Signals and Systems Using MATLAB
Luis F. Chaparro Department of Electrical and Computer Engineering University of Pitt...
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Signals and Systems Using MATLAB
Luis F. Chaparro Department of Electrical and Computer Engineering University of Pittsburgh
AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Academic Press is an imprint of Elsevier
Academic Press is an imprint of Elsevier 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA Elsevier, The Boulevard, Langford Lane, Kidlington, Oxford, OX5 1GB, UK c 2011 Elsevier Inc. All rights reserved. Copyright No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than R as may be noted herein). MATLAB is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks R does not warrant the accuracy of the text or exercises in this book. This books use or discussion of MATLAB software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical R approach or particular use of the MATLAB software. Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. Library of Congress CataloginginPublication Data Chaparro, Luis F. R Signals and systems using MATLAB / Luis F. Chaparro. p. cm. ISBN 9780123747167 1. Signal processing–Digital techniques. 2. System analysis. 3. MATLAB. I. Title. TK5102.9.C472 2010 621.382’2–dc22 2010023436 British Library CataloguinginPublication Data A catalogue record for this book is available from the British Library.
For information on all Academic Press publications visit our Web site at www.elsevierdirect.com Printed in the United States of America 10 11 12 13 9 8 7 6 5 4 3 2 1
To my family, with much love.
Contents
PREFACE .....................................................................................................................
xi
ACKNOWLEDGMENTS ................................................................................................ xvi
Part 1
Introduction
1
CHAPTER 0
From the Ground Up! .............................................................................
3
0.1
Signals and Systems and Digital Technologies ........................................
3
0.2
Examples of Signal Processing Applications ...........................................
5
0.3
0.2.1 CompactDisc Player ................................................................ 0.2.2 SoftwareDefined Radio and Cognitive Radio............................... 0.2.3 ComputerControlled Systems ................................................... Analog or Discrete? .............................................................................
5 6 8 9
0.4
0.3.1 ContinuousTime and DiscreteTime Representations .................. 0.3.2 Derivatives and Finite Differences ............................................. 0.3.3 Integrals and Summations......................................................... 0.3.4 Differential and Difference Equations ......................................... Complex or Real? ................................................................................
10 12 13 16 20
0.5
0.4.1 Complex Numbers and Vectors.................................................. 0.4.2 Functions of a Complex Variable ................................................ 0.4.3 Phasors and Sinusoidal Steady State .......................................... 0.4.4 Phasor Connection ................................................................... Soft Introduction to MATLAB ...............................................................
20 23 24 26 29
0.5.1 Numerical Computations .......................................................... 0.5.2 Symbolic Computations ............................................................ Problems............................................................................................
30 43 53
Part 2 CHAPTER 1
iv
Theory and Application of ContinuousTime Signals and Systems
63
ContinuousTime Signals .........................................................................
65
1.1
Introduction .......................................................................................
65
1.2
Classification of TimeDependent Signals...............................................
66
Contents
1.3
ContinuousTime Signals .....................................................................
67
1.4
1.3.1 Basic Signal Operations—Time Shifting and Reversal ................... 1.3.2 Even and Odd Signals .............................................................. 1.3.3 Periodic and Aperiodic Signals .................................................. 1.3.4 FiniteEnergy and Finite Power Signals ...................................... Representation Using Basic Signals.......................................................
71 75 77 79 85
1.4.1 1.4.2 1.4.3 1.4.4
1.5
Complex Exponentials .............................................................. UnitStep, UnitImpulse, and Ramp Signals ................................. Special Signals—the Sampling Signal and the Sinc ....................... Basic Signal Operations—Time Scaling, Frequency Shifting, and Windowing ....................................................................... 1.4.5 Generic Representation of Signals.............................................. What Have We Accomplished? Where Do We Go from Here?....................
85 88 100 102 105 106
Problems............................................................................................ 108
CHAPTER 2
ContinuousTime Systems ....................................................................... 117 2.1
Introduction ....................................................................................... 117
2.2
System Concept .................................................................................. 118
2.3
2.2.1 System Classification................................................................ 118 LTI ContinuousTime Systems .............................................................. 119
2.4
2.3.1 Linearity ................................................................................. 2.3.2 Time Invariance ....................................................................... 2.3.3 Representation of Systems by Differential Equations .................... 2.3.4 Application of Superposition and Time Invariance ....................... 2.3.5 Convolution Integral................................................................. 2.3.6 Causality ................................................................................ 2.3.7 Graphical Computation of Convolution Integral ........................... 2.3.8 Interconnection of Systems—Block Diagrams .............................. 2.3.9 BoundedInput BoundedOutput Stability ................................... What Have We Accomplished? Where Do We Go from Here?....................
120 125 130 135 136 143 145 147 153 156
Problems............................................................................................ 157
CHAPTER 3
The Laplace Transform ............................................................................ 165 3.1
Introduction ....................................................................................... 165
3.2
The TwoSided Laplace Transform ........................................................ 166
3.3
3.2.1 Eigenfunctions of LTI Systems................................................... 167 3.2.2 Poles and Zeros and Region of Convergence ............................... 172 The OneSided Laplace Transform ........................................................ 176 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5
Linearity ................................................................................. Differentiation ......................................................................... Integration .............................................................................. Time Shifting........................................................................... Convolution Integral.................................................................
185 188 193 194 196
v
vi
Contents
3.4
Inverse Laplace Transform ................................................................... 197
3.5
3.4.1 Inverse of OneSided Laplace Transforms ................................... 3.4.2 Inverse of Functions Containing e−ρs Terms ................................ 3.4.3 Inverse of TwoSided Laplace Transforms ................................... Analysis of LTI Systems .......................................................................
3.6
3.5.1 LTI Systems Represented by Ordinary Differential Equations ........ 214 3.5.2 Computation of the Convolution Integral .................................... 221 What Have We Accomplished? Where Do We Go from Here?.................... 226
197 209 212 214
Problems............................................................................................ 226
CHAPTER 4
Frequency Analysis: The Fourier Series .................................................. 237 4.1
Introduction ....................................................................................... 237
4.2
Eigenfunctions Revisited ..................................................................... 238
4.3
Complex Exponential Fourier Series ...................................................... 245
4.4
Line Spectra ....................................................................................... 248
4.5
4.4.1 Parseval’s Theorem—Power Distribution over Frequency ............. 248 4.4.2 Symmetry of Line Spectra ......................................................... 250 Trigonometric Fourier Series ................................................................ 251
4.6
Fourier Coefficients from Laplace .......................................................... 255
4.7
Convergence of the Fourier Series......................................................... 265
4.8
Time and Frequency Shifting................................................................ 270
4.9
Response of LTI Systems to Periodic Signals........................................... 273
4.9.1 Sinusoidal Steady State............................................................. 274 4.9.2 Filtering of Periodic Signals ....................................................... 276 4.10 Other Properties of the Fourier Series .................................................... 279 4.10.1 Reflection and Even and Odd Periodic Signals ............................. 4.10.2 Linearity of Fourier Series—Addition of Periodic Signals ............... 4.10.3 Multiplication of Periodic Signals ............................................... 4.10.4 Derivatives and Integrals of Periodic Signals ............................... 4.11 What Have We Accomplished? Where Do We Go from Here?....................
279 282 284 285 289
Problems............................................................................................ 290
CHAPTER 5
Frequency Analysis: The Fourier Transform ........................................... 299 5.1
Introduction ....................................................................................... 299
5.2
From the Fourier Series to the Fourier Transform .................................... 300
5.3
Existence of the Fourier Transform ....................................................... 302
5.4
Fourier Transforms from the Laplace Transform ..................................... 302
5.5
Linearity, Inverse Proportionality, and Duality ........................................ 304 5.5.1 5.5.2 5.5.3
Linearity ................................................................................. 304 Inverse Proportionality of Time and Frequency ............................ 305 Duality ................................................................................... 310
Contents
5.6
Spectral Representation ....................................................................... 313
5.7
5.6.1 Signal Modulation .................................................................... 5.6.2 Fourier Transform of Periodic Signals ......................................... 5.6.3 Parseval’s Energy Conservation................................................. 5.6.4 Symmetry of Spectral Representations........................................ Convolution and Filtering.....................................................................
313 317 320 322 327
5.8
5.7.1 Basics of Filtering .................................................................... 5.7.2 Ideal Filters ............................................................................. 5.7.3 Frequency Response from Poles and Zeros.................................. 5.7.4 Spectrum Analyzer................................................................... Additional Properties ..........................................................................
329 332 337 341 344
5.9
5.8.1 Time Shifting .......................................................................... 344 5.8.2 Differentiation and Integration .................................................. 346 What Have We Accomplished? What Is Next? ....................................... 350 Problems............................................................................................ 350
CHAPTER 6
Application to Control and Communications ........................................... 359 6.1
Introduction ....................................................................................... 359
6.2
System Connections and Block Diagrams ............................................... 360
6.3
Application to Classic Control............................................................... 363
6.4
6.3.1 Stability and Stabilization ......................................................... 369 6.3.2 Transient Analysis of First and SecondOrder Control Systems ..... 371 Application to Communications ............................................................ 377
6.5
6.4.1 AM with Suppressed Carrier ..................................................... 6.4.2 Commercial AM ....................................................................... 6.4.3 AM Single Sideband ................................................................. 6.4.4 Quadrature AM and FrequencyDivision Multiplexing .................. 6.4.5 Angle Modulation .................................................................... Analog Filtering..................................................................................
379 380 382 383 385 390
6.6
6.5.1 Filtering Basics........................................................................ 6.5.2 Butterworth LowPass Filter Design........................................... 6.5.3 Chebyshev LowPass Filter Design ............................................ 6.5.4 Frequency Transformations ...................................................... 6.5.5 Filter Design with MATLAB ...................................................... What Have We Accomplished? What Is Next? ........................................
390 393 396 402 405 409
Problems............................................................................................ 409
Part 3 CHAPTER 7
Theory and Application of DiscreteTime Signals and Systems
417
Sampling Theory ...................................................................................... 419 7.1
Introduction ....................................................................................... 419
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Contents
7.2
Uniform Sampling ............................................................................... 420
7.3
7.2.1 Pulse Amplitude Modulation ..................................................... 7.2.2 Ideal Impulse Sampling ............................................................ 7.2.3 Reconstruction of the Original ContinuousTime Signal ................ 7.2.4 Signal Reconstruction from Sinc Interpolation.............................. 7.2.5 Sampling Simulation with MATLAB ........................................... The NyquistShannon Sampling Theorem ..............................................
7.4
7.3.1 Sampling of Modulated Signals .................................................. 438 Practical Aspects of Sampling ............................................................... 439
7.5
7.4.1 SampleandHold Sampling ....................................................... 7.4.2 Quantization and Coding .......................................................... 7.4.3 Sampling, Quantizing, and Coding with MATLAB ........................ What Have We Accomplished? Where Do We Go from Here?....................
420 421 428 432 433 437
439 441 444 446
Problems............................................................................................ 447
CHAPTER 8
DiscreteTime Signals and Systems ......................................................... 451 8.1
Introduction ..................................................................................... 451
8.2
DiscreteTime Signals .......................................................................... 452
8.3
8.2.1 Periodic and Aperiodic Signals .................................................. 8.2.2 FiniteEnergy and FinitePower DiscreteTime Signals ................. 8.2.3 Even and Odd Signals .............................................................. 8.2.4 Basic DiscreteTime Signals ...................................................... DiscreteTime Systems ........................................................................
454 458 461 465 478
8.3.1 8.3.2
481
8.4
Recursive and Nonrecursive DiscreteTime Systems..................... DiscreteTime Systems Represented by Difference Equations ............................................................................... 8.3.3 The Convolution Sum ............................................................... 8.3.4 Linear and Nonlinear Filtering with MATLAB.............................. 8.3.5 Causality and Stability of DiscreteTime Systems ......................... What Have We Accomplished? Where Do We Go from Here?....................
486 487 494 497 502
Problems............................................................................................ 502
CHAPTER 9
The ZTransform ...................................................................................... 511 9.1
Introduction ....................................................................................... 511
9.2
Laplace Transform of Sampled Signals................................................... 512
9.3
TwoSided ZTransform ....................................................................... 515
9.4
9.3.1 Region of Convergence ............................................................. 516 OneSided ZTransform........................................................................ 521 9.4.1 9.4.2 9.4.3
Computing the ZTransform with Symbolic MATLAB ................... 522 Signal Behavior and Poles ......................................................... 522 Convolution Sum and Transfer Function ..................................... 526
Contents
9.5
9.4.4 Interconnection of DiscreteTime Systems................................... 537 9.4.5 Initial and Final Value Properties ............................................... 539 OneSided ZTransform Inverse ............................................................ 542
9.6
9.5.1 LongDivision Method .............................................................. 9.5.2 Partial Fraction Expansion ........................................................ 9.5.3 Inverse ZTransform with MATLAB............................................ 9.5.4 Solution of Difference Equations ................................................ 9.5.5 Inverse of TwoSided ZTransforms ............................................ What Have We Accomplished? Where Do We Go from Here?....................
542 544 547 550 561 564
Problems............................................................................................ 564
CHAPTER 10 Fourier Analysis of DiscreteTime Signals and Systems ........................... 571 10.1 Introduction ....................................................................................... 571 10.2 DiscreteTime Fourier Transform .......................................................... 572 10.2.1 Sampling, ZTransform, Eigenfunctions, and the DTFT ................. 10.2.2 Duality in Time and Frequency .................................................. 10.2.3 Computation of the DTFT Using MATLAB .................................. 10.2.4 Time and Frequency Supports ................................................... 10.2.5 Parseval’s Energy Result........................................................... 10.2.6 Time and Frequency Shifts........................................................ 10.2.7 Symmetry ............................................................................... 10.2.8 Convolution Sum ..................................................................... 10.3 Fourier Series of DiscreteTime Periodic Signals......................................
573 575 577 580 585 587 589 595 596
10.3.1 Complex Exponential Discrete Fourier Series .............................. 10.3.2 Connection with the ZTransform .............................................. 10.3.3 DTFT of Periodic Signals ........................................................... 10.3.4 Response of LTI Systems to Periodic Signals ............................... 10.3.5 Circular Shifting and Periodic Convolution .................................. 10.4 Discrete Fourier Transform ..................................................................
599 601 602 604 607 614
10.4.1 DFT of Periodic DiscreteTime Signals ........................................ 10.4.2 DFT of Aperiodic DiscreteTime Signals ...................................... 10.4.3 Computation of the DFT via the FFT .......................................... 10.4.4 Linear and Circular Convolution Sums ........................................ 10.5 What Have We Accomplished? Where Do We Go from Here?....................
614 616 617 622 628
Problems............................................................................................ 629
CHAPTER 11 Introduction to the Design of Discrete Filters .......................................... 639 11.1 Introduction ....................................................................................... 639 11.2 FrequencySelective Discrete Filters ...................................................... 641 11.2.1 Linear Phase ........................................................................... 641 11.2.2 IIR and FIR Discrete Filters ....................................................... 643
ix
x
Contents
11.3 Filter Specifications ............................................................................. 648 11.3.1 FrequencyDomain Specifications .............................................. 648 11.3.2 TimeDomain Specifications ...................................................... 652 11.4 IIR Filter Design.................................................................................. 653 11.4.1 Transformation Design of IIR Discrete Filters .............................. 11.4.2 Design of Butterworth LowPass Discrete Filters ......................... 11.4.3 Design of Chebyshev LowPass Discrete Filters ........................... 11.4.4 Rational Frequency Transformations .......................................... 11.4.5 General IIR Filter Design with MATLAB ..................................... 11.5 FIR Filter Design .................................................................................
654 658 666 672 677 679
11.5.1 Window Design Method ........................................................... 681 11.5.2 Window Functions ................................................................... 683 11.6 Realization of Discrete Filters ............................................................... 689 11.6.1 Realization of IIR Filters ............................................................ 690 11.6.2 Realization of FIR Filters ........................................................... 699 11.7 What Have We Accomplished? Where Do We Go from Here?.................... 701 Problems............................................................................................ 701
CHAPTER 12 Applications of DiscreteTime Signals and Systems ................................. 709 12.1 Introduction ....................................................................................... 709 12.2 Application to Digital Signal Processing ................................................. 710 12.2.1 Fast Fourier Transform ............................................................. 12.2.2 Computation of the Inverse DFT ................................................ 12.2.3 General Approach of FFT Algorithms ......................................... 12.3 Application to SampledData and Digital Control Systems ........................
711 715 716 722
12.3.1 OpenLoop SampledData System .............................................. 724 12.3.2 ClosedLoop SampledData System ............................................ 726 12.4 Application to Digital Communications .................................................. 729 12.4.1 Pulse Code Modulation ............................................................. 12.4.2 TimeDivision Multiplexing ....................................................... 12.4.3 Spread Spectrum and Orthogonal FrequencyDivision Multiplexing............................................................................ 12.5 What Have We Accomplished? Where Do We Go from Here?....................
APPENDIX
730 733 735 742
Useful Formulas ....................................................................................... 743
BIBLIOGRAPHY ........................................................................................................... 746 INDEX ......................................................................................................................... 749
Preface
In this book I have only made up a bunch of other men’s flowers, providing of my own only the string that ties them together. M. de Montaigne (1533–1592) French essayist Although it is hardly possible to keep up with advances in technology, it is reassuring to know that in science and engineering, development and innovation are possible through a solid understanding of basic principles. The theory of signals and systems is one of those fundamentals, and it will be the foundation of much research and development in engineering for years to come. Not only engineers will need to know about signals and systems—to some degree everybody will. The pervasiveness of computers, cell phones, digital recording, and digital communications will require it. Learning as well as teaching signals and systems is complicated by the combination of mathematical abstraction and concrete engineering applications. Mathematical sophistication and maturity in engineering are needed. Thus, a course in signals and systems needs to be designed to nurture the students’ interest in applications, but also to make them appreciate the significance of the mathematical tools. In writing this textbook, as in teaching this material for many years, the author has found it practical to follow Einstein’s recommendation that “Everything should be made as simple as possible, but not simpler,” and Melzak’s [47] dictum that “It is downright sinful to teach the abstract before the concrete.” The aim of this textbook is to serve the students’ needs in learning signals and systems theory as well as to facilitate the teaching of the material for faculty by proposing an approach that the author has found effective in his own teaching. We consider the use of MATLAB, an essential tool in the practice of engineering, of great significance in the learning process. It not only helps to illustrate the theoretical results but makes students aware of the computational issues that engineers face in implementing them. Some familiarity with MATLAB is beneficial but not required.
LEVEL The material in this textbook is intended for courses in signals and systems at the junior level in electrical and computer engineering, but it could also be used in teaching this material to mechanical engineering and bioengineering students and it might be of interest to students in applied mathematics. The “studentfriendly” nature of the text also makes it useful to practicing engineers interested in learning or reviewing the basic principles of signals and systems on their own. The material is organized so that students not only get a solid understanding of the theory—through analytic examples as well as software examples using MATLAB—and learn about applications, but also develop confidence and proficiency in the material by working on problems.
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Preface
The organization of the material in the book follows the assumption that the student has been exposed to the theory of linear circuits, differential equations, and linear algebra, and that this material will be followed by courses in control, communications, or digital signal processing. The content is guided by the goal of nurturing the interest of students in applications, and of assisting them in becoming more sophisticated mathematically. In teaching signals and systems, the author has found that students typically lack basic skills in manipulating complex variables, in understanding differential equations, and are not yet comfortable with basic concepts in calculus. Introducing discretetime signals and systems makes students face new concepts that were not explored in their calculus courses, such as summations, finite differences, and difference equations. This text attempts to fill the gap and nurture interest in the mathematical tools.
APPROACH In writing this text, we have taken the following approach: 1. The material is divided into three parts: introduction, theory and applications of continuoustime signals and systems, and theory and applications of discretetime signals and systems. To help students understand the connection between continuous and discretetime signals and systems, the connection between infinitesimal and finite calculus is made in the introduction part, together with a motivation as to why complex numbers and functions are used in the study of signals and systems. The treatment of continuous and discretetime signals and systems is then done separately in the next two parts; combining them is found to be confusing to students. Likewise, the author believes it is important for students to understand the connections and relevance of each of the transformations used in the analysis of signals and systems so that these transformations are seen as a progression rather than as disconnected methods. Thus, the author advocates the presentation of the Laplace analysis followed by the Fourier analysis, and the Ztransform followed by the discrete Fourier, and capping each of these topics with applications to communications, control, and filtering. The mathematical abstraction and the applications become more sophisticated as the material unfolds, taking advantage as needed of the background on circuits that students have. 2. An overview of the topics to be discussed in the book and how each connects with some basic mathematical concepts—needed in the rest of the book—is given in Chapter 0 (analogous to the ground floor of a building). The emphasis is in relating summations, differences, difference equations, and sequence of numbers with the calculus concepts that the students are familiar with, and in doing so providing a new interpretation to integrals, derivatives, differential equations, and functions of time. This chapter also links the theory of complex numbers and functions to vectors and to phasors learned in circuit theory. Because we strongly believe that the material in this chapter should be covered before beginning the discussion of signals and systems, it is not relegated to an appendix but placed at the front of the book where it cannot be ignored. A soft introduction to MATLAB is also provided in this chapter. 3. A great deal of effort has been put into making the text “student friendly.” To make sure that the student does not miss some of the important issues presented in a section, we have inserted wellthoughtout remarks— we want to minimize the common misunderstandings we have observed from our students in the past. Plenty of analytic examples with different levels of complexity are given to illustrate issues. Each chapter has a set of examples in MATLAB, illustrating topics presented in the text or special issues that the student should know. The MATLAB code is given so that students can learn by example from it. To help students follow the mathematical derivations, we provide extra steps whenever necessary and do not skip steps that are necessary in the understanding of a derivation. Summaries of important issues are boxed and concepts and terms are emphasized to help students grasp the main points and terminology. 4. Without any doubt, learning the material in signals and systems requires working analytical as well as computational problems. It is important to provide problems of different levels of complexity to exercise not only basic problemsolving skills, but to achieve a level of proficiency and mathematical sophistication. The problems at the end of the chapter are of different types, some to be done analytically, others using
Preface
MATLAB, and some both. The repetitive type of problem was avoided. Some of the problems explore issues not covered in the text but related to it. The MATLAB problems were designed so that a better understanding of the theoretical concepts is attained by the student working them out. 5. We feel two additional features would be beneficial to students. One is the inclusion of quotations and footnotes to present interesting ideas or historical comments, and the other is the inclusion of sidebars that attempt to teach historical or technical information that students should be aware of. The theory of signals and systems clearly connects with mathematics and a great number of mathematicians have contributed to it. Likewise, there is a large number of engineers who have contributed significantly to the development and application of signals and systems. All of them need to be recognized for their contributions, and we should learn from their experiences. 6. Finally, other features are: (1) the design of the index of the book so that it can be used by students to find definitions, symbols, and MATLAB functions used in the text; and (2) a list of references to the material.
CONTENT The core of the material is presented in the second and third part of the book. The second part of the book covers the basics of continuoustime signals and systems and illustrates their application. Because the concepts of signals and systems are relatively new to students, we provide an extensive and complete presentation of these topics in Chapters 1 and 2. The presentation in Chapter 1 goes from a very general characterization of signals to very specific classes that will be used in the rest of the book. One of the aims is to familiarize students with continuoustime as well as discretetime signals so as to avoid confusion in their processing later on—a common difficulty encountered by students. Chapter 1 initiates the representation of signals in terms of basic signals that will be easily processed later with the transform methods. Chapter 2 introduces the general concept of systems, in particular continuoustime systems. The concepts of linearity, time invariance, causality, and stability are introduced in this chapter, trying as much as possible to use the students’ background in circuit theory. Using linearity and time invariance, the computation of the output of a continuoustime system using the convolution integral is introduced and illustrated with relatively simple examples. More complex examples are treated with the Laplace transform in the following chapter. Chapter 3 covers the basics of the Laplace transform and its application in the analysis of continuoustime signals and systems. It introduces the student to the concept of poles and zeros, damping and frequency, and their connection with the signal as a function of time. This chapter emphasizes the solution of differential equations representing linear timeinvariant (LTI) systems, paying special attention to transient solutions due to their importance in control, as well as to steadystate solutions due to their importance in filtering and in communications. The convolution integral is dealt with in time and using the Laplace transform to emphasize the operational power of the transform. The important concept of transfer function for LTI systems and the significance of its poles and zeros are studied in detail. Different approaches are considered in computing the inverse Laplace transform, including MATLAB methods. Fourier analysis of continuoustime signals and systems is covered in detail in Chapters 4 and 5. The Fourier series analysis of periodic signals, covered in Chapter 4, is extended to the analysis of aperiodic signals resulting in the Fourier transform of Chapter 5. The Fourier transform is useful in representing both periodic and aperiodic signals. Special attention is given to the connection of these methods with the Laplace transform so that, whenever possible, known Laplace transforms can be used to compute the Fourier series coefficients and the Fourier transform—thus avoiding integration but using the concept of the region of convergence. The concept of frequency, the response of the system (connected to the location of poles and zeros of the transfer function), and the steadystate response are emphasized in these chapters. The ordering of the presentation of the Laplace and the Fourier transformations (similar to the Ztransform and the Fourier representation of discretetime signals) is significant for learning and teaching of the material.
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Preface
Our approach of presenting first the Laplace transform and then the Fourier series and Fourier transform is justified by several reasons. For one, students coming into a signals and systems course have been familiarized with the Laplace transform in their previous circuits or differential equations courses, and will continue using it in control courses. So expertise in this topic is important and the learned material will stay with them longer. Another is that a common difficulty students have in applying the Fourier series and the Fourier transform is connected with the required integration. The Laplace transform can be used not only to sidestep the integration but to provide a more comprehensive understanding of the frequency representation. By asking students to consider the twosided Laplace transform and the significance of its region of convergence, they will appreciate better the Fourier representation as a special case of Laplace’s in many cases. More importantly, these transforms can be seen as a continuum rather than as different transforms. It also makes theoretical sense to deal with the Laplace representation of systems first to justify the existence of the steadystate solution considered in the Fourier representations, which would not exist unless stability of the system is guaranteed, and stability can only be tested using the Laplace transform. The paradigm of interest is the connection of transient and steadystate responses that must be understood by students before they can understand the connections between Fourier and Laplace analyses. Chapter 6 presents applications of the Laplace and the Fourier transforms to control, communications, and filtering. The intent of the chapter is to motivate interest in these areas. The chapter illustrates the significance of the concepts of transfer function, response of systems, and stability in control, and of modulation in communications. An introduction to analog filtering is provided. Analytic as well as MATLAB examples illustrate different applications to control, communications, and filter design. Using the sampling theory as a bridge, the third part of the book covers the theory and illustrates the application of discretetime signals and systems. Chapter 7 presents the theory of sampling: the conditions under which the signal does not lose information in the sampling process and the recovery of the analog signal from the sampled signal. Once the basic concepts are given, the analogtodigital and digitaltoanalog converters are considered to provide a practical understanding of the conversion of analogtodigital and digitaltoanalog signals. Discretetime signals and systems are discussed in Chapter 8, while Chapter 9 introduces the Ztransform. Although the treatment of discretetime signals and systems in Chapter 8 mirrors that of continuoustime signals and systems, special emphasis is given in this chapter to issues that are different in the two domains. Issues such as the discrete nature of the time, the periodicity of the discrete frequency, the possible lack of periodicity of discrete sinusoids, etc. are considered. Chapter 9 provides the basic theory of the Ztransform and how it relates to the Laplace transform. The material in this chapter bears similarity to the one on the Laplace transform in terms of operational solution of difference equations, transfer function, and the significance of poles and zeros. Chapter 10 presents the Fourier analysis of discrete signals and systems. Given the accumulated experience of the students with continuoustime signals and systems, we build the discretetime Fourier transform (DTFT) on the Ztransform and consider special cases where the Ztransform cannot be used. The discrete Fourier transform (DFT) is obtained from the Fourier series of discretetime signals and sampling in frequency. The DFT will be of great significance in digital signal processing. The computation of the DFT of periodic and aperiodic discretetime signals using the fast Fourier transform (FFT) is illustrated. The FFT is an efficient algorithm for computing the DFT, and some of the basics of this algorithm are discussed in Chapter 12. Chapter 11 introduces students to discrete filtering, thus extending the analog filtering in Chapter 6. In this chapter we show how to use the theory of analog filters to design recursive discrete lowpass filters. Frequency transformations are then presented to show how to obtain different types of filters from lowpass prototype filters. The design of finiteimpulse filters using the window method is considered next. Finally, the implementation of recursive and nonrecursive filters is shown using some basic techniques. By using MATLAB for the design of recursive and nonrecursive discrete filters, it is expected that students will be motivated to pursue on their own the use of more sophisticated filter designs.
Preface
Finally, Chapter 12 explores topics of interest in digital communications, computer control, and digital signal processing. The aim of this chapter is to provide a brief presentation of topics that students could pursue after the basic courses in signals and systems.
TEACHING USING THIS TEXT The material in this text is intended for a twoterm sequence in signals and systems: one on continuoustime signals and systems, followed by a term in discretetime signals and systems with a lab component using MATLAB. These two courses would cover most of the chapters in the text with various degrees of depth, depending on the emphasis the faculty would like to give to the course. As indicated, Chapter 0 was written as a necessary introduction to the rest of the material, but does not need to be covered in great detail—students can refer to it as needed. Chapters 6 and 11 need to be considered together if the emphasis on applications is in filter design. The control, communications, and digital signal processing material in Chapters 6 and 12 can be used to motivate students toward those areas.
TO THE STUDENT It is important for you to understand the features of this book, so you can take advantage of them to learn the material: 1. Refer as often as necessary to the material in Chapter 0 to review or to learn the mathematical background; to understand the overall structure of the material; or to review or learn MATLAB as it applies to signal processing. 2. As you will see, the complexity of the material grows as it develops. The material in part three has been written assuming good understanding of the material in the first two. See also the connection of the material with applications in your own areas of interest. 3. To help you learn the material, clear and concise results are emphasized by putting them in boxes. Justification of these results is then given, complemented with remarks regarding issues that need a bit more clarification, and illustrated with plenty of analytic and computational examples. Important terms are emphasized throughout the text. Tables provide a good summary of properties and formulas. 4. A heading is used in each of the problems at the end of the chapters, indicating how it relates to specific topics and if it requires to use MATLAB to solve it. 5. One of the objectives of this text is to help you learn MATLAB, as it applies to signal and systems, on your own. This is done by providing the soft introduction to MATLAB in Chapter 0, and then by showing examples using simple code in each of the chapters. You will notice that in the first two parts basic components of MATLAB (scripts, functions, plotting, etc.) are given in more detail than in part three. It is assumed you are very proficient by then to supply that on your own. 6. Finally, notice the footnotes, the vignettes, and the historical sidebars that have been included to provide a glance at the background in which the theory and practice of signals and systems have developed.
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Acknowledgments
I would like to acknowledge with gratitude the support and efforts of many people who made the writing of this text possible. First, to my family—my wife Cathy, my children William, Camila, and Juan, and their own families—many thanks for their support and encouragement despite being deprived of my attention. To my academic mentor, Professor Eliahu I. Jury, a deep sense of gratitude for his teachings and for having inculcated in me the love for a scholarly career and for the theory and practice of signals and systems. Thanks to Professor William Stanchina, chair of the Department of Electrical and Computer Engineering at the University of Pittsburgh, for his encouragement and support that made it possible to dedicate time to the project. Sincere thanks to Seda Senay and Mircea Lupus, graduate students in my department. Their contribution to the painful editing and proofreading of the manuscript, and the generation of the solution manual (especially from Ms. Senay) are much appreciated. Equally, thanks to the publisher and its editors, in particular to Joe Hayton and Steve Merken, for their patience, advising, and help with the publishing issues. Thanks also to Sarah Binns for her help with the final editing of the manuscript. Equally, I would like to thank Professor James Rowland from the University of Kansas and the following reviewers for providing significant input and changes to the manuscript: Dimitrie Popescu, Old Dominion University; Hossein Hakim, Worcester Polytechnic Institute; Mark Budnik, Valparaiso University; Periasamy Rajan, Tennessee Tech University; and Mohamed Zohdy, Oakland University. Thanks to my colleagues Amro ElJaroudi and Juan Manfredi for their early comments and suggestions. Lastly, I feel indebted to the many students I have had in my courses in signals and systems over the years I have been teaching this material in the Department of Electrical and Computer Engineering at the University of Pittsburgh. Unknown to them, they contributed to my impetus to write a book that I felt would make the teaching of signals and systems more accessible and fun to future students in and outside the university.
RESOURCES THAT ACCOMPANY THIS BOOK A companion website containing downloadable MATLAB code for the worked examples in the book is available at:
http://booksite.academicpress.com/chaparro For instructors, a solutions manual and image bank containing electronic versions of figures from the book are available by registering at:
www.textbooks.elsevier.com
Also Available for Use with This Book – Elsevier Online Testing Webbased testing and assessment feature that allows instructors to create online tests and assignments which automatically assess student responses and performance, providing them with immediate feedback. Elsevier’s online testing includes a selection of algorithmic questions, giving instructors the ability to create virtually unlimited variations of the same problem. Contact your local sales representative for additional information, or visit http://booksite.academicpress.com/chaparro/ to view a demo chapter.
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1 PART
Introduction
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CHAPTER 0
From the Ground Up!
In theory there is no difference between theory and practice. In practice there is. Lawrence “Yogi” Berra, 1925 New York Yankees baseball player This chapter provides an overview of the material in the book and highlights the mathematical background needed to understand the analysis of signals and systems. We consider a signal a function of time (or space if it is an image, or of time and space if it is a video signal), just like the voltages or currents encountered in circuits. A system is any device described by a mathematical model, just like the differential equations obtained for a circuit composed of resistors, capacitors, and inductors. By means of practical applications, we illustrate in this chapter the importance of the theory of signals and systems and then proceed to connect some of the concepts of integrodifferential Calculus with more concrete mathematics (from the computational point of view, i.e., using computers). A brief review of complex variables and their connection with the dynamics of systems follows. We end this chapter with a soft introduction to MATLAB, a widely used highlevel computational tool for analysis and design. Significantly, we have called this Chapter 0, because it is the ground floor for the rest of the material in the book. Not everything in this chapter has to be understood in a first reading, but we hope that as you go through the rest of the chapters in the book you will get to appreciate that the material in this chapter is the foundation of the book, and as such you should revisit it as often as needed.
0.1 SIGNALS AND SYSTEMS AND DIGITAL TECHNOLOGIES In our modern world, signals of all kinds emanate from different types of devices—radios and TVs, cell phones, global positioning systems (GPSs), radars, and sonars. These systems allow us to communicate messages, to control processes, and to sense or measure signals. In the last 60 years, with the advent of the transistor, the digital computer, and the theoretical fundamentals of digital signal Signals and Systems Using MATLAB®. DOI: 10.1016/B9780123747167.000028 c 2011, Elsevier Inc. All rights reserved.
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CH A P T E R 0: From the Ground Up!
processing, the trend has been toward digital representation and processing of data, most of which are in analog form. Such a trend highlights the importance of learning how to represent signals in analog as well as in digital forms and how to model and design systems capable of dealing with different types of signals.
1948 The year 1948 is considered the birth year of technologies and theories responsible for the spectacular advances in communications, control, and biomedical engineering since then. In June 1948, Bell Telephone Laboratories announced the invention of the transistor. Later that month, a prototype computer built at Manchester University in the United Kingdom became the first operational storedprogram computer. Also in that year, many fundamental theoretical results were published: Claude Shannon’s mathematical theory of communications, Richard W. Hamming’s theory on errorcorrecting codes, and Norbert Wiener’s Cybernetics comparing biological systems with communication and control systems [51].
Digital signal processing advances have gone handinhand with progress in electronics and computers. In 1965, Gordon Moore, one of the founders of Intel, envisioned that the number of transistors on a chip would double about every two years [35]. Intel, the largest chip manufacturer in the world, has kept that pace for 40 years. But at the same time, the speed of the central processing unit (CPU) chips in desktop personal computers has dramatically increased. Consider the wellknown Pentium group of chips (the Pentium Pro and the Pentium I to IV) introduced in the 1990s [34]. Figure 0.1 shows the range of speeds of these chips at the time of their introduction into the market, as well as the number of transistors on each of these chips. In five years, 1995 to 2000, the speed increased by a factor of 10 while the number of transistors went from 5.5 million to 42 million.
MHz
2000
1000
0
FIGURE 0.1 The Intel Pentium CPU chips. (a) Range of CPU speeds in MHz for the Pentium Pro (1995), Pentium II (1997), Pentium III (1999), and Pentium IV (2000). (b) Number of transistors (in millions) on each of the above chips. (Pentium data taken from [34].)
Million transistors
4
1995
1997 Year (a)
1999
2000
40 30 20 10 1995
1996
1997 1998 Year (b)
1999
2000
0.2 Examples of Signal Processing Applications
Advances in digital electronics and in computer engineering in the past 60 years have permitted the proliferation of digital technologies. Digital hardware and software process signals from cell phones, highdefinition television (HDTV) receivers, radars, and sonars. The use of digital signal processors (DSPs) and more recently of fieldprogrammable gate arrays (FPGAs) have been replacing the use of applicationspecific integrated circuits (ASICs) in industrial, medical, and military applications. It is clear that digital technologies are here to stay. Today, digital transmission of voice, data, and video is common, and so is computer control. The abundance of algorithms for processing signals, and the pervasive presence of DSPs and FPGAs in thousands of applications make digital signal processing theory a necessary tool not only for engineers but for anybody who would be dealing with digital data; soon, that will be everybody! This book serves as an introduction to the theory of signals and systems—a necessary first step in the road toward understanding digital signal processing.
DSPs and FPGAs A digital signal processor (DSP) is an optimized microprocessor used in realtime signal processing applications [67]. DSPs are typically embedded in larger systems (e.g., a desktop computer) handling generalpurpose tasks. A DSP system typically consists of a processor, memory, analogtodigital converters (ADCs), and digitaltoanalog converters (DACs). The main difference with typical microprocessors is they are faster. A fieldprogrammable gate array (FPGA) [77] is a semiconductor device containing programmable logic blocks that can be programmed to perform certain functions, and programmable interconnects. Although FPGAs are slower than their applicationspecific integrated circuits (ASICs) counterparts and use more power, their advantages include a shorter time to design and the ability to be reprogrammed.
0.2 EXAMPLES OF SIGNAL PROCESSING APPLICATIONS The theory of signals and systems connects directly, among others, with communications, control, and biomedical engineering, and indirectly with mathematics and computer engineering. With the availability of digital technologies for processing signals, it is tempting to believe there is no need to understand their connection with analog technologies. It is precisely the opposite is illustrated by considering the following three interesting applications: the compactdisc (CD) player, softwaredefined radio and cognitive radio, and computercontrolled systems.
0.2.1 CompactDisc Player Compact discs [9] were first produced in Germany in 1982. Recorded voltage variations over time due to an acoustic sound is called an analog signal given its similarity with the differences in air pressure generated by the sound waves over time. Audio CDs and CD players illustrate best the conversion of a binary signal—unintelligible—into an intelligible analog signal. Moreover, the player is a very interesting control system. To store an analog audio signal (e.g., voice or music) on a CD the signal must be first sampled and converted into a sequence of binary digits—a digital signal—by an ADC and then especially encoded to compress the information and to avoid errors when playing the CD. In the manufacturing of a CD,
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CH A P T E R 0: From the Ground Up!
Speaker Laser DAC Sensor
Audio amplifier
FIGURE 0.2 When playing a CD, the CD player follows the tracks in the disc, focusing a laser on them, as the CD is spun. The laser shines a light that is reflected by the pits and bumps put on the surface of the disc and corresponding to the coded digital signal from an acoustic signal. A sensor detects the reflected light and converts it into a digital signal, which is then converted into an analog signal by the DAC. When amplified and fed to the speakers such a signal sounds like the originally recorded acoustic signal.
pits and bumps corresponding to the ones and zeros from the quantization and encoding processes are impressed on the surface of the disc. Such pits and bumps will be detected by the CD player and converted back into an analog signal that approximates the original signal when the CD is played. The transformation into an analog signal uses a DAC. As we will see in Chapter 7, an audio signal is sampled at a rate of about 44,000 samples/second (sec) (corresponding to a maximum frequency around 22 KHz for a typical audio signal) and each of these samples is represented by a certain number of bits (typically 8 bits/sample). The need for stereo sound requires that two channels be recorded. Overall, the number of bits representing the signal is very large and needs to be compressed and especially encoded. The resulting data, in the form of pits and bumps impressed on the CD surface, are put into a spiral track that goes from the inside to the outside of the disc. Besides the binarytoanalog conversion, the CD player exemplifies a very interesting control system (see Figure 0.2). Indeed, the player must: (1) rotate the disc at different speeds depending on the location of the track within the CD being read, (2) focus a laser and a lens system to read the pits and bumps on the disc, and (3) move the laser to follow the track being read. To understand the exactness required, consider that the width of the track and the high of the bumps is typically less than a micrometer (10−6 meters or 3.937 × 10−5 inches) and a nanometer (10−9 meters or 3.937 × 10−8 inches), respectively.
0.2.2 SoftwareDefined Radio and Cognitive Radio Softwaredefined radio and cognitive radio are important emerging technologies in wireless communications [43]. In softwaredefined radio (SDR), some of the radio functions typically implemented in hardware are converted into software [64]. By providing smart processing to SDRs, cognitive radio (CR) will provide the flexibility needed to more efficiently use the radio frequency spectrum and to make available new services to users. In the United States the Federal Communication Commission (FCC), and likewise in other parts of the world the corresponding agencies, allocates the bands for
0.2 Examples of Signal Processing Applications
different users of the radio spectrum (commercial radio and T V, amateur radio, police, etc.). Although most bands have been allocated, implying a scarcity of spectrum for new users, it has been found that locally at certain times of the day the allocated spectrum is not being fully utilized. Cognitive radio takes advantage of this. Conventional radio systems are composed mostly of hardware, and as such cannot be easily reconfigured. The basic premise in SDR as a wireless communication system is its ability to reconfigure by changing the software used to implement functions typically done by hardware in a conventional radio. In an SDR transmitter, software is used to implement different types of modulation procedures, while ADCs and DACs are used to change from one type of signal to another. Antennas, audio amplifiers, and conventional radio hardware are used to process analog signals. Typically, an SDR receiver uses an ADC to change the analog signals from the antenna into digital signals that are processed using software on a generalpurpose processor. See Figure 0.3. Given the need for more efficient use of the radio spectrum, cognitive radio (CR) uses SDR technology while attempting to dynamically manage the radio spectrum. A cognitive radio monitors locally the radio spectrum to determine regions that are not occupied by their assigned users and transmits in those bands. If the primary user of a frequency band recommences transmission, the CR either moves to another frequency band, or stays in the same band but decreases its transmission power level or modulation scheme to avoid interference with the assigned user. Moreover, a CR will search
Antenna
TRANSMITTER
Microphone Modulator
ADC
Antenna
DAC
RECEIVER Speaker Superheterodyne
ADC
Demodulator
DAC
FIGURE 0.3 Schematics of a voice SDR mobile twoway radio. Transmitter: The voice signal is inputted by means of a microphone, amplified by an audio amplifier, converted into a digital signal by an ADC, and then modulated using software, before being converted into analog by an DAC, amplified, and sent as a radio frequency signal via an antenna. Receiver: The signal received by the antenna is processed by a superheterodyne frontend, converted into a digital signal by an ADC before being demodulated and converted into an analog signal by a DAC, amplified, and fed to a speaker. The modulator and demodulator blocks indicate software processing.
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for network services that it can offer to its users. Thus, SDR and CR are bound to change the way we communicate and use network services.
0.2.3 ComputerControlled Systems The application of computer control ranges from controlling simple systems such as a heater (e.g., keeping a room temperature comfortable while reducing energy consumption) or cars (e.g., controlling their speed), to that of controlling rather sophisticated machines such as airplanes (e.g., providing automatic flight control) or chemical processes in very large systems such as oil refineries. A significant advantage of computer control is the flexibility computers provide—rather sophisticated control schemes can be implemented in software and adapted for different control modes. Typically, control systems are feedback systems where the dynamic response of a system is changed to make it follow a desirable behavior. As indicated in Figure 0.4, the plant is a system, such as a heater, car, or airplane, or a chemical process in need of some control action so that its output (it is also possible for a system to have several outputs) follows a reference signal (or signals). For instance, one could think of a cruisecontrol system in a car that attempts to keep the speed of the car at a certain value by controlling the gas pedal mechanism. The control action will attempt to have the output of the system follow the desired response, despite the presence of disturbances either in the plant (e.g., errors in the model used for the plant) or in the sensor (e.g., measurement error). By comparing the reference signal with the output of the sensor, and using a control law implemented in the computer, a control action is generated to change the state of the plant and attain the desired output. To use a computer in a control application it is necessary to transform analog signals into digital signals so that they can be inputted into the computer, while it is also necessary that the output of the computer be converted into an analog signal to drive an actuator (e.g., an electrical motor) to provide an action capable of changing the state of the plant. This can be done by means of ADCs and DACs. The sensor should also be able to act as a transducer whenever the output of the plant is
w (t) r (t) + −
ADC
y (t)
Digital computer
DAC
Plant
Clock
Sensor v (t)
FIGURE 0.4 Computercontrolled system for an analog plant (e.g., cruise control for a car). The reference signal is r(t) (e.g., desired speed) and the output is y(t) (e.g., car speed). The analog signals are converted to digital signals by an ADC, while the digital signal from the computer is converted into an analog signal (an actuator is probably needed to control the car) by a DAC. The signals w(t) and v(t) are disturbances or noise in the plant and the sensor (e.g., electronic noise in the sensor and undesirable vibration in the car).
0.3 Analog or Discrete?
of a different type than the reference. Such would be the case, for instance, if the plant output is a temperature while the reference signal is a voltage.
0.3 ANALOG OR DISCRETE? Infinitesimal calculus, or just plain calculus, deals with functions of one or more continuously changing variables. Based on the representation of these functions, the concepts of derivative and integral are developed to measure the rate of change of functions and the areas under the graphs of these functions, or their volumes. Differential equations are then introduced to characterize dynamic systems. Finite calculus, on the other hand, deals with sequences. Thus, derivatives and integrals are replaced by differences and summations, while differential equations are replaced by difference equations. Finite calculus makes possible the computations of calculus by means of a combination of digital computers and numerical methods—thus, finite calculus becomes the more concrete mathematics.1 Numerical methods applied to sequences permit us to approximate derivatives, integrals, and the solution of differential equations. In engineering, as in many areas of science, the inputs and outputs of electrical, mechanical, chemical, and biological processes are measured as functions of time with amplitudes expressed in terms of voltage, current, torque, pressure, etc. These functions are called analog or continuoustime signals, and to process them with a computer they must be converted into binary sequences—or a string of ones and zeros that is understood by the computer. Such a conversion is done in a way as to preserve as much as possible the information contained in the original signal. Once in binary form, signals can be processed using algorithms (coded procedures understood by computers and designed to obtain certain desired information from the signals or to change them) in a computer or in a dedicated piece of hardware. In a digital computer, differentiation and integration can be done only approximately, and the solution of differential equations requires a discretization process as we will illustrate later in this chapter. Not all signals are functions of a continuous parameter—there exist inherently discretetime signals that can be represented as sequences, converted into binary form, and processed by computers. For these signals the finite calculus is the natural way of representing and processing them. Analog or continuoustime signals are converted into binary sequences by means of an ADC, which, as we will see, compresses the data by converting the continuoustime signal into a discretetime signal or a sequence of samples, each sample being represented by a string of ones and zeros giving a binary signal. Both time and signal amplitude are made discrete in this process. Likewise, digital signals can be transformed into analog signals by means of a DAC that uses the reverse process of the ADC. These converters are commercially available, and it is important to learn how they work so that digital representation of analog signals is obtained
1 The
use of concrete, rather than abstract, mathematics was coined by Graham, Knuth, and Patashnik in Concrete Mathematics: A Foundation for Computer Science [26]. Professor Donald Knuth from Stanford University is the the inventor of the Tex and Metafont typesetting systems that are the precursors of Latex, the document layout system in which the original manuscript of this book was done.
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with minimal information loss. Chapters 1, 7, and 8 will provide the necessary information about continuoustime and discretetime signals, and show how to convert one into the other and back. The sampling theory presented in Chapter 7 is the backbone of digital signal processing.
0.3.1 ContinuousTime and DiscreteTime Representations There are significant differences between continuoustime and discretetime signals as well as in their processing. A discretetime signal is a sequence of measurements typically made at uniform times, while the analog signal depends continuously on time. Thus, a discretetime signal x[n] and the corresponding analog signal x(t) are related by a sampling process: x[n] = x(nTs ) = x(t)t=nTs
(0.1)
That is, the signal x[n] is obtained by sampling x(t) at times t = nTs , where n is an integer and Ts is the sampling period or the time between samples. This results in a sequence, {· · · x(−Ts ) x(0) x(Ts ) x(2Ts ) · · · } according to the sampling times, or equivalently {· · · x[−1] x[0] x[1] x[2] · · · } according to the ordering of the samples (as referenced to time 0). This process is called sampling or discretization of an analog signal. Clearly, by choosing a small value for Ts we could make the analog and the discretetime signals look very similar—almost indistinguishable—which is good, but this is at the expense of memory space required to keep the numerous samples. If we make the value of Ts large, we improve the memory requirements, but at the risk of losing information contained in the original signal. For instance, consider a sinusoid obtained from a signal generator: x(t) = 2 cos(2πt) for 0 ≤ t ≤ 10 sec. If we sample it every Ts1 = 0.1 sec, the analog signal becomes the following sequence: x1 [n] = x(t) t=0.1n = 2 cos(2πn/10) 0 ≤ n ≤ 100 providing a very good approximation to the original signal. If, on the other hand, we let Ts2 = 1 sec, then the discretetime signal becomes x2 [n] = x(t) t=n = 2 cos(2πn) = 2
0 ≤ n ≤ 10
See Figure 0.5. Although for Ts2 the number of samples is considerably reduced, the representation of the original signal is very poor—it appears as if we had sampled a constant signal, and we have thus lost information! This indicates that it is necessary to come up with a way to choose Ts so that sampling provides not only a reasonable number of samples, but, more importantly, guarantees that the information in the analog and the discretetime signals remains the same.
0.3 Analog or Discrete?
x1(0.1n)
2 1 0 −1 −2 0
2
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8
10
6
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(a) 2 1 x2 (n)
FIGURE 0.5 Sampling an analog sinusoid x(t) = 2 cos(2πt), 0 ≤ t ≤ 10, with two different sampling periods, (a) Ts1 = 0.1 sec and (b) Ts2 = 1 sec, giving x1 (0.1n) and x2 (n). The sinusoid is shown by dashed lines. Notice the similarity between the discretetime signal and the analog signal when Ts1 = 0.1 sec, while they are very different when Ts2 = 1 sec, indicating loss of information.
0 −1 −2 0
2
4 t(sec) (b)
ACM Closings, Jan. 2006−Dec. 2009
260 240
Dollars
220 200 180 160
FIGURE 0.6 Weekly closings of ACM stock for 160 weeks in 2006 to 2009. ACM is the trading name of the stock of the imaginary company, ACME Inc., makers of everything you can imagine.
140 120 100 20
40
60
80 Week
100
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As indicated before, not all signals are analog; there are some that are naturally discrete. Figure 0.6 displays the weekly average of the stock price of a fictitious company, ACME. Thinking of it as a signal, it is naturally discretetime as it does not come from the discretization of an analog signal. We have shown in this section the significance of the sampling period Ts in the transformation of an analog signal into a discretetime signal without losing information. Choosing the sampling period requires knowledge of the frequency content of the signal—this is an example of the relation between time and frequency to be presented in great detail in Chapters 4 and 5, where the Fourier representation of periodic and nonperiodic
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signals is given. In Chapter 7, where we consider the problem of sampling, we will use this relation to determine appropriate values for the sampling period.
0.3.2 Derivatives and Finite Differences Differentiation is an operation that is approximated in finite calculus. The derivative operator D[x(t)] =
x(t + h) − x(t) dx(t) = lim dt h h→0
(0.2)
measures the rate of change of an analog signal x(t). In finite calculus the forward finitedifference operator 1[x(nTs )] = x((n + 1)Ts ) − x(nTs )
(0.3)
measures the change in the signal from one sample to the next. If we let x[n] = x(nTs ), for a known Ts , the forward finitedifference operator becomes a function of n: 1[x[n]] = x[n + 1] − x[n]
(0.4)
The forward finitedifference operator measures the difference between two consecutive samples: one in the future x((n + 1)Ts ) and the other in the present x(nTs ). (See Problem 0.4 for a definition of the backward finitedifference operator.) The symbols D and 1 are called operators as they operate on functions to give other functions. The derivative and the finitedifference operators are clearly not the same. In the limit, we have that dx(t) 1[x(nTs )] t=nTs = lim Ts →0 dt Ts
(0.5)
Depending on the signal and the chosen value of Ts , the finitedifference operation can be a crude or an accurate approximation to the derivative multiplied by Ts . Intuitively, if a signal does not change very fast with respect to time, the finitedifference approximates well the derivative for relatively large values of Ts , but if the signal changes very fast one needs very small values of Ts . The concept of frequency of a signal can help us understand this. We will learn that the frequency content of a signal depends on how fast the signal varies with time; thus a constant signal has zero frequency while a noisy signal that changes rapidly has high frequencies. Consider a constant signal x0 (t) = 2 having a derivative of zero (i.e., such a signal does not change at all with respect to time or it is a zerofrequency signal). If we convert this signal into a discretetime signal using a sampling period Ts = 1 (or any other positive value), then x0 [n] = 2 and so 1[x0 [n]] = 2 − 2 = 0 coincides with the derivative. Consider then a signal x1 (t) = t with derivative 1 (this signal changes faster than x(t) so it has frequencies larger than zero). If we sample it using Ts = 1, then x1 [n] = n and the finite difference is 1[x1 [n]] = 1[n] = (n + 1) − n = 1
0.3 Analog or Discrete?
which again coincides with the derivative. Finally, we consider a signal that changes faster than x(t) and x1 (t) such as x2 (t) = t2 . Sampling x2 (t) with Ts = 1, we have x2 [n] = n2 and its forward finite difference is given by 1[x2 [n]] = 1[n2 ] = (n + 1)2 − n2 = 2n + 1 which gives as an approximation to the derivative 1[x2 [n]]/Ts = 2n + 1. The derivative of x2 (t) is 2t, which at 0 equals 0, and at 1 equals 2. On the other hand, 1[n2 ]/Ts equals 1 and 3 at n = 0 and n = 1, respectively, which are different values from those of the derivative. Suppose Ts = 0.01, so that x2 [n] = x2 (nTs ) = (0.01n)2 = 0.0001n2 . If we compute the difference for this signal we get 1[x2 (0.01n)] = 1[(0.01n)2 ] = (0.01n + 0.01)2 − 0.0001n2 = 10−4 (2n + 1) which gives as an approximation to the derivative 1[x2 (0.01n)]/Ts = 10−2 (2n + 1), or 0.01 when n = 0 and 0.03 when n = 1 which are a lot closer to the actual values of dx2 (t) t=0.01n = 2t t=0.01n = 0.02n dt The error now is 0.01 for each case instead of 1 as in the case when Ts = 1. Thus, whenever the rate of change of the signal is faster, the difference gets closer to the derivative by making Ts smaller. It becomes clear that the faster the signal changes, the smaller the sampling period Ts should be in order to get a better approximation of the signal and its derivative. As we will learn in Chapters 4 and 5 the frequency content of a signal depends on the signal variation over time. A constant signal has frequency zero, while a signal that changes very fast over time would have high frequencies. The higher the frequencies in a signal, the more samples would be needed to represent it with no loss of information, thus requiring that Ts be smaller.
0.3.3 Integrals and Summations Integration is the opposite of differentiation. To see this, suppose I(t) is the integration of a continuous signal x(t) from some time t0 to t (t0 < t), Zt I(t) = t0
x(τ )dτ
(0.6)
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CH A P T E R 0: From the Ground Up!
or the sum of the area under x(t) from t0 to t. Notice that the upper bound of the integral is t so the integrand depends on a dummy variable.2 The derivative of I(t) is Zt
dI(t) I(t) − I(t − h) 1 = lim = lim dt h h→0 h→0 h
x(τ )dτ
t−h
≈ lim
h→0
x(t) + x(t − h) = x(t) 2
where the integral is approximated as the area of a trapezoid with sides x(t) and x(t − h) and height h. Thus, for a continuous signal x(t), d dt
Zt
x(τ )dτ = x(t)
(0.7)
t0
or if using the derivative operator D[.], then its inverse D−1 [.] should be the integration operator. That is, the above equation can be written D[D−1 [x(t)]] = x(t).
(0.8)
We will see in Chapter 3 a similar relation between the derivative and the integral. The Laplace transform operators s and 1/s (just like D and 1/D) imply differentiation and integration in the time domain. Computationally, integration is implemented by sums. Consider, for instance, the integral of x(t) = t from 0 to 10, which we know is equal to Z10 t dt =
t2 10 = 50. 2 t=0
0
That is, the area of a triangle with a base of 10 and a height of 10. For Ts = 1, suppose we approximate the signal x(t) by pulses p[n] of width Ts = 1 and height nTs = n, or pulses of area n for n = 0, . . . , 9. This can be seen as a lowerbound approximation to the integral, as the total area of these pulses gives a result smaller than the integral. In fact, the sum of the areas of the pulses is given by " 9 # 9 9 0 X X X X p[n] = n = 0 + 1 + 2 + · · · 9 = 0.5 n+ k n=0
n=0
n=0
" = 0.5
9 X
n=0
2 The
n+
9 X
# (9 − n) =
n=0
9 2
9 X n=0
1=
k=9
10 × 9 = 45 2
integral I(t) is a function of t and as such the integrand needs to be expressed in terms of a socalled dummy variable τ that takes values from t0 to t in the integration. It would be confusing to let the integration variable be t. The variable τ is called a dummy variable because it is not crucial to the integration; any other variable could be used with no effect on the integration.
0.3 Analog or Discrete?
10 8 t t, n
6 4 2
FIGURE 0.7 Approximation of area under x(t) = t, t ≥ 0, 0 otherwise, by pulses of width 1 and height nTs , where Ts = 1 and n = 0, 1, . . .
0 0
2
4
6
8
10
t
The approximation of the area using Ts = 1 is very poor (see Figure 0.7). In the above, we used the fact that the sum is not changed whether we add the numbers from 0 to 9 or backwards from 9 to 0, and that doubling the sum and dividing by 2 would not change the final answer. The above sum can thus be generalized to N−1 X n=0
1 n= 2
"N−1 X
n+
n=0
N−1 X
#
(N − 1 − n) =
n=0
N−1 1X (N − 1) 2 n=0
N × (N − 1) = 2
(0.9)
a result that Gauss found out when he was a preschooler!3 To improve the approximation of the integral we use Ts = 10−3 , which gives a discretized signal nTs for 0 ≤ nTs < 10 or 0 ≤ n ≤ (10/Ts ) − 1. The area of the pulses is nTs2 and the approximation to the integral is then 4 −1 10 X
p[n] =
n=0
4 −1 10 X
n10−6
n=0
=
104 × (104 − 1) 106 × 2
= 49.995
3 Carl
Friedrich Gauss (1777–1855) was a German mathematician. He was seven years old when he amazed his teachers with his trick for adding the numbers from 1 to 100 [7]. Gauss is one of the most accomplished mathematicians of all times [2]. He is in a group of selected mathematicians and scientists whose pictures appear in the currency of a country. His picture was on the Mark, the previous currency of Germany [6].
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CH A P T E R 0: From the Ground Up!
which is a lot better result. In general, we have that the integral can be computed quite accurately using a very small value of Ts , indeed (10/T s )−1 X n=0
p[n] =
(10/T s )−1 X
nTs2
n=0
(10/Ts ) × ((10/Ts ) − 1) 2 10 × (10 − Ts ) = 2
= Ts2
which for very small values of Ts (so that 10 − Ts ≈ 10) gives 100/2 = 50, as desired. Derivatives and integrals take us into the processing of signals by systems. Once a mathematical model for a dynamic system is obtained, typically differential equations characterize the relation between the input and output variable or variables of the system. A significant subclass of systems (used as a valid approximation in some way to actual systems) is given by linear differential equations with constant coefficients. The solution of these equations can be efficiently found by means of the Laplace transform, which converts them into algebraic equations that are much easier to solve. The Laplace transform is covered in Chapter 3, in part to facilitate the analysis of analog signals and systems early in the learning process, but also so that it can be related to the Fourier theory of Chapters 4 and 5. Likewise for the analysis of discretetime signals and systems we present in Chapter 9 the Ztransform, having analogous properties to those from the Laplace transform, before the Fourier analysis of those signals and systems.
0.3.4 Differential and Difference Equations A differential equation characterizes the dynamics of a continuoustime system, or the way the system responds to inputs over time. There are different types of differential equations, corresponding to different systems. Most systems are characterized by nonlinear, timedependent coefficient differential equations. The analytic solution of these equations is rather complicated. To simplify the analysis, these equations are locally approximated as linear constantcoefficient differential equations. Solution of differential equations can be obtained by means of analog and digital computers. An electronic analog computer consists of operational amplifiers (opamps), resistors, capacitors, voltage sources, and relays. Using the linearized model of the opamps, resistors, and capacitors it is possible to realize integrators to solve a differential equation. Relays are used to set the initial conditions on the capacitors, and the voltage source gives the input signal. Although this arrangement permits the solution of differential equations, its drawback is the storage of the solution, which can be seen with an oscilloscope but is difficult to record. Hybrid computers were suggested as a solution—the analog computer is assisted by a digital component that stores the data. Both analog and hybrid computers have gone the way of the dinosaurs, and it is digital computers aided by numerical methods that are used now to solve differential equations. Before going into the numerical solution provided by digital computers, let us consider why integrators are needed in the solution of differential equations. A firstorder (the highest derivative present in the equation); linear (no nonlinear functions of the input or the output are present) with
0.3 Analog or Discrete?
1F
vi (t)
FIGURE 0.8 RC circuit.
i(t)
+ 1Ω
−
vi (t ) +
FIGURE 0.9 Realization of firstorder differential equation using (a) a differentiator and (b) an integrator.
vc(t )
+
−
dvc (t ) dt
vi (t ) −
dvc (t ) dt
∫ (·)dt
vc(t )
d(·) dt (a)
(b)
constantcoefficient differential equations obtained from a simple RC circuit (Figure 0.8) with a constant voltage source vi (t) as input and with resistor R = 1; and capacitor C = 1 F (with huge plates!) connected in series is given by vi (t) = vc (t) +
dvc (t) dt
(0.10)
with an initial voltage vc (0) across the capacitor. Intuitively, in this circuit the capacitor starts with an initial charge of vc (0), and will continue charging until it reaches saturation, at which point no more charge will flow (the current across the resistor and the capacitor is zero). Therefore, the voltage across the capacitor is equal to the voltage source–that is, the capacitor is acting as an open circuit given that the source is constant. Suppose, ideally, that we have available devices that can perform differentiation. There is then the tendency to propose that the differential equation (Eq. 0.10) be solved following the block diagram shown in Figure (0.9). Although nothing is wrong analytically, the problem with this approach is that in practice most signals are noisy (each device produces electronic noise) and the noise present in the signal may cause large derivative values given its rapidly changing amplitudes. Thus, the realization of the differential equation using differentiators is prone to being very noisy (i.e., not good). Instead of, as proposed years ago by Lord Kelvin,4 using differentiators we need to smooth out the process by using integrators, so that the voltage across the capacitor vc (t) is obtained by integrating both sides of Equation (0.10). Assuming that the source is switched on at time t = 0 and that the capacitor has an initial voltage vc (0), using the inverse relation between derivatives and integrals gives vc (t) =
Zt
[vi (τ ) − vc (τ )]dτ + vc (0)
t≥0
(0.11)
0
4 William
Thomson, Lord Kelvin, proposed in 1876 the differential analyzer, a type of analog computer capable of solving differential equations of order 2 and higher. His brother James designed one of the first differential analyzers [78].
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CH A P T E R 0: From the Ground Up!
which is represented by the block diagram in Figure 0.9(b). Notice that the integrator also provides a way to include the initial condition, which in this case is the initial voltage across the capacitor, vc (0). Different from the accentuating the effect of differentiators on noise, integrators average the noise, thus reducing its effects. Block diagrams like the ones shown in Figure 0.9 allow us to visualize the system much better, and are commonly used. Integrators can be efficiently implemented using operational amplifiers with resistors and capacitors.
How to Obtain Difference Equations Let us then show how Equation (0.10) can be solved using integration and its approximation, resulting in a difference equation. Using Equation (0.11) at t = t1 and t = t0 for t1 > t0 , we have that vc (t1 ) − vc (t0 ) =
Zt1
vi (τ )dτ −
t0
Zt1
vc (τ )dτ
t0
If we let t1 − t0 = 1t where 1t → 0 (i.e., a very small time interval), the integrals can be seen as the area of small trapezoids of height 1t and bases vi (t1 ) and vi (t0 ) for the input source and vc (t1 ) and vc (t0 ) for the voltage across the capacitor (see Figure 0.10). Using the formula for the area of a trapezoid we get an approximation for the above integrals so that vc (t1 ) − vc (t0 ) = [vi (t1 ) + vi (t0 )]
1t 1t − [vc (t1 ) + vc (t0 )] 2 2
from which we obtain 1t 1t 1t vc (t1 ) 1 + = [vi (t1 ) + vi (t0 )] + vc (t0 ) 1 − 2 2 2
Assuming 1t = T, we then let t1 = nT and t0 = (n − 1)T. The above equation can be written as vc (nT) =
T 2−T [vi (nT) + vi ((n − 1)T)] + vc ((n − 1)T) 2+T 2+T
n≥1
(0.12)
and initial condition vc (0) = 0. This is a firstorder linear difference equation with constant coefficients approximating the differential equation characterizing the RC circuit. Letting the input vc (t1) vc (t0 )
FIGURE 0.10 Approximation of area under the curve by a trapezoid.
t1
t0 Δt
t
0.3 Analog or Discrete?
be vi (t) = 1 for t ≥ 0, we have vc (nT) =
0 2T 2+T
+
2−T 2+T vc ((n − 1)T)
n=0 n≥1
(0.13)
The advantage of the difference equation is that it can be solved for increasing values of n using previously computed values of vc (nT), which is called a recursive solution. For instance, letting T = 10−3 , vi (t) = 1, and defining M = 2T/(2 + T), K = (2 − T)/(2 + T), we obtain n=0
vc (0) = 0
n=1
vc (T) = M
n=2
vc (2T) = M + KM = M(1 + K)
n=3
vc (3T) = M + K(M + KM) = M(1 + K + K 2 )
n=4
vc (4T) = M + KM(1 + K + K 2 ) = M(1 + K + K 2 + K 3 )
··· The values are M = 2T/(2 + T) ≈ T = 10−3 , K = (2 − T)/(2 + T) < 1, and 1 − K = M. The response increases from the zero initial condition to a constant value, which is the effect of the dc source—the capacitor eventually acts as an open circuit, so that the voltage across the capacitor equals that of the input. Extrapolating from the above results it seems that in the steadystate (i.e., when nT → ∞) we have5 vc (nT) = M
∞ X
Km =
m=0
M =1 1−K
Even though this is a very simple example, it clearly illustrates that very good approximations to the solution of differential equations can be obtained using numerical methods that are appropriate for implementation in digital computers. The above example shows how to solve a differential equation using integration and approximation of the integrals to obtain a difference equation that a computer can easily solve. The integral approximation used above is the trapezoidal rule method, which is one among many numerical methods used to solve differential equations. Also we will see later that the above results in the bilinear transformation, which connects the Laplace s variable with the z variable of the Ztransform, and that will be used in Chapter 11 in the design of discrete filters.
5 The
infinite sum converges if K < 1, which is satisfied in this case. If we multiply the sum by (1 − K) we get (1 − K)
∞ X m=0
Km =
∞ X
Km −
m=0
=1+
∞ X
K m+1
m=0 ∞ X m=1
Km −
∞ X
K` = 1
`=1
where we changed the variable in the second equation to ` = m + 1. This explains why the sum is equal to 1/(1 − K).
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CH A P T E R 0: From the Ground Up!
0.4 COMPLEX OR REAL? Most of the theory of signals and systems is based on functions of a complex variable. Clearly, signals are functions of a real variable corresponding to time or space (if the signal is twodimensional, like an image) so why would one need complex numbers in processing signals? As we will see later, timedependent signals can be characterized by means of frequency and damping. These two characteristics are given by complex variables such as s = σ + j (where σ is the damping factor and is the frequency) in the representation of analog signals in the Laplace transform, or z = rejω (where r is the damping factor and ω is the discrete frequency) in the representation of discretetime signals in the Ztransform. Both of these transformations will be considered in detail in Chapters 3 and 9. The other reason for using complex variables is due to the response of systems to pure tones or sinusoids. We will see that such response is fundamental in the analysis and synthesis of signals and systems. We thus need a solid grasp of what is meant by complex variables and what a function of these is all about. In this section, complex variables will be connected to vectors and phasors (which are commonly used in the sinusoidal steadystate analysis of linear circuits).
0.4.1 Complex Numbers and Vectors A complex number z represents any point (x, y) in a twodimensional plane by z = x + jy, where x = Re[z] (real part of z) is the coordinate √ in the x axis and y = Im[z] (imaginary part of z) is the coordinate in the y axis. The symbol j = −1 just indicates that z needs to have two components to represent a point in the twodimensional plane. Interestingly, a vector Ez that emanates from the origin of the complex plane (0, 0) to the point (x, y) with a length q Ez = x2 + y2 = z (0.14) and an angle θ = ∠Ez = ∠z
(0.15)
also represents the point (x, y) in the plane and has the same attributes as the complex number z. The couple (x, y) is therefore equally representable by the vector Ez or by a complex number z that can be written in a rectangular or in a polar form, z = x + jy = zejθ
(0.16)
where the magnitude z and the phase θ are defined in Equations (0.14) and (0.15). It is important to understand that a rectangular plane or a polar complex plane are identical despite the different representation of each point in the plane. Furthermore, when adding or subtracting complex numbers the rectangular form is the appropriate one, while when multiplying or dividing complex numbers the polar form is more advantageous. Thus, if complex numbers z = x + jy = zej∠z and v = p + jq = vej∠v are added analytically, we obtain z + v = (x + p) + j(y + q)
0.4 Complex or Real?
z +v
(x, y) y z
v
z
θ (a)
x
j = j 1, j 5, … −1 = j 2, j 6, …
FIGURE 0.11 (a) Representation of a complex number z by a vector (b) addition of complex numbers z and v; (c) integer powers of j; and (d) complex conjugate.
1 = j 0, j 4, …
(b) (x, y) z θ −θ z
−j = j 3, j 7, …
(x, −y) (c)
(d)
Using their polar representations requires a geometric interpretation: the addition of vectors (see Figure 0.11). On the other hand, the multiplication of z and v is easily done using their polar forms as zv = zej∠z vej∠v = zvej(∠z+∠v) but it requires more operations if done in the rectangular form—that is, zv = (x + jy)(p + jq) = (xp − yq) + j(xq + yp) It is even more difficult to obtain a geometric interpretation. Such an interpretation will be seen later on. Addition and subtraction as well as multiplication and division can thus be done more efficiently by choosing the rectangular and the polar representations, respectively. Moreover, the polar representation is also useful when finding powers of complex numbers. For the complex variable z = ze∠z , we have that zn = zn ejn∠z 1.5 for n integer or rational. For instance, if n = 10, then z10 = z10 ej10∠z √ , and if n = 3/2, then z = √ 3 j1.5∠z ( z) e . The powers of j are of special interest. Given that j = −1 then, we have
jn = (−1)n/2 =
(−1)m n = 2m, n even (−1)m j n = 2m + 1, n odd
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22
CH A P T E R 0: From the Ground Up!
so that j0 = 1, j1 = j, j2 = −1, j3 = −j, and so on. Letting j = 1ejπ/2 , we can see that the increasing powers of jn = 1ejnπ/2 are vectors with angles of 0 when n = 0, π/2 when n = 1, π when n = 2, and 3π/2 when n = 3. The angles repeat for the next four values, the four after that, and so on. See Figure 0.11. One operation possible with complex numbers that is not possible with real numbers is complex conjugation. Given a complex number z = x + jy = zej∠z its complex conjugate is z∗ = x − jy = ze−j∠z —that is, we negate the imaginary part of z or reflect its angle. This operation gives that (i)
z + z∗ = 2x
(ii)
z − z∗ = 2jy
(iii) zz∗ = z2 z = ej2∠z z∗
(iv)
or Re[z] = 0.5[z + z∗ ] or Im[z] = 0.5[z − z∗ ] √ or z = zz∗ or
∠z = −j0.5[log(z) − log(z∗ )]
(0.17)
The complex conjugation provides a different approach to the division of complex numbers in rectangular form. This is done by making the denominator a positive real number by multiplying both numerator and denominator by the complex conjugate of the denominator. For instance, z=
1 + j1 (1 + j1)(3 − j4) 7−j 7−j = = = 3 + j4 (3 + j4)(3 − j4) 9 + 16 25
Finally, the conversion of complex numbers from rectangular to polar needs to be done with care, especially when computing the angles. For instance, z = 1 +√j has a vector representing in the first quadrant of the complex plane, and its magnitude is z = 2 while the tangent of its angle θ is tan(θ ) = 1 or θ = π/4 radians. If z = −1 + j, the vector representing it is now in the second quadrant with the same magnitude as before, but its angle is now θ = π − tan−1 (1) = 3π/4 That is, we find the angle with respect to the negative real axis and subtract it from π. Likewise, if z = −1 − j, the magnitude does not change but the phase is now θ = π + tan−1 (1) = 5π/4 which can also be expressed as −3π/4. Finally, when z = 1 − j, the angle is −π/4 and the magnitude remains the same. The conversion from polar to rectangular form is much easier. Indeed, given a complex number in polar form z = zejθ its real part is x = z cos(θ ) (i.e., the projection of the vector corresponding√to z onto the real axis) and the imaginary part is y = z sin(θ ), so that z = x + jy. For instance, z = 2e3π/4 can be written as z=
√
√ 2 cos(3π/4) + j 2 sin(3π/4) = −1 + j
0.4 Complex or Real?
0.4.2 Functions of a Complex Variable Just like realvalued functions, functions of a complex variable can be defined. For instance, the logarithm of a complex number can be written as v = log(z) = log(zejθ ) = log(z) + jθ by using the inverse connection between the exponential and the logarithmic functions. Of particular interest in the theory of signals and systems is the exponential of complex variable z defined as v = e z = ex+jy = ex ejy It is important to mention that complex variables as well as functions of complex variables are more general than real variables and realvalued functions. The above definition of the logarithmic function is valid when z = x, with x a real value, and also when z = jy, a purely imaginary value. Likewise, the exponential function for z = x is a realvalued function.
Euler’s Identity One of the most famous equations of all times6 is 1 + ejπ = 1 + e−jπ = 0 due to one of the most prolific mathematicians of all times, Leonard Euler.7 The above equation can be easily understood by establishing Euler’s identity, which connects the complex exponential and sinusoids: ejθ = cos(θ ) + j sin(θ )
(0.18)
One way to verify this identity is to consider p the polar representation of the complex number cos(θ ) + j sin(θ ), which has a unit magnitude since cos2 (θ ) + sin2 (θ ) = 1 given the trigonometric identity cos2 (θ ) + sin2 (θ ) = 1. The angle of this complex number is sin(θ ) =θ ψ = tan−1 cos(θ ) Thus, the complex number cos(θ) + j sin(θ ) = 1ejθ which is Euler’s identity. Now in the case where θ = ±π the identity implies that e±jπ = −1, explaining the famous Euler’s equation. 6 A reader’s poll done by Mathematical
Intelligencer named Euler’s identity the most beautiful equation in mathematics. Another poll by Physics World in 2004 named Euler’s identity the greatest equation ever, together with Maxwell’s equations. Paul Nahin’s book Dr. Euler’s Fabulous Formula (2006) is devoted to Euler’s identity. It states that the identity sets “the gold standard for mathematical beauty” [73]. 7 Leonard Euler (1707–1783) was a Swiss mathematician and physicist, student of John Bernoulli, and advisor of Joseph Lagrange. We √ owe Euler the notation f (x) for functions, e for the base of natural logs, i = −1, π for pi, 6 for sum, the finite difference notation 1, and many more!
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24
CH A P T E R 0: From the Ground Up!
The relation between the complex exponentials and the sinusoidal functions is of great importance in signals and systems analysis. Using Euler’s identity the cosine can be expressed as cos(θ) = Re[ejθ ] =
ejθ + e−jθ 2
(0.19)
sin(θ) = Im[ejθ ] =
ejθ − e−jθ 2j
(0.20)
while the sine is given by
Indeed, we have ejθ = cos(θ ) + j sin(θ ) e−jθ = cos(θ ) − j sin(θ ) Adding them we get the above expression for the cosine, and subtracting the second from the first we get the given expression for the sine. The variable θ is in radians, or in the corresponding angle in degrees (recall that 2π radians equals 360 degrees). These relations can be used to define the hyperbolic sinusoids as e−α + eα = cosh(α) 2 e−α − eα j sin( jα) = = − sinh(α) 2 cos( jα) =
(0.21) (0.22)
from which the other hyperbolic functions are defined. Also, we obtain the following expression for the realvalued exponential: e−α = cosh(α) − sinh(α)
(0.23)
Euler’s identity can also be used to find different trigonometric identities. For instance, #2 " jθ + e−jθ 1 1 1 e = [2 + ej2θ + e−j2θ ] = + cos(2θ ) cos2 (θ) = 2 4 2 2 1 1 − cos(2θ ) 2 2 ejθ − e−jθ ejθ + e−jθ ej2θ − e−j2θ 1 sin(θ) cos(θ) = = = sin(2θ ) 2j 2 4j 2
sin2 (θ) = 1 − cos2 (θ) =
0.4.3 Phasors and Sinusoidal Steady State A sinusoid x(t) is a periodic signal represented by x(t) = A cos(0 t + ψ)
−∞ 0) or growth (if α < 0) in the signal, while the imaginary part indicates the frequency of the cosine in the signal. Again, the poles will be complex conjugate pairs since the signal d(t) is real valued. The conclusion is that the location of the poles (and to some degree the zeros), as indicated in the previous two cases, determines the characteristics of the signal. Signals are characterized by their damping and frequency and as such can be described by the poles of its Laplace transform.
If we were to add the different signals considered above, then the Laplace transform of the resulting signal would be the sum of the Laplace transform of each of the signals and the poles would be the aggregation of the poles from each. This observation will be important when finding the inverse Laplace transform, then we would like to do the opposite: To isolate poles or pairs of poles (when they are complex conjugate) and associate with each a general form of the signal with parameters that are found by using the zeros and the other poles of the transform. Figure 3.10 provides an example illustrating the importance of the location of the poles, and the significance of the σ and j axes.
187
CH A P T E R 3: The Laplace Transform
cos(5t)exp(− 0.5t)u(t)
cos(5t)u(t)
1
1
0
0
−1
−1 0
5 t
10
0
5 t
10
10
jΩ
188
0
−10 −2
−1
0
1
σ u(t)
exp(−0.5t)u(t) 1
1
0.5
0.5
0
0
5 t
10
0
0
5 t
10
FIGURE 3.10 For poles shown in the middle, possible signals are displayed around them anti–clockwise from bottom right. The pole s = 0 corresponds to a unitstep signal; the complex conjugate poles on the j axis correspond to a sinusoid; the pair of complex conjugate poles with a negative real part provides a sinusoid multiplied by an exponential; and the pole in the negative real axis gives a decaying exponential. The actual amplitudes and phases are determined by the other poles and by the zeros.
3.3.2 Differentiation For a signal f (t) with Laplace transform F(s) its onesided Laplace transform of its firstand secondorder derivatives are df (t) L u(t) = sF(s) − f (0−) (3.11) dt
3.3 The OneSided Laplace Transform
# df (t) d2 f (t) u(t) = s2 F(s) − sf (0−) − L t=0− dt dt2 "
(3.12)
In general, if f (N) (t) denotes an Nthorder derivative of a function f (t) that has a Laplace transform F(s), we have L[f (N) (t)u(t)] = sN F(s) −
N−1 X
f (k) (0−)sN−1−k
(3.13)
k=0
where f (m) (t) = dm f (t)/dtm is the mthorder derivative, m > 0, and f (0) (t) , f (t).
The Laplace transform of the derivative of a causal signal is Z∞ df (t) df (t) −st u(t) = e dt L dt dt
0−
This integral is evaluated by parts. Let w = e−st , then dw = −se−st dt, and let v = f (t) so that dv = [df (t)/dt]dt, and Z Z wdv = wv − vdw We would then have Z∞
df (t) −st e dt = e−st f (t) ∞ 0− − dt
0−
Z∞
f (t)(−se−st )dt
0−
Z∞ =s
f (t)e−st dt − f (0−)
0−
= sF(s) − f (0−) where e−st f (t)t=0− = f (0−) and e−st f (t)t→∞ = 0 since the region of convergence guarantees that lim f (t)e−σ t = 0
t→∞
For a secondorder derivative we have that " # 2 d f (t) df (1) (t) u(t) = L u(t) L dt2 dt = sL[f (1) (t)] − f (1) (0−) = s2 F(s) − sf (0−) −
df (t) t=0− dt
where we used the notation f (1) (t) = df (t)/dt. This approach can be extended to any higher order to obtain the general result shown above.
189
190
CH A P T E R 3: The Laplace Transform
Remarks n
The derivative property for a signal x(t) defined for all t is Z∞
dx(t) −st e dt = sX(s) dt
−∞
This can be seen by computing the derivative of the inverse Laplace transform with respect to t, assuming that the integral and the derivative can be interchanged. Using Equation (3.3): dx(t) 1 = dt 2πj 1 = 2πj
σZ+j∞
X(s) σ −j∞
dest ds dt
σZ+j∞
(sX(s))est ds
σ −j∞
or that sX(s) is the Laplace transform of the derivative of x(t). Thus, the twosided transform does not include initial conditions. The above result can be generalized to any order of the derivative as L[dN x(t)/dtN ] = sN X(s) n
Application of the linearity and the derivative properties of the Laplace transform makes solving differential equations an algebraic problem.
n Example 3.8 Find the impulse response of an RL circuit in series with a voltage source vs (t) (see Figure 3.11). The current i(t) is the output and the input is the voltage source vs (t). Solution To find the impulse response of the RL circuit we let vs (t) = δ(t) and set the initial current in the inductor to zero. According to Kirchhoff’s voltage law, vs (t) = L
di(t) + Ri(t) dt
i(0−) = 0 R
i(t) i(t)
FIGURE 3.11 Impulse response i(t) of an RL circuit with input vs (t).
vs(t)
+ −
L t
3.3 The OneSided Laplace Transform
which is a firstorder linear differential equation with constant coefficients, zero initial condition, and a causal input so that it is a linear timeinvariant system, as discussed before. Letting vs (t) = δ(t) and computing the Laplace transform of the above equation (using the linearity and the derivative properties of the transform and remembering the initial condition is zero), we obtain the following equation in the sdomain: di(t) L[ δ(t)] = L L + Ri(t) dt 1 = sLI(s) + RI(s) where I(s) is the Laplace transform of i(t). Solving for I(s) we have that I(s) =
1/L s + R/L
which as we have seen is the Laplace transform of i(t) =
1 −(R/L)t e u(t) L
Notice that i(0−) = 0 and that the response has the form of a decaying exponential trying to follow the input signal, a delta function. n
n Example 3.9 In this example we consider the duality between the time and the Laplace domains. The differentiation property indicates that computing the derivative of a function in the time domain corresponds to multiplying by s the Laplace transform of the function (assuming initial conditions are zero). We will illustrate in this example the dual of this—that is, when we differentiate a function in the sdomain its effect in the time domain is to multiply by −t. Consider the connection between δ(t), u(t), and r(t) (i.e., the unit impulse, the unit step, and the ramp, respectively), and relate it to the indicated duality. Explain how this property connects with the existence of multiple poles, real and complex, in general. Solution The relation between the signals δ(t), u(t), and r(t) is seen from 1 L[r(t)] = 2 s dr(t) 1 1 L u(t) = =s 2 = dt s s du(t) 1 L δ(t) = =s =1 dt s which also shows that a double pole at the origin, 1/s2 , corresponds to a ramp function r(t) = tu(t).
191
192
CH A P T E R 3: The Laplace Transform
The above results can be explained by looking for a dual of the derivative property. Multiplying by −t the signal x(t) corresponds to differentiating X(s) with respect to s. Indeed for an integer N > 1, dN X(s) = dsN
Z∞ x(t)
dN e−st dt dsN
0
Z∞ =
x(t)(−t)N e−st dt
0
Thus, if x(t) = u(t), X(s) = 1/s, then −tx(t) has Laplace transform dX(s)/ds = −1/s2 , or tu(t) and 1/s2 are Laplace transform pairs. In general, the Laplace transform of tN−1 u(t), for N ≥ 1, has N poles at the origin. What about multiple real (different from zero) and multiple complex poles? What are the corresponding inverse Laplace transforms? The inverse Laplace transform of 20 s/(s2 + 20 )2 having double complex poles at ±j0 , is t sin(0 t)u(t) Likewise, te−at u(t) has as Laplace transform 1/(s + a)2 . So multiple poles correspond to multiplication by t in the time domain. n
n Example 3.10 Obtain from the Laplace transform of x(t) = cos(0 t)u(t) the Laplace transform of sin(t)u(t) using the derivative property. Solution The causal sinusoid x(t) = cos(0 t)u(t) has a Laplace transform X(s) =
s2
s + 20
Then, dx(t) d cos(0 t) du(t) = u(t) + cos(0 t) dt dt dt
3.3 The OneSided Laplace Transform
= −0 sin(0 t)u(t) + cos(0 t)δ(t) = −0 sin(0 t)u(t) + δ(t) so that the Laplace transform of dx(t)/dt is given by sX(s) − x(0−) = −0 L[sin(0 t)u(t)] + L[δ(t)] Thus, the Laplace transform of the sine is sX(s) − x(0−) − 1 0 1 − sX(s) = 0 0 = 2 s + 20
L[sin(0 t)u(t)] = −
since x(0−) = 0 and X(s) = L[cos(o T)] given above.
n
Notice that whenever the signal is discontinuous at t = 0, as in the case of x(t) = cos(0 t)u(t), its derivative will include a δ(t) signal due to the discontinuity. On the other hand, whenever the signal is continuous at t = 0, for instance y(t) = sin(0 t)u(t), its derivative does not contain δ(t) signals. In fact, dy(t) = 0 cos(0 t)u(t) + sin(0 t)δ(t) dt = 0 cos(0 t)u(t) since the sine is zero at t = 0.
3.3.3 Integration The Laplace transform of the integral of a causal signal y(t) is given by t Z Y(s) L y(τ )dτ u(t) = s 0
This property can be shown by using the derivative property. Call the integral f (t) =
Zt
y(τ )dτ u(t)
0
Using the fundamental theorem of calculus, we then have that df (t) = y(t)u(t) dt
(3.14)
193
194
CH A P T E R 3: The Laplace Transform
and so df (t) L = sF(s) − f (0) dt
= Y(s) since f (0) = 0 (the integral over a point), then t Z Y(s) F(s) = L y(τ )dτ = s 0
n Example 3.11 Suppose that Zt
y(τ )dτ = 3u(t) − 2y(t)
0
Find the Laplace transform of y(t), a causal signal. Solution Applying the integration property gives Y(s) 3 = − 2Y(s) s s so that solving for Y(s) we obtain Y(s) =
3 2(s + 0.5)
corresponding to y(t) = 1.5e−0.5t u(t).
n
3.3.4 Time Shifting If the Laplace transform of f (t)u(t) is F(s), the Laplace transform of the timeshifted signal f (t − τ )u(t − τ ) is L[f (t − τ )u(t − τ )] = e−τ s F(s)
(3.15)
This indicates that when we delay (advance) the signal to get f (t − τ )u(t − τ ) ( f (t + τ )u(t + τ )) its corresponding Laplace transform is F(s) multiplied by e−τ s (eτ s ). This property is easily shown by a change of variable when computing the Laplace transform of the shifted signals.
3.3 The OneSided Laplace Transform
n Example 3.12 Suppose we wish to find the Laplace transform of the causal sequence of pulses x(t) shown in Figure 3.12. Let x1 (t) denote the first pulse (i.e., for 0 ≤ t < 1). x1(t)
x(t)
··· 0
1
t
0
1
2
3
t
FIGURE 3.12 Generic causal pulse signal.
Solution We have for t ≥ 0, x(t) = x1 (t) + x1 (t − 1) + x1 (t − 2) + · · · and 0 for t < 0. According to the shifting and linearity properties, we have X(s) = X1 (s) 1 + e−s + e−2s + · · · 1 = X1 (s) 1 − e−s Notice that 1 + e−s + e−2s + · · · = 1/(1 − e−s ), which is verified by crossmutiplying: [1 + e−s + e−2s + · · · ](1 − e−s ) = (1 + e−s + e−2s + · · · ) − (e−s + e−2s + · · · ) = 1 The poles of X(s) are the poles of X1 (s) and the roots of 1 − e−s = 0 (the s values such that e−s = 1, or sk = ±j2πk for any integer k ≥ 0). Thus, there is an infinite number of poles for X(s), and the partial fraction expansion method that uses poles to invert Laplace transforms, presented later, will not be useful. The reason this example is presented here, ahead of the inverse Laplace, is to illustrate that when we are finding the inverse of this type of Laplace function we need to consider the timeshift property, otherwise we would need to consider an infinite partial fraction expansion. n
n Example 3.13 Consider the causal fullwave rectified signal shown in Figure 3.13. Find its Laplace transform.
195
CH A P T E R 3: The Laplace Transform
1 0.8 x(t)
196
0.6 0.4 0.2 0
FIGURE 3.13 Fullwave rectified causal signal.
−0.2 −1
0
1
2
3 t
4
5
6
7
Solution The first period of the fullwave rectified signal can be expressed as x1 (t) = sin(2πt)u(t) + sin(2π(t − 0.5))u(t − 0.5) and its Laplace transform is X1 (s) =
2π(1 + e−0.5s ) s2 + (2π )2
And the train of these sinusoidal pulses x(t) =
∞ X
x1 (t − 0.5k)
k=0
will then have the following Laplace transform: X(s) = X1 (s)[1 + e−s/2 + e−s + · · · ] = X1 (s)
1 2π(1 + e−s/2 ) = 1 − e−s/2 (1 − e−s/2 )(s2 + 4π 2 )
n
3.3.5 Convolution Integral Because this is the most important property of the Laplace transform we provide a more extensive coverage later, after considering the inverse Laplace transform. The Laplace transform of the convolution integral of a causal signal x(t), with Laplace transforms X(s), and a causal impulse response h(t), with Laplace transform H(s), is given by L[(x ∗ h)(t)] = X(s)H(s)
(3.16)
3.4 Inverse Laplace Transform
If the input of an LTI system is the causal signal x(t) and the impulse response of the system is h(t), then the output y(t) can be written as Z∞ y(t) =
x(τ )h(t − τ )dτ
t≥0
0
and zero otherwise. Its Laplace transform is ∞ Z Z∞ Z∞ Y(s) = L x(τ )h(t − τ )dτ = x(τ )h(t − τ )dτ e−st dt 0
Z∞ = 0
0
0
∞ Z x(τ ) h(t − τ ) e−s(t−τ ) dt e−sτ dτ = X(s)H(s) 0
where the internal integral is shown to be H(s) = L[h(t)] (change variable to ν = t − τ ) using the causality of h(t). The remaining integral is the Laplace transform of x(t). The system function or transfer function H(s) = L[h(t)], the Laplace transform of the impulse response h(t) of an LTI system, can be expressed as the ratio H(s) =
L[y(t)] L[ output ] = L[ x(t)] L[ input ]
(3.17)
This function is called transfer function because it transfers the Laplace transform of the input to the output. Just as with the Laplace transform of signals, H(s) characterizes an LTI system by means of its poles and zeros. Thus, it becomes a very important tool in the analysis and synthesis of systems.
3.4 INVERSE LAPLACE TRANSFORM Inverting the Laplace transform consists in finding a function (either a signal or an impulse response of a system) that has the given transform with the given region of convergence. We will consider three cases: n n n
Inverse of onesided Laplace transforms giving causal functions. Inverse of Laplace transforms with exponentials. Inverse of twosided Laplace transforms giving anticausal or noncausal functions.
The given function X(s) we wish to invert can be the Laplace transform of a signal or a transfer function—that is, the Laplace transform of an impulse response.
3.4.1 Inverse of OneSided Laplace Transforms When we consider a causal function x(t), the region of convergence of X(s) is of the form {(σ , ) : σ > σmax , −∞ < < ∞}
197
198
CH A P T E R 3: The Laplace Transform
where σmax is the maximum of the real parts of the poles of X(s). Since in this section we only consider causal signals, the region of convergence will be assumed and will not be shown with the Laplace transform. The most common inverse Laplace method is the socalled partial fraction expansion, which consists in expanding the given function in s into a sum of components of which the inverse Laplace transforms can be found in a table of Laplace transform pairs. Assume the signal we wish to find has a rational Laplace transform—that is, X(s) =
N(s) D(s)
(3.18)
where N(s) and D(s) are polynomials in s with realvalued coefficients. In order for the partial fraction expansion to be possible, it is required that X(s) be proper rational, which means that the degree of the numerator polynomial N(s) is less than that of the denominator polynomial D(s). If X(s) is not proper, then we need to do long division until we obtain a proper rational function—that is, X(s) = g0 + g1 s + · · · + gm sm +
B(s) D(s)
(3.19)
where the degree of B(s) is now less than that of D(s)—so that we can perform partial expansion for B(s)/D(s). The inverse of X(s) is then given by x(t) = g0 δ(t) + g1
dδ(t) dm δ(t) −1 B(s) + · · · + gm + L dt dtm D(s)
(3.20)
The presence of δ(t) and its derivatives (called doublets, triplets, etc.) are very rare in actual signals, and as such the typical rational function has a numerator polynomial that is of lower degree than the denominator polynomial. Remarks n
n
Things to remember before performing the inversion are: n The poles of X(s) provide the basic characteristics of the signal x(t). n If N(s) and D(s) are polynomials in s with real coefficients, then the zeros and poles of X(s) are real and/or complex conjugate pairs, and can be simple or multiple. n In the inverse, u(t) should be included since the result of the inverse is causal—the function u(t) is an integral part of the inverse. The basic idea of the partial expansion is to decompose proper rational functions into a sum of rational components of which the inverse transform can be found directly in tables. Table 3.1 displays common onesided Laplace transform pairs, while Table 3.2 provides properties of the onesided Laplace transform.
We will consider now how to obtain a partial fraction expansion when the poles are real, simple and multiple, and in complex conjugate pairs, simple and multiple.
3.4 Inverse Laplace Transform
Table 3.1 OneSided Laplace Transforms Function of Time 1.
δ(t)
2.
u(t)
Function of s, ROC 1,
3.
r(t)
4.
e−at u(t), a > 0
5.
cos(0 t)u(t) sin(0 t)u(t)
6. 7.
e−at
cos(0 t)u(t), a > 0
8.
e−at
sin(0 t)u(t), a > 0
9.
e−at
cos(0 t + θ )u(t), a > 0
2A
1 N−1 u(t) (N−1)! t 1 N−1 e−at u(t) (N−1)! t
10. 11. 12.
2A N−1 e−at (N−1)! t
cos(0 t + θ )u(t)
whole splane
1 s , Re[s] > 0 1 , Re[s] > 0 s2 1 , s+a Re[s] > −a s , Re[s] > 0 s2 +20 0 , Re[s] > 0 s2 +20 s+a , Re[s] > −a (s+a)2 +20 0 , Re[s] > −a (s+a)2 +20 A∠−θ A∠θ s+a−j0 + s+a+j0 , Re[s] > −a 1 N an integer, Re[s] > 0 sN 1 N an integer, Re[s] > −a (s+a)N A∠θ A∠−θ + (s+a+j N , Re[s] > −a (s+a−j0 )N 0)
Table 3.2 Basic Properties of OneSided Laplace Transforms Causal functions and constants
αf (t), βg(t)
Linearity
αf (t) + βg(t)
αF(s) + βG(s)
Time shifting
f (t − α)
e−αs F(s)
Frequency shifting
eαt f (t)
F(s − α)
Multiplication by t
t f (t)
− dF(s) ds
Derivative
df (t) dt
sF(s) − f (0−) s2 F(s) − sf (0−) − f (1) (0)
Integral
d2 f (t) 2 R dt t 0 0− f (t )dt
Expansion/contraction
f (αt) α 6= 0
Initial value
f (0+) = lims→∞ sF(s)
Final value
limt→∞ f (t) = lims→0 sF(s)
Second derivative
αF(s), βG(s)
F(s) s 1 s α F α
Simple Real Poles If X(s) is a proper rational function X(s) =
N(s) N(s) = Q D(s) k (s − pk )
(3.21)
199
200
CH A P T E R 3: The Laplace Transform
where the {pk } are simple real poles of X(s), its partial fraction expansion and its inverse are given by X(s) =
X Ak X ⇔ x(t) = Ak epk t u(t) s − pk k
(3.22)
k
where the expansion coefficients are computed as Ak = X(s)(s − pk ) s=pk
According to Laplace transform tables the time function corresponding to Ak /(s − pk ) is Ak epk t u(t), thus the form of the inverse x(t). To find the coefficients of the expansion, say Aj , we multiply both sides of the Equation (3.22) by the corresponding denominator (s − pj ) so that X(s)(s − pj ) = Aj +
X Ak (s − pj ) k6=j
s − pk
If we let s = pj , or s − pj = 0, in the above expression, all the terms in the sum will be zero and we find that Aj = X(s)(s − pj ) s=pj
n Example 3.14 Consider the proper rational function X(s) =
s2
3s + 5 3s + 5 = + 3s + 2 (s + 1)(s + 2)
Find its causal inverse. Solution The partial fraction expansion is X(s) =
A1 A2 + s+1 s+2
Given that the two poles are real, the expected signal x(t) will be a superposition of two decaying exponentials, with damping factors −1 and −2, or x(t) = [A1 e−t + A2 e−t ]u(t) where as indicated above, A1 = X(s)(s + 1)s=−1 =
3s + 5 s=−1 = 2 s+2
3.4 Inverse Laplace Transform
and A2 = X(s)(s + 2)s=−2 =
3s + 5 s=−2 = 1 s+1
Therefore, X(s) =
1 2 + s+1 s+2
and as such x(t) = [2e−t + e−2t ]u(t) To check that the solution is correct one could use the initial or the final value theorems shown in Table 3.2. According to the initial value theorem, x(0) = 3 should coincide with 3s2 + 5s 3 + 5/s lim sX(s) = 2 =3 = lim s→∞ s→∞ s + 3s + 2 1 + 3/s + 2/s2 as it does. The final value theorem indicates that limt→∞ x(t) = 0 should coincide with 3s2 + 5s lim sX(s) = 2 =0 s→0 s + 3s + 2 as it does. Both of these validations seem to indicate that the result is correct.
n
Remarks The coefficients A1 and A2 can be found using other methods. For instance, n
We can compute X(s) =
A1 A2 + s+1 s+2
(3.23)
for two different values of s (as long as we do not divide by zero), such as s = 0 and s = 1, 5 1 = A1 + A2 2 2 8 1 1 s = 1 X(1) = = A1 + A2 6 2 3 s = 0 X(0) =
n
which gives a set of two linear equations with two unknows, and applying Cramer’s rule we find that A1 = 2 and A2 = 1. We crossmultiply the partial expansion given by Equation (3.23) to get X(s) =
3s + 5 s(A1 + A2 ) + (2A1 + A2 ) = s2 + 3s + 2 s2 + 3s + 2
Comparing the numerators, we have that A1 + A2 = 3 and 2A1 + A2 = 5, two equations with two unknowns, which can be shown to have as unique solutions A1 = 2 and A2 = 1, as before.
201
202
CH A P T E R 3: The Laplace Transform
Simple Complex Conjugate Poles The partial fraction expansion of a proper rational function X(s) =
N(s) (s + α)2 + 20
=
N(s) (s + α − j0 )(s + α + j0 )
(3.24)
with complex conjugate poles {s1,2 = −α ± j0 } is given by X(s) =
A∗ A + s + α − j0 s + α + j0
where A = X(s)(s + α − j0 )s=−α+j0 = Aejθ so that the inverse is the function x(t) = 2Ae−αt cos(0 t + θ)u(t)
(3.25)
Because the numerator and the denominator polynomials of X(s) have real coefficients, the zeros and poles whenever complex appear as complex conjugate pairs. One could thus think of the case of a pair of complex conjugate poles as similar to the case of two simple real poles presented above. Notice that the numerator N(s) must be a firstorder polynomial for X(s) to be proper rational. The poles of X(s), s1,2 = −α ± j0 , indicate that the signal x(t) will have an exponential e−αt , given that the real part of the poles is −α, multiplied by a sinusoid of frequency 0 , given that the imaginary parts of the poles are ±0 . We have the expansion X(s) =
A A∗ + s + α − j0 s + α + j0
where the expansion coefficients are complex conjugate of each other. From the pole information, the general form of the inverse is x(t) = Ke−αt cos(0 t + 8)u(t) for some constants K and 8. As before, we can find A as A = X(s)(s + α − j0 )s=−α+j0 = Aejθ and that X(s)(s + α + j0 )s=−α−j0 = A∗ can be easily verified. Then the inverse transform is given by h i x(t) = Ae−(α−j0 )t + A∗ e−(α+j0 )t u(t) = Ae−αt (ej(0 t+θ ) + e−j(0 t+θ ) )u(t) = 2Ae−αt cos(0 t + θ )u(t).
3.4 Inverse Laplace Transform
Remarks n
An equivalent partial fraction expansion consists in expressing the numerator N(s) of X(s), for some constants a and b, as N(s) = a + b(s + α), a firstorder polynomial, so that X(s) =
a s+α a + b(s + α) 0 = +b 0 (s + α)2 + 20 (s + α)2 + 20 (s + α)2 + 20
so that the inverse is a sum of a sine and a cosine multiplied by a decaying exponential. The inverse Laplace transform is a −αt −αt x(t) = e sin(0 t) + be cos(0 t) u(t) 0 which can be simplified, using the sum of phasors corresponding to sine and cosine, to s a2 a −αt −1 2 x(t) = +b e cos 0 t − tan u(t) 0 b 20 n
When α = 0 the above indicates that the inverse Laplace transform of X(s) =
a + bs s2 + 20
is s x(t) =
n
a2 a 2 cos t − tan−1 u(t) + b 0 0 b 20
which is transform of a cosine with a phase shift not commonly found in tables. When the frequency 0 = 0, we get that the inverse Laplace transform of X(s) =
a + b(s + α) a b = + (s + α)2 (s + α)2 s+α
(corresponds to a double pole at −α) is x(t) = lim
0 →0
a −αt e sin(0 t) + be−αt cos(0 t) u(t) 0
= [ate−αt + be−αt ]u(t) where the first limit is found by L’Hˆopital’s rule. Notice that when computing the partial fraction expansion of the double pole s = −α the expansion is composed of two terms, one with denominator (s + α)2 and the other with denominator s + α of which the sum gives a firstorder numerator and a secondorder denominator to satisfy the proper rational condition.
203
204
CH A P T E R 3: The Laplace Transform
n Example 3.15 Consider the Laplace function X(s) =
s2
2s + 3 2s + 3 = + 2s + 4 (s + 1)2 + 3
Find the corresponding causal signal x(t), then use MATLAB to validate your answer. Solution √ The poles are at −1 ± j 3, so that we expect that x(t) is a decaying exponential with √ a damping factor of −1 (the real part of the poles) multiplied by a causal cosine of frequency 3. The partial fraction expansion is of the form X(s) =
s2
2s + 3 a + b(s + 1) = + 2s + 4 (s + 1)2 + 3
so that 3 + 2s = (a + b) + bs, or b = 2 and a + b = 3 or a = 1. Thus, √ 3 1 s+1 X(s) = √ +2 (s + 1)2 + 3 3 (s + 1)2 + 3 which corresponds to √ √ 1 x(t) = √ sin( 3t) + 2 cos( 3t) e−t u(t) 3
The value x(0) = 2 and according to the initial value theorem the following limit should equal it: 2 + 3/s 2s2 + 3s = lim =2 lim sX(s) = 2 s→∞ s→∞ 1 + 2/s + 4/s2 s + 2s + 4 which is the case, indicating the result is probably correct (satisfying the initial value theorem is not enough to indicate the result is correct, but if it does not the result is wrong). We use the MATLAB function ilaplace to compute symbolically the inverse Laplace transform and plot the response using ezplot, as shown in the following script. %%%%%%%%%%%%%%%%% % Example 3.15 %%%%%%%%%%%%%%%%% clear all; clf syms s t w num = [0 2 3]; den = [1 2 4]; % coefficients of numerator and denominator subplot(121) splane(num,den) % plotting poles and zeros disp(’>>>>> Inverse Laplace Zeros Poles Residues Zeros > Poles > Residues > Inverse Laplace M
(3.29)
where x(t) is the input and y(t) is the output of the system, and initial conditions {y(k) (t), 0 ≤ k ≤ N − 1}
(3.30)
is obtained by inverting the Laplace transform Y(s) =
1 B(s) X(s) + I(s) A(s) A(s)
where Y(s) = L[y(t)], X(s) = L[x(t)], and A(s) =
N X
ak sk
k=0
B(s) =
M X `=0
b ` s`
aN = 1
(3.31)
3.5 Analysis of LTI Systems
I(s) =
N X
ak
k−1 X
sk−m−1 y(m) (0)
m=0
k=1
That is, I(s) depends on the initial conditions.
The notation y(k) (t) and x(`) (t) indicates the kth and the `th derivatives of y(t) and of x(t), respectively (it is to be understood that y(0) (t) = y(t) and likewise x(0) (t) = x(t) in this notation). The assumption N > M avoids the presence of δ(t) and its derivatives in the solution, which are realistically not possible. To obtain the complete response y(t) we compute the Laplace transform of Equation (3.29): # " N "M # N k−1 X X X X ak sk Y(s) = ak b` s` X(s) + s(k−1)−m y(m) (0) `=0
k=0

{z
A(s)
k=1
{z

}
B(s)
}
m=0
{z

I(s)
}
which can be written as A(s)Y(s) = B(s)X(s) + I(s)
(3.32)
by defining A(s), B(s), and I(s) as indicated above. Solving for Y(s) in Equation (3.32), we have Y(s) =
B(s) 1 X(s) + I(s) A(s) A(s)
and finding its inverse we obtain the complete response y(t). Letting H(s) =
1 B(s) and H1 (s) = A(s) A(s)
the complete response y(t) = L−1 [Y(s)] of the system is obtained by the inverse Laplace transform of Y(s) = H(s)X(s) + H1 (s)I(s)
(3.33)
y(t) = yzs (t) + yzi (t)
(3.34)
which gives
where zerostate response:
yzs (t) = L−1 [H(s)X(s)]
zeroinput response:
yzi (t) = L−1 [H1 (s)I(s)]
In terms of convolution integrals, Zt y(t) = 0
x(τ )h(t − τ )dτ +
Zt 0
i(τ )h1 (t − τ )dτ
(3.35)
215
216
CH A P T E R 3: The Laplace Transform
where h(t) = L−1 [H(s)] and h1 (t) = L−1 [H1 (s)], and N k−1 X X i(t) = L−1 [I(s)] = ak y(m) (0)δ (k−m−1) (t) k=1
m=0
where {δ (m) (t)} are mth derivatives of the impulse signal δ(t) (as indicated before, δ (0) (t) = δ(t)).
ZeroState and ZeroInput Responses Despite the fact that linear differential equations, with constant coefficients, do not represent linear systems unless the initial conditions are zero and the input is causal, linear system theory is based on these representations with initial conditions. Typically, the input is causal so it is the initial conditions not always being zero that causes problems. This can be remedied by a different way of thinking about the initial conditions. In fact, one can think of the input x(t) and the initial conditions as two different inputs to the system, and apply superposition to find the responses to these two different inputs. This defines two responses. One is due completely to the input, with zero initial conditions, called the zerostate solution. The other component of the complete response is due exclusively to the initial conditions, assuming that the input is zero, and is called the zeroinput solution. Remarks n
It is important to recognize that to compute the transfer function of the system H(s) =
n
Y(s) X(s)
according to Equation (3.33) requires that the initial conditions be zero, or I(s) = 0. If there is no polezero cancellation, both H(s) and H1 (s) have the same poles, as both have A(s) as denominator, and as such h(t) and h1 (t) might be similar.
Transient and SteadyState Responses Whenever the input of a causal and stable system has poles in the closed lefthand splane, poles in the jaxis being simple, the complete response will be bounded. Moreover, whether the response exists as t → ∞ can then be determined without using the inverse Laplace transform. The complete response y(t) of an LTI system is made up of transient and steadystate components. The transient response can be thought of as the system’s reaction to the initial inertia after applying the input, while the steadystate response is how the system reacts to the input away from the initial time when the input starts. If the poles (simple or multiple, real or complex) of the Laplace transform of the output, Y(s), of an LTI system are in the open lefthand splane (i.e., no poles on the j axis), the steadystate response is yss (t) = lim y(t) = 0 t→∞
3.5 Analysis of LTI Systems
In fact, for any real pole s = −α, α > 0, of multiplicity m ≥ 1, we have that X m N(s) −1 L = Ak tk−1 e−αt u(t) (s + α)m k=1
where N(s) is a polynomial of degree less or equal to m − 1. Clearly, for any value of α > 0 and any order m ≥ 1, the above inverse will tend to zero as t increases. The rate at which these terms go to zero depends on how close the pole(s) is (are) to the j axis: The farther away, the faster the term goes to zero. Likewise, complex conjugate pairs of poles with a negative real part also give terms that go to zero as t → ∞, independent of their order. For complex conjugate pairs of poles s1,2 = −α ± j0 of order m ≥ 1, we have # " m X N(s) = 2Ak tk−1 e−αt cos(0 t + ∠(Ak ))u(t) L−1 ((s + α)2 + 20 )m k=1
where again N(s) is a polynomial of degree less or equal to 2m − 1. Due to the decaying exponentials this type of term will go to zero as t goes to infinity. Simple complex conjugate poles and a simple real pole at the origin of the splane cause a steadystate response. Indeed, if the pole of Y(s) is s = 0 we know that its inverse transform is of the form Au(t), and if the poles are complex conjugates ±j0 the corresponding inverse transform is a sinusoid— neither of which is transient. However, multiple poles on the jaxis, or any poles in the righthand splane will give inverses that grow as t → ∞. This statement is clear for the poles in the righthand splane. For double or higherorder poles in the j axis their inverse transform is of the form " # m X N(s) L−1 = 2Ak tm−1 cos(0 + ∠(Ak ))u(t) (s2 + 20 )m k=1
which will continuously grow as t increases. In summary, when solving differential equations—with or without initial conditions—we have n The steadystate component of the complete solution is given by the inverse Laplace transforms of the partial fraction expansion terms of Y(s) that have simple poles (real or complex conjugate pairs) in the jaxis. n The transient response is given by the inverse transform of the partial fraction expansion terms with poles in the lefthand splane, independent of whether the poles are simple or multiple, real or complex. n Multiple poles in the j axis and poles in the righthand splane give terms that will increase as t increases.
n Example 3.22 Consider a secondorder (N = 2) differential equation, d2 y(t) dy(t) +3 + 2y(t) = x(t) 2 dt dt
217
218
CH A P T E R 3: The Laplace Transform
Assume the above equation represents a system with input x(t) and output y(t). Find the impulse response h(t) and the unitstep response s(t) of the system. Solution If the initial conditions are zero, computing the two or onesided Laplace transform of the two sides of this equation, after letting Y(s) = L[y(t)] and X(s) = L[x(t)], and using the derivative property of Laplace, we get Y(s)[s2 + 3s + 2] = X(s) To find the impulse response of this system (i.e., the system response y(t) = h(t)), we let x(t) = δ(t) and the initial condition be zero. Since X(s) = 1, then Y(s) = H(s) = L[h(t)] is H(s) =
1 1 A B = = + s2 + 3s + 2 (s + 1)(s + 2) s+1 s+2
We obtain values A = 1 and B = −1, and the inverse Laplace transform is then h(t) = e−t − e−2t u(t) which is completely transient. In a similar form we obtain the unitstep response s(t), by letting x(t) = u(t) and the initial conditions be zero. Calling Y(s) = S(s) = L[s(t)], since X(s) = 1/s, we obtain S(s) =
H(s) 1 A B C = 2 = + + s s(s + 3s + 2) s s+1 s+2
It is found that A = 1/2, B = −1, and C = 1/2, so that s(t) = 0.5u(t) − e−t u(t) + 0.5e−2t u(t) The steady state of s(t) is 0.5 as the two exponentials go to zero. Interestingly, the relation sS(s) = H(s) indicates that by computing the derivative of s(t) we obtain h(t). Indeed, ds(t) = 0.5δ(t) + e−t u(t) − e−t δ(t) − e−2t u(t) + 0.5e−2t δ(t) dt = [0.5 − 1 + 0.5]δ(t) + [e−t − e−2t ]u(t) = [e−t − e−2t ]u(t) = h(t)
n
Remarks n
Because the existence of the steadystate response depends on the poles of Y(s) it is possible for an unstable causal system (recall that for such a system BIBO stability requires all the poles of the system transfer function be in the open, lefthand splane) to have a steadystate response. It all depends on the input. Consider, for instance, an unstable system with H(s) = 1/(s(s + 1)), being unstable due to the pole at
3.5 Analysis of LTI Systems
s = 0; if the system input is x1 (t) = u(t) so that X1 (s) = 1/s, then Y1 (s) = 1/(s2 (s + 1)). There will be no steady state because of the double pole s = 0. On the other hand, X2 (s) = s/(s + 2)2 will give Y2 (s) = H(s)X2 (s) =
n
n
1 s 1 = 2 s(s + 1) (s + 2) (s + 1)(s + 2)2
which will give a zero steady state, even though the system is unstable. This is possible because of the polezero cancellation. The steadystate response is the response of the system away from t = 0, and it can be found by letting t → ∞ (even though the steady state can be reached at finite times, depending on how fast the transient goes to zero). In Example 3.22, the steadystate response of h(t) = (e−t − e−2t )u(t) is zero, while for s(t) = 0.5u(t) − e−t u(t) + 0.5e−2t u(t) it is 0.5. The transient responses are then h(t) − 0 = h(t) and s(t) − 0.5u(t) = −e−t u(t) + 0.5e−2t u(t). These transients eventually disappear. The relation found between the impulse response h(t) and the unitstep response s(t) can be extended to more cases by the definition of the transfer function—that is, H(s) = Y(s)/X(s) so that the response Y(s) is connected with H(s) by Y(s) = H(s)X(s), giving the relation between y(t) and h(t). For instance, if x(t) = δ(t), then Y(s) = H(s) × 1, with inverse the impulse response. If x(t) = u(t), then Y(s) = H(s)/s is S(s), the Laplace transform of the unitstep response, and so s(t) = dh(t)/dt. And if x(t) = r(t), then Y(s) = H(s)/s2 is ρ(s), the Laplace transform of the ramp response, and so ρ(t) = d2 h(t)/dt2 = ds(t)/dt.
n Example 3.23 Consider again the secondorder differential equation in the previous example, d2 y(t) dy(t) +3 + 2y(t) = x(t) 2 dt dt but now with initial conditions y(0) = 1 and dy(t)/dtt=0 = 0, and x(t) = u(t). Find the complete response y(t). Could we find the impulse response h(t) from this response? How could we do it? Solution The Laplace transform of the differential equation gives [s2 Y(s) − sy(0) −
dy(t) t=0 ] + 3[sY(s) − y(0)] + 2Y(s) = X(s) dt
Y(s)(s2 + 3s + 2) − (s + 3) = X(s) so we have that Y(s) = =
X(s) s+3 + (s + 1)(s + 2) (s + 1)(s + 2) 1 + 3s + s2 B1 B2 B3 = + + s(s + 1)(s + 2) s s+1 s+2
219
220
CH A P T E R 3: The Laplace Transform
after replacing X(s) = 1/s. We find that B1 = 1/2, B2 = 1, and B3 = −1/2, so that the complete response is y(t) = [0.5 + e−t − 0.5e−2t ]u(t)
(3.36)
Again, we can check that this solution satisfies the initial condition y(0) and dy(0)/dt (this is particularly interesting to see, try it!). The steadystate response is 0.5 and the transient [e−t − 0.5e−2t ]u(t). According to Equation (3.36), the complete solution y(t) is composed of the zerostate response, due to the input only, and the response due to the initial conditions only or the zeroinput response. Thus, the system considers two different inputs: One that is x(t) = u(t) and the other the initial conditions. If we are able to find the transfer function H(s) = Y(s)/X(s) its inverse Laplace transform would be h(t). However that is not possible when the initial conditions are nonzero. As shown above, in the case of nonzero initial conditions, we get that the Laplace transform is Y(s) =
X(s) I(s) + A(s) A(s)
where in this case A(s) = (s + 1)(s + 2) and I(s) = s + 3, and thus we cannot find the ratio Y(s)/X(s). If we make the second term zero (i.e., I(s) = 0), we then have that Y(s)/X(s) = H(s) = 1/A(s) and h(t) = e−t u(t) − e−2t u(t). n n Example 3.24 Consider an analog averager represented by y(t) =
1 T
Zt
x(τ )dτ
(3.37)
t−T
where x(t) is the input and y(t) is the output. The derivative of y(t) gives the firstorder differential equation dy(t) 1 = [x(t) − x(t − T)] dt T with a finite difference for the input. Let us find the impulse response of this analog averager. Solution The impulse response of the averager is found by letting x(t) = δ(t) and the initial condition be zero. Computing the Laplace transform of the two sides of the differential equation, we obtain sY(s) =
1 [1 − e−sT ]X(s) T
3.5 Analysis of LTI Systems
and substituting X(s) = 1, then H(s) = Y(s) =
1 [1 − e−sT ] sT
The impulse response is then h(t) =
1 [u(t) − u(t − T)]. T
n
3.5.2 Computation of the Convolution Integral From the point of view of signal processing, the convolution property is the most important application of the Laplace transform to systems. The computation of the convolution integral is difficult even for simple signals. In Chapter 2 we showed how to obtain the convolution integral analytically as well as graphically. As we will see in this section, it is not only that the convolution property of the Laplace transform gives an efficient solution to the computation of the convolution integral, but that it introduces an important representation of LTI systems, namely the transfer function of the system. A system, like signals, is thus represented by the poles and zeros of the transfer function. But it is not only the polezero characterization of the system that can be obtained from the transfer function. The system’s impulse response is uniquely obtained from the poles and zeros of the transfer function and the corresponding region of convergence. The way the system responds to different frequencies will be also given by the transfer function. Stability and causality of the system can be equally related to the transfer function. Design of filters depends on the transfer function. The Laplace transform of the convolution y(t) = [x ∗ h](t) is given by the product (3.38)
Y(s) = X(s)H(s) where X(s) = L[x(t)] and H(s) = L[h(t)]. The transfer function of the system H(s) is defined as H(s) = L[h(t)] =
Y(s) X(s)
(3.39)
H(s) transfers the Laplace transform X(s) of the input into the Laplace transform of the output Y(s). Once Y(s) is found, y(t) is computed by means of the inverse Laplace transform.
n Example 3.25 Use the Laplace transform to find the convolution y(t) = [x ∗ h](t) when (1) the input is x(t) = u(t) and the impulse response is a pulse h(t) = u(t) − u(t − 1), and (2) the input and the impulse response of the system are x(t) = h(t) = u(t) − u(t − 1). Solution n
The Laplace transforms are X(s) = L[u(t)] = 1/s and H(s) = L[h(t)] = (1 − e−s )/s, so that Y(s) = H(s)X(s) =
1 − e−s s2
221
222
CH A P T E R 3: The Laplace Transform
Its inverse is y(t) = r(t) − r(t − 1)
n
where r(t) is the ramp signal. This result coincides with the one obtained graphically in Example 2.12 in Chapter 2. In the second case, X(s) = H(s) = L[u(t) − u(t − 1)] = (1 − e−s )/s, so that (1 − e−s )2 1 − 2e−s + e−2s = s2 s2
Y(s) = H(s)X(s) = which corresponds to
y(t) = r(t) − 2r(t − 1) + r(t − 2) or a triangular pulse as we obtained graphically in Example 2.13 in Chapter 2.
n
n Example 3.26 To illustrate the significance of the Laplace approach in computing the output of an LTI system by means of the convolution integral, consider an RLC circuit in series with input a voltage source x(t) and as output the voltage y(t) across the capacitor (see Figure 3.17). Find its impulse response h(t) and its unitstep response s(t). Let LC = 1 and R/L = 2. Solution The RLC circuit is represented by a secondorder differential equation given that the inductor and the capacitor are capable of storing energy and their initial conditions are not dependent on each other. To obtain the differential equation we apply Kirchhoff’s voltage law (KVL) x(t) = Ri(t) + L
di(t) + y(t) dt
where i(t) is the current through the resistor, inductor and capacitor and where the voltage across the capacitor is given by 1 y(t) = C
Zt
i(σ )dσ + y(0)
0
R
FIGURE 3.17 RLC circuit with input a voltage source x(t) and output the voltage across the capacitor y(t).
+ x(t) −
L + C
y(t) −
3.5 Analysis of LTI Systems
with y(0) the initial voltage in the capacitor and i(t) the current through the resistor, inductor, and capacitor. The above two equations are called integrodifferential given that they are composed of an integral equation and a differential equation. To obtain a differential equation in terms of the input x(t) and the output y(t), we find the first and second derivative of y(t), which gives dy(t) 1 dy(t) = i(t) ⇒ i(t) = C dt C dt d2 y(t) 1 di(t) di(t) d2 y(t) = ⇒L = LC 2 dt C dt dt dt2 which when replaced in the KVL equation gives d2 y(t) dy(t) + y(t) (3.40) + LC dt dt2 which, as expected, is a secondorder differential equation with two initial conditions: y(0), the initial voltage in the capacitor, and i(0) = Cdy(t)/dtt=0 , the initial current in the inductor. To find the impulse response of this circuit, we let x(t) = δ(t) and the initial conditions be zero. The Laplace transform of Equation (3.40) gives x(t) = RC
X(s) = [LCs2 + RCs + 1]Y(s) The impulse response of the system is the inverse Laplace transform of the transfer function Y(s) 1/LC = 2 X(s) s + (R/L)s + 1/LC
H(s) =
If LC = 1 and R/L = 2, then the transfer function is H(s) =
1 (s + 1)2
which corresponds to the impulse response h(t) = te−t u(t) Now that we have the impulse response of the system, suppose then the input is a unitstep signal, x(t) = u(t). To find its response we consider the convolution integral Z∞ y(t) =
x(τ )h(t − τ )dτ
−∞
Z∞ =
u(τ )(t − τ )e−(t−τ ) u(t − τ )dτ
−∞
Zt =
(t − τ )e−(t−τ ) dτ
0
= [1 − e−t (1 + t)]u(t)
223
224
CH A P T E R 3: The Laplace Transform
which satisfies the initial conditions and attempts to follow the input signal. This is the unitstep response. In the Laplace domain, the above can be easily computed as follows. From the transfer function, we have that Y(s) = H(s)X(s) 1 1 = 2 (s + 1) s where we replaced the transfer function and the Laplace transform of x(t) = u(t). The partial fraction expansion of Y(s) is then Y(s) =
A B C + + s s + 1 (s + 1)2
and after obtaining that A = 1, C = −1, and B = −1, we get y(t) = s(t) = u(t) − e−t u(t) − te−t u(t) which coincides with the solution of the convolution integral. It has been obtained, however, in a much easier way. n
n Example 3.27 Consider the positive feedback system created by a microphone close to a set of speakers that are putting out an amplified acoustic signal (see Figure 3.18), which we considered in Example 2.18 in Chapter 2. Find the impulse response of the system using the Laplace transform, and use it to express the output in terms of a convolution. Determine the transfer function and show that the system is not BIBO stable. For simplicity, let β = 1, τ = 1, and x(t) = u(t). Connect the location of the poles of the transfer function with the unstable behavior of the system. Solution As we indicated in Example 2.18 in Chapter 2, the impulse response of a feedback system cannot be explicitly obtained in the time domain, but it can be done using the Laplace transform. The input–output equation for the positive feedback is y(t) = x(t) + βy(t − τ ) x (t)
y (t) +
FIGURE 3.18 Positive feedback created by closeness of a microphone to a set of speakers.
βy (t − τ)
Delay τ
×
β
3.5 Analysis of LTI Systems
If we let x(t) = δ(t), the output is y(t) = h(t) or h(t) = δ(t) + βh(t − τ ) and if H(s) = L[h(t)], then the Laplace transform of the above equation is H(s) = 1 + βH(s)e−sτ or solving for H(s): H(s) = =
1 1 = 1 − βe−sτ 1 − e−s ∞ X
e−sk = 1 + e−s + e−2s + e−3s + · · ·
k=0
after replacing the given values for β and τ . The impulse response h(t) is the inverse Laplace transform of H(s) or h(t) = δ(t) + δ(t − 1) + δ(t − 2) + · · · =
∞ X
δ(t − k)
k=0
If x(t) is the input, the output is given by the convolution integral Z∞ y(t) =
x(t − τ )h(τ )dτ =
∞ Z∞ X
δ(τ − k)x(t − τ )dτ
−∞ k=0
−∞
=
Z∞ X ∞
δ(τ − k)x(t − τ )dτ =
k=0−∞
∞ X
x(t − k)
k=0
and replacing x(t) = u(t), we get y(t) =
∞ X
u(t − k)
k=0
which tends to infinity as t increases. For this system to be BIBO stable, the impulse response h(t) must be absolutely integrable, which is not the case for this system. Indeed, Z∞ h(t)dt =
Z∞ X ∞
δ(t − k)dt
−∞ k=0
−∞
=
∞ Z∞ X k=0−∞
δ(t − k)dt =
∞ X
1→∞
k=0
The poles of H(s) are the roots of 1 − e−s = 0, which are the values of s such that e−sk = 1 = ej2πk or sk = ±j2πk. That is, there is an infinite number of poles on the j axis, indicating that the system is not BIBO stable. n
225
226
CH A P T E R 3: The Laplace Transform
3.6 WHAT HAVE WE ACCOMPLISHED? WHERE DO WE GO FROM HERE? In this chapter you have learned the significance of the Laplace transform in the representation of signals as well as of systems. The Laplace transform provides a complementary representation to the time representation of a signal, so that damping and frequency, poles and zeros, together with regions of convergence, conform a new domain for signals. But it is more than that—you will see that these concepts will apply for the rest of this part of the book. When discussing the Fourier analysis of signals and systems we will come back to the Laplace domain for computational tools and for interpretation. The solution of differential equations and the different types of responses are obtained algebraically with the Laplace transform. Likewise, the Laplace transform provides a simple and yet very significant solution to the convolution integral. It also provides the concept of transfer function, which will be fundamental in analysis and synthesis of linear timeinvariant systems. The common thread of the Laplace and the Fourier transforms is the eigenfunction property of LTI systems. You will see that understanding this property will provide you with the needed insight into the Fourier analysis, which we will cover in the next two chapters.
PROBLEMS 3.1. Generic signal representation and the Laplace transform The generic representation of a signal x(t) in terms of impulses is Z∞ x(t) =
x(τ )δ(t − τ )dτ
−∞
Considering the integral an infinite sum of terms x(τ )δ(t − τ ) (think of x(τ ) as a constant, as it is not a function of time t), find the Laplace transform of each of these terms and use the linearity property to find X(s) = L[x(t)]. Are you surprised at this result? 3.2. Impulses and the Laplace transform Given x(t) = 2[δ(t + 1) + δ(t − 1)] (a) Find the Laplace transform X(s) of x(t) and determine its region of convergence. (b) Plot x(t). (c) The function X(s) is complex. Let s = σ + j and carefully obtain the magnitude X(σ + j) and the phase ∠X(σ + j). 3.3. Sinusoids and the Laplace transform Consider the following cases involving sinusoids: (a) Find the Laplace transform of y(t) = sin(2π t)u(t) − sin(2π(t − 1))u(t − 1)) and its region of convergence. Carefully plot y(t). Determine the region of convergence of Y(s). (b) A very smooth pulse, called the raised cosine, x(t) is obtained as x(t) = 1 − cos(2π t)
0≤t≤1
Problems
and zero elsewhere. The raised cosine is used in communications to transmit signals with minimal interference. Find its Laplace transform and its corresponding region of convergence. (c) Indicate three possible approaches to finding the Laplace transform of cos2 (t)u(t). Use two of these approaches to find the Laplace transform. 3.4. Unitstep signals and the Laplace transform Find the Laplace transform of the reflection of the unitstep signal (i.e., u(−t)) and its region of convergence. Then use the result together with the Laplace transform of u(t) to see if you can obtain the Laplace transform of a constant or u(t) + u(−t) (assume u(0) = 0.5 so there is no discontinuity at t = 0). 3.5. Laplace transform of noncausal signal Consider the noncausal signal x(t) = e−t u(t + 1) Carefully plot it, and find its Laplace transform X(s) by separating x(t) into a causal signal and an anticausal signal, xc (t) and xac (t), respectively, and plot them separately. Find the ROC of X(s), Xc (s), and Xac (s). 3.6. Transfer function and differential equation The transfer function of a causal LTI system is 1 H(s) = 2 s +4 (a) Find the differential equation that relates the input x(t) and the output y(t) of the system. (b) Suppose we would like the output y(t) to be identically zero for t greater or equal to zero. If we let x(t) = δ(t), what would the initial conditions be equal to? 3.7. Transfer function The input to an LTI system is x(t) = u(t) − 2u(t − 1) + u(t − 2) If the Laplace transform of the output is given by Y(s) =
(s + 2)(1 − e−s )2 ) s2 (s + 1)2
determine the transfer function of the system. 3.8. Inverse Laplace transform—MATLAB Consider the following inverse Laplace transform problems for a causal signal x(t): (a) Given the Laplace transform s4 + 2s + 1 X(s) = 3 s + 4s2 + 5s + 2 which is not proper, determine the amplitude of the δ(t) and dδ(t)/dt terms in the inverse signal x(t). (b) Find the inverse Laplace transform of X(s) =
s2 − 3 (s + 1)(s + 2)
Can you use the initialvalue theorem to check your result? Explain.
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CH A P T E R 3: The Laplace Transform
(c) The inverse Laplace transform of X(s) =
3s − 4 s(s + 1)(s + 2)
should give a response of the form x(t) = [Ae−t + B + Ce−2t ]u(t) Find the values of A, B, and C. Use the MATLAB function ilaplace to get the inverse. 3.9. Steady state and transient Consider the following cases where we want to determine either the steady state, transient, or both. (a) Without computing the inverse of the Laplace transform X(s) =
1 s(s2 + 2s + 10)
corresponding to a causal signal x(t), determine its steadystate solution. What is the value of x(0)? Show how to obtain this value without computing the inverse Laplace transform. (b) The Laplace transform of the output of an LTI system is Y(s) =
1 s((s + 2)2 + 1)
What would be the steadystate response yss (t)? (c) The Laplace transform of the output of an LTI system is Y(s) =
e−s s((s − 2)2 + 1)
How would you determine if there is a steady state or not? Explain. (d) The Laplace transform of the output of an LTI system is Y(s) =
s+1 s((s + 1)2 + 1)
Determine the steadystate and the transient responses corresponding to Y(s). 3.10. Inverse Laplace transformation—MATLAB Consider the following inverse Laplace problems using MATLAB for causal signal x(t): (a) Use MATLAB to compute the inverse Laplace transform of X(s) =
s2 + 2s + 1 s(s + 1)(s2 + 10s + 50)
and determine the value of x(t) in the steady state. How would you be able to obtain this value without computing the inverse? Explain (b) Find the poles and zeros of X(s) =
(1 − se−s ) s(s + 2)
Find the inverse Laplace transform x(t) (use MATLAB to verify your result).
Problems
3.11. Convolution integral Consider the following problems related to the convolution integral: (a) The impulse response of an LTI system is h(t) = e−2t u(t) and the system input is a pulse x(t) = u(t) − u(t − 3). Find the output of the system y(t) by means of the convolution integral graphically and by means of the Laplace transform. (b) It is known that the impulse response of an analog averager is h(t) = u(t) − u(t − 1). Consider the input to the averager x(t) = u(t) − u(t − 1), and determine graphically as well as by means of the Laplace transform the corresponding output of the averager y(t) = [h ∗ x](t). Is y(t) smoother than the input signal x(t)? Provide an argument for your answer. (c) Suppose we cascade three analog averagers each with the same impulse response h(t) = u(t) − u(t − 1). Determine the transfer function of this system. If the duration of the support of the input to the first averager is M sec, what would be the duration of the support of the output of the third averager? 3.12. Deconvolution In convolution problems the impulse response h(t) of the system and the input x(t) are given and one is interested in finding the output of the system y(t). The socalled ”deconvolution” problem consists in giving two of x(t), h(t), and y(t) to find the other. For instance, given the output y(t) and the impulse response h(t) of the system, one wants to find the input. Consider the following cases: (a) Suppose the impulse response of the system is h(t) = e−t cos(t)u(t) and the output has a Laplace transform Y(s) =
4 s((s + 1)2 + 1)
What is the input x(t)? (b) The output of an LTI system is y(t) = r(t) − 2r(t − 1) + r(t − 2), where r(t) is the ramp signal. Determine the impulse response of the system if it is known that the input is x(t) = u(t) − u(t − 1). 3.13. Application of superposition One of the advantages of LTI systems is the superposition property. Suppose that the transfer function of a LTI system is s H(s) = 2 s +s+1 Find the unitstep response s(t) of the system, and then use it to find the response due to the following inputs: x1 (t) = u(t) − u(t − 1) x2 (t) = δ(t) − δ(t − 1) x3 (t) = r(t) x4 (t) = r(t) − 2r(t − 1) + r(t − 2) Express the responses yi (t) due to xi (t) for i = 1, . . . , 4 in terms of the unitstep response s(t). 3.14. Properties of the Laplace transform Consider computing the Laplace transform of a pulse ( 1 0≤t≤1 p(t) = 0 otherwise
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(a) Use the integral formula to find P(s), the Laplace transform of p(t). Determine the region of convergence of P(s). (b) Represent p(t) in terms of the unitstep function and use its Laplace transform and the timeshift property to find P(s). Find the poles and zeros of P(s) to verify the region of convergence obtained above. 3.15. Frequencyshift property Duality occurs between time and frequency shifts. As shown, if L[x(t)] = X(s), then L[x(t − t0 )] = X(s)e−t0 s . The dual of this would be L[x(t)e−αt ] = X(s + α), which we call the frequencyshift property. (a) Use the integral formula for the Laplace transform to show the frequencyshift property. (b) Use the above frequencyshift property to find X(s) = L[x(t) = cos(0 t)u(t)] (represent the cosine using Euler’s identity). Find and plot the poles and zeros of X(s). (c) Recall the definition of the hyperbolic cosine, cosh(0 t) = 0.5(e0 t + e−0 t ), and find the Laplace transform Y(s) of y(t) = cosh(0 t)u(t). Find and plot the poles and zeros of Y(s). Explain the relation of the poles of X(s) and Y(s) by connecting x(t) with y(t). 3.16. Poles and zeros Consider the pulse x(t) = u(t) − u(t − 1). (a) Find the zeros and poles of X(s) and plot them. (b) Suppose x(t) is the input of an LTI system with a transfer function H(s) = 1/(s2 + 4π 2 ). Find and plot the poles and zeros of Y(s) = L[y(t)] = H(s)X(s) where y(t) is the output of the system. (c) If the transfer function of the LTI system is G(s) =
∞ Y Z(s) 1 = X(s) s2 + (2kπ)2 k=1
and the input is the above signal x(t), compute the output z(t). 3.17. Poles and zeros—MATLAB The poles corresponding to the Laplace transform X(s) of a signal x(t) are p1,2 = −3 ± jπ/2 p3 = 0 (a) Within some constants, give a general form of the signal x(t). (b) Let X(s) =
1 (s + 3 − jπ/2)(s + 3 − jπ/2)s
From the location of the poles, obtain a general form for x(t). Use MATLAB to find x(t) and plot it. How well did you guess the answer? 3.18. Solving differential equations—MATLAB One of the uses of the Laplace transform is the solution of differential equations. (a) Suppose you are given the differential equation that represents an LTI system, y(2) (t) + 0.5y(1) (t) + 0.15y(t) = x(t)
t≥0
where y(t) is the output and x(t) is the input of the system, and y(1) (t) and y(2) (t) are first and secondorder derivatives with respect to t. The input is causal, (i.e., x(t) = 0 t < 0). What should the initial conditions be for the system to be LTI? Find Y(s) for those initial conditions.
Problems
(b) If y(1) (0) = 1 and y(0) = 1 are the initial conditions for the above differential equation, find Y(s). If the input to the system is doubled—that is, the input is 2x(t) with Laplace transform 2X(s)—is Y(s) doubled so that its inverse Laplace transform y(t) is doubled? Is the system linear? (c) Use MATLAB to find the poles and zeros and the solutions of the differential equation when the input is u(t) and 2u(t) with the initial conditions given above. Compare the solutions and verify your response in (b). 3.19. Differential equation, initial conditions, and impulse response—MATLAB The following function Y(s) = L[y(t)] is obtained applying the Laplace transform to a differential equation representing a system with nonzero initial conditions and input x(t), with Laplace transform X(s): X(s) s+1 Y(s) = 2 + 2 s + 2s + 3 s + 2s + 3 (a) Find the differential equation in y(t) and x(t) representing the system. (b) Find the initial conditions y0 (0) and y(0). (c) Use MATLAB to determine the impulse response h(t) of this system. Find the poles of the transfer function H(s) and determine if the system is BIBO stable. 3.20. Different responses—MATLAB Let Y(s) = L[y(t)] be the Laplace transform of the solution of a secondorder differential equation representing a system with input x(t) and some initial conditions, X(s) s+1 Y(s) = 2 + 2 s + 2s + 1 s + 2s + 1 (a) (b) (c) (d) (e)
Find the zerostate response (response due to the input only with zero initial conditions) for x(t) = u(t). Find the zeroinput response (response due to the initial conditions and zero input). Find the complete response when x(t) = u(t). Find the transient and the steadystate response when x(t) = u(t). Use MATLAB to verify the above responses.
3.21. Poles and stability The transfer function of a BIBOstable system has poles only on the open lefthand splane (excluding the j axis). (a) Let the transfer function of a system be H1 (s) =
1 Y(s) = X(s) (s + 1)(s − 2)
and let X(s) be the Laplace transform of signals that are bounded (i.e., the poles of X(s) are on the lefthand splane). Find limt→∞ y(t). Determine if the system is BIBO stable. If not, determine what makes the system unstable. (b) Let the transfer function be H2 (s) =
Y(s) 1 = X(s) (s + 1)(s + 2)
and X(s) be as indicated above. Find lim y(t)
t→∞
Can you use this limit to determine if the system is BIBO stable? If not, what would you do to check its stability?
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3.22. Poles, stability, and steadystate response The steadystate solution of stable systems is due to simple poles in the j axis of the splane coming from the input. Suppose the transfer function of the system is H(s) =
Y(s) 1 = X(s) (s + 1)2 + 4
(a) Find the poles and zeros of H(s) and plot them in the splane. Find then the corresponding impulse response h(t). Determine if the impulse response of this system is absolutely integrable so that the system is BIBO stable. (b) Let the input x(t) = u(t). Find y(t) and from it determine the steadystate solution. (c) Let the input x(t) = tu(t). Find y(t) and from it determine the steadystate response. What is the difference between this case and the previous one? (d) To explain the behavior in the case above consider the following: Is the input x(t) = tu(t) bounded? That is, is there some finite value M such that x(t) < M for all times? So what would you expect the output to be knowing that the system is stable? 3.23. Responses from an analog averager The input–output equation for an analog averager is given by
y(t) =
1 T
Zt
x(τ )dτ
t−T
where x(t) is the input and y(t) is the output. This equation corresponds to the convolution integral. (a) Change the above equation so that you can determine from it the impulse response h(t). (b) Graphically determine the output y(t) corresponding to a pulse input x(t) = u(t) − u(t − 2) using the convolution integral (let T = 1) relating the input and the output. Carefully plot the input and the output. (The output can also be obtained intuitively from a good understanding of the averager.) (c) Using the impulse response h(t) found above, use now the Laplace transform to find the output corresponding to x(t) = u(t) − u(t − 2). Let again T = 1 in the averager. 3.24. Transients for secondorder systems—MATLAB The type of transient you get in a secondorder system depends on the location of the poles of the system. The transfer function of the secondorder system is H(s) =
Y(s) 1 = 2 X(s) s + b1 s + b0
and let the input be x(t) = u(t). (a) Let the coefficients of the denominator of H(s) be b1 = 5 and b0 = 6. Find the response y(t). Use MATLAB to verify the response and to plot it. (b) Suppose then that the denominator coefficients of H(s) are changed to b1 = 2 and b0 = 6. Find the response y(t). Use MATLAB to verify the response and to plot it. (c) Explain your results above by relating your responses to the location of the poles of H(s). 3.25. Effect of zeros on the sinusoidal steady state To see the effect of the zeros on the complete response of a system, suppose you have a system with a transfer function H(s) =
Y(s) s2 + 4 = X(s) s((s + 1)2 + 1)
Problems
(a) Find and plot the poles and zeros of H(s). Is this BIBO stable? (b) Find the frequency 0 of the input x(t) = 2 cos(0 t)u(t) such that the output of the given system is zero in the steady state. Why do you think this happens? (c) If the input is a sine instead of a cosine, would you get the same result as above? Explain why or why not. 3.26. Zero steadystate response of analog averager—MATLAB The analog averager can be represented by the differential equation dy(t) 1 = [x(t) − x(t − T)] dt T where y(t) is the output and x(t) is the input. (a) If the input–output equation of the averager is 1 y(t) = T
Zt
x(τ )dτ
t−T
show how to obtain the above differential equation and that y(t) is the solution of the differential equation. (b) If x(t) = cos(π t)u(t), choose the value of T in the averager so that the output is y(t) = 0 in the steady state. Graphically show how this is possible for your choice of T. Is there a unique value for T that makes this possible? How does it relate to the frequency 0 = π of the sinusoid? (c) Use the impulse response h(t) of the averager found before, to show using Laplace that the steady state is zero when x(t) = cos(π t)u(t) and T is the above chosen value. Use MATLAB to solve the differential equation and to plot the response for the value of T you chose. (Hint: Consider x(t)/T the input and use superposition and time invariance to find y(t) due to (x(t) − x(t − T))/T.) 3.27. Partial fraction expansion—MATLAB Consider the following functions Yi (s) = L[yi (t)], i = 1, 2 and 3: Y1 (s) =
s+1 s(s2 + 2s + 4)
1 (s + 2)2 s−1 Y3 (s) = 2 s ((s + 1)2 + 9)
Y2 (s) =
where {yi (t), i = 1, 2, 3} are the complete responses of differential equations with zero initial conditions. (a) For each of these functions, determine the corresponding differential equation, if all of them have as input x(t) = u(t). (b) Find the general form of the complete response {yi (t), i = 1, 2, 3} for each of the {Yi (s) i = 1, 2, 3}. Use MATLAB to plot the poles and zeros for each of the {Yi (s)}, to find their partial fraction expansions, and the complete responses. 3.28. Iterative convolution integral—MATLAB Consider the convolution of a pulse x(t) = u(t + 0.5) − u(t − 0.5) with itself many times. Use MATLAB for the calculations and the plotting. (a) Consider the result for N = 2 of these convolutions—that is, y2 (t) = (x ∗ x)(t)
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Find Y2 (s) = L[y2 (t)] using the convolution property of the Laplace transform and find y2 (t). (b) Consider then the result for N = 3 of these convolutions—that is, y3 (t) = (x ∗ x ∗ x)(t) Find Y3 (s) = L[y3 (t)] using the convolution property of the Laplace transform and find y3 (t). (c) The signal x(t) can be considered the impulse response of an averager that ”smooths” out a signal. Letting y1 (t) = x(t), plot the three functions yi (t) for i = 1, 2, and 3. Compare these signals on their smoothness and indicate their supports in time. (For y2 (t) and y3 (t), how do their supports relate to the supports of the signals convolved?) 3.29. Positive and negative feedback There are two types of feedback, negative and positive. In this problem we explore their difference. (a) Consider negative feedback. Suppose you have a system with transfer function H(s) = Y(s)/E(s) where E(s) = C(s) − Y(s), and C(s) and Y(s) are the transforms of the feedback system’s reference c(t) and output y(t). Find the transfer function of the overall system G(s) = Y(s)/C(s). (b) In positive feedback, the only equation that changes is E(s) = C(s) + Y(s); the other equations remain the same. Find the overall feedback system transfer function G(s) = Y(s)/C(s). (c) Suppose that C(s) = 1/s, H(s) = 1/(s + 1). Determine G(s) for both negative and positive feedback. Find y(t) = L−1 [Y(s)] for both types of feedback and comment on the difference in these signals. 3.30. Feedback stabilization An unstable system can be stabilized by using negative feedback with a gain K in the feedback loop. For instance, consider an unstable system with transfer function H(s) =
2 s−1
which has a pole in the righthand splane, making the impulse response of the system h(t) grow as t increases. Use negative feedback with a gain K > 0 in the feedback loop, and put H(s) in the forward loop. Draw a block diagram of the system. Obtain the transfer function G(s) of the feedback system and determine the value of K that makes the overall system BIBO stable (i.e., its poles in the open lefthand splane). 3.31. Allpass stabilization Another stabilization method consists in cascading an allpass system with the unstable system to cancel the poles in the righthand splane. Consider a system with a transfer function H(s) =
s+1 (s − 1)(s2 + 2s + 1)
which has a pole in the righthand splane, s = 1, so it is unstable. (a) The poles and zeros of an allpass filter are such that if p12 = −σ ± j0 are complex conjugate poles of the filter, then z12 = σ ± j0 are the corresponding zeros, and for real poles p = −σ there is a corresponding z = σ . The orders of the numerator and the denominator of the allpass filter are equal. Write the general transfer function of an allpass filter Hap (s) = KN(s)/D(s). (b) Find an allpass filter Hap (s) so that when cascaded with H(s) the overall transfer function G(s) = H(s)Hap (s) has all its poles in the lefthand splane. (c) Find K of the allpass filter so that when s = 0 the allpass filter has a gain of unity. What is the relation between the magnitude of the overall system G(s) and that of the unstable filter H(s). 3.32. Halfwave rectifier—MATLAB In the generation of DC from AC voltage, the ”halfwave” rectified signal is an important part. Suppose the AC voltage is x(t) = sin(2π t)u(t).
Problems
(a) Carefully plot the halfwave rectified signal y(t) from x(t). (b) Let y1 (t) be the period of y(t) between 0 ≤ t ≤ 1. Show that y1 (t) can be written as y1 (t) = sin(2π t)u(t) + sin(2π(t − 0.5))u(t − 0.5) or y1 (t) = sin(2π t)[u(t) − u(t − 0.5)] Use MATLAB to verify this. Find the Laplace transform X1 (s) of x1 (t). (c) Express y(t) in terms of y1 (t) and find the Laplace transform Y(s) of y(t). 3.33. Polynomial multiplication—MATLAB When the numerator or denominator is given in a factorized form, we need to multiply polynomials. Although this can be done by hand, MATLAB provides the function conv that computes the coefficients of the polynomial resulting from the product of two polynomials. (a) Use help in MATLAB to find how conv can be used, and then consider two polynomials P(s) = s2 + s + 1 and Q(s) = 2s3 + 3s2 + s + 1 Do the multiplication of these polynomials by hand to find Z(s) = P(s)Q(s) and use conv to verify your results. (b) The output of a system has a Laplace transform Y(s) =
N(s) (s + 2) = 2 D(s) s (s + 1)((s + 4)2 ) + 9)
Use conv to find the denominator polynomial and then find the inverse Laplace transform using ilaplace. 3.34. Feedback error—MATLAB Consider a negative feedback system used to control a plant G(s) = 1/(s(s + 1)(s + 2)). The output y(t) of the feedback system is connected via a sensor with transfer function H(s) = 1 to a differentiator where the reference signal x(t) is also connected. The output of the differentiator is the feedback error e(t) = x(t) − v(t) where v(t) is the output of the feedback sensor. (a) Carefully draw the feedback system, and find an expression for E(s), the Laplace transform of the feedback error e(t). (b) Two possible reference test signals for the given plant are x(t) = u(t) and x(t) = r(t). Choose the one that would give a zero steadystate feedback error. (c) Use MATLAB to do the partial fraction expansions for the two error functions E1 (s), corresponding to when x(t) = u(t) and E2 (s) when x(t) = r(t). Use these partial fraction expansions to find e1 (t) and e2 (t), and thus verify your results obtained before.
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CHAPTER 4
Frequency Analysis: The Fourier Series
A Mathematician is a device for turning coffee into theorems. Paul Erdos (1913–1996) mathematician
4.1 INTRODUCTION In this chapter and the next we consider the frequency analysis of continuoustime signals and systems—the Fourier series for periodic signals in this chapter, and the Fourier transform for both periodic and aperiodic signals as well as for systems in Chapter 5. In these chapters we consider: n
n
Spectral representation—The frequency representation of periodic and aperiodic signals indicates how their power or energy is allocated to different frequencies. Such a distribution over frequency is called the spectrum of the signal. For a periodic signal the spectrum is discrete, as its power is concentrated at frequencies multiples of a socalled fundamental frequency, directly related to the period of the signal. On the other hand, the spectrum of an aperiodic signal is a continuous function of frequency. The concept of spectrum is similar to the one used in optics for light, or in material science for metals, each indicating the distribution of power or energy over frequency. The Fourier representation is also useful in finding the frequency response of linear timeinvariant systems, which is related to the transfer function obtained with the Laplace transform. The frequency response of a system indicates how an LTI system responds to sinusoids of different frequencies. Such a response characterizes the system and permits easy computation of its steadystate response, and will be equally important in the synthesis of systems. Eigenfunctions and Fourier analysis—It is important to understand the driving force behind the representation of signals in terms of basic signals when applied to LTI systems. For instance, the convolution integral that gives the output of an LTI system resulted from the representation of its input signal in terms of shifted impulses. Along with this result came the concept of the impulse response of an LTI system. Likewise, the Laplace transform can be seen as the representation of signals in terms of general eigenfunctions. In this chapter and the next we will see that complex
Signals and Systems Using MATLAB®. DOI: 10.1016/B9780123747167.000077 c 2011, Elsevier Inc. All rights reserved.
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CH A P T E R 4: Frequency Analysis: The Fourier Series
n
n
exponentials or sinusoids are used in the Fourier representation of periodic as well as aperiodic signals by taking advantage of the eigenfunction property of LTI systems. The results of the Fourier series in this chapter will be extended to the Fourier transform in Chapter 5. Steadystate analysis—Fourier analysis is in the steady state, while Laplace analysis considers both transient and steady state. Thus, if one is interested in transients, as in control theory, Laplace is a meaningful transformation. On the other hand, if one is interested in the frequency analysis, or steady state, as in communications theory, the Fourier transform is the one to use. There will be cases, however, where in control and communications both Laplace and Fourier analysis are considered. Application of Fourier analysis—The frequency representation of signals and systems is extremely important in signal processing and in communications. It explains filtering, modulation of messages in a communication system, the meaning of bandwidth, and how to design filters. Likewise, the frequency representation turns out to be essential in the sampling of analog signals—the bridge between analog and digital signal processing.
4.2 EIGENFUNCTIONS REVISITED As indicated in Chapter 3, the most important property of stable LTI systems is that when the input is a complex exponential (or a combination of a cosine and a sine) of a certain frequency, the output of the system is the input times a complex constant connected with how the system responds to the frequency at the input. The complex exponential is called an eigenfunction of stable LTI systems. If x(t) = ej0 t , −∞ < t < ∞, is the input to a causal and a stable system with impulse response h(t), the output in the steady state is given by y(t) = ej0 t H( j0 )
(4.1)
Z∞ H( j0 ) = h(τ )e−j0 τ dτ
(4.2)
where
0
is the frequency response of the system at 0 . The signal x(t) = ej0 t is said to be an eigenfunction of the LTI system as it appears at both input and output.
This can be seen by finding the output corresponding to x(t) = ej0 t by means of the convolution integral, Z∞ y(t) =
j0 t
h(τ )x(t − τ )dτ = e
0 j0 t
=e
Z∞ 0
H( j0 )
h(τ )e−j0 τ dτ
4.2 Eigenfunctions Revisited
where we let H( j0 ) equal the integral in the second equation. The input signal appears in the output modified by the frequency response of the system H( j0 ) at the frequency 0 of the input. Notice that the convolution integral limits indicate that the input started at −∞ and that we are considering the output at finite time t—this means that we are in steady state. The steadystate response of a stable LTI system is attained by either considering that the initial time when the input is applied to the system is −∞ and we reach a finite time t, or by starting at time 0 and going to ∞. The above result for one frequency can be easily extended to the case of several frequencies present at the input. If the input signal x(t) is a linear combination of complex exponentials, with different amplitudes, frequencies, and phases, or X x(t) = Xk ejk t k
where Xk are complex values, since the output corresponding to Xk ejk t is Xk ejk t H( jk ) by superposition the response to x(t) is X y(t) = Xk ejk t H( jk ) k
=
X
Xk H( jk )ej(k t+∠H( jk ))
(4.3)
k
The above is valid for any signal that is a combination of exponentials of arbitrary frequencies. As we will see in this chapter, when x(t) is periodic it can be represented by the Fourier series, which is a combination of complex exponentials harmonically related (i.e., the frequencies of the exponentials are multiples of the fundamental frequency of the periodic signal). Thus, when a periodic signal is applied to a causal and stable LTI system its output is computed as in Equation (4.3). The significance of the eigenfunction property is also seen when the input signal is an integral (a sum, after all) of complex exponentials, with continuously varying frequency, as the integrand. That is, if 1 x(t) = 2π
Z∞
X()ejt d
−∞
then using superposition and the eigenfunction property of a stable LTI system, with frequency response H( j), the output is 1 y(t) = 2π
Z∞
X()ejt H( j)d
−∞
=
1 2π
Z∞ −∞
X()H( j)e( jt+j∠H( j)) d
(4.4)
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The above representation of x(t) corresponds to the Fourier representation of aperiodic signals, which will be covered in Chapter 5. Again here, the eigenfunction property of LTI systems provides an efficient way to compute the output. Furthermore, we also find that by letting Y() = X()H( j) the above equation gives an expression to compute y(t) from Y(). The product Y() = X()H( j) corresponds to the Fourier transform of the convolution integral y(t) = x(t) ∗ h(t), and is connected with the convolution property of the Laplace transform. It is important to start noticing these connections, to understand the link between Laplace and Fourier analysis. Remarks n
n
n
Notice the difference of notation for the frequency representation of signals and systems used above. If x(t) is a periodic signal its frequency representation is given by {Xk }, and if aperiodic by X(), while for a system with impulse response h(t) its frequency response is given by H( j). When considering the eigenfunction property, the stability of the LTI system is necessary to ensure that H( j) exists for all frequencies. The eigenfunction property applied to a linear circuit gives the same result as the one obtained from phasors in the sinusoidal steady state. That is, if x(t) = A cos(0 t + θ ) =
Aejθ j0 t Ae−jθ −j0 t e + e 2 2
(4.5)
is the input of a circuit represented by the transfer function H(s) =
L[y(t)] Y(s) = X(s) L[x(t)]
then the corresponding steadystate output is given by Aejθ j0 t Ae−jθ −j0 t e H( j0 ) + e H(−j0 ) 2 2 = AH( j0 ) cos(0 t + θ + ∠H( j0 ))
yss (t) =
(4.6)
where, very importantly, the frequency of the output coincides with that of the input, and the amplitude and phase of the input are changed by the magnitude and phase of the frequency response of the system for the frequency 0 . The frequency response is H( j0 ) = H(s)s=j0 , and as we will see its magnitude is an even function of frequency, or H( j) = H(−j), and its phase is an odd function of frequency, or ∠H( j0 ) = −∠H(−j0 ). Using these two conditions we obtain Equation (4.6). The phasor corresponding to the input x(t) = A cos(0 t + θ ) is defined as a vector, X = A∠θ rotating in the polar plane at the frequency of 0 . The phasor has a magnitude A and an angle θ with respect to the positive real axis. The projection of the phasor onto the real axis, as it rotates at the given
4.2 Eigenfunctions Revisited
frequency, with time generates a cosine of the indicated frequency, amplitude, and phase. The transfer function is computed at s = j0 or Y X (ratio of the phasors corresponding to the output Y and the input X). The phasor for the output is thus H(s)s=j0 = H( j0 ) =
Y = H( j0 )X = Yej∠Y Such a phasor is then converted into the sinusoid (which equals Eq. 4.6): yss (t) = Re[Yej0 t ] = Y cos(0 t + ∠Y) A very important application of LTI systems is filtering, where one is interested in preserving desired frequency components of a signal and getting rid of lessdesirable components. That an LTI system can be used for filtering is seen in Equations (4.3) and (4.4). In the case of a periodic signal, the magnitude H( jk ) can be set ideally to one for those components we wish to keep and to zero for those we wish to get rid of. Likewise, for an aperiodic signal, the magnitude H( j) could be set ideally to one for those components we wish to keep and zero for those components we wish to get rid of. Depending on the filtering application, an LTI system with the appropriate characteristics can be designed, obtaining the desired transfer function H(s).
n
For a stable LTI with transfer function H(s) if the input is x(t) = Re[Aej(0 t+θ ) ] = A cos(0 t + θ) the steadystate output is given by y(t) = Re[AH( j0 )ej(0 t+θ ) ] = AH( j0 ) cos(0 t + θ + ∠H( j0 ))
(4.7)
where H( j0 ) = H(s)s=j0
n Example 4.1 Consider the RC circuit shown in Figure 4.1. Let the voltage source be vs (t) = 4 cos(t + π/4) volts. the resistor be R = 1, and the capacitor C = 1 F. Find the steadystate voltage across the capacitor. Solution This problem can be approached in two ways. n
Phasor approach. From the phasor circuit in Figure 4.1, by voltage division we have the following phasor ratio, where Vs is the phasor corresponding to the source vs (t) and Vc the phasor
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CH A P T E R 4: Frequency Analysis: The Fourier Series
1Ω
1 +
+ Vs
vs (t) − 1F
FIGURE 4.1 RC circuit and corresponding phasor circuit.
+ vc (t)
−
−
−j
+ Vc −
corresponding to vc (t): √ Vc −j −j(1 + j) 2 = = = ∠−π/4 Vs 1−j 2 2 Since Vs = 4∠π/4, then √ Vc = 2 2∠0 so that in the steady state, √ vc (t) = 2 2 cos(t) n
Eigenfunction approach. Considering the output is the voltage across the capacitor and the input is the voltage source, the transfer function is obtained using voltage division as H(s) =
Vc (s) 1/s 1 = = Vs (s) 1 + 1/s s+1
so that the system frequency response at the input frequency 0 = 1 is √ 2 H( j1) = ∠−π/4 2 According to the eigenfunction property the steadystate response of the capacitor is vc (t) = 4H( j1) cos(t + π/4 + ∠H( j1)) √ = 2 2 cos(t) which coincides with the solution found using phasors.
n
n Example 4.2 An ideal communication system provides as output the input signal with only a possible delay in the transmission. Such an ideal system does not cause any distortion to the input signal beyond
4.2 Eigenfunctions Revisited
the delay. Find the frequency response of the ideal communication system, and use it to determine the steadystate response when the delay caused by the system is τ = 3 sec, and the input is x(t) = 2 cos(4t − π/4). Solution The impulse response of the ideal system is h(t) = δ(t − τ ) where τ is the delay of the transmission. In fact, the output according to the convolution integral gives Z∞ y(t) = 0
δ(ρ − τ ) x(t − ρ)dρ = x(t − τ )  {z } h(ρ)
as expected. Let us then find the frequency response of the ideal communication system. According to the eigenvalue property, if the input is x(t) = ej0 t, then the output is y(t) = ej0 t H( j0 ) but also y(t) = x(t − τ ) = ej0 (t−τ ) so that comparing these equations we have that H( j0 ) = e−jτ 0 For a generic frequency 0 ≤ < ∞, we would get H( j) = e−jτ which is a complex function of , with a unity magnitude H( j) = 1, and a linear phase ∠H( j) = −τ . This system is called an allpass system, since it allows all frequency components of the input to go through with a phase change only. Consider the case when τ = 3, and that we input into this system x(t) = 2 cos(4t − π/4), then H( j) = 1e−j3 , so that the output in the steady state is y(t) = 2H( j4) cos(4t − π/4 + ∠H( j4)) = 2 cos(4(t − 3) − π/4) = x(t − 3) where we used H( j4) = 1e−j12 (i.e., H( j4) = 1 and ∠H( j4) = 12).
n
n Example 4.3 Although there are better methods to compute the frequency response of a system represented by a differential equation, the eigenfunction property can be easily used for that. Consider the RC
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circuit shown in Figure 4.1 where the input is vs (t) = 1 + cos(10,000t) with components of low frequency, = 0, and of large frequency, = 10,000 rad/sec. The output vc (t) is the voltage across the capacitor in steady state. We wish to find the frequency response of this circuit to verify that it is a lowpass filter (it allows lowfrequency components to go through, but filters out highfrequency components). Solution Using Kirchhoff’s voltage law, this circuit is represented by a firstorder differential equation, vs (t) = vc (t) +
dvc (t) dt
Now, if the input is vs (t) = ejt , for a generic frequency , then the output is vc (t) = ejt H( j). Replacing these in the differential equation, we have ejt = ejt H( j) +
dejt H( j) dt
= ejt H( j) + jejt H( j) so that H( j) =
1 1 + j
or the frequency response of the filter for any frequency . The magnitude of H( j) is 1 H( j) = √ 1 + 2 which is close to one for small values of the frequency, and tends to zero when the frequency values are large—the characteristics of a lowpass filter. For the input vs (t) = 1 + cos(10,000t) = cos(0t) + cos(10,000t) (i.e., it has a zero frequency component and a 10,000rad/sec frequency component) using Euler’s identity, we have that vs (t) = 1 + 0.5 ej10,000t + e−j10,000t and the steadystate output of the circuit is vc (t) = 1H( j0) + 0.5H( j10,000)ej10,000t + 0.5H(−j10,000)e−j10,000t ≈1+
1 cos(10,000t − π/2) ≈ 1 10,000
4.3 Complex Exponential Fourier Series
since H( j0) = 1 H( j10,000) ≈ H(−j10,000) ≈
1 −j = 4 j 10 10,000 1 j = −j 104 10,000
Thus, this circuit acts like a lowpass filter by keeping the DC component (with the low frequency = 0) and essentially getting rid of the highfrequency ( = 10,000) component of the signal. Notice that the frequency response can also be obtained by considering the phasor ratio for a generic frequency , which by voltage division is 1 Vc 1/j = = Vs 1 + 1/j 1 + j which for = 0 is 1 and for = 10,000 is approximately −j/10,000 (i.e., corresponding to H( j0) and H( j10,000) = H∗ ( j10,000)). n
Fourier and Laplace French mathematician JeanBaptisteJoseph Fourier (1768–1830) was a contemporary of Laplace with whom he shared many scientific and political experiences [2, 7]. Like Laplace, Fourier was from very humble origins but he was not as politically astute. Laplace and Fourier were affected by the political turmoil of the French Revolution and both came in close contact with Napoleon Bonaparte, French general and emperor. Named chair of the mathematics department of the Ecole Normale, Fourier led the most brilliant period of mathematics and science education in France. His main work was “The Mathematical Theory of Heat Conduction” where he proposed the harmonic analysis of periodic signals. In 1807 he received the grand prize from the French Academy of Sciences for this work. This was despite the objections of Laplace, Lagrange, and Legendre, who were the referees and who indicated that the mathematical treatment lacked rigor. Following Galton’s advice of “Never resent criticism, and never answer it,” Fourier disregarded these criticisms and made no change to his 1822 treatise in heat conduction. Although Fourier was an enthusiast for the Revolution and followed Napoleon on some of his campaigns, in the Second Restoration he had to pawn his belongings to survive. Thanks to his friends, he became secretary of the French Academy, the final position he held.
4.3 COMPLEX EXPONENTIAL FOURIER SERIES The Fourier series is a representation of a periodic signal x(t) in terms of complex exponentials or sinusoids of frequency multiples of the fundamental frequency of x(t). The advantage of using the Fourier series to represent periodic signals is not only the spectral characterization obtained, but in finding the response for these signals when applied to LTI systems by means of the eigenfunction property. Mathematically, the Fourier series is an expansion of periodic signals in terms of normalized orthogonal complex exponentials. The concept of orthogonality of functions is similar to the concept of
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CH A P T E R 4: Frequency Analysis: The Fourier Series
perpendicularity of vectors: Perpendicular vectors cannot be represented in terms of each other, as orthogonal functions provide mutually exclusive information. The perpendicularity of two vectors can be established using the dot or scalar product of the vectors, and the orthogonality of functions is established by the inner product, or the integration of the product of the function and its conjugate. Consider a set of complex functions {ψk (t)} defined in an interval [a, b], and such that for any pair of these functions, let’s say ψ` (t) and ψm (t), ` 6= m, their inner product is Zb
∗ ψ` (t)ψm (t)dt
=
a
0 ` 6= m 1 `=m
(4.8)
Such a set of functions is called orthonormal (i.e., orthogonal and normalized). A finiteenergy signal x(t) defined in [a, b] can be approximated by a series X xˆ (t) = ak ψk (t)
(4.9)
k
according to some error criterion. For instance, we could minimize the energy of the error function ε(t) = x(t) − xˆ (t) or Zb a
2 Zb X ak ψk (t) dt ε(t) dt = x(t) − 2
(4.10)
k
a
The expansion can be finite or infinite, and may not approximate the signal point by point. Fourier proposed sinusoids as the functions {ψk (t)} to represent periodic signals, and solved the quadratic minimization posed in Equation (4.10) to obtain the coefficients of the representation. For most signals, the resulting Fourier series has an infinite number of terms and coincides with the signal pointwise. We will start with a more general expansion that uses complex exponentials and from it obtain the sinusoidal form. In Chapter 5 we extend the Fourier series to represent aperiodic signals—leading to the Fourier transform that is in turn connected with the Laplace transform. Recall that a periodic signal x(t) is such that n n
It is defined for −∞ < t < ∞ (i.e., it has an infinite support). For any integer k, x(t + kT0 ) = x(t), where T0 is the fundamental period of the signal or the smallest positive real number that makes this possible. The Fourier series representation of a periodic signal x(t), of period T0 , is given by an infinite sum of weighted complex exponentials (cosines and sines) with frequencies multiples of the signal’s fundamental frequency 0 = 2π/T0 rad/sec, or x(t) =
∞ X k=−∞
Xk ejk0 t
0 =
2π T0
(4.11)
4.3 Complex Exponential Fourier Series
where the Fourier coefficients Xk are found according to 1 Xk = T0
t0Z+T0
x(t)e−jk0 t dt
(4.12)
t0
for k = 0, ±1, ±2, . . . , and any t0 . The form of Equation (4.12) indicates that the information needed for the Fourier series can be obtained from any period of x(t).
Remarks n
The Fourier series uses the Fourier basis {ejk0 t , k integer} to represent the periodic signal x(t) of period T0 . The Fourier basis functions are also periodic of period T0 (i.e., for an integer m, ejk0 (t+mT0 ) = ejk0 t ejkm2π = ejk0 t
n
as ejkm2π = 1). The Fourier basis functions are orthonormal over a period—that is, 1 T0
t0Z+T0
e
jk0 t
[e
( 1 k=` ] dt = 0 k= 6 `
j`0 t ∗
t0
(4.13)
That is, ejk0 t and ej`0 t are said to be orthogonal when for k 6= ` the above integral is zero, and they are normal (or normalized) when for k = ` the above integral is unity. The functions ejk0 t and ej`0 t are orthogonal since 1 T0
t0Z+T0
e t0
jk0 t
[e
1 ] dt = T0
j`0 t ∗
1 = T0
t0Z+T0
ej(k−`)0 t dt
t0 t0Z+T0
[cos((k − `)0 t) + j sin((k − `)0 t)] dt
t0
k 6= `
=0
The above integrals are zero given that the integrands are sinusoids and the limits of the integrals cover one or more periods of the integrands. The normality of the Fourier functions is easily shown when for k = ` the above integral is 1 T0
t0Z+T0
ej0t dt = 1
t0
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CH A P T E R 4: Frequency Analysis: The Fourier Series
n
The Fourier coefficients {Xk } are easily obtained using the orthonormality of the Fourier functions: First, we multiply the expression for x(t) in Equation (4.11) by e−j`0 t and then integrate over a period to get Z X Z x(t)e−j`0 t dt = Xk ej(k−`)0 t dt k
T0
=
X
T0
Xk T0 δ(k − `)
k
= X` T0
n
R given that when k = `, then T0 ej(k−`)0 t dt = T0 ; otherwise it is zero according to the orthogonality of the Fourier exponentials. This then gives us the expression for the Fourier coefficients {X` } in Equation (4.12). You need to recognize that the k and ` are dummy variables in the Fourier series, and as such the expression for the coefficients is the same regardless of whether we use ` or k. It is important to realize from the given Fourier series equations that for a periodic signal x(t), of period T0 , any period x(t), t0 ≤ t ≤ t0 + T0 provides all the necessary information in the timedomain characterizing x(t). In an equivalent way the coefficients and their corresponding frequencies {Xk , k0 } provide all the necessary information about x(t) in the frequency domain.
4.4 LINE SPECTRA The Fourier series provides a way to determine the frequency components of a periodic signal and the significance of these frequency components. Such information is provided by the power spectrum of the signal. For periodic signals, the power spectrum provides information as to how the power of the signal is distributed over the different frequencies present in the signal. We thus learn not only what frequency components are present in the signal but also the strength of these frequency components. In practice, the power spectrum can be computed and displayed using a spectrum analyzer, which will be described in Chapter 5.
4.4.1 Parseval’s Theorem—Power Distribution over Frequency Although periodic signals are infiniteenergy signals, they have finite power. The Fourier series provides a way to find how much of the signal power is in a certain band of frequencies. The power Px of a periodic signal x(t), of period T0 , can be equivalently calculated in either the time or the frequency domain: Z X 1 Px = x(t)2 dt = Xk 2 (4.14) T0 T0
k
4.4 Line Spectra
The power of a periodic signal x(t) of period T0 is given by Z 1 Px = x(t)2 dt T0 T0
Replacing the Fourier series of x(t) in the power equation we have that Z Z XX 1 1 ∗ j0 kt −j0 mt x(t)2 dt = Xk Xm e e dt T0 T0 m T0
T0
=
XX k
=
X
k
∗ Xk Xm
m
1 T0
Z
ej0 kt e−j0 mt dt
T0
Xk 2
k
after we apply the orthonormality of the Fourier exponentials. Even though x(t) is real, we let x(t)2 = x(t)x∗ (t) in the above equations, permitting us to express them in terms of Xk and its conjugate. The above indicates that the power of x(t) can be computed in either the time or the frequency domain giving exactly the same result. Moreover, considering the signal to be a sum of harmonically related components or X X x(t) = Xk ejk0 t = xk (t) k
k
the power of each of these components is given by Z Z Z 1 1 1 xk (t)2 dt = Xk ejk0 t 2 dt = Xk 2 dt = Xk 2 T0 T0 T0 T0
T0
T0
and the power of x(t) is the sum of the powers of the Fourier series components. This indicates that the power of the signal is distributed over the harmonic frequencies {k0 }. A plot of Xk 2 versus the harmonic frequencies k0 , k = 0, ±1, ±2, . . . , displays how the power of the signal is distributed over the harmonic frequencies. Given the discrete nature of the harmonic frequencies {k0 } this plot consists of a line at each frequency and as such it is called the power line spectrum (that is, a periodic signal has no power in nonharmonic frequencies). Since {Xk } are complex, we define two additional spectra, one that displays the magnitude Xk  versus k0 , called the magnitude line spectrum, and the phase line spectrum or ∠Xk versus k0 showing the phase of the coefficients {Xk } for k0 . The power line spectrum is simply the magnitude spectrum squared. A periodic signal x(t), of period T0 , is represented in the frequency by its Magnitude line spectrum : Phase line spectrum :
Xk  vs k0 ∠Xk vs k0
(4.15) (4.16)
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CH A P T E R 4: Frequency Analysis: The Fourier Series
The power line spectrum Xk 2 versus k0 of x(t) displays the distribution of the power of the signal over frequency.
4.4.2 Symmetry of Line Spectra For a realvalued periodic signal x(t), of period T0 , represented in the frequency domain by the Fourier coefficients {Xk = Xk ej∠Xk } at harmonic frequencies {k0 = 2πk/T0 }, we have that ∗ Xk = X−k
(4.17)
or equivalently that 1.
Xk  = X−k  (i.e., magnitude Xk  is even function of k0 )
2.
∠Xk = −∠X−k (i.e., phase ∠Xk is odd function of k0 )
(4.18)
Thus, for realvalued signals we only need to display for k ≥ 0 the Magnitude line spectrum: Plot of Xk  versus k0 Phase line spectrum: Plot of ∠Xk versus k0
For a real signal x(t), the Fourier series of its complex conjugate x∗ (t) is x (t) = ∗
" X
#∗ X` e
j`0 t
`
=
X `
X`∗ e−j`0 t =
X
∗ jk0 t X−k e
k
Since x(t) = x∗ (t), the above equation is equal to x(t) =
X
Xk ejk0 t
k ∗ = X , which means Comparing the Fourier series coefficients in the expressions, we have that X−k k that if Xk = Xk ej∠Xk , then
Xk  = X−k  ∠Xk = −∠X−k or that the magnitude is an even function of k, while the phase is an odd function of k. Thus, the line spectra corresponding to realvalued signals is given for only positive harmonic frequencies, with the understanding that for negative values of the harmonic frequencies the magnitude line spectrum is even and the phase line spectrum is odd.
4.5 Trigonometric Fourier Series
4.5 TRIGONOMETRIC FOURIER SERIES The trigonometric Fourier series of a realvalued, periodic signal x(t), of period T0 , is an equivalent representation that uses sinusoids rather than complex exponentials as the basis functions. It is given by x(t) = X0 + 2
∞ X
Xk  cos(k0 t + θk )
k=1
= c0 + 2
∞ X
[ck cos(k0 t) + dk sin(k0 t)]
0 =
k=1
2π T0
(4.19)
where X0 = c0 is called the DC component, and {2Xk  cos(k0 t + θk )} are the kth harmonics for k = 1, 2 . . .. The frequencies {k0 } are said to be harmonically related. The coefficients {ck , dk } are obtained from x(t) as follows: 1 ck = T0 1 dk = T0
t0Z+T0
x(t) cos(k0 t) dt
k = 0, 1, . . .
t0 t0Z+T0
x(t) sin(k0 t) dt
k = 1, 2, . . .
(4.20)
t0
The coefficients Xk = Xk ejθk are connected with the coefficients ck and dk by q Xk  = ck2 + d2k d θk = −tan−1 k ck The functions {cos(k0 t), sin(k0 t)} are orthonormal. ∗ , obtained in the previous section, we express the exponential Fourier Using the relation Xk = X−k series of a realvalued periodic signal x(t) as
x(t) = X0 +
∞ X [Xk ejk0 t + X−k e−jk0 t ] k=1
= X0 +
∞ h i X Xk ej(k0 t+θk ) + Xk e−j(k0 t+θk ) k=1
= X0 + 2
∞ X
Xk  cos(k0 t + θk )
k=1
which is the top equation in Equation (4.19).
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CH A P T E R 4: Frequency Analysis: The Fourier Series
Let us then show how the coefficients ck and dk can be obtained directly from the signal. Using the ∗ and the fact that for a complex number z = a + jb, then z + z∗ = (a + jb) + (a − relation Xk = X−k jb) = 2a = 2Re(z), we have that x(t) = X0 +
∞ X [Xk ejk0 t + X−k e−jk0 t ] k=1
= X0 +
∞ X [Xk ejk0 t + Xk∗ e−jk0 t ] k=1
= X0 +
∞ X
2Re[Xk ejk0 t ]
k=1
Since Xk is complex (verify this!), 2Re[Xk ejk0 t ] = 2Re[Xk ] cos(k0 t) − 2Im[Xk ] sin(k0 t) Now, if we let 1 ck = Re[Xk ] = T0
t0Z+T0
x(t) cos(k0 t) dt
k = 1, 2, . . .
t0
1 dk = −Im[Xk ] = T0
t0Z+T0
x(t) sin(k0 t) dt
k = 1, 2, . . .
t0
we then have x(t) = X0 +
∞ X
2Re[Xk ] cos(k0 t) − 2Im[Xk ] sin(k0 t)
k=1
= X0 + 2
∞ X
(ck cos(k0 t) + dk sin(k0 t))
k=1
and since the average X0 = c0 we obtain the second form of the trigonometric Fourier series shown in Equation (4.19). Notice that d0 = 0 and so it is not necessary to define it. The coefficients Xk = Xk ejθk are connected with the coefficients ck and dk by q Xk  = ck2 + d2k −1 dk θk = −tan ck This can be shown by adding the phasors corresponding to ck cos(k0 t) and dk sin(k0 t) and finding the magnitude and phase of the resulting phasor.
4.5 Trigonometric Fourier Series
Finally, since the exponential basis {ejk0 t } = {cos(k0 t) + j sin(k0 t)}, the sinusoidal bases cos(k0 t) and sin(k0 t) just like the exponential basis are periodic, of period T0 , and orthonormal.
n Example 4.4 Find the Fourier series of a raisedcosine signal (B ≥ A), x(t) = B + A cos(0 t + θ ) which is periodic of period T0 and fundamental frequency 0 = 2π/T0 . Call y(t) = B + cos(0 t − π/2). Find its Fourier series coefficients and compare them to those for x(t). Use symbolic MATLAB to compute the Fourier series of y(t) = 1 + sin(100t). Find and plot its magnitude and phase line spectra. Solution In this case we do not need to compute the Fourier coefficients since x(t) is already in the trigonometric form. From Equation (4.19) its dc value is B, and A is the coefficient of the first harmonic in the trigonometric Fourier series, so that X0 = B, X1  = A/2, and ∠X1 = θ. Likewise, using Euler’s identity we obtain that i A h j(0 t+θ ) e + e−j(0 t+θ ) 2 Aejθ j0 t Ae−jθ −j0 t =B+ e + e 2 2
x(t) = B +
which gives X0 = B Aejθ 2 ∗ = X1
X1 = X−1 If we let θ = −π/2 in x(t), we get
y(t) = B + A sin(0 t) Its Fourier series coefficients are Y0 = B and Y1 = Ae−jπ/2 /2 so that Y1  = Y−1  = A/2 and ∠Y1 = −∠Y−1 = −π/2. The magnitude and phase line spectra of the raised cosine (θ = 0) and of the raised sine (θ = −π/2) are shown in Figure 4.2. For both x(t) and y(t) there are only two frequencies—the dc frequency and 0 —and as such the power of the signal is concentrated at those two frequencies as shown in Figure 4.2. The difference between the line spectra of x(t) and y(t) is in the phase.
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CH A P T E R 4: Frequency Analysis: The Fourier Series
Xk 
Yk 
B A 2
B A 2
A 2
−Ω0
Ω0
k Ω0
−Ω0
Ω0
k Ω0
∠Yk
∠Xk
FIGURE 4.2 (a) Magnitude (top left) and phase (bottom left) line spectra of raised cosine and (b) magnitude (top right) and phase (bottom right) line spectra of raised sine.
A 2
π 2 Ω0
−Ω0
−Ω0
k Ω0 −
(a)
Ω0
π 2
k Ω0
(b)
Using symbolic MATLAB integration we can easily find the Fourier series coefficients, and the corresponding magnitude and phase are then plotted using stem to obtain the line spectra. Using our MATLAB function fourierseries the magnitude and phase of the line spectrum corresponding to the periodic raised sine y(t) = 1 + sin(100t) is shown in Figure 4.3. function [X, w] = fourierseries(x, T0, N) %%%%% % symbolic Fourier Series computation % x: periodic signal % T0: period % N: number of harmonics % X,w: Fourier series coefficients at harmonic frequencies %%%%% syms t % computation of N Fourier series coefficients for k = 1:N, X1(k) = int(x ∗ exp(−j ∗ 2 ∗ pi ∗ (k  1) ∗ t/T0), t, 0, T0)/T0; X(k) = subs(X1(k)); w(k) = (k−1) ∗ 2 ∗ pi/T0; % harmonic frequencies end
4.6 Fourier Coefficients from Laplace
2
y (t )
1.5 1 0.5 0 0
0.1
0.2 t (sec)
Magnitude line spectrum
Phase line spectrum
1 1
0.6
Yk 
Yk 
0.8
0
0.4 −1
0.2 0
−200
0 Ω (rad/sec)
200
−200
0 Ω (rad/sec)
200
FIGURE 4.3 Line spectra of Fourier series of y(t) = 1 + sin(100t) (top figure). Notice the even and the odd symmetries of the magnitude and the phase spectra. The phase is −π/2 at = 100 rad/sec. n
Remarks Just because a signal is a sum of sinusoids, which are always periodic, is not enough for it to have a Fourier series. The signal should be periodic. The signal x(t) = cos(t) − sin(πt) has components with periods T1 = 2π and T2 = 2 so that the ratio T1 /T2 = π is not a rational number. Thus, x(t) is not periodic and no Fourier series for it is possible.
4.6 FOURIER COEFFICIENTS FROM LAPLACE The computation of the Xk coefficients (see Eq. 4.12) requires integration that for some signals can be rather complicated. The integration can be avoided whenever we know the Laplace transform of a period of the signal as we will show. In general, the Laplace transform of a period of the signal exists over the whole splane, given that it is a finitesupport signal. In some cases, the dc coefficient cannot be computed with the Laplace transform, but the dc term is easy to compute directly.
255
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CH A P T E R 4: Frequency Analysis: The Fourier Series
For a periodic signal x(t), of period T0 , if we know or can easily compute the Laplace transform of a period of x(t), x1 (t) = x(t)[u(t0 ) − u(t − t0 − T0 )]
for any t0
Then the Fourier coefficients of x(t) are given by Xk =
1 L [x1 (t)]s=jk0 T0
0 =
2π fundamental frequency T0
(4.21)
This can be seen by comparing the equation for the Xk coefficients with the Laplace transform of a period x1 (t) = x(t)[u(t0 ) − u(t − t0 − T0 )] of x(t). Indeed, we have that 1 Xk = T0 1 = T0 =
t0Z+T0
x(t)e−jk0 t dt
t0 t0Z+T0
x(t)e−st dt s=jk0
t0
1 L [x1 (t)]s=jk0 T0
n Example 4.5 Consider the periodic pulse train x(t), of period T0 = 1, shown in Figure 4.4. Find its Fourier series. Solution Before finding the Fourier coefficients, we see that this signal has a dc component of 1, and that x(t) − 1 (zeroaverage signal) is well represented by cosines, given its even symmetry, and as such x(t)
2
··· −1.25
FIGURE 4.4 Train of rectangular pulses.
··· −0.75
−0.25
0.25
0.75 T0 = 1
1.25
t
4.6 Fourier Coefficients from Laplace
the Fourier coefficients will be real. Doing this analysis before the computations is important so we know what to expect. The Fourier coefficients are obtained directly using their integral formulas or from the Laplace transform of a period. Since T0 = 1, the fundamental frequency of x(t) is 0 = 2π rad/sec. Using the integral expression for the Fourier coefficients we have Z3/4
1 Xk = T0 =
−j0 kt
x(t)e
Z1/4 dt =
−1/4
2 πk
"
2e−j2πkt dt
−1/4
ejπk/2 − e−jπk/2 2j
# =
sin(πk/2) (πk/2)
which are real as we predicted. The Fourier series is then x(t) =
∞ X sin(πk/2) jk2πt e (πk/2)
k=−∞
To find the Fourier coefficients with the Laplace transform, let the period be x1 (t) = x(t) for −0.5 ≤ t ≤ 0.5. Delaying it by 0.25 we get x1 (t − 0.25) = 2[u(t) − u(t − 0.5)] with a Laplace transform e−0.25s X1 (s) =
2 (1 − e−0.5s ) s
so that X1 (s) = (2/s)[e0.25s − e−0.25s ], and therefore 1 L [x1 (t)] s=jk0 T0 2 2j sin(k0 /4) = jk0 T0
Xk =
and for 0 = 2π, T0 = 1, we get Xk =
sin(πk/2) πk/2
k 6= 0
Since the above equation gives zero over zero when k = 0 (i.e., it is undefined), the dc value is found from the integral formula as Z1/4 2dt = 1
X0 = −1/4
These Fourier coefficients coincide with the ones found before. The following script is used to find the Fourier coefficients with our function fourierseries and to plot the magnitude and phase line spectra.
257
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CH A P T E R 4: Frequency Analysis: The Fourier Series
%%%%%%%%%%%%%%%%% % Example 4.5Fourier series of train of pulses %%%%%%%%%%%%%%%%% clear all;clf syms t T0 = 1; m = heaviside(t) − heaviside(t − T0/4) + heaviside(t − 3 ∗ T0/4);x = 2 ∗ m [X,w] = fourierseries(x,T0,20); subplot(221); ezplot(x,[0 T0]); grid subplot(223); stem(w,abs(X)) subplot(224); stem(w,angle(X))
Notice that in this case: 1. The Xk Fourier coefficients of the train of pulses are given in terms of the sin(x)/x or the sinc function. This function was presented in Chapter 1. Recall that the sinc is n Even—that is, sin(x)/x = sin(−x)/(−x). ˆ n The value at x = 0 is found by means of L’Hopital’s rule because the numerator and the denominator of sinc are zero for x = 0, so d sin(x)/dx sin(x) = lim =1 x→0 x→0 x dx/dx lim
n
It is bounded, indeed −1 sin(x) 1 ≤ ≤ x x x
2. Since the dc component of x(t) is 1, once it is subtracted it is clear that the rest of the series can be represented as a sum of cosines: ∞ X
x(t) = 1 +
k=−∞,k6=0
=1+2
sin(πk/2) jk2πt e (πk/2)
∞ X sin(πk/2) k=1
(πk/2)
cos(2πkt)
This can also be seen by considering the trigonometric Fourier series of x(t). Since x(t) sin(k0 t) is odd, as x(t) is even and sin(k0 t) is odd, then the coefficients corresponding to the sines in the expansion will be zero. On the other hand, x(t) cos(k0 t) is even and gives nonzero Fourier coefficients. See Equations (4.20). 3. In general, the Fourier coefficients are complex and as such need to be represented by their magnitudes and phases. In this case, the Xk coefficients are realvalued, and in particular zero when kπ/2 = ±mπ, m an integer, or when k = ±2, ±4, . . .. Since the Xk values are real, the corresponding phase would be zero when Xk ≥ 0, and ±π when Xk < 0. In Figure 4.5 we show a period of the signal, and the magnitude and phase line spectra displayed only for positive values of frequency (with the understanding that the magnitude spectrum is even and the phase is odd functions of the frequency).
4.6 Fourier Coefficients from Laplace
Period 2
x(t)
1.5 1 0.5 0 0.2
0
0.6
0.4
1
0.8
t Magnitude line spectrum
1
Phase line spectrum
4
0.8
3 Xk 
Xk 
0.6 0.4
1
0.2 0
2
0
50 Ω (rad/sec)
0
100
0
50 Ω (rad/sec)
100
FIGURE 4.5 Period of train of rectangular pulses (top) and its magnitude and phase line spectra (bottom).
4. The Xk coefficients and its squares, related to the power line spectrum, are obtained using the fourierseries function (see Figure 4.5):
k
Xk2
Xk = X−k
0
1
1
1
0.64
0.41
2
0
0
3
−0.21
4
0
0
5
0.13
0.016
6
0
0
7
−0.09
0.041
0.008
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CH A P T E R 4: Frequency Analysis: The Fourier Series
Notice that about 11 of them (including the zero values), or the dc value and 5 harmonics, provide a very good approximation of the pulse train, and would occupy a bandwidth of approximately 10π rad/sec. The power contribution, as indicated by Xk2 after k = ±6, is n relatively small.
n Example 4.6 Find the Fourier series of the fullwave rectified signal x(t) =  cos(πt) shown in Figure 4.6. This signal is used in the design of dc sources. The rectification of an ac signal is the first step in this design. Solution The integral to find the Fourier coefficients is Z0.5 Xk =
cos(πt)e−j2πkt dt
−0.5
which can be computed by using Euler’s identity or any other method. We want to show that this can be avoided by using the Laplace transform. A period x1 (t) of x(t) can be expressed as x1 (t − 0.5) = sin(πt)u(t) + sin(π(t − 1))u(t − 1)
1 1 0.8
0.8
0.6
0.6
x1(t )
x (t)
260
0.4
0.4
0.2
0.2
0 0 −0.2 −2
−1
0 t (a)
1
2
FIGURE 4.6 (a) Fullwave rectified signal x(t) and (b) one of its periods x1 (t).
−0.6
−0.4
−0.2
0 t (b)
0.2
0.4
0.6
4.6 Fourier Coefficients from Laplace
Period 1 0.8 0.6 0.4 0.2 0 0
0.2
0.4
0.6
0.8
1
t Magnitude line spectrum
Phase line spectrum
4
0.6 3
0.4
<Xk
Xk
0.5
0.3 0.2
2 1
0.1 0
0
50 Ω (rad/sec)
0
100
0
50 Ω (rad/sec)
FIGURE 4.7 Period of fullwave rectified signal x(t) and its magnitude and phase line spectra.
and using the Laplace transform we have X1 (s)e−0.5s =
π [1 + e−s ] s2 + π 2
so that X1 (s) =
π [e0.5s + e−0.5s ] s2 + π 2
The Fourier coefficients are then Xk =
1 X1 (s)s=j0 k T0
where T0 = 1 and 0 = 2π, giving Xk = =
π 2 cos(2πk/2) ( j2πk)2 + π 2 2(−1)k π(1 − 4k2 )
100
261
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CH A P T E R 4: Frequency Analysis: The Fourier Series
since cos(πk) = (−1)k . The DC value of the fullwave rectified signal is X0 = 2/π. Notice that the Fourier coefficients are real given that the signal is even. The MATLAB script used in the previous example can be used again with the following modification for the generation of a period of x(t). The results are shown in Figure 4.7. %%%%%%%%%%%%%%%%% % Example 4.6Fourier series of fullwave rectified signal %%%%%%%%%%%%%%%% % period generation T0 = 1; m = heaviside(t) − heaviside(t − T0);x = abs(cos(pi ∗ t)) ∗ m
n
n Example 4.7 Computing the derivative of a signal enhances higher harmonics. To illustrate this consider the train of triangular pulses y(t) (Figure 4.8) with fundamental period T0 = 2. Let x(t) = dy(t)/dt. Find its Fourier series and compare Xk  with Yk  to determine which of these signals is smoother—that is, which one has lower frequency components. Solution A period of y(t), −1 ≤ t ≤ 1, is given by y1 (t) = r(t + 1) − 2r(t) + r(t − 1) with a Laplace transform
Y1 (s) =
1 s e − 2 + e−s 2 s
x (t ) =
y(t )
FIGURE 4.8 (a) Train of triangular pulses y(t) and (b) its derivative x(t). Notice that y(t) is a continuous function while x(t) is discontinuous.
1
1
··· −2
··· −1
dy(t ) dt
0
1
2
··· t
−2
··· −1
0 −1
(a)
(b)
1
2
t
4.6 Fourier Coefficients from Laplace
so that the Fourier coefficients are given by (T0 = 2, 0 = π ): Yk = =
1 1 [2 cos(πk) − 2] Y1 (s)s=jo k = T0 2( jπk)2 1 − cos(πk) 1 − (−1)k = π 2 k2 π 2 k2
k 6= 0
This is also equal to
sin(πk/2) Yk = 0.5 (πk/2)
2 (4.22)
using the identity 1 − cos(πk) = 2 sin2 (πk/2). By observing y(t) we deduce that its DC value is Y0 = 0.5. Let us then consider the periodic signal x(t) = dy(t)/dt (shown in Fig. 4.8(b)) with a dc value X0 = 0. For −1 ≤ t ≤ 1, its period is x1 (t) = u(t + 1) − 2u(t) + u(t − 1) and X1 (s) =
1 s e − 2 + e−s s
which gives the Fourier series coefficients (T0 = 2, (the period and the fundamental frequency are equal to the ones for y(t)) Xk =
sin2 (kπ/2) j kπ/2
(4.23)
Period
1 0.8 0.6 0.4 0.2 0 0
1 t
0.5
1.5
2
0.5
1 t
1.5
0.8
2
0.5
0.6
1.5
0.4
0
0.4
0.2
−0.5
0.2
0
−1
0
0
20 40 Ω (rad/sec)
60
(a)
0
20 40 Ω (rad/sec)
60
Yk 
1
0.6
Xk 
0.8 Yk 
Yk 
Period 1 0.5 0 −0.5 −1 0
x(t )
y(t )
since Xk = 12 X1 (s)s=jπk .
2
1 0.5
0
20 40 Ω (rad/sec)
60
0
0
20 40 Ω (rad/sec)
(b)
FIGURE 4.9 Magnitude and phase line spectra of (a) triangular signal y(t) (top left) and (b) its derivative x(t) (top right). Ignoring the dc values, the {Yk } decay faster to zero than the {Xk }, thus y(t) is smoother than x(t).
60
263
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CH A P T E R 4: Frequency Analysis: The Fourier Series
For k 6= 0 we have Yk  = Xk /(πk), so that as k increases the frequency components of y(t) decrease in magnitude faster than the corresponding ones of x(t). Thus, y(t) is smoother than x(t). The magnitude line spectrum Yk , ignoring its average, goes faster to zero than the magnitude line spectrum Xk , as seen in Figure 4.9. Notice that in this case y(t) is even and its Fourier coefficients Yk are real, while x(t) is odd and its Fourier coefficients Xk are purely imaginary. If we subtract the average of y(t), the signal y(t) can be clearly approximated as a series of cosines, thus the need for real coefficients in its complex exponential Fourier series. The signal x(t) is zeroaverage and as such it can be clearly approximated by a series of sines requiring its Fourier coefficients Xk to be imaginary. n n Example 4.8 Integration of a periodic signal, provided it has zero mean, gives a smoother signal. To see this, find and compare the magnitude line spectra of a sawtooth signal x(t), of period T0 = 2, and its integral Z y(t) =
x(t)dt
shown Figure 4.10. Solution Before doing any calculations it is important to realize that the integral would not exist if the dc is not zero. Using the following script we can compute the Fourier series coefficients of x(t) and y(t). A period of x(t) is x1 (t) = tw(t) + (t − 2)w(t − 1) 0 ≤ t ≤ 2 where w(t) = u(t) − u(t − 1) is a rectangular window.
y (t ) = ∫ x(t)dt
x(t)
0.5
1
FIGURE 4.10 (a) Sawtooth signal x(t) and (b) its integral y(t). Notice that x(t) is a discontinuous function while y(t) is continuous.
··· −2
··· −1
0
1
t ···
−2
··· −1
0
−1 (a)
(b)
1
2
t
4.7 Convergence of the Fourier Series
Period 1 0 −0.5 −1
0
0.5
1.5
1 t
2
0
0.5
1 t
0.4
4
0.3
1
0.3
3
0.2
2
0.2
0
0.1
−1
0
−2
0
20
40
60
Ω
Xk 
2
Yk 
0.4 Xk 
Xk 
y(t )
x(t )
0.5
Period
0.5 0.4 0.3 0.2 0.1 0
20
40
60
Ω
0
2
0
20
1
0.1 0
1.5
0
20
40
60
Ω
(a)
0
40
60
Ω
(b)
FIGURE 4.11 (a) Periods of the sawtooth signal x(t) and (b) its integral y(t) and their magnitude and phase line spectra.
%%%%%%%%%%%%%%%%% % Example 4.8Sawtooth signal and its integral %%%%%%%%%%%%%%%% syms t T0 = 2; m = heaviside(t) − heaviside(t − T0/2); m1 = heaviside(t − T0/2)  heaviside(t − T0); x = t ∗ m + (t − 2) ∗ m1; y = int(x); [X, w] = fourierseries(x, T0, 20); [Y, w] = fourierseries(y, T0, 20);
The signal y(t) is smoother than x(t); y(t) is a continuous function of time, while x(t) is discontinuous. This is indicated as well by the magnitude line spectra of the two signals. Ignoring the dc components, the {Yk } of y(t) decay a lot faster to zero than the {Xk } of x(t) (See Figure 4.11). As we will see in Section 4.10, computing the derivative of a periodic signal is equivalent to multiplying its Fourier series coefficients by j0 k, which emphasizes the higher harmonics. If the periodic signal is zeromean so that its integral exists, the Fourier coefficients of the integral can be found by dividing them by j0 k so that now the low harmonics are emphasized. n
4.7 CONVERGENCE OF THE FOURIER SERIES It can be said, without overstating it, that any periodic signal of practical interest has a Fourier series. Only very strange signals would not have a converging Fourier series. Establishing convergence is necessary because the Fourier series has an infinite number of terms. To establish some general
265
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CH A P T E R 4: Frequency Analysis: The Fourier Series
conditions under which the series converges, we need to classify signals with respect to their smoothness. A signal x(t) is said to be piecewise smooth if it has a finite number of discontinuities, while a smooth signal has a derivative that changes continuously. Thus, smooth signals can be considered special cases of piecewise smooth signals. The Fourier series of a piecewise smooth (continuous or discontinuous) periodic signal x(t) converges for all values of t. The mathematician Dirichlet showed that for the Fourier series to converge to the periodic signal x(t), the signal should satisfy the following sufficient (not necessary) conditions over a period: n Be absolutely integrable. n Have a finite number of maxima, minima, and discontinuities. The infinite series equals x(t) at every continuity point and equals the average 0.5[x(t + 0+) + x(t + 0−)] of the right limit x(t + 0+) and the left limit x(t + 0−) at every discontinuity point. If x(t) is continuous everywhere, then the series converges absolutely and uniformly.
Although the Fourier series converges to the arithmetic average at discontinuities, it can be observed that there is some ringing before and after the discontinuity points. This is called the Gibb’s phenomenon. To understand this phenomenon it is necessary to explain how the Fourier series can be seen as an approximation to the actual signal, and how when a signal has discontinuities the convergence is not uniform around them. It will become clear that the smoother the signal x(t) is, the easier it is to approximate it with a Fourier series with a finite number of terms. When the signal is continuous everywhere, the convergence is such that at each point t the series approximates the actual value x(t) as we increase the number of terms in the approximation. However, that is not the case when discontinuities occur in the signal. This is despite the fact that a minimum meansquare approximation seems to indicate that the approximation could give a zero error. Let
xN (t) =
N X
Xk ejk0 t
(4.24)
k=−N
be the Nthorder approximation of a periodic signal x(t), of fundamental frequency 0 , that minimizes the average quadratic error over a period 1 EN = T0
Z T0
x(t) − xN (t)2 dt
(4.25)
4.7 Convergence of the Fourier Series
with respect to the Fourier coefficients Xk . To minimize EN with respect to the coefficients Xk we set its derivative with respect to Xk to zero. Let ε(t) = x(t) − xN (t), so that dEN 1 = dXk T0 =−
Z 2ε(t) T0
1 T0
Z
dε∗ (t) dt dXk
2[x(t) − xN (t)]e−jk0 t dt
T0
=0 which after replacing xN (t) and using the orthogonality of the Fourier exponentials gives
Xk =
1 T0
Z
x(t)e−j0 kt dt
(4.26)
T0
corresponding to the Fourier coefficients of x(t) for −N ≤ k ≤ N. As N → ∞ the average error EN → 0. The only issue left is how xN (t) converges to x(t). As indicated before, if x(t) is smooth xN (t) approximates x(t) at every point, but if there are discontinuities the approximation is in an average fashion. The Gibb’s phenomenon indicates that around discontinuities there will be ringing, regardless of the order N of the approximation, even though the average quadratic error EN goes to zero as N increases. This phenomenon will be explained in Chapter 5 as the effect of using a rectangular window to obtain a finitefrequency representation of a periodic signal.
n Example 4.9 To illustrate the Gibb’s phenomenon consider the approximation of a train of pulses x(t) with zero mean and period T0 = 1 (see the dashed signal in Figure 4.12) with a Fourier series xN (t) with N = 1, . . . , 20. Solution We compute analytically the Fourier coefficients of x(t) and use them to obtain an approximation xN (t) of x(t) having a zero DC component and up to 20 harmonics. The dashedline plot in Figure 4.12 is x(t) and the solid–line plot is xN (t) when N = 20. The discontinuities of the pulse train cause the Gibb’s phenomenon. Even if we increase the number of harmonics there is an overshoot in the approximation around the discontinuities.
267
CH A P T E R 4: Frequency Analysis: The Fourier Series
1 0.5 x(t), xN (t )
268
FIGURE 4.12 Approximate Fourier series xN (t) of the pulse train x(t) (discontinuous) using the DC component and 20 harmonics. The approximate xN (t) displays the Gibb’s phenomenon around the discontinuities.
0 −0.5 −1 0
0.2
0.4
0.6
0.8
t (sec)
%%%%%%%%%%%%%%%%% % Example 4.9Simulation of Gibb’s phenomenon %%%%%%%%%%%%%%%% clf; clear all w0 = 2 ∗ pi; DC = 0; N = 20; % parameters of periodic signal % computation of Fourier series coefficients for k = 1:N, X(k) = sin(k ∗ pi/2)/(k ∗ pi/2); end X = [DC X]; % Fourier series coefficients % computation of periodic signal Ts = 0.001; t = 0:Ts:1 − Ts; L = length(t); x = [ones(1, L/4) zeros(1, L/2) ones(1, L/4)]; x = x − 0.5; % computation of approximate xN = X(1)∗ones(1,length(t)); for k = 2:N, xN = xN + 2 ∗ X(k) ∗ cos(2 ∗ pi ∗ (k − 1). ∗ t); % approximate signal plot(t, xN); axis([0 max(t) 1.1 ∗ min(xN) 1.1 ∗ max(xN)]) hold on; plot(t, x, ’r’) ylabel(’x(t), x N(t)’); xlabel(’t (sec)’);grid hold off pause(0.1) end
When you execute the above script, it pauses to display the approximation for an increasing number of terms in the approximation. At each of these values ringing around the discontinuities the Gibb’s phenomenon is displayed. n
4.7 Convergence of the Fourier Series
n Example 4.10 Consider the meansquare error optimization to obtain an approximation of the periodic signal x(t) shown in Figure 4.4 from Example 4.5. We wish to obtain an approximate x2 (t) = α + 2β cos(0 t), given that it is clear that x(t) has an average, and that once we subtract it from the signal the resulting signal is approximated by a cosine function. Minimize the meansquare error Z 1 E2 = x(t) − x2 (t)2 dt T0 T0
with respect to α and β to find these values. Solution To minimize E2 we set to zero its derivatives with respect to α and β to get Z Z 1 dE2 1 =− 2[x(t) − α − 2β cos(0 t)]dt = − 2[x(t) − α]dt = 0 dα T0 T0 T0
dE2 1 =− dβ T0 which, after getting rid of of the Fourier basis, gives
Z
T0
2[x(t) − α − 2β cos(0 t)] cos(0 t)dt = 0
T0 2 T0
of both sides of the above equations and applying the orthogonality
α=
1 T0
1 β= T0
Z x(t)dt T0
Z x(t) cos(0 t)dt T0
For the signal in Figure 4.4 we obtain α=1 β=
2 π
giving as approximation the signal x2 (t) = 1 +
4 cos(2πt) π
which at t = 0 gives x2 (0) = 2.27 instead of the expected 2; x2 (0.25) = 1 (because of the discontinuity at this point, this value is the average of 2 and 0, the values, respectively, before and after the discontinuity) instead of 2 and x2 (0.5) = −0.27 instead of the expected 0. n
269
270
CH A P T E R 4: Frequency Analysis: The Fourier Series
n Example 4.11 Consider the train of pulses in Example 4.5. Determine how many Fourier coefficients are necessary to get a representation containing 97% of the power of the periodic signal. Solution The desired 97% of the power of x(t) is 1 0.97 T0
Z
2
x (t)dt = 0.97
T0
0.25 Z
4dt = 1.94
−0.25
and so we need to find an integer N such that N X k=−N
N X sin(πk/2) 2 Xk  = (πk/2) = 1.94 2
k=−N
The value of N is found by trial and error, adding consecutive values of the magnitude squared of Fourier coefficients. Using MATLAB, it is found that for N = 5 (dc and 5 harmonics) the Fourier series approximation has a power of 1.93. Thus, 11 Fourier coefficients give a very good approximation to the periodic train of pulses, with about 97% of the signal power. n
4.8 TIME AND FREQUENCY SHIFTING Time shifting and frequency shifting are duals of each other. n
Timeshifting: A periodic signal x(t), of period T0 , remains periodic of the same period when shifted in time. If Xk are the Fourier coefficients of x(t), the Fourier coefficients for x(t − t0 ) are n o Xk e−jk0 t0 = Xk ej(∠Xk −k0 t0 ) (4.27) That is, only a change in phase is caused by the time shift. The magnitude spectrum remains the same.
n
Frequencyshifting: When a periodic signal x(t), of period T0 , modulates a complex exponential ej1 t : n The modulated signal x(t)ej1 t is periodic of period T0 if 1 = M0 for an integer M ≥ 1. n The Fourier coefficients Xk are shifted to frequencies k0 + 1 . n The modulated signal is realvalued by multiplying x(t) by cos(1 t).
If we delay or advance in time a periodic signal, the resulting signal is periodic of the same period. Only a change in the phase of the coefficients occurs to accommodate for the shift. Indeed, if x(t) =
X k
Xk ejk0 t
4.8 Time and Frequency Shifting
we then have that x(t − t0 ) =
X
x(t + t0 ) =
X
Xk ejk0 (t−t0 ) =
Xh
Xk ejk0 (t+t0 ) =
Xh
k
i Xk e−jk0 t0 ejk0 t
k
k
i Xk ejk0 t0 ejk0 t
k
so that the Fourier coefficients {Xk } corresponding to x(t) are changed to {Xk e∓jk0 t0 } for x(t ∓ t0 ). In both cases, they have the same magnitude Xk  but different phases. In a dual way, if we multiply the above periodic signal x(t) by a complex exponential of frequency 1 , ej1 t , we obtain a socalled modulated signal y(t) and its spectrum is shifted in frequency by 1 with respect to the spectrum of the periodic signal x(t). In fact, y(t) = x(t)ej1 t X = Xk ej(0 k+1 )t k
indicating that the harmonic frequencies are shifted by 1 . The signal y(t) is not necessarily periodic. Since T0 is the period of x(t), then y(t + T0 ) = x(t + T0 )ej1 (t+T0 ) and for it to be equal to y(t), then 1 T0 = 2πM, for an integer M 6= 0 or 1 = M0
M >> 1
which goes along with the condition that the modulating frequency 1 is chosen much larger than 0 . The modulated signal is then given by y(t) =
X
Xk ej(0 k+1 )t =
k
X
Xk ej0 (k+M)t =
X
k
X`−M ej0 `t
`
so that the Fourier coefficients are shifted to new frequencies 0 (k + M). To keep the modulated signal realvalued, one multiplies the periodic signal x(t) by a cosine of frequency 1 = M0 for M >> 1 to obtain a modulated signal y1 (t) = x(t) cos(1 t) X = 0.5Xk [ej(k0 +1 )t + ej(k0 −1 )t ] k
so that the harmonic components are now centered around ±1 .
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CH A P T E R 4: Frequency Analysis: The Fourier Series
n Example 4.12 To illustrate the modulation property using MATLAB consider modulating a sinusoid cos(20πt) with a train of square pulses x1 (t) = 0.5[1 + sign(sin(πt)] and with a sinusoid x2 (t) = cos(πt) Use our function fourierseries to find the Fourier series of the modulated signals and plot their magnitude line spectra. Solution The function sign is defined as sign(x(t)) =
−1 x(t) < 0 1 x(t) ≥ 0
(4.28)
That is, it determines the sign of the signal. Thus, 0.5[1 + sign sin(πt)] = u(t) − u(t − 1) equals 1 for 0 ≤ t ≤ 1, and 0 for 1 < t ≤ 2, which corresponds to a period of a train of square pulses. The following script allows us to compute the Fourier coefficients of the two modulated signals. %%%%%%%%%%%%%%%%% % Example 4.12Modulation %%%%%%%%%%%%%%%% syms t T0 = 2; m = heaviside(t) − heaviside(t − T0/2); m1 = heaviside(t) − heaviside(t − T0); x = m ∗ cos(20 ∗ pi ∗ t); x1 = m1 ∗ cos(pi ∗ t) ∗ cos(20 ∗ pi ∗ t); [X, w] = fourierseries(x, T0, 60); [X1, w1] = fourierseries(x1, T0, 60);
The modulated signals and their corresponding magnitude line spectra are shown in Figure 4.13. The Fourier coefficients of the modulated signals are now clustered around the frequency 20π.
4.9 Response of LTI Systems to Periodic Signals
0.5
0.5 x2(t)
1
x1(t)
1
0 −0.5
−0.5
−1
−1 0
4
2 t
0
0.2
2 t
4
0.2 X2k 
X1k 
FIGURE 4.13 (a) Modulated squarewave x1 (t) cos(20πt) and (b) cosine x2 (t) cos(20πt) and their respective magnitude line spectra.
0
0.1 0
0.1 0
0
50
100
150
0
50
100
150
Ω (b)
Ω (a)
n
4.9 RESPONSE OF LTI SYSTEMS TO PERIODIC SIGNALS The most important property of LTI systems is the eigenfunction property. Eigenfunction property: In steady state, the response to a complex exponential (or a sinusoid) of a certain frequency is the same complex exponential (or sinusoid), but its amplitude and phase are affected by the frequency response of the system at that frequency.
Suppose that the impulse response of an LTI system is h(t) and that H(s) = L[h(t)] is the corresponding transfer function. If the input to this system is a periodic signal x(t), of period T0 , with Fourier series x(t) =
∞ X
Xk ejk0 t
k=−∞
0 =
2π T0
(4.29)
then according to the eigenfunction property the output in the steady state is yss (t) =
∞ X k=−∞
[Xk H( jk0 )] ejk0 t
(4.30)
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CH A P T E R 4: Frequency Analysis: The Fourier Series
If we call Yk = Xk H( jk0 ) we have a Fourier series representation of yss (t) with Yk as its Fourier coefficients.
4.9.1 Sinusoidal Steady State If the input of a stable and causal LTI system, with impulse response h(t), is x(t) = Aej0 t , the output is Z∞ y(t) =
j0 t
h(τ )x(t − τ )dτ = Ae
0
Z∞
h(τ )e−j0 τ dτ
0
= Aej0 t H( j0 ) = AH( j0 )ej0 t+∠H( j0 )
(4.31)
The limits of the first integral indicate that the system is causal (the h(τ ) = 0 for τ < 0) and that the input x(t − τ ) is applied from −∞ (when τ = ∞) to t (when τ = 0); thus y(t) is the steadystate response of the system. If the input is a sinusoid—for example, x1 (t) = Re[x(t) = Aej0 t ] = A cos(0 t)
(4.32)
then the corresponding steadystate response is y1 (t) = Re[AH( j0 )ej0 t+∠H( j0 ) ] = AH( j0 ) cos(0 t + ∠H( j0 )).
(4.33)
As in the eigenfunction property, the frequency of the output coincides with the frequency of the input, however, the magnitude and the phase of the input signal is changed by the response of the system at the input frequency. The following script simulates the convolution of a sinusoid x(t) of frequency = 20π, amplitude 10, and random phase with the impulse response h(t) (a modulated decaying exponential) of an LTI system. The convolution integral is approximated using the MATLAB function conv. %%%%%%%%%%%%%%%%% % Simulation of Convolution %%%%%%%%%%%%%%%% clear all; clf Ts = 0.01; Tend = 2; t = 0:Ts:Tend; x = 10 ∗ cos(20 ∗ pi ∗ t + pi ∗ (rand(1, 1) − 0.5)); % input signal h = 20 ∗ exp(−10.ˆt). ∗ cos(40 ∗ pi ∗ t); % impulse response % approximate convolution integral y = Ts ∗ conv(x, h);
4.9 Response of LTI Systems to Periodic Signals
M = length(x); figure(1) x1 = [zeros(1, 5) x(1:M)]; z = y(1); y1 = [zeros(1, 5) z zeros(1, M − 1)]; t0 = −5 ∗ Ts:Ts:Tend; for k = 0:M − 6, pause(0.05) h0 = fliplr(h); h1 = [h0(M  k  5:M) zeros(1, M  k  1)]; subplot(211) plot(t0, h1, ’r’) hold on plot(t0, x1, ’k’) title(’Convolution of x(t) and h(t)’) ylabel(’x(τ ), h(tτ )’); grid; axis([min(t0) max(t0) 1.1*min(x) 1.1*max(x)]) hold off subplot(212) plot(t0, y1, ’b’) ylabel(’y(t) = (x ∗ h)(t)’); grid; axis([min(t0) max(t0) 0.1 ∗ min(x) 0.1 ∗ max(x)]) z = [z y(k + 2)]; y1 = [zeros(1, 5) z zeros(1, M  length(z))]; end
Figure 4.14 displays the last step of the convolution integral simulation. Notice that the steady state is attained in a very short time (around t = 0.5 sec). The transient changes every time that the script is executed due to the random phase.
x(τ ), h(t − τ)
10 5 0 −5 −10
0
0.5
1 τ
1.5
2
1.5
2
FIGURE 4.14 Convolution simulation: (a) input x(t) (solid line) and h(t − τ ) (dashed line), and (b) output y(t): transient and steadystate response.
y(t) = (x *h)(t)
(a) 0.5 0 −0.5 0
0.5
1 t (b)
275
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CH A P T E R 4: Frequency Analysis: The Fourier Series
If the input x(t) of a causal and stable LTI system, with impulse response h(t), is periodic of period T0 and has the Fourier series x(t) = X0 + 2
∞ X
Xk  cos(k0 t + ∠Xk )
k=1
0 =
2π T0
(4.34)
the steadystate response of the system is y(t) = X0 H( j0) cos(∠H( j0)) + 2
∞ X
Xk H( jk0 ) cos(k0 t + ∠Xk + ∠H( jk0 ))
(4.35)
k=1
where Z∞ H( jk0 ) = h(τ )e−jk0 τ dτ
(4.36)
0
is the frequency response of the system at k0 .
Remarks n
If the input signal x(t) is a combination of sinusoids of frequencies that are not harmonically related, the signal is not periodic, but the eigenfunction property still holds. For instance, if X x(t) = Ak cos(k t + θk ) k
and the frequency response of the LTI system is H( j), the steadystate response is X y(t) = Ak H( jk ) cos(k t + θk + ∠H( jk )) k n
It is important to realize that if the LTI system is represented by a differential equation and the input is a sinusoid, or combination of sinusoids, it is not necessary to use the Laplace transform to obtain the complete response and then let t → ∞ to find the sinusoidal steadystate response. The Laplace transform is only needed to find the transfer function of the system, which can then be used in Equation (4.35) to find the sinusoidal steady state.
4.9.2 Filtering of Periodic Signals According to Equation (4.35) if we know the frequency response of the system (Eq. 4.36), at the harmonic frequencies of the periodic input, H( jk0 ), we have that in the steady state the output of the system y(t) is as follows: n n
Periodic of the same period as the input. Its Fourier coefficients are those of the input Xk multiplied by the frequency response at the harmonic frequencies, H( jk0 ).
4.9 Response of LTI Systems to Periodic Signals
n Example 4.13 To illustrate the filtering of a periodic signal, consider a zeromean pulse train ∞ X
x(t) =
k=−∞,6=0
sin(kπ/2) j2kπt e kπ/2
as the driving source of an RC circuit that realizes a lowpass filter (i.e., a system that tries to keep the lowfrequency harmonics and get rid of the highfrequency harmonics of the input). The transfer function of the RC lowpass filter is H(s) =
1 1 + s/100
Solution The following script computes the frequency response of the filter at the harmonic frequencies H( jk0 ) (see Figure 4.15). %%%%%%%%%%%%%%%%% % Example 4.13 %%%%%%%%%%%%%%%% % Freq response of H(s)=1/(s/scale+1)  lowpass filter w0 = 2 * pi; % fundamental frequency of input M = 20; k = 0:M  1; w1 = k. * w0; % harmonic frequencies H = 1./(1 + j * w1/100); Hm = abs(H); Ha = angle(H); % frequency response subplot(211) stem(w1, Hm, ’filled’); grid; ylabel(’—H(jω)—’) axis([0 max(w1) 0 1.3]) subplot(212) stem(w1, Ha, ’filled’); grid axis([0 max(w1) 1 0]) ylabel(’¡H(j ω)’); xlabel(’w (rad/sec)’)
The response due to the pulse train can be found by finding the response to each of its Fourier series components and adding them. Approximating x(t) using N = 20 harmonics by 20 X
xN (t) =
k=−20,6=0
sin(kπ/2) j2kπt e kπ/2
Then the output voltage across the capacitor is given in the steady state, yss (t) =
20 X k=−20,6=0
H( j2kπ )
sin(kπ/2) j2kπt e kπ/2
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CH A P T E R 4: Frequency Analysis: The Fourier Series
Because the magnitude response of the lowpass filter changes very little in the range of frequencies of the input, the output signal is very much like the input (see Figure 4.15). The following script is used to find the response.
1 0.5 0 0
20
40
−1 0
20
40
60
80
100
60 80 Ω (rad/sec)
100
0 −0.5
(a)
x (t ), y (t )
H(j Ω)
% lowpass filtering % FS coefficients of input X(1) = 0; % mean value for k = 2:M  1, X(k) = sin((k − 1) ∗ pi/2)/((k − 1) ∗ pi/2); end % periodic signal Ts = 0.001; t1 = 0:Ts:1  Ts;L = length(t1); x1 = [ones(1, L /4) zeros(1, L /2) ones(1, L /4)]; x1 = x1 − 0.5; x = [x1 x1]; % output of filter t = 0:Ts:2 − Ts; y = X(1) ∗ ones(1, length(t)) ∗ Ha(1); plot(t, y); axis([0 max(t) − .6 .6]) for k = 2:M  1, y = y + X(k) ∗ Hm(k) ∗ cos(w0 ∗ (k − 1). ∗ t + Ha(k)); plot(t, y); axis([0 max(t) − .6 .6]); hold on plot(t, x, ’r’); axis([0 max(t) − 0.6 0.6]); grid ylabel(’x(t), y(t)’); xlabel(’t (sec)’) ; hold off pause(0.1) end
> 1 and 1 is the cutoff frequency of the lowpass filter) corresponds to the impulse response of a bandpass filter centered around 0 . Indeed, its Fourier transform is given by F[2hlp (t) cos(0 t)] = Hlp ( j( − 0 )) + Hlp ( j( + 0 ))
n
n
which is the frequency response of the lowpass filter shifted to new center frequencies 0 and −0 , making it a bandpass filter. A zerophase ideal lowpass filter Hlp ( j) = u( + 1 ) − u( − 1 ) has as impulse response a sinc function with a support from −∞ to ∞. This ideal lowpass filter is clearly noncausal as its impulse response is not zero for negative values of time t. To make it causal we could approximate its impulse response by a function h1 (t) = hlp (t)w(t) where w(t) = u(t + τ ) − u(t − τ ) is a rectangular window where the value of τ is chosen so that outside the window the values of the impulse response hlp (t) are very close to zero. Although the Fourier transform of h1 (t) is a very good approximation of the desired frequency response, the frequency response of h1 (t) displays ringing around the cutoff frequency 1 because of the rectangular window. Finally, we delay h1 (t) by τ to get a causal filter with linear phase. That is, h1 (t − τ ) has as its magnitude response H1 ( j) ≈ Hlp ( j) and its phase response is ∠H1 (j) = −τ . Although the above procedure is a valid way to obtain approximate lowpass filters with linear phase, they are not guaranteed to be rational and would be difficult to implement. Thus, other methods are used to design filters. Since ideal filters are not causal they cannot be used in realtime applications—that is when the input signal needs to be processed as it comes to the filter. Imposing causality on the filter restricts the frequency response of the filter in significant ways. According to the PaleyWiener integral condition, a causal
333
334
CH A P T E R 5: Frequency Analysis: The Fourier Transform
and stable filter with frequency response H( j) should satisfy the condition Z∞  log(H( j)) d < ∞ 1 + 2
(5.22)
−∞
n
To satisfy this condition, H( j) cannot be zero in any band of frequencies, because in such cases the numerator of the integrand would be infinite. The PaleyWiener integral condition is clearly not satisfied by ideal filters. So they cannot be implemented or used in actual situations, but they can be used as models for designing filters. That ideal filters are not realizable can be understood also by considering what it means to make the magnitude response of a filter zero in some frequency bands. A measure of attenuation is given by the loss function in decibels, defined as α() = −10 log10 H( j)2 = −20 log10 H( j) dB Thus, when H( j) = 1 and there is no attenuation the loss is 0 dB, and when H( j) = 10−5 for a large attenuation the loss is 100 dB. You quickly convince yourself that if a filter achieves a magnitude response of 0 at any frequency this would mean a loss or attenuation at that frequency of ∞ dBs! Values of 60 to 100 dB attenuation are considered extremely good, and to obtain that the signal needs to be attenuated by a factor of 10−3 to 10−5 . A curious term JND or “just noticeable difference” is used by experts in human hearing to characterize the smallest sound intensity that can be judged by a human as different. Such a value varies from 0.25 to 1 dB. To illustrate what is loud in the dB scale, consider the following cases: A sound pressure level higher than 130 dB causes pain; 110 dB is generated by an amplified rock band performance [65].
n Example 5.15 The Gibb’s phenomenon, which we mentioned when discussing the Fourier series of periodic signals with discontinuities, consists in ringing around these discontinuities. To see this, consider a periodic train of square pulses x(t) of period T0 displaying discontinuities at kT0 /2, for k = ±1, ±2, . . .. Show how the Gibb’s phenomenon is due to ideal lowpass filtering. Solution Choosing 2N + 1 of the Fourier series coefficients to approximate the signal x(t) is equivalent to passing x(t) through an ideal lowpass filter, 1 −c ≤ ≤ c H( j) = 0 otherwise having as impulse response a sinc function h(t). If the Fourier transform of the periodic signal x(t) of fundamental frequency 0 = 2π/T0 is X() =
∞ X k=−∞
2πXk δ( − k0 )
5.7 Convolution and Filtering
the output of the filter is the signal xN (t) = F −1 [X()H( j)] N X = F −1 2πXk δ( − k0 ) k=−N
or the inverse Fourier transform of X() multiplied by a lowpass filter with an ideal magnitude response of 1 for −c < < c where the cutoff frequency c is chosen so that N0 < c < (N + 1)0 . As such, xN (t) is the convolution xN (t) = [x ∗ h](t) where h(t) is the inverse Fourier transform of H( j), or a sinc signal of infinite support. The convolution around the discontinuities of x(t) causes ringing before and after them, and this ringing appears independent of the value of N. n
n Example 5.16 Obtain different filters from an RLC circuit (Figure 5.9) by choosing different outputs. Let the input be a voltage source with Laplace transform Vi (s). For simplicity, let R = 1 , L = 1 H, and C = 1 F, and assume the initial conditions to be zero. Solution n
Lowpass filter: Let the output be the voltage across the capacitor; by voltage division we have that VC (s) =
Vi (s)/s Vi (s) = 2 1 + s + 1/s s +s+1
so that the transfer function is Hlp (s) =
VC (s) 1 = 2 Vi (s) s +s+1
This is the transfer function of a secondorder lowpass filter. If the input is a dc source, so that its frequency is = 0, the inductor is a short circuit (its impedance would be 0) and +
vR (t ) R
+
FIGURE 5.9 RLC circuit for implementing different filters.
vi (t ) −
−
L +
vL(t )
− C
+ vC (t ) −
335
336
CH A P T E R 5: Frequency Analysis: The Fourier Transform
n
the capacitor is an open circuit (its impedance would be infinite), so that the voltage in the capacitor is equal to the voltage in the source. On the other hand, if the frequency of the input source is very high, then the inductor is an open circuit and the capacitor a short circuit (its impedance is zero) so that the capacitor voltage is zero. This is a lowpass filter. Notice that this filter has no finite zeros, and complex conjugate poles. Highpass filter: Suppose then that we let the output be the voltage across the inductor. Then again by voltage division the transfer function Hhp (s) =
n
is that of a highpass filter. Indeed, for a dc input (frequency zero) the impedance in the inductor is zero, so that the inductor voltage is zero, and for very high frequency the impedance of the inductor is very large so that it can be considered open circuit and the voltage in the inductor equals that of the source. This filter has the same poles of the lowpass filter (this is determined by the overall impedance of the circuit, which has not changed) and double zeros at zero. It is these zeros that make the frequency response for low frequencies be close to zero. Bandpass filter: Letting the output be the voltage across the resistor, its transfer function is Hbp (s) =
n
VL (s) s2 = 2 Vi (s) s +s+1
VR (s) s = 2 Vi (s) s +s+1
or the transfer function of a bandpass filter. For zero frequency, the capacitor is an open circuit so the current is zero and the voltage across the resistor is zero. Similarly, for very high frequency the impedance of the inductor is very large, or an open circuit, making the voltage across the resistor zero because again the current is zero. For some middle frequency the serial combination of the inductor and the capacitor resonates and will have zero impedance. At the resonance frequency, the current achieves its largest value and the voltage across the resistor does too. This behavior is that of a bandpass filter. This filter again has the same poles as the other two, but only one zero at zero. Bandstop filter: Finally, suppose we consider as output the voltage across the connection of the inductor and the capacitor. At low and high frequencies, the impedance of the LC connection is very high, or open circuit, and so the output voltage is the input voltage. At the resonance frequency r = 1 the impedance of the LC connection is zero, so the output voltage is zero. The resulting filter is a bandstop filter with the transfer function Hbs (s) =
s2 + 1 s2 + s + 1
Secondorder filters can then be easily identified by the numerator of their transfer functions. Secondorder lowpass filters have no zeros, and the numerator is N(s) = 1; bandpass filters have a zero at s = 0 so N(s) = s, and so on. We will see next that such a behavior can be easily seen from a geometric approach. n
5.7 Convolution and Filtering
5.7.3 Frequency Response from Poles and Zeros Given a rational transfer function H(s) = B(s)/A(s), to calculate its frequency response we let s = j and find the magnitude and phase for a discrete set of frequencies. This can be done using MATLAB. A geometric way to obtain an approximate magnitude and phase frequency responses is using the effects of zeros and poles on the frequency response of a system. Consider a function G(s) =
s−z s−p
with a zero z and a pole p, as shown in Figure 5.10. The frequency response corresponding to G(s) at some frequency 0 is found by letting s = j0 , or G(s)s=j0 =
j0 − z j0 − p
Representing j0 , z, and p, which are complex numbers, as vectors coming from the origin, then the E 0 ) = j0 − z (adding to Z( E 0 ) the vector corresponding to z gives a vector corresponding vector Z( E 0 ) = j0 − p goes from the pole p to to j0 ) goes from the zero z to j0 , and likewise the vector P( j0 . The argument 0 in the vectors indicates that the magnitude and phase of these vectors depend on the frequency at which we are finding the frequency response. As we change the frequency at which we are finding the frequency response, the lengths and the phases of these vectors change. Therefore, G( j0 ) =
E 0) E 0 ) j(∠Z( Z( Z( E 0 )−∠P( E 0 )) = e E E P(0 ) P(0 )
and the magnitude response is G( j0 ) =
E 0 ) Z( E 0 ) P(
(5.23)
and the phase response is E 0 ) − ∠P( E 0) ∠G( j0 ) = ∠Z(
(5.24)
splane j Ω0 →
P (Ω0)
FIGURE 5.10 Geometric interpretation of poles and zeros.
p
→
Z (Ω0)
z
337
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
E 0 ) and P( E 0 ), the ratio of these So that for 0 ≤ 0 < ∞, if we compute the length and the angle of Z( lengths gives the magnitude response and the difference of their angles gives the phase response. For a filter with a transfer function Q (s − zi ) H(s) = Q i (s k − pk ) E i () = j − zi and PE k () = j − pk , going from each of where zi , pk are zeros and poles of H(s) with vectors Z the zeros and poles to the frequency at which we are computing the magnitude and phase response in the j axis, gives Q E i () Z H( j) = H(s)s=j = Q i E k Pk () hP i Q P E E E i Zi () j i ∠(Zi ())− k ∠(Pk () (5.25) = Q e {z } PE k ()  j∠H( j)  k {z } e H( j)
n Example 5.17 Consider series RC circuit with a voltage source vi (t). Choose the output to obtain lowpass and highpass filters and use the poles and zeros of the transfer functions to determine their frequency responses. Let R = 1 , C = 1 F, and the initial conditions be zero. Solution n
Lowpass filter: Let the output be the voltage across the capacitor. By voltage division, we obtain that the transfer function of the filter is VC (s) 1/Cs H(s) = = Vi (s) R + 1/Cs For dc frequency, the capacitor behaves as an open circuit so that the output voltage equals the input voltage, and for very high frequencies the impedance of the capacitor tends to zero so that the voltage across the capacitor also goes to zero. This is a lowpass filter. Let R = 1 and C = 1 F, so 1 1 = H( j) = E 1 + j P() E Drawing a vector from the pole s = −1 to any point on the j axis gives P(), and for different frequencies we get =1
E P(0) = 1ej0 √ E P(1) = 2ejπ/4
=∞
E P(∞) = ∞ ejπ/2
=0
5.7 Convolution and Filtering
Since there are no zeros, the frequency response of this filter depends inversely on the behavior E of the pole vector P(). The frequency responses for these three frequencies are: H( j0) = 1ej0 H( j1) = 0.707e−jπ/4 H( j∞) = 0e−jπ/2
n
Thus, the magnitude response is unity at = 0 and it decays as the frequency increases. The phase is zero at = 0, −π/4 at = 1, and −π/2 at → ∞. The magnitude response is even and the phase response is odd. Highpass filter: Consider then the output being the voltage across the resistor. Again by voltage division we obtain the transfer function of this circuit as H(s) =
Vr (s) CRs = Vs (s) CRs + 1
Again, let C = R = 1, so the frequency response is H( j) =
E Z() j = E 1 + j P()
E E The vector Z() goes from zero at the origin s = 0 to j in the j axis, and the vector P() goes from the pole s = −1 to j in the j axis. The vectors and the frequency response, at three different frequencies, are given by =0
E P(0) = 1ej0
=1
E P(1) =
=∞
√
E Z(0) = 0ejπ/2
2ejπ/4
E P(∞) = ∞ ejπ/2
H( j0) =
E Z(1) = 1ejπ/2
E Z(∞) = ∞ ejπ/2
E Z(0) = 0ejπ/2 E P(0)
H( j1) =
E Z(1) = 0.707ejπ/4 E P(1)
H( j∞) =
E Z(∞) = 1ej0 E P(∞)
E Thus, the magnitude response is zero at = 0 (this is due to the zero at s = 0, making Z(0) =0 as it is right on top of the zero), and it grows to unity as the frequency increases (at very high frequency, the lengths of the pole and the zero vectors are alike and so the magnitude response is unity and the phase response is zero). n Remarks n
Poles create “hills” at frequencies in the j axis in front of the poles imaginary parts. The closer the pole is to the j axis, the narrower and higher the hill. If, for instance, the poles are on the j axis (this would correspond to an unstable and useless filter) the frequency response at the frequency of the poles will be infinity.
339
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
n
Zeros create “valleys” at the frequencies in the j axis in front of the zeros imaginary parts. The closer the zero is to the j axis (from its left or its right, as the zeros are not restricted by stability to be in the open lefthand splane) the closer the frequency response is to zero. If the zeros are on the j axis, the frequency response at the frequency of the zeros is zero. Thus, poles produce frequency responses that look like hills (or like the main pole in a circus) around the frequencies of the poles, and zeros make the frequency response go to zero in the form of valleys around the frequencies of the zeros.
n Example 5.18 Use MATLAB to find and plot the poles and zeros and the corresponding magnitude and phase frequency responses of: (a) A secondorder bandpass filter and a highpass filter realized using a series connection of a resistor, an inductor, and a capacitor, each with unit resistance, inductance, and capacitance. Let the input be a voltage source vs (t) and initial conditions be zero. (b) An allpass filter with a transfer function H(s) =
s2 − 2.5s + 1 s2 + 2.5s + 1
Solution Our function freq resp s computes and plots the poles and the zeros of the filter transfer function and the corresponding frequency response (the function requests the coefficients of its numerator and denominator in decreasing order of powers of s). (a) As from a Example 5.16, the transfer functions of the bandpass and highpass secondorder filters are s Hbp (s) = 2 s +s+1 Hhp (s) =
s2 s2 + s + 1
The denominator in the two cases is exactly the same since the values of R, L, and C remain the same for the two filters—the only difference is in the numerator. To compute the frequency response of these filters and to plot their poles and zeros, we used the following script, which uses two functions: freqresp s, which we give below, and splane, which plots the poles and zeros. The coefficients of the numerator and the denominator correspond to the coefficients, from the highest to the lowest order of s, of the transfer function. %%%%%%%%%%%%%%%%%%%%% % Example 5.18Frequency response %%%%%%%%%%%%%%%%%%%%% n = [0 1 0]; % numerator coefficients  bandpass % n = [1 0 0]; % numerator coefficients  highpass d = [1 1 1]; % denominator coefficients
5.7 Convolution and Filtering
wmax = 10; % maximum frequency [w, Hm, Ha] = freqresp s(n, d, wmax); % frequency response splane(n, d) % plotting of poles and zeros
The following is the function freqresp s used to compute the magnitude and phase response of the filter with the given numerator and denominator coefficients. function [w, Hm, Ha] = freqresp s(b, a, wmax) w = 0:0.01:wmax; H = freqs(b, a, w); Hm = abs(H); % magnitude Ha = angle(H) ∗ 180/pi; % phase in degrees
Bandpass filter: Letting the output of the filter be the voltage across resistor, we find that the transfer function has a zero at zero, so that the frequency response is zero at = 0. When goes to infinity, one of the two poles cancels the zero effect so that the other pole makes the frequency response tend to zero. n Highpass filter: When the output of the filter is the voltage across the inductor the filter is high pass. In this case there is a double zero at s = 0, and the poles are located as before. Thus, when = 0 the magnitude response is zero due to the double zeros at zero, and when goes to infinity the effect of two poles and the two zeros cancel out giving a constant magnitude response, which corresponds to a highpass filter. The results for the bandpass and the highpass filters are shown in Figure 5.11. Notice that the frequency response of the bandpass and the highpass filter is determined by the ’number’ of zeros at the origin. The ’location’ of zeros, like in the allpass filter we consider next, also determines the frequency response. (b) Allpass filter: The poles and the zeros of an allpass filter have the same imaginary parts, but the negative of its real part. At any frequency in the jaxis the lengths of the vectors from the poles equal the length of the vectors from the zeros to the frequency in the j axis. Thus the magnitude response of the filter is unity. The following changes to the above script are needed for the allpass filter: n
clear all clf n = [1 −2.5 1]; d = [1 2.5 1]; wmax = 10; freq resp s(n, d, wmax)
The results are shown in Figure 5.12.
n
5.7.4 Spectrum Analyzer A spectrum analyzer is a device that measures the spectral characteristics of a signal. It can be implemented as a bank of narrow band bandpass filters with fixed bandwidths covering the desired frequencies (see Figure 5.13). The power at the output of each filter is computed and displayed at the corresponding center frequency. Another possible implementation is using a bandpass filter with an adjustable center frequency, with the power in its bandwidth being computed and displayed [16].
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
Magnitude Response
Phase Response 50
0.8
∠H( j Ω)
H(j Ω)
1 0.6 0.4 0
0 −50
0.2 0
5 Ω
0
10
5 Ω
10
Poles/Zeros
jΩ
1 0 −1 −1
−0.5
0
0.5
σ (a)
Magnitude Response
Phase Response
1
150 ∠H( j Ω)
H( j Ω)
0.5
100 50
0
0
5 Ω
0
10
0
5 Ω
10
Poles/Zeros 1 jΩ
342
0 −1 −1
−0.5
0
0.5
σ (b)
FIGURE 5.11 Frequency response and poles/zeros location of (a) bandpass and (b) highpass RLC filters.
5.7 Convolution and Filtering
Phase Response
Magnitude Response 1 100 ∠H(j ω)
H(j ω)
0.8 0.6 0.4
−100
0.2 0
0
0
5 Ω
10
0
5 Ω
10
Poles/Zeros
1
jΩ
0.5 0 −0.5 −1
−2
0 σ
2
FIGURE 5.12 Frequency response and poles/zeros location of the allpass filter.
LPF
Power measurement
Px (0)
BPF1
Power measurement
Px (Ω1)
. . .
. . .
BPFN
Power measurement
x (t )
FIGURE 5.13 Bankoffilter spectrum analyzer. LPF stands for lowpass filter, and BPFi corresponds to bandpass filters, i = 1, . . . , N.
Px (ΩN)
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
If the input of the spectrum analyzer is x(t), the output of either the fixed or the adjustablebandpass filters in the implementations—assumed to have a very narrow bandwidth 1—would be 1 y(t) = 2π ≈
0 +0.51 Z
X()ejt d
0 −0.51
1 1 X(0 )ej0 t 2π
Computing the mean square of this signal we get Z 1 2 1 2 y(t) dt = X(0 )2 T 2π T
which is proportional to the power or the energy of the signal in 0 ± 1. A similar computation can be done at each of the frequencies of the input signal. Remarks n n
The bankoffilter spectrum analyzer is used for the audio range of the spectrum. Radio frequency spectrum analyzers resemble an AM demodulator. It usually consists of a single narrowband intermediate frequency (IF) bandpass filter fed by a mixer. The local oscillator sweeps across the desired band, and the power at the output of the filter is computed and displayed on a monitor.
5.8 ADDITIONAL PROPERTIES We consider now some additional properties of the Fourier transform, some of which look like those of the Laplace transform when s = j and some are different.
5.8.1 Time Shifting If x(t) has a Fourier transform X(), then x(t − t0 ) ⇔ X()e−jt0 x(t + t0 ) ⇔ X()ejt0
The Fourier transform of x(t − t0 ) is F[x(t − t0 )] =
Z∞
x(t − t0 )e−jt dt
−∞
Z∞ =
x(τ )e−j(τ +t0 ) dτ = e−jt0 X()
−∞
where we changed the variable to τ = t − t0 . Likewise for x(t + t0 ).
(5.26)
5.8 Additional Properties
It is important to realize that shifting in time does not change the frequency content of the signal— that is, the signal does not change when delayed or advanced. This is clear when we see that the magnitude of the two transforms, corresponding to the original and the shifted signals, is the same, X() = X()e±jt0  and the effect of the time shift is only in the phase spectrum. n Example 5.19 Consider the signal x(t) = A[δ(t − τ ) + δ(t + τ )] Find its Fourier transform X(). Use this Fourier pair and the duality property to verify the Fourier transform of a cos(0 t) obtained before. Solution Applying the timeshift property, we have X() = A[1e−jτ + 1ejτ ] = 2A cos(τ ) giving the Fourier transform pair x(t) = A[δ(t − τ ) + δ(t + τ )]
⇔
X() = 2A cos(τ )
Using the duality property, we then have X(t) = 2A cos(tτ )
2πx(−) = 2πA[δ(− − τ ) + δ(− + τ )]
⇔
Let τ = 0 , Use the evenness of δ(t) to get A cos(0 t)
⇔
πA[δ( + 0 ) + δ( − 0 )]
n
n Example 5.20 Consider computing the Fourier transform of y(t) = sin(0 t) using the Fourier transform of the cosine signal x(t) = cos(0 t) we just found. Solution Since y(t) = cos(0 t − π/2) = x(t − π/(20 )), applying the timeshifting property, we then get F[sin(0 t)] = F[x(t − π/20 )] = π[δ( − 0 ) + δ( + 0 )]e−jπ/(20 ) = πδ( − 0 )e−jπ/2 + π δ( + 0 )ejπ/2 = −jπδ( − 0 ) + jπ δ( + 0 )
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
after applying the sifting property of δ(). The above shows that this Fourier transform is different from the one for the cosine in the phase only. n
5.8.2 Differentiation and Integration If x(t), −∞ < t < ∞, has a Fourier tranform X(), then dN x(t) dtN Zt
x(σ )dσ
( j)N X()
(5.27)
X() + πX(0)δ() j
(5.28)
⇔
⇔
−∞
where Z∞ X(0) =
x(t)dt −∞
From the inverse Fourier transform given by 1 x(t) = 2π
Z∞
X()ejt d
−∞
we then have that Z∞
dx(t) 1 = dt 2π
X()
d ejt d dt
−∞
Z∞
1 = 2π
[X()j] ejt d
−∞
indicating that dx(t) ⇔ jX() dt and similarly for higher derivatives. The proof of the integration property can be done in two parts: 1. The convolution of u(t) and x(t) gives the integral—that is Zt −∞
x(τ )dτ =
Z∞ −∞
x(τ )u(t − τ )dτ = [x ∗ u](t)
5.8 Additional Properties
since u(t − τ ) as a function of τ equals u(t − τ ) =
1 τ t
We thus have that
Zt
F
x(τ )dτ = X()F[u(t)]
(5.29)
−∞
2. Since the unitstep signal is not absolutely integrable its Fourier transform cannot be found from the integral definition, and we cannot use its Laplace transform either because its ROC does not include the j axis. Let’s transform it into an absolutely integrable signal by subtracting 1/2 and multiplying the result by 2. This gives the sign signal: 1 t>0 sgn(t) = 2[u(t) − 0.5] = −1 t < 0 The derivative of sgn(t) is dsgn(t) = 2δ(t) dt and thus if S() = F[sgn(t)], S() =
2 j
using the derivative property. The linearity of the Fourier transform applied to the definition of sgn(t) gives F[sgn(t)] = 2F[u(t)] − 2πδ()
⇒ F[u(t)] =
1 + π δ() j
(5.30)
Replacing the Fourier transform of u(t) in Equation (5.29), we get t Z 1 F x(τ )dτ = X() + π δ() j −∞
=
X() + πX(0)δ() j
(5.31)
Remarks n
Just like in the Laplace transform where the operator s corresponds to the derivative operation in time of the signal, in the Fourier transform j becomes the corresponding operator for the derivative operation in time of the signal.
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
n
If X(0) (i.e., the dc value of X()) is zero, then the operator 1/( j) corresponds to integration in time of x(t), just like 1/s in the Laplace domain. For X(0) to be zero the integral of the signal from −∞ to ∞ must be zero.
n Example 5.21 Suppose a system is represented by a secondorder differential equation with constant coefficients: 2y(t) + 3
dy(t) d2 y(t) + = x(t) dt dt2
and that the initial conditions are zero. Let x(t) = δ(t). Find y(t). Solution Computing the Fourier transform of this equation, we get [2 + 3j + ( j)2 ]Y() = X() Replacing X() = 1 and solving for Y(), we have 1 2 + 3j + ( j)2 1 = ( j + 1)( j + 2) 1 −1 = + ( j + 1) ( j + 2)
Y() =
and the inverse Fourier transform of these terms gives y(t) = [e−t − e−2t ]u(t)
n Example 5.22 Find the Fourier transform of the triangular pulse x(t) = r(t) − 2r(t − 1) + r(t − 2) which is piecewise linear, using the derivative property. Solution A first derivative gives dx(t) = u(t) − 2u(t − 1) + u(t − 2) dt
n
5.8 Additional Properties
and a second derivative gives d2 x(t) = δ(t) − 2δ(t − 1) + δ(t − 2) dt2 Using the timeshift and the derivative properties, we get from the expression for the second derivative and letting X() be the Fourier transform of x(t): ( j)2 X() = 1 − 2e−j + e−j2 = e−j [ej − 2 + e−j ] so that X() =
2e−j [1 − cos()] 2
n
n Example 5.23 Consider the integral Zt y(t) =
x(τ )dτ − ∞ < t < ∞
−∞
where x(t) = u(t + 1) − u(t − 1). Find the Fourier transform Y() directly and from the integration property. Solution The integral is t < −1 0 y(t) = t + 1 −1 ≤ t < 1 2 t≥1 or y(t) = [r(t + 1) − r(t − 1) − 2u(t − 1)] +2u(t − 1)  {z } y1 (t)
The Fourier transform of y1 (t) is e s − e−s 2e−s Y1 () = − s2 s
= s=j
−2j sin() 2e−j + j 2
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
The Fourier transform of 2u(t − 1) is −2je−j / + 2π δ() so that Y() =
−2j sin() 2e−j 2e−j +j −j + 2π δ() 2
−2j sin() + 2π δ() 2 To use the integration property we first need X(), which is =
X() =
2 sin()
and according to the property, Y() = =
X() + πX(0)δ() j −2j sin() + 2π δ() 2
ˆ since X(0) = 2 (using L’Hopital’s rule). As expected, the two results coincide.
n
5.9 WHAT HAVE WE ACCOMPLISHED? WHAT IS NEXT? You should by now have a very good understanding of the frequency representation of signals and systems. In this chapter, we have unified the treatment of periodic and nonperiodic signals and their spectra, and consolidated the concept of frequency response of a linear timeinvariant system. Basic properties of the Fourier transform and important Fourier pairs are given in Tables 5.1 and 5.2. Two significant applications are in filtering and communications. We introduced the basics of filtering in this chapter and will expand on them in Chapter 6. The fundamentals of modulation provided in this chapter will be illustrated in Chapter 6 where we will consider their application in communications. Certainly the next step is to find out where the Laplace and the Fourier analyses apply, which will be done in Chapter 6. After that, we will go into discretetime signals and systems. We will show that sampling, quantization, and coding bridge the continuoustime and the digital signal processing, and that transformations similar to the Laplace and the Fourier transforms will permit us to do processing of discrete–time signals and systems.
PROBLEMS 5.1. Fourier series versus Fourier transform—MATLAB The connection between the Fourier series and the Fourier transform can be seen by considering what happens when the period of a periodic signal increases to a point at which the periodicity is not clear as only one period is seen. Consider a train of pulses x(t) with a period T0 = 2, and a period of x(t) is x1 (t) = u(t + 0.5) − u(t − 0.5). Let T0 be increased to 4, 8, and 16.
Problems
Table 5.1 Basic Properties of the Fourier Transform Time Domain
Frequency Domain
Signals and constants
x(t), y(t), z(t), α, β
X(), Y(), Z()
Linearity
αx(t) + βy(t)
Expansion/contraction in time
x(αt), α 6= 0
αX() + βY() 1 α X α
x(−t) R∞
Reflection Parseval’s energy relation
Ex =
Duality
X(t)
Time differentiation
dn x(t) dtn ,
X(−) 1 2π
R∞
2 −∞ x(t) dt
Ex =
n ≥ 1, integer
( j)n X()
2 −∞ X() d
2π x(−)
Integration
−jtx(t) Rt 0 0 −∞ x(t )dt
Time shifting
x(t − α)
dX() d X() j + π X(0)δ() e−jα X()
Frequency shifting
ej0 t x(t)
X( − 0 )
Modulation Periodic signals
x(t) cos(c t) P x(t) = k Xk ejk0 t
0.5[X( − c ) + X( + c )] P X() = k 2π Xk δ( − k0 )
Symmetry
x(t) real
X() = X(−)
Convolution in time
z(t) = [x ∗ y](t)
Z() = X()Y()
Windowing/multiplication
x(t)y(t)
1 2π [X
Cosine transform
x(t) even
Sine transform
x(t) odd
Frequency differentiation
∠X() = −∠X(−) ∗ Y]() R∞ X() = −∞ x(t) cos(t)dt, real R∞ X() = −j −∞ x(t) sin(t)dt, imaginary
Table 5.2 Fourier Transform Pairs Function of Time
Function of
1
δ(t)
1
2
δ(t − τ )
e−jτ
3
u(t)
4
u(−t)
5
sgn(t) = 2[u(t) − 0.5]
1 j + πδ() −1 j + πδ() 2 j
6
A, −∞ < t < ∞
2π Aδ()
7
Ae−at u(t),
8
Ate−at u(t),
9
e−at ,
a>0 a>0
a>0
A j+a A ( j+a)2 2a a2 +2
10
cos(0 t), −∞ < t < ∞
π [δ( − 0 ) + δ( + 0 )]
11
sin(0 t), −∞ < t < ∞
−jπ [δ( − 0 ) − δ( + 0 )]
12
A[u(t + τ ) − u(t − τ )], τ > 0
) 2Aτ sin(τ τ
13
sin(0 t) πt
14
x(t) cos(0 t)
u( + 0 ) − u( − 0 ) 0.5[X( − 0 ) + X( + 0 )]
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
(a) Find the Fourier series coefficient X0 for each of the values of T0 and indicate how it changes for the different values of T0 . (b) Find the Fourier series coefficients for x(t) and carefully plot the magnitude line spectrum for each of the values of T0 . Explain what is happening in these spectra. (c) If you were to let T0 be very large, what would you expect to happen to the Fourier coefficients? Explain. (d) Write a MATLAB script that simulates the conversion from the Fourier series to the Fourier transform of a sequence of rectangular pulses as the period is increased. The Fourier coefficients need to be multiplied by the period so that they do not become insignificant. Plot using stem the magnitude line spectrum for pulse sequences with periods T0 from 4 to 62. 5.2. Fourier transform from Laplace transform—MATLAB The Fourier transform of finitesupport signals, which are absolutely integrable or finite energy, can be obtained from their Laplace transform rather than doing the integral. Consider the following signals: x1 (t) = u(t + 0.5) − u(t − 0.5) x2 (t) = sin(2π t)[u(t) − u(t − 0.5)] x3 (t) = r(t + 1) − 2r(t) + r(t − 1) (a) Plot each of the signals. (b) Find the Fourier transforms {Xi ()} for i = 1, 2, and 3 using the Laplace transform. (c) Use MATLAB’s symbolic function fourier to compute the Fourier transform of the given signals. Plot the magnitude spectrum corresponding to each of the signals. 5.3. Fourier transform from Laplace transform of infinitesupport signals—MATLAB For signals with infinite support, their Fourier transforms cannot be derived from the Laplace transform unless they are absolutely integrable or the region of convergence of the Laplace transform contains the j axis. Consider the signal x(t) = 2e−2t . (a) Plot the signal x(t) for −∞ < t < ∞. (b) Use the evenness of the signal to find the integral Z∞ x(t)dt −∞
and determine whether this signal is absolutely integrable or not. (c) Use the integral definition of the Fourier transform to find X(). (d) Use the Laplace transform of x(t) to verify the above found Fourier transform. (e) Use MATLAB’s symbolic function fourier to compute the Fourier transform of x(t). Plot the magnitude spectrum corresponding to x(t). 5.4. Fourier and Laplace transforms—MATLAB Consider the signal x(t) = 2e−2t cos(2π t)u(t). (a) Use the fact this signal is bounded by the exponential ±2e−2t u(t) to show that the integral Z∞ x(t)dt −∞
Problems
is finite, indicating the signal is absolutely integrable and also finite energy. (b) Use the Laplace transform to find the Fourier transform X() of x(t). (c) Use the MATLAB function fourier to compute the magnitude and phase spectrum of X(). 5.5. Fourier transform of causal signals Any causal signal x(t) having a Laplace transform with poles in the openleft splane (i.e., not including the j axis) has, as we saw before, a region of convergence that includes the j axis, and as such its Fourier transform can be found from its Laplace transform. Consider the following signals: x1 (t) = e−2t u(t) x2 (t) = r(t) x3 (t) = x1 (t)x2 (t) (a) Determine the Laplace transform of the above signals (use properties of the Laplace transform) indicating the corresponding region of convergence. (b) Determine for which of these signals you can find its Fourier transform from its Laplace transform. Explain. (c) Give the Fourier transform of the signals that can be obtained from their Laplace transform. 5.6. Duality of Fourier transform There are some signals for which the Fourier transforms cannot be found directly by either the integral definition or the Laplace transform, and for those we need to use the properties of the Fourier transform, in particular the duality property. Consider, for instance, sin(t) t
x(t) =
or the sinc signal. Its importance is that it is the impulse response of an ideal lowpass filter. (a) Let X() = A[u( + 0 ) − u( − 0 ] be a possible Fourier transform of x(t). Find the inverse Fourier transform of X() using the integral equation to determine the values of A and 0 . (b) How could you use the duality property of the Fourier transform to obtain X()? Explain. 5.7. Cosine and sine transforms The Fourier transforms of even and odd functions are very important. The reason is that they are computationally simpler than the Fourier transform. Let x(t) = e−t and y(t) = e−t u(t) − et u(−t). (a) Plot x(t) and y(t), and determine whether they are even or odd. (b) Show that the Fourier transform of x(t) is found from Z∞ x(t) cos(t)dt
X() = −∞
which is a real function of , thus its computational importance. Show that X() is also even as a function of . (c) Find X() from the above equation (called the cosine transform). (d) Show that the Fourier transform of y(t) is found from Z∞ Y() = −j −∞
y(t) sin(t)dt
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
which is imaginary function of , thus its computational importance. Show that Y() is also odd as a function of . (e) Find Y() from the above equation (called the sine transform). Verify that your results are correct by finding the Fourier transform of z(t) = x(t) + y(t) directly and using the above results. (f) What advantages do you see to the cosine and sine transforms? How would you use the cosine and the sine transforms to compute the Fourier transform of any signal, not necessarily even or odd? Explain. 5.8. Time versus frequency—MATLAB The supports in time and in frequency of a signal x(t) and its Fourier transform X() are inversely proportional. Consider a pulse x(t) =
1 [u(t) − u(t − T0 )] T0
(a) Let T0 = 1 and T0 = 10 and find and compare the corresponding X(). (b) Use MATLAB to simulate the changes in the magnitude spectrum when T0 = 10k for k = 0, . . . , 4 for x(t). Compute X() and plot its magnitude spectra for the increasing values of T0 on the same plot. Explain the results. 5.9. Smoothness and frequency content—MATLAB The smoothness of the signal determines the frequency content of its spectrum. Consider the signals x(t) = u(t + 0.5) − u(t − 0.5) y(t) = (1 + cos(π t))[u(t + 0.5) − u(t − 0.5)] (a) Plot these signals. Can you tell which one is smoother? (b) Find X() and carefully plot its magnitude X() versus frequency . (c) Find Y() (use the Fourier transform properties) and carefully plot its magnitude Y() versus frequency . (d) Which one of these two signals has higher frequencies? Can you now tell which of the signals is smoother? Use MATLAB to decide. Make x(t) and y(t) have unit energy. Plot 20 log10 Y() and 20 log10 X() using MATLAB and see which of the spectra shows lower frequencies. 5.10. Smoothness and frequency—MATLAB Let the signals x(t) = r(t + 1) − 2r(t) + r(t − 1) and y(t) = dx(t)/dt. (a) Plot x(t) and y(t). (b) Find X() and carefully plot its magnitude spectrum. Is X() real? Explain. (c) Find Y() (use properties of Fourier transform) and carefully plot its magnitude spectrum. Is Y() real? Explain. (d) Determine from the above spectra which of these two signals is smoother. Use MATLAB to plot 20 log10 Y() and 20 log10 X() and decide. Would you say in general that computing the derivative of a signal generates high frequencies or possible discontinuities? 5.11. Integration and smoothing—MATLAB Consider the signal x(t) = u(t + 1) − 2u(t) + u(t − 1) and let Zt y(t) = −∞
x(τ )dτ
Problems
(a) Plot x(t) and y(t). (b) Find X() and carefully plot its magnitude spectrum. Is X() real? Explain. (Use MATLAB to do the plotting.) (c) Find Y() and carefully plot its magnitude spectrum. Is Y() real? Explain. (Use MATLAB to do the plotting.) (d) Determine from the above spectra which of these two signals is smoother. Use MATLAB to decide. Would you say that in general by integrating a signal you get rid of higher frequencies, or smooth out a signal? 5.12. Duality of Fourier transforms As indicated by the derivative property, if we multiply a Fourier transform by ( j)N , it corresponds to computing an Nth derivative of its time signal. Consider the dual of this property—that is, if we compute the derivative of X(), what would happen to the signal in the time? (a) Let x(t) = δ(t − 1) + δ(t + 1). Find its Fourier transform (using properties) X(). (b) Compute dX()/d and determine its inverse Fourier transform. 5.13. Periodic functions in frequency The duality property provides interesting results. Consider the signal x(t) = δ(t + T1 ) + δ(t − T1 ) (a) Find X() = F[x(t)] and plot both x(t) and X(). (b) Suppose you then generate a signal y(t) = δ(t) +
∞ X
[δ(t + kT0 ) + δ(t − kT0 )]
k=1
Find its Fourier transform Y() and plot both y(t) and Y(). (c) Are y(t) and the corresponding Fourier transform Y() periodic in time and in frequency? If so, determine their periods. 5.14. Sampling signal The sampling signal δTs (t) =
∞ X
δ(t − nTs )
n=−∞
will be important in the sampling theory later on. (a) As a periodic signal of period Ts , express δTs (t) by its Fourier series. (b) Determine then the Fourier transform 1() = F[δTs (t)]. (c) Plot δTs (t) and 1() and comment on the periodicity of these two functions. 5.15. Piecewise linear signals The derivative property can be used to simplify the computation of some Fourier transforms. Let x(t) = r(t) − 2r(t − 1) + r(t − 2) (a) Find and plot the second derivative with respect to t of x(t), or y(t) = d2 x(t)/dt2 . (b) Find X() from Y() using the derivative property. (c) Verify the above result by computing the Fourier transform X() directly from x(t) using the Laplace transform.
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CH A P T E R 5: Frequency Analysis: The Fourier Transform
5.16. Periodic signalequivalent representations Applying the time and frequency shifts it is possible to get different but equivalent Fourier transforms of periodic signals. Assume a period of a periodic signal x(t) of period T0 is x1 (t), so that x(t) =
X
x1 (t − kT0 )
k
and as seen in Chapter 4 the Fourier series coefficients of x(t) are found as Xk = X1 ( jk0 )/T0 , so that x(t) can also be represented as x(t) =
1 X X1 ( jk0 )ejk0 t T0 k
(a) Find the Fourier transform of the first expression given above for x(t) using the timeshift property. (b) Find the Fourier transform of the second expression for x(t) using the frequencyshift property. (c) Compare the two expressions and comment on your results. 5.17. Modulation property Consider the raisedcosine pulse x(t) = [1 + cos(π t)](u(t + 1) − u(t − 1))
(a) Carefully plot x(t). (b) Find the Fourier transform of the pulse p(t) = u(t + 1) − u(t − 1). (c) Use the definition of the pulse p(t) and the modulation property to find the Fourier transform of x(t) in terms of P() = F[p(t)]. 5.18. Solution of differential equations An analog averager is characterized by the relationship dy(t) = 0.5[x(t) − x(t − 2)] dt where x(t) is the input and y(t) the output. If x(t) = u(t) − 2u(t − 1) + u(t − 2): (a) Find the Fourier transform of the output Y(). (b) Find y(t) from Y(). 5.19. Generalized AM Consider the following generalization of amplitude modulation where instead of multiplying by a cosine we multiply by a periodic signal with harmonic frequencies much higher than those of the message. Suppose the carrier c(t) is a periodic signal with fundamental frequency 0 , let’s say c(t) =
6 X
2 cos(k0 t)
k=4
and that the message is a sinusoid of frequency 0 = 2π , or x(t) = cos(0 t). (a) Find the AM signal s(t) = x(t)c(t). (b) Determine the Fourier transform S(). (c) What would be a possible advantage of this generalized AM system? Explain.
Problems
5.20. Filter for halfwave rectifier Suppose you want to design a dc source using a halfwave rectified signal x(t) and an ideal filter. Let x(t) be periodic, T0 = 2, and with a period x1 (t) =
sin(π t) 0
0≤t≤1 1 < t ≤ 2,
(a) Find the Fourier transform X() of x(t), and plot the magnitude spectrum including the dc and the first three harmonics. (b) Determine the magnitude and cutoff frequency of an ideal lowpass filter H( j) such that when we have x(t) as its input, the output is y(t) = 1. Plot the magnitude response of the ideal lowpass filter. (For simplicity assume the phase is zero.) 5.21. Passive RLC filters—MATLAB Consider an RLC series circuit with a voltage source vs (t). Let the values of the resistor, capacitor, and inductor be unity. Plot the poles and zeros and the corresponding frequency responses of the filters with the output the voltage across the (a) Capacitor (b) Inductor (c) Resistor Indicate the type of filter obtained in each case. Use MATLAB to plot the poles and zeros, the magnitude, and the phase response of each of the filters obtained above. 5.22. AM modulation and demodulation A pure tone x(t) = 4 cos(1000t) is transmitted using an AM communication system with a carrier cos(10,000t). The output of the AM system is y(t) = x(t) cos(10,000t) At the receiver, to recover x(t) the sent signal y(t) needs first to be separated from the thousands of other signals. This is done with a bandpass filter with a center frequency equal to the carrier frequency, and the output of this filter then needs to be demodulated. (a) Consider an ideal bandpass filter H( j). Let its phase be zero. Determine its bandwidth, center frequency, and amplitude so we get as its output 10y(t). Plot the spectrum of x(t), 10y(t), and the magnitude frequency response of H( j). (b) To demodulate 10y(t), we multiply it by cos(10,000t). You need then to pass the resulting signal through an ideal lowpass filter to recover the original signal x(t). Plot the spectrum of z(t) = 10y(t) cos(10,000t) and from it determine the frequency response of the lowpass filter G( j) needed to recover x(t). Plot the magnitude response of G( j). 5.23. Ideal lowpass filter—MATLAB Consider an ideal lowpass filter H(s) with zero phase and magnitude response H( j) =
1 0
−π ≤ ≤ π otherwise
(a) Find the impulse response h(t) of the lowpass filter. Plot it and indicate whether this filter is a causal system or not.
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(b) Suppose you wish to obtain a bandpass filter G( j) from H( j). If the desired center frequency of G( j) is 5π , and its desired magnitude is 1 at the center frequency, how would you process h(t) to get the desired filter? Explain your procedure. (c) Use symbolic MATLAB to find h(t), g(t), and G( j). Plot H( j), h(t), g(t), and G( j). 5.24. Magnitude response from poles and zeros—MATLAB Consider the following filters with the given poles and zeros and dc constant: H1 (s) :
K = 1 poles
p1 = −1, p2,3 = −1 ± jπ; zeros
z1 = 1, z2,3 = 1 ± jπ
H2 (s) :
K = 1 poles
p1 = −1, p2,3 = −1 ± jπ; zeros
z1,3 = ±jπ
H3 (s) :
K = 1 poles
p1 = −1, p2,3 = −1 ± jπ; zero
z1 = 1
Use MATLAB to plot the magnitude responses of these filters and indicate the type of filters they are. 5.25. Different types of AM modulations—MATLAB Let the signal m(t) = sin(2π t)[u(t) − u(t − 1)] be the message or input to different types of AM systems with the output the following signals. Carefully plot m(t) and the following outputs in 0 ≤ t ≤ 1 and their corresponding spectra using MATLAB. Let the sampling period be Ts = 0.001. (a) y1 (t) = m(t) cos(20π t) (b) y2 (t) = [1 + m(t)] cos(20πt) 5.26. Windows—MATLAB The signal x(t) in Problem 5.17 is called a raisedcosine window. Notice that it is a very smooth signal and that it decreases at both ends. The rectangular window is the signal y(t) = u(t + 1) − u(t − 1). (a) Use MATLAB to compute the magnitude spectrum of x(t) and y(t) and indicate which is the smoother of the two by considering the presence of high frequencies as an indication of roughness. (b) When computing the Fourier transform of a very long signal it makes sense to break it up into smaller sections and compute the Fourier transform of each. In such a case, windows are used to smooth out the transition from one section to the other. Consider a sinusoid z(t) = cos(2π t) for 0 ≤ t ≤ 1000 sec. Divide the signal into two sections of duration 500 sec. Multiply the corresponding signal in each of the sections by a raisedcosine x(t) and rectangular y(t) windows of length 500 and compute using MATLAB the corresponding Fourier transforms. Compare them to the Fourier transform of the whole signal and comment on your results. Sample all the signals using Ts = 1/(4π) as the sampling period. (c) Consider the computation of the Fourier transform of the acoustic signal corresponding to a train whistle, which MATLAB provides as a sampled signal in “train.mat” using the discrete approximation of the Fourier transform. The frequency content of the whole signal (hard to find) would not be as meaningful as the frequency content of a smaller section of it as they change with time. Compute the Fourier transform of sections of 1000 samples by windowing the signal with the raisedcosine window (sampled with the same sampling period as the “train.mat” signal or Ts = 1/Fs where Fs is the sampling frequency given for “train.mat”). Plot the spectra of a few of these segments and comment on the change in the frequency content as time changes.
CHAPTER 6
Application to Control and Communications
Who are you going to believe? Me or your own eyes. Julius “Groucho” Marx (1890–1977) comedian and actor
6.1 INTRODUCTION Control and communications are areas in electrical engineering where the Laplace and the Fourier analyses apply. In this chapter, we illustrate how these transform methods and the concepts of transfer function, frequency response, and spectrum connect with the classical theories of control and communications. In classical control, the objective is to change the dynamics of a given system to be able to achieve a desired response by frequencydomain methods. This is typically done by means of a feedback connection of a controller to a plant. The plant is a system such as a motor, a chemical plant, or an automobile we would like to control so that it responds in a certain way. The controller is a system we design to make the plant follow a prescribed input or reference signal. By feeding back the response of the system to the input, it can be determined how the plant responds to the controller. The commonly used negative feedback generates an error signal that permits us to judge the performance of the controller. The concepts of transfer function, stability of systems, and different types of responses obtained through the Laplace transform are very useful in the analysis and design of classical control systems. A communication system consists of three components: a transmitter, a channel, and a receiver. The objective of communications is to transmit a message over a channel to a receiver. The message is a signal, for instance, a voice or a music signal, typically containing low frequencies. Transmission of the message can be done over the airwaves or through a line connecting the transmitter to the receiver, or a combination of the two—constituting channels with different characteristics. Telephone communication can be done with or without wires, and radio and television are wireless. The concepts of Signals and Systems Using MATLAB®. DOI: 10.1016/B9780123747167.000090 c 2011, Elsevier Inc. All rights reserved.
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frequency, bandwidth, spectrum, and modulation developed by means of the Fourier transform are fundamental in the analysis and design of communication systems. The aim of this chapter is to serve as an introduction to problems in classical control and communications and to link them with the Laplace and Fourier analyses. More indepth discussion of these topics can be found in many excellent texts in control and communications. The other topic covered in this chapter is an introduction to analog filter design. Filtering is a very important application of LTI systems in communications, control, and digital signal processing. The material in this chapter will be complemented by the design of discrete filters in Chapter 11. Important issues related to signals and system are illustrated in the design and implementation of filters.
6.2 SYSTEM CONNECTIONS AND BLOCK DIAGRAMS Control and communication systems consist of interconnection of several subsystems. As we indicated in Chapter 2, there are three important connections of LTI systems: Cascade Parallel Feedback
n n n
Cascade and parallel result from properties of the convolution integral, while the feedback connection relates the output of the overall system to its input. With the background of the Laplace transform we present now a transform characterization of these connections that can be related to the timedomain characterizations given in Chapter 2. The connection of two LTI continuoustime systems with transfer functions H1 (s) and H2 (s) (and corresponding impulse responses h1 (t) and h2 (t)) can be done in: n Cascade (Figure 6.1(a)): Provided that the two systems are isolated, the transfer function of the overall system is H(s) = H1 (s)H2 (s) n
Parallel (Figure 6.1(b)): The transfer function of the overall system is H(s) = H1 (s) + H2 (s)
n
(6.1)
(6.2)
Negative feedback (Figure 6.4): The transfer function of the overall system is H(s) =
n n
H1 (s) 1 + H2 (s)H1 (s)
Openloop transfer function: Ho` (s) = H1 (s). Closedloop transfer function: Hc` (s) = H(s).
(6.3)
6.2 System Connections and Block Diagrams
Cascading of LTI Systems Given two LTI systems with transfer functions H1 (s) = L[h1 (t)] and H2 (s) = L[h2 (t)] where h1 (t) and h2 (t) are the corresponding impulse responses of the systems, the cascading of these systems gives a new system with transfer function H(s) = H1 (s)H2 (s) = H2 (s)H1 (s) provided that these systems are isolated from each other (i.e., they do not load each other). A graphical representation of the cascading of two systems is obtained by representing each of the systems with blocks with their corresponding transfer function (see Figure 6.1(a)). Although cascading of systems is a simple procedure, it has some disadvantages: n n
It requires isolation of the systems. It causes delay as it processes the input signal, possibly compounding any errors in the processing.
Remarks n
Loading, or lack of system isolation, needs to be considered when cascading two systems. Loading does not allow the overall transfer function to be the product of the transfer functions of the connected systems. Consider the cascade connection of two resistive voltage dividers (Figure 6.2), each with a simple transfer function Hi (s) = 1/2, i = 1, 2. The cascade in Figure 6.2(b) clearly will not have as transfer function H(s) = H1 (s)H2 (s) = (1/2)(1/2) unless we include a buffer (such as an operational amplifier voltage
H1(s) +
X(s) H1(s)
y (t)
x (t )
Y(s)
X(s)
Y(s)
H2(s)
H 2(s) y(t )
x (t) (a)
(b)
FIGURE 6.1 (a) Cascade and (b) parallel connections of systems with transfer function H1 (s) and H2 (s). The input and output are given in the time or in the frequency domains. 1Ω
+ −
+ V0(s)
1Ω
1Ω
1Ω
1Ω
+
+ −
−
1Ω
V1(s)
V0(s)
1Ω
1Ω
+ V2(s) −
− (a)
(b)
FIGURE 6.2 Cascading of two voltage dividers: (a) using a voltage follower gives V1 (s)/V0 (s) = (1/2)(1/2) with no loading effect, and (b) using no voltage follower V2 (s)/V0 (s) = 1/5 6= V1 (s)/V0 (s) due to loading.
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Modulator
x (t )
FIGURE 6.3 Cascading of (a) an LTV and (b) an LTI system. The outputs are different, y1 (t) 6= y2 (t).
n
Modulator d dt
×
y2 (t )
x(t )
d dt
×
f (t )
y1(t )
f (t ) (a)
(b)
follower) in between (see Figure 6.2(a)). The cascading of the two voltage dividers without the voltage follower gives a transfer function H1 (s) = 1/5, as can be easily shown by doing mesh analysis on the circuit. The block diagrams of the cascade of two or more LTI systems can be interchanged with no effect on the overall transfer function, provided the connection is done with no loading. That is not true if the systems are not LTI. For instance, consider cascading a modulator (LTV system) and a differentiator (LTI) as shown in Figure 6.3. If the modulator is first, Figure 6.3(a), the output of the overall system is y2 (t) =
dx(t)f (t) dx(t) df (t) = f (t) + x(t) dt dt dt
while if we put the differentiator first, Figure 6.3(b), the output is y1 (t) = f (t)
dx(t) dt
It is obvious that if f (t) is not a constant, the two responses are very different.
Parallel Connection of LTI Systems According to the distributive property of the convolution integral, the parallel connection of two or more LTI systems has the same input and its output is the sum of the outputs of the systems being connected (see Figure 6.1(b)). The parallel connection is better than the cascade, as it does not require isolation between the systems, and reduces the delay in processing an input signal. The transfer function of the parallel system is H(s) = H1 (s) + H2 (s) Remarks n
n
Although a communication system can be visualized as the cascading of three subsystems—the transmitter, the channel, and the receiver—typically none of these subsystems is LTI. As we discussed in Chapter 5, the lowfrequency nature of the message signals requires us to use as the transmitter a system that can generate a signal with much higher frequencies, and that is not possible with LTI systems (recall the eigenfunction property). Transmitters are thus typically nonlinear or linear time varying. The receiver is also not LTI. A wireless channel is typically time varying. Some communication systems use parallel connections (see quadrature amplitude modulation (QAM) later in this chapter). To make it possible for several users to communicate over the same channel, a combination of parallel and cascade connections are used (see frequency division multiplexing (FDM) systems later in this chapter). But again, it should be emphasized that these subsystems are not LTI.
6.3 Application to Classic Control
x(t ) +
FIGURE 6.4 Negativefeedback connection of systems with transfer function H1 (s) and H2 (s). The input and the output are x(t) and y(t), respectively, and e(t) is the error signal.
e (t )
H1(s)
y (t )
−
H2(s)
Feedback Connection of LTI Systems In control, feedback connections are more appropriate than cascade or parallel connections. In the feedback connection, the output of the first system is fed back through the second system into the input (see Figure 6.4). In this case, like in the parallel connection, beside the blocks representing the systems we use adders to add/subtract two signals. It is possible to have positive or negativefeedback systems depending on whether we add or subtract the signal being fed back to the input. Typically, negative feedback is used, as positive feedback can greatly increase the gain of the system. (Think of the screeching sound created by an open microphone near a loudspeaker: the microphone continuously picks up the amplified sound from the loudspeaker, increasing the volume of the produced signal. This is caused by positive feedback.) For negative feedback, the connection of two systems is done by putting one in the feedforward loop, H1 (s), and the other in the feedback loop, H2 (s) (there are other possible connections). To find the overall transfer function we consider the Laplace transforms of the error signal e(t), E(s), and of the output y(t), Y(s), in terms of the Laplace transform of the input x(t), X(s), and the transfer functions H1 (s) and H2 (s) of the systems: E(s) = X(s) − H2 (s)Y(s) Y(s) = H1 (s)E(s) Replacing E(s) in the second equation gives Y(s)[1 + H1 (s)H2 (s)] = H1 (s)X(s) and the transfer function of the feedback system is then H(s) =
Y(s) H1 (s) = X(s) 1 + H1 (s)H2 (s)
(6.4)
As you recall, in Chapter 2 we were not able to find an explicit expression for the impulse response of the overall system and now you can understand why.
6.3 APPLICATION TO CLASSIC CONTROL Because of different approaches, the theory of control systems can be divided into classic and modern control. Classic control uses frequencydomain methods, while modern control uses timedomain methods. In classic linear control, the transfer function of the plant we wish to control is available;
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η(t )
η (t ) x(t ) +
e(t )
Hc(s)
G (s)
+
y(t )
x (t )
Hc(s)
G (s)
+
y(t)
− (a)
(b)
FIGURE 6.5 (a) Closed and (b) openloop control systems. The transfer function of the plant is G(s) and the transfer function of the controller is Hc (s).
let us call it G(s). The controller, with a transfer function Hc (s), is designed to make the output of the overall system perform in a specified way. For instance, in a cruise control the plant is the car, and the desired performance is to automatically set the speed of the car to a desired value. There are two possible ways the controller and the plant are connected: in openloop or in closedloop (see Figure 6.5).
OpenLoop Control In the openloop approach the controller is cascaded with the plant (Figure 6.5(b)). To make the output y(t) follow the reference signal at the input x(t), we minimize an error signal e(t) = y(t) − x(t) Typically, the output is affected by a disturbance η(t), due to modeling or measurement errors. If we assume initially no disturbance, η(t) = 0, we find that the Laplace transform of the output of the overall system is Y(s) = L[y(t)] = Hc (s)G(s)X(s) and that of the error is E(s) = Y(s) − X(s) = [Hc (s)G(s) − 1]X(s) To make the error zero, so that y(t) = x(t), it would require that Hc (s) = 1/G(s) or the inverse of the plant, making the overall transfer function of the system Hc (s)G(s) unity. Remarks Although openloop systems are simple to implement, they have several disadvantages: n
n
n
The controller Hc (s) must cancel the poles and the zeros of G(s) exactly, which is not very practical. In actual systems, the exact location of poles and zeros is not known due to measurement errors. If the plant G(s) has zeros on the righthand splane, then the controller Hc (s) will be unstable, as its poles are the zeros of the plant. Due to ambiguity in the modeling of the plant, measurement errors, or simply the presence of noise, the output y(t) is typically affected by a disturbance signal η(t) mentioned above (η(t) is typically random— we are going to assume for simplicity that it is deterministic so we can compute its Laplace transform).
6.3 Application to Classic Control
The Laplace transform of the overall system output is Y(s) = Hc (s)G(s)X(s) + η(s) where η(s) = L[η(t)]. In this case, E(s) is given by E(s) = [Hc (s)G(s) − 1]X(s) + η(s) Although we can minimize this error by choosing Hc (s) = 1/G(s) as above, in this case e(t) cannot be made zero—it remains equal to the disturbance η(t) and we have no control over this.
ClosedLoop Control Assuming y(t) and x(t) in the openloop control are the same type of signals, (e.g., both are voltages, or temperatures), if we feed back y(t) and compare it with the input x(t) we obtain a closedloop control. Considering the case of negativefeedback system (see Figure 6.5(a)), and assuming no disturbance (η(t) = 0), we have that E(s) = X(s) − Y(s) Y(s) = Hc (s)G(s)E(s) and replacing Y(s) gives E(s) =
X(s) 1 + G(s)Hc (s)
If we wish the error to go to zero in the steady state, so that y(t) tracks the input, the poles of E(s) should be in the open lefthand splane. If a disturbance signal η(t) (consider it for simplicity deterministic and with Laplace transform η(s)) is present (See Figure 6.5(a)), the above analysis becomes E(s) = X(s) − Y(s) Y(s) = Hc (s)G(s)E(s) + η(s) so that E(s) = X(s) − Hc (s)G(s)E(s) − η(s) or solving for E(s), E(s) =
η(s) X(s) − 1 + G(s)Hc (s) 1 + G(s)Hc (s)
= E1 (s) + E2 (s) If we wish e(t) to go to zero in the steady state, then poles of E1 (s) and E2 (s) should be in the open lefthand splane. Different from the openloop control, the closedloop control offers more flexibility in achieving this by minimizing the effects of the disturbance.
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Remarks A control system includes two very important components: n
n
Transducer: Since it is possible that the output signal y(t) and the reference signal x(t) might not be of the same type, a transducer is used to change the output so as to be compatible with the reference input. Simple examples of a transducer are: lightbulbs, which convert voltage into light; a thermocouple, which converts temperature into voltage. Actuator: A device that makes possible the execution of the control action on the plant, so that the output of the plant follows the reference input.
n Example 6.1: Controlling an unstable plant Consider a dc motor modeled as an LTI system with a transfer function G(s) =
1 s(s + 1)
The motor is not BIBO stable given that its impulse response g(t) = (1 − e−t )u(t) is not absolutely integrable. We wish the output of the motor y(t) to track a given reference input x(t), and propose using a socalled proportional controller with transfer Hc (s) = K > 0 to control the motor (see Figure 6.6). The transfer function of the overall negativefeedback system is H(s) =
Y(s) KG(s) = X(s) 1 + KG(s)
Suppose that X(s) = 1/s, or the reference signal is x(t) = u(t). The question is: What should be the value of K so that in the steady state the output of the system y(t) coincides with x(t)? Or, equivalently, is the error signal in the steady state zero? We have that the Laplace transform of the error signal e(t) = x(t) − y(t) is E(s) = X(s) [1 − H(s)] =
1 s+1 = s(1 + G(s)K) s(s + 1) + K
The poles of E(s) are the roots of the polynomial s(s + 1) + K = s2 + s + K, or √ s1,2 = −0.5 ± 0.5 1 − 4K For 0 < K ≤ 0.25 the roots are real, and complex for K > 0.25, and in either case in the lefthand splane. The partial fraction expansion corresponding to E(s) would be E(s) =
B1 B2 + s − s1 s − s2
x(t ) +
FIGURE 6.6 Proportional control of a motor.
−
Proportional controller e (t )
K
Motor G(s ) =
1 s (s +1)
y(t )
6.3 Application to Classic Control
for some values B1 and B2 . Given that the real parts of s1 and s2 are negative, their corresponding inverse Laplace terms will have a zero steadystate response. Thus, lim e(t) → 0
t→∞
This can be found also with the final value theorem, e(0) is sE(s)s=0 = 0 So for any K > 0, y(t) → x(t) in steady state. Suppose then that X(s) = 1/s2 or that x(t) = tu(t), a ramp signal. Intuitively, this is a much harder situation to control, as the output needs to be continuously growing to try to follow the input. In this case, the Laplace transform of the error signal is E(s) =
1 s2 (1 + G(s)K)
=
s+1 s(s(s + 1) + K)
In this case, even if we choose K to make the roots of s(s + 1) + K be in the lefthand splane, we have a pole at s = 0. Thus, in the steady state, the partial fraction expansion terms corresponding to poles s1 and s2 will give a zero steadystate response, but the pole s = 0 will give a constant steadystate response A where A = E(s)ss=0 = 1/K In the case of a ramp as input, it is not possible to make the output follow exactly the input command, although by choosing a very large gain K we can get them to be very close. n Choosing the values of the gain K of the openloop transfer function G(s)Hc (s) =
KN(s) D(s)
to be such that the roots of 1 + G(s)Hc (s) = 0 are in the open lefthand splane, is the rootlocus method, which is of great interest in control theory. n Example 6.2: A cruise control Suppose we are interested in controlling the speed of a car or in obtaining a cruise control. How to choose the appropriate controller is not clear. We consider initially a proportional plus integral (PI) controller Hc (s) = 1 + 1/s and will ask you to consider the proportional controller as an exercise. See Figure 6.7. Suppose we want to keep the speed of the car at V0 miles/hour for t ≥ 0 (i.e., x(t) = V0 u(t)), and that the model for a car in motion is a system with transfer function Hp (s) = β/(s + α)
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Pl Controller e (t )
x (t )
FIGURE 6.7 Cruise control system: reference speed x(t) = V0 u(t) and output speed of car v(t).
+
1 Hc(s) = 1 + s
Plant c(t )
Hp(s)
v (t )
−
with both β and α positive values related to the mass of the car and the friction coefficient. For simplicity, let α = β = 1. The question is: Can this be achieved with the PI controller? The Laplace transform of the output speed v(t) of the car is V(s) = =
Hc (s)Hp (s) X(s) 1 + Hc (s)Hp (s) V0 (s + 1) V0 = s(s2 + 2s + 1) s(s + 1)
The poles of V(s) are s = 0 and s = −1 on the lefthand splane. We can then write V(s) as V(s) =
B A + s+1 s
where A = V0 . The steadystate response is lim v(t) = V0
t→∞
since the inverse Laplace transform of the first term goes to zero due to its poles being in the lefthand splane. The error signal e(t) = x(t) − v(t) in the steady state is zero. The controlling signal c(t) (see Figure 6.7) that changes the speed of the car is Zt c(t) = e(t) +
e(τ )dτ
0
so that even if the error signal becomes zero at some point—indicating the desired speed had been reached—the value of c(t) is not necessarily zero. The values of e(t) at t = 0 and at steady–state can be obtained using the initial and the finalvalue theorems of the Laplace transform applied to V0 1 E(s) = X(s) − V(s) = 1− s s+1 The finalvalue theorem gives that the steadystate error is lim e(t) = lim sE(s) = 0
t→∞
s→0
coinciding with our previous result. The initial value is found as e(0) = lim sE(s) s→∞ = lim V0 1 − s→∞
1/s = V0 1 + 1/s
6.3 Application to Classic Control
The PI controller used here is one of various possible controllers. Consider a simpler and cheaper controller such as a proportional controller with Hc (s) = K. Would you be able to obtain the same results? Try it. n
6.3.1 Stability and Stabilization A very important question related to the performance of systems is: How do we know that a given causal system has finite zeroinput, zerostate, or steadystate responses? This is the stability problem of great interest in control. Thus, if the system is represented by a linear differential equation with constant coefficients the stability of the system determines that the zeroinput, the zerostate, as well as the steadystate responses may exist. The stability of the system is also required when considering the frequency response in the Fourier analysis. It is important to understand that only the Laplace transform allows us to characterize stable as well as unstable systems; the Fourier transform does not. Two possible ways to look at the stability of a causal LTI system are: n
n
When there is no input so that the response of the system depends on initial energy in the system. This is related to the zeroinput response of the system. When there is a bounded input and no initial condition. This is related to the zerostate response of the system.
Relating the zeroinput response of a causal LTI system to stability leads to asymptotic stability. An LTI system is said to be asymptotically stable if the zeroinput response (due only to initial conditions in the system) goes to zero as t increases—that is, yzi (t) → 0
(6.5)
t→∞
for all possible initial conditions. The second interpretation leads to the boundedinput boundedoutput (BIBO) stability, which we defined in Chapter 2. A causal LTI system is BIBO stable if its response to a bounded input is also bounded. The condition we found in Chapter 2 for a causal LTI system to be BIBO stable was that the impulse response of the system be absolutely integrable—that is Z∞
h(t)dt < ∞
(6.6)
0
Such a condition is difficult to test, and we will see in this section that it is equivalent to the poles of the transfer function being in the open lefthand splane, a condition that can be more easily visualized and for which algebraic tests exist. Consider a system being represented by the differential equation y(t) +
N X k=1
ak
M X dk y(t) d` x(t) = b x(t) + b 0 ` dt` dtk `=1
M 0 or K > 2, the feedback system will be stable.
n
6.3.2 Transient Analysis of First and SecondOrder Control Systems Although the input to a control system is not known apriori, there are many applications where the system is frequently subjected to a certain type of input and thus one can select a test signal. For instance, if a system is subjected to intense and sudden inputs, then an impulse signal might be the
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appropriate test input for the system; if the input applied to a system is constant or continuously increasing, then a unit step or a ramp signal would be appropriate. Using test signals such as an impulse, a unitstep, a ramp, or a sinusoid, mathematical and experimental analyses of systems can be done. When designing a control system its stability becomes its most important attribute, but there are other system characteristics that need to be considered. The transient behavior of the system, for instance, needs to be stressed in the design. Typically, as we drive the system to reach a desired response, the system’s response goes through a transient before reaching the desired response. Thus, how fast the system responds and what steadystate error it reaches need to be part of the design considerations.
FirstOrder Systems As an example of a firstorder system consider an RC serial circuit with a voltage source vi (t) = u(t) as input (Figure 6.9), and as the output the voltage across the capacitor, vc (t). By voltage division, the transfer function of the circuit is Vc (s) 1 H(s) = = Vi (s) 1 + RCs Considering the RC circuit, a feedback system with input vi (t) and output vc (t), the feedforward transfer function G(s) in Figure 6.9 is 1/RCs. Indeed, from the feedback system we have E(s) = Vi (s) − Vc (s) Vc (s) = E(s)G(s) Replacing E(s) in the second of the above equations, we have that Vc (s) G(s) 1 = = Vi (s) 1 + G(s) 1 + 1/G(s) so that the openloop transfer function, when we compare the above equation to H(s), is G(s) =
1 RCs
The RC circuit can be seen as a feedback system: the voltage across the capacitor is constantly compared with the input voltage, and if found smaller, the capacitor continues charging until its voltage coincides with it. How fast depends on the RC value. R vi (t ) +
FIGURE 6.9 Feedback modeling of an RC circuit in series.
e (t ) −
G(s) =
1 RCs
vc (t ) vi (t )
+ −
vc (t) + − C
6.3 Application to Classic Control
0.5 RC = 1 1 RC = 1 vc (t )
jΩ
0.8 0
0.6 RC = 10
0.4 RC = 10 −0.5
0.2 0
−1
−0.8
−0.6
−0.4
−0.2
0
0
10
20
σ (a)
30
40
50
t (b)
FIGURE 6.10 (a) Clustering of poles and (b) time responses of a firstorder feedback system for 1 ≤ RC ≤ 10.
For vi (t) = u(t), so that Vi (s) = 1/s, then the Laplace transform of the output is Vc (s) =
1 1/RC 1 1 = = − s(sRC + 1) s(s + 1/RC) s s + 1/RC
so that vc (t) = (1 − e−t/RC )u(t) The following MATLAB script plots the poles Vc (s)/Vi (s) and simulates the transients of vc (t) for 1 ≤ RC ≤ 10, shown in Figure 6.10. Thus, if we wish the system to respond fast to the unitstep input we locate the system pole far from the origin. %%%%%%%%%%%%%%%%%%%%% % Transient analysis %%%%%%%%%%%%%%%%%%%%% clf; clear all syms s t num = [0 1]; for RC = 1:2:10, den = [ RC 1]; figure(1) splane(num, den) % plotting of poles and zeros hold on vc = ilaplace(1/(RC ∗ s ˆ 2 + s)) % inverse Laplace figure(2) ezplot(vc, [0, 50]); axis([0 50 0 1.2]); grid hold on end hold off
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SecondOrder System A series RLC circuit with the input a voltage source, vs (t), and the output the voltage across the capacitor, vc (t), has a transfer function Vc (s) 1/Cs 1/LC = = 2 Vs (s) R + Ls + 1/Cs s + (R/L)s + 1/LC If we define 1 Natural frequency : n = √ CL r C Damping ratio : ψ = 0.5R L
(6.9) (6.10)
we can write Vc (s) 2n = 2 Vs (s) s + 2ψn s + 2n
(6.11)
A feedback system with this transfer function is given in Figure 6.11 where the feedforward transfer function is G(s) =
2n s(s + 2ψn )
Indeed, the transfer function of the feedback system is given by H(s) =
Vc (s) G(s) = Vs (s) 1 + G(s)
=
2n s2 + 2ψn s + 2n
The dynamics of a secondorder system can be described in terms of the parameters n and ψ, as these two parameters determine the location of the poles of the system and thus its response. We adapted the previously given script to plot the cluster of poles and the time response of the secondorder system. Assume n = 1 rad/sec and let 0 ≤ ψ ≤ 1 (so that the poles of H(s) are complex conjugate for 0 ≤ ψ < 1 and double real for ψ = 1). Let the input be a unitstep signal so that Vs (s) = 1/s. We then have: vs (t ) +
FIGURE 6.11 Secondorder feedback system.
e (t )
vc(t ) G (s)
−
6.3 Application to Classic Control
1
1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1
0.6 0.4 0.2 0
0
10
20
30
40
50
30
40
50
t (b) 2 1.5 −1
−0.5 σ (a)
0
vc (t)
jΩ
vc (t )
0.8
1 0.5 0 0
10
20 t (c)
FIGURE 6.12 √ (a) Clustering of poles and time responses vc (t) of secondorder feedback system for (b) 2/2 ≤ ψ ≤ 1 and √ (c) 0 ≤ ψ ≤ 2/2.
(a) If we plot the poles of H(s) as ψ changes from 0 (poles in j axis) to 1 (double real poles) the response y(t) in the steady state changes from a sinusoid shifted up by 1 to a damped signal. The locus of the poles is a semicircle of radius n = 1. Figure 6.12 shows this behavior of the poles and the responses. (b) As in the firstorder system, the location of the poles determines the response of the system. The system is useless if the poles are on the j axis, as the response is completely oscillatory and the input will never be followed. On the other extreme, the response of the system is slow when the poles become real. The√designer would have to choose a value in between these two for ψ. (c) For values of ψ between 2/2 to 1 the oscillation√is minimum and the response is relatively fast (see Figure 6.12(b)). For values of ψ from 0 to 2/2 the response oscillates more and more, giving a large steadystate error (see Figure 6.12(c)). n Example 6.4 In this example we find the response of an LTI system to different inputs by using functions in the control toolbox of MATLAB. You can learn more about the capabilities of this toolbox, or set of specialized functions for control, by running the demo respdemo and then using help to learn more about the functions tf, impulse, step, and pzmap, which we will use here.
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CH A P T E R 6: Application to Control and Communications
We want to create a MATLAB function that has as inputs the coefficients of the numerator N(s) and of the denominator D(s) of the system’s transfer function H(s) = N(s)/D(s) (the coefficients are ordered from the highest order to the lowest order or constant term). The other input of the function is the type of response t where t = 1 corresponds to the impulse response, t = 2 to the unitstep response, and t = 3 to the response to a ramp. The output of the function is the desired response. The function should show the transfer function, the poles, and zeros, and plot the corresponding response. We need to figure out how to compute the ramp response using the step function. Consider the following transfer functions: s+1 +s+1 s (b) H2 (s) = 3 s + s2 + s + 1
(a) H1 (s) =
s2
Determine the stability of these systems. Solution The following script is used to look at the desired responses of the two systems and the location of their poles and zeros. We consider the second system; you can run the script for the first system by putting % at the numerator and the denominator after H2 (s) and getting rid of % after H1 (s) in the script. The function response computes the desired responses (in this case the impulse, step, and ramp responses). %%%%%%%%%%%%%%%%%%% % Example 6.4  Control toolbox %%%%%%%%%%%%%%%%%%% clear all; clf % % H 1(s) % nu = [1 1]; de = [1 1 1]; %% H 2(s) nu = [1 0]; de = [1 1 1 1]; % unstable h = response(nu, de, 1); s = response(nu, de, 2); r = response(nu, de, 3); function y = response(N, D, t) sys = tf(N, D) poles = roots(D) zeros = roots(N) figure(1) pzmap(sys);grid if t == 3, D1 = [D 0]; % for ramp response end
6.4 Application to Communications
figure(2) if t == 1, subplot(311) y = impulse(sys,20); plot(y);title(’ Impulse response’);ylabel(’h(t)’);xlabel(’t’); grid elseif t == 2, subplot(312) y = step(sys, 20); plot(y);title(’ Unitstep response’);ylabel(’s(t)’);xlabe(’t’);grid else subplot(313) sys = tf(N, D1); % ramp response y = step(sys, 40); plot(y); title(’ Ramp response’); ylabel(’q(t)’); xlabel(’t’);grid end
The results for H2 (s) are as follows. Transfer function: s sˆ3 + sˆ2 + s + 1 poles = −1.0000 −0.0000 + 1.0000i −0.0000 − 1.0000i zeros = 0
As you can see, two of the poles are on the j axis, and so the system corresponding to H2 (s) is unstable. The other system is stable. Results for both systems are shown in Figure 6.13. n
6.4 APPLICATION TO COMMUNICATIONS The application of the Fourier transform in communications is clear. The representation of signals in the frequency domain and the concept of modulation are basic in communications. In this section we show examples of linear (amplitude modulation or AM) as well as nonlinear (frequency modulation or FM, and phase modulation or PM) modulation methods. We also consider important extensions such as frequencydivision multiplexing (FDM) and quadrature amplitude modulation (QAM). Given the lowpass nature of most message signals, it is necessary to shift in frequency the spectrum of the message to avoid using a very large antenna. This can be attained by means of modulation, which is done by changing either the magnitude or the phase of a carrier: A(t) cos(2πfc + θ (t))
(6.12)
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CH A P T E R 6: Application to Control and Communications
Impulse Response
0.5
1
0
0.8
−0.5
0
10
20
50 t
60
70
80
0
10
20
30
40
50 t
60
70
0.6
90 100
0.4
1 0
80
90 100
0
10
20
30
40
50 t
60
70
80
0.32
0.22
0.1
0.96
1
0.8
0.6
0.4
0.2
System: sys Pole : −0.5 − 0.866i Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/sec): 1
0.96
0.86 0.72
−1
90 100
0.44
−0.2
−0.8 0
0.58
0 86
0
−0.6
20
Pole−Zero Map 0.72
0.2
−0.4
Ramp Response
40 q (t)
40
Unit−Step Response
2 s (t)
30
Imaginary Axis
h (t)
1
0.58
−1
−0.8
0.44
−0.6 Real Axis
0 32
0.22
0.1
−0.2
−0.4
0
(a) Impulse Response Pole−Zero Map
0 −1 0
10
20
40
50 t
60
70
80
0.8
10
20
30
2
40
50 60 70 t Ramp Response
80
90 100
0.4 0
30
40
50 t
60
70
0.32
0.2
0.1
0 84
0.95
1.2
1
0.8
−1
−0.8
0.6
0.4
0.2
−0.2 −0.4
−1 20
0.44
0.2
−0.8
10
0.56
0.95
−0.6
1 0 0
0.7
0.6
0 −1 0
1
90 100
Unit−Step Response
1 s(t)
30
Imaginary Axis
h(t)
1
q(t)
378
80
−1.2
90 100
0.84
0.7
0.56
0.44
−0.6
0.32
−0.4
02
−0.2
01
0
0.2
Real Axis (b)
FIGURE 6.13 Impulse, unitstep, and ramp responses and poles and zeros for system with transfer function (a) H1 (s) and (b) H2 (s).
When A(t) is proportional to the message, for constant phase, we have amplitude modulation (AM). On the other hand, if we let θ(t) change with the message, keeping the amplitude constant, we then have frequency modulation (FM) or phase modulation (PM), which are called angle modulations.
6.4 Application to Communications
6.4.1 AM with Suppressed Carrier Consider a message signal m(t) (e.g., voice or music, or a combination of the two) modulating a cosine carrier cos(c t) to give an amplitude modulated signal s(t) = m(t) cos(c t)
(6.13)
The carrier frequency c >> 2πf0 where f0 (Hz) is the maximum frequency in the message (for music f0 is about 22 KHz). The signal s(t) is called an amplitude modulated with suppressed carrier (AMSC) signal (the last part of this denomination will become clear later). According to the modulation property of the Fourier transform, the transform of s(t) is S() =
1 [M( − c ) + M( + c )] 2
(6.14)
where M() is the spectrum of the message. The frequency content of the message is now shifted to a much larger frequency c (rad/sec) than that of the baseband signal m(t). Accordingly, the antenna needed to transmit the amplitude modulated signal is of reasonable length. An AMSC system is shown in Figure 6.14. At the receiver, we need to first detect the desired signal among the many signals transmitted by several sources. This is possible with a tunable bandpass filter that selects the desired signal and rejects the others. Suppose that the signal obtained by the receiver, after the bandpass filtering, is exactly s(t)—we then need to demodulate this signal to get the original message signal m(t). This is done by multiplying s(t) by a cosine of exactly the same frequency of the carrier in the transmitter (i.e., c ), which will give r(t) = 2s(t) cos(c t), which again according to the modulation property has a Fourier transform R() = S( − c ) + S( + c ) = M() +
1 [M( − 2c ) + M( + 2c )] 2
(6.15)
The spectrum of the message, M(), is obtained by passing the received signal r(t) through a lowpass filter that rejects the other terms M( ± 2c ). The obtained signal is the desired message m(t). The above is a simplification of the actual processing of the received signal. Besides the many other transmitted signals that the receiver encounters, there is channel noise caused by interferences from cos(Ωct ) m(t)
×
2cos(Ωct )
Channel
Transmitter
FIGURE 6.14 AMSC transmitter, channel, and receiver.
Bandpass filter
×
Receiver
Lowpass filter
ˆ ) m(t
379
380
CH A P T E R 6: Application to Control and Communications
equipment in the transmission path and interference from other signals being transmitted around the carrier frequency. This noise will also be picked up by the bandpass filter and a perfect recovery of m(t) will not be possible. Furthermore, the sent signal has no indication of the carrier frequency c , which is suppressed in the sent signal, and so the receiver needs to guess it and any deviation would give errors. Remarks n
n
n
The transmitter is linear but time varying. AMSC is thus called a linear modulation. The fact that the modulated signal displays frequencies much higher than those in the message indicates the transmitter is not LTI—otherwise it would satisfy the eigenfunction property. A more general characterization than c >> 2πf0 where f0 is the largest frequency in the message is given by c >> BW where BW (rad/sec) is the bandwidth of the message. You probably recall the definition of bandwidth of filters used in circuit theory. In communications there are several possible definitions for bandwidth. The bandwidth of a signal is the width of the range of positive frequencies for which some measure of the spectral content is satisfied. For instance, two possible definitions are: n The halfpower or 3dB bandwidth is the width of the range of positive frequencies where a peak value at zero or infinite frequency (lowpass and highpass signals) or at a center frequency (bandpass signals) is attenuated to 0.707, the value at the peak. This corresponds to the frequencies for which the power at dc, infinity, or center frequency reduces to half. n The nulltonull bandwidth determines the width of the range of positive frequencies of the spectrum of a signal that has a main lobe containing a significant part of the energy of the signal. If a lowpass signal has a clearly defined maximum frequency, then the bandwidth are frequencies from zero to the maximum frequency, and if the signal is a bandpass signal and has a minimum and a maximum frequency, its bandwidth is the maximum minus the minimum frequency. In AMSC demodulation it is important to know exactly the carrier frequency. Any small deviation would cause errors when recovering the message. Suppose, for instance, that there is a small error in the carrier frequency—that is, instead of c the demodulator uses c + 1—so that the received signal in that case has the Fourier transform ˜ R() = S( − c − 1) + S( + c + 1) =
1 [M( + 1) + M( − 1)] 2 +
1 [M( − 2(c + 1/2)) + M( + 2(c + 1/2)] 2
The lowpass filtered signal will not be the message.
6.4.2 Commercial AM In commercial broadcasting, the carrier is added to the AM signal so that information of the carrier is available at the receiver helping in the identification of the radio station. For demodulation, such information is not important, as commercial AM uses envelope detectors to obtain the message. By making the envelope of the modulated signal look like the message, detecting this envelope is all
6.4 Application to Communications
that is needed. Thus, the commercial AM signal is of the form s(t) = [K + m(t)] cos(c t) where the AM modulation index K is chosen so that K + m(t) > 0 for all values of t so that the envelope of s(t) is proportional to the message m(t). The Fourier transform is given by 1 [M( − c ) + M( + c )] 2 The receiver for this type of AM is an envelope receiver, which basically detects the message by finding the envelope of the received signal. S() = Kπ [δ( − c ) + δ( + c )] +
Remarks n
n
n
n
The advantage of adding the carrier to the message, which allows the use of a simple envelope detector, comes at the expense of increasing the power in the transmitted signal. The demodulation in commercial AM is called noncoherent. Coherent demodulation consists in multiplying—or mixing—the received signal with a sinusoid of the same frequency and phase of the carrier. A local oscillator generates this sinusoid. A disadvantage of commercial as well as suppressedcarrier AM is the doubling of the bandwidth of the transmitted signal compared to the bandwidth of the message. Given the symmetry of the spectrum, in magnitude as well as in phase, it becomes clear that it is not necessary to send the upper and the lower sidebands of the spectrum to get back the signal in the demodulation. It is thus possible to have upper and lowersideband AM modulations, which are more efficient in spectrum utilization. Most AM receivers use the superheterodyne receiver technique developed by Armstrong and Fessenden.2
n Example 6.5: Simulation of AM modulation with MATLAB For simulations, MATLAB provides different data files, such as “train.mat” (the extension mat indicates it is a data file) used here. Suppose the analog signal y(t) is a recording of a “choochoo” train, and we wish to use it to modulate a cosine cos(c t) to create an amplitude modulated signal z(t). Because the train y(t) signal is given in a sampled form, the simulation requires discretetime processing, and so we will comment on the results here and leave the discussion of the issues related to the code for the next chapters. The carrier frequency is chosen to be fc = 20.48 KHz. For the envelope detector to work at the transmitter we add a constant K to the message to ensure this sum is positive. The envelope of the AMmodulated signal should resemble the message. Thus, the AM signal is z(t) = [K + y(t)] cos(c t)
c = 2πfc
In Figure 6.15 we show the train signal, a segment of the signal, and the corresponding modulated signal displaying the envelope, as well as the Fourier transform of the segment and of its modulated 2 Reginald
Fessenden was the first to suggest the heterodyne principle: mixing the radiofrequency signal using a local oscillator of different frequency, resulting in a signal that could drive the diaphragm of an earpiece at an audio frequency. Fessenden could not make a practical success of the heterodyne receiver, which was accomplished by Edwin H. Armstrong in the 1920s using electron tubes.
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CH A P T E R 6: Application to Control and Communications
1 0.8 0.6 0.4 y(t )
0.2 0 −0.2 −0.4 −0.6 −0.8 −1 0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
t (sec) (a) 1
150 Z(Ω)
y(t )
0.5 0 −0.5 −1 0
100 50 0 −2.5
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
−2 −1.5
−1 −0.5
0
0.5
1
1.5
2
2.5 ×104
f (Hz)
t (sec) 2 Y(Ω)
1 z(t )
382
0 −1 −2 0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
500 400 300 200 100 0 −2.5 −2
−1.5 −1
−0.5
0
t (sec)
f (Hz)
(b)
(c)
0.5
1
1.5
2
2.5 ×104
FIGURE 6.15 Commercial AM modulation: (a) original signal, (b) part of original signal and corresponding AMmodulated signal, and (c) spectrum of the original signal, and of the modulated signal.
version. Notice that the envelope resembles the original signal. Also from the spectrum of the segment of the train signal its bandwidth is about 5 Khz, while the spectrum of the modulated segment displays the frequencyshifted spectrum plus the large spectrum at fc corresponding to the carrier. n
6.4.3 AM Single Sideband The message m(t) is typically a realvalued signal that, as indicated before, has a symmetric spectrum—that is, the magnitude and the phase of the Fourier transform M() are even and odd
6.4 Application to Communications
cos(Ωct ) m(t)
FIGURE 6.16 Upper sideband AM transmitter. c is the carrier frequency and B is the bandwidth in rad/sec of the message.
×
s(t)
BPF H( j Ω)
1
Ωc
Ωc + B
Ω
functions of frequency. When using AM modulation the resulting spectrum has redundant information by providing the upper and the lower sidebands. To reduce the bandwidth of the transmitted signal, we could get rid of either the upper or the lower sideband of the AM signal. The resulting modulation is called AM single sideband (AMSSB) (upper or lower sideband depending on which of the two sidebands is kept). This type of modulation is used whenever the quality of the received signal is not as important as the advantages of a narrowband and having less noise in the frequency band of the received signal. AMSSB is used by amateur radio operators. As shown in Figure 6.16, the upper sideband modulated signal is obtained by bandpass filtering the upper sideband in the modulated signal. At the receiver, bandpass filtering the received signal the output is then demodulated like in an AMSC system, and the result is then lowpass filtered using the bandwidth of the message.
6.4.4 Quadrature AM and FrequencyDivision Multiplexing Quadrature amplitude modulation (QAM) and frequency division multiplexing (FDM) are the precursors of many of the new communication systems. QAM and FDM are of great interest for their efficient use of the radio spectrum.
Quadrature Amplitude Modulation QAM enables two AMSC signals to be transmitted over the same frequency band, conserving bandwidth. The messages can be separated at the receiver. This is accomplished by using two orthogonal carriers, such as a cosine and a sine (see Figure 6.17). The QAMmodulated signal is given by s(t) = m1 (t) cos(c t) + m2 (t) sin(c t)
(6.16)
where m1 (t) and m2 (t) are the messages. You can think of s(t) as having a phasor representation that is the sum of two phasors perpendicular to each other (the cosine leading the sine by π/2); indeed, s(t) = Re[(m1 (t)ej0 + m2 (t)e−jπ/2 )ejc t ]. Since m1 (t)ej0 + m2 (t)e−jπ/2 = m1 (t) − jm2 (t) we could interpret the QAM signal as the result of amplitude modulating the real and the imaginary parts of a complex message m(t) = m1 (t) − jm2 (t).
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CH A P T E R 6: Application to Control and Communications
m1(t )
×
×
cos(Ωct)
cos(Ωct)
π shift 2 m2(t )
+
LPF
m˜ 1(t )
s(t ) r (t ) π shift 2
×
×
Transmitter
LPF
m˜ 2(t )
Receiver
FIGURE 6.17 QAM transmitter and receiver: s(t) is the transmitted signal and r(t) is the received signal.
To simplify the computation of the spectrum of s(t), let us consider the message m(t) = m1 (t) − jm2 (t) (i.e., a complex message) with spectrum M() = M1 () − jM2 () so that s(t) = Re[m(t)ejc t ] = 0.5[m(t)ejc t + m∗ (t)e−jc t ] where ∗ stands for complex conjugate. The spectrum of s(t) is then given by S() = 0.5[M( − c ) + M∗ ( + c )] = 0.5[M1 ( − c ) − jM2 ( − c ) + M∗1 ( + c ) + jM∗2 ( + c )] where the superposition of the spectra of the two messages is clearly seen. At the receiver, if we multiply the received signal (for simplicity assume it to be s(t)) by cos(c t), we get r1 (t) = s(t) cos(c t) = 0.25[m(t) + m∗ (t)] + 0.25[m(t)ej2c t + m∗ (t)e−j2c t ] which when passed through a lowpass filter, with the appropriate bandwidth, gives 0.25[m(t) + m∗ (t)] = 0.25[m1 (t) − jm2 (t) + m1 (t) + jm2 (t)] = 0.5m1 (t) Likewise, to get the second message we multiply s(t) by sin(c t) and pass the resulting signal through a lowpass filter.
FrequencyDivision Multiplexing Frequencydivision multiplexing (FDM) implements sharing of the spectrum by several users by allocating a specific frequency band to each. One could, for instance, think of the commercial AM or FM
6.4 Application to Communications
cos(Ω1t )
cos(Ω1t ) m1(t )
×
BPF
×
+
Channel
BPF
m1(t )
×
×
LPF
m2(t )
LPF
m3(t )
∧
cos(Ω3t)
cos(Ω3t ) m3(t )
∧
LPF
cos(Ω2t)
cos(Ω2t ) m2(t )
×
BPF
×
∧
FIGURE 6.18 FDM system: transmitter (left), channel, and receiver (right).
locally as an FDM sytem. In the United States, the Federal Communication Commission (FCC) is in charge of the spectral allocation. In telephony, using a bank of filters it is possible to also get several users in the same system—it is, however, necessary to have a similar system at the receiver to have a twoway communication. To illustrate an FDM system (Figure 6.18), consider we have a set of messages of known finite bandwidth (we could lowpass filter the messages to satisfy this condition) that we wish to transmit. Each of the messages modulate different carriers so that the modulated signals are in different frequency bands without interfering with each other (if needed a frequency guard could be used to make sure of this). These frequencymultiplexed messages can now be transmitted. At the receiver, using a bank of bandpass filters centered at the carrier frequencies in the transmitter and followed by appropriate demodulators recover the different messages (see FDM receiver in Figure 6.18). Any of the AM modulation techniques could be used in the FDM system.
6.4.5 Angle Modulation Amplitude modulation is said to be linear modulation, because as a system it behaves like a linear system. Frequency and phase, or angle, modulation systems on the other hand are nonlinear. The interest in angle modulation is due to the decreasing effect of noise or interferences on it, when compared with AM, although at the cost of a much wider bandwidth and greater complexity in implementation. The nonlinear behavior of angle modulation systems makes their analysis more difficult than that for AM. The spectrum of an FM or PM signal is much harder to obtain than that of an AM signal. In the following we consider the case of the socalled narrowband FM where we are able to find its spectrum directly. Professor Edwin H. Armstrong developed the first successful frequency modulation system— narrowband FM.3 If m(t) is the message signal, and we modulate a carrier signal of frequency 3 Edwind
H. Armstrong (1890–1954), professor of electrical engineering at Columbia University, and inventor of some of the basic electronic circuits underlying all modern radio, radar, and television, was born in New York. His inventions and developments form the backbone of radio communications as we know it.
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CH A P T E R 6: Application to Control and Communications
c (rad/sec) with m(t), the transmitted signal s(t) in angle modulation is of the form s(t) = A cos(c t + θ (t))
(6.17)
where the angle θ(t) depends on the message m(t). In the case of phase modulation, the angle function is proportional to the message m(t)—that is, θ(t) = Kf m(t)
(6.18)
where Kf > 0 is called the modulation index. If the angle is such that dθ(t) = 1 m(t) dt
(6.19)
this relation defines frequency modulation. The instantaneous frequency, as a function of time, is the derivative of the argument of the cosine or d[c t + θ (t)] dt dθ (t) = c + dt = c + 1 m(t)
(6.20)
IF(t) =
(6.21) (6.22)
indicating how the frequency is changing with time. For instance, if θ (t) is a constant—so that the carrier is just a sinusoid of frequency c and constant phase θ—the instantaneous frequency is simply c . The term 1 m(t) relates to the spreading of the frequency about c . Thus, the modulation paradox Professor E. Craig proposed in his book [17]: In amplitude modulation the bandwidth depends on the frequency of the message, while in frequency modulation the bandwidth depends on the amplitude of the message.
Thus, the modulated signals are PM : FM :
sPM (t) = cos(c t + Kf m(t)) sFM (t) = cos(c t + 1
Zt
(6.23) m(τ )dτ )
(6.24)
−∞
Narrowband FM In this case the angle θ(t) is small, so that cos(θ (t)) ≈ 1 and sin(θ (t)) ≈ θ (t), simplifying the spectrum of the transmitted signal: S() = F [cos(c t + θ (t))] = F [cos(c t) cos(θ (t)) − sin(c t) sin(θ (t))] ≈ F [cos(c t) − sin(c t)θ (t)]
(6.25)
6.4 Application to Communications
Using the spectrum of a cosine and the modulation theorem, we get S() ≈ π [δ( − c ) + δ( + c )] −
1 [2( − c ) − 2( + c )] 2j
(6.26)
where 2() is the spectrum of the angle, which is found to be (using the derivative property of the Fourier transform) 2() =
1 M() j
(6.27)
If the angle θ (t) is not small, we have wideband FM and its spectrum is more difficult to obtain. n Example 6.6: Simulation of FM modulation with MATLAB In these simulations we will concern ourselves with the results and leave the discussion of issues related to the code for the next chapter since the signals are approximated by discretetime signals. For the narrowband FM we consider a sinusoidal message m(t) = 80 sin(20πt)u(t), and a sinusoidal carrier of frequency fc = 100 Hz, so that for 1 = 0.1π the FM signal is Zt x(t) = cos(2πfc t + 0.1π
m(τ )dτ )
−∞
Figure 6.19 shows on the top left the message and the narrowband FM signal x(t) right below it, and on the top right their corresponding magnitude spectra M() and below X(). The narrowband FM has only shifted the frequency of the message. The instantaneous frequency (the derivative of the argument of the cosine) is IF(t) = 2πfc + 0.1πm(t) = 200π + 8π sin(20πt) ≈ 200π That is, it remains almost constant for all times. For the narrowband FM, the spectrum of the modulated signal remains the same for all times. To illustrate this we computed the spectrogram of x(t). Simply, the spectrogram can be thought of as the computation of the Fourier transform as the signal evolves with time (see Figure 6.19(c)). To illustrate the wideband FM, we consider two messages, m1 (t) = 80 sin(20πt)u(t) m2 (t) = 2000tu(t) giving FM signals, xi (t) = cos(2πfci t + 50π
Zt
mi (τ )dτ )
i = 1, 2
−∞
where fc1 = 2500 Hz and fc2 = 25 Hz. In this case, the instantaneous frequency is IFi (t) = 2πfci + 50πmi (t)
i = 1, 2
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CH A P T E R 6: Application to Control and Communications
10 M(Ω)
m (t )
50 0 −50 0
0.05
0.1
0.15
5
0 −500
0.2
0
500
0 f
500
0.1
1 x(Ω)
0.5 x (t )
0 −0.5 0
0.05
0.1 t (sec)
0.15
0.05
0 −500
0.2
(a)
(b)
500 450 400 350 Frequency
388
300 250 200 150 100 50 0 0.6/7 0.7/7 0.8/7 0.9/7
1/7
1.1/7 1.2/7 1.3/7
0.2
Time (c)
FIGURE 6.19 Narrowband frequency modulation: (a) message m(t) and narrowband FM signal x(t); (b) magnitude spectra of m(t) and x(t); and (c) spectrogram of x(t) displaying evolution of its Fourier transform with respect to time.
These instantaneous frequencies are not almost constant as before. The frequency of the carrier is now continuously changing with time. For instance, for the ramp message the instantaneous frequency is IF2 (t) = 50π + 105 tπ so that for a small time interval [0, 0.1] we get a chirp (sinusoid with timevarying frequency), as shown in Figure 6.20(b). Figure 6.20 display the messages, the FM signals, and their corresponding magnitude spectra and their spectrograms. These FM signals are broadband, occupying a band of frequencies much larger than the messages, and their spectrograms show that their spectra change with time. n
6.4 Application to Communications
200 150
m (t)
m (t)
50 0
100 50
−50 0.02
0.04
0.06
0.08
0
0.1
1 0.5 0 −0.5
0
0.02
0.04
0.06
0.08
0.1
0
0.02
0.04 0.06 t (sec)
0.08
0.1
1 0.5 0 −0.5
x (t )
x (t )
0
−1
−1 0
0.02
0.04
0.06 t (sec)
0.08
0.1
10 M (Ω)
M (Ω)
10 5 0 −200 −150 −100 −50
0
50
0 −1000
100 150 200
3
0.01
X(Ω)
X(Ω)
0.015
0.005
−500
0
500
1000
×10−3
2.5 2 1.5
0 −5000
0 f
1 −5000
5000
5000
5000
4500
4500
4000
4000
3500
3500 Frequency
Frequency
5
3000 2500 2000
5000
3000 2500 2000
1500
1500
1000
1000
500
500
0
0 f
0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
Time (a)
Time (b)
FIGURE 6.20 Wideband frequency modulation, from top to bottom, for (a) the sinusoidal message and for (b) the ramp message: messages, FMmodulated signals, spectra of messages, spectrum of FM signals, and spectrogram of FM signals.
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CH A P T E R 6: Application to Control and Communications
6.5 ANALOG FILTERING The basic idea of filtering is to get rid of frequency components of a signal that are not desirable. Application of filtering can be found in control, in communications, and in signal processing. In this section we provide a short introduction to the design of analog filters. Chapter 11 is dedicated to the design of discrete filters and to some degree that chapter will be based on the material in this section. According to the eigenfunction property of LTI systems (Figure 6.21) the steadystate response of an LTI system to a sinusoidal input—with a certain magnitude, frequency, and phase—is a sinusoid of the same frequency as the input, but with magnitude and phase affected by the response of the system at the frequency of the input. Since periodic as well as aperiodic signals have Fourier representations consisting of sinusoids of different frequencies, the frequency components of any signal can be modified by appropriately choosing the frequency response of the LTI system, or filter. Filtering can thus be seen as a way of changing the frequency content of an input signal. The appropriate filter for a certain application is specified using the spectral characterization of the input and the desired spectral characteristics of the output. Once the specifications of the filter are set, the problem becomes one of approximation as a ratio of polynomials in s. The classical approach in filter design is to consider lowpass prototypes, with normalized frequency and magnitude responses, which may be transformed into other filters with the desired frequency response. Thus, a great deal of effort is put into designing lowpass filters and into developing frequency transformations to map lowpass filters into other types of filters. Using cascade and parallel connections of filters also provide a way to obtain different types of filters. The resulting filter should be causal, stable, and have realvalued coefficients so that it can be used in realtime applications and realized as a passive or an active filter. Resistors, capacitors, and inductors are used in the realization of passive filters, while resistors, capacitors, and operational amplifiers are used in active filter realizations.
6.5.1 Filtering Basics A filter H(s) = B(s)/A(s) is an LTI system having a specific frequency response. The convolution property of the Fourier transform gives that (6.28)
Y() = X()H( j) where H( j) = H(s)s=j
Thus, the frequency content of the input, represented by the Fourier transform X(), is changed by the frequency response H( j) of the filter so that the output signal with spectrum Y() only has desirable frequency components. LTI system Ae j(Ω0t+θ )
FIGURE 6.21 Eigenfunction property of continuous LTI systems.
H(s)
AH( j Ω0)e j (Ω0t +θ +∠H( jΩ0)
6.5 Analog Filtering
Magnitude Squared Function The magnitudesquared function of an analog lowpass filter has the general form 1 1 + f (2 )
H( j)2 =
(6.29)
where for low frequencies f (2 ) ≈ 0 so that H( j)2 ≈ 1, and for high frequencies f (2 ) → ∞ so that H( j)2 → 0. Accordingly, there are two important issues to consider: n n
Selection of the appropriate function f (.). The factorization needed to get H(s) from the magnitudesquared function.
As an example of the above steps, consider the Butterworth lowpass analog filter. The Butterworth magnitudesquared response of order N is 1
HN ( j)2 = 1+
h
hp
i2N
(6.30)
where hp is the halfpower frequency of the filter. We then have that for > hp , then HN ( j) → 0. To find H(s) we need to factorize Equation (6.30). Letting S be a normalized variable S = s/hp , the magnitudesquared function (Eq. 6.30) can be expressed in terms of the S variable by letting S/j = /hp to obtain H(S)H(−S) =
1 1 + (−S2 )N
since H( j0 )2 = H( j0 )H∗ ( j0 ) = H( j0 )H(−j0 ). As we will see, the poles of H(S)H(−S) are symmetrically clustered in the splane with none on the j axis. The factorization then consists of assigning poles in the open lefthand splane to H(S), and the rest to H(−S). We thus obtain 1 1 D(S) D(−S)
H(S)H(−S) = so that the final form of the filter is
1 D(S) where D(S) has roots on the lefthand splane. A final step is the replacement of S by the unnormalized variable s, to obtain the final form of the filter transfer function: H(S) =
Butterworth lowpass filter : H(s) = H(S)S=s/ hp
(6.31)
Filter Specifications Although an ideal lowpass filter is not realizable (recall the PaleyWiener condition in Chapter 5) its magnitude response can be used as prototype for specifying lowpass filters. Thus, the desired magnitude is specified as 1 − δ2 ≤ H( j) ≤ 1 0 ≤ H( j) ≤ δ1
0 ≤ ≤ p (passband) ≥ s (stopband)
(6.32)
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CH A P T E R 6: Application to Control and Communications
H( jΩ)
α (Ω)
1
α min
1− δ 2
FIGURE 6.22 Magnitude specifications for a lowpass filter.
α max
δ1 Ωp
Ωs
Ω
0
Ωp
Ωs
Ω
for some small values δ1 and δ2 . There is no specification in the transition region p < < s . Also the phase is not specified, although we wish it to be linear at least in the passband. See Figure 6.22. To simplify the computation of the filter parameters, and to provide a scale that has more resolution and physiological significance than the specifications given above, the magnitude specifications are typically expressed in a logarithmic scale. Defining the loss function (in decibels, or dBs) as α() = −10 log10 H( j)2 = −20 log10 H( j)
dBs
(6.33)
an equivalent set of specifications to those in Equation (6.32) is 0 ≤ α() ≤ αmax α() ≥ αmin
0 ≤ ≤ p (passband) ≥ s (stopband)
(6.34)
where αmax = −20 log10 (1 − δ2 ) and αmin = −20 log10 (δ1 ). In the above specifications, the dc loss was 0 dB corresponding to a normalized dc gain of 1. In more general cases, α(0) 6= 0 and the loss specifications are given as α(0) = α1 , α2 in the passband and α3 in the stopband. To normalize these specifications we need to subtract α1 , so that the loss specifications are α(0) = α1 (dc loss) αmax = α2 − α1 (maximum attenuation in passband) αmin = α3 − α1 (minimum attenuation in stopband) Using {αmax , p , αmin , s } we proceed to design a magnitudenormalized filter, and then use α1 to achieve the desired dc gain. The design problem is then: Given the magnitude specifications in the passband (α(0), αmax , and p ) and in the stopband (αmin and s ) we then 1. Choose the rational approximation method (e.g., Butterworth). 2. Solve for the parameters of the filter to obtain a magnitudesquared function that satisfies the given specifications.
6.5 Analog Filtering
3. Factorize the magnitudesquared function and choose the poles on the lefthand splane, guaranteeing the filter stability, to obtain the transfer function HN (s) of the filter.
6.5.2 Butterworth LowPass Filter Design The magnitudesquared approximation of a lowpass Nthorder Butterworth filter is given by HN ( j0 )2 =
1 1 + / hp
2N
0 =
hp
(6.35)
where hp is the halfpower or −3dB frequency. This frequency response is normalized with respect to the halfpower frequency (i.e., the normalized frequency is 0 = / hp ) and normalized in magnitude as the dc gain is H( j0) = 1. The frequency 0 = / hp = 1 is the normalized halfpower frequency since HN ( j1)2 = 1/2. The given magnitudesquared function is thus normalized with respect to frequency (giving a unity halfpower frequency) and in magnitude (giving a unity DC gain for the lowpass filter). The approximation improves (i.e., gets closer to the ideal filter) as the order N increases. Remarks n
The halfpower frequency is called the −3dB frequency because in the case of the lowpass filter with a dc gain of 1, at the halfpower frequency hp the magnitudesquared function is H( jhp )2 =
H( j0)2 1 = . 2 2
(6.36)
In the logarithmic scale we have 10 log10 (H( jhp )2 ) = −10 log10 (2) ≈ −3 (dB) n
(6.37)
This corresponds to a loss of 3 dB. It is important to understand the significance of the frequency and magnitude normalizations typical in filter design. Having a lowpass filter with normalized magnitude, its dc gain is 1, if one desires a filter with a DC gain K 6= 1 it can be obtained by multiplying the magnitudenormalized filter by the constant K. Likewise, a filter H(S) designed with a normalized frequency, say 0 = / hp so that the normalized halfpower frequency is 1, is converted into a denormalized filter H(s) with a desired hp by replacing S = s/ hp in H(S).
Factorization To obtain a filter that satisfies the specifications and that is stable we need to factorize the magnitudesquared function. By letting S = s/hp be a normalized Laplace variable, then S/j = 0 = /hp and H(S)H(−S) =
1 1 + (−S2 )N
If the denominator can be factorized as D(S)D(−S) = 1 + (−S2 )N
(6.38)
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CH A P T E R 6: Application to Control and Communications
we let H(S) = 1/D(S)—that is, we assign to H(S) the poles in the lefthand splane so that the resulting filter is stable. The roots of D(S) in Equation (6.38) are S2N k =
ej(2k−1)π = ej(2k−1+N)π for integers k = 1, . . . , 2N e−jπN
after replacing −1 = ej(2k−1)π and (−1)N = e−jπN . The 2N roots are then Sk = ej(2k−1+N)π/(2N)
k = 1, . . . , 2N
(6.39)
Remarks n
n
n
Since Sk  = 1, the poles of the Butterworth filter are on a circle of unit radius. De Moivre’s theorem guarantees that the poles are also symmetrically distributed around the circle, and because of the condition that complex poles should be complex conjugate pairs, the poles are symmetrically distributed with respect to the σ axis. Letting S = s/hp be the normalized Laplace variable, then s = Shp , so that the denormalized filter H(s) has its poles in a circle of radius hp . No poles are on the j0 axis, as can be seen by showing that the angle of the poles are not equal to π/2 or 3π/2. In fact, for 1 ≤ k ≤ N, the angle of the poles are bounded below and above by letting 1 ≤ k and then k ≤ N to get 1 (2k − 1 + N)π π 1 π 1+ ≤ ≤ 3− 2 N 2N 2 N and for integers N ≥ 1 the above indicates that the angle will not be equal to either π/2 or 3π/2, or on the j0 axis. Consecutive poles are separated by π/N radians from each other. In fact, subtracting the angles of two consecutive poles can be shown to give ±π/N.
Using the above remarks and the fact that the poles must be in conjugate pairs, since the coefficients of the filter are realvalued, it is easy to determine the location of the poles geometrically.
n Example 6.7 A secondorder lowpass Butterworth filter, normalized in magnitude and in frequency, has a transfer function of 1 H(S) = √ 2 S + 2S + 1 We would like to obtain a new filter H(s) with a dc gain of 10 and a halfpower frequency hp = 100 rad/sec. The DC gain of H(S) is unity—in fact, when = 0, S = j0 gives H( j0) = 1. The halfpower frequency of H(S) is unity, indeed letting 0 = 1, then S = j1 and 1 1 H( j1) = = √ √ 2 j +j 2+1 j 2 so that H( j1)2 = H( j0)2 /2 = 1/2, or 0 = 1 is the halfpower frequency.
6.5 Analog Filtering
Thus, the desired filter with a dc gain of 10 is obtained by multiplying H(S) by 10. Furthermore, if we let S = s/100 be the normalized Laplace variable when S = j0hp = j1, we get that s = jhp = j100, or hp = 100, the desired halfpower frequency. Thus, the denormalized filter in frequency H(s) is obtained by replacing S = s/100. The denormalized filter in magnitude and frequency is then H(s) =
10 105 = √ √ (s/100)2 + 2(s/100) + 1 s2 + 100 2s + 104
n
Design For the Butterworth lowpass filter, the design consists in finding the parameters N, the minimum order, and hp , the halfpower frequency, of the filter from the constrains in the passband and in the stopband. The loss function for the lowpass Butterworth is α() = −10 log10 HN (/hp )2 = 10 log10 (1 + / hp
2N
)
The loss specifications are 0 ≤ α() ≤ αmax
0 ≤ ≤ p
αmin ≤ α() < ∞
≥ s
10 log10 (1 + p / hp
2N
At = p , we have that ) ≤ αmax
so that
p hp
2N
≤ 100.1αmax − 1
(6.40)
and similarly for = s , we have that 100.1αmin − 1 ≤
s hp
2N (6.41)
We then have that from Equation (6.40) and (6.41), the halfpower frequency is in the range p s ≤ hp ≤ 1/2N 0.1α − 1) (10 min − 1)1/2N
(100.1αmax
(6.42)
and from the log of the two extremes of Equation (6.42), we have that N≥
log10 [(100.1αmin − 1)/(100.1αmax − 1)] 2 log10 (s / p )
(6.43)
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Remarks n
n
n
According to Equation (6.43) when either n The transition band is narrowed (i.e., p → s ), or n The loss αmin is increased, or n The loss αmax is decreased the quality of the filter is improved at the cost of having to implement a filter with a high order N. The minimum order N is an integer larger or equal to the right side of Equation (6.43). Any integer larger than the minimum N also satisfies the specifications but increases the complexity of the filter. Although there is a range of possible values for the halfpower frequency, it is typical to make the frequency response coincide with either the passband or the stopband specifications giving a value for the halfpower frequency in the range. Thus, we can have either hp =
p − 1)1/2N
(6.44)
s − 1)1/2N
(6.45)
(100.1αmax
or hp =
n
n
(100.1αmin
as possible values for the halfpower frequency. The design aspect is clearly seen in the flexibility given by the equations. We can select out of an infinite possible set of values of N and of halfpower frequencies. The optimal order is the smallest value of N and the halfpower frequency can be taken as one of the extreme values. After the factorization, or the formation of D(S) from the poles, we need to denormalize the obtained transfer function HN (S) = 1/D(S) by letting S = s/ hp to get HN (s) = 1/D(s/ hp ), the filter that satisfies the specifications. If the desired DC gain is not unit, the filter needs to be denormalized in magnitude by multiplying it by an appropriate gain K.
6.5.3 Chebyshev LowPass Filter Design The normalized magnitudesquared function for the Chebyshev lowpass filter is given by HN (0 )2 =
1 2 (/ ) 1 + ε2 CN p
0 =
p
(6.46)
where the frequency is normalized with respect to the passband frequency p so that 0 = / p , N stands for the order of the filter, ε is a ripple factor, and CN (.) are the Chebyshev orthogonal4 polynomials of the first kind defined as cos(N cos−1 (0 )) 0  ≤ 1 0 CN ( ) = (6.47) −1 cosh(N cosh (0 )) 0  > 1 The definition of the Chebyshev polynomials depends on the value of 0 . Indeed, whenever 0  > 1, the definition based in the cosine is not possible since the inverse would not exist; thus the cosh(.) 4 Pafnuty
Chebyshev (1821–1894), a brilliant Russian mathematician, was probably the first one to recognize the general concept of orthogonal polynomials.
6.5 Analog Filtering
definition is used. Likewise, whenever 0  ≤ 1, the definition based in the hyperbolic cosine would not be possible since the inverse of this function only exists for values of 0 bigger or equal to 1 and so the cos(.) definition is used. From the definition it is not clear that CN (0 ) is an Nthorder polynomial in 0 . However, if we let θ = cos−1 (0 ) or 0 = cos(θ ) when 0  ≤ 1, we have that CN (0 ) = cos(Nθ ) and CN+1 (0 ) = cos((N + 1)θ) = cos(Nθ ) cos(θ ) − sin(Nθ ) sin(θ ) CN−1 (0 ) = cos((N − 1)θ) = cos(Nθ ) cos(θ ) + sin(Nθ ) sin(θ ) so that adding them we get CN+1 (0 ) + CN−1 (0 ) = 2 cos(θ ) cos(Nθ ) = 20 CN (0 ) This gives a threeterm expression for computing CN (0 ), or a difference equation CN+1 (0 ) + CN−1 (0 ) = 20 CN (0 )
N≥0
(6.48)
with initial conditions C0 (0 ) = cos(0) = 1 C1 (0 ) = cos(cos−1 (0 )) = 0 We can then see that C0 (0 ) = 1 C1 (0 ) = 0 C2 (0 ) = −1 + 202 C3 (0 ) = −30 + 403 .. . which are polynomials in 0 of order N = 0, 1, 2, 3, . . .. In Chapter 0 we gave a script to compute and plot these polynomials using symbolic MATLAB. Remarks n
n
Two fundamental characteristics of the CN (0 ) polynomials are: (1) they vary between 0 and 1 in the range 0 ∈ [−1, 1], and (2) they grow outside this range (according to their definition, the Chebyshev polynomials outside this range become cosh(.) functions, which are functions always bigger than 1). The first characteristic generates ripples in the passband, while the second makes these filters have a magnitude response that goes to zero faster than Butterworth’s. There are other characteristics of interest for the Chebyshev polynomials. The Chebyshev polynomials are unity at 0 = 1 (i.e., CN (1) = 1 for all N). In fact, C0 (1) = 1, C1 (1) = 1, and if we assume that CN−1 (1) = CN (1) = 1, we then have that CN+1 (1) = 1 according to the threeterm recursion. This indicates that the magnitudesquare function is HN ( j1)2 = 1/(1 + ε2 ) for any N.
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CH A P T E R 6: Application to Control and Communications
n
n
Different from the Butterworth filter that has a unit dc gain, the dc gain of the Chebyshev filter depends on the order of the filter. This is due to the property of the Chebyshev polynomial of being CN (0) = 0 if √ N is odd and 1 if N is even. Thus, the dc gain is 1 when N is odd, but 1/ 1 + ε2 when N is even. This is due to the fact that the Chebyshev polynomials of odd order do not have a constant term, and those of even order have 1 or −1 as the constant term. 0 Finally, the polynomials CN ( √ ) have N real roots between −1 and 1. Thus, the Chebyshev filter displays N/2 ripples between 1 and 1 + ε2 for normalized frequencies between 0 and 1.
Design The loss function for the Chebyshev filter is 2 (0 ) α(0 ) = 10 log10 1 + ε2 CN
0 =
p
(6.49)
The design equations for the Chebyshev filter are obtained as follows: n
Ripple factor ε and ripple width (RW): From CN (1) = 1, and letting the loss equal αmax at that normalized frequency, we have that p ε = 100.1αmax − 1 1 RW = 1 − √ 1 + ε2
(6.50)
0
n
Minimum order: The loss function at s is bigger or equal to αmin , so that solving for the Chebyshev polynomial we get after replacing ε, 0
0
CN (s ) = cosh(N cosh−1 (s )) .1αmin 0.5 10 −1 ≥ 10.1αmax − 1 0
where we used the cosh(.) definition of the Chebyshev polynomials since s > 1. Solving for N we get h i0.5 −1 100.1αmin −1 cosh 100.1αmax −1 N≥ (6.51) s cosh−1 p n
Halfpower frequency: Letting the loss at the halfpower frequency equal 3 dB and using that 100.3 ≈ 2, we obtain from Equation 6.49 the Chebyshev polynomial at that normalized frequency to be 0
CN (hp ) =
1 ε
0 = cosh N cosh−1 (hp )
6.5 Analog Filtering
0
where the last term is the definition of the Chebyshev polynomial for hp > 1. Thus, we get 1 1 hp = p cosh cosh−1 N ε
(6.52)
Factorization The factorization of the magnitudesquared function is a lot more complicated for the Chebyshev filter than for the Butterworth filter. If we let the normalized variable S = s/ p equal j0 , the magnitudesquared function can be written as H(S)H(−S) =
1 1 = 2 (S/j) D(S)D(−S) 1 + ε 2 CN
As before in the Butterworth case, the poles in the lefthand splane gives H(S) = 1/D(S), a stable filter. The poles of the H(S) can be found to be in an ellipse. They can be connected with the poles of the corresponding order Butterworth filter by an algorithm due to Professor Ernst Guillemin. The poles of H(S) are given by the following equations for k = 1, . . . , N, with N the minimal order of the filter: 1 −1 1 a = sinh N ε σk = −sinh(a) cos(ψk )
(real part)
0k
(imaginary part)
= ± cosh(a) sin(ψk )
(6.53)
where 0 ≤ ψk < π/2 (refer to Equation 6.39) are the angles corresponding to the Butterworth filters (measured with respect to the negative real axis of the splane). Remarks n
n
n
The dc gain of the Chebyshev filter is not easy to determine as in the Butterworth filter, as it depends on the order N. We can, however, set the desired dc value by choosing the appropriate value of a gain K so ˆ that H(S) = K/D(S) satisfies the dc gain specification. The poles of the Chebyshev filter depend now on the ripple factor ε and so there is no simple way to find them as it was in the case of the Butterworth. The final step is to replace the normalized variable S = s/ p in H(S) to get the desired filter H(s).
n Example 6.8 Consider the lowpass filtering of an analog signal x(t) = [−2 cos(5t) + cos(10t) + 4 sin(20t)]u(t) with MATLAB. The filter is a thirdorder lowpass Butterworth filter with a halfpower frequency hp = 5 rad/sec—that is, we wish to attenuate the frequency components of the frequencies 10 and 20 rad/sec. Design the desired filter and show how to do the filtering. The design of the filter is done using the MATLAB function butter where besides the specification of the desired order, N = 3, and halfpower frequency, hp = 5 rad/sec, we also need to indicate that
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CH A P T E R 6: Application to Control and Communications
the filter is analog by including an ’s’ as one of the arguments. Once the coefficients of the filter are obtained, we could then either solve the differential equation from these coefficients or use the Fourier transform, which we choose to do. Symbolic MATLAB is thus used to compute the Fourier transform of the input X(), and after generating the frequency response function H( j) from the filter coefficients, we multiply these two to get Y(), which is inversely transformed to obtain y(t). To obtain H( j) symbolically we multiply the coefficients of the numerator and denominator obtained from butter by variables ( j)n where n corresponds to the order of the coefficient in the numerator or the denominator, and then add them. The poles of the designed filter and its magnitude response are shown in Figure 6.23, as well as the input x(t) and the output y(t). The following script was used for the filter design and the filtering of the given signal. cos(10t ) − 2cos(5t ) + 4sin(20t ) 5 x (t )
1
5 4 3
0.9 0.8
H( jΩ)
0 −1 −2
0.6
σ
0
5
10 t
15
20
0
5
10 t
15
20
0.5 0.4
5 y (t )
0.3
−3 −4 −5 −6 −4 −2
0 −5
0.7
2 1 jΩ
400
0.2 0.1
0 −5
0 0
2
0
10
20
30
Ω
FIGURE 6.23 Filtering of an analog signal x(t) using a lowpass Butterworth filter. Notice that the output of the filter is approximately the sinusoid of 5 rad/sec in x(t), as the other two components have been attenuated.
%%%%%%%%%%%%%%%%%%% % Example 6.8  Filtering with Butterworth filter %%%%%%%%%%%%%%%%%%% clear all; clf syms t w x = cos(10 ∗ t) − 2 ∗ cos(5 ∗ t) + 4 ∗ sin(20 ∗ t); % input signal X = fourier(x); N = 3; Whp = 5; % filter parameters [b, a] = butter(N, Whp, ’s’); % filter design W = 0:0.01:30; Hm = abs(freqs(b, a, W)); % magnitude response in W % filter output n = N:−1:0; U = ( j ∗ w).ˆn num = b − conj(U’); den = a − conj(U’);
6.5 Analog Filtering
H = num/den; % frequency response Y = X ∗ H; % convolution property y = ifourier(Y, t); % inverse Fourier
n
n Example 6.9 In this example we will compare the performance of Butterworth and Chebyshev lowpass filters in the filtering of an analog signal x(t) = [−2 cos(5t) + cos(10t) + 4 sin(20t)]u(t) using MATLAB. We would like the two filters to have the same halfpower frequency. The magnitude specifications for the lowpass Butterworth filter are αmax = 0.1dB, p = 5rad/sec
(6.54)
αmin = 15dB, s = 10 rad/sec
(6.55)
and a dc loss of 0 dB. Once this filter is designed, we would like the Chebyshev filter to have the same halfpower frequency. In order to obtain this, we need to change the p specification for the Chebyshev filter. To do that we use the formulas for the halfpower frequency of this type of filter to find the new value for p . The Butterworth filter is designed by first determining the minimum order N and the halfpower frequency hp using the function buttord, and then finding the filter coefficients by means of the function butter. Likewise, for the design of the Chebyshev filter we use the function cheb1ord to find the minimum order and the cutoff frequency (the new p is obtained from the halfpower frequency). The filtering is implemented using the Fourier transform as before. There are two significant differences between the designed Butterworth and Chebyshev filters. Although both of them have the same halfpower frequency, the transition band of the Chebyshev filter is narrower, [6.88 10], than that of the Butterworth filter, [5 10], indicating that the Chebyshev is a better filter. The narrower transition band is compensated by a lower minimum order of five for the Chebyshev compared to the sixorder Butterworth. Figure 6.24 displays the poles of the Butterworth and the Chebyshev filters, their magnitude responses, as well as the input signal x(t) and the output y(t) for the two filters (the two perform very similarly). %%%%%%%%%%%%%%%%%%% % Example 6.9  Filtering with Butterworth and Chebyshev filters %%%%%%%%%%%%%%%%%%% clear all;clf syms t w x = cos(10 ∗ t) − 2 ∗ cos(5 ∗ t) + 4 ∗ sin(20 ∗ t); X = fourier(x); wp = 5;ws = 10;alphamax = 0.1;alphamin = 15; % filter parameters % butterworth filter [N, whp] = buttord(wp, ws, alphamax, alphamin, ’s’) [b, a] = butter(N, whp, ’s’) % cheby1 filter epsi = sqrt(10ˆ(alphamax/10) −1)
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CH A P T E R 6: Application to Control and Communications
Butterworth 1
0
cos(10t) − 2cos(5t) + 4sin(20t )
Butterworth Chebyshev
0.9
5
−5
x(t)
0.8
−8 −6 −4 −2 σ
0
2
Chebyshev 5
H( jΩ)
0.7
0.5
0
0.2
−5
0.1 −2
0 σ
2
0
5
10 t
15
20
0
5
10 t
15
20
0.4 0.3
−4
0 −5
0.6
0
5 y(t )
jΩ
5
jΩ
402
0 −5
0
10
20
30
Ω
FIGURE 6.24 Comparison of filtering of an analog signal x(t) using a lowpass Butterworth and Chebyshev filter with the same halfpower frequency.
wp = whp/cosh(acosh(1/epsi)/N) % recomputing wp to get same whp [N1, wn] = cheb1ord(wp, ws, alphamax, alphamin, ’s’); [b1, a1] = cheby1(N1, alphamax, wn, ’s’); % frequency responses W = 0:0.01:30; Hm = abs(freqs(b, a, W)); Hm1 = abs(freqs(b1, a1, W)); % generation of frequency response from coefficients n = N:−1:0; n1 = N1:−1:0; U = ( j ∗ w).ˆn; U1 = ( j ∗ w).ˆn1 num = b ∗ conj(U’); den = a ∗ conj(U’); num1 = b1 ∗ conj(U1’); den1 = a1 ∗ conj(U1’) H = num/den; % Butterworth LPF H1 = num1/den1; % Chebyshev LPF % output of filter Y = X ∗ H; Y1 = X ∗ H1; y = ifourier(Y, t) y1 = ifourier(Y1, t)
n
6.5.4 Frequency Transformations As indicated before, the design of an analog filter is typically done by transforming the frequency of a normalized prototype lowpass filter. The frequency transformations were developed by Professor Ronald Foster [72] using the properties of reactance functions. The frequency transformations for the
6.5 Analog Filtering
basic filters are given by: Low passlow pass : Low passhigh pass : Low passband pass :
s 0 0 S= s s2 + 20 S= s BW
S=
Low passband eliminating :
S=
s BW s2 + 20
(6.56)
where S is the normalized and s the final variables, while 0 is a desired cutoff frequency and BW is a desired bandwidth. Remarks n
n
The lowpass to lowpass (LPLP) and lowpass to highpass (LPHP) transformations are linear in the numerator and denominator; thus the number of poles and zeros of the prototype lowpass filter is preserved. On the other hand, the lowpass to bandpass (LPBP) and lowpass to bandeliminating (LPBE) transformations are quadratic in either the numerator or the denominator, so that the number of poles/zeros is doubled. Thus, to obtain a 2Nthorder bandpass or bandeliminating filter the prototype lowpass filter should be of order N. This is an important observation useful in the design of these filters with MATLAB. It is important to realize that only frequencies are transformed, and the magnitude of the prototype filter is preserved. Frequency transformations will be useful also in the design of discrete filters, where these transformations are obtained in a completely different way, as no reactance functions would be available in that domain.
n Example 6.10 To illustrate how the above transformations can be used to convert a prototype lowpass filter we use the following script. First a lowpass prototype filter is designed using butter, and then to this filter we apply the lowpass to highpass transformation with 0 = 40 (rad/sec) to obtain a highpass filter. Let then 0 = 6.32 (rad/sec) and BW = 10 (rad/sec) to obtain a bandpass and a bandeliminating filters using the appropriate transformations. The following is the script used. The magnitude responses are plotted with ezplot. Figure 6.25 shows the results. clear all; clf syms w N = 5; [b, a] = butter(N, 1, ’s’) % lowpass prototype omega0 = 40;BW = 10; omega1=sqrt(omega0); % transformation parameters % lowpass prototype n = N:−1:0; U = ( j ∗ w).ˆn; num = b ∗ conj(U’); den = a ∗ conj(U’); H = num/den; % lowpass to highpass
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CH A P T E R 6: Application to Control and Communications
LP−HP
1
1
0.8
0.8 H1( j Ω)
H(j Ω)
LP Prototype
0.6 0.4
0.6 0.4
0.2
0.2
0
0 0
5 Ω (a)
0
10
20
0.8
0.8 H3( j Ω)
1
0.6 0.4
30
40
0.4 0.2
0
0 20 Ω (c)
80
0.6
0.2
10
60
LP−BE
1
0
40 Ω (b)
LP−BP
H2(jΩ)
404
30
40
0
10
20 Ω (d)
FIGURE 6.25 Frequency transformations: (a) prototype lowpass filter, (b) lowpass to highpass transformation, (c) lowpass to bandpass transformation, and (d) lowpass to bandeliminating transformation.
U1 = (omega0/( j ∗ w)).ˆn; num1 = b ∗ conj(U1’); den1 = a ∗ conj(U1’); H1 = num1/den1; % lowpass to bandpass U2 = ((−wˆ2 + omega1ˆ2)/(BW ∗ j ∗ w)).ˆn num2 = b ∗ conj(U2’); den2 = a ∗ conj(U2’); H2 = num2/den2; % lowpass to bandeliminating U3 = ((BW ∗ j ∗ w)/(−wˆ2 + omega1ˆ2)).ˆn num3 = b ∗ conj(U3’); den3 = a ∗ conj(U3’); H3 = num3/den3
n
6.5 Analog Filtering
6.5.5 Filter Design with MATLAB The design of filters, analog and discrete, is simplified by the functions that MATLAB provides. Functions to find the filter parameters from magnitude specifications, as well as functions to find the filter poles/zeros and to plot the designed filter magnitude and phase responses, are available.
LowPass Filter Design The design procedure is similar for all of the approximation methods (Butterworth, Chebyshev, elliptic) and consists of both n n
Finding the filter parameters from loss specifications. Obtaining the filter coefficients from these parameters.
Thus, to design an analog lowpass filter using the Butterworth approximation, the loss specifications αmax and αmin , and the frequency specifications, p and s are first used by the function buttord to determine the minimum order N and the halfpower frequency hp of the filter that satisfies the specifications. Then the function butter uses these two values to determine the coefficients of the numerator and the denominator of the designed filter. We can then use the function freqs to plot the designed filter magnitude and phase. Similarly, this applies for the design of lowpass filters using the Chebyshev or the elliptic design methods. To include the design of lowpass filters using the Butterworth, Chebyshev (two versions), and the elliptic methods we wrote the function analogfil. function [b, a] = analogfil(Wp, Ws, alphamax, alphamin, Wmax, ind) %% % Analog filter design % Parameters % Input: loss specifications (alphamax, alphamin), corresponding % frequencies (Wp,Ws), frequency range [0,Wmax] and indicator ind (1 for % Butterworth, 2 for Chebyshev1, 3 for Chebyshev2 and 4 for elliptic). % Output: coefficients of designed filter. % Function plots magnitude, phase responses, poles and zeros of filter, and % loss specifications %%% if ind == 1,% Butterworth lowpass [N, Wn] = buttord(Wp, Ws, alphamax, alphamin, ’s’) [b, a] = butter(N, Wn, ’s’) elseif ind == 2, % Chebyshev lowpass [N, Wn] = cheb1ord(Wp, Ws, alphamax, alphamin, ’s’) [b, a] = cheby1(N, alphamax, Wn, ’s’) elseif ind == 3, % Chebyshev2 lowpass [N, Wn] = cheb2ord(Wp, Ws, alphamax, alphamin, ’s’) [b, a] = cheby2(N, alphamin, Wn, ’s’) else % Elliptic lowpass [N, Wn] = ellipord(Wp, Ws, alphamax, alphamin, ’s’) [b, a] = ellip(N, alphamax, alphamin, Wn, ’s’) end
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CH A P T E R 6: Application to Control and Communications
W = 0:0.001:Wmax; % frequency range for plotting H = freqs(b, a, W); Hm = abs(H); Ha = unwrap(angle(H)) % magnitude (Hm) and phase (Ha) N = length(W); alpha1 = alphamax ∗ ones(1, N); alpha2 = alphamin ∗ ones(1, N); % loss specs subplot(221) plot(W, Hm); grid; axis([0 Wmax 0 1.1 ∗ max(Hm)]) subplot(222) plot(W, Ha); grid; axis([0 Wmax 1.1 ∗ min(Ha) 1.1 ∗ max(Ha)]) subplot(223) splane(b, a) subplot(224) plot(W, −20 ∗ log10(abs(H))); hold on plot(W, alpha1, ’r’, W, alpha2, ’r’); grid; axis([0 max(W) −0.1 100]) hold off
n Example 6.11 To illustrate the use of analogfil consider the design of lowpass filters using the Chebyshev2 and the Elliptic design methods. The specifications for the designs are α(0) = 0, αmax = 0.1, αmin = 60 dB p = 10, s = 15 rad/sec
0
10 15 20 25 Ω
20 10 0 −10 −20 −4
−2
0
2
σ
100 80 60 40 20 0
5
10 15 20 25 Ω
0 in the sampling are: n The spectrum of the ideal sampled signal xs (t) is now weighted by the sinc function of the frequency response H( j) of the zeroorder hold filter. Thus, the spectrum of the sampled signal using the sampleandhold system will not be periodic and will decay as increases. n The reconstruction of the original signal x(t) requires a more complex filter than the one used in the ideal sampling. Indeed, the concatenation of the zeroorder hold filter with the reconstruction filter should be such that H(s)Hr (s) = 1, or that Hr (s) = 1/H(s). A circuit used for implementing the sampleandhold system is shown in Figure 7.11. In this circuit the switch closes every Ts seconds and remains closed for a short time 1. If the time constant rC > Ts when the switch opens 1 seconds later, the capacitor slowly discharges. The cycle repeats providing a signal that approximates the output of the sampleandhold system explained before. The DAC also uses a holder to generate an analog signal from the discrete signal coming out of the decoder into the DAC. There are different possible types of holders, providing an interpolation that will make the final smoothing of the signal a lot easier. The socalled zeroorder hold basically expands the sample value in between samples, providing a rough approximation of the discrete signal, which is then smoothed out by a lowpass filter to provide the analog signal.
7.4.2 Quantization and Coding Amplitude discretization of the sampled signal xs (t) is accomplished by a quantizer consisting of a number of fixed amplitude levels against which the sample amplitudes {x(nTs )} are compared. The output of the quantizer is one of the fixed amplitude levels that best represents x(nTs ) according to some approximation scheme. The quantizer is a nonlinear system. Independent of how many levels, or equivalently of how many bits are allocated to represent each level of the quantizer, there is a possible error in the representation of each sample. This is called the quantization error. To illustrate this, consider a 2bit or 22 level quantizer shown in Figure 7.12. The input of the quantizer are the samples x(nTs ), which are compared with the values in the bins [−21, −1], [−1, 0], [0, 1], and [1, 21], and depending on which of these bins the sample falls in it is replaced by the corresponding levels −21, −1, 0, or 1. The value of the quantization step 1 for the fourlevel quantizer is 1=
2 maxx(t) 22
(7.23)
441
442
CH A P T E R 7: Sampling Theory
∧
x (nTs ) 01
Δ −2Δ
−Δ
00 Δ 11
FIGURE 7.12 Fourlevel quantizer and coder.
10
2Δ
x (nTs )
−Δ −2Δ
That is, 1 is assigned so as to cover the possible peaktopeak range of values of the signal, or its dynamic range. To each of the levels a binary code is assigned. The code assigned to each of the levels uniquely represents the different levels [−21, −1, 0, 1]. As to the way to approximate the given sample to one of these levels, it can be done by rounding or by truncating. The quantizer shown in Figure 7.12 approximates by truncation—that is, if the sample k1 ≤ x(nTs ) < (k + 1)1, for k = −2, −1, 0, 1, then it is approximated by the level k1. To see the quantization, coding, and quantization error, let the sampled signal be x(nTs ) = x(t)t=nTS The given fourlevel quantizer is such that k1 ≤ x(nTs ) < (k + 1)1
⇒
xˆ (nTs ) = k1
k = −2, −1, 0, 1
(7.24)
where the sampled signal x(nTs ) is the input and the quantized signal xˆ (nTs ) is the output. Therefore, −21 ≤ x(nTs ) < −1
⇒
xˆ (nTs ) = −21
−1 ≤ x(nTs ) < 0
⇒
xˆ (nTs ) = −1
0 ≤ x(nTs ) < 1
⇒
xˆ (nTs ) = 0
1 ≤ x(nTs ) < 21
⇒
xˆ (nTs ) = 1
To transform the quantized values into unique binary 2bit values, one could use a code such as xˆ (nTs ) xˆ (nTs ) xˆ (nTs ) xˆ (nTs )
= −21 = −1 = 01 = 1
⇒ ⇒ ⇒ ⇒
10 11 00 01
which assigns a unique 2bit binary number to each of the four quantization levels. If we define the quantization error as ε(nTs ) = x(nTs ) − xˆ (nTs )
7.4 Practical Aspects of Sampling
and use the characterization of the quantizer given in Equation (7.24), we have then that the error ε(nTs ) is obtained from xˆ (nTs ) ≤ x(nTs ) ≤ xˆ (nTs ) + 1 by subtracting xˆ (nTs ) ⇒ 0 ≤ ε(nTs ) ≤ 1
(7.25)
indicating that one way to decrease the quantization error is to make the quantization step 1 very small. That clearly depends on the quality of the ADC. Increasing the number of bits of the ADC makes 1 smaller (see Equation (7.23) where the denominator is 2 raised to the number of bits), which will make the quantization error smaller. In practice, the quantization error is considered random, and so it needs to be characterized probabilistically. This characterization becomes meaningful only when the number of bits is large, and the input signal is not a deterministic signal. Otherwise, the error is predictable and thus not random. Comparing the energy of the input signal to the energy of the error, by means of the socalled signaltonoise ratio (SNR), it is possible to determine the number of bits that are needed in a quantizer to get a reasonable quantization error.
n Example 7.5 Suppose we are trying to decide between an 8 and a 9bit ADC for a certain application. The signals in this application are known to have frequencies that do not exceed 5 KHz. The amplitude of the signals is never more than 5 volts (i.e., the dynamic range of the signals is 10 volts, so that the signal is bounded as −5 ≤ x(t) ≤ 5). Determine an appropriate sampling period and compare the percentage of error for the two ADCs of interest. Solution The first consideration in choosing the ADC is the sampling period, so we need to get an ADC capable of sampling at fs = 1/Ts > 2fmax samples/sec. Choosing fs = 4fmax = 20 K samples/sec, then Ts = 1/20 msec/sample. Suppose then we look at an 8bit ADC, which means that the quantizer would have 28 = 256 levels so that the quantization step is 1 = 10/256 volts. If we use the truncation quantizer given above the quantization error would be 0 ≤ ε(nTs) ≤ 10/256 If we find that objectionable we can then consider a 9bit ADC, with a quantizer of 29 = 512 levels and the quantization step is 1 = 10/512 or half that of the 8bit ADC 0 ≤ ε(nTs ) ≤ 10/512 So that by increasing 1 bit we cut the quantization error in half from the previous quantizer (in practice, one of the 8 or 9 bits is used to determine the sign of the sampled value). Inputting a signal of constant amplitude 5 into the 9bit ADC gives a quantization error of [(10/512)/5] × 100% = (100/256)% ≈ 0.4% in representing the input signal. For the 8bit ADC it would correspond to a 0.8% error. n
443
CH A P T E R 7: Sampling Theory
7.4.3 Sampling, Quantizing, and Coding with MATLAB The conversion of an analog signal into a digital signal consists of three steps: sampling, quantization, and coding. These are the three operations an ADC does. To illustrate them consider a sinusoid x(t) = 4 cos(2πt). Its sampling period, according to the Nyquist sampling rate condition, is Ts ≤ π/max = 0.5 sec/sample as the maximum frequency of x(t) is max = 2π. We let Ts = 0.01 (sec/sample) to obtain a sampled signal xs (nTs ) = 4 cos(2πnTs ) = 4 cos(2πn/100), a discrete sinusoid of period 100. The following script is used to get the sampled x[n] and the quantized xq [n] signals and the quantization error ε[n] (see Figure 7.13).
4
4
2
2 x[n]
x(t )
Sampled signal
0 −2
0 −2
−4
−4 0
0.5 t (sec)
0
1
(a)
50 n
100
(b)
Quantized signal 4
4
3 e[n]
2 xq [n]
444
0 −2
2 1
−4
0 0
50 n (c)
100
0
50 n (d)
100
FIGURE 7.13 (a) Sinusoid, (b) sampled sinusoid using Ts = 0.01, (c) quantized sinusoid using four levels, and (d) quantization error.
7.4 Practical Aspects of Sampling
%%%%%%%%%%%%%%%%%%%%%%%%%%% % Sampling, quantization and coding %%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clf % analog signal t = 0:0.01:1; x = 4 ∗ sin(2 ∗ pi ∗ t); % sampled signal Ts = 0.01; N = length(t); n = 0:N − 1; xs = 4 ∗ sin(2 ∗ pi ∗ n ∗ Ts); % quantized signal Q = 2; % quantization levels is 2Q [d,y,e] = quantizer(x,Q); % binary signal z = coder(y,d)
The quantization of the sampled signal is implemented with the function quantizer which compares each of the samples xs (nTs ) with four levels and assigns to each the corresponding level. Notice the appproximation of the values given by the quantized signal samples to the actual values of the signal. The difference between the original and the quantized signal, or the quantization error, ε(nTs ), is also computed and shown in Figure 7.13. function [d,y,e] = quantizer(x,Q) % Input: x, signal to be quantized at 2Q levels % Outputs: y quantized signal % e, quantization error % d quantum % USE [d,y,e] = quantizer(x,Q) % N = length(x); d = max(abs(x))/Q; for k = 1:N, if x(k)> = 0, y(k) = floor(x(k)/d)*d; else if x(k) == min(x), y(k) = (x(k)/abs(x(k))) ∗ (floor(abs(x(k))/d) ∗ d); else y(k) = (x(k)/abs(x(k))) ∗ (floor(abs(x(k))/d) ∗ d + d); end end if y(k) == 2 ∗ d, y(k) = d; end end
445
446
CH A P T E R 7: Sampling Theory
The binary signal corresponding to the quantized signal is computed using the function coder which assigns the binary codes ’10’,’11’,’00’, and ’01’ to the four possible levels of the quantizer. The result is a sequence of 0s and 1s, each pair of digits sequentially corresponding to each of the samples of the quantized signal. The following is the function used to effect this coding. function z1 = coder(y,delta) % Coder for 4level quantizer % input: y quantized signal % output: z1 binary sequence % USE z1 = coder(y) % z1 = ’00’; % starting code N = length(y); for n = 1:N, y(n) if y(n) == delta z = ’01’; elseif y(n) == 0 z = ’00’; elseif y(n) == delta z = ’11’; else z = ’10’; end z1 = [z1 z]; end M = length(z1); z1 = z1(3:M) % get rid of starting code
7.5 WHAT HAVE WE ACCOMPLISHED? WHERE DO WE GO FROM HERE? The material in this chapter is the bridge between analog and digital signal processing. The sampling theory provides the necessary information to convert a continuoustime signal into a discretetime signal and then into a digital signal with minimum error. It is the frequency representation of an analog signal that determines the way in which it can be sampled and reconstructed. Analogtodigital and digitaltoanalog converters are the devices that in practice convert an analog signal into a digital signal and back. Two parameters characterizing these devices are the sampling rate and the number of bits each sample is coded into. The rate of change of a signal determines the sampling rate, while the precision in representing the samples determines the number of levels of the quantizer and the number of bits assigned to each sample. In the following chapters we will consider the analysis of discretetime signals, as well as the analysis and synthesis of discrete systems. The effect of quantization in the processing and design of systems
Problems
is an important problem that is left for texts in digital signal processing. We will, however, develop the theory of discretetime signals.
PROBLEMS 7.1. Sampling actual signals Consider the sampling of real signals. (a) Typically, a speech signal that can be understood over a telephone shows frequencies from about 100 Hz to about 5 KHz. What would be the sampling frequency fs (samples/sec) that would be used to sample speech without aliasing? How many samples would you need to save when storing an hour of speech? If each sample is represented by 8 bits, how many bits would you have to save for the hour of speech? (b) A music signal typically displays frequencies from 0 up to 22 KHz. What would be the sampling frequency fs that would be used in a CD player? (c) If you have a signal that combines voice and musical instruments, what sampling frequency would you use to sample this signal? How would the signal sound if played at a frequency lower than the Nyquist sampling frequency? 7.2. Sampling of bandlimited signals Consider the sampling of a sinc signal and related signals. (a) For the signal x(t) = sin(t)/t, find its magnitude spectrum X() and determine if this signal is band limited or not. (b) Suppose you want to sample x(t)). What would be the sampling period Ts you would use for the sampling without aliasing? (c) For a signal y(t) = x2 (t), what sampling frequency fs would you use to sample it without aliasing? How does this frequency relate to the sampling frequency used to sample x(t)? (d) Find the sampling period Ts to sample x(t) so that the sampled signal xs (0) = 1, otherwise xs (nTs ) = 0 for n 6= 0. 7.3. Sampling of timelimited signals—MATLAB Consider the signals x(t) = u(t) − u(t − 1) and y(t) = r(t) − 2r(t − 1) + r(t − 2). (a) Are either of these signals band limited? Explain. (b) Use Parseval’s theorem to determine a reasonable value for a maximum frequency for these signals (choose a frequency that would give 90% of the energy of the signals). Use MATLAB. (c) If we use the sampling period corresponding to y(t) to sample x(t), would aliasing occur? Explain. (d) Determine a sampling period that can be used to sample both x(t) and y(t) without causing aliasing in either signal. 7.4. Uncertainty in time and frequency—MATLAB Signals of finite time support have infinite support in the frequency domain, and a bandlimited signal has infinite time support. A signal cannot have finite support in both domains. (a) Consider x(t) = (u(t + 0.5) − u(t − 0.5))(1 + cos(2π t)). Find its Fourier transform X(). Compute the energy of the signal, and determine the maximum frequency of a bandlimited approximation signal xˆ (t) that would give 95% of the energy of the original signal. (b) The fact that a signal cannot be of finite support in both domains is expressed well by the uncertainty principle, which says that 1(t)1() ≥
1 4π
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where ∞ 0.5 R 2 2 dt t x(t) −∞ 1(t) = Ex measures the duration of the signal for which the signal is significant in time, and 0.5 ∞ R 2 X()2 d −∞ 1() = Ex measures the frequency support for which the Fourier representation is significant. The energy of the signal is represented by Ex . Compute 1(t) and 1() for the given signal x(t) and verify that the uncertainty principle is satisfied. 7.5. Nyquist sampling rate condition and aliasing Consider the signal x(t) = (a) (b) (c) (d)
sin(0.5t) 0.5t
Find the Fourier transform X() of x(t). Is x(t) band limited? If so, find its maximum frequency max . Suppose that Ts = 2π . How does s relate to the Nyquist frequency 2max ? Explain. What is the sampled signal x(nTs ) equal to? Carefully plot it and explain if x(t) can be reconstructed.
7.6. Antialiasing Suppose you want to find a reasonable sampling period Ts for the noncausal exponential x(t) = e−t (a) (b) (c) (d)
Find the Fourier transform of x(t), and plot X(). Is x(t) band limited? Find a frequency 0 so that 99% of the energy of the signal is in −o ≤ ≤ o . If we let s = 2π/Ts = 50 , what would be Ts ? Determine the magnitude and bandwidth of an antialiasing filter that would change the original signal into the bandlimited signal with 99% of the signal energy.
7.7. Sampling of modulated signals Assume you wish to sample an amplitude modulated signal x(t) = m(t) cos(c t) where m(t) is the message signal and c = 2π 104 rad/sec is the carrier frequency. (a) If the message is an acoustic signal with frequencies in a band of [0, 22] KHz, what would be the maximum frequency present in x(t)? (b) Determine the range of possible values of the sampling period Ts that would allow us to sample x(t) satisfying the Nyquist sampling rate condition. (c) Given that x(t) is a bandpass signal, compare the above sampling period with the one that can be used to sample bandpass signals.
Problems
7.8. Sampling output of nonlinear system The input–output relation of a nonlinear system is y(t) = x2 (t) where x(t) is the input and y(t) is the output. (a) The signal x(t) is band limited with a maximum frequency M = 2000π rad/sec. Determine if y(t) is also band limited, and if so, what is its maximum frequency max ? (b) Suppose that the signal y(t) is lowpass filtered. The magnitude of the lowpass filter is unity and the cutoff frequency is c = 5000π rad/sec. Determine the value of the sampling period Ts according to the given information. (c) Is there a different value for Ts that would satisfy the Nyquist sampling rate condition for both x(t) and y(t) and that is larger than the one obtained above? Explain. 7.9. Signal reconstruction You wish to recover the original analog signal x(t) from its sampled form x(nTs ). (a) If the sampling period is chosen to be Ts = 1 so that the Nyquist sampling rate condition is satisfied, determine the magnitude and cutoff frequency of an ideal lowpass filter H( j) to recover the original signal and plot them. (b) What would be a possible maximum frequency of the signal? Consider an ideal and a nonideal lowpass filter to reconstruct x(t). Explain. 7.10. CD player versus record player Explain why a CD player cannot produce the same fidelity of music signals as a conventional record player. (If you do not know what these are, ignore this problem, or get one to find out what they do or ask your grandparents about LPs and record players!) 7.11. Twobit analogtodigital converter—MATLAB Let x(t) = 0.8 cos(2π t) + 0.15, 0 ≤ t ≤ 1, and zero otherwise, be the input to a 2bit analogtodigital converter. (a) For a sampling period Ts = 0.025 sec determine and plot using MATLAB the sampled signal, x(nTs ) = x(t)t=nTS (b) The fourlevel quantizer (see Figure 1.2) corresponding to the 2bit ADC is defined as k1 ≤ x(nTs ) < (k + 1)1
→
xˆ (nTs ) = k1
k = −2, −1, 0, 1
(7.26)
where x(nTs ), found above, is the input and xˆ (nTs ) is the output of the quantizer. Let the quantization step be 1 = 0.5. Plot the input–output characterization of the quantizer, and find the quantized output for each of the sample values of the sampled signal x(nTs ). (c) To transform the quantized values into unique binary 2bit values, consider the following code: xˆ (nTs ) = −21
→
10
xˆ (nTs ) = −1
→
11
xˆ (nTs ) = 01
→
00
xˆ (nTs ) = 1
→
01
Obtain the digital signal corresponding to x(t).
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CHAPTER 8
DiscreteTime Signals and Systems
´ avu, ` It’s like dej all over again. Lawrence “Yogi” Berra (1925) Yankees baseball player
8.1 INTRODUCTION As you will see in this chapter, the basic theory of discretetime signals and systems is very much like that for continuoustime signals and systems. However, there are significant differences that need to be understood. Specifically in this chapter we will consider the following contrasting issues: n
n
n
Discretetime signals resulting from sampling of continuoustime signals are only available at uniform times determined by the sampling period; they are not defined inbetween sampling periods. It is important to emphasize the significance of sampling according to the Nyquist sampling rate condition since the characteristics of discretetime signals will depend on it. Given the knowledge of the sampling period, discretetime signals depend on an integer variable n, which unifies the treatment of discretetime signals obtained from analog signals by sampling and those that are naturally discrete. It will also be seen that the frequency in the discrete domain differs from the analog frequency. The radian discrete frequency cannot be measured, and depends on the sampling period used whenever the discretetime signals result from sampling. Although the concept of periodicity of discretetime signals coincides with that for continuoustime signals, there are significant differences. As functions of an integer variable, discretetime periodic signals must have integer periods. This imposes some restrictions that do not exist in continuoustime periodic signals. For instance, continuoustime sinusoids are always periodic as their period can be a positive real number; however, that will not be the case for discretetime sinusoids. It is possible to have discretetime sinusoids that are not periodic, even if they resulted from the uniform sampling of continuoustime sinusoids. Characteristics such as energy, power, and symmetry of continuoustime signals are conceptually the same for discretetime signals. Integrals are replaced by sums, derivatives by finite differences, and differential equations by difference equations. Likewise, one can define a set of basic signals
Signals and Systems Using MATLAB®. DOI: 10.1016/B9780123747167.000120 c 2011, Elsevier Inc. All rights reserved.
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n
just like those for continuoustime signals. However, some of these basic signals do not display the mathematical complications of their continuoustime counterparts. For instance, the discreteimpulse signal is defined at every integer value in contrast with the continuousimpulse response, which is not defined at zero. The discrete approximation of derivatives and integrals provides an approximation of differential equations, representing dynamic continuoustime systems by difference equations. Extending the concept of linear time invariance to discretetime systems, we obtain a convolution sum to represent LTI systems. Thus, dynamic discretetime systems can be represented by difference equations and convolution sums. A computationally significant difference with continuoustime systems is that the solution of difference equations can be recursively obtained, and that the convolution sum provides a class of systems that do not have a counterpart in the analog domain.
8.2 DISCRETETIME SIGNALS A discretetime signal x[n] can be thought of as a real or complexvalued function of the integer sample index n: x[.] : I → R (C) n
(8.1)
x[n]
The above means that for discretetime signals the independent variable is an integer n, the sample index, and that the value of the signal at n, x[n], is either a real or a complexvalue function. Thus, the signal is only defined at integer values n—no definition exists for values between the integers. Remarks n
n
It should be understood that a sampled signal x(nTs ) = x(t)t=nTs is a discretetime signal x[n] that is a function of n only. Once the value of Ts is known, the sampled signal only depends on n, the sample index. However, this should not prevent us in some situations from considering a discretetime signal obtained through sampling as a function of time t where the signal values only exist at discrete times {nTs }. Although in many situations discretetime signals are obtained from continuoustime signals by sampling, that is not always the case. There are many signals that are inherently discrete—think, for instance, of a signal consisting of the final values attained daily by the shares of a company in the stock market. Such a signal would consist of the values reached by the share in the days when the stock market opens. This signal is naturally discrete. A signal generated by a random number generator in a computer would be a sequence of real values and can be considered a discretetime signal. Telemetry signals, consisting of measurements—for example, voltages, temperatures, pressures—from a certain process, taken at certain times, are also naturally discrete.
n Example 8.1 Consider a sinusoidal signal x(t) = 3 cos(2πt + π/4)
−∞ < t < ∞
8.2 DiscreteTime Signals
Determine an appropriate sampling period Ts according to the Nyquist sampling rate condition, and obtain the discretetime signal x[n] corresponding to the largest allowed sampling period. Solution To sample x(t) so that no information is lost, the Nyquist sampling rate condition indicates that the sampling period should be Ts ≤
π π = = 0.5 max 2π
For the largest allowed sampling period Ts = 0.5, we obtain x[n] = 3 cos(2πt + π/4)t=0.5n = 3 cos(πn + π/4)
−∞ < n < ∞
which is a function of the integer n.
n
n Example 8.2 To generate the celebrated Fibonacci sequence of numbers, {x[n]}, we use the recursive equation x[n] = x[n − 1] + x[n − 2]
n≥2
x[0] = 0 x[1] = 1 which is a difference equation with zero input and two initial conditions. The Fibonacci sequence has been used to model different biological systems.1 Find the Fibonacci sequence. Solution The given equation allows us to compute the Fibonacci sequence recursively. For n ≥ 2, we find x[2] = 1 + 0 = 1 x[3] = 1 + 1 = 2 x[4] = 2 + 1 = 3 x[5] = 3 + 2 = 5 .. . where we are simply adding the previous two numbers in the sequence. The sequence is purely discrete as it is not related to a continuoustime signal. n 1 Leonardo of Pisa (also known as Fibonacci) in his book Liber Abaci described how his sequence could be used to model the reproduction of rabbits over a number of months assuming bunnies begin breeding when they are a few months old.
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8.2.1 Periodic and Aperiodic Signals A discretetime signal x[n] is periodic if It is defined for all possible values of n, −∞ < n < ∞. n There is a positive integer N, the period of x[n], such that n
(8.2)
x[n + kN] = x[n] for any integer k. Periodic discretetime sinusoids, of period N, are of the form 2π m n+θ x[n] = A cos N
−∞ < n < ∞
(8.3)
where the discrete frequency is ω0 = 2π m/N rad, for positive integers m and N, which are not divisible by each other, and θ is the phase angle.
The definition of a discretetime periodic signal is similar to that of continuoustime periodic signals, except for the period being an integer. That discretetime sinusoids are of the given form can be easily shown: Shifting the sinusoid in Equation (8.3) by a multiple k of the period N, we have 2πm x[n + kN] = A cos (n + kN) + θ N 2πm = A cos n + 2πmk + θ = x[n] N since we add to the original angle a multiple mk (an integer) of 2π, which does not change the angle. Remarks n
n
The units of the discrete frequency ω is radians. Moreover, discrete frequencies repeat every 2π (i.e., ω = ω + 2πk for any integer k), and as such we only need to consider the range −π ≤ ω < π . This is in contrast with the analog frequency , which has rad/sec as units, and its range is from −∞ to ∞. If the frequency of a periodic sinusoid is ω=
2π m N
for nondivisible integers m and N > 0, the period is N. If the frequency of the sinusoid cannot be written like this, the discrete sinusoid is not periodic.
n Example 8.3 Consider the sinusoids x1 [n] = 2 cos(πn − π/3) x2 [n] = 3 sin(3πn + π/2)
−∞ < n < ∞
8.2 DiscreteTime Signals
From their frequencies determine if these signals are periodic, and if so, determine their corresponding periods. Solution The frequency of x1 [n] can be written as ω1 = π =
2π 2
where m = 1 and N = 2, so that x1 [n] is periodic of period N1 = 2. Likewise, the frequency of x2 [n] can be written as ω2 = 3π =
2π 3 2
where m = 3 and N = 2, so that x2 [n] is also periodic of period N2 = 2, which can be verified as follows: x2 [n + 2] = 3 sin(3π(n + 2) + π/2) = 3 sin(3πn + 6π + π/2) = x[n] n n Example 8.4 What is true for continuoustime sinusoids—that they are always periodic—is not true for discretetime sinusoids. These sinusoids can be nonperiodic even if they result from uniformly sampling a continuoustime sinusoid. Consider the discrete signal x[n] = cos(n + π/4), which is obtained by sampling the analog sinusoid x(t) = cos(t + π/4) with a sampling period Ts = 1 sec/sample. Is x[n] periodic? If so, indicate its period. Otherwise, determine values of the sampling period, satisfying the Nyquist sampling rate condition, that when used in sampling x(t) result in periodic signals. Solution The sampled signal x[n] = x(t)t=nTs = cos(n + π/4) has a discrete frequency ω = 1 rad that cannot be expressed as 2πm/N for any integers m and N because π is an irrational number. So x[n] is not periodic. Since the frequency of the continuoustime signal x(t) is = 1 (rad/sec), then the sampling period, according to the Nyquist sampling rate condition, should be Ts ≤
π =π
and for the sampled signal x(t)t=nTs = cos(nTs + π/4) to be periodic of period N or cos((n + N)Ts + π/4) = cos(nTs + π/4)
is necessary that
NTs = 2kπ
for an integer k (i.e., a multiple of 2π). Thus, Ts = 2kπ/N ≤ π satisfies the Nyquist sampling condition at the same time that it ensures the periodicity of the sampled signal. For instance, if we
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wish to have a sinusoid with period N = 10, then Ts = 0.2kπ for k chosen so the Nyquist sampling rate condition is satisfied—that is, 0 < Ts = kπ/5 ≤ π
so that 0 < k ≤ 5.
From these possible values for k we choose k = 1 and 3 so that N and k are not divisible by each other and we get the desired period N = 10 (the values k = 2 and 4 would give 5 as the period, and k = 5 would give a period of 2 instead of 10). Indeed, if we let k = 1, then Ts = 0.2π satisfies the Nyquist sampling rate condition, and we obtain the sampled signal π 2π n+ x[n] = cos(0.2nπ + π/4) = cos 10 4 which according to its frequency is periodic of period 10. This is the same for k = 3.
n
When sampling an analog sinusoid x(t) = A cos(0 t + θ)
−∞ < t < ∞
of period T0 = 2π/ 0 , 0 > 0, we obtain a periodic discrete sinusoid, 2π Ts x[n] = A cos(0 Ts n + θ) = A cos n+θ T0
(8.4)
(8.5)
provided that Ts m = T0 N
(8.6)
for positive integers N and m, which are not divisible by each other. To avoid frequency aliasing the sampling period should also satisfy Ts ≤
π T0 = 0 2
(8.7)
Indeed, sampling a continuoustime signal x(t) using as sampling period Ts , we obtain x[n] = A cos(0 Ts n + θ) 2πTs = A cos n+θ T0 where the discrete frequency is ω0 = 2πTs /T0 . For this signal to be periodic we should be able to express this frequency as 2πm/N for nondivisible positive integers m and N. This requires that T0 N = Ts m be a rational number, or that mT0 = NTs
(8.8)
8.2 DiscreteTime Signals
which says that a period (m = 1) or several periods (m > 1) should be divided into N > 0 segments of duration Ts seconds. If the condition in Equation (8.6) is not satisfied, then the discretized sinusoid is not periodic. To avoid frequency aliasing the sampling period should be chosen so that Ts ≤
π T0 = 0 2
The sum z[n] = x[n] + y[n] of periodic signals x[n] with period N1 , and y[n] with period N2 is periodic if the ratio of periods of the summands is rational—that is, N2 p = N1 q where p and q are integers not divisible by each other. If so, the period of z[n] is qN2 = pN1 .
If qN2 = pN1 , we then have that z[n + pN1 ] = x[n + pN1 ] + y[n + pN1 ] = x[n] + y[n + qN2 ] = x[n] + y[n] = z[n] since pN1 and qN2 are multiples of the periods of x[n] and y[n].
n Example 8.5 The signal z[n] = v[n] + w[n] + y[n] is the sum of three periodic signals v[n], w[n], and y[n] of periods N1 = 2, N2 = 3, and N3 = 4, respectively. Determine if z[n] is periodic, and if so, determine its period. Solution Let x[n] = v[n] + w[n], so that z[n] = x[n] + y[n]. The signal x[n] is periodic since N2 /N1 = 3/2 is a rational number and 3 and 2 are nondivisible by each other, and its period is N4 = 3N1 = 2N2 = 6. The signal z[n] is also periodic since N4 6 3 = = N3 4 2 Its period is N = 2N4 = 3N3 = 12. Thus, z[n] is periodic of period 12, indeed z[n + 12] = v[n + 6N1 ] + w[n + 4N2 ] + y[n + 3N3 ] = v[n] + w[n] + y[n] = z[n] n
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n Example 8.6 Determine if the signal x[n] =
∞ X
ω0 =
Xm cos(mω0 n)
m=0
2π N0
is periodic, and if so, determine its period. Solution The signal x[n] consists of the sum of a constant X0 and cosines of frequency mω0 =
2πm N0
m = 1, 2, . . .
The periodicity of x[n] depends on the periodicity of the cosines. According to the frequency of the cosines, they are periodic of period N0 . Thus, x[n] is periodic of period N0 . Indeed x[n + N0 ] =
∞ X
Xm cos(mω0 (n + N0 ))
m=0
=
∞ X
Xm cos(mω0 n + 2πm) = x[n]
m=0
n
8.2.2 FiniteEnergy and FinitePower DiscreteTime signals For discretetime signals, we obtain definitions for energy and power similar to those for continuoustime signals by replacing integrals by summations. For a discretetime signal x[n], we have the following definitions: Energy:
εx =
∞ X
x[n]2
(8.9)
n=−∞
Power:
N X 1 x[n]2 N→∞ 2N + 1
Px = lim
(8.10)
n=−N
n n
x[n] is said to have finite energy or to be square summable if εx < ∞. x[n] is called absolutely summable if ∞ X n=−∞
n
x[n] is said to have finite power if Px < ∞.
x[n] < ∞
(8.11)
8.2 DiscreteTime Signals
n Example 8.7 A “causal” sinusoid, obtained from a signal generator after it is switched on, is
x(t) =
( 2 cos(0 t − π/4) t ≥ 0 0
otherwise
The signal x(t) is sampled using a sampling period of Ts = 0.1 sec to obtain a discretetime signal x[n] = x(t)t=0.1n = 2 cos(0.10 n − π/4)
n≥0
and zero otherwise. Determine if this discretetime signal has finite energy and finite power and compare these characteristics with those of the continuoustime signal x(t) when 0 = π and when 0 = 3.2 rad/sec (an upper approximation of π). Solution The continuoustime signal x(t) has infinite energy, and so does the discretetime signal x[n], for both values of 0 . Indeed, its energy is εx =
∞ X
x[n]2 =
n=−∞
∞ X
4 cos2 (0.10 n − π/4) → ∞
n=0
Although the continuoustime and the discretetime signals have infinite energy, they have finite power. That the continuoustime signal has finite power can be shown as indicated in Chapter 1. For the discretetime signal x[n], we have for the two frequencies: 1. For 0 = π, x1 [n] = 2 cos(πn/10 − π/4) = 2 cos(2πn/20 − π/4) for n ≥ 0 and zero otherwise. Thus, x[n] repeats every N0 = 20 samples for n ≥ 0, and its power is N N X X 1 1 x1 [n]2 = lim x1 [n]2 N→∞ 2N + 1 N→∞ 2N + 1
Px = lim
n=0
n=−N
N0 −1 N0 −1 1 1 X 1 X x1 [n]2 = x1 [n]2 < ∞ N N→∞ 2N + 1 N0 2N0 n=0 n=0  {z } power of period, n≥0
= lim
where we used the causality of the signal (x1 [n] = 0 for n < 0), and considered N periods of x1 [n] for n ≥ 0, and for each computed its power to get the final result. Thus, for 0 = π the discretetime signal x1 [n] has finite power and can be computed using a period for n ≥ 0.
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To find the power we use the trigonometric identity (or Euler’s equation) cos2 (θ ) = 0.5(1 + cos(2θ )), and so replacing x1 [n], we have NX NX 0 −1 0 −1 4 cos(0.2πn − π/2) 1+ Px = 0.5 2N0 =
N0 = 20
n=0
n=0
2 [20 + 0] = 1 40
where the sum of the cosine is zero, as we are adding the values of the periodic cosine over a period. 2. For 0 = 3.2, x2 [n] = 2 cos(3.2n/10 − π/4) for n ≥ 0 and zero otherwise. The signal now does not repeat periodically after n = 0, as the frequency 3.2/10 (which equals the rational 32/100) cannot be expressed as 2πm/N (which due to π is an irrational value) for integers m and N. Thus, in this case we do not have a close form for the power, so we can simply say that the power is N X 1 x2 [n] 2 N→∞ 2N + 1
Px = lim
n=−N
and conjecture that because the corresponding analog signal has finite power, so would x2 [n]. Thus, we use MATLAB to compute the power for both cases. %%%%%%%%%%%%%%%%%%%%%%% % Example 8.7  Power %%%%%%%%%%%%%%%%%%%%%%% clear all;clf n = 0:100000; x2 = 2∗cos(0.1∗n∗3.2 − pi/4); % nonperiodic for positive n x1 = 2∗cos(0.1∗n∗pi − pi/4); % periodic for positive n N = length(x1) Px1 = sum(x1.ˆ2)/(2∗N+1) % power of x1 Px2 = sum(x2.ˆ2)/(2∗N+1) % power of x2 P1 = sum(x1(1:20).ˆ2)/(20); %power in period of x1
The signal x1 [n] in the script has unit power and so does the signal x2 [n] when we consider n 100,001 samples. The two signals and their squared magnitudes are shown in Figure 8.1.
n Example 8.8 Determine if a discretetime exponential x[n] = 2(0.5)n
n≥0
and zero otherwise, has finite energy, finite power, or both.
2
2
1
1 x1[n]
x2[n]
8.2 DiscreteTime Signals
0 −1
−1
−2 −10
0
10
20
30
−2 −10
40
0
10
4
20
30
40
n (b)
n (a) x 22 [n] x 21[n]
3
x 21[n], x 22 [n]
FIGURE 8.1 (a) Signal x2 [n] (nonperiodic for n ≥ 0) and (b) signal x1 [n] (periodic for n ≥ 0). The arrows show that the values are not equal for x2 [n] and equal for x1 [n]. (c) The square of the signals differ slightly, suggesting that if x1 [n] has finite power so does x2 [n].
0
2 1 0 −10
0
10
20
30
40
n (c)
Solution The energy is given by εx =
∞ X
4(0.5)2n = 4
n=0
∞ X
(0.25)n =
n=0
4 16 = 1 − 0.25 3
thus, x[n] is a finiteenergy signal. Just as with continuoustime signals, a finiteenergy signal is a finitepower (actually zero power) signal. n
8.2.3 Even and Odd Signals Time shifting and scaling of discretetime signals are very similar to the continuoustime cases, the only difference being that the operations are now done using integers. A discretetime signal x[n] is said to be Delayed by N (an integer) samples if x[n − N] is x[n] shifted to the right N samples. n Advanced by M (an integer) samples if x[n + M] is x[n] shifted to the left M samples. n Reflected if the variable n in x[n] is negated (i.e., x[−n]). n
The shifting to the right or the left can be readily seen by considering where x[0] is attained. For x[n − N], this is when n = N (i.e., N samples to the right of the origin), or x[n] is delayed by N samples.
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Likewise, for x[n + M] the x[0] appears advanced by M samples, or shifted to the left (i.e., when n = −M). Negating the variable n flips over the signal with respect to the origin.
n Example 8.9 A triangular discrete pulse is defined as n 0 ≤ n ≤ 10 x[n] = 0 otherwise Find an expression for y[n] = x[n + 3] + x[n − 3] and z[n] = x[−n] + x[n] in terms of n and carefully plot them. Solution Replacing n by n + 3 and n − 3 in the definition of x[n], we get the advanced and delayed signals n + 3 −3 ≤ n ≤ 7 x[n + 3] = 0 otherwise and x[n − 3] =
n − 3 3 ≤ n ≤ 13 0 otherwise
so that when added, we get n+3 2n y[n] = x[n + 3] + x[n − 3] = n −3 0
−3 ≤ n ≤ 2 3≤n≤7 8 ≤ n ≤ 13 otherwise
Likewise, we have that n 0 z[n] = x[n] + x[−n] = −n 0 The results are shown in Figure 8.2.
1 ≤ n ≤ 10 n=0 −10 ≤ n ≤ −1 otherwise n
n Example 8.10 We will see that in the convolution sum we need to figure out how a signal x[n − k] behaves as a function of k for different values of n. Consider the signal k 0≤k≤3 x[k] = 0 otherwise
12
12
10
10
10
8
8
8
6 4
6 4
0
5 n
10
0 −5
15
6 4 2
2
2 0 −5
x [n − 3]
12
x [n + 3]
x [n]
8.2 DiscreteTime Signals
0
5 n
10
0 −5
15
0
5 n
10
15
y[n] = x [n + 3] + x[n − 3]
15
10
5
5 n (a)
0
10
12
12
10
10
8
8 x[−n]
x[n]
0 −5
6
6
4
4
2
2
0
−10
0 n
0
10
15
−10
0 n
10
z [n] = x [n] + x [−n]
12 10 8 6 4 2 0 −15
−10
−5
0 n (b)
5
FIGURE 8.2 Generation of (a) y[n] = x[n + 3] + x[n − 3] and (b) z[n] = x[n] + x[−n].
10
15
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CH A P T E R 8: DiscreteTime Signals and Systems
Obtain an expression for x[n − k] for −2 ≤ n ≤ 2 and determine in which direction it shifts as n increases from −2 to 2. Solution Although x[n − k], as a function of k, is reflected it is not clear if it is advanced or delayed as n increases from −2 to 2. If n = 0, −k −3 ≤ k ≤ 0 x[−k] = 0 otherwise For n 6= 0, we have that x[n − k] =
n−k n−3≤k≤n 0 otherwise
As n increases from −2 to 2 the supports of x[n − k] move to the right. For n = −2 the support of x[−2 − k] is −5 ≤ k ≤ −2, while for n = 0 the support of x[−k] is −3 ≤ k ≤ 0, and for n = 2 the support of x[2 − k] is −1 ≤ k ≤ 2, each shifted to the right. n We can thus use the above to define even and odd signals and obtain a general decomposition of any signal in terms of even and odd signals. Even and odd discretetime signals are defined as x[n] is even: x[n] is odd:
⇔ ⇔
x[n] = x[−n]
(8.12)
x[n] = −x[−n]
(8.13)
Any discretetime signal x[n] can be represented as the sum of an even and an odd component, x[n] =
1 1 x[n] + x[−n] + x[n] − x[−n] 2 {z } 2 {z } xe [n]
xo [n]
= xe [n] + xo [n]
(8.14)
The even and odd decomposition can be easily seen. The even component xe [n] = 0.5(x[n] + x[−n]) is even since xe [−n] = 0.5(x[−n] + x[n]) equals xe [n], and the odd component xo [n] = 0.5(x[n] − x[−n]) is odd since xo [−n] = 0.5(x[−n] − x[n]) = −xo [n]. n Example 8.11 Find the even and the odd components of the discretetime signal 4−n 0≤n≤4 x[n] = 0 otherwise
8.2 DiscreteTime Signals
Solution The even component of x[n] is given by xe [n] = 0.5 x[n] + x[−n] When n = 0 then xe [0] = 0.5 × 2x[0] = 4, when n > 0 then xe [n] = 0.5x[n], and when n < 0 then xe [n] = 0.5x[−n], giving 2 + 0.5n −4 ≤ n ≤ −1 4 n=0 xe [n] = 2 − 0.5n 1 ≤ n ≤ 4 0 otherwise The odd component xo [n] = 0.5 x[n] − x[−n] gives 0 when n = 0, 0.5x[n] for n > 0, and −0.5x[−n] when n < 0, or −2 − 0.5n −4 ≤ n ≤ −1 0 n=0 xo [n] = 2 − 0.5n 1 ≤ n ≤ 4 0 otherwise The sum of these two components gives x[n].
n
Remarks Expansion and compression of discretetime signals is more complicated than in the continuoustime signals. In the discrete domain, expansion and compression can be related to the change of the sampling period in the sampling. Thus, if a continuoustime signal x(t) is sampled using a sampling period Ts , by changing the sampling period to MTs for an integer M > 1, we obtain fewer samples, and by changing the sampling period to Ts /L for an integer L > 1, we increase the number of samples. For the corresponding discretetime signal x[n], increasing the sampling period would give x[Mn], which is called the downsampling of x[n] by M. Unfortunately, because the argument of discretetime signals must be integers, it is not clear what x[n/L] is unless the values for n are multiples of L (i.e., n = ±0, ±L, ±2L, . . .) without a clear definition when n takes other values. This leads to the definition of the upsampled signal x[n/L] n = ±0, ±L, ±2L, . . . xu [n] = (8.15) 0 otherwise To replace the zero entries with the values obtained by decreasing the sampling period we need to lowpass filter the upsampled signal. MATLAB provides the functions decimate and interp to implement the downsampling and upsampling without losing information due to possible frequency aliasing. In Chapter 10, we will continue the discussion of these operations including their frequency characterization.
8.2.4 Basic DiscreteTime Signals The representation of discretetime signals via basic signals is simpler than in the continuoustime domain. This is due to the lack of ambiguity in the definition of the impulse and the unitstep discretetime signals. The definitions of impulses and unitstep signals in the continuoustime domain are more abstract.
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DiscreteTime Complex Exponential Given complex numbers A = Aejθ and α = αejω0 , a discretetime complex exponential is a signal of the form x[n] = Aα n = Aαn ej(ω0 n+θ ) = Aαn [cos(ω0 n + θ) + j sin(ω0 n + θ)]
(8.16)
where ω0 is a discrete frequency in radians.
Remarks n
The discretetime complex exponential looks different from the continuoustime complex exponential. This can be explained by sampling the continuoustime complex exponential x(t) = Ae(−a+j0 )t (for simplicity we let A be real) using as sampling period Ts . The sampled signal is x[n] = x(nTs ) = Ae(−anTs +j0 nTs ) = A(e−aTs )n ej(0 Ts )n = Aα n ejω0 n
n
where we let α = e−aTs and ω0 = 0 Ts . Just as with the continuoustime complex exponential, we obtain different signals depending on the chosen parameters A and α. For instance, the real part of x[n] in Equation (8.16) is a real signal g[n] = Re[x[n]] = Aαn cos(ω0 n + θ )
n
where when α < 1 it is a damped sinusoid, and when α > 1 it is a growing sinusoid (see Figure 8.3). If α = 1 then the above signal is a sinusoid. It is important to realize that for α > 0 the real exponential x[n] = (−α)n = (−1)n α n = α n cos(πn)
n Example 8.12 Given the analog signal x(t) = e−at cos(0 t)u(t) determine the values of a > 0, 0 , and Ts that permit us to obtain a discretetime signal y[n] = α n cos(ω0 n)
n≥0
4
4
3
3 x2 [n] = 1.25n
x1[n] = (0.8)n
8.2 DiscreteTime Signals
2 1 0
2 1 0
−6
−4
−2
0 n
2
4
−6
6
−4
−2
0 n
2
4
6
4
4
2
2 y2 [n]
y1[n]
(a)
0 −2 −4
0 −2
−5
0 n
−4
5
−5
0 n
5
(b)
FIGURE 8.3 (a) Real exponential x1 [n] = 0.8n , x2 [n] = 1.25n , and (b) modulated exponential y1 [n] = x1 [n] cos(πn) and y2 [n] = x2 [n] cos(πn).
and zero otherwise. Consider the case when α = 0.9 and ω0 = π/2. Find a, 0 , and Ts that will permit us to obtain y[n] from x(t) by sampling. Plot x[n] and y[n] using MATLAB. Solution Comparing the sampled continuoustime signal x(nTs ) = (e−aTs )n cos((0 Ts )n)u[n] with y[n] we obtain the following two equations: α = e−aTs ω0 = 0 Ts with three unknowns (a, 0 , and Ts ), so there is no unique solution. According to the Nyquist sampling rate condition, Ts ≤
π max
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Assuming the maximum frequency is max = N0 for N ≥ 2 (since the signal is not band limited the maximum frequency is not known; to estimate it we could use Parseval’s result as indicated in Chapter 7, instead we are assuming that it is a multiple of 0 ), if we let Ts = π/N0 after replacing it in the above equations, we get α = e−aπ/N0 ω0 = 0 π/N0 = π/N If we want α = 0.9 and ω0 = π/2, we have that N = 2 and 20 log 0.9 π for any frequency 0 > 0. For instance, if 0 = 2π, then a = −4 log 0.9 and Ts = 0.25. Figure 8.4 displays the continuous and the discretetime signals generated using the above parameters. The following script is used. The continuoustime and the discretetime signals coincide at the sample times. a=−
%%%%%%%%%%%%%%%%%%%%% % Example 8.12 %%%%%%%%%%%%%%%%%%%%% a = −4∗log(0.9);Ts = 0.25; % parameters alpha = exp(−a∗Ts); n = 0:30; y = alpha.ˆn.∗cos(pi∗n/2); % discretetime signal t = 0:0.001:max(n)∗Ts; x = exp(−a∗t).∗cos(2∗pi∗t); % analog signal stem(n, y, ’r’); hold on plot(t/Ts, x); grid; legend(’y[n]’, ’x(t)’); hold off 1 y [n]
0.8
x (t )
0.6 0.4 x (t ), y [n]
468
0.2 0 −0.2 −0.4
FIGURE 8.4 Determination of parameters for a continuoustime signal x(t) that when sampled gives a desired discretetime signal y[n].
−0.6 −0.8 −1
0
5
10
15 t /Ts
20
25
30
n
8.2 DiscreteTime Signals
n Example 8.13 Show how to obtain the discretetime exponential x[n] = (−1)n for n ≥ 0 and zero otherwise, by sampling a continuoustime signal x(t). Solution Because the values of x[n] are 1 and −1, x[n] cannot be generated by sampling a real exponential signal e−at u(t); indeed, e−at > 0 for any values of a and t. The discrete signal can be written as x[n] = (−1)n = cos(πn) for n ≥ 0. If we sample an analog signal x(t) = cos(0 t)u(t) with a sampling period Ts , we get x[n] = x(nTs ) = cos(0 nTs ) = cos(πn)
n≥0
and zero otherwise. Thus, 0 Ts = π, giving Ts = π/ 0 . For instance, for 0 = 2π, then Ts = 0.5. n
DiscreteTime Sinusoids Discretetime sinusoids are a special case of the complex exponential. Letting α = ejω0 and A = Aejθ , we have according to Equation (8.16), x[n] = Aα n = Aej(ω0 n+θ ) = A cos(ω0 n + θ ) + jA sin(ω0 n + θ )
(8.17)
so the real part of x[n] is a cosine, while the imaginary part is a sine. As indicated before, discrete sinusoids of amplitude A and phase shift θ are periodic if they can be expressed as A cos(ω0 n + θ) = A sin(ω0 n + θ + π/2)
−∞ < n < ∞
(8.18)
where w0 = 2πm/N rad is the discrete frequency for integers m and N > 0, which are not divisible by each other. Otherwise, discretetime sinusoids are not periodic. Because ω is given in radians, it repeats periodically with 2π as the period—that is, ω = ω + 2πk
k integer
(8.19)
To avoid this ambiguity, we will let −π < ω ≤ π as the possible range of discrete frequencies. This is possible since ω − 2πk when ω > 2π, for some k > 0 integer ω= ω − 2π 0 ≤ ω ≤ 2π
(8.20)
See Figure 8.5. Thus, sin(3πn) equals sin(πn), and sin(1.5πn) equals sin(−0.5πn) = −sin(0.5πn).
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CH A P T E R 8: DiscreteTime Signals and Systems
ω = π /2, 5π /2, 9π /2 · · · = π /2
ω = π , 3π , 5π ··· = π = −π
FIGURE 8.5 Discrete frequencies ω.
ω = 0, 2π , 4π , ··· = 0
ω = 3π /2, 7π /2, 11π /2 , ··· = −π /2
n Example 8.14 Consider the following four sinusoids: (a) x1 [n] = sin(0.1πn) (b) x2 [n] = sin(0.2πn) (c) x3 [n] = sin(0.6πn) (d) x4 [n] = sin(0.7πn) Find if they are periodic, and if so, determine their periods. Are these signals harmonically related? Use MATLAB to plot these signals from n = 0, . . . , 40. Comment on which of these signals resemble sampled analog sinusoids. Solution To find if they are periodic, rewrite the given signals as follows indicating that the signals are periodic of periods 20, 10, 10, and 20: 2π (a) x1 [n] = sin(0.1πn) = sin n 20 2π (b) x2 [n] = sin(0.2πn) = sin n 10 2π (c) x3 [n] = sin(0.6πn) = sin 3n 10 2π (d) x4 [n] = sin(0.7πn) = sin 7n 20 If we let ω1 = 2π/20, the frequencies of x2 [n], x3 [n], and x4 [n] are 2ω1 , 6ω1 , and 7ω1 , respectively; thus they are harmonically related. Also, one could consider the frequencies of x1 [n] and x4 [n] harmonically related (i.e., the frequency of x4 [n] is seven times that of x1 [n]), and likewise the frequencies of x2 [n] and x3 [n] are also harmonically related, with the frequency of x3 [n] being
1
1
0.5
0.5 x2 [n]
x1[n]
8.2 DiscreteTime Signals
0 −0.5
−0.5
−1
−1 0
10
20 n (a)
30
40
1
1
0.5
0.5 x4 [n]
x3 [n]
0
0
−0.5
−1
−1 10
20 n (c)
30
40
10
20 n (b)
30
40
0
10
20 n (d)
30
40
0
−0.5
0
0
FIGURE 8.6 Periodic signals xi [n], (a) i = 1, (b) i = 2, (c) i = 3, and (d) i = 4, given in Example 8.14.
three times that of x2 [n]. When plotting these signals using MATLAB, the first two resemble analog sinusoids but not the other two. See Figure 8.6. n Remarks n n
The discretetime sine and cosine signals, as in the continuoustime case, are out of phase π/2 radians. The discrete frequency ω is given in radians since n, the sample index, does not have units. This can also be seen when we sample a sinusoid using a sampling period Ts so that cos(0 t)t=nTs = cos(0 Ts n) = cos(ω0 n)
n
where we defined ω0 = 0 Ts , and since 0 has rad/sec as units and Ts has seconds as units, then ω0 has radians as units. The frequency of analog sinusoids can vary from 0 (dc frequency) to ∞. Discrete frequencies ω as radian frequencies can only vary from 0 to π. Negative frequencies are needed in the analysis of realvalued signals; thus −∞ < < ∞ and −π < ω ≤ π. A discretetime cosine of frequency 0 is constant for all n, and a discretetime cosine of frequency π varies from −1 to 1 from sample to sample, giving the largest variation possible for the discretetime signal.
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DiscreteTime UnitStep and UnitSample Signals The unitstep u[n] and the unitsample δ[n] discretetime signals are defined as ( 1 n≥0 u[n] = 0 n= −ad, y(i) = 1; end end
Finally, the following function can be used to compute the even and the odd decomposition of a discretetime signal. The MATLAB function flliplr reflects the signal as needed in the generation of the even and the odd components. function [ye, yo] = evenodd(y) % even/odd decomposition % NOTE: the support of the signal should be % symmetric about the origin % y: analog signal % ye, yo: even and odd components yr = fliplr(y); ye = 0.5∗(y + yr); yo = 0.5∗(y − yr);
The results are shown in Figure 8.7. The discretetime signal is given as 0 n ≤ −21 −20 ≤ n ≤ −6 0.45n + 9 y[n] = −0.45n + 3 −7 ≤ n ≤ 0 3 1 ≤ n ≤ 19 0 n ≥ 20 n
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CH A P T E R 8: DiscreteTime Signals and Systems
ze [n]
7 6 5
5 4 3 2 1 0 −1
4 y [n]
−30
−20
−10
0 n (b)
10
20
30
−30
−20
−10
0 n (c)
10
20
30
3 2
2
1
1
zo [n]
478
0 −1
0 −1
−30
−20
−10
0 n (a)
10
20
30
−2
FIGURE 8.7 (a) Discretetime signal, and (b) even and (c) odd components.
8.3 DISCRETETIME SYSTEMS Just as with continuoustime systems, a discretetime system is a transformation of a discretetime input signal x[n] into a discretetime output signal y[n]—that is, y[n] = S{x[n]}
(8.26)
Just as we were when we studied the continuoustime systems, we are interested in dynamic systems S{.} having the following properties: Linearity Time invariance Stability Causality
n n n n
A discretetime system S is said to be Linear: If for inputs x[n] and v[n] and constants a and b, it satisfies the following n Scaling: S{ax[n]} = aS{x[n]} n Additivity: S{x[n] + v[n]} = S{x[n]} + S{v[n]} or equivalently if superposition applies—that is,
n
S{ax[n] + bv[n]} = aS{x[n]} + bS{v[n]} n
(8.27)
Timeinvariant: If for an input x[n] with a corresponding output y[n] = S{x[n]}, the output corresponding to a delayed or advanced version of x[n], x[n ± M], is y[n ± M] = S{x[n ± M]} for an integer M.
8.3 DiscreteTime Systems
n Example 8.19 A squareroot computation system. The input–output relation characterizing a discretetime system is nonlinear if there are nonlinear terms that include the input x[n], the output y[n], or both (e.g., a square root of x[n], products of x[n] and y[n], etc.). Consider the development of an iterative algorithm to compute the square root of a positive real number α. If the result of the algorithm is y[n] as n → ∞, then y2 [n] = α and likewise y2 [n − 1] = α, thus y[n] = 0.5(y[n − 1] + y[n − 1]). Replacing y[n − 1] = α/y[n − 1] in this equation, the following difference equation, with some initial condition y[0], can be used to find the square root of α: α y[n] = 0.5 y[n − 1] + n>0 y[n − 1] Find recursively the solution of this difference equation. Use the results of finding the square roots of 4 and 2 to show the system is nonlinear. Solve the difference equation and plot the results for α = 4, 2 with MATLAB. Solution The given difference equation is first order, nonlinear (expanding it you get the product of y[n] with y[n − 1] and y2 [n − 1], which are nonlinear terms) with constant coefficients. This equation can be solved recursively for n > 0 by replacing y[0] to get y[1], and use this to get y[2] and so on—that is, α y[1] = 0.5 y[0] + y[0] α y[2] = 0.5 y[1] + y[1] α y[3] = 0.5 y[2] + y[2] .. . For instance, let y[0] = 1 and α = 4 (i.e., we wish to find the square root of 4), y[0] = 1
4 = 2.5 y[1] = 0.5 1 + 1 4 y[2] = 0.5 2.5 + = 2.05 2.5 .. .
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CH A P T E R 8: DiscreteTime Signals and Systems
y1[n]
2
1 square root of 2
0
0
1
2
3
4
5
6
7
8
9
5
6
7
8
9
5
6
7
8
9
n (a)
y2 [n]
4
FIGURE 8.8 Nonlinear system: (a) square root of 2, (b) square root of 4 compared with twice the square root of 2, and (c) sum of previous responses with response when computing square root of 2 + 4. Figure (b) shows scaling does not hold and (c) shows that additivity does not hold either. The system is nonlinear.
2 square root of 4
0
0
1
2
3
4 n (b)
4 y3 [n]
480
2 square root of 6
0
0
1
2
3
4 n (c)
which is converging to 2, the square root of 4 (see Figure 8.8). Thus, as indicated before, when n → ∞, then y[n] = y[n − 1] = Y, for some value√Y, which according to the difference equation satisfies the relation Y = 0.5Y + 0.5(4/Y) or Y = 4 = 2. Suppose then that the input is α = 2, half of what it was before. If the system is linear, we should get half the previous output according to the scaling property. That is not the case, however. For the same initial condition y[0] = 1, we obtain recursively for α[n] = 2u[n − 1]: y[0] = 1 y[1] = 0.5[1 + 2] = 1.5 2 y[2] = 0.5 1.5 + = 1.4167 1.5 .. . This solution is clearly not half of the previous one. Moreover, as n → √ ∞, we expect y[n] = y[n − 1] = Y, for Y that satisfies the relation Y = 0.5Y + 0.5(2/Y) or Y = 2 = 1.4142, so that √ the solution is tending to 2 and not to 2 as it would if the system were linear. Finally, if we add
8.3 DiscreteTime Systems
the signals in the above two cases and compare the resulting signal with the one we obtain when finding the square root of the previous two values of α, 2 and 4, they do not coincide. The additive condition is not satisfied either, so the system is not linear. n
8.3.1 Recursive and Nonrecursive DiscreteTime Systems Depending on the relation between the input x[n] and the output y[n] two types of discretetime systems of interest are: n Recursive system: y[n] = −
N−1 X
ak y[n − k] +
M−1 X
bm x[n − m]
initial conditions y[−k], k = 1, . . . , N − 1 n
n≥0
m=0
k=1
(8.28)
This system is also called infiniteimpulse response (IIR). Nonrecursive system: y[n] =
M−1 X
bm x[n − m]
(8.29)
m=0
This system is also called finiteimpulse response (FIR).
The recursive system is analogous to a continuoustime system represented by a differential equation. For this type of system the discretetime input x[n] and the discretetime output y[n] are related by an (N − 1)thorder difference equation. If such a difference equation is linear, with constant coefficients, zero initial conditions, and the input is zero for n < 0, then it represents a linear and timeinvariant system. For these systems the output at a present time n, y[n], depends or recurs on previous values of the output {y[n − k], k = 1, . . . , N − 1}, and thus they are called recursive. We will see that these systems are also called infiniteimpulse response or IIR because their impulse responses are typically of infinite length. On the other hand, if the output y[n] does not depend on previous values of the output, but only on weighted and shifted inputs {bm x[n − m], m = 0, . . . , M − 1}, the system is called nonrecursive. We will see that the impulse response of nonrecursive systems is of finite length; as such, these systems are also called finite impulse response or FIR. n Example 8.20 Movingaverage discrete filter: A thirdorder movingaverage filter (also called a smoother since it smooths out the input signal) is an FIR filter for which the input x[n] and the output y[n] are related by y[n] =
1 (x[n] + x[n − 1] + x[n − 2]) 3
Show that this system is linear time invariant.
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CH A P T E R 8: DiscreteTime Signals and Systems
Solution This is a nonrecursive system that uses a present sample, x[n], and two past values, x[n − 1] and x[n − 2], of the input to get an average, y[n], at every n. Thus, its name, movingaverage filter. Linearity: If we let the input be ax1 [n] + bx2 [n], and assume that {yi [n], i = 1, 2} are the corresponding outputs to {xi [n], i = 1, 2}, the filter output is 1 [(ax1 [n] + bx2 [n])+(ax1 [n − 1] + bx2 [n − 1])+(ax1 [n − 2] + bx2 [n − 2])] = ay1 [n] + by2 [n] 3 That is, the system is linear. Time invariance: If the input is x1 [n] = x[n − N], the corresponding output to it is 1 1 (x1 [n] + x1 [n − 1] + x1 [n − 2]) = (x[n − N] + x[n − N − 1] + x[n − N − 2]) 3 3 = y[n − N] That is, the system is time invariant.
n
n Example 8.21 Autoregressive discrete filter: The recursive discretetime system represented by the firstorder difference equation (with initial condition y[−1]) y[n] = ay[n − 1] + bx[n]
n ≥ 0, y[−1]
is also called an autoregressive (AR) filter. “Autoregressive” refers to the feedback in the output—that is, the present value of the output y[n] depends on its previous value y[n − 1]. Find recursively the solution of the difference equation and determine under what conditions the system represented by this difference equation is linear and time invariant. Solution Let’s first discuss why the initial condition is y[−1]. The initial condition is the value needed to compute y[0], which according to the difference equation y[0] = ay[−1] + bx[0] is y[−1] since x[0] is known. Assume that the initial condition is y[−1] = 0, and that the input is x[n] = 0 for n < 0 (i.e., the system is not energized for n < 0). The solution of the difference equation when the input x[n] is not defined can be found by a repetitive substitution of the input–output relationship. Thus, replacing y[n − 1] = ay[n − 2] + bx[n − 1] in the difference equation, and then replacing
8.3 DiscreteTime Systems
y[n − 2] = ay[n − 3] + bx[n − 2], and so on, we obtain y[n] = a(ay[n − 2] + bx[n − 1]) + bx[n] = a(a(ay[n − 3] + bx[n − 2])) + abx[n − 1] + bx[n] = ··· = · · · a3 bx[n − 3] + a2 bx[n − 2] + abx[n − 1] + bx[n] until we reach x[0]. The solution can be written as
y[n] =
n X
bak x[n − k]
(8.30)
k=0
which we will see in the next section is the convolution sum of the impulse response of the system and the input. To see that Equation (8.30) is actually the solution of the given difference equation, we need to show that when replacing the above expression for y[n] in the right term of the difference equation we obtain the left term y[n]. Indeed, we have that
ay[n − 1] + bx[n] = a
"n−1 X
# k
ba x[n − 1 − k] + bx[n]
k=0
=
n X
bam x[n − m] + bx[n] =
m=1
n X
bam x[n − m] = y[n]
m=0
where the dummy variable in the sum was changed to m = k + 1, so that the limits of the summation became m = 1 when k = 0, and m = n when k = n − 1. The final equation is identical to y[n]. To establish if the system represented by the difference equation is linear, we use the solution Eq. (8.30) with input x[n] = αx1 [n] + βx2 [n], where the outputs {yi [n], i = 1, 2} correspond to inputs {xi [n], i = 1, 2}, and α and β are constants. The output for x[n] is n X
bak x[n − k] =
k=0
n X
bak αx1 [n − k] + βx2 [n − k]
k=0
=α
n X k=0
So the system is linear.
bak x1 [n − k] + β
n X k=0
bak x2 [n − k] = αy1 [n] + βy2 [n]
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The time invariance is shown by letting the input be v[n] = x[n − N], n ≥ N, and zero otherwise. The corresponding output according to Equation (8.30) is n X
bak v[n − k] =
k=0
n X
bak x[n − N − k]
k=0
=
n−N X
n X
bak x[n − N − k] +
k=0
bak x[n − N − k] = y[n − N]
k=n−N+1
since the summation n X
bak x[n − N − k] = 0
k=n−N+1
given that x[−N] = · · · = x[−1] = 0 is assumed. Thus, the system represented by the above difference equation is linear and time invariant. As in the continuoustime case, however, if the initial condition y[−1] is not zero, or if x[n] 6= 0 for n < 0, the system characterized by the difference equation is not LTI. n
n Example 8.22 Autoregressive moving average filter: The recursive system represented by the firstorder difference equation y[n] = 0.5y[n − 1] + x[n] + x[n − 1]
n ≥ 0, y[−1]
is called the autoregressive moving average given that it is the combination of the two systems discussed before. Consider two cases: n
n
Let the initial condition be y[−1] = −2, and the input be x[n] = u[n] first and then x[n] = 2u[n]. Let the initial condition be y[−1] = 0, and the input be x[n] = u[n] first and then x[n] = 2u[n].
Determine in each of these cases if the system is linear. Find the steadystate response—that is, lim y[n]
n→∞
8.3 DiscreteTime Systems
Solution For an initial condition y[−1] = −2 and x[n] = u[n], we get recursively y[0] = 0.5y[−1] + x[0] + x[−1] = 0 y[1] = 0.5y[0] + x[1] + x[0] = 2 y[2] = 0.5y[1] + x[2] + x[1] = 3 ... Let us then double the input (i.e., x[n] = 2u[n]) and call the response y1 [n]. As the initial condition remains the same (i.e., y1 [−1] = −2), we get y1 [0] = 0.5y1 [−1] + x[0] + x[−1] = 1 y1 [1] = 0.5y1 [0] + x[1] + x[0] = 4.5 y1 [2] = 0.5y1 [1] + x[2] + x[1] = 6.25 ... It is clear that the y1 [n] is not 2y[n]. Due to the initial condition not being zero, the system is nonlinear. If the initial condition is set to zero, and the input x[n] = u[n], the response is y[0] = 0.5y[−1] + x[0] + x[−1] = 1 y[1] = 0.5y[0] + x[1] + x[0] = 2.5 y[2] = 0.5y[1] + x[2] + x[1] = 3.25 ... If we double the input (i.e., x[n] = 2u[n]) and call the response y1 [n], y1 [−1] = 0, we obtain y1 [0] = 0.5y1 [−1] + x[0] + x[−1] = 2 y1 [1] = 0.5y1 [0] + x[1] + x[0] = 5 y1 [2] = 0.5y1 [1] + x[2] + x[1] = 6.5 ...
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For the zero initial condition, it is clear that y1 [n] = 2y[n] when we double the input. One can also show that superposition holds for this system. For instance, if we let the input be the sum of the previous inputs, x[n] = u[n] + 2u[n] = 3u[n], and let y12 [n] be the response when the initial condition is zero, y12 [0] = 0, we get y12 [0] = 0.5y12 [−1] + x[0] + x[−1] = 3 y12 [1] = 0.5y12 [0] + x[1] + x[0] = 7.5 y12 [2] = 0.5y12 [1] + x[2] + x[1] = 9.75 ... showing that y12 [n] is the sum of the responses when the inputs are u[n] and 2u[n]. Thus, the system represented by the given difference equation with a zero initial condition is linear. Although when the initial condition is −2 or 0, and x[n] = u[n] we cannot find a closed form for the response, we can see that the response is going toward a final value or a steadystate response. Assuming that as n → ∞ we have that Y = y[n] = y[n − 1], and since x[n] = x[n − 1] = 1, according to the difference equation the steadystate value Y is found from Y = 0.5Y + 2
or Y = 4
independent of the initial condition. Likewise, when x[n] = 2u[n], the steadystate solution Y is obtained from Y = 0.5Y + 4 or Y = 8, independent of the initial condition. n Remarks n
n
Like in the continuoustime system, to show that a discretetime system is linear and time invariant an explicit expression relating the input and the output is needed. Although the solution of linear difference equations can be obtained in the time domain, just like with differential equations, we will see in the next chapter that their solution can also be obtained using the Ztransform, just like the Laplace transform being used to solve linear differential equations.
8.3.2 DiscreteTime Systems Represented by Difference Equations As we saw before, a recursive discretetime system is represented by a difference equation y[n] = −
N−1 X k=1
ak y[n − k] +
M−1 X
bm x[n − m]
n≥0
m=0
initial conditions y[−k], k = 1, . . . , N − 1
(8.31)
If the system is discretetime, the difference equation naturally characterizes the dynamics of the system. On the other hand, the difference equation could be the approximation of a differential equation representing a continuoustime system that is being processed discretely. For instance, to approximate a secondorder differential equation by a difference equation, we could approximate
8.3 DiscreteTime Systems
the first derivative of a continuoustime function vc (t) as dvc (t) vc (t) − vc (t − Ts ) ≈ dt Ts and its second derivative as d dvdtc (t) d2 vc (t) d(vc (t) − vc (t − Ts ))/Ts ) = ≈ dt2 dt dt ≈
vc (t) − 2vc (t − Ts ) + vc (t − 2Ts ) Ts2
to obtain a secondorder difference equation when t = nTs . Choosing a small value for Ts provides an accurate approximation to the differential equation. Other transformations can be used. In Chapter 0 we indicated that approximating integrals by the trapezoidal rule gives the bilinear transformation, which can also be used to change differential into difference equations. Just as in the continuoustime case, the system being represented by the difference equation is not LTI unless the initial conditions are zero and the input is causal. The Ztransform will, however, allow us to find the complete response of the system even when the initial conditions are not zero. When the initial conditions are not zero, just like in the continuous case, these systems are incrementally linear. The complete response of a system represented by the difference equation can be shown to be composed of a zeroinput and a zerostate responses—that is, if y[n] is the solution of the difference Equation (8.31) with initial conditions not necessarily equal to zero, then y[n] = yzi [n] + yzs [n]
(8.32)
The component yzi [n] is the response when the input x[n] is set to zero, thus it is completely due to the initial conditions. The response yzs [n] is due to the input, as we set the initial conditions equal to zero. The complete response y[n] is thus seen as the superposition of these two responses. The Ztransform provides an algebraic way to obtain the complete response, whether the initial conditions are zero or not. It is important, as in the continuoustime system, to differentiate the zeroinput and the zerostate responses from the transient and the steadystate responses.
8.3.3 The Convolution Sum Let h[n] be the impulse response of an LTI discretetime system, or the output of the system corresponding to an impulse δ[n] as input and initial conditions (if needed) equal to zero. Using the generic representation of the input x[n] of the LTI system, x[n] =
∞ X k=−∞
x[k]δ[n − k]
(8.33)
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the output of the system is given by either of the following two equivalent forms of the convolution sum: y[n] =
∞ X
x[k]h[n − k]
k=−∞
=
∞ X
x[n − m]h[m]
(8.34)
m=−∞
The impulse response h[n] of a discretetime system is due exclusively to an input δ[n]; as such, the initial conditions are set to zero. In some cases there are no initial conditions, as in the case of nonrecursive systems. Now, if h[n] is the response due to δ[n], by time invariance the response to δ[n − k] is h[n − k]. By superposition, the response due to x[n] with the generic representation X x[n] = x[k]δ[n − k] k
is the sum of responses due to x[k]δ[n − k], which is x[k]h[n − k] (x[k] is not a function of n), or X y[n] = x[k]h[n − k] k
i.e., the convolution sum of the input x[n] with the impulse response h[n] of the system. The second expression of the convolution sum in Equation (8.34) is obtained by a change of variable m = n − k. Remarks n
The output of nonrecursive or FIR systems is the convolution sum of the input and the impulse response of the system. The input–output expression of an FIR system is y[n] =
N−1 X
bk x[n − k]
(8.35)
k=0
and its impulse response is found by letting x[n] = δ[n], which gives h[n] =
N−1 X
bk δ[n − k] = b0 δ[n] + b1 δ[n − 1] + · · · + bN−1 δ[n − (N − 1)]
k=0
so that h[n] = bn for n = 0, . . . , N − 1, and zero otherwise. Replacing the bk coefficients in Equaiton (8.35) by h[k] we find that the output can be written as y[n] =
N−1 X k=0
h[k]x[n − k]
8.3 DiscreteTime Systems
n
or the convolution sum of the input and the impulse response. This is a very important result, indicating that the output of FIR systems is obtained by means of the convolution sum rather than difference equations, which gives great significance to the efficient computation of the convolution sum. Considering the convolution sum as an operator—that is, y[n] = [h ∗ x][n] =
∞ X
x[k]h[n − k]
k=−∞
it is easily shown to be linear. Indeed, whenever the input is ax1 [n] + bx2 [n], and {yi [n]} are the outputs corresponding to {xi [n]} for i = 1, 2, then we have that [h ∗ (ax1 + bx2 )][n] =
X X X (ax1 [k] + bx2 [k])h[n − k] = a x1 [k]h[n − k] + b x2 [k]h[n − k] k
k
k
= a[h ∗ x1 ][n] + b[h ∗ x2 ][n] = ay1 [n] + by2 [n] as expected, since the system was assumed to be linear when the expression for the convolution sum was obtained. We will then have that if the output corresponding to x[n] is y[n], given by the convolution sum, then the output corresponding to a shifted version of the input, x[n − N], should be y[n − N]. In fact, if we let x1 [n] = x[n − N], the corresponding output is [h ∗ x1 ][n] =
X
x1 [n − k]h[k] =
k
X
x[n − N − k]h[k]
k
= [h ∗ x][n − N] = y[n − N]
n
Again, this result is expected given that the system was considered time invariant when the convolution sum was obtained. From the equivalent representations for the convolution sum we have that X X [h ∗ x][n] = x[k]h[n − k] = x[n − k]h[k] k
k
= [x ∗ h][n]
n
n
n
indicating that the convolution commutes with respect to the input x[n] and the impulse response h[n]. Just as with continuoustime systems, when conecting two LTI discretetime systems (with impulse responses h1 [n] and h2 [n]) in cascade or in parallel, their respective impulse responses are given by [h1 ∗ h2 ][n] and h1 [n] + h2 [n]. See Figure 8.9 for block diagrams. There are situations when instead of giving the input and the impulse response to compute the output, the information that it is available is, for instance, the input and the output and we wish to find the impulse response of the system, or we have the output and the impulse response and wish to find the input. This type of problem is called deconvolution. We consider this problem later in this chapter after considering causality, and in Chapter 9 where we show that it can be easily solved using the Ztransform. The computation of the convolution sum is typically difficult. It is made easier when the Ztransform is used, as we will see. MATLAB provides the function conv which greatly simplifies the computation.
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x [n]
h1[n]
y [n]
h2[n]
x[n]
FIGURE 8.9 (a) Cascade and (b) parallel connections of LTI systems with impulse responses h1 [n] and h2 [n]. Equivalent systems on the right. Notice the interchange of systems in the cascade connection.
x[n]
h2[n]
(h1 ∗ h2) [n]
y [n]
y [n]
h1[n]
(a) h1[n] +
x[n]
y [n]
x [n]
h1[n] + h2[n]
y [n]
h2[n] (b)
n Example 8.23 Consider a movingaveraging filter where the input is x[n] and the output is y[n]:
y[n] =
1 (x[n] + x[n − 1] + x[n − 2]) 3
Find the impulse response h[n] of this filter. Then, (a) Let x[n] = u[n]. Find the output of the filter y[n] using the input–output relation and the convolution sum. (b) If the input of the filter is x[n] = A cos(2πn/N)u[n], determine the values of A and N, so that the steadystate response of the filter is zero. Explain. Use MATLAB to verify your results. Solution (a) If the input is x[n] = δ[n], the output of the filter is y[n] = h[n], or the impulse response of the system. No initial conditions are needed. We thus have that
h[n] =
1 (δ[n] + δ[n − 1] + δ[n − 2]) 3
so that h[0] = 1/3 as δ[0] = 1 but δ[−1] = δ[−2] = 0; likewise, h[1] = h[2] = 1/3 so that the coefficients of the filter equal the impulse response of the filter at n = 0, 1, and 2.
8.3 DiscreteTime Systems
Now if x[n] is the input to the filter according to the convolution sum, its output is y[n] =
n X
x[n − k]h[k] = h[0]x[n] + h[1]x[n − 1] + h[2]x[n − 2]
k=0
1 x[n] + x[n − 1] + x[n − 2] 3 Notice that the lower bound of the sum is set by the impulse response being zero for n < 0, while the upper bound is set by the input being zero for n < 0 (i.e., if k > n, then n − k < 0 and x[n − k] = 0). The convolution sum coincides with the input–output equation. This holds for any FIR filter. =
For any input x[n], let us then find a few values of the convolution sum to see what happens as n grows. If n < 0, the arguments of x[n], x[n − 1], and x[n − 2] are negative giving zero values, and so the output is also zero (i.e., y[n] = 0, n < 0). For n ≥ 0, we have 1 1 x[0] + x[−1] + x[−2] = x[0] 3 3 1 1 y[1] = x[1] + x[0] + x[−1] = (x[0] + x[1]) 3 3 1 1 x[2] + x[1] + x[0] = (x[0] + x[1] + x[2]) y[2] = 3 3 1 1 y[3] = x[3] + x[2] + x[1] = (x[1] + x[2] + x[3]) 3 3 y[0] =
··· Thus, if x[n] = u[n], then we have that y[0] = 1/3, y[1] = 2/3, and y[n] = 1 for n ≥ 2. (b) Notice that for n ≥ 2, the output is the average of the present and past two values of the input. Thus, when the input is x[n] = A cos(2πn/N), if we let N = 3 and A be any real value, the input repeats every three samples and the local average of three of its values is zero, giving y[n] = 0 for n ≥ 2; thus the steadystate response will be zero. The following MATLAB script uses the function conv to compute the convolution sum when the input is either x[n] = u[n] or x[n] = cos(2πn/3)u[n]. %%%%%%%%%%%%%%%%%% % Example 8.23  Convolution sum %%%%%%%%%%%%%%%%%% x1 = [0 0 ones(1, 20)] % unitstep input n = −2:19; n1 = 0:19; x2 = [0 0 cos(2∗pi∗n1/3)]; % cosine input h = (1/3)∗ones(1, 3); % impulse response y = conv(x1, h); y1 = y(1:length(n)); % convolution sums y = conv(x2, h); y2 = y(1:length(n));
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1
1
0.8
0.8
0.6
0.6
y1[n]
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CH A P T E R 8: DiscreteTime Signals and Systems
0.4 0.2
0.4 0.2
0
0 0
5
10
15
0
5
0
5
n
15
10
15
0.3 y2 [n]
0.5
FIGURE 8.10 Convolution sums for a movingaveraging filter with input x1 [n] = u[n] and x2 [n] = cos(2π n/3)u[n].
10 n
0.4
1
x2 [n]
492
0
−0.5
0.2 0.1 0
−1 0
5
10
−0.1
15
n
n
Notice that each of the input sequences has two zeros at the beginning so that the response can be found at n ≥ −2. Also, when the input is of infinite support, like when x[n] = u[n], we can only approximate it as a finite sequence in MATLAB, and as such the final values of the convolution obtained from conv are not correct and should not be considered. In this case, the final two values of the convolution results are not correct and are not considered. The results are shown in Figure 8.10. n n Example 8.24 Consider an autoregressive system represented by a firstorder difference equation y[n] = 0.5y[n − 1] + x[n]
n≥0
Find the impulse response h[n] of the system and then compute the response of the system to x[n] = u[n] − u[n − 3] using the convolution sum. Verify results with MATLAB. Solution The impulse response h[n] can be found recursively. Letting x[n] = δ[n], y[n] = h[n], and initial condition y[−1] = h[−1] = 0, we have h[0] = 0.5h[−1] + δ[0] = 1 h[1] = 0.5h[0] + δ[1] = 0.5 h[2] = 0.5h[1] + δ[2] = 0.52 h[3] = 0.5h[2] + δ[3] = 0.53 ···
8.3 DiscreteTime Systems
from which a general expression for the impulse response is obtained as h[n] = 0.5n u[n]. The response to x[n] = u[n] − u[n − 3] using the convolution sum is then given by y[n] =
∞ X k=−∞
∞ X
x[k]h[n − k] =
(u[k] − u[k − 3])0.5n−k u[n − k]
k=−∞
Since as functions of k, u[k]u[n − k] = 1 for 0 ≤ k ≤ n, zero otherwise, and u[k − 3]u[n − k] = 1 for 3 ≤ k ≤ n, zero otherwise (in the two cases, draw the two signals as functions of k and verify this is true), y[n] can be expressed as " n # n X X n −k −k y[n] = 0.5 0.5 − 0.5 u[n] k=0
k=3
n= 0.95, noise(m) = −1.5; else noise(m) = 0; end end
8.3 DiscreteTime Systems
2 y1[n]
1 0 −1 −2
0
20
40
60
80
100 120 140 160 180 200 n
(a) 2
2
1 z2 [n]
z1[n]
1 0
−1
−1 −2
0
0
20
40
60
80
100 120 140 160 180 200 n
−2
0
20
40
60
80
(b)
100 120 140 160 180 200 n
(c)
FIGURE 8.13 Top figure (a): noisy signal (dashed blue line) and clean signal (solid line). The clean signal (dashed line) is superposed on the denoised signal (solid blue line) in the bottom plots. The solid line in plot (b) is the result of median filtering, and the solid line in plot (c) is the result of the averager. x = [2∗cos(pi∗n(1:100)/256) zeros(1, 100)]; y1 = x + noise; % linear filtering z2 = averager(15, y1); % nonlinear filtering  median filtering z1(1) = median([0 0 y1(1) y1(2) y1(3)]); z1(2) = median([0 y1(1) y1(2) y1(3) y1(4)]); z1(N − 1) = median([y1(N − 3) y1(N − 2) y1(N − 1) y1(N) 0]); z1(N) = median([y1(N − 2) y1(N − 1) y1(N) 0 0]); for k = 3:N − 2, z1(k) = median([y1(k − 2) y1(k − 1) y1(k) y1(k + 1) y1(k + 2)]); end
Although the theory of nonlinear filtering is beyond the scope of this book, it is good to remember that in cases like this when linear filters do not seem to do well, there are other methods to use.
8.3.5 Causality and Stability of DiscreteTime Systems As with continuoustime systems, two additional independent properties of discretetime systems are causality and stability. Causality relates to the conditions under which computation can be performed in real time, while stability relates to the usefulness of the system.
Causality In many situations signals need to be processed in real time—that is, the processing must be done as the signal comes into the system. In those situations, the system must be causal. In many other
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situations, realtime processing is not required as the data can be stored and processed without the requirements of real time. Under such circumstances causality is not necessary. A discretetime system S is causal if: Whenever the input x[n] = 0, and there are no initial conditions, the output is y[n] = 0. n The output y[n] does not depend on future inputs.
n
Causality is independent of the linearity and timeinvariance properties of a system. For instance, the system represented by the input–output equation y[n] = x2 [n] where x[n] is the input and y[n] is the output is nonlinear but time invariant. According to the above definition it is a causal system: The output is zero whenever the input is zero, and the output depends on the present value of the input. Likewise, an LTI system can be noncausal, as can be seen in the following discretetime system that computes the moving average of the input: y[n] =
1 (x[n + 1] + x[n] + x[n − 1]). 3
The input–output equation indicates that at the present time n to compute y[n] we need a present value x[n], a past value x[n − 1], and a future value x[n + 1]. Thus, the system is LTI but noncausal since it requires future values of the input. n
n
n
An LTI discretetime system is causal if the impulse response of the system is such that h[n] = 0
n n). According to this equation the output depends on inputs {x[0], . . . , x[n]}, which are past and present values of the input. n Example 8.25 So far we have considered the convolution sum as a way of computing the output y[n] of an LTI system with impulse response h[n] for a given input x[n]. But it actually can be used to find either of these three variables given the other two. The problem is then called deconvolution. Assume the input x[n] and the output y[n] of a causal LTI system are given. Find equations to compute recursively the impulse response h[n] of the system. Consider finding the impulse response h[n] of a causal LTI system with input x[n] = u[n] and output y[n] = δ[n]. Use the MATLAB function deconv to find h[n]. Solution If the system is causal and LTI, the input x[n] and the output y[n] are connected by the convolution sum y[n] =
n X
h[n − m]x[m] = h[n]x[0] +
m=0
n X
h[n − m]x[m]
m=1
To find h[n] from given x[n] and y[n], under the condition that x[0] 6= 0, the above equation can be rewritten as " # n X 1 y[n] − h[n − m]x[m] h[n] = x[0] m=1
so that the impulse response of the causal LTI can be found recursively as follows: 1 y[0] x[0] 1 h[1] = y[1] − h[0]x[1] x[0] 1 h[2] = y[2] − h[0]x[2] − h[1]x[1] x[0]
h[0] =
.. .
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For the given case where y[n] = δ[n] and x[n] = u[n], we get, according to the above, 1 y[0] = 1 x[0] 1 h[1] = y[1] − h[0]x[1] = 0 − 1 = −1 x[0] 1 y[2] − h[0]x[2] − h[1]x[1] = 0 − 1 + 1 = 0 h[2] = x[0] 1 h[3] = y[3] − h[0]x[3] − h[1]x[2] − h[2]x[3] = 0 − 1 + 1 − 0 = 0 x[0]
h[0] =
.. . and, in general, h[n] = δ[n] − δ[n − 1]. The length of the convolution y[n] is the sum of the lengths of the input x[n] and of the impulse response h[n] minus one. Thus, length of h[n] = length of y[n] − length of x[n] + 1 When using deconv we need to make sure that the length of y[n] is always larger than that of x[n]. If x[n] is of infinite length, like when x[n] = u[n], this would require an even longer y[n], which is not possible. However, MATLAB can only provide a finitesupport input, so we make the support of y[n] larger. In this example we have found analytically that the impulse response h[n] is of length 2, so if the length of y[n] is chosen so that length y[n] is larger than the length of x[n] by one, we get the correct answer (case (a) in the script below); otherwise we do not (case (b)). Run the two cases to verify this (get rid of % symbol to run case (b)). %%%%%%%%%%%%%%%%%%%%%% % Example 8.25  Deconvolution %%%%%%%%%%%%%%%%%%%%%% clear all x = ones(1, 100); y = [1 zeros(1, 100)]; % (a) correct h % y = [1 zeros(1, 99)]; % (b) incorrect h [h, r] = deconv(y, x)
n
BoundedInput BoundedOutput Stability Stability characterizes useful systems. A stable system provides wellbehaved outputs for wellbehaved inputs. Boundedinput boundedoutput (BIBO) stability establishes that for a bounded (that is what is meant by wellbehaved) input x[n] the output of a BIBOstable system y[n] is also bounded. This means that if there is a finite bound M < ∞ such that x[n] < M for all n (you can think of it as an envelope [−M, M] inside which the input is in for all time), the output is also bounded (i.e., y[n] < L for L < ∞ and all n).
8.3 DiscreteTime Systems
An LTI discretetime system is said to be BIBO stable if its impulse response h[n] is absolutely summable, X h[k] < ∞ (8.39) k
Assuming that the input x[n] of the system is bounded, or that there is a value M < ∞ such that x[n] < M for all n, the output y[n] of the system represented by a convolution sum is also bounded, or X ∞ X ∞ x[n − k]h[k] ≤ x[n − k]h[k] y[n] ≤ k=−∞ k=−∞ ≤M
∞ X
h[k] ≤ MN < ∞
k=−∞
provided that
P∞
k=−∞ h[k]
< N < ∞, or that the impulse response be absolutely summable.
Remarks n
n
n
Nonrecursive or FIR systems are BIBO stable. Indeed, the impulse response of such a system is of finite length and thus absolutely summable. For a recursive or IIR system represented by a difference equation, to establish stability we need to find the system impulse response h[n] and determine whether it is absolutely summable or not. A much simpler way to test the stability of an IIR system will be based on the location of the poles of the Ztransform of h[n], as we will see in Chapter 9.
n Example 8.26 Consider an autoregressive system y[n] = 0.5y[n − 1] + x[n] Determine if the system is BIBO stable. Solution As shown in Example 8.24, the impulse response of the system is h[n] = 0.5n u[n]. Checking the BIBO stability condition, we have ∞ X n=−∞
Thus, the system is BIBO stable.
h[n] =
∞ X n=0
0.5n =
1 =2 1 − 0.5 n
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CH A P T E R 8: DiscreteTime Signals and Systems
8.4 WHAT HAVE WE ACCOMPLISHED? WHERE DO WE GO FROM HERE? As you saw in this chapter the theory of discretetime signals and systems is very similar to the theory of continuoustime signals and systems. Many of the results in the continuoustime theory are changed by swapping integrals for sums and differential equations for difference equations. However, there are significant differences imposed by the way the discretetime signals and systems are generated. For instance, the discrete frequency can be considered finite but circular, and it depends on the sampling time. Discrete sinusoids, as another example, are not necessarily periodic. Thus, despite the similarities there are also significant differences between the continuoustime and the discretetime signals and systems. Now that we have a basic structure for discretetime signals and systems, we will continue developing the theory of linear timeinvariant discretetime systems using transforms. Again, you will find a great deal of similarity but also some very significant differences. In the next chapters, carefully notice the relation that exists between the Ztransform and the Fourier representations of discretetime signals and systems, not only with each other but with the Laplace and Fourier transforms. There is a great deal of connection among all of these transforms, and a clear understanding of this would help you with the analysis and synthesis of discretetime signals and systems.
PROBLEMS 8.1. Discrete sequence—MATLAB Consider the following formula x[n] = x[n − 1] + x[n − 3]
n≥3
x[0] = 0 x[1] = 1 x[2] = 2 Find the rest of the sequence for 0 ≤ n ≤ 50 and plot it using the MATLAB function stem. 8.2. Finiteenergy signals—MATLAB Given the discrete signal x[n] = 0.5n u[n]: (a) Use MATLAB to plot the signal x[n] for n = −5 to 200. (b) Is this a finiteenergy discretetime signal? That is compute the infinite sum ∞ X
x[n]2
n=−∞
Hint: Show that ∞ X n=0
αn =
1 1−α
Problems
or equivalently that ∞ X
(1 − α)
αn = 1
n=0
provided α < 1. (c) Verify your results by using symbolic MATLAB to find an expression for the above sum. 8.3. Periodicity of sampled signals—MATLAB Consider an analog periodic sinusoid x(t) = cos(3π t + π/4) being sampled using a sampling period Ts to obtain the discretetime signal x[n] = x(t)t=nTs = cos(3π Ts n + π/4). (a) Determine the discrete frequency of x[n]. (b) Choose a value of Ts for which the discretetime signal x[n] is periodic. Use MATLAB to plot a few periods of x[n], and verify its periodicity. (c) Choose a value of Ts for which the discretetime signal x[n] is not periodic. Use MATLAB to plot x[n] and choose an appropriate length to show the signal is not periodic. (d) Determine under what condition the value of Ts makes x[n] periodic. 8.4. Even and odd decomposition and energy—MATLAB Suppose you sample the analog signal x(t) =
1−t 0
0≤t≤1 otherwise
with a sampling period Ts = 0.25 to generate x[n] = x(t)t=nTs . (a) Use MATLAB to plot x[−n] for an appropriate interval. (b) Find xe [n] = 0.5[x[n] + x[−n]] and plot it carefully using MATLAB. (c) Find xo [n] = 0.5[x[n] − x[−n]] and plot it carefully using MATLAB. (d) Verify that xe [n] + xo [n] = x[n] graphically. (e) Compute the energy of x[n] and compare it to the sum of the energies of xe [n] and xo [n]. 8.5. Signal representation in terms of u[n]—MATLAB We have shown how any discretetime signal can be represented as a sum of weighted and shifted versions of δ[n]. Given that δ[n] = u[n] − u[n − 1] it should be possible to represent any signal as a combination of unitstep functions. Consider a discretetime ramp r[n] = nu[n], which in terms of δ[n] is written as r[n] =
∞ X
r[k]δ[n − k]
k=−∞
Replace r[k] = ku[k] and use δ[n] = u[n] − u[n − 1] to show that r[n] can be expressed in terms of u[n] as r[n] =
∞ X
u[n − k]
k=1
Does this equation make sense? Use MATLAB to plot the obtained r[n] to help you answer this.
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8.6. Generation of periodic discretetime signals—MATLAB Periodic signals can be generated by obtaining a period and adding shifted versions of this period. Suppose we wish to generate a train of triangular pulses. A period of the signal is x[n] = r[n] − 2r[n − 1] + r[n − 2] where r[n] = nu[n] is the discretetime ramp signal. (a) Carefully plot x[n]. (b) Let y[n] =
∞ X
x[n − 2k]
k=−∞
and carefully plot it. Indicate the period N of y[n]. (c) Write a MATLAB script to generate and plot the periodic signal y[n]. 8.7. Expansion and compression of discretetime signals—MATLAB Consider the discretetime signal x[n] = cos(2π n/7). (a) The discretetime signal can be compressed by getting rid of some of its samples (downsampling). Consider the downsampling by 2. Write a MATLAB script to obtain and plot z[n] = x[2n]. Plot also x[n] and compare it with z[n]. What happended? Explain. (b) The expansion for discretetime signals requires interpolation, and we will see it later. However, a first step of this process is the socalled upsampling. Upsampling by 2 consists in defining a new signal y[n] such that y[n] = x[n/2] for n even, and y[n] = 0 otherwise. Write a MATLAB script to perform upsampling on x[n]. Plot the resulting signal y[n] and explain its relation with x[n]. (c) If x[n] resulted from sampling a continuoustime signal x(t) = cos(2π t) using a sampling period Ts and with no frequency aliasing, determine Ts . How would you sample the analog signal x(t) to get the downsampled signals z[n]? That is, choose a value for the sampling period Ts to get z[n] directly from x(t). Can you choose Ts to get y[n] from x(t) directly? Explain. 8.8. Absolutely summable and finiteenergy discretetime signals—MATLAB Suppose we sample the analog signal x(t) = e−2t u(t) using a sample period Ts = 1. (a) Expressing the sampled signal as x(nTs ) = x[n] = α n u[n], what is the corresponding value of α? Use MATLAB to plot x[n]. (b) Show that x[n] is absolutely summable—that is, show the following sum is finite: ∞ X
x[n]
n=−∞
(c) If you know that x[n] is absolutely summable, could you say that x[n] is a finiteenergy signal? Use MATLAB to plot x[n] and x2 [n] in the same plot to help you decide. (d) In general, for what values of α are signals y[n] = α n u[n] finite energy? Explain. 8.9. Discretetime periodic signals Determine whether the following discretetime sinusoids are periodic or not. If periodic, determine its period N0 . x[n] = 2 cos(πn − π/2) y[n] = sin(n − π/2) z[n] = x[n] + y[n] v[n] = sin(3π n/2)
Problems
8.10. Periodicity of discretetime signals Consider periodic signals x[n], of period N1 = 4, and y[n], of period N2 = 6. What would be the period of z[n] = x[n] + y[n] v[n] = x[n]y[n] w[n] = x[2n] 8.11. Periodicity of sum and product of periodic signals—MATLAB If x[n] is periodic of period N1 > 0 and y[n] is periodic of period N2 > 0: (a) What should be the condition for the sum of x[n] and y[n] to be periodic? (b) What would be the period of the product x[n]y[n]? (c) Would the formula N1 N2 gcd(N1, N2) (gcd(N1 , N2 ) stands for the greatest common divisor of N1 and N2 ) give the period of the sum and the product of the two signals x[n] and y[n]? (d) Use MATLAB to plot the signals x[n] = cos(2π n/3)u[n], and y[n] = (1 + sin(6π n/7))u[n], their sum and product, and to find their periods and to verify your analytic results. 8.12. Echoing of music—MATLAB An effect similar to multipath in acoustics is echoing or reverberation. To see the effects of an echo in an acoustic signal consider the simulation of echoes on the “handel.mat” signal y[n]. Pretend that this piece is being played in a round theater where the orchestra is in the middle of two concentric circles and the walls on one half side are at a radial distances of 17.15 meters (corresponding to the inner circle) and 34.3 meters (corresponding to the outer circle) on the other side (yes, an usual theater!) from the orchestra. The speed of sound is 343 meters/sec. Assume that the recorded signal is the sum of the original signal y[n] and attenuated echoes from the two walls so that the recorded signal is given by r[n] = y[n] + 0.8y[n − N1 ] + 0.6y[n − N2 ] where N1 is the delay caused by the closest wall and N2 is the delay caused by the farther wall. The recorder is at the center of the auditorium where the orchestra is and we record for 1.5 seconds. (a) Find the values of the two delays N1 and N2 . Give the expression for the discretetime recorded signal r[n]. The sampling frequency Fs of “handel.mat” is given when you load it in MATLAB. (b) Simulate the echo signal. Plot r[n]. Use sound to listen to the original and the echoed signals. 8.13. Envelope modulation—MATLAB In the generation of music by computer, the process of modulation is extremely important. When playing an instrument, the player typically does it in three stages: (1) rise time or attack, (2) sustained time, and (3) decay time. Suppose we model these three stages as an envelope continuoustime signal given by e(t) =
1 1 [r(t) − r(t − 3)] − [r(t − 20) + r(t − 30)] 3 0.1
where r(t) is the ramp signal. (a) For a simple tone x(t) = cos(2π/T0 t), the modulated signal is y(t) = x(t)e(t). Find the period T0 so that 100 cycles of the sinusoid occur for the duration of the envelope signal. (b) Simulate in MATLAB the modulated signal using the value of T0 = 1 and a simulation sampling time of 0.1T0 . Plot y(t) and e(t) (discretized with the sampling period 0.1T0 ) and listen to the modulated signal using sound.
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CH A P T E R 8: DiscreteTime Signals and Systems
8.14. LTI of ADCs An ADC can be thought of as composed of three subsystems: a sampler, a quantizer, and a coder. (a) The sampler, as a system, has as input an analog signal x(t) and as output a discretetime signal x(nTs ) = x(t)t=nTs where Ts is the sampling period. Determine whether the sampler is a linear system or not. (b) Sample x(t) = cos(0.5π t)u(t) and x(t − 0.5) using Ts = 1 to get y(nTs ) and z(nTs ), respectively. Plot x(t), x(t − 0.5), and y(nTs ) and z(nTs ). Is z(nTs ) a shifted version of y(nTs ) so that you can say the sampler is time invariant? Explain. 8.15. LTI of ADCs (part 2) A twobit quantizer of an ADC has as input x(nTs ) and as output xˆ (nTs ), such that if k1 ≤ x(nTs ) < (k + 1)1
→
xˆ (nTs ) = k1
k = −2, −1, 0, 1
(a) Is this system time invariant? Explain. (b) Suppose that the value of 1 in the quantizer is 0.25, and the sampled signal is x(nTs ) = nTs , Ts = 0.1 and −5 ≤ n ≤ 5. Use the sampled signal to determine whether the quantizer is a linear system or not. Explain. (c) From the results in this and the previous problem, would you say that the ADC is an LTI system? Explain. 8.16. Rectangular windowing system—MATLAB A window is a signal w[n] that is used to highlight part of another signal. The windowing process consists in multiplying an input signal x[n] by the window signal w[n], so that the output is y[n] = x[n]w[n] There are different types of windows used in signal processing. One of them is the socalled rectangular window, which is given by w[n] = u[n] − u[n − N]
(a) Determine whether the rectangular windowing system is linear. Explain. (b) Suppose x[n] = nu[n]. Plot the output y[n] of the windowing system (with N = 6). (c) Let the input be x[n − 6]. Plot the corresponding output of the rectangular windowing system, and indicate whether the rectangular windowing system is time invariant. 8.17. Impulse response of an IIR system—MATLAB A discretetime IIR system is represented by the difference equation y[n] = 0.15y[n − 2] + x[n]
n≥0
where x[n] is the input and y[n] is the output. (a) To find the impulse response h[n] of the system, let x[n] = δ[n], y[n] = h[n], and the initial conditions be zero, y[n] = h[n] = 0, n < 0. Find recursively the values of h[n] for values of n ≥ 0. (b) As a second way to do it, replace the relation between the input and the output given by the difference equation to obtain a convolution sum representation that will give the impulse response h[n]. What is h[n]? (c) Use the MATLAB function filter to get the impulse response h[n] (use help to learn about the function filter).
Problems
8.18. FIR filter—MATLAB An FIR filter has a nonrecursive input–output relation
y[n] =
5 X
kx[n − k]
k=0
Find and plot using MATLAB the impulse response h[n] of this filter. Is this a causal and stable filter? Explain. Find and plot the unitstep response s[n] for this filter. If the input x[n] for this filter is bounded, i.e., x[n] < 3, what would be a minimum bound M for the output (i.e., y[n] ≤ M)? (e) Use the MATLAB function filter to compute the impulse response h[n] and the unitstep response s[n] for the given filter and plot them.
(a) (b) (c) (d)
8.19. LTI and convolution sum—MATLAB The impulse response of a discretetime system is h[n] = (−0.5)n u[n]. (a) If the input of the system is x[n] = δ[n] + δ[n − 1] + δ[n − 2], use the linearity and time invariance of the system to find the corresponding output y[n]. (b) Find the convolution sum corresponding to the above input, and show that your solution coincides with the output y[n] obtained above. (c) Use the MATLAB function conv to find the output y[n] due to the given input x[n]. Plot x[n], h[n], and y[n] using MATLAB. 8.20. Steady state of IIR systems—MATLAB Suppose an IIR system is represented by a difference equation y[n] = ay[n − 1] + x[n] where x[n] is the input and y[n] is the output. (a) If the input x[n] = u[n] and it is known that the steadystate response is y[n] = 2, what would be a for that to be possible (in steady state x[n] = 1 and y[n] = y[n − 1] = 2 since n → ∞). (b) Writing the system input as x[n] = u[n] = δ[n] + δ[n − 1] + δ[n − 2] + · · · then according to the linearity and time invariance, the output should be y[n] = h[n] + h[n − 1] + h[n − 2] + · · · Use the value for a found above, that the initial condition is zero (i.e., y[−1] = 0) and that the input is x[n] = u[n], to find the values of the impulse response h[n] for n ≥ 0 using the above equation. The system is causal. (c) Use the MATLAB function filter to compute the impulse response h[n] and compare it with the one obtained above. 8.21. Causal systems and realtime processing Systems that operate under realtime conditions need to be causal—that is, they can only process present and past inputs. When no realtime processing is needed the system can be noncausal.
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CH A P T E R 8: DiscreteTime Signals and Systems
(a) Consider the case of averaging an input signal x[n] under realtime conditions. Suppose you are given two different filters, • y[n] =
N−1 1 X x[n − k] N k=0
• y[n] =
1 N
N−1 X
x[n − k]
k=−N+1
Which one of these would you use and why? (b) If you are given a tape with the data, which of the two filters would you use? Why? Would you use either? Explain. 8.22. IIR versus FIR systems A significant difference between IIR and FIR discretetime systems is stability. Consider an IIR filter with the difference equation y1 [n] = x[n] − 0.5y1 [n − 1] where x[n] is the input and y1 [n] is the output. Then consider an FIR filter y2 [n] = x[n] + 0.5x[n − 1] + 3x[n − 2] + x[n − 5] where x[n] is the input and y2 [n] is the output. (a) Since to check the stability of these filters we need their impulse responses, find the impulse responses h1 [n] corresponding to the IIR filter by recursion, and h2 [n] corresponding to the FIR filter. (b) Use the impulse response h1 [n] to check the stability of the IIR filter. (c) Use the impulse response h2 [n] to check the stability of the FIR filter. (d) Since the impulse response of a FIR filter has a finite number of nonzero terms, would it be correct to say that FIR filters are always stable? Explain. 8.23. Unitstep versus impulse response—MATLAB The unitstep response of a discretetime LTI system is s[n] = 2[(−0.5)n − 1]u[n] Use this information to find (a) The impulse response h[n] of the discretetime LTI system. (b) The response of the LTI system to a ramp signal x[n] = nu[n]. Use the MATLAB function filter and superposition to find it. 8.24. Convolution sum—MATLAB A discretetime system has a unitimpulse response h[n]. (a) Let the input to the discretetime system be a pulse x[n] = u[n] − u[n − 4]. Compute the output of the system in terms of the impulse response. (b) Let h[n] = 0.5n u[n]. What would be the response of the system y[n] to x[n] = u[n] − u[n − 4]? Plot the output y[n]. (c) Use the convolution sum to verify your response y[n]. (d) Use the MATLAB function conv to compute the response y[n] to x[n] = u[n] − u[n − 4]. Plot both the input and output. 8.25. Discrete envelope detector—MATLAB Consider an envelope detector that would be used to detect the message sent in an AM system. Consider the envelope detector as a system composed of the cascading of two systems: one which computes the
Problems
absolute value of the input, and a second one that lowpass filters its input. A circuit that is used as an envelope detector consists of a diode circuit that does the absolute value operation, and an RC circuit that does the lowpass filtering. The following is an implementation of these operations in the discretetime system. Let the input to the envelope detector be a sampled signal, x(nTs ) = p(nTs ) cos(2000π nTs ) where p(nTs ) = u(nTs ) − u(nTs − 20Ts ) + u(nTs − 40Ts ) − u(nTs − 60Ts ) where two pulses of duration 20Ts and amplitude equal to one. (a) Choose Ts = 0.01, and generate 100 samples of the input signal x(nTs ) and plot it. (b) Consider then the subsystem that computes the absolute value of the input x(nTs ) and compute and plot 100 samples of y(nTs ) = x(nTs ). (c) Let the lowpass filtering be done by a moving averager of order 15—that is, if y(nTs ) is the input, then the output of the filter is z(nTs ) =
14 1 X y(nTs − kTs ) 15 k=0
Implement this filter using the MATLAB function filter, and plot the result. Explain your results. (d) Is this a linear system? Come up with an example using the script developed above to show that the system is linear or not.
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CHAPTER 9
The ZTransform
I was born not knowing and have had only a little time to change that here and there. Richard P. Feynman, (1918–1988) Professor and Nobel Prize physicist
9.1 INTRODUCTION Just as with the Laplace transform for continuoustime signals and systems, the Ztransform provides a way to represent discretetime signals and systems, and to process discretetime signals. Although the Ztransform can be related to the Laplace transform, the relation is operationally not very useful. However, it can be used to show that the complex zplane is in a polar form where the radius is a damping factor and the angle corresponds to the discrete frequency ω in radians. Thus, the unit circle in the zplane is analogous to the j axis in the Laplace plane, and the inside of the unit circle is analogous to the lefthand splane. We will see that once the connection between the Laplace plane and the zplane is established, the significance of poles and zeros in the zplane can be obtained like in the Laplace plane. The representation of discretetime signals by the Ztransform is very intuitive—it converts a sequence of samples into a polynomial. The inverse Ztransform can be achieved by many more methods than the inverse Laplace transform, but the partial fraction expansion is still the most commonly used method. Using the onesided Ztransform, for solving difference equations that could result from the discretization of differential equations, but not exclusively, is an important application of the Ztransform. As it was the case with the Laplace transform and the convolution integral, the most important property of the Ztransform is the implementation of the convolution sum as a multiplication of polynomials. This is not only important as a computational tool but also as a way to represent a discrete system by its transfer function. Filtering is again an important application, and as before, the Signals and Systems Using MATLAB®. DOI: 10.1016/B9780123747167.000132 c 2011, Elsevier Inc. All rights reserved.
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CH A P T E R 9: The ZTransform
localization of poles and zeros determines the type of filter. However, in the discrete domain there is a greater variety of filters than in the analog domain.
9.2 LAPLACE TRANSFORM OF SAMPLED SIGNALS The Laplace transform of a sampled signal x(t) =
X
x(nTs )δ(t − nTs )
(9.1)
n
is given by X(s) =
X
x(nTs )L[δ(t − nTs )]
n
=
X
x(nTs )e−nsTs
(9.2)
n
By letting z = esTs , we can rewrite Equation (9.2) as Z[x(nTs )] = L[xs (t)] z=esTs X = x(nTs )z−n
(9.3)
n
which is called the Ztransform of the sampled signal.
Remarks The function X(s) in Equation (9.2) is different from the Laplace transforms we considered before: n
Letting s = j, X() is periodic of period 2π/Ts (i.e., X( + 2π/Ts ) = X() for an integer k). Indeed, X( + 2π/Ts ) =
X
x(nTs )e−jn(+2π/Ts )Ts =
n n
X
x(nTs )e−jn(Ts +2π ) = X()
n
X(s) may have an infinite number of poles or zeros—complicating the partial fraction expansion when finding its inverse. Fortunately, the presence of the {e−nsTs } terms suggests that the inverse should be done using the timeshift property of the Laplace transform instead, giving Equation (9.1).
n Example 9.1 To see the possibility of an infinite number of poles and zeros in the Laplace transform of a sampled signal, consider a pulse x(t) = u(t) − u(t − T0 ) sampled with a sampling period Ts = T0 /N. Find the Laplace transform of the sampled signal and determine its poles and zeros. Solution The sampled signal is x(nTs ) =
1 0
0 ≤ nTs ≤ T0 or 0 ≤ n ≤ N otherwise
9.2 Laplace Transform of Sampled Signals
with Laplace transform
X(s) =
N X
e−nsTs =
n=0
1 − e−(N+1)sTs 1 − e−sTs
The poles are the sk values that make the denominator zero—that is, e−sk Ts = 1 = e j2πk
k integer, −∞ < k < ∞
or sk = −j2πk/Ts for any integer k, an infinite number of poles. Similarly, one can show that X(s) has an infinite number of zeros by finding the values sm that make the numerator zero, or e−(N+1)sm Ts = 1 = e j2πm
m integer, −∞ < m < ∞
or sm = −j2πm/((N + 1)Ts ) for any integer m. Such a behavior can be better understood when we consider the connection between the splane and the zplane. n
The History of the ZTransform The history of the Ztransform goes back to the work of the French mathematician De Moivre, who in 1730 introduced the characteristic function to represent the probability mass function of a discrete random variable. The characteristic function is identical to the Ztransform. Also, the Ztransform is a special case of the Laurent’s series, used to represent complex functions. In the 1950s the Russian engineer and mathematician Yakov Tsypkin (1919–1997) proposed the discrete Laplace transform, which he applied to the study of pulsed systems. Then Professor John Ragazzini and his students Eliahu Jury and Lofti Zadeh at Columbia University developed the Ztransform. Ragazzini (1912–1988) was chairman of the Department of Electrical Engineering at Columbia University. Three of his students are well recognized in electrical engineering for their accomplishments: Jury for the Ztransform, nonlinear systems, and the inners stability theory; Zadeh for the Ztransform and fuzzy set theory; and Rudolf Kalman for the Kalman filtering. Jury was born in Iraq, and received his doctor of engineering science degree from Columbia University in 1953. He was professor of electrical engineering at the University of California, Berkeley, and at the University of Miami. Among his publications, Professor Jury’s “Theory and Application of the Ztransform,” is a seminal work on the theory and application of the Ztransform.
Remarks n
The relation z = esTs provides the connection between the splane and the zplane: z = esTs = e(σ +j)Ts = eσ Ts e jTs
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Letting r = eσ Ts and ω = Ts , we have that z = re jω
n
which is a complex variable in polar form, with radius 0 ≤ r < ∞ and angle ω in radians. The variable r is a damping factor and ω is the discrete frequency in radians, so the zplane corresponds to circles of radius r and angles −π ≤ ω < π. Let us see how the relation z = e sTs maps the splane into the zplane. Consider the strip of width 2π/Ts across the splane shown in Figure 9.1. The width of this strip is related to the Nyquist condition establishing that the maximum frequency of the analog signals we are considering is M = s /2 = π/Ts where s is the sampling frequency and Ts is the sampling period. If Ts → 0, we would be considering the class of signals with maximum frequency approaching ∞—that is, all signals. The relation z = e sTs maps the real part of s = σ + j, Re(s) = σ , into the radius r = eσ Ts ≥ 0, and the analog frequencies −π/Ts ≤ ≤ π/Ts into −π ≤ ω < π, according to the frequency connection ω = Ts . Thus, the mapping of the j axis in the splane, corresponding to σ = 0, gives a circle of radius r = 1 or the unit circle.
n
The righthand splane, σ > 0, maps into circles with radius r > 1, and the lefthand splane, σ < 0, maps into circles of radius r < 1. Points A, B, and C in the strip are mapped into corresponding points in the zplane as shown in Figure 9.1. So the given strip in the splane maps into the whole zplane— similarly for other strips of the same width. Thus, the splane, as a union of these strips, is mapped onto the same zplane. The mapping z = e sTs can be used to illustrate the sampling process. Consider a bandlimited signal x(t) with maximum frequency π/Ts with a spectrum in the band [−π/Ts π/Ts ]. According to the relation z = e sTs the spectrum of x(t) in [−π/Ts π/Ts] is mapped into the unit circle of the zplane from [−π, π ) on the unit circle. Going around the unit circle in the zplane, the mapped frequency response repeats periodically just like the spectrum of the sampled signal. z = e sTs
jΩ
j
π Ts
A
−j
π Ts
r
B
ω σ
B C
A
C
splane
zplane
FIGURE 9.1 Mapping of the Laplace plane into the zplane. Slabs of width 2π/Ts in the lefthand splane are mapped into the inside of a unit circle in the zplane. The righthand side of the slab is mapped outside the unit circle. The jaxis in the splane is mapped into the unitcircle in the zplane. The whole splane as a union of these slabs is mapped onto the same zplane.
9.3 TwoSided ZTransform
9.3 TWOSIDED ZTRANSFORM Given a discretetime signal x[n], −∞ < n < ∞, its twosided Ztransform is X(z) =
∞ X
x[n]z−n
(9.4)
n=−∞
defined in a region of convergence (ROC) in the zplane.
Considering the sampled signal x(nTs ) a function of n in Equation (9.3), we obtain the twosided Ztransform. Remarks n
n
The Ztransform can be thought of as the transformation of the sequence {x[n]} into a polynomial X(z) (possibly of infinite degree in positive and negative powers of z) where to each x[n0 ] we attach a monomial z−n0 . Thus, given a sequence of samples {x[n]} its Ztransform simply consists in creating a polynomial with coefficients x[n] corresponding to z−n . Given a Ztransform as in Equation (9.4), its inverse is easily obtained by looking at the coefficients attached to the z−n monomials for positive as well as negative values of the sample value n. Clearly, this inverse is not in a closed form. We will see ways to compute these later in this chapter. The twosided Ztransform is not useful in solving difference equations with initial conditions, just as the twosided Laplace transform was not useful either in solving differential equations with initial conditions. To include initial conditions in the transformation it is necessary to define the onesided Ztransform. The onesided Ztransform is defined for a causal signal, x[n] = 0 for n < 0, or for signals that are made causal by multiplying them with the unitstep signal u[n]: X1 (z) = Z(x[n]u[n]) =
∞ X
x[n]u[n]z−n
(9.5)
n=0
in a region of convergence R1 . The twosided Ztransform can be expressed in terms of the onesided Ztransform as follows: X(z) = Z x[n]u[n] + Z x[−n]u[n] z − x[0]
(9.6)
The region of convergence of X(z) is R = R1 ∩ R2 where R1 is the region of convergence of Z x[n]u[n] and R2 is the region of convergence of Z x[−n]u[n] z .
The onesided Ztransform coincides with the twosided Ztransform whenever the discretetime signal x[n] is causal (i.e., x[n] = 0 for n < 0). If the signal is noncausal, multiplying it by u[n] makes it
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causal. To express the twosided Ztransform in terms of the onesided Ztransform we separate the sum into two and make each into a causal sum: X(z) =
∞ X
x[n]z−n =
n=−∞
∞ X
x[n]u[n]z−n +
0 X
x[n]u[−n]z−n − x[0]
n=−∞
n=0
∞ X = Z x[n]u[n] + x[−m]u[m]zm − x[0] m=0
= Z x[n]u[n] + Z x[−n]u[n] z − x[0] where the inclusion of the additional term x[0] in the sum from −∞ to 0 is compensated by subtracting it, and in the same sum a change of variable (m = −n) gives a onesided Ztransform in terms of positive powers of z, as indicated by the notation Z x[−n]u[n] z .
9.3.1 Region of Convergence The infinite summation of the twosided Ztransform must converge for some values of z. For X(z) to converge it is necessary that X X X −n x[n]z ≤ x[n]r −n e jωn  = x[n]r −n  < ∞ X(z) = n n n Thus, the convergence of X(z) depends on r. The region in the zplane where X(z) converges, its ROC, connects the signal and its Ztransform uniquely. As with the Laplace transform, the poles of X(z) are connected with its region of convergence. The poles of a Ztransform X(z) are complex values {pk } such that X(pk ) → ∞ while the zeros of X(z) are the complex values {zk } that make X(zk ) = 0
n Example 9.2 Find the poles and the zeros of the following Ztransforms: (a) X1 (z) = 1 + 2z−1 + 3z−2 + 4z−3 (b) X2 (z) =
(z−1 − 1)(z−1 + 2)2 √ z−1 (z−2 + 2z−1 + 1)
9.3 TwoSided ZTransform
Solution To see the poles and the zeros more clearly let us express X1 (z) as a function of positive powers of z: z3 (1 + 2z−1 + 3z−2 + 4z−3 ) z3 N1 (z) z3 + 2z2 + 3z + 4 = = z3 D1 (z)
X1 (z) =
There are three poles at z = 0, the roots of D1 (z) = 0, and the zeros are the roots of N1 (z) = z3 + 2z2 + 3z + 4 = 0. Likewise, expressing X2 (z) as a function of positive powers of z, X2 (z) =
z3 (z−1 − 1)(z−1 + 2)2 √ z3 (z−1 (z−2 + 2z−1 + 1))
(1 − z)(1 + 2z)2 N2 (z) = √ 2 D2 (z) 1 + 2z + z √ The poles of X2 (z) are the roots of D2 (z) = 1 + 2z + z2 = 0, while the zeros of X2 (z) are the roots of N2 (z) = (1 − z)(1 + 2z)2 = 0. n =
The region of convergence depends on the support of the signal. If it is finite, the ROC is very much the whole zplane; if it is infinite, the ROC depends on whether the signal is causal, anticausal, or noncausal. Something to remember is that in no case does the ROC include any poles of the Ztransform.
ROC of FiniteSupport Signals The ROC of the Ztransform of a signal x[n] of finite support [N0 , N1 ] where −∞ < N0 ≤ n ≤ N1 < ∞, X(z) =
N1 X
x[n]z−n
(9.7)
n=N0
is the whole zplane, excluding the origin z = 0 and/or z = ±∞ depending on N0 and N1 .
Given the finite support of x[n] its Ztransform has no convergence problem. Indeed, for any z 6= 0 (or z 6= ±∞ if positive powers of z occur in Equation (9.40)), we have X(z) ≤
N1 X
x[n]z−n  ≤ (N1 − N0 + 1) max x[n] max z−n  < ∞
n=N0
The poles of X(z) are either at the origin of the zplane (e.g., when N0 ≥ 0) or there are no poles (e.g., when N1 ≤ 0). Thus, only when z = 0 or z = ±∞ would X(z) go to infinity. The ROC is the whole zplane excluding these values.
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n Example 9.3 Find the Ztransform of a discretetime pulse 1 x[n] = 0
0≤n≤9 otherwise
Determine the region of convergence of X(z). Solution The Ztransform of x[n] is X(z) =
9 X
1 z−n =
n=0
z10 − 1 1 − z−10 = 1 − z−1 z9 (z − 1)
(9.8)
That this sum equals the term on the right can be shown by multiplying the left term by the denominator 1 − z−1 and verifying the result is the same as the numerator in negative powers of z. In fact, (1 − z−1 )
9 X n=0
1 z−n =
9 X
1 z−n −
n=0
9 X
1 z−n−1
n=0
= (1 + z
−1
+ · · · + z−9 ) − (z−1 + · · · + z−9 + z−10 ) = 1 − z−10
Since x[n] is a finite sequence there is no problem with the convergence of the sum, although X(z) in Equation (9.8) seems to indicate the need for z 6= 1 (z = 1 makes the numerator and denominator zero). From the sum, if we let z = 1, then X(1) = 10, so there is no need to restrict z to be different from 1. This is caused by the pole at z = 1 being canceled by a zero. Indeed, the zeros zk of X(z) (see Eq. 9.8) are the roots of z10 − 1 = 0, which are zk = e j2πk/10 for k = 0, . . . , 9. Therefore, the zero when k = 0, or z0 = 1, cancels the pole at 1 so that Q9 (z − e jπk/5 ) X(z) = k=1 9 z That is, X(z) has nine poles at the origin and nine zeros around the unit circle except at z = 1. Thus, the whole zplane excluding the origin is the region of convergence of X(z) . n
ROC of InfiniteSupport Signals Signals of infinite support are either causal, anticausal, or a combination of these or noncausal. Now for the Ztransform of a causal signal xc [n] (i.e., xc [n] = 0, n < 0) Xc (z) =
∞ X n=0
xc [n]z−n =
∞ X
xc [n]r −n e−jnω
n=0
to converge we need to determine appropriate values of r, the damping factor. The frequency ω has no effect on the convergence. If R1 is the radius of the farthestout pole of Xc (z), then there is
9.3 TwoSided ZTransform
an exponential Rn1 u[n] such that xc [n] < MRn1 for n ≥ 0 for some value M > 0. Then, for X(z) to converge we need that Xc (z) ≤
∞ X
xc [n]r −n  < M
n=0
∞ X R1 n R1 . As indicated, this ROC does not include any poles of Xc (z)—it is the outside of a circle containing all the poles of Xc (z). Likewise, for an anticausal signal xa [n], if we choose a radius R2 that is smaller than the radius of all the poles of Xa (z), the region of convergence is z = r < R2 . This ROC does not include any poles of Xa (z)—it is the inside of a circle that does not contain any of the poles of Xa (z). If the signal x[n] is noncausal, it can be expressed as x[n] = xc [n] + xa [n] where the supports of xa [n] and xc [n] can be finite or infinite or any possible combination of these two. The corresponding ROC of X(z) = Z{x[n]} would then be 0 ≤ R1 < z < R2 < ∞ This ROC is a torus surrounded on the inside by the poles of the causal component, and in the outside by the poles of the anticausal component. If the signal has finite support, then R1 = 0 and R2 = ∞, coinciding with the result for finitesupport signals. For the Ztransform X(z) of an infinitesupport signal: n A causal signal x[n] has a region of convergence z > R1 where R1 is the largest radius of the poles of X(z)—that is, the region of convergence is the outside of a circle of radius R1 . n An anticausal signal x[n] has as region of convergence the inside of the circle defined by the smallest radius R2 of the poles of X(z), or z < R2 . n A noncausal signal x[n] has as region of convergence R1 < z < R2 , or the inside of a torus of inside radius R1 and outside radius R2 corresponding to the maximum and minimum radii of the poles of Xc (z) and Xa (z), which are the Ztransforms of the causal and anticausal components of x[n].
n Example 9.4 The poles of X(z) are z = 0.5 and z = 2. Find all the possible signals that can be associated with it according to different regions of convergence. Solution Possible regions of convergence are: n
{R1 : z > 2}—the outside of a circle of radius 2, we associate X(z) with a causal signal x1 [n].
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n
n
{R2 : z < 0.5}—the inside of a circle of radius 0.5, an anticausal signal x2 [n] can be associated with X(z). {R3 : 0.5 < z < 2}—a torus of radii 0.5 and 2, a noncausal signal x3 [n] can be associated with X(z).
Three different signals can be connected with X(z) by considering three different regions of convergence. n
n Example 9.5 Find the regions of convergence of the Ztransforms of the following signals: n 1 u[n] 2 n 1 (b) x2 [n] = − u[−n − 1] 2 (a) x1 [n] =
Determine then the Ztransform of x1 [n] + x2 [n]. Solution The signal x1 [n] is causal, while x2 [n] is anticausal. The Ztransform of x1 [n] is X1 (z) =
∞ n X 1 n=0
2
z−n =
z 1 = −1 1 − 0.5z z − 0.5
provided that 0.5z−1  < 1 or that its region of convergence is R1 : z > 0.5. The region R1 is the outside of a circle of radius 0.5. The signal x2 [n] grows as n decreases from −1 to −∞, and the rest of its values are zero. Its Ztransform is found as ∞ −m −1 n X X 1 1 −n X2 (z) = − z =− zm + 1 2 2 n=−∞ m=0
=−
∞ X
2m zm + 1 =
m=0
−1 z +1= 1 − 2z z − 0.5
with a region of convergence of R2 : z < 0.5. Although the signals are clearly different, their Ztransforms are identical. It is the corresponding regions of convergence that differentiate them. The Ztransform of x1 [n] + x2 [n] does not exist given that the intersection of R1 and R2 is empty. n
9.4 OneSided ZTransform
Remarks The uniqueness of the Ztransform requires that the Ztransform of a signal be accompanied by a region of convergence. It is possible to have identical Ztransforms with different regions of convergence, corresponding to different signals.
n Example 9.6 Let c[n] = α n , 0 < α < 1, be a discretetime signal (it is actually an autocorrelation function related to the power spectrum of a random signal). Determine its Ztransform. Solution To find its twosided Ztransform C(z) we consider its causal and anticausal components. First, Z(c[n]u[n]) =
∞ X
α n z−n =
n=0
1 1 − αz−1
with the region of convergence of αz−1  < 1 or z > α. For the anticausal component, Z(c[−n]u[n])z =
∞ X
α n zn =
n=0
1 1 − αz
with a region of convergence of αz < 1 or z < 1/α. Thus, the twosided Ztransform of c[n] is (notice that the term for n = 0 was used twice in the above calculations, so we need to subtract it) 1 1 z z + −1= − −1 1 − αz 1 − αz z−α (z − 1/α) (α − 1/α)z = (z − α)(z − 1/α)
C(z) =
with a region of convergence of 1 α < z < α For instance, for α = 0.5, we get C(z) =
−1.5z (z − 0.5)(z − 2)
0.5 < z < 2
n
9.4 ONESIDED ZTRANSFORM In most situations where the Ztransform is used the system is causal (its impulse response is h[n] = 0 for n < 0) and the input signal is also causal (x[n] = 0 for n < 0). In such cases the onesided Ztransform is very appropriate. Moreover, as we saw before, the twosided Ztransform can be expressed
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in terms of onesided Ztransforms. Another valid reason to study the onesided Ztransform in more detail is its use in solving difference equations with initial conditions. Recall that the onesided Ztransform is defined as X1 (z) = Z(x[n]u[n]) =
∞ X
x[n]u[n]z−n
(9.9)
n=0
in a region of convergence R1 . Also recall that the computation of the twosided Ztransform using the onesided Ztransform is given in Equation (9.6).
9.4.1 Computing the ZTransform with Symbolic MATLAB Similar to the computation of the Laplace transform, the computation of the Ztransform can be done using the symbolic toolbox of MATLAB. The following is the necessary code for computing the Ztransform of h1 [n] = 0.9u[n] h2 [n] = u[n] − u[n − 10] h3 [n] = cos(ω0 n)u[n] h4 [n] = hsigna1 [n]h2 [n] The results are shown at the bottom. (As in the continuous case, in MATLAB the heaviside function is the same as the unitstep function.) %%%%%%%%%%%%%%%%%%%%% % Ztransform computation %%%%%%%%%%%%%%%%%%%%% syms n w0 h1 = 0.9. ˆ n; H1 = ztrans(h1) h2 = heaviside(n)  heaviside(n10); H2 = ztrans(h2) h3 = cos(w0 ∗ n) ∗ heaviside(n); H3 = ztrans(h3) H4 = ztrans(h1 ∗ h3) H1 = 10/9/(10/9 ∗ z  1) ∗ z H2 = 1 + 1/z + 1/z ˆ 2 + 1/z ˆ 3 + 1/z ˆ 4 + 1/z ˆ 5 + 1/z ˆ 6 + 1/z ˆ 7 + 1/z ˆ 8 + 1/z ˆ 9 H3 = (z  cos(w0)) ∗ z/(z ˆ 2  2 ∗ z ∗ cos(w0) + 1) H4 = 10/9 ∗ (10/9 ∗ z  cos(w0))*z/(100/81 ∗ z ˆ 2  20/9 ∗ z ∗ cos(w0) + 1)
The function iztrans computes the inverse Ztransform. We will illustrate its use later on.
9.4.2 Signal Behavior and Poles In this section we will use the linearity property of the Ztransform to connect the behavior of the signal with the poles of its Ztransform.
9.4 OneSided ZTransform
The Ztransform is a linear transformation, meaning that Z(ax[n] + by[n]) = aZ(x[n]) + bZ(y[n])
(9.10)
for signals x[n] and y[n] and constants a and b.
To illustrate the linearity property as well as the connection between the signal and the poles of its Ztransform, consider the signal x[n] = α n u[n] for real or complex values α. Its Ztransform will be used to compute the Ztransform of the following signals: n n
x[n] = cos(ω0 n + θ)u[n] for frequency 0 ≤ ω0 ≤ π and phase θ. x[n] = α n cos(ω0 n + θ)u[n] for frequency 0 ≤ ω0 ≤ π and phase θ.
Show how the poles of the corresponding Ztransform connect with the signals. The Ztransform of the causal signal x[n] = α n u[n] is X(z) =
∞ X n=0
α n z−n =
∞ X
(αz−1 )n =
n=0
1 z = 1 − αz−1 z−α
ROC: z > α
(9.11)
Using the last expression in Equation (9.11) the zero of X(z) is z = 0 and its pole is z = α, since the first value makes X(0) = 0 and the second makes X(α) → ∞. For α real, be it positive or negative, the region of convergence is the same, but the poles are located in different places. See Figure 9.2 for α < 0. If α = 1 the signal x[n] = u[n] is constant for n ≥ 0 and the pole of X(z) is at z = 1e j0 (the radius is r = 1 and the lowest discrete frequency ω = 0 rad). On the other hand, when α = −1 the signal is x[n] = (−1)n u[n], which varies from sample to sample for n ≥ 0; its Ztransform has a pole at z = −1 = 1e jπ (a radius r = 1 and the highest discrete frequency ω = π rad). As we move the pole toward the center of the zplane (i.e., α → 0), the corresponding signal decays exponentially for 0 < α < 1, and is a modulated exponential of αn (−1)n u[n] = αn cos(πn)u[n] for −1 < α < 0. When α > 1 the signal becomes either a growing exponential (α > 1) or a growing modulated exponential (α < −1).
zplane
−1
FIGURE 9.2 Region of convergence (shaded area) of X(z) with a pole at z = α, α < 0 (same ROC if pole is at z = −α).
α
×
1
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CH A P T E R 9: The ZTransform
For a real value α = αe jω0 for ω0 = 0 or π, x[n] = α n u[n]
X(z) =
⇔
1 z = −1 z − α 1 − αz
ROC: z > α
and the location of the pole of X(z) determines the behavior of the signal: n When α > 0, then ω0 = 0 and the signal is less and less damped as α → ∞. n When α < 0, then ω0 = π and the signal is a modulated exponential that grows as α → −∞.
To compute the Ztransform of x[n] = cos(ω0 n + θ )u[n], we use Euler’s identity to write x[n] as " x[n] =
# e j(ω0 n+θ ) e−j(ω0 n+θ ) + u[n] 2 2
Applying the linearity property and using the above Ztransform when α = e jω0 and its conjugate α ∗ = e−jω0 , we get 1 X(z) = 2
"
e jθ e−jθ + 1 − e jω0 z−1 1 − e−jω0 z−1
#
1 2 cos(θ ) − 2 cos(ω0 − θ )z−1 = 2 1 − 2 cos(ω0 )z−1 + z−2 =
cos(θ) − cos(ω0 − θ )z−1 1 − 2 cos(ω0 )z−1 + z−2
(9.12)
Expressing X(z) in terms of positive powers of z, we get X(z) =
z(z cos(θ) − cos(ω0 − θ )) z(z cos(θ ) − cos(ω0 − θ )) = z2 − 2 cos(ω0 )z + 1 (z − e jω0 )(z − e−jω0 )
(9.13)
which is valid for any value of θ. If x[n] = cos(ω0 n)u[n], then θ = 0 and the poles of X(z) are a complex conjugate pair on the unit circle at frequency ω0 radians. The zeros are at z = 0 and z = cos(ω0 ). When x[n] = sin(ω0 n)u[n] = cos(ω0 n − π/2)u[n], then θ = −π/2 and the poles are at the same location as those for the cosine, but the zeros are at z = 0 and z = cos(ω0 + π/2)/ cos(π/2) → ∞, so there is only one finite zero at zero. For any other value of θ, the poles are located in the same place but there is a zero at z = 0 and another at z = cos(ω0 − θ )/ cos(θ ). For simplicity, we let θ = 0. If ω0 = 0, one of the double poles at z = 1 is canceled by one of the zeros at z = 1, resulting in the poles and the zeros of Z([u[n]). Indeed, the signal when ω0 = 0 and θ = 0 is x[n] = cos(0n)u[n] = u[n]. When the frequency ω0 > 0 the poles move along the unit circle from the lowest (ω0 = 0 rad) to the highest (ω0 = π rad) frequency.
9.4 OneSided ZTransform
The Ztransform pairs of a cosine and a sine are, respectively, cos(ω0 n)u[n]
⇔
sin(ω0 n)u[n]
⇔
z(z − cos(ω0 )) (z − e jω0 )(z − e−jω0 ) z sin(ω0 ) (z − e jω0 )(z − e−jω0 )
ROC : z > 1
(9.14)
ROC : z > 1
(9.15)
The Ztransforms for these sinusoids have identical poles 1e±jω0 , but different zeros. The frequency of the sinusoid increases from the lowest (ω0 = 0 rad) to the highest (ω0 = π rad)) as the poles move along the unit circle from 1 to −1 in its lower and upper parts.
Consider then the signal x[n] = r n cos(ω0 n + θ)u[n], which is a combination of the above cases. As before, the signal is expressed as a linear combination " # e jθ (re jω0 )n e−jθ (re−jω0 )n x[n] = + u[n] 2 2 and it can be shown that its Ztransform is X(z) =
z(z cos(θ ) − r cos(ω0 − θ )) (z − re jω0 )(z − re−jω0 )
(9.16)
The Ztransform of a sinusoid is a special case of the above (i.e., when r = 1). It also becomes clear that as the value of r decreases toward zero, the exponential in the signal decays faster, and that whenever r > 1, the exponential in the signal grows making the signal unbound. The Ztransform pair r n cos(ω0 n + θ )u[n]
z(z cos(θ) − r cos(ω0 − θ))
⇔
(z − re jω0 )(z − re−jω0 )
(9.17)
shows how complex conjugate pairs of poles inside the unit circle represent the damping indicated by the radius r and the frequency given by ω0 in radians.
Double poles are related to the derivative of X(z) or to the multiplication of the signal by n. If X(z) =
∞ X
x[n]z−n
n=0
its derivative with respect to z is ∞
dX(z) X dz−n = x[n] dz dz n=0
= −z−1
∞ X n=0
nx[n]z−n
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Or the pair nx[n]u[n]
⇔
−z
dX(z) dz
(9.18)
For instance, if X(z) = 1/(1 − αz−1 ) = z/(z − α), we find that α dX(z) =− dz (z − α)2 That is, the pair nα n u[n]
⇔
αz (z − α)2
indicates that double poles correspond to multiplication of x[n] by n.
The above shows that the location of the poles of X(z) provides basic information about the signal x[n]. This is illustrated in Figure 9.3, where we display the signal and its corresponding poles.
9.4.3 Convolution Sum and Transfer Function The most important property of the Ztransform, as it was for the Laplace transform, is the convolution property. The output y[n] of a causal LTI system is computed using the convolution sum y[n] = [x ∗ h][n] =
n X
x[k]h[n − k] =
k=0
n X
h[k]x[n − k]
(9.19)
k=0
where x[n] is a causal input and h[n] is the impulse response of the system. The Ztransform of y[n] is the product Y(z) = Z{[x ∗ h][n]} = Z{x[n]}Z{h[n]} = X(z)H(z)
(9.20)
and the transfer function of the system is thus defined as H(z) =
Z[output y[n]] Y(z) = X(z) Z[input x[n]]
(9.21)
That is, H(z) transfers the input X(z) into the output Y(z).
Remarks n
The convolution sum property can be seen as a way to obtain the coefficients of the product of two polynomials. Whenever we multiply two polynomials X1 (z) and X2 (z), of finite or infinite order, the coefficients of the resulting polynomial can be obtained by means of the convolution sum. For instance, consider X1 (z) = 1 + a1 z−1 + a2 z−2
1 1 0 −1
h1(n)
Imaginary part
9.4 OneSided ZTransform
−2
0 Real part
2
0.5 0
0
5
10 n
15
20
0
5
10 n
15
20
0
5
10 n
15
20
0
5
10 n
15
20
1
1 h2(n)
Imaginary part
(a)
0 −1
−2
0 Real part
0 −1
2
1
1 2
0 −1
−2
0 Real part
h3(n)
Imaginary part
(b)
0 −1
2
50
1 h4(n)
Imaginary part
(c)
2
0 −1 −3
−2
−1
0 1 Real part
2
0 −50
3 (d)
FIGURE 9.3 Effect of pole location on the inverse Ztransform: (a) if the pole is at z = 1 the signal is u[n], constant for n ≥ 0; (b) if the pole is at z = −1 the signal is a cosine of frequency π continuously changing, constant amplitude; (c, d) when poles are complex, if inside the unit circle the signal is a decaying modulated exponential, and if outside the unit circle the signal is a growing modulated exponential.
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CH A P T E R 9: The ZTransform
and X2 (z) = 1 + b1 z−1 Their product is X1 (z)X2 (z) = 1 + b1 z−1 + a1 z−1 + a1 b1 z−2 + a2 z−2 + a2 b1 z−3 = 1 + (b1 + a1 )z−1 + (a1 b1 + a2 )z−2 + a2 b1 z−3 The convolution sum of the two sequences [1 a1 a2 ] and [1 b1 ], formed by the coefficients of X1 (z) and X2 (z), is [1 (a1 + b1 ) (a2 + b1 a1 ) a2 ], which corresponds to the coefficients of the product of the polynomials X1 (z)X2 (z). Also notice that the sequence of length 3 (corresponding to the firstorder polynomial X1 (z)) and the sequence of length 2 (corresponding to the secondorder polynomial X2 (z)) when convolved give a sequence of length 3 + 2 − 1 = 4 (corresponding to the thirdorder polynomial X1 (z)X2 (z)). A finiteimpulse response or FIR filter is implemented by means of the convolution sum. Consider an FIR with an input–output equation
n
y[n] =
N−1 X
bk x[n − k]
(9.22)
k=0
where x[n] is the input and y[n] is the output. The impulse response of this filter is (let x[n] = δ[n] and set initial conditions to zero, so that y[n] = h[n]) h[n] =
N−1 X
bk δ[n − k]
k=0
giving h[n] = bn for n = 0, . . . , N − 1, and accordingly, we can write Equation (9.22) as y[n] =
N−1 X
h[k]x[n − k]
k=0
which is the convolution of the input x[n] and the impulse response h[n] of the FIR filter. Thus, if X(z) = Z(x[n]) and H(z) = Z(h[n]), then Y(z) = H(z)X(z) and
n n
y[n] = Z −1 [Y(z)]
The length of the convolution of two sequences of lengths M and N is M + N − 1. If one of the sequences is of infinite length, the length of the convolution is infinite. Thus, for an infiniteimpulse response (IIR) or recursive filters the output is always of infinite length for any input signal, given that the impulse response of these filters is of infinite length.
9.4 OneSided ZTransform
n Example 9.7 Consider computing the output of an FIR filter, y[n] =
1 x[n] + x[n − 1] + x[n − 2] 2
for an input x[n] = u[n] − u[n − 4] using the convolution sum, analytically and graphically, and the Ztransform. Solution The impulse response is h[n] = 0.5(δ[n] + δ[n − 1] + δ[n − 2]), so that h[0], h[1], h[2] are, respectively, 0.5, 0.5, and 0.5, and h[n] is zero otherwise. Convolution sum formula: The equation y[n] =
n X
h[k]x[n − k]
k=0
= x[0]h[n] + x[1]h[n − 1] + · · · + x[n]h[0] n ≥ 0 with the condition that in each entry the arguments of x[.] and h[.] add to n ≥ 0, gives y[0] = x[0]h[0] = 0.5 y[1] = x[0]h[1] + x[1]h[0] = 1 y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = 1.5 y[3] = x[0]h[3] + x[1]h[2] + x[2]h[1] + x[3]h[0] = x[1]h[2] + x[2]h[1] + x[3]h[0] = 1.5 y[4] = x[0]h[4] + x[1]h[3] + x[2]h[2] + x[3]h[1] + x[4]h[0] = x[2]h[2] + x[3]h[1] = 1 y[5] = x[0]h[5] + x[1]h[4] + x[2]h[3] + x[3]h[2] + x[4]h[1] + x[5]h[0] = x[3]h[2] = 0.5 and the rest are zero. In the above computations, we notice that the length of y[n] is 4 + 3 − 1 = 6 since the length of x[n] is 4 and that of h[n] is 3. Graphical approach: The convolution sum is given by either y[n] =
n X
x[k]h[n − k]
k=0
=
n X
h[k]x[n − k]
k=0
Choosing one of these equations, let’s say the first one, we need x[k] and h[n − k], as functions of k, for different values of n. Multiply them and then add the nonzero values. For instance, for n = 0 the sequence h[−k] is the reflection of h[k]; multiplying x[k] by h[−k] gives only one value different from zero at k = 0, or y[0] = 1/2. For n = 1, the sequence h[1 − k], as a function of k,
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530
CH A P T E R 9: The ZTransform
x [k]
n = −1 h[−1 −k ]
y [−1] = 0
−3
−2
−1
0
1 (a)
2
3
4
5
k
x[k] n=2 h [2 − k] y[2] = x[0]h[2] + x[1]h[1] + x [2]h [0]
−3
−2
−1
0
1 (b)
2
3
4
5
k
x [k ] n=5 h [5 − k ] y[5] = x[3]h [2] −3
−2
−1
0
1 (c)
2
3
4
5
k
FIGURE 9.4 Graphical approach: convolution sum for (a) n = −1, (b) n = 2, and (c) n = 5 with corresponding outputs y[−1], y[2], and y[5]. Both x[k] and h[n − k] are plotted as functions of k for a given value of n. The signal x[k] remains stationary, while h[n − k] moves linearly from left to right. Thus, the convolution sum is also called a linear convolution.
is h[−k] shifted to the right one sample. Multiplying x[k] by h[1 − k] gives two values different from zero, which when added gives y[1] = 1, and so on. For increasing values of n we shift to the right one sample to get h[n − k], multiply it by x[k], and then add the nonzero values to obtain the output y[n]. Figure 9.4 displays the graphical computation of the convolution sum for n = −1, n = 2 and n = 5. Convolution sum property: We have X(z) = 1 + z−1 + z−2 + z−3 H(z) =
1 1 + z−1 + z−2 2
9.4 OneSided ZTransform
1
h[n]
x [n]
1
0.5
0
0.5
0 −2
0
2
4
6
−2
8
0
n (a)
2
4
6
8
n (b)
1.5
y [n]
1
0.5
0 −2
0
2
4
6
8
n (c)
FIGURE 9.5 Convolution sum for an averager FIR: (a) x[n], (b) h[n], and (c) y[n]. The output y[n] is of length 6 given that x[n] is of length 4 and h[n] is the impulse response of a secondorder FIR filter of length 3.
and according to the convolution sum property, Y(z) = X(z)H(z) =
1 (1 + 2z−1 + 3z−2 + 3z−3 + 2z−4 + z−5 ) 2
Thus, y[0] = 0.5, y[1] = 1, y[2] = 1.5, y[3] = 1.5, y[4] = 1, and y[5] = 0.5, just as before. In MATLAB the function conv is used to compute the convolution sum giving the results shown in Figure 9.5, which coincide with the ones obtained in the other approaches. n
n Example 9.8 Consider an FIR filter with impulse response h[n] = δ[n] + δ[n − 1] + δ[n − 2] Find the filter output for an input x[n] = cos(2πn/3)(u[n] − u[n − 14]). Use the convolution sum to find the output, and verify your results with MATLAB.
531
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CH A P T E R 9: The ZTransform
Solution Graphical approach: Let us use the formula y[n] =
n X
x[k]h[n − k]
k=0
which keeps the more complicated signal x[k] as the unchanged signal. The term h[n − k] is h[k] reversed for n = 0, and then shifted to the right for n ≥ 1. The output is zero for negative values of n, and for n ≥ 0 we have y[0] = 1 y[1] = 0.5 y[n] = 0
2 ≤ n ≤ 13
y[14] = 0.5 y[15] = −0.5 The first value is obtained by reflecting the impulse response to get h[−k], and when multiplied by x[k] we only have the value at k = 0 different from zero, therefore y[0] = x[0]h[0] = 1. As we shift the impulse response to the right to get h[1 − k] for n = 1 and multiply it by x[k], we get two values different from zero; when added they equal 0.5. The result for 2 ≤ n ≤ 13 is zero because we add three values of −0.5, 1 and −0.5 from the cosine. These results are verified by MATLAB as shown in Figure 9.6. (the cosine does not look like a sampled cosine given that only three values are used per period). Convolution property approach: By the convolution property, the Ztransform of the output y[n] is Y(z) = X(z)H(z) = X(z)(1 + z−1 + z−2 ) = X(z) + X(z)z−1 + X(z)z−2 The coefficients of Y(z) can be obtained by adding the coefficients of X(z), X(z)z−1 , and X(z)z−2 : z0 z−1 z−2 z−3 z−4 z−5 z−6 z−7 z−8 z−9 z−10 z−11 z−12 z−13 z−14 z−15 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5
Adding these coefficients vertically, we obtain Y(z) = 1 + 0.5z−1 + 0z−2 + · · · + 0z−13 + 0.5z−14 − 0.5z−15 = 1 + 0.5z−1 + 0.5z−14 − 0.5z−15 Notice from this example that n
The convolution sum is simply calculating the coefficients of the polynomial product X(z)H(z).
9.4 OneSided ZTransform
1.5 1
0.5
h [n]
x [n]
1 0.5
0 −0.5
0
−1
0
15
10
5
0
5
n (a)
10
15
n (b)
1
y [n]
0.5 0 −0.5 −1 −2
0
2
4
6
8
10
12
14
16
18
n (c)
FIGURE 9.6 Convolution sum for FIR filter: (a) x[n], (b) h[n], and (c) y[n]. n
The length of the convolution sum = length of x[n] + length of h[n] −1 = 14 + 3 − 1 = n 16—that is, Y(z) is a polynomial of order 15.
n Example 9.9 The convolution sum of noncausal signals is more complicated graphically than that of the causal signals we showed in the previous examples. Let 1 h1 [n] = δ[n + 1] + δ[n] + δ[n − 1] 3 be the impulse response of a noncausal averager FIR filter, and x[n] = u[n] − u[n − 4] be the input. Compute the filter output using the convolution sum. Solution Graphically, it is a bit confusing to plot h1 [n − k], as a function of k, to do the convolution sum. Using the convolution and the timeshifting properties of the Ztransform we can view the computation more clearly.
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CH A P T E R 9: The ZTransform
According to the convolution property the Ztransform of the output of the noncausal filter is Y1 (z) = X(z)H1 (z) = X(z)[zH(z)]
(9.23)
where we let 1 H1 (z) = Z[h1 [n]] = (z + 1 + z−1 ) 3 1 −1 −2 =z (1 + z + z ) = zH(z) 3 where H(z) = (1/3)Z[δ[n] + δ[n − 1] + δ[n − 2]] is the transfer function of a causal filter. Let Y(z) = X(z)H(z) be the Ztransform of the convolution sum y[n] = [x ∗ h][n] of x[n] and h[n], both of which are causal and can be computed as before. According to Equation (9.23), we then have Y1 (z) = zY(z) or y1 [n] = [x ∗ h1 ][n] = y[n + 1]. n Let x1 [n] be the input to a noncausal LTI system, with an impulse response h1 [n] such that h1 [n] = 0 for n < N1 < 0. Assume x1 [n] is also noncausal (i.e., x1 [n] = 0 for n < N0 < 0). The output y1 [n] = [x1 ∗ h1 ][n] has a Ztransform of Y1 (z) = X1 (z)H1 (z) = [zN0 X(z)][zN1 H(z)] where X(z) and H(z) are the Ztransforms of a causal signal x[n] and of a causal impulse response h[n]. If we let y[n] = [x ∗ h][n] = Z −1 [X(z)H(z)] then y1 [n] = [x1 ∗ h1 ][n] = y[n + N0 + N1 ]
Remarks n
The impulse response of an IIR system, represented by a difference equation, is found by setting the initial conditions to zero, therefore, the transfer function H(z) also requires a similar condition. If the initial conditions are not zero, the Ztransform of the total response Y(z) is the sum of the Ztransforms of the zerostate and the zeroinput responses—that is, its Ztransform is of the form Y(z) =
X(z)B(z) I0 (z) + A(z) A(z)
(9.24)
and it does not permit us to compute the ratio Y(z)/X(z) unless the component due to the initial conditions is I0 (z) = 0.
9.4 OneSided ZTransform
n
It is important to remember the relations H(z) = Z[h[n]] =
Y(z) Z[y[n]] = X(z) Z[x[n]]
where H(z) is the transfer function and h[n] is the impulse response of the system, with x[n] as the input and y[n] as the output.
n Example 9.10 Consider a discretetime IIR system represented by the difference equation y[n] = 0.5y[n − 1] + x[n]
(9.25)
with x[n] as the input and y[n] as the output. Determine the transfer function of the system and from it find the impulse and the unitstep responses. Determine under what conditions the system is BIBO stable. If stable, determine the transient and steadystate responses of the system. Solution The system transfer function is given by H(z) =
Y(z) 1 = X(z) 1 − 0.5z−1
and its impulse response is h[n] = Z −1 [H(z)] = 0.5n u[n] The response of the system to any input can be easily obtained by the transfer function. If the input is x[n] = u[n], we have Y(z) = H(z)X(z) = =
−1 1 − 0.5z−1
1 (1 − 0.5z−1 )(1 − z−1 ) 2 + 1 − z−1
so that the total solution is y[n] = −0.5n u[n] + 2u[n] From the transfer function H(z) of the LTI system, we can test the stability of the system by finding the location of its poles—very much like in the analog case. An LTI system is BIBO stable if and only if the impulse response of the system is absolutely summable—that is, X h[n] ≤ ∞ n
535
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CH A P T E R 9: The ZTransform
An equivalent condition is that the poles of H(z) are inside the unit circle. In this case, h[n] is absolutely summable, indeed ∞ X
0.5n =
n=0
1 =2 1 − 0.5
On the other hand, H(z) =
1 z = −1 1 − 0.5z z − 0.5
has a pole at z = 0.5, inside the unit circle. Thus, the system is BIBO stable. As such, its transient and steadystate responses exist. As n → ∞, y[n] = 2 is the steadystate response, and −0.5n u[n] is the transient solution. n
n Example 9.11 An FIR system has the input–output equation y[n] =
1 [x[n] + x[n − 1] + x[n − 2]] 3
where x[n] is the input and y[n] is the output. Determine the transfer function and the impulse response of the system, and from them indicate whether the system is BIBO stable or not. Solution The transfer function is 1 [1 + z−1 + z−2 ] 3 z2 + z + 1 = 3z2
H(z) =
and the corresponding impulse response is h[n] =
1 [δ[n] + δ[n − 1] + δ[n − 2]] 3
The impulse response of this system only has three nonzero values, h[0] = h[1] = h[2] = 1/3, and the rest of the values are zero. As such, h[n] is absolutely summable and the filter is BIBO stable. FIR filters are always BIBO stable given their impulse responses will be absolutely summable, due to their final support, and equivalently because the poles of the transfer function of these system are at the origin of the zplane, very much inside the unit circle. n Nonrecursive or FIR systems: The impulse response h[n] of an FIR or nonrecursive system y[n] = b0 x[n] + b1 x[n − 1] + · · · + bM x[n − M]
9.4 OneSided ZTransform
has finite length and is given by h[n] = b0 δ[n] + b1 δ[n − 1] + · · · + bM δ[n − M] Its transfer function is H(z) =
Y(z) X(z)
= b0 + b1 z−1 + · · · + bM z−M =
b0 zM + b1 zM−1 + · · · + bM zM
with all its poles at the origin z = 0 (multiplicity M), and as such the system is BIBO stable. Recursive or IIR systems: The impulse response h[n] of an IIR or recursive system y[n] = −
N X
ak y[n − k] +
M X
bm x[n − m]
m=0
k=1
has (possible) infinite length and is given by h[n] = Z −1 [H(z)] # " P M −m m=0 bm z −1 =Z P −k 1+ N k=1 ak z B(z) = Z −1 A(z) =
∞ X
h[`]δ[n − `]
`=0
where H(z) is the transfer function of the system. If the poles of H(z) are inside the unit circle, or A(z) 6= 0 for z ≥ 1, the system is BIBO stable.
9.4.4 Interconnection of DiscreteTime Systems Just like with analog systems, two discretetime LTI systems with transfer functions H1 (z) and H2 (z) (or with impulse responses h1 [n] and h2 [n]) can be connected in cascade, parallel, or feedback. The first two forms result from properties of the convolution sum. The transfer function of the cascading of the two LTI systems is H(z) = H1 (z)H2 (z) = H2 (z)H1 (z)
(9.26)
537
538
CH A P T E R 9: The ZTransform
x[n]
y[n] H1 (z)
x [n] =
H2(z)
y[n] H2(z)
x [n] =
H1(z)
y[n] H1(z)H2(z)
(a) H1(z)
y [n]
x [n]
x[n]
y[n] H1(z) + H2(z)
=
+ H2(z) (b) x [n] +
e[n]
y[n] H1(z)
x [n] =
− w[n]
H1(z) 1 + H1(z)H2 (z)
y [n]
H2(z) (c)
FIGURE 9.7 Connections of LTI systems: (a) cascade, (b) parallel, and (c) negative feedback.
showing that there is no effect on the overall system if we interchange the two systems (see Figure 9.7(a)). Recall that such a property is only valid for LTI systems. In the parallel system, as in Figure 9.7(b), both systems have the same input and the output is the sum of the output of the subsystems. The overall transfer function is H(z) = H1 (z) + H2 (z)
(9.27)
Finally, the negative feedback connection of the two systems shown in Figure 9.7(c) gives in the feedforward path Y(z) = H1 (z)E(z)
(9.28)
where Y(z) = Z[y[n]] is the Ztransform of the output y[n] and E(z) = X(z) − W(z) is the Ztransform of the error function e[n] = x[n] − w[n]. The feedback path gives that W(z) = Z[w[n]] = H2 (z)Y(z) Replacing W(z) in E(z), and then replacing E(z) in Equation (9.28), we obtain the overall transfer function H(z) =
Y(z) H1 (z) = X(z) 1 + H1 (z)H2 (z)
(9.29)
9.4 OneSided ZTransform
9.4.5 Initial and Final Value Properties In some control applications and to check a partial fraction expansion, it is useful to find the initial or the final value of a discretetime signal x[n] from its Ztransform. These values can be found as shown in the following box. If X(z) is the Ztransform of a causal signal x[n], then x[0] = lim X(z)
Initial value:
z→∞
lim x[n] = lim (z − 1)X(z)
Final value:
n→∞
z→1
(9.30)
The initial value results from the definition of the onesided Ztransform—that is, X x[n] = x[0] lim X(z) = lim x[0] + z→∞ z→∞ zn n≥1
To show the final value, we have that (z − 1)X(z) =
∞ X
x[n]z−n+1 −
n=0
= x[0]z +
∞ X
x[n]z−n
n=0 ∞ X
[x[n + 1] − x[n]]z−n
n=0
and thus the limit lim (z − 1)X(z) = x[0] +
z→1
∞ X
(x[n + 1] − x[n])
n=0
= x[0] + (x[1] − x[0]) + (x[2] − x[1]) + (x[3] − x[2]) · · · = lim x[n] n→∞
given that the entries in the sum cancel out as n increases, leaving x[∞].
n Example 9.12 Consider a negativefeedback connection of a plant with a transfer function G(z) = 1/(1 − 0.5z−1 ) and a constant feedback gain K (see Figure 9.8). If the reference signal is a unit step, x[n] = u[n], determine the behavior of the error signal e[n]. What is the effect of the feedback, from the error point of view, on an unstable plant G(z) = 1/(1 − z−1 )?
539
540
CH A P T E R 9: The ZTransform
x[n] +
e [n]
G(z)
y[n]
−
FIGURE 9.8 Negativefeedback system with plant G(z).
w [n]
K
Solution For G(z) = 1/(1 − 0.5z−1 ), the Ztransform of the error signal is E(z) = X(z) − W(z) = X(z) − KG(z)E(z) and for X(z) = 1/(1 − z−1 ), E(z) =
X(z) 1 = −1 1 + KG(z) (1 − z )(1 + KG(z))
The initial value of the error signal is then e[0] = lim E(z) = z→∞
1 1+K
since G(∞) = 1. The steadystate or final value of the error is (z − 1)X(z) z→1 z→1 1 + KG(z) z(z − 1) 1 = lim = z→1 (z − 1)(1 + KG(z)) 1 + 2K
lim e[n] = lim (z − 1)E(z) = lim
n→∞
since G(1) = 2. If we want the steadystate error to go to zero, then K must be large. In that case, the initial error is also zero. If G(z) = 1/(1 − z−1 ) (i.e., the plant is unstable), the initial value of the error function remains the same, e[0] = 1/(1 + K), but the steady state error goes to zero since G(1) → ∞. n Tables 9.1 and 9.2 provide a list of oneside Ztransforms and the basic properties of the onesided Ztransform.
9.4 OneSided ZTransform
Table 9.1 OneSided ZTransforms Function of Time 1.
δ[n]
2.
u[n]
3.
nu[n]
4.
n2 u[n]
5.
α n u[n], α < 1
6.
nα n u[n], α < 1
7.
cos(ω0 n)u[n]
8.
sin(ω0 n)u[n]
9.
α n cos(ω0 n)u[n], α < 1
10.
α n sin(ω0 n)u[n], α < 1
Function of z, ROC 1, whole zplane 1 , z > 1 1 − z−1 z−1 , z > 1 (1 − z−1 )2 −1 −1 z (1 + z ) , z > 1 (1 − z−1 )3 1 , z > α 1 − αz−1 −1 αz , z > α (1 − αz−1 )2 1 − cos(ω0 )z−1 , z > 1 1 − 2 cos(ω0 )z−1 + z−2 sin(ω0 )z−1 , z > 1 1 − 2 cos(ω0 )z−1 + z−2 −1 1 − α cos(ω0 )z , z > 1 1 − 2α cos(ω0 )z−1 + z−2 −1 α sin(ω0 )z , z > α 1 − 2α cos(ω0 )z−1 + z−2
Table 9.2 Basic Properties of OneSided ZTransform Causal signals and constants
αx[n], βy[n]
αX(z), βY(z)
Linearity
αX(z) + βY(z)
Convolution sum
αx[n] + βy[n] P (x ∗ y)[n] = k x[n]y[n − k]
Time shifting—causal
x[n − N] N integer
z−N X(z)
Time shifting—noncausal
x[n − N]
z−N X(z) + x[−1]z−N+1
x[n] noncausal, N integer
X(z)Y(z)
+ x[−2]z−N+2 + · · · + x[−N]
Time reversal
x[−n]
X(z−1 )
Multiplication by n
n x[n]
−z
Multiplication by n2
n2 x[n]
z2
Finite difference
x[n] − x[n − 1]
Accumulation
Pn
Initial value
x[0]
Final value
lim x[n]
k=0 x[k]
n→∞
dX(z) dz
d2 X(z) dX(z) +z dz2 dz (1 − z−1 )X(z) − x[−1] X(z) 1 − z−1 lim X(z)
z→∞
lim (z − 1)X(z)
z→1
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CH A P T E R 9: The ZTransform
9.5 ONESIDED ZTRANSFORM INVERSE Different from the inverse Laplace transform, which was done mostly by the partial fraction expansion, the inverse Ztransform can be done in different ways. For instance, if the Ztransform is given as a finiteorder polynomial, the inverse can be found by inspection. Indeed, if the given Ztransform is X(z) =
N X
x[n]z−n
n=0
= x[0] + x[1]z−1 + x[2]z−2 + · · · + x[N]z−N
(9.31)
by the definition of the Ztransform, x[k] is the coefficient of the monomial z−k for k = 0, 1, . . . , N; thus the inverse Ztransform is given by the sequence {x[0], x[1], . . . , x[n]}. For instance, if we have a Ztransform X(z) = 1 + 2z−10 + 3z−20 the inverse is a sequence x[n] = δ[n] + 2δ[n − 10] + 3δ[n − 20] so that x[0] = 1, x[10] = 2, x[20] = 3, and x[n] = 0 for n 6= 0, 10, 20, respectively. In this case it makes sense to do this because N is finite, but if N → ∞, this way of finding the inverse Ztransform might not be very practical. In that case, the longdivision method and the partial fraction expansion method, which we consider next, are more appropriate. In this section we will consider the inverse of onesided Ztransforms, and in the next section we consider the inverse of twosided transforms.
9.5.1 LongDivision Method When a rational function X(z) = B(z)/A(z), having as ROC the outside of a circle of radius R (i.e., x[n] is causal), is expressed as X(z) = x[0] + x[1]z−1 + x[2]z−2 + · · · then the inverse is the sequence {x[0], x[1], x[2], . . .}, or x[n] = x[0]δ[n] + x[1]δ[n − 1] + x[2]δ[n − 2] + · · ·
To find the inverse we simply divide the polynomial B(z) by A(z) to obtain a possible infiniteorder polynomial in negative powers of z−1 . The coefficients of this polynomial are the inverse values. The disadvantage of this method is that it does not provide a closedform solution, unless there is a clear connection between the terms of the sequence. But this method is useful when we are interested in finding some of the initial values of the sequence x[n].
9.5 OneSided ZTransform Inverse
n Example 9.13 Find the inverse Ztransform of X(z) =
1 1 + 2z−2
z >
√ 2
Solution We can perform the long division to find the x[n] values, or equivalently let X(z) = x[0] + x[1]z−1 + x[2]z−2 + · · · and find the {x[n]} samples so that the product X(z)(1 + 2z−2 ) = 1. Thus, 1 = (1 + 2z−2 )(x[0] + x[1]z−1 + x[2]z−2 + · · · ) = x[0] + x[1]z−1 + x[2]z−2 + x[3]z−3 + · · · + 2x[0]z−2 + 2x[1]z−3 + · · · and comparing the terms on the two sides of the equality gives x[0] = 1 x[1] = 0 x[2] + 2x[0] = 0 ⇒ x[2] = −2 x[3] + 2x[1] = 0 ⇒ x[3] = 0 x[4] + 2x[2] = 0 ⇒ x[4] = (−2)2 .. . So the inverse Ztransform is x[0] = 1 and x[n] = (−2)log2 (n) for n > 0 and even, and zero otherwise. Notice that this sequence grows as n → ∞. Another possible way to find the inverse is to use the geometric series equation ∞ X k=0
αn =
1 α < 1 1−α
with −α = 2z−2 (notice that α = 2/z2 < 1 or z > X(z) =
√
2, the given ROC). Therefore,
1 = 1 + (−2z−2 )1 + (−2z−2 )2 + (−2z−2 )3 + · · · 1 + 2z−2
but this method is not as general as the long division.
n
543
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CH A P T E R 9: The ZTransform
9.5.2 Partial Fraction Expansion The basics of partial fraction expansion remain the same for the Ztransform as for the Laplace transform. A rational function is a ratio of polynomials N(z) and D(z) in z or z−1 : X(z) =
N(z) D(z)
The poles of X(z) are the roots of D(z) = 0 and the zeros of X(z) are the roots of the equation N(z) = 0. Remarks n
n
n
n
The basic characteristic of the partial fraction expansion is that X(z) must be a proper rational function, or that the degree of the numerator polynomial N(z) must be smaller than the degree of the denominator polynomial D(z) (assuming both N(z) and D(z) are polynomials in either z−1 or z). If this condition is not satisfied, we perform long division until the residue polynomial is of a degree less than that of the denominator. It is more common in the Ztransform than in the Laplace transform to find that the numerator and the denominator are of the same degree—this is because δ[n] is not as unusual as the analog impulse function δ(t). The partial fraction expansion is generated, from the poles of the proper rational function, as a sum of terms of which the inverse Ztransforms are easily found in a Ztransform table. By plotting the poles and the zeros of a proper X(z), the location of the poles provides a general form of the inverse within some constants that are found from the poles and the zeros. Given that the numerator and the denominator polynomials of a proper rational function X(z) can be expressed in terms of positive or negative powers of z, it is possible to do partial fraction expansions in either z or z−1 . We will see that the partial fraction expansion in negative powers is more like the partial fraction expansion in the Laplace transform, and as such we will prefer it. Partial fraction expansion in positive powers of z requires more care.
n Example 9.14 Consider the nonproper rational function X(z) =
2 + z−2 1 + 2z−1 + z−2
(the numerator and the denominator are of the same degree in powers of z−1 ). Determine how to obtain an expansion of X(z) containing a proper rational term to find x[n]. Solution By division we obtain X(z) = 1 +
1 − 2z−1 1 + 2z−1 + z−2
9.5 OneSided ZTransform Inverse
where the second term is proper rational as the denominator is of a higher degree in powers of z−1 than the numerator. The inverse Ztransform of X(z) will then be x[n] = δ[n] + Z −1
1 − 2z−1 1 + 2z−1 + z−2
The inverse of the proper rational term is done as indicated in this section.
n
n Example 9.15 Find the inverse Ztransform of 1 + z−1 (1 + 0.5z−1 )(1 − 0.5z−1 ) z(z + 1) = z > 0.5 (z + 0.5)(z − 0.5)
X(z) =
by using the negative and the positive powers of z expressions. Solution Clearly, X(z) is proper if it is considered a function of negative powers z−1 (in z−1 , the numerator is of degree 1 and the denominator of degree 2), but it is not proper if it is considered a function of positive powers z (the numerator and the denominator are both of degree 2). It is, however, unnecessary to perform long division to make X(z) proper when it is considered as a function of z. One simple approach is to consider X(z)/z as the function. We wish to find its partial fraction expansion—that is, X(z) z+1 = z (z + 0.5)(z − 0.5)
(9.32)
which is proper. Thus, whenever X(z), as a function of z terms, is not proper it is always possible to divide it by some power in z to make it proper. After obtaining the partial fraction expansion then the z term is put back. Consider then the partial fraction expansion in z−1 terms, 1 + z−1 (1 + 0.5z−1 )(1 − 0.5z−1 ) A B = + 1 + 0.5z−1 1 − 0.5z−1
X(z) =
Given that the poles are real—one at z = −0.5 and the other at z = 0.5—from the Ztransform table, we get that a general form of the inverse is x[n] = [A (−0.5)n + B 0.5n ]u[n]
545
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CH A P T E R 9: The ZTransform
The A and B coefficients can be found (by analogy with the Laplace transform partial fraction expansion) as A = X(z)(1 + 0.5z−1 )z−1 =−2 = −0.5 B = X(z)(1 − 0.5z−1 )z−1 =2 = 1.5 so that x[n] = [−0.5(−0.5)n + 1.5(0.5)n ]u[n] Consider then the partial fraction expansion in positive powers of z. From Equation (9.32) the proper rational function X(z)/z can be expanded as X(z) z+1 = z (z + 0.5)(z − 0.5) C D = + z + 0.5 z − 0.5 The values of C and D are obtained as follows: C=
X(z) (z + 0.5)z=−0.5 = −0.5 z
D=
X(z) (z − 0.5)z=0.5 = 1.5 z
We then have that 1.5z −0.5z + z + 0.5 z − 0.5 which according to the table (if entries are in negative powers of z, convert them into positive powers of z) we get X(z) =
x[n] = [−0.5(−0.5)n + 1.5(0.5)n ]u[n] which coincides with the above result. Two simple checks on our result are given by the initial and the final value results. For the initial value, x[0] = 1 = lim X(z) z→∞
and lim x[n] = lim (z − 1)X(z) = 0
n→∞
z→1
Both of these check. It is important to recognize that these two checks do not guarantee that we did not make mistakes in computing the inverse, but if the initial or the final values were not to coincide with our results, our inverse would be wrong. n
9.5 OneSided ZTransform Inverse
9.5.3 Inverse ZTransform with MATLAB Symbolic MATLAB can be used to compute the inverse onesided Ztransform. The function iztrans provides the sequence that corresponds to its argument. The following script illustrates the use of this function. %%%%%%%%%%%%%%%%%% % Inverse Ztransform %%%%%%%%%%%%%%%%%% syms n z x1 = iztrans((z ∗ (z + 1))/((z + 0.5) ∗ (z  0.5))) x2 = iztrans((2  z)/(2 ∗ (z  0.5))) x3 = iztrans((8  4 ∗ z ˆ (1))/(z ˆ (2) + 6 ∗ z ˆ (1) + 8))
The above gives the following results: x1 = 3/2 ∗ (1/2) ˆ n  1/2 ∗ (1/2) ˆ n x2 = 2 ∗ charfcn[0](n) + 3/2 ∗ (1/2) ˆ n x3 = 3 ∗ (1/4) ˆ n + 4 ∗ (1/2) ˆ n
Notice that the Ztransform can be given in positive or negative powers of z, and that when it is nonproper the function charfcn[0] corresponds to δ[n].
Partial Fraction Expansion with MATLAB Several numerical functions are available in MATLAB to perform partial fraction expansion of a Ztransform and to obtain the corresponding inverse. In the following we consider the cases of single and multiple poles.
(1) Simple Poles Consider finding the inverse Ztransform of
X(z) =
z(z + 1) (1 + z−1 ) = (z − 0.5)(z + 0.5) (1 − 0.5z−1 )(1 + 0.5z−1 )
z > 0.5
The MATLAB function residuez provides the partial fraction expansion coefficients or residues r[k], the poles p[k], and the gain k corresponding to X(z) when the coefficients of its denominator and of its numerator are inputted. If the numerator or the denominator is given in a factored form (as is the case of the denominator above) we need to multiply the terms to obtain the denominator polynomial. Recall that multiplication of polynomials corresponds to convolution of the polynomial coefficients. Thus, to perform the multiplication of the terms in the denominator, we use the MATLAB function conv to obtain the coefficients of the product. The convolution of the coefficients [1 − 0.5] of p1 (z) = 1 − 0.5z−1 and [1 0.5] of p2 (z) = 1 + 0.5z−1 gives the denominator coefficients. By means of the MATLAB function poly we can obtain the polynomials in the numerator and denominator from the zeros and poles. These polynomials are then multiplied as indicated before to obtain the numerator with coefficients {b[k]}, and the denominator with coefficients {a[k]}.
547
548
CH A P T E R 9: The ZTransform
To find the poles and the zeros of X(z), given the coefficients {b[k]} and {a[k]} of the numerator and the denominator, we use the MATLAB function roots. To get a plot of the poles and the zeros of X(z), the MATLAB function zplane, with inputs the coefficients of the numerator and the denominator of X(z), is used (conventionally, an ’x’ is used to denote poles and an ’o’ for zeros). Two possible approaches can now be used to compute the inverse Ztransform x[n]. We can compute the inverse (below we call it x1 [n] to differentiate it from the other possible solution, which we call x[n]) by using the information on the partial fraction expansion (the residues r[k]) and the corresponding poles. An alternative is to use the MATLAB function filter, which considers X(z) as a transfer function, with the numerator and the denominator defined by the b and a vectors of coefficients. If we assume the input is a delta function of Ztransform unity, the function filter computes as output the inverse Ztransform x[n] (i.e., we have tricked filter to give us the desired result). The following script is used to implement the generation of the terms in the numerator and the denominator to obtain the corresponding coefficients, plot them, and find the inverse in the two different ways indicated above. For additional help on the functions used here use help. %%%%%%%%%%%%%%%%%%%%%%%%%%%% % Two methods for inverse Ztransform %%%%%%%%%%%%%%%%%%%%%%%%%%%% p1 = poly(0.5); p2 = poly(0.5); % generation of terms in denominator a = conv(p1,p2) % denominator coefficients z1 = poly(0); z2 = poly(1); % generation of terms in numerator b = conv(z1,z2) % numerator coefficients z = roots(b) % zeros of X(z) [r,p,k] = residuez(b,a) % partial fraction expansion, poles and gain zplane(b,a) % plot of poles and zeros d = [1 zeros(1,99)]; % impulse delta[n] x = filter(b,a,d); % x[n] computation from filter n = 0:99; x1 = r(1) ∗ p(1). ˆ n + r(2) ∗ p(2). ˆ n; % x[n] computation from residues a = 1.0000 b= 1 1 0 z= 0 1 r = 1.5000 0.5000 p = 0.5000 0.5000
0 0.2500
Figure 9.9 displays the plot of the zeros and the poles and the comparison between the inverses x1 [n] and x[n] for 0 ≤ n ≤ 99, which coincide sample by sample.
9.5 OneSided ZTransform Inverse
1
1.5
1
0.8
0.6 x[n], x1[n]
Imaginary part
0.5
0
0.4 −0.5
0.2
−1
−1.5
0 −1
−0.5
0 Real part (a)
0.5
1
0
5
10 n (b)
15
20
FIGURE 9.9 (a) Poles and zeros of X(z) and (b) inverse Ztransforms x[n] and x1 [n] found using filter and the residues.
(2) Multiple Poles Whenever multiple poles are present one has to be careful in interpreting the MATLAB results. First, use help to get more information on residuez and how the partial fraction expansion for multiple poles is done. Notice from the help file that the residues are ordered the same way the poles are. Furthermore, the residues corresponding to the multiple poles are ordered from the lowest to the highest order. Also notice the difference between the partial fraction expansion of MATLAB and ours. For instance, consider the Ztransform X(z) =
az−1 (1 − az−1 )2
z > a
with inverse x[n] = nan u[n]. Writing the partial fraction expansion as MATLAB does gives X(z) =
r1 r2 + −1 1 − az (1 − az−1 )2
r1 = −1, r2 = 1
(9.33)
549
550
CH A P T E R 9: The ZTransform
where the second term is not found in the Ztransforms table. To write it so that each of the terms in the expansion are in the Ztransforms table, we need to obtain values for A and B in the expansion X(z) =
Bz−1 A + −1 1 − az (1 − az−1 )2
(9.34)
so that Equations (9.33) and (9.34) are equal. We find that A = r1 + r2 , while B − Aa = −r1 a or B = ar2 . With these values we find the inverse to be x[n] = [(r1 + r2 )an + nr2 an ]u[n] = nan u[n] as expected. To illustrate the computation of the inverse Ztransform from the residues in the case of multiple poles, consider the transfer function X(z) =
0.5z−1 . 1 − 0.5z−1 − 0.25z−2 + 0.125z−3
The following script is used. %%%%%%%%%%%%%%%%%%%%%%%%%%%% % Inverse Ztransform  multiple poles %%%%%%%%%%%%%%%%%%%%%%%%%%%% b = [0 0.5 0 0 ]; a = [1 0.5 0.25 0.125] [r,p,k] = residuez(b,a) % partial fraction expansion, poles and gain zplane(b,a) % plot of poles and zeros n = 0:99; xx = p(1). ˆ n; yy = xx. ∗ n; x1 = (r(1) + r(2)). ∗ xx + r(2). ∗ yy + r(3) ∗ p(3). ˆ n; % inverse computation
The poles and the zeros and the inverse Ztransform are shown in Figure 9.10—there is a double pole at 0.5. The residues and the corresponding poles are r = 0.2500 0.5000 0.2500 p = 0.5000 0.5000 0.5000
Computationally, our method and MATLAB’s are comparable but the inverse transform in our method is found directly from the table, while in the case of MATLAB’s you need to change the expansion to get it into the forms found in the tables.
9.5.4 Solution of Difference Equations In this section we will use the shifting in time property of the Ztransform in the solution of difference equations with initial conditions. You will see that the partial fraction expansion used to find the inverse Ztransform is like the one used in the inverse Laplace transform.
9.5 OneSided ZTransform Inverse
2
0.6
1.5
0.5
1 0.4
0.3 2
0
2
x[n]
Imaginary part
0.5
0.2 −0.5 0.1 −1 0
−1.5
−2
−1
−0.5
0 Real part
0.5
−0.1
1
0
2
4
6
8
10
n
(a)
(b)
FIGURE 9.10 (a) Poles and zeros of X(z) and (b) inverse Ztransform x[n].
If x[n] has a onesided Ztransform X(z), then x[n − N] has the following onesided Ztransform: Z[x[n − N]] = z−N X(z) + x[−1]z−N+1 + x[−2]z−N+2 + · · · + x[−N]
(9.35)
Indeed, we have that Z(x[n − N]) =
∞ X n=0
= z−N
x[m]z−(m+N)
m=−N ∞ X
x[m]z−m +
m=0 −N
=z
∞ X
x[n − N]z−n =
−1 X
x[m]z−(m+N)
m=−N −N+1
X(z) + x[−1]z
+ x[−2]z−N+2 + · · · + x[−N]
where we first let m = n − N and then separated the sum into two, one corresponding to the Ztransform of x[n] multiplied by z−N (the delay on the signal) and a second sum that corresponds to initial values {x[i], −N ≤ i ≤ −1}.
551
552
CH A P T E R 9: The ZTransform
Remarks n
n
If the signal is causal, so that {x[i], −N ≤ i ≤ −1} are all zero, we then have that Z(x[n − N]) = z−N X(z), indicating that the operator z−1 is a delay operator. Thus, x[n − N] has been delayed N samples and its Ztransform is then simply X(z) multiplied by z−N . The shifting in time property is useful in the solution of difference equations, especially when it has nonzero initial conditions as we will see next. On the other hand, if the initial conditions are zero, either the onesided or the twosided Ztransforms could be used.
The analog of differential equations are difference equations, which result directly from the modeling of a discrete system or from discretizing differential equations. The numerical solution of differential equations requires that these equations be converted into difference equations since computers cannot perform integration. Many methods are used to solve differential equations with different degrees of accuracy and sophistication. This is a topic of numerical analysis, outside the scope of this text, and thus only simple methods are illustrated here.
n Example 9.16 A discretetime IIR system is represented by a firstorder difference equation y[n] = ay[n − 1] + x[n]
n≥0
(9.36)
where x[n] is the input of the system and y[n] is the output. Discuss how to solve it using recursive methods and the Ztransform. Obtain a general form for the complete solution y[n] in terms of the impulse response h[n] of the system. For input x[n] = u[n] − u[n − 11], zero initial conditions, and a = 0.8, use the MATLAB function filter to find y[n]. Plot the input and the output. Solution In the time domain a unique solution is obtained by using the recursion given by the difference equation. We would need an initial condition to compute y[0], indeed y[0] = ay[−1] + x[0] and as x[0] is given, we need y[−1] as the initial condition. Given y[0], recursively we find the rest of the solution: y[1] = ay[0] + x[1] y[2] = ay[1] + x[2] y[3] = ay[2] + x[3] ··· where at each step the needed output values are given by the previous step of the recursion. However, this solution is not in closed form.
9.5 OneSided ZTransform Inverse
To obtain a closedform solution, we use the Ztransform. Taking the onesided Ztransform of the two sides of the equation, we get Z(y[n]) = Z(ay[n − 1]) + Z[x[n]) Y(z) = a(z−1 Y(z) + y[−1]) + X(z) Solving for Y(z) in the above equation, we obtain Y(z) =
X(z) ay[−1] + −1 1 − az 1 − az−1
(9.37)
where the first term depends exclusively on the input and the second depends exclusively on the initial condition. If the input x[n] and the initial condition y[−1] are given, we can then find the inverse Ztransform to obtain the complete solution y[n] of the form y[n] = yzs [n] + yzi [n] where the zerostate response yzs [n] is due exclusively to the input x[n] and zero initial conditions, and the zeroinput response yzi [n] is the response due to the initial condition y[−1] with zero input. In this simple case we can obtain the complete solution for any input x[n] and any initial condition y[−1]. Indeed, by expressing 1/(1 − az−1 ) as its Ztransform sum—that is, ∞
X 1 ak z−k = −1 1 − az k=0
Equation (9.37) becomes Y(z) =
∞ X
k −k
X(z)a z
+ ay[−1]
k=0
∞ X
ak z−k
k=0
= X(z) + aX(z)z−1 + a2 X(z)z−2 + · · · + ay[−1](1 + az−1 + a2 z−2 + · · · ) Using the timeshift property we then get the complete solution, y[n] = x[n] + ax[n − 1] + a2 x[n − 2] + · · · + y[−1]a(1 + aδ[n − 1] + a2 δ[n − 2] + · · · ) =
∞ X k=0
ak x[n − k] + ay[−1]
∞ X
ak δ[n − k]
k=0
for any input x[n], initial condition y[−1], and a.
(9.38)
553
CH A P T E R 9: The ZTransform
To solve the difference equation with a = 0.8, x[n] = u[n] − u[n − 11], and zero initial condition y[−1] = 0, using MATLAB, we use the following script. %%%%%%%%%%%%%%%% % Example 9.16 %%%%%%%%%%%%%%%% N = 100; n = 0:N  1; x = [ones(1,10) zeros(1,N  10)]; den=[1 0.8]; num = [1 0]; y = filter(num, den,x)
The function filter requires that the initial conditions are zero. The results are shown in Figure 9.11. Let us now find the impulse response h[n] of the system. For that, let x[n] = δ[n] and y[−1] = 0, then we have y[n] = h[n] or Y(z) = H(z). Thus, we have
H(z) =
1 so that h[n] = an u[n] 1 − az−1
5
x[n]
4 3 2 1 0
0
5
10
15
20
25
30
15
20
25
30
n (a) 6 4 y [n]
554
2
0
0
5
10 n (b)
FIGURE 9.11 (a) Solution of the firstorder difference equation with (b) input.
9.5 OneSided ZTransform Inverse
You can then see that the first term in Equation (9.38) is a convolution sum, and the second is the impulse response multiplied by ay[−1]—that is, y[n] = yzs [n] + yzi [n] =
∞ X
h[k]x[n − k] + ay[−1]h[n] n
k=0
n Example 9.17 Consider a discretetime system represented by a secondorder difference equation with constant coefficients y[n] − a1 y[n − 1] − a2 y[n − 2] = x[n] + b1 x[n − 1] + b2 x[n − 2]
n≥0
where x[n] is the input, y[n] is the output, and the initial conditions are y[−1] and y[−2]. Use the Ztransform to obtain the complete solution. Solution Applying the onesided Ztransform to the two sides of the difference equation, we have Z(y[n] − a1 y[n − 1] − a2 y[n − 2]) = Z(x[n] + b1 x[n − 1] + b2 x[n − 2]) Y(z) − a1 (z−1 Y(z) + y[−1]) − a2 (z−2 Y(z) + y[−1]z−1 + y[−2]) = X(z)(1 + b1 z−1 + b2 z−2 ) where we used the linearity and the timeshift properties of the Ztransform. It was also assumed that the input is causal, x[n] = 0 for n < 0, so that the Ztransforms of x[n − 1] and x[n − 2] are simply z−1 X(z) and z−2 X(z). Rearranging the above equation, we have Y(z)(1 − a1 z−1 − a2 z−2 ) = (y[−1](a1 + a2 z−1 ) + a2 y[−2]) + X(z)(1 + b1 z−1 + b2 z−2 ) and solving for Y(z), we have Y(z) =
X(z)(1 + b1 z−1 + b2 z−2 ) y[−1](a1 + a2 z−1 ) + a2 y[−2]) + 1 − a1 z−1 − a2 z−2 1 − a1 z−1 − a2 z−2
where again the first term is the Ztransform of the zerostate response, due to the input only, and the second term is the Ztransform of the zeroinput response, which is due to the initial conditions alone. The inverse Ztransform of Y(z) will give us the complete response. n Remarks As we saw in Chapter 8, if either the initial conditions are not zero or the input is not causal, the system is not linear time invariant (LTI). However, the timeshift property allows us to find the complete response in that case. We can think of two inputs applied to the system: one due to the initial conditions and the other due to the regular input. By using superposition, we obtain the zerostate and the zeroinput responses, which add to the total response.
555
556
CH A P T E R 9: The ZTransform
Just as with the Laplace transform, the steadystate response of a difference equation y[n] +
N X
ak y[n − k] =
k=1
M X
bm x[n − m]
m=0
is due to simple poles of Y(z) on the unit circle. Simple or multiple poles inside the unit circle give a transient, while multiple poles on the unit circle or poles outside the unit circle create an increasing response.
n Example 9.18 Solve the difference equation y[n] = y[n − 1] − 0.25y[n − 2] + x[n]
n≥0
with zero initial conditions and x[n] = u[n]. Solution The Ztransform of the terms of the difference equation gives Y(z) = =
X(z) + 0.25z−2
1 − z−1
1 z3 = (1 − z−1 )(1 − z−1 + 0.25z−2 ) (z − 1)(z2 − z + 0.25)
z > 1
Y(z) has three zeros at z = 0, a pole at z = 1, and a double pole at z = 0.5. The partial fraction expansion of Y(z) is of the form Y(z) =
A B(1 − 0.5z−1 ) + Cz−1 + 1 − z−1 (1 − 0.5z−1 )2
(9.39)
where the terms of the expansion can be found in the Ztransforms table. Within some constants, the complete response is y[n] = Au[n] + [B(0.5)n + Cn(0.5)n ] u[n] The steady state is then yss [n] = A (corresponding to the pole on the unit circle z = 1) since the other two terms, corresponding to the double pole z = 0.5 inside the unit circle, make up the transient response. The value of A is obtained as A = Y(z)(1 − z−1 ) z−1 =1 = 4 To find the complete response y[n] we find the constants in Equation (9.39). Notice in Equation (9.39) that the expansion term corresponding to the double pole z = 0.5 has as numerator a firstorder polynomial, with constants B and C to be determined, to ensure that the term is proper
9.5 OneSided ZTransform Inverse
rational. That term equals B(1 − 0.5z−1 ) + Cz−1 B Cz−1 = + (1 − 0.5z−1 )2 1 − 0.5z−1 (1 − 0.5z−1 )2 which is very similar to the expansion for multiple poles in the inverse Laplace transform. Once we find the values of B and C, the inverse Ztransforms are obtained from the Ztransforms table. A simple method to obtain the coefficients B and C is to first obtain C by multiplying the two sides of Equation (9.39) by (1 − 0.5z−1 )2 to get Y(z)(1 − 0.5z−1 )2 = B(1 − 0.5z−1 ) + Cz−1 and then letting z−1 = 2 on both sides to find that C=
Y(z)(1 − 0.5z−1 )2 z−1 =2 = −0.5 z−1
The B value is then obtained by choosing a value for z−1 that is different from 1 or 0.5 to compute Y(z). For instance, assume you choose z−1 = 0 and that you have found A and C, then Y(z)z−1 =0 = A + B = 1 from which B = −3. The complete response is then y[n] = 4u[n] − 3(0.5)n − 0.5n(0.5)n u[n] n
n Example 9.19 Find the complete response of the difference equation y[n] + y[n − 1] − 4y[n − 2] − 4y[n − 3] = 3x[n]
n≥0
y[−1] = 1 y[−2] = y[−3] = 0 x[n] = u[n] Determine if the discretetime system corresponding to this difference equation is BIBO stable or not, and the effect this has in the steadystate response. Solution Using the timeshifting and linearity properties of the Ztransform, and replacing the initial conditions, we get Y(z)[1 + z−1 − 4z−2 − 4z−3 ] = 3X(z) + [−1 + 4z−1 + 4z−2 ]
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Letting A(z) = 1 + z−1 − 4z−2 − 4z−3 = (1 + z−1 )(1 + 2z−1 )(1 − 2z−1 ) we can write Y(z) = 3
X(z) −1 + 4z−1 + 4z−2 + A(z) A(z)
z > 2
(9.40)
To determine whether the steadystate response exists or not let us first consider the stability of the system associated with the given difference equation. The transfer function H(z) of the system is computed by letting the initial conditions be zero (this makes the second term on the right of the above equation zero) so that we can get the ratio of the Ztransform of the output to the Ztransform of the input. If we do that then H(z) =
Y(z) 3 = X(z) A(z)
Since the poles of H(z) are the zeros of A(z), which are z = −1, z = −2, and z = 2, then the impulse response h[n] = Z −1 [H(z)] will not be absolutely summable, as required by the BIBO stability, because the poles of H(z) are on and outside the unit circle. Indeed, a general form of the impulse response is h[n] = [C + D (2)n + E (−2)n ]u[n] where C, D, and E are constants that can be found by doing a partial fraction expansion of H(z). Thus, h[n] will grow as n increases and it would not be absolutely summable—that is, the system is not BIBO stable. Since the system is unstable, we expect the total response to grow as n increases. Let us see how we can justify this. The partial fraction expansion of Y(z), after replacing X(z) in Equation (9.40), is given by 2 + 5z−1 − 4z−3 (1 − z−1 )(1 + z−1 )(1 + 2z−1 )(1 − 2z−1 ) B1 B2 B3 B4 = + + + −1 −1 −1 1−z 1+z 1 + 2z 1 − 2z−1
Y(z) =
B1 = Y(z)(1 − z−1 )z−1 =1 = −
1 2
B2 = Y(z)(1 + z−1 )z−1 =−1 = −
1 6
B3 = Y(z)(1 + 2z−1 )z−1 =−1/2 = 0 B4 = Y(z)(1 − 2z−1 )z−1 =1/2 =
8 3
9.5 OneSided ZTransform Inverse
so that 1 8 n n y[n] = −0.5 − (−1) + 2 u[n] 6 3 which as expected will grow as n increases — there is no steadystate response. In a problem like this the chance of making computational errors is large, so it is important to figure out a way to partially check your answer. In this case we can check the value of y[0] using the difference equation, which is y[0] = − y[−1] + 4y(−2) + 4y(−3) + 3 = − 1 + 3 = 2, and compare it with the one obtained using our solution, which gives y[0] = −3/6 − 1/6 + 16/6 = 2. They coincide. Another way to partially check your answer is to use the initial and the final values theorems. n
Solution of Differential Equations The solution of differential equations requires converting them into difference equations, which can then be solved in a closed form by means of the Ztransform.
n Example 9.20 Consider an RLC circuit represented by the secondorder differential equation d2 vc (t) dvc (t) + + vc (t) = vs (t) dt2 dt where the voltage across the capacitor vc (t) is the output and the source vs (t) = u(t) is the input. Let the initial conditions be zero. Find the voltage across the capacitor vc (t). Solution The Laplace transform of the output is found from the differential equation as Vs (s) 1 + s + s2 1 1 = 2 = s(s + s + 1) s((s + 0.5)2 + 3/4)
Vc (s) =
where the final equation is obtained after replacing Vs (s) = 1/s. The solution of the differential equation is of the general form √ vc (t) = [A + Be−0.5t cos( 3/2t + θ )]u(t) for constants A, B, and θ. To convert the differential equation into a difference equation we approximate the first derivative as dvc (t) vc (t) − vc (t − Ts ) ≈ dt Ts
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and the second derivative as d dvdtc (t) d(vc (t) − vc (t − Ts ))/Ts ) d2 vc (t) = ≈ dt2 dt dt vc (t) − 2vc (t − Ts ) + vc (t − 2Ts ) ≈ Ts2 which when replaced in the differential equation, and computing the resulting equation for t = nTs , gives 1 1 2 1 1 + + 1 vc (nTs ) − + vc ((n − 1)Ts ) + vc ((n − 2)Ts ) = vs (nTs ) Ts2 Ts Ts2 Ts Ts2 Although we know that we need to choose a very small value for Ts to get a good approximation to the exact result, for simplicity let us first set Ts = 1, so that the difference equation is 3vc [n] − 3vc [n − 1] + vc [n − 2] = vs [n]
n≥0
For zero initial conditions and unitstep input, we can recursively compute this equation to get vc [0] = 1/3 vc [n] = 1
n→∞
A closedform solution can be obtained using the Ztransform, giving (assuming zero initial conditions) [3 − 3z−1 + z−2 ]Vc (z) =
1 1 − z−1
so that Vc (z) =
z3 (z − 1)(3z2 − 3z + 1)
√ from which we obtain that there is a triple zero at z = 0, and poles at 1 and −0.5 ± j 3/6. The partial fraction expansion will be of the form Vc (z) =
A B B∗ + + √ √ 1 − z−1 1 + (0.5 + j 3/6)z−1 1 + (0.5 − j 3/6)z−1
Since the complex poles are inside the unit circle, the steadystate response is due to the input that has a single pole at 1 (i.e., the steady state is limn→∞ vc [n] = A = 1). We first use the symbolic MATLAB functions ilaplace and ezplot to find the exact solution of the differential equation. We then sample the input signal using a sampling period Ts = 0.1 sec and use the approximations of the first and second derivatives to obtain the difference equation, which is computed using filter. The results are shown in Figure 9.12. The exact solution of the
9.5 OneSided ZTransform Inverse
1.2
1
y (t ), y[n]
0.8
0.6
0.4
0.2
0 0
1
2
3
4
5 t, nTs
6
7
8
9
10
FIGURE 9.12 Solution of the differential equation d2 vc (t)/dt2 + dvc (t)/dt + vc (t) = vs (t) (solid line). Solution of the difference equation approximating the differential equation for Ts = 0.1 (dotted line). Exact and approximate solutions are very close.
differential equation is well approximated by the solution of the difference equation obtained by approximating the first and second derivatives. %%%%%%%%%%%%%%%% % Example 9.20 %%%%%%%%%%%%%%%% syms s vc = ilaplace(1/(s ˆ 3 + s ˆ 2 + s)); % exact solution ezplot(vc,[0,10]);grid; hold on % plotting of exact solution Ts = 0.1; % sampling period a1 = 1/Tsˆ2 + 1/Ts + 1;a2 = 2/Ts ˆ 2  1/Ts;a3 = 1/Ts ˆ 2; % coefficients a = [1 a2/a1 a3/a1];b = 1; t = 0:Ts:10; N = length(t); vs=ones(1,N); % input vca = filter(b,a,vs);vca = vca/vca(N); % solution
n
9.5.5 Inverse of TwoSided ZTransforms When finding the inverse of a twosided Ztransform, or a noncausal discretetime signal, it is important to relate the poles to the causal and the anticausal components. The region of convergence plays
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a very important role in making this determination. Once this is done, the inverse is found by looking for the causal and the anticausal partial fraction expansion components in a Ztransforms table. The coefficients of the partial fraction expansion are calculated like those in the case of causal signals. n Example 9.21 Consider finding the inverse Ztransform of X(z) =
2z−1 (1 − z−1 )(1 − 2z−1 )2
1 < z < 2
which corresponds to a noncausal signal. Solution The function X(z) has two zeros at z = 0, a pole at z = 1, and a double pole at z = 2. For the region of convergence to be a torus of internal radius 1 and outer radius 2, we need to associate with the pole at z = 1 the region of convergence R1 : z > 1 corresponding to a causal signal, and with the pole at z = 2, we associate a region of convergence R2 : z < 2 associated with an anticausal signal. Thus, we have 1 < z < 2 = R1 ∩ R2 The partial fraction expansion is then done so that A B Cz−1 X(z) = + + −1 −1 (1 − 2z−1 )2 1 −{zz }  1 − 2z {z } R1 :z>1
R2 :z 1 so that the corresponding impulse response h1 [n] = Z −1 [H(z)] is causal with a general form h1 [n] = B0 δ[n] + [B1 (0.5)n + B2 ]u[n]
n
The pole at z = 1 makes this filter unstable, as its impulse response is not absolutely summable. R2 : z < 0.5 for which the corresponding impulse response h2 [n] = Z −1 [H(z)] is anticausal with a general form h2 [n] = B0 δ[n] − (B1 (0.5)n + B2 )u[−n − 1]
n
The region of convergence R2 does not include the unit circle and so the impulse response is not absolutely summable (H(z) cannot be defined at z = 1 because it is not in the region of convergence R2 ). The impulse response h2 [n] grows as n becomes smaller and negative. R3 : 0.5 < z < 1, which gives a twosided impulse response h2 [n] = Z −1 [H(z)] of general form h3 [n] = B0 δ[n] + B1 ((0.5)n u[n] −B2 u[−n − 1] {z } {z }  causal anticausal Again, this filter is not stable.
n
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9.6 WHAT HAVE WE ACCOMPLISHED? WHERE DO WE GO FROM HERE? Although the history of the Ztransform is originally connected with probability theory, for discretetime signals and systems it can be connected with the Laplace transform. The periodicity in the frequency domain and the possibility of an infinite number of poles and zeros makes this connection not very useful. Defining a new complex variable in polar form provides the definition of the Ztransform and the zplane. As with the Laplace transform, poles of the Ztransform characterize discretetime signals by means of frequency and attenuation. One and twosided Ztransforms are possible, although the onesided version can be used to obtain the twosided one. The region of convergence makes the Ztransform have a unique relationship with the signal, and it will be useful in obtaining the discrete Fourier representations in Chapter 10. Dynamic systems represented by difference equations use the Ztransform for representation by means of the transfer function. The onesided Ztransform is useful in the solution of difference equations with nonzero initial conditions. As in the continuoustime case, filters can be represented by difference equations. However, discrete filters represented by polynomials are also possible. These nonrecursive filters give significance to the convolution sum, and will motivate us to develop methods that efficiently compute it. You will see that the reason to present the Ztransform before the Fourier representation of discretetime signals and systems is to use their connection, thereby simplifying calculations.
PROBLEMS 9.1. Mapping of splane into the zplane The poles of the Laplace transform X(s) of an analog signal x(t) are p1,2 = −1 ± j1 p3 = 0 p4,5 = ±j1 There are no zeros. If we use the transformation z = e sTs with Ts = 1: (a) Determine where the given poles are mapped into the zplane. (b) How would you determine if these poles are mapped inside, on, or outside the unit circle in the zplane? Explain. (c) Carefully plot the poles and the zeros of the analog and the discretetime signals in the Laplace and the zplanes. 9.2. Mapping of zplane into the splane Consider the inverse relation given by z = e sTs —that is, how to map the zplane into the splane. (a) Find an expression for s in terms of z from the relation z = e sTs . (b) Consider the mapping of the unit circle (i.e., z = 1e jω , −π ≤ ω < π). Obtain the segment in the splane resulting from the mapping. (c) Consider the mapping of the inside and the outside of the unit circle. Determine the regions in the splane resulting from the mappings.
Problems
(d) From the above results, indicate the region in the splane to which the whole zplane is mapped into. Since ω = ω + 2π, is this mapping unique? Explain. 9.3. Ztransform and ROCs Consider the noncausal sequence s[n] = s1 [n] + s2 [n] where s1 [n] = u[n] is causal and s2 [n] = −u[−n] is anticausal. This signal is the signum, or sign function, that extracts the sign of a realvalued signal—that is, −1 x[n] < 0 0 x[n] = 0 s[n] = sgn(x[n]) = 1 x[n] > 0 (a) Find the Ztransforms of s1 [n] and s2 [n], indicating the corresponding ROC. (b) Determine the Ztransform S(z). 9.4. Ztransform and ROC Given the anticausal signal x[n] = −α n u[−n] (a) Determine the Ztransform X(z), and carefully plot the ROC when α = 0.5 and α = 2. For which of the two values of α does X(e jω ) exist? (b) Find the signal that corresponds to the derivative dX(z)/dz. Express it in terms of α. 9.5. Significance of ROC Consider a causal signal x1 [n] = u[n] and an anticausal signal x2 [n] = −u[−n − 1]. (a) Find the Ztransforms X1 (z) and X2 (z) and carefully plot their ROCs. If the ROCs are not included with the Ztransforms, would you be able to tell which is the correct inverse? Explain. (b) Determine if it is possible to find the Ztransform of x1 [n] + x2 [n]. 9.6. Fibonacci sequence generation—MATLAB Consider the Fibonacci sequence generated by the difference equation f [n] = f [n − 1] + f [n − 2]
n≥0
with initial conditions f [−1] = 1, f [−2] = −1. (a) Find the Ztransform of f [n], or F(z). (b) Find the poles φ1 and φ2 and the zeros of F(z) and plot them. How are the poles connected? How are they related to the “golden ratio”? (c) The Fibonacci difference equation has zero input, but its response is a sequence of everincreasing integers. Obtain a partial fraction expansion of F(z) and find f [n] in terms of the poles φ1 and φ2 , and show that the result is always integer. Use MATLAB to implement the inverse in term of the poles. 9.7. Laplace and Ztransforms of sampled signals An analog pulse x(t) = u(t) − u(t − 1) is sampled using a sampling period Ts = 0.1. (a) Obtain the discretetime signal x(nTs ) = x(t)t=nTs and plot it as a function of nTs . (b) If the sampled signal is represented as an analog signal as xs (t) =
N−1 X n=0
determine the value of N in the above equation.
x(nTs )δ(t − nTs)
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(c) Compute the Laplace transform of the sampled signal (i.e., Xs (s) = L[xs (t]). (d) Determine the Ztransform of x(nTs ), or X(z). (e) Indicate how to transform Xs (s) into X(z) 9.8. Computation of Ztransform—MATLAB Consider a discretetime pulse x[n] = u[n] − u[n − 10]. (a) Plot x[n] as a function of n and use the definition of the Ztransform to find X(z). (b) Use the Ztransform of u[n] and properties of the Ztransform to find X(z). Verify that the expressions obtained above for X(z) are identical. (c) Find the poles and the zeros of X(z) and plot them in the zplane. Use MATLAB to plot the poles and zeros. 9.9. Computation of Ztransform A causal exponential x(t) = 2e−2t u(t) is sampled using a sampling period Ts = 1. The corresponding discretetime signal is x[n] = 2e−2n u[n]. (a) Express the discretetime signal as x[n] = 2α n u[n] and give the value of α. (b) Find the Ztransform X(z) of x[n] and plot its poles and zeros in the zplane. 9.10. Computation of Ztransform Consider the signal x[n] = 0.5(1 + [−1]n )u[n]. (a) Plot x[n] and use the definition of the Ztransform to obtain its Ztransform, X(z). (b) Use the linearity property and the Ztransforms of u[n] and [−1]n u[n] to find the Ztransform X(z) = Z[x[n]]. (c) Determine and plot the poles and the zeros of X(z). 9.11. Solution of difference equations with Ztransform Consider a system represented by the firstorder difference equation y[n] = x[n] − 0.5y[n − 1] where y[n] is the output and x[n] is the input. (a) Find the Ztransform Y(z) in terms of X(z) and the initial condition y[−1]. (b) Find an input x[n] 6= 0 and an initial condition y[−1] 6= 0 so that the output is y[n] = 0 for n ≥ 0. Verify you get this result by solving the difference equation recursively. (c) For zero initial conditions, find the input x[n] so that y[n] = δ[n] + 0.5δ[n − 1]. 9.12. Transfer function, stability, and impulse response—MATLAB Consider a secondorder discretetime system represented by the difference equation y[n] − 2r cos(ω0 )y[n − 1] + r 2 y[n − 2] = x[n]
n≥0
where r > 0 and 0 ≤ ω0 ≤ 2π , y[n] is the output, and x[n] is the input. (a) Find the transfer function H(z) of this system. (b) Find the value of ω0 and determine the values of r that would make the system stable. Use the MATLAB function zplane to plot the poles and the zeros for r = 0.5 and ω0 = π/2 radians. (c) Let ω0 = π/2. Find the corresponding impulse response h[n] of the system. What other value of ω0 would get the same impulse response? 9.13. Generation of discretetime sinusoid—MATLAB Given that the Ztransform of a discretetime cosine A cos(ω0 n)u[n] is A(1 − cos(ω0 )z−1 ) 1 − 2 cos(ω0 )z−1 + z−2
Problems
(a) Use the given Ztransform to find a difference equation for which the output y[n] is a discretetime cosine A cos(ω0 n) and the input is x[n] = δ[n]. What should you use as initial conditions? (b) Verify your algorithm by generating a signal y[n] = 2 cos(πn/2)u[n] by implementing your algorithm in MATLAB. Plot the input and the output signals x[n] and y[n]. (c) Indicate how to change your previous algorithm to generate a sine function y[n] = 2 sin(πn/2)u[n]. Use MATLAB to find y[n], and to plot it. 9.14. Inverse Ztransform and poles and zeros When finding the inverse Ztransform of functions with z−1 terms in the numerator, the fact that z−1 can be thought of as a delay operator can be used to simplify the computation. Consider
X(z) =
1 − z−10 1 − z−1
(a) Use the Ztransform of u[n] and the properties of the Ztransform to find x[n]. (b) If we consider X(z) a polynomial in negative powers of z, what would be its degree and the values of its coefficients? (c) Find the poles and the zeros of X(z) and plot them on the zplane. Is there a pole or zero at z = 1? Explain. 9.15. Initial conditions and steady state Consider a secondorder system represented by the difference equation y[n] = 0.25y[n − 2] + x[n] where x[n] is the input and y[n] is the output. (a) For the zeroinput case (i.e., when x[n] = 0), find the initial conditions y[−1] and y[−2] so that y[n] = 0.5n u[n]. (b) Suppose the input is x[n] = u[n]. Without solving the difference equation can you find the corresponding steady state yss [n]? Explain how and give the steadystate output. Verify by inverse Ztransform that the steadystate response yss [n] is the one obtained. 9.16. Initial conditions and impulse response A secondorder system has the difference equation y[n] = 0.25y[n − 2] + x[n] where x[n] is the input and y[n] is the output. (a) Find the input x[n] so that for zero initial conditions, the output is given as y[n] = 0.5n u[n]. (b) If x[n] = δ[n] + 0.5δ[n − 1] is the input to the above difference equation, find the impulse response of the system. 9.17. Convolution sum and product of polynomials The convolution sum is a fast way to find the coefficients of the polynomial resulting from the multiplication of two polynomials. (a) Suppose x[n] = u[n] − u[n − 3]. Find its Ztransform X(z), a secondorder polynomial in z−1 . (b) Multiply X(z) by itself to get a new polynomial Y(z) = X(z)X(z) = X2 (z). Find Y(z). (c) Graphically show the convolution of x[n] with itself and verify that the result coincides with the coefficients of Y(z).
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9.18. Inverse Ztransform Find the inverse Ztransform of 8 − 4z−1 X(z) = −2 z + 6z−1 + 8 and determine x[n] as n → ∞. Assume x[n] is causal. 9.19. Ztransform properties and inverse transform Sometimes the partial fraction expansion is not needed in finding the inverse Ztransform—instead the properties of the transform can be used. Consider the function z+1 F(z) = 2 z (z − 1) (a) Determine whether F(z) is a proper rational function as a function of z and of z−1 . (b) Verify that F(z) can be written as F(z) =
z−2 z−3 + 1 − z−1 1 − z−1
Find the inverse Ztransform f [n] using the above expression. 9.20. Inverse Ztransform—MATLAB We are interested in the unitstep solution of a system represented by the difference equation y[n] = y[n − 1] − 0.5y[n − 2] + x[n] + x[n − 1] (a) Find an expression for Y(z). (b) Do a partial expansion of Y(z). (c) Find the inverse Ztransform y[n] and verify your results using MATLAB. 9.21. Pade´ approximation Suppose we are given a finitelength sequence h[n] (it could be part of an infinitelength impulse response from a discrete system that has been windowed) and would like to obtain a rational approximation for it. This means that if H(z) = Z[h[n]], a rational approximation of it would be H(z) = B(z)/A(z), from which we get H(z)A(z) = B(z) Letting
B(z) =
M−1 X
bk z−k
k=0
A(z) = 1 +
N−1 X
ak z−k
k=1
for some choice of M and N, equations from H(z)A(z) = B(z) should allow us to find the M + N − 1 coefficients {ak bk }.
Problems
(a) Find a matrix equation that would allow us to find the coefficients of B(z) and A(z). (b) Let h[n] = 0.5n (u[n] − u[n − 101]) be the sequence we wish to obtain a rational approximation and let B(z) = b0 while A(z) = a0 + a1 z−1 . Find the equations to solve for the coefficients {b0 , a0 , a1 }. 9.22. Prony’s rational approximation—MATLAB The Pade´ approximants provide an exact matching of M + N − 1 values of h[n] where M and N are, respectively, the orders of the numerator and the denominator of the rational approximation. But there is no method for choosing the numerator and the denominator orders, M and N. Also, there is no guarantee on how well the rest of the signal is matched. Prony’s rational approximation considers how well the rest of the signal is approximated when finding the approximation. Let h[n] = 0.9n u[n] be the exact impulse response for which we wish to find a rational approximation. Take the first 100 values of this signal as the impulse response.1 (a) Assume the order of the numerator and the denominator are equal, M = N = 1. Use the MATLAB function prony to obtain the rational approximation, and then use filter to verify that the impulse response of the rational approximation is close to the given 100 values. Plot the error between h[n] and the impulse response of the rational approximation for the first 200 samples. (b) Plot the poles and the zeros of the rational approximation and compare them to the poles and the zeros of H(z) = Z(h[n]. (c) Suppose that h[n] = (h1 ∗ h2 )[n]—that is, the convolution of h1 [n] = 0.9n u[n] and h2 [n] = 0.8n u[n]. Use again prony to find the rational approximation when the first 100 values of h[n] are available. Use conv from MATLAB to compute h[n]. Compare the impulse response of the rational approximation to h[n]. Plot the poles and the zeros of H(z) = Z(h[n]) and of the rational approximation. (d) Consider the h[n] given above, and perform the Prony approximation using orders M = N = 3. Explain your results. Plot the poles and the zeros. 9.23. MATLAB partial fraction expansion Consider the partial fraction expansion that MATLAB uses. (a) Find the inverse Ztransform of a/(1 − az−1 )2 . (b) Suppose that the partial fraction expansion given by MATLAB is X(z) =
−1 1 + −1 1 − 0.5z (1 − 0.5z−1 )2
Determine the inverse x[n]. 9.24. MATLAB partial fraction expansion Consider finding the inverse Ztransform of X(z) =
2z−1 (1 − z−1 )(1 − 2z−1 )2
z > 2
MATLAB does the partial fraction expansion as X(z) =
A B C + + −1 −1 1−z 1 − 2z (1 − 2z−1 )2
while we do it in the following form: X(z) =
D E Fz−1 + + −1 −1 1−z 1 − 2z (1 − 2z−1 )2
Show that the two partial fraction expansions give the same result. 1 Gaspar
de Prony (1765–1839) was a French mathematician and engineer, while Henri Pad´e (1863–1953) was a French mathematician interested in rational approximations.
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9.25. Prony method and ZTransform—MATLAB Consider finding the Ztransform of a noncausal signal h[n] = 0.5n u[n + 1] using the Prony approximation. (a) Use the prony function to find a rational approximation for h[n] (i.e., the Ztransform H(z) = B(z)/A(z)). Use a first order for the numerator and the denominator. (b) Separate the signal into its causal and anticausal components, and use prony to find the rational approximation of the causal and then add the anticausal component to correct the above result.
CHAPTER 10
Fourier Analysis of DiscreteTime Signals and Systems
Diligence is the mother of good luck. Benjamin Franklin (1706–1790) Printer, inventor, scientist, and diplomat I am a great believer in luck, and I find the harder I work, the more I have of it. President Thomas Jefferson (1743–1826) Main author of the U.S. Declaration of Independence
10.1 INTRODUCTION In this chapter we will consider the Fourier representation of discretetime signals and systems. Similar to the connection between the Laplace and the Fourier transforms of continuoustime signals and systems, if the region of convergence of the Ztransform of a signal or of the transfer function of a discrete system includes the unit circle, then the discretetime Fourier transform (DTFT) of the signal or the frequency response of the system is easily found. Duality in time and frequency is used whenever signals and systems do not satisfy this condition. We can thus obtain the Fourier representation of most discretetime signals and systems. Two computational disadvantages of the DTFT are that the direct DTFT is a function of a continuously varying frequency, and the inverse DTFT requires integration. These disadvantages can be removed by sampling in frequency the DTFT, resulting in the socalled discrete Fourier transform (DFT) (notice the difference in the naming of these two related frequency representations). An interesting connection determines their computational feasibility: a discrete–time signal has a periodic continuousfrequency transform—the DTFT—while a periodic discretetime signal has a periodic and discretefrequency transform—the DFT. As we will discuss in this chapter, any periodic or aperiodic signal can be represented by the DFT, a computationally feasible transformation where both time and frequency are discrete and no integration is required, and that can be implemented very efficiently by the Fast Fourier Transform (FFT) algorithm.
Signals and Systems Using MATLAB®. DOI: 10.1016/B9780123747167.000144 c 2011, Elsevier Inc. All rights reserved.
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In this chapter, we will see that a great deal of the Fourier representation of discretetime signals and characterization of discrete systems can be obtained from our knowledge of the Ztransform. To obtain the DFT, which is of great significance in digital signal processing, we will proceed in an opposite direction as in the continuoustime analysis. First, we consider the Fourier representation of aperiodic signals and then that of periodic discretetime signals, and finally use this representation to obtain the DFT.
10.2 DISCRETETIME FOURIER TRANSFORM The discretetime Fourier transform (DTFT) of a discretetime signal x[n], X X(e jω ) = x[n]e−jωn −π ≤ω α, and the ROC of the second term is z < 1/α. Thus, the region of convergence of X(z) is ROC: α < z
1, can be downsampled to generate a discretetime signal xd [n] = x[Mn] with Xd (e jω ) =
n
1 X(e jω/M ) M
(10.17)
which is an expanded version of X(e jω ). A signal x[n] is upsampled to generate a signal xu [n] = x[n/L] for n = ±kL, k = 0, 1, 2, . . . , and zero otherwise. The DTFT of xu [n] is X(e jLω ), or a compressed version of X(e jω ).
10.2 DiscreteTime Fourier Transform
n Example 10.4 Consider the frequency response of an ideal lowpass filter, H(e jω ) =
1 −π/2 ≤ ω ≤ π/2 0 −π ≤ ω < −π/2 and π/2 < ω ≤ π
which is the DTFT of an impulse response h[n]. Determine h[n]. Suppose that we downsample h[n] with a factor of M = 2. Find the downsampled impulse response hd [n] = h[2n] and its corresponding frequency response Hd (e jω ). Solution The impulse response h[n] corresponding to the ideal lowpass filter is found to be 1 h[n] = 2π
Zπ/2
e jωn dω =
0.5 n=0 sin(πn/2)/(πn) n = 6 0
−π/2
The downsampled impulse response is given by hd [n] = h[2n] =
0.5 n=0 sin(πn)/(2πn) = 0 n = 6 0
or hd [n] = 0.5δ[n], with a DTFT of Hd (e jω ) = 0.5 for −π < ω ≤ π (i.e., an allpass filter). This agrees with the downsampling theory, which gives that Hd (e jω ) =
1 1 H(e jω/2 ) = , 2 2
−π ≤ ω < π
That is, H(e jω ) multiplied by 1/M = 1/2 and expanded by M = 2.
n
n Example 10.5 A discrete pulse is given by x[n] = u[n] − u[n − 4]. Suppose we downsample x[n] by a factor of M = 2, so that the length 4 of the original signal is reduced to 2, giving xd [n] = x[2n] = u[2n] − u[2n − 4] = u[n] − u[n − 2] Find the corresponding DTFTs for x[n] and xd [n], and determine how they are related. Solution The Ztransform of x[n] is X(z) = 1 + z−1 + z−2 + z−3
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with the whole zplane (except for the origin) as its region of convergence. Thus, the DTFT of x[n] is h 3 i 3 1 1 3 X(e jω ) = e−j( 2 ω) e j( 2 ω) + e j( 2 ω) + e−j( 2 ω) + e−j( 2 ω) ω 3 3ω = 2e−j( 2 ω) cos + cos 2 2 The Ztransform of the downsampled signal (M = 2) is Xd (z) = 1 + z−1 and the DTFT of xd [n] is h 1 i 1 1 Xd (e jω ) = e−j( 2 ω) e j( 2 ω) + e−j( 2 ω) ω 1 = 2e−j( 2 ω) cos 2 Clearly, this is not equal to 0.5X(e jω/2 ). This is caused by aliasing: The maximum frequency of x[n] is not π/M = π/2 and so Xd (e jω ) is the sum of superposed and shifted X(e jω ). Suppose we pass x[n] through an ideal lowpass filter H(e jω ) with cutoff frequency π/2. Its output would be a signal x1 [n] with a maximum frequency of π/2, and downsampling it with M = 2 would give a signal with a DTFT of 0.5X1 (e jω/2 ). n
n Example 10.6 Discuss the effects of downsampling a discrete signal that is not band limited versus the case of one that is. Consider a unit rectangular pulse of length N = 10. Downsample it by a factor of M = 2, and compute and compare the DTFTs of the pulse and its downsampled version. Do a similar procedure to a sinusoid of discrete frequency π/4 and comment on the results. Explain the difference between the above two cases. Use the MATLAB function decimate (lowpass filtering is used to avoid aliasing followed by downsampling) to perform similar operations and comment on the differences with downsampling. Use the MATLAB function interp to interpolate (upsampling with smoothing by a lowpass filter) the downsampled signals. See Figure 10.3 for illustrations of downsampler and upsampler and decimator and interpolator. Solution As indicated, when we downsample a discretetime signal x[n] by a factor of M, in order not to have aliasing in frequency the signal must be band limited to π/M. If the signal satisfies this condition, the spectrum of the downsampled signal is an expanded version of the spectrum of x[n]. To illustrate this in the following script we downsample by a factor of M = 2 first a signal that is not band limited to π/2, and then another that is.
10.2 DiscreteTime Fourier Transform
Downsampler x [n] M
↓
x [M n]
Decimator x [n] H (z)
x 1 [n]
M
↓
x 1 [M n]
LPF Upsampler
FIGURE 10.3 Top: Downsampler and decimator. Bottom: upsampler and interpolator.
x [n] L
↑
Interpolator x [n/L]
x [n] L
↑
x˜ [n]
x [n/L] H (z) LPF
%%%%%%%%%%%%%%%%%%%%%%%%% % Example 10.6downsampling and decimation %%%%%%%%%%%%%%%%%%%%%%%%% x = [ones(1,10) zeros(1,100)];Nx = length(x); n1 = 0:19; % first signal % Nx = 200;n = 0:Nx  1; x = cos(pi ∗ n/4); % second signal y = x(1:2:Nx  1); % downsampling with M = 2 X = fft(x);Y = fft(y); % ffts of original and downsampled signals L = length(X);w = 0:2 ∗ pi/L:2 ∗ pi  2 ∗ pi/L;w1 = (w  pi)/pi; % frequency range z = decimate(x,2,‘fir’); % decimation with M = 2 Z = fft(z); % fft of decimated signal %%%%%%%%%%%%%%%%%%%%%%%%% % interpolation %%%%%%%%%%%%%%%%%%%%%%%%% s = interp(y,2);
As shown in Figure 10.4, the rectangular pulse is not band limited to π/2 since it has frequency components beyond π/2, while the sinusoid is band limited. The DTFT of the downsampled rectangular pulse (a narrower pulse) is not an expanded version of the DTFT of the pulse, while the DTFT of the downsampled sinusoid is an expanded version. The MATLAB function decimate uses an FIR lowpass filter to smooth out x[n] to a frequency of π/2 before downsampling. In the case of the sinusoid, which satisfies the downsampling condition, the downsampling and the decimation provide the same results, but not for the rectangular pulse. The original discretetime signal can be recovered by interpolation. This procedure is composed of upsampling followed by lowpass filtering. The MATLAB function interp is used to that effect. If we use the downsampled signal as input to this function, we obtain slightly better results for the sinusoid than for the pulse when comparing the interpolated signal to the original signal. The results are shown in Figure 10.5. The error s[n] − x[n] is shown also. The signal s[n] is the interpolation of the downsampled signal y[n]. n
10.2.5 Parseval’s Energy Result Just like in the case of continuoustime signals, the energy or power of a discretetime signal x[n] can be equally computed in either the time or the frequency domain.
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10
X (e jω )
x [n]
1 0.5 0
0
5
10
y [n]
Y (e jω ) 0
5
10
0.5
−0.5
0
0.5
−0.5
0 ω/π
0.5
−0.5
0
0.5
−0.5
0
0.5
−0.5
0 ω/π
0.5
10
Z (e jω )
z [n]
0
5 0 −1
15
1 0.5 0
−0.5
10
0.5 0
5 0 −1
15
1
0
5
10 n
5 0 −1
15 (a)
100 X (e jω )
x [n]
1 0 −1 0
5
10
100 Y (e jω )
y [n]
50 0 −1
15
1 0 −1 0
5
10
50 0 −1
15
1
100 Z (e jω )
z [n]
586
0 −1 0
5
10 n
50 0 −1
15 (b)
FIGURE 10.4 Downsampling of (a) nonbandlimited and (b) bandlimited discretetime signals. The signals x[n] correspond to the original signals, while y[n] and z[n] are their downsampled and decimated signals, respectively. The corresponding magnitude spectra are shown. Notice the difference between the downsampled and the decimated signals, they are identical when the signals are bandlimited, slightly different otherwise.
Interpolated Original
0
2
4
6
8
10 n
12
14
16
s [n], x [n]
1 0.8 0.6 0.4 0.2 0
−0.2 −0.4 2
4
6
8
10 n
12
14
16
Interpolated Original
0 −0.5 −1 0
0
0
1 0.5
18
s [n]−x [n]
s [n]−x [n]
s [n], x [n]
10.2 DiscreteTime Fourier Transform
2
4
6
8 n
10
12
14
16
0.1 0.05 0 −0.05 −0.1 −0.15
18
0
2
4
6
(a)
8 n
10
12
14
16
(b)
FIGURE 10.5 Interpolation of (a) nonbandlimited and (b) bandlimited discretetime signals. The interpolated signal is compared to the original signal, and the interpolation error is shown. The errors signals show that the original signal can be recovered almost exactly when the signal satisfies the bandlimiting condition, not otherwise.
If the DTFT of a finiteenergy signal x[n] is X(e jω ), we have that the energy of the signal is given by ∞ X
1 Ex = x[n] = 2π n=−∞ 2
Zπ
X(e jω )2 dω
(10.18)
−π
The magnitude square X(e jω )2 has the units of energy per radian, and so it is called an energy density. When X(e jω )2 is plotted against frequency ω, the plot is called the energy spectrum of the signal, or how the energy of the signal is distributed over frequencies.
10.2.6 Time and Frequency Shifts Shifting in time does not change the frequency content of a signal. Thus, the magnitude of the signal DTFT is not affected, only the phase is. Indeed, if x[n] has a DTFT X(e jω ), then the DTFT of x[n − N] for some integer N is X F(x[n − N]) = x[n − N]e−jωn n
=
X
x[m]e−jω(m+N) = e−jωN X(e jω )
m
If x[n] has a DTFT X(e jω ) = X(e jω )e jθ (ω) where θ (ω) is the phase, the shifted signal x1 [n] = x[n − N] has a DTFT of X1 (e jω ) = X(e jω )e−jωN = X(e jω )e−j(ωN−θ (ω))
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In a dual way, when we multiply a signal by a complex exponential e jω0 n for some frequency ω0 , the spectrum of the signal is shifted in frequency. So if x[n] has a DTFT X(e jω ), the modulated signal x[n]e jω0 n has as DTFT X(e j(ω−ω0 ) ). Indeed, the DTFT of x1 [n] = x[n]e jω0 n is X X X1 (e jω ) = x1 [n]e−jωn = x[n]e−j(ω−ω0 )n = X(e j(ω−ω0 ) ) n
n
The following pairs illustrate the duality in time and frequency shifts: if the DTFT of x[n] is X(e jω ) then x[n − N] ⇔ X(e jω )e−jωN x[n]e jω0 n ⇔ X(e j(ω−ω0 ) )
(10.19)
Remark The signal x[n]e jω0 n was called modulated because x[n] modulates the complex exponential or discretetime sinusoids. It can be written as x[n] cos(ω0 n) + jx[n] sin(ω0 n) n Example 10.7 The DTFT of x[n] = cos(ω0 n), −∞ < n < ∞, cannot be found from the Ztransform or from the sum defining the DTFT as x[n] is not a finiteenergy signal. Use the frequencyshift and the timeshift properties to find the DTFTs of x[n] = cos(ω0 n) and y[n] = sin(ω0 n). Solution Using Euler’s identity we have that x[n] = cos(ω0 n) =
e jω0 n + e−jω0 n 2
and so the DTFT of x[n] is given by X(e jω ) = F[0.5e jω0 n ] + F[0.5e−jω0 n ] = F[0.5]ω−ω0 + F[0.5]ω+ω0 = π[δ(ω − ω0 ) + δ(ω + ω0 )] Since y[n] = sin(ω0 n) = cos(ω0 n − π/2) = cos(ω0 (n − π/(2ω0 )) = x[n − π/(2ω0 )] we have that according to the timeshift property its DTFT is given by Y(e jω ) = X(e jω )e−jωπ/(2ω0 ) h i = π δ(ω − ω0 )e−jωπ/(2ω0 ) + δ(ω + ω0 )e−jωπ/(2ω0 )
10.2 DiscreteTime Fourier Transform
h i = π δ(ω − ω0 )e−jπ/2 + δ(ω + ω0 )e jπ/2 = −jπ[δ(ω − ω0 ) − δ(ω + ω0 )] Thus, the frequency content of the cosine and the sine is concentrated at the frequency ω0 . Although the sinusoids are infiniteenergy signals they have finite power and their spectra can be measured with a spectrum analyzer, which displays how the power is distributed over the frequencies. n
10.2.7 Symmetry When plotting or displaying the spectrum of a realvalued discretetime signal it is important to know that it is only necessary to show the magnitude and the phase spectra for frequencies [0 π], since the magnitude and the phase of X(e jω ) are even and odd functions of ω, respectively. This can be shown by considering a realvalued discretetime signal x[n], with inverse DTFT given by 1 x[n] = 2π
Zπ
X(e jω )e jωn dω
−π
and its complex conjugate is 1 x [n] = 2π ∗
Zπ
jω
X (e )e ∗
−jωn
1 dω = 2π
−π
Zπ
0
0
X∗ (e−jω )e jω n dω0
−π
Since x[n] = x∗ [n], as x[n] is real, comparing the above integrals we have that X(e jω ) = X∗ (e−jω ) X(e jω )e jθ (ω) = X(e−jω )e−jθ (−ω) Re[X(e jω )] + jIm[X(e jω )] = Re[X(e−jω )] − jIm[X(e−jω )] or that the magnitude is an even function of ω—that is, X(e jω ) = X(e−jω )
(10.20)
and that the phase is an odd function of ω, or θ(ω) = −θ (−ω)
(10.21)
Likewise, the real and the imaginary parts of X(e jω ) are also even and odd functions of ω: Re[X(e jω )] = Re[X(e−jω )] Im[X(e jω )] = −Im[X(e−jω )]
(10.22)
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n Example 10.8 For the signal x[n] = α n u[n], 0 < α < 1, find the magnitude and the phase of its DTFT X(e jω ). Solution The DTFT of x[n] is X(e jω ) =
1 1 jω = z=e −1 1 − αz 1 − αe−jω
since the Ztransform has a region of convergence z > α that includes the unit circle. Its magnitude is 1 X(e jω ) = p (1 − α cos(ω))2 + α 2 sin2 (ω) which is an even function of ω given that cos(ω) = cos(−ω) and sin2 (−ω) = (− sin(ω))2 = sin2 (ω). The phase is given by α sin(ω) θ(ω) = − tan−1 1 − α cos(ω) which is an odd function of ω. As functions of ω, the numerator is odd and the denominator is even, so that the argument of the inverse tangent is odd, which is in turn odd. n
n Example 10.9 For a discretetime signal x[n] = cos(ω0 n + φ)
−π ≤φ 0 for −π ≤ ω < π , the phase is ∠H2 (e jω ) = −4ω, which is a line through the origin with slope −4 (i.e., a linear phase) The following script is used to compute the frequency responses of the two filters using fft, their wrapped phases using angle, and then unwrapping them using unwrap. Figure 10.7 displays the magnitude responses of the two filters, as well as their wrapped and unwrapped phases.
0
jω
ω
N). On the other hand, if the length L < N the first period of y˜ [n] does not coincide with y[n] because of superposition of shifted versions of it (this corresponds to time aliasing, the dual of frequency aliasing, which occurs in time sampling). Assuming y[n] is of finite length N and that L ≥ N, as the dual of sampling in time we then have that y˜ [n] =
∞ X
y[n + rL] ⇔ Y[k] = Y(e j2πk/L ) =
r=−∞
N−1 X
y[n]e−j2πnk/L
k = 0, . . . , L − 1
(10.48)
n=0
The equation on the right is the DFT of y[n]. The inverse DFT is the Fourier series representation of y˜ [n] (normalized with respect to L) or its first period L−1
y[n] =
1X Y[k]e j2πnk/L L
0≤n≤L−1
(10.49)
k=0
where Y[k] = Y(e j2πk/L ). Thus, instead of the frequency aliasing that samplingintime causes, we have timealiasing whenever the length N of y[n] is greater than the chosen L in the samplinginfrequency. In practice, the generation of the periodic extension y˜ [n] is not needed—we just need to generate a period that either coincides with y[n] when L = N, or when L > N that coincides with y[n] with a sequence of L − N zeros attached to it (i.e., y[n] is padded with zeros). To avoid time aliasing we do not consider choosing L < N. If the signal y[n] is a very long signal, in particular if N → ∞, it does not make sense to compute its DFT, even if we could. Such a DFT would give the frequency content of the whole signal and since an infinitelength signal could have all types of frequencies its DFT would just give no valuable information. A possible approach to obtain, over time, the frequency content of a signal with a large time support is to window it and compute the DFT of each of these segments. Thus, when y[n] is of infinite length, or its length is much larger than the desired or feasible length L, we use a window WL [n] of length L, and represent y[n] as the superposition X y[n] = ym [n] where ym [n] = y[n]WL [n − mL] (10.50) m
10.4 Discrete Fourier Transform
Therefore, by the linearity of the DFT, we have the DFT of y[n] is X X Y[k] = DFT(ym [n]) = Ym [k] m
(10.51)
m
where each Ym [k] provides a frequency characterization of the windowed signal or the local frequency content of the signal. Practically, this would be more meaningful than finding the DFT of the whole signal. Now we have frequency information corresponding to segments of the signal and possibly evolving over time. The DFT of an aperiodic signal x[n] of finite length N is found as follows: Choose an integer L ≥ N that is the length of the DFT to be the period of a periodic extension x˜ [n] having x[n] as a period with padded zeros if necessary. n Find the normalized Fourier series representation of x˜ [n],
n
x˜ [n] =
L−1 1X˜ X[k]ej2πnk/L L
0≤n≤L−1
(10.52)
0≤k≤L−1
(10.53)
k=0
where ˜ X[k] =
L−1 X
x˜ [n]e−j2πnk/L
n=0 n
Then, ˜ n X[k] = X[k] for 0 ≤ k ≤ L − 1 is the DFT of x[n]. n x[n] = x˜ [n]W[n] where W[n] = u[n] − u[n − L] is a rectangular window of length N, is the IDFT of X[k]. The IDFT x[n] is defined for 0 ≤ n ≤ L − 1.
10.4.3 Computation of the DFT via the FFT Although we now have discrete frequencies and use only summations to compute the direct and the inverse DFT, there are still several issues that should be understood when computing these transforms. Assuming that the given signal is finite length, or it is made finite length by windowing, we have: n
n
Efficient computation with the FFT algorithm: A very efficient computation of the DFT is done by means of the FFT algorithm, which takes advantage of some special characteristics of the DFT, as we will discuss in Chapter 12. It should be understood that the FFT is not another transformation but an algorithm to efficiently compute DFTs. For now, we will consider the FFT as a black box that for an input x[n] (or X[k]) gives as output the DFT X[k] (or IDFT x[n]). Causal aperiodic signals: If the given signal x[n] is causal of length N—that is, the samples {x[n], n = 0, 1, . . . , N − 1} are given—we can proceed to obtain {X[k], k = 0, 1, . . . , N − 1} or the DFT of x[n] by means of an FFT of length L = N. To compute an L > N DFT we simply attach L − N zeros at the end of the
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n
above sequence and obtain L values corresponding to the DFT of x[n] of length L (why this could be seen as a better version of the DFT of x[n] is discussed below in frequency resolution). Noncausal aperiodic signals: When the given signal x[n] is noncausal of length N—that is, the samples {x[n], n = −n0 , . . . , 0, 1, . . . , N − n0 − 1} are given—we need to recall that a periodic extension of x[n] or x˜ [n] was used to obtain its DFT. This means that we need to create a sequence of N values corresponding to the first period of x˜ [n]—that is, x[0] x[1] · · · x[N − n0 − 1] x[−n0 ] x[−n0 + 1] · · · x[−1]  {z }  {z } causal samples noncausal samples where as indicated the samples x[−n0 ] x[−n0 + 1] · · · x[−1] are the values that make x[n] noncausal. If we wish to consider zeros after x[N − n0 − 1] to be part of the signal, so as to obtain a better DFT transform as we discuss later in frequency resolution, we simply attach zeros between the causal and noncausal components—that is, x[0] x[1] · · · x[N − n0 − 1] 0 0 · · · 0 0 x[−n0 ] x[−n0 + 1] · · · x[−1]  {z }  {z } causal samples noncausal samples
n
n
to compute an L > N DFT of the noncausal signal. The periodic extension x˜ [n] represented circularly instead of linearly would clearly show the above sequence. Periodic signals: If the signal x[n] is periodic of period N we will then choose L = N (or a multiple of N) and calculate the DFT X[k] by means of the FFT algorithm. If we use a multiple of the period (e.g., L = MN for some integer M > 0), we need to divide the obtained DFT by the value M. For periodic signals we cannot choose L to be anything but a multiple of N as we are really computing the Fourier series of the signal. Likewise, no zeros can be attached to a period (or periods when M > 1) to improve the frequency resolution of its DFT—by attaching zeros to a period we distort the signal. Frequency resolution: When the signal x[n] is periodic of period N, the DFT values are normalized Fourier series coefficients of x[n] that only exist for the harmonic frequencies {2πk/N}, as no frequency components exist for any other frequencies. On the other hand, when x[n] is aperiodic, the number of possible frequencies depend on the length L chosen to compute its DFT. In either case, the frequencies at which we compute the DFT can be seen as frequencies around the unit circle in the zplane. In both cases one would like to have a significant number of frequencies in the unit circle so as to visualize the frequency content of the signal well. The number of frequencies considered is related to the frequency resolution of the DFT of the signal. n If the signal is aperiodic we can improve the frequency resolution of its DFT by increasing the number of samples in the signal without distorting the signal. This can be done by padding the signal with zeros (i.e., attaching zeros to the end of the signal). These zeros do not change the frequency content of the signal (they can be considered part of the aperiodic signal) but permit us to increase the available frequency components of the signal.
10.4 Discrete Fourier Transform
On the other hand, the harmonic frequencies of a periodic signal of period N are fixed to 2πk/N for 0 ≤ k < N. In such a case we cannot pad the given period of the signal with an arbitrary number of zeros, because such zeros are not part of the periodic signal. As an alternative, to increase the frequency resolution of a periodic signal we consider more periods which give the same harmonic frequencies as for one period, but add zeros in between the harmonic frequencies when considering more than one period. Frequency scales: When computing the DFT we obtain a sequence of complex values X[k] for k = 0, 1, . . . , N − 1 corresponding to an input signal x[n] of length N. Since each of the k values corresponds to a discrete frequency 2πk/N the k = 0, 1, . . . , N − 1 scale is converted into a discrete frequency scale [0 2π(N − 1)/N] (rad) (the last value is always smaller than 2π to keep the periodicity in frequency of X[k]) by multiplying the integer scale {0 ≤ k ≤ N − 1} by 2π/N. Subtracting π to this frequency scale we obtain discrete frequencies [−π π − 2π/N] (rad) where again the last frequency does not coincide with π in order to keep the periodicity of 2π of the X[k]. We obtain a normalized discretefrequency scale by dividing the above scale by π so as to obtain a nonunits normalized scale of [0 1 − 2(N − 1)/N] or [−1 1 − 2/N]. Finally, if the signal is the result of sampling and we wish to display the analog frequency, we then use the relation where Ts is the sampling period and fs is the sampling frequency: n
n
=
ω ω ωfs (Hz) = ωfs (rad/sec) or f = = Ts 2πTs 2π
(10.54)
giving analog scales [0 πfs ] (rad/sec) and [0 fs /2] (Hz).
n Example 10.20 Consider the DFT computation via the FFT of a causal signal x[n] = (sin(πn/32))(u[n] − u[n − 34]) and of its advanced version x[n + 16]. To improve its frequency resolution compute FFTs of length N = 512. Explain the difference between computing the FFTs of the causal and the noncausal signals. Solution As indicated above, when computing the FFT of a causal signal the signal is simply inputted into the function. However, to improve the frequency resolution of the FFT we attach zeros to the signal. These zeros make it possible to have additional values of the frequency response of the signal, with no effect on the frequency content of the signal. For the noncausal signal x[n + 16], we need to recall that the DFTs of an aperiodic signal were computed by extending the signal into a periodic signal with an arbitrary period N, which exceeds the length of the signal. Thus, the periodic extension of x[n + 16] can be obtained by creating an input vector consisting of x[n], n = 0, . . . , 16; N − 33 zeros (N being the length of the FFT and 33 the length of the signal) to improve the frequency resolution, and x[n], n = −16, . . . , −1. In either case, the output of the FFT is available as an array of length N = 512 values. This array X[k], k = 0, . . . , N − 1 can be understood as values of the spectrum at frequencies 2πk/N—that is,
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from 0 to 2π(N − 1)/N rad (notice that this value is close to 2π but always smaller than 2π as needed to display a period of X[k]). We can change this scale into other frequency scales—for instance, if we wish a scale that considers positive as well as negative frequencies, to the above scale we subtract π, and if we wish a normalized scale [−1 1), we simply divide the previous scale by π. When shifting to a [−π π) or [−1 1) frequency scale, the spectrum also needs to be shifted accordingly—this is done using the fftshift function. To understand this change recall that X[k] is also periodic of period N. The following script is used to compute the DFT of x[n] and x[n + 16] given above. The results are shown in Figure 10.14. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Example 10.20FFFT computation of causal and % noncausal signals %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clf N = 512; % order of the FFT n = 0:N  1; % causal signal x = [ones(1,33) zeros(1,N  33)]; x = x. ∗ sin(pi ∗ n/32); % zeropadding X = fft(x); X = fftshift(X); % fft and its shifting to [1 1] frequency scale w = 2 ∗ [0:N  1]./N  1; % normalized frequencies n1 = [9:40]; % time scale % noncausal signal xnc = [zeros(1,3) x(1:33) zeros(1,3)]; % noncausal signal x = [x(17:33) zeros(1,N33) x(1:16)]; % periodic extension and zeropadding X = fft(x); X = fftshift(X); n1 = [19:19]; % time scale
n
n Example 10.21 Consider improving the frequency resolution of a periodic sampled signal y(nTs ) = 4 cos(2πf0 nTs ) − cos(2πf1 nTs )
f0 = 100 Hz, f1 = 4f0
where the sampling period is Ts = 1/(3f1 ) sec/sample. Solution In the case of a periodic signal, the frequency resolution of its FFT cannot be improved by attaching zeros. The length of the FFT must be the period or a multiple of the period of the signal. The following script illustrates how the FFT of the given periodic signal can be obtained by using 4 or 12 periods. As the number of periods increases the harmonic components appear in each case at exactly the same frequencies, and only zeros in between these fixed harmonic frequencies result from increasing the number of periods. However, the magnitude frequency response is increasing
10.4 Discrete Fourier Transform
x [n]
1 0.5 0 −10
−5
0
5
10
15 n
10
<X(e jω )
X (e jω )
25
30
35
40
4
15
5 0 −1
20
−0.5
0 ω /π
0.5
2 0 −2 −4 −1
1
−0.5
0 ω /π
0.5
1
(a)
x1 [n]
1
0.5
0 −20
−10
−15
−5
0 n
15
20
2 <X1 (e jω )
X1 (e jω )
20 15 10
0 −2
5 0 −1
10
4
25
FIGURE 10.14 Computation of the FFT of (a) a causal signal and (b) a noncausal signal. Notice that as expected the magnitude responses are equal—only the phase responses change.
5
−0.5
0
0.5
−4 −1
1
ω /π
−0.5
0 ω /π
0.5
1
(b)
as the number of periods increases. Thus, we need to divide by the number of periods used in computing the FFT. Since the signal is sampled, it is of interest to have the frequency scale of the FFTs in hertz, so we convert the discrete frequency ω (rad) into f (Hz) according to f =
ω ωfs = 2πTs 2π
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where fs = 1/Ts is the sampling rate given in samples/second. The results are shown in Figure 10.15. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Example 10.21Improving frequency resolution of FFFT of periodic signals %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% f0 = 100; f1 = 4 ∗ f0; % frequencies in Hz of signal Ts = 1/(3 ∗ f1); % sampling period t = 0:Ts:4/f0; % time for 4 periods y = 4 ∗ cos(2 ∗ pi ∗ f0 ∗ t)  cos(2 ∗ pi ∗ f1 ∗ t); % sampled signal (4 periods) M = length(y); Y = fft(y,M); Y = fftshift(Y)/4; % fft using 4 periods, shifting and normalizing t1 = 0:Ts:12/f0; % time for 12 periods y1 = 4 ∗ cos(2 ∗ pi ∗ f0 ∗ t1)  cos(2 ∗ pi ∗ f1 ∗ t1); % sampled signal (12 periods) Y1 = fft(y1);Y1 = fftshift(Y1)/12; % fft using 12 periods, shifting and normalizing w = 2 ∗ [0:M  1]./M  1;f = w/(2 ∗ Ts); % frequency scale (4 periods) N = length(y1); w1 = 2 ∗ [0:N  1]./N  1;f = w/(2 ∗ Ts); % frequency scale (12 periods)
n
10.4.4 Linear and Circular Convolution Sums The most important property of the DFT is the convolution property, which permits the computation of the linear convolution sum very efficiently by means of the FFT. Consider the convolution sum that gives the output y[n] of a discretetime LTI system with impulse response h[n] and input x[n]: X x[m]h[n − m] y[n] = m
In frequency, y[n] is the inverse DTFT of the product Y(e jω ) = X(e jω )H(e jω ) Assuming that x[n] has a finite length M and that h[n] has a finite length K, then y[n] has a finite length N = M + K − 1. If we choose a period L ≥ N for the periodic extension y˜ [n] of y[n], we would obtain the frequencysampled periodic sequence Y(e jω )ω=2πk/L = X(e jω )H(e jω )ω=2πk/L or the DFT of y[n] as the product of the DFTs of x[n] and h[n]: Y[k] = X[k]H[k] for k = 0, 1, . . . , L − 1 We then obtain y[n] as the inverse DFT of Y[k]. It should be noticed that the Llength DFT of x[n] and of h[n] requires that we pad x[n] with L − M zeros and h[n] with L − K zeros, so that both X[k] and H[k] have the same length L and can be multiplied at each k.
10.4 Discrete Fourier Transform
4 x (nTs)
2 0 −2 −4 −6
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
nTs 25
4 2
15
<X (f )
X (f )
20
10
0 −2
5 0 −600 −400 −200
0
−4 −600 −400 −200
200 400 600
f (Hz)
0 200 400 600 f (Hz)
(a) 4 x (nTs)
2 0 −2 −4 −6
0
0.02
0.04
0.06
0.08
0.1
0.12
nTs 25
4
20
2
15
<X (f )
X (f )
FIGURE 10.15 Computation of the FFT of a periodic signal using (a) 4 and (b) 12 periods to improve the frequency resolution of the FFT. Notice that both magnitude and phase responses look alike, but when we use 12 periods these spectra look sharper due to the increase in the number of frequency components added.
10
−2
5 0 −600 −400 −200
0
0 200 400 600 f (Hz)
−4 −600 −400 −200
(b)
0 200 400 600 f (Hz)
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Given x[n] and h[n] of lengths M and K, the linear convolution sum y[n] of length N = M + K − 1 can be found by following these three steps: n Compute DFTs X[k] and H[k] of length L ≥ N for x[n] and h[n]. n Multiply them to get Y[k] = X[k]H[k]. n Find the inverse DFT of Y[k] of length L to obtain y[n]. Although it seems computationally more expensive than performing the direct computation of the convolution sum, the above approach implemented with the FFT can be shown to be much more efficient.
The above procedure could be implemented by a circular convolution sum in the time domain, although in practice it is not done due to the efficiency of the implementation with FFTs. A circular convolution uses circular rather than linear representation of the signals being convolved. The periodic convolution sum introduced before is a circular convolution of fixed length—the period of the signals being convolved. When we use the DFT to compute the response of an LTI system the length of the circular convolution is given by the possible length of the linear convolution sum. Thus, if the system input is a finite sequence x[n] of length M and the impulse response of the system h[n] has a length K, then the output y[n] is given by a linear convolution of length M + K − 1. The length L ≥ M + K − 1 of the DFT Y[k] = X[k]H[k] corresponds to a circular convolution of length L of the x[n] and h[n] padded with zeros so that both have length L. In such a case the circular and the linear convolutions coincide. If x[n] of length M is the input of an LTI system with impulse response h[n] of length K, then Y[k] = X[k]H[k] ⇔ y[n] = (x ⊗L h)[n]
(10.55)
where X[k], H[k], and Y[k] are, respectively, DFTs of length L of the input, the impulse response, and the output of the LTI system, and ⊗L stands for the circular convolution of length L. If L is chosen so that L ≥ M + K − 1, the circular and the linear convolution sums coincide—that is, y[n] = (x ⊗L h)[n] = (x ∗ h)[n]
(10.56)
Remark If we consider the periodic expansions of x[n] and h[n] with period L = M + K − 1, we can use their circular representations and implement the circular convolution as shown in Figure 10.16. Since the length of the linear convolution or convolution sum, M + K − 1, coincides with the length of the circular convolution, the two convolutions coincide. Given the efficiency of the FFT algorithm in computing the DFT, the convolution is typically done using the DFT as indicated above.
n Example 10.22 To illustrate the connection between the circular and the linear convolution, compute using MATLAB the circular convolution of a pulse signal x[n] = u[n] − u[n − 21] of length N = 20 with itself for different values of its length. Determine the length for which the circular convolution coincides with the linear convolution of x[n] with itself.
10.4 Discrete Fourier Transform
n =0
y [2] y [3]
y [1] x [6] x [7]
FIGURE 10.16 Circular convolution of length L = 8 of x[n] and y[n]. The signal x[k] is stationary with a circular representation given by the inside circle, while y[n − k] is represented by the outside circle and rotated in the clockwise direction. The shown circular convolution sum corresponds to n = 0.
y [4]
x [5]
x [0]
x [4]
x [1] x [3] x [2]
y [5]
y [7] y [6] Circular convolution (L = 20)
20
20
15
15 y (n)
z (n)
Linear convolution
10
10 5
5 0
y [0]
0
10
20
30
0
40
0
10
20
(a)
20
15
15 y2(n)
20
10 5 0
40
Circular convolution (L = 49)
Circular convolution (L = 30)
y1(n)
FIGURE 10.17 Circular versus linear convolutions: (a) Plot corresponds to linear convolution. (b) and (c) Plots are circular convolutions wih L < 2N − 1. (d) Plot is circular convolution with L > 2N − 1 coinciding with the linear convolution.
30 (b)
10 5
0
10
20
30 n (c)
40
0
0
10
20
30
40
n (d)
Solution We know that the length of the linear convolution z[n] = (x ∗ x)[n] is N + N − 1 = 39. If we use the function circonv2 shown below to compute the circular convolution of x[n] with itself with length N < 2N − 1, for instance L = 20 as shown in Figure 10.17(b), the result will not equal the linear convolution. Likewise, if the circular convolution is of length N + 10 = 30 < 2N − 1, only part of the result resembles the linear convolution (see Figure 10.17(c)). If we let the length of the circular convolution be 2N + 9 = 49 > 2N − 1, the result is identical to the linear convolution (see Figure 10.17(d)). The script is given as follows.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Example 10.22Linear and circular convolution %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clf N = 20; x = ones(1,N); % linear convolution z = conv(x,x);z = [z zeros(1,10)]; % circular convolution y = circonv2(x,x,N); y1 = circonv2(x,x,N + 10); y2 = circonv2(x,x,2 ∗ N + 9); Mz = length(z); My = length(y); My1 = length(y1);My2 = length(y2); y = [y zeros(1,Mz  My)]; y1 = [y1 zeros(1,Mz  My1)]; y2 = [y2 zeros(1,MzMy2)];
The function circonv2 has as inputs the signals to be convolved and the desired length of the circular convolution. It computes and multiplies the FFTs of the signals and then finds the inverse FFT to obtain the circular convolution. If the desired length of the circular convolution is larger than the length of each of the signals, the signals are padded with zeros to make them the length of the circular convolution. The following is the code for this function. function xy = circonv2(x,y,N) M = max(length(x),length(y)) if M>N disp(‘Increase N’) end x = [x zeros(1,N  M)]; y = [y zeros(1,N  M)]; % circular convolution X = fft(x,N); Y = fft(y,N); XY = X. ∗ Y; xy = real(ifft(XY,N));
n
n Example 10.23 A significant advantage of using the FFT for computing the DFT is in filtering. Assume that the signal to filter consists of the MATLAB file “laughter.mat,” multiplied by 5, to which a signal that continuously changes between −0.3 and 0.3 is added. We wish to recover the original “laughter.mat” signal using a filter. Use the MATLAB function fir1 to design the required filter. Solution Noticing that since the disturbance 0.3(−1)n is a signal of frequency π, we need a lowpass filter with a wide bandwidth so as to get rid of the disturbance while trying to keep the frequency components of the desired signal. The following script is used to design the desired lowpass filter, and to implement the filtering. To compare the results obtained with the FFT we use the function conv to find the output of the filter in the time domain. The results are shown in Figure 10.18.
10.4 Discrete Fourier Transform
x [n], x1[n]
1 0.5 0 −0.5 −1
0
10
20
30
40
50 n (a)
60
70
80
90
100
1
y [n]
0.5 0 −0.5 −1
0
20
40
60
80
100
120
140
n
(b)
1
0.6
0.8
H (e jω )
0.8 0.4 0.2
0.4 0.2
0 −0.2
0.6
0
10
20 n
30
0
40
0
1
ω
2
3
120
140
(c) 4
×10−16
2 ε [n]
FIGURE 10.18 FIR filtering of disturbed signal. Comparison of results using conv and fft functions. (a) Actual and noisy signal (plotted as a continuous signal using black lines); (b) denoised signal (notice the delay caused by the FIR filter); (c) impulse response and magnitude response of lowpass FIR filter; and (d) error signal [n] = y[n] − y1 [n] between output from conv and fftbased filtering.
h [n]
1
0 −2
0
20
40
60
80 n
(d)
100
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Example 10.23Filtering using convolution and FFT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clf N = 100; n = 0:N  1; load laughter x = 5 ∗ y(1:N)’; x1 = x + 0.3 ∗ (1).ˆn; % desired signal plus disturbance h = fir1(40,0.9); [H,w] = freqz(h,1); % lowpass FIR filter design % filtering using convolution y = conv(x,h); % convolution of signal and impulse response of FIR % computing using FFT M = length(x) + length(h)  1; % circular and linear convolutions equal X = fft(x,M); H = fft(h,M); Y = X. ∗ H; y1 = ifft(Y); % output of filtering
n
10.5 WHAT HAVE WE ACCOMPLISHED? WHERE DO WE GO FROM HERE? In this chapter we have considered the Fourier representation of discretetime signals and systems. Just as with the Laplace and the Fourier transforms, in the continuous case there is a large class of discretetime signals and impulse responses of systems for which we are able to find their discretetime Fourier transform from their Ztransforms. For signals that are not absolutely summable, the timefrequency duality and other properties of the transform are used to find their DTFTs. Properties of the DTFT are very similar to those of the Ztransform. Although theoretically useful, the DTFT is computationally not feasible, due to the continuity of the frequency variable and to the integration required in the inverse transformation. It is the Fourier series of discretetime signals that makes the Fourier representation computationally feasible. In Table 10.1, the DTFT of common signals and some DTFT properties are given. The Fourier series coefficients constitute a periodic sequence of the same period as the signal; thus both are periodic. Moreover, the Fourier series and its coefficients are obtained as sums, and the frequency used is discretized. Thus, they can be obtained by computer. To take advantage of this, the spectrum of an aperiodic signal resulting from the DTFT is sampled so that in the time domain there is a periodic repetition of the original signal. For finitesupport signals we can then obtain a periodic extension that gives the discrete Fourier transform or DFT. The significance of this result is that we have frequency representations of discretetime signals that are computed algorithmically. Table 10.2 displays properties of the discrete Fourier series and of the Discrete Fourier Transform (DFT). What remains then is to take a look at the algorithm used for those computations or the fast Fourier transform (FFT). We will do that in Chapter 12, where we will show that this algorithm efficiently computes the DFT and makes the convolution sum a more feasible procedure.
Problems
Table 10.1 DTFT of Common Signals and DTFT Properties DiscreteTime Fourier Transforms DiscreteTime Signal
DTFT X(e jω ), Periodic of Period 2π
1.
δ[n]
1, − π ≤ ω < π
2.
A
2π Aδ(ω), − π ≤ ω < π
3.
e jω0 n
2πδ(w − ω0 ), − π ≤ ω < π
4.
α n u[n], α < 1
5.
nα n u[n], α < 1
1 , −π ≤ω 1−αe−jω −jω αe , −π ≤ (1−αe−jω )2
6.
cos(ω0 n)u[n]
π [δ(ω − ω0 ) + δ(ω + ω0 )] , − π ≤ ω < π
7.
sin(ω0 n)u[n]
−jπ [δ(ω − ω0 ) + δ(ω + ω0 )] , − π ≤ ω < π
8.
α n , α < 1
1−α 2 , 1−2α cos(ω)+α 2
9.
u[n + N/2] − u[n − N/2]
sin(ω(N+1)/2) , sin(ω/2)
10.
α n cos(ω0 n)u[n]
11.
α n sin(ω0 n)u[n]
1−α cos(ω0 )e−jω , 1−2α cos(ω0 )e−jω +α 2 e−2jω −jω α sin(ω0 )e , 1−2α cos(ω0 )e−jω +α 2 e−2jω