Circuits, Systems and Signal Processing

S. C. Dutta Roy

doi:10.1007/978-981-10-6919-2

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Circuits, Systems and Signal Processing

Roy Tamonob Roy

2018

https://doi.org/10.1007/978-981-10-6919-2

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323 pages

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The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Suhash Chandra Dutta Roy Circuits, Systems and Signal Processing A Tutorials Approach Circuits, Systems and Signal Processing Suhash Chandra Dutta Roy Circuits, Systems and Signal Processing A Tutorials Approach 123 Suhash Chandra Dutta Roy Department of Electrical Engineering Indian Institute of Technology Delhi New Delhi, Delhi India ISBN 978-981-10-6918-5 ISBN 978-981-10-6919-2 (eBook) https://doi.org/10.1007/978-981-10-6919-2 Library of Congress Control Number: 2017962026 © Springer Nature Singapore Pte Ltd. 2018 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional afﬁliations. Printed on acid-free paper This Springer imprint is published by Springer Nature The registered company is Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore Dedicated to My Parents Shri Suresh Chandra Dutta Roy who told me ‘high quality steel cannot be made without burning iron’ and said, ‘Suhash you will surely shine’ and Shrimati Suruchi Bala Dutta Roy who must have given me the genes to keep the ﬁre on and says, ‘Son, I am with you, even though you lost me at nine’ Preface Starting from 1962, I have written and published a large number of articles in IETE Journals. At various points of time, starting from the early 80s, I have been requested by students, as well as teachers and researchers, to publish a book of collected reprints, appropriately edited and sequenced. Of late, this request has intensiﬁed, and the demand for reprints of some of the tutorial papers I wrote, has increased considerably, not only from India, but also from abroad, because of my ﬁve video courses related to Circuits, Systems and Signal Processing (CSSP) successfully uploaded by NPTEL on the YouTube. I thought it would be a good idea to venture into such a project at this time. This is a book for you, students, teachers and researchers in the subjects related to CSSP. As you would notice, I have written this book in a conversational style to make you feel at ease while reading it. I have also injected some wit and humour at appropriate places to make it enjoyable. This book is divided into four parts, dealing with Signals and Systems, Passive Circuits, Active Circuits and Digital Signal Processing. An appendix has also been added to give simple derivations of mathematics used throughout this book. In each chapter, I have added some examples so that the students may appreciate the fundamentals treated in the chapter and apply them to practical cases. This book contains chapters based on only articles of tutorial nature and those containing educational innovations. Purely research papers are excluded. The details of the parts are as follows: Part I on Signals and Systems comprises six chapters on basic concepts in signals and systems, state variable characterization, some fundamental issues involving the impulse function and partial fraction expansion of rational functions in s and z, having repeated poles. Part II on Analysis of Passive Circuits consists of 16 chapters on circuit analysis without transforms, transient response of RLC networks, circuits which are deceptive in appearance, resonance, many faces of the single tuned circuit, analysis and design of the parallel-T RC network, perfect transformer, capacitor charging through a lamp, difference equations and resistive net- works, a third-order driving point synthesis problem, an example of LC vii viii Preface driving point synthesis, low-order Butterworth ﬁlters, band-pass/band-stop ﬁlter design by frequency transformation and optimum passive differentiators. Part III on Active Circuits contains eight chapters on BJT biasing, analysis of a high-frequency transistor stage, transistor Wien bridge oscillator, anal- ysis of sinusoidal oscillators, triangular to sine wave converter and the Wilson current mirror. Part IV Digital Signal Processing comprises eight chapters on the ABCD’s of digital signal processing, second-order band-pass and band-stop ﬁlters, all-pass digital ﬁlters, FIR lattice and fast Fourier transform. In the Appendix, simpliﬁed treatment of some apparently difﬁcult topics is presented. These are roots of a polynomial, Euler’s relation, approximation of the square root of the sum of two squares, solution to cubic and quartic equations, solving second-order linear differential equations and Chebyshev polynomials. I hope students and teachers will beneﬁt from these chapters. It is only then that I shall feel adequately rewarded. For any mistakes/confusions/ clariﬁcations, please feel free to contact me on email at s.c.dutta. [email protected]. I take such mail on top priority, I assure you. Happy learning! New Delhi, India Suhash Chandra Dutta Roy Acknowledgements I owe so much to so many that it is not possible to acknowledge all of them in this limited space. To the ones whose credit I have not been able to mention, I ask for forgiveness. To the successive three Presidents, Shri R. K. Gupta, Shrimati Smriti Dagur and Lt Gen (Dr.) AKS Chandele, PVSM, AVSM (retd), I am indebted for their enthusiasm and encouragement; To the successive Publication Committee Chairpersons and Members of the Governing Council, I would like to say a big thank you for supporting my proposal of writing this book; To Shrimati Sreelatha Menon, Former Managing Editor of IETE, for her constant advice and suggestions for taking the project forward; To Shrimati Sandeep Kaur Mangat, I would like to express my deep gratitude for successfully leading this project to a conclusion, with her immense patience, and love and passion for work, much beyond her duty, and for carrying out all the necessary hard work; To all the workers of the Publications Section and the Secretariat in general, I have no words to convey my heartfelt appreciation for their diligent work; To my students—real and virtual, I would like to express my love and best wishes for their raising interesting questions in the class as well as outside, and through innumerable emails, which led to the polishing and refreshing of the contents; To my wife and lifelong companion, Shrimati Sudipta Dutta Roy, I cannot say enough to express my love and adoration for being with me in all weathers—in rain and sunshine—and in all the moments of happiness and depression, particularly during the years that I devoted to compiling and editing this book; To my two sons, Sumantra and Shoubhik, and their families, I would like to say ‘I have been and shall always be with you, even though many a times I could not pay due attention to family duties and responsibilities’; Finally, to my grandson Soham, is due a special word. He stood by my table patiently and asked me to hurry up the work so that he gets his share of time to play with me. It is because of this that this book has seen the light of the day in such a short time. Suhash Chandra Dutta Roy ix About the Book This book, I claim, is unique in character. Such a book has never been written. This book is unique, because it is innovative all throughout. It is innovative in Titles of Chapters, Abstracts, Headings of Articles, Subhead- ings, References and Problems. I have injected wit and humour, so as to retain the interest of the readers and to ignite their imagination. I shall not write more about this book. Read and you will know. I do not wish to receive any compliments, whatsoever. What I wish to receive is your frank and blunt criticism, pointing out the deﬁciencies. I promise I shall take them seriously and make my best efforts to take care of them in the next edition. Happy reading! xi Contents Part I Signals and Systems 1 Basic Concepts in Signals and Systems . . . . . . . . . . . . . . . . . . 3 Linear System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Elementary Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Time Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Impulse Response and Convolution . . . . . . . . . . . . . . . . . . . . . . . 6 LTI System Response to Exponential Signals . . . . . . . . . . . . . . . 7 The Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Linear System Response to Periodic Excitation . . . . . . . . . . . . . . 10 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Spectral Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Concluding Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2 The Mysterious Impulse Function and its Mysteries . . . . . . . . 17 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Impulse Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 How Do You Solve Differential Equations Involving an Impulse Function? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... 19 How Do You Solve Difference Equations Involving an Impulse Function? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Fourier Transform of the Unit Step Function . . . . . . . . . . . . . . . 21 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3 State Variables—Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Why State Variables? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 What is a State? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Deﬁnitions and Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Standard Form of Linear State Equations . . . . . . . . . . . . . . . . . . 27 How Do You Choose State Variables in a Physical System? . . . 28 How Do You Choose State Variables When the System Differential Equation is Given? . . . . . . . . . . . . . . . . . . . . . . . . . . 29 xiii xiv Contents How Does One Solve Linear Time-Invariant State Equations? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 An Advice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 4 State Variables—Part II . . . . . . . . . . . . . . . . . . . . . . . . . . .... 37 Properties of the Fundamental Matrix . . . . . . . . . . . . . . . . . .... 37 The Fundamental State Transition Equation . . . . . . . . . . . . .... 39 Procedures for Evaluating the Fundamental Matrix: Described in Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... 39 By Exponential Series Expansion . . . . . . . . . . . . . . . . . . .... 39 By Solution of the Homogeneous Differential Equations using Classical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 By Evaluating the Inverse Laplace Transform of (sI–A) . . . . . 40 State Transition Flow Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Concluding Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Appendix on Review of Matrix Algebra . . . . . . . . . . . . . . . . . . . 46 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 5 Carry Out Partial Fraction Expansion of Functions with Repeated Poles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 The Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Another Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 6 A Very Simple Method of Finding the Residues at Repeated Poles of a Rational Function in z−1 . . . . . . . . . . . 53 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 The Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Part II Passive Circuits 7 Circuit Analysis Without Transforms . . . . . . . . . . . . . . . . . . . 61 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Some Nomenclatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Force-Free Response: General Considerations . . . . . . . . . . . . . . . 63 Force-Free Response of a Simple RC Circuit . . . . . . . . . . . . . . . 63 Force-Free Response of a Simple RL Circuit . . . . . . . . . . . . . . . 65 First-Order Circuits with More Than One Energy Storage Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Force-Free Response of a Second-Order Circuit . . . . . . . . . . . . . 65 Root Locus of the Second-Order Circuit . . . . . . . . . . . . . . . . . . . 68 Natural Frequencies of Circuits with a Forcing Function . . . . . . . 68 Contents xv Concept of Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Relation Between Impedance and Natural Frequencies . . . . . . . . 69 Forced Response to an Exponential Excitation . . . . . . . . . . . . . . 70 Forced Response Due to DC . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Forced Response to a Sinusoidal Excitation . . . . . . . . . . . . . . . . 70 Basic Elements and Their V–I Relationships for Sinusoidal Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 An Example of the Use of Phasors and Impedances . . . . . . . . . . 72 Back to Complete Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Step Response of an RL Circuit . . . . . . . . . . . . . . . . . . . . . . . . . 75 Sinusoidal Response of a Series RC Circuit . . . . . . . . . . . . . . . . 75 Response of an RC Circuit to an Exponential Excitation . . . . . . 76 Step Response of an RLC Circuit . . . . . . . . . . . . . . . . . . . . . . . . 78 Sinusoidal Response of an RLC Circuit . . . . . . . . . . . . . . . . . . . 78 Pulse Response of an RC Circuit . . . . . . . . . . . . . . . . . . . . . . . . 80 Impulse Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 8 Transient Response of RLC Networks Revisited . . . . . . . . . . . 83 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Example Circuit and the Differential Equation . . . . . . . . . . . . . . 83 Analytical Solution of the Differential Equation . . . . . . . . . . . . . 84 Evaluating the Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Overdamped Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Underdamped Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Critically Damped Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Concluding Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 9 Appearances Can Be Deceptive: A Circuit Paradox . . . . . . . . 89 The Illusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 AC Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 DC Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 10 Appearances Can Be Deceptive: An Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 The Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Establishing I2(0−): One Possibility . . . . . . . . . . . . . . . . . . . . . . . 93 Establishing I2 (0−): Another Possibility . . . . . . . . . . . . . . . . . . . 93 Solve the Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 11 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Q: A Figure of Merit for Coils and Capacitors . . . . . . . . . . . . . . 95 Series Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Parallel Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 xvi Contents Impedance/Admittance Variation with Frequency: Universal Resonance Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Bandwidth of Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Other Types of Resonant Circuits . . . . . . . . . . . . . . . . . . . . . . . . 100 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Some Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12 The Many Faces of the Single-Tuned Circuit . . . . . . . . . . . . . 105 Notations: First Things First . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 The Possible Conﬁgurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 The Low-Pass Conﬁguration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 The High-Pass Conﬁguration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 The Band-pass Conﬁguration . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 The Band-stop Conﬁguration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 13 Analyzing the Parallel-T RC Network . . . . . . . . . . . . . . . . . . . 111 Mesh Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Node Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Two-Port Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Analysis by Miller’s Equivalence . . . . . . . . . . . . . . . . . . . . . . . . 113 Splitting the T’S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Yet Another Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 14 Design of Parallel-T Resistance–Capacitance Networks For Maximum Selectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Network Conﬁguration and Simpliﬁcation . . . . . . . . . . . . . . . . . . 118 Null Condition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Selectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Linearity of the Selectivity Curve . . . . . . . . . . . . . . . . . . . . . . . . 121 Selectivity of an Ampliﬁer Using the General Parallel-T RC Network in the Negative Feedback Line . . . . . . . . . . . . . . . . . . . 123 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 15 Perfect Transformer, Current Discontinuity and Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 Analysis of the General Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . 126 Contents xvii Condition for Continuity of Currents Under Perfect Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 16 Analytical Solution to the Problem of Charging a Capacitor Through a Lamp . . . . . . . . . . . . . . . . . . . . . . . . . . 131 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 The Circuit and the Differential Equation . . . . . . . . . . . . . . . . . . 131 Solution of the Differential Equation . . . . . . . . . . . . . . . . . . . . . . 132 Energy Dissipated in the Lamp . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 17 Difference Equations, Z-Transforms and Resistive Ladders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Solution by Difference Equation Approach . . . . . . . . . . . . . . . . . 136 Z-Transform Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Resistance Between Any Two Arbitrary Nodes of an Inﬁnite Ladder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Concluding Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 18 A Third-Order Driving Point Synthesis Problem . . . . . . . . . . 141 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Is Z(s) at All Realizable? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Alternative Realization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 A Problem for the Student. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 19 Interference Rejection in a UWB System: An Example of LC Driving Point Synthesis . . . . . . . . . . . . . . 147 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 The Four Canonical Realizations . . . . . . . . . . . . . . . . . . . . . . . . . 148 Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 Effect of Losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 20 Low-Order Butterworth Filters: From Magnitude to Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Butterworth Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Basis of the Alternative Method . . . . . . . . . . . . . . . . . . . . . . . . . 152 Application to Low Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 First-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Second-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 xviii Contents Third-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Fourth-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Fifth-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Sixth-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Seventh-Order Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Application to Chebyshev Filters . . . . . . . . . . . . . . . . . . . . . . . . . 156 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 21 Band-Pass/Band-Stop Filter Design by Frequency Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Band-Pass Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Band-Stop Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Concluding Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 22 Optimum Passive Differentiators . . . . . . . . . . . . . . . . . . . . . . . 165 Optimal Transfer Function and Its Realizability . . . . . . . . . . . . . 166 Second-order Optimal and Suboptimal Differentiators . . . . . . . . . 170 Third-order Suboptimal Passive Differentiator . . . . . . . . . . . . . . . 171 Optimal RC Differentiators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Suboptimal RC Differentiator . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Part III Active Circuits 23 Ampliﬁer Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 24 Appearances Can Be Deceptive: The Case of a BJT Biasing Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Analysis of N2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 An Example of Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 25 BJT Biasing Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 The Generalized Circuits and Special Cases . . . . . . . . . . . . . . . . 191 Bias Stability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Contents xix An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Design of N1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Design of N2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Design of N3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 Performances of N1, N2 and N3 . . . . . . . . . . . . . . . . . . . . . . . . . . 196 Design and Performance of N4 . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Using the Total Change Formula . . . . . . . . . . . . . . . . . . . . . . . . . 197 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 26 Analysis of a High-Frequency Transistor Stage . . . . . . . . . . . 199 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Two Port Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 27 Transistor Wien Bridge Oscillator . . . . . . . . . . . . . . . . . . . . . . 203 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 Circuit 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 Circuit 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 Circuit 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Practical Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 Discussions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 28 Analysing Sinusoidal Oscillator Circuits: A Different Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 An Op-Amp Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Transistor Version of the Wien Bridge Oscillator . . . . . . . . . . . . 214 Another Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Concluding Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 29 Triangular to Sine-Wave Converter . . . . . . . . . . . . . . . . . . . . . 217 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 30 Dynamic Output Resistance of the Wilson Current Mirror . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 xx Contents Part IV Digital Signal Processing 31 The ABCDs of Digital Signal Processing––PART 1 . . . . . . . . 229 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 The Basic Digital Signal Processor . . . . . . . . . . . . . . . . . . . . . . . 230 The Sampling Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 Quantization Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Z-Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 Transfer Function of a Digital Signal Processor . . . . . . . . . . . . . 236 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 32 The ABCDs of Digital Signal Processing–PART 2 . . . . . . . . . 241 Realization of Digital Signal Processors . . . . . . . . . . . . . . . . . . . 241 The Discrete Fourier Transform. . . . . . . . . . . . . . . . . . . . . . . . . . 242 The Fast Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 Applications of FFT to Compute Convolution and Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ......... 247 Application of FFT to Find the Spectrum of a Continuous Signal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 Concluding Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 33 On Second-Order Digital Band-Pass and Band-Stop Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Design for Arbitrary Pass-band Tolerance . . . . . . . . . . . . . . . . . . 257 Realization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 34 Derivation of Second-Order Canonic All-Pass Digital Filter Realizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 Derivation of the Structure of Fig. 34.2 . . . . . . . . . . . . . . . . . . . . 263 Derivation of the Structure of Fig. 34.3 . . . . . . . . . . . . . . . . . . . . 263 Alternative Derivation of the Structure of Fig. 34.2 . . . . . . . . . . 264 Yet Another Derivation of the Structure of Fig. 34.2 . . . . . . . . . 265 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 35 Derivation of the FIR Lattice Structure . . . . . . . . . . . . . . . . . . 267 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Concluding Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Contents xxi 36 Solution to a Problem in FIR Lattice Synthesis . . . . . . . . . . . 271 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Conventional Synthesis Procedure . . . . . . . . . . . . . . . . . . . . . . . . 272 Linear Phase Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Nonlinear Phase FIR Function with hN (N ) = ±1 . . . . . . . . . . . . 274 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 37 FIR Lattice Structures with Single-Multiplier Sections . . . . . . 279 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Realization 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Realization 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Realization 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Realization 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 38 A Note on the FFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 Derivation of the Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Alternative Derivation for M(N) . . . . . . . . . . . . . . . . . . . . . . . . . 285 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 Appendix: Some Mathematical Topics Simpliﬁed . . . . . . . . . . . . . . 287 About the Author Suhash Chandra Dutta Roy I was fortunate enough to be educated at the Calcutta University, where great professors and researchers like C. V. Raman, S. N. Bose, M. N. Saha and S. Radhakrishnan taught and made a name, globally, not only for them, but also for the uni- versity and the country. Immediately after my master's, I had to take up a job in a West Bengal Government Research Institute. I soon discovered that many things are done at the Institute, but not research (like an American Drug Store!). I there- fore quit and shifted to the newly established University of Kalyani as a lecturer. There I spent a few years, and after my Ph.D., I left the country for taking up an Assistant Professorship at the University of Minnesota. A few years passed by, but I felt increasingly guilty that I was not doing anything for my motherland. I therefore returned to the country as an Associate Professor at IIT Delhi where I served for more than four decades as professor, head of the department and dean. The last two positions were imposed upon me, and I suffered because they cut down the time available for research and interaction with my dear students. I formally retired at the age of 60, but continued to teach and carry out research, as an Emeritus Fellow, followed by INSA Senior Scientist and INSA Honorary Scientist. Since my term as an Emeritus Fellow was over, I did not xxiii xxiv About the Author take any money from IIT Delhi and served voluntarily, simply for the love of teaching and research. I ﬁnally quit IIT Delhi 12 years after retirement. I am now settled in a DDA Flat at Hauz Khas, where I live happily with my wife, but still continue to do research. We both are reasonably healthy, because of strict diet and exercise, including pranayama. I have been extraordinarily lucky to have had gems of Ph.D. students, 30 of them, who have done exemplary work in research and innovation. In fact, I consider myself as shining from reflected glory. I have also been lucky to have received high recognition through Fellowship of IEEE, Distinguished Fellowship of IETE and Fellowship of all the relevant national academies. I have been awarded some prestigious national awards, including the Shanti Swarup Bhatnagar Prize. Over and above all the awards and recognition, however, what I value most is the love and affection of my students. I now spend time giving professional lectures and also delivering sermons on innovations in teaching and research, in general, and also on how to improve their standards in the country. I have the hobby of listening to Hindustani classical music and researching on its masters. I also love spending quality time with my grandson, Soham. Reading political history and detective stories are my other hobbies. I also read poetry and compose some poems and short stories, for my own pleasure. In short, I have lived a complete life, with nothing to complain about. Part I Signals and Systems This part contains six chapters based on the same number of articles. Although they may appear to be disjointed, in reality, they are not. For example, the second chapter is about an impulse function, which often appears as a signal and in input–output characterization of systems. State variable characterization, dealt with in the following two parts, relates to a special kind of system, viz. linear system. State variable was a hot topic when the corresponding articles were written. They are still used, mostly in control systems. The ﬁfth part relates to a rational function in s which is the Laplace transform of the output of a system. A simple method is presented, in contrast to usual textbook methods, for partial fraction expansion of a function with repeated poles. The last part relates to the same topic in Digital Signal Pro- cessing where s is replaced by the variable z. All throughout here, as well as everywhere in the book, my effort has been to simplify life so that you can sail in gentle waters and do not have to shed tears. Tears are costly and should not be allowed to flow just like that! Reserve them for practical life where you would have ample opportunities to shed them. All the examples here, at the end of each chapter, have some twists and turns; you have to unwind them to be able to ﬁgure out how you should proceed. Do not think along difﬁcult lines because difﬁcult problems always have simple solutions. This is true not only here but in life in general. Do not give up. If a path does not lead you to the goal, return and choose an alternative path. Sometimes, when you go some way in the latter, it appears that the ﬁrst path gives you the solution! There are no straightforward rules; you have to ﬁnd your own path. All the time, do not allow your mind to be polluted by complicated rules and procedures. All that is complicated is useless. This is also true for life in general. But enough of this philosophical discourse. Coming down to the problem at hand, you should learn the fundamentals carefully and completely. Once you make them your own, no application will appear difﬁcult. I have done this throughout my life, in teaching as well as research, and I have received excellent rewards, well beyond my expectations. 2 Part I: Signals and Systems Help your classmates whenever they are in difﬁculty in solving problems, or otherwise. This helps in strengthening the bonds and also in clearing your own understanding at places where they were fuzzy. The problems I have set have a different character as compared to the usual textbook ones. Some have intentionally designed wrong results. You will have to correct them before going ahead. Do workout all the problems. Intentionally, I have not given the correct answers. Consult your classmates; if a majority gets the same result, then you can be conﬁdent that your solution is correct. A good teacher is born and not made. Identify one or two such good teachers and consult them in case of difﬁculties. Don’t give up if they turn you down on one pretext or other. After all, your teacher should realize that teaching–learning is an interactive process, and unless there is a strong and loving relationship among the three components, viz. teacher, student and subject, teaching–learning becomes routine, monotonous and repelling. Consult the video courses, fully downloadable from the YouTube, I have created ﬁve of them, but there are many more from IITs through the NPTEL programme. Choose the best few and follow them. It is not that these are all faultless, but faults are few and far between. If you ﬁnd a mistake, conﬁrm with an email to the concerned teacher and clarify. I have been doing this every day since my ﬁrst video course appeared in the YouTube. Finally, I wish you happy reading and happy learning. Write to me at the email address given earlier, if you have a question. I enjoy interacting through emails as most of the teachers also do. Basic Concepts in Signals and Systems 1 In this chapter, I shall tell you all you wanted Keywords to know about signals and systems but were Linear systems Signals Amplitude afraid to ask your teacher. Starting with the and phase spectra Response to periodic deﬁnition of linear systems, some elementary signals Non-periodic signals Fourier series signals are introduced, which is followed by and transforms Energy and power signals the notion of time-invariant systems. Signals Spectral density and systems are then coupled through impulse response, convolution and response to expo- nential signals. This naturally leads to Fourier series representation of periodic signals and Linear System consequently, signal representation in the frequency domain in terms of amplitude and The concept of linear systems plays an important phase spectra. Does this sound difﬁcult? role in the analysis and synthesis of most prac- I shall simplify it to the extent possible, do tical systems, be it communication, control or not worry. Linear system response to periodic instrumentation. Consider a system S which signals, discussed next, is then easy to under- produces an output y when an input x is applied stand. To handle non-periodic signals, Fourier to it (both y and x are usually functions of time). transform is introduced by viewing a non- We shall denote this symbolically as periodic function as the limiting case of a periodic one, and its application to linear x S y ! system analysis is illustrated. The concepts of energy and power signals and the correspond- Then S is said to be linear if it obeys two ing spectral densities are then introduced. principles, viz. principle of superposition and principle of homogeneity. The former implies that if x1 ! y1 (note that we have omitted S above the arrow: this is implied) and x2 ! y2, then x1 + x2 ! y1 + y2. The principle of homo- Source: S. C. Dutta Roy, “Basic Concepts in Signals and geneity implies that if x ! y, then ax ! ay where Systems,” IETE Journal of Education, vol. 40, pp. 3–11, a is an arbitrary constant. Note, in passing, that a January–June 1999. could be zero, i.e. zero input should lead to zero Lectures delivered at Trieste, Italy, in the URSI/ICTP output in a linear system. Combining the two Basic Course on Telecommunication Science, January 1989. © Springer Nature Singapore Pte Ltd. 2018 3 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_1 4 1 Basic Concepts in Signals and Systems principles, we can now formally deﬁne a linear surprise you, because life, in general, is nonlinear system as one in which and we make it simple by approximating it by a linear one. Otherwise, life would have been so x1;2 ! y1;2 ) ax1 þ bx2 ! ay1 þ by2 ; ð1:1Þ complicated that it would not be worth living. Enjoyment would have been out of question! where the notation ) is used to mean ‘implies’. Also, in many situations, a nonlinear system is As an example, consider the system described ‘incrementally’ linear, i.e. the system is linear if by the well-known equation of a straight line an increment Dx in x is considered as the input and the corresponding increment Dy in y is y ¼ mx þ c ð1:2Þ considered as the output. Both Eqs. 1.2 and 1.4 are descriptions of such incrementally linear It may seem surprising but Eq. 1.2 does not systems. A transistor ampliﬁer is a highly non- describe a linear system unless c = 0, simply linear system, but it behaves as a linear one if the because zero input does not lead to zero output. input is an AC signal superimposed on a much Another way of demonstrating this is to apply ax larger DC bias. as input; then the output is y0 ¼ max þ c 6¼ ay ¼ max þ ca ð1:3Þ Elementary Signals By the same token, the dynamic system described by A signal, in the context of electrical engineering, is a time-varying current or voltage. An arbitrary dy signal can be decomposed into some elementary þ 5y ¼ 5x þ 6 ð1:4Þ dt or ‘basic’ signals, which, by themselves also occur frequently in nature. These are (i) the is not linear, because (x = 0, y = 0) does not exponential signal eat where a may be real, satisfy the equation. imaginary or complex, (ii) the unit step function Another, and a bit more subtle example, is u(t) and (iii) the unit impulse function, d(t). shown in Fig. 1.1. Is this system linear? Obvi- When a is purely imaginary in eat, we get a ously x = 0 leads to z = 0 but then this is only a particularly important situation, because if necessary condition for a linear system. Is it a = jx and x is real, then sufﬁcient? To test this, apply x = x0; then z = z0. Now apply x = −x0; the output is still z0 instead ejxt ¼ cos xt þ j sin xt ð1:5Þ of –z0. The obvious conclusion is that the system is nonlinear. Thus sinusoidal signals, cos xt and sin xt, Almost all practical systems are nonlinear, which are so important in the study of commu- which are usually much more difﬁcult to handle nications, are special cases of the exponential than linear systems. Hence, we make our life signal. The quantity x, as is well known, is the comfortable by approximating (or idealizing?) a frequency in radians/s, while f = x/(2p) is the nonlinear system by a linear one. This should not frequency in cycles/s or Hz. y = yo (x/x0)2 z = zo (y/y0)1/2 x y z Fig. 1.1 A subtle example of a nonlinear system Elementary Signals 5 The unit step function, shown in Fig. 1.2, is representation of d(t). Two important properties deﬁned by of d(t) are that 0 t\0 xðtÞdðtto Þ ¼ xðto Þdðtto Þ ð1:10Þ uð t Þ ¼ ð1:6Þ 1 t[0 and Note that it is discontinuous at t = 0. The unit Z1 impulse function d(t) is related to u(t) through xðsÞdðt sÞds ¼ xðtÞ ð1:11Þ Zt 1 uð t Þ ¼ dðsÞds ð1:7Þ Equation 1.11 easily follows from Eqs. 1.9 1 and 1.10, and represents the ‘Sifting’ or ‘Sam- pling’ property of the impulse function. or duðtÞ dðtÞ ¼ ð1:8Þ Time Invariance dt Obviously, it exists only at t = 0, and the At this point, we need to introduce another value there is inﬁnitely large, but concept, viz. that of time invariance of a system. A system S is time invariant if a time shift in the Z1 Z0 þ input signal causes the same time shift in the dðsÞds ¼ dðsÞds ¼ 1; ð1:9Þ output signal, i.e. if x(t) ! y(t) implies x(t – t0) ! y(t − t0). 1 0 Both Eqs. 1.2 and 1.4 are descriptions of i.e. the area under the plot of d(t) versus t is unity. time-invariant systems. On the other hand, This is called the strength of the impulse; for y(t) = tx(t) represents a time-varying system. example, the strength of the impulse Kd(t) is K. Most of the practical systems we encounter are Obviously, there is some formal difﬁculty with time-invariant systems. regard to the deﬁnition of d(t), but we shall not Systems which are linear and time invariant enter into this debate here. d(t) can be viewed as (LTI) are particularly simple to analyze in terms the limit of the rectangular pulse shown in of their impulse response or frequency response Fig. 1.3 as D tends to 0; Fig. 1.3 also shows the function, as will be demonstrated in what follows. u(t) 1 (t) 0 1/ 1 t 0 t t 0 0 Fig. 1.2 The unit step function Fig. 1.3 A limiting view of d(t) 6 1 Basic Concepts in Signals and Systems Impulse Response and Convolution h(t) * x(t)); in fact, this is what equivalence of Eqs. 1.16 and 1.17 implies), associative (i.e. Consider an LTI system whose response to a unit x(t) * [h1(t) * h2(t)] = [x(t) * h1(t)] * h2(t)); this impulse function is h(t), i.e. is useful in the analysis of cascade connection of systems) and distributive (i.e. x(t) * [h1(t) + dð t Þ ! hð t Þ ð1:12Þ h2(t)] = x(t) * h1(t) + x(t) * h2(t)); this is useful in the analysis of parallel systems). By time invariance, therefore As an example of application of the convo- lution integral, consider the RC network shown dðt sÞ ! hðt sÞ ð1:13Þ in Fig. 1.4, where both x(t) and y(t) are voltages, and the capacitor is uncharged before application By homogeneity, if we multiply the left-hand of x(t) (an alternate way of expressing this is to side of Eq. 1.13 by x(s)ds, the right-hand side say that C is initially relaxed). When x(t) = d(t), should also get multiplied by x(s)ds, i.e. the current in the circuit is i(t) = d(t)/R. This xðsÞdðt sÞds ! xðsÞhðt sÞds ð1:14Þ impulse of current charges the capacitor to a voltage By superposition, if we integrate the left-hand side of Eq. 1.14, we should do the same for the Z0 þ 1 dðsÞ 1 right-hand side, i.e. ds ¼ ð1:18Þ C R RC 0 Z1 Z1 xðsÞdðt sÞds ! xðsÞhðt sÞds at t = 0+. For t > 0+, d(t) = 0; hence the capaci- 1 1 tor charge decays exponentially; so does the ð1:15Þ voltage across it, according to 1 t=ðRCÞ But, by Eq. 1.11, the left-hand side of yðtÞ ¼ e ð1:19Þ Eq. 1.15 is simply x(t), so the right-hand side RC should be y(t). Thus, if the unit impulse response Thus, the impulse response of the RC network h(t) of an LTI system is known, then one can ﬁnd is the output of the system due to an arbitrary excitation x(t) as 1 t=T hðtÞ ¼ e uðtÞ; ð1:20Þ Z1 T yðtÞ ¼ xðsÞhðt sÞds ð1:16Þ where T = RC is called the time constant of the 1 network. Now, suppose the input is changed to a Z1 unit step voltage, i.e. x(t) = u(t). Then the ¼ xðt sÞhðsÞds; ð1:17Þ response is, by Eq. 1.17, 1 where the second form follows simply, through a change of variable. The integral Eqs. 1.16 or 1.17 R is called the convolution integral and the opera- + + tion of convolution is symbolically denoted as x(t) i(t) C y(t) y ð t Þ ¼ x ð t Þ hð t Þ – – It is a simple matter to prove that convolution operation is commutative (i.e. x(t) * h(t) = Fig. 1.4 An RC network Impulse Response and Convolution 7 Z1 part (cost xt) or imaginary part (sin xt), then the 1 s=T output will be H(jx) ejxt or Re [H(jx) ejxt] or Im yðtÞ ¼ e uðsÞuðt sÞds ð1:21Þ T [H(jx)ejxt], respectively. For example, if 1 H(jx) = |H(jx)| ej∠H(jx) and the input is cos xt, Zt then the output shall be |H(jx)| cos (xt + 1 ∠H(jx)). H(jx) varies with frequency, and the ¼ es=T ds ¼ 1 et=T uðtÞ; ð1:22Þ T plots of |H(jx)| and ∠H(jx) versus x are known 0 as magnitude and phase responses, respectively. where the lower limit arises due to the factor u(s) Since the principle of superposition holds in a linear system, the response to a linear combina- and the upper limit arises as a consequence of the P factor u(t − s) in the integrand. tion of exponential signals, i ai esi t , will be of P the form i ai Hðsi Þesi t . It is precisely this fact which motivated Fourier to explore if an arbitrary LTI System Response to Exponential signal could be represented as a superposition of Signals exponential signals. As is now well known, this can indeed be done by a Fourier series for a Let x(t) = est where s, as you will see later, is the periodic signal and by the Fourier transform for a complex frequency r + j x, be applied to a general, not necessarily periodic, signal. system with impulse response h(t); then by Eq. 1.17, the response is The Fourier Series Z1 yðtÞ ¼ hðsÞ esðtsÞ ds ð1:23Þ Consider a linear combination of the exponential 1 signal ejx0 t with its harmonically related expo- nential signals ejkx0 t k = 0, ±1, ±2, … : Z1 X 1 ¼e st hðsÞ ess ds ð1:24Þ xðtÞ ¼ ak ejkx0 t ð1:27Þ 1 k¼1 In this, k = 0 gives a constant term or dc, in ¼ H ðsÞest ; ð1:25Þ electrical engineering language; ejx0 t is the smallest frequency term, with a frequency x0 and where period T = 2p/x0, and is called the fundamental Z1 frequency. The term ej2x0 t has a frequency 2x0, HðsÞ ¼ hðsÞ ess ds ð1:26Þ while ej2x0 t has a frequency −2x0; the period of either term is T/2, and both the terms represent 1 what is known as the second harmonic. A similar is called the system function or transfer function interpretation holds for the general term ejkx0 t ; of the system and is a function of s only. A signal which has a period T/|k|. Note that we take the for which the output differs from the input only frequency as positive or negative, but the period by a scaling factor (perhaps complex) is called is taken as positive. Obviously, the summation the eigenfunction of the system, and the scaling Eq. 1.27 is periodic with a period equal to T, in factor is called the eigenvalue of the system. which there are |k| periods of the general term Obviously, est is an eigen function of an LTI ejkx0 t but only one period of the fundamental. system, and H(s) is its eigenvalue. What about a given periodic function x(t) with When s = jx, H represents the frequency a period T, i.e. x(t + mT) = x(t), m = 0, ±1, ±2, response of the system, i.e. if x(t) = ejxt or its real … ? Can it be decomposed into the form 8 1 Basic Concepts in Signals and Systems Eq. 1.27? It turns out that under certain condi- T2 t T=2 which virtually becomes −s/2 tions which are satisﬁed by all but a few excep- t +s/2, because x(t) = 0 at other values of tional cases, one can indeed do so. To determine t within the chosen interval. Hence ak’s, multiply both sides of Eq. 1.27 by ejnx0 t and integrate over the interval 0 to T. Obviously, Zs=2 RT A this results in an integral 0 ejðknÞx0 t dt on the ak ¼ ejkx0 t dt T right-hand side, which is zero if k 6¼ n, and T if s=2 RT k = n. Thus, an = (1/T) 0 xðtÞ ejnx0 t dt or 2A kx0 s ð1:29Þ ¼ sin Z kx0 2 T 1 sin kx20 s ak ¼ xðtÞ ejkx0 t dt ð1:28Þ ¼ sA kx0 s T 0 2 ak represents the weight of the kth harmonic and This is of the form sA sin x/x, where x = kx0s/2. is called the spectral coefﬁcient of x(t). ak is, in Note that ak is real, and can be positive, zero general, complex. A plot of |ak| versus k will or negative. Hence, separate amplitude and phase consist of discrete lines at k = 0, ±1, ±2, …, it spectrum plots are not necessary; a single dia- resembles a spectrum as observed on a spectro- gram sufﬁces and is shown in Fig. 1.6. Note that scope and is called the amplitude spectrum. ak has a maximum value at DC, i.e. k = 0, the Similarly, one can draw a phase spectrum. value being sA (this checks with direct calcula- It is obvious from Eq. 1.27 that x(t) could be tion from Fig. 1.5). The envelope of the spec- written as the summation of a sine and a cosine trum is of the form sin x/x and exhibits damped series, and that the corresponding coefﬁcients RT oscillations with zeros at x = p (i.e. kx0 = 2p/s), could be found from 0 xðtÞ cos kxt dt 2p (i.e. kx0 = p/s), … Further, the sketch is RT and 0 xðtÞ sin kxt dt: It is, however, much more symmetrical about x = 0, because sin x/x is an convenient to handle the exponential form of the even function. The spectrum consists of discrete Fourier series as given in Eq. 1.27. lines, two adjacent lines being separated by 2p T ¼ As an example of application of the Fourier x0 rad/s. series, consider the pulse stream shown in Some important points emerge from the Fig. 1.5. sketch of Fig. 1.6. As T increases, the lines get At this point, note that in Eq. 1.28, the lower closer and ultimately when T ! ∞, corre- and the upper limit of integration are not sponding to a single pulse, the spectrum becomes important so long as their difference is T. This is continuous and will be characterized by the Rt þT sin xs=2 so because t00 ejðknÞx0 t is independent of t0. function sA xs=2 , where kx0 has become the In the example under consideration, it is obvi- continuous variable x. This, as we shall see, is ously convenient to choose the interval the Fourier transform of the single pulse. Second, since the lines concentrated in the lower frequency range are of higher amplitude, x(t) most of the energy of the periodic wave of Fig. 1.5 must be conﬁned to lower frequencies. A Third, as s decreases, the spectrum spreads out, i.e. there is an inverse relationship between pulse width and frequency spread. t Since the energy of the periodic wave is –T –T/2 /2 0 /2 T/2 T mostly conﬁned to the lower frequency range, a convenient measure of bandwidth of the signal is Fig. 1.5 A rectangular pulse stream from zero frequency to the frequency of the ﬁrst The Fourier Series 9 ak TA –3p –2p –p p 2p 3p x 6p - 4p - 2p - –w w0 2p 4p 6p w = 2x/t 0 T T T T T T 0 Fig. 1.6 Spectrum of x(t) of Fig. 1.5 zero crossing, i.e. the bandwidth in Hz, B, can be A periodic signal that is of great importance in taken as 1/s. digital communication is the impulse train If x(t) of Eq. 1.27 is the voltage across or the current through a one-ohm resistor, then X 1 RT xðtÞ ¼ dðt kTÞ ð1:32Þ the average power dissipated is T1 0 jxðtÞj2 dt. k¼1 If one writes |x(t)|2 = x(t) x(t), where bar denotes complex conjugate, and substitutes for as shown in Fig. 1.7. If this is expanded in x(t) and x(t) from Eq. 1.27, there results the Fourier series following: X 1 X 1 X 1 xðtÞ ¼ ak ejkx0 t ; ð1:33Þ 2 jxðtÞj ¼ ak an e jðknÞx0 t ð1:30Þ k¼1 k¼1 n¼1 RT As we have already seen, 0 ejðknÞx0 t dt is x(t) zero if k 6¼ n and equals T when k = n. Thus, the average power becomes 1 ZT X 1 1 jxðtÞj2 dt ¼ jak j2 ð1:31Þ T t 0 k¼1 3T 2T T 0 T 2T 3T This is known as Parseval’s theorem. Fig. 1.7 A periodic impulse train 10 1 Basic Concepts in Signals and Systems where xo = 2p/T, then As a simple example, consider the RC net- work of Fig. 1.4; we have already derived its ZT=2 impulse response as 1 1 ak ¼ dðtÞejkx0 t dt ¼ ð1:34Þ T T 1 t=ðRCÞ T=2 hðtÞ ¼ e uðtÞ ð1:37Þ RC The spectrum is sketched in Fig. 1.8 so that What is the bandwidth of this signal? The amplitude is a constant at all frequencies, unlike Z1 1 tðjx þ RC1 Þ the spectrum of Fig. 1.6. Hence, the bandwidth is HðjxÞ ¼ e dt ð1:38Þ inﬁnite. This agrees with our observation about RC 0 bandwidth and pulse duration, because Fig. 1.7 is the degenerate form of Fig. 1.5 with s ! 0 and 1 ¼ ð1:39Þ A ! ∞. jxRC þ 1 When excited by the periodic impulse train of Linear System Response to Periodic Fig. 1.7, the response y(t) can be found in two Excitation ways. First, since dðtÞ ! hðtÞ; From the discussion on LTI system response to exponential signals, it follows that a linear sys- it follows that tem, excited by the periodic signal of Eq. 1.27, will produce an output signal which is also X 1 X 1 periodic with the same period, and is given by dðt kTÞ ! hðt kTÞ k¼1 k¼1 X 1 yðtÞ ¼ ak Hðjkx0 Þejkx0 t ; ð1:35Þ so that k¼1 1 X 1 where yðtÞ ¼ eðtkTÞ=ðRCÞ uðt kTÞ ð1:40Þ RC k¼1 Z1 HðjxÞ ¼ hðtÞejxt dt ð1:36Þ Should you try to sketch this waveform, you would realize how messy it looks; also not much 1 information about the effect of the RC network and h(t) is the unit impulse response. will be obvious from this sketch. On the other hand, the Fourier series method gives, from Eqs. 1.35, 1.34 and 1.39. X 1 1=T yðtÞ ¼ ejkx0 t ð1:41Þ ak jkx0 RC þ 1 k¼1 1/T P Let this be written as 1k¼1 bk e jkx0 t ; then the sketch of |bk| versus x = kx0 looks like that shown in Fig. 1.9. Comparing this with Fig. 1.8, w 2w 0 w0 0 w0 2w 0 we note that the RC network attenuates higher frequencies as compared to lower ones and hence Fig. 1.8 Spectrum of the impulse train of Fig. 1.7 acts as a low-pass ﬁlter. The bandwidth of the Linear System Response to Periodic Excitation 11 bk Now, turn to the frequency domain. If the signal bandwidth is taken as B = 1/s Hz, then obviously for ﬁdelity, the RC ﬁlter must pass all frequencies up to 1/s Hz with as little attenuation as possible. Thus, Bf must be at least equal to B = 1/s. Since the attenuation is 3 dB instead of zero at Bf and the input spectrum is not limited to 2w 0 w0 0 w0 w = kw 0 B, the pulse shape will be distorted. For reduced 2w 0 distortion, we need to increase Bf and we expect Fig. 1.9 Spectrum of output y(t) given in Eq. 1.41 good results if Bf B i.e. RC s. ﬁlter Bf is deﬁned as the frequency at which the | The Fourier Transform H(jx)| falls down by 3 dB as compared to its dc value, i.e. Now consider a non-periodic function x(t) which pﬃﬃﬃ exists in the range −T/2 t T/2, and is zero jHðj2pBf Þj ¼ Hðj0Þ= 2 ð1:42Þ outside this range. Consider a periodic extension xp(t) of x(t), as shown in Fig. 1.11. xp(t) can be Combining this with Eq. 1.39 gives Bf = expanded in Fourier series as 1/(2pRC). What would be the response of the RC ﬁlter to X 1 the rectangular pulse stream of Fig. 1.5? This xp ðtÞ ¼ ak ejkx0 t ; ð1:43Þ will of course depend on the relative values of T, k¼1 s and Bf. First let us conﬁne ourselves to the time where x0 = 2p/T and domain. If the product RC is comparable to T, ZT=2 then the output will consist of overlapping pul- 1 ak ¼ xp ðtÞejkx0 t dt ð1:44Þ ses, and will retain very little similarity to the T T=2 input. Let, therefore, RC T; then depending on s, the response during one period will be of ZT=2 the form shown in Fig. 1.10. It is obvious that for 1 ¼ xðtÞejkx0 t dt ð1:45Þ ﬁdelity, i.e. if the output is to closely resemble T the input, we require RC s < T. T=2 Fig. 1.10 Response of RC y(t) y(t) network in one period of Fig. 1.5 t t 0 t 0 t RC ≥ t RC << t 12 1 Basic Concepts in Signals and Systems x(t) Z1 XðjxÞ ¼ F ½xðtÞ ¼ xðtÞejxt dt ð1:51Þ 1 t -T 0 T and the inverse Fourier transform as 2 2 xp (t) Z1 1 1 xðtÞ ¼ F ½XðjxÞ ¼ XðjxÞejxt dx 2p t 1 –T -T 0 T T ð1:52Þ 2 2 Without entering into the question of exis- Fig. 1.11 A non-periodic function x(t) and its periodic extension xp(t) tence, we simply state below the conditions, named after Dirichlet, under which x(t) is Fourier transformable. These are because for |t| T/2, xp(t) = x(t). Also, since x(t) = 0 for |t| > T/2, we can write R1 (1) jxðtÞjdt\1 Z1 1 1 (2) ﬁnite number of maxima and minima within ak ¼ xðtÞejkx0 t dt ð1:46Þ T any ﬁnite interval, and 1 (3) ﬁnite number of ﬁnite discontinuities within If we deﬁne any ﬁnite interval. Z1 Referring to Eqs. 1.26 or 1.36, it should be XðjxÞ ¼ xðtÞejxt dt ð1:47Þ obvious that the impulse response h(t) and the 1 frequency response H(jx) are Fourier transform pairs. Explicitly then from Eq. 1.46, we get HðjwÞ ¼ F ½hðtÞ ð1:53Þ 1 1 ak ¼ Xðjkx0 Þ ¼ Xðjkx0 Þx0 ð1:48Þ As an example of Fourier transformation, T 2p consider the rectangular pulse shown in Thus, Eq. 1.43 can be written as Fig. 1.12. Notice that this is the limiting form of 1 X 1 xp ðtÞ ¼ Xðjkx0 Þx0 ejkx0 t ð1:49Þ 2p k¼1 x(t) Now let T tend to inﬁnity; then xp(t) tends to x(t), kx0 tends to x, a continuous variable, x0 tends to dx, and the summation becomes an integral. Thus, Eq. 1.49 becomes Z1 1 xðtÞ ¼ XðjxÞejxt dx ð1:50Þ t 2p –t /2 0 t /2 1 Combining Eqs. 1.47 and 1.48, we now for- mally deﬁne the Fourier transform of x(t) as Fig. 1.12 A rectangular pulse The Fourier Transform 13 the periodic function of Fig. 1.5 with T ! ∞. Let t − s = n; then the integral inside the Applying Eq. 1.51, we get bracket becomes e−jxs H(jx), so that Zs=2 Z1 sinðxs=2Þ XðjxÞ ¼ A ejxt dt ¼ sA ð1:54Þ YðjxÞ ¼ HðjxÞxðsÞejxt ds; ð1:59Þ ðxs=2Þ s=2 1 This, as will be easily recognized, is the lim- i.e. iting form of Fig. 1.6, and is the envelope of the same ﬁgure. This veriﬁes the observation made YðjxÞ ¼ HðjxÞXðjxÞ ð1:60Þ in the discussion on Fourier series. In words, this amounts to saying that the The Fourier transform has many important spectrum of the output of a linear system is properties, the most important in the context of simply the product of the spectrum of the input analysis of linear systems being that it converts a signal and the frequency response of the system. convolution in the time domain to a multiplica- The output in the time domain, y(t) can be simply tion in the frequency domain, i.e. if found by taking the inverse Fourier transform of Z1 Y(jx). yðtÞ ¼ xðtÞ hðtÞ ¼ xðsÞhðt sÞds ð1:55Þ To illustrate the application of Eq. 1.60, 1 consider a linear system having the impulse response then, assuming that y(t), x(t) and h(t) are Fourier transformable, and F [y(t)] = Y(jx), we get hðtÞ ¼ ea t uðtÞ; a [ 0 ð1:61Þ YðjxÞ ¼ XðjxÞHðjxÞ ð1:56Þ which is excited by an input signal The proof of Eq. 1.56 is simple and proceeds xðtÞ ¼ eb t uðtÞ; b [ 0 ð1:62Þ as follows: By direct integration, it is easily shown that YðjxÞ ¼ F ½yðtÞ 2 3 1 1 Z1 Z1 HðjxÞ ¼ and XðjxÞ ¼ ð1:63Þ 4 a þ jx b þ jx ¼ xðsÞhðt sÞds5 ejxt dt 1 1 Thus ð1:57Þ 1 YðjxÞ ¼ ð1:64Þ Interchange the order of integration and notice ða þ jxÞðb þ jxÞ that x(s) does not depend on t; the result is 2 3 To determine y(t), one may write Z1 Z1 YðjxÞ ¼ xðsÞ4 hðt sÞ ejxt dt5ds A B YðjxÞ ¼ þ ð1:65Þ 1 1 ða þ jxÞ ðb þ jxÞ ð1:58Þ 14 1 Basic Concepts in Signals and Systems and ﬁnd A and B as Depending on the nature of the signal, the spectral density is also to be qualiﬁed as power or energy. 1 Consider an energy signal x(t). Using the facts A ¼ B ¼ ð1:66Þ ba that jxðtÞj2 ¼ xðtÞxðtÞ and F ½xðtÞ ¼ XðjxÞ and combining with the inversion integral Eq. 1.52, it so that is not difﬁcult to show that 1 1 1 Z1 Z1 yðtÞ ¼ F 1 1 b a a þ jx b þ jx E¼ 2 jxðtÞj dt ¼ jXðjxÞj2 dx ð1:72Þ ð1:67Þ 2p 1 1 1 at ¼ e ebt uðtÞ ð1:68Þ The rightmost expression in Eq. 1.72 shows ba that |X(jx)|2/(2p) has the dimension of energy per Things are of course, different if a = b; then unit radian frequency, i.e. |X(jx)|2 has the one goes back to Eq. 1.64 and uses the property dimension of energy per unit Hz. For this reason, that if F [x(t)] = X(jx) then F [tx(t)] = j dX(jx)/ |X(jx)|2 is referred to as the energy density dx. Accordingly if a = b, then spectrum of the signal x(t). Incidentally, Eq. 1.72 is known as the Parseval’s relation (cf. Eq. 1.31). yðtÞ ¼ t ea t uðtÞ ð1:69Þ For a periodic signal, which is a power signal, we have already seen in Eq. 1.31 that the average P power P is given by 1 2 k¼1 jak j ; where |ak| is Spectral Density the amplitude of the kth harmonic. If we deﬁne, in similarity with Eq. 1.72, In using Fourier transform to calculate the energy Z1 or power of a signal, the notion of spectral den- P¼ Sx ðf Þdf ð1:73Þ sity is an important one. The total energy and 1 average power of a signal x(t) are deﬁned as ZT Z1 then Sx(f) qualiﬁes as the power per unit Hz and 2 2 is called the power spectral density. In terms of E ¼ lim jxðtÞj dt ¼ jxðtÞj dt ð1:70Þ T!1 |ak|, it is easily seen that T 1 X 1 and Sx ðf Þ ¼ jak j2 dðf kfo Þ; ð1:74Þ k¼1 ZT 1 P ¼ lim jxðtÞj2 dt; ð1:71Þ where f0 = x0/(2p) is the fundamental frequency. T!1 2T T respectively. A signal x(t) is called an energy signal if 0 < E < ∞ and a power signal if Concluding Comments 0 < P < ∞. A given signal x(t) can be either an energy signal or a power signal but not both. For more on the impulse function, see the next A periodic signal (e.g. the one of Fig. 1.5) is chapter. In this chapter, we have talked about the usually a power signal, while a non-periodic signal fundamentals of signals and systems and their (e.g. the one of Fig. 1.12) is usually an energy relationship. Fourier series and Fourier transform signal. Power and energy signals are mutually and the concepts of spectra––amplitude as well exclusive because the former has inﬁnite energy as phase––and the concepts of energy, power and while the latter has zero average power. their density functions have also been introduced. Problems 15 P:5 Determine the impulse response of a system Problems consisting of a cascade of two systems characterized by the impulse response You have to think carefully. These are designed also carefully, as the problems are designed to f ðtÞ ¼ eat uðtÞ test your grasp of the fundamentals and the ease with which you can ﬁnd a clue. and P:1 Sketch (−1), (−½), (0), (½), (1), (1 − t) gðtÞ ¼ ebt uðtÞ; P:2 The impulse response of a system is h(t − t0). Find an expression for the output y(t), if the where a > 0, b > 0 and (i) a 6¼ b input is t2x(t1/2). (ii) a = b. P:3 Determine the Fourier transform of the function tu(t). P:4 Determine and sketch the spectrum of the following function: Bibliography 8 < 0; t\0 1. A.V. Oppenheim, A.S. Willsky, I.T. Young, Signals xðtÞ ¼ 1; 0\t\ T20 : and systems. Prentice Hall (1983) dðt TÞ; t[0 The Mysterious Impulse Function and its Mysteries 2 Some fundamental issues relating to this Introduction mysterious but fascinating impulse function in continuous as well as discrete time domain The issues considered in this chapter are usually are discussed ﬁrst. These are: deﬁnition and treated in a routine manner in courses on circuit relation to the unit step function, dimension of theory, signals and systems and digital signal impulse response, solution of differential and processing. That problems may arise without a difference equations involving the impulse thorough understanding of the concepts involved function and Fourier transform of the unit step are usually not clearly brought out, or avoided. function. Conceptual understanding is empha- These problems are usually related to the sized at every point. This is very important, occurrence of the impulse function dðtÞ in the particularly for beginners like you. So read the continuous time domain and dðnÞ in the discrete chapter carefully and grasp the contents. This time domain in various situations. The chapter will help you to understand the later course on makes an attempt to clarify the concepts involved signals and systems, control, DSP, etc. in a few of such problem cases and the manner in which they can be solved. In particular, starting Keywords with the deﬁnitions of dðtÞ and dðnÞ and their Impulse function Unit step function relationship with the unit step functions u(t) and Impulse response Differential equation u(n), respectively, we discuss the following Difference equation Fourier transform issues: dimension of impulse response, solution Signals and systems Digital signal of differential and difference equations involving the impulse function, and Fourier transforms of processing u(t) and u(n). The last discussion is a concluding one. The chapter is written in a tutorial style so as to be appealing to students and teachers alike. The chapter also asks some questions, the answers to which are left as open problems. Source: S. C. Dutta Roy, “Some Fundamental Issues Related to the Impulse Function,” IETE Journal of Education, Vol 57, pp 2–8, January–June 2016. © Springer Nature Singapore Pte Ltd. 2018 17 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_2 18 2 The Mysterious Impulse Function and its Mysteries Definitions starting time is t = 0, then the value of the pulse, rectangular or triangular, is 0 at t = 0−, 0, as well The unit impulse function dðtÞ in the continuous as 0+, which is not the case with dðtÞ. time domain is deﬁned as The impulse function may also be looked upon as the limiting value of some distributions, dðtÞ ¼ 0; t 6¼ 0 and dðtÞ ! 1; t ¼ 0 ð2:1Þ p e.g. ða pÞ1 expðt2 =a2 Þ or a 1 sinc ðx=aÞ, the integral of which from −∞ to +∞ is unity, as such that a ! 0. However, this point of view is not rele- Zb vant in the context of this chapter, and will not be dðtÞ dt ¼ 1; ð2:2Þ pursued further. The function dðtÞ is best visualized by relating a it to the unit step function u(t), deﬁned by where a 0− and b 0+. The adjective ‘unit’ refers to Eq. 2.2 and the area under the function, uðtÞ ¼ 1; t 0 þ and uðtÞ ¼ 0; t 0 ð2:3Þ which is also known as the strength of the Clearly, the differentiation of u(t) will give impulse. An impulse A dðtÞ will have an area dðtÞ, satisfying both Eqs. 2.1 and 2.2. Hence, the A under the integral and will be called an impulse relationship between the two functions can be of strength A. The deﬁnition must include formally written as Eqs. 2.1 and 2.2 together. Often Eq. 2.2 is not paid attention to, but it must be understood that Zt inﬁnity is incomprehensible in a ﬁnite world. If dðtÞ ¼ duðtÞ=dt and uðtÞ ¼ dðtÞ dt; ð2:4Þ Eq. 2.2 was not coupled to Eq. 2.1, then dðtÞ a would not be admissible in the realm of func- tions, particularly for engineering analysis and where a 0−. Unlike dðtÞ, u(t) can be generated design. Even otherwise, the impulse function is a in the laboratory by a switch in series with a mathematical anomaly. From a purely mathe- voltage source. matical point of view, dðtÞ is not strictly a On the other hand, in the discrete time function because any extended real function that domain, there is no uncertainty or confusion is zero everywhere except at a single point must about the unit impulse function dðnÞ, which is have a total integral equal to zero. There is a deﬁned as fascinating history behind the acceptance of dðtÞ, not as a function, but as a distribution or a gen- dðnÞ ¼ 1; n ¼ 0 and dðnÞ ¼ 0; n 6¼ 0 ð2:5Þ eralized function by mathematicians. However, it would be too much of a diversion to go into the The unit step function u(n) is deﬁned as details in this chapter. Instead, we refer to an excellent tutorial article by Balakrishnan [1] on uðnÞ ¼ 1; n 0 and uðnÞ ¼ 0; n\0 ð2:6Þ the subject. For our purposes here, Eqs. 2.1 and 2.2 as the deﬁnition of dðtÞ will sufﬁce. Thus, the relationships between dðnÞ and u It must be understood that dðtÞ is only a (n) are mathematical concept and is introduced to facil- itate analysis and design. It cannot be generated dðnÞ ¼ uðnÞ uðn 1Þ and uðnÞ Xn in the laboratory. Textbooks usually give a visual ¼ dðn kÞ ð2:7Þ aid of a rectangular pulse of duration s and height k¼0 1=s or a triangular pulse of width 2s and height 1=s. By allowing s ! 0, one can visualize dðtÞ. In order to distinguish between the continuous However, this limiting approach may fail to give and discrete time domains, dðnÞ should more correct results in system analysis, because if the appropriately be called the ‘unit sample Definitions 19 function’. However, in conformity with the HðsÞ ¼ L½yðtÞ=L½xðtÞjzero initial conditions ; ð2:10Þ common usage, we shall continue to call dðnÞ as the unit impulse function, it being implied that where L stands for the Laplace transform and the the argument of d will make it clear which initial conditions are on y(t) and its derivative(s), domain we are referring to. if the order of the system is more than one. It is known that the h(t) and H(s) are related to each other by Impulse Response HðsÞ ¼ L½hðtÞ evaluated under zero initial conditions In the continuous time domain, the impulse ð2:11Þ response h(t) of a system is generally interpreted as the response of the system to an excitation Textbooks usually omit the condition under d(t). What is the dimension of h(t)? To answer which h(t) is to be evaluated. It must be this question, look at the ﬁrst relation in Eq. 2.4; emphasized that h(t) is the impulse response if u(t) is a voltage, then dðtÞ has the dimension of under zero-state condition. volts/second, which we can neither generate nor Similarly, for the discrete time domain, the apply to a system. However, for a linear system, transfer function is deﬁned as differentiation of the unit step response will give h(t). This is how we can measure h(t) in the HðzÞ ¼ Z ½yðnÞ=Z ½xðnÞjzero initial conditions ð2:12Þ laboratory. Obviously, h(t) will have the dimen- sion of (second)−1. and Now look at the convolution relation: HðzÞ ¼ Z ½hðnÞ evaluated under zero initial conditions; Zþ 1 ð2:13Þ yðtÞ ¼ xðsÞhðt sÞds; ð2:8Þ 1 where Z stands for the z-transform. where y(t) is the response of a system to an excitation x(t). Clearly, if both x(t) and y(t) are voltages or currents, then h(t) will have the How Do You Solve Differential dimension of (second)−1. However, if x(t) is a Equations Involving an Impulse voltage and y(t) is a current, then h(t) will have Function? the dimension of (ohm-second)−1. Similarly, if x (t) is a current and y(t) is a voltage, then the We shall illustrate the kind of difﬁculties that dimension of h(t) will be ohm/second. arise in solving differential equations involving In the discrete time domain, h(n) is dimen- an impulse function, with a typical problem. sionless, as are all signals, because the process- Consider the following differential equation: ing is concerned only with numbers. The y00 þ 2y0 þ 2y ¼ dðtÞ; ð2:14Þ convolution relation X 1 where prime denotes differentiation with respect yðnÞ ¼ xðnÞ hðnÞ ¼ xðkÞhðnkÞ ð2:9Þ to t. Let the given initial conditions be k¼a yð0Þ ¼ 1 and y0 ð0Þ ¼ 2 ð2:15Þ does not involve any dimensions. In the context of the continuous time domain, Let us try to solve Eq. 2.14 by using Laplace the transfer function is deﬁned as transforms, as is usually done. In order to take 20 2 The Mysterious Impulse Function and its Mysteries account of dðtÞ on the right-hand side of constants in the former from the initial condi- Eq. 2.14, we take the Laplace transform of y(t) as tions. The method gives correct results if there is no impulse function in the excitation function, Z1 but not in the present case. Why? The question is YðsÞ ¼ yðtÞ expðstÞdt ð2:16Þ left to you as an open problem. 0 Applying the ﬁrst method, we ﬁrst ﬁnd the zero-input solution, i.e. the solution of Eq. 2.14 Then, the Laplace transformation of Eq. 2.14 with the right-hand side equal to zero. It can be gives easily veriﬁed that the solution is of the form s2 YðsÞsyð0Þy0 ð0Þ þ 2 ½sYðsÞyð0Þ þ 2YðsÞ ¼ 1 yzi ðtÞ ¼ A cos ðt þ hÞ; ð2:22Þ ð2:17Þ where A and h are constants to be evaluated from Combining Eqs. 2.17 and 2.15 and simplify- the initial conditions given by Eq. 2.15. Carrying ing, we get out the steps, we get p YðsÞ ¼ ðs þ 1Þ=ðs2 þ 2s þ 2Þ ð2:18Þ A¼ 5 and h ¼ arctan 2 ð2:23Þ Using the standard table of Laplace trans- Now consider the zero-state solution where forms, inversion of Eq. 2.18 gives the initial conditions are to be put equal to zero. Obviously, this can be done by the Laplace yðtÞ ¼ ½expðtÞ cos t uðtÞ ð2:19Þ transform method by putting y(0−) = 0 and y′(0−) = 0 in Eq. 2.17. Then, Eq. 2.17 gives Differentiation of Eq. 2.19 gives Yzs ðsÞ ¼ 1=ðs2 þ 2s þ 2Þ ð2:24Þ y0 ðtÞ ¼ ½expðtÞ cos tdðtÞ exp ðtÞ:½cos t þ sin t uðtÞ ð2:20Þ On inversion, Eq. 2.24 gives The solutions Eqs. 2.19 and 2.20 are valid for yzs ðtÞ ¼ ½expðtÞ sin t uðtÞ ð2:25Þ t 0+. In particular, at t = 0+, Combining Eqs. 2.22, 2.23 and 2.25, we get 0 the total solution as yð0 þ Þ ¼ 1 and y ð0 þ Þ ¼ 1 ð2:21Þ p Note that while y(0 +) = y(0−), y′(0+) ¼ 6 y′(0−). yðtÞ ¼ 5 cosðt þ arctan 2Þ þ ½expðtÞ sin t uðtÞ This discontinuity in initial condition is typical, ð2:26Þ whenever an impulse function ﬁgures in the right-hand side of a differential equation. Also note Note that the ﬁrst term does not involve u(t) that if Eqs. 2.19 and 2.20 are applied at t = 0−, the because the zero-input solution is valid for all values are: y(0−) = 0 and y′(0−) = 0, because both t. Differentiating Eq. 2.26, we get u(t) and dðtÞ are zero at t = 0−. p Can we get a solution which is valid at t = 0− y0 ðtÞ ¼ 5 sin ðt þ arctan 2Þ also? The answer is yes, if we apply the super- þ ½expðtÞ sin t dðtÞ½expðtÞðsin t þ cos tÞuðtÞ position of ‘zero-input’ and ‘zero-state’ solu- ð2:27Þ tions. This method is seldom emphasized in circuit theory courses. Instead, one uses the Putting t = 0− in Eqs. 2.26 and 2.27 and superposition of the complementary function and noting that both u(t) and dðtÞ are zero at t = 0−, the particular solution, and then evaluates the we get the same values as in Eq. 2.15. How Do You Solve Differential Equations Involving an Impulse Function? 21 On the other hand, putting t = 0+ in Eqs. 2.26 Combining Eqs. 2.31 with 2.29 and simpli- and 2.27 and noting that u(0+) = 1 and fying gives dð0 þ Þ ¼ 0; we get the same values as in Eq. 2.21. YðzÞ ¼ 6ðz1 1Þ=ð1 þ z 1 6z2 Þ ð2:32Þ Expanding Eq. 2.32 in partial fractions and taking the inverse transform gives How Do You Solve Difference Equations Involving an Impulse yðnÞ ¼ ½4:8ð3Þn þ 1:2ð2Þn uðnÞ ð2:33Þ Function? This solution is strictly valid for n 0. Do the types of difﬁculties discussed in the pre- However, it can be easily modiﬁed to take care of vious section arise in the discrete time domain n < 0 by adding the terms dðn þ 1Þ and also? The answer is yes, if z-transform method is dðn þ 2Þ to the right-hand side of Eq. 2.33. applied blindly because the resulting solution is Consider now the zero-input, zero-state strictly valid for n 0. However, the situation method of solution. The zero-input solution is can be corrected by adding the necessary terms of the form representing the initial conditions for n < 0. As in the continuous time domain, the complemen- yzi ðnÞ ¼ K1 ð3Þn þ K2 ð2Þn ; ð2:34Þ tary function and particular solution method does not work here. Why? The answer to this question where K1 and K2 are to be evaluated from is again left to you as an open problem. On the Eq. 2.29. The result is other hand, as in the continuous time case, if the difference equation is solved in the n- domain by yzi ðnÞ ¼ ½5:4ð3Þn þ 1:6 ð2Þn ð2:35Þ the zero-input and zero-state method, correct This part of the solution is valid for all n. solution is obtained, with initial conditions taken The zero-state solution can be obtained by the z- into account. As in the previous case, we shall transform technique by putting y(−1) = y(−2) = 0 illustrate these facts with an example. in Eq. 2.31. Inverting the resulting Y(z), we get Let it be required to solve the difference equation yzs ðnÞ ¼ ½0:6 ð3Þn þ 0:4 ð2Þn uðnÞ ð2:36Þ yðnÞ þ yðn1Þ6yðn2Þ ¼ dðnÞ ð2:28Þ which is valid for n 0. The total solution is the sum of Eqs. 2.35 and 2.36, i.e. subject to the initial conditions yðnÞ ¼ ½5:4ð3Þn þ 1:6 ð2Þn þ ½0:6 ð3Þn þ 0:4 ð2Þn uðnÞ yð1Þ ¼ 1 and yð2Þ ¼ 1 ð2:29Þ ð2:37Þ If we use the z-transform method blindly and Clearly, Eq. 2.37 will give the correct values take the z-transform of y(n) as for y(−1) and y(−2). Also, for n 0, Eq. 2.37 X 1 gives the same result as Eq. 2.33, as expected. YðzÞ ¼ yðnÞzn ð2:30Þ n¼0 Fourier Transform of the Unit Step then the z-transform of Eq. 2.28 gives Function YðzÞ þ z1 YðzÞ þ yð1Þ 6 z2 YðzÞ For ready reference, recall that the Fourier ð2:31Þ þ z 1 yð1Þ þ yð2Þ ¼ 1 transform of a continuous time function y(t) is deﬁned as 22 2 The Mysterious Impulse Function and its Mysteries Zþ 1 Zþ 1 YðjxÞ ¼ yðtÞ exp ðjxtÞdt ð2:38Þ F ½uðtÞ ¼ uðtÞ expðjxtÞdt; 1 1 Zþ 1 ð2:42Þ while that of the discrete time function y(n) is ¼ expðjxtÞdt deﬁned as 0 X 1 ¼ ½expðjxtÞ=ðjxÞ10 Y ½exp ðjxÞ ¼ yðnÞ exp ðjxnÞ ð2:39Þ n¼1 where F stands for the Fourier transform. Since exp(−j∞) is not known, we evaluate Eq. 2.42 Note that in either case, the transform is a indirectly. It is easy to show that continuous function of x, although y(n) is a discrete variable. In Eq. 2.38, x ¼ 2pf , where F ½expðatÞ uðtÞ ¼ 1=ða þ jxÞ ð2:43Þ f is the frequency in Hz, while in Eq. 2.39, x ¼ 2pf =fs , where fs is the sampling frequency It is tempting to put a = 0 in Eq. 2.43 and in Hz. Consequently, in Eq. 2.38, x has the conclude that F[u(t)] = 1/(jx), but this is part of dimension of (second)−1, while in Eq. 2.39, x is the story. Hidden here is the fact that Eq. 2.43 dimensionless. has a real part as well as an imaginary part, i.e. In order to be useful, any transformation must be reversible, i.e. one must be able to recover the 1=ða þ jxÞ ¼ ½a=ða2 þ x2 Þ ½jx=ða2 þ x2 Þ original signal from the transformed one in a ð2:44Þ unique way. The Fourier transform is no excep- tion for which the inverse Fourier transforms is While 1/( jx) takes care of the imaginary part when a = 0, the real part becomes 1=x2 when Zþ 1 a = 0, and it tends to inﬁnity when x ! 0: yðtÞ ¼ ½1=ð2pÞ YðjxÞ expðjxtÞdx ð2:40Þ Hence, there is an impulse function at x ¼ 0: (It 1 would be wrong to say that the real part becomes of the form 0/0 when both a and x are zero, corresponding to Eq. 2.38 and because the denominator becomes 02 so that the Z1 real part becomes 1/0, i.e. inﬁnite). To determine yðnÞ ¼ ½1ð2pÞ Y ½expðjxÞ exp ðjxnÞdx the strength of the impulse, we ﬁnd the area under a=ða2 þ x2 Þ: This is given by 1 ð2:41Þ Zþ 1 ½a=ða2 þ x2 Þdx ¼ ½arctan ðx=aÞj1 1 ¼ p corresponding to Eq. 2.40. The limits of the 1 integral in Eq. 2.41 cover a range of 2p because Y[exp(jx)] is periodic with a period of 2p. That ð2:45Þ both Eqs. 2.38 and 2.40 or Eqs. 2.39 and 2.41 Thus, the real part in Eq. 2.44 becomes pdðxÞ cannot be deﬁnitions is often not appreciated by when both a and x tend to zero. Finally, students. therefore, Now, we turn to the main item of discussion, i.e. the Fourier transforms of u(t) and u(n). F ½uðtÞ ¼ pdðxÞ þ ½1=jxÞ ð2:46Þ Applying the deﬁnition, we get Fourier Transform of the Unit Step Function 23 For ﬁnding the Fourier transform of u(n), if confusions have been clariﬁed. That the com- we apply the deﬁnition Eq. 2.39 blindly, then it plementary function, and particular solution is tempting to conclude that the required trans- method does not work if the excitation contains form is 1=½1 expðjxÞ. However, as in the an impulse function does not appear to have been case of u(t), this is a part of the story; hidden here recorded earlier. The reason why it does not work is the fact that u(n) has an average value of ½. To has been left as an open problem for you. show that 1=½1 expðjxÞ is not F[u(n)], con- The reason why misconceptions and confu- sider the sequence sions persist in the minds of students and teachers lies in the kind of rote learning that is sgn ðnÞ ¼ 1=2 for n\0 and þ 1=2 for n 0 emphasized and practiced in the class. Also, even ð2:47Þ if the teacher is knowledgeable and wishes to explain the concepts and subtle points, he/she It is not difﬁcult to show, by applying (2.39) ﬁnds no time to do this because of the pressure to that F ½sgnðnÞ ¼ 1=½1 expðjxÞ. Clearly, ‘cover’ the syllabus, which invariably is overloaded. uðnÞ ¼ sgnðnÞ þ ð1=2Þ ð2:48Þ Thus, F[u(n)] will have an additional term Problems corresponding to the average value of ½. To ﬁnd this term, consider the following inverse P:1 Solve the equation z-transform: y000 þ 3y00 þ 3y0 þ y ¼ dðtÞ; X þ1 Zþ p F 1 ½p dðx þ 2pkÞ ¼ ½1ð2pÞ p 1 given p X þ1 dðx þ 2pkÞ expðjxnÞdx yð0Þ ¼ 0; y0 ð0Þ ¼ 0; y00 ð0Þ 1 ð2:49Þ P:2 Solve the equation In the range of integration, only the k = 0 term yðn4Þ þ yðn3Þ þ yðn1Þ ¼ dðn1Þ will have an effect, which gives the right-hand side of Eq. 2.49 as ½. Since the Fourier trans- given form gives a one-to-one correspondence between yð0Þ ¼ yð1Þ ¼ yð2Þ ¼ yð3Þ ¼ 1: the n and x domains, we conclude that P:3 Solve the equation F½uðnÞ ¼ 1=½1 expðjxÞ X1 ð2:50Þ dyðtÞ þp dðx þ 2pkÞ þ q0 yðtÞ ¼ Aeat uðtÞ 1 dt given that Conclusion yð0Þ ¼ y0 In this chapter, some fundamental and conceptual P:4 Solve the equation issues, relating to the unit impulse function, in both continuous and discrete time domains, are y0 þ q0 y ¼ q0 x0 þ x presented and some possible misconceptions and 24 2 The Mysterious Impulse Function and its Mysteries Acknowledgement The author acknowledges the help given that received from Professor Yashwant V Joshi in the prepa- ration of this chapter. yð0Þ ¼ 0 P:5 Solve the equation Reference yðnÞ þ q0 yðn1Þ ¼ xðnÞ; 1. V. Balakrishnan, All about the Dirac delta function(?). given that Resonance 8(8), 48–58 (2003) yð1Þ ¼ y0 : State Variables—Part I 3 Here, we introduce state variables, which were linear and time-invariant systems only. Linearity the hot topics in the middle of 1960s. Later, implies that if inputs x1(t) and x2(t) produce they have seeped into and made deep impact outputs y1(t) and y2(t), respectively, then an input on signals and systems, circuit theory, con- ax1(t) + bx2(t), where a and b are arbitrary con- trols, etc. You better get familiar with them stants, should lead to the output ay1(t) +by2(t). and make friends with them as early as This involves the two principles of homogeneity possible. and superposition. Time invariance implies that if an input x(t) produces the output y(t), then the delayed input x(t − s) should produce an output Keywords y(t − s), delayed by the same amount s. For state State variables Standard forms Choice of variable characterization of systems which do not state variables Solution of state equations obey linearity and/or time invariance, the reader is referred to the literature listed under references. This discussion on state variables is organized Why State Variables? in two parts. In Part I, we introduce the concept of state; clarify deﬁnitions and symbols; present The state variable approach provides an extre- the standard form of linear state equations; dis- mely powerful technique for the study of system cuss how state variables are chosen in a given theory and has led to many results of far-reaching physical system, or a differential equation importance. This chapter is intended to serve as description of the same and elaborate on the an introduction to the basic concepts and tech- methods of solving the state equations. In Part II niques involved in the state variable characteri- of this presentation, in the next chapter, we shall zation of systems. The discussion is restricted to deal with the properties of the fundamental matrix and procedures for evaluation of the same, and dwell upon the state transition flow graph method in considerable details. The references Source: S. C. Dutta Roy, “State Variables—Part I,” for the whole chapter will be given in Part II, IETE Journal of Education, vol. 38, pp. 11–18, January– which will also include an appendix on the March 1997. essentials of matrix algebra. This chapter and the next one are based on some notes prepared for some students at the University of Minnesota in USA as early as 1965. © Springer Nature Singapore Pte Ltd. 2018 25 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_3 26 3 State Variables—Part I What is a State? state is described by a set of numbers, called state variables, which contains sufﬁcient information Consider an electrical network containing resis- regarding the past history of the system so that tances, capacitances and inductances, whose the future behaviour can be computed. For response to an excitation e(t) is r(t). Since the example, in an electrical network, the inductor network is known, a complete knowledge of currents and capacitor voltages constitute the e(t) over the time interval −∞ to t is sufﬁcient to state variables. determine the output r(t) over the same time To represent a system of more than two or interval. However, if e(t) is known over the time three state variables, it is convenient to use the interval t0 to t, as is usually the case (with t0 vector space notation of modern algebra. A sys- taken as equal to zero), then the currents through tem deﬁned by n independent states xi(t), i = 1, the inductors and the voltages across the capac- 2, … n, (corresponding to an nth order system), itors in the network must be known at some time can be represented by the state vector t1, t0 t1 t (usually, t1 = t0, hence the name 2 3 ‘initial conditions’), in order to determine r x1 ðtÞ 6 x2 ðtÞ 7 (t) over the interval t0 to t. These currents and 6 7 xð t Þ ¼ 6 . 7 ð3:1Þ voltages constitute the ‘state’ of the network at 4 .. 5 time t1. In this sense, the state of the network is xn ðtÞ related to its memory; for a purely resistive net- work (zero memory), only the present input is Consider the multiple-input, multiple-output required to determine the present output. (or multivariable) system shown in Fig. 3.1. It Next, consider a general type of linear system, can be described by a set of simultaneous which is described by a set of linear differential differential equations relating the state vector equations for t t0. The complete solution of x(t) given by Eq. 3.1, the excitation vector1 these equations will involve arbitrary constants 2 3 which can be determined from ‘initial’, ‘given’ u1 ðtÞ or ‘boundary’ conditions at time t = t0 (or at 6 u2 ðtÞ 7 6 7 t = t1 where t0 t1 t). The boundary con- uðtÞ ¼ 6 . 7; ð3:2Þ 4 .. 5 ditions can thus be termed as the ‘state’ of the um ðtÞ system at t = t0. Heuristically, the state of a system separates the future from the past, so that the state contains all relevant information con- cerning the past history of the system, which is u1(t) y1(t) required to determine the response for any input. u2(t) y2(t) . . . . The evolution of an excited system from one . . . . n-th Order . . state to another may be visualized as a process of . . . . System . . . . . . state transition. . . . . . . . . . . . . um(t) yp(t) Definitions and Symbols Fig. 3.1 Schematic representation of a multivariable system The state of a dynamical system may now be formally deﬁned as the minimal amount of information necessary at any time such that this 1 information, together with the input or the exci- Note that the symbol u(t) is usually reserved for the unit step function. We have used u(t) to denote the excitation tation function and the equations describing the function for conformity with the control literature. We dynamics of the system, characterize completely shall use the symbol l(t) for the unit step function when the future state and output of the system. The the occasion arises. Definitions and Symbols 27 where m is the number of inputs, and the D response vector 2 3 . + y1 ðtÞ u B + Σ x x C + ? y 6 y2 ðtÞ 7 + 6 7 yðtÞ ¼ 6 . 7; ð3:3Þ 4 .. 5 A yp ðtÞ Fig. 3.2 Block diagram representation of the state where p is the number of outputs, through the equations system parameters, as discussed next. Example 1 Consider the network shown in Fig. 3.3. We identify the inductor current i1 and Standard Form of Linear State the capacitor voltage v2 as the state variables. Equations There are two excitations, e1(t) and e2(t). Let the desired responses be i1 and i2. Thus, for this In general, the equations of a dynamical system system, we identify can be written in the following functional forms xðtÞ ¼ f½xðt0 Þ; uðt0 ; tÞ ð3:4Þ i1 e1 i x¼ ; u¼ and y¼ 1 ð3:8Þ v2 e2 i2 yðtÞ ¼ g½xðt0 Þ; uðt0 ; tÞ ð3:5Þ for t t0. If the system can be described by a A system of two ﬁrst-order differential equa- set of ordinary linear differential equations, then tions can be written for this electric circuit as the state equations can be written as follows: _ ¼ AðtÞxðtÞ þ BðtÞuðtÞ xðtÞ ð3:6Þ Lðdi1 =dtÞ ¼ e1 ðv2 þ e2 Þ ¼ e1 e2 v2 ð3:9Þ yðtÞ ¼ CðtÞxðtÞ þ DðtÞuðtÞ; ð3:7Þ i2 ¼ Cðdv2 =dtÞ ¼ i1 ðv2 þ e2 Þ=R ð3:10Þ where the dot above the symbol x denotes differentiation with respect to time and A(t), B(t), Equations 3.9 and 3.10 can be rearranged as C(t) and D(t) are, in general, time-varying real follows: matrices. Equations 3.6 and 3.7 are the standard forms of linear state equations, as will be illus- ðdi1 =dtÞ ¼ ð1=LÞv2 þ ðl=LÞe1 þ ð1=LÞe2 trated in the examples to follow. If the system ð3:11Þ is time invariant, then the matrices A(t), B(t), C(t) and D(t) are constants. A general block ðdv2 =dtÞ ¼ ð1=C Þi1 þ ½1=ðRCÞv2 þ ½ 1=ðRC Þe2 diagram representation for the state equations is given in Fig. 3.2. ð3:12Þ Referring to Eqs. 3.1 to 3.7, we observe that These two equations can be combined into the the dimensions of the matrices A, B, C and D are following single matrix equation: n n, n m, p n and p m, respectively. d i1 0 1=L i1 1=L 1=L e1 ¼ þ ð3:13Þ dt v2 1=C 1=ðRCÞ v2 0 1=ðRCÞ e2 28 3 State Variables—Part I L C How Do You Choose State Variables in a Physical System? i1 + v2 – + + e1(t) R e2(t) The choice of state variables in Example 1 was – –i very simple. We chose the capacitor voltage and 2 the inductor current as the two state variables. This should not lead one to believe that, in Fig. 3.3 Circuit for Example 1 general, an electrical network with l inductors and c capacitors would require l + c state vari- ables. This is because not all capacitor voltages A comparison of Eq. 3.13 with Eq. 3.6 yields can be speciﬁed independently when there are the identiﬁcation capacitor loops in the network. Similarly, not all inductor currents can be speciﬁed independently 0 1=L when there are inductance cut-sets2 in the net- A¼ and 1=C 1=ðRCÞ work. An allowable set of state variables is the ð3:14Þ 1=L 1=L set of all capacitor voltages, diminished by a B¼ 0 1=ðRCÞ subset equal to the number of capacitor loops, plus the set of inductor currents, diminished by a The equation for the output vector y is derived subset equal to the number of inductance by recognizing the fact that cut-sets. Thus to ﬁnd the state equations, we must eliminate, from the set of equations governing the network, all non-state variables, that is all i1 ¼ i1 ð3:15Þ resistance voltages and currents, and those and capacitor voltages and inductor currents which correspond to certain members of capacitance i2 ¼ i1 ðv2 =RÞðe2 =RÞ ð3:16Þ loops and inductance cut-sets. These topological circuit constraints, which Equations 3.15 and 3.16 may be combined reduce the total number of state variables from into the following single matrix equation: the total number of storage elements, apply to other systems also. However, the state variables of a non-electrical system can perhaps be detec- i1 1 0 i1 0 0 e1 ¼ þ ted more easily by drawing the analogous elec- i2 1 1=R v2 0 1=R e2 trical system and by ﬁnding the all capacitor ð3:17Þ loops and all inductance cut-sets in this analo- gous representation. Equation 3.17 is of the form Eq. 3.7 with Example 2 Consider the network shown in Fig. 3.4 in which there is one all capacitor loop 1 0 0 0 C¼ and D ¼ and one all inductance node. Thus, although 1 1=R 0 1=R there are six energy storage elements, we shall ð3:18Þ require only 6 − 1 − 1 = 4 state variables. This reduction occurs because we can assign arbitrary In this chapter, we shall restrict the term initial voltages to two capacitors only (the initial ‘state equation’ to denote the set of ﬁrst-order voltage on the remaining capacitor is determined differential equations given by Eq. 3.6 only, by Kirchoff’s voltage law); similarly, we can because the other equation, viz. Eq. 3.7, assign arbitrary initial currents in two inductors relating the output to the input and the state vectors, can be easily obtained by algebraic 2 A cut-set is a set of branches, which, when removed, means. splits the network into two unconnected parts. How Do You Choose State Variables in a Physical System? 29 C3 detailed exposition of this topic, see Kuh and i1 i2 Rohrer [1]. R1 L1 L2 + L3 + + V1 V1 C1 C2 V2 – – – How Do You Choose State Variables R2 When the System Differential Equation is Given? Fig. 3.4 Circuit for Example 2 When the physical system is given, the natural choice of state variables is the quantities associ- only (the third inductor current is determined by ated with the energy storage elements. Suppose, Kirchoff’s current law). Let us, therefore, choose however, that the only given information about v1, v2, i1 and i2 as the state variables. We can then the system is a single differential equation write the following equations: involving the output of the system, y, and its ﬁrst 3 9 k derivatives, e.g. ðvi v1 Þ=R1 ¼ C1 ðdv1 =dtÞ þ i1 þ C3 ½dðv1 v2 Þ=dt > > = C2 ðdv2 =dtÞ ¼ i2 þ C3 ½dðv1 v2 Þ=dt v1 L1 ðdi1 =dtÞ ¼ v2 þ L2 ðdi2 =dtÞ > F y; y_ ; yð2Þ ; . . .; yðkÞ þ uðtÞ ¼ 0 ð3:20Þ > ; ¼ L3 ½dði1 i2 Þ=dt þ R2 ði1 i2 Þ ð3:19Þ If Eq. 3.20 can be rearranged to the form yðkÞ ¼ f y; y_ ; yð2Þ ; . . .; yðk 1Þ þ uðtÞ ð3:21Þ By manipulating these equations, one can express the quantities (dv1/dt), (dv2/dt), (di1/dt) then a natural, but quite arbitrary choice of state and (di2/dt) in terms of v1, v2, i1, i2 and vi, and variables would be the following: hence obtain the state formulation. This reduc- tion to a canonical form is left to you as an x1 ¼ y; x2 ¼ y_ ; x3 ¼ yð2Þ ; . . .; xk ¼ yðk1Þ exercise. ð3:22Þ In the preceding example, it was easy to identify the all-capacitance loop and the It follows that the state variables obey the all-inductance node. In a complicated network, following differential equations: however, difﬁculties may arise in counting the number of such subsets. To organize the count- x_ 1 ¼ x2 ing in such a situation, ﬁrst, open circuit all x_ 2 ¼ x3 elements in the network excepting the capacitors ... ð3:23Þ and count the number of independent loops by x_ k1 ¼ xk the usual topological rule, namely (Nl)c = Nb − x_ k ¼ f ðx1 ; x2 ; . . .; xk Þ þ uðtÞ Nj + Ns, where Nb is the number of branches, Nj is the number of nodes, and Ns is the number of For linear systems, f(x1, x2,…, xk) will be of separate parts. Next, short circuit all elements of the form the network excepting the inductors and ﬁnd the number of independent nodes by the formula f ¼ a1 x 1 þ a2 x 2 þ þ ak x k ð3:24Þ (Nn)l = Nj − Ns. Then, the minimal number of state variables is N = Nl+c − (Nl)c − (Nn)l, where Nl+c is the total number of energy storage ele- ments in the network. It should also become apparent now that the choice of state variables for 3 The symbol y(k) stands for the kth derivative of y with a given physical system is not unique. For a more respect to t. 30 3 State Variables—Part I Hence, the state equation will be of the form where b is a constant to be determined, and leave Eq. 3.6 with the other state variables unchanged. Then the 2 3 dynamic equations become 0 1 0 0 0 0 6 0 0 1 0 0 0 7 6 7 x_ 1 ¼ x2 ; x_ 2 ¼ x3 ; . . .; x_ k2 ¼ xk1 ; ð3:32Þ 6 0 0 0 1 0 0 7 A¼6 6 7 6 7 7 and 4 0 0 0 0 0 1 5 a1 a2 a3 a4 a5 ak x_ k1 ¼ yðk1Þ ¼ xk þ bu ð3:25Þ The last equation must satisfy Eq. 3.27, so which is of dimension k k, and that 2 3 0 ð2Þ x_ k ¼ xk1 bu_ ¼ yðkÞ bu_ 607 6 7 ¼ a1 ðxk þ buÞ a2 xk1 ak x1 B¼6 07 6.7 ð3:26Þ 4 .. 5 þ ðb1 bÞu_ þ b2 u 1 ð3:33Þ which has the dimension k 1. The term involving u_ disappears if we choose A more complicated situation arises when the b1 = b. Then differential equation also involves the derivatives x_ k ¼ a1 xk a2 xk l ; ak xl þ ðb2 a1 b1 Þu of the input. For example, consider the equation ð3:34Þ ðk1Þ y ð k Þ þ a1 y þ þ ak1 y_ þ ak y ¼ b1 u_ ðtÞ þ b2 uðtÞ ð3:27Þ and the state equation becomes of the form Eq. 3.6 with If we make the same choice of the state 2 3 variables as in the previous case, i.e. 0 1 0 0 0 6 0 0 1 0 0 7 6 7 A¼6 6 7 7 x1 ¼ y; x2 ¼ y_ ; . . .; xk ¼ yðk 1Þ ð3:28Þ 4 0 0 0 0 1 5 ak ak1 a1 we would obtain the following dynamic equations: ð3:35Þ x_ 1 ¼ x2 ; x_ 2 ¼ x3 ; . . .; x_ k1 ¼ xk ð3:29Þ which is of dimension k k, and and the remaining equation for x_ k would be, from Eq. 3.27, 2 3 0 6 0 7 x_ k ¼ a1 xk a2 xk l ; ak x1 þ bl u_ þ b2 u 6 .. 7 B¼6 6 . 7 7 ð3:36Þ ð3:30Þ 4 5 b b2 a1 b 1 which is not in the normal form because of the presence of the u_ term. which is of dimension k 1. For a more general To avoid this difﬁculty, let us choose procedure for determining the normal form of linear differential equations, you are referred to xk ¼ yðk lÞ bu; ð3:31Þ Schwartz and Friedland [2]. How Does One Solve Linear Time-Invariant State Equations? 31 How Does One Solve Linear We expect the solution to be of the same form Time-Invariant State Equations? as Eq. 3.41, but involving matrix exponential functions. Consider the familiar ﬁrst-order differential equation Laplace Transform Method Taking the Laplace transform of Eq. 3.42 gives ðdx=dtÞ ¼ ax þ bu; ð3:37Þ where x(t) and u(t) are scalar functions of time. sXðsÞxð0Þ ¼ AXðsÞ þ BUðsÞ ð3:43Þ The solution of this equation can be obtained by using either the integrating factor or the Laplace A block diagram representation of Eq. 3.43 is transform method. Using the latter method, we shown in Fig. 3.5. The matrix solution of this get, from Eq. 3.37, equation is given by sX ðsÞxð0Þ ¼ aX ðsÞ þ bU ðsÞ ð3:38Þ XðsÞ ¼ ðsIAÞ1 xð0Þ þ ðsIAÞ1 BUðsÞ; or, ð3:44Þ ðsaÞX ðsÞ ¼ xð0Þ þ bU ðsÞ ð3:39Þ where I is the identity matrix or, 2 3 1 0 0 0 X ðsÞ ¼ ½xð0Þ=ðs aÞ þ ½bU ðsÞ=ðsaÞ ð3:40Þ 6 0 1 0 0 7 I¼6 4 7 ð3:45Þ 5 or, 0 0 0 1 Zt of dimension n n and ( )−1 indicates matrix 1 at aðt sÞ xðtÞ ¼ L X ðsÞ ¼ e xð0Þ þ e buðsÞds inversion. Clearly, the solution is determined by 0 the properties of the matrix (sI − A), aptly called ð3:41Þ the characteristic matrix of the system. If we write where the last term on the right-hand side is easily recognized as the familiar convolution LðsÞ ¼ sIA ð3:46Þ integral. The same result could also have been then obtained by using the integrating factor 2at . Now, consider the vector differential equation L1 ðsÞ ¼ ðsIAÞ1 ¼ La ðsÞ=DðsÞ; ð3:47Þ ðdx=dtÞ ¼ Ax þ Bu ð3:42Þ Fig. 3.5 Block diagram x(o) representation of (3.43) + U(s) Input B + å 1/S X(s) + State output vector A 32 3 State Variables—Part I where La(s) is the adjoint of the characteristic The Laplace transform method of solution matrix and requires evaluation of the adjoint of a matrix. Also, in general, it is difﬁcult to obtain the DðsÞ ¼ det ðsIAÞ ð3:48Þ required inverse transformation. Before we proceed to discuss a second This determinant is a polynomial of degree method of solution, we digress a little in order to n and is called the characteristic polynomial of introduce the important concept of the transfer the matrix A or of the system characterized by matrix. Taking the Laplace transform of Eq. 3.7 the matrix A. The roots of this polynomial are and substituting the value of X(s) from Eq. 3.44, called by various different names, some of them we get being eigenvalues, natural modes, natural fre- quencies and characteristic roots. As the ele- YðsÞ ¼ CðsIAÞ1 xð0Þ þ ½CðsIAÞ1 B þ DXðsÞ ments in La(s) are the co-factors of L(s), these will be of order (n − 1). ð3:52Þ In the event that the entries in the adjoint With zero initial conditions, this becomes matrix have a common factor, this will also appear in Δ(s). This will permit the cancellation of the common factor in L−1(s). This cancellation YðsÞ ¼ ½CðsIAÞ1 B þ DXðsÞ ¼ HðsÞXðsÞ; should be done before evaluating the inverse ð3:53Þ Laplace transform of L−1(s). The time domain solution follows by taking where the inverse Laplace transform of the expression HðsÞ ¼ 2½CðsIAÞ1 B þ D 3 ½1=DðsÞLa ðsÞxð0Þ þ ½1=DðsÞLa ðsÞBUðsÞ H11 ðsÞ H12 ðsÞ H1n ðsÞ ð3:49Þ 6 H21 ðsÞ H22 ðsÞ H2n ðsÞ 7 ¼64 7 5 As in the case of ordinary expressions in Hp1 ðsÞ Hp2 ðsÞ Hpn ðsÞ Laplace variable s, we expand each of the terms ð3:54Þ in (3.49) into a partial fraction and take the inverse Laplace transform. For example, for the Thus, the ith component Yi(s) of the transform force-free system, assuming that the roots of Y(s) of the output vector may be written as Δ(s) = 0 are all distinct, we have h i Yi ðsÞ ¼Hi1 ðsÞX1 ðsÞ þ Hi2 ðsÞX2 ðsÞ xðtÞ ¼ Ll ðsI AÞ1 xð0Þ ð3:55Þ þ þ Hin ðsÞXn ðsÞ ¼ L1 f½X1 =ðs s1 Þ þ ½X2 =ðs s2 Þ þ þ ½Xn =ðs sn Þg Clearly, Hij(s), the (i, j)th element of H(s), is the transfer function between Xj(s) and Yi(s) and ¼ X1 es1 t þ X2 es2 t þ þ Xn esn t ð3:50Þ hij ðtÞ ¼ L1 ½Hij ðsÞ ð3:56Þ where is the response at the ith output terminal due to an unit impulse at the jth input terminal. Xi ¼ lim ½ðs si ÞLa ðsÞ=DðsÞxð0Þ ð3:51Þ We have thus found a general expression for s!sl the transfer functions of a system in terms of the The case in which the roots are not all distinct matrices appearing in the normal form charac- can be handled by the same technique as in the terization. The matrix H(s) is called the transfer ordinary expressions in the Laplace variable s. matrix of the system. How Does One Solve Linear Time-Invariant State Equations? 33 Example 3 Consider again the circuit in Fig. 3.5 The transfer matrix is given by Eq. 3.61 with L = (1/2) Henry, C = 1 Farad and R = (1/3) Thus, the response i2(t) to the input ohm. Then from Eqs. 3.14 and 3.18, A, B, C and e1(t) = d(t) is the inverse transform of H21(s), D matrices become giving h21 ðtÞ ¼ 4e2t 2et ð3:62Þ 0 2 2 2 A¼ B¼ 1 3 0 3 which checks with the result of a direct calcula- 1 0 0 0 tion. It may also be noted that only H22(s) has a C¼ and D ¼ 1 3 0 3 numerator of second degree, so that only ð3:57Þ h22(t) involves an impulsive term. This is exactly what we expect, since e2(t) = d(t) will result in an impulse of current through C. Therefore By Series Expansion We next consider a solution of the vector dif- s 2 LðsÞ ¼ ðsIAÞ ¼ ð3:58Þ ferential equation by series expansion. First, 1 sþ3 consider the homogeneous equation and 1 s þ 3 2 ðsIAÞ1 ¼ s2 þ 3s þ 2 1 s ð3:59Þ ½1=ðs þ 2Þ þ ½2=ðs þ 1Þ½2=ðs þ 2Þ þ ½2=ðs þ 1Þ ¼ ½1=ðs þ 2Þ þ ½1=ðs þ 1Þ½2=ðs þ 2Þ þ ½1=ðs þ 1Þ For the force-free case, the time domain x_ ¼ Ax ð3:63Þ solution is Performing a Taylor series expansion about the origin of the state space, we get e2t þ 2et 2e2t 2et xð t Þ ¼ x(0) e2t þ et 2e2t et xðtÞ ¼ E0 þ E1 t þ E2 t2 þ ð3:64Þ ð3:60Þ where Ei’s are column vectors, whose elements are constants. In order to determine these vectors, HðsÞ ¼ CðsIAÞ1 B þ D 1 1 0 s þ 3 2 2 2 0 0 ¼ 2 þ s þ 3s þ 2 1 3 1 s 0 3 0 3 ð3:61Þ 1 2ðs þ 3Þ 2s ¼ 2 s þ 3s þ 2 2s ð3s2 þ 2sÞ 34 3 State Variables—Part I we successively differentiate Eq. 3.64 and set Substituting this expression in Eq. 3.6, we t = 0; thus obtain _ xð0Þ ¼ E0 ; xð0Þ ¼ E1 ; xð2Þ ð0Þ ¼ 2E2 ; . . . ðd/=dtÞq þ /ðdq=dtÞ ¼ A/q þ B u ð3:70Þ ð3:65Þ But However, from Eq. 3.63, we note that d/=dt ¼ d eAt =dt ¼ AeAt ¼ A/ ð3:71Þ ð2Þ _ xð0Þ _ ¼ Axð0Þ; x ð0Þ ¼ A xð0Þ ¼ A½Axð0Þ so that ¼ A2 xð0Þ; . . . ð3:66Þ /ðdq=dtÞ ¼ B u ð3:72Þ where A2 implies the matrix multiplication or A A. Continuing the process, we obtain the following series solution for x(t): dq=dt ¼ /1 ðtÞB uðtÞ ð3:73Þ xðtÞ ¼ I þ At þ ðA2 t2 =2!Þ þ ðA3 t3 =3!Þ þ xð0Þ or ð3:67Þ Zt This inﬁnite series deﬁnes the matric expo- qðtÞ ¼ /1 ðsÞBuðsÞ ds ð3:74Þ nential function eAt which may be shown to be 0 convergent for all square matrices A. Therefore Thus, the particular integral is given by At xð t Þ ¼ e xð 0Þ ð3:68Þ Zt Comparing this result with Eq. 3.41, we note the xp ðtÞ ¼ /ðtÞ /1 ðsÞBuðsÞ ds ð3:75Þ equivalence of the unforced part of the solution. 0 The function eAt contains a great deal of informa- and the complete solution is tion about the system behaviour and, as such, is called the fundamental or the transition matrix, Zt denoted by /ðtÞ. It possesses all the important xðtÞ ¼ /ðtÞxð0Þ þ /ðtÞ/1 ðsÞBuðsÞ ds properties of the scalar exponential function. In 0 particular, the derivative of the function yields the function itself pre-multiplied by a constant. Using ð3:76Þ this property and the method of variation of parameters, we now proceed to determine the This expression is analogous to the familiar forced part of the solution, i.e. the particular complete solution of a scalar ﬁrst-order differ- integral of the original equation given by Eq. 3.6. ential equation, given in Eq. 3.41. The proper- By analogy with the particular solution of the ties of the fundamental matrix /ðtÞ allow a scalar equation given by Eq. 3.37, we let the further simpliﬁcation of Eq. 3.76. These prop- particular integral of the vector differential erties and methods for evaluation of /ðtÞ will be equation be discussed in details in Part II, which will also include a bibliography and an appendix on xp ðtÞ ¼ /ðtÞqðtÞ ð3:69Þ matrix algebra. An Advice 35 L3 + C1 C3 u (t ) C2 R L – R1 L2 + u (t ) L1 R2 Fig. P.1 Circuit for Problem P.1 – Fig. P.3 Circuit for Problem P.3 + R C2 u(t) C1 C3 R – L3 L1 R Fig. P.2 Circuit for Problem P.2 + u (t ) L2 C L4 – An Advice Fig. P.4 Circuit for Problem P.4 The best way to comprehend a topic like this is to follow the examples given here as well as those P:3. Write the state equations for the circuit in the next chapter and in addition, solve as many shown in Fig. P.3. examples as possible, from the reference books, P:4. Write the state equations for the circuit cited below. shown in Fig. P.4. P:5. Solve the state equations for the circuit in Fig. P.4. Problems I believe, as I mentioned on earlier occasion, that they are not difﬁcult. References P:1. Write the state equations for the circuit in 1. E.S. Kuh, R.A. Rohrer, The state variable approach to network theory, in Proc IRE, vol 53 (July 1965), Fig. P.1. pp. 672–686 P:2. Write the state equations for the circuit in 2. R.J. Schwartz, B. Friedland, Linear Systems (McGraw Fig. P.2. Hill, 1965) State Variables—Part II 4 In the ﬁrst part of this discussion on state the same. In particular, we dwell upon the state variables, which hopefully you have grasped, transition flow graph method in considerable we presented the basic concepts of state details. We provide a bibliography at the end and variables and state equations, and some include an appendix on the essentials of matrix methods for solution of the latter. In this algebra. second and concluding part, we dwell upon Equations and ﬁgures occurring in Part I and the properties and evaluation of the funda- referred to here are not reproduced, for brevity, mental matrix. An appendix on matrix algebra and it is suggested that you make it convenient to is also included. Several examples have been have Part I at hand for ready reference. given to illustrate the techniques. This is the last part, be assured! Properties of the Fundamental Matrix Keywords Fundamental matrix Fundamental state One important property of the fundamental equation Evaluating fundamental state matrix /(t), namely that matrix State transition flow graphs Review of matrix algebra d/=dt ¼ A/ has already been mentioned. In Part I of this discussion, we introduced the Consider now the solution to the homoge- basic concepts of state variables and state equa- neous vector differential equation, given by tions, and discussed several methods of solution Eq. 3.68 in Part I: of the latter. In this second and the concluding part, we deal with the properties of the funda- xðtÞ ¼ /ðtÞxð0Þ ð4:1Þ mental matrix and procedures for evaluation of Or, explicitly Source: S. C. Dutta Roy, “State Variables—Part II,” IETE Journal of Education, vol. 38, pp. 99–107, April– June 1997. © Springer Nature Singapore Pte Ltd. 2018 37 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_4 38 4 State Variables—Part II 2 3 2 32 3 Consider the evaluation of the transient x1 ðtÞ /11 ðtÞ /12 ðtÞ /1n ðtÞ x1 ð0Þ 6 x2 ðtÞ 7 6 6 x2 ð0Þ 7 response of a time-invariant system at various 6 7 /21 ðtÞ /22 ðtÞ /2n ðtÞ 7 76 7 6 .. 7 ¼ 6 6 7 4 ... 5 4 5 values of time t1 and t2 while the initial time was 4 . 5 t0. At t = t1, xn ðtÞ /n1 ðtÞ /n2 ðtÞ /nn ðtÞ xn ð0Þ ð4:2Þ xðt1 Þ ¼ /ðt1 t0 Þxðt0 Þ ð4:7Þ We can determine the elements of the funda- At t = t2, if the initial time is considered as t1, then mental matrix by setting to zero all initial con- ditions except one, evaluating the output of the xðt2 Þ ¼ /ðt2 t1 Þxðt1 Þ ð4:8Þ states, and repeating the procedure. For example, ¼ /ðt2 t1 Þ/ðt1 t0 Þxðt0 Þ setting On the other hand, considering the initial time x1 ð0Þ ¼ 1; x2 ð0Þ ¼ x3 ð0Þ ¼ ¼ xn ð0Þ ¼ 0 as t0, we get ð4:3Þ xðt2 Þ ¼ /ðt2 t0 Þxðt0 Þ ð4:9Þ we obtain Comparing Eqs. 4.8 and 4.9, we get x1 ðtÞ ¼ /11 ðtÞ; x2 ðtÞ ¼ /21 ðtÞ; . . .; xn ðtÞ ¼ /nl ðtÞ /ðt2 t0 Þ ¼ /ðt2 t1 Þ/ðt1 t0 Þ ð4:10Þ ð4:4Þ This important relationship justiﬁes the name Thus, in general, /ij(t) is the transient response ‘state transition matrix’ for /(t). Clearly, of the ith state due to unit initial condition of the /(t2 − t0) is a sequence of state transitions. Since jth state, when zero initial conditions apply to all the relation must hold at t2 = t0, we obtain other states. This property will be utilized later to determine the state equations of any system /ðt2 t0 Þ ¼ /ðt2 t2 Þ ¼ I represented by an analog computer simulation. ð4:11Þ So far, in solving the state equations, the ini- ¼ /ðt0 t1 Þ/ðt1 t0 Þ tial conditions were taken at t = 0. If, instead, the initial time is taken as t = t0, then the funda- and mental matrix becomes /1 ðt1 t0 Þ ¼ /ðt0 t1 Þ ð4:12Þ /ðt t0 Þ ¼ eAðtt0 Þ ð4:5Þ Now setting the initial time of interest to zero, and the complementary solution changes to we get xðtÞ ¼ /ðtt0 Þxðt0 Þ ð4:6Þ /1 ðt1 Þ ¼ /ðt1 Þ ð4:13Þ Furthermore, we have only considered Or, in general time-invariant systems. For a time-varying sys- tem, we must use the fundamental matrix /(t, t0) /1 ðtÞ ¼ /ðtÞ ð4:14Þ which depends on both the present time t and the This relation can be used to simplify the initial time t0, and not just on the time difference general solution, given by Eq. 3.76 of the state t − t0. Also the matrices A, B, C and D become equation. functions of time t. The Fundamental State Transition Equation 39 The Fundamental State Transition By Exponential Series Expansion Equation Example 4 For the network in Fig. 3.3 with the For an initial time t = t0, the general solution of speciﬁed element values, as in Example 3, the the vector differential equation given by Eq. 3.6, A matrix was shown to be as found in Eq. 3.76, modiﬁes as follows: 0 2 Zt A¼ ð4:18Þ 1 3 xðtÞ ¼ /ðtt0 Þ xðt0 Þ þ /(t)/1 ðsÞBuðsÞds t0 Thus ð4:15Þ 0 2 0 2 2 6 A2 ¼ ¼ ð4:19Þ Using Eqs. 4.10 and 4.14, we may write the 1 3 1 3 3 7 fundamental state transition equation of a 2 6 0 2 6 14 time-invariant system as A ¼ 3 ¼ ð4:20Þ 3 7 1 3 7 15 Zt and so on. Hence xðtÞ ¼ /ðtt0 Þ xðt0 Þ þ /(t - s)BuðsÞds t0 1 0 0 2 2 6 t2 /ðtÞ ¼ þ tþ 0 1 1 3 3 7 2! ð4:16Þ 6 14 t3 þ þ 7 15 3! For a time-varying system, the more general " # t 2 t3 t2 t3 1 þ 0:t 2 2! þ 6 3! 0 2t þ 6 2! 14 3! þ form of the transition matrix /(t, t0) must be used ¼ 2 t 3 7t2 t 3 0 þ t 3t2! þ 7 3! þ 1 3t þ 2! 15 3! þ and then the state transition equation becomes ð4:21Þ Zt xðtÞ ¼ /ðt; t0 Þ xðt0 Þ þ /(t; s)BðsÞuðsÞds lf the calculation is continued, each series in Eq. 4.21 turns out to be the expansion of the sum t0 of two exponentials, and the fundamental matrix ð4:17Þ becomes 2t The solution in either Eqs. 4.16 or 4.17 e þ 2et 2e2t 2et /ðtÞ ¼ ð4:22Þ consists of the unforced natural response due to e2t þ et 2e2t et the initial conditions plus a matrix convolu- tion integral containing the matrix of the inputs This method is effective only in very simple u (s). cases. By Solution of the Homogeneous Procedures for Evaluating Differential Equations using Classical the Fundamental Matrix: Described Methods in Steps The fundamental matrix /(t) = eAt can be eval- Example 5 Consider the network in Fig. 3.3 uated by various methods which are now dis- again. The homogeneous differential equation for cussed through some examples. this system is 40 4 State Variables—Part II 2t x_ ¼ Ax ð4:23Þ x1 ðtÞ e þ2et 2e2t 2et x1 ð0Þ ¼ x2 ðtÞ e2t þet 2e2t et x2 ð0Þ or, ð4:31Þ x_ 1 0 2 x1 ¼ ð4:24Þ Since x(t) = /(t) x(0), the square matrix in x_ 2 1 3 x2 Eq. 4.31 is /(t). This result checks with the one obtained by the previous method. This matrix equation represents two simulta- This method also works out well for simple neous differential equations, namely systems but, for higher order systems, it becomes laborious and time consuming. x_ 1 ¼ 2x2 ð4:25Þ x_ 2 ¼ x1 3x2 Eliminating x2, we obtain By Evaluating the Inverse Laplace Transform of (sI–A) ð2Þ x1 ¼ 2x2 ¼ 2x1 þ 6x2 Comparing the ﬁrst equation in Eqs. 3.50 and ¼ 2x1 þ 6ð_x1 =2Þ ¼ 2x1 3_x1 4.1, we obtain the important relationship or /ðtÞ ¼ L1 ½sIAÞ1 ð4:32Þ ð2Þ x1 þ 3x1 þ 2x1 ¼ 0 ð4:26Þ This relation can be used to determine the fundamental matrix, as illustrated in the follow- The general solution for this homogeneous ing example. second-order differential equation is Example 6 Consider the circuit in Fig. 3.3 once 2t t x1 ðtÞ ¼ C1 e þ C2 e ; ð4:27Þ more. For this, where C1 and C2 are arbitrary constants. From s 0 0 2 ðsIAÞ ¼ Eqs. 4.25 and 4.27, we get 0 s 1 3 s 2 ¼ ð4:33Þ x2 ðtÞ ¼ C1 e2t þ ðC2 =2Þet ð4:28Þ 1 s þ 3 We have to determine the matrix that multi- The characteristic function of the system is plies the initial state vector x(0) to yield x(t). Thus, we must express C1 and C2 in terms of the DðsÞ ¼ dctðsIAÞ ¼ ðs þ 1Þðs þ 2Þ ð4:34Þ components of x(0). From Eqs. 4.27 and 4.28, we ﬁnd that which has two real distinct roots. Thus C1 þ C2 ¼ x1 ð0Þ adjoint of ðsI AÞ ð4:29Þ ðsIAÞ1 ¼ ð4:35Þ ðs þ 1Þðs þ 2Þ C1 þ ðC2 =2Þ ¼ x2 ð0Þ 1 s þ 3 2 ¼ ð4:36Þ Solving Eq. 4.29, we get ðs þ 1Þðs þ 2Þ 1 s 2 C1 ¼ x1 ð0Þ þ 2x2 ð0Þ 1 s þ2 1 þ s þ2 2 ð4:30Þ ¼ s þ1 1 s þ1 2 ð4:37Þ sþ1 sþ2 sþ1 þ sþ2 1 2 C2 ¼ 2x1 ð0Þ2x2 ð0Þ Substitution of these expressions for the con- Evaluating the inverse transform of each of stants C1 and C2 into Eqs. 4.27 and 4.28 gives the terms, we ﬁnd Procedures for Evaluating the Fundamental Matrix: Described in Steps 41 simpliﬁes the necessary steps. The state transi- /ðtÞ ¼ L1 ½ðsIAÞ1 ð4:38Þ tion flow graph provides such a method; in 2et e2t 2et þ 2e2t addition, it intuitively illustrates the physical ¼ ð4:39Þ et e2t et þ 2e2t foundation of the vector state transition equation. which agrees with Eq. 4.31. Next, consider the system response to a unit State Transition Flow Graphs step in place of e2(t), while e1(t) = 0. To simplify matters, assume that the initial conditions are The state transition flow graphs differ from zero. Then Mason’s signal flow graphs in that, while the latter precludes consideration of initial conditions, the Zt former does include all initial conditions. /11 ðt sÞ /12 ðt sÞ 0 xð t Þ ¼ ds Analog computers are considered obsolete /21 ðt sÞ /22 ðt sÞ 1 t0 nowadays, but for the purpose of developing ð4:40Þ state transition flow graphs, consider, for a moment, how systems are simulated on an ana- since1 l(t – t0) = 1 for t > t0. Thus log computer. For linear systems, we require 2 3 only the following computing abilities: Rt 6 /12 ðt sÞds 7 6t 7 (i) Multiply a machine variable by a positive xðtÞ ¼ 6 R0t 7 ð4:41Þ 4 5 or a negative constant coefﬁcient (In /22 ðt sÞds t0 practice, this can be done by potentiome- ters and ampliﬁers). Combining Eqs. 4.40 with 4.41, we get (ii) Sum up two or more machine variables (This is accomplished by summing Zt ampliﬁers). x1 ðtÞ ¼ eðtsÞ e2ðtsÞ ds (iii) Produce the time integral of a machine t0 ð4:42Þ variable (This is done by an integrating 1 1 ampliﬁer). ¼ eðtt0 Þ þ e2ðtt0 Þ 2 2 The mathematical descriptions of these three Zt basic analog computations are x2 ðtÞ ¼ eðtsÞ þ 2e2ðtsÞ ds ð4:43Þ 9 t0 ð1Þ x2 ðtÞ ¼ ax1 ðtÞ > > ð2Þ x3 ðtÞ ¼ x0 ðtÞx1 ðtÞ x2 ðtÞ = ¼ eðtt0 Þ e2ðtt0 Þ ; R Rt ð3Þ x2 ðtÞ ¼ x1 ðtÞdt ¼ x1 ðtÞdt þ x2 ð0Þ > > ; In this method, we must determine the adjoint 0 of a matrix (sI − A), the roots of the character- ð4:44Þ istic function D(s) and the inverse transform of each term in (sI − A) matrix. This process will respectively, where a is a constant, positive or demand an increasingly tedious calculation for negative. higher order systems. Taking the Laplace transform of each of these In view of the shortcomings of the three equations, we get methods illustrated in the preceding examples, it 9 would be worthwhile to develop a method of ð1Þ X2 ðsÞ ¼ aX1 ðsÞ = determining the state transition matrix which ð2Þ X3 ðsÞ ¼ X0 ðsÞX1 ðsÞ X2 ðsÞ ð4:45Þ ; ð3Þ X2 ðsÞ ¼ X1sðsÞ þ x2 ð0Þ s Recall that l(t) stands for the unit step function. 1 42 4 State Variables—Part II The analog computer representations of where Eq. 4.44 and the flow graph representations of Eq. 4.45 are shown in Fig. 4.1 on the left- and Mk ðsÞ ¼ gain of the kth forward path; ð4:49aÞ right-hand sides, respectively. Note that the rep- DðsÞ ¼ system determinant or characteristic function resentations of operations (1) and (2) are identi- ¼ 1ðsum of all individual loop gainsÞ cal to the corresponding flow graph notation but þ ðsum of gain products of all possible the integral operation is different. A state transition of flow graph is deﬁned as a combinations of two non - touching loops; i:e:; cause and effect representation of a set of system two loops having no common nodeÞ equations in normal form using the flow graphs of ðsum of gain products of all possible basic computer elements such as those shown in combinations of three non - touching loopsÞ þ : Fig. 4.1. The only dynamic (or storage) element in ð4:49bÞ the computer is the integrator. The output of each integrator (or each node point in a signal flow graph Dk ðsÞ ¼ the value of DðsÞfor that part of the graph not in which a branch with transmittance s−1 ends) can touching the kth forward path: therefore be considered as a state variable (or ð4:49cÞ transform of the state variable). These constitute a set of state variables (or their transforms). There- The method can best be illustrated by con- fore, if we can present the system under investi- sidering an example. gation by an analog computer diagram or a state Example 7 Consider, for a change, another transition flow graph, we can easily determine the second-order system, described by the following state vector x(t) or its transform X(s). state equations: The state transition equation and its Laplace transform are given by 0 1 0 x_ ¼ xþ uðtÞ ð4:50Þ Zt 2 3 1 xðtÞ ¼ /ðtt0 Þxðt0 Þ þ /ðt sÞBuðsÞds Taking the Laplace transform of both sides, t0 we get ð4:46Þ X1 ðsÞ 0 1 X1 ðsÞ XðsÞ ¼ ðsIAÞ1 xðt0 Þ þ ðsIAÞ1 BUðsÞ ¼ 1 X2 ðsÞ s 2 3 X2 ðsÞ ð4:47Þ ð4:51Þ 0 x1 ðt0 Þ þ UðsÞ þ Now if we can determine Eq. 4.47 directly 1 x2 ðt0 Þ from the state transition flow graph, then we can avoid the matrix inversion, and by taking the The state transition flow graph diagram can be inverse Laplace transform of each element, we drawn as shown in Fig. 4.2, where the initial can obtain Eq. 4.46 directly. The system equa- time has been assumed to be t0. There are only tion is obtained in the form of Eq. 4.47 by two loops in the graph of gains −2s−2 and −3s−1 applying Mason’s gain formula to the state which touch at a node. Thus, the system char- transition flow graph. This formula relates the acteristic function is gain Mxy(s) between an independent node x and a dependent node y in the following manner: 3 2 ðs þ 1Þðs þ 2Þ DðsÞ ¼ 1 þ þ ¼ ð4:52Þ s s2 s2 Xy ðsÞ X Mk ðsÞDk ðsÞ Mxy ðsÞ ¼ ¼ ; ð4:48Þ Now, applying Mason’s gain formula Xx ðsÞ k DðsÞ Eq. 4.48, we get State Transition Flow Graphs 43 α (1) x1(t) α x2(t) x1(s) x2(s) x1(t) x0(s) + 1 (2) x0(t) + ∑ x3(t) 1 x1(s) x3(s) - -1 x2(s) x2(t) x2(o) 2(0) -1 (3) s 1 -1 1 s -1 s x1(t) ∫ x2(t) x1(s) x2(s) x1(s) x2(s) Fig. 4.1 Time and frequency domain representations of basic operations in an analog computer U11 ðsÞ ¼ gain from node x1 ðt0 Þ to X1 ðsÞ s X2 ðsÞ=U ðsÞ ¼ ð4:58Þ ð4:53Þ ðs þ 1Þðs þ 2Þ ðs þ 3Þ ¼ ðs þ 1Þðs þ 2Þ Thus U12 ðsÞ ¼ gain from x2 ðt0 Þ to X1 ðsÞ " sþ3 1 # 1 ð4:54Þ X1 ðsÞ ðs þ 1Þðs þ 2Þ ðs þ 1Þðs þ 2Þ x1 ðt0 Þ ¼ ¼ 2 s ðs þ 1Þðs þ 2Þ X2 ðsÞ ðs þ 1Þðs þ 2Þ ðs þ 1Þðs þ 2Þ x2 ðt0 Þ " # 1 U21 ðsÞ ¼ gain from node x1 ðt0 Þ to X2 ðsÞ þ ðs þ 1Þðs þ 2Þ UðsÞ s 2 ð4:55Þ ðs þ 1Þðs þ 2Þ ¼ ðs þ 1Þðs þ 2Þ ð4:59Þ and The state transition equation is obtained by taking the inverse Laplace transform of U22 ðsÞ ¼ gain from node x2 ðt0 Þ to X2 ðsÞ Eq. 4.59 s ð4:56Þ ¼ ðs þ 1Þðs þ 2Þ x1 ðtÞ 2et e2t et e2t ¼ The relation between X1(s) and U(s) is given x2 ðtÞ 2et þ 2e2t et þ 2e2t (" #) by the gain, from node U(s) to X1(s) and is equal x1 ðt0 Þ ðs þ 1 þ þ L1 1Þðs 2Þ to s UðsÞ x2 ðt0 Þ ðs þ 1Þðs þ 2Þ 1 ð4:60Þ X1 ðsÞ=U ðsÞ ¼ ð4:57Þ ðs þ 1Þðs þ 2Þ The fundamental matrix in this result checks The relation between X2(s) and U(s) is given by with the one found out earlier in Eq. 4.39 using the gain from node U(s) to X2(s) and is equal to the inverse matrix process. The extra beneﬁt that 44 4 State Variables—Part II Fig. 4.2 State transition flow x2(to) x1(to) graph of the system of Example 7 - s-1 s1 U(s) 1 s -1 s-1 x2(s) x1(s) -3 -2 has been derived from the state transition flow The state variables are chosen as graph is that the effect of the input signal has been included. It is a distinct advantage that we x1 ¼ w; x2 ¼ w_ and x3 ¼ wð2Þ ð4:64Þ have avoided the necessity of evaluating the integral Then, the set of ﬁrst-order differential equa- tions describing the system is Zt 9 /ðt sÞBuðsÞ ds ð4:61Þ x_ 1 ¼ x2 = x_ 2 ¼ x3 ð4:65Þ t0 ; x_ 3 ¼ 2x2 3x3 þ uðtÞ For example, consider the case where the system is subjected to a unit step applied at In matrix form, this set of equations becomes t = t0, i.e. u (t) = l(t − t0) and the initial condi- 2 3 2 3 tions are zero. From Eq. 4.60, we easily obtain 0 1 0 0 x_ ¼ 4 0 0 1 5x þ 4 0 5uðtÞ ð4:66Þ est0 ) 0 2 3 1 x1 ðtÞ ¼ L1 ðsð1=sÞ þ 1Þðs þ 2Þ ¼ e 1 2 ðtt0 Þ þ 12 e2ðtt0 Þ st x2 ðtÞ ¼ L1 ðs þe1Þðs0þ 2Þ ¼ eðtt0 Þ e 2ðtt0 Þ The state transition flow graph of this system ð4:62Þ is shown in Fig. 4.3. The characteristic function of the system is These check with the earlier obtained results [see Eqs. 4.42 and 4.43]. 3 2 DðsÞ ¼ 1 3s1 2s1 s1 ¼ 1 þ þ 2 The dominant advantage of the signal flow s s graph method of obtaining the state transition ðs þ 1Þðs þ 2Þ ¼ ð4:67Þ equation is that it does not become more difﬁcult or s2 tedious as the order of the system increases. This assertion will be clear from the next example, Using Mason’s gain formula, we obtain where a third-order system is considered. s1 D1 ðsÞ s2 D2 ðsÞ X1 ðsÞ ¼ x1 ðt0 Þ þ x2 ðt0 Þ Example 8 Consider a third-order system repre- DðsÞ DðsÞ sented by the differential equation s3 D3 ðsÞ s3 D4 ðsÞ þ x3 ðt0 Þ þ UðsÞ DðsÞ DðsÞ d3 w d2 w dw 3 þ3 2 þ2 ¼ uðtÞ ð4:63Þ ð4:68Þ dt dt dt State Transition Flow Graphs 45 x3(to) x2(to) x1(to) s-1 s-1 s-1 U(s) 1 s-1 s-1 s-1 x3(s) x2(s) x1(s) -3 -2 Fig. 4.3 State transition flow graph of the system of Example 8 Now 21 3 2 UðsÞ 3 sþ3 1 s sqðsÞ sqðsÞ sqðsÞ 9 6 sþ3 1 7 6 1 6 UðsÞ 7 7 D1 ðsÞ ¼ 1 ð3s1 2s2 Þ ¼ DðsÞ > > xðtÞ ¼ L1 4 0 qðsÞ qðsÞ 5xðt0 Þ þ L 4 qðsÞ 5 = D2 ðsÞ ¼ 1 ð3s1 Þ ¼ 1 þ 3=s 0 2s 1 sUðsÞ ð4:69Þ qðsÞ qðsÞ qðsÞ D3 ðsÞ ¼ 1 > > ; ð4:71Þ D4 ðsÞ ¼ 1 Therefore where q(s) = s2 + 3 s + 2. For an input step of magnitude u(t0), we get 2 3 lðtÞ 2 lðsÞ 3 2es þ 12 e2s 2 lðsÞ 1 es þ 12 e2s 6 7 xðtÞ ¼ 4 0 2es e2s es e2s 5xðt0 Þ 0 2es 4e2s es þ 2e2s 2 3 3 ð4:72Þ 4 lðsÞ þ 12 s 12 e2s þ 2es 6 s 7 þ4 2 lðsÞ e 1 þ 12 e2s 5uðt0 Þ; s 2s e e where s = t − t0 and l(t) stands for the unit step 1 ðs þ 3Þ function. X1 ðsÞ ¼ x1 ðt0 Þ þ x2 ðt0 Þ s ðs þ 1Þðs þ 2Þ 1 1 þ x3 ðt0 Þ þ UðsÞ sðs þ 1Þðs þ 2Þ sðs þ 1Þðs þ 2Þ Concluding Discussion ð4:70Þ As stated in the introduction in Part I, the aim Continuing the process, we obtain the state of this presentation was to introduce the reader transition equation as given by Eq. 4.72. to the fundamentals of state variable 46 4 State Variables—Part II characterization of linear systems. We started Dij ¼ ð1Þi þ j Mij ð4:74Þ with the concept of state and discussed the choice of state variables for a given system. Next, we attempted to solve the state equation and we Operations discovered the importance of the state transition or the fundamental matrix. Special methods were Addition shown to be convenient for evaluating the fun- Two matrices of the same size are added by damental matrix in simple cases. In the general summing the corresponding elements, i.e. if case of a high-order system, one must take resort to the state transition flow graph, the properties AþB ¼ C ð4:75Þ and applications of which have been briefly explained. then The references cited below should be useful to readers interested in further exploration of state cij ¼ aij þ bij ð4:76Þ variables in characterization, analysis and syn- thesis of systems. Multiplication of a matrix by a scalar In the appendix, we have given a short, but If b is a scalar quantity and comprehensive review of matrix algebra for ready reference on the notations, operations, bA ¼ Ab ¼ C ¼ ½Cij ð4:77Þ properties and types of matrices. then the new matrix C has the elements Appendix on Review of Matrix Cij ¼ baij ð4:78Þ Algebra Multiplication of matrices The product AB may be formed if the number Notations of columns in A is equal to the number of rows in A matrix A of dimension m n is a rectangular B. Such matrices are said to be conformable in the array of mn elements arranged in m rows and order stated. Thus, if A is m p and B is p n, n columns as follows: then AB = C is an m n matrix deﬁned by 2 3 a11 a12 a1n 6 a21 X p 6 a22 a2n 7 7 ¼ aij Cij ¼ aij bkj ð4:79Þ A ¼4 5 ð4:73Þ k¼1 am1 am2 amn Matrix multiplication is associative [i.e. (AB) The elements aij are called scalars; they may C = A(BC)], distributive with respect to addition be real or complex numbers, or functions of real [i.e. A (B + C) = AB + AC] but not, in general, or complex variables. commutative [i.e. AB 6¼ BA in general]. In the A square matrix has the same number of rows product AB, A is said to pre-multiply B; an and columns and can be associated with a equivalent statement is that B post-multiplies A. determinant, usually written as det A or |A|. Transpose of a matrix A minor of an n n determinant, denoted by The transpose of A, denoted as AT, is a matrix Mij, is the (n − 1) (n − 1) determinant formed formed by interchanging the rows and columns by crossing out the ith row and the jth column. of A, i.e. The corresponding co-factor Dij is Appendix on Review of Matrix Algebra 47 Inverse and transpose of a product ½aij T ¼ ½aji ¼ AT ð4:80Þ It can be shown that Adjoint of a matrix ðA BÞ1 ¼ B1 A1 The adjoint of A, denoted as Adj A, is a ð4:86Þ matrix formed by replacing each element in A by ðABÞT ¼ BT AT its co-factor and then taking the transpose of the result, i.e. Cayley–Hamilton theorem The matrix (A − k I) is called the character- Adj A ¼ ½Dij T ð4:81Þ istic matrix of A. If A is a square matrix, then the equation Conjugate matrix is formed The conjugate of A, denoted as A, detðAkIÞ ¼ uðkÞ ¼ 0 ð4:87Þ by replacing each element of A by its complex conjugate, i.e. is called its characteristic equation. Cayley– ¼ ½aij Hamilton theorem states that every square matrix A ð4:82Þ A satisﬁes its characteristic equation, i.e. u (A) = 0 Inversion The inverse of A, denoted by A−1 is deﬁned Types as that matrix which, when pre-multiplied or Real and complex post-multiplied by A, gives the unit or identity A is real or complex according to whether matrix (see Eq. 4.74) I, i.e. aij’s are real or complex. Diagonal scalar and unit or identity AA1 ¼ A1 A ¼ I ð4:83Þ A is diagonal if aij = 0, i 6¼ j. A is a scalar if A is diagonal and diagonal elements are all equal, It can be shown that i.e. A = kI. A is unit or identity matrix I if A is diagonal and all the diagonal elements are unity. adj A ½Dij T A1 ¼ ¼ ð4:84Þ Hermitian and Skew-Hermitian det A D A is Hermitian if AT = A, i.e. aij = aij . Obviously, the diagonal elements of a Hermitian A are real numbers. Properties Obviously, A is skew-Hermitian if AT = A. Equality the diagonal elements in this case are either zero The matrices A and B are said to be equal if or pure imaginaries. and only if aij = bij for all i and j. Orthogonal and unitary Equivalents A is orthogonal AT = A−1. A is unitary if A is The matrices A and B are said to be equiva- both Hermitian and orthogonal, i.e. if A T = A−1. lent if and only if non-singular matrices P and Positive deﬁnite Q exist such that A real symmetric A is positive deﬁnite if X n B ¼ PAQ ð4:85Þ xT Ax ¼ aij xi xj [ 0 ð4:88Þ i¼j¼1 Rank The rank of A is deﬁned as the dimension of for all non-trivial x. A necessary and sufﬁcient the largest square sub-matrix in A whose deter- condition is that the characteristic roots of A are minant does not vanish. positive. 48 4 State Variables—Part II Singular P:1. Determine the fundamental state transition A is singular if det A = 0, i.e. no inverse matrix for P.1 circuit of the previous exists. chapter. Symmetric P:2. Same for P.2 circuit of the previous chapter. A is symmetric if it is square and A = AT, i.e. P:3. Same for P.3 circuit of previous chapter. aij = aji. P:4. Draw the state transition flow graph for the Vector, column or row circuit of P.4 of the previous chapter. A is a column vector if it has one column and P:5. Same for P.3 circuit of the previous chapter. a row vector if it has one row. Zero A is zero if aij = 0. Bibliography Problems 1. L.A. Zadeh, C.A. Desoer, Linear System Theory: A State Space Approach (McGraw-Hill, 1963) 2. S. Seeley, Dynamic System Analysis (Reinhold Pub- These problems are slightly more difﬁcult than lishing Co, 1964) those in the previous chapters. But have no fear—if you have followed the contents of this chapter, then these should not be an issue. You will sail through comfortably. Carry Out Partial Fraction Expansion of Functions with Repeated Poles 5 This chapter aims to simplify partial fraction For ﬁnding the constants K0 ; K1 ; . . .Kn1 , expansion with repeated poles––presented most of the textbooks (see, e.g. [1]) recommend a here are some techniques which should make procedure based on differentiation of the function this topic considerably easier. F1 ðsÞ ¼ ðs s0 Þn F ðsÞN ðsÞ=DðsÞ ð5:3Þ The general formula is, in fact Keywords Partial fraction expansion Repeated poles Kr ¼ 1 dr F1 ðsÞ r ¼ 0 to n 1 ð5:4Þ New method r! dsr s¼s0 For even a moderate value of n, this can As you are well aware, the function become quite tedious. Several alternative proce- dures have therefore been suggested in the liter- NðsÞ ature—mostly in journals on circuits, systems FðsÞ ¼ ð5:1Þ ðs s0 Þn DðsÞ and controls—and a few of them have been mentioned in some recent textbooks. While can be expanded in partial fractions as follows: making a critical survey of all these procedures, K0 K1 K2 Kn1 N1 ðsÞ FðsÞ ¼ n þ þ þ þ þ ð5:2Þ ðs s0 Þ ðs s0 Þn1 ðs s0 Þ n2 ðs s0 Þ DðsÞ Source: S. C. Dutta Roy, “Carry Out Partial Fraction Expansion of Functions with Repeated Poles without Tears,” Students’ Journal of the IETE, vol. 26, pp. 129– 131, October 1985. © Springer Nature Singapore Pte Ltd. 2018 49 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_5 50 5 Carry Out Partial Fraction Expansion of Functions … the author has come to the conclusion that a On long division, we get procedure given in [2] and credited to Professor Leonard O. Goldstone of the Polytechnic Insti- tute of Brooklyn, is the best. The method will be described here with a slight modiﬁcation; it is hoped that you will readily appreciate its merits as compared to the differentiation or any other procedure you may have come across so far, and will adopt it in your future work. The Method Thus Look at Eq. 5.2 and note that if s s0 is replaced 1 GðpÞ ¼ p2 þ p 1 p=4 ð5:8Þ by 1p,1 then it becomes 2 4 2p þ 1 Now restoring s, we get Gð pÞ ¼ K0 pn þ K1 pn1 þ þ Kn1 p þ G1 ð pÞ; ð5:5Þ 1 1 1 GðpÞ ¼ FðsÞ ¼ 2 2 þ 4 4 ð5:9Þ ðs þ 1Þ sþ1 sþ3 where for convenience, we have called F ðsÞ ¼ Fðs0 þ 1=pÞ as Gð pÞ and N1 ðs0 þ 1=pÞ= which completes the expansion. Dðs0 þ 1=pÞ as G1 ð pÞ. Thus, if one transforms F(s) to G(p) and carries out a long division to extract, in the quotient, all powers of p from n to Another Example 1 then the constants K0 ; K1 ; . . .Kn1 will auto- matically be found out! The remaining function Consider, next, a more complicated example. G1(p) should then be transformed back to the Let s-variable, and expanded for other simple or multiple poles. 2s2 þ s þ 1 FðsÞ ¼ ð5:10Þ ðs þ 1Þðs þ 2Þ2 ðs þ 3Þ3 An Example Putting s þ 3 ¼ 1=p and simplifying, we get Now consider a speciﬁc example for illustration. 2s2 þ s þ 1 ¼ ð2=p2 Þð11=pÞ þ 16 ð5:11Þ Let and sþ2 FðsÞ ¼ ð5:6Þ ðs þ 1Þ2 ðs þ 3Þ ðs þ 1Þðs þ 2Þ2 ¼ ð1=p3 Þð4=p2 Þ þ ð5=pÞ2 ð5:12Þ Putting s þ 1 ¼ 1=p and simplifying gives We have found the transformed form of p3 þ p2 (s + 1) (s + 2)2 separately because we shall need GðpÞ ¼ ð5:7Þ 2p þ 1 it later, while expanding in terms of the multiple pole at s = −2. Combining Eqs. 5.10, 5.11 and 5.12, we get, after elementary simpliﬁcations 1 This is the modiﬁcation; Another Example 51 16p6 11p5 þ 2p4 7q3 28q2 þ 85 4 q FðsÞ ¼ GðpÞ ¼ ð5:13Þ G2 ðqÞ ¼ G1 ðpÞ ¼ ð5:17Þ 2p3 þ 5p2 4p þ 1 q þ 1 The long division proceeds as follows: Carrying out the long division gives - q + 1) 7 q 3 - 28q 2 + 85 4 ( q -7 q 2 + 21q 3 2 7q - 7q 85 -21q 2 + q 4 -21q 2 + 21q 1 4 q Thus 1 q G2 ðqÞ ¼ 7q2 þ 21q þ 4 ð5:18Þ q þ 1 From Eqs. 5.14 and 5.18, we get 29 2 85 Thus FðsÞ ¼ 8p3 p p 2 41 q 29 2 85 2 7q þ 21q þ 4 ð5:19Þ GðpÞ ¼ 8p3 p p þ G1 ðpÞ; ð5:14Þ q þ 1 2 4 where Finally substituting p ¼ 1=ðs þ 3Þ and q ¼ 1=ðs þ 2Þ in Eq. 5.19 gives the desired partial 225 3 141 2 85 4 p 2 p þ 4 p fraction expansion: G1 ðpÞ ¼ ð5:15Þ 2p3 þ 5p2 4p þ 1 8 29=2 85=4 FðsÞ ¼ 3 2 Expansion in terms of the repeated pole at ðs þ 3Þ ðs þ 3Þ sþ3 ð5:20Þ s = −3 is complete. The remainder function is 7 21 1 2 þ þ 4 now to be expanded in terms of the repeated pole ðs þ 2Þ sþ2 sþ1 at s = −2. In order to accomplish this, G1(p) has to be transformed back to a function of s, which we shall call F2(s); the job is not difﬁcult because Problems dividing both numerator and denominator of Eq. 5.15 by p3, and taking help of Eq. 5.12, we P:1. Expand the function simply get sþ3 225 141 85 2 F ðsÞ ¼ 4 2 ðs þ 3Þ þ 4 ðs þ 3Þ ðs þ 1Þ3 ðs þ 2Þ2 G1 ðpÞ ¼ F2 ðsÞ ¼ 2 ðs þ 1Þðs þ 2Þ ð5:16Þ P:2. Do the same for Now, put s + 2 = 1/q in Eq. 5.16 and call the sþ2 F ðsÞ ¼ resulting function as G2(q). After a bit of sim- ðs þ 1Þ4 pliﬁcation, the following result is obtained: 52 5 Carry Out Partial Fraction Expansion of Functions … P:3. Do the same for P:5. Do the same for sþ3 sþ3 F ðsÞ ¼ F ðsÞ ¼ ðs þ 1Þ3 ðs þ 2Þ4 s2 ðs þ 1Þ2 P:4. Do the same for References 3 ðs þ 1Þ ðs þ 2Þ F ðsÞ ¼ s4 1. M.E. van Valkenburg, Network Analysis (Prentice-Hall of India, New Delhi, pp. 186–187) (1974) and ﬁnd f(t). Note: f(t) will contain a 2. F.F. Kuo, Network Analysis and Synthesis (Wiley, New York, pp. 153–154) d-function. Beware! A Very Simple Method of Finding the Residues at Repeated Poles 6 of a Rational Function in z−1 If you have followed the last chapter carefully, Y ðzÞ ¼ P1 ðzÞ=½ð1 pz1 Þq Q1 ðzÞ this one would be a cakewalk! The two discussions are similar except for the variables. ¼ ½A1 =ð1 pz1 Þ þ ½A2 =ð1 pz1 Þ2 þ . . . A very simple method is given for ﬁnding the þ ½Aq =ð1 pz1 Þq þ Y1 ðzÞ; residues at repeated poles of a rational function ð6:1Þ in z−1. Compared to the multiple differentiation formula given in most textbooks, and several where Ai’s are the residues, i = 1 to q, and other alternatives, this method appears to be Y1(z) contains terms due to other poles. Text- the simplest and the most elegant. It requires books (see, e.g. [1]) usually give the following only a long division preceded by a small formula for ﬁnding Ai’s: amount of processing of the given function. Ai ¼ f1=½ðq iÞ!ðpÞqi g jdqi =dðz1 Þqi ½ð1pz1 ÞL Y ðzÞj1 pz ¼ 1: Keywords ð6:2Þ Partial fraction expansion Repeated poles New method This expression, involving multiple differen- tiations, is indeed formidable, and students invariably make mistakes in calculation. In a recent paper [1, 2], three alternative methods Introduction were outlined. These are Let Y(z), a proper rational function in z−1, have a (1) Multiply both sides of Eq. 6.1 by pole at z = p, where p may be real or complex, of ½ð1pz1 Þq Q1 ðzÞ, simplify the right-hand multiplicity q. Then, Y(z) can be expanded in side, equate the coefﬁcients of powers of z−1 partial fractions as follows: on both sides to get a set of linear equations in the unknown constants, and solve them. Source: S. C. Dutta Roy, “A Very Simple Method of Finding the Residues at Repeated Poles of a Rational Function in z−1,” IETE Journal of Education, vol. 56, pp 68–70, July–December 2015. © Springer Nature Singapore Pte Ltd. 2018 53 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_6 54 6 A Very Simple Method of Finding the Residues at Repeated Poles … (2) Put arbitrary speciﬁc values of z−1 on both and let, in general, sides, like 0; 14 ; 1; 2 etc., and solve the F ðzÞjð1=pÞz1 ¼ x ¼ F 0 ð xÞ ð6:7Þ resulting set of linear algebraic equations. (3) Obtain Aq as Then, Eq. 6.1 becomes Aq ¼ ð1 pz Þ 1 q Y ðzÞj1 pz ¼1 ð6:3Þ Y 0 ð xÞ ¼ P01 ð xÞ=½xq Q01 ð xÞ ¼ ½ðA1 =pÞ=x þ ½ðA2 =p2 Þ=x2 þ . . . þ ½ðAq =pq Þ=xq þ Y10 ð xÞ: and ﬁnd the rational function, ð6:8Þ Y2 ðzÞ ¼ Y ðzÞ½Aq =ð1pz1 Þq ð6:4Þ Multiply both sides by xq, so that Eq. 6.8 becomes Clearly, Y2(z) will have a multiple pole of order q − 1 at z = p. Now, Aq−1 can be obtained as P01 ð xÞ=Q01 ð xÞ ¼ A01 xq1 þ A02 xq 2 þ . . . þ A0q þ xq Y10 ð xÞ; ð6:9Þ Aq1 ¼ ð1pz1 Þq1 Y2 ðzÞj1 pz ¼ 1 ð6:5Þ where The process can now be repeated till all the Ai’s are obtained and the remainder function A0i ¼ Ai =pi : ð6:10Þ Yq(z) can then be handled. Here, we present yet another method, which Now make a long division of P1′(x) by Q1′(x), does not appear to be known to teachers and starting with the lowest powers. Then, the quo- students, and to the best of knowledge of the tients will give all the required residues and the author, it has not appeared in any literature. The remainder gives the numerator of the function method is based on a small amount of prepro- Y1′(x) which can then be analysed for the resi- cessing of Eq. 6.1 followed by a long division; dues at the other poles. besides elegance, it can claim to be the simplest We shall now illustrate the method by an of all known methods. example. The method presented here is an adaptation of a similar one for Laplace transforms, given in the previous chapter and Kuo’s book [3] which was Example published long back, in 1966, and a tutorial paper on the method [1] appeared in 1985, the method Let did not ﬁgure in any textbook so far, and is not known to teachers and students. The author has Y ðzÞ ¼ ½1 ð1=8Þz1 =½1 ð1=2Þz1 3 ½1 ð1=4Þz1 g used this method routinely in his courses on ¼ fA1 =½1 ð1=2Þz1 3 g þ fA2 =½1 ð1=2Þz1 2 g network theory and signals and systems and has þ fA3 =½1 ð1=2Þz1 g þ fA4 =½1 ð1=4Þz1 g: found it to be well received by students. ð6:11Þ Equation 6.11 can be rewritten as The Method Y ðzÞ ¼ 4ð8 z1 Þ=½ð2 z1 Þ3 ð4 z1 Þ In the method under discussion, ﬁrst make a change of variable from z−1 to ¼ ½8A1 =ð2 z1 Þ3 þ ½4A2 =ð2 z1 Þ2 þ ½2A3 =ð2 z1 Þ þ ½4A4 =ð4 z1 Þ: x ¼ ð1=pÞz1 ð6:6Þ ð6:12Þ Example 55 Put x ¼ 2 z1 and multiply both sides of Conclusion Eq. 6.12 by x3. The result is A very simple method, claimed to be the simplest 4ð6 þ xÞ=ð2 þ xÞ ¼ 8A1 þ 4A2 x þ 2A3 x 2 and the most elegant, has been presented for þ ½4A4 x3 =ð2 þ xÞ: ﬁnding the residues at multiple poles of a rational function in z−1. ð6:13Þ Now, make a long division of 24 + 4x by Problems 2 + x as shown below. All problems concern partial fraction expansion and ﬁnding the residues. 1 2 5 þ 4z 0:9z P:1. F ðzÞ ¼ ð10:6z1 Þ2 ð1 þ 0:5z1 Þ 1 2 þ 0:48z P:2. F ðzÞ ¼ 41:2z ð10:4z1 Þ3 4 5 4z 3:2z 5:2z þ 4:5z3 2 Comparing the result with the right-hand side P:3. F ðzÞ ¼ ðz0:5Þ2 ðz þ 2:1Þ2 ðz3Þðz þ 4:5Þ of Eq. 6.13, we get Clue: First convert this F(z) into a rational A1 ¼ 3=2; A2 ¼ 1; A3 ¼ 1; and A4 ¼ 1=2: function in z−1 ð6:14Þ þ 4:5z5 2 P:4. F ðzÞ ¼ ðz 2z þ 0:6Þ3 ðz2:1Þ Substituting these values in Eq. 6.11 and using the inversion formula Clue: Same as in P.3 4 3 2 1 4z 3:2z 5:2z 4:5z 5 Z 1 ½1=ð1 pz1 Þq ¼ ½1=ðq 1Þ!ðn þ 1Þðn þ 2Þ. . . P:5. F ðzÞ ¼ ðz1 0:5Þ2 1 ðz þ 2:1Þ2 ðz1 3Þðz1 þ 4:5Þ ðn þ q 1Þpn uðnÞ; ð6:15Þ Clue: First convert the denominator factors into (1 − az−1) forms. we ﬁnally get, after simpliﬁcation, yðnÞ ¼ Z 1 Y ðzÞ ¼ f½ð3n2 þ 5n þ 6Þ=4ð1=2Þn ð1=2Þð1=4Þn guðnÞ ð6:16Þ References u(n) being the unit step function. 1. S.C. Dutta Roy, Comments on fair and square computation of inverse z-transforms of rational func- tions. IEEE Trans. Educ. 58(1), 56–57 (Feb 2015) 2. S.C. Dutta Roy, Carry out partial fraction expansion of rational functions with multiple poles—without tears. Stud. J. IETE. 26, 129–31 (Oct 1985) 3. F.F. Kuo, in Network Analysis and Synthesis (Wiley, New York, 1966), pp. 153–154 Part II Passive Circuits This is the largest of the four parts and contains 16 chapters. As in Part I, the topics are interrelated here also. We deal with passive circuits and two dis- tinct aspects of it, viz. analysis and synthesis. Most curricula today empha- size only the ﬁrst aspect, viz. analysis, which deals with the problem of ﬁnding the response of a given circuit to a given excitation. The synthesis aspect deals with designing a circuit to perform in a speciﬁed way; by far, synthesis is more exciting than analysis. Analysis, however difﬁcult it may be, is always possible. However, a synthesis problem may or may not have a solution. In real life, it is synthesis that is required more than analysis. Synthesis is an art and not a science. The beauty of synthesis is that if one solution exists, then there exists an indeﬁnite number of solutions. For analysis, the solution is unique. Synthesis, therefore, facilitates choice. From among a variety of solutions, you can select the one that is the best for your situation. The ﬁrst 11 chapters of this part are concerned with analysis. Circuit analysis can be performed with ease in the case of linear circuits with the help of transforms—Fourier or Laplace—which transports the problem from the time domain to the frequency domain. In the latter, there is no differentiation or integration; instead, there are only algebraic manipulations, viz. addition, multiplication and division. That is the reason why you always prefer to work in the frequency domain. However, frequency domain analysis has its own demerits, as you have observed in Chap. 2. A differential (or difference) equation cannot be solved for all times by transforming it to an algebraic equation. There comes the difﬁculty of initial conditions which do not match. So time domain solutions are comprehensive; they do not give rise to such anomalies. The other reason why time domain cannot be divorced once for all is that most undergraduate curricula treat transform techniques later, may be in the second year. On the other hand, a basic Electrical Engineering course is taught in the very ﬁrst year. Hence, one has to work in the time domain. This is how it was and is still is at IIT Delhi. Your curriculum would be no exception. I faced a difﬁculty while teaching the ﬁrst-year course on basic Electrical Engineering. Circuit analysis forms a large part of the course, and I found that most textbooks either bring in the Laplace transform there 58 Part II: Passive Circuits itself or avoid it by introducing an artiﬁcial excitation est. Then follows the concept of poles, zeroes and inversion of a function in s. When the student goes to the second year, it is difﬁcult for him or her to accept Laplace transform in place of est excitation. To remove this difﬁculty, I prepared a set of notes by remaining solely in the time domain and discarded the prescribed textbook. Students found these notes very useful and my other colleagues asked for them when they were assigned to teach this course. This is the genesis of Chap. 7. Chapter 8 deals most comprehensively with the RLC circuit analysis in the time domain. The three cases of damping, viz. under-damping, critical damping and over-damping, are thoroughly analysed, and simple methods are devised for ﬁnding the response easily. Chapter 11 deals in the same circuit with Fourier transforms. Many types of resonances that may occur by taking the output across various elements or a combination of them are treated in Chap. 12. Chapters 9 and 10 deal with two problems which counteract the popular belief that seeing is believing. Chapter 9 gives a circuit paradox which I hope you will ﬁnd interesting, while in Chap. 10, I talk of the problem of initial values in an inductor. The parallel-T RC network is an important circuit and ﬁnds application in many situations. Primarily, it is a null network, but it can be used to make a selective ampliﬁer, a measuring instrument or a frequency discriminator. Chapter 13 presents several different methods for analysing this network and gives some simple methods to avoid the messy mesh or node equations. Chapter 14 goes deeper into the performance of the network with regard to selectivity and spells out design conditions to achieve maximum selectivity. A perfect transformer does not exist in practice, but the concept facilitates design and applications of this important component. There arise some peculiar phenomena, like current discontinuity from t=0− to t=0+, and degeneracy. These are of great theoretical interest and help you to understand the limitations of the device. Read this with attention to realize that what you take for granted with this simple device is not in general practical. Resistive ladders are very commonly treated in the ﬁrst course of circuits. Inﬁnite ladders are particularly important because they bring in some irra- tional numbers and functions in their performance characteristics. Such networks can be analysed by step-by-step analysis and clever reasoning rather than mesh and node equation, which would be inﬁnite in number. Difference equation formulation, however, gives an easier method, and z-transforms can be easily applied to them. This is what forms the contents of Chap. 17. Chapters 18 and 19 deal with synthesis. In Chap. 18, we start with the driving-point synthesis of a relatively simple function, viz. one of the third order. Chapter 19 shows how the practical problem of interference rejection in an ultra-wideband system reduces to the synthesis of an LC driving-point function. We work out several alternative solutions to facilitate a choice of the best one. A ﬁlter design problem has to be solved in several steps. First, from the speciﬁcation of cut-off frequency(ies) and pass- and stop-band tolerances, Part II: Passive Circuits 59 you choose a magnitude function. Then by the process of analytic continu- ation, ﬁnd the corresponding transfer function. Finally, realize the transfer function by using inductors and capacitors. The second step involves ﬁnding the poles and zeroes of the transfer function. Chapter 20 gives some shortcuts for moving from the magnitude function to the transfer function, by simple coefﬁcient matching. This saves a lot of calculation. Also, normally you consider the Butterworth type because of its simplicity. Design of band-pass (BP) and band-stop (BS) ﬁlters usually proceeds from a corresponding normalized low-pass (LP) transfer function. After realizing the LP function, you apply the frequency transformation technique to obtain the actual BP/BS components. There exists some confusion regarding this particular transformation. Chapter 21 attempts to remove these confusions. Chapter 22 deals with passive differentiators. It gives, starting with the well-known RC differentiators, other circuits which give improved linearity of the frequency response. This contains a number of innovations and the new circuits are truly new, not available in textbooks. Circuit Analysis Without Transforms 7 Is it simpler? In most cases, it is. Remember curriculum. At this stage, you are not exposed to the difﬁculty you faced by working solely in Laplace and Fourier transforms, and hence you the time domain, in the previous chapter cannot appreciate how they simplify circuit (Chap. 2), in solving a differential equation analysis. This chapter is an attempt to show that with impulsive excitation? Except for these circuits can indeed be completely analysed odd cases, time domain analysis is usually without the help of transform techniques. simpler. In this chapter, we discuss how linear We ﬁrst introduce the concepts of natural circuits can be completely analysed without response, forced response, transient response and using Laplace or Fourier transforms. Is this steady-state response, and deal with typical analysis simpler than that using transform examples of force-free response. The concepts of techniques? You should judge for yourself to impedance, admittance, poles and zeros are then realize. introduced through the artiﬁce of est excitation. It is shown that the natural frequencies of a circuit depend upon the kind of forcing function, and Keywords that they are related to the poles and zeros of an Circuit analysis Differential equation impedance function. Time domain Force-free response Next, we deal with forced response to expo- First-order circuit Second-order circuit nential excitation, and in particular to sinusoidal Damping Root locus Impedance excitation; introduce the concept of phasors; and Natural response Natural frequencies demonstrate how steady-state sinusoidal DC and sinusoidal excitation Pulse response response can be found out only through phasors Impulse response and impedances. Several examples of complete response are worked out. The chapter concludes Analysis of electrical circuits forms part of a core with an introduction to the impulse and an course for all engineering students which is example of impulse response of a circuit. usually offered in the very ﬁrst semester of the Throughout the chapter, the emphasis is on understanding through examples, rather than on intricate theories. Once you learn the technique through simple, heuristic and common sense Source: S. C. Dutta Roy, “Circuit Analysis without arguments, detailed justiﬁcation through trans- Transforms,” IETE Journal of Education, vol. 39, form techniques, an exposure to which will come pp. 111–127, April–June 1998. © Springer Nature Singapore Pte Ltd. 2018 61 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_7 62 7 Circuit Analysis Without Transforms to you later in the curriculum, will make you a d2 i R di 1 master of the same. þ þ i¼0 ð7:5Þ dt 2 L dt LC Let us start with an example. In general, for a second-order equation like Eq. 7.5, ic(t) will be of the form An Example ic ðtÞ ¼ A1 il ðtÞ þ A2 i2 ðtÞ; ð7:6Þ Consider the circuit in Fig. 7.1, where we wish to ﬁnd i(t) due to an excitation by a current source where A1 and A2 are constants. The particular is(t). Note that C may have an initial voltage integral ip(t) of Eq. 7.4 is any function which vc(0) = V, say, and L may have an initial current satisﬁes the equation as it is. For example, if i(0) = I, say, it being assumed that is(t) has been is(t) = est, then ip(t) = Kest where, by substitution switched on at t = 0. By KCL, we get in Eq. 7.4, we get dvc ðtÞ is ðtÞ ¼ C þ iðtÞ ð7:1Þ R 1 1 dt K s þ sþ 2 ¼ ð7:7Þ L LC LC Also or diðtÞ vc ðtÞ ¼ L þ RiðtÞ ð7:2Þ 1=ðLCÞ dt K¼ ð7:8Þ s2 þ RL s þ LC 1 Substituting Eq. 7.2 in Eq. 7.1 gives In general, therefore, the complete solution to d2 i di Eq. 7.4 is given by is ðtÞ ¼ LC 2 þ RC þ i; ð7:3Þ dt dt iðtÞ ¼ A1 i1 ðtÞ þ A2 i2 ðtÞ þ ip ðtÞ ð7:9Þ where we have dropped the argument t from i(t) for brevity. Equation 7.3 can be written more It is emphasized that ip(t) will not contain any succinctly as unknown constants. The unknown constants here are A1 and A2, which are to be evaluated from the d2 i Rdi 1 is ðtÞ þ þ i¼ ð7:4Þ two initial conditions. The condition i(0) = I dt2 Ldt LC LC gives This is a second-order differential equation with constant coefﬁcients and as you are aware, I ¼ A1 i1 ð0Þ þ A2 i2 ð0Þ þ ip ð0Þ ð7:10Þ the solution will consist of two parts—the com- The other condition vc(0) = V gives, from plementary function and the particular integral. Eq. 7.2, The complementary function ic(t) is the solution to the homogeneous equation di V ¼ Rið0Þ þ L ð7:10aÞ dt t¼0 L or, combining with Eq. 7.9. i(t) + V ¼ RI þ L½A1 i01 ð0Þ þ A2 i02 ð0Þ þ i0p ð0Þ; ð7:11Þ C vc(t) R iS (t) ˗ where the symbol f′(0) has been used for df/dt|t = 0. In Eqs. 7.10 and 7.11, everything else is known except A1 and A2, and hence they can Fig. 7.1 An RLC circuit excited by a current source be found out. Some Nomenclatures 63 Some Nomenclatures not necessarily the same, neither are the steady-state response and forced response. In the previous example, the term ic(t) of Eq. 7.6 arises due to initial energy in the capacitor as Force-Free Response: General well as the inductor. However, Eq. 7.6 only Considerations gives the form of the response due to the initial conditions; the values of the constants depend First, let us study some force-free cases. Obvi- upon the particular solution ip(t) and its deriva- ously, the equation to be solved will be a tive at t = 0. homogeneous equation of order determined by This part of the response is called the natural the number of energy storage elements. In the response. Had the circuit been force free, i.e. if example given earlier in the chapter (page 62), is(t) = 0, then obviously A1 and A2 would be the order was two because of one L and one C. In determined from the initial conditions only. general, our equation will be of the form Hence, although the form of the natural response remains the same in force free as well as forced dn x dn1 x dx n þ an1 n1 þ þ a1 þ a0 ¼ 0; cases, the actual values will be different. dt dt dt The part ip(t) of the total response depends ð7:12Þ solely on the forcing function and the parameters and the architecture of the circuit; it is indepen- where x is a voltage or a current and an–1, an–2,… dent of the initial conditions. ip(t) is called the a1 and a0 are constants determined by the forced response and is usually of the same form parameters of the circuit. as the forcing function. An example has already Although there are various methods of solving been given for is(t) = est. As another example, if Eq. 7.12, we ﬁnd it convenient to assume a is(t) = Is, a dc, then ip(t) is also a dc, Ip, whose solution of the form x = Aest, substitute it in value is obtained from Eq. 7.4 as Ip = Is, because Eq. 7.12, and obtain the following algebraic equation in s: dIp d2 I p ¼ 0 and ¼ 0: sn þ an 1 sn 1 þ þ a1 s þ a0 ¼ 0 ð7:13Þ dt dt2 This is called the characteristic equation of The superposition of natural response (in form the circuit and its roots are called the charac- only) and the forced response gives the total or teristic roots or natural frequencies of the circuit. complete response. Since Eq. 7.13 has n roots—s1, s2, … sn, natu- The natural response of the circuit, in the rally, our desired solution will be of the form force-free case, usually decays with time and becomes negligible as t ! ∞ the term ‘usually’ x ¼ A1 e s 1 t þ A 2 e s 2 t þ . . . þ A n e s n t ; ð7:14Þ indicates the practical situation where dissipation is invariably present. In the dissipationless case, where the constants A1, A2, , An, are to be it is possible that the natural response does not determined from the n initial conditions––one on decay with time. The forced response can be each energy storage element. maintained indeﬁnitely by an appropriate forcing We now consider some speciﬁc examples. function, but for some forcing functions, (like e–at, a > 0), it may also decay with time. That part of the total response which decays with time Force-Free Response of a Simple RC is called the transient response. The value of the Circuit response as t ! ∞ (or in practice, after the transient part has become negligible) is called the Consider the situation depicted in Fig. 7.2, where steady state response. It must be emphasized that the switch S is in position a for a long time so the transient response and natural response are that C is fully charged to the voltage V. At t = 0, 64 7 Circuit Analysis Without Transforms S with the fact that the circuit has only one a b energy storage element, viz. C. Hence, our i(t) solution is t=0 + iðtÞ ¼ Aet=ðRCÞ ð7:18Þ – V R To evaluate A, we take help of the fact that i C (0) R = V. Here, the argument of i is to be interpreted as 0+, i.e. immediately after the switch has been shifted from a to b [note that i Fig. 7.2 A simple RC circuit (0−) = 0]. Hence the switch is shifted to the position b. We then V ¼A ð7:19Þ have a force-free RC circuit (R and C are in series R or parallel?) with an initial condition on C, as and ﬁnally shown in Fig. 7.3. KVL around the loop in Fig. 7.3 gives V t=ðRCÞ iðtÞ ¼ e ð7:20Þ Zt R 1 Ri þ idt V ¼ 0 ð7:15Þ C A plot of Eq. 7.20 is shown in Fig. 7.4. The 0 product RC has the dimension of time, is denoted The sign is negative on V because V and by T, and is called the time constant of the cir- i oppose each other, i.e. i tends to charge C to a cuit. It is the time after which the current decays voltage whose polarity is opposite to that of to e1 ¼ 0:368 times the initial value. At t = 5T, V. Equation 7.15 is an integral equation and can the current value is VR e5 ﬃ 0:0067 VR which is be converted to a homogeneous differential only 0.67% of the initial value and hence can be equation by differentiation. The result is: neglected. It is therefore said that the current has a life time of 5T. di 1 Physically, the current i represents the rate of R þ i¼0 ð7:16Þ dt C decay of charge in the capacitor. As t ! ∞, the charge tends to zero and the current also tends to Assuming i = Aest, we get the characteristic zero. equation 1 sþ ¼0 ð7:17Þ RC i(t) V Thus, there is only one characteristic root or R natural frequency at s ¼ RC 1 ; this is consistent V eR i(t) + C V R – t T Fig. 7.3 Circuit in Fig. 7.2 at t 0, with i(0) R = V Fig. 7.4 Plot of Eq. 7.20 Force-Free Response of a Simple RL Circuit 65 Force-Free Response of a Simple RL Circuit Now consider the circuit in Fig. 7.5, where the switch S is in the position a for a long time, so that a steady current V/r is established in the inductor, and then at t = 0, S is shifted to position b. We wish to ﬁnd the current i(t), t 0+. The differential equation is, by KVL, di L þ Ri ¼ 0 ð7:21Þ dt and by following a similar procedure as in the previous example, we get V Rt=L iðtÞ ¼ e ð7:22Þ r Fig. 7.6 More than one C or L but still ﬁrst-order Here, the time constant is T = L/R. First-Order Circuits with More Than One Energy Storage Element Both of the circuits in the previous two examples are ﬁrst-order circuits, because they are governed Fig. 7.7 Second order circuits by a ﬁrst-order differential equation. It is not necessary that there should be only one energy storage element for a ﬁrst-order circuit. There can be more than one capacitor, for example, in an Force-Free Response RC circuit, but they must be connected in such a of a Second-Order Circuit fashion that the capacitors can be combined into one. As an example, the circuits in Fig. 7.6 are Consider a charged capacitor C, with an initial ﬁrst-order circuits, but not the ones in Fig. 7.7. voltage V0 to be connected across a series RL combination at t = 0 as shown in Fig. 7.8. Assume the inductor to be initially relaxed. r Application of KVL gives a Zt S di 1 L þ Ri þ idt V0 ¼ 0 ð7:23Þ b dt C t=0 0 V Differentiating both sides gives L R d2 i Rdi i þ þ ¼0 ð7:24Þ dt2 L dt LC Fig. 7.5 A simple RL circuit 66 7 Circuit Analysis Without Transforms i (t ) Hence, the solution becomes V0 L R iðtÞ ¼ ðes1 t es2 t Þ ð7:32Þ Lðs1 s2 Þ C Case I: 2 + vC (t) – R 1 [0 2L LC Fig. 7.8 A second-order circuit In this case, the roots s1 and s2 are real, neg- which is the same as Eq. 7.5. The characteristic ative and distinct and the solution is equation is at ) iðtÞ ¼ V0 e2bL ðebt ebt Þ R 1 ; ð7:33Þ s2 þ sþ ¼0 ð7:25Þ at ¼ V0 e sinh bt L LC bL which has the roots where s ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ R R2 1 R R 2 1 a¼ þ and b ¼ ¼ ð7:34Þ s1;2 ¼ ð7:26Þ 2L 4L 2 LC 2L 2L LC Case II: The most general solution is, therefore, R 2 1 \0 iðtÞ ¼ A1 es1 t þ A2 es2 t ð7:27Þ 2L LC Clearly, the nature of the solution will depend In this case, the roots will be complex con- R 2 1 jugates of each other: upon whether 2L LC is >, = or <0. Accord- ingly, we shall have three cases. But ﬁrst let us s1;2 ¼ a jx; ð7:35Þ evaluate A1 and A2. The condition i(0) = 0 gives, from Eq. 7.27, where a is deﬁned in Eq. 7.34 and A1 þ A 2 ¼ 0 ð7:28Þ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 R2 x¼ 2 ¼ x2n a2 ð7:36Þ Also, from Eq. 7.23, putting t = 0, we get LC 4L xn being given by di L ¼ V0 ð7:29Þ dt t¼0 x2n ¼ 1=ðLC Þ ð7:37Þ which, combined with Eq. 7.27, gives Hence, the complete solution is s1 A1 þ s2 A2 ¼ V0 =L ð7:30Þ iðtÞ ¼ eat ½A1 ejxt þ A2 ejxt ð7:38Þ Solving Eqs. 7.28 and 7.30, we get which can be simpliﬁed to the form V0 A1 ¼ ¼ A2 ð7:31Þ iðtÞ ¼ Aeat sinðxt þ hÞ ð7:39Þ Lðs1 s2 Þ Force-Free Response of a Second-Order Circuit 67 This is a damped sinusoid with the damping there are oscillations, while case I is called the coefﬁcient a (per second), natural frequency of overdamped case because the response decays oscillation x and initial phase h. A and h are to be monotonically with time after reaching the ﬁrst determined, as in the earlier case, from i(0) = 0 maximum. In the background of this nomencla- and ddti ¼ VL0 . Applied to Eq. 7.39, it gives ture, the case under consideration should be t¼0 called the critically damped case. From the the- ory of differential equations, the solution to this h¼0 ð7:40Þ case will be since s1 = s2 = −a, and iðtÞ ¼ eat ðA1 þ A2 tÞ ð7:43Þ V0 A¼ ð7:41Þ While this is the general form, i(0) = 0 indi- xL cates that A1 = 0. Hence, Thus, ﬁnally, iðtÞ ¼ A2 t eat ð7:44Þ V0 at iðtÞ ¼ e sin xt ð7:42Þ xL The other initial condition, viz. ddti ¼ V0 =L t¼0 Note that Eq. 7.39 is the most general form; it gives has reduced to Eq. 7.42 because we took the V0 initial current in the inductor as zero. A plot of A2 ¼ ð7:44aÞ L Eq. 7.42 is shown in Fig. 7.9. Case III: Hence, ﬁnally, V0 at R 2 1 iðtÞ ¼ te ð7:44bÞ ¼ L 2L LC a plot of which is shown in Fig. 7.10. Note that In this case, the roots are real and equal. i(t) = 0 at t = 0 as well as at t = ∞, and that there is Case II is called the underdamped case, because a maximum at t = 1/a, the maximum value being V0 i(t) IM ¼ ð7:45Þ aLe V 0 V -a t 0e wL i(t) wL V 0 a Le V - 0 2p t wL 1 w a Fig. 7.9 Plot of Eq. 7.42 Fig. 7.10 Plot of Eq. 7.44b 68 7 Circuit Analysis Without Transforms qﬃﬃﬃ It is worth mentioning here that the plot of (iii) when R ¼ 2 L C, both the roots coalesce at Eq. 7.33 (for the overdamped case) will be sim- ilar to that of Fig. 7.10 except that the maximum s = −a, i.e. on the negative real axis (point P in Fig. 7.11); and will be smaller and the decay will be slower. qﬃﬃﬃ (iv) when R increases beyond 2 CL , we get the overdamped case, with both roots on the Root Locus of the Second-Order negative real axis, one (s1) going towards Circuit the origin and the other (s2) moving towards −∞. The roots of the characteristic equation of the circuit considered in the preceding section move in the complex plane s = r + jx as R is varied. Referring to Eq. 7.26, we see that Natural Frequencies of Circuits with a Forcing Function (i) when R = 0 (note R cannot be negative), the roots are purely imaginary Consider the example in Fig. 7.1 again, we have pﬃﬃﬃﬃﬃﬃ seen that the natural response satisﬁes Eq. 7.5, s1;2 ¼ j= LC ¼ jxn ð7:46Þ which is same as Eq. 7.25. The natural response pﬃﬃﬃﬃﬃﬃﬃﬃﬃ is therefore of the form Eq. 7.27 with the natural (ii) When 0 \ R\ 2 L=C , the roots are frequencies given by Eq. 7.26. Now suppose complex: instead of a current source is(t), we connect a voltage source vs(t) across C, as shown in s1;2 ¼ a jx; ð7:47Þ Fig. 7.12. The differential equation now becomes where di L þ Ri ¼ vs ðtÞ ð7:48Þ dt a2 þ x2 ¼ x2n ; ð7:47aÞ so that the complementary function ic(t) satisﬁes i.e. roots move in a circle, starting from + the following homogeneous equation jxn and −jxn, towards the negative real axis; the centre of the circle is at the origin di L þ Ri ¼ 0 ð7:49Þ of the complex plane and its radius is xn, dt as shown in Fig. 7.11; The characteristic equation is, therefore, s þ ðR=LÞ ¼ 0 ð7:50Þ jω jωn jω i(t) ωn P σ L α + vs (t) C R ˗ – jωn Fig. 7.11 Root locus of the series RLC circuit Fig. 7.12 Circuit of Fig. 7.1 driven by a voltage source Natural Frequencies of Circuits with a Forcing Function 69 and the natural frequency is concerned, they perform the same role as that of a resistance. Impedances can be combined s0 ¼ R=L ð7:51Þ exactly as resistances. For example, consider a series combination of R, L and C carrying a This is quite different from Eq. 7.26 and leads current est, then the voltage drop across the us to the conclusion that the natural frequencies combination is of a circuit under a forcing function depend upon the nature of the latter. We next consider a 1 st v¼ R þ sL þ e ð7:56Þ general technique based on the concept of sC impedance for ﬁnding the natural frequencies. But before that, note that s1, 2 of Eq. 7.26 can be Thus, the ratio of v to i is called open-circuit natural frequencies because is(t) = 0 in Fig. 7.1 makes C look at an open 1 ZðsÞ ¼ R þ sL þ ð7:57Þ circuit to the left, while s0 of Eq. 7.51 qualiﬁes to sC be called a short-circuit natural frequency because vs(t) = 0 makes C look at a short circuit. which is simply the addition of the individual impedances. For the circuit in Fig. 7.1, the impedance seen by is(t) is Concept of Impedance sC ðR þ sLÞ 1 ZðsÞ ¼ ð7:58Þ R þ sL þ sC 1 Suppose a current est passes through an inductor L, then the voltage developed across it will be di Relation Between Impedance v¼L ¼ sLest ð7:52Þ and Natural Frequencies dt The ratio of v to i is Equation 7.58 can be simpliﬁed to as follows: ZL ¼ sL ð7:53Þ 1 ðs þ R=LÞ ZðsÞ ¼ ð7:59Þ C s2 þ s RL þ LC 1 and is called the impedance of the inductor. Similarly, if the voltage across a capacitor C is and can be rewritten as est, then the current through it is given by 1 ðs s0 Þ dv ZðsÞ ¼ ; ð7:60Þ i ¼ C ¼ sCest ð7:54Þ C ðs s1 Þðs s2 Þ dt where s0 is the same as the natural frequency so that the ratio of v to i in this case is with voltage excitation, given in Eq. 7.51, and s1 1 and s2 are the natural frequencies with current Zc ¼ ð7:55Þ excitation, given by Eq. 7.26. These observations sC hold in general, too. That is, the open-circuit Zc is called the impedance of the capacitor. natural frequencies are the value of s at which Z For a resistance R, the ratio of v to i is simply R, (s) ! ∞; these values are called the poles of Z independent of the form of v or i. It follows that (s). Similarly, the short-circuit natural frequen- Zc and ZL have the same dimension as that of R, cies are those values of s at which Z(s) = 0; these and that as far as an exponential excitation is values are given the name zeros. Clearly, both 70 7 Circuit Analysis Without Transforms jω 2Ω 1 R2 X j - 5e -3t + + LC 4L2 1H 0.5 F v vC (t) – – 3Ω σ R Fig. 7.14 An example of exponential excitation - L Since Z(s) = v/i when v or i is of the form est or Aest, it follows that R - X 2L vðtÞ ¼ iðtÞZ ðsÞjs ¼ 3 ð7:62Þ ¼ 5e3t 2 ¼ 10e3t Fig. 7.13 Pole-zero plot for the circuit in Fig. 7.1 with If we are interested in ﬁnding the voltage R2 < 4L/C across the capacitor, then we can use potential division, i.e. poles and zeros can, in general, be complex quantities. In the s = r + jx plane, a zero is 2ðs þ 3Þ ðs þ 2Þðs þ 1Þ indicated by a small circle, while a pole is indi- vc ðtÞ ¼ vðtÞ þ 3Þ ¼ 0 ð7:63Þ cated by a cross and the picture so obtained is 2 þ ðs þ2ðs2Þðs þ 1Þs¼3 called the pole-zero plot. For the circuit in Fig. 7.1, the pole-zero plot for a typical under- damped case is shown in Fig. 7.13. Forced Response Due to DC An admittance is deﬁned as the reciprocal of an impedance and is denoted by Y. The poles and A DC can be considered as an exponential zeros of Y are obviously the zeros and poles, excitation with s = 0. Then, the impedances of respectively, of Z. R, L and C become R, 0 and ∞, respectively. Hence for DC excitation, an inductor acts as a short circuit and a capacitor behaves as an open Forced Response to an Exponential circuit. Excitation We shall illustrate the calculation of forced Forced Response to a Sinusoidal response to an exponential excitation by an Excitation example. Consider the circuit in Fig. 7.14. We are interested in ﬁnding the voltage v. The The forced response to an excitation of the form impedance faced by the current source is est is of the same form, except for multiplication by a constant. The forced response to dc is also a ðs þ 3Þ 0:5s 1 dc. If s = jx, then the forced response will be of ZðsÞ ¼ 2 þ s þ 3 þ 0:5s 1 the form Aeixx where A can be a complex ð7:61Þ 2ðs þ 3Þ quantity. Let A = |A|ejh; then, symbolically, we ¼ 2þ ðs þ 2Þðs þ 1Þ can write Forced Response to a Sinusoidal Excitation 71 Imaginary part ejxt ! j Ajejðxt þ hÞ ð7:64Þ Since ejxt is the superposition of cos xt and b j sin xt, and since we are considering only linear b circuits, it follows that a Imðejxt Þ ¼ sinxt ! Im½j Ajejðxt þ hÞ ð7:65Þ ¼ j Ajsinðxt þ hÞ θ Real part and a Reðejxt Þ ¼ cosxt ! Re½j Ajejðxt þ hÞ Fig. 7.15 Phasor representation ð7:66Þ ¼ j Ajcosðxt þ hÞ or This suggests a methodology for ﬁnding the Vm cosðxt þ hÞ ! jBjcosðxt þ /Þ ð7:70Þ forced response of a general circuit to sinusoidal excitation. Suppose, we have a voltage excitation Thus, one can divorce the time dependence of the form and work only in terms of A = |A|ejh and impe- pﬃﬃﬃ dances to ﬁnd B = |B|ej/. It is conventional to vðtÞ ¼ Vm cosðxt þ hÞ ¼ 2Vcosðxt þ hÞ; work in terms of rms rather than peak values, so ð7:67Þ that the given excitation will be represented by pﬃﬃﬃ a ¼ A= 2 instead of A and the response where V is the rms value of the voltage. We can pﬃﬃﬃ write obtained will then be in the form b ¼ B= 2 rather than B. Each of these quantities is called a vðtÞ ¼ Re½Vm ejh ejxt ¼ Re½Aejxt ; ð7:68Þ phasor and can be represented as a vector in the complex plane as shown in Fig. 7.15. where A is the complex quantity Vm eih. We can From the convention just discussed, obviously then ﬁnd the response of the circuit to the a current 10 cos xt will be represented by the pﬃﬃﬃ exponential Aejxt by using impedance concepts. phasor ð10= 2Þej0 ; i.e. cos xt has been taken to In this case, the impedance of R, L and C will be deﬁne the reference direction for angle. (It must ZR = R be mentioned that there is nothing sacred about ZL ¼ jxL ¼ jXL ð7:69Þ this convention. In fact, we shall, as illustrated later, sometime ﬁnd it convenient to use sinxt as the reference phasor for angles.) 1 ZC ¼ ¼ jXC As examples, the current 10 cos (xt + p/6) will jxC pﬃﬃﬃ be represented by the phasor ð10= 2Þejp=6 while XL = xL and Xc = –1/(xC) are called the the current 10 sin (xt + p/6) will be represented reactances of L and C, respectively. by the phasor p10ﬃﬃ2 e j p6 p2 ¼ p10ﬃﬃ2 ejp=3 because Suppose the response is Bejxt where B is another complex quantity |B|ej/. Then sinðxt þ p=6Þ ¼ cosðxt þ p=6 p=2Þ ð7:71Þ Aejxt ! Bejxt Phasors can be added or subtracted like vec- Re½Aejxt ! Re½Bejxt tors. Suppose, for example, we wish to ﬁnd v = pﬃﬃﬃ v1 − v2 where v1 ¼ 100 2 cos ð100ptp=6Þ 72 7 Circuit Analysis Without Transforms Im B Basic Elements and Their V–I Relationships for Sinusoidal Excitation 200 pﬃﬃﬃ Note that if a current i ¼ 2I cos xt flows pﬃﬃﬃ through a resistance, the voltage across it is 2IR p /3 0 cos xt. Thus, the current and the voltage phasors Rc are Iej0 and IRej0. Thus, the voltage and current -p / 6 are in phase. 100 A If the same current flows through an induc- pﬃﬃﬃ tance L, the voltage across it is 2 LIx sin pﬃﬃﬃ xt ¼ 2x LI cos ðxt þ p=2Þ. The phasor cor- responding to this is xL I ejp/2 = jxLI. The impedance of the inductor jxL is C therefore simply the ratio of voltage to current phasor. Also note that the voltage leads the D current by 90° or the current lags the voltage by 90°. Fig. 7.16 Phasor addition Similarly for a capacitor C, with sinusoidal pﬃﬃﬃ excitation, the impedance is jxC1 and the voltage and v2 ¼ 200 2 cosð100pt þ p=3Þ. The phasors 1 ¼ across it lags the current by 90°. corresponding to these two voltage are V In general, if a current phasor I flows that an 100e jp=6 and V2 ¼ 200e þ jp=3 , which are shown impedance Z(jx), then the voltage phasor across in Fig. 7.16 by the vectors OA and OB, respec- the latter is tively. The vector OC represents the negative of OB and OD and the vector sum of OA and OC ¼ IZ V ð7:75Þ represents the phasor corresponding to v1 − v2. One can ﬁnd the phasor corresponding to OD by which is often referred to as Ohm’s law for measurements or by geometrical formulas. It is, sinusoidal excitation. Phasors, like actual however, most convenient to compute this by voltages/currents, obey KCL and KVL, pro- converting both OA and OB to rectangular vided, of course, the circuit is linear, for which coordinates. Thus all currents and voltages will be of the same frequency. 1 ¼ 100 cos p=6 j100 sin p=6 V pﬃﬃﬃ ð7:72Þ ¼ 50 3 j50 2 ¼ 200 cos p=3 j200 sin p=3 V An Example of the Use of Phasors pﬃﬃﬃ ð7:73Þ ¼ 100 þ j100 3 and Impedances pﬃﬃﬃ Hence Let a current source iðtÞ ¼ 10 2 cos pﬃﬃﬃ pﬃﬃﬃ ð1000t þ p=4Þ be connected across a parallel 1 V V 2 ¼ ð50 3 100Þ jð50 þ 100 3Þ combination of R, L and C as shown in Fig. 7.17. ð7:74Þ It is required to determine the voltage v(t) across the combination and the currents through the whose magnitude M and phase h can be easily individual branches. found out. The quantity v1 − v2 will then be The admittance (=1/impedance) of the com- pﬃﬃﬃ M 2 cos ð100pt þ hÞ. bination is An Example of the Use of Phasors and Impedances 73 Thus 9 C = 0.2 mF + IR ¼ p10ﬃﬃ A > i(t) = L = 1 mH 2 pﬃﬃ v R=1Ω IL ¼ 10= 2 ¼ 10ﬃﬃ jp=2 p e jxL pﬃﬃﬃ 2 pﬃﬃﬃ pﬃﬃﬃ > ; – Ic ¼ ð10= 2ÞjxC ¼ j10 2 ¼ 10 2ejp=2 iR iC iL ð7:80Þ Fig. 7.17 A parallel RLC circuit with sinusoidal These various phasors are shown in Fig. 7.18. excitation In the time domain, the expressions would be 9 Y ¼ R1 þ jxL 1 þ jxC = 9 vðtÞ ¼ 10 cos 1000t >> ¼ 1 j 100010 1 3 þ j 1000 2 10 3 = ; iR ðtÞ ¼ 10 cos 1000t ¼ 1þj ð7:81Þ iL ðtÞ ¼ 10 sin 1000t > > ; ð7:76Þ ic ðtÞ ¼ 20 sin 1000t Hence, the impedance is 1 1j Back to Complete Response Z¼ ¼ ð7:77Þ 1þj 2 We now return to the complete response of a The current phasor is given circuit to a given excitation. Consider ﬁrst a simple RL circuit to which a sinusoidal voltage 10 source Vm cos xt is connected in series at t = 0, I ¼ 10ejp=4 ¼ p ﬃﬃﬃ ð1 þ jÞ ð7:78Þ 2 as shown in Fig. 7.19. The differential equation is Hence, the voltage phasor will be di L þ Ri ¼ Vm cos xt ð7:82Þ ¼ IZ ¼ 10 V 10 pﬃﬃﬃ ð1 jÞð1 þ jÞ ¼ pﬃﬃﬃ ð7:79Þ dt 2 2 2 whose complete solution is iðtÞ ¼ ic ðtÞ þ ip ðtÞ; ð7:83Þ Im IC = 10 2e j p /2 t=0 R I = 10e j p /2 + i(t) Vm cos ωt Re – V = 10 / 2 I R = 10 / 2 L 10 IL = e - j p /2 2 Fig. 7.18 Phasor diagram for the circuit in Fig. 7.17 Fig. 7.19 A series RL circuit with sinusoidal excitation 74 7 Circuit Analysis Without Transforms where To obtain A, one appeals to the initial condi- tion. Let i(0) = 0; then from Eq. 7.91, ic ðtÞ ¼ AeRt=L ð7:84Þ Vm 1 xL A ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ cos tan ð7:92Þ and R2 þ x2 L2 R ip ðtÞ ¼ Im cosðxt þ hÞ ð7:85Þ As is evident from Fig. 7.20. The particular solution parameters Im and h 1 xL R cos tan ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð7:93Þ can be easily obtained from phasor analysis. The R R þ x2 L2 2 voltage phasor is Vm jo Hence, ﬁnally ¼p V ﬃﬃﬃ e ð7:86Þ 2 Vm 1 xL R Rt=L iðtÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ cos xt tan pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ e ð7:94Þ R2 þ x2 L2 R R2 þ x2 L2 The impedance is The ﬁrst term in square brackets is the forced Z ¼ R þ jxL ð7:87Þ response, while the second term is the natural Hence, the current phasor is response. Note that the natural response form depends on R and L only, but the actual value is Vmﬃﬃ p affected by the forcing function. Also note that I ¼ Vm 1 2 ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ej tan ðxL=RÞ the natural response here is also the transient part R þ jxL 2ðR2 þ x2 L2 Þ while the forced response is also the steady-state ð7:88Þ response. In general, if the forcing function is not Hence sinusoidal, one ﬁnds the particular solution by assuming it to be of the same form as the forcing Vm Im ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð7:89Þ function, and substituting it in the complete dif- ðR þ x2 L2 Þ 2 ferential equation to determine the unknown constant. and xL h ¼ tan1 ð7:90Þ R R2 Thus + w2 L2 Vm ωL iðtÞ ¼ AeRt=L þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ R2 þ x2 L2 ð7:91Þ θ xL cos xt tan1 R R Fig. 7.20 Computing cos tan1 xL R Back to Complete Response 75 S t=0 t=0 R R i(t) V i(t) + L v(t) =Vm sin t + – vC (t) C Fig. 7.21 An RL circuit excited by a DC voltage source – Fig. 7.23 A series RC circuit with sinusoidal excitation We now consider several other examples of complete response. and ﬁnally V Step Response of an RL Circuit iðtÞ ¼ ð1 eRt=L Þ ð7:98Þ R Consider the circuit in Fig. 7.21, where S was A plot of i is shown in Fig. 7.22. open for a long time and switched on at t = 0. Hence, there is no initial current in L, i.e. i(0+) = i(0−) = 0. ( continuity of current in an Sinusoidal Response of a Series RC inductor). From the differential equation Circuit di L þ Ri ¼ V; t [ 0 þ ð7:95Þ Consider the situation shown in Fig. 7.23, where dt vc(0) = 0. The impedance of the series RC circuit it follows that for est excitation is iðtÞ ¼ ic ðtÞ þ ip ðtÞ 1 ZðsÞ ¼ R þ ð7:99Þ ð7:96Þ sC V ¼ AeRt=L þ R and since the excitation is a voltage source, the natural frequency will be the zero of Z(s), because the natural frequency is the zero of Z occurring at s = −1/(CR). Also, the forced current (s) = R + sL and the forced response is a con- response can be found from the voltage phasor stant whose value is determined by Eq. 7.95 itself. Note that i = constant means that di/dt = 0 pﬃﬃﬃ ¼ ðVm = 2Þej0 V ð7:100Þ so that the particular solution is V/R. To ﬁnd A, take help of the fact that i(0+) = 0; thus (where we have taken sin xt as the reference phasor, instead of cos xt; this does not change A ¼ V=R ð7:97Þ the result) and the impedance for ejxt excitation; 1 ZðjxÞ ¼ R þ i jxC rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ V 1 1 ¼ R2 þ 2 2 \ tan1 R x C xCR ð7:101Þ t 1 Note that instead of writing e tan 1=ðxCRÞ , we Fig. 7.22 Step response of an RL circuit have indicated ∠tan−1 1/(xCR); these are 76 7 Circuit Analysis Without Transforms interchangeable notations. Hence, the current phasor will be 1 + w2 I ¼ V ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Vm 1 l C2 ﬃ \ tan1 R2 ZðjxÞ 2 R2 þ 1 xCR x2 C 2 θ ð7:102Þ ωCR Thus, the forced response is Fig. 7.24 Computing sin tan1 xCR 1 Vm 1 ip ðtÞ ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sin xt þ tan1 R2 þ x21C2 xCR Response of an RC Circuit ð7:103Þ to an Exponential Excitation and the complete response is In the circuit shown in Fig. 7.25, let vc(0) = V and the current source be exponentially Vm 1 iðtÞ ¼ Aet=ðRCÞ þ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sin xt þ tan1 decaying, i.e. R2 þ x2 C2 1 xCR ð7:104Þ is ðtÞ ¼ Ieat ; ð7:108Þ To evaluate A, note that vc(0−) = vc(0+) has been assumed to be zero so that i(0+) = v(0)/ R = 0. Hence t=0 Vm 1 + A ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sin tan1 ð7:105Þ R xCR iS (t) C v(t) R2 þ x21C2 – Vm 1 ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð7:106Þ R þ x2 C 2 2 1 1 þ x 2 C 2 R2 Fig. 7.25 Current excited parallel RC circuit from Fig. 7.24. Hence, ﬁnally Vm 1 et=ðRCÞ iðtÞ ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sin xt þ tan1 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð7:107Þ R2 þ x21C2 xCR 1 þ x2 C2 R2 Response of an RC Circuit to an Exponential Excitation 77 where a is real, positive and not equal to 1/(CR). This gives the same value of K as in The impedance to est excitation is Eq. 7.113. Hence, we have the complete response of the circuit as R sC 1 1 1 ZðsÞ ¼ ¼ ð7:109Þ I 1 R þ sC C s þ CR 1 1 vðtÞ ¼ AetðCRÞ þ eat ð7:115Þ C CR 1 a and therefore the natural frequency is given by To ﬁnd A, put v(0) = V; this gives 1 s¼ ð7:110Þ I CR V ¼ Aþ 1 ð7:116Þ C CR a the pole of Z(s). The forced response will be of the form vp(t) = Ke−at and referring to the dif- or, ferential equation I dv v A¼V 1 ð7:117Þ C þ ¼ is ðtÞ ð7:111Þ C CR a dt R This can be substituted in Eq. 7.115 to get the we get complete solution. The result is K CKa þ ¼I ð7:112Þ I R vðtÞ ¼ 1 ðeat eCR Þ þ VeCR 1 1 C CR a or ð7:118Þ I I 1 K¼1 ¼ 1 ð7:113Þ R Ca C CR a We have written v(t) in this form to illustrate the case when a ¼ CR 1 . In general, by series expansion, we can write " # I n o CR 1 et ð Þ1 CRa 1 1 vðtÞ ¼ e Vþ C CR a " ( 1 2 )# ð7:119Þ CR I 1 t2 CR a Vþ 1 t 1 ¼e a þ þ C CR a CR 2! Note that an easier method for calculating K would have been to refer to the fact that Z When a ! CR 1 , Eq. 7.119 gives (s) = v/is(t) for est excitation so that for Ie−at at I excitation, we would have vðtÞ ¼ e Vþ t ð7:120Þ C I 1 vp ðtÞ ¼ Ieat ZðaÞ ¼ eat ð7:114Þ This result can also be derived from the fact C CR 1 a that if the characteristic root of a ﬁrst-order 78 7 Circuit Analysis Without Transforms equation (s = −a) is also contained in the forcing equation, it is clear that the forced response is exponential function, then the particular solution zero. Then, the total response is is of the form Kte−at. Substituting this in Eq. 7.111 gives iðtÞ ¼ A1 es1 t þ A2 es2 t ; ð7:126Þ 1 where A1 and A2 are to be found from the initial CK½tðaÞeat þ eat þ Kteat ¼ Ieat R conditions on L and C. Depending on the value ð7:121Þ of R, we can have three cases. For real and distinct s1,2 (overdamped case), Eq. 7.126 is the in which the ﬁrst and the third terms on the general form. For s1 = s2 (critically damped left-hand side cancel, giving case), the solution becomes I iðtÞ ¼ ðA1 þ A2 tÞes1 t ð7:127Þ K¼ ð7:122Þ C For complex s1,2 = −a ± jx (underdamped Hence, v(t) is of the form case), the solution can be put in the form I iðtÞ ¼ Aeat sinðxt þ hÞ ð7:128Þ vðtÞ ¼ A þ t eat ð7:123Þ C In each case, there are two constants to be Putting t = 0 in this gives determined from the initial conditions. Let L and C be initially relaxed. Then i(0) = 0 and vc(0) = 0. A¼V ð7:124Þ The ﬁrst condition gives, for the overdamped case, A1 þ A 2 ¼ 0 ð7:128aÞ Step Response of an RLC Circuit while the second gives In the second-order circuit of Fig. 7.26, as we have seen, the natural response will be of the vL ð0Þ ¼ Við0ÞRvc ð0Þ ¼ V; ð7:128bÞ form A1 es1 t þ A2 es2 t ; where i.e. qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ) R R2 di s1;2 ¼ 2L 4L 2 LC 1 ð7:125Þ L ¼V ð7:128cÞ dt t ¼ 0 D a jx or while the forced response will be a constant. From the circuit as well as from this differential V s 1 A2 þ S 2 A2 ¼ ð7:129Þ L Hence, A1 and A2 can be determined. The S other two cases can be similarly dealt with. t=0 R i(t) V + Sinusoidal Response of an RLC VL L Circuit – + C vc – Consider the circuit in Fig. 7.27, where the switch has been in the closed position for a long time so Fig. 7.26 An RLC circuit with step excitation that v(0) = 0 and i(0) = 0 (why don’t we write v Sinusoidal Response of an RLC Circuit 79 The complete solution is, therefore, R = 1Ω pﬃﬃﬃ t=2 3 + vðtÞ ¼ Ae cos t þ h þ sinðt p=4Þ iS (t) = 2 C = 1F v ð7:134Þ 1 t=0 sin t – 2 L = 1H To determine A and h, take help of v(0) = 0 and i(0) = 0. The ﬁrst condition gives Fig. 7.27 RLC circuit excited by a sinusoidal current 0 ¼ A cos h sin p=4 ð7:135Þ (0) = i(0) = 0?) Since the excitation is a current or generator, the poles of Z(s) or the zeros of Y(s) will 1 determine the natural frequencies. We have A cos h ¼ pﬃﬃﬃ ð7:136Þ 2 1 1 s2 þ s þ 1 YðsÞ ¼ sC þ ¼ sþ ¼ The second condition says that R þ sL sþ1 sþ1 ð7:130Þ dv C ¼ is ð0Þ ið0Þ ð7:137Þ The natural frequencies are, therefore, the dt t¼0 roots of Substituting for v(t) from Eq. 7.134 and s þ s þ 1 ¼ 0; 2 ð7:130aÞ simplifying, we get pﬃﬃﬃ i.e. 2 pﬃﬃﬃ A ¼ pﬃﬃﬃ ð7:138Þ 1 3 3 sin h þ cos h s1;2 ¼ j ð7:131Þ 2 2 Combining Eq. 7.136 and Eq. 7.138, we get –t/2 The natural response is thus of the form Ae pﬃﬃ pﬃﬃﬃ cos 3 t þ h . The forced response phasor, V, is 1 2 2 pﬃﬃﬃ ¼ pﬃﬃﬃ ð7:139Þ given by the product of the current phasor cor- 2 cos h 3 sin h þ cos h responding to i(t) and the impedance at s = j1. Hence or pﬃﬃﬃ 3 sin h ¼ cos h ð7:140Þ ¼ 1 \0 1 þ j1 V 2 j1 ð7:132Þ or 1 1 ¼ \0 ð1 j1Þ ¼ pﬃﬃﬃ \ p=4; 2 2 p h¼ ð7:141Þ 6 where we have again taken sin xt as the refer- ence phasor, without any loss of accuracy. Hence, from Eq. 7.136, Hence, the forced response will be rﬃﬃﬃ 2 vP ¼ sinðtp=4Þ ð7:133Þ A¼ ð7:142Þ 3 80 7 Circuit Analysis Without Transforms Finally, therefore, the solution for v is v(t) rﬃﬃﬃ pﬃﬃﬃ 2 t=2 3 p V vðtÞ ¼ e cos tþ þ sinðt p=4Þ 3 2 6 ð7:143Þ The ﬁrst term is the transient response while 0 T t the second term represents the steady-state response. Fig. 7.29 Pulse excitation of the circuit in Fig. 7.28: A gate Pulse Response of an RC Circuit iðtÞ ¼ i1 ðtÞi2 ðtÞ Vh 1 tT i ð7:147Þ For the circuit in Fig. 7.28, where v(t) is a pulse ¼ e RC uðtÞ e RC uðt TÞ R as shown in Fig. 7.29, we can write A sketch of i(t) is shown in Fig. 7.30. vðtÞ ¼ V ½uðtÞuðtT Þ; ð7:144Þ What about the voltage across the capacitor? It can be found by integrating i(t) from 0 to t. where u(t) is the unit step function. We know the Physically, it is not difﬁcult to argue that C charges response to Vu(t) to be according to the relation V 1 i1 ðtÞ ¼ e RC uðtÞ; ð7:145Þ vc ðtÞ ¼ V½1et=ðCRÞ ð7:148Þ R for 0 < t T. At t = T, the voltage source where the multiplication by u(t) indicates that becomes zero, and current direction reverses. i1(t) = 0 for t < 0. The response to the delayed Hence, C discharges through R from the value V step Vu(t − T) will be, from the principle of time [1 − e−T/(CR)] to zero exponentially. Hence, invariance (which states that if the excitation is vc(t) will vary with time as shown in Fig. 7.31. delayed, so will the response be), The expression for vc(t) is V tT i2 ðtÞ ¼ e RC uðt TÞ; ð7:146Þ R Current where multiplication by u(t − T) indicates that i2(t) = 0 for t < T. Hence, by superposition, the complete response will be i(t) i2 (t) V R i1 (t) 0 t R i(t ) T + v(t) – V C - -i2 (t) R Fig. 7.28 An RC circuit excited by the pulse of Fig. 7.29 Fig. 7.30 Pulse response of the circuit in Fig. 7.28 Pulse Response of an RC Circuit 81 vC (t) Z0 þ T - V (1 - e CR ) dðtÞdt ¼ 1 ð7:150Þ V 0 It follows that d(t − a) is a unit impulse occurring at t = a and that Za þ dðt aÞdt ¼ 1 ð7:151Þ T t a Fig. 7.31 Plot of vc(t) for the circuit in Fig. 7.28 If the range of integral does not include the value of t at which the impulse occurs, then the integral will be zero, e.g. 8 < V 1 eCR1 ; 0 t T Z1 Z0 vc ðtÞ ¼ ð7:149Þ : V 1 eCRT etT CR t T dðtÞdt ¼ 0 ¼ dðtÞdt ð7:152Þ 0þ 1 Also Impulse Response f ðtÞdðt aÞ ¼ f ðaÞdðt aÞ ð7:153Þ Consider the pulse shown in Fig. 7.32 whose area is unity. Let T be decreased to T/2, and the and height increased to 2/T so that the area is still Za þ unity. Let T ! 0 and the height ! ∞ in such f ðtÞdðt aÞdt ¼ f ðaÞ ð7:154Þ manner that the area is still unity. This limiting a condition of the pulse, whose duration is zero and the height is inﬁnite such that the area under Impulse is, of course, a hypothetical function, it is unity, is called a unit impulse. A unit but is a useful concept in analysing circuits and impulse function is denoted by d(t), and since systems. Let an impulse Q(t) be applied to an RC inﬁnite height is not a determinate quantity, we circuit as shown in Fig. 7.33 with v(0−) = 0. deﬁne d(t) by the integral Then, the inﬁnite current Q(t) flows through C and establishes a voltage Z0 þ þ 1 Q vð0 Þ ¼ QdðtÞ dt ¼ ð7:155Þ C C 0 f(t) At t 0+, Q(t) acts as an open circuit; hence 1/T v(t) decays with time according to Q t=ðCRÞ vðtÞ ¼ e ð7:156Þ C T t Note that here v(0−) 6¼ v(0+), this is so because an inﬁnite amount of current flows Fig. 7.32 A pulse which becomes a unit impulse when through C for the short duration 0− to 0+. T!0 82 7 Circuit Analysis Without Transforms i i1 + R R Q (t) C v R + – v – ~ i2 C L = CR 2 Fig. 7.33 An RC circuit excited by an impulsive current source Fig. 7.34 Circuit for P.1 Similarly when a voltage excitation /d(t) is P:2. Find the transfer impedance ZT ðsÞ ¼ VðsÞ IðsÞ for applied across a series RL combination with no the circuit of P.1 and sketch its poles and initial current in L, a current i(0+) = //L is zeros. established. Here also i(0−) 6¼ i(0+) in an P:3. Write the differential equation governing inductor. But for these exceptions, which are, of the circuit in Fig. 7.33 in the text. Clue: course, hypothetical ones, the inductor current Chap. 8. and capacitor voltage cannot change P:4. Determine the transfer function ZT ðsÞ ¼ VðsÞ Is ðsÞ instantaneously. for the circuit in Fig. 7.27 of the text and sketch is poles and zeros. Problems P:5. Working in the frequency domain, ﬁnd the unit step response of the circuit in Fig. 7.26 P:1. Write the differential equation for the circuit in the text. Find the inverse transform and shown below for i, i1 as well as i2 verify that the response in the time domain, (Fig. 7.34). as determine in the text, is identical. Transient Response of RLC Networks Revisited 8 As compared to the conventional approach of Aest is shown to be suitable for the overdamping trial solutions for solving the differential and underdamping cases while for critical equation governing the transient response of damping, the trial solution assumes the form RLC networks, we present here a different (A1 + A2t)est because of repeated roots of the approach which is totally analytical. We also characteristic equation. To a beginner, it is not show that the three cases of damping, viz. clear why one has to use trial solutions, and why overdamping, critical damping and under- they have to be of these speciﬁc forms, but since damping, can be dealt with in a uniﬁed they work, he (or she) accepts the solutions manner from the general solution. Won’t faithfully (This illustrates the principle of the end you appreciate my innovations? Please do justifying the means!). This chapter presents a and encourage me. different approach to the solution of the differ- ential equation, which is totally analytical. Also, it is shown that the three cases, particularly, the Keywords critical damping one, need not be treated sepa- Transient response of RLC network rately and that a uniﬁed treatment is possible Overdamped case Underdamped case from the general solution of the differential Critically damped case equation. Introduction Example Circuit and the Differential Equation In dealing with transients in RLC networks, most textbooks on circuit theory [1, 2] derive the For a clear exposition and understanding of the governing second-order differential equation and analytical approach, we shall consider the simple then treat the three different cases of damping, series RLC circuit shown in Fig. 8.1, where the viz. overdamping, critical damping and under- switch is closed at t = 0 with i (0−) = 0 and vC damping separately. A trial solution of the form (0−) = V. KVL gives, for t > 0. Zt di 1 Source: S. C. Dutta Roy, “Transient Response of RLC Ri þ L þ idt V ¼ 0: ð8:1Þ dt C Networks Revisited,” IETE Journal of Education, vol. 0 44, pp. 207–211, October–December 2003. © Springer Nature Singapore Pte Ltd. 2018 83 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_8 84 8 Transient Response of RLC Networks Revisited t=0 R y0 þ ðs þ 2aÞy þ ðs2 þ 2as þ x20 Þi ¼ 0 ð8:7Þ Since s is our choice, let s2 þ 2as þ x20 ¼ 0 ð8:8Þ + C i(t) – L Note that this is precisely the result we would have obtained if we tried the solution i = Aest on Eq. 8.2, and as is well known, Eq. 8.8 is called the characteristic equation of the system, having the two roots Fig. 8.1 The simple RLC circuit s1;2 ¼ a b; ð8:9Þ Differentiating Eq. 8.1 and dividing both sides where by L, we get qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ b¼ a2 x20 ð8:10Þ i00 þ 2ai0 þ x20 i ¼ 0; ð8:2Þ Combining Eqs. 8.7 and 8.8, we have where prime (′) denotes differentiation with respect to t, y0 þ ðs þ 2aÞy ¼ 0; ð8:11Þ 2a ¼ R=L and x20 ¼ 1=ðLCÞ ð8:3Þ which can be written in the form dy=y ¼ ðs þ 2aÞdt ð8:12Þ Analytical Solution of the Differential Integrating Eq. 9.12 and simplifying, we get Equation y ¼ k1 eðs þ 2aÞt ; ð8:13Þ As already mentioned, the conventional approach to solving Eq. 8.2 is to try the solution i = Aest, where k1 is a constant, as are all the k1’s in what with little or, at the best, heuristic justiﬁcation. follows. Combining Eqs. 8.13 with 8.4, we get We take a different approach here, by introducing the following ﬁrst-order equation in i: the new variable i0 si ¼ k1 eðs þ 2aÞt ð8:14Þ y ¼ i0 si; ð8:4Þ Unlike Eq. 8.11, however, Eq. 8.14 is not a where s is a constant to be chosen shortly. From homogeneous equation. That should not be a Eq. 8.4, we have cause for worry because we can use the i0 ¼ y þ si ð8:5Þ well-known method of integrating factor, which in this case is e–st. Multiplying both sides of so that Eq. 8.14 by this factor, the result can be put in the form i00 ¼ y0 þ si0 ¼ y0 þ sy þ s2 i ð8:6Þ ðiest Þ0 ¼ k1 e2ðs þ aÞt ð8:15Þ Substituting Eqs. 8.5 and 8.6 in Eq. 8.2 and simplifying, we get Integrating Eq. 8.15 and simplifying, we get Analytical Solution of the Differential Equation 85 i ¼ k2 est þ k2 eðs þ 2aÞt ð8:16Þ Also, referring to Eq. 8.1 and putting t = 0+, we get Now s has two possible values given by Eq. 8.9. If we choose s = s1 = −a + b then −(s + 2a) = −a – b = s2. Similarly, if we choose ðdi=dtÞ0 þ ¼ V=L ð8:21Þ s = s2, then −(s + 2a) = −a + b = s1. Thus in Combining this with Eq. 8.20 and simplify- either case, our solution is of the form ing, we get i ¼ k4 es1 t þ k5 es2 t ð8:17Þ k4 ¼ V=½Lðs1 s2 Þ ð8:22Þ From Eqs. 8.20, 8.22 and 8.9, the general Evaluating the Constants solution can be written in the form Now we go back to the example circuit. To V at bt i¼ e ðe ebt Þ ð8:23Þ evaluate the constants in Eq. 8.17, we invoke the 2bL two initial conditions, viz. i(0−) = 0 and vC(0−) = V. Because of continuity of inductor This single equation, as will be shown here, is currents and capacitor voltages, we have adequate for considering all the cases of damping. ið0 þ Þ ¼ ið0 Þ ¼ 0 and vC ð0 þ Þ ¼ vC ð0 Þ ¼ V ð8:18Þ Overdamped Case From Eq. 8.17 and the ﬁrst condition in Eq. 8.18, we get The RLC circuit is overdamped if a > x0, i.e. from Eq. 8.3, R > 2√(L/C). Consequently, b is k4 þ k5 ¼ 0 ð8:19Þ real and we can write Eq. 8.23 in the form so that V at i¼ e sinh bt ð8:24Þ bL i ¼ k4 ðes1 t es2 t Þ ð8:20Þ Fig. 8.2 Current response for various cases of damping 86 8 Transient Response of RLC Networks Revisited This gives i = 0 at t = 0, as it should; the while its zero crossings occur at intervals of same holds at t = ∞ also because b< a. Hence, p/xd. The maxima and minima occur at times there must be a maximum at some value of t, satisfying the equation di/dt = 0. Carrying out say t. Differentiating Eq. 8.24 with respect to the required algebra, we get maxima at t and putting the result to zero gives, after simpliﬁcation, t2n þ 1 ¼ ð1=xd Þ½tan1 ðxd =aÞ þ 2np; n ¼ 0; 1; 2; . . . ð8:31Þ tb ¼ ð1=bÞ tanh1 ðb=aÞ ð8:25Þ and minima at The maximum value of i (= i) is obtained by combining Eqs. 8.24 and 8.25; using the identity t2n ¼ ðl=xd Þ½tan1 ðxd =aÞ þ ð2nlÞp; ð8:32Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ n ¼ 1; 2; 3; . . . sinh h ¼ 1= 1 þ coth2 h ð8:26Þ Combining Eqs. 8.30, 8.31 and the identity to simplify the result, we get pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 sin h ¼ 1= 1 þ cot2 h; ð8:33Þ ib ¼ V C=Leða=bÞ tanh ðb=aÞ ð8:27Þ and simplifying, we get the value of the ﬁrst The general nature of variation of the current maximum current as is shown in Fig. 8.2. Using the series pﬃﬃﬃﬃﬃﬃﬃﬃﬃ ða=x Þ tan1 ðx =aÞ imin;1 ¼ V C=Le d d ð8:34Þ tanh1 h ¼ h þ ðh3 =3Þ þ ðh5 =5Þ þ . . .to 1; jhj\1; ð8:28Þ Similarly, the value of the ﬁrst minimum current can be found as It is easily shown that as b increases, tb pﬃﬃﬃﬃﬃﬃﬃﬃﬃ ða=x Þ½tan1 ðx =aÞ þ p increases and ib decreases. The time tb is the imin;1 ¼ V C=Le d d ð8:35Þ smallest and the current ib is the highest when b = 0, but we shall talk about this limiting situ- All successive maxima (as well as minima) ation later. differ from each other in magnitude by the factor e2ap=xd : Underdamped Case Critically Damped Case The RLC circuit is underdamped if a < x0, i.e. R < 2√(L/C). In this case, b is imaginary. Let For this case, b = 0, i.e. a = x0 or R = 2√(L/C), qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ and as already mentioned, in the conventional b ¼ j x20 a2 ¼ jxd ð8:29Þ approach, one notes that here s1 = s2 = −a, and to obtain two independent solutions, one tries out Putting this value in Eq. 8.23 and simplifying, a solution of the form (A1 + A2t)est. One we get approach [3] for justifying this trial solution is to V at assume a solution of the form f(t) est; substitute it i¼ e sin xd t ð8:30Þ in Eq. 8.2; take help of the fact that s + a = 0; xd L and hence obtain f″(t) = 0, so that f(t) has to be of Obviously, the response is oscillatory but the form A1 + A2t. Here, we obtain the solution damped, as shown in Fig. 8.2. The envelope of directly from the general solution Eq. 8.23 as the the damped sine wave deceases exponentially, limiting case of b ! 0. Thus Critically Damped Case 87 V at ebt ebt Concluding Comments i¼ e lim : ð8:36Þ 2L b!0 b A totally analytical approach is given here for Using L’Hospital’s rule, this becomes dealing with transients in second-order networks and systems, which does not require trial solu- V at tions. It deserves to be mentioned here that our i¼ te ð8:37Þ method is an improvement over that using the L operator concept, as given in some books [4]. In A plot of Eq. 8.37 is shown in Fig. 8.2. It is the latter, one uses the operator D for d/dt and D2 easily shown that the maximum occurs at the for d2/dt2 so that Eq. 8.2 can be written as time t = t0 where ðD2 þ 2aD þ x20 Þi ¼ 0 ð8:43Þ t0 ¼ 1=a; ð8:38Þ One then treats (D2 + 2aD + x20) as an alge- the maximum value being braic expression, ﬁnds its roots s1 and s2 (same as those given by Eq. 8.9) and rewrites Eq. 8.43 as pﬃﬃﬃﬃﬃﬃﬃﬃﬃ i0 ¼ ðV=eÞ C=L; ð8:39Þ ðDs1 ÞðDs2 Þi ¼ 0 ð8:44Þ where use has been made of the fact that The conceptual difﬁculty of a student arises on R = 2√(L/C). Using the inﬁnite series Eq. 8.28 two counts—ﬁrst, in accepting D2 for d2/dt2 and and the following: second, in treating (D2 + 2aD + x20) as an alge- braic expression in D, i.e. in treating D as a sinh h ¼ h þ ðh3 =3!Þ þ ðh5 =5!Þ þ . . .: to 1; jhj\1; variable instead of an operator. Of course, here also, as in the trial solution method, the end can ð8:40Þ be used to justify the means as follows: it is easily shown that Eqs. 8.38 and 8.39 are the limiting values of Eqs. 8.25 and 8.27, respec- ðD s1 ÞðD s2 Þi ¼ ðD s1 Þ ddti s2 i tively, for b ! 0. 2 ¼ ddt2i s2 ddti s1 ddti þ s1 s2 i ¼ i00 þ 2ai0 þ x20 i; The critically damped case could also have been treated ab initio. The procedure is the same ð8:45Þ up to Eq. 8.15, at which stage we take account of which is the same as the left-hand side of the fact that s1 = s2 = −a. Equation 8.15 then Eq. 8.2. The method of solution is similar to ours becomes in that one solves two ﬁrst-order differential equations, viz. the homogeneous equation ðiest Þ0 ¼ k1 ð8:41Þ (D − s1)y = 0 ﬁrst, and then the nonhomoge- neous equation (D − s2)i = y. Here also, no Integrating Eq. 8.41 gives special care is needed to deal with the critically damped case in which s1 = s2 = −a. i ¼ ðk1 t þ k2 Þest ð8:42Þ k1 and k2 can now be evaluated from the ini- Problems tial conditions to arrive at precisely the same result as Eq. 8.37. P:1. Arrange an RLC circuit to give a band-stop response in as many ways as you can. 88 8 Transient Response of RLC Networks Revisited L1 L3 C2 P:4. Draw as many third-order circuits as pos- sible and comment on the nature of the + frequency response. + v/H L2 C1 C3 R v o /H P:5. Take any one circuit of P.4 and determine – – its poles and zeros. Fig. 8.3 For P.3 References P:2. Find at least two circuits using a parallel LC 1. W.H Hayt, Jr, J. Kemmerly, Engineering Circuit and a series LC circuit to obtain a band-stop Analysis (McGraw-Hill, 1978) response. Find the poles and zeroes of the 2. M.E Van Valkenburg, Network Analysis (Prentice Hall transfer function. of India, 1974) P:3. Obtain the differential equation governing 3. A.B. Carlson, Circuits (John Wiley, 1996) 4. D.K. Cheng, Analysis of Linear Systems (Addison the circuit Fig. 8.3. Wesley, 1959) Appearances Can Be Deceptive: A Circuit Paradox 9 How can a paradox be deceptive? What – gmRc, where gm = gm3 = gm4. Since Q1 as well appears to be an obvious conclusion may not as Q2 basically act as emitter followers, it is be correct after all! This is illustrated with the logical to assume that v3 ≅ v1 and v4 ≅ v2 so that help of a differential ampliﬁer circuit, whose the overall gain of the circuit is the same as Ad. gain is actually half of what it appears to be. However convincing this logic may be, things This paradox is indeed deceptive. See for turn out to be quite different in practice. In fact, yourself and decide if you wish to agree or the differential gain of the overall ampliﬁer is not. half of Ad! Let us see how. Keywords AC Analysis Paradox DC analysis First, recall that the gain of the simple follower (all gains are referred to the mid-band situation, of course) in Fig. 9.2, ignoring the effects of rx and r0 in the hybrid-p equivalent circuit, is [1–3]. v0 ðb þ 1ÞRE ¼ ð9:1Þ vi rp þ ðb þ 1ÞRE The Illusion ﬃ l, if rp ðb þ 1ÞRE ð9:2Þ Consider the differential ampliﬁer circuit shown in Fig. 9.1, where the symbols v0–v4 are used for The emitter follower Q1 has a load of rp3 in small signal ac voltages. As has been proved in the differential mode (in this mode the node standard textbooks [1–3], the gain of the internal E acts as virtual ac ground); hence its gain is differential ampliﬁer comprising of matched transistors Q3 and Q4 is Ad = v0/(v3 – v4) ≅ v3 ðb1 þ 1Þrp3 ¼ ; ð9:3Þ v1 rp1 þ ðb1 þ 1Þrp3 where the subscripts on the parameters refer to Source: S. C. Dutta Roy, “Appearances can be Deceptive: A Circuit Paradox,” Students’ Journal of the corresponding transistors. Now the IETE, vol. 37, pp. 79–81, July–September 1996. © Springer Nature Singapore Pte Ltd. 2018 89 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_9 90 9 Appearances Can Be Deceptive: A Circuit Paradox Fig. 9.1 The circuit under +VCC consideration RC RC + – Q2 + + Q1 v0 v2 v1 – – Q3 Q4 + + v3 v4 – – E IEE R EE b1 b V T b V T VT rp1 rp1 ¼ ¼ 1 ¼ 1 ¼ ð9:4Þ ¼ ; ð9:6Þ gm1 IC1 b1B1 IB1 b1 þ 1 and, similarly, where VT stands for the thermal voltage (kT/q ≅ 25 mV at room temperature) and capital I with V T VT VT capital subscript stands for DC current. Substi- rp3 ¼ ¼ ¼ ð9:5Þ IB3 IE1 ðb1 þ 1ÞIB1 tuting Eq. 9.6 in Eq. 9.3 gives v3 1 ¼ ð9:7Þ v1 2 Fig. 9.2 AC equiva- lent of a simple emitter Hence, each of the emitter followers Q1 and follower circuit Q2 has a gain of 1/2 instead of 1, and the actual differential gain of the overall circuit is Q + vi v0 v0 gm R c ¼ ¼ ; ð9:8Þ – + v1 v2 2ðv3 v4 Þ 2 v0 where gm = gm3 = gm4, as mentioned earlier. RE – DC Analysis 91 DC Analysis Thus, The result obtained above can be corroborated by b3 ðb1 þ 1ÞIC1 b2 VBE3 VBE4 ¼ VT ln analysing the DC characteristics of the circuit. b1 b4 ðb2 þ 1Þ=IC2 We wish to establish a relationship between ð9:15Þ V0 D VC3 VC4 ¼ ðVCC IC3 RC ÞðVCC IC4 RC Þ If the transistors are matched, as is the case with IC fabrication, then Eq. 9.15 becomes ¼ ðIC4 IC3 ÞRC ð9:9Þ IC1 VBE3 VBE4 ¼ VT ln ¼ VBE1 VBE2 IC2 and ð9:16Þ Vi D VB1 VB2 ð9:10Þ From Eqs. 9.12 and 9.16, we get Recall the basic current–voltage relationship IC3 of an active transistor Vi ¼ 2ðVBE3 VBE4 Þ ¼ 2VT ln ð9:17Þ IC4 Ic ﬃ Is exp ðVBE =VT Þ ð9:11Þ Solving for IC4/IC3 from Eq. 9.17 gives Now, IC4 Vi ¼ exp ð9:18Þ IC3 2VT Vi ¼VBE1 þ VBE3 þ VE ðVBE2 þ VBE4 þ VE Þ ¼ðVBE1 VBE2 Þ þ ðVBE3 VBE4 Þ so that ð9:12Þ Vi IC4 IC3 exp 2V T 1 Also, assuming IS1 = IS2 = IS3 = IS4 = IS, we ¼ ð9:19Þ IC4 þ IC3 Vi exp 2V þ1 have T IC3 b IB3 Also, note that IC4 + IC3 = IEE = constant. VBE3 ¼ VT ln ¼ VT ln ¼ 3 IS IS Hence from Eqs. 9.19 and 9.9, we get Vi b IE1 b ðb þ 1ÞIC1 V0 ¼ IEE R; tanh ð9:20Þ ¼ VT ln 3 ¼ VT ln ¼ 3 1 ð9:13Þ 4VT IS b1 IS The differential gain, evaluated at V1 = 0, is Similarly, b ðb þ 1ÞIC2 dV0 2 Vi 1 ¼ IEE RC sech VBE4 ¼ VT ln 4 2 b2 IS ð9:14Þ dVi Vi ¼0 4VT 4VT Vi ¼0 92 9 Appearances Can Be Deceptive: A Circuit Paradox IEE RC P:4. What happens when b1, b2 & b3 all ! ∞ in ¼ ð9:21Þ Eq. 9.15 of text? Comment on the result. 4VT P:5 What happens when RE is replaced by a For Vi = 0, IC3 = IC4 = IEE/2; hence the gain current generator? is Acknowledgements Acknowledgement is due to my IC3 g m RC students in the EE204 N class, to whom I had posed this RC ¼ ; ð9:22Þ paradox as a challenge during the semester commencing 2VT 2 January 1996. Special mention must be made of Ankur Srivastava, Atul Saroop and Ram Sadhwani whose where gm = gm3 = gm4. This is exactly the same enthusiastic participation in the resolution of the paradox as the result derived under AC analysis. made it an enjoyable experience for me. References Problems 1. S.G. Burns, P.R. Bond, Principles of Electronic P:1. What happens when REE ! 1 in Fig. 9.1 Circuits (West Publishing Co, St Paul, USA, 1987) of text. What about REE = 1? 2. A.S. Sedra, K.C. Smith, Microelectronic Circuits P:2. What happens when RE ! 1 in Fig. 9.2 of (Sanders College Publishing, Fort Worth, USA, 1992) text. What about RE = 1? 3. J. Millman, A. Grabel, Microelectronics, 2nd edn. (McGraw Hill, New York, 1987) P:3. Approximate exp. function in Eq. 9.19 of text by the ﬁrst two terms and ﬁnd IC4 in terms of IC3. Appearances Can Be Deceptive: An Initial Value Problem 10 An initial value problem is posed and solved Establishing I2(0−): One Possibility in a systematic way, illustrating the fact that what meets the eye may not be the truth! To investigate the problem, let us examine how i2(0−) can be established in the circuit. One pos- sible way is shown in Fig. 10.2, where the switch is Keyword closed at t = −∞ with i1(−∞−) = 0 and Initial value problem i2(−∞−) = I2, an arbitrary value. Since for t > −∞, the circuit is equivalent to a series com- bination of V, R and L1, i1 increases exponentially The Problem from zero towards the asymptotic value V/R, reachable, at t = 0. All this time, can i2 change Consider the circuit shown in Fig. 10.1 in which from the value I2? This is not possible, because any the switch is closed from t = −∞ and is opened change in i2 requires a corresponding change in the at t = 0. The problem that is posed here is: what voltage across it (v2), but since L2 is is i2(0−) (This forms part of Problem 5.9 in [1])? short-circuited, v2 is forced to remain zero. Thus, i2 One way of looking at the problem is to remains I2 throughout, and I2 can be arbitrary! The realize that at t = 0−, L2 behaves as a short circuit current through the short circuit keeps on changing and therefore, there are two short circuits in from −I2 at t = −∞− to (V/R) −I2 at t = 0−. parallel. Since i1(0−) = V/R, is i2(0−) = V/(2R)? The answer, as we shall demonstrate here, is; no, not necessarily. In fact, we show that i2(0−) can Establishing I2 (0−): Another have an arbitrary value I2. Possibility Figure 10.3 shows another method of establish- ing i2(0−). The switch is shorted at t = −∞ with i1(−∞−) = i2(−∞−) = V1/R. The current i1 remains constant at V1/R (why?) and so does i2 at the same value, the current through the short Source: S. C. Dutta Roy, “Appearances can be Deceptive: An Initial Value Problem,” IETE Journal of Education, vol. 45, pp. 31–32, January–March 2004. © Springer Nature Singapore Pte Ltd. 2018 93 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_10 94 10 Appearances Can Be Deceptive: An Initial Value Problem L1 i1 (t) L1 i1 ð0 Þ þ L2 i2 ð0 Þ ¼ L1 i1 ð0 þ Þ þ L2 i2 ð0 þ Þ: ð10:1Þ R i2(t) L2 S Since i1(0+) = i2(0+), i1(0−) = V/R and + t=0 i2(0−) = I2, we get V i1 ð0 þ Þ ¼ ðVL1 =R þ L2 I2 Þ=ðL1 þ L2 Þ: ð10:2Þ Fig. 10.1 The circuit under consideration For t 0+, i1 = i2 so that differential equa- tion governing the circuit becomes t= ∞ L1 Ri1 þ ðL1 þ L2 Þ ðdi1 =dtÞ ¼ V: ð10:3Þ S i1 (t) This equation has a solution of the form i 1(t) –I2 R i1 ðtÞ ¼ A þ BeRt=ðL1 þ L2 Þ : ð10:4Þ i2(t) L2 + Using the boundary conditions given by V Eq. 10.2 and i1(∞) = V/R, one can ﬁnd out the – constants A and B. The ﬁnal result is: Fig. 10.2 One possible way of establishing i2 (0−) V L2 I2 VL2 =R L Rtþ L i1 ðtÞ ¼ þ e 1 2 ð10:5Þ R L 1 þ L2 L1 Note that the circuit of Fig. 10.1 is an exam- i1 (t) ple of a situation where, upon switching, inductor currents are not continuous, i.e. i1(0−) 6¼ i1(0+) R and i2(0−) 6¼ i2(0+). i2(t) L2 S t= ∞ + Problems V1 P:1. What happens when the switch in Fig. 10.1 − of text is closed for t = 0 to t = T0 and then Fig. 10.3 Another possible way of establishing i2(0 ) opened? P:2. What happens when the switch in Fig. 10.1 circuit being zero all the time. Since V1 is arbi- is shifted to be across L1? trary, we see that i2 (0−) can also be arbitrary. P:3. Determine the response of the circuit of Fig. 10.1 when L2 is replaced by a capacitor C. P:4. Same when C of P.3 is shifted to be across L1. Solve the Circuit P:5. Same when C as well as the switch are shifted to be across L1. We have established that i2(0−) in the circuit of Fig. 10.1 can have an arbitrary value I2. Hence, for solving for i1 in this circuit, I2 also needs to Reference be speciﬁed (in [1], I2 is not speciﬁed). The solution proceeds by invoking the principle of 1. F F Kuo, Network Analysis and Synthesis (Wiley, 1966), p. 129 conservation of flux, viz. Resonance 11 In this chapter, we discuss the basic concepts power factor in unity. It is also clear that this of resonance in electrical circuits, and its situation can arise only when the inductive and characterization, and illustrate its application capacitive reactances (or susceptances) cancel by an example. Several problems have been each other. Such a situation is known as reso- added at the end for the students to work out. nance, which plays a very important role in Do work them out. impedance matching, ﬁltering, measurements, and many other applications. Two types of resonance are distinguished, viz. Keywords series and parallel. A cancellation of reactances Resonance Figure of merit for coils and in series is referred to as series resonance, while capacitors Q Series resonance Parallel if susceptances in parallel cancel, we call it par- resonance Impedance Admittance allel resonance. In either case, the condition of Bandwidth resonance is usually associated with extremum (maximum or minimum) of impedance or admittance magnitude, and voltage/current. A one-port network containing resistors and inductors has the property that the current lags behind the voltage. For a resistor–capacitor Q: A Figure of Merit for Coils one-port, on the other hand, current leads the and Capacitors voltage. If a one-port contains inductors as well as capacitors in addition to the inevitable dissi- Dissipation, as already mentioned, is an inevi- pative element, viz. resistors, then the circuit table phenomenon in nature, in general, and in may be inductive at some frequencies, capaci- electric circuits, in particular. In other words, you tive at some other frequencies and, most inter- cannot make a pure inductor or capacitor in estingly, purely resistive at one or more practice; there will always be some losses. The frequencies. In the last situation, the current less the losses are, the better is the reactive ele- obviously is in phase with the voltage and the ment. A ﬁgure of merit, Q, is deﬁned for reactive Source: S. C. Dutta Roy, “Resonance,” Students’ Journal of the IETE, vol. 36, pp. 169–178, October–December 1995. © Springer Nature Singapore Pte Ltd. 2018 95 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_11 96 11 Resonance elements in terms of energy stored and dissipated in it, as follows: maximum energy stored per cycle Q ¼ 2p : energy disspated per cycle ð11:1Þ Consider a coil having an inductance L and a series resistance R, through which a current1 iðtÞ ¼ I sin xt flows. Then, the maximum energy Fig. 11.1 A series RLC circuit stored in a cycle is ð1=2ÞLIm2 , while the average power dissipated is ð1=2ÞRIm2 : But power is the phasor Vg ∠0°. The current through the circuit, energy per unit time, so that energy dissipated per represented by the phasor I ¼ jIj\h is given by cycle is (average power) (time duration of one cycle) = ð1=2ÞRIm2 f . Thus, the Q of an inductor is Vg I¼ ; ð11:4Þ R þ jxL þ 1 jxC 2pfL xL Q¼ ¼ : ð11:2Þ R R where the denominator represents the total impe- dance ZðjxÞ. Note that R includes the generator Similarly, for a capacitor, usually represented internal resistance, coil losses and any external by a pure capacitance C in parallel with a resis- resistance which may have been inserted. Taking tance R, you can show (Problem 1) that the Q is the magnitude and phase of Eq. 11.4, we have given by Vg jI j ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ: ð11:5Þ Q ¼ xCR: ð11:3Þ 1 2 R2 þ xL xC As you can see from Eqs. 11.2 and 11.3, Q is a linear function of frequency. While this is very xL xC 1 nearly the case in the case of an air capacitor, it is h ¼ tan1 ð11:6Þ not so in the case of a coil or other types of R capacitors. For a coil, the Q usually increases at Note that at x ¼ 0, i.e. DC, at which the low frequencies, attains a maximum at some capacitor acts as an open circuit, the current is frequency, and then decreases. This happens zero; the same is true at x ¼ 1, at which the because of skin effect, due to which the resistance inductor acts as an open circuit. In between, the increases with frequency, and at sufﬁciently high current will show a maximum at the frequency frequencies, this increase is at a more rapid rate x0 at which the second term in the denominator than the linear increase of frequency. of Eq. 11.5 vanishes, i.e. 1 1 xo L ¼ or xo ¼ pﬃﬃﬃﬃﬃﬃ ð11:7Þ Series Resonance xo C LC Consider the circuit of Fig. 11.1 in which a series The maximum value of the current, denoted RLC circuit is excited by a sinusoidal voltage by Io is generator of variable frequency, represented by Vg Io ¼ ð11:8Þ R 1 As a matter of notations, we shall use small i or v for instantaneous value, subscript m for maximum value, A sketch of |I| versus x is shown in capital I or V for phasor representation, and |I| or |V| for Fig. 11.2a, while Fig. 11.2b shows a sketch of the rms value. Series Resonance 97 the corresponding phase h. It is easy to argue Parallel Resonance from Eq. 11.6 that the phase is +p/2 at dc, decreasing to 0° at x ¼ xo and then to –p/2 as Figure 11.4 shows a parallel RLC circuit excited x ! 1. This is a reflection of the fact that the by a sinusoidal current generator of varying fre- impedance ZðjxÞ is capacitive for x\xo quency. The voltage across the circuit is given by purely resistive at x ¼ xo and inductive for the product of Ig and total impedance, i.e. x [ xo . At the frequency xo , the current and Ig voltage are in phase, and by deﬁnition, reso- V¼ ; ð11:12Þ G þ jxC þ 1 jxL nance occurs. Taking the current as the reference, the phasor diagram for the series resonant circuit is shown in where G = 1/R. Note the similarity of this with Fig. 11.3 for three frequencies, viz. (i) x\xo , Eq. 11.4; this is, in fact, expected because of the (ii) x ¼ xo and (iii) x [ xo . In each of these duality of the circuits of Figs. 11.1 and 11.4. diagrams, Thus, all the results derived earlier will apply here also. For completeness, we summarize I below the main results: VR ¼ IR; VL ¼ jxLI and VC ¼ ð11:9Þ jxC (i) If Ig = Ig ∠0°, V = |V| ∠/, then At x\xo ; 1=xC [ xL so that |VC| > |VL|. Ig Consequently, the resultant of VR and VC + VL jVj ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð11:13Þ 1 2 which gives Vg lagging behind I, i.e. I leads Vg. G2 þ xC xL At x [ xo the reverse is the case, while at x ¼ xo , VC = −VL and Vg and VR are the same. It is interesting to observe that at resonance, xC xL 1 / ¼ tan1 ð11:14Þ G xo L VL ¼ jxo LIo ¼ j Vg ¼ jQVg : ð11:10Þ R (ii) |V| is a maximum, equal to IgR at the res- onance frequency xo given by Eq. 11.7 where Q refers to that of the series RL combi- (iii) V leads Ig for x\xo lags Ig for x [ xo nation at resonance. Similarly, at resonance and is in phase with Ig at x ¼ xo (iv) At resonance, IL = –jQ Ig and Ic = jQ Ig, Io VC ¼ ¼ jxo LIo ¼ jQVg ; ð11:11Þ where Q ¼ xRo L ¼ xo CR (Problem 3). jxo C where use has been made of Eq. 11.7.2 In most The phasor diagram can be easily constructed, practical situations, Q is required to be greater and is left to you as an exercise (Problem 4). than unity; thus the voltage across the capacitor or the inductor at resonance is greater than the input voltage, i.e. the resonant circuit can be used Impedance/Admittance Variation as a voltage magniﬁer. with Frequency: Universal Resonance Curves From the similarity between series and parallel resonance equations, it follows that the two cases can be described by a single equation of the form 1 ; 2 Equation 11.11 can also be written as VC = Vg/(jo CR). H ðjxÞ ¼ ð11:15Þ Obviously, the Q of the series RC circuit at resonance is l/ 1 þ j xb xy 1 (o CR), in contrast to Eq. 11.3 for a parallel RC circuit (Problem 2). 98 11 Resonance (a) (b) Fig. 11.2 Variation of a magnitude and b phase of current in a series RLC circuit (a) (b) (c) Fig. 11.3 Phasor diagrams for the series RLC circuit at frequencies a below, b at and c above the resonance frequency 1 AðjxÞ ¼ aH ðjxÞ ¼ : ð11:16Þ j aþ a xb xc 1 The interpretation of AðjxÞ is also given in Table 11.1, where the subscript zero refers to the value at resonance. Fig. 11.4 A parallel RLC circuit Note that in either series or parallel resonance, pﬃﬃﬃﬃﬃ the resonance frequency is xo ¼ 1= bc; where the interpretation of the symbols for the sﬃﬃﬃ sﬃﬃﬃ pﬃﬃﬃﬃﬃ b bx two kinds of resonance are given in Table 11.1. xb ¼ x bc ¼ ð11:17Þ Equation 11.15 can be written in a normalized c c xo form as follows: Table 11.1 Interpretation of the symbols in Eq. 11.15 for series and parallel resonances Type of resonance H ðjxÞ a b c AðjxÞ Series Y ðjxÞ R L C Y ðjxÞ ¼ I ðIjx Þ Yo o Parallel Z ðjxÞ G C L Z ðjxÞ ¼ V ðVjx Þ Zo o Impedance/Admittance Variation with Frequency: Universal … 99 and similarly, A more useful form of the universal resonance rﬃﬃﬃ curve can be obtained if the behaviour at or near c x xc ¼ ð11:18Þ resonance is only of concern. Deﬁne the frac- b xo tional deviation of source frequency from the resonance frequency as introducing these in Eq. 11.16, we get 1 x xo AðjxÞ ¼ qﬃﬃ : ð11:19Þ d¼ ; ð11:23Þ j xo 1þ a b x c xo xo x i.e. Now, for series resonance, x ¼ xo ð1 þ dÞ ð11:24Þ sﬃﬃﬃ rﬃﬃﬃﬃ 1 b 1 L L xo L Then Eq. 11.22 can be written as ¼ ¼ pﬃﬃﬃﬃﬃﬃ ¼ ¼ Q: ð11:20Þ a c R C R LC R 1 AðjxÞ ¼ : ð11:25Þ Similarly for parallel resonance, 1 þ jQdð2 þ dÞ=ð1 þ dÞ sﬃﬃﬃ Universal resonance curves are obtained by 1 b R plotting |A| and ∠A versus d with Q as a ¼ ¼ Q: ð11:21Þ a c xo L parameter. Note that d = 0 corresponds to x ¼ x0 or x ¼ x=x0 ¼ 1 (Fig. 11.5). Thus, Eq. 11.19 becomes Equation 11.25 is an exact expression; for d 1, it can be approximated by 1 AðjxÞ ¼ : ð11:22Þ 1 þ jQ x xxo 1 xo AðjxÞ ﬃ : ð11:26Þ 1 þ j2Qd This expression is independent of the type of resonance or the actual element values used in the circuit, and is, therefore, applicable to all Bandwidth of Resonance resonant circuits of the form of Fig. 11.1 or Fig. 11.4. A family of curves can be drawn for The sharpness of resonance, as we shall see, is the variation of |A| and ∠A with the normalized determined by Q. A measure of the sharpness is frequency x ¼ xxo , taking Q as a parameter. the bandwidth, deﬁned as the band of frequencies around xo at which the magnitude of AðjxÞ is no Because of universal applicability, these are pﬃﬃﬃ called universal resonance curves. less than 1= 2 (a) (b) Fig. 11.5 Showing the universal resonance curves 100 11 Resonance Figure 11.5 shows the bandwidth, B, as Other Types of Resonant Circuits B ¼ x2 ¼ x1 ; ð11:27Þ The parallel resonant circuit of Fig. 11.4 cannot where x2 is called the ‘upper half power fre- be made in practice because you cannot make a quency’ and x1 is called the ‘lower half power pure inductance (in contrast, almost pure capac- frequency’. The nomenclature ‘half power’ is itances can be readily obtained). A practical cir- derived from the following consideration. Con- cuit is shown in Fig. 11.6, where R is the sider, for example series resonance in which winding resistance of the coil, usually much less AðjxÞ ¼ I ðjxÞ=Io . The power dissipated in R at than the inductive reactance xL at or around pﬃﬃﬃ2 resonance. If this is true, then the total impedance x1;2 will be Io = 2 R ¼ Io2 R 2, which is half of the circuit is of that dissipated at xo . A similar interpretation can be derived for parallel resonance. 1 To determine x1;2 turn to Eq. 11.22 and notice Z ðjxÞ ¼ jxC þ R þ1jxL that jAðjxÞj ¼ p1ﬃﬃ2 implies R þ jxL ¼ ð11:32Þ 1 þ jxRC x2 LC x xo Q ¼ 1 ð11:28Þ jxL xo x ﬃ : 1 þ jxRC x2 LC Solving 11.28, we get four solutions for x, The last expression can be written as two of which are negative while the other two are positive. Obviously, the latter are the acceptable 1 ones. These are given by (Problem 7) ZðjxÞ ¼ RC ð11:33Þ L þ jxC þ 1 jxL 2sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 3 2 1 1 which is of the form of Eq. 11.15, and hence all x1;2 ¼ xo 4 1 þ 5: ð11:29Þ 2Q 2Q the results we have derived will apply. Notice that at resonance, the impedance attains a maxi- mum value, given by Thus xo L xo L 1 B ¼ x2 x1 ¼ : ð11:30Þ Zo ﬃ ¼ Q RC R xo C ð11:34Þ Thus, higher Q leads to a lower B and hence a ðxo LÞ 2 xo L 2 ¼ ¼R ¼ RQ2 ; sharper resonance. R R In a general selective response curve of the type of Fig. 11.5, one often deﬁnes the ‘sharp- ness of resonance’ or ‘selectivity’ as the ratio (frequency of maximum response)/(bandwidth). Applied to the two cases of resonance discussed here, this quantity, as is evident from Eq. 11.3, is identical with Q. Also note, from Eq. 11.29, that x1 x2 ¼ x20 ð11:31Þ which shows that the resonance curve is geo- metrically symmetrical about xo , i.e. the response at x ¼ xo =a is the same as that at a xo . This is also obvious from Eq. 11.22. Fig. 11.6 A practical parallel resonant circuit Other Types of Resonant Circuits 101 where Q is again xo L=R because R and L are in where series. The condition xL R assumed for the 1 1 approximations is usually taken to mean Q 10, Z ðjxÞ ¼ ¼ : ð11:36Þ jxC þ R þ1jxL Y ðjxÞ as a rule of thumb. Another interesting resonant circuit is shown in Fig. 11.7, which has the property that if Winding a coil of Q 10 at 1000 kHz does pﬃﬃﬃﬃﬃﬃﬃﬃﬃ not pose a problem at all; thus from the results R1 ¼ R2 ¼ L=C , then the impedance is R at all derived under ‘other types of resonant circuits’ frequencies, i.e. this becomes an all-pass reso- discussed earlier, nant circuit (Problem 8). 1 xo ¼ 2p 106 ﬃ pﬃﬃﬃﬃﬃﬃ ð11:37Þ LC An Example and In concluding this discussion, we illustrate the L analysis and design of a resonant circuit of Zðjxo Þ ¼ Zo ﬃ ¼ RQ2 ð11:38Þ RC practical importance. The fractional deviation in frequency corre- Example 1 The voltage induced in a radio sponding to x2 = 2p 1050 103 r/s is receiver aerial may be approximated by a voltage generator of internal resistance 2000 X and an 1050 1000 50 emf containing equal amplitudes of the frequen- d¼ ¼ ¼ 0:05 1; 1000 1000 cies 1000 and 1050 kHz. It is desired to tune the ð11:39Þ receiver to the ﬁrst frequency with the second frequency discriminated at least by a factor of 2. so that we can apply Eq. 11.26. Thus Design a resonant circuit of the form of Fig. 11.6 for the purpose. Z ðjx2 Þ 1 ¼ : ð11:40Þ Zo 1 þ j0:1Q Solution: The overall circuit is shown in Fig. 11.8. By superposition, the voltage V will be The condition of the problem demands that the sum of the voltages developed due to the two sources applied independently. Let us, therefore, Vg Vg consider the response due to a source Vg ∠0° of 1 þ R Y þ 1 þ R Y ðjx Þ 2: ð11:41Þ g o g 2 frequency x; then Combining Eqs. 11.40 and 11.41 and simpli- Vg Z ðjxÞ Vg fying we get V¼ ¼ ; ð11:35Þ Rg þ Z ðjxÞ 1 þ Rg Y ðjxÞ pﬃﬃﬃ 0:1Q 3: ð11:42Þ 1 þ Zo =Rg Let 0:1Q ¼ 2: ð11:43Þ 1 þ Zo =Rg Then Zo ¼ :05 Q 1: ð11:44Þ Rg Fig. 11.7 A resonant circuit having resistances in both parallel branches 102 11 Resonance Fig. 11.8 Circuit for Example 1 Thus, we must have Q > 20. Let Q = 40; then 4. With V as the reference phasor, draw the phasor diagrams for the circuit of Fig. 11.4 Zo ¼ Rg ¼ 2000 X ð11:45Þ for x\xo ; x ¼ xo and x [ xo where pﬃﬃﬃﬃﬃﬃ xo ¼ 1= LC . From Eq. 11.38, therefore 5. In a series resonant circuit, derive an expression for the voltage across the capac- R ¼ Zo =Q2 ¼ 2000=16; 000 ¼ 1:25 X ð11:46Þ itor. Find the frequency at which this voltage is a maximum; ﬁnd also this maximum value. Also, from Q = Q ¼ xo L=R, we have 6. Derive an expression for the bandwidth on RQ 1:25 40 25 the basis of expression Eq. 11.26. L¼ ¼ H ¼ lH: ð11:47Þ 7. Verify Eq. 11.29. xo 2p 106 p 8. Analyze the circuit of Fig. 11.7 and derive Hence, ﬁnally, from Eq. 11.37, the conditions for all-pass resonance. 9. A medium wave broadcast receiver spans 1 1 the range 570–1560 kHz, with tuning C¼ ¼ 2 F xo L 4p 10 25 2 12 p 10 6 accomplished by a series resonant circuit 1 using an air variable capacitor, ranging from ¼ lF ð11:48Þ 100p 3 to 500 pF. What value of inductance is needed? At 570 kHz, the capacitor voltage and the design is complete. is desired to be 10 times the signal picked up by the aerial. What is the total resistance in the tuned circuit? What will be the signal Some Problems multiplication factor at 1560 kHz? What are the bandwidths of the circuit when tuned at If you have understood this chapter, then you 570 and 1560 kHz? should be able to work out the following prob- 10. A signal generator produces a fundamental lems. Try them. at 1 kHz of 1 V amplitude and its second and third harmonics at 0.5 and 0.3 V 1. Show that for a capacitor, represented by an amplitudes respectively. It is required to equivalent circuit consisting of a pure suppress each harmonic to less than 1% of capacitance C in parallel with a resistance the fundamental. Design a suitable circuit R, the Q is given by Q ¼ xCR. for the purpose. 2. Show that for a series RC circuit, the Q is 11. Determine the frequency of resonance for given by Q ¼ 1=ðxCRÞ. the circuit of Fig. 11.7 exactly, and the 3. Show that for a parallel RL circuit, the Q is value of the impedance at resonance. Also given by Q ¼ R=ðxLÞ. ﬁnd the condition for maximum impedance. Some Problems 103 12. A 1 V, 1 MHz, 2500 X internal resistance Bibliography source is to deliver maximum power to a load of 1 X. Explain how this can be 1. F.F. Kuo, Network Analysis and Synthesis (Wiley, achieved using resonance. New York, 1966) The Many Faces of the Single-Tuned Circuit 12 It is shown that the simple single-tuned circuit Notations: First Things First is capable of performing a variety of ﬁltering functions and that it can be analyzed graph- In the analysis of the various conﬁgurations of ically for obtaining the relevant performance Fig. 12.1, we shall adopt the following notations, parameters. which are more or less standard, and illustrated in Fig. 12.2. Keywords Single-turned circuit Low-pass High-pass p1, p*1 = complex conjugate poles of the network function Band-pass −a =−R/(2L) = real part of either pole b = {[1/(LC)] − [R2/(4L2)]}1/2 = imaginary part Figure 12.1 shows the simple single-tuned RLC of either pole series circuit under consideration. In the usual f = (R/2)(L/C)1/2 = damping factor textbooks, the circuit is analyzed for its beha- h = tan−1(b/a) = cos−1f viour at and near resonance, where resonance xn = l/(LC)1/2 = (a2 + b2)1/2 = undamped natu- implies the in-phase condition of V1 and I. That it ral frequency is capable of performing a variety of ﬁltering xm = frequency of maximum response functions, depending on how the output is cho- M1(x), M2(x) = vectors drawn from the poles to sen, is not generally discussed. Also, except for an arbitrary point jx Kuo [1] who treated one of the conﬁgurations w = angle between M1(x) and M2(x) graphically, it is hard to ﬁnd a reference in which x2,1 = upper and lower 3-dB cutoff frequencies graphical construction is used to ﬁnd the major performance indices of the circuit. In this chap- ter, we bring out the versatility of the circuit and * denotes complex conjugate extend the graphical analysis of [1] to other conﬁgurations. The Possible Configurations The possible conﬁgurations in which the circuit of Source: S. C. Dutta Roy, “The Many Faces of the Single Fig. 12.1 can be used differ from each other in the Tuned Circuit,” IETE Journal of Education, vol. 41, location of the output and are shown in Fig. 12.3. pp. 101–104, July-December 2000. Let Hx denote the transfer function V2/V1 of © Springer Nature Singapore Pte Ltd. 2018 105 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_12 106 12 The Many Faces of the Single-Tuned Circuit Fig. 12.1 The circuit under consideration Fig. 12.3, where x = a, b, c, d, e or f. Then, the variety of transfer functions obtained is as follows. Ha ðsÞ ¼ x2n =ðs2 þ 2fxn s þ x2n Þ ð12:1aÞ Hb ðsÞ ¼ s2 =ðs2 þ 2fxn s þ x2n Þ ð12:1bÞ Hc ðsÞ ¼ 2fxn s=ðs2 þ 2fxn s þ x2n Þ ð12:1cÞ Hd ðsÞ ¼ ðs2 þ x2n Þ=ðs2 þ 2fxn s þ x2n Þ ð12:1dÞ He ðsÞ ¼ ð2fxn s þ x2n Þ=ðs2 þ 2fxn s þ x2n Þ ð12:1eÞ Fig. 12.2 Illustrating the notations used in the chapter graphics gives much more physical meaning than Hf ðsÞ ¼ ðs2 þ 2fxn sÞ=ðs2 þ 2fxn s þ x2n Þ is possible by using routine algebra and calculus. ð12:1fÞ The ﬁrst four transfer functions are basically The Low-Pass Configuration those of low-pass, high-pass, band-pass and band-stop ﬁlters, respectively. However, if f is small, Figure 12.1a represents a low-pass conﬁguration Ha, as well as Hb, may be used as band-pass ﬁlters. with unity DC response and zero response at The transfer functions He and Hf represent mixed inﬁnite frequency. In between these two type ﬁlters and are not of much interest; hence, they extremes, the response may be monotonically will not be considered any further in this chapter. decreasing or may show a maximum. By putting We shall analyze each conﬁguration in a s ¼ jx in (1a) and taking the magnitude, we get conventional manner and follow it up with qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ graphical constructions and interpretations. While jHðjxÞj ¼ x2n = ðx2n x2 Þ2 þ 4f2 x2n x2 only the ﬁnal results are of interest, the use of ð12:2Þ The Low-Pass Configuration 107 Fig. 12.3 Various possible (a) R L (b) R conﬁgurations of the + + single-tuned circuit C + + V1 C V2 V1 V2 – – L – – (c) (d) R L C + + + + C V1 R V2 V1 V2 – – L – – (e) L (f) C C + + L + + V1 V1 V2 V2 – R – R – – By differentiating Eq. 12.2 with respect to x and putting the result to zero shows that the maximum response occurs at the frequency xma, where qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ xma ¼ b a ¼ xn 1 2f2 2 2 ð12:3Þ Obviously, the maximum will exist if pﬃﬃﬃ f\1= 2, which is equivalent to b > a or h > p/ pﬃﬃﬃ 4. For f ¼ f\1= 2, the response will be max- pﬃﬃﬃ imally flat (MF), while for f [ 1= 2, the response will lie below the MF curve, as shown Fig. 12.4 Magnitude response of the low-pass conﬁgu- in Fig. 12.4. The maximum response is obtained ration of Fig 12.1 for various values of f by combining Eqs. 12.2 and 12.3 and is given by qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ equation obtained by equating Eq. 12.2 to Hma ¼ 1=ð2f 1 f2 Þ ð12:4Þ pﬃﬃﬃ Hma = 2. After a considerable amount of algebra, pﬃﬃﬃ we get If Hma is greater than 2, which occurs for f < 0.383, then there will exist two 3 dB cutoff qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ1=2 frequencies x2a and x1a, otherwise there will x2a ; x1a ¼ xn 1 2f2 2f 1 f2 exist only one upper cut off frequency x2a. These frequencies can be calculated by solving the ð12:5Þ 108 12 The Many Faces of the Single-Tuned Circuit Note that is easily shown from the right-angled triangle O- qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ jxma-C that xma is indeed given by Eq. 12.3. x2a ; x1a ¼ x2n 1 8f2 þ 8f4 6¼ x2ma ð12:6Þ Now if an arc is drawn with A as the centre and pﬃﬃﬃ Ap1 = 2b as the radius, then its intersection in contrast to what Kuo claims in [1], p. 236. with the positive jx-axis occurs at jx2a, x2a being At this point, it is instructive to bring in Kuo’s the upper cutoff frequency, because the value of x graphical analysis of the conﬁguration of there is p/4, being half of p/2, the angle subtended Fig. 12.1a. Note that the area of the shaded tri- at the centre A by the arc from p1 to p*1. It can be angle in Fig. 12.2 is given by ba as well as (1/2)| veriﬁed by considering the right-angled triangle M1||M2| sin x so that O-jx2a-A that x2a is indeed given by Eq. 12.5. Kuo [1], at this point, claimed that the lower jM1 jjM2 j ¼ 2ba= sin w ð12:7Þ cutoff frequency x1a is given by x2ma/x2a, which, as we have shown in Eq. 12.6, is not true. How- Now the magnitude response of Ha can be ever, a graphical construction is also possible for written as ﬁnding x1a, as shown by Martinez [2]. Draw an arc with B as the centre and Bp1 as the radius. The jHa j ¼ x2n =jM1 jjM2 j ¼ ½x2n =ð2baÞ sin w point of its intersection with the positive jx-axis is ð12:8Þ jx1a, because the value of x there is (p/2) + (p/4) pﬃﬃﬃ i.e. sin x is again 1= 2. From the right-angled Thus, the variation of |Ha| with x is consoli- triangle O-jx1a-B, it can now be veriﬁed that x1a dated in the variation of the angle w with x. Now is indeed given by Eq. 12.5. refer to Fig. 12.5, where p1Ap*1B is the so-called peaking circle of radius b, centred at C. Its intersection with the jx-axis previously occurs at The High-Pass Configuration jxma (xma being the frequency of maximum response) because the value of w at this point is The conﬁguration of Fig. 12.1b is the dual of the maximum, equal to p/2, being the angle Fig. 12.1a, because its dc response is zero while subtended by a diameter on the circumference. It at inﬁnite frequency, Hb becomes unity. By fol- lowing the same procedure as in the low-pass case, it can be shown that the maximum response occurs at the frequency xmb, where qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ xmb ¼ xn = 1 2f2 ð12:9Þ Comparing Eqs. 12.3 and 12.9, we see that xma xmb = x2n i.e. xma and xmb are geometri- cally symmetrical about xn. In fact, xmb can be obtained by the graphical construction shown in Fig. 12.6. Mark the points −xma and −xn, on the negative real axis. Join −xma to jxn, thereby creating the angle x. Construct the same angle x at −xn. The intersection of the new line with the positive jx-axis gives jxmb, because tan u ¼ xn =xma ¼ xmb =xn ð12:10Þ Fig. 12.5 Graphical analysis of the low-pass conﬁgura- tion of Fig. 12.1 The High-Pass Configuration 109 x xn Hc ðjxÞ ¼ 2fxn 2fxn þ jxn xn x ð12:13Þ It is clear that the magnitude characteristic will have true geometric symmetry about xn, which is also the frequency of maximum response. The maximum response, in this case, is unity, and by the usual procedure, the 3 dB cutoff frequencies are obtained as qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 x2c ; x1c ¼ xn 1þf f ð12:14Þ Fig. 12.6 Graphical construction for obtaining xmb from x Clearly, x2c ; x1c ¼ x2n ; which is a reflection of the geometric symmetry of the transfer func- tion Hc(jx). A graphical construction for ﬁnding x2c and It is easily shown that Hmb, the maximum x1c is shown in Fig. 12.7. With the origin as the response of |Hb(jx)| is the same as Hma, as given centre and Op1 (=xn) as the radius, draw the by Eq. 12.4. circle p1Ap*1B. It cuts the positive jx-axis at C Continuing the analysis, the 3 dB cutoff which is jxn. Since OD is fxn, the distance DC frequencies x2b and x1b can be obtained by pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ must be xn 1 þ f2 . With D as the centre and equating |Hb(jx)| to Hmb 2. After consider- DC as the radius, draw the circle ECFG. Now the able algebra, one obtains the following pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ result: distance OE is xn 1 þ f2 þ f , while the pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ1=2 distance OF is xn 1 þ f2 f . Hence, x2b ; x1b ¼ xn = 1 2f2 2f 1 f2 drawing two arcs with OE and OF as radii will cut the positive jx-axis at jx2c and jx1c, respec- ð12:11Þ tively, as shown in Fig. 12.7. Comparing Eqs. 12.5 and 12.11, we note that x2b x1a ¼ x2a x1b ¼ x2n ð12:12Þ Thus, following the same procedure as depicted in Fig. 12.6, one can ﬁnd x2b and x1b graphically from Fig. 12.5. The Band-pass Configuration The conﬁguration of Fig. 12.1c, characterized by the transfer function Hc of (1c), is a true band-pass ﬁlter because of its response at dc, as well as Fig. 12.7 Graphical analysis of the bandpass conﬁgura- inﬁnite frequency, are zero. By writing Hc(jx) as tion of Fig. 12.1c 110 12 The Many Faces of the Single-Tuned Circuit The Band-stop Configuration Problems By looking at the circuit of Fig. 12.1d or its P:1. In a single-tuned circuit, if the output is transfer function Hd, it is clear that it will give taken across the series combination of R and null transmission at the frequency xn and that the L, what kind of frequency response will you dc and inﬁnite frequency responses would be obtain? Sketch it. unity. To determine the 3 dB cutoff frequencies P:2. If L and C are in parallel in an RLC circuit x2d and x1d, we let and the output is taken across L, what will be the frequency response? Sketch it. ðx2 x2n Þ2 1 P:3. You require two poles in the frequency ¼ ð12:15Þ ðx2 x2n Þ2 þ 4f 2 x2 x2n 2 response and are supplied with three reactive elements. Draw an appropriate circuit and and obtain the same values as those given by sketch its frequency response. Comment on Eq. 12.14. Hence, the graphical construction of the d.c and inﬁnite frequency responses. Fig. 12.7 works for this circuit also except that P:4. Same as above except that you require two xn is now the frequency of rejection and not of nulls in the frequency response. maximum response. P:5. If you require three nulls, what is the min- imum number of reactances needed? Draw an appropriate circuit and comment on the Conclusion d.c. and inﬁnite frequency responses. Also, comment on the height of the peaks. We have shown in this chapter, how a relatively simple circuit like that of series RLC combination can be used to illustrate various circuit concepts like poles, zeros and their effects on the fre- References quency response; ﬁltering of various kinds; 1. F.F. Kuo, Network Analysis and Synthesis (Wiley, graphical analysis; geometric symmetry, etc. 1966) This should be of interest to teachers as well as 2. J.R. Martinez, Graphical solution for 3 dB points. students of circuit theory. Electron. Eng., 48–51 (January 1967) Analyzing the Parallel-T RC Network 13 Following a review of the various alternative [16, 17], FM detection [18], and sine-wave methods available for analyzing the parallel-T generation [19]. Various methods are available RC network, we present yet another concep- in the literature for analyzing this network, a tually elegant method. This discussion illus- review of which is given in this chapter. This is trates the famous saying of Ramakrishna followed by yet another method, which is con- Paramhansa: As many religions, as many ceptually elegant and is believed to be new. ways. Don’t just grab one method; learn all For illustrating the various methods of anal- of them and decide for yourself which one you ysis, we have used the symmetrical conﬁguration ﬁnd to be the simplest. shown in Fig. 13.1b, for simplicity. It should, however, be emphasized that the method of analysis does not depend upon the composition Keywords of the individual arms, each of which could as Parallel-T-network Mesh analysis well be a general RLC impedance. Node analysis Two-port method Splitting the parallel-T Mesh Analysis The parallel-T RC network shown in Fig. 13.1a has fascinated many circuits researchers, includ- The network of Fig. 13.1b has been redrawn in ing me [1–9]. It has many applications, foremost Fig. 13.2 in order to clearly indicate one choice among them being notch ﬁltering [10, 11], active of four independent meshes. Following standard band-pass ﬁltering [12, 13], measurements procedure, the four mesh equations can be writ- [14, 15], compensation in control systems ten in the matrix form as follows: Source: S. C. Dutta Roy, “Analyzing the Parallel-T RC Network,” IETE Journal of Education, vol. 44, pp. 111– 116, July–September 2003. © Springer Nature Singapore Pte Ltd. 2018 111 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_13 112 13 Analyzing the Parallel-T RC Network 2 32 3 2 3 R þ ½1=ð2sCÞ 1=ð2sCÞ R 0 I1 VI 6 1=ð2sCÞ R þ RL þ ½1=ð2sCÞ R RL 7 6 I2 7 6 0 7 6 76 7 ¼ 6 7: 4 R R 2R þ ½2=ðsCÞ 1=ðsCÞ 5 4 I3 5 4 0 5 0 RL 1=ðsCÞ ðR=2Þ þ RL þ ½1=ðsCÞ I4 0 ð13:1Þ (a) C1 C22 I2 ¼ Vi D12 =D and I4 ¼ Vi D14 =D ð13:3Þ R1 R2 so that from Eq. 13.2, the voltage transfer func- + + tion is obtained as Vi R3 C3 V0 RL – – T ¼ V0 =Vi ¼ ðD12 D14 ÞRL =D ¼ ðM14 M12 ÞRL =D; ð13:4Þ (b) where Mij is the minor of Dij i.e. Dij = (–1)i+j Mij. i1 C C i2 Evaluating D and its minors, and simpliﬁcation of Eq. 13.4 will require pages of calculations. R R However, if it is done with care, one ends up + + with the following expression: Vi R/2 2C V0 RL – – T ¼ ðp2 þ 1Þ=½p2 þ 2ð2 þ rÞp þ 1 þ 2r; ð13:5Þ where we have used the following notations: Fig. 13.1 a The general parallel-T-network; b The sym- metrical form p ¼ sCR and r ¼ R=RL : ð13:6Þ C I3 Node Analysis C I4 R R Refer to Fig. 13.1b again where all node voltages + + have been identiﬁed. The node equations for Vi I1 2C I2 V0 RL R/2 – V1, V2 and V0 can be written in the following – matrix form: 2 32 3 Fig. 13.2 Redrawn form of Fig. 13.lb for mesh analysis 2ðG þ sCÞ 0 G V1 4 0 2ðG þ sCÞ sC 54 V 2 5 2G 3 sC sC þ G þ GL V0 Since GVi V0 ¼ ðI2 I4 ÞRL ; ð13:2Þ ¼ 4 sCVi 5; 0 we need to evaluate I2 and I4 only. If D denotes ð13:7Þ the determinant of the 4 4 mesh impedance matrix in Eq. 13.2 and Dij its cofactors, then where Node Analysis 113 G ¼ 1=R and GL ¼ 1=RL : ð13:8Þ Finally, the transfer function is Denoting the determinant of the 3 3 matrix on T ¼ y21 =ðy22 þ GL Þ the left of Eq. 13.7 by D, we get ¼ ðp2 þ 1Þ=½p2 þ 4p þ 1 þ 2RGL ðp þ 1Þ; ð13:14Þ V0 ¼ ðGD013 þ sCD023 Þ=Vi =D0 ð13:9Þ which is identical to Eq. 13.5 because RGL = r. so that 0 T ¼ ðGM13 0 sCM23 Þ=D0 : ð13:10Þ Analysis by Miller’s Equivalence 0 The evaluation of D, M13 and M23 is a much easier task than that in mesh analysis. It is not Refer to Fig. 13.1b again. By Miller’s theorem, difﬁcult to show that simplifying the resulting this circuit is equivalent to that shown in expression gives the same result as in Eq. 13.5. Fig. 13.3a where Two-Port Method Y1 ¼ i1 =Vi and Y2 ¼ i2 =V0 : ð13:15Þ The network of Fig. 13.1b is the parallel con- Since Y1 occurs across the input voltage nection of two T-networks, viz. A: R–2C–R and source, it does not affect the transfer function. B: C–R/2–C, which is terminated in RL. The We, therefore, have to ﬁnd Y2 only. Obviously, z-parameters of T-networks are as follows: i2 ¼ sCðV0 V2 Þ: ð13:16Þ z11A ¼ z22A ¼ R þ ½1=ð2sCÞ; To ﬁnd V2, write the node equation at this z12A ¼ z21A ¼ 1=ð2sCÞ; node as follows: ð13:11Þ z11B ¼ z22B ¼ ðR=2Þ þ ½1=ðsCÞ; and z12B ¼ z21B ¼ R=2: ð2sC þ 2GÞV2 ¼ sCVi þ sCV0 ð13:17Þ The corresponding y-parameters can be found or, from the conversion formulas [20] as V2 ¼ sCðVi þ V0 Þ=½2ðsC þ GÞ: ð13:18Þ y11A ¼ y22A ¼ ð2p þ 1Þ=½2Rðp þ 1Þ; Combining Eq. 13.15 with Eqs. 13.16 and y12A ¼ y21A ¼ 1=½2Rðp þ 1Þ; 13.18, we get y11B ¼ y22B ¼ pðp þ 2Þ=½2Rðp þ 1Þ; and y12B ¼ y21B ¼ p2 =½2Rðp þ 1Þ: Y2 ¼ p½ð1 T 1 Þp þ 2=½2Rðp þ 1Þ: ð13:19Þ ð13:12Þ Although Y2 involves T which we wish to ﬁnd, one should not be worried. As you would see, we shall Thus, the overall y-parameters are the following: ﬁnd T in terms of T–1 and then by cross multiplying and simplifying, we shall ﬁnd an explicit expres- y11 ¼ y22 ¼ ðp2 þ 4p þ 1Þ=½2Rðp þ 1Þ and y12 sion for T. By applying Thevenin’s theorem to the ¼ y21 ¼ ðp2 þ 1Þ=½2Rðp þ 1Þ: left of the XX′ line in Fig. 13.3a, we get the ð13:13Þ equivalent circuit shown in Fig. 13.3b. Thus 114 13 Analyzing the Parallel-T RC Network X (a) (b) i1 R i2 R R R + + + + Vi 2p + 1 Vi Y1 2C Y2 RL V0 2p + 1 Y2 + GL V0 X Fig. 13.3 a Miller’s equivalent of the network of Fig. 13.1b; b Equivalent circuit of Fig. 13.3a obtained by using Thevenin’s theorem T ¼ V0 =Vi Simplifying Eq. 13.22 and taking the ratio T gives the same result as Eq. 13.5. ¼ ½1=ð2p þ 1Þ: ½1=ðY2 þ GL Þ=f½1=ðY2 þ GL Þ : þ R þ ½R=ð2p þ 1Þg ð13:20Þ Yet Another Method Combining Eqs. 13.20 with 13.19, and simpli- Look at Fig. 13.4a again. Instead of applying fying, we get Thevenin’s theorem, let us apply ladder analysis method [20] starting from RL and going to the T ¼ 1=fp½ðlT 1 Þp þ 2 þ 2ðp þ 1ÞRGL þ 2p þ 1g: left, and again starting from RL and going to the ð13:21Þ right. In order not to clutter Fig. 13.4a, we have redrawn the circuit in Fig. 13.5, where all branch Now cross multiply and simplify. The ﬁnal result currents have been identiﬁed. At node V0, is the same as Eq. 13.5. I1 þ I2 ¼ IL ¼ V0 GL : ð13:23Þ Splitting the T’S Going to the left, we get Let the C–R/2–C T-network in Fig. 13.1b be V1 ¼ RI1 þ V0 ; ð13:24aÞ separated at the input side, turned through 180° and be terminated in another voltage source Vi. I3 ¼ 2sCV1 ¼ 2pI1 þ 2pGV0 ; ð13:24bÞ The result is shown in Fig. 13.4a, which would be completely equivalent to Fig. 13.1b because I5 ¼ I1 þ I3 ¼ ð2p þ 1ÞI1 þ 2pGV0 ð13:24cÞ potentials at all the nodes have been preserved. and Now apply Thevenin’s theorem to the left of XX′ and to the right of YY′ to get the equivalent circuit Vi ¼ RI5 þ V1 ¼ 2ðp þ lÞRI1 þ ð2p þ lÞV0 : shown in Fig. 13.4b [8]. Next, write the node ð13:24dÞ equation at the load as follows: V0 ½Vi =ð2p þ 1Þ V0 V0 ½Vi p=ðp þ 2Þ From the last equation, we have þ þ R þ ½R=ð2p þ 1Þ RL ðR=pÞ þ ½R=ðp þ 2Þ I1 ¼ ½Vi ð2p þ lÞV0 =½2ðp þ 1ÞR: ð13:25Þ ¼ 0: ð13:22Þ Yet Another Method 115 Fig. 13.4 a Spread out X Y (a) version of Fig. 13.1b by V1 V0 V2 splitting the two T’s; b Equivalent circuit of R R C C Fig. 13.4a obtained by two + applications of Thevenin’s Vi 2C + RL R/2 Vi theorem – – X Y (b) R/(2p + 1) V0 R/(p + 2) R C + + Vi /(2p + 1) RL – pVi /(p + 2) – Fig. 13.5 Fig 13.4a circuit I5 I1 I2 I6 redrawn to illustrate the new V1 V0 V2 method R R C C + Vi 2C + RL R/2 Vi – I3 IL I4 – Now start from V0 and go to the right. This is Now combine Eqs. 13.23, 13.25 and 13.27 to get what we get: ðp2 þ 1ÞVi ðp2 þ 4p þ 1ÞV0 ¼ 2ðp þ lÞRGL V0 : V2 ¼ ½I2 =ðsCÞ þ V0 ¼ ðR=pÞI2 þ V0 ; ð13:26aÞ ð13:28Þ I4 ¼ 2GV2 ¼ ð2=pÞI2 þ 2GV0 ; ð13:26bÞ Simplifying Eq. 13.28 gives the same result as I6 ¼ I4 þ I2 ¼ ½ð2=pÞ þ 1I2 þ 2GV0 ð13:26cÞ Eq. 13.5. and Conclusion Vi ¼ ½1=ðsC ÞI6 þ V2 ¼ ð2R=pÞ½ð1=pÞ þ 1I2 þ ½ð2=pÞ þ 1V0 : In this chapter, we have discussed six different ð13:26dÞ methods for analyzing the parallel-T RC net- work. Of these, mesh analysis requires more The last equation gives effort than any other method. The node analysis comes next in terms of computational effort. The I2 ¼ ½p2 Vi pðp þ 2ÞV0 =½2ðp þ 1ÞR: ð13:27Þ efforts needed in the two-port method and the 116 13 Analyzing the Parallel-T RC Network method using Miller’s equivalence are compa- 4. S.C. Dutta Roy, The deﬁnition of Q of RC networks. rable and can be bracketed to occupy the joint Proc. IEEE. 52, 44, (1964) 5. D.G.O. Morris & S.C. Dutta Roy, Q and selectivity. third position in terms of decreasing computa- Proc. IEEE. 53, 87–89, (1965) tional effort. Splitting the T’s is common to the 6. S.C. Dutta Roy, Dual input null networks, Proc. last two methods—one using Thevenin’s theo- IEEE. 55, 221–222, (1967) rem, and the other using ladder analysis tech- 7. S.C. Dutta Roy & N. Choudhury, An application of dual input networks. Proc. IEEE. 58, 847–848, nique. Both are conceptually elegant and require (1970) almost the same amount of effort. They, there- 8. S.C. Dutta Roy, A quick method for analyzing fore, qualify for the joint fourth position in the parallel ladder networks. Int. J. Elect. Eng. Educ. 13, list; of these, the last method does not seem to 70–75, (1976) 9. S.C. Dutta Roy, Miller’s theorem revisited Circ. Syst. have appeared earlier in the literature and is Signal Process. 19, 487–499, (2000) therefore believed to be new. 10. L. Stanton, Theory and applications of the parallel- T resistance capacitance frequency selective network. Proc. IRE. 34, 447–456 (1946) 11. A.E. Hastings, Analysis of the resistance capacitance Problems parallel-T network and applications. Proc. IRE. 34, 126–129 (1946) P:1. What kind of transfer function do you get if, 12. H. Fleischer, Low frequency feedback ampliﬁers, in in Fig. 13.1a R3 ! ∞ and C3 = 0? Vacuum Tube Ampliﬁers, ed. by G.E. Valley Jr., H. Wallman, McGrawHill, (1948, Chapter 10) P:2. Same, if in Fig. 13.1b, R/2 ! ∞ and 13. C.K. Battye, A low frequency selective ampliﬁer. 2C = 0 in the shunt branches? J. Sci. Inst. 34, 263–265 (1957) P:3. Find the two-port parameters of the circuit 14. W.N. Tuttle, Bridged-T and parallel-T null networks of P.1 and hence the transfer function. for measurements at rf. Proc. IRE. 28, 23–30 (1940) 15. K. Posel, Recording of pressure step functions of low P:4. Same for the circuit of P.2. amplitude by means of composite dielectric capaci- P:5. Apply Miller to P.3 and P.4 and verify that tance transducer in parallel-T network. Amer. Rocket you get the same transfer functions. Soc. J. 21, 1243–1251 (1961) 16. A.B. Rosenstein, J. Slaughter, Twin T compensation using root locus method. AlEE Trans, Part II (Applications and Industry) 81, 339–350 (1963) 17. A.C. Barker, A.B. Rosenstein, s-plane synthesis of References the symmetrical twin-T network. IEEE. Trans. Appl. Indus. 83, 382–388 (1964) 18. J.R. Tillman, Linear frequency discriminator. Wirel. 1. S.C. Dutta Roy, A twin-tuned RC network. Ind. Eng. 23, 281–286 (1946) J. Phys. 36, 369–378 (1962) 19. A.P. Bolle, Theory of twin-T RC networks and their 2. S.C. Dutta Roy, On the design of parallel-T resistance applications to oscillators. J. Brit. IRE. 13, 571–587 capacitance networks for maximum selectivity. (1953) J. Inst. Telecommun. Eng. 8, 218–233, (1962) 20. F.F. Kuo, Network Analysis and Synthesis (John 3. S.C. Dutta Roy, Parallel-T RC networks: limitations of Wiley, 1966, Chapter 9) design equations and shaping the transmission char- acteristic. Ind. J. Pure Appl. Phys. 1, 175–181, (1963) Design of Parallel-T Resistance– Capacitance Networks For Maximum 14 Selectivity A simple analysis is presented for obtaining an Keywords expression for the transfer function and hence Parallel-T RC network Selectivity the selectivity, QT, of a general parallel-T Selective ampliﬁer resistance–capacitance network. The maxi- mum value of QT obtainable by a suitable choice of elements is shown to be 12. A design Introduction procedure for approaching this maximum value is given.An expression for the selectiv- In the low-frequency range, an inductance– ity, QA, of an ampliﬁer using a general capacitance-tuned circuit is seldom used as a parallel-T resistance–capacitance network in frequency-selective network because of the follow- the negative feedback line has been deduced ing disadvantages: (a) large physical size of the and the advantages of having an increased QT inductor requires space and makes the equipment explained. Parallel-T RC is an important bulky, (b) an inductor of large value is expensive and network and you cannot do without it, if you (c) the value of Q obtainable is low. A resistance– wish to remain in circuit design.An expression capacitance network is a better choice for all these has been given for estimating the departure considerations. Of all such networks, the parallel-T from linearity of the amplitude response RC network is the most extensively used one. characteristic at a particular frequency. This Much work has been done on the symmetrical is used to ﬁnd an optimum value of QT for conﬁguration of the parallel-T RC network and a best performance of the network as an F.M. fairly impressive list of references is available on discriminator at low frequencies. It is shown its theory and applications. The general asym- that the required value of QT is very near its metrical conﬁguration of the network has maximum value. received less attention, important work in this line being due to Stanton [1], Wolf [2] and Oono [3]. Stanton [1] has given an expression for the transfer function of a general parallel-T RC net- work in which the components occur as the ratios ða series arm resistance or reactance) Source: S. C. Dutta Roy, “Design of Parallel-T Resistance– ðtotal series arm resistance or reactance) Capacitance Networks for Maximum Selectivity,” Journal of the Institution of Telecommunication Engineers, vol. 8, ð14:1Þ pp. 218–223, September 1962. © Springer Nature Singapore Pte Ltd. 2018 117 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_14 118 14 Design of Parallel-T Resistance–Capacitance Networks For … and By itself, the parallel-T RC network behaves as a rejection ﬁlter. A resonance characteristic, ðtotal series arm reactance) similar to that of a tuned ampliﬁer, can be : ðtotal series arm resistance) obtained by using it as the feedback network of an ampliﬁer which has an odd number of stages. A simpler expression is deduced in this chapter Fleischer [7] has shown that by using a sym- by assuming the series arm impedances to be metrical conﬁguration of the network, the maxi- arbitrary multiples of the shunt arm impedances. mum value of Q obtainable is approximately Go/ Using Morris’s [4] deﬁnition of Q of resistance– 4, where Go is the open-loop gain of the ampli- capacitance networks, an expression is deduced ﬁer. In this chapter, an expression has been for the selectivity, QT, of the general parallel-T deduced for the selectivity, QA, of an ampliﬁer network. It is shown that the maximum value of using a general parallel-T network in the negative QT is 12, which is in conformity with Wolf’s [2] feedback line and the advantages of having an result. A design procedure is then suggested for increased value of QT are explained. such networks, which is more general than that given by Wolf [2]. Oono’s [3] work is an extension of that of Stanton [1] for the case when Network Configuration the effects of the source and the load impedances and Simplification are not negligible. Throughout this chapter, however, the source impedance has been A parallel-T RC network with completely arbi- assumed to be negligible and the load impedance trary values of the elements is shown in has been assumed to be inﬁnite. Fig. 14.1, where R is a resistance parameter, C a In its application in an F.M. discriminator [5, capacitance parameter and m1, m2, m3, n1, n2 and 6], it is desired that the amplitude transfer func- n3 are numerical constants. In the conventional tion should have a linear variation with fre- symmetrical conﬁguration, quency. An expression is given in this chapter for estimating the departure from linearity of the m1 ¼ m2 ¼ n1 ¼ n2 ¼ 1; m3 ¼ 2=k and amplitude transfer characteristic at a particular n3 ¼ 1=ð2kÞ; frequency. This is used to ﬁnd an optimum value of QT for best performance of a single parallel-T where the parameter k controls the selectivity of network in the above application. It is shown that the transfer characteristic and has a value of unity in the frequency range of interest, the required for maximum Q equal to 14. value of QT is very near to its maximum value. Without any loss of generality, we can assume m3 = n3 = 1. Each of the two tees in Fig. 14.1 can be converted to an equivalent pi-network. m1C m2C For the two networks in Fig. 14.2 to be equiva- lent, the elements should be related as follows: ZA ¼ R=Z2 ; ZB ¼ R=Z3 ; ZC ¼ R=Z1 ; ð14:1Þ n1R n2R m3C where R = Z1Z2 + Z2Z3 + Z3Z1. Employing Eq. 14.1 IN n3R OUT and denoting by subscripts 1 and 2 the equivalent pi-elements of the R-C-R and C-R-C tees respectively, we have Fig. 14.1 A General Parallel-T network Network Configuration and Simplification 119 (a) Z1 Z2 (b) ZB Z3 ZA ZC Fig. 14.2 a A Tee-network, b A Pi-network 9 (n1 + n2 )R n1n2CR 2 ZA1 ¼ n1 R þ ðn1 þ n2 Þ=ðn2 pCÞ > > > > ZB1 ¼ ðn1 þ n2 ÞR þ pn1 n2 CR2 > > > > > > ZC1 ¼ n2 R þ ðn1 þ n2 Þ=ðn1 pCÞ > > = ZA2 ¼ Rðm1 þ m2 Þ=m1 þ 1=ðpm1 CÞ ; > > ¼ 1=ðm1 m2 p C RÞ2 2 > > ZB2 > > 1 > > m1m2 þ ðm1 þ m2 Þ=ðpCm1 m2 Þ > > - C > > m1m2w C R 2 2 (m1 + m2 ) and ZC2 ¼ Rðm1 þ m2 Þ=m2 þ 1=ðpm2 CÞ ; ð14:2Þ Fig. 14.3 Showing the ZB arm of the pi-equivalent of the network of Fig. 14.1 where p = jx, x being the angular frequency. The pi-equivalent of the network of Fig. 14.1 Without any loss of generality, we can let will then have its elements given by x20 ¼ 1=ðC 2 R2 Þ. Then ZA1 ZA2 ZB1 ZB2 m1 m2 ðn1 þ n2 Þ ¼ m1 m2 n1 n2 =ðm1 þ m2 Þ ¼ 1: ZA ¼ ; ZB ¼ ; ZC ZA1 þ ZA2 ZB1 þ ZB2 ð14:4Þ ZC1 ZC2 ¼ : ð14:3Þ ZC1 þ ZC2 Equation 14.4 shows that if x0 is ﬁxed, then the number of arbitrary numerical constants is reduced to two only. Null Condition For zero transmission, since ZA and ZC cannot be Transfer Function zero with ordinary circuit elements, we must have ZB = a. At this point, it is convenient to Let xCR = x; then the rejection frequency is have a look at the elements composing the ZB given by x = 1 and from Eq. 14.2, arm as shown in Fig. 14.3. We note that this is 9 simply an anti-resonant circuit having inﬁnite ZB1 ¼ Rfðn1 þ n2 Þ þ jn1 n2 xg > > > impedance at a frequency given by ZB2 ¼ fR=ðm1 m2 xÞgfð1=xÞ þ jðm1 þ m2 Þg = : ZC1 ¼ fR=ðn1 xÞgfn1 n2 x jðn1 þ n2 Þg > > > ; 1 m1 þ m2 ¼ fR=ðm2 xÞgfðm1 þ m2 Þx jg x20 ¼ ¼ 2 2 : and ZC2 C 2 R2 m 1 m2 ðn1 þ n2 Þ C R m1 m2 n1 n2 ð14:5Þ 120 14 Design of Parallel-T Resistance–Capacitance Networks For … Table 14.1 Examples of n1 m1 QT m2 n2 design for maximum QT 1.0 0.95 0.487 0.026 40.04 1.0 0.90 0.475 0.053 20.11 1.0 0.85 0.459 0.081 13.51 1.0 0.80 0.444 0.111 10.25 1.0 0.75 0.429 0.143 8.33 Combining Eqs. 14.3, 14.4 and 14.5 and Using Morris’s [4] deﬁnition of Q of a resis- simplifying, we get tance–capacitance network, we have from 9 Eq. 14.8 1 þ jm1 m2 n1 n2 x > ZB ¼ R > m1 m2 ð1 x2 Þ = QT ¼ l=ðn1 þ 1=m1 Þ: ð14:9Þ : ð14:6Þ m 1 m 2 n1 n2 x j > > ZC ¼ R ; Thus QT can be increased by decreasing n1 m2 ð1 þ m1 n1 Þx and increasing m1. The extent to which this can For zero source and inﬁnite load impedances, be done depends, however, on m2 and n2 also, the transfer function of the network is given by because these must remain positive as m1 and n1 are changed. At this stage, therefore, we require T ¼ ZC =ðZB þ ZC Þ: the expressions for m2 and n2 in terms of m1 and n1. They can be easily obtained from Eq. 14.4 as Substituting for ZB and ZC from Eq. 14.6 and simplifying, we get n1 m1 m21 n1 þ 1 m2 ¼ and n2 ¼ : m 1 n1 þ 1 2 m1 ðn1 m1 Þ 1 T¼ : ð14:7Þ ð14:10Þ 1 jðn1 þ 1=m1 Þ=ðx 1=xÞ Thus for m2 and n2 to be positive, (n1 − m1) must This expression for the transfer function of a remain positive. Under this restriction, QT will general parallel-T network is much simpler to have a maximum value of 12 when m1 = n1 = 1. handle than that of Stanton [1], as it contains only two numerical constants. In Eq. 14.7, the term varying with frequency occurs as (x − 1/x); Design it thus follows that both the amplitude and phase transfer characteristics of the network will be At m1 = n1 = 1, Eq. 14.10 gives m2 = 0 and symmetrical about x = 1 when plotted on a log n2 = a, so that the corresponding arms are (x) scale. effectively open circuited and the output is zero at all frequencies. Even with ﬁnite elements of moderate values, however, QT can be made to Selectivity approach this maximum value, as will be evident from the following example. Let n1 = 1.0 and Equation 14.7 can be written as m1 = 0.9; then QT = 0.475. In the conventional 1 x2 symmetrical case, QT = 0.250 so that the T¼ ð14:8Þ improvement is as much as 90%. Also from 1 þ jðn1 þ 1=m1 Þx x2 Eq. 14.10, m2 = 0.053 and n2 = 20.11. For a Design 121 rejection frequency of 1000 c/s., we can choose or, C = 0.01 lF and R = 16 KX. Then the series 1=jT j2 ¼ 1 þ 1= Q2T y2 ð14:11Þ resistances required are n1R = l6 KX and n2R = 321.7 KX and the series capacitances required where y = x − 1/x. Differentiating Eq. 14.11 are m1C = 0.009 lF, and m2C = 530 llF. Thus with respect to y gives elements of reasonable values can be used to approach the maximum selectivity. Table 14.1 djTj jTj3 shows some typical examples of design for ¼ 2 3: ð14:12Þ dy QT y improved QT. Differentiating again, we get ! Linearity of the Selectivity Curve d2 jT j 3jT j3 jT j2 ¼ 2 4 1 : dy2 QT y Q2T y2 Detection of a frequency-modulated signal is usually carried out by ﬁrst converting it into an Combining this with Eq. 14.11, we have amplitude-modulated signal by a device called a discriminator and then applying the A.M. signal d 2 jT j 3jT j5 ¼ : ð14:13Þ to an ordinary A.M. detector. The circuit dy2 Q2T y4 arrangement of the discriminators used in the high-frequency range may be looked upon as Again, consisting of two channels, each containing an djT j d jT j dy inductance–capacitance circuit. The two LC cir- ¼ dx dy dx cuits are tuned to two different frequencies f1 and f2 such that (f1 * f2) is slightly greater than d 2 jT j d jT j d 2 y d2 jT j dy 2 ) ¼ þ : twice the peak deviation and (f1 + f2)/2 is equal dx2 dy dx2 dy2 dx to the carrier frequency of the F.M. wave to be detected. The difference between the rectiﬁed Substituting the values of d|T|/dy and d2|T|/dy2 outputs of the two channels then varies linearly from Eqs. 14.12 and 14.13, we have with frequency in the frequency range of interest. ( ) In the low-frequency range, the two tuned cir- d2 jTj jTj3 d2 y 3jTj2 dy 2 ¼ 2 3 : ð14:14Þ cuits are replaced by two parallel-T RC networks dx2 QT y dx2 y dx [5, 6] whose rejection frequencies are chosen in the same manner as f1 and f2 in the Also, high-frequency circuit. A single parallel-T net- dy d2 y work can also be used as a discriminator if it can ¼ 1 þ 1=x2 and 2 ¼ 2=x3 : ð14:15Þ be so designed that a linear relation exists dx dx between the amplitude transfer function (|T|) and Combining Eq. 14.14 with Eqs. 14.11 and 14.15 the frequency (x) in the frequency range of gives interest. It will be shown that this condition is approximately satisﬁed when the network d 2 jT j 1 ¼ 3=2 selectivity is nearly equal to its maximum dx2 Q2T y3 1 þ Q21y2 value. 8 T 2 9 From Eqs. 14.7 and 14.9, we can write <2 3 1 þ x12 = þ : :x3 y 1 þ 1 ; 1 2 QT y 2 jT j ¼ 1=2 f1 þ 1=ðQ2T y2 Þg For a perfectly linear curve, the ﬁrst differential coefﬁcient is a constant and the second 122 14 Design of Parallel-T Resistance–Capacitance Networks For … differential coefﬁcient is zero. Thus, the value of It is natural to suggest that x0 should be cho- d2|T|/dx2 (neglecting sign) is a measure of the sen to be somewhere near the centre of the band departure from linearity, the least value corre- 0 < x < 1 so that with the carrier frequency sponding to maximum linearity. From the above, coincident with x0, a frequency deviation of the we see that d2|T|/dx2 is a function of both x and order of 50 per cent of the carrier frequency can QT so that for a particular value of QT, the lin- be detected. Since, however, |T| ! 1 as x ! 0, earity varies from point to point. there will be a considerable deviation from lin- In the particular application considered, the earity at very low frequencies. We thus choose x0 frequency range of interest is 0 < x < 1. In this to be nearer to 1 than to 0. Let x0 = 0.55; then the range, y is negative and the expression within the required value of QT is 0.485, which is very near second bracket can be made zero, i.e. perfect to its maximum value. The improvement in lin- linearity can be attained at a single frequency by earity as QT approaches this value will be evident suitably choosing QT. If the normalized value of from Fig. 14.4, where the magnitude of the this frequency is denoted by x0 and y0 = x0–1/x0, amplitude transfer function has been plotted in the required value of QT is given by the band 0 x 1 for various values of QT. The curve for QT = 0.495 is appreciably linear 1=2 ð1=y0 Þ over the range 0.2 < x < 1. QT ¼ : ð14:16Þ 1:5x0 ð1 þ 1=x20 Þ þ y0 3 Fig. 14.4 Showing the 1.0 selectivity curves of the parallel-T network for QT = 0.250, 0.350 and 0.495 0.8 0.6 QT =0 .49 0.3 0 0.2 5 50 5 0.4 T 0.2 0 0.2 0.4 0.6 0.8 1.0 x Selectivity of an Amplifier Using the General Parallel-T RC … 123 Selectivity of an Amplifier Using Thus with the network considered previously, the General Parallel-T RC Network in the Negative Feedback Line QA ¼ 0:475ðG0 þ 1Þ In a low-frequency selective ampliﬁer, a while with the conventional symmetrical parallel-T RC network is used in the negative network, feedback line. If the open-loop gain of the QA ¼ 0:25ðG0 þ 1Þ: ampliﬁer is G0, then the gain with feedback is G ¼ G0 =ð1 þ G0 TÞ: For the same open-loop gain of 50 (say), the values of QA in these cases are respectively 24.20 Combining this with Eq. 14.7, we get and 12.75 while for the same QA of value 12.75, the ampliﬁer with the asymmetrical network need have a gain of 26 only. 1þ j QT 1x x 2 G ¼ G0 G0 þ 1 þ j QT 1x x 2 8 91=2 Conclusion 2 > < 1þ 1x x 1 > = 2 QT ) jGj ¼ G0 2 : In situations where a continuous adjustment of :ðG0 þ 1Þ2 þ 1 x 2 > > ; the rejection frequency is desired, a general QT 1x conﬁguration will, of course, be of limited applicability, as the elements of the same kind The resonant gain is G0; the gain is 3 dB. below are neither equal nor simply related. But for a this value at frequencies given by |G| = 2−1/2 G0 ﬁxed rejection frequency, a general network with which on simpliﬁcation reduces to the following: proper asymmetry will deﬁnitely be a better 2 choice. Also in its application as an F.M. dis- 1 x ¼ G20 þ 2G0 1: criminator in the low-frequency range, a value of QT 1 x2 QT nearly equal to its maximum value is required. Thus, the design procedure given in the The solutions of this equation are chapter will be of much use in these situations. ( 1=2 ) 1 1 1 x1;2 ¼ þ 4G00 ; 2G00 Q2T QT Problems where P:1. Determine the transfer function of Fig. 14.1 circuit if m1 = 0 and comment on the kind G00 ¼ ðG20 þ 2G0 1Þ2 : 1 of ﬁltering it can do. P:2. Same if m2 = 0. Thus the selectivity of the ampliﬁer is P:3. Same if m3 = 0. P:4. Same if n1 = n2 = 0 QA ¼ 1=ðx2 x1 Þ ¼ G00 QT : P:5. Same if n3 = ∞. For G0 > 20, G00 ’G0 þ 1 to within an error of less than 0.25% so that Acknowledgments The author is indebted to Prof. J. N. Bhar, D.Sc., F.N.I., and to Dr. A. K. Choudhury, M.Sc., D. Phil., for their kind help and advice in the preparation QA ’ ðG0 þ 1ÞQT : of this chapter. 124 14 Design of Parallel-T Resistance–Capacitance Networks For … References 4. D. Morris, Q as a mathematical parameter. Electron. Eng. 306 (1954) 5. J.R. Tillman, Linear frequency discriminator. Wirel. 1. L. Stanton, Theory and applications of parallel-T Engr. 23, 281 (1946) resistance capacitance frequency selective network. 6. Paul T. Stine, Parallel-T discriminator design tech- Proc. IRE. 34, 447 (1946) nique. Proc. Natl. Elec. Conf. IX, 26 (1950) 2. A. Wolf, Note on a parallel-T resistance capacitance 7. H. Fleischer, in Vacuum Tube Ampliﬁers, ed. by G.E. network, Proc. I.R.E. 34, 659 (1946) Valley Jr. and H. Wallman (McGraw-Hill, 1948), 3. Y. Oono, Design of parallel-T resistance capacitance Chap. 10, p. 394 network. Proc. I.R.E. 43, 617 (1953) Perfect Transformer, Current Discontinuity and Degeneracy 15 That on connecting a source in the primary It is not common to ﬁnd an analysis of coupled circuit of a perfectly coupled transformer, the coils with initial currents in textbooks on circuit currents in both the primary and secondary theory. Somehow, in the large number of books coils may be discontinuous does not appear to consulted by the author, it is always assumed that have been widely discussed in the literature. the coils are initially relaxed and imperfectly In this discussion, we present an analysis of coupled. The only exception happens to be the the general circuit and show that in general, book by Kuo [1], where the circuit shown in the currents will be discontinuous, except for Fig. 15.1 has been analyzed with due regard to speciﬁc combinations of the initial currents in initial conditions. It has been shown that when the two coils. Although unity coupling coef- M 2 < L1L2, i.e. when the coefﬁcient of coupling pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬁcient cannot be realized in practice, a k ¼ M= L1 L2 \1, the currents i1 and i2 must be perfectly coupled transformer is a useful continuous at t = 0. On the other hand, if k = 1, concept in circuit analysis and synthesis, and then for the speciﬁc case i1(0–) = i2(0–) = 0, the the results presented here should be of interest currents are discontinuous, with to students as well as teachers of circuit theory. i1 ð0 þ Þ ¼ VL2 =ðR1 L2 þ R2 L1 Þ ð15:1Þ and Keywords i2 ð0 þ Þ ¼ VM=ðR1 L2 þ R2 L1 Þ ð15:2Þ Perfect transform Current discontinuity Degeneracy Kuo is, however, silent on what happens when the coils are not initially relaxed. The speciﬁc question is the following: Is k = 1 necessary as well as sufﬁcient for the currents to be discon- tinuous? We show, in this chapter, that this con- dition is necessary but not sufﬁcient. In other words, even for k = 1, the currents may display Source: S. C. Dutta Roy, “Perfect Transformer, Current continuity. First, we demonstrate this through an Discontinuity and Degeneracy”, IETE Journal of example. We next consider a more general circuit Education, vol. 43, pp. 135–138, July–September 2002. and analyze it to obtain expressions for i1(0+) and © Springer Nature Singapore Pte Ltd. 2018 125 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_15 126 15 Perfect Transformer, Current Discontinuity and Degeneracy R1 voltage source is generalized to v(t) instead of a M battery. We assume that v(t) does not contain t=0 + impulses. The loop equations now become V – I1 I2 R2 L1 L2 vðtÞ ¼R1 i1 ðtÞ þ L1 i01 ðtÞ þ Mi02 ðtÞ Zt 1 ð15:5Þ þ i1 ðtÞ dt þ v1 ð0Þ C1 Fig. 15.1 The circuit analyzed by Kuo [1] and used in 0 the example of this chapter with speciﬁc values 0 ¼Mi01 ðtÞ þ R2 i2 þ L2 i02 ðtÞ Zt i2(0+) in terms of circuit parameters, source value 1 ð15:6Þ þ i2 ðtÞ dt þ v2 ð0Þ at t = 0+, and currents and voltages in the circuit C2 0 at t = 0−. We then derive the condition for cur- rent continuity in a perfectly coupled transformer. The sum of the last two terms on the right-hand Finally, we consolidate the main results of the side of Eq. 15.5 represents v1(t) = q1(t)/C1, where chapter and make some concluding remarks. q1(t) denotes the charge on C1. Similarly, v2(t) = q2(t)/C2. Integrals of v1(t), v2(t) as well as v(t) from t = 0− to t = 0+ will be zero because none of An Example them contains impulses. Thus, if we integrate Eqs. 15.5 and 15.6 from t = 0− to t = 0+, For the sake of completeness and for ready ref- we get erence, we include, briefly, the analysis and results of Kuo for the circuit shown in Fig. 15.1 0 ¼ L1 ½i1 ð0 þ Þ in Appendix A. Let i1 ð0Þ þ M ½i2 ð0 þ Þ i2 ð0Þ ð15:7Þ 9 L1 ¼ 4H; L2 ¼ 1H; M ¼ 2H = 0 ¼ M ½i1 ð0 þ Þ i1 ð0Þ þ L2 ½i2 ð0 þ Þ i2 ð0Þ R1 ¼ 8X; R2 ¼ 3X; V ¼ 6V ð15:3Þ ; ð15:8Þ i1 ð0Þ ¼ 0 and ið0Þ ¼ 1A From Eqs. 15.25 and 15.26, then, we get These are the same as in Kuo’s circuit, as given in Eqs. 15.20 and 15.21. Note that Eqs. 15.7 and i1 ð0 þ Þ ¼ 0 15.8 imply that the principle of conservation of ð15:4Þ i2 ð0 þ Þ ¼ 1A flux applies to each coil individually, i.e. the flux in either coil at t = 0− is the same as that at t = 0+. Hence, the currents are continuous despite k = 1. Also note that the generalized circuit does not This counterexample is sufﬁcient to prove that change the conclusion arrived at in Kuo’s circuit, k = 1 is only a necessary but not a sufﬁcient viz. that if k < 1, then the currents in the two coils condition for current discontinuity. must be continuous. For the case k = 1, Eq. 15.6 gives, at t = 0+, the following equation: Analysis of the General Circuit 0 ¼R2 i2 ð0 þ Þ þ ðM=L1 Þ½L1 i01 ð0 þ Þ ð15:9Þ We now consider the general circuit shown in þ Mi02 ð0 þ Þ þ v2 ð0Þ; Fig. 15.2 which includes an initially charged capacitor in each loop and, in addition, the which can be rewritten as Analysis of the General Circuit 127 R1 Condition for Continuity of Currents M Under Perfect Coupling t=0 + v(t) If the currents are to be continuous, then it – I1 L1 I2 R2 L2 sufﬁces to equate Eq. 15.14 to i1(0−) or C1 C2 Eq. 15.15 to i2(0−) because from Eq. 15.7 or Eq. 15.8, i1(0+) = i(0−) guarantees that – v1(t) + +v2(t) – i2(0+) = i2(0−), and vice versa. Equating Eq. 15.15 to i2(0−) gives the following condition Fig. 15.2 A more general circuit than that shown in Fig. 15.1 with a generalized source v(t), and initially for continuity: charged capacitors in each loop R1 Mi1 ð0Þ M½vð0 þ Þ v1 ð0Þ L1 v2 ð0Þ L1 i01 ð0 þ Þ þ Mi02 ð0 þ Þ ¼ ðL1 =MÞ i2 ð0Þ ¼ R2 L1 ð15:10Þ ½R2 i2 ð0 þ Þ þ v2 ð0Þ ð15:16Þ Now, putting t = 0+ in Eq. 15.5 and substituting with i1(0−) arbitrary. In other words, for every from Eq. 15.10, we get i1(0−), there exists one i2(0−) for the currents to be continuous and vice versa. For all other com- vð0þ Þ¼R1 i01 ð0þ Þ binations of i1(0−) and i2(0−), the currents will be ðL1 =MÞ½R2 i2 ð0þ Þþv2 ð0Þþv1 ð0Þ: discontinuous. It is, of course, implied that other ð15:11Þ conditions, viz., v(0+), v1(0−) and v2(0−), do not change. Should that be the case, it is clear that the Combining Eq. 15.11 with 15.7, we get the fol- relationship between i2(0−) and i1(0−), as given lowing two simultaneous equations in i1(0+) and by Eq. 15.16, is a straight line with a slope of i2(0+): R1M/(R2L1) and an intercept of R2 L1 i1 ð0 þ Þ i2 ð0 þ Þ R1 M M½vð0 þ Þ v1 ð0Þ þ L1 v2 ð0Þ vð0 þ Þ v1 ð0Þ þ ðL1 =MÞv2 ð0Þ ð15:17Þ ¼ R2 L1 R1 ð15:12Þ on the i2(0−) axis. For the example considered earlier, the slope is 4/3 while the intercept is M M −1 A. i1 ð0 þ Þ þ i2 ð0 þ Þ ¼ i1 ð0Þ þ i2 ð0Þ L1 L1 ð15:13Þ Solving Eqs. 15.12 and 15.13 gives, ﬁnally, R2 ½L1 i1 ð0Þ þ Mi2 ð0Þ þ L2 ½vð0 þ Þ v1 ð0Þ þ Mv2 ð0Þ i1 ð0 þ Þ ¼ ð15:14Þ R1 L2 þ R2 L1 R1 ½L2 i2 ð0Þ þ Mi1 ð0Þ þ M½vð0 þ Þ v1 ð0Þ þ L1 v2 ð0Þ i2 ð0 þ Þ ¼ ð15:15Þ R1 L2 þ R2 L1 128 15 Perfect Transformer, Current Discontinuity and Degeneracy Concluding Remarks by the two examples in [1] (pp. 124–126). For k < 1 in the circuit shown in Fig. 15.1, the sys- We have shown in this chapter that in an tem has two natural frequencies, although it has imperfectly coupled transformer, the currents in three inductances L1, L2 and M. They are not the two coils are always continuous. For perfect physically connected at a junction, but in the coupling, on the other hand, the currents are equivalent1 circuit shown in Fig. 15.3, we do always discontinuous except for speciﬁc combi- have a junction of L1 − M, M and L2 − M. This nations of the two initial currents. More speciﬁ- is not an ‘effective’ junction in the sense of [2], cally, for each initial current in one coil, there but we may call it an ‘equivalent’ junction. It is exists a particular value of the initial current in no wonder, therefore, that we get the the other coil, for which the currents will be second-order system, instead of the third-order continuous. These combinations lie on a straight one. An alternative way of justifying the result is line, when one current is plotted against the to note that we can specify only two initial other. These conclusions are valid for any com- conditions for the system, the initial current in bination of resistors and capacitors in the two M being dependent on those in L1 and L2. loops, with or without initial charges in the However, despite the degeneracy, there is no capacitors. It is obvious, however, that including discontinuity in the currents! another inductor in either or both loops makes When k = 1, further degeneracy sets in, not the coupling imperfect, and the currents will then because we cannot specify two initial currents, be always continuous. but (in our opinion) because M is completely In this context, the following two observations speciﬁed if L1 and L2 are speciﬁed. As the second made by Seshu and Balabanian [2] are of example of [2] (pp. 125–126) demonstrates, the interest: system now has only one natural frequency and behaves like the ﬁrst-order system. Despite this (1) ‘If idealized R, L and C branches, voltage ‘double’ degeneracy, however, the currents are generators, and current generators are arbi- not always discontinuous, as demonstrated in this trarily connected together, the system may chapter analytically and by an example. It is clear not have the maximum possible order. It is that a deeper examination of the case is needed to only when such degeneracies are present that resolve the issue in terms of physical concepts.2 discontinuities in inductance currents and capacitance voltages are encountered. No existence theorems have been proved by Problems mathematicians for these cases …’ (p. 103). (2) ‘… it may be expected that inductance cur- P:1. Suppose R1 in Fig. 15.1 circuit is shunted rents will be discontinuous when there are by a capacitor C. Investigate the disconti- junctions or effective junctions… at which nuity in this circuit. only inductances and current generators are P:2. Same, with C shifted to be across R2. present’ (p. 104). P:3. Same, with C shifted to be in series with R2. P:4. Same, with C in series with R1. By an effective junction in the second obser- vation, the authors mean ‘a junction at which 1 This ‘equivalent’ circuit implies only mathematical only inductances and current generators would equivalence (of the loop equations) but not physical meet if we suitably interchanged series connected equivalence, because the two coils have no common two terminal networks or shorted some branches. point. 2 Thus an effective junction is the same as a cut Notably, in [2], there are no examples or discussions on set’ (p. 104, footnote). initial conditions in coupled coils. In the only example in which a coupled coil appears (pp 110–112), inductor The ﬁrst observation regarding degeneracy is junctions are created through additional inductors in each clearly demonstrated in the case of coupled coils circuit and a current generator in the secondary circuit. Problems 129 Fig. 15.3 A mathematical L2 – M L1 – M equivalent circuit for two coupled coils: coupling is not always good! M M L1 ≡ L2 P:5. Same, with C1 in series with R1, C2 in series which, along with Eq. 15.20 or 15.21, clearly with R2, and a voltage output taken across indicates that if L1L2 > M2, i.e. k < l, then the R2. currents are continuous at t = 0. On the other hand, if k = 1, then they need not be. In fact, in this case, Eq. 15.19 gives at t = 0+: Appendix R2 i2 ð0 þ Þ ¼ ðM=L1 Þ L1 i01 ð0 þ Þ þ Mi02 ð0 þ Þ ð15:23Þ Kuo’s analysis and results for the circuit are shown in Fig. 15.1. which, substituted in Eq. 15.18 with t = 0+, The loop equations for the circuit shown in yields Fig. 15.1 are V ¼ R1 i1 ð0 þ Þ ðL1 =MÞR2 i2 ð0 þ Þ ð15:24Þ VuðtÞ ¼ L1 i01 ðtÞ þ R1 i1 ðtÞ þ Mi02 ðtÞ ð15:18Þ Combining this with Eq. 15.20, one can solve 0 ¼ Mi01 ðtÞ þ R2 i2 ðtÞ þ L2 i02 ðtÞ ð15:19Þ for i1(0+) and i2(0+). The results are3 Integrating Eqs. 15.18 and 15.19 from t = 0− VL2 þ R2 ½L1 i1 ð0Þ þ Mi2 ð0Þ to t = 0+, we get i1 ð0 þ Þ ¼ R1 L2 þ R2 L1 ð15:25Þ L1 ½il ð0 þ Þ i1 ð0Þ þ M ½i2 ð0 þ Þ i2 ð0Þ ¼0 VM þ R1 ½Mi1 ð0Þ þ L2 i2 ð0Þ i2 ð0 þ Þ ¼ ð15:20Þ R1 L2 þ R2 L1 ð15:26Þ M ½i1 ð0 þ Þ il ð0Þ þ L2 ½i2 ð0 þ Þ i2 ð0Þ ¼0 ð15:21Þ References Combining Eqs. 15.19 and 15.21 gives 1. F.F. Kuo, Network Analysis and Synthesis (John Wiley, New York, 1966), pp. 123–126 ðL1 L2 2. S. Seshu, N. Balabanian, Linear Network Analysis M 2 Þ½i1 ð0 þ Þ i1 ð0Þ½i2 ð0 þ Þ i2 ð0Þ (John Wiley, New York, 1963), pp. 101–112 ¼0 ð15:22Þ 3 Kuo [1], at this point, assumes i1(0–) = i2(0–), presum- ably, as an example. We give general results in Eqs. 15.25 and 15.26. Analytical Solution to the Problem of Charging a Capacitor Through 16 a Lamp An analytical solution is presented for the R varies with the current i flowing through it. problem of charging a capacitor through a They solved the resulting differential equation by lamp, by assuming a polynomial relationship applying numerical techniques and found a close between the resistance of the lamp and the ﬁt between these results and the experimental current flowing through it. The total energy ones. The aim of this chapter is to present an dissipated in the lamp is also easily calculated analytical, rather than numerical solution to the thereby. An example of an available practical problem. For this purpose, we assume a poly- case is used to illustrate the theory. nomial relationship between R(i) and i. The total energy dissipated in the lamp is also easily cal- culated thereby. The experimental data of RV are Keywords used to illustrate the validity of the theory. Capacitor charging Differential equation Energy The Circuit and the Differential Equation Introduction The circuit under consideration is shown in Fig. 16.1, which obeys the integral equation The charging of a capacitor from a battery Zt through a resistance is a standard topic in the 1 iRðiÞ þ i dt ¼ V ð16:1Þ undergraduate curriculum of Physics or Engi- C neering in the theory as well as laboratory clas- 0 ses. A 2006 paper by Ross and Venugopal [1] Differentiating Eq. 16.1, we get (hereafter referred to as RV) deals with an interesting variation of this topic in which the di dR i resistor is replaced by a lamp, whose resistance R þi þ ¼ 0; ð16:2Þ dt dt C where, for brevity, the dependence of R on i is not shown explicitly. Assuming, as in RV, that Source: S. C. Dutta Roy, “Analytical Solution to the the thermal relaxation time of the lamp ﬁlament Problem of Charging a Capacitor through a Lamp,” is much less than the time during which the IETE Journal of Education, vol. 47, pp. 145–147, July– September 2006. © Springer Nature Singapore Pte Ltd. 2018 131 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_16 132 16 Analytical Solution to the Problem of Charging a Capacitor … LAMP C dR ¼ R0 ða1 þ 2a2 i þ 3a3 i2 Þ ð16:7Þ R(i ) di i Combining Eq. 16.4 with Eqs. 16.6 and 16.7, we get, on simpliﬁcation, V t=0 ½ð1=iÞ þ 2a1 þ 3a2 i þ 4a3 i2 di ¼ dt=ðR0 CÞ: Fig. 16.1 The basic charging circuit ð16:8Þ Integrating both sides of Eq. 16.8 gives current in the ﬁlament changes signiﬁcantly, we can write ln i þ 2a1 i þ ð3=2Þa2 i2 þ ð4=3Þa3 i3 dR dR di ¼ ½t=ðR0 CÞ þ K: ð16:9Þ ¼ : ð16:3Þ dt di dt To evaluate the integration constant K, we Combining Eqs. 16.2 and 16.3, we get note that at t = 0, i = i0 = V/(R0). Putting this initial condition in Eq. 16.9, we get the value of dR di i K as the left hand side of Eq. 16.9 with i replaced Rþi þ ¼ 0: ð16:4Þ di dt C by i0. Finally, therefore, the equation for the current becomes Solution of the Differential Equation t ¼ R0 C½Inði0 =iÞ þ 2a1 ði0 iÞ þ ð3=2Þa2 ði20 i2 Þ þ ð4=3Þa3 ði30 i3 Þ: As illustrated in Fig. 3 of RV, the variation of R ð16:10Þ (i) with i is approximately linear, except at high values of i. In general, we can assume R and i to Equation 16.10 is transcendental in i and for a obey a polynomial relationship of the form given t, it has to be solved numerically. A better ! strategy would be to compute t for various values X N of i in the range of interest and then to plot the R ¼ R0 1 þ ak i k ; ð16:5Þ variation of i with t, as we shall do in the k¼1 example to follow. where N will depend upon the required accuracy. For most practical situations, N = 2 or 3 sufﬁces. We shall consider here a third order polynomial, Energy Dissipated in the Lamp but if required, the treatment can be extended to The energy dissipated in the lamp is given by any order. Let, therefore, Z1 R ¼ R0 ð1 þ a1 i þ a2 i2 þ a3 i3 Þ ð16:6Þ E¼ RðiÞi2 dt ð16:11Þ Then 0 Energy Dissipated in the Lamp 133 Combining Eq. 16.11 with Eq. 16.6, substituting A plot of Eq. 16.14 is shown in Fig. 16.2, which, for dt from Eq. 16.8, changing the limits of the as predicted by RV, is virtually indistinguishable integral (from t = 0 to i = i0 and t = ∞ to i = 0) from that given in Fig. 4 of their paper. and simplifying, we get Zt0 E¼ CR20 i þ 3a1 i2 þ 4a2 þ 2a21 i3 þ 5ða3 þ a1 a2 Þi4 þ 6a1 a3 þ 3a22 i5 þ 7a2 a3 i6 þ 4a23 i7 di 0 2 2 i0 a21 4 a22 6 a23 8 ¼ CR0 þ a1 i0 þ a2 þ 3 i þ ða3 þ a1 a2 Þi0 þ a1 a3 þ 5 i þ a2 a3 i0 þ i0 7 2 2 0 2 0 2 ð16:12Þ The total energy dissipated in the lamp for this case is given by Example 2 i0 a21 4 E¼ CR20 þ a1 i 0 þ i 0 ; 3 ð16:15Þ 2 2 We use the experimental data given in RV to illustrate the application of the theory presented which is calculated as 2.772 J. here. As mentioned earlier, Fig. 3 of RV shows that the variation of R(i) with i is predominantly linear. By considering the two points (0.03 A, Conclusion 10 X) and (0.07 A, 20 X) in this ﬁgure, we get RðiÞ ¼ 2:5ð1 þ 100iÞ ð16:13Þ It is shown that if the functional dependence of the lamp resistance on current is known in the form of With C = 0.154 F and i0 = 0.15 A (as given in a polynomial relationship, then the charging Fig. 4 of RV), Eq. 16.10 becomes, for this case, process of a series capacitor can be analytically determined. It is then also easy to determine the t ¼ 0:385ð28:1 ln i 200iÞ: ð16:14Þ energy dissipated in the lamp during the charging process. It is easily shown that the discharging of 0 a charged capacitor through a lamp also follows 10 Eq. 16.2 and hence the theory presented here also applies to the discharging process. –1 Current (i) 10 Problems P:1. In the circuit of Fig. 16.1, add an inductor L in series. Write the differential equation and –2 10 solve it. 0 2 4 6 8 10 12 P:2. Let, in Fig. 16.1, C be shifted to be across Time (t ) the lamp. Obtain the differential equation Fig. 16.2 Variation of i with t for the example and solve it. 134 16 Analytical Solution to the Problem of Charging a Capacitor … P:3. Can you solve Eq. 16.10 analytically? After Acknowledgements The author thanks Professor Jaya- all, it is a cubic equation, and can be solved deva for his help in the preparation of Fig. 16.2. by Cardan’s method. Try it. P:4. What happens if Eq. 16.6 has another term Reference a 4t 4? P:5. Repeat the example in the text with an extra 1. R. Ross, P. Venugopal, On the problem of (dis) charg- term 10i2 in Eq. 16.13. ing a capacitor through a lamp. Am. J. Phys. 74, 523– 525 (2006) Difference Equations, Z-Transforms and Resistive Ladders 17 It is shown that the semi-inﬁnite and inﬁnite equations. KCL (Kirchoff’s Current Law), KVL resistive ladder networks composed of identi- (Kirchoff’s Voltage Law) and Ohm’s law should cal resistors can be conveniently analyzed by be adequate for dealing with such networks. Yet, the use of difference equations or z-trans- there are situations where the use of difference forms. Explicit and simple expressions are equations and/or frequency domain techniques obtained for the input resistance, node volt- offers signiﬁcant advantages over conventional ages and the resistance between two arbitrary methods. This chapter is concerned with one nodes of the network. such situation, viz. a semi-inﬁnite or inﬁnite resistive ladder network. The semi-inﬁnite resistive ladder network Keywords shown in Fig. 17.1 is often posed as a problem Inﬁnite networks Resistive ladders [1, 2] to undergraduate students for ﬁnding the Difference equations Z-transforms input resistance Ri = V0/I0. The solution is easily found by noting that the resistance looking to the right of nodes 1 and ground should also be Ri. Thus, Introduction R2i RRi R2 ¼ 0 ð17:1Þ Difference equations and z-transforms are tech- which gives the quadratic equation niques for dealing with discrete time signals and systems, of which the former is in time domain R2i RRi R2 ¼ 0: ð17:2Þ and the latter is in the frequency domain. Anal- ysis of a purely resistive network does not nor- Noting that Ri must be positive, we get mally require any tool in the frequency domain, pﬃﬃﬃ neither does the network process discrete time Ri ¼ R 1 þ 5 =2 ¼ RU; ð17:3Þ signals so as to require the use of difference where U is the so-called ‘golden ratio’. What about the potential vn at node n, 1 Source: S. C. Dutta Roy, “Difference Equations, n < ∞? Solution of this problem appears in [3] Z-Transforms and Resistive Ladders,” IETE Journal of in the form of an integral obtained by using the Education, vol. 52, pp. 11–15, January–June 2011. concept of discrete Fourier transform. It will be © Springer Nature Singapore Pte Ltd. 2018 135 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_17 136 17 Difference Equations, Z-Transforms and Resistive Ladders 0 I0 1 2 n 1 n n+ V0 ... ... ... to infinity Fig. 17.1 The semi-inﬁnite resistive ladder. Each resistance is of value R shown in this chapter that the solution can be analyzed a semi-inﬁnite ladder in which the obtained in a simpler form by using the theory of resistors in the successive sections differ by a difference equations or by application of the z- factor of b. He showed that by choosing b ap- p transform technique. In the process, we have also propriately, one can obtain the golden ratio, 2 considered the inﬁnite resistive ladder of and some other irrational numbers in a Fig. 17.2 and have calculated the resistance non-geometric context. Parera-Lopez [7] made offered to a battery connected between two some generalizations of [5, 6]. Denardo et al. [8] arbitrary nodes of this inﬁnite ladder. presented some numerical and laboratory exper- Besides [3], there exists a substantial volume iments on ﬁnite N-section ladders and showed of literature on the subject of semi-inﬁnite and that the convergence of the input resistance to inﬁnite resistive ladders. Some of the prominent RU is exponential and rapid. For example, for ones, which are of educational and pedagogic N 5, the deviation from RU occurs only in the interest, will be reviewed here. Lavatelli [4] fourth place of decimal, while for N 7, the considered an inﬁnite balanced ladder i.e. one in deviation occurs in ﬁfth place of decimal. which the lower ground line of Fig. 17.2 is Bapeswara Rao [9] related the ﬁnite resistance replaced by a chain of resistors. He gave a dif- ladder to the effective resistance between the ference equation formulation for the resistance centre and a vertex of an N-sided polygon of between two arbitrary nodes. resistors. Our treatment here in Part IV has been Besides these papers of pedagogic interest, inspired by his work and follows the same line of there have appeared many scholarly papers on analysis. Srinivasan [5] considered the inﬁnite networks in IEEE and other professional semi-inﬁnite ladder with different values of series journals, the most prominent author being and shunt resistors and showed that when they Zemanian (see, e.g. [10, 11] and the references are equal, the input resistance is RU, as in cited there). Reference [10] deserves special Eq. 17.3. He also showed that the successive mention because it is a tutorial paper addressed convergents of the continued fraction form of the to undergraduate students in a rather unique and input resistance are related to the Fibonacci enjoyable style. Zemanian’s book [12] gives a sequence. As an extension of [5], Thomson [6] comprehensive treatment of the subject with the necessary mathematical rigour. 1 0 +1 Solution by Difference Equation to ... ... ... to infinity I0 Approach infinity Consider, in Fig. 17.1, the nodes n − 1, n and n + 1, n > 0. By writing KCL at node n and Fig. 17.2 Inﬁnite resistive ladder driven by a current simplifying, we get source I0 at node 0. Each resistance has a value R Solution by Difference Equation Approach 137 3vn vn1 vn þ 1 ¼ 0: ð17:4Þ we have Z ½dðnÞ ¼ 1; Z ½vn þ 1 ¼ zV ðzÞ; This is a difference equation of order 2, and ð17:11Þ assuming a solution of the form kn, we get the Z ½vn1 ¼ z1 V ðzÞ: characteristic equation Thus taking the z-transform of both sides of k2 3k þ 1 ¼ 0: ð17:5Þ 17.9 and simplifying, we get The solution of Eq. 17.5 are VðzÞ z1 ¼ I0 R 1 3z1 þ z2 pﬃﬃﬃ z1 k1; 2 ¼ 3 5 =2: ð17:6Þ ¼ ; ð17:12Þ ð1 az Þð1 z1 =aÞ 1 Note that k1k2 = 1; for convenience, we shall where a is the same as that given by 17.7. call k1 as a so that k2 = a−1. Thus, the general Expending 17.12 in partial fractions and using solution for vn is p the fact that a 1=a ¼ 5, we get pﬃﬃﬃ vn ¼ Aan þ Ban ; a ¼ 3 þ 5 =2: ð17:7Þ VðzÞ 1 1 1 ¼ pﬃﬃﬃ : ð17:13Þ I0 R 5 1 z1 =a 1 az1 The constants A and B are evaluated from the boundary conditions v0 = V0 and v∞ = 0, the The pole at z = a is outside the unit circle latter being dictated by physical considerations. while that at z = 1/a is inside the unit circle. The The second condition forces A to be zero while physical situation demands that the sequence vn the ﬁrst one makes B = V0. Thus, ﬁnally, should decrease on both sides of n = 0 and tend h pﬃﬃﬃ in to zero when n ! ∞. Hence, the ﬁrst term in v n ¼ V0 3 5 =2 : ð17:8Þ 17.13 represents the z-transform of the right-sided sequence {v0, v1, … to ∞} with |z| < a as the region of convergence, while the Z-Transform Solution second term represents the z-transform of the left-sided sequence {v−1, v−2, … to ∞} with To apply the z-transform technique [13], it is |z| < a as the region of convergence. Thus, the instructive to consider the inﬁnite ladder of inversion of Eq. 17.13 gives Fig. 17.2, with a current generator I0 connected vn 1 between node 0 and ground. Then the difference ¼ pﬃﬃﬃ ½an uðnÞ þ an uðn 1Þ; ð17:14Þ I0 R 5 equation 17.4 is modiﬁed to the following: where u(n) is the unit step function, having the 3vn vn1 vn þ 1 ¼ I0 dðnÞ; ð17:9Þ value unity for n 0 and zero otherwise. More where d(n) = 1 for n = 0 and zero otherwise. explicitly, Deﬁning the z-transform in the usual manner, i.e. pﬃﬃﬃ!n I0 R 3 5 X 1 vn ¼ pﬃﬃﬃ ; n 0; ð17:15Þ n 5 2 Z ½vn ¼ VðzÞ ¼ vn z ; ð17:10Þ n¼1 138 17 Difference Equations, Z-Transforms and Resistive Ladders pﬃﬃﬃ!n The only difference between 17.18 and 17.4 is I0 R 3 þ 5 vn ¼ pﬃﬃﬃ ; n\0 ð17:16Þ that the right-hand side in the former is not zero. 5 2 Hence we shall have a constant term, represent- ing the particular solution of Eq. 17.18, in addi- This gives the complete solution for the inﬁ- tion to the solution of the form given by nite ladder of Fig. 17.2. The resistance seen by Eq. 17.7. It is easily seen from Eq. 17.18 that the current generator I0 is this constant term is I0. Thus the solution of pﬃﬃﬃ Eq. 17.18 is R1 ¼ RRi Ri ¼ R= 5 ð17:17Þ pﬃﬃﬃ p in ¼ Aan þ Ban þ I0 ; a ¼ 3 þ 5 =2: so that v0 ¼ I0 R= 5; this veriﬁes that Eq. 17.15 gives correct results for the semi-inﬁnite ladder, ð17:19Þ as derived independently in Eq. 17.8. Also, as expected, vn = v−n, and both tend to zero as The constant A and B have to be determined n ! ∞. from the boundary conditions that hold at nodes m and m + r. Since the network is perfectly symmetrical with respect to an imaginary vertical Resistance Between Any Two line at the middle, the voltages at nodes Arbitrary Nodes of an Infinite Ladder m + r and m are, respectively, +V0/2 and −V0/2. Thus We now consider another relevant problem in the inﬁnite ladder, viz. that of ﬁnding the resistance i1 ¼ ir ¼ V0 =ð2RT Þ: ð17:20Þ offered to a source connected between any two Combining Eqs. 17.19 and 17.20, we get two arbitrary nodes m and m + r. Let the source be a simultaneous equations in A and B, the solution voltage generator V0 and let a set of r + 1 mesh of which gives currents be formulated as shown in Fig. 17.3, where the last mesh includes V0 and the network ½V0 =ð2RT Þ I0 1 r to the left of node m and that to the right of node ðA; BÞ ¼ a ;a : ð17:21Þ 1 þ ar1 m + r have been replaced by an equivalent p resistance RT ¼ RjjRi ¼ ð 5 1Þ=2. Consider Thus, ﬁnally, the nth mesh, 1 < n < r. Writing KVL around this mesh gives the equation ½V0 =ð2RT Þ I0 n1 n þ r in ¼ I0 þ a þ a : 1 þ ar1 3in in1 in þ 1 ¼ I0 : ð17:18Þ ð17:22Þ I0 m m +1 m V0 /2 in 1 i1 in in+1 ir RT ... ... RT Fig. 17.3 Circuit for determining the resistance between any two arbitrary nodes m and m + r. Each unmarked resistance has a value R Resistance Between Any Two Arbitrary Nodes of an Infinite Ladder 139 Application of KVL around the (r + 1)th mesh earlier, the use of z-transforms is believed to be gives new and instructive. The explicit formulas for the node voltages and the resistance between two X r arbitrary nodes also appear to be new. V0 ¼ ðI0 in ÞR: ð17:23Þ n¼1 Combining Eqs. 17.22 and 17.23 gives Problems ½V0 =ð2RT Þ þ I0 Xr P:1. Suppose in Fig. 17.1, the ladder is termi- V0 ¼ r1 an1 þ an þ r : 1þa n¼1 nated at the third node on the right. What impedance does I0 face? This is easy! ð17:24Þ P:2. Suppose, in Fig. 17.2, the current generator is replaced by a voltage generator and the Clearly, ladder is terminated in node X r X r 1 ar marked + n and −n. What current will flow an1 ¼ an þ r ¼ : ð17:25Þ from the generator? This is super-easy! n¼1 n¼1 1a P:3. Suppose, in Fig. 17.2, each shunt resistors or is replaced by a capacitor C. What is the Using Eq. 17.25 in Eq. 17.24 and simplify- input impedance? This is not so easy, but ing, we get not difﬁcult too! " pﬃﬃﬃ!n # P:4. Same as P.4, but each series resistor is 2R 3 5 Rr ¼ pﬃﬃﬃ 1 : ð17:26Þ replaced by a capacitor C. What is the input 5 2 impedance? Same level of difﬁculty as in P.3. It is easily veriﬁed by direct calculation that P:5. Same as P.5, but each shunt resistor is Eq. 17.26 give correct results for r = 1, 2 and 3, replaced by an inductance L. which are, respectively, pﬃﬃﬃ R1 ¼ Rð1 1= 5Þ; Acknowledgments This work was supported by the pﬃﬃﬃ Indian National Science Academy through the Honorary R2 ¼ Rð3 5Þ; ð17:27Þ Scientist scheme. pﬃﬃﬃ R3 ¼ 8Rð1 2= 5Þ: For a semi-inﬁnite ladder, the condition of References symmetry no longer holds and the appropriate boundary conditions have to be used at both the 1. E.M. Purcell, in Electricity and Magnetism, Berkeley Physics Course—Vol. 2, 2nd edn. (New York, end meshes 1 and r. For examples, if m = 1, then McGraw-Hill, 1985), pp. 167–168 the boundary conditions are i1 = ir = V0/(RT + 2. F.W. Sears, M.W. Zemansky, in College Physics, R); for m = 0, the resistance would be R + (the World Students, 5th edn. (Reading, MA, Addison-Wesley, 1980) value for m = 1); for m = 2, the boundary con- 3. R.M. Dimeo, Fourier transform solution to the ditions are i1 = ir = V0/(RT + 2/3) and so on. semi-inﬁnite resistance ladder. American J. Phys. 68(7), 669–670 (2000) 4. L. Lavatelli, The resistive net and difference equa- tion. American J. Phys. 40(9), 1246–1257 (1972, Concluding Discussion September) 5. T.P. Srinivasan, Fibonacci sequence, golden ratio and a network of resistors. American J. Phys. 60(5), We have used difference equations and z-trans- 461–462 (1992) form to analyze semi-inﬁnite and inﬁnite resis- 6. D. Thompson, Resistor networks and irrational tive ladders. While the former has been used numbers. American J. Phys. 65(1), 88 (1997) 140 17 Difference Equations, Z-Transforms and Resistive Ladders 7. J.J. Parera-Lopez, T-iterated electrical networks and 11. A.H. Zemanian, Inﬁnite electrical networks. Proc. numerical sequences. American J. Phys. 65(5), 437– IEEE 64(1), 1–17 (1976) 439 (1997) 12. A.H. Zemanian, Transﬁniteness for graphs, electrical 8. B. Denardo, J. Earwood, V. Sazonava, Experiments networks and random walks (Birkhauser, Boston, with electrical resistive networks. American J. Phys. MA, 1996) 67(11), 981–986 (1999, November) 13. S.K. Mitra, in Digital Signal Processing—A Com- 9. V.V. Bapeswara Rao, Analysis of doubly excited puter Based Approach, 3rd edn, Chapter 6 (New symmetric ladder networks. American J. Phys. 68(5), York, McGraw-Hill, 2006) 484–485 (2000) 10. A.H. Zemanian, Inﬁnite electrical networks: a reprise. IEEE Trans. Circuits Sys. 35(11), 1346– 1358 (1988) A Third-Order Driving Point Synthesis Problem 18 Minimal realizations of an interesting multiplying constant, which in this case is unity, third-order impedance function are discussed. is a hidden speciﬁcation. The solution, based on an elegant algebraic The order of the impedance function, deﬁned identity, illustrates several basic concepts of as the degree of the numerator or denominator, driving point function synthesis. whichever is higher, being three, we shall natu- rally require three reactive elements. The fourth element must then be a resistance. Can all the Keyword three reactive elements be of the same kind, viz. Driving point synthesis Third-order either inductance or capacitance? Having a pole impedance function at the origin (s = 0) obviously excludes an all inductor solution, because an RL impedance cannot have such a pole. How about all reactive elements being capacitances? That is, how about Introduction an RC realization of Eq. 18.1? Note that Eq. 18.1 has a pole at s = ∞ and we know that Consider the impedance function an RC impedance cannot have such a pole. Also note that Z(s) poles are at s = 0 and ð s þ aÞ ð s þ bÞ ð s þ c Þ s = −(a + b + c) while its zeros are at s = −a, Z ðsÞ ¼ ; ð18:1Þ sðs þ a þ b þ cÞ −b and −c. Since a + b + c > a, b as well as c, we conclude that poles and zeros of Z(s) do not where a, b and c are arbitrary non-negative real alternate. This alternation of poles and zeros is an quantities. The problem is to have a minimal essential requirement of RC or RL impedances. realization of Eq. 18.1, i.e. a realization which Hence we conclude that Eq. 18.1 can neither be uses no more than four elements. Why four? RL nor RC; if at all realizable, it must be RLC. Apparently, there are three speciﬁcations, namely a, b and c, but then you should realize that the Is Z(s) at All Realizable? The question that arises at this stage is the fol- Source: S. C. Dutta Roy, “A Third-Order Driving Point lowing: Is Z(s) at all realizable? It is known that Synthesis Problem,” Students’ Journal of the IETE, vol. Z(s) will be realizable if it is a positive real 36, pp. 179–183, October–December 1995. function (PRF) [1], i.e. if (i) Z(s) is real for s real © Springer Nature Singapore Pte Ltd. 2018 141 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_18 142 18 A Third-Order Driving Point Synthesis Problem and (ii) Re Z(s) 0 for Re s 0. There are is twice the residue of Z(s) at s ¼ jx1 , and many ways of testing for a PRF, but one Z3(s) is the remaining function to be tested. The pre-processing or simpliﬁcation that should ﬁrst term in Eq. 18.6 represents a parallel con- invariably be carried out is to look for poles and nection between an inductance K1 =x21 and a zeros on the jx-axis, including s = 0 and s = ∞, capacitance 1/K1. and to remove them. This step is the testing for a If instead of a pole, one ﬁnds one or more PRF is known as the ‘Foster preamble’. In par- zeros of Z(s) on the jx-axis, then one removes ticular, if Z(s) has a pole at the origin (s = 0), them from Y(s) = 1/Z(s) which will have a pole then one can write at those points. Here, removal of a pole at s = 0, s = ∞ and s ¼ jx1 corresponds to the removal K0 Z ðsÞ ¼ þ Z1 ðsÞ; ð18:2Þ of an inductance, capacitance and a series con- s nection of inductance and capacitance, respec- tively, all in parallel with the remaining where admittance function to be tested. K0 ¼ sZ ðsÞjs¼0 : ð18:3Þ It can be shown that if the original function was PR, then so is the remainder function after is the residue of Z(s) at the pole at s = 0 and removal of any pole or zero on the jx-axis. This, Z1(s) is the remaining function to be tested. The in fact, gives validity of the Foster preamble! But term K0/s obviously represents a capacitor of then, how are we simplifying the testing? Note value 1/K0. Naturally K0 has to be positive, that Z1(s) of Eq. 18.2 as well as Z2(s) of Eq. 18.4 otherwise no further testing is needed. In fact if Z will be one order less than Z(s), while Z3(s) of (s) is PRF, then it can be shown that its residue at Eq. 18.6 will have an order reduction by two. all poles on the jx-axis have to be real and Hence, indeed, the remainder functions are positive, but not necessarily vice versa. simpliﬁed. If, instead of the origin, Z(s) has a pole at In the present case of Z(s) given by Eq. 18.1, s = ∞, then one can write we have a pole s = 0 due to the factor s in the denominator, and also at s = ∞ because the Z ðsÞ ¼ K1 s þ Z2 ðsÞ; ð18:4Þ degree of the numerator is one greater than that of the denominator (it cannot be more than one or where less than one, see [1]). Let us remove them. The residues are, from Eqs. 18.3 and 18.5, Z ðsÞ K1 ¼ Lim ð18:5Þ s!1 s K0 ¼ abc=ða þ b þ cÞ and K1 ¼ 1: ð18:8Þ is the residue of Z(s) at the pole at s = ∞ and If we remove the pole at s = ∞ ﬁrst, the Z2(s) is the remaining function to be tested. Here remainder function is K∞s represents an inductor of value K∞. Finally, if Z(s) has poles at s ¼ jx1 , then one can write Z10 ðsÞ ¼ Z ðsÞ s ð s þ aÞ ð s þ bÞ ð s þ c Þ K1 s ¼ s Z ðsÞ ¼ þ Z3 ðsÞ; ð18:6Þ sðs þ a þ b þ cÞ ð18:9Þ s2 þ x21 sðab þ bc þ caÞ þ abc ¼ : where sðs þ a þ b þ c Þ s2 þ x21 Z ðsÞ This step, as explained earlier, leads to the K1 ¼ 2 ð18:7Þ partial realization of Fig. 18.1a and reduces the s s ¼x2 1 order from three to two. As is obvious from Is Z(s) at All Realizable? 143 Eq. 18.9, Z10 ðsÞ does not (and cannot) have a pole which can be easily veriﬁed. Thus Z20 ðsÞ has no at s = ∞, but it retains the pole at s = 0 of Z obvious defect for positive realness. Not only (s) with the same residue. If we now remove this that, because Z20 ðsÞ has a zero at s = ∞, its pole from Z10 ðsÞ, we have a remainder function reciprocal Y20 ðsÞ has a pole at s = ∞ which can be removed. The corresponding residue is abc Z20 ðsÞ ¼ Z10 ðsÞ : ð18:10Þ sða þ b þ c Þ 0 Y20 ðsÞ On simpliﬁcation, this reduces to the K12 ¼ Lims!1 s following: ða þ b þ cÞðs þ a þ b þ cÞ ¼ Lims!1 s ð a þ bÞ ð b þ c Þ ð c þ aÞ ða þ b þ cÞðab þ bc þ caÞ abc Z20 ðsÞ ¼ : ð18:13Þ ða þ b þ c Þðs þ a þ b þ c Þ ð18:11Þ aþbþc The partial realization resulting from this step ¼ ; ð18:14Þ ð a þ bÞ ð b þ c Þ ð c þ aÞ is shown in Fig. 18.1b. Note also that the order of Z20 ðsÞ is one, which is one less than that of where in Eq. 18.13, we have used the identity Z10 ðsÞ, as expected. Eq. 18.12 in conjunction with Eq. 18.11. If we In order to proceed further with the testing, it remove this pole from Y20 ðsÞ which corresponds 0 is necessary to ensure that the numerator constant to a capacitance of value K12 in parallel, we are of Eq. 18.11 is positive. It is indeed so, because left with the following remainder function of the algebraic identity ða þ b þ cÞs ða þ b þ cÞðab þ bc þ caÞ Y30 ðsÞ ¼ Y20 ðsÞ : ða þ bÞðb þ cÞðc þ aÞ ð18:12Þ ¼ ða þ bÞðb þ cÞðc þ aÞ þ abc; ð18:15Þ Fig. 18.1 Various steps in a +b + c the testing of Z(s) for positive realness, leading to a (a) 1 (b) abc 1 complete realization through Foster preamble only Z(s) Z1¢ (s) Z(s) Z2¢ (s) a +b + c 1 abc (c) Z(s) (a +b)(b + c)(c + a) (a +b + c)2 a +b + c (a +b)(b + c)(c + a) 144 18 A Third-Order Driving Point Synthesis Problem Simpliﬁcation of Eq. 18.15 leads to Y10 ðsÞ 0 K11 ¼ Lims!1 s ða þ b þ c Þ2 sðs þ a þ b þ c Þ Y30 ðsÞ ¼ ð18:16Þ ¼ Lims!1 ða þ bÞðb þ cÞðc þ aÞ s½sðab þ bc þ caÞ þ abc 1 which is a positive constant, equivalent to a ¼ : ab þ bc þ ca resistance of value (a + b) (b + c) (c + a)/(a + b + c)2. The realization obtained at this stage is ð18:17Þ shown in Fig. 18.1c, which is, in fact, a complete This removal means partial realization realization. Nothing is left to test anymore! 0 through a capacitance of value K11 in parallel, We have, therefore, shown that Z(s) is PR and 0 leaving a remainder Y2 ðsÞ, as shown in in the process, which involved only the Foster preamble, we have solved the synthesis problem. Fig. 18.2b, where sðs þ a þ b þ c Þ s Y2 ðsÞ ¼ sðab þ bc þ caÞ þ abc ab þ bc þ ca Alternative Realization s½ða þ b þ cÞðab þ bc þ caÞ abc ¼ : It is known that solution to a synthesis problem, ðab þ bc þ caÞ½sðab þ bc þ caÞ þ abc if it exists, is never unique [1]. Can we, in the ð18:18Þ present case, ﬁnd another realization? Let us see. As in the previous section, let us ﬁrst remove Once again, because of Eq. 18.12, the coefﬁ- the pole at s = ∞, leaving the remainder Z10 ðsÞ cient of s in the numerator of Eq. 18.18 is posi- given by Eq. 18.9. Instead of removing the pole tive, and we can re-write Y2 ðsÞ as at s = 0 from Z10 ðsÞ, note that Z10 ðsÞ has a zero at s = ∞. Let us, therefore, consider the admittance s ð a þ bÞ ð b þ c Þ ð c þ aÞ Y2 ðsÞ ¼ : Y10 ðsÞ ¼ l=Z10 ðsÞ and remove its pole at s = ∞. ðab þ bc þ caÞ½sðab þ bc þ caÞ þ abc The residue is ð18:19Þ Fig. 18.2 Various steps in (a) the alternative realization of Z 1 (b) 1 (s) 1 Z(s) Z(s) Y2 (s ) Z1¢ (s) ab +bc + ca (a + b )(b + c )(c + a ) 1 abc (ab + bc + ca ) (c) 1 Z(s) (ab + bc + ca )2 ab + bc + ca (a + b )(b + c )(c + a ) Alternative Realization 145 Y2 ðsÞ has a zero at the origin, which can be A word of caution must be sounded here. That removed as the pole of Z2 ðsÞ ¼ 1=Y2 ðsÞ: In fact, continued fraction expansion works here is a we can easily see that matter of luck; it may not work in a general RLC case. Even in this case, you may try continued ðab þ bc þ caÞ2 fraction expansion starting with the lowest Z2 ðsÞ ¼ ða þ bÞðb þ cÞðc þ aÞ powers and soon get frustrated! ð18:20Þ abcðab þ bc þ caÞ þ : sða þ bÞðb þ cÞðc þ aÞ A Problem for the Student The second term corresponds to the pole at the origin. Also, observe that this decomposition Can you ﬁnd out another alternative minimal corresponds to a series combination of a capaci- realization? If this proves tough, try relaxing on tance and resistance value indicated in Fig. 18.2c. the minimal requirement—ﬁrst with three reac- The synthesis is complete and as you can see, this tances and more than one resistance and later is different from the network of Fig. 18.1c. with more than three reactances and more than It is interesting to observe that the alternative one resistance. No more! Isn’t life simple? realization can be mechanized through continued fraction expansion starting with the highest Acknowledgments Acknowledgement is made to S. powers, as follows: Tirtoprodjo who ﬁrst posed the problem [2] and to S. Erfani et al. who gave the solution of Fig. 18.1, although s3 þ s2 ða þ b þ cÞ þ sðab þ bc þ caÞ þ abc in a cryptic form [3]. Z ðsÞ ¼ s2 þ sða þ b þ c Þ ð18:21Þ ( References s ðab þ bc þ caÞ2 ¼ s þ 1= þ 1= ab þ bc þ ca ða þ bÞðb þ cÞðc þ aÞ 1. F.F. Kuo, Network Analysis and Synthesis (Wiley, New York, 1966). Chapter 10 sða þ bÞðb þ cÞðc þ aÞ þ 1= : 2. S. Tirtoprodjo, On the lighter side. IEEE CAS abcðab þ bc þ caÞ Magazine, 5(1), 25 (1983, March) ð18:22Þ 3. S. Erfani et al., On the lighter side—Solution to the march puzzle. IEEE CAS Magazine 5(2), 22 (1983) Interference Rejection in a UWB System: An Example of LC Driving 19 Point Synthesis Synthesis of an LC driving point function is extensive tables are available in textbooks [1] one of the initial topics in the study of network and handbooks [2]. In particular, one-port or synthesis. This chapter gives a practical driving point synthesis, one of the starting topics example of application of such synthesis in in the subject, appears to be of little use in the design of a notch ﬁlter for interference practice. This chapter deals with a recent appli- rejection in an ultra wide-band (UWB) system. cation of LC driving point synthesis, which may The example can used to motivate students to be used to enhance the motivation of students to learn network synthesis with all seriousness, learn the subject with all seriousness. and not merely as a matter of academic The example is taken from a 2009 paper [3] exercise. dealing with an integrated double-notch ﬁlter and implemented with 0.13 lm CMOS technology, for rejection of interference in an ultra wide-band Keywords (UWB) system. The problem, translated to net- LC driving point synthesis Notch ﬁlter work synthesis language, is to design a ﬁlter to UWB systems Network synthesis reject the frequencies around f1 and f2 and pass those around fp, where f1 < fp < f2. The authors of [3] set the design values as x1 = 2p 2.4 109 rad/s, x2 = 2p 5.2 109 rad/s and xp = 2p 4.8 109 rad/s, where x = 2pf, and Introduction suggested and designed the network shown in Fig. 19.1 for this purpose. The current generator The subject of network synthesis is considered and the shunt resistance in Fig. 19.1 represent the by most students as moderately difﬁcult, mathe- equivalent circuit of an ampliﬁer and the LC matical and mainly of academic interest. The network has a driving point impedance Z(s), underlying reason is that it encounters very few which is required to have series resonance (and practical applications, except in the design of hence zero impedance) at x1 and x2, thus ﬁlters, which is a two-port network, for which shunting out all the current from the load at these Source: S. C. Dutta Roy, “Interference Rejection in a frequencies, and parallel resonance (and hence UWB System: An Example of LC Driving Point inﬁnite impedance) at xp, thus passing all the Synthesis,” IETE Journal of Education, vol. 50, current through the load at this frequency. Thus, pp. 55–58, May–August 2009. © Springer Nature Singapore Pte Ltd. 2018 147 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_19 148 19 Interference Rejection in a UWB System: An Example … basically, one requires to design an impedance K1 ¼ 1 ¼ L1 ; ð19:3bÞ Z(s) of the form and s2 þ x21 s2 þ x22 ZðsÞ ¼ ; ð19:1Þ s s2 þ x2p Kp ¼ x2p x21 x22 x2p =x2p ) C2 ¼ 1=Kp ; L2 ¼ Kp =x2p : where without any loss in generality, the scaling ð19:3cÞ constant is assumed to be unity. In this chapter, we shall treat Eq. 19.1 as the function to be These results are reproduced in Table 19.1, in synthesized, and derive the form of Fig. 19.1, as which the capacitors are given as C1 x21 and well as the other alternative canonic forms, along C2 x21 , for later convenience. with their element values. We shall then compare Foster II form is obtained by the PFE of Y(s) the various networks on the basis of the = 1/Z(s); the results are as follows: required total inductance, total capacitance and grounded and ungrounded capacitors, which are K1 s K2 s important considerations for integrated circuit Y ðsÞ ¼ þ 2 ; þ x1 s þ x22 s22 implementation. K1 ¼ x2p x21 = x22 x2p ; ð19:4aÞ The Four Canonical Realizations K2 ¼ x22 x2p = x22 x21 ; The network of Fig. 19.1 is easily recognized as the L1 ¼ 1=K1 ; C1 ¼ 1= x21 L1 ; L2 ¼ 1=K2 ; 2 Foster I realization of Eq. 19.1 [4]. As is well C2 ¼ 1= x2 L2 : known, there are four basic structures for the ð19:4bÞ canonical synthesis of an LC driving point synthe- sis, viz. Foster I, Foster II, Cauer I and Cauer II [4]. The elements in Eq. 19.4b refer to the net- Foster I form is obtained by partial fraction work shown in Fig. 19.2, and the values are expansion (PFE) of Z(s), given by shown in Table 19.1. For Cauer I network, we make a continued K0 Kp s Z ðsÞ ¼ þ K1 s þ 2 ; ð19:2Þ fraction expansion (CFE) of Eq. 19.1 starting s s þ x2p with the highest powers. The quotients of the CFE give the following element values with where with reference to Fig. 19.1, reference to the structure shown in Fig. 19.3. These values are also shown in Table 19.1. K0 ¼ x21 x22 =x2p ) C1 ¼ 1=K0 ; ð19:3aÞ 1 1 L1 ¼ 1; C1 ¼ ¼ 2; x21 2 þ x2 xp x3 2 C1 L1 x43 x23 x2p x21 x22 L2 ¼ ; C2 ¼ x23 x2p x21 x22 x23 x22 x21 Z(s) L2 C2 ð19:5Þ The Cauer II realization will be of the form shown in Fig. 19.4, and the element values are Fig. 19.1 Foster I network connected to a current obtained from the quotients of the CFE of generator and a shunt resistance, which represent the Eq. 19.1 starting with the lowest powers. These equivalent circuit of an ampliﬁer The Four Canonical Realizations 149 Table 19.1 I Comparison of the four canonical structures Parameter Expressions and values of network L1 L2 L1 + L2 x21 C1 x21 C2 x21 (C1 + C2) (Num. (Num. value) value) ! ! Foster I 1 x2 x22 2.130 x2p x21 2.775 1 12 1 x2p L2 xp x2p x22 (0.130) (0.852) (1.923) Foster II ðx22 x21 Þ ðx22 x21 Þ 1.630 1/L1 x21 3.429 ðx2p x21 Þ ðx22 x2p Þ (3.268) x22 L2 0.306) (1.324) (0.161) Cauer I 1 x43 2.378 x21 x23 0.852 x23 x2p x21 x22 x23 x22 L2 (1.378) (0.590) (0.262) x23 ¼ x21 þ x22 x2p Cauer II x24 x24 9.810 x2p x21 0.877 x2p x24 x2p x22 x24 L2 (1.130) (8.680) (0.852) (0.025) x2 x2 x24 ¼ x21 x22 122 xp Note The expressions are slightly modiﬁed versions of those in the text values are given below and are also shown in Table 19.1. L1 L2 L1 ¼ x24 =x2p ; x24 ¼ x21 þ x22 x21 x22 =x2p ; Z(s) C1 C2 L2 ¼ x24 = x24 x2p ; C2 ¼ 1= x24 L2 ; ð19:6aÞ Fig. 19.2 Foster II form of Z(s) and L1 L2 C1 ¼ x2p = x21 x22 : ð19:6bÞ Z(s) C1 C2 Comparison Fig. 19.3 Cauer I form of Z(s) Using the same speciﬁcations as given in [3], and mentioned in the Introduction, we have C1 C2 computed the numerical values of the elements for the various structures. These are shown in Table 19.1 inside brackets below the corre- Z(s) L1 L2 sponding algebraic expression. Note that no powers of 10 are involved in the expressions for C1 x21 and C2 x21 because of multiplica- tion of the capacitors by x21 . Also, for Fig. 19.4 Cauer II form of Z(s) 150 19 Interference Rejection in a UWB System: An Example … computational convenience, some of the alge- Problems braic expressions given in Table 19.1 are also slightly modiﬁed versions of the formulas given P:1. Can you ﬁnd an alternative network to the in the text. C1, L1, L2, C2 conﬁguration in Fig. 19.1? A look at the total capacitance (Ct) and total P:2. Could we do with third-order impedances inductance (Lt) values in Table 19.1 show that for C1, L1 combination as will as L2, C2 Cauer realizations have considerably smaller Ct combination in Fig. 19.1. What will this as compared to the Foster realizations, with circuit perform as? Cauer I having the lowest value and Cauer II P:3. Suppose there are two frequencies which having a marginal increase over the same. The have to be rejected. Draw the necessary reverse is the case with respect to Lt with Foster I circuit conﬁguration. having the lowest value and Foster II having a P:4. Same as P.3 except that three frequencies marginal increase over it. Another point to be have to be rejected. Draw an alternation noted is that both Foster II and Cauer I networks circuit also. have both capacitors connected to ground, which P:5. Draw another alternative circuit for P.4 and is, in general, a desirable feature in integrated compare the two. circuits. Acknowledgements The work was supported by the Indian National Science Academy through their Honorary Scientist scheme. The author acknowledges the help of Effect of Losses Dr. Sumantra Dutta Roy for his help in the preparation of the diagrams. In practice, all reactive elements are lossy, i.e. all inductors have a series resistance and all capacitors have a shunt conductance. How- References ever, the losses in inductors dominate over those in the capacitors. A practical scheme for 1. L. Weinberg, Network Analysis and Synthesis (McGrawHill, New York, 1962) effective compensation of the losses for the 2. W.K. Chen (ed.), in Passive, Active and Digital network in Fig. 19.1 has been given in [3] Filters, Volume 3 of Handbook of Circuits and Filters using a single negative resistance realized with (Boca Raton, CRC Press, 2009) active devices. Analysis of the effects of los- 3. A. Vallese, A. Bevilacqua, C. Sandner, M. Tiebout, A. Gerosa, A. Neviani, Analysis and design of an ses on the notch depths and maximum output integrated notch ﬁlter for the rejection of interference for the four structures will be a worthwhile in UWB systems. IEEE J. Solid-State Circuits 44, project for the students, and a comparison 331–343 (2009) may reveal the superiority of one network 4. F.F. Kuo, Network Analysis and Synthesis (Wiley, New York, 1966). Chapter 11 over the others. Low-Order Butterworth Filters: From Magnitude to Transfer Function 20 A simple method is given for obtaining the HN(s) has all its poles on the left half of the unit transfer function of Butterworth ﬁlters of circle centered at s = 0, at equal angular intervals orders 1 to 6. of p/N, with none occurring on the jx-axis. This gives rise to the property that if we write Keywords HN ðsÞ ¼ 1=BN ðsÞ ð20:2Þ Butterworth ﬁlters Transfer functions Magnitude Orders of ﬁlter Chebyshev ﬁlter then, BN(s), the so-called Butterworth polyno- mial, has symmetrical coefﬁcients, i.e. BN(s) is of the form BN ðsÞ ¼ 1 þ b1 s þ b2 s2 þ þ b2 sN2 þ b1 sN1 þ sN Butterworth Filters ð20:3Þ Butterworth ﬁlter is the most elegant of all ﬁlters In standard textbooks (see, e.g. [1–3]), the in more than one way. The magnitude squared procedure prescribed for ﬁnding BN(s) is to locate function of a low-pass Butterworth ﬁlter, having all the roots of a normalized 3 dB cutoff frequency of 1 rad/sec, is given by BN ðsÞBN ðsÞ ¼ 1 þ ðs2 ÞN ð20:4Þ and then to take the left half-plane ones for BN(s), jHN ðjxÞj2 ¼ 1= 1 þ x2N ; ð20:1Þ i.e. where N is the order of the ﬁlter. As is well Y N known, it has a monotonically decaying response BN ð s Þ ¼ ðs sk Þ; Re sk \0; k ¼ 1 to N in the whole frequency range. It is maximally flat k¼1 at dc, and the corresponding transfer function ð20:5Þ Source: S. C. Dutta Roy, “Low Order Butterworth Filters: From Magnitude to Transfer Function,” Journal of the IETE, vol. 37, pp. 221–225, October–December 1996. © Springer Nature Singapore Pte Ltd. 2018 151 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_20 152 20 Low-Order Butterworth Filters: From Magnitude to Transfer … It may be noted that Eq. 20.4 is simply the so that denominator of Eq. 20.1 with x2 replaced by BN ðsÞ ¼ mN ðsÞnN ðsÞ ð20:10Þ −s2, a process known as ‘analytic continuation’. An explicit formula for sk is [3]: Combining Eqs. 20.4, 20.8 and 20.9, we get sk ¼ sin½ð2k 1Þp=ð2N Þ þ jcos½ð2k1Þp=ð2N Þ; m2N n2N ¼ 1 þ ðlÞN s2N ð20:11Þ k ¼ 1 to N ð20:6Þ or, 2 2 Obviously, these roots occur in symmetry 1 þ b2 s2 þ b1 s þ b3 s3 þ with respect to the real as well as the imaginary ¼ 1 þ ð1ÞN s2N ð20:12Þ axes. It is also clear that if N is odd, then there is a root of BN(s) at s = −1, i.e. (s + 1) is a factor of Obviously, the constant terms and the BN(s); the other (N − 1) roots will be complex coefﬁcients of s2N on both sides of Eq. 20.12 are and will occur in conjugate pairs. Consequently, identically equal. The coefﬁcients of s2 ; s4 ; . . .; one can write s2N2 must then each be zero. This, combined 8 > > Q2 N=2 > < s þ 2s sin½ð2k 1Þp=ð2N Þ þ 1 ; N even BN ð s Þ ¼ k¼1 ð20:7Þ > > Q 2 ðN1Þ=2 > : ðs þ 1Þ s 2s sin½ð2k 1Þp=ð2N Þ þ 1 ; N odd k¼1 When N is large, one can use Eq. 20.7 for with the symmetry of the coefﬁcients, makes it ﬁnding BN(s) in quickest possibly way. However, easy to ﬁnd BN(s) for low orders. when N is low, an alternative trick can be applied, A further simpliﬁcation can be obtained for and that is the subject matter of this chapter. the odd-order case. It follows from Eq. 20.11 that if N is odd, then (1 − s2) will be a factor of BN(s) BN(−s), which contributes (1 + s) to Basis of the Alternative Method BN(s) and (1 − s) to BN(−s) as factors. The other factor of BN(s) BN(−s) will be Since the roots of BN(s) are strictly in the left half ð1 þ s2 þ s4 þ þ s2N2 Þ. Hence, the proce- of the s-plane, BN(s) qualiﬁes as a strict Hurwitz dure simpliﬁes to that of ﬁnding the coefﬁcients polynomial. We can write of the polynomial BN ð s Þ ¼ 1 þ b 2 s 2 þ þ b 1 s þ b 3 s 3 þ CN1 ðsÞ ¼1 þ c1 s þ c2 s2 þ þ c2 sN3 ¼ mN ðsÞ þ nN ðsÞ; þ c1 sN2 þ sN1 ; ð20:8Þ ð20:13Þ where mN and nN are the even and odd parts of BN(s). By deﬁnition where mN ðsÞ ¼ mN ðsÞ and nN ðsÞ ¼ nN ðsÞ; CN1 ðsÞCN1 ðsÞ ¼ 1 þ s2 þ s4 þ þ s2N2 ð20:9Þ ð20:14Þ Basis of the Alternative Method 153 Notice that CN−1(s) is also written with coef- Second-Order Case ﬁcient symmetry. This follows from the fact that all real polynomial factors of a polynomial with For N = 2, we have from Eq. 20.3, symmetrical coefﬁcients retain the symmetry property. From Eqs. 20.13 and 20.14, we have B2 ð s Þ ¼ 1 þ b 1 s þ s 2 ð20:19Þ 2 2 and hence, from Eq. 20.12, 1 þ c2 s2 þ c1 s þ c 3 s3 þ ¼ 1 þ s2 þ s4 þ þ s2N2 ð20:15Þ 2 1 þ s2 b21 s2 ¼ 1 þ s4 ð20:20Þ 0 2N−2 Again, the coefﬁcients of s and s are identical on both sides, while the coefﬁcients of Equating the coefﬁcients of s2 on both sides gives s2 ; s4 ; . . .; s2N 4 , each equated to unity, will form a set of equations for determining the ci 2 b21 ¼ 0 ð20:21Þ coefﬁcients. or We shall now apply the method to all the pﬃﬃﬃ possible low orders; in the process, the difﬁcul- b1 ¼ 2 ð20:22Þ ties encountered for high orders will become obvious. It will also be clear that one cannot Hence, apply the method blindly, because of the occur- pﬃﬃﬃ rence of multiple solutions for the coefﬁcients, B2 ðsÞ ¼ 1 þ 2s þ s2 ð20:23Þ and the consequent need for identifying the correct solutions. An obvious constraint is that all coefﬁcients must be positive, but, as will be Third-Order Case shown, this alone is not enough. As already pointed out, (1 + s) will be a factor of B3(s), so that Application to Low Orders B3 ðsÞ ¼ ð1 þ sÞC2 ðsÞ ð20:24Þ First-Order Case where For N = 1, we have from Eq. 20.11, C2 ðsÞ ¼ 1 þ c1 s þ s2 ð20:25Þ m21 n21 ¼ 1 s2 ð20:16Þ From Eq. 20.15, then, we get 2 Consequently, 1 þ s2 c21 s2 ¼ 1 þ s2 þ s4 ð20:26Þ m1 ¼ 1 and n1 ¼ s ð20:17Þ Equating the coefﬁcients of s2 on both sides gives so that 2 c21 ¼ 1 ð20:27Þ B1 ðsÞ ¼ 1 þ s ð20:18Þ or, This is, of course, a trivial case. c1 ¼ 1 ð20:28Þ 154 20 Low-Order Butterworth Filters: From Magnitude to Transfer … so that Which ones are admissible? To test this, we bring in the strict Hurwitz character of B4(s), B3 ð s Þ ¼ ð 1 þ s Þ 1 þ s þ s 2 ð20:29Þ which demands that the ratio m4(s)/n4(s) should be an LC driving point function of the fourth ¼ 1 þ 2s þ 2s2 þ s3 ð20:30Þ order. Let us try rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ pﬃﬃﬃ b2 ¼ 2 2 and b1 ¼ 2 2 2 Fourth-Order Case We can write For N = 4, we have from Eq. 20.8 and the symmetry of coefﬁcients, pﬃﬃﬃ m4 ðsÞ s4 þ 2 2 s2 þ 1 ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃ ; B4 ð s Þ ¼ 1 þ b1 s þ b2 s 2 þ b 1 s 3 þ s 4 ð20:31Þ n4 ð s Þ 2 2 2 ðs3 þ sÞ and from Eq. 20.12, we get " pﬃﬃﬃ 2 # 1 1 2 s þ1 2 2 ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃ þs ; 1 þ b2 s2 þ s4 b1 s þ b1 s3 ¼ 1 þ s8 2 2 2 s3 þ s ð20:32Þ ð20:38Þ Equating the coefﬁcients of s and s on both 6 4 p Since ð1 2Þ is negative, Eq. 20.38 cannot sides of Eq. 20.32 gives rise to the following set qualify as an LC driving point function. Hence, of nonlinear equations: the acceptable solutions are the ones with positive signs in Eqs. 20.36 and 20.37. We therefore get 2b2 ¼ b21 ð20:33Þ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ pﬃﬃﬃ pﬃﬃﬃ and B4 ðsÞ ¼1 þ 2 2 þ 2 s þ 2 þ 2 s2 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ b22 þ 2 ¼ 2b21 ð20:34Þ pﬃﬃﬃ 3 þ 2 2 þ 2 s þ s4 Note that because of symmetry, equating the ð20:39Þ coefﬁcients of s2 on both sides of Eq. 20.32 gives the same result as Eq. 20.33. Combining Eqs. 20.33 and 20.34 gives ¼ 1 þ 2:6131 s þ 3:4142s2 þ 2:6131 s3 þ s4 ð20:40Þ b22 4b2 þ 2 ¼ 0 ð20:35Þ This case illustrates the difﬁculty we Solving this quadratic, we get the following encounter with multiple solutions. Finding the two solutions for b2: acceptable solution requires a Hurwitz test, pﬃﬃﬃ which, in this particular case, has not proved to b2 ¼ 2 2 ð20:36Þ be difﬁcult. Note that for the Hurwitz test, one can also use continued fraction expansion, rather Consequently, from Eq. 20.33 than partial fraction expansion. rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ b1 ¼ 2 2 2 ð20:37Þ Fifth-Order Case Both of these values of b2 and b1 are posi- As in the third-order case, (1 + s) is a factor of tive and are candidates for belonging to B4(s). B5(s) and we can write Application to Low Orders 155 p p p B5 ðsÞ ¼ ð1 þ sÞC4 ðsÞ; ð20:41Þ ¼ð1 þ ð 5 þ 1Þs þ ð 5 þ 3Þs2 þ ð 5 þ 3Þs3 p þ ð 5 þ 1Þs4 þ s5 Þ; ¼ ðl þ sÞ l þ c1 s þ c2 s2 þ c1 s3 þ s4 ; ð20:42Þ ð20:52Þ where ¼1 þ 3:2361 s þ 5:2361 s2 þ 5:2361 s3 2 2 1 þ c2 s2 þ s4 c1 s þ c1 s3 þ 3:2361 s4 þ s5 ¼ 1 þ s2 þ s4 þ s6 þ s8 ð20:43Þ ð20:53Þ Equating the coefﬁcients of s2 and s4 on both sides gives Sixth-Order Case 2c2 c21 ¼1 ð20:44Þ For N = 6, and B6 ðsÞ ¼ 1 þ b1 s þ b2 s2 þ b3 s3 þ b2 s4 þ b1 s5 þ s6 c22 þ2 2c21 ¼1 ð20:45Þ ð20:54Þ Combining Eqs. 20.44 and 20.45 give the and following quadratic equation in c2: 2 2 1 þ b2 s 2 þ b2 s 4 þ s 6 b1 s þ b3 s 3 þ b1 s 5 c22 4c2 þ 3 ¼ 0 ð20:46Þ ¼ 1 þ s12 It is easy to see that the solutions for c2 are ð20:55Þ c2 ¼ 3; 1 ð20:47Þ Equating the coefﬁcients of s2, s4 and s6 on both sides give the following equations: Correspondingly, from Eq. 20.44, 2b22 ¼ b21 ; ð20:56Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p c1 ¼ 2c2 1 ¼ 5; 1 ð20:48Þ 2b2 þ b22 ¼ 2b1 b3 ; ð20:57Þ Try c2 = c1 = 1; then and C4 ðsÞ ¼ 1 þ s þ s2 þ s3 þ s4 ð20:49Þ 2 þ 2b22 ¼ 2b21 þ b23 ð20:58Þ The ratio of even to odd parts of C4(s) is From Eq. 20.56, we get s4 þ s2 þ 1 1 ¼ sþ 3 ; ð20:50Þ pﬃﬃﬃﬃﬃﬃﬃ s3 þ s s þs b1 ¼ 2b2 ð20:59Þ which is obviously not an LC driving point and from Eqs. 20.58 and 20.59, we get function. Hence, the acceptable solutions are: p p c2 = 3 and c1 ¼ 5 giving b3 ¼ 2j1 b2 j; ð20:60Þ p p p B5 ðsÞ ¼ ð1 þ sÞ 1 þ 5s þ 3s2 þ 5s3 þ s4 where the magnitude sign is included to indicate ð20:51Þ that b3 must be positive. Combining Eqs. 20.57, 156 20 Low-Order Butterworth Filters: From Magnitude to Transfer … 20.59 and 20.60 give the following cubic equa- Equating the coefﬁcients of like powers of tion in b2: s on both sides gives b32 12b22 þ 36b2 16 ¼ 0 ð20:61Þ 2c2 c21 ¼ 1; ð20:69Þ A cubic equation is analytically solvable [4], although not as easily as the quadratic equation. c22 þ 2c2 2c1 c3 ¼ 1; ð20:70Þ For the general cubic equation and ax þ bx þ cx þ d ¼ 0 3 2 ð20:62Þ 2 þ 2c22 c23 2c21 ¼ 1 ð20:71Þ all the three roots are real if b − 3ac > 0, while 2 From Eq. 20.69, we get if b2 − 3ac < 0, then there is only one real root, the other two being complex conjugates of each pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ other. As can be easily veriﬁed, the second case c1 ¼ 2c2 1 ð20:72Þ is valid for Eq. 20.61; hence, our job is simply to ﬁnd the real root of Eq. 20.61. Trial and error while Eq. 20.71 gives seems to be the best policy at this stage, and after a few trials, we get qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ c3 ¼ 3 4c2 þ 2c22 ð20:73Þ b2 ¼ 7:4641 ð20:63Þ Substituting these values in Eq. 20.70, we get as a reasonably good solution. The correspond- ing values of the other two coefﬁcients are qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ obtained from Eqs. 20.59 and 20.60 as c22 þ 2c2 1 ¼ 2 ð2c2 1Þ 3 4c2 þ 2c22 b1 ¼ 3:8637 and b3 ¼ 9:1416 ð20:64Þ ð20:74Þ Hence, ﬁnally, Squaring both sides and simplifying, we get the following quartic equation in c2: B6 ðsÞ ¼1 þ 3:8637 s þ 7:4671 s2 þ 9:1416 s3 c42 12c32 þ 42c22 44c2 þ 13 ¼ 0 ð20:75Þ þ 7:4671 s4 þ 3:8637 s5 þ s6 ð20:65Þ A cubic equation was bad enough; this is worse! Fortunately, however, a quartic equation Seventh-Order Case can also be solved analytically [4]; however, the effort involved does not justify proceeding fur- Since the order is odd, we have ther. Using Eq. 20.7 appears to be a much better proposition, not just for N = 7, but for all higher B7 ðsÞ ¼ ð1 þ sÞC6 ðsÞ; ð20:66Þ orders. ¼ ð1 þ sÞ 1 þ c1 s þ c2 s2 þ c3 s3 þ c2 s4 þ c1 s5 þ s6 ; ð20:67Þ Application to Chebyshev Filters where What about ﬁlters other than Butterworth? For 6 2 2 example, Chebyshev? Does the procedure pre- 1 þ c2 s þ c2 s þ s 2 4 c1 s þ c 3 s3 þ c 1 s5 sented here offer any simplicity? Let us ¼ 1 þ s2 þ s4 þ s6 þ s8 þ s10 examine. ð20:68Þ Application to Chebyshev Filters 157 For the Chebyshev low-pass ﬁlter, the nor- Thus, malized magnitude squared function is given by b0 ¼ 1 and b1 ¼ e ð20:82Þ jHN ðjxÞj2 ¼ 1= 1 þ e2 TN2 ðxÞ ; ð20:76Þ and where TN is the Chebyshev polynomial of the D1 ðsÞ ¼ 1 þ es ð20:83Þ ﬁrst kind, of order N. For the ﬁrst few orders, we For the second-order case, we get have 2 2 T1 ð x Þ ¼ x b0 þ b2 s2 b21 s2 ¼ 1 þ e2 2x2 1 jx2 ¼s2 T2 ðxÞ ¼ 2x2 ð20:84Þ ð20:77Þ T3 ðxÞ ¼ 4x3 3x1 On simpliﬁcation, Eq. 20.84 becomes T4 ðxÞ ¼ 8x4 8x2 þ 1 b20 þ 2b2 b21 s2 þ b22 s4 Let ¼ 1 þ e2 þ 4e2 s2 þ 4e2 s4 ð20:85Þ HN ðsÞ ¼ 1=DN ðsÞ ¼ 1=½mN ðsÞ þ nN ðsÞ; Equating the coefﬁcients of like powers of ð20:78Þ s gives pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ b0 ¼ 1 þ e 2 ð20:86Þ ¼ 1= b0 þ b1 s þ b2 s2 þ þ bN sN ; ð20:79Þ b2 ¼ 2e ð20:87Þ where the usual notations have been used. Note that neither b0 nor bN is ﬁxed, as in the Butter- and worth case; there is no coefﬁcient symmetry either. The equation for ﬁnding the coefﬁcients is 2b0 b2 b21 ¼ 4e2 ð20:88Þ 2 2 Combining Eqs. 20.86–20.88, we get b0 þ b2 s 2 þ b1 s þ b3 s 3 þ ¼ 1 þ e2 TN2 ðs=jÞ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð20:80Þ b1 ¼ 2 e 1 þ e2 e ð20:89Þ Although the right-hand side of Eq. 20.80 Thus, involves j, in actual practice, CN2 ðxÞ shall involve rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ only x2 ; x4 ; . . . so that x2 has to be replaced by pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ D2 ðsÞ ¼ 1 þ e2 þ 2 e 1 þ e e s þ 2es2 2 −s2; hence, only real coefﬁcients would be encountered. ð20:90Þ For the ﬁrst-order case, we have Now consider the third-order case, for which we b20 b21 s2 ¼ 1 e2 s2 ð20:81Þ get, from Eqs. 20.77 and 20.80. 2 2 2 b0 þ b2 s2 b1 s þ b3 s3 ¼ 1 þ e2 4x3 3x jx2 ¼s2 2 ¼ 1 þ e2 ðsÞ2 4s2 3 ð20:91Þ ¼ 1 9e2 s2 24e2 s4 16e2 s6 158 20 Low-Order Butterworth Filters: From Magnitude to Transfer … Equating the coefﬁcients of like powers of s on Butterworth ﬁlter and other types of ﬁlters, even both sides, we get of a lower order, have been pointed out. b0 ¼ 1; b3 ¼ 4e ð20:92Þ Problems 2b2 b21 ¼ 9e2 ð20:93Þ P:1. Without ﬁnding poles and zeroes, can you and formulate a procedure and give the equa- tions for ﬁnding an Nth-order Butterworth b22 8eb1 ¼ 24e2 ; ð20:94Þ polynomial? No, no, I am not asking you to solve these equations, because that will be where, in Eqs. 20.93 and 20.94, the values given too must to ask for. With the kind of in Eq. 20.92 have been utilized. Combining training I have given to you, I believe you Eqs. 20.93 and 20.94 give the following cubic should be able to do it. equation for b2: P:2. For the third-order case, ﬁnd the zeroes of the third-order Butterworth polynomial. Do b32 þ 48e2 b2 128e2 ¼ 0 ð20:95Þ not bring poles and zeroes into the scene. They pollute and hamper you intellectual According to the theory of cubic equations development! Of course, you substitute the [4], this also has only one real root, which values of the coefﬁcients from the text. obviously depends on e. P:3. Same as P.1 for order = 4. One should, at this point, be convinced that P:4. Same as P.2 for order = 5. the applicability of the technique presented in P:5. Same as P.2 for order = 7. this chapter: to the Chebyshev ﬁlter, or for that matter, to any other kind of ﬁlter would be lim- ited. Even for the Butterworth case, the limit appears to be set by the sixth order. References 1. M.E. Van Valkenburg, Introduction to Modern Net- work Synthesis (Wiley, New York, 1964) Conclusion 2. N. Balabanian, in Network Synthesis (Englewood Cliffs, NJ, Prentice Hall, Inc, 1958) 3. S. Karni, Network Theory: analysis and Synthesis A simple method is presented for ﬁnding the (Allyn and Bacon Inc, Boston, 1966) Butterworth polynomials of orders one to six, 4. S. Neumark, Solution of Cubic and Quartic Equations and its limitations for higher orders of (Pergamon Press, London, 1965) Band-Pass/Band-Stop Filter Design by Frequency Transformation 21 Given the speciﬁcations of a band-pass ﬁlter x0 s x0 S¼ þ ; ð21:1Þ (BPF) or a band-stop ﬁlter (BSF), the same B x0 s can be translated to those of a normalized low-pass ﬁlter (LPF) by frequency transfor- where S = R + jX is the LPF complex frequency mation. Once the latter is designed, one can variable, s ¼ r þ jx is the BPF complex fre- realize the BPF/BSF by using the same quency variable, transformation in a reverse manner. The process of translation to the normalized LPF B ¼ xp2 xp1 ð21:2Þ is usually not explained in details in standard textbooks, and in some of them, the process is the bandwidth and has even been wrongly stated or illustrated. pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x0 ¼ xp1 xp2 ¼ xs1 xs2 ð21:3Þ This chapter clariﬁes this important step in BPF/BSF design. is the centre frequency of the BPF response, which is geometrically symmetrical about x0 . Keywords Similarly, the BSF response of Fig. 21.1c can be Band-Pass Band-Stop Frequency obtained from the LPF response of Fig. 21.1a through the transformation. transformation x0 s x0 S ¼ 1= þ ; ð21:4Þ B x0 s Introduction where, again Eqs. 21.2 and 21.3 are valid, but B does not have the interpretation of bandwidth. As is well known the normalized LPF response As in the BPF case, the BSF characteristic is also of Fig. 21.1a can be transformed to the BPF geometrically symmetrical about x0 . response of Fig. 21.1b by the transformation Given Fig. 21.1a, it is easy to obtain the characteristics of Fig. 21.1b or Fig. 21.1c by using Eqs. 21.1 or 21.4 as the case may be, but given a BPF or BSF response, how does one go Source: S. C. Dutta Roy, “Band-Pass/Band-Stop Filter Design by Frequency Transformation,” IETE Journal of to the normalized LPF response? In particular, Education, vol. 45, pp. 145–149, July–September 2004. how does one ﬁnd the edge of the stop-band Xs, © Springer Nature Singapore Pte Ltd. 2018 159 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_21 160 21 Band-Pass/Band-Stop Filter Design by Frequency Transformation (a) (b) (c) Fig. 21.1 a Normalized LPF characteristic. b BPF response obtained through Eq. 21.1. c BSF response obtained through Eq. 21.4 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ in Fig. 21.1a? Also, obviously xp1 xp2 may not If xp1 xp2 ¼ x0s1 x0s2 then no modiﬁcation of pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ be equal to xs1 xs2 in the given speciﬁcations. the characteristics is necessary. However, if this How does one proceed? These questions are is not the case, then two cases may arise: either not answered or not adequately explained in textbooks [1–3]. Some textbooks [4, 5] in fact I. xp1 xp2 \x0s1 x0s2 have given wrong answers/illustrations (see II. xp1 xp2 [ x0s1 x0s2 Appendix). The purpose of this chapter is to clarify these important points. In case I, reduce x0s2 to xs2 ¼ xp1 xp2 =x0s1 and rename x0s1 as xs1 , as shown in Fig. 21.2a. In case II, increase x0s1 to xs1 ¼ xp1 xp2 =x0s2 and Band-Pass Case rename x0s2 as xs2 , as shown in Fig. 21.2b. In both cases adjustments have been made to Let the BPF speciﬁcations be: guarantee geometric symmetry, thus facilitating the application of Eq. 21.1. 1 magnitude dp ; for xp1 x p2 ; and Now comes the question of ﬁnding Xs. Note 0 magnitude ds ; for 0 x x0s1 and x0s2 that S = ±jXs should correspond to s ¼ jxs1 as x 1: well as s ¼ jxs2 , where the signs may or may Band-Pass Case 161 (a) (b) Fig. 21.2 Adjustments in given BPF response to ensure that xp1 xp2 ¼ xs1 xs2 not correspond to each other. Putting S = jXs and xp2 xp1 Xs ¼ ð21:6Þ s ¼ jxs1 in Eq. 21.1 and simplifying, we get xs2 xs1 xs2 xs1 Xs ¼ ð21:5Þ xp2 xp1 Example Since a positive value of Xs has been obtained, the correspondence of S = jXs to s ¼ As an example, let us design a maximally flat jxs1 is validated. Similarly, one can show that BSF to satisfy the speciﬁcations shown in S = jXs also corresponds to s ¼ jxs2 and that Fig. 21.4a, where magnitude refers to that of the S = −jXs corresponds to s ¼ jxs1 as well as transfer function V2/11 of the network shown in s ¼ jxs2 . Fig. 21.4b. Since here fs10 fs20 ¼ 6 ðkHzÞ2 [ fp1 fp2 ¼ 5 ðkHzÞ2 , f denoting x=2p, we adjust fs10 to Band-Stop Case fs1 ¼ 5=3 kHz, and set fs2 ¼ fs20 ¼ 3kHz. Thus, by Eq. 21.6, we get The adjustments needed in the band-stop case are illustrated in Fig. 21.3. It is easily shown that, here (a) (b) Fig. 21.3 Adjustments in given BSF response to ensure that xp1 xp2 ¼ xs1 xs2 . a xp1 xp2 \x0s1 x0s2 ; xs1 ¼ xp1 xp2 =x0s2 ; x0s2 ¼ xs2 ; b xp1 xp2 \x0s1 x0s2 ; xs2 ¼ xp1 xp2 =x0s2 ; x0s1 ¼ xs1 162 21 Band-Pass/Band-Stop Filter Design by Frequency Transformation (a) (b) Fig. 21.4 a BSF speciﬁcation, b desired network fp2 fp1 Xs ¼ ¼ 2:25 ð21:7Þ fs2 fs1 Thus, the normalized LPF to be designed has the speciﬁcations shown in Fig. 21.5 with a network of the form of Fig. 21.4b but with a terminating resistance of 1 ohm. Since a maxi- mally flat design is needed, the order required is given by pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ log10 104 1 N ¼ 5:67887 ð21:8Þ log10 2:25 Thus, a sixth-order Butterworth ﬁlter is nee- ded. Taking values from standard Tables [1], we get the complete normalized LPF shown in Fig. 21.5 Characteristics of the normalized LPF corre- Fig. 21.6. To convert it to a de-normalized BSF sponding to the adjusted BSF response of Fig. 21.4a with a termination of 1 K, the following replacements are to be made: Fig. 21.6 Normalized Butterworth ﬁlter satisfying the speciﬁcations of Fig. 21.5 Example 163 (1) each inductance Li by a parallel combination Appendix of inductance RLi B=x20 and capacitance 1=ðRLi BÞ; Temes and Lapatra [4] recommend adjusting (2) each capacitance Ci by a series combination either of the pass-band edges to obtain geometric of inductance R=ðCi BÞ and capacitance symmetry. This is obviously not advisable Ci B=ðx20 RÞ; and because admitting part of the transition band into (3) resistance 1 ohms by R. the pass-band allows undesirable frequencies to be passed, along with the noise at these fre- p quencies, thus deteriorating the signal to noise where R = 1000 ohms, x0 ¼ 2p 5 103 radians/sec and B = 8p 103 radians/sec. ratio. On the other hand, adjusting the stop-band by pushing some of the transition band into the stop-band not only attenuates undesired fre- Concluding Comments quencies to a greater extent, but also improves the signal to noise ratio. This chapter attempts to supply the important Karni [5], in an example to illustrate the steps needed in designing a BPF/BSF through design of a BPF, computes the stop-band edge of frequency transformation. As noted in the intro- the normalized LPF as the ratio of the upper duction, these steps are either wrongly stop-band edge of the BPF to the bandwidth. stated/illustrated, as discussed in the Appendix, This is obviously wrong! or not explained in details in standard textbooks. References Problems 1. F.F. Kuo, in Network Analysis and Synthesis (Wiley, 1996) P:1. Determine the low-pass transfer function 2. H. Ruston, J. Bordogna, in Electric Networks: func- corresponding to a BPF having the same tions, Filters, Analysis (McGraw-Hill, 1966) speciﬁcation as those given in the example 3. M.E. Van Valkenburg, in Introduction to Modem in the text. Network Synthesis (Wiley, 1964) 4. G.C. Temes, J.W. Lapatra, in Introduction to Circuit P:2. Design the BPF corresponding to the above. Synthesis and Design (McGraw-Hill, 1977), pp. 556– P:3. Design the LPF corresponding to P.1. 557 P:4. What will happen if geometric symmetry is 5. S. Karni, in Network Theory–Analysis and Synthesis ignored in the BPF design? (Allyn and Bacon, 1966), p. 379 P:5. Same as above except that the design is to be for a BSF. Optimum Passive Differentiators 22 A general, nth order, the transfer function (TF) This chapter is complementary to [1], which is derived, whose time-domain response deals with optimum passive integrators. Follow- approximates optimally that of an ideal differ- ing, a parallel approach, optimum differentiators entiator, optimality criterion chosen being the of order n have been suggested here, the opti- maximization of the ﬁrst n derivatives of the mality criterion being the maximum possible ramp response at t = 0+. It is shown that values for the ﬁrst n derivatives of the ramp transformerless, passive, unbalanced realizabil- response at t = 0+. ity is ensured for n < 3, but for n > 3, the TF is We show that transformerless RLC unbal- unstable. For n = 3, the TF is not realizable, anced realization is possible only for n < 3, and however, near optimum results can be obtained that for n 3, the optimum transfer function by perturbation of the pole locations. Optimum (TF) is unstable. Near optimum results can be TFs are also derived for the additional con- achieved for n = 3 by perturbation of the pole straint of inductorless realizability. It is shown locations. RLC realizations for n 2 give, in that TFs for n 2 are not realizable. For all n, general, a damped oscillatory output around the however, near optimum results can be achieved ideal differentiated value. However, one can by small perturbations of the pole locations; this reduce the amplitude of these oscillations such is illustrated in this chapter for n = 2. Network that the output is within a prescribed limit of realizations, for a variety of cases, are also tolerance. In this chapter, we assume the toler- given. ance as ±5% of the ideal differentiated value. We will also derive an nth-order TF with an additional constraint of RC realizability, and Keywords show that for n > 2, optimal RC realizations are Differentiators Networks Optimization not possible; however, near optimum results can be achieved by small perturbations of the pole locations for all n. © Springer Nature Singapore Pte Ltd. 2018 165 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_22 166 22 Optimum Passive Differentiators Network realizations are given for the fol- With an input vin(t) = t u(t), the output will be lowing cases: n = 3, suboptimal, RLC; n = 2, optimal RLC; n = 2, suboptimal RLC (Oscilla- vn ðtÞ ¼ L1 Vn ð s Þ ¼ L 1 ð1=s2 ÞHn ðsÞ : tions limited to ±5% of the ideal differentiated ð22:4bÞ value); n = 2, suboptimal, RC. The optimality criteria chosen are: Optimal Transfer Function and Its Realizability ðiÞ vn ðtÞjt¼1 ¼ 1 ð22:5aÞ The ideal differentiator has a normalized TF maximum possible mð1Þ þ ð2Þ þ n ð0 Þ; mn ð0 Þ; ðiiÞ mð3Þ þ ðnÞ þ n ð0 Þ. . .; mn ð0 Þ: H 0 ðsÞ ¼ s ð22:1Þ ð22:5bÞ ðiÞ so that when the input voltage vin(t) is t u(t), i.e. a where vn ðtÞ: denotes the ith derivative of ramp function, the output voltage vout(t) = u(t), a vn(t), and unit step function. (iii) highest possible order zero, at s = 0, of The TF given by Eq. 22.1 is not realizable and most common approximation used is L½1 vn ðtÞ ð22:5cÞ H1 ðsÞ ¼ sT=ðsT þ 1Þ: ð22:2Þ Criterion (ii) is chosen to minimise the rise which can be realized by an RC or an RL net- time of vn(t); the reason for the choice of criterion work, shown in Figs. 22.1a, b, respectively. (iii) will be discussed later. We shall, in the When driven by vin(t) = t u(t), the output is sequel, impose the condition of a grounded transformerless network as part of realization v1 ðtÞ ¼ Tð1 et=T ÞuðtÞ: ð22:3Þ constraints. Combining Eqs. 22.4a, 22.4b and 22.5a with the ﬁnal value theorem gives v1(t) rises exponentially to the true (ideal) value of differentiation (=T u(t)), with a time constant T, a0 ¼ 0 and a1 ¼ b0 ð22:6Þ and reaches the ideal value only at t = ∞. How- ever, in practice, the time taken for the output to Using Eq. 22.6, Hn(s) and vn(s) become reach ±10% of the ideal value determines the usefulness of a differentiator. We shall choose a an sn þ an1 sn1 þ þ a2 s2 þ b0 s Hn ðsÞ ¼ tighter tolerance of ±5% to compare the perfor- sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 mance of various differentiators in this chapter. ð22:7aÞ The approximation will improve with decreasing T, in Eq. 22.3, but it also reduces the and output level. We, therefore, derive an optimal TF, of general order n, which does not suffer from this 1 an sn þan1 sn1 þ þa2 s2 þb0 s Vn ðsÞ ¼ 2 n disadvantage. Let, the required nth-order TF be s s þbn1 sn1 þ þb2 s2 þb1 sþb0 ð22:7bÞ an sn þ an1 sn1 þ þ a1 s þ a0 Hn ðsÞ ¼ n : s þ bn1 sn1 þ þ b1 s þ b0 By the initial value theorem, we have ð22:4aÞ vn ð0Þ ¼ Lim s Vn ðsÞ ¼ 0 ð22:7cÞ s!1 Optimal Transfer Function and Its Realizability 167 (a) (b) (c) (d) (e) (f) (g) (h) (a) (b) (c) (d) (e) (f) (g) (h) Fig. 22.1 Differentiator Networks 168 22 Optimum Passive Differentiators ðiÞ ð2Þ If we let Vn,1(s) D Lmn ðtÞ; then by the differen- To maximize vn ð0Þ; (criterion Eq. 22.5b), tiation theorem of Laplace transform, under the Fialkow–Gerst condition, an−1 bn−1, we have to choose Vn;i ðsÞ ¼ sVn;i1 ðsÞ vi1 n ð0Þ; i 1 ð22:8Þ an1 ¼ bn1 ð22:12Þ Thus, which gives Vn;1 ðsÞ sVn ðsÞ vn ð0Þ vð2Þ n ð0Þ ¼ 0 ð22:13Þ 1 an sn þ an1 sn1 þ þ a2 s2 þ b0 s ¼ s sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 From Eqs. 22.11a and 22.12, we get ð22:9aÞ Vn;2 ðsÞ ðan2 bn2 Þsn2 þ þ ða2 b2 Þs2 þ ðb0 b1 Þs b0 ¼ and sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 ð22:14Þ vð1Þ n ð0Þ ¼ Lim s Vn;1 ðsÞ ¼ an ð22:9bÞ s!1 Again, using Eq. 22.8, we have Vn;3 ðsÞ ¼ sVn2 ðsÞ mð2Þ n ð0Þ From criterion Eq. 22.5b, and the Fialkow– ðiÞ ðan2 bn2 Þsn1 þ þ ða2 b2 Þs3 þ ðb0 b1 Þs2 b0 s Gerst condition, we note that vn ð0Þ ¼ an ¼ 1: ¼ sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 Thus Eqs. 22.7a and 22.9a become, respectively, ð22:15aÞ sn þ an1 sn1 þ þ a2 s2 þ b0 s Hn ðsÞ ¼ and sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 ð22:10aÞ vð3Þ n ð0Þ ¼ Lim s Vn;3 ðsÞ ¼ an2 bn2 s!1 ð22:15bÞ 1 sn þ an1 sn1 þ þ a2 s2 þ b0 s Vn;1 ðsÞ ¼ s s þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 n By arguments similar to those already used, ð22:10bÞ ð3Þ the maximum value of vn ð0Þ is obtained when Again we have, from Eqs. 22.8 and 22.10b an2 ¼ bn2 ð22:16Þ n1 Vn;2 ðsÞ ¼ ðan1 bn1 Þs ðan2 bn2 Þsn2 þ þ ða2 b2 Þs2 þ ðb0 b1 Þs b0 for which þ sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 ð22:11aÞ vð3Þ n ð0Þ ¼ 0 ð22:17Þ so that This yields vð2Þ n ð0Þ ¼ Lim s Vn;2 ðsÞ ¼ an1 bn1 Vn;3 ðsÞ s!1 ðan3 bn3 Þsn2 þ þ ða2 b2 Þs3 þ ðb0 b1 Þs2 b0 s ð22:11bÞ ¼ sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ b0 ð22:18Þ Optimal Transfer Function and Its Realizability 169 Proceeding in this manner, till we maximize and ﬁnally, the optimum TF is ðnÞ vn ð0Þ we get sn þ sn1 þ sn2 þ þ s2 þ s Hn ðsÞ ¼ a1 ¼ bi ; i ¼ 1; 2; . . .; ðn1Þ ð22:19Þ sn þ sn1 þ sn2 þ þ s2 þ s þ 1 ð22:25Þ Combining these with Eq. 22.6 gives sðsn 1Þ ¼ ð22:26Þ sn þ 1 1 sn þ bn1 sn1 þ þ b2 s2 þ b0 s Hn ðsÞ ¼ n s þ bn1 sn1 þ þ b2 s2 þ b0 s þ b0 The poles of Hn(s) are located on the unit ð22:20aÞ circle, at and Sr ¼ cj2pr=ðn þ 1Þ r ¼ 1; 2; . . .; n ð22:27Þ 1 sn1 þbn1 sn2 þ þb2 sþb0 where r = 0 is excluded because s = 1 is a pole Vn ðsÞ ¼ s sn þbn1 sn1 þ þb2 s2 þb0 sþb0 as well as a zero. For stability, the poles should ð22:20bÞ be in the left half of the s-plane, i.e. We may also write Hn(s) of Eq. 22.20a as p=2 2pr=ðn þ 1Þ 3p=2 ð22:28Þ Nn ðsÞ Nn ðsÞ Equation 22.28 is violated for n = 4. The TF Hn ðsÞ ¼ ¼ ; ð22:21Þ for n = 3 is Dn ðsÞ Nn ðsÞ þ Dn ð0Þ s3 þ s2 þ s where Nn(s) and Dn(s) denote, respectively, the H3 ðsÞ ¼ ð22:29Þ numerator and the denominator polynomials of s3 þ s2 þ s þ 1 Eq. 22.20a. It can be easily shown that the poles of H3(s), Let at s = ±j, do not have purely imaginary residues, qðtÞDð1 vn ðtÞÞ uðtÞ ð22:22Þ hence H3(s) cannot be realized [3]. Had H3(s) been realizable, the ramp response would denote the deviation of vn(t) from the ideal output have been given by u(t). Had vn(t) been the ideal output; q(t) as well 1 2 vs ðtÞ ¼ L 1=s H 3 ðsÞ as Q(s) = Lq(t) would be zero. Since this is not 12 1 1 the case in practice, we impose condition ¼ 1 et cosðt þ p=4Þ uðtÞ 2 2 Eq. 22.5c, i.e. Q(s) should have a zero of the highest possible order at s = 0. Now ð22:30Þ 1 A plot of v3(t) is shown in Fig. 22.2 (curve f). Q ðsÞ ¼ Vn ðsÞ ð22:23Þ s Clearly, v3(t) is of little use due to the undamped oscillations. Substituting for Vn(s) from Eq. 22.20b and We thus conclude that the only optimum, simplifying, it is easy to observe that Q(s) will passive, grounded and transformerless network have a zero of the highest order (=n − 1), at realizable approximations of Eq. 22.1 are s = 0 if bn1 ¼ bn2 ¼ ¼ b2 ¼ b0 ¼ 1 ð22:24Þ 170 22 Optimum Passive Differentiators Fig. 22.2 Ramp responses of various differentiators with (e = 0.601), f third-order, optimal RLC (unrealizable), ﬁnal value normalised to unity. a Ideal case, b ﬁrst-order g third-order, suboptimal RLC (e = 0.5), h third-order, RC, c second-order, suboptimal RC (e = 0.01), d sec- suboptimal RLC (e = 0.71) ond-order, optimal RLC, e second-order, suboptimal RLC H1 ðsÞ ¼ s=ðs þ 1Þ ð22:31Þ fully compensated by appropriate reduction in the series resistance R. The ramp response of the second-order differentiator is given by H2 ðsÞ ¼ ðs2 þ sÞ=ðs2 þ s þ 1Þ ð22:32Þ ms ðtÞ ¼ L1 1=s2 H2 ðsÞ n pﬃﬃﬃ hpﬃﬃﬃﬃﬃﬃﬃﬃ io ¼ 1 2= 3 et=2 cos 3=2 t þ p=6 Equation 22.31 is the same as Eq. 22.2 with uðtÞ T normalized to unity. ð22:33Þ and is plotted in Fig. 22.2, curve d. The damped oscillations exhibited by v2(t) re- Second-order Optimal duce the utility of H2(s) vis-a-vis H1(s). How- and Suboptimal Differentiators ever, by shifting the poles of H2(s), we may bring down the amplitude of the oscillations to achieve Dividing the numerator and denominator of the desired tolerance limits of the output. If we Eq. 22.32 by s, two simple network realizations take of second-order passive differentiators, as shown in Figs. 22.1c, d are obtained. Realization of H20 ðsÞ¼ðs2 þsÞ= s2 þð1þeÞsþ1 ;where0\e\1 Fig. 22.1c may be preferred to that of Fig. 22.1d ð22:34aÞ since the effect of losses in the inductor L can be Second-order Optimal and Suboptimal Differentiators 171 then where " # 1 2e 2 1 1 2 a ¼ tan and b ¼ ð1 e2 =4Þ2 m2 ðtÞ ¼ 1 1=2 eð1þeÞt=2 cosðbt þaÞ uðtÞ; 2þe ð3þeÞ ð22:36bÞ ð22:34bÞ H20 ðtÞ can be realized using the Fialkow-Gerst Where technique [3], one such realization being shown 1 in Fig. 22.1g. 1e 2 1 1 a ¼ tan and b ¼ ð3 2e e2 Þ2 =2 As is clear from Eq. 22.36a, v03 ðtÞ also gives 3þe damped oscillations around the ideal differenti- ð22:34cÞ ated value. However, we can decrease the oscil- lations by increasing e. In particular, e = 1 gives It may be seen that damping increases as we critical damping [4], i.e. no oscillations and increase e from 0 to 1. In particular, H20 ðsÞ e¼0 ¼ v03 ðtÞje¼1 ¼ 1 ¼ v1 ðtÞ. The optimum value of e so H2 ðsÞ and m02 ðtÞ e¼0 ¼ v2 ðtÞ: Critical damping is that v03 ðtÞ may reach the ideal differentiated value achieved for e = 1 [4], and v02 ðtÞje¼1 ¼ 1 ¼ v1 ðtÞ. within a tolerance of ±5%, in the shortest pos- Thus for critical damping, v02 ðtÞ coincides with sible time, is found to be e = 0.71. Curves g and v1(t) and the rise time of v02 ðtÞ is maximal. The h in Fig. 22.2. show v03 ðtÞ for e = 0.5 and for rise time of v02 ðtÞ decreases with the decrease of e = 0.71, respectively. damping (i.e. with the decrease of e). The opti- mum value of e, such that v02 ðtÞ may reach the ideal differentiated value, within a tolerance of Optimal RC Differentiators ±5%, in the shortest possible time is found to be e = 0.601, and the response under this condition From Eq. 22.21, one can write the nth-order is shown by curve e in Fig. 22.2. differentiator TF as Nn ðsÞ Third-order Suboptimal Passive Hn ðsÞ ¼ ð22:37Þ Nn ðsÞ þ bo Differentiator For RC realizability, the roots of the denom- The third-order TF given by Eq. 22.29 is not inator polynomial should be distinct and located realizable due to its poles at s = ±j. We may, on the negative real axis of the s-plane, i.e. however, realize a network by shifting the poles slightly to the left in the s-plane. The TF will no Y n Nn ðsÞ þ bo ¼ ðs þ ri Þ; ð22:38Þ longer remain optimal, and we call this a sub- i¼1 optimal realization. The suboptimal TF will be where sðs2 þ s þ 1Þ H30 ðsÞ ¼ ; where 0\e\ 1 ðs þ 1Þðs2 þ es þ 1Þ 0\r1 \r2 \ \rn ð22:39Þ ð22:35Þ Equating the constant terms and the coefﬁ- The ramp response of Eq. 22.35 is given by cients of s on both sides of Eq. 22.38, we get " # Y n et ð1eÞ et=2 bo ¼ ri ð22:40Þ v03 ðtÞ¼ 1 e 1 cosðbtþaÞ uðtÞ; 2e bð2eÞ 2 i¼1 ð22:36aÞ 172 22 Optimum Passive Differentiators and where ! Xn 1 Yn bo ¼ ri ð22:41Þ a ¼ 2ð1 þ eÞ and b ¼ 2ð2e þ e2 Þ1=2 ð22:47bÞ r j¼1 j i¼1 Also, Combining Eqs. 22.40 and 22.41 gives v02RC ðtÞ e¼0 Dv2RC ðtÞ ¼ 1 ð1 þ tÞe2t uðtÞ X n ð22:48Þ 1=ri ¼ 1 ð22:42Þ i¼1 A plot of v02RC ðtÞ, for e = 0.01, is shown in For optimum results, we must maximize bo, as Fig. 22.2. (curve c). As the plots for v2RC ðtÞ and shown in the appendix. Following, a procedure v02RC ðtÞ do not differ by more than 1% in the time similar to one used in Section 5 of [1]; maxi- range shown, these are not shown separately in mization of bo yields Fig. 22.2. A realization of H20 RC ðsÞ; using the F-G tech- r1 ¼ r2 ¼ r3 ¼ ¼ rn ¼ n ð22:43Þ nique [3] is shown in Fig. 22.1h. and bo ¼ nn ð22:44Þ Conclusion Thus nth-order optimal RC differentiator is The problem of obtaining an optimum approxima- tion to the ideal differentiator by passive, transfor- ðs þ nÞn nn merless, unbalanced network has been investigated. HnRC ðsÞ ¼ ð22:45Þ ðs þ nÞn The following conclusions have been arrived at: (a) RLC, optimal differentiators are not realiz- able for order n 3; however, suboptimal Suboptimal RC Differentiator RLC differentiators, for n = 3 can be realized by pole perturbation technique. HnRC ðsÞ is not realizable for n 2 as the TFs (b) RC, optimal differentiators are not realizable have multiple poles at s = −n for n 2. To for order n 2; however, suboptimal RC make the poles distinct, we perturb them from differentiators for all n can be realized by location −n, so that HnRC ðsÞ becomes realizable pole perturbation. for all n. The same methodology as suggested in (c) RC differentiators, of all orders, reach the Section 6 of [1] can be followed; then the ideal value of differentiation only at t = ∞. second-order suboptimal TF becomes (d) Although the response of optimal RLC dif- s2 þ 4s ferentiators approaches the ideal value faster H20 RC ðsÞ ¼ 2 ð22:46Þ than that of RC differentiators, the former s þ 4ð1 þ eÞs þ 4 exhibit damped/undamped oscillations. This which gives, for a ramp input restricts the use of optimal RLC differentia- tors. However, the amplitude of oscillations 1 can be limited to any desired tolerance by m02RC ðtÞ¼ 1 eat ðaþb1Þebt ðab1Þebt uðtÞ; 2b pole perturbation, such that the RLC differ- ð22:47aÞ entiators give better performance than the RC differentiators. We have chosen a tolerance of Conclusions 173 ±5% of the ideal differentiated value and the e ¼ 0:71 results obtained are given in Table 22.1. A variety of differentiators using operational RLC, suboptimal n = 3 2 þ es þ 1Þ ampliﬁers are known. But the constraints of the þs þs active device viz. offset voltages and currents, 2 H30 ðsÞ ¼ ðs þ s1Þðs ﬁnite gain-bandwidth product, ﬁnite dynamic 2 range and slew rate limiting, etc. lead to further problems. The proposed optimum passive dif- 0.576 Table 22.1 Values of s5, the normalised time taken by various differentiators to give output voltage within ±5% of the ideal output voltage ferentiators, which are free from aforesaid limi- tations, may be successfully employed in areas, where the frequency spectrum of the signal is e ¼ 0:601 relatively wide or where simple and reliable cir- RLC, suboptimal n = 2 cuits with minimum power consumption are a necessity. A few such applications are the hom- H20 ðsÞ ¼ s2 sþ esþ þs 1 ing devices of an underwater torpedo, (where the 2 dynamic range requirement is large of the order of 80 dB) and guidance system of a long-range 0.587 guided missile (where the signal spectrum is wide, up to about 60 MHz). e ¼ 0:01 Problems þ eÞs þ 4 RC, suboptimal n = 2 P:1. Apply a square pulse of duration 1 s and s þ 4s height 1 V to a ﬁrst-order differentiator. H20 RC ðsÞ ¼ s2 þ 4ð1 2 Find the output v0/H and sketch it. P:2. Determine the transfer function of a sub- optimal differentiator of order 4 and obtain 0.7 the output for a ramp function. P:3. Obtain the transfer function for an optimal RC differentiator of order 2 and ﬁnd its H1 ðsÞ ¼ s þs 1 RC, optimal support for a unit step input. n=1 P:4. Same as P.3 except that the input is a ramp 1.0 function. P:5. Same as P.3 except that the input is an H0(s) = s (unrealisable) impulse function. Ideal case Appendix 0.0 In this section, we examine the nature of bo and substantiate the assertions made in Section 5 that differentiators higher the value of bo, better is the approxima- tion. For the ﬁrst-order case, the TF and the TF of the corresponding ramp response are, respectively, Case s5 174 22 Optimum Passive Differentiators bo s H1 ðsÞ ¼ ð22:49Þ s þ bo 1 D GðsÞ ð22:56Þ and s v1 ðtÞ ¼ ð1 ebo t ÞuðtÞ ð22:50Þ where Clearly, higher the bo, closer is v1(t) to u sn1 þ bn1 sn2 þ þ b3 s2 þ b2 s þ bo GðsÞ ¼ (t) which is the ideal ramp response. Maximum sn þ bn1 sn1 þ þ b2 s2 þ b1 s þ bo value of bo can be unity in H1(s) [F-G conditions ð22:57aÞ of realizability of H1(s)]. 1 þ ðb2 =bo Þs þ ðb3 =bo Þs2 þ For the second-order case, the TF and the ¼ ð22:57bÞ ramp response are, respectively 1 þ ðb1 =bo Þs þ ðb2 =bo Þs2 þ s 2 þ bo s Equation 22.56 shows that vn(t) = L−1 H2 ðsÞ ¼ 2 ð22:51Þ Vn(s) can be interpreted as the unit step response s þ bo s þ bo of the low-pass function G(s). Equation 22.55, and together with the initial and ﬁnal value theorems of Laplace transforms shows that vn(t) rises from " # ebo t=2 1 bo a value zero at t = 0 to unity at t = ∞. To enable m2 ðtÞ¼ 1 cos xo ttan uðtÞ; us make vn(t) achieve unity value in as short a ð4bo Þ1=2 2xo time as possible, we must choose b0 such that the ð22:52Þ rise time sr, of vn(t) is as small as possible. Using Elmore’s formula [2], with the assumption that where the plot of vn(t) is monotonic, (whereby Elmore’s 1=2 xo ¼ bo b2o =4 ð22:53Þ formula can be applied), we get 1 2 1 bo sr ¼ 2p b1 b22 2bo ðb2 b3 Þ 2 As cos xo t tan1 2x 1; smaller the o bo bo t=2 value e 1=2 ; closer is v2(t) to u(t). Increase of ð22:58Þ ð4bo Þ bo decreases ebo t=2 faster (i.e. exponentially) sr, decreases monotonically with the increase of than (4 − bo)1/2. Thus higher the bo, smaller is bo t=2 bo. Thus bo should be as large as possible. The the value of e and consequently closer is assumption of vn(t) being monotonic has impli- ð4bo Þ1=2 the v2(t) to the ideal value u(t). cations as mentioned in the Appendix of [1]. For the general case, a semi-rigorous argu- ment can be forwarded as follows. As References sn þ bn1 sn1 þ þ b2 s2 þ bo s Hn ðsÞ ¼ n s þ bn1 sn1 þ þ b2 s2 þ b1 s þ bo 1. S.C. Dutta Roy, Optimum passive integrators, in IEE Proceedings, part G, (vol. 130, No. 5, pp. 196–200), ð22:54Þ Oct 1983 2. W.C. Elmore, Transient response of damped linear and network with particular regard to wide band ampli- ﬁers. J. Appl. Phys. 19, 55–63 (1948) 1 sn1 þbn1 sn2 þ þb3 s2 þb2 sþbo 3. N. Balabanian, Network Synthesis (Prentice Hall, 1958) Vn ðsÞ ¼ 4. M.E. Van Valkenburg, Network analysis (Prentice s sn þbn1 sn1 þ þb2 s2 þb1 sþbo Hall of India, 1983) ð22:55Þ Part III Active Circuits Passive circuits have their own limitations and can do very little when ampliﬁcation, oscillation and other essential function of practical circuits can offer. Part III therefore concentrates on active circuits, which are combina- tions of passive circuits with active devices. Vacuum tubes are the things of the past and are seldom used, except in broadcast applications. We therefore treat circuits with transistors and operational ampliﬁers as the active devices. Ampliﬁer fundamentals are presented in Chap. 23; this material was the ﬁrst broadcast in India on actual educational materials and was done from studios of Space Application Centre at Ahmedabad, under the ‘Teacher in the Sky’ experiment of IETE. Judged by the positive feedback from students, it was a great success. Again, that appearances can be deceptive occurs in active circuits also. This is illustrated in Chap. 24 with the BJT biasing circuit as an example. BJT biasing is dealt with, comprehensively, in Chap. 25, and it is proved that bias stability is the best in the four resistor circuit. A high-frequency tran- sistor stage, consisting of emitter feedback, is analysed in detail in Chap. 26, using the hybrid equivalent circuit of the transistor, which was carefully avoided till then in most textbooks. Transistor Wien Bridge oscillator is treated comprehensively in Chap. 27, where various circuits and their merits and demerits are enumerated. In contrast to the hybrid parameter, I used the h-parameter equivalent circuit of the transistor because the former was not known till then. The usual analysis of the oscillator, as given in textbooks, is to use the Nyquist criterion Ab=1. In Chap. 28, I formulate several simpler, in fact much simpler, methods for doing the same, without the difﬁculty of identifying A and b, which is not easy even for experienced researchers. The triangular to sine wave converter is discussed in Chap. 29 with step-by-step logical anal- ysis. The Wilson current mirror, presented in Chap. 30, is a versatile circuit and is used as an essential component in various analog ICs. The dynamic resistance is calculated easily. That completes our journey through active circuits. I hope it will be a smooth one, without getting lost in the rather complicated equivalent circuits. Amplifier Fundamentals 23 This chapter presents the fundamentals of a ampliﬁer is an essential component. The Public bipolar junction transistor ampliﬁer and Address system used at large gatherings, like includes the following aspects: choice of political rallies and music concerts, is another Q point, classes of operation, incremental very common example. Under this topic, we equivalent circuit, frequency response, cas- shall discuss the essential features of an ampliﬁer cading, broadbanding and pulse testing. The along with the analysis of a typical circuit. emphasis is on understanding the fundamen- In order to introduce the subject, consider a tal, rather than rigorous analysis or elaborate typical single-transistor ampliﬁer circuit, shown in design procedure. Fig. 23.1, in which the transistor is connected in the common emitter conﬁguration. The phrase ‘common emitter’, incidentally, implies that the Keywords emitter terminal is common between the input and Ampliﬁer Transistor characteristics the output. In the circuit of Fig. 23.1, we have CE conﬁguration Biasing Hybrid P shown an n-p-n transistor, whose dc equivalent circuit collector-to-emitter voltage, VCE, is determined by the supply voltage VCC, the collector resistance RC and the emitter resistance RE through the equation The term ‘ampliﬁer’ stands for any device which ampliﬁes or magniﬁes a weak signal so as to VCE ¼ VCC IC RC IE RE ð23:1Þ make it detectable and useful. An ampliﬁer is perhaps the most important electronic circuit and IC and IE are the dc collector and emitter was the motivation or the leading reason behind currents, which are, of course, approximately the invention of the triode and the transistor. An equal, because the DC base current IB is much ampliﬁer is also a part of our daily life. The smaller than IC (IB = IC/b, b * 50). IB is deter- Radio, the Television and the Stereo are common mined by the relation (see Fig. 23.2) examples of electronic equipment where the R2 VCC ¼ IB ðR1 jjR2 Þ þ VBE þ IE RE ; R 1 þ R2 ð23:2Þ S.C Dutta Roy, “Ampliﬁer Fundamentals,” Students’ Journal of the IETE, vol. 35, pp. 143–150, July– where VBE is the dc base-to-emitter voltage and December 1994. is of the order of 0.7 V for a silicon transistor. © Springer Nature Singapore Pte Ltd. 2018 177 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_23 178 23 Amplifier Fundamentals virtually short-circuited. If CE is not there (opened), then the signal voltage developed across RE would cause a negative feedback, because the actual signal input to the transistor would have been the voltage across R2 minus the voltage across RE. The gain is thereby reduced to approximately—RL =RE ; RL ¼ RC jjR0L . Now let us come back to the question of choosing the Q point. Figure 23.3 shows the transistor characteristics along with a plot of Eq. 23.1 given by the line ABC. The usable region of the characteristic is bounded by the Fig. 23.1 A single-stage CE ampliﬁer maximum ratings of the transistor viz VCEmax, ICmax, the maximum collector dissipation IC and VCE determine the dc operating point or (VCEIC), PDmax, along with the saturation line the so-called Q point of the transistor. More will near VCE = 0, and the cutoff line iB = 0. If VCE be said about this later. and IC are such that the Q point is near B, then CC1 and CC2 in Fig. 23.1 are coupling signal excursions like that shown will lead to capacitors, their values being so chosen that they faithful variations in the collector current, as act as short circuits at the signal frequency. Had shown on the left side of the ﬁgure. This is the CC1 not been there (short-circuited), R2 would linear range of operation and goes under the have been virtually short-circuited (R1 R2, name of Class A operation in the literature. usually) and the transistor could not have been Obviously, Class A operation cannot extend to a biased properly. Similarly, if CC2 is not there point close to A during the positive excursions of (short-circuited), the DC through RC would have iB, or to a point close to C during the negative found a path through the load resistor R0L , so that excursions of iB, because of the distortion due to the Q point would shift with variations in the saturation and cutoff, respectively. If the Q point load. The capacitor CE across RE is chosen to be is the same as the point marked VCC in Fig. 23.3, large so that at the signal frequency, RE is then current will flow in the collector circuit only during the positive excursions of the signal. No current will flow during the negative excursions (this condition can be achieved by opening R1 in Fig. 23.1). Obviously, the circuit will act as a half-wave rectiﬁer. In order to reproduce the positive, as well as the negative portions of the signal, we need another transistor to take care of the negative half. These two transistors are con- ﬁgured in the so-called push–pull operation, as shown in Fig. 23.4, which, incidentally, uses a complementary symmetric pair of transistors (n-p-n and p-n-p) in a transformerless conﬁgu- ration. The usual arrangement of push–pull uses a centre-tapped transformer at the input and another such transformer at the output. The important point is that the transistors in Fig. 23.4 operate under what is deﬁned as the Class B Fig. 23.2 DC equivalent of the base-emitter circuit condition. Naturally, the Class B condition 23 Amplifier Fundamentals 179 Fig. 23.3 Transistor characteristics, showing load line and the bounds of operation condition. A further improvement in efﬁciency is possible if the transistor is biased beyond cutoff under quiescent conditions; the collector current will then flow for only a part of the positive half cycle (assuming an n-p-n transistor). This, of course, leads to excessive distortion, which can be avoided by using a parallel reso- nant circuit in the collector, tuned to the signal frequency. Then the voltage developed across the tuned circuit will mostly consist of the signal component. This is called the Class C mode of operation, and can be used to achieve practical Fig. 23.4 Complementary symmetry push–pull class B efﬁciencies of about 85%. ampliﬁer We shall now conﬁne our attention to small signal Class A operation so that the analysis is possible through an incremental or ac equivalent produces more distortion than the Class A circuit. The adjective ‘incremental’ refers to the operation. Nevertheless, Class B is invariably condition that signal components of current and used in low-frequency (audio) large signal or voltage in the transistor are small perturbations to power ampliﬁers, because of a drastic improve- the total current and voltage respectively. Under ment in efﬁciency (from a maximum of 25% in this assumption, the ampliﬁer behaves as a linear Class A to a maximum of 78.5% in Class B) system and DC and AC analyses can be done arising out of the reduced (to zero, ideally) power separately, the latter being carried out with the dissipation under the quiescent or no-signal equivalent circuit. 180 23 Amplifier Fundamentals Fig. 23.5 Hybrid-p equivalent circuit of transistor We shall now consider the circuit of Fig. 23.1 again, and carry out its analysis. Although vari- ous equivalent circuits have been proposed in the literature, the most versatile one viz the hybrid p, as shown in Fig. 23.5, shall be used here. We have used somewhat simpliﬁed notation as compared to what is used in most textbooks. Fig. 23.6 Simpliﬁed hybrid-p equivalent circuit of the While B, E and C stand for the base, emitter and transistor collector respectively, B′ is used to denote the internal base. The difference between B and B′ is has also been ignored. By inspection, you see the occurrence of the ‘base-spreading’ resistor, rx that the gain is which is usually an order of magnitude smaller vo than rp, the base-emitter resistor. Typically, Ao ¼ ¼ gm RL ð23:3Þ vi rx = 100 Ω, rp = 1 K, r = 4 MΩ, ro = 80 K, Cp = 100 pF, C = 3 pF and Co = 1 pF, while gm Next, consider the low-frequency response for is determined by the quiescent collector current which we ignore internal capacitances but not the IC according to the relation external ones. There are three such capacitances viz CC1, CC2 and CE, and it becomes rather gm ¼ ½IC ðin mAÞ=26 mhos involved to consider the effects of all three simultaneously. We, therefore, consider them Except at high frequencies, or in rigorous one by one. Suppose CE, CC2 ! ∞; then the analysis at other frequencies, we can ignore rx. equivalent circuit becomes that shown in Also, notice that ro will come across the load Fig. 23.8. With Rp ¼ R1 kR2 krp , we have the which is normally much smaller than 80 K; low-frequency gain hence it can be ignored, along with Co which is basically the stray capacitance. Normally, r can vo vo V also be ignored in comparison to the impedance AL ðsÞ ¼ ¼ ; vi V vi of C at the frequencies at which it counts. Hence, r also qualiﬁes to be omitted from consideration, Rp ¼ gm RL ; ð23:4Þ thus leading to the much simpliﬁed form of Rp þ sC1C1 Fig. 23.6. First, we consider midband frequencies at Putting s = jx and Rp CC1 = l/x1 we get which the effects of all capacitances—internal (Cp, C) as well as external (CC1, CC2 and CE) can Ao AL ðjxÞ ¼ ð23:5Þ be ignored, the former acting as open circuits 1 j xx1 while the latter act as short circuits. Then, the equivalent circuit of Fig. 23.1 becomes that This shows that with increasing frequency, the shown in Fig. 23.7, where the source resistance gain rises from zero to the midband value Ao and 23 Amplifier Fundamentals 181 Fig. 23.7 Equivalent circuit of Fig. 23.1 at midband frequencies and 1 1 x001 ¼ 1 þ gm þ RE ; ð23:8Þ RE C E rp gm ð23:9Þ CE Fig. 23.8 Low-frequency equivalent circuit with CE, CC2 ! ∞ The form of Eq. 23.9 arises because pﬃﬃﬃ reaches the value Ao = 2 when x = x1. In terms gm rp ¼ bð¼ hfe Þ 1 of decibels, this is equivalent to saying that the gain reaches 3 dB below the midband value at Usually, x001 xz . x = x1. Hence, x1 is called the low-frequency The question now arises: how to determine the cutoff point. low-frequency 3 dB cutoff (xL) when none of the In a similar manner, we can calculate the three capacitances can be considered as effect of CC2 with CC1, CE ! ∞ as giving rise to short-circuits. A guideline for the designer is that an expression of the form Eq. 23.5 with x1 one of the capacitances should be used to control replaced by x01 ¼ 1= CC2 ðRC þ R0L Þ . The effect xL while the other two should be so chosen that the of CE with CC1, CC2 ! ∞ is a bit more critical frequencies due to them are an order of involved, and it can be shown that the gain is magnitude less than the desired xL. For example, proportional to with CC1 ! ∞, CE = 200 lF, CC2 = 10 lF, RE = 100 Ω, Rc = R′L = 2 K, Rp rp = 1 K and 1 jxz =x b = gm rp = 100, the gain is of the form ; ð23:6Þ 1 jx001 =x ðconstantÞð1 jx=xz Þ ; ð23:10Þ where ð1 jx=x01 Þð1 jx=x001 Þ 1 where xz, = 50 r/s, x′1 = 25 r/s and x001 ¼ 509. xz ¼ ð23:7Þ ðCE RE Þ 1 r/s. The value of xL is then determined by x001 and is given by fL = 509.1/(2p) = 81.02 Hz. 182 23 Amplifier Fundamentals Fig. 23.9 High frequency equivalent circuit of the ampliﬁer of Fig. 23.1 Finally, we consider the high-frequency The effect of Rx, as expected, is to reduce the response of the ampliﬁer, for which the equiva- midband gain by the factor rp/(Rx + rp). Also lent circuit is shown in Fig. 23.9. Notice that we putting s = jx, and denoting CT ðrp k Rx Þ by have no longer ignored the effect of rp or of the 1/x2, we see that the HF 3-dB cutoff is given by source internal resistance Rs. The reason is that the small capacitor Cl reflects at the input (across 1 x2 ¼ ð23:14Þ Cp) as a much larger capacitance CT ðrp k Rx Þ To get an idea of x2, let, for a typical tran- CM ¼ Cl ð1 þ gm RL Þ; ð23:11Þ sistor ampliﬁer, which is approximately the midband gain times RS ¼ 900 X; rx ¼ 100 X: rp ¼ 1 K; Cl ¼ 4 pF Cl. This is known as the ‘Miller effect’. Cp ¼ 31 pF; gm ¼ 58 m - mhos and RL ¼ 2 K The total capacitance ð23:15Þ CT ¼ Cp þ CM ð23:12Þ Then may have a reactance which is comparable to CT ¼ 31 þ 4 ð1 þ 58 103 2 103 Þ Rx = rx + Rs at high frequencies. With Miller 500 pF effect taken into account, the equivalent circuit simpliﬁes to that shown in Fig. 23.10, where we ð23:16Þ have ignored R1 k R2 in comparison to rp. rp k Rx ¼ 1 K k 1 K ¼ 500 X ð23:17Þ Obviously, the gain is given by and vo V gm RL ðrp k 1=sCT Þ x2 1 AH ðsÞ ¼ ¼ f2 ¼ ¼ Hz V vi Rx þ ðrp k 1=sCT Þ 2p 2p 500 1012 500 gm RL rp ¼ 636 kHz ¼ ð23:13Þ sCT rp Rx þ Rx þ rp ð23:18Þ gm RL rp 1 ¼ Rx þ rp 1 þ sCT ðrp k Rx Þ Fig. 23.10 Simpliﬁed equivalent of Fig. 23.9 using Fig. 23.11 Frequency response of the gain of circuit of Miller effect Fig. 23.1 23 Amplifier Fundamentals 183 Fig. 23.12 A cascade of three stages Fig. 23.13 Showing how stages interact with each other As we can see from the analysis, the gain of the typical ampliﬁer of Fig. 23.1, called an RC cou- Fig. 23.14 Effect of cascading a number of stages pled ampliﬁer for obvious reasons, will have a band-pass characteristic as shown in Fig. 23.11. Aon ¼ Ano ð23:19Þ The gain of a single-stage ampliﬁer may not be adequate for the speciﬁc application. Hence, pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ one uses multistage ampliﬁers by cascading x1n ¼ x1 = 21=n 1 ð23:20Þ several stages as shown in Fig. 23.12. Analysis pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ of such multistage ampliﬁers proceeds stage by x2n ¼ x2 21=n 1 ð23:21Þ stage by considering the Thevenin equivalent of the previous stage as constituting the source, In a given ampliﬁer, how does one increase while the succeeding stage constitutes the load. the x2 and decrease x1? A number of such This is illustrated in Fig. 23.13 for analysis of compensation techniques are available and we Stage#2 of the circuit in Fig. 23.12. In general, shall discuss, qualitatively, one example of each. therefore, cascading does not lead to a multipli- A simple philosophy of increasing x2 is to use a cation of gains. load whose impedance increases with frequency, Suppose that we have succeeded in cascading so that the fall in gain due to the factor 1 + jx/x2 in a number of identical non-interacting stages, the denominator can be partially compensated. each having a midband gain Ao and low-and Such a load is shown in Fig. 23.15. high-frequency 3-dB cutoff at x1 and x2 A similar philosophy can be applied to com- respectively. The term ‘non-interacting’ here pensate for low-frequency fall off, by using a means that each stage has an input impedance load which increases with decreasing frequency. which is much higher than the output impedance One such load is shown in Fig. 23.16. of the previous stage, i.e. Zi, n+1 Z0, n. Under We conclude this discussion by pointing out this condition, what happens to the overall x1 that in order to test a given ampliﬁer for its low- and x2? This is illustrated qualitatively in and high-frequency responses, it is convenient to Fig. 23.14. Obviously, x1 increase and x2 use a pulse as the input. As shown in Fig. 23.17, decreases, i.e. the overall bandwidth decreases the response to a pulse will be a gradually rising while the gain increases. It can be easily shown waveform, which after reaching a maximum, that when n such stages are cascaded, the overall does not stay there, but sags a little before set- parameters are given by tling down to the zero value. The rise time, 184 23 Amplifier Fundamentals Fig. 23.15 HF compensation Fig. 23.17 Pulse response of an RC coupled ampliﬁer between the joint of R1A and R1B and Fig. 23.16 L-F ground. Derive the necessary equations for compensation the biasing condition of the transistor. P:2. At frequencies at which r 1/(xC) and r0 (1/xC0), ﬁnd an expression for the frequency response of a transistor, assuming a load RL and a source of resistance RS. P:3. Consider a two-stage cascaded ampliﬁer with source of resistance RS and load RL. Find an expression for the overall gain, if the stages interact with each other. P:4. In Fig. 23.15, a capacitor C is connected from the joint of L and RC and ground. Find an expression for the high-frequency gain. P:5. In Fig. 23.16, C is neither open-nor short-circuit. Derive an expression for the deﬁned as the time required for the waveform to low-frequency gain. rise from 10 to 90% of the ﬁnal value, can be related to x2, while the amount of sag can be For further information on ampliﬁers, see [1] related to x1. Reference Problems 1. A.S. Sedra, K.C. Smith, Microelectronic Circuits P:1. Suppose in Fig. 23.1, R1 is split into R1A (Sanders College Publishing, Fortworth, 1992) and R1B and a resistor R1C is connected Appearances Can Be Deceptive: The Case of a BJT Biasing Circuit 24 It is shown that bias stability is the best with Keywords the four resistor circuit. A two-resistor BJT Bias Bias stability 2, 3 and 4 resistor biasing circuit, which appears to be an attrac- biasing tive alternative to the familiar four resistor circuit, is shown to have serious limitations. It is also shown that even when augmented by one or two resistors, these limitations are only partially overcome and that the bias stability Introduction that can be achieved thereby is poorer than that of the four resistor circuit. Any student of Electronics should be familiar with the four resistors BJT biasing circuit shown in Fig. 24.1, to be called N1, hereafter, and any standard textbook would give the derivation of the following expression for the collector current IC (see, e.g. [1–3]): ½VCC R1 =ðR1 þ R2 Þ VBE þ ICBO ½1 þ ð1bÞ½ðRE þ R1 k R2 Þ IC ¼ ; ð24:1Þ RE þ f½RE þ ðR1 k R2 Þ=bg where the symbols VBE, ICBO and b have their usual meanings. Stabilization of IC against vari- ations of these three parameters due to tempera- ture change and/or replacement of transistor demands that Source: S. C. Dutta Roy, “Appearances can be Deceptive: The Case of a BJT Biasing Circuit,” Students’ Journal of the IETE, vol. 37, pp. 3–6, January–June 1996. © Springer Nature Singapore Pte Ltd. 2018 185 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_24 186 24 Appearances Can Be Deceptive … R1 k R2 bRE ð24:2Þ and VCC R1 =ðR1 þ R2 Þ VBE þ ICBO ½RE þ ðR1 k R2 Þ; ð24:3Þ where the usual assumption of b 1 has been made. It would be recognized that Eq. 24.3 is a conservative condition because what we require is that the left-hand side should be much greater than both of the two terms on the right-hand side. Fig. 24.2 N2—an apparently attractive alternative to N1 It is also easy to see that the midband gain of the circuit is −gmRc if RE is bypassed for AC by a stability achieved thereby compares poorly with capacitor, as shown in Fig. 24.1. that of N1. Another circuit for BJT biasing, usually mentioned as an exercise for the student in textbooks, is that of Fig. 24.2, to be designated Analysis of N2 as N2, for brevity. As compared to N1, N2 looks attractive because it uses only two, instead of Application of Kirchoff’s voltage law to the loop four resistors. It is believed that bias stabilization formed by VCC, RC, RF and the base–emitter occurs in N2 due to feedback through RF. We junction gives shall examine this circuit critically in this chapter and demonstrate that it has serious limitations, as VCC ¼ ðIC þ IB ÞRC þ IB RF þ VBE ð24:4Þ compared to N1. We also show that even with the addition of one or two resistors, these limitations Also, the base and collector currents are can only be overcome partially, and that the bias related to the following equation IC ¼ bIB þ ðb þ 1ÞICBO ð24:5Þ Combining Eqs. 24.4 and 24.5, we get VCC VBE þ ICBO ½1 þ ð1=bÞðRC þ RF Þ IC ¼ RC þ ½ðRC þ RF Þ=b ð24:6Þ Bias stability now demands that RF bRC ð24:7Þ and VCC VBE þ ICBO ðRC þ RF Þ ð24:8Þ under the usual assumption that b 1 and the conservatism mentioned earlier. The midband Fig. 24.1 N1—the familiar four resistor BJT biasing gain of the circuit can be easily derived to be −gm circuit Analysis of N2 187 (RF k RC ) under the assumption of gm (1/RF), which is usually satisﬁed. The limitations of the circuit are best brought out through a numerical example, as worked out next. An Example of Design Let the transistor Q have the following parameters VBE ¼ 0:6 V; ICBO ¼ 10 nA and b ¼ 100 ð24:9Þ Fig. 24.3 N3—modiﬁed form of N2 to avoid saturation at 25 °C, and let the Q-point be of Q VCE ¼ 4 V and IC ¼ 4 mA ð24:10Þ with VCC = 12 V. Also, let the midband gain for a gain of −160, we need RF1 = 2 K. RF2 has required be −160. The gm of the transistor is to be chosen to satisfy the base current IB = 0.04 mA. From the relation gm ¼ 4 mA/25 mV ¼ 0:16 f ð24:11Þ IB ¼ ðVCC VBE Þ=ðRF2 þ RF1 Þ ð24:16Þ so that for a gain of 160, we need we calculate RF2 as 83 K. Thus, our RF is 85 K, RC k RF ¼ 1K ð24:12Þ which, to our disappointment, does not satisfy Eq. 24.7, because b RC ¼ 200 K. To satisfy The speciﬁcation on VCE determines RC as Eq. 24.7, RF should not exceed 20 K, taking the thumb rule of 1:10 for the sign ‘’ to be satis- RC ﬃ 2K ð24:13Þ ﬁed. Clearly, N3 does not achieve bias stability! What should we do now? Use one more From Eqs. 24.12 and 24.13, we get resistor? Let us see. This additional resistor can be either from the RF ¼ 2K ð24:14Þ base to ground, as shown in Fig. 24.4, or from the emitter to ground, as shown in Fig. 24.5. These two Note that 1/RF equals 0:001 f which is indeed circuits will be designated as N4 and N5, respectively. much smaller than gm ¼ 0:16 f thus validating Consider N5 ﬁrst. It is not difﬁcult to realize that the midband gain formula, but the problem arises since the same current passes through RC and RE, the elsewhere. With RF = 2 K, the base current is expression for IC will be the same as Eq. 24.6 except IB ¼ ðVCE VBE Þ=RF ¼ 1:7 mA ð24:15Þ that RC + RE will take the place of RC. Thus, for bias stability, we need RF b ðRE þ RC Þ and since IB Since bIB has a value of 170 mA, clearly, the is still given by Eq. 24.16, RF does not change. transistor will be saturated! Hence, this modiﬁcation is of no use. What is the remedy? In [2], it is suggested that For N4, given in Fig. 24.4, the currents in the we split RF into two parts and use a bypass various branches, as indicated, can be easily capacitor, as shown in Fig. 24.3. This circuit, to established. Kirchoff’s voltage law can be used be called N3, has a gain −gm (RC k RF1 ) so that to write the following equation: 188 24 Appearances Can Be Deceptive … With b 1, bias stability requirements now become RF bRC ð24:19Þ and VCC R1 =ðR1 þ RC þ RF Þ VBE þ ICBO ½R1 k ðRC þ RF Þ ð24:20Þ Since RC will have to be 2 K to satisfy the Q- point and RF1 will also have to be 2 K to satisfy Fig. 24.4 N4—a modiﬁcation of N3 the gain requirement, the question that arises is the following: is it possible to choose RF2 < 83 K? Note that ½IB þ ðVBE =R1 ÞRF ¼ VCE VBE ð24:21Þ Putting numerical values, this gives ð3:4=RF Þð0:6=R1 Þ ¼ 0:04 103 ð24:22Þ We want to have RF 20 K. If RF is chosen, arbitrarily, as 17 K, Eq. 24.22 gives R1 = 3.75 K. Now look at the other requirement, given by Eq. 24.20. Putting numerical values, the left-hand side is calculated as 1.978 V while the right-hand side is greater than 0.6 V. Hence, Eq. 24.20 is not satisﬁed. In fact, it can be shown Fig. 24.5 N5—another modiﬁcation of N3 that the highest value of the left-hand side of Eq. 24.20 under the constraint of Eq. 24.22 and VCC ¼ ½IC þ IB þ ðVBE =R1 ÞRC RF 20 K occurs when RF = 20 K, and that ð24:17Þ this value is only 2.075. Hence, we conclude that þ ½IB þ ðVBE =R1 ÞRF þ VBE N4, like N5, is also not of much use in stabilizing Combining this with Eq. 24.5, and solving for the transistor Q-point. IC gives ½VCC R1 =ðR1 þ RC þ RF Þ VBE þ ICBO ½1 þ ð1=bÞ½ðR1 jjRC þ RF Þ IC ¼ ð24:18Þ ½R1 RC =ðR1 þ RC þ RF Þ þ f½R1 jjðRC þ RF Þ=bg Conclusion 189 Conclusion P:2. In Fig. 24.1 circuit, a resistor RL is con- nected between the collector and ground. The preceding example clearly demonstrates the Comment on the bias stability of the circuit. limitations of N2, and its modiﬁed versions—N3, Justify. N4 and N5, in stabilizing the Q-point of a BJT. P:3. In Fig. 24.3 circuit, the capacitor is there On the other hand, one can easily show that N1 and is neither a short circuit nor an open with RC = RE = 1 K and R1 = R2 = 20 K gives circuit. Deriver an expression for the low-frequency gain. R1 k R2 ¼ 10 K; bRE ¼ 100 K; P:4. In Fig. 24.4 circuit, the capacitor is there VCC R1 =ðR1 þ R2 Þ ¼ 6 V; and and is neither a short circuit nor an open VBE þ ICBO ðRE þ R1 k R2 Þ ﬃ 0:6V circuit. Derive an expression for the high-frequency gain. so that both Eqs. 24.2 and 24.3 are approximately P:5. What happens when the capacitor is shifted satisﬁed. We conclude therefore that N1 is the to have a position between the collector and best choice for stabilizing the Q-point of a BJT. ground? Carry out the necessary analysis for the low-frequency gain. Problems You may have to couple these with the previous References chapter 1. J. Millman, A. Grabel, Microelectronics (McGraw- P:1. In Fig. 24.1 circuit, the capacitor is neither Hill, New York, 1987) 2. S.G. Burns, P.R. Bond, Principles of Electronic open nor short. Find an expression for the Circuits (West Publishing Company, St. Paul, 1987) low-frequency gain, using, of course, the 3. A.S. Sedra, K.C. Smith, Microelectronic Circuits hybrid p equivalent circuit. (Sanders College Publishing, Fortworth, 1992) BJT Biasing Revisited 25 The familiar four resistor circuit for biasing a choice of a few additional resistors and then bipolar junction transistor (BJT) is generalized transformed to a different topology. The latter is through simple reasoning, and transformed to shown to yield, as special cases, three alternative yield a different topology. Three alternative four resistor circuits, to be called N2, N3 and N4, four resistor circuits are derived as special which do not appear to have been widely known cases of the transformed generalized circuit, in the literature in the context of biasing a BJT. which do not appear to have been widely From a detailed and careful analysis, it is shown known in the literature. A detailed and careful that the bias stability parameters achieved in all analysis reveals that the bias stability param- the four circuits—N1, N2, N3 and N4—are com- eters of all alternative circuits are comparable parable. An illustrative example of bias design is to those of the conventional circuit. An worked out to demonstrate this fact. illustrative example is included for demon- strating this fact. The Generalized Circuits and Special Keywords Cases BJT Biasing Bias stability Design Let a resistor connected between nodes X and Y be denoted by R(X, Y). In Fig. 25.1a, there are ﬁve nodes—A, B, C, E and G—and a most Introduction general biasing circuit would be the one in which every node is connected to every other node by a Figure 25.1a shows the familiar BJT biasing resistor. Several exceptions are to be made, circuit for linear class A ampliﬁcation, and is to however. The resistors R(A, G), R(A, E), R(C, E) be called N1 in the sequel. It uses four resistors, and R(C, G) are not necessary for biasing and of which RE1 is usually by-passed for AC [1–3]. cause additional loss of power. If these four In this chapter, N1 is generalized by a proper resistors are excluded, then the generalized biasing circuit looks like that shown in Fig. 25.1b. If the two p-networks BCA and BEG are Source: S. C. Dutta Roy, “BJT Biasing Revisited,” IETE converted into T’s, then the transformed circuit Journal of Education, vol. 46, pp. 27–33, January– March 2005. becomes that shown in Fig. 25.1c. Note that © Springer Nature Singapore Pte Ltd. 2018 191 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_25 192 25 BJT Biasing Revisited A +VCC IC + I 1 RC1 (a) (b) (c) D R2 IC RC2 I1 A A +VCC +VCC IB C R2 Rc2 IC B Q E C C R1 I1 IB B Q B Q IC + IB RE1 E E F R1 RBE C RE1 RE2 IC + I 1 G G G Fig. 25.1 a N1—the familiar four resistor BJT biasing circuit; b a generalized BJT biasing circuit; c alternative generalized BJT biasing circuit obtained by transformation of the circuit of (b) since p to T conversion does not involve any gives the circuit of Fig. 25.4, henceforth to subtraction operation, all resistances in be referred to as N4. Fig. 25.1c are positive. Hence, both circuits can claim to be general canonic biasing circuits for Note that (i) for the convenience of reference, the BJT. Four special cases of four resistor cir- we have designated the circuits such that Ni cuits can be derived from the two generalized refers to Fig i, and that (ii) N2 cannot be derived circuits as follows. from Fig. 25.1b. Because of the latter, the circuit of Fig. 25.1c, therefore seems to have an edge (1) Let R(B, C) = R(B, E) = ∞ in Fig. 25.1b or over that of Fig. 25.1b in terms of topological R(A, D) = R(F, G) = 0 in Fig. 25.1c; then generality. Also, note that another four resistor we get the conventional circuit N1 of circuit can be obtained by setting R(A, B) = R(B, Fig. 25.1a. G) = ∞ or R(D, C) = R(E, F) = 0 in Fig. 25.1c. (2) Let R(E, F) = R(F, G) = 0 in Fig. 25.1c; However, in the resulting circuit, RC1 and RE2 then we get the circuit of Fig. 25.2, hence- carry the same DC; hence for biasing purposes, forth referred to as N2. they can be combined into one resistance and the (3) Let R(B, C) = R(B, G) = ∞ in Fig. 25.1b or circuit thereby behaves as a three resistor one. It R(A, D) = R(E, F) = 0 in Fig. 25.1c; then we has been found that such a circuit has a poorer obtain the circuit of Fig. 25.3, which we bias stability than N1, and will not, therefore, be shall refer to as N3. considered further. (4) Let R(A, B) = R(B, E) = ∞ in Fig. 25.1b or In N2, it is of advantage to by-pass RC1 for R(D, C) = R(F, G) = 0 in Fig. 25.1c; this AC, as shown, by connecting a large capacitor The Generalized Circuits and Special Cases 193 A A +V• +VCC RC1 I1 + IC RC1 I1 + I C C D IC I1 R2 I1 R2 IC RC2 B Q IC IB C E I1 - I B R1 B Q RE1 IB E R1 G G Fig. 25.4 N4—yet another alternative BJT biasing circuit Fig. 25.2 N2—an alternative BJT biasing circuit A Bias Stability Analysis +VCC RC2 IC The generalized circuit of Fig. 25.1c will only be R2 analyzed because it gives all the four special cases. For this purpose, all the resistors have C been named and the currents in all of them have B Q been identiﬁed. Let the transistor Q be charac- E terized by the parameters VBE, b and ICBO, where the symbols have their usual meanings. The R1 currents IC and IB are related by the following equation RE2 IC ¼ bIB þ ðb þ 1ÞICBO : ð25:1Þ G By applying Kirchoff’s voltage law, one obtains Fig. 25.3 N3—another alternative BJT biasing circuit VCC ðIC þ I1 Þ ðRC1 þ RE2 Þ ¼ I1 R2 þ ðI1 IB ÞR1 ; between nodes D and G, so that the gain is ð25:2Þ determined by RC2 only. It may be noted that this form of the circuit ﬁnds use in multistage and ampliﬁers for power supply decoupling, where RC1 is common to all the stages. For single-stage ðI1 IB ÞR1 ¼ VBE þ ðIC þ IB ÞRE1 ð25:3Þ biasing, however, N2 does not appear to have been used. As compared to N1, N2 trades an From Eq. 25.3, I1 can be obtained in terms of additional resistor at the collector side for that at IB and IC. Substituting this value in Eq. 25.2, the emitter side. In N3 as well as N4, the emitter replacing IB in terms of IC and ICBO from resistance is by-passed for ac, as in N1. Eq. 25.1, and carrying out some algebraic 194 25 BJT Biasing Revisited manipulations, one obtains the following and/or replacement of transistor, where the expression for IC: changes are not inﬁnitesimal. One can ﬁnd, from Eq. 25.4, the net change ΔIC = IC2 − IC1 and V1 VBE þ ICBO r1 divide by IC1 to determine the fractional (or IC ¼ ; ð25:4Þ r2 þ ðr1 =bÞ percentage) variation of IC. In most textbooks, however, this procedure is not followed because where the resulting expression is considered to be ‘very formidable and not too informative’ [2, p. 412]. V1 ¼ VCC R1 =ðR1 þ R2 þ RC1 þ RE2 Þ; ð25:5aÞ Instead, they consider the change of IC due to r1 ¼ RE1 þ R1 k ðR2 þ RC1 þ RE2 Þ; ð25:5bÞ each parameter separately, holding the other two constant. We shall also follow the same proce- r2 ¼ RE1 þ ½R1 ðRC1 þ RE2 Þ=ðR1 þ R2 þ RC1 þ RE2 Þ; dure to start with, and then show that considering ð25:5cÞ all the changes simultaneously is not as difﬁcult as it is made out to be. and we have made the simplifying, but practical To follow the conventional procedure, let d, assumption that b 1 so that the factor [1 + (1/ dv and dI denote the partial fractional changes in b)] can be approximated by unity. IC due to changes in b, VBE and ICBO, respec- Bias stability is achieved if IC can be made tively. When all the three parameters vary insensitive to variations in VBE, b and ICBO. Note simultaneously, and each d is small (<0.1), the that IC is independent of RC2, but of course, RC2 total fractional change in IC, to be denoted by dT, has an important effect on VCE. Also, note that is estimated as the sum of d, dv, and dI. RE2 always occurs in combination with RC1 in the From Eq. 25.4, one can easily derive expres- form RC1 + RE2. This is to be expected because, sions for the fractional deviations. The results as is clear from Fig. 25.1c, the same current are: IC + I1 flows in them. As mentioned earlier, both resistors are not necessary; one can make either Db=b1 RC1 = 0 or RE2 = 0 without any loss of general- db ¼ ; ð25:9Þ 1 þ b2 ðr2 =r1 Þ ity. This fact is also reflected in the circuits N1– N4. DVBE dv ¼ ; ð25:10Þ Referring to Eq. 25.4, we observe that bias V1 VBE1 þ ICBO r1 stability demands the following conditions to be met: and DICBO r2 =r1 1=b; ð25:6Þ d1 ¼ ð25:11Þ ICBO1 þ ½ðV1 VBE1 Þ=r1 V1 VBE ; ð25:7Þ It is clear that the resistance r1, given by and Eq. 25.5b determines dv and dI, while d is determined by the ratio r2/r1, which can be V1 ICBO r1 ð25:8Þ obtained from Eqs. 25.5b and 25.5c as As will be evident from the practical designs r2 RE1 ðR1 þR2 þRC1 þRE2 ÞþR1 ðRC1 þRE2 Þ ¼ worked out later in the chapter, usually VBE r1 RE1 ðR1 þR2 þRC1 þRE2 ÞþR1 ðRC1 þRE2 ÞþR1 R2 ICBO r1 so that satisfying Eq. 25.7 automatically ð25:12Þ satisﬁes Eq. 25.8. To obtain a quantitative measure of bias sta- The values of V1, r1 and r2/r1 for the four bility, consider the case in which the parameter circuits are given in Table 25.1. In practical cir- set (VBE, b, ICBO) changes from (VBE1, b1, ICBO1) cuits, RC1, RE1 and RE2 will be of the same orders to (VBE2, b2, ICBO2) due to temperature variation of magnitude (≅1 K) while R1 and R2 will be one Bias Stability Analysis 195 order higher. It can, therefore, be observed that r1 a negligible shunting effect. Let the various cir- and r2/r1 are comparable for all the four circuits, cuit and transistor parameters be as follows: which makes them comparable in terms of bias stability performance. In particular, if RC1 of N2 VCE ¼ 4 V; IC ¼ 4 mA; VCC ¼ 12 V; VBE is the same as RE2 of N3, then N2 and N3 will ¼ 0:6 V; ICBO ¼ 10 nA have identical behaviour. and b ¼ 100; When all the three parameters vary simulta- ð25:14Þ neously, as is usually the case in practice, one can easily show, using Eq. 25.4, that The last two quantities being measured at 25 ° C. Also, let the gain required be −160, so that the DIC DVBE þ ICBO r1 required RC2 = 160/gm = 160/(40 IC) = 1 K. d0 ¼ ¼ 1þ IC1 V1 VBE þ ICBO r1 1 Db Db 1þ 1þ 1 Design of N1 b1 b1 þ ðr1 =r2 Þ ð25:13Þ Since b = 100 1, we require RC2 + RE1 ≅ ( VCC − VCE)/IC = 2 K. Thus RE1 = 1 K. From This expression does indeed look formidable, Eq. 25.6 and Table 25.1, it is required to have but is not difﬁcult for computation once the RE1 =ðRE1 þ R1 k R2 Þ 1=b; with numerical numerical values are available. Also, to compare values substituted, this translates to the competing circuits, all that changes are the R1 k R2 99 K. Let R1 and R2 be arbitrarily values of r1 and r2/r1. chosen as 20 K each. Then V1 becomes 6 V so In the next section, an illustrative example of that Eq. 25.7 is satisﬁed. Also, r1 is calculated as design is worked out for absolute as well as 11 K so that ICBO r1 = 11 10−5; thus Eq. 25.8 comparative performances of the four circuits. is also satisﬁed. The design is summarized in column 2 of Table 25.2. An Example Design of N2 For a fair comparison of the four circuits, one should design each circuit for the same Q point The gain requirement ﬁxes RC2 as 1 K. From and the same gain. First, consider N1, N2 and N3, Eqs. 25.6–25.8 and Table 25.1, the requirements in all of which, the gain is approximately of bias stability become −gmRC2. Since the Q points are the same, one should have identical RC2 in each circuit. This R2 99RC1 ensures that the output resistance is also equal, and VCC R1 =ðR1 þ RC1 þ R2 Þ 0:6; 108 R1 under the usual assumption of r0 of the BJT k ðRC1 þ R2 Þ: being much greater than RC2. The input resis- ð25:15Þ tance in each circuit is approximately r in the usual situation of base biasing resistances having Also, for this circuit, Table 25.1 Values of r1 Circuit r1 r2/r1 and r2/r1 for N1–N4 N1 RE1 þ R1 k R2 RE1 =ðRE1 þ R1 k R2 Þ N2 R1 k ðRC1 þ R2 Þ RC1/(RC1 + R2) N3 R1 k ðRE2 þ R2 Þ RE2/(RE2 + R2) N4 RE1 þ R1 k ðR2 þ RC1 Þ RE1 ðR1 þ R2 þ RC1 Þ þ R1 RC1 RE1 ðR1 þ R2 þ RC1 Þ þ R1 RC1 þ R1 R2 196 25 BJT Biasing Revisited Table 25.2 Bias stability of example designs Circuit N1 N2 N3 N4 Design parameters RC2 = RE1 = 1 K RC2 = RC1 = 1 K RC2 = RC1 = 1 K RE1 = 0.896 K R2 = R1 = 20 K R1 = 10 K R1 = 10 K R1 = 32.5 K R2 = 9 K R2 = 9 K R2 = 20 K RC1 = 1.05 K V1(volts) 6 6 Same 7.28 r1(K) 11 5 Values 12.76 r2(K) 1 0.5 As 1.43 db 0.0342 0.0313 Those 0.0281 dv 0.0463 0.0463 Listed 0.0374 dI 0.0208 0.0095 For 0.0195 dT 0.1013 0.0871 Circuit 0.0850 d0 0.1034 0.0883 N2 0.0863 VCE ¼ VCC ðI1 þ IC ÞRC1 IC RC2 ; ð25:16Þ Performances of N1, N2 and N3 and Let b change from b1 = 100 to b2 = 150 and let I1 ¼ IB þ ðVBE =R1 Þ ﬃ ðIC =bÞ þ ðVBE =R1 Þ: the temperature change from 25 to 125 °C. As is ð25:17Þ well known, VBE decreases with temperature at the rate of 2.5 mV/°C, and ICBO doubles for Combining Eqs. 25.16 and 25.17 and substi- every 10 °C rise in temperature. Thus, VBE tuting numerical values (note RC2 = 1 K) gives changes by ΔVBE = –250 mV while the corre- the condition sponding change in ICBO is ΔICBO = (210 − 1) ICBO = 10.23 lA. ð4=RC1 Þð0:6=R1 Þ ¼ 4:04 103 ð25:18Þ For each of the three designs, the values of d, dv and dI can now be calculated from Eqs. 25.9– Let R1 = 10 K; then RC1 is obtained from 25.11 and Table 25.1. These values are given Eq. 25.18 as 0.976 K ≅ 1 K. Now from Table 25.2 along with other necessary informa- Eq. 25.15, we should have R2 99 K. With tion. Note that no partial fractional deviation is R2 = 9 K, V1, which is the left-hand side of more than 0.1 so that in each circuit, the total Eq. 25.15, becomes 6 V, and ICBO r1, which is fractional deviation can be obtained by summing the second expression on the right-hand side of the three partial ones. The circuits N2 and N3 Eq. 25.15, becomes 5 10−8; thus Eq. 25.15 is cause a smaller change in IC than N1, although no satisﬁed. The design is complete and is given in special care was taken to show N2 and N3 in a column 3 of Table 25.2. brighter light than N1. However, no generaliza- tion can be made from this speciﬁc design; it is possible that with a redesign of N1, balance can Design of N3 be tilted in its favour. All that can be said, and it has been said earlier, is that bias stabilities that As mentioned earlier, the design of N2 will also can be achieved by the three circuits are work for N3 if RE2 is taken as 1 K. comparable. Design and Performance of N4 197 Design and Performance of N4 as in the other designs, the value of ICBO r1 is negligible compared to 0.6 so that Eq. 25.8 is N4 is different from N1, N2 and N3 in that it had satisﬁed. DC as well as AC feedback through R2. The ac The complete design along with values of d, equivalent circuit is shown in Fig. 25.5, from dv, dI and dT are given in Table 25.2. which the gain can be calculated as Vo ð1=R2 Þ gm Using the Total Change Formula ¼ ð25:19Þ Vi ð1=RC1 Þ þ ð1=R2 Þ For the example under consideration, the total By Miller’s theorem, the input impedance change formula Eq. 25.13 was used for each would be R1 krp kR2 =ð1 þ jgainjÞ. While the design, and the values of d0 are found to be shunting effect of R1 can be ignored, that of R2/ 0.1034, 0.0883, 0.0883 and 0.0863 for N1, N2, N3 (1 + |gain|) cannot; in fact, the latter will, in and N4, respectively. Clearly, d0 > dT. Thereby practice, be one order smaller than r ! The output showing that the estimate of total fractional impedance is approximately RC1 k R2 =ð1 þ change on the basis of partial changes is an jgainj1 Þ ﬃ RC1 if R2 RC1 as is usually the optimistic one. case. With a gain requirement of −160, and R2 arbitrarily chosen as 20 K, Eq. 25.19 gives Conclusion RC1 = 1.05 K. Referring to Fig. 25.4, we see that I1 = (VCE − VBE)/R2 = 0.17 mA. Hence, voltage The BJT biasing circuit has been generalized and drop across RE1 is VCC − VCE − (IC + I1)RC1 = transformed to yield three alternative four resistor 3.62 V; since the current through RE1 is circuits, whose bias stability performance is 4.04 mA, we get RE1 = 0.896 K. Finally, comparable to that of the commonly used four R1 = (VBE + voltage drop across RE1)/ resistor circuit. It has been shown that the com- (I1 − IB) = 32.5 K. monly used performance measure dT obtained by With the above design, r1 and r2/r1 are cal- summing the partial fractional changes is an culated from Table 25.1 as 12.76 K and 0.112; optimistic one. It has also been shown that the thus Eq. 25.6 is satisﬁed. Also, V1 here becomes calculation of the total fractional change d0 poses 7.28 V so that Eq. 25.7 is also satisﬁed. Finally, no problem even though the formula looks for- midable. Another general guideline in designing a bias circuit that has been revealed in our R2 designs is that once V1 VBE has been estab- lished, V1 ICBO r1 need not be checked + + because VBE is usually a few orders greater than Vi R1 rx gm Vi RC1 Vo ICBO r1. - The circuits N1, N2 and N3 have similar per- - formance in terms of gain, input and output impedances, while for the same gain and output impedance, N4 has one order lower input Fig. 25.5 AC equivalent circuit of N4 impedance. 198 25 BJT Biasing Revisited Problems References P:1. Replace the dotted capacitor C by a ﬁrm 1. J. Millman, A. Grabel, Microelectronics connection. Choose C such that its impe- (McGraw-Hill, New York, 1987) dance is comparable to RE1. What happens 2. S.G. Burns, P.R. Bond, Principles of Electronic Circuits (West Publishing House, St Paul, 1987) to the biasing? Analyze. 3. A.S. Sedra, K.C. Smith, Microelectronic Circuits P:2. What if RBE is absent in Fig. 25.1b? (Oxford University Press, New York, 1998) P:3. What if RBE = 0 in Fig. 25.1b? P:4. Let REZ = 0 in Fig. 25.1c. What is the effect on biasing? P:5. What if C in Fig. 25.2 is not too large to become a short circuit at AC? Analysis of a High-Frequency Transistor Stage 26 It is shown that, contrary to popular belief, circuit is not valid for output impedance calcu- classical two-port network theory is adequate lations. There exist several other methods for for an exact analysis of a general carrying out the analysis: the classical node or high-frequency transistor stage, including mesh analysis, analysis using feedback concepts, emitter feedback, almost by inspection. driving point impedance technique [3], and the recently proposed open and short circuit tech- nique [2]. Of these, the last one appears attrac- Keywords tive, and is based on the calculation of two Two-port analysis High-frequency stage simpler gain functions and a driving point impedance. The purpose of this chapter is to show that classical two-port network theory is adequate for Introduction analyzing the circuit exactly, almost by inspec- tion. The method has been tested in the under- Consider the high-frequency ampliﬁer circuit graduate classes and has been well received. shown in Fig. 26.1, which includes an un-bypassed emitter resistance RE. The capacitor Cl is traditionally singled out as the troublesome Two Port Analysis element, but for which the analysis would have been much simpler. In most textbooks on elec- Let as indicated in Fig. 26.1, tronics, therefore, the circuit is unilateralized through application of Miller’s theorem, and Rx ¼ Rs þ rz and Zp ¼ 1=ðgp þ sCp Þ ¼ 1=Yp simpliﬁed by assuming a resistive load, and ð26:1Þ ignoring the reflected Miller admittance on the load side [1]. These assumptions, as one readily where gp = 1/rp. We shall carry out the analysis appreciates, are not always valid; further, as in several steps. First, consider the two-port pointed out by Yeung [2], the Miller equivalent shown in Fig. 26.2a. By inspection, its y-matrix is yp 0 Source: S. C. Dutta Roy, “Analysis of a High Frequency ½ya ¼ ð26:2Þ gm 0 Transistor Stage,” Students’ Journal of the IETE, vol. 29, pp. 5–7, January 1988 © Springer Nature Singapore Pte Ltd. 2018 199 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_26 200 26 Analysis of a High-Frequency Transistor Stage N (a) (b) IS IL + Cm 1 V Zp gm V 2 1 2 RS rx Cm + RX Cp V rp gmV ZL VL + - VS (c) Cm (d) - Zp + RE 1 V Zp gm V 2 1 RE 2 Fig. 26.1 Equivalent circuit incorporated along with the (e) Cm actual components + V Zp gm V Next, consider the two-port of Fig. 26.2b for 1 2 which, again by inspection, RE sCl sCl ½yb ¼ ð26:3Þ sCl sCl Fig. 26.2 Steps in derivation Now, connect the two two-ports of Fig. 26.2a, If we connect the two-ports of Fig. 26.2c, d in b in parallel, as in Fig. 26.2c; the y-matrix of this series, the two-port of Fig. 26.2e results, whose two-port is the sum of Eqs. 26.2 and 26.3, i.e., z-matrix is obtained by adding 26.6 and 26.7, i.e., " # yp þ sCl sCl RE þ 1 yp þ gm RE þ 1 yp þ g m ½yc ¼ ð26:4Þ ½ze ¼ gm sCl sCl RE þ sCl gm RE þ yp þ sCl sCl ðyp þ gm Þ sCl ðyp þ gm Þ The determinant of Eq. 26.4 is ð26:8Þ j yjc ¼ sCl ðyp þ gm Þ ð26:5Þ Adding Rx in series at the input port in Fig. 26.2e gives the two-port N, indicated by Hence the z-matrix of the two-port of dashed outline in Fig. 26.1. The z-matrix of N is Fig. 26.2c is therefore the same as given in Eq. 26.8 except for an increase of z11 by Rx. Hence 1 y22c y12c " # ½zc ¼ jyj y y11c R x þ RE þ 1 RE þ 1 " c 21c 1 1 # ½zN ¼ yp þ g m sCl gm y p þ gm yp þ sCl y p þ gm y p þ gm RE þ sCl ðyp þ gm Þ RE þ sCl ðyp þ gm Þ ¼ gm þ sCl yp þ sCl ð26:6Þ sCl ðyp þ gm Þ sCl ðyp þ gm Þ ð26:9Þ Next consider the two-port of Fig. 26.2d. Its Now postulate the currents IS and IL as shown z-matrix is given by in Fig. 26.1. Then RE RE Vs ¼ Is z11N þ IL z12N ð26:10Þ ½zd ¼ ð26:7Þ RE RE Two Port Analysis 201 VL ¼ Is z21N þ IL z22N ð26:11Þ high-frequency transistor stage having an un-bypassed emitter resistor and a general load. It should be clear that the effect of rl could be VL ¼ IL ZL ð26:12Þ taken account of by replacing sCl in Eq. 26.9 by gl + sCl and that the effect of a parallel ro, Co Solving for IL from Eqs. 26.11 and 26.12, we combination across the current generator gm get V could be taken account of by putting y22 a = go + sCo, instead of zero, in Eq. 26.2 and con- IL ¼ Is z21N =ðz22N þ ZL Þ ð26:13Þ tinuing the analysis. Substituting this in Eqs. 26.11 and 26.12 gives the voltage transfer function H(s) = VL/Vs Problems and the input impedance Zin = Vs/Is as P:1. Rederive the equations by assuming gm H ðsÞ ¼ ZL z21N =ðZ11N ZL þ jzjN Þ ð26:14Þ ! ∞ in Fig. 26.1. P:2. What happens if Cl = 0 in Fig. 26.1? Zin ¼ ðz11N ZL þ jzjNÞ=ðz22N þ ZL Þ ð26:15Þ P:3. What happens if Cl ! ∞ in Fig. 26.1? P:4. Now, it’s the turn of RE. What happens The output impedance Zout faced by the load when RE = 0 in Fig. 26.1? is precisely 1/y22N, i.e. P:5. What happens when RE ! ∞ in Fig. 26.1? Zout ¼ jzjN =z11N ð26:16Þ Combining Eqs. 26.14–26.16 with Eq. 26.9 References gives the desired expressions. 1. J. Millman, Microelectronics (McGraw-Hill, New York, 1979) 2. K.S. Yeung, An open and short circuit technique for Conclusion analyzing electronic circuits. IEEE Trans. Educ. E-30, 55–56 (1987) It has been demonstrated that simple two-port 3. R.D. Kelly, Electronic circuit analysis and design by techniques are adequate for analyzing, exactly driving point impedance techniques. IEEE Trans. Educ. E-13, 154–167 (1970) and almost by inspection, a general Transistor Wien Bridge Oscillator 27 Three possible circuits of transistor Wien working into an inﬁnite impedance load, it has a bridge oscillator, derived from analogy with transfer function given by the corresponding vacuum tube circuit, are vout 1 described. Approximate formulas for the fre- ¼ quency of oscillation and the voltage gain vin 1þ C1 R1 þ C2 R2 þ j xC2 R1 xC11 R2 C1 R 2 required for maintenance of oscillations are ð27:1Þ deduced. A practical circuit using two OC71 transistors is given. The frequency of oscilla- where x = 2 pf denotes the angular frequency of tion is found to agree fairly well with that the driving source. The phase shift produced by calculated from theory. The relative merits of the network is zero at a frequency xo, where the different forms have also been discussed. 1=2 1 xo ¼ ð27:2Þ Keywords R1 R2 C1 C2 Transistor Oscillator Wien bridge Figure 27.2 shows the circuit of a vacuum tube oscillator using the network of Fig. 27.1. It consists of a two-stage ampliﬁer with positive feedback provided through the Wien network. Introduction Under open loop conditions, the ampliﬁer has a flat gain and a phase shift of 360° over the fre- The RC network shown in Fig. 27.1 is a quency range of interest. Thus the circuit will degenerated form of the Wien bridge and will, oscillate at a frequency given by Eq. 27.2 pro- henceforward, be referred to as the Wien net- vided that the open loop gain Ao of the ampliﬁer work. Driven by an ideal voltage generator and satisﬁes the inequality C2 R1 Ao 1 þ þ C1 R2 A common emitter (CE) transistor ampliﬁer is Source: S. C. Dutta Roy, “Transistor Wien Bridge Oscillator,” Journal of the Institution of analogous to a common cathode vacuum tube Telecommunication Engineers, vol. 8, pp. 186–196, ampliﬁer in that both give voltage ampliﬁcation July 1962 with a phase reversal. A transistor circuit, © Springer Nature Singapore Pte Ltd. 2018 203 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_27 204 27 Transistor Wien Bridge Oscillator maintenance of oscillations. In view of the R1 C1 low input impedance of a CE ampliﬁer, the gain of the ﬁrst stage will be small; the vin R2 C2 vout second stage will then have to provide the necessary voltage gain. (4) In an oscillator circuit, it is desirable that the frequency of oscillation should be controlled Fig. 27.1 Degenerated Wien bridge network by varying the passive elements only. In the above circuit, since both the series and the shunt arms arc supplemented by transistor H .T . + impedances, the frequency will be largely RL controlled by the transistor impedance R1 C1 CL parameters, which are again functions of the OUT transistor operating point. V1 V2 These difﬁculties may be obviated by using a current dual of the Wien network as the feedback R2 C2 + + network [1, 2]. In this chapter, it is shown that - - H .T . – circuits analogous to that of Fig. 27.2 can be designed to overcome some or all the difﬁculties listed above. In all, three circuits have been dis- Fig. 27.2 Circuit of a vacuum tube Wien bridge cussed, using 4, 3 and 2 transistors. In the anal- oscillator ysis of these circuits, a transistor ampliﬁer has been assumed to be capable of representation by analogous to that of Fig. 27.2, will, however, the equivalent circuit shown in Fig. 27.3 in have the following drawbacks. which the effect of the collector capacitance has been assumed to be negligible. The expressions (1) A transistor ampliﬁer in the CE mode has a for the voltage gain (A) and the output impedance high output impedance so that the behaviour (Zi) are: at the output terminals is essentially that of a current generator. The series arm of the Wien v2 h21 ZL A¼ ¼ ð27:3Þ network will thus be supplemented by the v1 h11 þ Dk ZL output impedance of the equivalent voltage generator. In contrast, in Fig. 27.2, V2 acts as a voltage generator with an internal impe- v1 h11 þ Dk ZL Zi ¼ ¼ ð27:4Þ dance rpRL/(rp + RL) (rp = plate resistance of i1 1 þ h22 ZL V2) which is usually small compared to R1. (2) The input impedance of a CE transistor i1 i2 ampliﬁer is low and causes a considerable + - loading of the shunt arm of the Wien net- h11 1/h22 h12v2 work. In contrast, the input impedance of a Zg vacuum tube is very high, ideally inﬁnite, so v1 v2 ZL that the Wien network in Fig. 27.2 works h21i1 vin into an open circuit load. (3) Since the shunt arm is heavily loaded and the impedance of the series arm is increased, a Fig. 27.3 Low frequency equivalent circuit of a transis- large voltage gain will be required for tor ampliﬁer Introduction 205 Table 27.1 Low Parameter Conﬁguration frequency h-parameters of OC71 at Vc = −2 V, CB CE CC Ic = 3 mA h11 (X) 17 800 800 h12 8 10−1 5.4 10−4 1 h21 −0.979 47 −47 h22 ðfÞ 1.6 10−6 80 10−6 80 10−6 Δk 8.1 10−1 0.0386 47.06 where effect of the biasing resistors is negligible and that the coupling and decoupling condensers behave Dk ¼ h11 h22 h12 h21 as short circuits at the frequency of oscillation. The two-stage ampliﬁer T1T2 gives an output voltage which is in phase with the input voltage. A practical circuit, using two OC71 transis- The ﬁrst stage has a load approximately equal to tors, is given. Each transistor is maintained at the the input impedance of the second and as such following operating point: collector to emitter has a low gain. In view of the high input impe- voltage, Vc = −2 V; collector current, Ic = 3 dance of the CC stage T3, T2 has a load mA. The h-parameters at the above operating approximately equal to RL2 and can be designed point are given in Table 27.1. The frequency of to have a high gain. As the output impedance of oscillation is found to agree fairly well with that T3 is low, the Wien network is effectively sup- calculated from theory. plied by a voltage generator. The network has a load equal to the input impedance of the CC stage T4 and as such, if R2 is not too high, the Circuit 1 loading may be considered to be negligible. T4 has a low output impedance and is not loaded A common collector (CC) transistor ampliﬁer is when connected to the input of T1. The CC analogous to a cathode follower circuit. It has a stages produce no phase shift. Thus the circuit high input and a low output impedance while the will oscillate at a frequency at which the phase gain is slightly less than unity. Thus the high shift through the network is zero. impedance output of a CE stage can he trans- An approximate solution for this circuit can be formed into a low impedance one by cascading a obtained as follows. Let us assume that the CC stage to it. The problem of loading of the transistors T1–T4 are identical. If the effective shunt arm of the Wien network can be solved in a load impedance of T2 is not very large, then its similar way. The transistor analogue of the Wien input impedance is approximately h11e, where the bridge oscillator of Fig. 27.2 will then look like subscript e is used to mean a common emitter that shown in Fig. 27.4. Only the A.C. equivalent parameter. The effective load impedance of T1 is circuit has been drawn on the assumption that the T1 T2 T3 T4 RL1 RL2 C1 R1 Re1 C2 R2 Re2 Fig. 27.4 Circuit 1 using four transistors (2 CE and 2 CC) 206 27 Transistor Wien Bridge Oscillator RL1 h11e A3 ’ 1. Thus if the voltage at the input of T1 is ðRL Þ1 ’ v1, then that at the input of the Wien network is RL1 þ h11e A1A2v1. The voltage gain of this stage is, by formula Since Re2 is shunted by the input impedance of Eq. 27.3, T1 which is small ð’ h11e Þ the input impedance of T4, (Zi)4, will not, in general, be negligible com- h21e ðRL Þ1 pared with the impedance of the shunt arm of the A1 ¼ h11e þ ðDh Þe ðRL Þ1 Wien network. (Zi)4, however, can be calculated as follows. The load impedance of T4 is Now, (RL)1 < h11e and from Table 27.1, (Dk)e 1. Thus Re2 h11e ðRL Þ4 ’ Re2 þ h11e h21e h21e RL1 A1 ’ ðRL Þ1 ¼ ð27:5Þ h11e RL1 þ h11e Thus from Eq. 27.4, The effective load impedance of T2 is h11e þ ðDk Þe ðRL Þ4 ðZi Þ4 ¼ RL1 ðZi Þ3 1 þ h22e ðRL Þ4 ðRL Þ2 ¼ RL2 þ ðZi Þ3 Now, (RL)4 < h11e so that from Table 27.1, where (Zi)3 is the input impedance of T3. From h22e(RL)4 1. Therefore, Eq. 27.4, Re2 h11e ðZi Þ4 ’ h11e þ ðDk Þ ð27:7Þ h11e þ ðDh Þe Re1 Re2 þ h11e ðZi Þs ’ 1 þ h22e Re1 Let The approximation involved in the above R2 ðZi Þ4 equation is that the loading of Re1 by the Wien R02 ¼ ð27:8Þ network and the following stage is negligible. R2 þ ðZi Þ4 Substituting values from Table 27.1 and assum- ing Re1 = 1 kX, we get ðZi Þ3 ’ 48 kX. Now, RL2 Then from Eq. 27.1, the voltage at the input will be of the order of 3 kX or less so that we can of T4 is put (Zi)3 RL2 and get ðRL Þ2 ’ RL2 . Thus the A1 A2 v 1 gain of the second stage is ð27:9Þ C1 R1 þ C2 R02 1þ C1 R02 þ j xC2 R1 xC11 R0 2 h21e RL2 A2 ’ ð27:6Þ The gain of T4 is, by formula Eq. 27.3, h11e þ ðDk Þe RL2 h21e ðRL Þ4 The gain of the third stage is A4 ¼ h11e þ ðDk Þe ðRL Þ4 h21e Re2 A3 ’ h11e þ ðDk Þe Re1 Assuming Re2 = 1 kX and substituting for the parameters from Table 27.1, we get A4 ’ 1. Assuming Re1 = 1 kX and putting the values Thus the output voltage of T4 is given by of the parameters from Table 27.1, we get Eq. 27.9; but, this is equal to v1 so that Circuit 1 207 C1 R1 þ C2 R02 1 Total negative feedback may be applied by A1 A2 ¼ 1 þ 0 þ j xC 2 R 1 0 C1 R2 xC1 R2 connecting a suitable resistance from the output ð27:10Þ of T2, T3 or T4 to the ﬁrst emitter, the biasing resistance connected to it being partly or wholly Equating the imaginary parts on either side of unbypassed. If a suitable non-linear element is Eq. 27.10 gives the frequency of oscillation as used for the feedback resistance, amplitude sta- bilization of the oscillator output will result. 1=2 The use of negative feedback raises the input 1 xo ¼ 0 ð27:11Þ impedance of T1 and thus reduces the loading of C1 C2 R1 R2 Re2. As a result, the input impedance of T4 Combining Eqs. 27.7 and 27.8 with increases and the loading of the shunt arm of the Eq. 27.11, we have Wien network decreases. If sufﬁcient negative feedback can be used, then a1 ! 0 and x0 ! xn. " ( )#1=2 The output of the oscillator is to be taken from R2 ðRe2 þh11e Þ xo ¼xn 1þ the third emitter, as the output impedance of T3 is h11e ðh11e þRe2 ÞþðDk Þe Re2 h11e small (’ 64 X for OC71 transistor if RL2 = 2 ð27:12Þ 2 kX and Re1 = 1 kX). The output impedance of T4 is also small, but a load connected at this point where xn denotes the angular frequency at will reduce (Zi)4 and as such R02 . Thus a variation which the phase shift through the isolated net- in the load impedance will result in a change in work is zero. The quantity within the second the frequency of oscillation. bracket in Eq. 27.12 may be looked upon as a correction factor, a1. If R2 < 3 kX and Re2 = 1 kX, then for the OC71 transistor, a1 < 0 14 and Circuit 2 we can write A common base transistor ampliﬁer can give 1 voltage ampliﬁcation with zero phase shift. Thus xo ’ xn 1 þ a1 2 the two-stage CE ampliﬁer in Fig. 27.4 can be replaced by a single stage CB ampliﬁer as shown The condition for maintenance of oscillations in Fig. 27.5. As in the previous case, only the A. is obtained by equating the real parts on either C. equivalent circuit has been drawn on the same side of Eq. 27.10. Combining this with Eqs. 27.5 assumptions as made before. and 27.6 gives Let the voltage at the input of T1 be v1. As in 2 h21e RL1 RL2 the previous case, the input impedance of T2 will be high compared to RL so that the gain of T1 is ðRL1 þ h11e Þ h11e þ ðDk Þe RL2 C 2 R1 h21b RL ¼ 1þ þ ð1 þ a1 Þ ð27:13Þ A1 ’ C 1 R2 h11b þ ðDk Þ RL b Normally, the left-hand side of eq. 27.13 will Substituting values from Table 27.1, we note be far in excess of the right-hand side, so that the that h21b ’ 1 and that if RL < 3 kX then output waveform will be distorted. A good (Dk)bRL < 2 43. Thus, to a ﬁrst approximation, waveform can be obtained by inserting a suitable we can neglect (Dk)bRL compared to h11b and get resistance Rf at the point marked X in Fig. 27.4. It is, however, better to reduce the gain by negative RL A1 ’ ð27:14Þ feedback. Local negative feedback may be h11b applied through unbypassed emitter resistance. 208 27 Transistor Wien Bridge Oscillator Fig. 27.5 Circuit 2 using T1 three transistors (1 CB and 2 T2 T3 CC) RL C1 R1 Re1 C2 R2 Re2 The gain of T2 is approximately unity so that the wo ¼ wn ð1 þ a2 Þ1=2 input to the Wien network is A1v1. The input impedance of T3, (Zi)3, will be small compared where with that of T2 because of the heavy loading of Re2 by the input circuit of T1. The load impedance of R2 T3 is ’ h11b and from Table 27.1, h22eh11b 1; a2 ¼ h11e þ ðDk Þe h11b thus The correction factor a2 in this case is quite ðZi Þ3 ’ h11e þ ðDk Þe h11b large. For the OC71 transistor, if R2 = 3.2 kX, then a2 = 2. Equating the real parts on either side Let of Eq. 27.17 gives the condition for maintenance of oscillations. Combining this with Eqs. 27.14 R2 h11e þ ðDk Þe h11b R02 ¼ ð27:15Þ and 27.16, we get R2 þ h11e þ ðDk Þe h11b h21e RL C 2 R1 k ¼ 1þ þ ð1 þ a2 Þ Then the output of the Wien network will be h11e þ ðD Þe h11b C 1 R2 given by ð27:18Þ A 1 v1 v3 ¼ As in the previous case, the left-hand side of C1 R1 þ C2 R02 1þ C1 R02 þ j xC2 R1 xC11 R0 Eq. 27.18 will usually be in excess of the 2 right-hand side. A suitable resistance may be The gain of T3 is used between the emitters of T3 and T1 to get a good waveform. Alternatively, negative feedback h21e h11b may be applied by connecting a suitable resis- A3 ’ ð27:16Þ h11e þ ðDk Þe h11b tance from the output of T1, T2 or T3 to the base of T1, the biasing resistances at this point being The output of T3 is given by v4 = A3v3. But partly or wholly unbypassed. As in the previous v4 = v1; thus case, negative feedback reduces the correction factor, a2. C2 R1 1 A 1 A3 ¼ 1 þ þ þ j xC2 R1 As the output impedance of T3 and the input C1 R02 xC1 R02 impedance of T1 are both small, the output ð27:17Þ voltage may be taken from the third emitter, if the load impedance is not too small. Alterna- Equating the imaginary parts on either side of tively, the output may be taken from the second Eq. 27.17 gives the frequency of oscillation as emitter as in the previous case. Circuit 3 209 T1 h21b ZL C1 R1 T2 A1 ¼ h11b þ ðDk Þb ZL RL C2 R2 Re Now ZL < RL and RL is of the order of 3 kX. Thus (Dk)bZL h11b and since h21b ’ 1, Fig. 27.6 Circuit 3 using two transistors (1 CB and 1 ZL A1 ’ ð27:23Þ CC) h11b The voltage across R1 is A1v1, v1 being the Circuit 3 voltage at the input of T1. The output voltage of the Wien network is, therefore, A further simpliﬁcation of the Wien bridge oscillator circuit is possible if we omit the tran- A 1 v1 sistor T2 in Fig. 27.5. The resulting circuit is v2 ¼ C1 R1 þ C2 R02 shown in Fig. 27.6. In view of the high output 1þ C1 R02 þ j xC2 R1 xC11 R0 2 impedance of T1, the series arm of the Wien ð27:24Þ network will also be supplemented by an extra impedance in this circuit. The voltage gain of T2 is As in circuit 2, the effective load impedance of T2 is approximately h11b and since h22ch11b 1, h21c h11b A2 ’ ð27:25Þ its input impedance is h11c þ ðDk Þc h11b ðZi Þ2 ’ h11c þ ðDk Þc h11b The output of T2 is A2v2 = v1. Combining this with Eqs. 27.22–27.25, we get Let h21c RL RL ¼ D 1þ ð27:26Þ R2 ðZi Þ2 R2 h11c þ ðDk Þc h11b Z1 þ Z2 R02 ¼ ¼ R2 þ ðZi Þ2 1 þ R2 = h11c þ ðDk Þc h11b where D denotes the denominator of the ð27:19Þ right-hand side of Eq. 27.24. Now from 1 Eqs. 27.20 and 27.21, Z1 ¼ R 1 þ ð27:20Þ jxC1 R02 D Z1 þ Z2 ¼ ð27:27Þ and jxC2 R02 þ 1 R02 Combining 27.26 and 27.27 gives Z2 ¼ ð27:21Þ jxC2 R02 þ 1 h21c RL C2 R1 þRL ¼1þ þ The effective load impedance of T1 is h h11c þðD Þc h11b C1 R02 1 RL ðZ1 þ Z2 Þ þj xC2 ðR1 þRL Þ ZL ¼ ð27:22Þ xC1 R02 R L þ Z1 þ Z2 ð27:28Þ so that the voltage gain of T1 is 210 27 Transistor Wien Bridge Oscillator Equating the imaginary parts on either side of because even if (ZL)2 = 1 kX, h22c(ZL)2 = 80 Eq. 27.28 and substituting for R02 from Eq. 27.19 10−3 1. The frequency of oscillation will be gives the frequency of oscillation as given by " #1=2 ( ) 1 þ R2 = h11c þ ðDk Þc h11b 1þ R2 xo ¼ xn h11c þ ðDk Þc Re ðRf þ h11b Þ=ðRe þ Rf þ h11b Þ 1 þ ðRL =R1 Þ xo ¼ xn 1 þ ðRL =R1 Þ ð27:29Þ ð27:32Þ Equation 27.29 shows that xo can be made The condition of oscillation is modiﬁed to the equal to xn by choosing following: R1 R2 ¼ RL h11c þ ðDk Þc h11b ð27:30Þ h21c RL C R R n o ¼ 1þ 2 þ 1 1þ L h ðRe þ Rf þ h11b Þ h11b ðDk Þc þ 11cRe ðR C1 R2 R1 f þ h11b Þ Equating the real parts on either side of 8 9 < R2 = Eq. 27.28 gives the condition for maintenance of 1þ : h þ ðDk Þc Re ðRf þ h11b Þ; oscillations as 11c ðRe þ Rf þ h11b Þ h21c RL C2 R1 RL k ¼ 1þ þ 1þ This can be solved to ﬁnd the appropriate h11c þ ðD Þc h11b C R2 R1 ( 1 ) value of Rf. It is, however, more convenient to R2 put a variable resistance for Rf and to adjust it 1þ h11e þ ðDk Þe h11b experimentally. The output voltage is taken from the emitter of ð27:31Þ T2 from the same considerations as stated in the Here also the left-hand side of Eq. 27.31 will previous case. be in excess of the right-hand side, and the gain can be reduced by the same methods as employed in circuit 2. If degeneration is used, Practical Circuit then the input impedance of T1 will be raised and the output impedance lowered. The former A practical version of the two-transistor circuit is reduces the loading of the shunt arm and the shown in Fig. 27.7a. Each transistor is main- latter reduces the impedance adding to the series tained at the operating point at which the arm of the Wien network. If sufﬁcient negative parameters of Table 27.1 apply. This was done feedback can be applied, then xo can be made to for comparing the actual frequency with that approach xn very closely. calculated from theory. By choosing a smaller For simplicity’s sake, let us suppose that the value of Ic, the circuit could be designed to work gain is reduced by inserting a resistance Rf on a 6 V. battery. A 9 V battery could also be between the emitters of T2 and T1. Then the load used for establishing the required operating impedance of T2 is point, but the biasing resistors required are so small that besides drawing a large power from ðRf þ h11b ÞRe the battery, their effect on the A.C. operation ðZL Þ2 ’ Re þ Rf þ h11b becomes quite appreciable. A slightly lower value of resistance was used Therefore at the collector of T2 than that at the collector of ðZi Þ2 ’ h21e þ ðDk Þe ðZL Þ2 T1 to establish a slight difference of potential between the two emitters. Practical Circuit 211 (a) (b) 2.2 K 30 K 30 K 2.1 K 25 m T1 C1 R1 - + (OC71) 25 m + - T2 - T1 (OC71) 1M 12 V + + - C2 - 100 m 1.11 K 10 K 25 m 1.11 K 1K - R2 19 K + 19 K 25 m + - 9K 25 m + Fig. 27.7 a Practical oscillator circuit using two transistors; b arrangement for negative feedback Table 27.2 Comparing R1 (kX) C1 (lF) R2 (kX) C2 (lF) R4 (X) fa (c/s) fc (c/s) the actual frequency with that calculated from theory 40 0.106 89 0.105 140 115 113 0.97 0.106 10.4 0.105 395 215 212 0.97 0.106 1.48 0.105 415 755 730 1.4 0.022 1.48 0.0208 375 3365 3400 0.97 0.011 1.48 0.0095 410 7223 7580 0.97 0.0065 1.48 0.00642 385 11,312 12,000 In Fig. 27.7a, the gain is shown to be reduced The dependence of frequency of oscillation by inserting a variable resistance Rf in the posi- on the transistor operating point is most pro- tive feedback line. Thus for this circuit, for- nounced in circuit 2, because the correction mula 27.32 will be applicable. The arrangement factor a2 is usually greater than unity. The for reducing the gain by negative feedback is correction factor a1 in circuit 1 is generally less shown in Fig. 27.7b. than unity while that in circuit 3 can be made a The values of the Wien network components minimum by choosing R1, R2 and RL such that (R1, C1, R2 and C2), the feedback resistance (Rf). Eq. 27.30 is satisﬁed. Condition Eq. 27.30 the actual frequency of oscillation (fa) and the cannot, however, be maintained in the lower frequency calculated from Eq. 27.32 (fc) are audio range because of the large values of shown in Table 27.2. In calculating fc for the ﬁrst condensers required. two cases, the effects of the biasing resistances The change in the frequency of oscillation due were also taken into account. It will be seen that to a given change of load impedance will be the fa agrees fairly well with fc in the frequency range highest in circuit 3 and the least in circuit 1. shown. The lower limit of frequency in either of the three circuits considered will be set by the maxi- mum value of the coupling and bypass capacitors Discussions that can be used while the high frequency limit will be set primarily by the collector capacitance. From economic considerations, circuit 3 should Note that OC71 is obsolete. So do not search be preferred as it uses the least number of tran- for one in the market. Instead wire of a circuit sistors and other components. with a commonly available tansistor. 212 27 Transistor Wien Bridge Oscillator When this paper was written, [3] was our Bible P:5. What if there is no negative feedback? for transistor circuits. Also see [4] for an early form of transistor oscillator. References Problems 1. D.E. Hooper, A.E. Jackets, Current derived resistance capacitance oscillators using junction transistor. Elec- P:1. Analyze the circuit of Fig. 27.3 with tron. Eng. 28, 333 (1956) hybrid-p parameter. h-parameters are not 2. R. Hutchins, Selective RC ampliﬁer using transistors. used anymore. Do you know the reason? Electron. Eng. 33, 84 (1961) 3. R.F. Shea, Principles of Transistor Circuits (Wiley, P:2. Same for the circuit of Fig. 27.4. 1953), p. 336 P:3. Same for the circuit of Fig. 27.5. 4. P.G. Sulzer, Low distortion transistor audio oscillator. P:4. Same for the circuit of Fig. 27.6. Electronics 26, 171 (1953) Analysing Sinusoidal Oscillator Circuits: A Different Approach 28 Conventionally, in analysing sinusoidal oscil- the ampliﬁer and the feedback networks load lator circuits, one uses the Berkhausen’s each other and the identiﬁcation of A and b poses criterion, viz. Ab = 1 in a positive feedback a problem. In this chapter, we propose a different ampliﬁer whose gain without feedback is approach which does not require such identiﬁ- A and whose feedback factor is b. However, cation. In fact, we do not use feedback concepts the identiﬁcation of A and b poses problems at all. Instead, we assume a voltage at an arbi- because of mutual loading of the ampliﬁer and trary node and come back to the same node the feedback network. A different approach is through the feedback loop. This results in the presented here which does not require such so-called characteristic equation of the oscillator. identiﬁcation. The method is based on assum- By putting s = jx and equating the real and ing a voltage at an arbitrary node and coming imaginary parts of the equation to zero, we get back to it through the feedback loop. the condition for, and the frequency of oscillation. Keywords Sinusoidal oscillator Different approach An Op-Amp Oscillator Introduction Consider the Wien bridge RC oscillator, shown in Fig. 28.1, using an op-amp as the gain ele- In most textbooks on analog electronic circuits, ment. Here, A refers to the gain between the sinusoidal oscillator circuits are analysed by nodes N1 and N2, and clearly, using the Berkhausen’s criterion, viz Ab = 1 in a A ¼ 1 þ ðR2 =R1 Þ ð28:1Þ positive feedback ampliﬁer, where A is the gain of the ampliﬁer without feedback and b is the which is independent of the feedback network feedback factor. However, except where the because the input impedance of the op-amp tends ampliﬁer is nearly ideal, e.g. in op-amp circuits, to inﬁnity and the output impedance tends to zero. Source: S. C. Dutta Roy, “Analyzing Sinusoidal Oscillator Circuits: A Different Approach,” IETE Journal of Education, vol. 45, pp. 9–12, January– March 2004. © Springer Nature Singapore Pte Ltd. 2018 213 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_28 214 28 Analysing Sinusoidal Oscillator Circuits: A Different Approach R2 +VCC R1 RC R3 R - R C R1 + N2 • C N1 R C • C R R2 R4 • • Fig. 28.1 An op-amp Wien bridge oscillator Fig. 28.2 Transistor Wien bridge oscillator Because of the same reason, the transfer function of the b network is, by inspection, gm1V1 gm2V2 b¼½R=ððsCRþ1ÞÞ=fRþ ½ð1=sC Þ rp1 C + + þR=½ðsCRþ1Þg¼sCR= s2 C2 R2 þ3sCRþ1 : V1 V2 R R C RC rp2 - - ð28:2Þ Putting the values of A and b from Eqs. 28.1 and 28.2 in Ab = 1, and simplifying, we get the Fig. 28.3 AC equivalent circuit of Fig. 28.2 characteristic equation as s2 C 2 R2 þ ½2ððR2 =R1 ÞsCR þ 1 ¼ 0: ð28:3Þ usual meanings. To analyse this circuit, we start at V1, and return to V1 through the feedback Now putting s = jx in Eq. 28.3 and equating loop. Note that its real and imaginary parts, we get the condition of oscillation as V2 ¼ gm1 V1 R0c ; ð28:6Þ R2 ¼ 2R1 ð28:4Þ where and the frequency of oscillation as R0c ¼ Rc jjrp2 : ð28:7Þ x0 ¼ 1=ðRC Þ: ð28:5Þ The current generator gm2 V2 in parallel with R can be converted to a voltage source −gm2 V2R in series with R. Then, one can ﬁnd V1 as Transistor Version of the Wien Bridge Oscillator R0 =ð1 þ sCR0 Þ V1 ¼ gm2 V2 R ; R þ ½1=ðsC Þ þ ½R0 =ðð1 þ sCR0 Þ Now consider the transistorized version of the ð28:8Þ Wien bridge oscillator shown in Fig. 28.2. Assume that the shunting effects of R1, R2, R3 where and R4 are negligible that the coupling and bypass capacitances behave as short circuits and R0 ¼ Rjjrp1 : ð28:9Þ that the transistor internal capacitances behave as open circuits at the frequency of oscillation. Combining Eq. 28.8 with Eq. 28.6, cancelling Then, the AC equivalent circuit becomes that V1 from both sides and simplifying, we get the shown in Fig. 28.3, where the symbols have their following characteristic equation: Transistor Version of the Wien Bridge Oscillator 215 Cm L s2 C2 RR0 þ sCðR þ 2R0 gm1 gm2 RR0 R0c Þ þ 1 ¼ 0 ð28:10Þ + V1 rp Cp RC C C Putting s = jx in this equation and equating - g m V1 the real and imaginary parts, we get the condition of oscillation as Fig. 28.5 AC equivalent circuit of Fig. 28.4 gm1 gm2 RR0 R0c ¼ R þ 2R0 ð28:11Þ and the frequency of oscillation as Cm pﬃﬃﬃﬃﬃﬃﬃﬃ x0 ¼ 1= C RR0 : ð28:12Þ L + gmV1 RC C rp C V1 Cp - Another Example As another example, consider the transistor Colpitt’s oscillator shown in Fig. 28.4. At the Fig. 28.6 Redrawn form of Fig. 28.5 frequencies at which the Colpitt’s circuit is used, the transistor internal capacitances also may have to be considered and we shall do so. Again, we Z1 Z2 + assume the coupling and bypass capacitances to - behave as shorts and ignore the shunting effects g mV 1 Z 1 Z3 V1 of R1 and R2. Then, the ac equivalent circuit + becomes that shown in Fig. 28.5, which is - redrawn in Fig. 28.6 in a form more suitable for analysis. The current generator gmV1 and its shunting elements Rc and C can be replaced by a Fig. 28.7 Simpliﬁed version of Fig. 28.6 Thevenin equivalent, and the resulting circuit is shown in Fig. 28.7, where Z1 ¼ Rc =ðsRc C þ lÞ; Z2 ¼ sL= s2 LCl þ 1 and Z3 ¼ rp =½srp ðC þ Cp Þ þ 1: ð28:13Þ +VCC From Fig. 28.7, we get V1 ¼ gm V1 Z1 Z3 =ðZ1 þ Z2 þ Z3 Þ: ð28:14Þ RC R1 Cancelling V1 from both sides, combining with Eq. 28.13, and simplifying, one gets the • following characteristic equation: L s3 L Cm 2C þCp þC C þCp R2 • C C þs2 L Cm Gc þgp þgm þGc C þCp þgp C þs 2C þCp þLGC gp þ Gc þgp þgm ¼ 0: ð28:15Þ Fig. 28.4 Transistor Colpitt’s oscillator 216 28 Analysing Sinusoidal Oscillator Circuits: A Different Approach Putting s = jx in Eq. 28.15 and equating the real and imaginary parts on both sides give the real and imaginary parts, we get the frequency frequency of, as well as the condition for oscillation as given by oscillation. 2C þ Cp þ LGc gp x20 ¼ L Cl ðð2C þ Cp Þ þ CðC þ Cp Þ Problems G c þ gp þ gm ¼ ; L Cl ðGc þ gp þ gm Þ þ Gc ðC þ Cp Þ þ gp C P:1. What happens when series RC is inter- changed with parallel RC in Fig. 28.1? ð28:16Þ Derive the necessary equations, and justify where the second part gives the condition of your conclusions. oscillation. P:2. Suppose in Fig. 28.1, the series R is absent, what will happen? Oscillations? Justify your answer with necessary equations. Concluding Comments P:3. What happens in Fig. 28.2 if the capacitor marked ∞ is not inﬁnite? Again, justify Rather than undertaking the involved task of your answer with equations. identifying A and b in a sinusoidal oscillator P:4. In Fig. 28.3, if rp1 and rp2 are inﬁnitely circuit, we show that it is easier and less prone to large, what will happen? Justify. mistake to start at a convenient node voltage and P:5. If in Fig. 28.1, the two C’s are replaced by return to the same through the feedback two L’s and C is replaced by a single L, loop. The characteristic equation is thus what would happen? Justify you answer obtained; putting s = jx in it and equating the with necessary derivations. Triangular to Sine-Wave Converter 29 This chapter describes how a triangular wave zero. If Vi is a symmetrical triangular wave with a is converted into a sine wave by using a peak value of Vp = V4′, then the output shall be piecewise linear transfer characteristic. A de- an approximation to a sine wave with a peak tailed analysis of the basic circuit is given, and value of V4′ as shown in Fig. 29.3. If Vp exceeds its actual implementation in an available IC V4′, it is obvious that the resulting sine wave chip is briefly discussed. shall have a clipped top and bottom (Fig. 29.3). On the other hand, if Vp < V4′, then we get a reduced amplitude sine wave of poorer quality as Keyword compared to the case when Vp = V4′. Conversion of waves The basic electronic circuit utilized to achieve the transfer characteristics (Fig. 29.2) is shown in Fig. 29.4, where V1 < V2 < V3 < V4, these being reference voltages derived from the power supply Introduction through an appropriate resistive voltage divider network. The circuit is functionally symmetrical Given a symmetrical triangular wave as shown in about the centre line. The upper half of the circuit Fig. 29.1, is it possible to convert it into a sine realizes the characteristic for Vi > 0, while the wave by an electronic circuit? The answer turns lower half takes care of the part Vi < 0. Because out to be in the afﬁrmative. Such a converter is, of symmetry, it sufﬁces to consider only the part in fact, available as an analog IC chip, whose for Vi > 0. transfer or input–output characteristic consists of To keep life simple, assume that all the diodes nine symmetrical, piecewise linear segments, as are ideal, i.e. they act as short circuits. We shall shown in Fig. 29.2. The central segment has a see later that in the actual chip, this is approxi- slope of unity, while the slope of the succeeding mately ensured by a pnp–npn transistor combi- segments is in decreasing order as we go to the nation. Suppose 0 < Vi < V1; then none of the right or to the left. The last two segments, viz. diodes conduct and Vo = Vi. This is the situation those for Vi > V4′ and Vi < –V4′, have a slope of for the central part of the characteristic in Fig. 29.2. When Vi is increased such that V1 Vi < V2′, where V2′ is the input needed to make Vo = V2′, diode D1 conducts and the equivalent Source: S. C. Dutta Roy, “Triangular to Sine Wave Converter,” Students’ Journal of the IETE, vol. 31, pp. 90–94, April 1990. © Springer Nature Singapore Pte Ltd. 2018 217 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_29 218 29 Triangular to Sine-Wave Converter Fig. 29.1 A symmetrical Voltage triangular wave and the approximate sine wave that Approximate can be obtained by shaping it Vp sinewave Time -Vp Triangular wave Fig. 29.2 Transfer V0 characteristics of the triangular to sine-wave V4 converter V3 V2 V1 -V4¢ -V3¢ -V2¢ -V1 0 V1 V2¢ V3¢ V4¢ Vi -V1 -V2 -V3 -V4 V4 V1 V2 V3 V4 0 D1 D2 D3 D4 V4 R1 R2 R3 Vp < V 4¢ Vp = V 4¢ Vp > V 4¢ Fig. 29.3 Shape of the output waveform for various values of Vp in relation to V4′ Vi Vo Ri circuit is shown in Fig. 29.5a. To ﬁnd Vo, apply KCL at the node which gives R1 R2 R3 D¢1 D¢2 D¢3 D 4¢ ðVi Vo ÞGi ¼ ðVo V1 ÞG1 ; ð29:lÞ where Gx = l/Rx, x = i, 1, 2, 3. Solving for Vo -V1 -V2 -V3 -V4 gives Fig. 29.4 The basic circuit of the triangular to sine-wave Vo ¼ ðV1 G1 þ Vi Gi Þ=ðG1 þ Gi Þ ð29:2Þ converter Introduction 219 V1 V20 ¼ V2 ðl þ Ri =R1 ÞV1 ðRi =R1 Þ ð29:3Þ R1 V1 £ Vi < V2 Note that V2′ > V2, which is of course expected. Vi Vo When V20 Vi \V30 ; where V3′ is the input R1 needed to make Vo = V3, both of the diodes D1 (a) and D2 conduct and the equivalent circuit is V1 V2 shown in Fig. 29.5b. Again applying KCL and simplifying, we get R1 R2 V ¢2 £ Vi < V3 Vo ¼ ðVi Gi þ V1 G1 þ V2 G2 Þ=ðGi þ G1 þ G2 Þ ð29:4Þ V1 Vo R1 (b) This describes the third segment in Fig. 29.2, whose slope is Gi =ðGi þ G1 þ G2 Þ ¼ ðR1 jjR2 Þ= V1 V2 V3 ½ðR1 jjR3 Þ þ Ri : To determine V3′, put Vi = V3′ and Vo = V3 in Eq. 29.4 and solve for V3′. The R1 R2 R3 V ¢3 £ Vi < V4 result is V1 V0 V30 ¼ V3 ½1 þ ðRi =R1 Þ þ ðRi =R2 ÞV1 Ri =R1 V2 Ri =R2 R1 ð29:5Þ (c) V1 V2 V3 V4 When Vi is further increased so that V3′ Vi < V4′, where V4′ is the input needed to make Vo = V4′, diodes D1, D2 and D3 conduct and the R1 R2 R3 V ¢4 £ Vi equivalent circuit is shown in Fig. 29.5c. From this, one can solve for Vo as V1 V0 R1 Vo ¼ðVi G1 þ V1 G1 þ V2 G2 þ V3 G3 Þ= ð29:6Þ Fig. 29.5 Equivalent circuits for various input voltage ðGi þ G1 þ G2 þ G3 Þ ranges This characterizes the fourth segment of Fig. 29.2, which has a slope of This describes the second segment of the characteristic in Fig. 29.2. Whose slope is Gi =ðGi þ G1 þ G2 þ G3 Þ Gi =ðG1 þ Gi Þ ¼ R1 =ðR1 þ Ri Þ\1: To determine ¼ ðR1 jjR2 jjR3 Þ=½ðR1 jjR2 jjR3 Þ þ Ri : V2′, put Vo = V2 and Vi = V2′ in Eq. 29.2; this gives, on simpliﬁcation. By putting V1 = V4′ and Vo = V4 in Eq. 29.6, one can obtain V4′ as V40 ¼ V4 ½1 þ ðRi =R1 Þ þ ðR1 =R2 Þ þ ðR1 =R3 ÞV1 Ri =R1 V2 Ri =R2 V3 R1 =R3 ð29:7Þ 220 29 Triangular to Sine-Wave Converter Finally, when V1 V4′, all four diodes D1′, Fig. 29.6 The resistive +10 V D2, D3 and D4 conduct, the equivalent circuit is voltage divider network for generating the refer- shown in Fig. 29.5d and Vo settles at V4. This ence voltages 5.2 K corresponds to the last segment of the charac- teristic in Fig. 29.2. V4 For negative input voltages, a similar analysis 200 W can be performed with the part of the circuit below the centre line in Fig. 29.4 and it can be V3 shown that the characteristic shown in Fig. 29.2 in the third quadrant is realized thereby. 375 W In the Intersil 8038 chip implementation of the V2 circuit shown in Fig. 29.4, the resistance values used are Ri = l K, R1 = 10 K, R2 = 2.7 K and 330 W R3 = 0.8 K. The voltages V1, V2, V3 and V4 are derived from the +10 V, −10 V supplies through V1 the resistive network shown in Fig. 29.6. It is 1.6 K readily calculated that V4 = 2.469 V, V3 = 2.180 V, V2 = 1.637 V and V1 = 1.159 V. -V1 The implementation of the diode is done in a clever way such that (i) the Thevenin impedance 330 W of each reference voltage is transformed to an -V2 insigniﬁcant value, and (ii) the voltage drop across the conducting diode is virtually reduced 375 W to zero. The actual circuit for the R1, D1, V1 and R1, D1′, −V1 legs of Fig. 29.4 is shown in -V3 Fig. 29.7. The diode D1 is realized by the com- plementary pnp (Q2)–npn (Q1) emitter follower 200 W pair. If Q1 and Q2 are matched, then their -V4 base-emitter drops will be equal and opposite. Thus, the voltage at the emitter of Q2 will be 5.2 K V1–VBE, Q1−VBE, Q2 = V1. Also, because of the 33 K resistor in the emitter lead of Q1, it will –10 V present an impedance of 33 K multiplied by its beta (bQ1 *100) to the source V1. This impe- dance will therefore be of the order 3300 K 0.34, 0.33 and 0.32 X, respectively. The scheme and should not affect the potential divider shown shown in Fig. 29.4 is therefore realized to a high in Fig. 29.6 at all! On the other hand, the degree of accuracy. effective Thevenin impedance of the V1 source, Since all parameters are known, we can now viz. [(1.6 + 0.33 + 0.375 + 0.2 + 5.2)|| (0.33 + calculate the input voltages at the various break 0.375 + 0.2 + 5.2)] K ≅ 3.4 K will be trans- points shown in Fig. 29.2 from Eqs. 29.3, 29.5 formed to 3.4 K/Q1 at the emitter of Q1, and to and 29.7. These are, respectively, V2′ = 1.685 V, 3.4 (K/Q1)/Q2 at the emitter of Q2. Assuming V3′ = 2.483 V and V4′ = 3.269 V. The circuit can bQ1 = bQ2 = 100, the resulting impedance redu- therefore be made optimum use of if the trian- ces to 0.34 X only. Similarly, the impedances gular wave peak voltage is 3.269 V; then a sine presented to D2, D3 and D4 can be calculated as wave with a peak voltage of 2.469 V is obtained. Introduction 221 V+ Problems R1 P:1. Convert a sine waveform to a triangular Q1 V1 (10 K) waveform. Q2 P:2. Same, but a square waveform. 33 k V- P:3. Same, but the desired waveform is shown in R1 the ﬁgure below. Vi V0 (1 K) V+ amplitude 33 k Q¢2 R1 Time Q¢1 -V1 (10 K) Fig P.3. V- P:4. Suppose the diodes in Fig. 29.4 are all nonideal, but identical. What will happen? Fig. 29.7 Realization of diodes D1 and D1′ P:5. Given a square wave going positive as well as negative. How would you generate a chain of positive impulses? Why should one bother about generating a sine wave by conversion of a triangular wave? Instead, why should not one use an LC or RC sinusoidal oscillator? The reason is that it is very Bibliography difﬁcult to obtain a wide range variable fre- quency sinusoidal oscillator. In contrast, one can 1. S. Soclof, Applications of Analog Integrated Circuits. easily generate a 100: 1 frequency sweep with a (Prentice Hall, 1985) voltage-controlled triangular wave oscillator. The 2. A.B. Grebene, Analog Integrated Circuit Design. resulting wave can then be shaped to a sine wave (Van Nostrand Reinhold, 1972) by a triangular to sine-wave converter as described in this chapter. Dynamic Output Resistance of the Wilson Current Mirror 30 A simple derivation is given for the dynamic also do not prove this result. Soclof [5] attempted output resistance of the Wilson current mirror, a simple proof but his result is higher by a factor which forms a basic building block in many of 2 due to a mistake in the assumed current analog integrated circuits. distributions. In view of the importance of the Wilson current mirror and in view of the fact that Soclof’s books [5, 6] are the most comprehensive Keywords texts available on the subject, we present here a Current mirror Wilson circuit Dynamic simple and correct analysis leading to the result output resistance claimed by Wilson [1] and others. Introduction Derivation The Wilson current mirror, shown in Fig. 30.1b, We adopt here the same approach as that of Soclof is a basic building block in many analog inte- [5] and represent Q3 by an ideal transistor Q03 in grated circuits. As compared to the simple cur- parallel with its dynamic collector-to-emitter rent mirror shown in Fig. 30.1a, it has the conductance g0 ¼ 1=r0 , as shown in Fig. 30.2. advantage of achieving base current cancellation, Let the output voltage change by a small amount so that I0 = I1, even if the base currents of the ΔV0 and let the consequent change in the output transistors (all assumed identical) are not negli- current be ΔI0. If ΔI0 can be determined in terms gible as compared to their respective collector of ΔV0 and transistor parameters, then the currents. Further, its dynamic output resistance is dynamic output conductance can be calculated as greater than that of the simple current mirror by a factor of b/2. This has been mentioned by Wilson g00 ¼ 1=r00 ¼ D I0 =DV0 : [1] but not proved. Grebene [2] follows Wilson, but refers to a hybrid-p equivalent circuit anal- Due to the incremental change ΔV0, let the ysis made by Davidse [3]. Gray and Meyer [4] collector current of Q2 change by ΔI; since Q1 and Q2 are matched and have the same base-to-emitter voltages, the collector current of Q1 will also change by the same amount ΔI. Assuming that the current I1 remains a constant, i.e. DI1 ¼ 0, KCL Source: S. C. Dutta Roy, “Dynamic Output Resistance of the Wilson Current Mirror,” Students’ Journal of the dictates that the base current of Q3 must change IETE, vol. 31(4) 1990 and 32(1), pp. 165–168, 1991 © Springer Nature Singapore Pte Ltd. 2018 223 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_30 224 30 Dynamic Output Resistance of the Wilson Current Mirror Fig. 30.1 a A simple current (a) (b) mirror b Wilson current V+ mirror I1 I0 I1 I0 Q3 R1 Q1 Q2 Q1 Q2 v v+ DI0 incremental base currents of Q1 as well as Q2, each of which is ΔI/b. Thus DI1 = 0 bDI DI 0 ¼ D I þ 2 DI=b ð30:2Þ g0 g0DV0 Q3 DI Combining Eqs. 30.1 and 30.2, we get DI DV0 D I ¼ D I0 =½2ð1 þ 1=bÞ ð30:3Þ DI DI 2DI/b Q3 Q1 Q2 Now applying KCL at the collector of Q03 , we DI/b DI/b have D I0 ¼ g0 DV0 bD I ð30:4Þ Fig. 30.2 Incremental equivalent circuit of the Wilson Combining Eqs. 30.3 and 30.4 and simplify- current mirror ing, we get D I0 =DV0 ¼ g00 ¼ g0 =f1 þ b=½2ð1 þ 1=bÞg by −ΔI. This causes Q03 collector current to ð30:5Þ change by −b ΔI. The increment of current through g0 will be g0 ΔV0, because the Assuming b 2, as is the case in practice, diode-connected transistor Q2 offers a negligible this reduces to dynamic resistance compared to that of Q3 (= r0). Now consider the dotted rectangular box in g00 ¼ 2g0 =b ð30:6Þ Fig. 30.2 (representing Q3); it has two currents entering, viz. −Δ I and Δ I0, and one current so that the equivalent dynamic output resistance leaving, viz, ΔI′, so that by KCL, becomes DI0 ¼ D I ¼ D I 0 ð30:1Þ r00 ¼ br0 =2 ð30:7Þ Assuming all transistors to have the same b, This agrees with the claim of Wilson [1] and we see that DI 0 supplies ΔI, and also the others. Derivation 225 It may be mentioned here that Soclof’s P:5. What happens when g0 end is shifted from equivalent circuit assumed DI ¼ DI0 , which, as emitter to base? is obvious, violates KCL and makes DI 0 ¼ 0! Problems References P:1. Search out the literature for other current 1. G.R. Wilson, A monolithic junction FET-NPN oper- mirrors. Make a list of them and enumerate ational ampliﬁer. IEEE J. Solid State Circ.SC-3, 341– 348 (December 1968) their merits and demerits as compared to the 2. A.B. Grebene, Bipolar and MOS Analog Integrated Wilson current mirror. Hint: Consult the Circuit Design. (John Wiley, 1984) references given above, and if you cannot 3. J. Davidse, Integration of Analogue Electronic Cir- ﬁnd them, ask me. cuits. (Academic Press, 1979) 4. P.R. Gray, R.G. Meyer, Analysis and Design of P:2. Justify, with derivations, what happens Analog Integrated Circuits. (John Wiley, 1984) when Q1 and Q2 emitters are connected to 5. S. Soclof, Analog Integrated Circuits. Prentice (Hall, current generators. 1985) P:3. Why is DI1 6¼ 0 in Fig. 30.2. What happens 6. S. Soclof, Applications of Analog Integrated Circuit. (Prentice Hall, 1985) when DI1 6¼ 0? P:4. Why is g0 connected from the collector to emitter in Fig. 30.2? Part IV Digital Signal Processing The ﬁeld of Digital Signal Processing (DSP) has fascinated me for over three decades, since I ﬁrst met it in the early 1970s. This is reflected in the eight chapters of Part IV. The ﬁrst two articles, on which Chaps. 31 and 32 are based, were written while I was teaching at the University of Leeds, during 1972–1973. DSP was at its infant state at that time, and students had difﬁculty in understanding the basic concepts. That is why, I innovated the title as ‘The ABCD’s of Digital Signal Processing’. In these articles, I described DSP from common-sense arguments. These became popular with the students I taught, almost instantly. I hope beginning students of DSP will like them, even now. The article on second-order digital ﬁlters, described in Chap. 33, was again inspired by inquisitive questions from students in the class, and I took pains to give simple derivations of band-pass and band-stop ﬁlters, touching on the limits of selectivity attainable by them. Chapters 34 through 37 were also inspired by student’s queries in the class. Chapter 34 topic was in fact solved by four M.Tech. students of IIT Delhi, with clues from me, and it was satisfying to note that they could come up with simple derivations of this important element of DSP, viz. all pass ﬁlters. I have always encouraged my teacher colleagues at IIT Delhi and elsewhere to involve students in their research by throwing challenges on unsolved problems to students in the class. This article was a result of such a challenge. The FIR lattice structure is usually ﬁrst described and then analysed to ﬁnd the performance parameters. I took upon myself the task of viewing this as a synthesis problem and succeeded. This forms the content of Chap. 35. A special problem arose during the course of this development, and I solved it in the article on which Chap. 36 is based. FIR lattice, as described in the textbooks, uses twice the minimum number of multipliers, as required in a canonic realization. Long back, Johnson had answered the question with afﬁrmative, but surprisingly, his paper, although published in IEEE Trans- 228 Part IV: Digital Signal Processing actions on Audio and Electroacoustics, went completely ignored by later workers, even the famous ones. I took upon myself the job of giving due credit to this unsung hero in the article on which Chap. 36 is based. In FFT signal flow graphs, there are some redundant multipliers. Also, there are identical multipliers which can be combined. Taking all these into account, the minimum possible number of multipliers is found out in Chap. 38. That completes Part IV and the main contents of the book. The ABCDs of Digital Signal Processing––PART 1 31 In this chapter, the basic concepts of digital Introduction signal processing will be introduced, leading to a mathematical description of a digital The ﬁrst textbook on digital signal processing, by signal processor in terms of, ﬁrst, a difference Rader and Gold (1969), came out in 1969, and equation and, second, a z-domain transfer included the following sentence in its preface: function. In the process, the effects of sam- ‘The ﬁeld of digital signal processing is too new pling and quantization will be briefly touched to allow us to predict subsequent developments’. upon. Implementation of a processor by Today, more than four decades later, one cannot special purpose hardware and discrete Fourier certainly claim the ﬁeld to be new, particularly in transform technique will be discussed. The view of the phenomenal progress made in the fast Fourier transform (FFT) will be intro- techniques of digital signal processing, leading to duced and several of its applications will be dramatic improvements in system efﬁciency, and presented, along with the pitfalls and incorrect its many applications in very diverse ﬁelds like usage of the technique. biomedical engineering, geophysics, acoustics, radar and sonar, radio astronomy, etc. These advances, in turn, have been stimulated by fan- Keyword Fundamentals of DSP Difference equation tastic advances in integrated circuit technology Z-transform Sampling Quantization and computer hardware, in terms of volume, cost DFT FFT and its pitfalls and speed. One way of gauging the progress of a ﬁeld is to look at the available literature related to it. By as early as the 1970s, there were more than ten textbooks [1–10], four collections of signiﬁcant papers [11–14], a large number of special issues of professional research journals [15] and numerous journal conference articles reporting on new techniques, or improvements on known ones, or novel applications of digital signal pro- Source: S. C. Dutta Roy, “The ABCD’s of Digital Signal Processing (Part 1),” Students’ Journal of the IETE, vol. cessing [16]. 21, pp. 3–12, January 1980. On a subject as vast as the literature explosion suggests, it is not easy to decide as to what © Springer Nature Singapore Pte Ltd. 2018 229 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_31 230 31 The ABCDs of Digital Signal Processing––PART 1 x*(nT) x (nT) y(nT) x(t) LPF1 A/D Processor y(T) Extra- y*(nT) y(t) LPF2 polator D/A Fig. 31.1 Block diagram of a digital signal processor whose input and output are continuous signals should be included in the ‘ABCD’s’ because as with. Let us then consider a continuous signal x the subject advances, so do the ‘ABCD’s’. The (t), which is to be processed so as to facilitate the choice and organization of topics in this chapter extraction of a desired information which, we have been greatly influenced by discussions with will assume, is again a continuous signal. In some members of potential readers and, of other words, we wish to transform the given course, by personal preferences. Starting with the signal x(t) into another signal y(t) which is, in basic concepts involved in sampling and quan- some sense, more desirable than the original. For tization, a mathematical description will be given example, x(t) may, typically, be the desired sig- of the digital signal processor in terms of a dif- nal contaminated by an undesired interference ference equation and then in terms of a z-domain and our aim in processing may be to get rid of the transfer function. Implementation of the proces- latter. As another example, we may wish to sor by special purpose hardware as well as by the enhance or estimate some component or param- discrete Fourier transform (DFT) technique will eter of a signal. Whatever our aim may be for the be discussed. In the latter, two basic forms of the processing, if the processor is to be digital, x(t) fast Fourier transform (FFT) algorithm will be must be converted ﬁrst to a discrete form x*(nT). introduced. The presentation will conclude with A conceptually simple way of doing this is to some applications of the FFT, along with its have a switch in series which closes for a very pitfalls and potential incorrect usage. short time s after every T seconds, where s T, Throughout the presentation, efforts will be as shown in Fig. 31.1. Ignore the box marked made to keep the mathematics as simple as LPF, for the time being; also do not worry as to possible, and rigorous proofs and derivations will what should the value of T be. The resulting be avoided as much as practicable. System signal x*(nT) would appear as a series of narrow design aspects, starting from a given speciﬁca- pulses at times t = 0, T, 2T, …, the height of each tion, will not be dealt with at all. This topic, pulse being equal to the value of the continuous along with others, will form the subject matter of signal at that instant of time. This sampled signal a future chapter. x*(nT) is then suitably coded or quantized in the analog-to-digital converter (A/D). The output of the A/D converter, x(nT), is a series of coded The Basic Digital Signal Processor numbers, coming out every T seconds (ignoring the small but non-zero A/D conversion time, as While digital signals and systems can be compared to T). The signal x(nT) is now in a designed without reference to continuous sys- form which can be processed by digital hard- tems, it is intuitively appealing, and often easier ware, indicated by the box labelled PRO- to understand, to build the theory of digital signal CESSOR in Fig. 31.1. The processor thus processing starting from continuous signals and accepts a time series, i.e. series of numbers systems which most engineers are more familiar appearing at equal intervals of T seconds at its The Basic Digital Signal Processor 231 input, and performs some operations on them to ﬁltering and the digital signal processor is then produce another time series y(nT). These opera- more commonly known as a digital ﬁlter. tions, in the most general linear digital signal The output of the processor in Fig. 31.1, y processor, can be described by the following (nT), is another sequence of numbers, which can linear1 difference equation: be fed to a decoder, the digital-to-analog con- verter, to produce pulses of short duration, whose X M X N amplitudes are in proportion with the value of y yðnTÞ ¼ ai xðnT iTÞ þ bj yðnT jTÞ; (nT). Continuous output can be obtained by i¼0 j¼1 passing this pulse sequence through an extrapo- ð31:1Þ lator or data reconstructor, which can be a simple zero-order hold, described by where ai’s and bj’s are suitable constants. x(nT) and y(nT) are, respectively, the input and output y1 ðnT þ tÞ ¼ y1 ðnT Þ; 0 t T; ð31:2Þ of the processor at the instant nT, while x(nT – iT) and y(nT – jT) represent, respectively, the input at i.e. the value between two sampling instants is some past instant nT – iT and the output at another held at the value of the immediately preceding past instant nT – jT. At the instant nT, therefore, sample.2 The output of the extrapolator, y1(t), the computation of the output y(nT) requires the will contain some high-frequency ripples, which past M inputs and N outputs; the processor may be removed, if desired, by passing the signal therefore should have a memory in which these through a low-pass ﬁlter, LPF2. This ﬁlter is, past input and output numbers can be stored. The however, a simple, inexpensive one and typi- constants ai and bj are also to be stored, of course. cally, a single R–C section serves the purpose. What the processor does after receiving the pre- Before we conclude this section, a discussion of sent input x(nT), in fact, is then to recall the the block LPF1, is in order. The choice of T or the constants, past inputs and past outputs and per- sampling frequency fs = 1/T should be in confor- form the computation speciﬁed in Eq. 31.1, to mity with the sampling theorem, i.e. fs should be deliver the output y(nT). The computation does greater than 2fh, where fh is the highest frequency require some small but non-zero time, but this content of the signal x(t) to be processed. If this is must be smaller than T in order that the processor not the case, distortion occurs and x*(nT) will not be may be ready to receive the next input with a a true discrete representation of x(t); another way of clean state. We shall, in the following discus- saying this is that it would not be possible to recover sions, ignore this computation time and continue x(t) from x*(nT). The sampling frequency should to call the output due to x(nT) as y(nT). therefore be sufﬁciently high. However, for other Before we proceed further, it is wise to recall reasons (e.g. coefﬁcient quantization error, to be the following sources of nonidealness in our discussed later), the sampling frequency should not signal processing system in Fig. 31.1. (i) s, the be too high. For a signal contaminated with duration for which the sampling switch remains high-frequency noise (impulse noise), the mini- closed, (ii) A/D conversion time, and (iii) pro- mum required sampling frequency may be inordi- cessor computation time. The ﬁrst two ideally nately high. In such cases, it would be advantageous should be zero, and the third should be at least to ﬁlter the continuous signal x(t), before sampling, less than T, if not signiﬁcantly so. by passing it through an inexpensive analog By choosing the ai’s, bj’s, M and N in low-pass ﬁlter, LPF1 as shown in Fig. 31.1. LPF1 Eq. 31.1, one can achieve a variety of process- may also typically be a single R–C section ing. If the emphasis is on shaping the spectrum of x(t) in a desired fashion, we call the processing as 2 It can be shown that the zero-order hold has a lowpass frequency response with linear phase characteristic. Higher-order holds are not generally used because their 1 We shall conﬁne our attention to linear digital signal implementation is difﬁcult, their phase response is not processing only. linear and they introduce more delay to the signal. 232 31 The ABCDs of Digital Signal Processing––PART 1 The Sampling Process 1 X 1 SðtÞ ¼ ejkxst ; ð31:4Þ T k¼1 As already mentioned, a sampler may be viewed as a switch which closes for a short duration s, where xs ¼ 2p=T ¼ 2pfs . Thus, after every T seconds, where s T, as shown in Fig. 31.2a. Implicit, of course, is also the 1 X 1 assumption that the sampler can open and close x ðnTÞ ¼ xðtÞejkxs t ð31:5Þ T k¼1 instantaneously. A typical input x(t) and output x* (nT) of the sampler are shown in Fig. 31.2b, c, respectively. A considerable simplicity in If one now takes the Fourier transform of both understanding and analysis is achieved if s is sides and uses the notations X ðxÞ and XðxÞ to assumed to be zero, i.e. if the sampler is assumed represent the spectra of x(t) and x*(nT) respec- to be an ideal impulse sampler whose output is a tively, then one obtains. series of impulses appearing at the sampling 1 X X ðxÞ ¼ Xðx þ kxs Þ ð31:6Þ instants, as shown in Fig. 31.2d, the strength of T k¼1 the impulse at t = nT being equal to x(t)|t = nT. Then one can write Thus the spectrum of the sampled signal is obtained by superimposing, on the original signal X 1 x ðnTÞ ¼ xðtÞ dðt nTÞD xðtÞSðtÞ; say spectrum, the same spectrum shifted on the fre- n¼1 quency scale by ks, where k takes all integer ð31:3Þ values, positive and negative, except zero. This is pictorially shown in Fig. 31.3 for a hypothetical What Eq. 31.3 says in essence is that the X ðxÞ which is bandlimited such that XðxÞ ¼ 0 sampled sequence x*(nT) is obtained by multi- for jxj [ xs =2. Note that X ðxÞ is the same as plying the continuous signal x(t) by an impulse XðxÞ in the band jxj [ xs =2, called the base train occurring at 2T, – T, 0, T, 2T,…; we have band, and it is possible to recover XðxÞ from named this impulse train as another function S(t). X ðxÞ by using a low-pass ﬁlter whose trans- Obviously, S(t) is periodic with period T and can mission characteristic is shown by dotted lines. be expanded in Fourier series. When this is done, An example of a situation where XðxÞ is not one ﬁnds bandlimited and has considerable amplitude Fig. 31.2 The sampling x*(n ) process a sampler, b a (a) (c) continuous signal, c its sampled version and d idealized sampled version t x(t) T t T (b) x(t) (d) x*(nT)|t = 0 t t The Sampling Process 233 xw wh w -ws -wh O ws 2 2 x* w LPF RESPONSE w -3ws -ws –ws -ws O ws ws ws 3ws 2 2 2 2 Fig. 31.3 Showing the spectrum of the sampled signal in relation to that of the continuous signal beyond xs/2, called the Nyquist frequency, is frequency, xh, contained in the signal. This, in shown in Fig. 31.4. An alternative equivalent essence, is the well-known sampling theorem. description of this situation is that the Nyquist Practical signal are not, however, bandlimited frequency is lower than the highest frequency xh and the signal spectrum recovered by passing contained in XðxÞ. In this case obviously, XðxÞ X ðxÞ through an ideal low-pass ﬁlter will be does not keep its identity in X ðxÞ and no ﬁlter different from XðxÞ due to spillover from the can recover XðxÞ from X ðxÞ. Thus, to keep the adjacent bands. The error so introduced is called information content of the signal intact in the aliasing or folding error. Further, an ideal ﬁlter, sampled version, we must choose the sampling with the brickwall characteristic, is not realizable frequency xs to be at least twice the highest in practice and this introduces an additional error. To keep these two errors within the tolerable limits, the sampling frequency is often required to be sufﬁciently high. xw Another source of error is the fact that s is not zero in practice; the nature and extent of its contribution to distortion has been discussed in Shapiro (1978) and will not be considered here. w -ws O ws Quantization Errors3 2 2 -wh wh In this section, we would like to point out an x* w inherent limitation on the accuracy of digital signal processors. This limitation arises due to -wh -ws O ws wh w 2 2 3 This example is reprinted by kind permission of John Fig. 31.4 Aliasing error Wiley & Sons Inc. 234 31 The ABCDs of Digital Signal Processing––PART 1 the fact that all digital systems operate with a To begin with, the numbers we are dealing ﬁnite number of bits, or a ﬁnite word length. with must be represented in binary notation in Rather than going into a detailed theory, we order to be stored, manipulated and operated prefer to illustrate the various errors, resulting upon by digital hardware. Consider the coefﬁ- from ﬁnite word length, through a simple cient 0.81, it can be written as example (Peled and Liu [7]). Suppose our pro- 1 1 1 1 cessor is required to implement the following 0:81 ¼ ð Þ1 þ ð Þ2 þ ð Þ5 þ ð Þ6 þ 2 2 2 2 difference equation. i.e. in base 2, 0.81 can be represented as yðnÞ ¼xðnÞxðn2Þ 0.11001110101 … An inﬁnite number of bits are þ 1:2727922yðn1Þ0:81yðn2Þ; needed to represent this coefﬁcient exactly. Since all practical memory circuits have a ﬁnite number of bits for each word, the inﬁnite binary string yðnÞ ¼xðnÞxðn2Þ must be modiﬁed. If one uses a memory with a þ 1:2727922yðn1Þ0:81yðn2Þ; 6-bit word length, a simple way to store our ð31:7Þ number will be to keep only the 6 most signiﬁ- cant bits, that is, 0.11001 as the approximate where for brevity, x(nT) and y(nT) have been value for 0.81. However, 0.11001 in base 2 represented as x(n) and y(n), respectively. A pos- represents the number 0.78125, thus introducing sible basic arrangement is shown in Fig. 31.5. an error of 0.02875 in this coefﬁcient. Similarly, It consists of a memory for storing the coef- 1.2727922 has a 6-bit base 2 representation of ﬁcients; a set of data registers for storing the 1.01000 or l.25, resulting in an error of input and output samples; an arithmetic unit to 0.0227922. Obviously, ±1 or 0 can be repre- perform the computation according to Eq. 31.7 sented exactly; ﬁnally, therefore, the equation and a control unit (not shown) for providing the that the processor actually implements is timing signals. yðnÞ ¼ xðnÞxðn2Þ þ 1:25yðn1Þ0:78125yðn2Þ ð31:8Þ MEMORY FOR CO-EFFICIENTS The resulting error is called the coefﬁcient 0.81 quantization error. 1.2727922 Another source of error is the quantization of the input data in the A/D converter. Suppose x(t) in Fig. 31.1 is sinusoidal and consider the following DATA REGISTERS input segment… 0.2955, 0.5564, 0.8912, 0.9320… Suppose the A/D converter yields 8 bits x(n) and let the data registers in Fig. 31.5 be of the same capacity. Truncated to 8 bits, the above input data x(n 2) segment becomes, in binary form, ARITHMETIC x(n) UNIT 0:0100101; 0:1000111; 0:111010; 0:111011 y(n 1) corresponding to the values… 0.2890625, y(n 2) 0.5546875, 0.890625, 0.9265,… which, obvi- ously, differ from the actual samples of the sinusoidal signal. The resulting error is called the Fig. 31.5 Implementation of Eq. 31.7. Control unit is input quantization error. not shown Quantization Errors 235 The third source of error is due to the limited Also, to keep things simple, let us consider a accuracy with which arithmetic operation can be causal signal, i.e. let x(t) = 0, t < 0. Further, performed. In computing the term—0.78125y since a delta function exists only when its argu- (n − 2) in Eq. 31.8, for example, the product of a ment is zero, we could rewrite Eq. 31.9 in the 6-bit number (−0.78125) and an 8-bit number form [y (n − 2)] will give 14 signiﬁcant bits. This must be shortened to 8 bits so that the result will ﬁt in X 1 X ðnTÞ ¼ xðnTÞdðt nTÞ ð31:10Þ the 8-bit data register. The error thus committed n¼0 is known as the round-off error. Further and more importantly, the previously computed out- If we take Laplace transform (LT) on both put samples are used via Eq. 31.8 to compute sides and call the LT of x*(nT) as x*(s), we get later output samples, and this has a cumulative effect. X 1 X ðsÞ ¼ xðnTÞesnT ð31:11Þ What are the overall effects of these errors? n¼0 Unless carefully analysed and accounted for, the results can be very disappointing. For example, To get rid of the transcendental function est, coefﬁcient quantization error may convert a let us replace it by z, i.e. let stable processor into an unstable one. The arith- metic round-off errors can result in low-level z ¼ esT ð31:12Þ limit cycles and overflow oscillations. Further, let X ðsÞ , X ðzÞ; then Z-Transform X 1 XðzÞ ¼ X ðsÞD xðnTÞzn ð31:13Þ n¼0 Recall that the input–output relation of a digital signal processor is expressed by a linear differ- The variable z need not necessarily be thought ence equation of the form of Eq. 31.1. It is well of as esT; it could be interpreted as an ordinary known that the solution of such an equation is variable whose exponent (ignoring the negative greatly simpliﬁed by using the z-transform (just sign) represents the position of the particular as the solution of linear differential equations, pulse in the sequence {x(nT)}. When viewed in which can be used to characterize linear contin- the latter light, X(z) is a ‘generating function’ and uous systems, is greatly simpliﬁed by using may be treated without identiﬁcation with a Laplace transforms). Further, a better under- Laplace transform. standing of the digital signal processor, particu- The inﬁnite summation Eq. 31.13 deﬁnes the larly its frequency domain or spectral behaviour, z-transform of the sequence {x(nT)} or more is obtained from its z-transformed description. concisely {x(n)}. (Note the use of {.} to represent Consider the sampled signal x*(nT) described a sequence and the dropping of T for brevity). in Eq. 31.3, reproduced below in a slightly dif- Thus, formally, ferent, but equivalent form X 1 X 1 ZfxðnÞg ¼ XðzÞD xðnÞzn ð31:14Þ X ðnTÞ ¼ xðtÞdðt nTÞ ð31:9Þ n¼0 n¼1 236 31 The ABCDs of Digital Signal Processing––PART 1 We shall not go into the details of existence, In the discrete case, by analogy with Eq. (19), convergence and other mathematical properties we deﬁne convolution of two sequences {x1(n)} of the z-transform here, but it is better to and {x2(n)} as another sequence {x(n)} such that remember that the series Eq. 31.14 converges outside a circle in the z-plane whose radius X n xðnÞ ¼ x1 ðrÞx2 ðn rÞ equals the n-th root of maximum x(n) in {x (n)}. r¼0 Given X(z), one can recover {x(n)}, in gen- ð31:20Þ Xn eral, by applying the inversion integral and ¼ x1 ðn rÞx2 ðrÞ Cauchy’s residue theorem; however, for rational r¼0 X (z), as is usually the case, a long division is adequate. As an example, if The discrete convolution can be given a graphical interpretation, analogous to continuous z 1 convolution, but we would not discuss this here. XðzÞ ¼ ¼ ð31:15Þ Instead, we now state the third important prop- z k 1 kz1 erty of z-transforms, viz. that if Eq. 31.20 is true, then then so is, X ðzÞ ¼ 3 þ kzl þ k2 z2 þ ð31:16Þ X ðzÞ ¼ X1 ðzÞX2 ðzÞ ð31:21Þ and obviously x(n) = kn. This result is exactly analogous of the Laplace Three important properties of z-transforms transform of the convolution of two continuous will now be stated without proof. First, the functions. The proofs of Eqs. 31.17, 31.18 and z-transform is a linear operation, i.e. if 31.21 follow easily from the deﬁnition of Z fxi ðnÞg ¼ Xi ðzÞ; then z-transform given in Eq. 31.14. " # X P X P Z fai xi ðnÞg ¼ ai xi ðzÞ ð31:17Þ i¼1 i¼1 Transfer Function of a Digital Signal Processor The second property concerns the z-transform of a shifted sequence, viz. Consider the digital signal processor described in Eq. 31.1 once again and let Z[{x(n)}; {y Z fxðnmÞg ¼ zm X ðzÞ ð31:18Þ (n)}] = X(z); Y(z). If one takes the z-transform of both sides and uses the shifting property of The third concerns the z-transform of a con- z-transforms mentioned in the preceding section, volution of two sequences. Before we state this, it is easy to see that however, let us understand what we mean by the convolution of two sequences. In the continuous X M X N domain, convolution x(t) of two functions YðzÞ ¼ ai XðzÞzi þ bj YðzÞzj ð31:22Þ x1(t) and x2(t) is deﬁned by i¼0 j¼0 xðtÞ ¼ x1 ðtÞ x2 ðtÞ This can be put in the form Zt P M ¼ x1 ðtÞx2 ðt TÞdT ai z1 YðzÞ ð31:19Þ HðzÞ ¼ ¼ i¼0 ð31:23Þ 0 XðzÞ P N Zt 1 bj z1 ¼ x1 ðt TÞx2 ðTÞdT j¼0 0 The quantity H(z), deﬁned ‘as the ratio of z-transform of output sequence to the z-transform Transfer Function of a Digital Signal Processor 237 of the input sequence’, obviously is a character- X M istic of the processor only and is an adequate yðnÞ ¼ hðrÞxðn rÞ ð31:28Þ representation of it. It is, by analogy with con- r¼0 tinuous systems, called the z-domain transfer Note that this is of the same form as Eq. 13.1 function or simply the transfer of the digital with bj’s equal to zero, and ar = h (r). Under this signal processor. Note that condition, the transfer function H (z) given in Y ðzÞ ¼ H ðzÞ when X ðzÞ ¼ 1 ð31:24Þ Eq. 31.23 becomes a polynomial in z−1. A pro- cessor which is not FIR is of the Inﬁnite Impulse What does X(z) = 1 signify? We can write Response (IIR) type. For this, at least one bj in Eq. 31.1 is non-zero and H (z) in Eq. 31.23 is a X ðzÞ ¼ 1 þ 0:zl þ 0:z2 ð31:25Þ rational function in z−1. There are two other terms which are very If one compares Eq. 31.25 with Eq. 31.14, it commonly used in digital signal processing ter- is obvious that XðzÞ ¼ 1 corresponds to an input minology; these are non-recursive and recursive. sequence. A very common mistake that has been perpetu- ated in the literature is to identify FIR with x ð nÞ ¼ 1 for n ¼ 0 ¼ 0 otherwise ð31:26Þ non-recursive and IIR with recursive. As pointed out by Gold and Jordan [17], the terms recursive i:e:xðnÞ ¼ fl; 0; 0; . . .g. This is called the unit and non-recursive should be used only to describe pulse and we see that it plays the same role as the the method of realization. A realization in which impulse function dðtÞ in a continuous system. no past values of the output have to be called back The inverse z-transform of H(z), denoted by to compute the present output is called fhðnÞg, is obviously the output sequence of the non-recursive; if one or more past values of the processor when the input is a unit pulse. fhðnÞg output are required for computing the present is called the impulse response of the digital sig- value of the output, the realization is called nal processor. recursive. Obviously, FIR processors are realized From Eq. 31.23 and the z-transform property most conveniently in non-recursive form, while for convolution, it should also be apparent that recursive form is to be preferred for IIR proces- for a general input fxðnÞg, the output sequence sors. But, as shown by Gold and Jordan (loc. cit.), fyðnÞg should be given by the convolution of the FIR processors can be realized recursively, and input sequence with the impulse response, i.e. IIR processors can be realized non-recursively. The transfer function Eq. 31.23 can be X N yðnÞ ¼ xðrÞhðn rÞ ð31:27Þ expressed as r¼0 ðz1 a1 Þðz1 a2 Þ. . .ðz1 aM Þ HðzÞ ¼ A ; X n ðz1 b1 Þðz1 b2 Þ. . .ðz1 bN Þ ¼ xðn rÞhðrÞ ð31:29Þ r¼0 Suppose we have a digital signal processor, where A is a real constant, and a0 s and b0 s are characterized by an impulse response fhðnÞg, either real or complex if they are complex they where h (n) = 0, n > M. Such a processor is said occur in conjugate pairs, a0 s are called zeros and to be of the Finite Impulse Response (FIR) type. b0 s are called poles of the digital signal processor For this, Eq. 31.27 becomes in the z−1 plane. A digital signal processor is 238 31 The ABCDs of Digital Signal Processing––PART 1 (a) j Im Z (b) (c) H (e jwT ) Avg H (e jwT ) wT ws Re Z 2 ws w O |Z| = 1 ws ws 2 Fig. 31.6 Pole-zero sketch of Eqs. 31.30 or 31.31, b Magnitude response, c Phase response (To be continued) stable if the poles in the z−1 plane all lie outside magnitude response is symmetrical while the the unit circle, which is equivalent to having all phase response is antisymmetrical around the poles inside the unit circle with z-plane; this Nyquist frequency xs ; =2 and that both responses comes from the correspondence of the jx-axis in repeat after every xs radians. the s-plane to the unit circle in the z−1 plane, see Eq. 31.12. The frequency response of the digital signal processor is obtained by putting z ¼ ejx T Problems in H (z). As an example, let a digital signal processor be described by the difference equation P:1. What happens to the spectrum if the impulses in Fig. 31.2d are replaced by thin yðnÞ ¼ xðnÞxðnlÞ0:8yðn2Þ ð31:30Þ rectangular pulses? P:2. If the base spectrum in Fig. 31.3 is a full The transfer function of the system is sinusoid form—xh and xh [ x2s ; what will happen to the sampled spectrum? 1 z1 P:3. Why are all powers of z in z-transform HðzÞ ¼ negative? What is the meaning of positive 1 þ 0:81z2 1 z1 powers? Comment on their realizability in ¼ ð31:31Þ real time and virtual time. ð1 þ j0:9z1 Þð1 j0:9z1 Þ P:4. Can a z-transform with negative powers of zðz 1Þ z have a numerator of degree higher than ¼ ; ðz þ j0:9Þðz j0:9Þ that of the denominator? What will be its inverse transform? where the last form has been used to facilitate a P:5. Can you realize a difference equation with pole-zero sketch in the z-plane, as shown in term like x(n + 1), x(n + 2) …? Fig. 31.6a. Putting z ¼ ejxt in Eq. 31.31 and evaluating the amplitude and phase, one can obtain the plots shown in Figs. 31.6b, c. It can References also be done graphically by drawing the vectors shown in Fig. 31.6a for a particular frequency. 1. A.V. Oppenheim, R.W. Schafer, Digital Signal Obviously, our signal processor described by Processing. (Prentice-Hall, 1975) Eq. 31.30 represents a band-pass ﬁlter in the 2. L.R. Rabiner, B. Gold, Theory and Applications of baseband. It should be mentioned that the Digital Signal Processing. (Prentice Hall, 1975) References 239 3. W.D. Stanley. Digital Signal Processing. (Reston, on Communications, IEEE Transaction on Comput- 1975) ers, Bell System International Journal of Circuit 4. M.H. Ackroyd, Digital Filters. (Butterworth, 1973) Theory and Applications, Proceedings IEEE, IEEE, 5. E.O. Brigham, The Fast Fourier Transform. Journal on Electronic Circuits and Systems, Elec- (Prentice-Hall, 1974) tronics Letters, Radio Electronic Engineer, Journal 6. K. Steiglitz, An Introduction to Discrete Systems. on Acoustical Society of America., and many others. (Wiley, 1974) Conferences which devote a signiﬁcant portion of 7. A. Peled, B. Liu, Digital Signal Processing.(Wiley, time to digital signal processing papers are IEEE 1976) International Conference on ASSP, IEEE Interna- 8. S.A. Tretter, Introduction to Discrete Time Signal tional Conference on CAS, Allerton, Asilomar, Processing. (Wiley, 1976) Midwest Symposium., European Conference on Cir- 9. R.E. Bogner, A.G. Constantinides, Introduction to cuit Theory and Design, NATO Special Conferences, Digital Filtering. (Wiley Interscience, 1975) Summer Schools on Circuit Theory held at Prague, 10. D. Childers, A. Durling, Digital Filtering and Signal etc. etc Processing. (West Pub. Co., 1975) 17. B. Gold, K.L. Jordan, Digital Signal Processing. 11. A.V. Oppenheim (ed.), Papers on Digital Signal (McGraw-Hill, 1968) Processing. (MIT Press, 1969) 12. L.R. Rabiner, C. Rader (eds.), Digital Signal Pro- cessing. (IEEE Press, 1972) 13. B. Liu, (ed.), Digital Filters and the Fast Fourier Bibliography Transform. (Dowden Hutchinson Ross, 1975) 14. A.V. Oppenheim et al. (eds.), Selected Papers in Digital Signal Processing II. (IEEE Press, 1976) 18. B. Gold, C. Rader, Digital Processing of Signals. 15. See e.g. IEEE Transactions on Audio and Electroa- (McGraw-Hill, 1969) coustics; June 1967, September 1968, June 1969, 19. H.D. Helms, et al. (eds.), Literature in Digital Signal June 1970, December 1970, October 1972, June Processing. (IEEE Press, 1976) 1973, June 1975. IEEE Transactions on Circuit 20. L. Shapiro, Sampling Theory in Digital Processing. Theory: November 1971, July 1973. IEEE Transac- Electron. Eng. 45–50 (May 1978) tions on Circuits and Systems: March 1975. Pro- 21. B. Gold, K. Jordan, A Note on Digital Filter ceedings of lEEE: July 1972, October 1972, April Synthesis Proceedings on IEEE 56, October 1968. l975. IEEE Transactions on Computers: July 1972, pp. 1717–1718 May 1974. IEEE Transactions on Communication 22. J.W. Cooley, J.W. Tukey, An algorithm for the Technology: December 1971 Machine Calculation of Complex Fourier Series. 16. Digital signal processing papers appear. in Proceed- Math. Comput. 19, 297–301 (April 1965) ings of IEEE, IEEE Transaction on Acoustics, 23. W.M. Gentlemen, G. Sande, Fast Fourier Trans- Speech and Signal Processing (formerly Audio and forms–for Fun and Proﬁt. in 1966 Fall Joint Electroacoustics), IEEE Transaction on Circuits and Computer Conference on AFIPS Proceedings, Systems (formerly Circuit Theory), IEEE Transaction pp. 563–578 The ABCDs of Digital Signal Processing–PART 2 32 Here, we deal with the realizations of DSP’s can be represented by a variety of equivalent DFT, FFT, application of FFT to compute realization diagrams or structures. When imple- convolution and correlation, and application mented in a general-purpose computer, the of FFT to ﬁnd the spectrum of a continuous structure may be thought of as the representation signal. of a computational algorithm, from which a computer program is derived. When imple- mented by special-purpose hardware, it is often Keyword convenient to think of the structure as specifying DSP realization DFT FFT and its a hardware conﬁguration. applications Convolution Correlation Corresponding to the basic operations required Picket fence effect for implementation of a digital signal processor, the basic elements required to represent a difference equation pictorially are an adder, a delay and a constant multiplier, the commonly used symbols for Realization of Digital Signal which are shown in Fig. 32.1. Physically, Fig. 32.1a Processors represents a means for adding together two sequences, Fig. 32.1b represents a means for mul- It should be clear from what has been discussed tiplying a sequence by a constant and Fig. 32.1c so far that a digital signal processor may be represents a means for storing the previous value of a realized by use of the storage registers, arithmetic sequence. The representation used for a single unit and the control unit of a general-purpose sample delay arises from the fact that the z-transform computer. Alternatively, special digital hardware of x(n – 1) is simply z–1 times the z-transform of x(n). may be designed to perform the required com- As an example of the representation of a dif- putations; this would result in a special-purpose ference equation in terms of these elements, processor (e.g. for radar or sonar signals) that consider the second-order equation would more or less be committed to a speciﬁc job. In either case, the digital signal processor yðnÞ ¼ b1 yðn1Þ þ b2 yðn2Þ þ axðnÞ: ð32:1Þ A realization structure for Eq. 32.2 is shown Source: S. C. Dutta Roy, “The ABCDs of Digital Signal in Fig. 32.2. In terms of a computer program, Processing (Part 2),” Students’ Journal of the IETE, vol. Fig. 32.2 shows explicitly that storage must be 21, pp. 60–70, April 1980. provided for the variables y(n – 1) and y(n – 2) Revised version of the text of a seminar series. and also the constants b1, b2 and a. Further, © Springer Nature Singapore Pte Ltd. 2018 241 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_32 242 32 The ABCDs of Digital Signal Processing-PART 2 x2 (n ) a1 a2 (a) HðzÞ ¼ ð32:3Þ 1 c1 z1 1 þ c2 z1 x1(n) x1(n) x2(n) a1 a11 ¼ 1 þ ; ð32:4Þ 1 c1 z 1 þ c2 z1 (b) a where c1 – c2 = b1, c1 c2 = b2, a1 a2 = a = x(n) a (n) a1 + a2 and a1c2 – a2 c2 = 0. Each ﬁrst-order (c) form x(n) z-1 x(n 1) constant 1 c1 z1 Fig. 32.1 Basic elements used in the realization diagram of a digital signal processor can be implemented by using a delay, two con- stant multiplications and an adder. To implement form Eq. 32.1, one needs to cascade two a (n) y(n) ﬁrst-order realizations, as shown in Fig. 32.3a, while form Eq. 32.4 requires parallel connection z -1 of two ﬁrst-order forms, as shown in Fig. 32.3b. b1 Obviously, the three realizations of Figs. 32.1, y(n 1) 32.3a, b differ in computational algorithm and z -1 hardware requirements; what is more important, however, is that the quantization errors also, in b2 y(n 2) general, differ from structure to structure. We shall not, however, explore this point further in Fig. 32.2 A realization of Eq. 32.1 this chapter, but shall be content with the knowledge that a digital signal processor can be implemented by various equivalent structures, Fig. 32.2 shows that to compute an output sam- and one should choose the one which is the ple y(n), one must form the products b1y(n – l) optimum under the given set of constraints. and b2y(n – 2), ax(n), and add them together. In We shall not also discuss here methods of terms of special digital hardware, Fig. 32.2 ﬁnding out the difference equation to suit a par- indicates that we must provide storage for the ticular set of speciﬁcations; this we shall reserve variables and constants, as well as means for for a future article. The rest of this chapter will multiplication and addition. Thus, diagrams such deal with DFT technique in relation to digital as Fig. 32.2 serve to depict the complexity of a signal processing. digital signal processor algorithm and the amount of hardware required to realize the processor. As mentioned earlier, a variety of structures The Discrete Fourier Transform can be derived to implement a given difference equation. In case of Eq. 32.1, for example, the As we have already seen, one can implement a transfer function is digital signal processor, described by Eq. 31.1, in a general-purpose computer or by a HðzÞ ¼ a=ð1b1 z1 b2 z2 Þ: ð32:2Þ special-purpose hardware. Another way of implementing a digital signal processor is based Let b1 and b2 be such that the poles of H(z) are on the fact that the output sequence {y(n)} is the real. Then, one could write H(z) in the following convolution of the input sequence {x(n)} with two equivalent forms: the impulse response sequence {h(n)} of the The Discrete Fourier Transform 243 Fig. 32.3 a Cascade (a) realization. b Parallel a1 a2 realization x(n) y(n) C1 z -1 C2 z -1 (b) a1 x(n) C1 z -1 y(n) a11 C2 z -1 processor, as given by Eq. 31.27. This equation (DFT) has been found most suitable. The signal is reproduced here for convenience as follows: processing operation then simply boils down to the following sequence of computations: X n yðnÞ ¼ hðrÞxðn rÞ r¼0 (i) Compute the DFT of {x(n)}. ð32:5Þ (ii) Compute the DFT of {h(n)}. Xn ¼ hðn rÞxðrÞ: (iii) Multiply the two. r¼0 (iv) Compute the inverse DFT (IDFT) of the product. Recall that in the continuous signal case, the convolution of the input signal x(t) with the Let, for simplicity, the notation xk be used for impulse response h(t) gives the output y(t) and x(k) x(kT), and consider a sequence {xk} of that in the frequency domain, this amounts to a length N, i.e. k = 0, 1, 2, … N – 1. Then, the multiplication of the transforms (Laplace or DFT of {xk} is deﬁned by Fourier) of x(t) and h(t) to give the transform of y X N 1 (t). y(t) can then be obtained by the inverse Ar ¼ xk ej2prk=N ; r ¼ 0; 1; 2; . . . N 1: transform operation. A similar operation can be k¼0 performed with discrete time systems if we have ð32:6Þ a suitable transform. As we have already seen, the z-transform does provide such a vehicle; The DFT is thus also a sequence {Ar} of however, for numerical computation, a modiﬁed length N. The xk’s may be complex numbers; the version of it, called the discrete Fourier transform 244 32 The ABCDs of Digital Signal Processing-PART 2 Ar ’s are almost always complex. For notational would require N2 complex multiplications; in convenience, let contrast, application of FFT can reduce this number to (N/2) log2N. For example, for W ¼ ej2p=N ; ð32:7Þ N = 512, the ratio (N/2) log2N N2 becomes so that less than 1%. This drastic reduction in compu- X N 1 tation time through FFT has made the FFT an Ar ¼ xk W rk ; r ¼ 0; 1; . . .N 1: ð32:8Þ important tool in many signal processing k¼0 applications. The DFT, given by Eq. 32.8 and its inverse, If one compares Eq. 32.6 with the continuous given by Eq. 32.10 are of the same form so that Fourier transform A(x) of a signal x(t), viz. any algorithm capable of computing one may be Z1 used for computing the other by simply exchanging the roles of xk and Ar, and making AðxÞ ¼ xðtÞej2pft dt ð32:9Þ appropriate scale factor and sign changes. There 1 are two basic forms of FFT; the ﬁrst, due to Cooley and Tukey [1], is known as decimation in then one way of interpreting the DFT is that it time, while the other, obtained by reversing the gives the N-point discrete spectrum of the N- roles of xk and Ar, gives the form called deci- point time series {x(kT)} at the frequency points mation in frequency, and was proposed by r/(NT); r = 0, 1, … N – 1; the fundamental fre- Gentleman and Sande [2]. Clearly, they should quency, obviously, is fo = 1/(NT). be equivalent; it is, however, worth distinguish- The inverse DFT (IDFT) of the complex ing between them and discussing them sequence {Ar}, r = 0, 1 … N – 1, is given by separately. Let N be even and the sequence {xk} be 1XN 1 xk ¼ Ar W rk ; k ¼ 0; 1; . . . N 1: decomposed as N r¼0 ð32:10Þ fxk g ¼ fuk g þ fvk g; ð32:12Þ where That this exists and is unique can be easily established by substituting Eq. 32.8 in Eq. 32.10 uk ¼ x2k and carrying out some elementary manipulations. k ¼ 0; 1; 2; . . . N=21 ð32:13Þ vk ¼ x2k þ 1 : Since ej is periodic with a period 2p, it follows from Eq. 32.8 and Eq. 32.10 that Thus {uk} contains the even-numbered points Ar ¼ Ar þ mN ; m ¼ 0; 1; 2; . . . and {vk} contains the odd-numbered points of ð32:11Þ {xk} and each has N/2 points. The DFTs of {uk} xk ¼ xk þ mN : and {vk} are, therefore, i.e. both DFT and IDFT yield sequences which are periodic, with periods Nfo = T−1 = fs and NT P N=21 Br ¼ uk ej2prk=ðN=2Þ respectively. k¼0 P N=21 ¼ uk ej4prk=N r ¼ 0; 1; 2; . . .N=2 1 k¼0 The Fast Fourier Transform P N=21 Cr ¼ vk ej4prk=N : k¼0 The fast Fourier transform (FFT) is a highly ð32:14Þ efﬁcient method for computing the DFT of a time series. A direct computation from Eq. 32.8 The Fast Fourier Transform 245 The DFT we want is x0 = u0 B0 A0 x2 = u1 B1 W0 X N 1 DISCRETE A1 Ar ¼ xk ej2prk=N FOURIER W1 k¼0 x4 = u2 TRANSFORM B2 (N = 4) A2 X N=21 X N=21 W2 ¼ x2k ej4rpk=N þ x2k þ 1 ej2prð2k þ 1Þ=N ; x6 = u3 B3 A3 k¼0 k¼0 W3 r ¼ 0; 1; 2; . . .N 1 x1 = v0 C0 A4 j2pr=N W4 ¼ Br þ e Cr ; 0 r\N=2: x3 = v1 C1 DISCRETE A5 ð32:15Þ FOURIER W5 x5= v2 TRANSFORM C2 (N = 4) A6 because Br and Cr are deﬁned for r = 0 to (N/2) – W6 x7 = v3 C3 1. Further, Br and Cr are periodic with period N/2 A7 W7 so that Fig. 32.4 Illustrating the ﬁrst step in decimation in time Br þ N=2 ¼ Br and Cr þ N=2 ¼ Cr : ð32:16Þ form of FFT for N = 8 Thus computation of {Br} and {Cr} reduces to the task of ﬁnding the DFTs of four sequences, each of N/ Ar þ N=2 ¼ Br þ ej2pðr þ N=2Þ =N Cr 4 samples. These reductions can be continued as ¼ Br ej2pr=N Cr ; 0 r\N=2: long as each sequence has an even number of ð32:17Þ samples. Thus if N = 2n, one can make n such reductions by applying Eq. 32.13 and Eq. 32.18, Finally, using Eqs. 32.7, 32.15 and 32.17, we ﬁrst for N, then for N/2 and so on, and ﬁnally for get a two-point function. The DFT of a one-point function is, of course, the sample itself. The Ar ¼ Br þ W r Cr successive reduction of an 8-point DFT, which 0 r\N=2 ð32:18Þ began in Fig. 32.4, is continued in Figs. 32.5 and Ar þ N=2 ¼ Br W r Cr: 32.6. In Fig. 32.6, the operation has been com- pletely reduced to complex multiplications and A direct calculation of Br and Cr, from additions. The number of summing nodes is Eq. 32.14 requires (N/2)2 complex multiplica- (8) (3) = 24 and 24 complex additions are, tions each. Another N such multiplications are therefore, required; the number of complex required to compute Ars from Eq. 32.18, thus multiplications needed are also 24 = (no. of making a total of 2(N/2)2 + N = N2/2 + N, stages) (no. of multiplications in each stage) = which is less than N2 if N > 2. This is illustrated (3) (8). Half of these multiplications are easily in Fig. 32.4 by a signal flow diagram for N = 8, eliminated by noting that W7 = −W3, W6 = −W2, where we have used the fact that WN/2 = –1, so W5 = −W1 and W4 = −W0. Thus, in general, that –Wr = Wr + N/2, N log2 N complex additions and, at most, (1/2) The DFTs of {uk} and {vk}, k = 0, l, … (N/2) N log2 N complex multiplications are required for –1, can now be computed through a similar the computation of an N-point DFT, when N is a decomposition if N/2 is even; thus the power of 2. 246 32 The ABCDs of Digital Signal Processing-PART 2 Fig. 32.5 Illustrating two x0 = u0 B0 A0 steps of decimation in DFT W0 W0 frequency form of FFT for (N = 2) x4 = u2 A1 N=8 B1 W1 W2 x2 = u1 B2 A2 W4 W2 x6 = u3 A3 W5 B3 W3 x1 = v0 C0 A4 W0 W4 x5 = v2 A5 C1 W5 W2 x3 = v1 C2 A6 W4 W6 x7 = v3 A7 W6 C3 W7 Fig. 32.6 Illustrating x0 A0 decimation in time form of W0 W0 W0 FFT for N = 8 x4 A1 W4 W2 W1 x2 A2 W0 W4 W2 x6 A3 W4 W6 W3 x1 A4 W0 W4 W0 x5 A5 W4 W2 W5 x3 A6 W0 W4 W6 x7 A7 W4 W6 W7 When N is not a power of 2, but has a factor of ðiÞ Each of these sequences has a DFT Br , and p, one can develop equations analogous to 32.13 the DFT of {xk} can be computed from p simpler through 32.18 by forming p different sequences, DFT’s. Further simpliﬁcation occurs if N has ðiÞ fuk g ¼ fxpk þ i g; i = 0 to p – 1, each having N/ additional prime factors. p samples. For example, if N = 15 having a In the decimation in frequency form of FFT, factor 3, we can form three sequences as follows: the sequence {xk}, k = 0, l, …, N – 1 and N even, is decomposed as ð0Þ fuk g ¼ fx0 ; x3 ; x6 ; x9 ; x12 g ð1Þ uk ¼ xk k ¼ 0; 1; . . .; N=21 fuk g ¼ fx1 ; x4 ; x7 ; x10 ; x13 g ð32:19Þ ð32:20Þ vk ¼ xk þ N=2 ð2Þ fuk g ¼ fx2 ; x5 ; x8 ; x11 ; x14 g: The Fast Fourier Transform 247 i.e. {uk} is composed of the ﬁrst N/2 points and x0 A0 {vk} is composed of the last N/2 points of {xk}. x1 A2 Then one can write DFT (N = 4) x2 A4 X N=21 Ar ¼ ½uk ej2prk=N þ vk e2prðk þ N=2Þ=N x3 A8 k¼0 W0 X N=21 x4 ¼ ðuk þ ejpr vk Þe2prk=N ; W0 A1 W1 k¼0 x5 A3 W1 DFT r ¼ 0; 1; . . .N1: x6 W2 (N = 4) A5 W2 ð32:21Þ W3 x7 A7 W3 Consider the even-numbered and odd-numbered points of the DFT separately; let Fig. 32.7 Illustrating the ﬁrst step in decimation in frequency form of FFT for N = 8 Rr ¼ A2r 0 r\N=2 ð32:22Þ Sr ¼ A2r þ 1 : Fig. 32.2 by two 2-point DFTs, and each of the 2-point DFTs by two 1-point transforms, these It is this step that may be called the decima- last being equivalency operations. These steps tion in frequency. Note that for computing Rr, are indicated in Figs. 32.8 and 32.9. Eq. 32.21 becomes There are many variations and modiﬁcations of the two basic FFT schemes, which we would X N=21 not discuss here. Rr ¼ A2r ¼ ðuk þ vk Þej2rpk=ðN=2Þ : k¼0 ð32:23Þ Applications of FFT to Compute which we recognize as the N/2-point DFT of the Convolution and Correlation sequence {uk + vk}. Similarly, It may be recalled that our motivation for intro- X N=21 ducing the DFT and FFT was to convert the Sr ¼ A2r þ 1 ¼ ½uk þ vk ejpð2r þ 1Þ ej2pð2r þ 1Þk=N convolution relation Eq. 32.5, viz. k¼0 X n X N=21 yn ¼ hr xnr j2pk=N j2prk=ðN=2Þ ¼ ðuk vk Þe e : r¼0 k¼0 ð32:25Þ X n ð32:24Þ ¼ hnr xr r¼0 which we recognize as the N/2-point DFT of the sequence fðuk vk Þej2pk=n g. into a product form, through the DFT. To this Thus, the DFT of an N–sample sequence {xk}, end, assume that both the impulse response {hn}, N even, can be computed as the N/2 point DFT of and the input {xn} are band limited to 1/2T Hz. a simple combination of the ﬁrst N/2 and the last Then the output {yn} is also frequency band N/2 samples of {xk} for even-numbered points, limited. Also, if both {hn} and {xn} are deﬁned and a similar DFT of a different combination of for the range 0 n N – I, then {yn} is deﬁned the same samples of {xk} for the odd-numbered for the range 0 n 2 N – 1. For example, if points. This is illustrated in Fig. 32.7 for N = 8. {hn} = {h0, h1} and {xn} = {x0, x1}, then As was the case with decimation in time, we fyn g ¼ fh0 x0 ; h0 x1 þ h1 x0 ; h1 x1 g. Let the DFT’s can replace each of the DFTs indicated in of {xn} and {hn} be {Ar} and {Hr} respectively. 248 32 The ABCDs of Digital Signal Processing-PART 2 x0 A0 DFT (N = 2) x1 A4 W0 x2 A2 – W0 DFT W2 (N = 2) x3 A6 – W2 W0 x4 A1 W0 DFT W1 (N = 2) x5 A5 – W1 W2 W0 x6 A3 – W2 – W0 DFT W3 W2 (N = 2) x7 A7 – W3 – W2 Fig. 32.8 Illustrating two steps of decimation in time form of FFT for N = 8 x0 A0 W0 x1 A4 W0 W0 x2 A2 – W0 W2 W0 x3 A3 – W2 W0 W0 x4 A1 – W0 W1 W0 x5 A5 – W1 W0 W2 W0 x6 A3 – W2 – W0 W3 W2 W0 x7 A7 – W3 – W2 W0 Fig. 32.9 Illustrating decimation in frequency form of FFT for N = 8 Applications of FFT to Compute Convolution and Correlation 249 Then, the nth sample in the IDFT of the product arises due to the fact that the DFT assumes both {ArHr} is {xn} and {hn} to be periodic. Further, fy0n g is of length N instead of 2 N –1. Note that if we X N 1 extend both {xn} and {hn} to a length 2 N by y0n ¼ 1=N Ar Hr W rn ; r¼0 ð32:26Þ adding N zeros to each, i.e. if we change {xn} to f^xn g = {x0, x1 … xN – 1, 0, … 0} and similarly n ¼ 0; 1. . . N1: for {hn}, then the perturbation term becomes zero. Further, the sequence {yn} will be N + N – Substituting in Eq. 32.26, 1 = 2 N – 1 terms long, i.e. y2N– 2 will be the last, X N 1 non-zero term in {yn}. As an example, let N = 4, Ar ¼ xk W rk ; i.e. k¼0 ð32:27Þ fxn g ¼ fx0 ; x1 ; x2 ; x3 g X N 1 ð32:30Þ Hr ¼ rl hl W : fhn g ¼ fh0 ; h1 ; h2 ; h3 g: l¼0 The true convolution of {xn} with {hn} gives and carrying out some elementary manipulations, it is not difﬁcult to show that Eq. 32.26 simpliﬁes y 0 ¼ x 0 h0 to y 1 ¼ x 0 h1 þ x 1 h0 X N 1 y 2 ¼ x 0 h2 þ x 1 h1 þ x 2 h0 y0n ¼ xk hnk y 3 ¼ x 0 h3 þ x 1 h2 þ x 2 h1 þ x 3 h0 ð32:31Þ k¼0 ð32:28Þ y 4 ¼ x 1 h3 þ x 2 h2 þ x 3 h1 X n X N 1 ¼ xk hnk þ xk hN þ nk y 5 ¼ x 2 h3 þ x 3 h2 k¼0 k¼n þ 1 y 6 ¼ x 3 h3 : On the other hand, the DFT procedure, lead- ¼ yn þ perturbation term: ð32:29Þ ing to Eq. 32.28 gives The last form is obtained by comparison with X 3 Eq. 32.25, while the last term in Eq. 32.28 rep- y0n ¼ xk hnk ; ð32:32Þ resents the ‘cyclical’ part of the convolution, k¼0 arising out of the periodicity of DFT; and IDFT; so that h is the cyclical variable passing from ho to hN – 1 as k passes from n to n + 1. The convolution can y00 ¼ x0 h0 þ x1 h1 þ x2 h2 þ x3 h3 be made cyclical in x instead of h by ¼ x0 h0 þ ðx1 h3 þ x2 h2 þ x3 h1 Þ inter-changing x and h in Eq. 32.28. y01 ¼ x0 h1 þ x1 h0 þ ðx2 h3 þ x3 h2 Þ ð32:33Þ The procedure outlined for implementing a y02 ¼ x0 h2 þ x1 h1 þ x2 h0 þ ðx3 h1 Þ digital signal processor, viz. taking the DFTs of 0 y 3 ¼ x 0 h3 þ x 1 h2 þ x 2 h1 þ x 3 h0 : {xn} and {hn}, multiplying them, and taking IDFT of the product, does not, therefore, give the where the perturbation terms are bracketed. Also, desired output sequence {yn} unless the pertur- fy0n g consists of only 4 terms. Now let bation term in 32.28 can be made zero. This term 250 32 The ABCDs of Digital Signal Processing-PART 2 f^xn g ¼ fx0 ; x1 ; x2 ; x3 ; 0; 0; 0; 0g be corrected, as demonstrated by the example, by ð32:34Þ adding zeros to both {xn} and {hn} and thereby fh^n g ¼ fh0 ; h1 ; h2 ; h3 ; 0; 0; 0; 0g: increase their lengths sufﬁciently so that no overlap occurs in the resultant convolution. Then the DFT procedure gives We now state formally the steps for comput- ing convolution by DFT: X 7 y0n ¼ xk hnk ; ð32:35Þ k¼0 (i) Let {xn} be deﬁned for so that 0 n M1 y00 ¼ x 0 h0 y01 ¼ x 0 h1 þ x 1 h0 and {hn} be deﬁned for y02 ¼ x 0 h2 þ x 1 h1 þ x 2 h0 0 n P1 y03 ¼ x 0 h3 þ x 1 h2 þ x 2 h1 þ x 3 h0 ð32:36Þ y04 ¼ x 1 h3 þ x 2 h2 þ x 3 h1 (ii) Select N such that y05 ¼ x 2 h3 þ x 3 h2 N P þ M1 y06 ¼ x 3 h3 : N ¼ 2k By comparing with Eq. 32.31, we see that (iii) Form the new sequences f^xn g and f^ hn g such that fy0n g ¼ fyn g n ¼ 0; 1; 2; . . .7: xn ; 0 n M 1 ^xn ¼ 0; M n N 1 Thus, the modiﬁcation does give correct hn ; 0 n P 1 results. ^ hn ¼ 0; P n N 1 Before stating this simple remedy in formal terms, we would like to emphasize that blind use ^ r g and fH (iv) Compute the DFTs fA ^ r g of f^xn g of FFT for computing the convolution of two ^ and fhn g by FFT. sequences will lead to incorrect results, because the DFT introduces a periodic extension of both (v) Compute data and processor impulse response. This results fB ^rH ^ r g ¼ fA ^rg in cyclic or periodic convolution, rather than the desired noncyclic or aperiodic convolution. If ^ r g by FFT; the result (vi) Find the IDFT of fB {xn} and {hn} contain N samples each, then the is {yn}. true convolution should result in 2 N – 1 samples for {yn}. If DFT is used, then {Ar} and {Hr} each This technique is referred to as select-saving. consist of N samples, so does {ArHr} and hence Next, we consider the application of FFT to its IDFT. Hence, {yn′} found by DFT is not the compute the cross-correlation sequence {Rxy(k)} same as {yn} because of folding (or aliasing or of two given sequences {xn} and {yn}, each of cycling) occurring in the time domain. This can length N, where Applications of FFT to Compute Convolution and Correlation 251 X N 1 Application of FFT to Find Rxy ðkÞ D 1=N xn ynk ð32:37Þ the Spectrum of a Continuous Signal n¼0 The DFT, as we have seen, is speciﬁcally con- and the auto-correlation sequence {Rxx(k)} of a cerned with the analysis and processing of dis- sequence {xn}, where crete periodic signals, and that it is a zero-order X N 1 approximation of the continuous Fourier trans- Rxx ðkÞ D 1=N xn xnk : ð32:38Þ form. It is, therefore, tempting to apply the DFT n¼0 directly to provide, through FFT, a numerical spectral analysis of sampled versions of contin- Note that the essential difference between uous signals. This would be a perfectly valid convolution, as given by Eq. 32.25, and corre- application, if the continuous, signal is periodic, lation, as given by Eqs. 32.37 and 32.38 is that band limited and sampled in accordance with the one of the sequences is reversed in direction for sampling theorem. Deviations from these cause one operation as compared with the other. Thus, errors, and most of the problems in using the if FFT is to be used to compute correlation, the DFT to approximate the CFT (C for continuous) same kind of precautions, as discussed for con- are caused by a misunderstanding of what this volution, are to be exercised. The procedure here, approximation involves. is based on the fact that if There are, essentially, three phenomena, which contribute to errors in relating the DFT to DFTfxn g ¼ fAr g the CFT. The ﬁrst, called aliasing, has already and been discussed (Part 1). The solution to this DFTfyn g ¼ fBr g problem is to ensure that the sampling rate is high enough to avoid any spectral overlap. This then requires some prior knowledge of the nature of ( the spectrum, so that the appropriate sampling X N 1 rate may be chosen. In absence of such prior DFT xn ynk g ¼ Ar Br ; ð32:39Þ knowledge, the signal must be preﬁltered to n¼0 ensure that no components higher than the fold- where bar denotes complex conjugate. Thus ing frequency appear. applied to Eqs. 32.37 and 32.38, one obtains The second problem is that of leakage, arising due to the practical requirement of observing the fRxy ðkÞg ¼ IDFTfAr Br =Ng signal over a ﬁnite interval. This is equivalent to ð32:40Þ multiplying the signal by a window function. The ¼ IDFTfSxy ðrÞg simplest window is a rectangular function as shown in Fig. 32.10a, and its effect on the spectrum of a sine signal, shown in Fig. 32.10b, fRxx ðkÞg ¼ IDFTf1 Ar =2 =Ng ð32:41Þ is displayed in Fig. 32.10c. Note that there ¼ IDFTfSxx ðrÞg occurs a spreading or leakage of the spectral components away from the correct frequency; where {Sxy(r)} and {Sxx(r)} are the cross-power this results in an undesirable modiﬁcation of the spectrum sequence and auto power spectrum total spectrum. sequences respectively. 252 32 The ABCDs of Digital Signal Processing-PART 2 (a) w(t) |D(t)| t t (b) w(t) |w(t)| t t (c) w(t) |D(t) * w(t)| t t Fig. 32.10 Illustrating ‘leakage’ due to ﬁnite observation time The leakage effect cannot always be isolated window in the time domain is relatively small, as from the aliasing effect because leakage may also compared to other continuously varying weight lead to aliasing if the highest frequency of the windows, e.g. the Hamming window. composite spectrum moves beyond the folding The third problem in relating the DFT to the CFT frequency. This possibility is particularly signif- is the picket fence effect, resulting from the inability icant in the case of a rectangular window, of the DFT to observe the spectrum as a continuous because the tail of the window spectrum does not function, since the computation of the spectrum is converge rapidly. limited to integer multiples of the fundamental fre- The solution to the leakage problem is to choose quency fo = 1/(NT). In a sense, the observation of a window function that minimizes the spreading. the spectrum with the DFT is analogous to looking One example is the so-called ‘raised cosine’ win- at it through a sort of ‘picket fence’ since we can dow in which a raised cosine wave is applied to the observe the exact behaviour only at discrete points. ﬁrst and last 10 per cent of the data and a weight of It is possible that a major peak lies between two of unity is applied in between. Since only 20% of the the discrete transform lines, and this will go unde- terms in the time series is given a weight other than tected without some additional processing. unity, the computation required to apply this Application of FFT to Find the Spectrum of a Continuous Signal 253 One procedure for reducing the picket fence are always ‘better than one’. ‘… And three are effect is to vary the number of points N in a time better still’, the proverb continues; this third ‘ar- period by adding zeros at the end of the original row’ is provided by the charge transfer devices, record, while maintaining the original record intact. which can perform analog as well as digital signal This process artiﬁcially changes the period, which, processing. As compared to the digital signal in turn, changes the locations of the spectral lines processors we have talked about, the charge without altering the continuous form of the original transfer processing has the distinct advantage of spectrum. In this manner, spectral components not requiring an A/D conversion, and hence is less originally hidden from view may be shifted to expensive, more versatile and more accurate. points where they may be observed. Problems Concluding Comments P:1. Draw the equivalent of Fig. 32.6 for The aim of this chapter was to introduce the basic N = 16. You have to take an A3 paper with concepts involved in digital signal processing, 90° turn around. including an introduction to the FFT and its P:2. Draw the FFT diagram for N = 16 using applications. We went through the sampling decimation in frequency. process carefully, and pointed out the various P:3. What is the minimum possible number of errors introduced by quantization. A brief dis- non-trivial multipliers in Fig. 32.8? cussion on structures was included to facilitate an P:4. What is better? DIT or DIF? understanding of the implementation of a digital P:5. What is a possible remedy for eliminating signal processor in a general-purpose computer leakage altogether? Is it practicable? or by a special-purpose hardware. Two basic forms of FFT were introduced, and two of the most important applications of the FFT were discussed. It was pointed out that correct appli- References cation of FFT requires a much more than casual understanding of the periodic extension intro- 1. J.W. Cooley, J.W. Tukey An algorithm for the duced by the DFT process. machine calculation of complex fourier series. Math. Comput. 19, 297–301 (April 1965) In conclusion, it is worth mentioning that 2. W.M. Gentleman, G. Sande, Fast Fourier digital signal processing is not an answer to all Transforms-for Fun and Proﬁt. in 1966 Fall Joint signal processing problems. Digital and analog Computer Conference of AFIPS Proceedings, techniques form ‘two arrows in the quiver’, which pp. 563–578 On Second-Order Digital Band-Pass and Band-Stop Filters 33 The chapter deals with the derivation, design, is that of a normalized digital band-pass ﬁlter limitations and realization of second-order (BPF) whose centre frequency x0 and 3-dB digital band-pass (BP) and band-stop (BS) ﬁl- bandwidth B are given by ters with independent control of the centre frequency and the bandwidth in the BP case, 2a x0 ¼ cos1 b and B ¼ cos1 : ð33:2Þ and rejection frequency and the difference 1 þ a2 between the pass-band edges in the BS case. Thus x0 and b are independently controllable by varying a and b, provided one can realize Eq. 33.1 Keywords by only two multipliers of the same values. Such a Digital ﬁlter Band-stop Band-pass realization using the lattice structure has been given Second-order ﬁlters in [1]. The complement of Eq. 33.1, obtained by subtracting H1 (z) from unity, is 1þa 1 2bz1 þ z2 H2 ðzÞ ¼ : ð33:3Þ 2 1 bð1 þ aÞz1 þ az2 It represents a band-stop ﬁlter (BSF) whose rejection frequency x0 and the parameter B = x2 Introduction – x1, where x2 and x1 are the two 3 dB fre- quencies, are also given by Eq. 33.2. Hence, the The second-order transfer function BPF realization can also be used for realizing the BSF. 1a 1 z2 While one appreciates the elegance of the H1 ðzÞ ¼ ð33:1Þ transfer function Eq. 33.1, the question of how it 2 1 bð1 þ aÞz1 þ az2 was conceived of has not been answered in textbooks. Another question that arises is the following: Is the a-controllability of the band- width valid for any arbitrary pass-band toler- ance? Yet another relevant question concerns the Source: S. C. Dutta Roy, “On Second-Order Digital realization of a canonic structure with multipliers Band-Pass and Band-Stop Filters,” IETE Journal of a and b. Are structures other than that given in Education, vol. 49, pp. 59–63, May–August 2008. © Springer Nature Singapore Pte Ltd. 2018 255 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_33 256 33 On Second-Order Digital Band-Pass and Band-Stop Filters Eq. 33.1 possible? This chapter presents answers cos x0 ¼ 2r cos h=ð1 þ r 2 Þ: ð33:8Þ to all these questions. Comparing with Eq. 33.2, we, therefore, have Derivation 2r cos h=ð1 þ r 2 Þ ¼ b: ð33:9Þ For a BPF, we argue that the response at x = 0 Also, combining Eqs. 33.7 and 33.8, we note and p should both be zero. With z = ejx, x = 0 that the maximum magnitude will be unity if and p translate to z = 1 and –1, respectively. Hence, the numerator polynomial of the transfer K ¼ ð1r 2 Þ=2; ð33:10Þ function must be of the form K(1 – z−2), K being a constant. Also, we know that real poles and, combining Eq. 33.7 with Eq. 33.10, we can severely limit the selectivity of a BPF. Hence, we write let the poles be at re±jh, where r is close to but less than unity for high selectivity. The denom- 1 jH1 ðejx Þj2 ¼ h i2 : ð33:11Þ inator polynomial of the BPF transfer function is, ð1 þ r2 Þ cos x2r cos h 1þ ð1r 2 Þ sin x therefore, ð1rejh z1 Þð1rejh z1 Þ Now, suppose, instead of the range pﬃﬃﬃ ¼ 12r cos hz1 þ r 2 z2 : ð33:4Þ 1= 2\ jHðejx Þj \1; the pass-band is deﬁned pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ by 1= 1 þ e2 \ jHðejx Þj \1; where e < 1 is The required transfer function is, therefore, arbitrary. Then from Eq. 33.11, the pass-band edge frequencies x2 and x1 will satisfy the Kð1 z2 Þ equation H1 ðzÞ ¼ : ð33:5Þ 1 2r cos hz1 þ r 2 z2 ð1 þ r 2 Þcos x2r cos h ¼ ð1r 2 Þsin x: where K will be chosen to normalize the maxi- ð33:12Þ mum magnitude to unity. The frequency response of Eq. 33.5 is given by We let (at the moment arbitrarily, but justiﬁed by later results) 2jK sin x H1 ðejx Þ ¼ : ½ð1 r 2 Þ cos x 2r cos hð1 r2 Þ sin x ð1 þ r 2 Þcos x1 2r cosh ¼ ð1r 2 Þsin x1 ; ð33:6Þ ð33:13Þ The magnitude squared function can be writ- and ten as ð1 þ r 2 Þcosx2 2r cos h ¼ ð1r 2 Þsin x2 : 2 4K 2 jH1 ðe Þj ¼ h jx i2 : ð33:14Þ ð1 þ r 2 Þ cos x2r cos h sin x þ ð1 r 2 Þ2 Subtracting Eq. 33.14 from Eq. 33.13 gives ð33:7Þ Maximum value of |H1(ejx)|2 is reached when ð1 þ r 2 Þ ðcosx1 cosx2 Þ the ﬁrst term in the denominator vanishes, i.e. at ¼ ð1r 2 Þ ðsinx1 þ sinx2 Þ: ð33:15Þ x = x0, where Derivation 257 Or, Or, x2 þ x1 x2 x1 cos x1 þ cos x2 4rcos h=ð1 þ r 2 Þ ¼ 2b ð1 þ r 2 Þ sin sin 2 2 ¼ 2cos x0 : x2 þ x1 x2 x1 ¼ eð1 r 2 Þ sin cos : ð33:23Þ 2 2 ð33:16Þ Thus, cosx1 and cosx2 are approximately This gives arithmetically symmetrical about cosx0. x2 x1 eð1 r 2 Þ tan ¼ : ð33:17Þ 2 1 þ r2 Design for Arbitrary Pass-band Tolerance Using the relationships Combining Eqs. 33.17 and 33.21, and letting be cos 2 h ¼ 2cos2 h1 and sec2 h ¼ 1 þ tan2 h; denote the bandwidth for an arbitrary pass-band ð33:18Þ tolerance speciﬁed by e, we get we get from Eq. 33.17, after simpliﬁcation, Be eð1 aÞ tan ¼ : ð33:24Þ 2 1þa ð1 þ r 2 Þ2 e2 ð1 r 2 Þ2 cosðx2 x1 Þ ¼ : ð1 þ r 2 Þ2 þ e2 ð1 r 2 Þ2 Thus, given e and Be, one can choose a from ð33:19Þ e tan B2e a¼ : ð33:25Þ The 3-dB bandwidth B is obtained as x2 – x1 e þ tan B2e with e = 1. Hence, Equation 33.25 puts a constraint on the 2r 2 speciﬁcations of e and Be. Since 0 < Be < p, tan cos b ¼ : ð33:20Þ 1 þ r4 (Be/2) > 0; also a = r2 is a real positive quantity. Hence, e and Be must satisfy Comparing Eq. 33.20 with Eq. 33.2, we get Be e [ tan : ð33:26Þ r ¼ a: 2 ð33:21Þ 2 Substituting Eqs. 33.9, 33.10 and 33.21 in This is quite logical because, with two Eq. 33.5, it becomes identical with Eq. 33.1. parameters e and b, one cannot satisfy three Note, in passing, that for small pass-band tol- speciﬁcations, viz. e, Be and x0. The constraint erance (e ! 0) or small bandwidth (x2 – x1 ! 0) Eq. 33.26 is shown graphically in Fig. 33.1 in or both addition of Eqs. 33.13 and 33.14 gives the form of a plot of e ¼ tan B2e : No speciﬁcation point which lies below the curve can be met by ð1 þ r 2 Þðcos x1 þ cos x2 Þ 4r cos h: ð33:22Þ the second-order BPF characterized by Eq. 33.1. 258 33 On Second-Order Digital Band-Pass and Band-Stop Filters 2.0 a bð1 þ aÞz1 þ z2 A2 ðzÞ ¼ : ð33:27Þ 1 bð1 þ aÞz1 þ az2 1.5 by noting that H1;2 ðzÞ ¼ ð1=2Þ½1 A2 ðzÞ: ð33:28Þ e 1.0 Figure 33.2 shows the implementation of Eq. 33.28, while Fig. 33.3 shows the realization of A2(z) with a lattice structure using the two 0.5 multipliers a and b. We give here another realization of A2(z), starting from that of the transfer function 0 0 p/4 p/2 3p/4 p d2 þ d1 z1 þ z2 Be A2 ðzÞ ¼ : ð33:29Þ 1 þ d1 z1 þ d2 z2 Fig. 33.1 Plot of e ¼ tan B2e : Speciﬁcation points below as given in [1] and reproduced in Fig. 33.4. For the curve cannot be met by the second-order ﬁlter Eq. 33.28, d1 ¼ bð1 þ aÞ and d2 ¼ a: ð33:30Þ Realization The part marked by nodes A, B and C in In [1], the realizations of Eq. 33.1 and 33.3 have Fig. 33.4 can be modiﬁed to the equivalent been derived from that of the all-pass ﬁlter conﬁguration shown in Fig. 33.5, where the two Fig. 33.2 Implementation of X(z)H1(z) Eq. 33.28 1 1/2 X(z) A2(z) X(z)H2(z) Fig. 33.3 Realization of A2(z) of Eq. 33.27, as given in X1(z) a b [1] 1 1 A2(z)X1(z) z 1 z 1 Realization 259 Fig. 33.4 Implementation of 1 1 z z Y1 A2(z) of Eq. 33.29, as given in X1 [1] A d1 1 B d2 C Fig. 33.5 Equivalent A A representation of the part ABC of Fig. 33.4 d1= b (1+a) b 1+a B C B C d2 = a 1 multipliers a and b have been separated out. each other. Design equations have been derived Replacing the part ABC of Fig. 33.4 by part and the limitations of the design have been (b) of Fig. 33.5 gives an alternative to the lattice pointed out for arbitrary pass-band tolerance. An structure of Fig. 33.3. Whether other alternative alternative canonic realization structure has also structures are possible or not is left as an open been presented, in which a and b are the only problem for the reader. two multipliers. Conclusion Problems A derivation has been given of the elegant P:1. Suppose b = 0 in Eq. 33.3. What kind of second-order band-pass/band-stop ﬁlter transfer ﬁlter do you get? function, in which the two parameters a and b P:2. In Eq. 33.3, investigate what happens when control the centre frequency and the difference (i) b = +1 and (ii) b = –1. between the pass-band edges, independently of 260 33 On Second-Order Digital Band-Pass and Band-Stop Filters P:3. Look at Eq. 33.19. Find cos (x2 + x1) and Reference ﬁnd the product cos (x2 + 1) cos (x2 – x1). Interpret the result. 1. S.K. Mitra, Digital Signal Processing—A Computer P:4. What happens when b = 0 in Eq. 33.27. Based Approach, Second Edition (McGraw-Hill, New P:5. What happens when (i) b = +1 and York, 2000) (ii) b = –1 in Eq. 33.27? Derivation of Second-Order Canonic All-Pass Digital Filter Realizations 34 This chapter deals with the derivation of two 34.3, which realize, respectively, the following canonic all-pass digital ﬁlter realizations, ﬁrst transfer functions: proposed by Mitra and Hirano. In contrast to d1 þ z1 their derivation, which uses a three-pair A1 ðzÞ ¼ ; ð34:1Þ 1 þ d1 z1 approach, our derivation is much simpler because we use a two-pair approach, in which d1 d2 þ d1 z1 þ z2 only four, instead of nine parameters have to A2 ðzÞ ¼ ; ð34:2Þ be chosen. 1 þ d1 z1 þ d1 d2 z2 and Keywords d2 þ d1 z1 þ z2 Canonical All-pass Digital ﬁlter B2 ðzÞ ¼ 1 þ d1 z1 þ d2 z2 ð34:3Þ Realizations Note that the transformed forms of these structures [3] will also give canonic realizations Introduction of the same transfer functions; these are not being considered in this chapter. All-pass digital ﬁlters have been recognized as The derivation of the ﬁrst-order structure was basic building blocks of many digital signal given in [2] by using the two-pair approach, i.e. processors [1]. Any arbitrary order all-pass ﬁlter by assuming a multiplier-less two-pair with the can be realized by cascading ﬁrst- and single multiplier d1 as its termination, as shown second-order ones only. Mitra and Hirano [2] in Fig. 34.4. Using the two-pair relationship proposed the canonic ﬁrst- and second-order conﬁgurations shown in Figs. 34.1, 34.2 and Y1 t11 t12 X1 ¼ ; ð34:4Þ Y2 t21 t22 X2 and the terminating constraint: X2 ¼ d1 Y2 ; ð34:5Þ Source: S. C. Dutta Roy, P. Uday Kiran, Bhargav R. Vyas,Tarun Aggarwal and D. G. Senthil Kumar, “Derivation of Second-Order Canonic All-Pass Digital Filter Realizations”, IETE Journal of Education, vol. 47, pp. 153–157, October–December 2006. © Springer Nature Singapore Pte Ltd. 2018 261 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_34 262 34 Derivation of Second-Order Canonic All-Pass … X1 z –1 –1 d1 Y1 Fig. 34.1 Canonical realization of Eq. 34.1 Fig. 34.2 Canonical 1 1 Y1 X1 z z realization of Eq. 34.2 1 d1 d2 X2 Y2 z 1 Fig. 34.3 Canonical realization of Eq. 34.3 X1 z 1 Y1 1 d1 d2 X1 one obtains Y2 Y1 t11 d1 ðt11 t12 t12 t21 Þ A1 ðzÞ ¼ ¼ : ð34:6Þ X1 1 d1 t22 Digital d1 2-pair Comparing Eq. 34.6 with Eq. 34.1, we see Y1 that various choices are possible, of which the set X2 t11 ¼ z1 ; t22 ¼ z1 ; t12 ¼ 1 þ z1 ; and Fig. 34.4 Digital two-pair terminated in multiplier d1 t21 ¼ 1z1 ð34:7Þ Introduction 263 gives the structure of Fig. 34.1, while inter- Again, obvious choices of t12 and t21 are as changing the expressions of t12 and t21 in follows: Eq. 34.7 gives the transposed form of Fig. 34.1. If we follow the same procedure for deriving t12 ¼ 1z2 ð34:12Þ the structures for Eqs. 34.2 and 34.3, we have to start with a multiplier-less three-pair, two of whose and pairs will be terminated in d1 and d2. As given in [2], analysis for the required t-parameters of the t21 ¼ d2 þ z1 þ d2 z2 : ð34:13Þ 3 3 t-matrix becomes quite involved. The pur- pose of this chapter is to present much simpler From Eqs. 34.4, 34.9, 34.12 and 34.13, we derivations of the structures of Figs. 34.2 and get 34.3, by using the two-pair approach only. Y1 ¼ z2 X1 þ ð1z2 ÞX2 ; ð34:14Þ and Derivation of the Structure of Fig. 34.2 Y2 ¼ ðd2 þ z1 þ d2 z2 ÞX1 ðz1 þ d2 z2 ÞX2 : ð34:15Þ With the aim of deriving the structure of Fig. 34.2, we start with same constrained two-pair shown in Equations 34.14 and 34.15 can be rewritten in Fig. 34.4, where the two-pair is no longer the following forms multiplier-less. Instead, it contains one multiplier (d2) and two delays. Following the steps of Y1 ¼ z2 ðX1 X2 Þ þ X2 ; ð34:16Þ Eqs. 34.4, 34.5 and 34.6, we now have to match the right-hand sides of Eqs. 34.6 and 34.2, i.e. and t11 d1 ðt11 t22 t12 t21 Þ 1 d1 d2 þ d1 z þ z 2 Y2 ¼ z1 ðX1 X2 Þ þ d2 X1 þ z2 ðX1 X2 Þ ¼ ð34:17Þ 1 d1 t22 1 þ d1 z1 þ d1 d2 z2 ð34:8Þ It is easily veriﬁed that Fig. 34.2 is a real- An obvious set of simple choices is the ization of these two equations, where, for con- following: venience, the locations of the signals Y2 and X2 are also indicated. t11 ¼ z2 ; t22 ¼ ðz1 þ d2 z2 Þ; ð34:9Þ t11 t22 t12 t21 ¼ ðd2 þ z1 Þ: Derivation of the Structure From Eq. 34.9, we get of Fig. 34.3 t12 t21 ¼ t11 t22 þ d2 þ z1 In contrast to the derivation of [2], which again ¼ z2 ðz1 þ d2 z2 Þ þ d2 þ z1 ; uses the three-pair approach, ab initio, for ð34:10Þ Eq. 34.3, we derive the structure of Fig. 34.3 from that of Fig. 34.2 by an elementary manip- which can be simpliﬁed to the following: ulation. Let, in Eq. 34.3, d2 = d1 p2. Then Eq. 34.3 becomes identical in form to Eq. 34.2 t12 t21 ¼ ð1z2 Þðd2 þ z1 þ d2 z2 Þ ð34:11Þ with d2 replaced by p2. In the resulting diagram, the part with p2, Y2 and d1 is reproduced in 264 34 Derivation of Second-Order Canonic All-Pass … (a) (b) Then from Eqs. 34.19–34.21, we get d1 z2 þ d1 z1 d1 z2 d1 t12 t21 ¼ ; 1 þ d1 z1 1 þ d1 z1 1 þ d1 z1 P2 d2 ð34:22Þ d1 Y2 which, on simpliﬁcation, gives Fig. 34.5 a Part of modiﬁed Fig. 34.2 and b its equivalent d1 ð1 z2 Þð1 þ d1 z1 þ z2 Þ t12 t21 ¼ 1 þ d1 z1 1 þ d1 z1 Fig. 34.5a. Shifting d1 to the two inputs of ð34:23Þ summing point gives the equivalent diagram of Fig. 34.5b. Replacing the latter in the original Subject to the constraint of Eq. 34.23, various diagram gives the conﬁguration of Fig. 34.3. choices for t12 and t21 are possible. By some trial and error, we have found the following choices to yield a canonic solution: Alternative Derivation of the Structure of Fig. 34.2 d1 ð1 z2 Þ t12 ¼ ; ð34:24Þ 1 þ d1 z1 We now ask the question: If, in Fig. 34.4, we replace d1 by d2, can we get another canonic and realization of Eq. 34.2? In effect, then, we should have ð1 þ d1 z1 þ z2 Þ z2 t21 ¼ 1 ¼ 1þ 1 þ d1 z 1 þ d1 z1 z2 þ d1 z1 þ d1 d2 t11 d2 ðt11 t22 t12 t21 Þ ð34:25Þ ¼ d1 d2 z2 þ d1 z1 þ 1 1 d2 t22 ð34:18Þ We now have the following basic equations: In order to have the two denominators in z2 þ d1 z1 d1 ð1 z2 Þ Y1 ¼ X 1 þ X2 ; ð34:26Þ Eq. 34.18 of the same form, we now divide both 1 þ d1 z1 1 þ d1 z1 the numerator and denominator of the left-hand side of Eq. 34.18 by (1 + d1z−1). Then, we and identify z2 d1 z2 Y2 ¼ 1þ X 1 X2 d1 z2 1 þ d1 z1 1 þ d1 z1 t22 ¼ : ð34:19Þ 1 þ d1 z1 ð34:27Þ Correspondingly, for the modiﬁed numerator, A systematic procedure for obtaining the we get realization diagram is depicted in Fig. 34.6. Part (a) of the ﬁgure shows how t11X1 is obtained with d1 X2 = 0, while part (b) of the same ﬁgure shows t11 t22 t12 t21 ¼ ; ð34:20Þ 1 þ d1 z1 how the same hardware can realize t12X2 with X1 = 0. Superimposing parts (a) and (b), we get and the part (c) with the solid lines, which give the output Y1. To obtain the output Y2, note the value z2 þ d1 z1 t11 ¼ ; ð34:21Þ of the signal at node A, as indicated, and that just 1 þ d1 z1 adding X1 to it gives Y2 according to Eq. 34.26. Alternative Derivation of the Structure of Fig. 34.2 265 (a) X1 X1z -2 - X 2d1z -2 (b) 1 + d1z -1 1 + d1z -1 1 + d1z -1 X1 z 1 t11X1 z 1 t12X2 1 z 1 z 1 1 d1 d1 X2 X1d1z -1 X 2d1 X2 1 + d1z -1 1 + d1z -1 1 + d1z -1 X1z -2 X 2d1z -2 (c) 1 + d1z -1 1 + d1z -1 X1 z 1 A Y1 z 1 1 d1 X2 d2 Y2 Fig. 34.6 Steps in the realization of Eqs. 34.25 and 34.26 Finally, multiplying Y2 by d2 gives X2, as indi- under this constraint, one possible set of choices cated by the broken lines in Fig. 34.6c. This for the t-parameters is the following: conﬁguration is indeed identical with that of Fig. 34.2. t11 ¼ d1 d2 ; t22 ¼ ðd1 þ d1 d2 z1 Þ; t12 Besides the transposed structures of Figs. 34.2 ¼ ð1d1 d2 Þ and t21 ¼ d1 þ ð1 þ d1 d2 Þz1 ; and 34.3, are there other canonical possibilities? ð34:28Þ We leave this as an open question to the reader. and that the resulting realization is the transpose of that of Fig. 34.2. Yet Another Derivation of the Structure of Fig. 34.2 Conclusion Another question that arises at this point is the following: If, in Fig. 34.4, we replace d1 by z−1, In this chapter, we have derived canonic real- and aim for a two-pair containing one delay and izations of second-order all-pass digital ﬁlters by the multipliers d1 and d2, do we get a new a procedure, which is much simpler than that of structure? It is left to the readers to verify that the original three-pair approach of [2]. We have 266 34 Derivation of Second-Order Canonic All-Pass … also derived the realization of Eq. 34.3 from that to Eq. 34.1). Find the overall transfer for Eq. 34.2 by an elementary manipulation of function and ﬁnd its characteristics. Fig. 34.2; this is drastically simpler as compared P:5. Do the same for two second-order ones, to the repetition of the three-pair approach, as having different d1 and d2 (refer to done in [2]. Eq. 34.2). Problems P:1. Write down the transfer function of a References third-order all-pass ﬁlter and draw its structure. 1. P.A. Regalia, S.K. Mitra, P.P. Vaidyanathan, The digital all-pass network: a versatile signal P:2. Do the same for a fourth-order transfer processing building block. Proc. IEEE 76, 19–37 function. (1988) P:3. When two digital 2-points are cascaded, 2. S.K Mitra, K. Hirano, Digital all-pass networks. how do you ﬁnd the overall parameters. Do IEEE Trans. Circ. Sys. 21, 688–700 (September 1974) this in terms of t-parameter. 3. A.V. Oppenheim, R.W. Schafer, J.R. Buck, P:4. Two ﬁrst-order all-pass transfer functions Discrete-Time Signal Processing (Prentice Hall, New are cascaded. They have different d1s (refer Jersey, 2000), p. 363 Derivation of the FIR Lattice Structure 35 A simple derivation is presented for the FIR and then derive the recursion formula for the lattice structure, based on the digital two-pair coefﬁcients of the lower order transfer functions concept. Go ahead, read it and judge for yourself whether it is simple or not! Xi ðzÞ Hi ðzÞ ¼ X0 ðzÞ X i ¼ 1þ aðiÞ n n z ; i ¼ N 1 to 1: Keywords Lattice structure Realization n¼1 ð35:2Þ In the process, one ﬁnds the multipliers as ðiÞ k i ¼ ai ð35:3Þ Introduction and also the relationship In discussing the FIR lattice structure, it is usual in Xi0 ðzÞ textbooks on digital signal processing (see, e.g. Hi0 ðzÞ ¼ ¼ zi Hi ðz1 Þ: ð35:4Þ [1–3]) to assume the conﬁguration of Fig. 35.1a, X0 ðzÞ where each section is of the form shown in Fig. 35.1b, for realizing the transfer function i.e. the two transfer functions Hi(z) and Hi0 (z) are a pair of mirror image polynomials. Speciﬁcally, XN ðzÞ XN with Hi(z) given by Eq. 35.2. HN ðzÞ ¼ ¼ 1þ anðNÞ zn ð35:1Þ X0 ðzÞ n¼1 X i1 ðiÞ Hi0 ðzÞ ¼ zi þ ain zn : ð35:5Þ n¼0 Even Mitra [1], who introduced the concept of digital two-pair [4] and used the same to derive IIR lattice structures, did not use it to derive FIR lattice structure. However, Vaidyanathan [5], a Source: S. C. Dutta Roy, “Derivation of the FIR Lattice former student of Mitra, used this approach to Structure”, IETE Journal of Education, vol. 45, pp. 211– derive a variety of FIR lattice structures for the 212, October–December 2004. © Springer Nature Singapore Pte Ltd. 2018 267 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_35 268 35 Derivation of the FIR Lattice Structure (a) X1 X2 XN 1 XN X0 #1 #2 #N X 1¢ X 2¢ X N¢ - 1 X N¢ (b) Xi Xi 1 ki ki X i¢- 1 X i¢ z 1 Fig. 35.1 a The general cascaded FIR lattice structure. b The ith section of (a) so-called ‘lossless bounded real (LBR)’ transfer 0 Hi ðzÞ ¼ t11 Hi1 ðzÞ þ t12 Hi1 ðzÞ; ð35:9Þ functions. We present here a simple derivation of the Hi0 ðzÞ ¼ t21 Hi1 ðzÞ þ t22 Hi1 0 ðzÞ: ð35:10Þ FIR lattice structure of Fig. 35.1 using the Observe that in Eq. 35.9, t11 has to be unity in two-pair approach, for a general transfer function order to satisfy the requirement of Eq. 35.2 that of the form Eq. 35.1 with the only constraints of the constant term in Hi(z) should be unity. Also, Eqs. 35.2 and 35.4, and no others. for satisfying the requirement that Hi(z) should be a polynomial of order i in z−1, t12 must be of the form kiz−1. Thus, we have Derivation t11 ¼ 1 and t12 ¼ ki z1 : ð35:11Þ Consider the ith stage of the FIR lattice, shown in Fig. 35.1b, and let it be characterized by the We next put the constraint of Eq. 35.4 and transmission matrix get, from Eqs. 35.9 and 35.11, " # ðiÞ ðiÞ t11 ðiÞ t12 Hi0 ðzÞ ¼ zi Hi ðz1 Þ T ¼ ðiÞ ðiÞ : ð35:6Þ t21 t22 ¼ zi Hi1 ðz1 Þ þ ki zHi1 0 ðz1 Þ ¼ zi Hi1 ðz1 Þ þ ki zði1Þ Hi1 0 ðz1 Þ For simplicity, we shall drop the superscript (i) in the following discussion. Equation 35.6 ¼ z1 zði1Þ Hi1 ðz1 Þ þ ki zði1Þ Hi1 0 ðz1 Þ implies that ¼ ki Hi1 ðzÞ þ z1 Hi1 0 ðzÞ 0 Xi ðzÞ ¼ t11 Xi1 ðzÞ þ t12 Xi1 ðzÞ ð35:7Þ ð35:12Þ Comparing the last line of Eq. 35.12 with Xi0 ðzÞ ¼ t21 Xi1 ðzÞ þ t22 Xi1 0 ðzÞ: ð35:8Þ Eq. 35.10, we observe that In terms of the transfer functions Hi(z) and t21 ¼ ki and t22 ¼ z1 ð35:13Þ Hi0 ðzÞ; Eqs. 35.7 and 35.8 translate to the following: Derivation 269 The structure resulting from Eqs. 35.11 and Problems 35.13 is precisely that of Fig. 35.1b. To obtain the coefﬁcients of Hi − 1(z) from P:1. What happens if the lower bound slanting those of Hi(z), one follows the same procedure as arrows in Fig. 35.1b points in the opposite in [1]. The result is direction? What if the upper bound slanting arrow points in the opposite direction? ðiÞ ðiÞ an ki ain What if both do the same? aði1Þ n ¼ ; n ¼ i 1 to 1; 1 ki2 P:2. Suppose all arrows in Fig. 35.1a are reversed. What kind of transfer function do i ¼ N 1 to 1 we get? ð35:14Þ P:3. Suppose x1, x1′ arrows point in the opposite direction. What kind of transfer function would you get? Concluding Comments P:4. Re-derive the equations in terms of trans- mission parameters. A simple derivation has been presented for the P:5. Besides t- and transmission parameters, FIR lattice structure on the basis of the constraint what other parameter are meaningful in the that the two transfer functions obtained at the context of digital two-pairs? How are they output of any section bear a mirror image rela- related t- and transmission parameters? tionship to each other. It is not difﬁcult to appreciate that other lattice structures are possible to derive by assuming some other relationship between Hi(z) and Hi0 ðzÞ, e.g. References Hi0 ðzÞ ¼ zi Hi ðz1 Þ: ð35:15Þ 1. S.K. Mitra, Digital Signal Processing—A Computer-Based Approach (McGrawHill, New York, These structures, when carefully derived, differ 2001) 2. A.V. Oppenheim, R.W. Schafer, Discrete-Time Signal from those of Fig. 35.1b in one or more of the Processing, Englewood Cliffs (Prentice Hall, NJ, following aspects: the position of the delay branch; 1989) positions of the multipliers; relative signs of the two 3. J.G. Proakis, D.G. Manolakis, Introduction to Digital multipliers, i.e. one multiplier may be the negative Signal Processing (Macmillan, New York, 1989) 4. S.K. Mitra, R.J. Sherwood, Digital ladder networks. of the other; and nonuniformity of the signs of the IEEE Trans. Audio Electroacoust. AU-21, 30–36 multipliers from one section to the next. Examples (February 1973) of such structures can be found in [5]. Detailed 5. P.P. Vaidyanathan, Passive cascaded lattice structures derivation of these structures for the general transfer for low-sensitivity FIR design, with applications to ﬁlter banks. IEEE Trans. Circ. Syst. CAS-33, 1045– function of Eq. 35.1 and their recurrence relations 1064 (November 1986) will be presented in a later chapter. Solution to a Problem in FIR Lattice Synthesis 36 In FIR lattice synthesis, if at any but the last which is known to be realizable by the lattice stage, a lattice parameter becomes ±1, then structure of Fig. 36.1a, where each ki block has the synthesis fails. A linear phase transfer the structure shown in Fig. 36.1b, provided that function is an example of this situation. This at no stage in the synthesis procedure except the chapter, written in a tutorial style, is con- last one, one encounters a parameter ki = ±1. For cerned with a simple solution to this problem, example, if hN(N) = ±1, one cannot obtain a demonstrated through simple examples, rather lattice by the usual procedure. This problem was than detailed mathematical analysis, some of not adequately addressed to in the literature (see which is available in (Dutta Roy in IEE e.g. [3–5]), and was solved in [1] for the linear Proc-Vis Image Signal Process 147:549–552, phase case with rigorous mathematical analysis 2000 [1]). and proofs. However, the solution for the non- linear phase case with hN(N) = ±1 was given in [1] in terms of parallel lattices, which, in general, Keywords is neither delay canonic nor multiplier canonic. FIR ﬁlters Lattice synthesis In this chapter, written in a tutorial style, we present the essence of [1] through several simple examples for easy comprehension by the stu- dents. We also give a canonic solution to the nonlinear phase case through a tapped lattice Introduction structure. Note that although two multipliers have Consider the FIR transfer function been shown in Fig. 36.1b, each lattice section X N should also be realized by a single multiplier HN ðzÞ ¼ 1 þ hN ðnÞzn ð36:1Þ structure. Unfortunately, however, such a struc- n¼1 ture does not exist as yet. However, we shall refer to the structure of Fig. 36.1 as multiplier canonic, even if we show two multipliers in each lattice section. Source: S. C. Dutta Roy, ‘Solution to a Problem in FIR Lattice Synthesis’, IETE Journal of Education, vol. 43, pp. 33–36, January–March 2002 (Corrections on p. 219, October–December 2002). © Springer Nature Singapore Pte Ltd. 2018 271 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_36 272 36 Solution to a Problem in FIR Lattice Synthesis (a) x(n) y(n) 0 1 2 N 1 N 0.38889 0.38462 0.2 x(n) y (n) k1 k1 kN Fig. 36.2 Lattice structure for the transfer function given (b) i 1 i by Eq. 36.6 i 1 i ki ki ki Example 1 Let z 1 H3 ðzÞ ¼ 1 þ 0:5 z1 þ 0:3 z2 0:2 z3 ð36:6Þ Fig. 36.1 a General FIR lattice structure b Composition of the ith block in (a) Here, h3(1) = 0.5, h3(2) = 0.3 and h3(3) = −0.2. Hence k3 = h3(3) = −0.2. From Eq. 36.5, Conventional Synthesis Procedure h2 ð1Þ ¼ ½h3 ð1Þ k3 h3 ð2Þ=ð1 k32 Þ ¼ 0:53846 ð36:7Þ We ﬁrst review the conventional FIR lattice synthesis procedure [2]. By analysis of Fig. 36.1, and it is easily shown that if Hi(z) is the transfer function from node 0 to node i, then h2 ð2Þ ¼ ½h3 ð2Þ k3 h3 ð1Þ=ð1 k32 Þ ¼ 0:38462 ð36:8Þ Hi ðzÞ ¼ Hi1 ðzÞ þ ki zi Hi1 ðz1 Þ; i ¼ 1 ! N ð36:2Þ Hence, Also, Hi(z) is of the form H2 ðzÞ ¼ 1 þ 0:53846 z1 þ 0:38462 z2 ð36:9Þ X i so that k2 = 0.38462. Apply Eq. 36.5 again to get Hi ðzÞ ¼ 1 þ hi ðnÞzn ð36:3Þ n¼1 h1 ð1Þ ¼ ½h2 ð1Þ k2 h2 ð1Þ=ð1 k22 Þ ð36:10Þ The lattice parameters (ki) of Fig. 36.1 are ¼ h2 ð1Þ=ð1 þ k2 Þ ¼ 0:38889 given by Thus ki ¼ hi ðiÞ; i ¼ 1 ! N ð36:4Þ H1 ðzÞ ¼ 1 þ 0:38889 z1 ð36:11Þ To obtain Hi−1(z) from Hi(z), i = N ! 2, one giving k1 = 0.38889. The synthesis is thereby uses the following recursion formula: complete and the resulting structure is shown in Fig. 36.2. hi1 ðnÞ ¼½hi ðnÞ ki hi ði nÞ=ð1 ki2 Þ i¼N!2 ð36:5Þ Linear Phase Transfer Function Obviously, if ki = ±1, then the synthesis fails! As is well known, there are four different types of We now illustrate the conventional procedure linear phase transfer functions, viz. (1) symmet- by an example. rical impulse response of even length; Linear Phase Transfer Function 273 (2) symmetrical impulse response of odd length; Nth-order case, N odd, which will require (N−1)/2 (3) asymmetrical impulse response of even non-trivial lattice sections to begin with, followed length; and (4) asymmetrical impulse response of by the same number of simple delays, and ending odd length. We shall consider each of these cases in one unity parameter lattice section. through simple examples. Example 3: Illustrating case 2 Example 2: Illustrating Case 1 Let Let 4 ðzÞ ¼ 1 þ h4 ð1Þz1 þ h4 ð2Þz2 þ h4 ð1Þz3 þ z4 1 2 3 H5 ðzÞ ¼1 þ h5 ð1Þz þ h5 ð2Þz þ h5 ð2Þz ð36:16Þ 4 5 þ h5 ð1Þz þz Here we have k4 = 1; also Eq. 36.16 can be ð36:12Þ rewritten as Here k5 = 1. Note that Eq. 36.12 can be rewritten as H4 ðzÞ ¼ 1 þ h4 ð1Þz1 þ ð1=2Þh4 ð2Þz2 H5 ðzÞ ¼ 1 þ h5 ð1Þz1 þ h5 ð2Þz2 þ ð1Þz4 1 þ h4 ð1Þz þ ð1=2Þh4 ð2Þz2 þ ð1Þz5 1 þ h5 ð1Þz þ h5 ð2Þz2 ð36:17Þ ð36:13Þ Combining this with Eq. 36.2 with i = 4, we get Comparing this with Eq. 36.2 with i = 5, we H3 ðzÞ ¼ 1 þ h4 ð1Þz1 þ ð1=2Þh4 ð2Þz2 ¼ H2 ðzÞ note that ð36:18Þ H4 ðzÞ ¼ 1 þ h5 ð1Þz1 þ h5 ð2Þz2 ¼ H2 ðzÞ which implies that k3 = 0 and k2 = (1/2)h4(2). ð36:14Þ Finally, as in Eq. 36.15, we have because the order of the polynomial is 2. This order k1 ¼ h1 ð1Þ ¼ h2 ð1Þ=ð1 þ k2 Þ reduction means that k4 = k3 = 0. Also k2 = h5(2), ð36:19Þ and by the formula in Eq. 36.10, we get ¼ h4 ð1Þ=½1 þ ð1=2Þh4 ð2Þ k1 ¼ h1 ð1Þ ¼ h2 ð1Þ=ð1 þ k2 Þ The resulting lattice is shown in Fig. 36.4. In ¼ h5 ð1Þ=½1 þ h5 ð2Þ ð36:15Þ general, for an Nth-order transfer function with N even, we shall require N/2 non-trivial lattice sections, (N/2)−1 simple delays, and a unity The synthesis is now complete and the result- parameter lattice section. ing structure is shown in Fig. 36.3. Note that only two non-trivial lattice parameters are needed for Example 4: Illustrating case 3 the synthesis of a ﬁfth-order transfer function; this Let is what it should be, because there are only two independent parameters in the transfer function H5 ðzÞ ¼ 1 þ h5 ð1Þz1 þ h5 ð2Þz2 h5 ð2Þz3 36.12. This can be easily generalized to the h5 ð1Þz4 z5 ð36:20Þ x(n) y(n) y(n) k1 k2 k5 = 1 x(n) k1 k2 k4 = 1 z 1 z 1 z 1 Fig. 36.3 Synthesis of Eq. 36.12 Fig. 36.4 Synthesis of Eq. 36.16 274 36 Solution to a Problem in FIR Lattice Synthesis x(n) y(n) k1 = h4(1) k4 = 1 z 1 z 1 Fig. 36.5 Synthesis of Eq. 36.22 Here k5 = −1 and we can rewrite Eq. 36.20 as Nonlinear Phase FIR Function H5 ðzÞ ¼ 1 þ h5 ð1Þz1 þ h5 ð2Þz2 with hN(N) = –1 þ ð1Þz5 1 þ h5 ð1Þz þ h5 ð2Þz2 Let ð36:21Þ X N 1 Comparing with Example 2, we see that the HN ðzÞ ¼ 1 þ hN ðnÞzn zN ð36:25Þ lower order polynomial is the same in this case n¼1 also. Hence, the realization of Fig. 36.2 is valid for this case also, except that the last lattice where N may be even or odd. We ﬁrst consider section will have k5 = −1. the case of even N and illustrate our new pro- cedure with examples. Example 5: Illustrating case 4 Example 6 Let Let H4 ðzÞ ¼ 1 þ h4 ð1Þz1 h4 ð1Þz3 z4 ð36:22Þ H4 ðzÞ ¼ 1 þ h4 ð1Þz1 þ h4 ð2Þz2 þ h4 ð3Þz3 þ z4 ð36:26Þ Note that because of asymmetry, h4(2) is identi- cally zero. Here, we have k4 = −1 and we can write This can be decomposed as follows: 1 H4 ðzÞ ¼ 1 þ h4 ð1Þz þ ð1Þz4 ½1 þ h4 ð1Þz H4 ðzÞ ¼ 1 þ h4 ð3Þz1 þ h4 ð2Þz2 þ h4 ð3Þz3 þ z4 ð36:23Þ þ ½h4 ð1Þ h4 ð3Þz1 Thus, ð36:27Þ H3 ðzÞ ¼ 1 þ h4 ð1Þz1 ¼ H1 ðzÞ ð36:24Þ The ﬁrst transfer function (within square brackets) is linear phase and is the same as i.e. k3 = k2 = 0 and k1 = h4(1). The synthesis is, Eq. 36.16 with h4(1) replaced by h4(3). The therefore, complete and the resulting structure is second transfer function in Eq. 36.27 (within shown in Fig. 36.5. In general, for N even and curly brackets) can be realized by tapping the k1 asymmetric impulse response, there will be (N/2) block after the delay z−1, multiplying it by [h4(1) −1 non-trivial lattice sections, N/2 simple delays −h4(3)], and adding it to the main output, as and a last lattice section with the parameter −1. shown in Fig. 36.6. Nonlinear Phase FIR Function with hN(N) = ±1 275 Fig. 36.6 Synthesis of x(n) Eq. 36.26: k1 ¼ h4 ð3Þ=½1 þ ð1=2Þh4 ð2Þ k1 y(n) k2 k4 = 1 and k2 ¼ ð1=2Þh4 ð2Þ k1 z 1 z 1 [h4(1) h4 (3)] Note that if each lattice can be realized by a H4 ðzÞ ¼ 1 h4 ð3Þz1 þ h4 ð3Þz3 z4 single multiplier structure; then the realization of Fig. 36.6 will be a delay, as well as multiplier þ ½h4 ð3Þ þ h4 ð1Þz1 þ h4 ð2Þz2 canonic. Unfortunately, however, such a struc- ð36:29Þ ture does not exist; this is still an unsolved problem. For higher order transfer functions, one The ﬁrst of these transfer functions is linear would require more tappings at the end of delays phase with k4 = −1, k3 = k2 = 0 and k1 = −h4(3). and the multiplier coefﬁcients have to be appro- The second transfer function is realized by taking priately chosen. The next example illustrates tappings after the ﬁrst and second delays, as both of these points, although the order of the shown in Fig. 36.7. The three outputs are then transfer function is the same. combined. The multipliers a and b are found from the following equation: Example 7 Let z1 a þ k1 þ z1 z1 b ¼ ½h4 ð3Þ þ h4 ð1Þz1 þ h4 ð2Þz2 ð36:30Þ H4 ðzÞ ¼ 1 þ h4 ð1Þz1 þ h4 ð2Þz2 þ h4 ð3Þz3 z4 ð36:28Þ This gives Here k4 = −1 and the necessary decomposition is b ¼ h4 ð2Þ and a ¼ h4 ð3Þ þ h4 ð1Þ h4 ð2Þh4 ð3Þ ð36:31Þ as follows: Fig. 36.7 Synthesis of x(n) Eq. 36.28: a and b are given by Eq. 36.31 k1 S y(n) k4 = 1 k1= h4 (3) z 1 1 1 z z b a S 276 36 Solution to a Problem in FIR Lattice Synthesis x(n) The ﬁrst transfer function is linear phase, k1 k2 antisymmetrical and can be realized by the pro- k1 k2 k5 = 1 cedure already illustrated. We shall have z 1 z 1 z 1 z 1 k5 ¼ 1; k4 ¼ k3 ¼ 0; k2 ¼ h5 ð3Þ and a b k1 ¼ h5 ð4Þ=½1 h5 ð3Þ S S ð36:36Þ y(n) The second transfer function in Eq. 36.35 is Fig. 36.8 Synthesis of Eq. 36.33: k1 ¼ h5 ð4Þ= realized by tappings after the ﬁrst and second ½1 þ h5 ð3Þ; k2 ¼ h5 ð3Þ; a ¼ h5 ð1Þ h5 ð4Þ bk1 and b ¼ h5 ð2Þ þ h5 ð3Þ delays. The ﬁnal realization is the same as that shown in Fig. 36.8 with k values given by Eq. 36.36 and Example 8 Let a ¼ h5 ð1Þ þ h5 ð4Þ bk1 and ð36:37Þ b ¼ h5 ð2Þ þ h5 ð3Þ H5 ðzÞ ¼1 þ h5 ð1Þz1 þ h5 ð2Þz2 ð36:32Þ þ h5 ð3Þz3 þ h5 ð4Þz4 þ z5 Here k5 = 1 and we can rewrite Eq. 36.32 as Conclusion H5 ðzÞ ¼ 1 þ h5 ð4Þz1 þ h5 ð3Þz2 þ h5 ð3Þz3 þ h5 ð4Þz4 þ z5 We have demonstrated, through simple exam- þ ½h5 ð1Þ h5 ð4Þz1 þ ½h5 ð2Þ h5 ð3Þz2 ples, how an FIR lattice transfer function P ð36:33Þ HN ðzÞ ¼ 1 þ N1 n¼1 hN ðnÞz n zN can be realized for both linear and nonlinear phase cases The ﬁrst transfer function has already been with canonic delays and multipliers. The proce- realized in Example 2, except that here h5(4) dure is expected to be useful to students and takes the place of h5(1) and h5(3) takes the place teachers, as well as designers concerned with of h5(2). The second transfer function in digital signal processing. Eq. 36.33 can be realized by tapping the signals after the ﬁrst and the second delays. The proce- dure is the same as in Example 7 and is not repeated here. The ﬁnal result is shown in Problems Fig. 36.8. P:1. What happens in Fig. 36.1b if one k1 is in the reverse direction? Example 9 P:2. Obtain a lattice structure for Let H5 ðzÞ ¼1 þ h5 ð1Þz1 þ h5 ð2Þz2 ð36:34Þ þ h5 ð3Þz3 þ h5 ð4Þz4 z5 H0 ðzÞ ¼ 1 þ 0:5 z1 þ 0:3 z2 þ 0:5 z3 þ z4 The necessary decomposition is as follows: P:3. Obtain a lattice if z−5 in Eq. 36.12 has a H5 ðzÞ ¼ 1 h5 ð4Þz1 h5 ð3Þz2 þ h5 ð3Þz3 þ h5 ð4Þz4 z 5 negative sign. þ ½h5 ð1Þ þ h5 ð4Þz1 þ ½h5 ð2Þ þ h5 ð3Þz2 P:4. Obtain a lattice for Eq. 36.16 with ð36:35Þ h4(2) = 0. Problems 277 P:5. Can you decompose Eq. 36.26 in any 2. A.V. Oppenheim, R.W. Schafer, Discrete Time Signal fashion other than Eq. 36.27? Give as many Processing (Prentice Hall, New Jersey, 1989) 3. M. Bellanger, Digital Processing of Signals (Wiley, as possible if you can. Hoboken, 1984) 4. J.G. Proakis, D.G. Manolakis, Digital Signal Process- ing (McMillan, Basingstoke, 1992) 5. L.B. Jackson, Digital Filters and Signal Processing (Kluwer, Alphen aan den Rijn, 1989) References 1. S.C. Dutta Roy, Synthesis of FIR lattice structures. IEE Proc-Vis Image Signal Process 147, 549–552 (2000) FIR Lattice Structures with Single-Multiplier Sections 37 An alternative derivation is given for the linear This structure uses two identical multipliers in prediction FIR lattice structures with each section, and is attributed to Itakura and single-multiplier sections. As compared to the Saito [1]. Most textbooks on Digital Signal previous approaches, this method is believed to Processing (DSP) refer to this structure and its be conceptually simpler and more straightfor- variations in details. They assume, rather than ward. derive, the structure and then analyze it for ﬁnding the transfer function as well as some recurrence formulas. A simple derivation of the Keywords structure has recently been given in [2], and an FIR lattice Single-multiplier realization alternative lattice structure in which the two transfer functions HN(z) and HN′(z) are comple- mentary to each other, i.e. HN′(z) = z– N HN(−z−1), has been given in [3]. Introduction There exists a corresponding lattice structure for all-pass IIR transfer functions, which also The linear prediction FIR lattice ﬁlter, shown in uses two multipliers per section. This structure Fig. 37.1, realizes the transfer function has been derived by Mitra by using his two-pair XN or multiplier extraction approach [4]. Using the YðzÞ HN ðzÞ ¼ ¼ 1þ hn zn ð37:1Þ same approach, Mitra also derived a modiﬁed IIR XðzÞ n¼1 lattice structure which uses a single multiplier per section, and is therefore canonic in multipliers. and its mirror image transfer function As it is well known, the basic all-pass IIR structure can be used with additional multipliers Y 0 ðzÞ HN0 ðzÞ ¼ ¼ zN HN ðz1 Þ: ð37:2Þ and summers to realize any arbitrary IIR transfer XðzÞ function [5]. A question arises as to whether a single-multiplier structure is possible for the FIR lattice also. The answer was given by Makhoul, as early as 1978 [6]. He derived two structures, called LF2 and LF3, each section of which Source: S. C. Dutta Roy, “FIR Lattice Structures with Single-Multiplier Sections,” IETE Journal of Education, contains three multipliers, and then converted vol. 47, pp. 119–122, July–September 2006. them to single-multiplier ones by a clever © Springer Nature Singapore Pte Ltd. 2018 279 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_37 280 37 FIR Lattice Structures with Single-Multiplier Sections Fig. 37.1 The basic f0(n) f1(n) f2(n) fN (n) two-multiplier structure of [1] = y(n) k1 k2 kN x(n) = f(n) z 1 k1 z 1 k2 z 1 kN g0(n) g1(n) g2(n) gN(n) = y ¢(n) manipulation. Nine years after the appearance of single-multiplier realizations, we require one or [6], Dognata and Vaidyanathan [7] gave an more additional variables involving a linear alternative derivation of the single-multiplier FIR combination of X1 and z−1 X2 without any mul- lattice by the multiplier extraction approach, and tiplication. The possibilities for a third variable another 2 years later, Krishna [8] indicated that are X1 + z−1 X2; X1 − z1 X2; −X1 + z−1 X2; and the same structure could also be arrived at by the −X1 − z−1 X2. We shall now investigate some of eigen-decomposition approach. these cases. Surprisingly, no mention could be found of Makhoul’s work in any of the large number of textbooks scanned by the author. The purpose of Realization 1 this chapter is, ﬁrst, to bring this ﬁne piece of work to the attention of teachers and students of In terms of the variable set (X1, z−1 X2, X1 + z−1 DSP, and, second, to present an alternative, class X2), Eqs. 37.4 and 37.5 can be rewritten as tested and conceptually simpler procedure for deriving Makhoul’s single-multiplier structures. Y1 ¼ ð1km ÞX1 þ km ðX1 þ z1 X2 Þ ð37:6Þ and Derivation Y2 ¼ ð1km Þz1 X2 þ km ðX1 þ z1 X2 Þ: ð37:7Þ Consider the mth section of the two-multiplier structure, shown in Fig. 37.2 with signals trans- Dividing both sides of Eqs. 37.6 and 37.7 by formed to the z-domain. Also, for simplicity, let (1 − km), we get Y1 km Fm1 ðzÞ ¼ X1 ðzÞ; Gm1 ðzÞ ¼ X2 ðzÞ; ð37:3aÞ ¼ X1 þ ðX1 þ z1 X2 Þ; ð37:8Þ 1 km 1 km and Fm ðzÞ ¼ Y1 ðzÞ; and Gm ðzÞ ¼ Y2 ðzÞ; ð37:3bÞ Fm -1 (z )[= X1 ] Fm (z )[ = Y1] By inspection of Fig. 37.2, we get km 1 Y1 ¼ X1 þ km z X2 ; ð37:4Þ and 1 km z 1 Y2 ¼ km X1 þ z X2 : ð37:5Þ Gm -1(z )[ = X 2 ] Gm (z )[ = Y2 ] The input variables occurring in Eqs. 37.4 and 37.5 are X1 and z−1 X2. Clearly, in order to obtain Fig. 37.2 The mth section of Fig. 37.1: Is m = 1? Derivation 281 and Since the procedure for other realizations is similar, we shall, for brevity, only give the main Y2 km ¼ z1 X2 þ ðX1 þ z1 X2 Þ ð37:9Þ results for three other cases. 1 km 1 km The resulting structure involves only one Realization 2 multiplier, as shown in Fig. 37.3a. However, each output is now scaled by the factor Taking the variable set as (X1, z−1 X2, X1 − z−1 1/(1 − km). Y1 and Y2 can, of course, be recov- X2), Eqs. 37.4 and 37.5 can be rewritten as ered by multiplying the outputs by the factor (1 − km). If each stage of Fig. 37.1 is thus con- Y1 ¼ ð1 þ km ÞX1 km ðX1 z1 X2 Þ; ð37:10Þ verted into a single-multiplier one, then all output multipliers can be clubbed into a single multiplier and QN of value ð1 km Þ at the input of the overall m¼1 Y2 ¼ ð1 þ km Þz1 X2 þ km ðX1 z1 X2 Þ: ð37:11Þ lattice, thus reducing the total number of multi- pliers from 2N to N + l. Note that instead of Single-multiplier realization is obtained by lumping the multipliers at the input, one can also dividing both sides of Eqs. 37.10 and 37.11 by distribute them appropriately in order to prevent (1 + km), and is shown in Fig. 37.3b which is the overflow and/or minimize quantization errors. same as LF2/l(b) of [6]. The multiplier needed at The total number of multipliers may still be Q N the input of the lattice is ð1 þ km Þ in this case. much less than 2N, as required in the lattice of m¼1 Fig. 37.1. One should also keep in mind that considerations of overflow and quantization errors may dictate the use of additional Realization 3 lumped/distributed scaling in the structure of Fig. 37.1 too. Also, observe that Fig. 37.3a is the For the same variable set as in realization 1, same as the one-multiplier lattice form LF2/l(a) Eqs. 37.4 and 37.5 can also be rewritten as of Makhoul [6]. follows: (a) (b) Y1/(1+km) Y1/(1 km ) X1 X1 1 1 km/(1 km ) km/(1 + km) 1 z 1 z X2 X2 Y2/(1 km ) Y2/(1 + km) Y1/(km 1) (d) Y1/(km + 1) X1 (c) X1 1/(km 1) 1/(km + 1) 1 z 1 1 1 z X2 X2 Y2/(km – 1) Y2/(km + 1) Fig. 37.3 Four single-multiplier realizations of the lattice section of Fig. 37.2 282 37 FIR Lattice Structures with Single-Multiplier Sections Y1 ¼ ðX1 þ z 1 X2 Þ þ ðkm lÞz1 X2 ; ð37:12Þ be written or in the new editions of the books which exist. It is also hoped that the conceptually and simpler approach presented here for deriving the single-multiplier FIR lattice would appeal to Y2 ¼ ðX1 þ z1 X2 Þ þ ðkm 1Þz1 X1 : ð37:13Þ students and teachers of DSP. The corresponding single-multiplier realiza- tion is obtained by dividing both sides of Problems Eqs. 37.12 and 37.13 by (km − 1) and is shown in Fig. 37.3c, which is the same as LF3/l(a) of P:1. In Fig. 37.1, if all arrows are reversed, what [6]. The multiplier needed at the input N of the kind of transfer function is obtained? Q N P:2. What if only the lower line arrows are lattice is ðkm 1Þ: m¼1 reversed in Fig. 37.1? P:3. Write the equations in Fig. 37.2 with only the lattice arrows reversed. Realization 4 P:4. In Fig. 37.3a and b, again reverse all the arrows, and comment on the transfer func- Taking the same variable set as in realization 2, tion so obtained. we can also modify Eqs. 37.4 and 37.5 as P:5. Do the same for Fig. 37.3c and d. follows: Y1 ¼ ðX1 z1 X2 Þ þ ð1 þ km Þz1 X2 ; ð37:14Þ Acknowledgments The author thanks R. Vishwanath for helpful discussions. and Y2 ¼ ðX1 z 1 X2 Þ þ ð1 þ km ÞX1 : ð37:15Þ References Single-multiplier realization is obtained 1. F. Itakura, S. Saito, Digital Filtering Techniques for by dividing both sides of Eqs. 37.14 and 37.15 Speech Analysis and Synthesis, in Proc 7th lnt Cong Acoust, (Budapest, Hungary, 1971) pp. 261–264 by (1 + km), and is shown in Fig. 37.3d, which is 2. S.C. Dutta Roy, R. Vishwanath, Derivation of the FIR the same as LF3/l(b) of [6]. The multiplier lattice. IETE J. Educ. 45, 211–212 (October–Decem- needed at the input of the lattice is now ber 2004) Q N 3. S.C. Dutta Roy, R. Vishwanath, Another FIR lattice ð1 þ km Þ: structure, Int. J. Circ. Theor. Appl. 33, 347–351, m¼1 (July–August 2005) As can be easily veriﬁed, the other input 4. S.K. Mitra, Digital Signal Processing: A Computer variable sets like (X1, z−1 X2, −X1 ± z−1 X2), Based Approach (McGrawHill, New York, 2001) 5. A.H. Gray, J.D. Markel, Digital lattice and ladder ﬁlter (−X1, z−1 X2, −X1 ± z−1 X2), etc., give simple synthesis. IEEE Trans. Audio Electroacoust. AU-21, variations of the four structures shown in 491–500 (December 1973) Fig. 37.3. 6. J. Makhoul, A class of all-zero lattice digital ﬁlters: properties and applications. IEEE Trans. Acoust. Speech Sig. Process. ASSP-26, 304–314, (August 1978) Conclusion 7. Z. Doganata, P.P. Vaidyanathan, On one-multiplier implementations of FIR lattice structures. IEEE Trans. As indicated in the introduction, Makhoul’s work Circ. Syst. CAS-34, 1608–1609 (December 1987) 8. H. Krishna, An eigen-decomposition approach to [6] has not received adequate recognition in one-multiplier realizations of FIR lattice structures. textbooks on DSP. It is hoped that this chapter IEEE Trans. Circ. Syst. CAS-36, 145–146, (January will facilitate inclusion of this work in books to 1989) A Note on the FFT 38 This chapter gives a formula for the exact predicted by the N log2 N formula. As another number of non-trival multipliers required in example, the FFT diagrams for a 32-point the basic N-point FFT algorithms, where N is sequence, in both DIT and DIF forms show the an integral power of 2. Now proceed further, number of non-trivial multipliers to be 49, but not too far! instead of 160, as predicted by the N log2 N for- mula. This reduction is effected by using the butterfly simpliﬁcation and the facts that WN0 ¼ 1 Keywords N=2 and WN ¼ 1; where WN = exp (−j2p/N). It FFT Computation Number of multipliers is, therefore, of interest to ﬁnd out the actual number of non-trivial multipliers needed in a general N-point FFT, where N = 2q, q being a positive integer. This note gives a formula for this purpose. Introduction In the usual presentation of the Fast Fourier Transform (FFT) in textbooks, it is mentioned Derivation of the Formula that the basic N-point FFT algorithms reduce the number of multipliers from N2 to N log2 N ([1], Consider the DIT algorithm of an N-point FFT, p 287), if N is a power of 2. If one looks at the incorporating butterfly simpliﬁcation. The last actual FFT diagrams, either decimation-in-time (qth) stage of the computation will have N/2 (DIT) or decimation-in-frequency (DIF), of an multipliers, of which WN0 is trivial and the others 8-point sequence, as shown in Fig. 38.1a and b, ðN=2Þ1 are WN1 ; WN2 ; . . .; WN : Thus the number of respectively, one ﬁnds that the actual number of multipliers at this stage is [(N/2) −1]. The pre- non-trivial multipliers is 5, instead of 24, as ceding stage [(q−1)th] will have two groups of multipliers, each having (N/4) members. In each group, there will be a WN0 multiplier. Hence, the number of multipliers at the [(q−1)th] stage is 2[(N/4)−1]. Similarly, at the (q−2)th stage, the Source: S. C. Dutta Roy, “A Note on the FFT,” IETE Journal of Education, vol. 46, pp. 61–63, April–June 2005. © Springer Nature Singapore Pte Ltd. 2018 283 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2_38 284 38 A Note on the FFT (a) x(0) X(0) x(4) X(1) –1 x(2) X(2) –1 W8 2 x(6) X(3) –1 –1 x(1) X(4) –1 W81 x(5) X(5) –1 –1 W8 2 x(3) X(6) –1 –1 W8 2 W83 x(7) X(7) –1 –1 –1 (b) x(0) X(0) x(1) X(4) –1 x(2) X(2) –1 W8 2 x(3) X(6) –1 –1 x(4) X(1) –1 W81 x(5) X(5) –1 –1 W8 2 x(6) X(3) –1 –1 3 2 W8 W8 x(7) X(7) –1 –1 –1 Fig. 38.1 a Decimation-in-time FFT flow diagram for a 8-point sequence, showing only the non-trivial multipliers; b Decimation-in-frequency FFT flow diagram for a 8-point sequence, again showing only the non-trivial multipliers number of multipliers is 4[(N/8) −1] and so on, second stage contributes to (N/4) multipliers. till we reach the second stage where there are The ﬁrst stage has (N/2) multipliers, each of (N/4) groups of multipliers, with two multipliers value WN0 : Thus, the total number of non-trivial in each group, one of them being WN0 : Hence the multipliers in an N-point FFT becomes Derivation of the Formula 285 N N N Recurrence Relation M ðN Þ ¼20 1 þ 21 1 þ 22 1 2 4 8 N A recurrence formula for M(N) can be derived as þ þ ð2 1Þ 4 follows. For a 2N-point FFT, Eq. 28.2 gives " " " " qth stage jðq1Þth stage j ðq2Þth stagej2nd stage M ð2N Þ ¼N ½log2 ð2N Þ2 þ 1 N N N ¼ 1 þ 2 þ 4 ¼N ½log2 2 þ log2 N2 þ 1 2 2 2 ð38:3Þ N N ¼N ðI þ log2 N2Þ þ 1 þ þ 2 4 ¼N ðlog2 N1Þ þ 1: N N ¼ ðq 1Þ 1 þ 2 þ 4 þ þ 2 4 N Also, ¼ ðlog2 N 1Þ 1 þ 2 þ 2 þ þ 2q2 2 2 N 1 2q1 2M ðN Þ ¼ N ðlog2 N 2Þ þ 2: ¼ ðlog2 N 1Þ 2 1 2 N N From Eqs. 38.3 and 38.4, we get ¼ ðlog2 N 1Þ þ 1 : 2 2 ð38:1Þ M ð2N Þ ¼ 2M ðN Þ þ N1; Finally, therefore, which is the required recurrence formula. N M ðN Þ ¼ ðlog2 N 2Þ þ 1: ð38:2Þ 2 Alternative Derivation for M(N) The same number of multipliers arises in DIF An alternative derivation of the formula for also, by recognizing that the index of stages will M(N) follows by noting that the actual number of be reversed in Eq. 38.1. Table 38.1 shows a multipliers after using the butterfly simpliﬁcation comparison of the actual number of non-trivial is (N/2) log2 N, in which the number of WN0 multipliers M(N) and the number predicted by the N log2 N formula. multipliers is, using the DIT, 1 at the qth stage; 2 at theðq1Þth stage; Table 38.1 Number of multipliers in FFT 4 at the ðq2Þth stage; N M(N) N log2 N . . .. . .. . .. . .. . .. . .. . .. . .; 2 0 2 ðN=4Þ at the 2nd stage; and 4 1 8 ðN=2Þ at the 1st stage: 8 5 24 16 17 64 Hence, the total number of WN0 multipliers is 32 49 160 N 64 129 384 1þ2þ4þ þ 2 128 321 896 ¼ 1 þ 2 þ 22 þ 2q l 256 769 2048 1 2q ¼ 512 1793 3584 12 ¼ N1: 286 38 A Note on the FFT Thus, P:3. What if the number of points is 15? I mean is DIF. N P:4. What if the number of points is 6? DIF M ðN Þ ¼ log2 N ðN 1Þ 2 ð38:4Þ again. N ¼ ðlog2 N 2Þ þ 1; P:5. Same as P.4 for DIT. 2 which is the same as Eq. 38.2. Reference Problems 1. A.V. Oppenheim, R.W. Schafer, Digital Signal Pro- cessing (Prentice Hall, New Jersey, 1975) P:1. What looks simpler? DIT or DIF? Why? P:2. Draw the DIF diagram for a 16-point FFT. Appendix: Some Mathematical Topics Simplified In the Appendix, I give some simple, common Chebyshev was out-and-out a mathematician. sense methods for deriving mathematical for- Little did he know that his polynomials would be mulas frequently used in CSSP. Appendix A.1 found so useful by ﬁlter designers. They do gives a semi-analytical method for ﬁnding the appear to be complicated to students, but a roots of a polynomial. Euler’s relation forms the reading of Appendices A.6 and A.7 would show basis of complex numbers. A fresh look at it that you can derive Chebyshev polynomial forms the content of Appendix A.2. The square identities with ease and compute the coefﬁcients root of the sum of two squares is required in of Chebyshev polynomials without difﬁculty. ﬁnding the magnitude of complex quantity. An As in Section I, all chapters in this section end up approximation appears in Appendix A.3. with ﬁve carefully designed problems. Each prob- It is well known that algebraic equations of lem requires a thorough understanding of the con- order more than 2 are difﬁcult to solve. For third tents of the corresponding chapter. Work them out and fourth orders, analytical solutions exist, but carefully and the joy of ﬁnding the clue can perhaps are difﬁcult to implement. For still higher orders, be compared with the joy you derive when you get a numerical methods have to be resorted to. For piece of your most favourite food. Learning is, by cubic and quartic equations, simpliﬁed proce- all accounts, feeding yourself. You can never dures are given in Appendix A.4. Appendix A.5 overeat, and if you think you have done so, it will gives many ways of solving an ordinary linear cause no uncomfortable feeling. Learning is con- second order differential equation. A simple suming food for your intellectual development. The method has been presented in Appendix A.5 for more you learn, the more you would like to learn. this purpose. Happy learning, dear students! © Springer Nature Singapore Pte Ltd. 2018 287 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 A.1: A Semi-analytical Method for Finding the Roots of a Polynomial A systematic method, which combines graphical, ing, in connection with the determination of analytical, and numerical techniques, is pre- poles and zeros of transfer functions, and in sented for ﬁnding the roots of a polynomial testing for stability of a given system. It is well P0(s) of any degree. Real roots are ﬁrst found by known that analytical solutions are possible only a simple graphical method, and then the purely for polynomials of degree 4 or less, and for imaginary roots are found by the Hurwitz test. higher degrees, numerical methods have to be When all the real and purely imaginary roots are resorted to. removed from P0(s), the remainder polynomial In this section, we present a semi-analytical P2(s) will have only complex conjugate roots and method, consisting of a combination of graphi- hence will be of even degree. When this degree is cal, analytical, and numerical techniques for 2, the roots are obvious. For P2(s) of degree 4, a ﬁnding the roots of a polynomial P0(s) of any variation of a previously published analytical degree. When P0(s) contains only one or two method, combined with a graphical display, is pairs of complex conjugate roots, besides those presented which is easier to apply. When the on the real and imaginary axes, it is shown that a degree of P2(s) is greater than 4, only numerical combination of graphical and analytical methods methods have to be used. sufﬁces. When P0(s) contains three or more pairs of complex conjugate roots, numerical methods have to be resorted to, after extracting all the real Keywords and imaginary axis roots from P0(s). Hurwitz test • Polynomial roots • Quartic poly- nomial • Solution of algebraic equations Roots on the Real Axis Let Introduction P0 ðsÞ ¼ sN þ aN 1 sN1 þ ðA:1:1Þ The problem of ﬁnding the roots of a given þ a2 s 2 þ a1 s þ a0 polynomial arises in all ﬁelds of science and engineering, particularly in electrical engineer- where the coefﬁcients are real and a0 6¼ 0. (If a0 = 0, then there is a root at s = 0, and the poly- nomial degree is reduced by one.) Let s = r + jx, where r and x are real and can be positive or negative. If P0(s) contains real roots, all on the Source: S. C. Dutta Roy, “A Semi-Analytical Method for negative r axis, then all ai’s will be positive, Finding the Roots of a Polynomial,” IETE Journal of while the existence of one or more positive real Education, vol. 55, pp. 90–93, July–December 2014. roots will be indicated by one or more ai’s being © Springer Nature Singapore Pte Ltd. 2018 289 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 290 A.1: A Semi-analytical Method for Finding the Roots of a Polynomial negative or missing. Such real roots can be indicates that there are real roots at s = −2 and s = simply obtained by plotting jP0 ðrÞj versus r −3. for one or both sides of r = 0, as appropriate. In general, if P0(s) of Eq. A.1.1 contains real We plot the magnitude rather than the value roots at s = ri ; i ¼ 1; 2; . . .; M; M N, where ri because visually the zero crossings from negative can be positive as well as negative, then to positive values, or vice versa, of P0(r) are not P0(s) can be written as as appealing (or perhaps not as accurate) as the Y position of the nulls, similar to those occurring in P0 ðsÞ ¼ ðs ri ÞP1 ðsÞ ðA:1:3Þ the magnitude response of null networks. i¼1M As an example, consider the 10th-degree polynomial where P1(s) does not have any real roots. In the case of Eq. A.1.2, the continued product term P0 ðsÞ ¼ s10 þ 8s9 þ 31s8 þ 87s7 þ 188s6 þ 317s5 simpliﬁes to (s2 + 5s + 6) and P1(s), obtained by dividing Eq. A.1.2 by this quadratic becomes þ 428s4 þ 452s3 þ 372s2 þ 204s þ 72: ðA:1:2Þ P1 ðsÞ ¼ s8 þ 3s7 þ 10s6 þ 19s5 þ 33s4 þ 38s3 þ 40s2 þ 24s þ 12: Since there are no negative coefﬁcients, we need to plot jP0 ðrÞj only for negative values of r. ðA:1:4Þ This plot is shown in Fig. A.1.1 which clearly 16 14 12 10 8 6 |P 0 (s)|, dB 4 2 0 2 4 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 Fig. A.1.1 Plot of jP0 ðrÞj, in dB, versus r for the example of Eq. A.1.2 A.1: A Semi-analytical Method for Finding the Roots of a Polynomial 291 Roots on the Imaginary Axis Also, the plot of jP1 ðjxÞj2 versus x2, as given in Fig. A.1.2, shows nulls at x2 = 2 and 3, thus As is well known [1], roots on the imaginary axis conﬁrming Eq. A.1.7. are revealed by performing the Hurwitz test. It P2(s) of Eq. A.1.5 can be obtained by long consists of performing a continued fraction division of P1(s) by D(s). For the example case, expansion (CFE), starting with the highest pow- this process gives ers, of the odd rational function (even part of the polynomial)/(odd part of the polynomial) or its P2 ðsÞ ¼ s4 þ 3s3 þ 5s2 þ 9s þ 2 ðA:1:8Þ reciprocal, depending on which one has a pole at inﬁnity. The existence of jx-axis roots makes the This will have two pairs of complex conjugate CFE end prematurely, and the last divisor con- roots. How to ﬁnd them will be discussed in the tains all these roots. (Note, in passing, that if the next section. coefﬁcient of any quotient in the CFE is negative, then the polynomial has roots in the right half plane; this is important in stability testing.) Complex Conjugate Roots In the present case of P1(s), if the CFE men- tioned above does end prematurely, and if the In general, P2(s) will have (N − M − 2Q)/2 pairs last divisor is D(s), then we can write of complex conjugate roots. If this number is 1, then P2(s) is a quadratic and its roots are easily P1 ðsÞ ¼ DðsÞP2 ðsÞ ðA:1:5Þ found. If P2(s) is of degree 4, as in Eq. A.1.8, the method given in [2] or [3] may be followed. where D(s) is of the form However, a confusion is likely to arise about Y signs in following this procedure. A variation of DðsÞ ¼ ðs2 þ x2k Þ: ðA:1:6Þ this procedure will now be given, which avoids k¼1Q this confusion and also does not require the analytical solution of the ‘resolvent’ cubic Note that we have taken D(s) to be an even equation. Let polynomial because a possible root at s = 0 can be taken out either at the beginning or while P2 ðsÞ ¼ s4 þ a3 s3 þ a2 s2 þ a1 s þ a0 : ðA:1:9Þ ﬁnding the real roots. P2(s) is of degree N − M − 2Q and contains only complex conjugate roots. If We express the right-hand side of Eq. A.1.9 the degree of D(s) is high, it may not be possible as the difference of two squares, rather than the to ﬁnd its roots analytically. In such a case, put s2 product of two quadratics, as in [2] and [3], as = S. The resulting polynomial in S will have follows: roots only on the negative real axis of the com- plex variable S, and hence the graphical proce- s4 þ a3 s3 þ a2 s2 þ a1 s þ a0 dure used in Sect. “Roots on the Real Axis” can ðA:1:10Þ be used. ¼ ðs2 þ as þ bÞ2 ðcs þ dÞ2 Clearly, Hurwitz test could also be avoided by plotting |P1(jx)|2 versus x2 which will show where a, b, c, and d are constants to be deter- nulls at x2 = x2k . For the example of Eq. A.1.4, mined. Equating the coefﬁcients of powers of CFE of the even part/odd part ends prematurely s on both sides of Eq. A.1.10, we get the fol- at the fourth step, and the last divisor is lowing set of four equations: 2a ¼ a3 ðA:1:11Þ DðsÞ ¼ s4 þ 5s2 þ 6 ¼ ðs2 þ 3Þ ðs2 þ 2Þ: ðA:1:7Þ 292 A.1: A Semi-analytical Method for Finding the Roots of a Polynomial 16 14 12 10 8 6 |P 1 (jw)|2 4 2 0 2 4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 w2 Fig. A.1.2 Plot of jP1 ðjxÞj2 versus x2 for Eq. A.1.4 while Eqs. A.1.13 and A.1.15 give a2 þ 2b c2 ¼ a2 ðA:1:12Þ d ¼ ½ða3 y=2Þ a1 =ð2cÞ: ðA:1:17Þ 2ðab cdÞ ¼ a1 ðA:1:13Þ Finally, Eqs. A.1.14 and A.1.15 give b2 d 2 ¼ a0 : ðA:1:14Þ d2 ¼ ðy2 =4Þa0 : ðA:1:18Þ Equation A.1.11 gives Now combine Eqs. A.1.16–A.1.18; after a ¼ a3 =2: ðA:1:15Þ simpliﬁcation, we get the following cubic equa- tion in y: To solve for b, c, and d, we have found it convenient to express c and d in terms of 2b, FðyÞ ¼ y3 a2 y2 þ ða1 a3 4a0 Þy which, for reasons to be made clear a little later, will be denoted by y. Then from Eqs. A.1.12 and þ ð4a0 a2 a3 a0 a23 a1 Þ ¼ 0: A.1.15, we get ðA:1:19Þ c2 ¼ a2 þ y þ ða23 =4Þ ðA:1:16Þ A.1: A Semi-analytical Method for Finding the Roots of a Polynomial 293 60 50 40 30 |F(y)| 20 10 0 0 1 2 3 4 5 6 y Fig. A.1.3 Plot of |F(y)| versus y It is interesting to note that although the For the example of Eq. A.1.8, Eq. A.1.19 approaches are slightly different, Eq. A.1.19 is becomes the same as the ‘resolvent’ cubic Eq. A.1.12 of [3]. This is not unexpected though, because B + FðyÞ ¼ y3 5y2 þ 4y þ 6 ¼ 0: ðA:1:24Þ b of [3] is the same as 2b in the approach adopted here. This is why 2b was denoted by y earlier in The plot of jFðyÞj versus y, as shown in this section. Fig. A.1.3, reveals only one real root at y1 = 3. F(y), being a cubic polynomial, must have at Substituting this value in Eqs. A.1.20–A.1.23, least one real root. Instead of following the along with the values of ai’s, we get a = b = ±1.5 analytical procedure of [3], we can plot jFðyÞj and c = d = ±0.5 (these are coincidences and not versus y to get the real root(s) from the null true in general). Substituting these values in location(s). If y1 is a real root, then our ﬁnal Eq. A.1.10 and factorizing, we get solution will be as follows: P2 ðsÞ ¼ ðs2 þ 2s þ 2Þðs2 þ s þ 1Þ: ðA:1:25Þ a ¼ a3 =2 ðA:1:20Þ If P2(s) is of degree 6 or more, there exists no b ¼ y1 =2 ðA:1:21Þ graphical or analytical method for ﬁnding the roots, qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ and one has to take help of numerical methods. c ¼ a2 þ y1 þ ða23 =4Þ ðA:1:22Þ d ¼ ½ða3 y1 =2Þ a1 =ð2cÞ ðA:1:23Þ 294 A.1: A Semi-analytical Method for Finding the Roots of a Polynomial Conclusions P:3. Solve s4 + 2s2 + 3s + 4 = 0 by any method. P:4. Write P(s) = s5 + as4 + bs3 + cs2 + d = In this section, a semi-analytical method has (s + e)(s + fs + g)(s2 + hs + j) and by 2 been presented for ﬁnding the real and purely solving for P(s) = 0. imaginary roots of an arbitrary polynomial. After P:5. Find the roots of (s + 1)4 = 0. removing the factors corresponding to these two types of roots, the remaining polynomial will Any resemblance to Butterworth polynomial only have complex conjugate roots. If there is roots? Show the roots graphically. only one pair of such roots, then the roots are obtained by solving a quadratic equation. For two pairs of complex conjugate roots, a variation Acknowledgements This work was supported by the Indian National Science Academy through the Honorary of an earlier published method is given, which is Scientist programme. The author thanks Professor Y. V. easier to apply. If there are more than two pairs Joshi for his help in the preparation of this manuscript. of complex conjugate roots, then there is no alternative but to use numerical methods. References 1. E.A. Guillemin, Synthesis of Passive Net- Problems works (Wiley, Hoboken, NJ, 1964) 2. M. Abramowitz, I.A. Stegun (eds.), Hand- P:1. Read Ref. 3, and try solving a cubic equa- book of Mathematical Functions (Dover, tion s3 + as2 + bs + c = 0 by decomposing New York, 1965) the LHS as (s2 + ds + e) (s + f) = 0 and 3. S.C. Dutta Roy, On the solution of quadratic eliminating all constants except one. What and cubic equations. IETE J. Educ. 47(2), do you get in this remaining constant? 91–95 (2006) P:2. Solve s3 + 2s2 + 4s + 1 = 0 by any method. A.2: A Fresh Look at the Euler’s Relation A direct proof of the Euler’s relation ej = cos h + Integrating Eq. A.2.3 and using the initial j sin h, is presented. It is direct in the sense that condition y(0) = 1, it follows that y(h) = ejh. unlike the existing proofs, it does not presume Both of these proofs presume that there is a any connection between ejh and the trigonometric connection between ejh and the trigonometric cos h functions cosh and sinh. and sin h. Presented here is a proof which does not do so, and in this sense it can be considered as a direct proof. Keywords Euler’s formula • Proof The Proof Euler’s relation Let the real and imaginary parts of the complex ejh ¼ cos h þ j sin h ðA:2:1Þ quantity ejh be denoted by f(h) and g(h), respectively; then is usually proved in mathematics and circuit theory texts [1, 2] by appealing to the inﬁnite ejh ¼ f ðhÞ þ jgðhÞ: ðA:2:4Þ h series expansions of ej , cosh and sinh. Another way [3, 4] of showing the truth of the formula is Differentiating Eq. A.2.4 and denoting d()/dh based on the observation that if by ()′, one obtains yðhÞ ¼ cos h þ j sin h ðA:2:2Þ f 0 ðhÞ þ jg0 ðhÞ ¼ ejh ¼ jf ðhÞ gðhÞ ðA:2:5Þ then Equating the real and imaginary parts on the two sides of Eq. A.2.5 gives dyðhÞ=yðhÞ ¼ jdh: ðA:2:3Þ f 0 ðhÞ g0 ðhÞ and g0 ðhÞ ¼ f ðhÞ ðA:2:6Þ Differentiating one equation in Eq. A.2.6 and combining with the other yields the following differential equations for f(h) and g(h): Source: S. C. Dutta Roy, “A Fresh Look at the Euler’s f 00 ðhÞ þ f ðhÞ ¼ 0 and g00 ðhÞ þ gðhÞ ¼ 0 ðA:2:7Þ Relation,” Students’ Journal of the IETE, vol. 22, pp. 1– 2, January 1981. © Springer Nature Singapore Pte Ltd. 2018 295 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 296 A.2: A Fresh Look at the Euler’s Relation Each of these equations describes a simple Substituting for hf in either of the equations in harmonic motion; the solutions for f(h) and g(h) Eq. A.2.11 gives are, therefore K¼l ðA:2:14Þ f ðhÞ ¼ Kf cosðh þ hf Þ and gðhÞ Thus, ﬁnally, ¼ Kg cosðh þ hg Þ ðA:2:8Þ f ðhÞ ¼ cos h and gðhÞ ¼ sin h ðA:2:15Þ where Kf, hf, Kg and hg are constants. Putting in Eq. A.2.4 gives the initial conditions f(0) = 1 and and from equations A.2.4 and A.2.15, Eq. A.2.1 g(0) = 0. Substituting these in Eq. A.2.8, there follows. results Q.E.D. Kf ¼ 1= cos hf and hg ¼ ð2r þ lÞp=2; Problems ðA:2:9Þ r ¼ 0; 1; 2; . . . P:1. Solve for h: ðcos hÞ4 ¼ 1. From Eqs. A.2.8 and A.2.9, one obtains P:2. Repeat for ðsin hÞ4 ¼ 1. h P:3. Solve for h: ejN ¼ 1, N > 1. Any relation f ðhÞ ¼ cos ðh þ hf Þ= cos hf and gðhÞ ¼ K sin h with the solutions of P.1 and P.2? Any rela- ðA:2:10Þ tion with Chebyshev? Maybe, maybe not. h h P:4. Solve for h : ej þ ejN ¼ 1, again N > 1. where K = ±Kg. Substituting Eq. A.2.10 in N N Eq. A.2.6 results in the following two equations: P:5. Solve for h: ejh 1 þ ejh 2 ¼ 1; N1 6¼ N2 ; N1;2 [ 1: sinðh þ hf Þ= cos hf ¼ K sin h and References cosðh þ hf Þ= cos hf ¼ K cos h: ðA:2:11Þ Dividing the ﬁrst equation in Eq. A.2.11 by 1. H. Sohon, Engineering Mathematics (Van the second gives Nostrand, New York, 1953), p. 65 2. W.H. Hayt Jr., J.E. Kemmerly, Engineering tanðh þ hf Þ ¼ tan h; ðA:2:12Þ Circuit Analysis (McGraw-Hill, New York, 1978), pp. 747–748 which is satisﬁed only if 3. W.H. Hayt Jr., J.E. Kemmerly, Engineering Circuit Analysis (McGraw-Hill, New York, hf ¼ 2pp; p ¼ 0; 1; 2; . . . ðA:2:13Þ 1962), p. 283 4. A.G. Beged-Dov, Another look at Euler’s relation. IEEE Trans. Educ. E-9, 44 (1966) A.3: Approximating the Square Root of the Sum of Two Squares pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ It is shown that ðx2 þ y2 Þ; with x > y, can be loss of accuracy, we propose, in this section, an approximated by x + y2/(2x) for 0 y/x 1/2, approximation for S for ~ S; consisting of two and by 0.816 (x + 0.722 y) for 1/2 y/x 1 to expressions valid for the ranges 0 y/x 0.5 within a relative error of 0.64%. This should be and 0.5 y/x l, and show that the maximum useful in computations, but more so in analytical percentage relative error e, deﬁned by developments involving such expressions. e ¼ ðS e SÞ=S 100 ðA:3:3Þ Keywords Square root • Sum of two squares is thereby reduced to a value of 0.64 only e S When dealing with complex numbers, as in cir- being the approximated value. cuit analysis or FFT computation, it is often required to calculate the value of Derivation pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ S ¼ ðx2 þ y2 Þ ðA:3:1Þ In view of Eq. A.3.2, we can write where, without loss of generality, it may be pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ S¼x ð1 þ t2 Þ ðA:3:4Þ assumed that where 0\y\x ðA:3:2Þ 0\t ¼ y=x\1 ðA:3:5Þ As is well known, evaluating the square root is somewhat tedious and time consuming. To The problem is therefore to approximate speed up the processing time, without signiﬁcant pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð1 þ t2 Þ; a plot of which is shown in Fig. A.3.1 (not to scale). Note that the latter part of the graph (for t > t1 say) can be approximated by a straight line, a possible candidate for which is indicated as a + bt. Also, when t is small, one can pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ approximate ð1 þ t2 Þ by 1 + t2/2, a plot of Source: S. C. Dutta Roy, “Approximating the Square Root of the Sum of Two Squares,” Students’ Journal of the which is also shown in Fig. A.3.1 (in a slightly IETE, vol. 32, pp. 11–13, April–June 1991. exaggerated form). In order to obtain a uniformly © Springer Nature Singapore Pte Ltd. 2018 297 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 298 A.3: Approximating the Square Root of the Sum of Two Squares 2 e3% 1+ t2 1 +ct e2 % a +bt 1 + t2/2 e¢1% e 1% 1 0 t1 t01 t2 t02 1 t pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Fig. A.3.1 Showing the variation (not to scale) of ð1 þ t2 Þ; ð1 þ t2 =2Þ and a þ bt ‘good’ approximation over the entire range of t, Now, according to Eq. A.3.7b, we have we assume that qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 þ t2 =2 0 t t1 ð1 þ t2 Þ ﬃ ðA:3:6a; bÞ ð1 þ t12 Þ ða þ b t1 Þ ð1 þ t12 Þ a þ bt t1 t 1 hpﬃﬃﬃ i pﬃﬃﬃ ¼ 2 ða þ bÞ = 2 ðA:3:8Þ and that the following relative errors are equal: (i) e1′ at t = t1 computed by using Eq. A.3.6a,b which can be simpliﬁed to the following: (ii) e1 at t = t1 computed by using Eq. A.3.6a,b (iii) e3 at t = 1, and b ¼ ab ðA:3:9Þ (iv) e2 at t = t2, which is the maximum value of e where in the range t01 t t02 (see Fig. A.3.1). pﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ b¼ 2 ð1 þ t12 Þ = ð1 þ t12 Þ 2t1 ðA:3:10Þ We therefore have the three equations To determine e2, note that in the range t01 e0t ¼ e2 ðA:3:7aÞ t t02. e1 ¼ e3 ðA:3:7bÞ h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃi pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ e ¼ 100 að1 þ btÞ ð1 þ t2 Þ = ð1 þ t2 Þ ðA:3:11Þ e2 ¼ e3 ðA:3:7cÞ The maximum of e occurs when de/dt = 0; for determining the unknown quantities t1, a and b. carrying out the differentiation and simplifying Once these are known, the maximum relative error gives the rather simple result em ð¼ e1 ¼ e01 ¼ e2 ¼ e3 Þ will also be known. A.3: Approximating the Square Root of the Sum of Two Squares 299 t2 ¼ b ðA:3:12Þ Numerical experimentation with this seem- ingly hopeless equation for t1 reveals that the Putting this in Eq. A.3.11, we get solution is, surprisingly, t1 ≅ 0.5. Further reﬁnement shows that t1 ≅ 0.5035, and a simple qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ calculation shows that under this condition, em = e2 ¼ 100 a ð1 þ b Þ 12 ðA:3:13Þ 0.64. Finally, we calculate b and a from Eqs. A.3.10 and A.3.15 as Now using Eq. A.3.7c, we get b ¼ 0:722 and a ¼ 0:816 ðA:3:19Þ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ a =ð1 þ b2 Þ 1 ¼ 1 að1 þ bÞ= 2 ðA:3:14Þ This can be simpliﬁed to the following; Concluding Comments qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ We have shown that a ¼ 2= ð1 þ b2 Þ þ ð1 þ bÞ= 2 ðA:3:15Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x þ y2 =ð2xÞ 0 y=x 0:5 ðx2 þ y2 Þ ﬃ 0:816ðx þ 0:722yÞ 0:5 y=x 1 Thus if b is known, which in turn requires that t1 is known, we can compute a from Eq. A.3.15 ðA:3:20a; bÞ and b from Eq. A.3.9. To ﬁnd t1, we use Eq. A.3.7a, which dictates that to within a relative error of 0.64%. This repre- sents a uniformly ‘good’ approximation, and qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ should be useful in speeding up the computation ð1 þ t1 =2Þ ð1 þ t12 Þ 2 ð1 þ t12 Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ of ðx2 þ y2 Þ, but more so in analytical devel- qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ opments involving such expressions. As an ¼ a þ b t2 ð1 þ t22 Þ ð1 þ t22 Þ example, let it be required to ﬁnd out if the ðA:3:16Þ equation pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Combining this with Eqs. A.3.9 and A.3.12, ð1 þ x2 Þ þ x2 þ x ¼ 2 ðA:3:21Þ and simplifying, we get has a real root in the range 0.5 x 1 and if qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ so, what is its approximate value. Using Eq. ð1 þ t12 =2Þ= ð1 þ t12 Þ ¼ a ð1 þ b2 Þ ðA:3:17Þ A.3.20a,b, Eq. A.3.21 becomes Now substituting the values of a and b from 0:816 ð1 þ 0:722xÞ þ x2 þ x2 ¼ 0 ðA:3:22Þ Eqs. A.3.15 and A.3.10 respectively, and sim- plifying, we get 1 þ t12 =2 2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðA:3:18Þ ð1 þ t1 Þ 1 þ ð1 t1 Þ= ½4ð1 þ t1 Þ 2ð1 þ t1 Þ f2ð1 þ t12 Þg 2 2 300 A.3: Approximating the Square Root of the Sum of Two Squares Solving this quadratic gives one value of x as P:3. Will the answer to P.2. be 0.816 0.5528. Putting this value in Eq. A.3.21, the (x + 0.722y) (f + 0.722g)? left-hand side becomes 2.001, which differs from P:4. What will be the approximation of pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ the right-hand side by 0.05% only. Note that the ðx2 þ y2 Þ if y x? exact solution of Eq. A.3.21 will require the P:5. Repeat this for P.1. solution of a quadratic equation. Problems pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ P:1. Can you approximate ða þ bxyÞðx2 þ y2 Þ? Take b a. P:2. How about approximating pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðx2 þ y2 Þðf 2 þ g2 Þ? A.4: On the Solution of Quartic and Cubic Equations A method of solving a quartic equation, which to ﬁnd the pass-band edges. Abramowitz and does not require extracting the roots of complex Stegun’s Handbook [1], which has been consid- numbers is explained in detail. In the process, the ered as ‘The Bible’ by scientists and engineers solution of a cubic equation has also been pre- for ages, was consulted, but to the author’s sur- sented, with the same degree of simplicity. prise, the correct solution could not be obtained. On deeper examination, it was found that there is a typographical mistake in signs in the last Keywords equation on page 17 and that the opening state- Quartic equation • Cubic equation • Solution ment on page 18 is ambiguous. Also, the method requires handling square roots of complex num- bers. A number of other such references [2–7] and internet sources [8, 9] were also consulted Introduction and it was found that they were either sketchy or had typographical mistakes or required ﬁnding When faced with a problem in mathematics, the square and cube roots of complex numbers. electrical engineers–students, faculty, researchers We present here a solution to the problem which and practitioners alike–usually consult mathe- does not require messy calculations with com- matics handbooks and encyclopaedias for a plex numbers. In the process, we also deal with quick solution. While trying to design a dualband the solution of a general cubic equation of the band-pass ﬁlter by using frequency transforma- form tion of a normalized low-pass ﬁlter, the author was confronted with the problem of solving a y 3 þ b2 y 2 þ b1 y þ b0 ¼ 0 ðA:4:2Þ quartic equation of the form with the same kind of simplicity, as compared to z4 þ a3 z3 þ a2 z2 þ a1 z þ a0 ¼ 0; ðA:4:1Þ the solutions given in [1–9] and also in [10] and [11]. The treatment is based on simpliﬁcation and consolidation of a monograph [12] by S Neumark, a British aeronautical engineer, whose work does not appear to have been Source: S. C. Dutta Roy, “On the Solution of Quartic and appreciated or even referred to in the literature. Cubic Equations,” IETE Journal of Education, vol. 47, We illustrate the procedures by examples whose pp. 91–95, April–June 2006. solutions are known beforehand. © Springer Nature Singapore Pte Ltd. 2018 301 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 302 A.4: On the Solution of Quartic and Cubic Equations Solution to the Quartic This is the so-called ‘resolvent’ cubic and checks with the equation given in [1], page 17. If Equation A.4.1 can be written as y = y1 satisﬁes Eq. A.4.12, then from Eqs. A.4.3, A.4.9 and A.4.10, the roots of the quartic Eq. ðz2 þ Az þ BÞðz2 þ a z þ bÞ ¼ 0; ðA:4:3Þ A.4.1 are obtained by solving the following two quadratic equations: where ! rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ A þ a ¼ a3 ; ðA:4:4Þ a3 a3 y1 2a1 y1 y21 z þ 2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ z þ a0 2 2 y21 4a0 2 4 B þ b þ Aa ¼ a2 ; ðA:4:5Þ ¼ 0: Ab þ aB ¼ a1 ðA:4:6Þ ðA:4:13Þ and From Eq. A.4.11, the second term in the Bb ¼ a0 : ðA:4:7Þ coefﬁcient of z in Eq. A.4.13 can be written as qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ a2 43 þ y1 a2 : Choosing the positive sign Let here leads to the mistake in [1] as pointed out in B þ b ¼ y: ðA:4:8Þ the Introduction. Choosing the negative sign gives the correct results, as demonstrated in the Then from Eqs. A.4.7 and A.4.8, we get a Example worked out later. Hence we get the quadratic equation in B or b, the solution of simpliﬁed form of Eq. A.4.13 as: which gives rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a3 a23 y1 y21 y y2 4a0 y y2 4a0 z þ 2 þ y 1 a2 z þ a0 B¼ and b ¼ 2 4 2 4 2 2 ðA:4:9Þ ¼ 0: ðA:4:14Þ From Eqs. A.4.4, A.4.6 and A.4.9 (hence- forth, we take the positive sign for B and the We next consider the solution of the cubic negative sign for b, without any loss of gener- equation A.4.12, and shall do so with general ality), we get coefﬁcients, as in Eq. A.4.2. a3 a3 y 2a1 A; a ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ : ðA:4:10Þ 2 2 y2 4a0 Solution of the Cubic Equation Finally, substituting for A, a, B and b in Consider Eq. A.4.2. In the literature, it is the Eq. A.4.5, and simplifying gives usual practice to derive a ‘depressed’ cubic i.e. another cubic equation in which the y2 term is a23 ða3 y 2a1 Þ2 missing. To this end, we include the ﬁrst two yþ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ a2 : ðA:4:11Þ 3 4 4 y2 4a0 terms of Eq. A.4.2 in y þ b32 . Then Eq. A.4.2 can be written as Simpliﬁcation of Eq. A.4.11 results in the 2 following cubic equation in y: b2 3 b2 b3 yþ b1 y þ b0 2 ¼ 0: y3 a2 y2 þ ða1 a3 4a0 Þy 3 3 27 ðA:4:15Þ ða21 þ a0 a23 4a0 a2 Þ ¼ 0: ðA:4:12Þ A.4: On the Solution of Quartic and Cubic Equations 303 We next supplement y in the second term by Since a cubic equation is required to have one +b2/3; then Eq. A.4.15 becomes real root, Eq. A.4.22 will be applicable only when b22 − 3b1 > 0. We shall consider the other 2 b2 3 b b case, i.e. b22 − 3b1 < 0 shortly. yþ 2 b1 y þ 2 þ b0 3 Notice that the left-hand side of Eq. A.4.21 is 3 3 b32 b2 b22 the third order Chebyshev polynomial in x which þ b1 27 3 3 oscillates between −1 and +1 for 1 x þ 1; ¼ 0: and is monotonically increasing or decreasing for ðA:4:16Þ |x| > 1, as shown in Fig. A.4.1. This ﬁgure will only give real root(s), and clearly, it sufﬁces to Let consider R 0, because changing the sign of R, simply leads to the roots changing their signs. It b2 is also obvious from the ﬁgure that for R < 1, yþ ¼ kx: ðA:4:17Þ 3 there will be three real roots, of which one is positive and the other two are negative. When Usually, k is taken as unity in the literature. R > 1, there will be only one real (positive) root. With a general k, Eq. A.4.16 can be simpliﬁed to When R = 1, there will be a double root at −1/2 the following: and a single root at +1. For R < 1, the real positive root x = x1 occurs 27k3 x3 9kðb22 3b1 Þx þ ð27b0 þ 2b32 9b1 b2 Þ between the point A, where x = √3/2 and x = 1. ¼ 0: Since x < 1, we can write ðA:4:18Þ x1 ¼ cos h; ðA:4:23Þ Dividing both sides by (27k3/4), we get where 0\h\p=6. Equation A.4.21 then 4ðb22 3b1 Þ becomes 4x3 xþ 3k2 cos 3h ¼ R: ðA:4:24Þ 4ð27b0 þ 2b32 9b1 b2 Þ ¼ 0: ðA:4:19Þ 27k3 Hence Now comes the brilliant idea of forcing the y1 ¼ cos ½ðcos1 RÞ=3: ðA:4:25Þ coefﬁcient of x as −3 by choosing pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Equation A.4.21 can now be rewritten in the b22 3b1 2 form k¼ ðA:4:20Þ 3 4x3 3xR ¼ ðxx1 Þ ð4x2 þ c1 x þ c0 Þ ¼ 0: Using the negative sign in Eq. A.4.20, Eq. ðA:4:26Þ A.4.19 becomes Comparing coefﬁcients, we get c1 = 4x1 and 4x3 3x ¼ R; ðA:4:21Þ c0 = R/x1 = 4x21 − 3. Thus the other two real roots of Eq. A.4.21 are the solutions of the quadratic where 27b0 þ 2b32 9b1 b2 equation R¼ : ðA:4:22Þ 2ðb22 3b1 Þ3=2 4x2 þ 4x1 x þ 4x21 3 ¼ 0 ðA:4:27Þ 304 A.4: On the Solution of Quartic and Cubic Equations 5 4 3 2 1 0 R A 1 2 3 4 5 1 5 x Fig. A.4.1 Plot of R = 4x3 − 3x so that pﬃﬃﬃ x2;3 ¼ ½ðcosh aÞ=2Þ jð 3=2Þ sinh a ðA:4:31Þ pﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x2;3 ¼ ðx1 =2Þ 3=2 1 x2 Finally, there remains the case b22 − 3b1 < 0. pﬃﬃﬃ 1 In this case, choose k such that the coefﬁcient of ¼ ½ðcos hÞ=2Þ 3=2 sin h p x in Eq. A.4.19 becomes +3, i.e. let ¼ cos h : ðA:4:28Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 3 2 3b1 b22 k¼ ðA:4:32Þ Now consider the case R > 1. Figure A.4.1 3 shows that there is only one real (positive) root at Then Eq. A.4.21 becomes x = x1 > 1. We can therefore write x1 ¼ cos h a: ðA:4:29Þ 27b0 þ 2b32 9b1 b2 4x3 þ 3x ¼ R ¼ ðA:4:33Þ 2jb22 3b1 j3=2 Substituting this in Eq. A.4.21 and simplify- ing, we get Notice that for uniformity with the previous case, we have modiﬁed the denominator of R. x1 ¼ cos h ½ðcos h1 RÞ=3: ðA:4:30Þ Here also, as in the previous case, we need to solve only for positive R; for negative R, all roots Equations A.4.26 and A.4.27 are applicable will change sign. The plot of the left-hand side of here also, so that the two complex roots are given Eq. A.4.33 is shown in Fig. A.4.2, from which it by is clear that there is only one real root x1 which is A.4: On the Solution of Quartic and Cubic Equations 305 15 10 5 R 0 5 10 15 1.5 1 0.5 0 0.5 1 1.5 x Fig. A.4.2 Plot of R = 4x3 + 3x positive for positive R. Since x1 can have any value, we let Example x1 ¼ sinh b: ðA:4:34Þ Let the equation to be solved be Then Eq. A.4.33 gives z4 5z3 þ 5z2 þ 5z 6 ¼ 0: ðA:4:38Þ x1 ¼ sinh sinh1 R =3 : ðA:4:35Þ As can be easily veriﬁed, Eq. A.4.38 has the roots −1, +1 and +5. Let us see what our pro- As in the earlier case, the cubic equation cedure gives. From Eq. A.4.12, the resolvent Eq. A.4.33 can be factorized as cubic becomes ðxx1 Þð4x2 þ 4x1 x þ 4x21 þ 3Þ ¼ 0 ðA:4:36Þ y3 5y2 y þ 5 ¼ 0: ðA:4:39Þ so that the other two roots which are complex As can be easily veriﬁed, the roots of Eq. conjugates, are given by A.4.39 are −1, +1 and 5, but let us follow the pﬃﬃﬃ procedure as given here. From Eqs. A.4.22 and x2;3 ¼ ½ðsinh bÞ=2Þ j 3=2 cosh b: A.4.39, R is calculated as −10/(7 √7). Since | R| < 1, we get, by applying Table A.4.1, ðA:4:37Þ x1;2;3 ¼ 0:756; 0:189; 0:945: ðA:4:40Þ Table A.4.1 gives a summary of the proce- dure for solving a general cubic equation. 306 A.4: On the Solution of Quartic and Cubic Equations Table A.4.1 Procedure for solving a cubic equation Equation to be solved: y3 + b2y2 + b1y + b0 = 0 27b0 þ 2b32 9b1 b2 Compute R ¼ 2jb22 3b1 j3=2 – 3b1 > 0 b22 b22 – 3b1 < 0 0<R<1 R>1 Any R h ¼ ðcos 1 RÞ=3 a ¼ ðcosh 1 RÞ=3 b = (sinh −1R)/3 x1 = cos h x1 ¼ cosh h x1 = sinh b pﬃﬃﬃ pﬃﬃﬃ x2;3 ¼ cos p3 R x2;3 ¼ ðcosh a j 3 sinh aÞ=2 x2;3 ¼ ðsinh b j 3 cosh bÞ=2 1=2 yi ¼ b2 þ 2b22 3b1 xi =3; i ¼ 1; 2; 3 Note The solutions are valid for positive R. For negative R, compute the solutions for |R|, and then negate the values of x before substitution in the equation for Yi Correspondingly, pﬃﬃﬃ For example, the quartic equation y1;2;3 ¼ 54 7x1;2;3 =3 ¼ 1; 1; 5: pﬃﬃﬃﬃﬃ ðA:4:41Þ z4 2z2 þ 3z0:5 ¼ 0 ðA:4:43Þ These values check with the solutions has the following resolvent cubic: obtained by inspection of Eq. A.4.39. Selecting y1 = 1 and using Eq. A.4.14, the quadratics then y3 þ 2y2 þ 2y þ 1 ¼ 0 ðA:4:44Þ become Equation A.4.44 has a real root at y = −1 and pﬃﬃﬃ z 4z þ 3 ¼ 0 and z z2 ¼ 0: 2 2 ðA:4:42Þ a pair of complex roots at y ¼ 1 þ j 3 =2. Using either of the complex roots, it can show These give z = −1, 1, 2 and 3 as expected. The that we get the correct roots as obtained by using same results are obtained had we selected y1 = y1 = −1. The real root is preferred because it −1 or 5. It can be veriﬁed that a reversal of signs reduces the computational effort, considerably. in the constant term of Eq. A.4.14 gives wrong results. This example also demonstrates that the opening sentence on p. 18 of [1] is ambiguous, Problems because it does not say what is to be done if all the three real roots give real coefﬁcients in the P:1. Can you solve a sixth order equation by quadratic equations. As the present example decomposing as in Eq. A.4.3? shows, any of them can be used. A question that P:2. How about an eight order? arises at this point is the following: Would it give P:3. Could you solve a cubic equation by start- the correct answer if we choose a complex, ing with trigonometry right from the start? instead of a real root for y1? The answer is yes. P:4. Solve x4 + x2 + 1 = 0 by any method. A.4: On the Solution of Quartic and Cubic Equations 307 P:5. Will the trigonometric approach work for 4. C.E. Pearson, Handbook of Applicable P.4? Yes or no answer will not do. You Mathematics (Van Nostrand, 1974) must have the necessary mathematical 5. W. Gellet et al., The VNR Concise Ency- support to justify your answer. clopaedia of Mathematics (Van Nostrand, 1975) 6. E.W. Weisstein, CRC Concise Encyclopaedia Acknowledgement The author thanks his former student of Mathematics (Chapman and Hall, 1999) and current colleague, Professor Jayadeva, for many 7. I.N. Bronshtein et al., Handbook of Mathe- helpful discussions on this topic during their evening matics (Springer, Berlin, 2000) walks in the corridors of IIT Delhi. 8. http://www.sosmath.com/algebra/factor/ fac12/fac12.html: The quartic formula References 9. http://mathforum.org/dr.math/faq/cubic. equations.html: Cubic and quartic equations 1. M. Abramowitz, I.A. Stegun, Handbook of 10. http://www.sosmath.com/algebra/factor/ Mathematical Functions (Dover, 1965) fac1/fac1l.html: The cubic formula 2. G.A. Korn, J.M. Korn, Mathematical 11. http://mathforum.org/dr.math/faq/cubic. Handbook for Scientists and Engineers equations.html: Cubic equations—another (McGraw-Hill, 1968) solution 3. R.S. Burington, Handbook of Mathematical 12. S. Neumark, Solution of Cubic and Quartic Tables and Formulas (McGraw-Hill, 1973) Equations (Pergamon, 1965) A.5: Many Ways of Solving an Ordinary Linear Second Order Differential Equation with Constant Coefficients There are many different ways of solving an the following ordinary linear second order dif- ordinary linear second order differential equation ferential equation: with constant coefﬁcients. Some of them are available in textbooks while some others are y00 þ 2ay0 þ x20 y ¼ 0; ðA:5:1Þ scattered in journal publications. A comprehen- sive survey of these methods is presented in this where, in the usual situation, the prime denotes section, along with the essential steps in each differentiation with respect to time t, and a and method and the relevant references. x0 are constants, subject to the initial conditions ‘As many faiths, as many ways’—Shri Ramakrishna Paramhansa yð0Þ ¼ y0 and y0 ð0Þ ¼ p0 : ðA:5:2Þ For example, when a capacitor C is charged to Keywords a voltage V and discharged through an induc- tance L in series with a resistance R, the current ODE • Solution y in the circuit obeys Eq. A.5.1 with a = R/(2L) and x20 = 1/(LC) [1]. Many techniques exist in the literature for solving Eq. A.5.1, of which the Introduction following are commonly available in one text- book or the other; (1) Laplace transform method; There are many problems in electrical engineer- (2) assuming an exponential solution; and ing and physics where one is required to solve (3) operator method. Several other methods have appeared in the literature, mostly in journals, some of which are quite simple, innovative, and/or of pedagogical interest. We present here a survey of all these techniques, along with the essential steps in each method and the relevant reference(s). Source: S. C. Dutta Roy, “Many Ways of Solving an Ordinary Linear Second Order Differential Equation with Constant Coefficients,” IETE Journal of Education, vol. 48, pp. 73–76, April–June 2007. © Springer Nature Singapore Pte Ltd. 2018 309 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 310 A.5: Many Ways of Solving an Ordinary Linear Second Order Differential Equation … Laplace Transform Method Putting these values in Eq. A.5.8 and simpli- fying, we get Taking the Laplace transform of Eq. A.5.1 and denoting the Laplace transform of y(t) by Y(s), yðtÞ ¼ y0 cos x0 t þ ðp0 =x0 Þ sin x0 t; ðA:5:10Þ we get which can be put in the form s YðsÞsy0 p0 þ 2a½sYðsÞ 2 y0 þ x20 YðsÞ ¼ 0: qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðA:5:3Þ yðtÞ ¼ y20 þ ðp0 =x0 Þ2 cos x0 t þ tan1 ½p0 =ðy0 x0 Þ ðA:5:11Þ On simpliﬁcation and factorization, this gives ðs þ 2aÞy0 þ p0 YðsÞ ¼ ; ðA:5:4Þ Assuming an Exponential ðs s1 Þðs s2 Þ Solution: Why Not Do It With where Trigonometric Functions? Because of Approximation qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Errors s1;2 ¼ a b and b ¼ a2 x20 : ðA:5:5Þ In this method, we assume a solution of the form Expanding Eq. A.5.4 in partial fractions and Aest for a 6¼ x0 and (A + Bt)est for a = x0 In the taking the inverse Laplace transform, we get ﬁrst case, putting the assumed solution in Eq. A.5.1 gives the so-called characteristic equation yðtÞ ¼ A1 es1 t þ A2 es2 t ; ðA:5:6Þ s2 þ 2as þ x20 þ 0 ðA:5:12Þ where 2ay0 þ p0 þ s1;2 y0 which has the roots given by Eq. A.5.5. Thus A1;2 ¼ ðA:5:7Þ 2b both es1 t and es2 t are solutions of Eq. A.5.1 and the general solution is the same as that given by Combining Eqs. A.5.5–A.5.7 and simplify- Eq. A.5.6. It is easily shown that for satisfying ing, one obtains the initial conditions given in Eq. A.5.2, A1, 2 are the same as given in Eq. A.5.7 so that the yðtÞ ¼ ðeat =bÞ ½ðy0 a þ p0 Þ sinh bt þ y0 b cosh bt: required solution is given by Eq. A.5.8. As ðA:5:8Þ mentioned earlier, this expression can handle all the three cases of relative values of a and x0, but As shown in [1], this expression is adequate for it is instructive to pursue the assumed solution considering all the three cases, viz. (i) overdamp- (A + Bt)est for the case a = x0 a bit further. As ing: a > x0; (ii) critical damping: a = x0; and (iii) pointed out in [1], a heuristic justiﬁcation for this underdamping: a < x0. We shall not therefore assumed solution comes from the argument that pursue these cases separately, at this stage. in this case, one can try a general solution of the It is interesting to observe that expression form f(t)est. instead of Aest, where f(t) is to be Eq. A.5.8 is also adequate for considering the determined. Substituting this trial solution in undamped case i.e. a = 0. Under this condition, Eq. A.5.1 and simplifying, we get b ¼ jx0 ; sinh bt ¼ j sin x0 t and cosh bt f 00 ðtÞ þ ðs2 þ 2as þ x20 Þf 0 ðtÞ þ 2ðs þ aÞf ðtÞ ¼ 0: ¼ cos x0 t: ðA:5:13Þ ðA:5:9Þ A.5: Many Ways of Solving an Ordinary Linear Second Order Differential Equation … 311 Since from Eq. A.5.12, s = −a = −x0 in this is of the same form as Eq. A.5.6. The rest of the case, Eq. A.5.13 reduces to procedure is the same as in the previous section. The critical damping case poses no problem f 00 ðtÞ ¼ 0; ðA:5:14Þ with this method. Under critical damping, Eq. A.5.18 becomes i.e. ðD þ aÞ ðD þ aÞy ¼ 0 ðA:5:22Þ f ðtÞ ¼ A þ Bt and yðtÞ ¼ ðA þ BtÞest : ðA:5:15Þ Evaluating A and B from the initial conditions, which can be solved by following the same steps we ﬁnally get, for this critically damped case, as in the case a 6¼ x0, and results in the same expression as Eq. A.5.16. yðtÞ ¼ ½y0 þ ðp0 þ y0 aÞt eat : ðA:5:16Þ Operator Method Solution by Change of Variable In this method, we deﬁne the operator D ¼ ddt As pointed out in [1], for the beginner student, 2 who has not been exposed to Laplace transforms, and D2 ¼ ddt2 so that Eq. A.5.1 becomes the conventional approach is to use either an assumed solution or the operator method. In ðD2 þ 2aD þ x20 Þy ¼ 0: ðA:5:17Þ either case, the student has conceptual difﬁculty in accepting why a solution should be assumed, We then treat the quadratic operator and that too of a particular type, or why with (D2 + 2aD + x20) as an algebraic expression and 2 2 factorize it to obtain the following changed form D ¼ ddt ; D2 y is ddt2y and not ddyt , and how of Eq. A.5.17: ðD2 þ 2aD þ x20 Þ can be treated as a polynomial and factorized. To obviate these difﬁculties, we ðDs1 ÞðDs2 Þy ¼ 0; ðA:5:18Þ proposed in [1] a change of variable from y to z with where s1,2 arc given by Eq. A.5.5. Now let z ¼ y0 sy ðA:5:23Þ ðDs2 Þy ¼ z ðA:5:19Þ where s is an unknown constant. Obtain y″ from so that Eq. A.5.18 becomes the following ﬁrst Eq. A.5.23 and substitute in Eq. A.5.1; the result order homogeneous equation in z: is ðDs1 Þz ¼ 0: ðA:5:20Þ z0 þ ðs þ 2aÞz þ ðs2 þ 2as þ x20 Þy ¼ 0 ðA:5:24Þ The solution of Eq. A.5.20 is Now choose s such that the y term vanishes in zðtÞ ¼ K1 e : s1 t ðA:5:21Þ Eq. A.5.24; this gives the same equation as Eq. A.5.12 with the possible values of s as given where K1 is a constant. Now combine equa- in Eq. A.5.5. Taking either value of s and solving tions A.5.19 and A.5.21 and solve the resulting the ﬁrst order homogeneous equation in z, we get non-homogeneous ﬁrst order differential equa- tion by the integrating factor method. The result zðtÞ ¼ K1 eðs þ 2aÞt : ðA:5:25Þ 312 A.5: Many Ways of Solving an Ordinary Linear Second Order Differential Equation … Putting this value in Eq. A.5.23 and solving D2m y ¼ b2 y: ðA:5:30Þ for y gives Since two operations by Dm is equivalent to yðtÞ ¼ K2 est þ K3 eðs þ 2aÞt : ðA:5:26Þ multiplication by b2, we see that where K2,3 are constants. Note that in Eq. A.5.25 Dm y ¼ by ðA:5:31Þ taking either s = s1 or s = s2 makes no difference, because If we take positive sign in Eq. A.5.31, and solve the resulting ﬁrst order non-homogeneous ðs1;2 þ 2aÞ ¼ ða b þ 2aÞ equation, we get y1 ðtÞ ¼ A1 es1 t ; while taking the ðA:5:27Þ ¼ a b ¼ s2;1 : negative sign gives y2 ðtÞ ¼ A2 es2 t :Thus the gen- eral solution is the same as Eq. A.5.6. For the Thus the solution is of the same form as critical damping case, let z = Dmy [4]; then the Eq. A.5.6. equation to be solved is Dmz = 0; further pro- It is to be recognized that the clue to the cedure is similar to that in the operator method of method is provided by the operator method but it Sect. 4. has the advantage that the student has no difﬁculty in comprehending this solution. Also, note that when s1 = s2 = −a (critical damping case), Eq. LKM 2: Another Change A.5.25 becomes zðtÞ ¼ K1 eat . Putting this value of Variable Method in Eq. A.5.23, solving for y and evaluating the constants lead to the same result as Eq. A.5.16. This method is due to Greenberg [5] and is based We next discuss some less known methods on a change of variable such that the ﬁrst dif- (LKM) for solving Eq. A.5.1, which do not ferential coefﬁcient term is eliminated; simulta- appear to be included in textbooks, but deserve neously, it also removes the difﬁculty in to be. analyzing the critical damping case. We let y ¼ zeat : ðA:5:32Þ LKM 1: Modified Operator Method Substituting this in Eq. A.5.1 and simplifying, we get This method is due to Garrison [2, 3] and starts with rewriting Eq. A.5.1 as z00 ¼ b2 z; ðA:5:33Þ b being the same as in Eq. A.5.5. Now choose d2 d þ 2a y ¼ x20 y ðA:5:28Þ the trial solution z ¼ ekt ; putting this in Eq. dt2 dt A.5.33 given k2 = b2 or k = ±b so that the Add a2y to both sides to get general solution for z becomes d2 d zðtÞ ¼ A1 ebt þ A2 ebt : ðA:5:34Þ þ 2a þ a y ¼ a2 x20 y: ðA:5:29Þ 2 dt2 dt Combining Eq. A.5.34 with Eq. A.5.32, we As in the operator of Sect. 4, the operator on get the same solution as Eq. A.5.6. For the crit- the left-hand side of Eq. A.5.29 can be factored ical damping case, b = 0, so that Eq. A.5.33 gives z″ = 0 or z = A + Bt, and combined with as ddt þ a ddt þ a . Denoting either factor by Eq. A.5.32, we get the same solution as Dm, Eq. A.5.29 combined with Eq. A.5.5 gives Eq. A.5.15. A.5: Many Ways of Solving an Ordinary Linear Second Order Differential Equation … 313 Instead of a trial solution, one could also and eAt is the so-called fundamental matrix. The make a change of variable from z to x = z′ − sz as latter can be calculated by using any of the in the method discussed in Sect. 4. Then well-known techniques [8]. The result, for this z00 ¼ x0 þ sz0 ¼ x0 þ sx þ s2 z. Substituting this in case, is Eq. A.5.33 and choosing s such that the z-term is eat a sinh bt þ b cosh bt sinh bt eAt ¼ absent in the result leads to s = ±b and x0 þ sx ¼ b x20 sinh bt b cosh bt a sinh bt 0; which has the solution x = K1e−st. Finally, ðA:5:41Þ solving the equation z′ − sz = K1e−st, we get z (t) = K2est + K3e−st, which is of the same form as Combining Eq. A.5.41 with Eqs. A.5.39 and Eq. A.5.34, irrespective of whether A.5.35 gives the desired y(t), which is the same s ¼ þ b or s ¼ b. as that given by Eq. A.5.8. State Variable Method Conclusion This method involves more efforts than in any A comprehensive survey has been presented here other method discussed so far, but is of peda- of the methods for solving Eq. A.5.1, some of gogical interest, when the state variables are ﬁrst which are well known and some are less known. introduced to undergraduate students [6]. We let Conceptually, the method based on change of variables given in Sect. 5 appears to be the x1 ¼ y and x2 ¼ y0 : ðA:5:35Þ simplest and easily comprehensible by the beginner. The less known methods given in Then we can write Sects. 6 and 7 are quite instructive and should ﬁnd a place in textbooks. The method based on x01 ¼ x2 and x02 ¼ x20 x1 2ax2 : ðA:5:36Þ state variables has a pedagogical value for introducing state variables to the beginner, rather The two equations in Eq. A.5.36 can be than for solving Eq. A.5.1. written in the familiar0 matrix from x ¼ Ax ðA:5:37Þ It should be mentioned here that except for the methods of Sects. 6 and 7, all other methods are where applicable for solving higher order ordinary lin- ear differential equations with constant coefﬁ- x 0 1 cients [9]. x¼ 1 and A ¼ : ðA:5:38Þ x2 x20 2a Problems The solution for Eq. A.5.37, as is well known [7], is P:1. Suppose, for Eq. A.5.1, the initial condi- x ¼ e x0 ; At ðA:5:39Þ tions Eq. A.5.2 are given at 0−. How would you modify the solution? where P:2. Instead of 0 on the RHS of Eq. A.5.1, let there be a constant. How do you ﬁnd a y solution? x0 ¼ 0 ðA:5:40Þ p0 P:3. Repeat with RHS = f(y). P:4. Repeat with RHS = f(x). 314 A.5: Many Ways of Solving an Ordinary Linear Second Order Differential Equation … P:5. Suppose the middle term on the LHS of this contribution, (x2 − a2) should be Eq. A.5.1 is 2ayy′. Can you ﬁnd a solution? replaced by (a2 − x2)) 6. D.S. Zrnic, Additional remarks on the equa- tion €x þ 2ax þ x2 x ¼ 0. Am. J. Phys. 41, 712 References (1973) (Note that in Eq. (3) of this contribu- tion, the sign of the (1, 2) element of the matrix should be positive) 1. S.C. Dutta Roy, Transients in RLC networks 7. S.C. Dutta Roy, An introduction to the state revisited. IETE J. Educ. 44, 207–211 (2003) variable characterization of linear systems— 2. J.D. Garrison, On the solution of the equation part I. IETE J. Educ. 38, 11–18 (1997) for damped oscillation. Am. J. Phys. 42, 694– 8. S.C. Dutta Roy, An introduction to the state 695 (1974) variable characterization of linear systems— 3. J.D. Garrison, Erratum: on the solution of the part II. IETE J. Educ. 38, 99–107 (1997) equation for damped oscillation. Am. J. Phys. 9. S.C. Dutta Roy, Solution of an Ordinary 43, 463 (1975) Linear Differential Equation With Constant 4. S. Balasubramanian, R. Fatchally, Comment Coefﬁcients, unpublished manuscript. That on the solution of the equation for damped gives me an idea. I should try to publish this oscillation. Am. J. Phys. 44, 705 (1976) manuscript as soon as possible. 5. H. Greenberg, Further remarks concerning the solution of the equation €x þ 2ax þ x2 x ¼ 0. Am. J. Phys. 44, 1135– 1136 (1976) (Note that in Eqs. (5) and (6) of A.6: Proofs of Two Chebyshev Polynomial Identities Useful in Digital Filter Design Alternate proofs of two Chebyshev polynomial where Ti is the i-th degree Chebyshev polyno- identities, which are useful in the design of mial of the ﬁrst kind. As shown by Shenoi and low-pass recursive digital ﬁlters, are presented. Agrawal [2], these identities are useful in the As compared to those provided by Yip [1], our design of recursive low-pass digital ﬁlters. In proofs appear to be simpler and are direct, rather proving Eq. A.6.1, Yip ﬁrst proved the identity than inductive. 1 Tn ðxÞTm ðxÞ ¼ c Tðm þ nÞ ðxÞ þ Tjmnj ðxÞ 2 Keywords ðA:6:3Þ Chebyshev polynomial • Identities • Application in DSP and then substituted m = n = N. In proving In 1980, Yip provided proofs of the following Eq. A.6.2, Yip used the method of induction. We two Chebyshev polynomial identities: present here simpler proofs of Eqs. A.6.1 and A.6.2, and in the latter case, we give a direct, T2N ðxÞ þ 1 ¼ 2 ½TN ðxÞ2 ðA:6:1Þ rather than inductive proof, based solely on the properties of trigonometric functions. and T2N þ 1 ðxÞ þ 1 ¼ ð1xÞ Proof of the First Identity " # X N N þ1 ðA:6:2Þ 2 i ð1Þ TN1 ðxÞ þ ð1Þ 2; Letting x = cos h, we have i¼0 Ti ðxÞ ¼ cos ih ðA:6:4Þ Using Eq. A.6.4 and the trigonometric formula cos 2/ ¼ 2 cos2 /l ðA:6:5Þ Source: S. C. Dutta Roy, “Proofs of Two Chebyshev Equation A.6.1 follows easily by putting Polynomial Identities Useful in Digital Filter Design” / = Nh. Journal of the IETE, vol. 28, p. 605, November 1982. (Corrections in vol. 29, p. 132, March 1983). © Springer Nature Singapore Pte Ltd. 2018 315 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 316 A.6: Proofs of Two Chebyshev Polynomial Identities Useful in Digital Filter Design Proof of the Second Identity Now using the trigonometric identity CþD jC Dj Using Eq. A.6.4, the right-hand side of Eq. cos C þ cos D ¼ 2 cos cos ðA:6:11Þ 2 2 A.6.2 becomes and Eq. A.6.5 in Eq. A.6.10, the latter RHS ¼ ð1 þ cos hÞ simpliﬁes to " #2 X N i N þ1 RHS ¼ 2 cos2 ½ð2N þ 1Þh=2 ðA:6:12Þ 2 ð1Þ cosðN iÞh þ ð1Þ i¼0 Using Eq. A.6.5 once again gives ðA:6:6Þ RHS ¼ cos ð2N þ 1Þh þ 1 Putting ðA:6:13Þ ¼ T2N þ 1 ðxÞ þ l ¼ LHS 2 cosðNiÞh ¼ ejðNiÞh þ e þ jðNiÞh ðA:6:7Þ which completes the proof. in Eq. A.6.6, the ﬁrst term within its square brackets, to be denoted by F for brevity, becomes Problems X N i X N i F ¼ ejNh ejh þ ejNh eh P:1. Can these identities be proved with other i¼0 i¼0 kinds of polynomial. For example, a second h N þ 1 i h N þ 1 i ejNh 1 ejh ejNh 1 ejh order one? Try it and let me know. ¼ þ P:2. Could the second identity be proved by 1 þ ejh 1 þ ejh bringing in Euler again? That is, replacing ðA:6:8Þ cosh by Re ejh ? By routine simpliﬁcation of Eq. A.6.8, we get P:3. Prove Eq. A.6.5 with Euler’s identity. P:4. Write T2N(x) as a polynomial in TN(x). cos Nh þ cosðN þ 1Þh ð1ÞN þ 1 ð1 þ cos hÞ P:5. Can you prove Eq. A.6.3 without invoking F¼ induction? That is, directly? 1 þ cos h ðA:6:9Þ Combining Eq. A.6.6 with Eq. A.6.9, and References simplifying gives 1. P.C. Yip, On a conjecture for the design of RHS ¼ ½cos Nh þ cos ðN þ 1Þh2 =ð1 þ cos hÞ low-pass recursive ﬁlters. IEEE Trans. ðA:6:10Þ ASSP-28, 6, 768 (1980) 2. K. Sbenoi, B.P. Agrawal, On the design of recursive low-pass digital ﬁlters. IEEE Trans. ASSP-28, 1, 79–84 (1980) A.7: Computation of the Coefficients of Chebyshev Polynomials A simple derivation is given of a closed form tational complexity of the elliptic functions. Next formula for the computation of the coefﬁcients of to them in the category of optimum ﬁlters comes Chebyshev polynomials, which, as is well the Chebyshev ﬁlter. For a normalized Cheby- known, are required for the design of equal ripple shev low-pass ﬁlter with cutoff at 1 rad/s, the ﬁlters. A modiﬁcation of the formula is also magnitude squared function is given by given for facilitating fast computation. 1 jHðjxÞj2 ¼ ðA:7:1Þ 1 þ e2 Cn2 ðxÞ Keywords Chebyshev polynomials • Computation • where Cn(x) is the Chebyshev polynomial, Coefﬁcients deﬁned by cosðn cos1 xÞ; x\1; Cn ðxÞ ¼ ðA:7:2Þ Introduction coshðn cosh1 xÞ; x [ 1: Chebyshev polynomials are required in the Cn(x) is usually computed by the recursion design of ﬁlters in which the pass-band or the relation stop-band is desired to have equal ripple char- acteristic [1, 2]. As is well known, elliptic ﬁlters, Cn þ 1 ðxÞ ¼ 2xCn ðxÞCn1 ðxÞ ðA:7:3Þ in which both pass- and stop-bands are equal ripple, are the optimum ones. However, their with the initial conditions design is rather involved because of the compu- C0 ðxÞ ¼ 1 and C1 ðxÞ ¼ x: ðA:7:4Þ Tables for low order Cn(x) are available in textbooks. However when n is high, one starts from the two highest n for which entries exist in the Table and then uses Eq. A.7.3 recursively. Clearly, this computation is time consuming. Nguyen [3] derived a recursive formula for the Source: S. C. Dutta Roy, “Computation of the Coeffi- cients of Chebyshev Polynomials”, IETE Journal of coefﬁcients and formulated some rules for cut- Education, vol. 49, pp. 19–21, January–April 2008. ting down on the computation time. Johnson and © Springer Nature Singapore Pte Ltd. 2018 317 S. C. Dutta Roy, Circuits, Systems and Signal Processing, https://doi.org/10.1007/978-981-10-6919-2 318 A.7: Computation of the Coefficients of Chebyshev Polynomials Johnson [4] derived the following closed form If (−l)i is deleted from Eq. A.7.9, then we get pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃn representation of Cn(x): the expansion for x þ x2 1 : Substituting bX n=2c these in Eq. A.7.8, we observe that the odd i- n Cn ðxÞ ¼ ð1Þk xn2k ð1 x2 Þk ðA:7:5Þ terms will cancel. Hence if we let k = i/2, then k¼0 2k Eq. A.7.8 becomes where bn=2c is the integer part of (n/2). They n=2c bX n k obtained this result by expressing Cn(x) of Eq. Cn ðxÞ ¼ xn2k x2 1 A.7.2 as k¼0 2k bX n=2c n k Cn ðxÞ ¼ Re expðjn cos1 xÞ : ðA:7:6Þ ¼ ð1Þ k xn2k 1 x2 k¼0 2k An alternative derivation of Eq. A.7.5 was ðA:7:10Þ given by Cole [5] by treating Eq. A.7.3 as a difference equation and applying z-transform to The last form in Eq. A.7.10 is the same as it. Eq. A.7.5. In this section, we give a simple derivation of Eq. A.7.5 and a modiﬁcation of this formula which directly gives the coefﬁcients of xn−2r, r = Simplification of Eq. A.7.10 0 to bn=2c, and facilitates faster computation as compared to the existing methods. We can write Eq. A.7.10 as bX n=2c k n 2 k Derivation Cn ðxÞ ¼ x n ð1Þ x 1 x2 k¼0 2k Let ðA:7:11Þ cos1 x ¼ h ðA:7:7Þ The term [x−2(1 − x2)]k in Eq. A.7.11 can be expressed as k so that X k 2 k k 2 k k 1 h jh n jh n i ðx 1Þ ¼ ð1Þ ð1 x Þ ¼ ð1Þ 1 r Cn ðxÞ ¼ cos nh ¼ ejnh þ ejnh ¼ e þ e r¼0 k 2 2 1 h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃn pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃn i X k ¼ x þ j 1 x2 þ x j 1 x2 ðx2 Þr ¼ ð1Þ k ð1Þr x2r : 2 r 1 h pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃn pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃn i r¼0 ¼ x x2 1 þ x þ x2 1 : ðA:7:12Þ 2 ðA:7:8Þ Substituting the last form in Eq. A.7.12 for k Note that this is the same as equation Eq. ½ðx2 ð1x2 Þ in Eq. A.7.11, we get A.7.12 in [5], derived by using z-transforms. n=2c bX X k Using the Binomial theorem, we get n k r Cn ðxÞ ¼ ð1Þ xn2r : n 2k r pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃn X k¼0 r¼0 n x x2 1 ¼ ðA:7:13Þ i¼0 i pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃi where (n/2) has the usual signiﬁcance. Equation ð1Þi xni x2 1 ðA:7:9Þ Eq. A.7.13 can also be written as A.7: Computation of the Coefficients of Chebyshev Polynomials 319 bX n=2c 3 X 7 7 7 7 7 Cn ðxÞ ¼ an2r x n2r ; ðA:7:14Þ r ¼ 0 : a7 ¼ ¼ þ þ þ k¼0 2k 0 2 4 6 r¼0 ¼ 1 þ 21 þ 35 þ 7 ¼ 64; X3 where 7 k 7 1 r ¼ 1 : a5 ¼ ¼ k¼0 2k 1 2 1 n=2c bX n k 7 2 7 3 an2r ¼ ð1Þr : ðA:7:15Þ þ þ 2k r 4 1 6 1 k¼0 ¼ ð21 1 þ 35 2 þ 7 3Þ ¼ 112; X3 Equations A.7.14 and A.7.15 constitute the 7 k 7 2 7 3 r ¼ 2 : a3 ¼ ¼ þ simpliﬁed formula for computation. In using k¼0 2k 2 4 2 6 2 these, note the following additional simplifying ¼ ð35 1 þ 7 3Þ ¼ 56; features: (1) the quantities n n n n and ; ; ;... are required 0 2 4 2bn=2c for each coefﬁcient and can be pre-calculated and X3 7 k r ¼ 3 : a1 ¼ k 2k 3 stored; (2) ¼ 0 for k\r; and k¼0 r 7 3 k k ¼ ¼ 7 1 ¼ 7: (3) ¼ ¼ 1: 6 3 0 k We now illustrate the computation with two Thus examples. C7 ðxÞ ¼ 64x7 112x5 þ 56x3 7x: ðA:7:18Þ Examples Next, consider the example of n = 8. With the Consider the case of n = 7. From Eqs. A.7.14 experience of the previous example, we can and A.7.15, we get directly write X 3 4 X C7 ðxÞ ¼ a72r x72r ; ðA:7:16Þ a82r ¼ 8 ð1Þr k r¼0 k¼0 2k r 8 0 8 1 8 2 where ¼ ð1Þr þ þ 0 r 2 r 4 r X3 þ 8 3 þ 8 4 7 k a72r ¼ ð1Þr : ðA:7:17Þ 6 r 8 r 2k r k¼0 r 0 1 2 ¼ ð1Þ þ 28 þ 70 r r r For various values of r, the coefﬁcients are 3 4 calculated as follows: þ 28 þ : r r ðA:7:19Þ 320 A.7: Computation of the Coefficients of Chebyshev Polynomials For various values of r, Eq. A.7.19 gives Problems r ¼ 0 : a8 ¼ 1 þ 28 þ 70 þ 28 þ 1 ¼ 128; r ¼ 1 : a6 ¼ ð28 1 þ 70 2 þ 28 3 þ 4Þ P:1. Any other method that you can ﬁnd out for ¼ 256; deriving Eq. A.7.6? P:2. Compute C15(x). r ¼ 2 : a4 ¼ ð70 1 þ 28 3 þ 6Þ ¼ 160; P:3. Repeat for C16(x). r ¼ 3 : a2 ¼ ð28 l þ 4Þ ¼ 32; P:4. Compare Eq. A.7.18 with a Butterworth polynomial of the same order. What differ- and ences do you observe? P:5. What about Legendre polynomials? Are r ¼ 4 : a0 ¼ 1: you not familiar with Legendre, a cousin of Butterworth? Read Kuo and thy will come Thus to know. C8 ðxÞ ¼ 128x8 256x6 þ 160x4 32x2 þ 1: ðA:7:20Þ References Equations A.7.18 and A.7.20 agree with those calculated by using any other method. 1. A. Budak, Passive and Active Network Analysis and Synthesis (Houghton Mifﬁn, 1974) Conclusion 2. H. Lam, Analog and Digital Filters (Prentice Hall, 1979) A method has been presented for rapid calcula- 3. T.V. Nguyen, A triangle of coefﬁcients for tion of the coefﬁcients of Chebyshev polynomi- Chebyshev polynomials, in Proceedings of als of high order. The method should be useful in IEEE, vol. 72 (July 1984), pp. 982–983 designing high order equal ripple ﬁlters. 4. D.E. Johnson, J.R. Johnson, Mathematical Methods in Engineering Physics (Ronals Press, 1965) 5. J.D. Cole, A new derivation of a closed from expression for Chebyshev polynomials of any order. IEEE Trans. Educ. 32, 390–392 (1989)

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J.W. Cooley, J.W. Tukey An algorithm for the machine calculation of complex fourier series. Math. Comput. 19, 297-301 (April 1965)
W.M. Gentleman, G. Sande, Fast Fourier Transforms-for Fun and Profit. in 1966 Fall Joint Computer Conference of AFIPS Proceedings, pp. 563-578
S.K. Mitra, Digital Signal Processing-A Computer Based Approach, Second Edition (McGraw-Hill, New York, 2000)
260 33 On Second-Order Digital Band-Pass and Band-Stop Filters References
P.A. Regalia, S.K. Mitra, P.P. Vaidyanathan, The digital all-pass network: a versatile signal processing building block. Proc. IEEE 76, 19-37 (1988)
S.K Mitra, K. Hirano, Digital all-pass networks. IEEE Trans. Circ. Sys. 21, 688-700 (September 1974)
A.V. Oppenheim, R.W. Schafer, J.R. Buck, Discrete-Time Signal Processing (Prentice Hall, New Jersey, 2000), p. 363
266 34 Derivation of Second-Order Canonic All-Pass …
S.K. Mitra, Digital Signal Processing-A Computer-Based Approach (McGrawHill, New York, 2001)
A.V. Oppenheim, R.W. Schafer, Discrete-Time Signal Processing, Englewood Cliffs (Prentice Hall, NJ, 1989)
J.G. Proakis, D.G. Manolakis, Introduction to Digital Signal Processing (Macmillan, New York, 1989)
S.K. Mitra, R.J. Sherwood, Digital ladder networks. IEEE Trans. Audio Electroacoust. AU-21, 30-36 (February 1973)
P.P. Vaidyanathan, Passive cascaded lattice structures for low-sensitivity FIR design, with applications to filter banks. IEEE Trans. Circ. Syst. CAS-33, 1045- 1064 (November 1986)
S.C. Dutta Roy, Synthesis of FIR lattice structures. IEE Proc-Vis Image Signal Process 147, 549-552 (2000)
A.V. Oppenheim, R.W. Schafer, Discrete Time Signal Processing (Prentice Hall, New Jersey, 1989)
M. Bellanger, Digital Processing of Signals (Wiley, Hoboken, 1984)
J.G. Proakis, D.G. Manolakis, Digital Signal Process- ing (McMillan, Basingstoke, 1992)
L.B. Jackson, Digital Filters and Signal Processing (Kluwer, Alphen aan den Rijn, 1989)
Problems 277
F. Itakura, S. Saito, Digital Filtering Techniques for Speech Analysis and Synthesis, in Proc 7th lnt Cong Acoust, (Budapest, Hungary, 1971) pp. 261-264
S.C. Dutta Roy, R. Vishwanath, Derivation of the FIR lattice. IETE J. Educ. 45, 211-212 (October-Decem- ber 2004)
S.C. Dutta Roy, R. Vishwanath, Another FIR lattice structure, Int. J. Circ. Theor. Appl. 33, 347-351, (July-August 2005)
S.K. Mitra, Digital Signal Processing: A Computer Based Approach (McGrawHill, New York, 2001)
A.H. Gray, J.D. Markel, Digital lattice and ladder filter synthesis. IEEE Trans. Audio Electroacoust. AU-21, 491-500 (December 1973)
J. Makhoul, A class of all-zero lattice digital filters: properties and applications. IEEE Trans. Acoust. Speech Sig. Process. ASSP-26, 304-314, (August 1978)
Z. Doganata, P.P. Vaidyanathan, On one-multiplier implementations of FIR lattice structures. IEEE Trans. Circ. Syst. CAS-34, 1608-1609 (December 1987)
H. Krishna, An eigen-decomposition approach to one-multiplier realizations of FIR lattice structures. IEEE Trans. Circ. Syst. CAS-36, 145-146, (January 1989) 282 37 FIR Lattice Structures with Single-Multiplier Sections
A.V. Oppenheim, R.W. Schafer, Digital Signal Pro- cessing (Prentice Hall, New Jersey, 1975)
286 References
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M. Abramowitz, I.A. Stegun (eds.), Hand- book of Mathematical Functions (Dover, New York, 1965)
S.C. Dutta Roy, On the solution of quadratic and cubic equations. IETE J. Educ. 47(2), 91-95 (2006)
H. Sohon, Engineering Mathematics (Van Nostrand, New York, 1953), p. 65
W.H. Hayt Jr., J.E. Kemmerly, Engineering Circuit Analysis (McGraw-Hill, New York, 1978), pp. 747-748
W.H. Hayt Jr., J.E. Kemmerly, Engineering Circuit Analysis (McGraw-Hill, New York, 1962), p. 283
A.G. Beged-Dov, Another look at Euler's relation. IEEE Trans. Educ. E-9, 44 (1966)
A.2: A Fresh Look at the Euler's Relation References 1. M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions (Dover, 1965)
G.A. Korn, J.M. Korn, Mathematical Handbook for Scientists and Engineers (McGraw-Hill, 1968)
R.S. Burington, Handbook of Mathematical Tables and Formulas (McGraw-Hill, 1973)
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http://www.sosmath.com/algebra/factor/ fac12/fac12.html: The quartic formula
http://mathforum.org/dr.math/faq/cubic. equations.html: Cubic and quartic equations 10. http://www.sosmath.com/algebra/factor/ fac1/fac1l.html: The cubic formula
http://mathforum.org/dr.math/faq/cubic. equations.html: Cubic equations-another solution
S. Neumark, Solution of Cubic and Quartic Equations (Pergamon, 1965)
P:5. Suppose the middle term on the LHS of Eq. A.5.1 is 2ayy′. Can you find a solution?
S.C. Dutta Roy, Transients in RLC networks revisited. IETE J. Educ. 44, 207-211 (2003)
J.D. Garrison, On the solution of the equation for damped oscillation. Am. J. Phys. 42, 694- 695 (1974)
J.D. Garrison, Erratum: on the solution of the equation for damped oscillation. Am. J. Phys. 43, 463 (1975)
S. Balasubramanian, R. Fatchally, Comment on the solution of the equation for damped oscillation. Am. J. Phys. 44, 705 (1976)
H. Greenberg, Further remarks concerning the solution of the equation € x þ 2ax þ x 2 x ¼ 0. Am. J. Phys. 44, 1135- 1136 (1976) (Note that in Eqs. (5) and (6) of this contribution, (x 2 -a 2 ) should be replaced by (a 2 -x 2 ))
D.S. Zrnic, Additional remarks on the equa- tion € x þ 2ax þ x 2 x ¼ 0. Am. J. Phys. 41, 712 (1973) (Note that in Eq. (3) of this contribu- tion, the sign of the (1, 2) element of the matrix should be positive)
S.C. Dutta Roy, An introduction to the state variable characterization of linear systems- part I. IETE J. Educ. 38, 11-18 (1997)
S.C. Dutta Roy, An introduction to the state variable characterization of linear systems- part II. IETE J. Educ. 38, 99-107 (1997)
S.C. Dutta Roy, Solution of an Ordinary Linear Differential Equation With Constant Coefficients, unpublished manuscript. That gives me an idea. I should try to publish this manuscript as soon as possible.
P.C. Yip, On a conjecture for the design of low-pass recursive filters. IEEE Trans. ASSP-28, 6, 768 (1980)
K. Sbenoi, B.P. Agrawal, On the design of recursive low-pass digital filters. IEEE Trans. ASSP-28, 1, 79-84 (1980)
A.6: Proofs of Two Chebyshev Polynomial Identities Useful in Digital Filter Design References
A. Budak, Passive and Active Network Analysis and Synthesis (Houghton Miffin, 1974)
H. Lam, Analog and Digital Filters (Prentice Hall, 1979)
T.V. Nguyen, A triangle of coefficients for Chebyshev polynomials, in Proceedings of IEEE, vol. 72 (July 1984), pp. 982-983
D.E. Johnson, J.R. Johnson, Mathematical Methods in Engineering Physics (Ronals Press, 1965)
J.D. Cole, A new derivation of a closed from expression for Chebyshev polynomials of any order. IEEE Trans. Educ. 32, 390-392 (1989)

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