(k.Deergha Rao) signals and systems.pdf

K. Deergha Rao

Signals and Systems

K. Deergha Rao

Signals and Systems

K. Deergha Rao Department of Electronics and Communication Engineering Vasavi College of Engineering (Afﬁliated to Osmania University) Hyderabad, Telangana, India

ISBN 978-3-319-68674-5 https://doi.org/10.1007/978-3-319-68675-2

ISBN 978-3-319-68675-2

(eBook)

Library of Congress Control Number: 2017958547

Mathematics Subject Classiﬁcation (2010): 94A12; 94A05; 93C55; 93C20; 35Q93

© Springer International Publishing AG, part of Springer Nature 2018 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional afﬁliations.

trademarks, service marks, etc.

Printed on acid-free paper

This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer International Publishing AG part of Springer Nature. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To My Parents Dalamma and Boddu, My Beloved Wife Sarojini, and My Mentor Prof. M.N.S. Swamy

Preface

The signals and systems course is not only an important element for undergraduate electrical engineering students but the fundamentals and techniques of the subject are essential in all the disciplines of engineering. Signals and systems analysis has a long history, with its techniques and fundamentals found in broad areas of applica- tions. The signals and systems is continuously evolving and developing in response to new problems, such as the development of integrated circuits technology and its applications.

In this book, many illustrative examples are included in each chapter for easy understanding of the fundamentals and methodologies of signals and systems. An attractive feature of this book is the inclusion of MATLAB-based examples with codes to encourage readers to implement exercises on their personal computers in order to become conﬁdent with the fundamentals and to gain more insight into signals and systems. In addition to the problems that require analytical solutions, MATLAB exercises are introduced to the reader at the end of some chapters.

This book is divided into 8 chapters. Chapter 1 presents an introduction to signals and systems with basic classiﬁcation of signals, elementary operations on signals, and some real-world examples of signals and systems. Chapter 2 gives time-domain analysis of continuous time signals and systems, and state-space representation of continuous-time LTI systems. Fourier analysis of continuous-time signals and sys- tems is covered in Chapter 3. Chapter 4 deals with the Laplace transform and analysis of continuous-time signals and systems, and solution of state-space equa- tions of continuous-time LTI systems using Laplace transform. Ideal continuous- time (analog) ﬁlters, practical analog ﬁlter approximations and design methodolo- gies, and design of special class ﬁlters based on pole-zero placement are discussed in Chapter 5. Chapter 6 discusses the time-domain representation of discrete-time signals and systems, linear time-invariant (LTI) discrete-time systems and their properties, characterization of discrete-time systems, and state-space representation of discrete-time LTI systems. Representation of discrete-time signals and systems in frequency domain, representation of sampling in frequency domain, reconstruction of a band-limited signal from its samples, and sampling of discrete-time signals are

vii

viii

Preface

detailed in Chapter 7. Chapter 8 describes the z-transform and analysis of LTI discrete-time systems, the solution of state-space equations of discrete-time LTI systems using z-transform, and transformations between the continuous-time sys- tems and discrete-time systems.

The salient features of this book are as follows:

(cid:129) Provides introductory and comprehensive exposure to all aspects of signal and

systems with clarity and in an easy way to understand.

(cid:129) Provides an integrated treatment of continuous-time signals and systems and

discrete-time signals and systems.

(cid:129) Several fully worked numerical examples are provided to help students under-

stand the fundamentals of signals and systems.

(cid:129) PC-based MATLAB m-ﬁles for the illustrative examples are included in

this book.

This book is written at introductory level for undergraduate classes in electrical engineering and applied sciences that are the prerequisite for upper level courses, such as communication systems, digital signal processing, and control systems.

Hyderabad, India

K. Deergha Rao

Contents

1.4

Introduction … … … … … … … … … … … … … … . 1.1 What is a Signal? … … … … … … … … … … … . . 1.2 What is a System? … … … … … … … … … … … . Elementary Operations on Signals … … … … … … … . . 1.3 Time Shifting … … … … … … … … … … . 1.3.1 Time Scaling … … … … … … … … … … . . 1.3.2 Time Reversal … … … … … … … … … … . 1.3.3 Classiﬁcation of Signals … … … … … … … … … … Continuous-Time and Discrete-Time Signals … … … 1.4.1 Analog and Digital Signals … … … … … … … . 1.4.2 Periodic and Aperiodic Signals … … … … … … . 1.4.3 Even and Odd Signals … … … … … … … … . 1.4.4 Causal, Noncausal, and Anticausal Signal … … … . . 1.4.5 Energy and Power Signals … … … … … … … . 1.4.6 Deterministic and Random Signals … … … … … . 1.4.7 Basic Continuous-Time Signals … … … … … … … … . The Unit Step Function … … … … … … … … 1.5.1 The Unit Impulse Function … … … … … … … . 1.5.2 The Ramp Function … … … … … … … … … 1.5.3 The Rectangular Pulse Function … … … … … … 1.5.4 The Signum Function … … … … … … … … . . 1.5.5 The Real Exponential Function … … … … … … . 1.5.6 The Complex Exponential Function … … … … … 1.5.7 The Sinc Function … … … … … … … … … . 1.5.8 Generation of Continuous-Time Signals Using MATLAB … … … … … … … … … … … . . Typical Signal Processing Operations … … … … … … … Correlation … … … … … … … … … … … 1.7.1 1.7.2 Filtering … … … … … … … … … … … . . 1.7.3 Modulation and Demodulation … … … … … … .

1.5

1.6

1.7

1 1 1 1 2 2 3 5 5 5 6 9 12 13 20 20 20 21 22 22 23 23 24 24

28 30 30 31 31

Contents

1.8

1.7.4 Transformation … … … … … … … … … … 1.7.5 Multiplexing and Demultiplexing … … … … … . . Some Examples of Real-World Signals and Systems … … … . Audio Recording System … … … … … … … . . 1.8.1 Global Positioning System … … … … … … … . 1.8.2 Location-Based Mobile Emergency 1.8.3 Services System … … … … … … … … … … Heart Monitoring System … … … … … … … . . 1.8.4 1.8.5 Human Visual System … … … … … … … … . 1.8.6 Magnetic Resonance Imaging … … … … … … . . 1.9 Problems … … … … … … … … … … … … … . . 1.10 MATLAB Exercises … … … … … … … … … … … Further Reading … … … … … … … … … … … … … . .

2.1 2.2

2 Continuous-Time Signals and Systems … … … … … … … . . The Representation of Signals in Terms of Impulses … … … . Continuous-Time Systems … … … … … … … … … . . Linear Systems … … … … … … … … … … 2.2.1 Time-Invariant System … … … … … … … … . 2.2.2 Causal System … … … … … … … … … … . 2.2.3 Stable System … … … … … … … … … … . 2.2.4 2.2.5 Memory and Memoryless System … … … … … . . Invertible System … … … … … … … … … . . 2.2.6 2.2.7 Step and Impulse Responses … … … … … … … The Convolution Integral … … … … … … … … … … Some Properties of the Convolution Integral … … … 2.3.1 Graphical Convolution … … … … … … … … . 2.3.2 Computation of Convolution Integral 2.3.3 Using MATLAB … … … … … … … … … . . Interconnected Systems … … … … … … … … Periodic Convolution … … … … … … … … . .

2.3

2.3.4 2.3.5 Properties of Linear Time-Invariant Continuous-Time System … … … … … … … … … … … … … … . LTI Systems With and Without Memory … … … … 2.4.1 Causality for LTI Systems … … … … … … … . 2.4.2 Stability for LTI Systems … … … … … … … . . 2.4.3 2.4.4 Invertible LTI System … … … … … … … … . . Systems Described by Differential Equations … … … … … Linear Constant-Coefﬁcient Differential Equations … . . 2.5.1 The General Solution of Differential Equation … … . . 2.5.2 Linearity … … … … … … … … … … … . . 2.5.3 Causality … … … … … … … … … … … . . 2.5.4 Time-Invariance … … … … … … … … … … 2.5.5 Impulse Response … … … … … … … … … . 2.5.6 Solution of Differential Equations Using 2.5.7 MATLAB … … … … … … … … … … … .

2.4

2.5

31 32 32 32 33

33 34 36 36 37 39 40

41 41 42 42 43 48 49 49 49 49 49 50 58

70 74 76

77 77 77 77 79 82 82 85 86 86 87 88

Contents

2.5.8

Determining Impulse Response and Step Response for a Linear System Described by a Differential Equation Using MATLAB … … … …

2.6

93 95

2.7 2.8

Block-Diagram Representations of LTI Systems Described by Differential Equations … … … … … … … . Singularity Functions … … … … … … … … … … . . State-Space Representation of Continuous-Time LTI Systems … … … … … … … … … … … … … State and State Variables … … … … … … … . . 2.8.1 State-Space Representation of Single-Input 2.8.2 Single-Output Continuous-Time LTI Systems … … . . State-Space Representation of Multi-input Multi-output Continuous-Time LTI Systems … … … 104 2.9 Problems … … … … … … … … … … … … … . . 105 2.10 MATLAB Exercises … … … … … … … … … … … 109 Further Reading … … … … … … … … … … … … … . . 110

98 98

2.8.3

3 Frequency Domain Analysis of Continuous-Time

Signals and Systems … … … … … … … … … … … … . 111 3.1 Complex Exponential Fourier Series Representation

of the Continuous-Time Periodic Signals … … … … … … . 111 Convergence of Fourier Series … … … … … … . . 113 3.1.1 Properties of Fourier Series … … … … … … … . . 113 3.1.2 3.2 Trigonometric Fourier Series Representation … … … … … . 128

3.2.1

Symmetry Conditions in Trigonometric Fourier Series … … … … … … … … … … . . 129

3.3 The Continuous Fourier Transform for Nonperiodic

3.3.3

Signals … … … … … … … … … … … … … … . . 133 Convergence of Fourier Transforms … … … … … . 135 3.3.1 Fourier Transforms of Some Commonly Used 3.3.2 Continuous-Time Signals … … … … … … … … 136 Properties of the Continuous-Time Fourier Transform … … … … … … … … … … … . . 139 3.4 The Frequency Response of Continuous-Time Systems … … … 159 Distortion During Transmission … … … … … … . 160 3.5 Some Communication Application Examples … … … … … . 162 Amplitude Modulation (AM) and Demodulation Amplitude Modulation … … … … … … … … . . 162 Single-Sideband (SSB) AM … … … … … … … . 164 Frequency Division Multiplexing (FDM) … … … … . 164 3.6 Problems … … … … … … … … … … … … … … 164 Further Reading … … … … … … … … … … … … … . . 170

3.5.2 3.5.3

3.5.1

3.4.1

4 Laplace Transforms … … … … … … … … … … … … . 171 The Laplace Transform … … … … … … … … … … . 171 Deﬁnition of Laplace Transform … … … … … … 171 4.1.1

4.1

xii

Contents

4.2 4.3 4.4

4.5 4.6

4.7

4.8

4.1.2 4.1.3 4.1.4

The Unilateral Laplace Transform … … … … … . . 172 Existence of Laplace Transforms … … … … … … 172 Relationship Between Laplace Transform and Fourier Transform … … … … … … … … . 172 4.1.5 Representation of Laplace Transform in the S-Plane … . 173 Properties of the Region of Convergence … … … … … … 174 The Inverse Laplace Transform … … … … … … … … . 176 Properties of the Laplace Transform … … … … … … … 178 4.4.1

Laplace Transform Properties of Even and Odd Functions … … … … … … … … … … . 182 Differentiation Property of the Unilateral Laplace Transform … … … … … … … … … . 183 Initial Value Theorem … … … … … … … … . . 186 4.4.3 4.4.4 Final Value Theorem … … … … … … … … . . 187 Laplace Transforms of Elementary Functions … … … … … 187 Computation of Inverse Laplace Transform Using Partial Fraction Expansion … … … … … … … … … . . 194 4.6.1

Partial Fraction Expansion of X(s) with Simple Poles … … … … … … … … … … … … . . 195 Partial Fraction Expansion of X(s) with Multiple Poles … … … … … … … … … … … … . . 195

4.6.2

4.4.2

Inverse Laplace Transform by Partial Fraction Expansion Using MATLAB … … … … … … … … … 201 Analysis of Continuous-Time LTI Systems Using the Laplace Transform … … … … … … … … … … . 202 Transfer Function … … … … … … … … … . . 202 4.8.1 Stability and Causality … … … … … … … … . 204 4.8.2 LTI Systems Characterized by Linear Constant 4.8.3 Coefﬁcient Differential Equations … … … … … . . 207 Solution of linear Differential Equations Using Laplace Transform … … … … … … … … … . 210 Solution of Linear Differential Equations Using Laplace Transform and MATLAB … … … … … . . 216 System Function for Interconnections of LTI Systems … … … … … … … … … … 217

4.8.4

4.8.5

4.8.6

4.9

Block-Diagram Representation of System Functions in the S-Domain … … … … … … … … … … … … 218

4.10 Solution of State-Space Equations Using Laplace

Transform … … … … … … … … … … … … … . 220 4.11 Problems … … … … … … … … … … … … … . . 222 4.12 MATLAB Exercises … … … … … … … … … … … 225 Further Reading … … … … … … … … … … … … … . . 225

Contents

xiii

5 Analog Filters … … … … … … … … … … … … … … 227 5.1 Ideal Analog Filters … … … … … … … … … … … . 227 5.2 Practical Analog Low-Pass Filter Design … … … … … … . 232 Filter Speciﬁcations … … … … … … … … … . 232 Butterworth Analog Low-Pass Filter … … … … … . 233 Chebyshev Analog Low-Pass Filter … … … … … . . 237 Elliptic Analog Low-Pass Filter … … … … … … . 245 Bessel Filter … … … … … … … … … … … 248 Comparison of Various Types of Analog Filters … … . . 249 Design of Analog High-Pass, Band-Pass, and Band-Stop Filters … … … … … … … … … 252 5.3 Effect of Poles and Zeros on Frequency Response … … … … 264

5.2.1 5.2.2 5.2.3 5.2.4 5.2.5 5.2.6 5.2.7

5.3.1

5.3.2

Effect of Two Complex System Poles on the Frequency Response … … … … … … … … . 264 Effect of Two Complex System Zeros on the Frequency Response … … … … … … … … . 264

5.4 Design of Specialized Analog Filters by Pole-Zero

Placement … … … … … … … … … … … … … . . 265 Notch Filter … … … … … … … … … … … . 266 5.4.1 5.5 Problems … … … … … … … … … … … … … … 267 Further Reading … … … … … … … … … … … … … . . 269

6.2

6.1

6 Discrete-Time Signals and Systems … … … … … … … … . . 271 The Sampling Process of Analog Signals … … … … … … 271 Impulse-Train Sampling … … … … … … … … 271 6.1.1 Sampling with a Zero-Order Hold … … … … … . . 272 6.1.2 6.1.3 Quantization and Coding … … … … … … … . . 274 Classiﬁcation of Discrete-Time Signals … … … … … … . 276 Symmetric and Anti-symmetric Signals … … … … . 276 6.2.1 Finite and Inﬁnite Length Sequences … … … … … 276 6.2.2 Right-Sided and Left-Sided Sequences … … … … . 277 6.2.3 Periodic and Aperiodic Signals … … … … … … . 277 6.2.4 6.2.5 Energy and Power Signals … … … … … … … . 279 Discrete-Time Systems … … … … … … … … … … . 281 Classiﬁcation of Discrete-Time Systems … … … … 282 6.3.1 6.3.2 Impulse and Step Responses … … … … … … … 286 Linear Time-Invariant Discrete-Time Systems … … … … … 286 Input-Output Relationship … … … … … … … . . 286 6.4.1 Computation of Linear Convolution … … … … … 288 6.4.2 Computation of Convolution Sum 6.4.3 Using MATLAB … … … … … … … … … . . 291 Some Properties of the Convolution Sum … … … … 291 Stability and Causality of LTI Systems in Terms of the Impulse Response … … … … … . . 295

6.4.4 6.4.5

6.4

6.3

Contents

xiv

6.5

6.6

6.7

Characterization of Discrete-Time Systems … … … … … . . 297 Non-Recursive Difference Equation … … … … … 298 6.5.1 Recursive Difference Equation … … … … … … . 298 6.5.2 Solution of Difference Equations … … … … … … 299 6.5.3 Computation of Impulse and Step Responses 6.5.4 Using MATLAB … … … … … … … … … . . 304 Sampling of Discrete-Time Signals … … … … … … … . 305 Discrete-Time Down Sampler … … … … … … . . 306 6.6.1 6.6.2 Discrete-Time Up-Sampler … … … … … … … . 306 State-Space Representation of Discrete-Time LTI Systems … . . 307 6.7.1

6.7.2

State-Space Representation of Single-Input Single-Output Discrete-Time LTI Systems … … … . . 307 State-Space Representation of Multi-input Multi-output Discrete-Time LTI Systems … … … … 309 6.8 Problems … … … … … … … … … … … … … . . 310 6.9 MATLAB Exercises … … … … … … … … … … … 312 Further Reading … … … … … … … … … … … … … . . 312

7 Frequency Domain Analysis of Discrete-Time Signals

and Systems … … … … … … … … … … … … … … . 313 7.1 The Discrete-Time Fourier Series … … … … … … … … . 313 Periodic Convolution … … … … … … … … … 314

7.1.1

7.2 Representation of Discrete-Time Signals and Systems

7.2.4

in Frequency Domain … … … … … … … … … … … 316 Fourier Transform of Discrete-Time Signals … … … . 316 7.2.1 Theorems on DTFT … … … … … … … … … . 317 7.2.2 Some Properties of the DTFT of a Complex 7.2.3 Sequence x(n) … … … … … … … … … … . . 320 Some Properties of the DTFT of a Real Sequence x(n) … … … … … … … … … … . . 322 7.3 Frequency Response of Discrete-Time Systems … … … … . . 332 Frequency Response Computation Using MATLAB … . . 338 7.4 Representation of Sampling in Frequency Domain … … … … 344 Sampling of Low-Pass Signals … … … … … … . . 346 7.5 Reconstruction of a Band-Limited Signal from Its Samples … … 347 7.6 Problems … … … … … … … … … … … … … … 349 Further Reading … … … … … … … … … … … … … . . 351

7.3.1

7.4.1

8 The z-Transform and Analysis of Discrete Time

LTI Systems … … … … … … … … … … … … … … . 353 Deﬁnition of the z-Transform … … … … … … … … … 353 8.1 Properties of the Region of Convergence for 8.2 the z-Transform … … … … … … … … … … … … 355 Properties of the z-Transform … … … … … … … … … 360 z-Transforms of Some Commonly Used Sequences … … … . . 365 The Inverse z-Transform … … … … … … … … … … 371

8.3 8.4 8.5

Contents

8.5.1 Modulation Theorem in the z-Domain … … … … . . 372 Parseval’s Relation in the z-Domain … … … … … 372 8.5.2 8.6 Methods for Computation of the Inverse z-Transform … … … 374

8.6.1

8.6.2

8.6.3

8.6.4

8.6.5

8.6.6

Cauchy’s Residue Theorem for Computation of the Inverse z-Transform … … … … … … … . 374 Computation of the Inverse z-Transform Using the Partial Fraction Expansion … … … … … 375 Inverse z-Transform by Partial Fraction Expansion Using MATLAB … … … … … … … … … . . 379 Computation of the Inverse z-Transform Using the Power Series Expansion … … … … … . 380 Inverse z-Transform via Power Series Expansion Using MATLAB … … … … … … … … … . . 383 Solution of Difference Equations Using the z-Transform … … … … … … … … . 383

8.7

8.8

Analysis of Discrete-Time LTI Systems in the z-Transform Domain … … … … … … … … … … … 385 Transfer Function … … … … … … … … … . . 385 8.7.1 Poles and Zeros of a Transfer Function … … … … . 386 8.7.2 Frequency Response from Poles and Zeros … … … . 388 8.7.3 Stability and Causality … … … … … … … … . 389 8.7.4 8.7.5 Minimum-Phase, Maximum-Phase, and

Mixed-Phase Systems … … … … … … … … . . 395 Inverse System … … … … … … … … … … 395 8.7.6 All-Pass System … … … … … … … … … … 397 8.7.7 8.7.8 All-Pass and Minimum-Phase Decomposition … … . . 399 One-Sided z-Transform … … … … … … … … … … . 401 8.8.1

Solution of Difference Equations with Initial Conditions … … … … … … … … … . . 404 Solution of State-Space Equations Using z-Transform … … … 405

8.9 8.10 Transformations Between Continuous-Time Systems

and Discrete-Time Systems … … … … … … … … … . 408 8.10.1 Impulse Invariance Method … … … … … … … . 409 8.10.2 Bilinear Transformation … … … … … … … … 411 8.11 Problems … … … … … … … … … … … … … . . 413 8.12 MATLAB Exercises … … … … … … … … … … … 416 Further Reading … … … … … … … … … … … … … . . 417

Index … … … … … … … … … … … … … … … … … 419

Chapter 1 Introduction

1.1 What is a Signal?

A signal is deﬁned as any physical quantity that carries information and varies with time, space, or any other independent variable or variables. The world of science and engineering is ﬁlled with signals: speech, television, images from remote space probes, voltages generated by the heart and brain, radar and sonar echoes, seismic vibrations, signals from GPS satellites, signals from human genes, and countless other applications.

1.2 What is a System?

A system is deﬁned mathematically as a transformation that maps an input signal x(t) into an output signal y(t) as illustrated in Figure 1.1. This can be denoted as

y tð Þ ¼ ℜ x tð Þ

(cid:2)

ð1:1Þ

where ℜ is an operator.

For example, a communication system itself is a combination of transmitter, channel, and receiver. A communication system takes speech signal as input and transforms it into an output signal, which is an estimate of the original input signal.

1.3 Elementary Operations on Signals

In many practical situations, signals related by a modiﬁcation of the independent variable t are to be considered. The useful elementary operations on signals including time shifting, time scaling, and time reversal are discussed in the following subsections.

1 Introduction

( )

Figure 1.1 Schematic representation of a system

x(t)

x(t-2)

-2

(a)

(b)

x(t+2)

(c)

-4

-2

Figure 1.2 Illustration of time shifting

1.3.1 Time Shifting

Consider a signal x(t). If it is time shifted by t0, the time-shifted version of x(t) is represented by x(t (cid:3) t0). The two signals x(t) and x(t (cid:3) t0) are identical in shape but time shifted relative to each other. If t0 is positive, the signal x(t) is delayed (right shifted) by t0. If t0 is negative, the signal is advanced (left shifted) by t0. Signals related in this fashion arise in applications such as sonar, seismic signal processing, radar, and GPS. The time shifting operation is illustrated in Figure 1.2. If the signal x (t) shown in Figure 1.2(a) is shifted by t0 ¼ 2 seconds, x(t (cid:3) 2) is obtained as shown in Figure 1.2(b), i.e., x(t) is delayed (right shifted) by 2 seconds. If the signal is advanced (left shifted) by 2 seconds, x(t þ 2) is obtained as shown in Figure 1.2(c), i.e., x(t) is advanced (left shifted) by 2 seconds.

1.3.2 Time Scaling

The compression or expansion of a signal is known as time scaling. The time-scaling operation is illustrated in Figure 1.3. If the signal x(t) shown in Figure 1.3(a) is

1.3 Elementary Operations on Signals

-4

-2

x(t)

-2

(a)

x(2t)

-2

(b)

-1

-2

x(t/2)

-2

-4

-2

Figure 1.3 Illustration of time scaling

x(t)

x(-t)

-2

(a)

(b)

Figure 1.4 Illustration of time reversal

compressed in time by a factor 2, x(2t) is obtained as shown in Figure 1.3(b). If the signal x(t) is expanded by a factor of 2, x(t/2) is obtained as shown in Figure 1.3(c).

1.3.3 Time Reversal

The signal x((cid:3)t) is called the time reversal of the signal x(t). The x((cid:3)t) is obtained from the signal x(t) by a reﬂection about t ¼ 0. The time reversal operation is illustrated in Figure 1.4. The signal x(t) is shown in Figure 1.4(a), and its time reversal signal x((cid:3)t) is shown in Figure 1.4(b).

1 Introduction

Example 1.1 Consider the following signals x(t) and xi(t), i ¼ 1,2,3. Express them using only x(t) and its time-shifted, time-scaled, and time-inverted version.

x(t)

-2

Solution

x(t)

x(t-2)

-2

x(-t)

x1 tð Þ ¼ x t (cid:3) 2

ð Þ þ x (cid:3)t þ 2

x(-t)

-2

1.4 Classiﬁcation of Signals

x2 tð Þ ¼ x t (cid:3) 2

ð Þ þ x (cid:3)t (cid:3) 2

(cid:1)

(cid:3)

t 2

(cid:3) 2

x3 tð Þ ¼ 2x

1.4 Classiﬁcation of Signals

Signals can be classiﬁed in several ways. Some important classiﬁcations of signals are:

1.4.1 Continuous-Time and Discrete-Time Signals

Continuous-time signals are deﬁned for a continuous of values of the independent variable. In the case of continuous-time signals, the independent variable t is continuous as shown Figure 1.5(a).

Discrete-time signals are deﬁned only at discrete times, and for these signals, the independent variable n takes on only a discrete set of amplitude values as shown in Figure 1.5(b).

1.4.2 Analog and Digital Signals

An analog signal is a continuous-time signal whose amplitude can take any value in a continuous range. A digital signal is a discrete-time signal that can only have a discrete set of values. The process of converting a discrete-time signal into a digital signal is referred to as quantization.

e d u t i l

p m A

3.5

2.5

1.5

0.5

0 −0.5

8 Time

10 12 14 16

e d u t i l

p m A

3.5

2.5

1.5

0.5

0 −0.5

10 8 6 Time index n

Figure 1.5 (a) Continuous-time signal, (b) discrete-time signal

1 Introduction

1.4.3 Periodic and Aperiodic Signals

A signal x(t) is said to be periodic with period T(a positive nonzero value), if it exhibits periodicity, i.e., x(t þ T) ¼ x(t), for all values of t as shown in Figure 1.6(a). Periodic signal has the property that it is unchanged by a time shift of T.

A signal that does not satisfy the above periodicity property is called an aperiodic

signal. The signal shown in Figure 1.6(b) is an example of an aperiodic signal.

Example 1.2 For each of the following signals, determine whether it is periodic or aperiodic. If periodic, ﬁnd the period. (i) x(t) ¼ 5 sin(2πt) (ii) x(t) ¼ 1 þ cos(4t þ 1) (iii) x(t) ¼ e(cid:3)2t ð (iv) x tð Þ ¼ ej 5tþπ (v) x tð Þ ¼ ej 5tþπ

Þe(cid:3)2t

Solution (i) It is periodic signal, period ¼ 2π (ii) It is periodic, period ¼ 2π 4 (iii) It is aperiodic, (iv) It is periodic. period ¼ 2π 5 (v) Since x(t) is a complex exponential multiplied by a decaying exponential, it is

2π ¼ 1:

¼ π 2

aperiodic.

Example 1.3 If a continuous-time signal x(t) is periodic, for each of the following signals, determine whether it is periodic or aperiodic. If periodic, ﬁnd the period. (i) x1(t) ¼ x(2t) (ii) x2(t) ¼ x(t/2)

Solution Let T be the period of x(t). Then, we have

ð x tð Þ ¼ x t þ T

x(t)

0.05

0.1 Time (sec)

(a)

0.15

0.2

Time

(b)

e d u t i l

p m A

1 0.5 0 −0.5 −1

Figure 1.6 (a) Periodic signal, (b) aperiodic signal

1.4 Classiﬁcation of Signals

(i) For x1(t) to be periodic,

x 2tð

ð x 2t þ T

Þ ¼ x 2t þ T ð (cid:4) (cid:4) Þ ¼ x 2 t þ T 2 (cid:5) (cid:4)

¼ x1

t þ T 2

(cid:5)

Since x1 tð Þ ¼ x1 t þ T 2

(cid:6)

(cid:7)

, x1(t) is periodic with fundamental period T 2.

As x1(t) is compressed version of x(t) by half, the period of x1(t) is also com-

pressed by half.

(ii) For x2(t) to be periodic,

Þ ¼ x x t=2ð (cid:3)

(cid:1)

t 2

þ T

(cid:1)

(cid:3)

þ T

t 2 (cid:4) 1 2 ¼ x2 t þ 2T ð

¼ x

t þ 2T

(cid:5)

Since x2(t) ¼ x2(t þ 2T ), x2(t) is periodic with fundamental period 2T. As x2(t) is expanded version of x(t) by two, the period of x2(t) is also twice the period of x(t).

Proposition 1.1 Let continuous-time signals x1(t) and x2(t) be periodic signals with fundamental periods T1 and T2, respectively. The signal x(t) that is a linear combi- nation of x1(t) and x2(t) is periodic if and only if there exist integers m and k such that mT1 ¼ kT2 and

T 1 T 2

¼ k m

¼ rational number

ð1:2Þ

The fundamental period of x(t) is given by mT1 ¼ kT2 provided that the values of m and k are chosen such that the greatest common divisor (gcd) between m and k is 1.

Example 1.4 For each of the following signals, determine whether it is periodic or aperiodic. If periodic, ﬁnd the period. (i) x(t) ¼ 2 cos(4πt) þ 3 sin(3πt) (ii) x(t) ¼ 2 cos(4πt) þ 3 sin(10t)

Solution (i) Let x1(t) ¼ 2 cos(4πt) and x2(t) ¼ 3 sin(3πt).

The fundamental period of x1(t) is

T 1 ¼ 2π 4π

¼ 1 2

1 Introduction

The fundamental period of x2(t) is

T 2 ¼ 2π 3π

¼ 2 3

The ratio T 1 T 2 The fundamental period of the signal x(t) is 4T1 ¼ 3T2 ¼ 2 seconds.

4 is a rational number. Hence, x(t) is a periodic signal.

¼ 3

¼ 1=2 2=3

(ii) Let x1(t) ¼ 2 cos(4πt) and x2(t) ¼ 3 sin(10t).

The fundamental period of x1(t) is

T 1 ¼ 2π 4π

¼ 1 2

The fundamental period of x2(t) is

T 2 ¼ 2π 10

The ratio T 1 T 2

¼ 1=2 π=5

¼ 5

2π is not a rational number. Hence, x(t) is an aperiodic signal.

Example 1.5 Consider the signals

x1 tð Þ ¼ cos

þ 2 sin

(cid:4) (cid:5) 8πt 5

(cid:4) (cid:5) 2πt 5 Þ

x2 tð Þ ¼ sin πtð

Determine whether x3(t) ¼ x1(t)x2(t) is periodic or aperiodic. If periodic, ﬁnd the

period.

Solution Decomposing signals x1(t) and x2(t) into sums of exponentials gives (cid:3) e(cid:3)j 8πt=5 x1 tð Þ ¼ 1

Þ þ ej 8πt=5

Þ þ 1

2 e(cid:3)j 2πt=5

2 ej 2πt=5

x2 tð Þ ¼ ej πtð

(cid:3) e(cid:3)j πtð 2j

Then,

ð ej 7πt=5

ð e(cid:3)j 3πt=5

x3 tð Þ ¼ 1 4j þ e(cid:3)j 3πt=5 2

Þ þ 1 4j ð (cid:3) e(cid:3)j 13πt=5 2 It is seen that all complex exponentials are powers of ej(π/5). Hence, it is periodic.

Þ (cid:3) 1 4j þ ej 3πt=5 2

Þ (cid:3) 1 4j

ð e(cid:3)j 7πt=5

ð ej 3πt=5

Þ (cid:3) ej 13πt=5

Period is 2π π=5

¼ 10 seconds:

1.4 Classiﬁcation of Signals

x(t)

-b

(a)

-A

(b)

Figure 1.7 (a) Even signal, (b) odd signal

1.4.4 Even and Odd Signals

The continuous-time signal is said to be even when x((cid:3)t) ¼ x(t). The continuous- time signal is said to be odd when x((cid:3)t) ¼ (cid:3)x(t). Odd signals are also known as nonsymmetrical signals. Examples of even and odd signals are shown in Figure 1.7 (a) and Figure 1.7(b), respectively.

Any signal can be expressed as sum of its even and odd parts as

x tð Þ ¼ xe tð Þ þ xo tð Þ

The even part and odd part of a signal are

xe tð Þ ¼ x tð Þ þ x (cid:3)t xo tð Þ ¼ x tð Þ (cid:3) x (cid:3)t

ð1:3Þ

ð1:3aÞ

ð1:3bÞ

Some important properties of even and odd signals are:

(i) Multiplication of an even signal by an odd signal produces an odd signal. Proof Let y(t) ¼ xe(t)xo(t)

ð y (cid:3)t

Hence, y(t) is an odd signal.

Þ ¼ xe (cid:3)t

ð Þxo (cid:3)t ¼ (cid:3)xe tð Þxo tð Þ ¼ (cid:3)y tð Þ

ð1:4Þ

(ii) Multiplication of an even signal by an even signal produces an even signal. Proof Let y(t) ¼ xe(t)xe(t)

ð y (cid:3)t

Hence, y(t) is an even signal.

ð Þxe (cid:3)t

Þ ¼ xe (cid:3)t ¼ xe tð Þxe tð Þ ¼ y tð Þ

ð1:5Þ

1 Introduction

(iii) Multiplication of an odd signal by an odd signal produces an even signal. Proof Let y(t) ¼ xo(t)xo(t)

ð y (cid:3)t

ð Þxo (cid:3)t Þ Þ (cid:3)xo tð Þ ð

Þ ¼ x0 (cid:3)t ¼ (cid:3)xo tð Þ ð ¼ xo tð Þxo tð Þ ¼ y tð Þ

Hence, y(t) is an even signal. It is seen from Figure 1.7(a) that the even signal is symmetric about the vertical

axis, and hence

(cid:3)b

xe tð Þdt ¼ 2

xe tð Þdt

From Figure 1.6(b), it is also obvious that

(cid:3)b

xo tð Þdt ¼ 0

ð1:6Þ

ð1:7Þ

Eqs. (1.6) and (1.7) are valid for no impulse or its derivative at the origin. These

properties are proved to be useful in many applications. Example 1.6 Find the even and odd parts of x(t) ¼ ej2t.

Solution From Eq. (1.3),

ej2t ¼ xe tð Þ þ xo tð Þ

where

xe tð Þ ¼ x tð Þ þ x (cid:3)t Þ xo tð Þ ¼ x tð Þ (cid:3) x (cid:3)t

¼ ej2t þ e(cid:3)j2t 2 ¼ ej2t (cid:3) e(cid:3)j2t 2

¼ cos 2tð

¼ j sin 2tð

Þ:

Example 1.7 If xe(t) and xo(t) are the even and odd parts of x(t), show that ð1

ð1

x2 tð Þdt ¼

e tð Þdt þ x2

o tð Þdt x2

(cid:3)1

ð1:8Þ

1.4 Classiﬁcation of Signals

Solution ð1

x2 tð Þdt ¼

(cid:3)1

ð1

(cid:3)1 ð1

(cid:3)1

Þ2dt

xe tð Þ þ xo tð Þ ð1

e tð Þdt þ 2 x2

(cid:3)1

ð1

xe tð Þxo tð Þdt þ

e tð Þdt þ x2

o tð Þdt x2

(cid:3)1

ð1

(cid:3)1

o tð Þdt x2

Since 2

ð1

(cid:3)1

xe tð Þxo tð Þdt ¼ 0

Example 1.8 For each of the following signals, determine whether it is even, odd, or neither (Figure 1.8)

Solution By deﬁnition a signal is even if and only if x(t) ¼ x((cid:3)t), while a signal is odd if and only if x(t) ¼ (cid:3)x((cid:3)t). (a) It is readily seen that x(t) 6¼ x((cid:3)t) for all t and x(t) 6¼ (cid:3)x((cid:3)t) for all t; thus x(t) is

neither even nor odd.

(b) Since x(t) is symmetric about t ¼ 0, x(t) is even. (c) Since x(t) ¼ (cid:3)x((cid:3)t), x(t) is odd in this case.

x(t)

-4

-2

(a)

(b)

-2

x(t)

-2

(c)

Figure 1.8 Signals of example 1.8

x(t)

1 Introduction

x(t)

(a)

(b)

(c)

Figure 1.9 (a) Causal signal, (b) noncausal signal, (c) anticausal signal

1.4.5 Causal, Noncausal, and Anticausal Signal

A causal signal is one that has zero values for negative time, i.e., t < 0. A signal is noncausal if it has nonzero values for both the negative and positive times. An anticausal signal has zero values for positive time, i.e., t > 0. Examples of causal, noncausal, and anticausal signals are shown in Figure 1.9(a), 1.9(b), and 1.9(c), respectively.

Example 1.9 Consider the following noncausal continuous-time signal. Obtain its realization as causal signal.

Solution

x(t)

-1

-0.5

0.5

x(t)

0.5

-1

1.5

1.4 Classiﬁcation of Signals

1.4.6 Energy and Power Signals

A signal x(t) with ﬁnite energy, which means that amplitude ! 0 as time ! 1, is said to be energy signal, whereas a signal x(t) with ﬁnite and nonzero power is said to be power signal. The instantaneous power p(t) of a signal x(t) can be expressed by

The total energy of a continuous-time signal x(t) can be deﬁned as

p tð Þ ¼ x2 tð Þ

for a complex valued signal

ð1

(cid:3)1

E ¼

x2 tð Þdt

ð1

(cid:3)1

E ¼

x tð Þ

j2dt

ð1:9Þ

ð1:10aÞ

ð1:10bÞ

Since the power is the time average of energy, the average power is deﬁned as

x2 tð Þdt. The signal x(t) expressed by Eq. (1.11), which is

P ¼ limT!1

1 T

T=2

(cid:3)T=2

shown in Figure 1.10(a), is an example of energy signal.

(

x tð Þ ¼

0 < t (cid:4) 1 1 < t (cid:4) 2

ð1:11Þ

The energy of the signal is given by ð

ð1

E ¼

(cid:3)1

x2 tð Þdt ¼

t2dt þ

1 dt ¼ 1 3

þ 1 ¼ 4 3

The signal x(t) shown in Figure 1.10(b) is an example of a power signal. The signal is periodic with period 2. Hence, averaging x2(t) over inﬁnitely large time interval is the same as averaging over one period, i.e., 2. Thus, the average power P is

P ¼ 1 2

(cid:3)1

x2 tð Þdt ¼ 1 2

(cid:3)1

4t2dt ¼ 4 3

Figure 1.10(a) Energy signal

x(t)

Time

Figure 1.10(b) Power signal

1 Introduction

x(t)

-2

-4

-2

Thus, an energy signal has ﬁnite energy and zero average power, whereas a power

signal has ﬁnite power and inﬁnite energy.

Example 1.10 Compute energy and power for the following signals, and determine whether each signal is energy signal, power signal, or neither. (i) x(t) ¼ 4sin(2πt), (cid:3)1 < t < 1. (ii) x(t) ¼ 2e(cid:3)2|t|, (cid:3)1 < t < 1. 2ﬃﬃ p t > 1 t t (cid:4) 1:

(iii) x tð Þ ¼

8 <

0 (iv) x(t) ¼ e(cid:3)at for real value of a (v) x(t) ¼ cos (t) (vi) xðtÞ ¼ e j 2tþ

π 4

Solution (i)

E ¼

ð1

j2dt ¼

(cid:3)1 1 (cid:3) cos 4πt ð 2 ð1

j2dt Þ

4 sin 2πt ð j (cid:10)

dt (cid:3) 8

(cid:3)1

ð cos 4πt

Þdt

x tð Þ (cid:9)

(cid:3)1

ð1

¼ 16

¼ 1

(cid:3)1 ð1

(cid:3)1

1 2

P ¼ limT!1

¼ limT!1

1 T

T=2

(cid:3)T=2

T=2

(cid:3)T=2

16 sin 2ð2πtÞdt

x2ðtÞdt ¼ limT!1

1 T

(cid:9)

1 (cid:3) cos ð4πtÞ 2

(cid:10)

(cid:3)T=2 ð

T=2

1 T

(cid:3)T=2

1 2

dt (cid:3) 16 limT!1

T=2

(cid:3)T=2

1 T

cos ð4πtÞ 2

¼ 16 limT!1

¼ 8

1.4 Classiﬁcation of Signals

The energy of the signal is inﬁnite, and its average power is ﬁnite; x(t) is a power

signal. (ii) x(t) ¼ 2e(cid:3)2|t|

ð1

E ¼

x tð Þ

j2dt ¼

(cid:11) (cid:11)

2e(cid:3)2 tj j

(cid:11) (cid:11)2

(cid:3)1 ð

(cid:3)1

ð1

e4tdt þ 4

e(cid:3)4tdt

(cid:12)

(cid:3)1 (cid:13) 0 (cid:3)1

e4t

(cid:12)

þ 4 4

e(cid:3)4t

(cid:13)1 0

þ 4 4

¼ 2

¼ 4

¼ 4 4 ¼ 4 4

P ¼ limT!1

1 T

x2 tð Þdt ¼ limT!1

1 T

T=2

(cid:3)T=2 ð

T=2

(cid:11) (cid:11)

(cid:3)T=2 ð

T=2

2e(cid:3)2 tj j

(cid:11) (cid:11)2

e(cid:3)4tdt

1 T (cid:12)

0 (cid:13)T=2 0

e(cid:3)4t

1 T

e4tdt þ 4 limT!1

(cid:3)T=2 (cid:13)

e4t

0 (cid:3)T=2

1 (cid:3) e(cid:3)2T

þ 4 4 þ 4 4

(cid:13)

limT!1

(cid:13)

e(cid:3)2T (cid:3) 1

(cid:12)

1 T

¼ 4 limT!1

1 T

(cid:12)

limT!1

¼ 4 4 ¼ 4 4

1 T 1 T ¼ 0 þ 0 ¼ 0

limT!1

The energy of the signal is ﬁnite, and its average power is zero; x(t) is an energy

signal.

(iii) x tð Þ ¼

8 <

2ﬃﬃ p t > 1 t

t (cid:4) 1:

ð1

E ¼

x tð Þ j

(cid:3)1 ¼ 4 ln t½ (cid:2)1 1 ¼ 1

j2dt ¼

ð1

4 t

1 Introduction

x2 tð Þdt ¼ lim T!1

1 T

4 t

(cid:5)

¼ 4 lim T!1

(cid:9) (cid:10) T 2

(cid:3) 1 T

(cid:5)

ln 1½ (cid:2)

T=2

1 (cid:4)

1 T

T=2

(cid:3)T=2

P ¼ lim T!1

1 T

¼ 4 lim T!1

(cid:4)

1 T

(cid:4)

1 T 0

¼ 4 lim T!1

B B @

1T=2

ln t½ (cid:2)

(cid:5)

(cid:9) (cid:10) T ln 2 (cid:9) (cid:10) 1 T 2 T

C C A

Using L’Hospital’s rule, we see that the power of the signal is zero. That is (cid:5)

(cid:4)

P ¼ 4 lim T!1

(cid:12) (cid:13) ln T 2 T

(cid:4) (cid:5) 2 T 1

¼ 0

¼ 4 lim T!1

The energy of the signal is inﬁnite and its average power is zero; x(t) is neither

energy signal nor power signal. (iv) x(t) ¼ e(cid:3)at for real value of a

ð1

(cid:3)1

e(cid:3)at

j2dt ¼ 1,

x2 tð Þdt ¼ limT!1

(cid:5)

(cid:4)

eaT 2aT

¼ lim T!1

T=2

(cid:3)T=2

1 T (cid:5)

(cid:3) lim T!1

e(cid:3)2at dt (cid:4)

(cid:5)

e(cid:3)aT 2aT

ð1

E ¼

x tð Þ j

(cid:3)1

P ¼ limT!1 (cid:4)

¼ lim T!1

(cid:4)

eaT 2aT

j2dt ¼ ð

T=2

1 T (cid:3)T=2 eaT (cid:3) e(cid:3)aT 2aT (cid:5)

(cid:3) 0

Using L’Hospital’s rule, we see that the power of the signal is inﬁnite. That is,

P ¼ lim T!1

(cid:5)

(cid:4)

eaT 2aT

(cid:4) (cid:5) eaT 2

¼ lim T!1

¼ 1

The energy of the signal is inﬁnite and its average power is inﬁnite; x(t) is neither

energy signal nor power signal. (v) x(t) ¼ cos(t)

ð1

(cid:3)1

E ¼

x tð Þ

j2dt ¼

ð1

(cid:3)1

cos 2 tð Þdt ¼ 1,

1.4 Classiﬁcation of Signals

T=2

(cid:3)T=2 ð T=2

(cid:3)T=2

1 T 1 T 1 T

P ¼ limT!1

¼ limT!1

¼ 1 2

T=2

(cid:3)T=2

1 T

cos 2 tð Þ dt

x2 tð Þdt ¼ limT!1 (cid:9)

(cid:10)

1 þ cos 2tð 2

1 2

dt þ limT!1

dt ð

1 T

T=2

(cid:3)T=2

cos 2tð 2

The energy of the signal is inﬁnite and its average power is ﬁnite; x(t) is a power

signal.

(vi) x tð Þ ¼ ej 2tþπ

ð1

, x tð Þ j

j ¼ 1. ð1

E ¼

x tð Þ

(cid:3)1

P ¼ limT!1

j2dt ¼ ð

T=2

(cid:3)T=2

1 T

dt ¼ 1,

(cid:3)1

x2 tð Þdt ¼ limT!1

T=2

(cid:3)T=2

1 T

1 dt ¼ limT!11 ¼ 1

The energy of the signal is inﬁnite and its average power is ﬁnite; x(t) is a power

signal.

Example 1.11 Consider the following signals, and determine the energy of each signal shown in Figure 1.11. How does the energy change when transforming a signal by time reversing, sign change, time shifting, or doubling it?

x(t)

(t)

-2

( )

-2

( )

(t)

Figure 1.11 Signals of example 1.11

Solution

xðtÞ ¼

Ex ¼

x1ðtÞ ¼

Ex1

x2ðtÞ ¼

Ex2

x3ðtÞ ¼

1 Introduction

t2dt ¼ t3 3

(cid:11) (cid:11) (cid:11) (cid:11)

¼ 8 3

(

0 < t (cid:4) 2

otherwise

0 ð1

jxðtÞj2dt ¼

(cid:3)1 0 ( t (cid:3)2 < t (cid:4) 0

(cid:3)2

t2dt ¼ t3 3

(cid:11) (cid:11) (cid:11) (cid:11)

(cid:3)2

¼ 8 3

t2dt ¼ t3 3

(cid:11) (cid:11) (cid:11) (cid:11)

¼ 8 3

otherwise

0 ð1

jx1ðtÞj2dt ¼

(cid:3)1 ( (cid:3)t

ð1

0 < t (cid:4) 2

otherwise ð

jx2ðtÞj2dt ¼

(cid:3)1 ( ðt (cid:3) 2Þ

2 < t (cid:4) 4

otherwise ð

jx3ðtÞj2dt ¼ (cid:5)(cid:11) (cid:11) 4 (cid:11) (cid:11) 2

(cid:3) 2t2 þ 4t

0 < t (cid:4) 2

0 ð1

(cid:3)1

otherwise

jx4ðtÞj2dt ¼

4t2dt ¼ 4

(cid:11) (cid:11) (cid:11) (cid:11)

t3 3

¼ 32 3

Ex3

ð1

(cid:3)1 (cid:4) ¼ t3 3

¼ 8 3 (

x4ðtÞ ¼

Ex4

ðt (cid:3) 2Þ2dt ¼

ðt2 (cid:3) 4t þ 4Þdt

The time reversal, sign change, and time shifting do not affect the signal energy. Doubling the signal quadruples its energy. Similarly, it can be shown that the energy of k x(t) is k2Ex.

Proposition 1.2 The sum of two sinusoids of different frequencies is the sum of the power of individual sinusoids regardless of phase. Proof Let us consider a sinusoidal signal x(t) ¼ Acos(Ωt + θ). The power of x(t) is given by

1.4 Classiﬁcation of Signals

P ¼ limT!1

T=2

(cid:3)T=2

1 T

x2ðtÞdt ¼ limT!1

1 T

A2cos 2ðΩt þ θÞdt

A2½1 þ cos 2ð2Ωt þ 2θÞ(cid:2)dt

dt þ

T=2

(cid:3)T=2

cos 2Ωt þ 2θ

Þdt

T=2

(cid:3)T=2 ð T=2

¼ limT!1

1 2T (cid:3)T=2 ” ð T=2 A2 2T ½T þ 0(cid:2) ¼ A2 2

(cid:3)T=2

¼ A2 2T

Thus, a sinusoid signal of amplitude A has a power A2

its frequency Ω and phase θ.

Now, consider the following two sinusoidal signals:

2 regardless of the values of

ð1:12Þ

x1 tð Þ ¼ A1 cos Ω1t þ θ1 x2 tð Þ ¼ A2 cos Ω2t þ θ2 xs tð Þ ¼ x1 tð Þ þ x2 tð Þ

ð ð

Þ Þ

Let

The power of the sum of the two sinusoidal signals is given by

Ps ¼ limT!1

¼ limT!1

1 T

þ limT!1

ðT 2

(cid:3)T 2 ð T=2

x2 s

ðtÞ

(cid:3)T=2 ðT 2

(cid:3)T 2 ðT 2

1 T

(cid:3)T 2 2A1A2 T

½A1cos ðΩ1t þ θ1Þ þ A2cos ðΩ2t þ θ2Þ(cid:2)2dt

1cos 2ðΩ1t þ θ1Þdt

2cos 2ðΩ2t þ θ2Þdt ðT 2

(cid:3)T 2

cos ðΩ1t þ θ1Þcos ðΩ2t þ θ2Þdt

The ﬁrst and second integrals on the right-hand side are the powers of the two

sinusoidal signals, respectively, and the third integral becomes zero since

cos Ω1t þ θ1

Þ cos Ω2t þ θ2

Hence,

½ Þ ¼ cos Ω1 þ Ω2 ð

þ cos Ω1 (cid:3) Ω2 ½ ð

Þt þ θ1 þ θ2 (cid:2) Þ ð ð Þt þ θ1 (cid:3) θ2

Ps ¼ A2 1 2

þ A2 2 2

(cid:2)

ð1:13Þ

1 Introduction

It can be easily extended to sum of any number of sinusoids with distinct

frequencies

1.4.7 Deterministic and Random Signals

For any given time, the values of deterministic signal are completely speciﬁed as shown in Figure 1.12(a). Thus, a deterministic signal can be described mathemati- cally as a function of time. A random signal takes random statistically characterized random values as shown in Figure 1.12(b) at any given time. Noise is a common example of random signal.

1.5 Basic Continuous-Time Signals

1.5.1 The Unit Step Function

The unit step function is deﬁned as

(cid:14) u tð Þ ¼ 1 0

t > 0 t < 0

ð1:14Þ

which is shown in Figure 1.13.

It should be noted that u(t) is discontinuous at t ¼ 0.

e d u t i l

p m A

1 0.5 0 −0.5 −1 0

0.05

0.1 Time (sec)

0.15

0.2

0.8

0.6

0.4

0.2

100

150

(a)

(b)

Figure 1.12 (a) Deterministic signal, (b) random signal

Figure 1.13 Unit step function

u(t)

1.5 Basic Continuous-Time Signals

1.5.2 The Unit Impulse Function

The unit impulse function also known as the Dirac delta function, which is often referred as delta function is deﬁned as

δ tð Þ ¼ 0, t 6¼ 0 ð1

δ tð Þ ¼ 1:

(cid:3)1

ð1:15aÞ

ð1:15bÞ

The delta function shown in Figure 1.14(b) can be evolved as the limit of the

rectangular pulse as shown in Figure 1.14(a).

δ tð Þ ¼ lim Δ!0

pΔ tð Þ

ð1:16Þ

As the width Δ ! 0, the rectangular function converges to the impulse function

δ(t) with an inﬁnite height at t ¼ 0, and the total area remains constant at one.

Some Special Properties of the Impulse Function (cid:129) Sampling property

If an arbitrary signal x(t) is multiplied by a shifted impulse function, the product is

given by

x tð Þδ t (cid:3) t0 ð

Þ ¼ x t0ð Þδ t (cid:3) t0

ð1:17aÞ

implying that multiplication of a continuous-time signal and an impulse function produces an impulse function, which has an area equal to the value of the continuous-time function at the location of the impulse. Also, it follows that for t0 ¼ 0,

x tð Þδ tð Þ ¼ x 0ð Þδ tð Þ

ð1:17bÞ

(cid:129) Shifting property

ð1

(cid:3)1

ð x tð Þδ t (cid:3) t0

Þdt ¼ x t0ð Þ

ð1:18Þ

Figure 1.14 (a) Rectangular pulse, (b) unit impulse

pΔ(t)

1/Δ

d (t)

−Δ/2

Δ/2

(a)

(b)

(cid:129) Scaling property

ð δ at þ b

Þ ¼ 1 aj j

(cid:5)

(cid:4) δ t þ b a

1 Introduction

ð1:19Þ

(cid:129) The unit impulse function can be obtained by taking the derivative of the unit step

function as follows:

δ tð Þ ¼ du tð Þ dt

ð1:20Þ

(cid:129) The unit step function is obtained by integrating the unit impulse function as

follows:

ð1:21Þ

ð1:22aÞ

ð1:22bÞ

u tð Þ ¼

(cid:3)1

δ tð Þdt

1.5.3 The Ramp Function

The ramp function is deﬁned as

which can also be written as

(cid:14) r tð Þ ¼ t 0

t > 0 t < 0

r tð Þ ¼ tu tð Þ

The ramp function is shown in Figure 1.15.

1.5.4 The Rectangular Pulse Function

The continuous-time rectangular pulse function is deﬁned as

Figure 1.15 The ramp function

1.5 Basic Continuous-Time Signals

Figure 1.16 The rectangular pulse function

Figure 1.17 The signum function

x(t)

x(t)=

-1

(cid:14) x tð Þ ¼ 1 0

tj j (cid:4) T 1 tj j > T 1

which is shown in Figure 1.16.

1.5.5 The Signum Function

The signum function also called sign function is deﬁned as

sgn tð Þ ¼

8 <

1 0 (cid:3)1

t > 0 t ¼ 0 t < 0

which is shown in Figure 1.17.

1.5.6 The Real Exponential Function

A real exponential function is deﬁned as

ð1:23Þ

ð1:24Þ

x tð Þ ¼ Aeσt

ð1:25Þ

where both A and σ are real. If σ is positive, x(t) is a growing exponential signal. The signal x(t) is exponentially decaying for negative σ. For σ ¼ 0, the signal x(t) is equal to a constant. Exponentially decaying signal and exponentially growing signal are shown in Figure 1.18(a) and (b), respectively.

1 Introduction

x(t)

(a)

x(t)

0 ( b)

Figure 1.18 Real exponential function. (a) Decaying, (b) growing

1.5.7 The Complex Exponential Function

A real exponential function is deﬁned as

x tð Þ ¼ Ae σþjΩ

Þt

x tð Þ ¼ AeσtejΩt

Hence

Using Euler’s identity

ejΩt ¼ cos Ωtð

Þ þ j sin Ωtð

Substituting Eq. (1.27) in Eq. (1.26a), we obtain

x tð Þ ¼ Aeσt cos Ωtð

Þ þ j sin Ωtð

Þ Þ

ð1:26Þ

ð1:26aÞ

ð1:27Þ

ð1:28Þ

Real sine function and real cosine function can be expressed by the trigonometric

identities as cos Ωtð

Þ ¼ ejΩtþe(cid:3)jΩt

and sin Ωtð

Þ ¼ ejΩt(cid:3)e(cid:3)jΩt

1.5.8 The Sinc Function

The continuous-time sinc function is deﬁned as

Sinc tð Þ ¼ sin πtð πt

ð1:28Þ

which is shown in Figure 1.19

1.5 Basic Continuous-Time Signals

Figure 1.19 The sinc function

Example 1.12 State whether the following signals are causal, anticausal, or noncausal. (a) x(t) ¼ e(cid:3)2tu(t) (b) x(t) ¼ tu(t) (cid:3) t(u(t (cid:3) 1) þ e(1(cid:3)t)u(t (cid:3) 1)) (c) x(t) ¼ et cos (2πt)u(1 (cid:3) t)

Solution (a)

x(t)

0.25

It is causal since x(t) ¼ 0 for t < 0 (b)

x(t)

It is causal since x(t) ¼ 0 for t < 0

(c)

(c )

u(1-t)

1 Introduction

It is non causal since

for t<0

Example 1.13 Determine and plot the even and odd components of the following continuous-time signal

x tð Þ ¼ tu t þ 2

ð Þ (cid:3) tu t (cid:3) 1

Solution

-2

x(t)

-2

x tð Þ ¼ tu t þ 2 ð ð Þ ¼ (cid:3)tu (cid:3)t þ 2

ð x (cid:3)t

ð Þ (cid:3) tu t (cid:3) 1

ð Þ þ tu (cid:3)t þ 1

x(-t)

-2

xeðtÞ ¼ xðtÞ þ xð(cid:3)tÞ (cid:1) uðt þ 2Þ (cid:3) uðt (cid:3) 1Þ (cid:3) uð(cid:3)t þ 2Þ þ uð(cid:3)t þ 1Þ

¼ 1 2

(cid:3)

1.5 Basic Continuous-Time Signals

-2

)(

-0.5

-1

xo tð Þ ¼ x tð Þ (cid:3) x (cid:3)t 2 ð t u t þ 2 ð

¼ 1 2

ð Þ (cid:3) u t (cid:3) 1

ð Þ þ u (cid:3)t þ 2

ð Þ (cid:3) u (cid:3)t þ 1

-2

-1

-0.5

Example 1.14 Simplify the following expressions:

(cid:1)

(cid:3) δ tð Þ sin t (a) t2þ3 (b) 4þjt δ t (cid:3) 1 ð Þ 3(cid:3)jt Ð 1 ð (cid:3)1 4t (cid:3) 3 (c)

ð ÞÞδ t (cid:3) 1

Þdt

Solution (cid:1)

(cid:3)

δ tð Þ ¼ 0

sin 0ð Þ (a) 0þ3 (b) 4þjt ð δ t (cid:3) 1 3(cid:3)jt Ð 1 ð (cid:3)1 4 1ð Þ (cid:3) 3 (c)

Þ ¼ 4þj ð δ t (cid:3) 1 3(cid:3)j ð ÞÞδ t (cid:3) 1

Þ Þdt ¼

Ð 1 ð (cid:3)1 1δ t (cid:3) 1

Þdt ¼ 1:

1 Introduction

1.6 Generation of Continuous-Time Signals Using

MATLAB

An exponentially damped sinusoidal signal can be generated using the following MATLAB command:

x(t) ¼ A ∗ sin (2 ∗ pi ∗ f0 ∗ t + θ) ∗ exp ((cid:3)a ∗ t)where a is positive for

decaying exponential.

Example 1.15 Write a MATLAB program to generate the following exponentially damped sinusoidal signal.

x(t) ¼ 5 sin (2πt)e(cid:3)0.4t

(cid:3) 10 (cid:4) t (cid:4) 10

Solution The following MATLAB program generates the exponentially damped sinusoidal signal as shown in Figure 1.20.

MATLAB program to generate exponentially damped sinusoidal signal

clear all;clc; x =inline(‘5sin(2pi1t).exp(-.4t)’,‘t’); t = (-10:.01:10);

plot(t,x(t));

xlabel (‘t (seconds)’); ylabel (‘Ámplitude’);

250

200

150

100

-50

-100

-150

-200

e d u t i l

p m Á

-250

-10

-8

-6

-4

-2

0 t (seconds)

Figure 1.20 Exponentially damped sinusoidal signal with exponential parameter a ¼ 0.4.

1.6 Generation of Continuous-Time Signals Using MATLAB

e d u t i l

p m Á

1.5

0.5

-0.5

-1

-1.5

-2

-5

-4

-3

-2

-1

t (seconds)

Figure 1.21 Unit step function

Example 1.16 Generate unit step function over [(cid:3)5,5] using MATLAB

Solution The following MATLAB program generates the unit step function over [(cid:3)5,5] as shown in Figure 1.21.

MATLAB program to generate unit step function over [(cid:3)5,5]

clear all;clc; u=inline(‘(t>=0)’,‘t’); t=-5:0.01:5; plot(t,u(t)) xlabel (‘t (seconds)’); ylabel (‘Ámplitude’) axis([-5 5 -2 2])

Example 1.17 Generate the following rectangular pulse function rect(t) using MATLAB:

(cid:14)

(cid:6) (cid:7) t 10

rect

¼ 1, (cid:3)5 < t < 5 elsewhere

Solution The following MATLAB program generates the rectangular pulse func- tion as shown in Figure 1.22.

1 Introduction

1.5

0.5

-0.5

-1

-1.5

e d u t i l

p m Á

-2

-10

-8

-6

-4

-2

t (seconds)

Figure 1.22 Rectangular pulse function

MATLAB program to generate rectangular pulse function

clear all;clc; u=inline(‘(t>=-5)& (t<5)’,‘t’); t=-10:0.01:10; plot(t,u(t)) xlabel (‘t (seconds)’); ylabel (‘Ámplitude’) axis([-10 10 -2 2])

1.7 Typical Signal Processing Operations

1.7.1 Correlation

Correlation of signals is necessary to compare one reference signal with one or more signals to determine the similarity between them and to determine additional infor- mation based on the similarity. Applications of cross correlation include cross- spectral analysis, detection of signals buried in noise, pattern matching, and delay measurements.

1.7 Typical Signal Processing Operations

1.7.2 Filtering

Filtering is basically a frequency domain operation. Filter is used to pass certain band of frequency components without any distortion and to block other frequency components. The range of frequencies that is allowed to pass through the ﬁlter is called the passband, and the range of frequencies that is blocked by the ﬁlter is called the stopband. A low-pass ﬁlter passes all low-frequency components below a certain speciﬁed frequency Ωc, called the cutoff frequency, and blocks all high-frequency components above Ωc. A high-pass ﬁlter passes all high-frequency components above a certain cutoff frequency Ωc and blocks all low-frequency components below Ωc. A band-pass ﬁlter passes all frequency components between two cutoff frequencies Ωc1 and Ωc2 where Ωc1 < Ωc2 and blocks all frequency components below the frequency Ωc1 and above the frequency Ωc2. A band-stop ﬁlter blocks all frequency components between two cutoff frequencies Ωc1 and Ωc2 where Ωc1 < Ωc2 and passes all frequency components below the frequency Ωc1 and above the frequency Ωc2. Notch ﬁlter is a narrow band-stop ﬁlter used to suppress a particular frequency, called the notch frequency.

1.7.3 Modulation and Demodulation

Transmission media, such as cables and optical ﬁbers, are used for transmission of signals over long distances; each such medium has a bandwidth that is more suitable for the efﬁcient transmission of signals in the high-frequency range. Hence, for transmission over such channels, it is necessary to transform the low-frequency signal to a high-frequency signal by means of a modulation operation. The desired low-frequency signal is extracted by demodulating the modulated high-frequency signal at the receiver end.

1.7.4 Transformation

The transformation is the representation of signals in the frequency domain, and inverse transform converts the signals from the frequency domain back to the time domain. The transformation provides the spectrum analysis of a signal. From the knowledge of the spectrum of a signal, the bandwidth required to transmit the signal can be determined. The transform domain representations provide additional insight into the behavior of the signal and make it easy to design and implement algorithms, such as those for ﬁltering, convolution, and correlation.

1 Introduction

1.7.5 Multiplexing and Demultiplexing

Multiplexing is used in situations where the transmitting media is having higher bandwidth, but the signals have lower bandwidth. Thus, multiplexing is the process in which multiple signals, coming from different sources, are combined and trans- mitted over a single channel. Multiplexing is performed by multiplexer placed at the transmitter end. At the receiving end, the composite signal is separated by demul- tiplexer performing the reverse process of multiplexing and routes the separated signals to their corresponding receivers or destinations.

In electronic communications, the two basic forms of multiplexing are time- division multiplexing (TDM) and frequency-division multiplexing (FDM). In time-division multiplexing, transmission time on a single channel is divided into non-overlapped time slots. Data streams from different sources are divided into units with same size and interleaved successively into the time slots. In frequency-division multiplexing (FDM), numerous low-frequency narrow bandwidth signals are com- bined for transmission over a single communication channel. A different frequency is assigned to each signal within the main channel. Code-division multiplexing (CDM) is a communication networking technique in which multiple data signals are combined for simultaneous transmission over a common frequency band.

1.8 Some Examples of Real-World Signals and Systems

1.8.1 Audio Recording System

An audio recording system shown in Figure 1.23(a) takes an audio or speech as input and converts the audio signal into an electrical signal, which is recorded on a magnetic tape or a compact disc. An example of recorded voice signal is shown in Figure 1.23(b).

Audio Recording System

Audio output signal

(a)

(b)

Figure 1.23 (a) Audio recording system, (b) the recorded voice signal “don’t fail me again”

1.8 Some Examples of Real-World Signals and Systems

1.8.2 Global Positioning System

The satellite-based global positioning system (GPS) consists of a constellation of 24 satellites at high altitudes above the earth. Figure 1.24 shows an example of the GPS used in air, sea, and land navigation. It requires signals at least from four satellites to ﬁnd the user position (X, Y, and Z) and clock bias from the user receiver. The measurements required in a GPS receiver for position ﬁnding are the ranges, i.e., the distances from GPS satellites to the user. The ranges are deduced from measured time or phase differences based on a comparison between the received and receiver- generated signals. To measure the time, the replica sequence generated in the receiver is to be compared to the satellite sequence.

The correlator in the user GPS receiver determines which codes are being received, as well as their exact timing. When the received and receiver-generated sequences are in phase, the correlator supplies the time delay. Now, the range can be obtained by multiplying the time delay by the velocity of light. For example, assuming the time delay as 3 ms (equivalent to 3 blocks of the C/A code of satellite 12), the correlation of satellite 12 producing a peak after 3 ms [Rao06] is shown in Figure 1.25.

1.8.3 Location-Based Mobile Emergency Services System

Mobile emergency services (MES) refer to the use of mobile positioning technology to pinpoint mobile users for purposes of providing enhanced wireless emergency dispatch services (including ﬁre, ambulance, and police) to mobile phone users. In this emergency service system, user should have assisted GPS-enabled mobile handset unit. Network service providers will support “Mobile Location Protocol

Figure 1.24 A pictorial representation of GPS positioning

Figure 1.25 The correlation of satellite 12 producing a peak

160

140

120

100

−20

1 Introduction

1000 2000 3000 4000 5000 6000 7000 8000 9000

(MLP).” The MLP serves as the interface between a location server and a location services (LCS) client.

Whenever user requires an emergency service, he will dial the speciﬁed number for emergency calling. Dialing of emergency service number will generate an “emergency location immediate service (ELIS).”

ELIS is used to retrieve the position of a mobile subscriber that is involved in an emergency call or has initiated an emergency service in some other way. The service consists of the following messages: emergency location immediate request (ELIR) and emergency location immediate answer (ELIA).

When user has dialed the emergency number, emergency location immediate

request is sent to network service provider.

After receiving the emergency location immediate request from the user, network service provider extracts the position information and sends emergency location immediate answer to the mobile user, and service provider asks him to select the service from ambulance, police, and ﬁre services. Mobile user selects the service, which he actually needs.

The service provider would ﬁnd the nearest emergency service center and send an emergency location report to that center. Whenever an emergency location report is received, a mark will appear on the corresponding digital map. This mark will indicate the user’s location. A schematic block diagram of location-based mobile emergency service system and tracking a mobile user are shown in Figure 1.26 (a) and (b), respectively.

1.8.4 Heart Monitoring System

In cardiac cells of the human body, a small electrical current is produced by the movement of sodium (Naþ) and potassium (Kþ) ions. The electrical potential

1.8 Some Examples of Real-World Signals and Systems

Figure 1.26 (a) Schematic block diagram (b) tracking a mobile user of location-based mobile emergency service system

Figure 1.27 One cycle of ECG signal

generated by these ions is known as an electrocardiogram (ECG) signal. The ECG signal is used by physicians to analyze heart conditions. The ECG signal is very small (normally 0.0001 to 0.003 volt). These signals are within the frequency range of 0.05 to 100 Hz. A typical one cycle ECG tracing of a normal heartbeat consists of a P wave, a QRS complex, and a T wave as shown in Figure 1.27. A small U wave is normally visible in 50 to 75% of ECGs.

The processing of ECG signal yields information, such as amplitude and timing, required for a physician to analyze a patient’s heart condition. Detection of R-peaks and computation of R-R interval of an ECG record are important requirements of comprehensive computed as (cid:6) Heart rate ¼

analysis (cid:7) (cid:5) 60.

arrhythmia 1 RR interval in seconds

systems. Heart

rate

An ECG signal with variations in heart rate is shown in Figure 1.28.

1 Introduction

e d u t i l

p m A

e d u t i l

p m A

e d u t i l

p m A

e d u t i l

p m A

0.5 0 −0.5

1 0 −1

0.4 0.2 0

100 50 0

Recorded ECG signal

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

Filtered ECG signal

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

ECG signal R peaks

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

Heart rate

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

2 (cid:2)104

Figure 1.28 An ECG signal with variations in heart rate

1.8.5 Human Visual System

The human visual system (HVS) can widely perform a number of image processing operations in a manner superior to anything we are currently able to execute with computers. To perform such signal processing operations, we have to understand the way HVS works.

When the reﬂection from an object (light ray) is observed by the eye, ﬁrst, it passes through the cornea, eventually through the aqueous humor, the iris, the lens, the vitreous humor, and ﬁnally reaching the retina. The retina consists photosensitive cells called cones and rods, which are responsible to convert the incident light energy into neural signals that are carried to human brain by the optic nerve (Figure 1.29).

1.8.6 Magnetic Resonance Imaging

When an oscillating strong magnetic ﬁeld is applied at a certain frequency on a certain part of the human body, the hydrogen atoms in the body emit radio-frequency waves to form image of the particular part of the body, which is captured by the MRI machine. An MRI imaging system and a MRI image with brain tumor are shown in Figure 1.30(a) and (b), respectively.

1.9 Problems

Figure 1.29 Human visual system

Figure 1.30 (a) MRI imaging system, (b) MRI image with brain tumor

1.9 Problems

Classify the following continuous-time signals as periodic or aperiodic. If

periodic, determine the period.

(cid:7)

(cid:6)

πt

(cid:6) (cid:7) þ 2 sin π 2t p ﬃﬃﬃ Þ þ sin 2 Þ (cid:6) Þ þ cos 6t þ π 3 (cid:7)

(cid:7)

(cid:6)

(i) x tð Þ ¼ cos 2π 3 t ð (ii) x tð Þ ¼ cos 2πt (iii) x tð Þ ¼ 1 2 cos 2tð (cid:3) 1 2 (iv) x tð Þ ¼ 1 þ sin 4tð (v) x(t) ¼ ej(4t + π/5) (vi) x tð Þ ¼ cos 2t þ π 4 (vii) x(t) ¼ cos(2πt)u(t) (viii) x(t) ¼ cos2(t)

(cid:6)

A periodic signal x1(t) has a period 2, and another periodic signal x2(t) has a the signal y

frequency and period for

period 3. Find the fundamental (t) ¼ x1(t) þ x2(t).

1 Introduction

Classify the following continuous-time signals as even or odd signals or neither even nor odd. Determine power and energy for each incase of power or energy signal. (i) x(t) ¼ (1þ t2)cos2(5t) (ii) x(t) ¼ u(t) (iii) x(t) ¼ tu(t) (iv) x(t) ¼ tsin(2t) (v) x(t) ¼ t + cos(2t) (vi) x(t) ¼ e(cid:3)2tsin(2t))
Consider the following continuous-time signal:

x tð Þ ¼ 2 sin

(cid:5) Þ

(cid:4)

ð 2π t (cid:3) T 10

Determine the values of T for which the signal is

(i) An even function (ii) An odd function

Classify the following continuous-time signals as power or energy signals or neither. Determine power and energy for each incase of power or energy signal. (i) x(t) ¼ sin(2πt)cos(πt) (ii) x(t) ¼ tu(t) (iii) x(t) ¼ e(cid:3)3tu(t) (iv) x(t) ¼ e(cid:3)j3t
Determine energy for each of the following signals and comment on the results.

x(t)

(a)

x(t)

(b)

-1

(c)

1 (d)

1.10 MATLAB Exercises

What is the energy of the signal x(t) ¼ cx(at (cid:3) b), where a 6¼ 0?
Verify that e(cid:3)ct is neither energy nor a power signal for a complex value of c

with nonzero real part.

Show that the energy of x(t) (cid:6) y(t) is Ex + Ey, if x(t) and y(t) are orthogonal.
Derive an expression for the power of the following continuous-time signal x

(t) ¼ A1cos(Ω1t + θ1) þ A2cos(Ω2t + θ2) for Ω1 ¼ Ω2.

Determine the power of the signal x(t) ¼ AejΩt.
Find power for each of the following signals:

(i) x(t) ¼ (5 þ 3sin(2t))cos(5t) (ii) x(t) ¼ 5 cos(5t) cos (10t) (iii) x(t) ¼ 2sin(5t) cos (10t)

Find odd and even components for each of the following signals:

(a) x(t) ¼ u(t) (b) x(t) ¼ e(cid:3)atu(t)

Evaluate the following expressions:

ð1

ð (cid:7)

(i)

cos (cid:6)

2 (cid:3)1 (ii) e2t cos 50 π t ð1

t (cid:3) 5

ÞÞδ 2t (cid:3) 3

Þdt

ð Þ δ t þ 1 (cid:4) (cid:5) π 50 π t

e2t cos

ð δ t þ 1

Þdt

ð Þδ t (cid:3) 1

Þdt

t þ cos 2πt ð e(cid:3)t dδ tð Þ dt e(cid:3)tδ t (cid:3) 1 ð

Þdt

(iii)

(iv)

(v)

(vi)

ð1 (cid:3)1

(cid:3)1

1.10 MATLAB Exercises

Use MATLAB to generate the continuous-time signal shown in Figure 1.p1.1.
Generate and plot each of

the following continuous-time signals using

MATLAB: (i) x(t) ¼ 10 sin(2πt) cos (πt (cid:3) 4) for (cid:3)10 (cid:4) t (cid:4) 10 (ii) x(t) ¼ 2e(cid:3)0.1t sin(2πt) for (cid:3)5 (cid:4) t (cid:4) 5

1 Introduction

-1

-2

-3

-4

e d u t i l

p m Á

-5 -10

-8

-6

-4

-2

0 t (seconds)

Figure p1.1 Signal of MATLAB exercise 1

Further Reading

Pierce, J.R., Noll, A.M.: Signals: The Science of Telecommunications. American Library, New

Delhi (1960)

Lathi, B.P.: Linear Systems and Signals, 2nd edn. Oxford University Press, New York (2005)
Mandal, M., Asif, A.: Continuous and Discrete Time Signals and Systems. Cambridge Univer-

sity Press, Cambridge (2007)

Chapter 2 Continuous-Time Signals and Systems

This chapter presents time-domain analysis of continuous-time systems. It develops representation of signals in terms of impulses. The notions of linearity, time- invariance, causality, stability, memorability, and invertibility are introduced. It has shown that the input-output relationship for linear time-invariant (LTI) contin- uous systems is described in terms of a convolution integral. The differential equation representation of LTI continuous systems and classical solutions of differ- ential equations are also presented. Next, block-diagram representation of LTI continuous-time systems is introduced. Furthermore, a brief discussion on singular- ity functions is provided. Finally, the state-space representation of continuous-time LTI systems is described.

2.1 The Representation of Signals in Terms of Impulses

Consider pulse or staircase approximation bx tð Þ to continuous-time signal x(t) as shown in Figure 2.1. Then, the approximation signal can be expressed as sum of all these pulse signals. Deﬁne

δΔ tð Þ ¼

8 <

1 Δ, 0,

0 < t < Δ

otherwise

ð2:1Þ

Since δΔ(t)Δ ¼ 1, bx tð Þ can be expressed as

bx tð Þ ¼

k¼(cid:2)1

x kΔð

Þ δΔ t (cid:2) kΔ ð

ÞΔ

ð2:2Þ

2 Continuous-Time Signals and Systems

Figure 2.1 Representation of a signal in terms of impulses

x(t)

…

As Δ approaches zero, the above approximation bx tð Þ can be written as

bx tð Þ ¼ limΔ!0

k¼(cid:2)1

x kΔð

Þ δΔ t (cid:2) kΔ ð

ÞΔ

ð2:3Þ

Also, as Δ ! 0, the summation approaches an integral, and the pulse approaches

unit impulse. Therefore, Eq. (2.3) can be rewritten as

bx tð Þ ¼

(cid:2)1

x τð Þδ t (cid:2) τ ð

Þ dτ

ð2:4Þ

Thus, a continuous-time signal can be represented as weighted superposition of shifted impulses. Here the superposition is integration due to nature of the continuous-time input. The weight x(τ) dτ on the impulse δ(t (cid:2) τ) is determined from the value of the input signal x(t) at the time of occurrence of each impulse.

2.2 Continuous-Time Systems

2.2.1 Linear Systems

Let x1(t) and x2(t) are the inputs applied to a system characterized by the transfor- mation operator ℜ[] and y1(t) and y2(t) are the system outputs. A linear system should satisfy the principles of homogeneity and superposition. Hence, the following equations hold for a linear system

Principle of homogeneity:

y1 tð Þ ¼ ℜ x1 tð Þ

(cid:3),

y2 tð Þ ¼ ℜ x2 tð Þ

(cid:3),

ℜ ax1 tð Þ

(cid:3) ¼ ay1 tð Þ,

ℜ bx2 tð Þ

(cid:3) ¼ by2 tð Þ

ð2:5aÞ

ð2:5bÞ

ð2:6aÞ

ð2:6bÞ

2.2 Continuous-Time Systems

Principle of superposition:

Linearity:

ℜ x1 tð Þ ½

(cid:3) þ ℜ x2 tð Þ ½

(cid:3) ¼ y1 tð Þ þ y2 tð Þ

ℜ ax1 tð Þ

(cid:3) þ ℜ bx2 tð Þ ½

(cid:3) ¼ ay1 tð Þ þ by2 tð Þ

ð2:7Þ

ð2:8Þ

where a and b are arbitrary constants.

2.2.2 Time-Invariant System

A system is time invariant if the behavior and characteristics of the system are ﬁxed over time. A system is time invariant if a time shift in the input signal results in an identical time shift in the output signal. For example, a time-invariant system should produce y(t (cid:2) t0) as the output when x(t (cid:2) t0) is the input. Mathematically it can be speciﬁed as

y t (cid:2) t0 ð

Þ ¼ ℜ x t (cid:2) t0 ½ ð

(cid:3)

ð2:9Þ

Example 2.1 Check for linearity and time-invariance of the following system

y tð Þ ¼ tx tð Þ

Solution

Linearity: Let x1(t) and x2(t) be two distinct inputs applied to the system, then

y1 tð Þ ¼ ℜ x1 tð Þ

(cid:3) ¼ tx1 tð Þ, y2 tð Þ ¼ ℜ x2 tð Þ

(cid:3) ¼ tx2 tð Þ

If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then

y tð Þ ¼ tax1 tð Þ þ tbx2 tð Þ ¼ ay1 tð Þ þ by2 tð Þ

Hence, the system is linear.

Time-invariance:

y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼ tx tð Þ

The output y(t) of the system delayed by t0 can be written as

y t (cid:2) t0 ð

Þ ¼ t (cid:2) t0 ð

Þx t (cid:2) t0

2 Continuous-Time Signals and Systems

For example, for an input x1(t) ¼ x(t (cid:2) t0), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ

(cid:3) ¼ tx1 tð Þ ¼ tx t (cid:2) t0

y t (cid:2) t0 ð

Þ 6¼ y1 tð Þ

Hence, it is a time variant system.

Example 2.2 Check for linearity and time-invariance of the following system:

y tð Þ ¼ sin x tð Þ

Solution

Linearity:

Let x1(t) and x2(t) be two distinct inputs applied to the system, then

y1 tð Þ ¼ ℜ x1 tð Þ

(cid:3) ¼ sin x1 tð Þ

Þ, y2 tð Þ ¼ ℜ x2 tð Þ

(cid:3) ¼ sin x2 tð Þ

If an input equal to sum of the inputs ax1(t), bx2(t),x(t) ¼ ax1(t) þ bx2(t) is applied,

then

y tð Þ ¼ sin ax1 tð Þ

Þ þ sin bx2 tð Þ ð

Þ 6¼ ay1 tð Þ þ by2 tð Þ

Hence, the system is nonlinear.

Time-invariance:

y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼ sin x tð Þ Þ

The output y(t) of the system delayed by t0 can be written as

y t (cid:2) t0 ð

Þ ¼ sin x t (cid:2) t0 ð

Þ Þ

For example, for an input x1(t) ¼ x(t (cid:2) t0), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ Þ ¼ y1 tð Þ y t (cid:2) t0 ð

(cid:3) ¼ sin x1 tð Þ

Þ ¼ sin x t (cid:2) t0 ð

Hence, it is a time-invariant system.

Example 2.3 Determine if the following continuous-time systems are linear or nonlinear:

(i) dy tð Þ

dt þ 2ty tð Þ ¼ t2x tð Þ

(ii) 2y(t) þ 3 ¼ x(t)

(iii) y tð Þ ¼

x τð Þdτ

(cid:2)1

(iv) dy tð Þ

dt þ 3y tð Þ ¼ x tð Þ dx tð Þ

2.2 Continuous-Time Systems

Solution (i) Let x1(t) and x2(t) be two distinct inputs applied to the system, then

dy1 tð Þ dt dy2 tð Þ dt

þ 2ty1 tð Þ ¼ t2x1 tð Þ

þ 2ty2 tð Þ ¼ t2x2 tð Þ

If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then

dy1 tð Þ dt

þ 2aty1 tð Þ þ b

dy2 tð Þ dt

þ 2bty2 tð Þ ¼ at2x1 tð Þ þ bt2x2 tð Þ

Hence, the system is linear.

(ii)

y tð Þ ¼ x tð Þ (cid:2)

3 2

Let x1(t) and x2(t) be two distinct (cid:3) ¼ x1 tð Þ (cid:2) 3

2 , y2 tð Þ ¼ ℜ x2 tð Þ

y1 tð Þ ¼ ℜ x1 tð Þ

inputs applied to the system, (cid:3) ¼ x2 tð Þ (cid:2) 3 2

then

If an input equal to sum of the inputs ax1(t), bx2(t),x(t) ¼ ax1(t) þ bx2(t) is applied,

then

y tð Þ ¼ ax1 tð Þ (cid:2)

3 2

þ bx2 tð Þ (cid:2)

3 2

6¼ ay1 tð Þ þ by2 tð Þ

Hence, the system is nonlinear.

(iii) Let x1(t) and x2(t) be two distinct inputs applied to the system, then x2 τð Þdτ

x1 τð Þdτ, y2 tð Þ ¼ ℜ x2 tð Þ

y1 tð Þ ¼ ℜ x1 tð Þ

(cid:3) ¼

(cid:2)1

If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then ð

y tð Þ ¼

ax1 τð Þdτ þ

bx2 τð Þdτ ¼ ay1 tð Þ þ by2 tð Þ

(cid:2)1

Hence, the system is linear.

(iv) Let x1(t) and x2(t) be two distinct inputs applied to the system, then

dy1 tð Þ dt dy2 tð Þ dt

þ 3y1 tð Þ ¼ x1 tð Þ

þ 3y2 tð Þ ¼ x2 tð Þ

dx1 tð Þ dt dx2 tð Þ dt

2 Continuous-Time Signals and Systems

If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then

dy1 tð Þ dt

þ 3ay1 tð Þ þ b

dy2 tð Þ dt

þ 3by2 tð Þ 6¼ a2x1 tð Þ

dx1 tð Þ dt

þ b2x2 tð Þ

dx2 tð Þ dt

The system is nonlinear.

Example 2.4 Determine if the following continuous-time systems are time invariant or time variant:

(i) y(t) ¼ x((cid:2)t),

(ii) y tð Þ ¼

x τð Þdτ,

(cid:2)1 (iii) y(t) ¼ x(4t) (iv) y tð Þ ¼ 2 þ sin tð Þ ð

Þx tð Þ vð Þy tð Þ ¼ dx tð Þ dt

Solution

(i)

y(t) ¼ ℜ[x(t)] ¼ x((cid:2)t)

The output y(t) of the system delayed by t0 can be written as

y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼ x (cid:2)t ð

y t (cid:2) t0 ð

Þ ¼ x (cid:2) t (cid:2) t0

Þ ¼ x (cid:2)t þ t0 ð

For example, for an input x1(t) ¼ x(t (cid:2) t0), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ Þ 6¼ y1 tð Þ y t (cid:2) t0 ð

(cid:3) ¼ x1 (cid:2)t

Þ ¼ x (cid:2)t (cid:2) t0 ð

Hence, it is a time-varying system.

(ii) Let x(t) ¼ δ(t), then y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼

(cid:2)3

δ tð Þdt ¼ 1

Now, for an input x1(t) ¼ x(t (cid:2) 6), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ Þ 6¼ y1 tð Þ y t (cid:2) 6 ð

(cid:3) ¼

3 (cid:2)3

δ t (cid:2) 6 ð

Þdt ¼ 0

Hence, it is a time-varying system.

(iii) The output y(t) of the system delayed by t0 can be written as

y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼ x 4tð

y t (cid:2) t0 ð

Þ ¼ x 4 t (cid:2) t0 ð ð

Þ ¼ x 4t (cid:2) 4t0 ð

2.2 Continuous-Time Systems

For example, for an input x1(t) ¼ x(t (cid:2) t0), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ Þ 6¼ y1 tð Þ y t (cid:2) t0 ð

(cid:3) ¼ x1 4tð

Þ ¼ x 4t (cid:2) t0 ð

Hence, it is a time-varying system.

(iv)

y(t) ¼ ℜ[x(t)] ¼ (2 þ sin (t))x(t)

The output y(t) of the system delayed by t0 can be written as

y t (cid:2) t0 ð

Þ ¼ 2 þ sin t (cid:2) t0

Þx t (cid:2) t0

For example, for an input x1(t) ¼ x(t (cid:2) t0), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ Þ 6¼ y1 tð Þ y t (cid:2) t0 ð

(cid:3) ¼ 2 þ sin tð Þ

Þx t (cid:2) t0

Hence, it is a time-varying system.

(v)

y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼

dx tð Þ dt

The output y(t) of the system delayed by t0 can be written as

y t (cid:2) t0 ð

Þ ¼

dx t (cid:2) t0 dt

For example, for an input x1(t) ¼ x(t (cid:2) t0), the output y1(t) can be written as

y1 tð Þ ¼ ℜ x1 tð Þ

(cid:3) ¼

y t (cid:2) t0 ð

Þ ¼ y1 tð Þ

dx t (cid:2) t0 dt

Hence, it is a time-invariant system.

Example 2.5 Consider an LTI system with the response y(t) as shown in Figure 2.2 to the input signal x(t) ¼ u(t) (cid:2) u(t (cid:2) 2).

Figure 2.2 Response y(t) to the input x(t)

y(t)

2 Continuous-Time Signals and Systems

(t)

-2

Figure 2.3 Response y1(t) to the input x1(t)

Figure 2.4 Response y2(t) to the input x2(t)

(t)

-2

Determine and sketch the response of the system to the following inputs:

(i) x1(t) ¼ x(t) (cid:2) x(t (cid:2) 4) (ii) x2(t) ¼ x(t) þ x(t þ 2)

Solution

(i) x1(t) ¼ x(t) (cid:2) x(t (cid:2) 4)

Since it is an LTI system, the response y1(t) to the input x1(t) is given by y1(t) ¼

y(t) (cid:2) y(t (cid:2) 4) as depicted in Figure 2.3.

(ii) x2(t) ¼ x(t) þ x(t þ 2)

Since it is an LTI system, the response y2(t) to the input x2(t) is given by y2(t) ¼ y

(t) þ y(t þ 2) as depicted in Figure 2.4.

2.2.3 Causal System

The causal system generates the output depending upon present and past inputs only. A causal system is non-anticipatory.

2.3 The Convolution Integral

2.2.4 Stable System

When the system produces bounded output for bounded input, then the system is called bounded-input and bounded-output stable. If the signal is bounded, then its magnitude will always be ﬁnite.

2.2.5 Memory and Memoryless System

The output of a memory system at any speciﬁed time depends on the inputs at that speciﬁed time and at other times. Such systems have memory or energy storage elements. The system is said to be static or memoryless if its output depends upon the present input only.

2.2.6

Invertible System

A system is said to be invertible if the input can be recovered from its output. Otherwise the system is noninvertible system.

2.2.7 Step and Impulse Responses

If the input to the system is unit impulse input δ(t), the system output is called the impulse response and denoted by h(t):

h tð Þ ¼ ℜ δ tð Þ

(cid:3)

ð2:10Þ

If the input to the system is a unit step input u(t), then the system output is called

the step response s(t);that is,

s tð Þ ¼ ℜ u tð Þ

(cid:3)

ð2:11Þ

2.3 The Convolution Integral

The output of a system for an input expressed as weighted superposition as in Eq. (2.4) is given by

y tð Þ ¼ ℜ x tð Þ

(cid:3) ¼ ℜ

x τð Þδ t (cid:2) τ ð

Þdτ

ð2:12Þ

(cid:2) ð

(cid:2)1

(cid:3)

2 Continuous-Time Signals and Systems

From the linearity property of the system, Eq. (2.12) can be rewritten as

y tð Þ ¼

(cid:2)1

x τð Þℜ δ t (cid:2) τ ½

(cid:3)dτ

ð2:13Þ

For a time-invariant system, ℜ[δ(t (cid:2) τ)] ¼ h(t (cid:2) τ). Hence, we obtain

y tð Þ ¼

(cid:2)1

x τð Þh t (cid:2) τ ð

Þdτ

ð2:14Þ

Thus, the output y(t) of a linear time-invariant system to an arbitrary input x(t) is obtained in terms of the unit impulse input δ(t). Eq. (2.14) is referred to as the convolutional integral and is denoted by the symbol * as

y tð Þ ¼ x tð Þ∗h tð Þ ¼

(cid:2)1

x τð Þh t (cid:2) τ ð

Þdτ

ð2:15Þ

2.3.1 Some Properties of the Convolution Integral

2.3.1.1 The Commutative Property

x1 tð Þ∗x2 tð Þ ¼ x2 tð Þ∗x1 tð Þ

ð2:16Þ

Proof This property can be proved by a change of variable.

By the deﬁnition of the convolution integral

x1 tð Þ∗x2 tð Þ ¼

(cid:2)1 x1 τð Þx2 t (cid:2) τ

Þdτ

Let V ¼ t (cid:2) τ so that τ ¼ t (cid:2) V, and dτ ¼ (cid:2)dV:

Then

ð2:17Þ

(cid:2)1 1 x1 t (cid:2) V ð

x1 tð Þ∗x2 tð Þ ¼ (cid:2) Ð 1 (cid:2)1 x1 t (cid:2) V ð ¼ ¼ x2 tð Þ∗x1 tð Þ

Þx2 Vð ÞdV

ð2:18Þ

2.3.1.2 The Distributive Property

x1 tð Þ∗ x2 tð Þ þ x3 tð Þ

(cid:3) ¼ x1 tð Þ∗x2 tð Þ þ x1 tð Þ∗x3 tð Þ

ð2:19Þ

2.3 The Convolution Integral

Proof By the deﬁnition of the convolution integral

x1 tð Þ∗ x2 tð Þ þ x3 tð Þ

(cid:3) ¼

(cid:4) x2 t (cid:2) τ ð

x1 τð Þ

Þdτ

Þ þ x3 t (cid:2) τ ð ð

(cid:2)1

x1 τð Þx2 t (cid:2) τ ð

Þdτ þ

x1 τð Þx3 t (cid:2) τ ð

Þdτ

(cid:2)1 1

(cid:2)1

ð2:20Þ

¼ x1 tð Þ∗x2 tð Þ þ x1 tð Þ∗x3 tð Þ

2.3.1.3 The Associative Property

x1 tð Þ∗x2 tð Þ ½

(cid:3)∗x3 tð Þ ¼ x1 tð Þ∗ x2 tð Þ∗x3 tð Þ

(cid:3)

ð2:21Þ

Proof The left-hand side of the property can be expressed by

x1 tð Þ∗x2 tð Þ

(cid:3)∗x3 tð Þ ¼

(cid:2)1

x1 τ1ð

Þx2 t (cid:2) τ1 ð

Þdτ1∗x3 tð Þ

ð2:22Þ

where x1(t) * x2(t) is expressed as a convolution integral. Expanding the second convolution gives

x1 tð Þ∗x2 tð Þ

(cid:3)∗x3 tð Þ ¼

(cid:5)

(cid:2)1

(cid:6)

x1 τ1ð

Þx2 τ2 (cid:2) τ1

Þdτ1

x3 t (cid:2) τ2

Þdτ2

ð2:23Þ

Reversing the order of integration gives

x1 tð Þ∗x2 tð Þ

e∗x3 tð Þ ¼

(cid:2)1

x1 τ1ð

Þx2 τ2 (cid:2) τ1

Þ x3 t (cid:2) τ2 ð

Þdτ1dτ2

ð2:24Þ

Similarly, the right-hand side of the property can be written as

x1 tð Þ∗ x2 tð Þ∗x3 tð Þ

(cid:7)

(cid:8)

e ¼ x1 tð Þ∗ Ð Ð 1 1 (cid:2)1

(cid:2)1 x2 τ2ð

Þdτ2

Þ x3 t (cid:2) τ2 ð Þx2 t (cid:2) τ1 (cid:2) τ2

Þx2 τ2ð

(cid:2)1 x1 τ1ð

ð2:25Þ

Þ dτ1 dτ2

Now, it is to be shown that ð

x1 τ1ð

Þx2 τ2 (cid:2) τ1

Þx3 t (cid:2) τ2 ð

Þ dτ1dτ2 ¼

(cid:2)1

x1 τ1ð

Þx2 τ2ð

(cid:2)1 (cid:2)1 x2 t (cid:2) τ1 (cid:2) τ2 ð

Þdτ1 dτ2

ð2:26Þ

In the right hand τ1 integration, let v ¼ τ1 + τ2 and dτ1 ¼ dv.

2 Continuous-Time Signals and Systems

Then ð

(cid:2)1

x1 τ1ð

Þx2 τ2 (cid:2) τ1

Þ∗x3 t (cid:2) τ2 ð

Þdτ1dτ2 ¼

x1 V (cid:2) τ2

Þx2 τ2ð

(cid:2)1

(cid:2)1 x3 t (cid:2) V ð

Þ dV dτ2

ð2:27Þ

Next, let u ¼ V (cid:2) τ2 and(cid:2)dτ2 ¼ du Then ð

x1 τ1ð

Þx2 τ2 (cid:2) τ1

Þx3 t (cid:2) τ2 ð

Þdτ1dτ2 ¼ (cid:2)

(cid:2)1

x1 uð Þx2 V (cid:2) u ð

(cid:2)1

(cid:2)1 x3 t (cid:2) V

1 ÞdV d u

(cid:2)1

x1 τ1ð

Þx2 τ2 (cid:2) τ1

Þx3 t (cid:2) τ2 ð

Þdτ1dτ2 ¼

x1 uð Þx2 V (cid:2) u ð

(cid:2)1

(cid:2)1 x3 t (cid:2) V ð

ÞdV d u

ð2:28Þ

ð2:29Þ

The above right-hand side and left-hand side integrals are the same except for

change of the variables. Hence, the associative property is proved.

2.3.1.4 Convolution with an Impulse

Proof By deﬁnition

x tð Þ∗δ tð Þ ¼ x tð Þ

ð2:30Þ

x tð Þ∗δ tð Þ ¼

(cid:2)1

x τð Þδ t (cid:2) τ ð

Þ dτ

ð2:31Þ

Since δ(t (cid:2) τ) is an impulse at τ ¼ t and by sampling property of the impulse,

(cid:2)1 x τð Þδ t (cid:2) τ

Þdτ ¼ x τð Þjτ¼t ¼ x tð Þ

ð2:32Þ

Hence

x tð Þ∗δ tð Þ ¼ x tð Þ

2.3 The Convolution Integral

2.3.1.5 Convolution with Delayed Input and Delayed Impulse Response

If y(t) ¼ x(t) * h(t), then

x t (cid:2) t1 ð

Þ∗h t (cid:2) t2 ð

Þ ¼ y t (cid:2) t1 (cid:2) t2

ð2:33Þ

Proof By the convolution integral, we have ð

and

y tð Þ ¼ x tð Þ∗h tð Þ ¼

x t (cid:2) t1 ð

Þ∗h t (cid:2) t2 ð

Þ ¼

(cid:2)1

x τð Þh t (cid:2) τ ð

Þ dτ

ð2:34Þ

(cid:2)1

x τ (cid:2) t1 ð

Þh t (cid:2) τ (cid:2) t2

Þ dτ

ð2:35Þ

Let τ (cid:2) t1 ¼ υ. Then τ ¼ υ + t1, and Eq. (2.35) becomes

x t (cid:2) t1 ð

Þ∗h t (cid:2) t2 ð

Þ ¼

(cid:2)1

x υð Þh t (cid:2) t1 (cid:2) t2 (cid:2) υ

Þ dυ

ð2:36Þ

It is observed that replacing t by t (cid:2) t1 (cid:2) t2 in Eq. (2.34), we obtain Eq. (2.36).

Thus, it is proved that

x t (cid:2) t1 ð

Þ∗h t (cid:2) t2 ð

Þ ¼ y t (cid:2) t1 (cid:2) t2

Example 2.6 Determine the continuous-time convolution of x(t) and h(t) for the following:

(i) x(t) ¼ u(t)

h(t) ¼ u(t)

(ii) x(t) ¼ u(t (cid:2) a)

h(t) ¼ u(t (cid:2) b)

(iii) x(t) ¼ u(t þ 1) (cid:2) u(t (cid:2) 1)

h(t) ¼ u(t þ 1) (cid:2) u(t (cid:2) 1)

(iv) x(t) ¼ e(cid:2)(t(cid:2)2)u(t (cid:2) 2)

h(t) ¼ u(t þ 2)

(v)

h(t)

(t-1)

x(t)

2 Continuous-Time Signals and Systems

(vi) x(t) ¼ u(t)

h(t) ¼ e(cid:2)tu(t)

(vii) x(t) ¼ 2(u(t) (cid:2) u(t (cid:2) 2))

h(t) ¼ e(cid:2)t/2u(t)

Solution

(i) y tð Þ ¼

(cid:2)1 x τð Þh t (cid:2) τ Ð (cid:2)1 u τð Þu t (cid:2) τ ( t > 0 t < 0

¼ tu tð Þ

Þ ¼ x t (cid:2) 1 ð Ð t

(

Þdτ ¼

Þ 0 1dτ, 0,

t > 0 t < 0

(ii) u t (cid:2) a ð

Þ∗u t (cid:2) b ð

Þ ¼ u tð Þ∗δ t (cid:2) a ¼ u tð Þ∗u tð Þ ¼ u tð Þ∗u tð Þ

Þ∗ u tð Þ∗δ t (cid:2) b Þ ð Þ∗δ t (cid:2) b ð

Þ∗ δ t (cid:2) a ð Þ∗δ t (cid:2) a (cid:2) b

Since u(t)*u(t) ¼ tu(t),

u t (cid:2) a ð

Þ∗u t (cid:2) b ð

Þ ¼ tu tð Þ ð

Þ∗δ t (cid:2) a (cid:2) b

(iii) x tð Þ∗h tð Þ ¼ u t þ 1

Þu t (cid:2) a (cid:2) b (cid:3)∗ u t þ 1 ð Þ

¼ t (cid:2) a (cid:2) b

Þ (cid:2) u t (cid:2) 1 ð Þ∗u t þ 1 ð

¼ u t þ 1 ð

Þ (cid:2) u t þ 1 ð

(cid:2) u t (cid:2) 1 ð

Þ∗u t þ 1 ð

Þ þ u t (cid:2) 1 ð

¼ u t þ 1 ð

Þ∗u t þ 1 ð

Þ (cid:2) 2u t þ 1 ð Þ∗u t (cid:2) 1 ð

þu t (cid:2) 1 ð

Þ (cid:2) u t (cid:2) 1 ð

(cid:3) Þ

Þ∗u t (cid:2) 1 ð

Þ Þ∗u t (cid:2) 1 ð Þ∗u t (cid:2) 1 ð

¼ t þ 2 ð

Þu t þ 2

Þ (cid:2) 2tu tð Þ þ t (cid:2) 2

Þu t (cid:2) 2

as shown in Figure 2.5

Figure 2.5 The convolution of x(t) and h(t)

x(t)*h(t)

-2

2.3 The Convolution Integral

(iv)

y tð Þ ¼

Þ ¼ x t (cid:2) 1 Þ ð Þdτ Þu t (cid:2) τ þ 2

(cid:2)1 x τð Þδ t (cid:2) τ (cid:2) 1 ð (cid:2)1 e(cid:2) τ(cid:2)2 1 Þu τ (cid:2) 2 ( Ð

tþ2 2

e(cid:2) τ(cid:2)2 ð

Þdτ,

t > 0 t < 0

Letting τ1 ¼ τ – 2,

(

tþ2 2

eτ1 dτ1 0

(

2 (cid:2) e(cid:2)t,

t > 0 t < 0

y tð Þ ¼

(cid:2)1

x τð Þδ t (cid:2) τ (cid:2) 1

Þ ¼ x t (cid:2) 1 ð

(v)

Hence, y(t) is a shifted version of x(t) as shown in Figure 2.6.

(vi)

y tð Þ ¼

Þdτ Þu t (cid:2) τ ð

(cid:2)1 x τð Þh t (cid:2) τ (cid:2)1 u τð Þe(cid:2) t(cid:2)τ 0 e(cid:2) t(cid:2)τ Þdτ, (cid:9) (cid:9) t Þ=2 0 ¼ 1 (cid:2) e(cid:2)t ð

t > 0,

¼ ¼ e(cid:2) t(cid:2)τ ð

Þdτ

Þ,

t > 0,

(vii)

y tð Þ ¼ 0

y tð Þ ¼

t < 0 Ð

y tð Þ ¼

2 (cid:4) t (cid:4) 0,

Þ=2dτ, (cid:11) ,

t 0 2e(cid:2) t(cid:2)τ (cid:10) ¼ 4 1 (cid:2) e(cid:2)t=2 0 2e(cid:2) t(cid:2)τ Þ=2

¼ 4e(cid:2) t(cid:2)τ ¼ 4e(cid:2)t=2 e (cid:2)1 ð

Þ=2dτ, t (cid:4) 2, (cid:11)(cid:9) (cid:10) (cid:9)2 0 ¼ 4 e(cid:2) t(cid:2)2 t (cid:4) 2,

Þ,

Þ=2 (cid:2) e(cid:2)t=2

(cid:11)

t (cid:4) 2,

y tð Þ ¼ 0 t (cid:5) 0

Example 2.7 Consider LTI system with the impulse response h(t); for an input x(t), the output y(t) is as shown in Figure 2.7.

Figure 2.6 The shifted version of x(t)

y(t)

2 Continuous-Time Signals and Systems

Figure 2.7 Response y(t) to the input x(t)

y(t)

y(t-2)

Figure 2.8 Response y(t (cid:2) 2) to the input x(t (cid:2) 2)

Figure 2.9 Response y1(t) to the input x1(t)

(t)

Determine the output of the system for an input x1(t) ¼ x(t) (cid:2) x(t (cid:2) 2))

Solution Since the system is LTI, for an input x(t (cid:2) 2), the output is y(t (cid:2) 2) as shown in Figure 2.8.

The output y1(t) for the input x1(t) ¼ x(t) (cid:2) x(t (cid:2) 2) is given by y1(t) ¼ y(t) (cid:2) y(t (cid:2) 2), which is shown in Figure 2.9.

Example 2.8 Consider a LTI system with input and output related through the equation

y tð Þ ¼

(cid:2)1

e(cid:2) t(cid:2)τ ð

Þx τ (cid:2) 3

Þ dτ

2.3 The Convolution Integral

(i) Determine the impulse response h(t) of the system. (ii) Determine the output y(t) of the system for the input

x(t) ¼ u(t þ 1) (cid:2) u(t (cid:2) 3).

Solution

(i)

Let τ1 ¼ τ (cid:2) 3, then

y tð Þ ¼

(cid:2)1

e(cid:2) t(cid:2)τ ð

Þx τ (cid:2) 3

Þ dτ

e(cid:2) t(cid:2)3(cid:2)τ1 ð

Þx τ1ð

Þ dτ1

Hence, h(t) ¼ e(cid:2)(t (cid:2) 3)u(t (cid:2) 3) ð

(ii)

y tð Þ ¼

(cid:2)1

e(cid:2) t(cid:2)3(cid:2)τ1 ð

Þ u t (cid:2) τ1 þ 1 ½

Þ (cid:2) u t (cid:2) τ1 (cid:2) 3

(cid:3) dτ1

e(cid:2) t(cid:2)3(cid:2)τ1 ð

Þ u t (cid:2) τ1 þ 1 ½

Þ (cid:2) u t (cid:2) τ1 (cid:2) 3

(cid:3) dτ1

x(t (cid:2) τ) and h(τ) are shown in Figure 2.10. Using Figure 2.10, y(t) can be written as

e(cid:2) τ1(cid:2)3 ð

Þdτ1 ¼ 1 (cid:2) e(cid:2) t(cid:2)2

Þ,

2 < t (cid:5) 6,

t (cid:5) 2,

e(cid:2) τ1(cid:2)3 ð

Þdτ1 ¼ e(cid:2) t(cid:2)6

(cid:4) Þ 1 (cid:2) e(cid:2)4

(cid:12)

t > 6:

< :

0, ð

tþ1

y tð Þ ¼

t 2 3

t-3

t+1

Figure 2.10 x(t (cid:2) τ) and h(τ)

2 Continuous-Time Signals and Systems

2.3.2 Graphical Convolution

An understanding of graphical interpretation of convolution is very useful in com- puting the convolution of more complex signals. The stepwise procedure for graph- ical convolution is as follows: Step 1: Make x(τ) ﬁxed. Step 2: Invert h(τ) about the vertical axis (t ¼ 0) to obtain h((cid:2)τ). Step 3: Shift the h((cid:2)τ) along the τ axis by t0 seconds so that the shifted h((cid:2)τ) is

representing h(t0 (cid:2) τ).

Step 4: The area under the product of x(τ) and h(t0 (cid:2) τ) is y(t0), the value of

convolution at t ¼ t0.

Step 5: Repeat steps 3 and 4 for different values of positive and negative to obtain y

(t) for all values of t.

Example 2.9 Graphically determine the continuous-time convolution of h(t) and x(t) for the following:

(

0 (cid:5) t (cid:5) 4

otherwise

0 (cid:5) t (cid:5) 4

otherwise

(i) x tð Þ ¼

h tð Þ ¼

Solution

To compute y(t) ¼ x(t) * h(t), ﬁrst h((cid:2)τ) is to be obtained by inverting h(τ) about the vertical axis. Then, the product of x(τ) and h(t (cid:2) τ) is formed, point by point, and this product is integrated to compute y(t). Thus, the overlap area between the rectangles forming x(τ) and h(t (cid:2) τ) is y(t).

Clearly, y(0) ¼ 0 because there is no overlap between the rectangles forming x(τ) and h(t (cid:2) τ) at t ¼ 0. For 0 < t < 8, there is overlap between the rectangles forming x (τ) and h(t (cid:2) τ). For t (cid:4) 8, there is no overlap, and hence, y(8) ¼ 0. These are illustrated in Figure 2.11 with the ﬁnal result for y(t). The shaded portion represents the overlap area of the product x(τ) and h(t (cid:2) τ).

2.3 The Convolution Integral

(0)= ( )h(0 − ) = 0

h( − )| = 0 1

( )

-4

h(1 − )

( )

(1)= ( )h(1 − ) = 1

-4

-3

h(2 − )

( )

(2)= ( )h(2 − ) = 2

-4

-2

h(3 − )

( )

(3)= ( )h(3 − ) = 3

-4

-1

h(4 − )

( )

(4)= ( )h(4 − ) = 4

-4

-1

Figure 2.11 Steps in the convolution and the ﬁnal result

2 Continuous-Time Signals and Systems

h(5 − )

( )

(5)= ( )h(5 − ) = 3

-4

-1

h(2 − )

( )

(2)= ( )h(2 − ) = 2

-4

-2

h(3 − )

( )

(3)= ( )h(3 − ) = 3

-4

-1

h(4 − )

( )

(4)= ( )h(4 − ) = 4

-4

-1

h(5 − )

( )

(5)= ( )h(5 − ) = 3

-4

-1

Figure 2.11 (continued)

2.3 The Convolution Integral

h(6 − )

( )

(6)= ( )h(6 − ) = 2

-4

-1

h(7 − )

( )

(7)= ( )h(7 − ) = 1

-4

-1

h(8 − )

( )

(8)= ( )h(8 − ) = 0

-4

-1

y(t)

Figure 2.11 (continued)

2 Continuous-Time Signals and Systems

Example 2.10 Determine graphically y(t) ¼ x(t) * h(t) for the following x(t) and h(t) shown.

-1

Solution

-1

-3

-1

There is no overlap area between x(τ) and h((cid:2)τ) at t ¼ 0, y(0) ¼ 0. For 0 < t < 6, there is overlap between the rectangles forming x(τ) and h((cid:2)τ). For t (cid:4) 6, there is no overlap, and hence, y(6) ¼ 0. These are illustrated in Figure 2.12 with the ﬁnal result for y(t). The shaded portion represents the overlap area of the products x(τ) and h(t (cid:2) τ).

Example 2.11 Consider the RC circuit shown in Figure 2.13. Determine the Vout(t) for Vin(t) ¼ u(t (cid:2) 1) (cid:2) u(t (cid:2) 2), and assume the time constant RC ¼ 1 sec. Assume the capacitor is initially discharged. Solution The impulse response of the RC low-pass ﬁlter is

V out tð Þ ¼

h tð Þ ¼ e(cid:2)tu tð Þ (cid:2)1 V in τð Þh t (cid:2) τ ð

Þ ¼ V in tð Þ∗h tð Þ

2.3 The Convolution Integral

h(1 − )

( )

(1)= ( )h(1 − ) = 1

-2

-1

3 h(2 − )

( )

-1

h(1 − )

(2)= ( )h(2 − ) = 2

-1

h(2 − )

( )

h(3 − )

(3)= ( )h(3 − ) = 1 − 1=0

-1

h(3 − )

( )

h(4 − )

(4)= ( )h(4 − ) = − 2

-1

h(4 − )

Figure 2.12 Steps in the convolution and the ﬁnal result

2 Continuous-Time Signals and Systems

(

)

h( 5 − )

( 5) = (

) h( 5 − ) = − 1

-1

(

)

h( 6 − )

( 6) = (

) h( 6 − ) = 0

-1

y(t)

-2

h( 6 − )

Figure 2.12 (continued)

The steps involved in the convolution are illustrated in Figure 2.14.

V out tð Þ ¼ 0,

t < 1

2.3 The Convolution Integral

Figure 2.13 RC circuit

Figure 2.14 Illustration of steps in the convolution

V out tð Þ ¼

¼ e(cid:2) t(cid:2)τ ð Ð

1 (cid:5) t (cid:5) 2, (cid:11)

Þdτ, (cid:10)

1 e(cid:2) t(cid:2)τ (cid:9) (cid:9) t 1 ¼ 1 (cid:2) e(cid:2) t(cid:2)1 1 e(cid:2) t(cid:2)τ Þdτ, (cid:9) (cid:10) (cid:9)2 1 ¼ e(cid:2) t(cid:2)2

2 (cid:5) t,

¼ e(cid:2) t(cid:2)τ ð

1 (cid:5) t (cid:5) 2,

Þ (cid:2) e(cid:2) t(cid:2)1

(cid:11)

2 (cid:5) t:

which is shown in Figure 2.15.

Example 2.12 If (t) ¼ x(t) * h(t), then show that

y(2t) ¼ 2x(2t) * h(2t)

y 2tð

Þ ¼

(cid:2)1

x 2t (cid:2) τ ð

Þh τð Þdτ

, we have

Solution

Letting τ1 ¼

y 2tð

Þ ¼

2 Ð

(cid:2)1 x 2t (cid:2) 2τ1 ð Þ∗h 2tð Þ

¼ 2x 2tð

Þh 2τ1 ð

Þ2dτ1 ¼ 2

(cid:2)1 x 2t (cid:2) 2τ1 ð

Þh 2τ1 ð

Þdτ1

Example 2.13 If x(t) and h(t) are odd signals, then show that

y(t) ¼ x(t) * h(t) is an even signal.

2 Continuous-Time Signals and Systems

e d u t i l

p m A

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.5

1.5

2.5

3.5

Time

Figure 2.15 Times versus Vout(t)

Solution y(t) ¼ x(t) * h(t)

y (cid:2)t ð

Þ∗h (cid:2)t Þ ¼ x (cid:2)t ð ð Þ Ð 1 (cid:2)1 x (cid:2) t (cid:2) τ ð ð (cid:2)1 x (cid:2)t þ τ ð

Þdτ

Þ Þh (cid:2)τð

Þh (cid:2)τð Þdτ

Since x(t) and h(t) are odd signals, Ð

y (cid:2)t ð

Þ ¼

(cid:2)1 x t (cid:2) τ ð

Þh τð Þdτ

Hence, y(t) is even because y(t) ¼ y((cid:2)t).

¼ y tð Þ

Example 2.14 Consider an LTI system with the impulse response h(t) ¼ e(cid:2)tu(t). Find the system response for the input x(t) ¼ sin2tu(t).

Solution

y tð Þ ¼

(cid:2)1 x τð Þh t (cid:2) τ ð sin 2τð sin 2τð

Þdτ Þe(cid:2) t(cid:2)τ ð Þe(cid:2) t(cid:2)τ ð

Þdτ

1 0 1 0 h (cid:10)

¼ sin 2τð

Þe(cid:2) t(cid:2)τ ð

¼ sin 2tð

Þu tð Þ (cid:2) u tð Þ

Þdτ (cid:11) (cid:9) (cid:9) t τ¼0 (cid:2) Ð 1 0 2 cos 2τð

0 2 cos 2τð Þe(cid:2) t(cid:2)τ ð

Þdτ

u tð Þ

Þe(cid:2) t(cid:2)τ ð

2.3 The Convolution Integral

Hence, Ð

1 0

sin 2τð

Þe(cid:2) t(cid:2)τ ð

Þdτ ¼ sin 2tð (cid:10) Þu tð Þ (cid:2) u tð Þ 2 cos 2τð

¼ sin 2tð

Þu tð Þ (cid:2) 2 cos 2tð ð

0 2 cos 2τð Þu tð Þ (cid:2) u tð Þ (cid:11) (cid:9) Ð (cid:9) t τ¼0 (cid:2) Ð 1 0 4 sin 2τð

Þe(cid:2) t(cid:2)τ ð 0 4 sin 2τð Þe(cid:2) t(cid:2)τ ð

Þe(cid:2) t(cid:2)τ ð

Þu tð Þ (cid:2)

Þ þ e(cid:2)t

Þdτ

Þe(cid:2) t(cid:2)τ ð Þdτ

The above equation can be rewritten as

Therefore,

sin 2τð

Þe(cid:2) t(cid:2)τ ð

Þdτ ¼ sin 2tð ½

Þ (cid:2) 2 cos 2tð

Þ þ e(cid:2)t

(cid:3)u tð Þ

y tð Þ ¼

sin 2τð

Þe(cid:2) t(cid:2)τ ð

Þdτ ¼

1 5

sin 2tð

Þ (cid:2) 2 cos 2tð

Þ þ 2e(cid:2)t

(cid:3)u tð Þ

Example 2.15 If the response of an LTI system to input x(t) is the output y(t), dt is dy show that the response of the system to dx dt , and using this result determines the impulse response of an LTI system having the response y(t) ¼ sin2t for an input x(t) ¼ e(cid:2)4tu(t).

Solution

y tð Þ ¼ x tð Þ∗h tð Þ Ð 1 (cid:2)1 h τð Þx t (cid:2) τ

Þdτ

Differentiating both sides with respect to t,

(cid:2)1

h τð Þ

dx dt

¼ h tð Þ∗dx dt

dy dt dy dt

t (cid:2) τ

Þdτ

dt ¼ 2 sin 2t, and for given x tð Þ, dx

For given y(t), dy From sampling property impulse function, it is known that x(t) δ (t) ¼ x(0)δ(t). Since e(cid:2)4tδ(t) ¼ e(cid:2)0δ(t) ¼ δ(t), dx

dt ¼ (cid:2)4e(cid:2)4t þ e(cid:2)4tδ tð Þ:

dt can be written as

dx dt

¼ (cid:2)4e(cid:2)4t þ δ tð Þ

Let x1(t) ¼ 4e(cid:2)4tu(t), then by homogeneity, the corresponding output

y1 tð Þ ¼ 4y tð Þ ¼ 4 sin 2t

2 Continuous-Time Signals and Systems

Let x2 tð Þ ¼ dx

dt ¼ (cid:2)4e(cid:2)4t þ δ tð Þ the corresponding output

y2 tð Þ ¼ 2 sin 2t

As it is LTI system, if (x1(t) þ x2(t)) is the input to the system, the corresponding

output is(y1(t) þ y2(t))

since x1(t) þ x2(t) ¼ 4e(cid:2)4t (cid:2) 4e(cid:2)4t + δ(t) ¼ δ(t), the impulse response h

(t) ¼ y1(t) þ y2(t) ¼ 4 sin 2t þ 2 sin 2t

Example 2.16 Consider a continuous-time LTI system with the unit step response s(t):

(i) Deduce that the response y(t) of the system to the input x(t) is

and also show that

y tð Þ ¼

(cid:2)1

dx τð Þ dt

s t (cid:2) τ ð

Þdτ

x tð Þ ¼

(cid:2)1

dx τð Þ dt

u t (cid:2) τ ð

Þdτ:

(ii) Determine the response of an LTI system with step response

(cid:10)

s tð Þ ¼ e(cid:2)2t (cid:2) e(cid:2)t þ 1

(cid:11)

u tð Þ

to an input x(t) ¼ etu(t).

Solution

(i) The step response s(t) is

s tð Þ ¼ h tð Þ∗u tð Þ Ð 1

(cid:2)1 h τð Þu t (cid:2) τ ð Ð (cid:2)1 h τð Þdτ

Þdτ

Consider the following equivalence.

x(t)

h(t)

y(t)

x(t)

h(t)

y(t)

From the above equivalence, we obtain

2.3 The Convolution Integral

h(t)

y(t)

Thus,

(cid:14)

h τð Þdτ

(cid:13) ð

∗

(cid:2)1

∗s tð Þ

dx tð Þ dt dx tð Þ dt

y tð Þ ¼

Since y(t) ¼ x(t) ∗ h(t) and if h(t) ¼ δ(t) and y(t) ¼ x(t) as x(t) ∗ δ(t) ¼ x(t),thus,

(cid:14)

(cid:16)

y tð Þ ¼

dx tð Þ dt

∗

(cid:13) ð

(cid:2)1

h τð Þdτ

becomes

putting

h(t) ¼ δ(t)

x tð Þ ¼ dx tð Þ dt (cid:15)

∗

t (cid:2)1

δ τð Þdτ (cid:16)

(cid:15) Ð

Since

(ii)

t (cid:2)1

x tð Þ ¼

δ τð Þdτ dx tð Þ dt x tð Þ ¼ etu tð Þ dx tð Þ dt

¼ etu tð Þ þ δ tð Þet

¼ u tð Þ

∗u tð Þ

Since

δ tð Þet ¼ δ tð Þe0 ¼ δ tð Þ ¼ etu tð Þ þ δ tð Þ

dx tð Þ dt

(cid:11)

u tð Þ

yðtÞ ¼

s tð Þ ¼ e(cid:2)2t (cid:2) e(cid:2)t þ 1 dxðtÞ dt 1 (cid:2)1 1

∗sðtÞ dxðτÞ dt

sðt (cid:2) τÞdτ

(cid:10)

Ð Ð Ð

¼ ¼ ¼

(cid:2)1 feτuðτÞ þ δðτÞgfe(cid:2)2ðt(cid:2)τÞ (cid:2) e(cid:2)ðt(cid:2)τÞ þ 1guðt (cid:2) τÞdτ 0 etfe(cid:2)2ðt(cid:2)τÞ (cid:2) e(cid:2)ðt(cid:2)τÞ þ 1gdτ þ fe(cid:2)2t (cid:2) e(cid:2)t þ 1guðtÞ 0 fe(cid:2)2tþ3τ (cid:2) e(cid:2)tþ2τ þ eτgdτ þ sðtÞ 1 1 e(cid:2)2tðe3t (cid:2) 1Þ (cid:2) e(cid:2)tðe(cid:2)2t (cid:2) 1Þ þ ðe(cid:2)t (cid:2) 1Þ þ sðtÞ 2 3 5 1 6 3

et (cid:2) 1 þ sðtÞ

e(cid:2)2t þ

e(cid:2)t þ

1 2

¼ (cid:2)

Example 2.17 Consider h(t) be the triangular pulse and x(t) be the unit impulse train as shown in Figure 2.16. Determine y(t) ¼ x(t) * h(t) for T ¼ 2.

2 Continuous-Time Signals and Systems

h (t)

x (t)

-1

-2T

-T

Figure 2.16 x(t) and h(t) of Example 2.17

y(t)

-3

-2

-1

Figure 2.17 Time versus y(t)

Solution

x tð Þ ¼

δ t (cid:2) nT ð

y tð Þ ¼ x tð Þ∗h tð Þ ¼

n¼(cid:2)1 h tð Þ∗δ t (cid:2) nT ð

Þ ¼

n¼(cid:2)1

h t (cid:2) nT ð

which is shown in Figure 2.17.

2.3.3 Computation of Convolution Integral Using MATLAB

MATLAB provides a function conv() that performs a discrete-time convolution of two discrete-time sequences. A new function convint() that uses conv() to numeri- cally integrate the continuous-time convolution is as follows.

2.3 The Convolution Integral

function[y,ty]=convint(x,tx,h,th) %Inputs: %x is the input signal vector %tx is the times of the samples in x %h is the impulse response vector %th is times of the samples in h %outputs: %y is the output signal vector, %length(y)=length(x)+length(h)-1 %ty is the time of the samples in y dt=tx(2)-tx(1); y=conv(x,h)*dt; ty=(tx(1)+th(1))+[0:(length(y)-1)]*dt;

The computation of convolution of continuous-time signals using MATLAB is

illustrated through the following numerical examples.

Example 2.18 (i) Verify the result of Example 2.9 using MATLAB. (ii) Verify the result of Example 2.10 using MATLAB.

Solution (i) The following MATLAB program 2.1 is used to compute the convo-

lution of x(t) and h(t) of Example 2.9.

Program 2.1

clc; clear all; close all; tx=[0:0.01:4]; x=ones(1,length(tx)); th=[0:0.01:4]; h=ones(1,length(th)); [y ty]=convint(x,tx,h,th); figure; plot(ty,y); xlabel(‘Time’) ylabel(‘Amplitude’); axis([0 8 0 4]);

The output y(t) ¼ x(t) * h(t) of the above program is shown in Figure 2.18. It is

observed to be the same as that shown in Figure 2.11. Thus, it is veriﬁed.

(ii) The following MATLAB program 2.2 is used to compute the convolution of x(t)

and h(t) of Example 2.10.

2 Continuous-Time Signals and Systems

3.5

2.5

1.5

0.5

e d u t i l

p m A

4 Time

Figure 2.18 Time versus y(t)

e d u t i l

p m A

1.5

0.5

-0.5

-1

-1.5

-2

Figure 2.19 Time versus y(t)

2.3 The Convolution Integral

Program 2.2

clc; clear all; close all; tx=[0:0.01:3]; tx1=[0:0.01:1]; x=[zeros(1,length(tx1)) ones(1,(length(tx)-length(tx1)))]; th1=[-1:0.01:1]; th2=[1.01:0.01:3]; h=[ones(1,length(th1)) -1*ones(1,length(th2))]; th=[-1:0.01:3]; [y ty]=convint(x,tx,h,th); figure; plot(ty,y); xlabel(‘Time’) ylabel(‘Amplitude’); axis([0 6 -2 2]);

The output y(t) ¼ x(t) * h(t) of tie above program is shown in Figure 2.19 It is

observed to be the same as that shown in Figure 2.12. Thus, it is veriﬁed

Example 2.19 Consider the RC circuit of Example 2.11 with time constant RC ¼ 1 3 sec . Determine the Vout(t) using MATLAB for Vin(t) ¼ (u(t (cid:2) 3) (cid:2) u (t (cid:2) 5)). Assume the capacitor is initially discharged.

Solution The impulse response of the RC circuit is given by

h tð Þ ¼

(cid:2)t=

1 RC

RC u tð Þ ¼ 3e(cid:2)3tu tð ÞV out tð Þ ¼

(cid:2)1

V in τð Þh t (cid:2) τ ð

Þ ¼ V in tð Þ∗h tð Þ

The following MATLAB program 2.3 is used to compute the convolution of vin(t)

and h(t).

Program 2.3

clc; clear all; close all; tx=[0:0.01:5]; tx1=[0:0.01:3]; x=[zeros(1,length(tx1)) ones(1,(length(tx)- length(tx1)))]; th=[0:0.01:5]; h =(3)* exp(-3*th); [y ty]=convint(x,tx,h,th); figure; plot(ty,y); xlabel(‘Time’) ylabel(‘Amplitude’);

2 Continuous-Time Signals and Systems

e d u t i l

p m A

0.12

0.1

0.08

0.06

0.04

0.02

Time

Figure 2.20 Time versus Vout(t)

(a)

(b)

Figure 2.21 (a) Cascade connection of two systems. (b) Equivalent system

The output Vout(t) ¼ Vin(t) * h(t) of the above program is shown in Figure 2.20.

2.3.4

Interconnected Systems

2.3.4.1 Cascade Connection of Systems

The system shown in Fig. 2.21 is formed by connecting two systems in cascade. The impulse responses of the systems are given by h1(t) and h2(t), respectively. Let y(t) be the output of the ﬁrst system. By the deﬁnition of convolution

y1 tð Þ ¼ x tð Þ∗h1 tð Þ

Then, the output of the overall system y(t) is given by

y tð Þ ¼ y1 tð Þ∗h2 tð Þ ¼ x tð Þ∗h1 tð Þ

(cid:3)∗h2 tð Þ

ð2:37Þ

ð2:38Þ

2.3 The Convolution Integral

By the associativity property of convolution, Eq. (2.38) can be rewritten as

y tð Þ ¼ y1 tð Þ∗h2 tð Þ ¼ x tð Þ∗ h1 tð Þ∗h2 tð Þ

(cid:3)

ð2:39Þ

Hence, the impulse response of the overall system is given by

h tð Þ ¼ h1 tð Þ∗h2 tð Þ ¼ ¼

(cid:2)1 h1 τð Þ h2 t (cid:2) τ (cid:2)1 h2 τð Þ h1 t (cid:2) τ

Þdτ Þdτ

ð2:40Þ

2.3.4.2 Parallel Connection of Two LTI Systems

The system shown in Fig. 2.22 is formed by connecting two systems in parallel. The impulse responses of the systems are given by h1(t) and h2(t), respectively. Let y1(t) and y2(t) be the outputs of the ﬁrst system and second system, respectively. By the deﬁnition of convolution

y1 tð Þ ¼ x tð Þ∗h1 tð Þ y2 tð Þ ¼ x tð Þ∗h2 tð Þ

ð2:41Þ ð2:42Þ

Then, the output of the overall system y(t) is given by

y tð Þ ¼ y1 tð Þ þ y2 tð Þ ¼ x tð Þ∗h1 tð Þ þ x tð Þ∗h2 tð Þ

ð2:43Þ

By the distributive property of convolution, Eq. (2.43) can be rewritten as

y tð Þ ¼ y1 tð Þ þ y2 tð Þ ¼ x tð Þ∗ h1 tð Þ þ h2 tð Þ

(cid:3)

Hence, the impulse response of the overall system is given by

h tð Þ ¼ h1 tð Þ þ h2 tð Þ

ð2:44Þ

(a)

(b)

Figure 2.22 (a) Parallel connection of two systems. (b) Equivalent system

2 Continuous-Time Signals and Systems

Example 2.20 An LTI system consists of two subsystems in cascade. The impulse responses of the subsystems are, respectively, given by

h1 tð Þ ¼ e(cid:2)3tu tð Þ; h2 tð Þ ¼ e(cid:2)tu tð Þ

Find the impulse response of the overall system.

Solution The overall impulse response of the system is given by

Þu t (cid:2) τ ð

Þdτ

h tð Þ ¼ h1 tð Þ∗h2 tð Þ 1

(cid:2)1 e(cid:2)3τu τð Þe(cid:2) t(cid:2)τ 0 e(cid:2)3τe(cid:2) t(cid:2)τ 0 e(cid:2)2τdτ (cid:11)

Þdτ

e(cid:2)t (cid:2) e(cid:2)2t

u tð Þ

¼ e(cid:2)t (cid:10) 1 2

2.3.5 Periodic Convolution

If the signals x1(t) and x2(t) are periodic with common period T, it can be easily shown that the convolution of x1(t) and x2(t) does not converge. In such a case, the periodic convolution of x1(t) and x2(t) is deﬁned as

y tð Þ ¼ x1 tð Þ

x2 tð Þ ¼

x1 τð Þx2 t (cid:2) τ ð

Þdτ

ð2:45Þ

Example 2.21 Let y(t) be the periodic convolution of x1(t) and x2(t). Show that

Solution (i) y t þ T

Þ ¼

y tð Þ ¼ y t þ T

x1 τð Þx2 t þ T (cid:2) τ ð

Þdτ

Since x2(t) is periodic with period T, x2(t þ T (cid:2) τ) ¼ x2(t (cid:2) τ)

0 x1 τð Þx2 t þ T (cid:2) τ ð

Þdτ ¼

0 x1 τð Þx2 t (cid:2) τ

Þdτ

y t þ T ð

Þ ¼ y tð Þ:

2.4 Properties of Linear Time-Invariant Continuous-Time System

2.4 Properties of Linear Time-Invariant Continuous-Time

System

2.4.1 LTI Systems With and Without Memory

The output y(t) of a memoryless system depends only on the present input x(t). If the system is LTI, then the relationship between the y(t) and x(t) for a memoryless system is

y tð Þ ¼ cx tð Þ

ð2:46Þ

where c is an arbitrary constant. Since the output of a continuous-time system can be written as

y tð Þ ¼

(cid:2)1

h τð Þx t (cid:2) τ ð

Þdτ

the corresponding impulse response is h(t) ¼ cδ(t).

Thus, a continuous-time system is memoryless if and only if

h tð Þ ¼ cδ tð Þ

ð2:47Þ

2.4.2 Causality for LTI Systems

The output of a continuous-time system can be written as

y tð Þ ¼

(cid:2)1

h τð Þx t (cid:2) τ ð

Þdτ

Since the impulse response h(τ) ¼ 0 for τ < 0 for a causal continuous-time system, the output of a causal system can be expressed by the following convolution integral:

y tð Þ ¼

h τð Þx t (cid:2) τ ð

Þdτ

ð2:48Þ

2.4.3 Stability for LTI Systems

A continuous-time system is BIBO stable if and only if the impulse response is absolutely integrable, that is,

2 Continuous-Time Signals and Systems

(cid:2)1

h τð Þdτ < 1

ð2:49Þ

Example 2.22 Check stability of continuous-time system having the following impulse responses: (i) h(t) ¼ e(cid:2)tu(t) (ii) h(t) ¼ e(cid:2)t cos (2t)u(t) (iii) h(t) is periodic and nonzero

Solution

(i)

(cid:2)1

h τð Þ

jdτ ¼

e(cid:2)τdτ ¼ 1

Indicating that h(t) is absolutely integrable and, hence, h(t) is the impulse

response of a stable system,

(ii)

(cid:2)1

h τð Þ

jdτ ¼

e(cid:2)τ cos 2τð

jdτ

Since e(cid:2)τ|cos (2τ)| is exponentially decaying for 0 (cid:5) t (cid:5) 1 , h(t) is absolutely

summable, and hence, h(t) is the impulse response of a stable system.

(iii) If h(t) is periodic with period T, then

(cid:2)1

h τð Þ j

jdτ ¼ N

T=2

(cid:2)T=2

h τð Þ

jdτ

where N ! 1

Since h(t) is nonzero

h τð Þ

jdτ ! 1, hence, h(t) is absolutely summable and,

hence, h(t) is the impulse response of an unstable system.

(cid:2)1

Example 2.23 Determine if each of the following system is causal or stable: (i) h(t) ¼ e(cid:2)tu(t (cid:2) 1) (ii) h(t) ¼ e(cid:2)tu((cid:2)t þ 1) (iii) h(t) ¼ e(cid:2)2tu(t þ 10) (iv) h(t) ¼ te(cid:2)tu(t) (v) h(t) ¼ e+tu((cid:2)t (cid:2) 1) (vi) h(t) ¼ e(cid:2)2|t|

Solution (i) Causal because h(t) ¼ 0 for t < 0. Stable because

(ii) Not causal because h(t) 6¼ 0 for t < 0. Unstable because ð

(iii) Not causal because h(t) 6¼ 0 for t < 0. Stable because

(cid:2)1

h τð Þ jdτ < 1. ð 1 h τð Þ

jdτ ¼ 1.

(cid:2)1 h τð Þ

jdτ < 1.

(cid:2)1

2.4 Properties of Linear Time-Invariant Continuous-Time System

(iv) Causal because h(t) ¼ 0 for t < 0. Stable because

h τð Þ

jdτ < 1.

(v) Not causal because h(t) 6¼ 0 for t < 0. Stable because

(vi) Not causal because h(t) 6¼ 0 for t < 0. Unstable because

(cid:2)1

ð (cid:2)1

h τð Þ

jdτ < 1.

j ð (cid:2)1

h τð Þ

jdτ ¼ 1.

2.4.4

Invertible LTI System

A system is invertible if its input x(t) can be recovered from its output y(t) ¼ x(t) * h (t). The cascade of an LTI system having impulse response h(t) with a LTI inverse system having impulse response g(t) ¼ h(cid:2)1(t) is shown in Figure 2.23.

The process of recovering x(t) from x(t) * h(t) is called deconvolution as it

corresponds to reverse of the convolution operation.

The overall impulse response of the invertible system shown in Figure 2.23 is the convolution of h(t) and g(t). It is required that the output of the invertible system is equivalent to the input:

implying that

(cid:10)

x tð Þ∗ h tð Þ∗h(cid:2)1 tð Þ

(cid:11)

¼ x tð Þ

h tð Þ∗h(cid:2)1 tð Þ ¼ δ tð Þ

ð2:50Þ

ð2:51Þ

As an example, it is veriﬁed that the inverse system for a continuous-time

integrator is a differentiator as follows:

(cid:2) ð

(cid:2)1

d dt

(cid:3)

x τð Þdτ

¼ x tð Þ

ð2:52Þ

Hence, the input-output relation for the inverse system shown in Figure 2.24 is

x tð Þ ¼

dy tð Þ dt

ð2:53Þ

x(t)

h(t)

y(t)

x(t)

(t)

Figure 2.23 Cascade connection of an LTI system and its inverse

2 Continuous-Time Signals and Systems

x(t)

y(t)

x(t)

Figure 2.24 Input-output relation for the inverse system

Example 2.24

(i) An echo of an auditorium can be modeled as a LTI system with an impulse

response consisting of a train of impulses:

h tð Þ ¼

k¼0

hkδ t (cid:2) kT ð

The inverse LTI system with impulse response g(t) is

where g(t) is also an impulse train that is modeled as

y tð Þ∗g tð Þ ¼ x tð Þ

g tð Þ ¼

k¼0

δ t (cid:2) kT ð

Obtain the relationship between hkand gk.

Solution

y(t) ¼ x(t) * h(t) and x(t) ¼ g(t) *y(t), then

However,

g tð Þ∗h tð Þ ¼

g tð Þ∗h tð Þ ¼ δ tð Þ:

δ t (cid:2) τ (cid:2) kT ð

hmδ τ (cid:2) mT

Þdτ

gkhmδ t (cid:2) τ (cid:2) kT

m¼0 Þδ t (cid:2) m þ k ð

ÞT

1 (cid:2)1 X1

k¼0 X1

k¼0

m¼0

Let n ¼ m þ k, then m ¼ n-k, and g(t) * h(t) can be rewritten as

g tð Þ∗h tð Þ ¼

n¼0

k¼0

gkhn(cid:2)k

δ t (cid:2) nT ð

2.4 Properties of Linear Time-Invariant Continuous-Time System

Hence,

Implying that

k¼0

gkhn(cid:2)k ¼

(

n ¼ 0, n 6¼ 0:

g0h0 ¼ 1, g0h1 þ g1h0 ¼ 0, g0h2 þ g1h1 þ g2h0 ¼ 0,

and so on, solution of the above equations leads to

g0 ¼

g1 ¼

(cid:2)g0h1 h0

1 h0 (cid:2)h1 h0h0

(cid:2)h1 h2 0 !

g0h2 þ g1h1 þ g2h0 ¼ (cid:2)

1 h0

(cid:2)

1 h0

h2 (cid:2)

h1 h2 0

¼ (cid:2)

1 h0

h2 h0

(cid:2)

h2 1 h2 0

(ii) Consider the following echo generation model characterized by

y tð Þ ¼ x tð Þ þ ay t (cid:2) T

where 0 < a < 1 and T is delay.

Construct the corresponding inverse system and obtain its impulse response.

Solution Assuming y(t) ¼ 0 for t < 0 and x(t) ¼ 0 for t < 0, the impulse response h(t) of the echo generation system is given by

hðtÞ ¼

k¼0

akδðt (cid:2) kTÞ

Thus, h0 ¼ 1, h1 ¼ a, hi ¼ 0 The inverse system has to obtain x(t) from the output y(t). Hence, the inverse

for i > 2.

system is characterized by

x tð Þ ¼ y tð Þ (cid:2) ay t (cid:2) T

and represented as depicted in Figure 2.25.

2 Continuous-Time Signals and Systems

Figure 2.25 An inverse system

y(t)

x(t)

Delay T

-a

The impulse response g(t) of the inverse system is given by

g tð Þ ¼

k¼0

(cid:2)að

Þkδ t (cid:2) kT ð

Hence, g0 ¼ 1, g1 ¼ (cid:2)a.

Example 2.25 Check y(t) ¼ x(2t) for causality and invertibility.

Solution

At time t ¼ 1

y tð Þ ¼ x 2tð

y 1ð Þ ¼ x 2ð Þ

indicating that the value of y(t) at time t ¼ 1 depends on x(t) at a time t ¼ 2. Therefore, y(t) ¼ x(2t) is not causal.

y(t) is invertible;

x tð Þ ¼ y t=2ð

2.5 Systems Described by Differential Equations

2.5.1 Linear Constant-Coefﬁcient Differential Equations

A general Nth-order linear constant-coefﬁcient differential equation is given by

n¼0

dny tð Þ dtn ¼

k¼0

dnx tð Þ dtn

ð2:54Þ

where coefﬁcients an and bn are real constants. The order N refers to the highest derivative of y(t) in Eq. (2.54). For example, consider the RC circuit considered in Example 2.11, the input x(t) and the output y(t) ¼ vo (t). If the current ﬂowing through the RC circuit is i(t), using Kirchhoff’s voltage law, we write

2.5 Systems Described by Differential Equations

which can be rewritten as

(cid:2)x tð Þ þ Ri tð Þ þ

1 c

i tð Þdt ¼ 0

Ri tð Þ þ

1 c

i tð Þdt ¼ x tð Þ

ð2:55Þ

ð2:56Þ

Since i tð Þ ¼ c dv0 tð Þ

dt ¼ c dy tð Þ

, substituting i tð Þ ¼ c dy tð Þ dt

following ﬁrst-order constant-coefﬁcient differential equation

in Eq. (2.56), we obtain the

dy tð Þ dt

1 RC

y tð Þ ¼

1 RC

x tð Þ

ð2:57Þ

relating the voltage across the capacitor y(t) and the input x(t).

Example 2.26 Find the differential equation relating the current y(t) and the input voltage x(t) for the RLC circuit shown in Figure 2.26 assuming R ¼ 3 Ohms, L ¼ 1 Henry, and C ¼ 1

2 Farad:

Solution Using Kirchhoff’s voltage law, we write the following loop equation for the given RLC circuit:

(cid:2)x tð Þ þ Ry tð Þ þ L

dy tð Þ dt

1 c

y tð Þdt ¼ 0

For R ¼ 3, L ¼ 1, and C ¼ 1

2 , the above equation becomes

þ 3y tð Þ þ 2

y tð Þdt ¼ x tð Þ

dy tð Þ dt

Differentiating this equation, we obtain

d2y tð Þ dt2 þ 3

dy tð Þ dt

þ 2y tð Þ ¼

dx tð Þ dt

Figure 2.26 RLC circuit

2 Continuous-Time Signals and Systems

Figure 2.27 Operational ampliﬁer circuit

Example 2.27 Find the differential equation relating the input voltage Vi tð Þ and the output voltage Vo tð Þ for the operational ampliﬁer circuit shown in Figure 2.27.

Solution

‘Ir1 ¼ Ir2 þ Ic1

;

Ir2 ¼ Ic2

; V2 ¼ V0

Rewriting the current node equations, we get

Vi (cid:2) V1 r1

V1 (cid:2) V0 r2

þ c1

d dt

V1 (cid:2) V0

which can be rewritten as

r1r2c1

dV1 dt

(cid:2) r1r2c1

þ r1 þ r2

ÞV1 (cid:2) r1V0 ¼ Vir2

dV0 dt V1 (cid:2) V0 r2

¼ c2

dV0 dt

;

V1 ¼ r2c2

dV0 dt

þ V0

Substituting the above equation for V1, the input-output relation can be written as

c2c1r2r1

d2V0 dt2 þ c2 r1 þ r2 ð

dV0 dt

þ V0 ¼ Vi

which is rewritten as

d2V0 dt2 þ

r1 þ r2 r1r2c1

dV0 dt

V0 r1r2c1c2

Vi r1r2c1c2

2.5 Systems Described by Differential Equations

2.5.2 The General Solution of Differential Equation

The general solution of Eq. (2.54) for a particular input x(t) is given by

y tð Þ ¼ yc tð Þ þ yp tð Þ

ð2:58Þ

where yc(t) is called the complementary solution and yp(t) is called the particular solution. The complementary solution yc(t) is obtained by setting x(t) ¼ 0 in Eq. (2.54). Thus yc(t) is the solution of the following homogeneous differential equation

n¼0

dny tð Þ dtn ¼ 0

ð2:59Þ

Example 2.28 Consider the RC circuit of Example 2.11 with time constant RC ¼ 1 sec. Determine the voltage across the capacitor for an input x(t) ¼ e(cid:2)2tu(t). Assume the capacitor is initially discharged.

Solution As the time constant RC ¼ 1, the input x(t) and the output y tð Þ ¼ V0 tð Þ of the RC circuit are related by

dy tð Þ dt

þ y tð Þ ¼ e(cid:2)2tu tð Þ

y 0ð Þ ¼ 0

The particular solution for the exponential input is of the form

yp tð Þ ¼ Ae(cid:2)2t

t > 0

Substituting yp(t) in the above differential equation, we get

(cid:2)2Ae(cid:2)2t þ Ae(cid:2)2t ¼ e(cid:2)2t

t > 0

Solving for A, we obtain A ¼ 1 and

yp tð Þ ¼ (cid:2)e(cid:2)2t

To obtain complementary solution, let us assume

Substituting this into

yc tð Þ ¼ Bekt

dyc tð Þ dt

þ yc tð Þ ¼ 0

yields

Thus, k ¼ (cid:2)1 and

Now,

2 Continuous-Time Signals and Systems

Bkekt þ Bekt ¼ 0 ÞBekt ¼ 0

k þ 1 ð

yc tð Þ ¼ Be(cid:2)t

y tð Þ ¼ yc tð Þ þ yp tð Þ ¼ Be(cid:2)t (cid:2) e(cid:2)2t

at t ¼ 0 y(0) ¼ B – 1.

Since the capacitor is initially discharged, y(0) ¼ 0, and we obtain B ¼ 1. Hence, the voltage across the capacitor is given by

(cid:10) y tð Þ ¼ e(cid:2)t (cid:2) e(cid:2)2t

(cid:11)

u tð Þ

2.5.3 Linearity

The system speciﬁed by Eq. (2.54) is linear only if all of the initial conditions are zero.

For instance, in the Example 2.11, if the capacitor is not assumed to be discharged

initially, then y 0ð Þ ¼ V0 0ð Þ 6¼ 0

A linear system has the property that zero input produces zero output. However, if we let x(t) ¼ 0, then

y tð Þ ¼ yc tð Þ ¼ y 0ð Þe(cid:2)t

ð2:60Þ

Thus, this system is nonlinear if y(0) 6¼ 0. If the capacitor is assumed to be discharged initially, then y 0ð Þ ¼ V0 0ð Þ ¼ 0: Then for x(t) ¼ 0,

y tð Þ ¼ yc tð Þ ¼ 0

ð2:61Þ

the system is linear

2.5.4 Causality

A linear system described by Eq. (2.54) is causal when it is initially relaxed. It implies that if x(t) ¼ 0 for t (cid:5) t0, then y(t) ¼ 0 for t (cid:5) t0, thus, the response for t > to with the initial conditions

2.5 Systems Described by Differential Equations

y t0ð Þ ¼

dy t0ð Þ dt

… ¼

dN(cid:2)1y t0ð Þ dtN(cid:2)1 ¼ 0 (cid:9) (cid:9) (cid:9) (cid:9) t¼t0

dny t0ð Þ dtn ¼

dny tð Þ dtn

2.5.5 Time-Invariance

For a linear causal system, initial rest also implies time-invariance.

For example, consider the system described by

dy tð Þ dt

þ y tð Þ ¼ x tð Þ

y 0ð Þ ¼ 0

Let y1(t) be the response to an input x1(t) and

so that

and

x1 tð Þ ¼ 0 t (cid:5) 0

dy1 tð Þ dt

þ y1 tð Þ ¼ x1 tð Þ

y1 0ð Þ ¼ 0

ð2:62aÞ

ð2:62bÞ

ð2:63Þ

ð2:64Þ

ð2:65Þ

ð2:66Þ

Now, let x2(t) ¼ x1(t (cid:2) τ) and y2 (t) be the corresponding response. From

Eq. (2.64), we get

x2 tð Þ ¼ 0

t (cid:5) τ

dy2 tð Þ dt

þ y2 tð Þ ¼ x2 tð Þ

τð Þ ¼ 0

Then y2(t) should satisfy

and

From Eq. (2.65), we write

dy1 t (cid:2) τ ð dt

þ y1 t (cid:2) τ ð

Þ ¼ x1 t (cid:2) τ ð

Þ ¼ x2 tð Þ

ð2:67Þ

ð2:68Þ

ð2:69Þ

2 Continuous-Time Signals and Systems

By letting y2(t) ¼ y1(t (cid:2) τ), we obtain from Eq. (2.66)

τð Þ ¼ y1

τ (cid:2) τ

Þ ¼ y1 0ð Þ ¼ 0:

Eqs. (2.68) and (2.69) are satisﬁed and thus the system is time invariant.

2.5.6

Impulse Response

The impulse response h(t) of the continuous-time LTI system described by Eq. (2.54) satisﬁes the differential equation

n¼0

dnh tð Þ dtn þ

n¼0

dnδ tð Þ dtn

ð2:70Þ

with the initial rest condition.

For example, let us consider the RC circuit of Example 2.11 with the time

constant RC ¼ 1 sec described by the following differential equation:

dy tð Þ dt

þ y tð Þ ¼ x tð Þ

The impulse response h(t) should satisfy the differential equation

dh tð Þ dt

þ h tð Þ ¼ δ tð Þ

Then, the complimentary solution hc(t) satisﬁes

(cid:11)

dhc tð Þ dt

þ hc tð Þ ¼ 0

To obtain complementary solution, let us assume

Substituting this into

yields

yc tð Þ ¼ Bekt

(cid:11)

dhc tð Þ dt

þ hc tð Þ ¼ 0

Bkekt þ Bekt ¼ 0 ÞBekt ¼ 0

k þ 1

2.5 Systems Described by Differential Equations

Thus, k ¼ (cid:2)1 and

hc tð Þ ¼ Be(cid:2)tu tð Þ

The particular solution hp(t) is zero since hp(t) cannot contain δ(t). Thus,

To ﬁnd the constant B, substituting h(t) ¼ Be(cid:2)tu(t)into

h tð Þ ¼ Be(cid:2)tu tð Þ

dh tð Þ dt

þ h tð Þ ¼ δ tð Þ

yields

(cid:2)Be(cid:2)tuðtÞ þ Be(cid:2)t duðtÞ dt Be(cid:2)t duðtÞ dt

¼ Be(cid:2)tδðtÞ ¼ δðtÞ

þ Be(cid:2)tuðtÞ ¼ δðtÞ

such that B ¼ 1 and hence

h tð Þ ¼ e(cid:2)tu tð Þ:

Example 2.29 Consider the RL circuit shown in Figure 2.28:

(i) Determine the impulse response. (ii) Determine the step response. Solution Writing the loop equation using Kirchhoff’s voltage law assuming i(t) is the current ﬂowing through the circuit, we obtain

(cid:2)x tð Þ þ Ri tð Þ þ L

di tð Þ dt

¼ 0

But i tð Þ ¼ y tð Þ R

Figure 2.28 RL circuit

2 Continuous-Time Signals and Systems

Substituting i tð Þ ¼ y tð Þ

R in the above equation, we get

which can be rewritten as

L R

dyðtÞ dt

þ yðtÞ ¼ xðtÞ

dy tð Þ dt

R L

y tð Þ ¼

R L

x tð Þ

(i) The impulse response h(t) should satisfy the differential equation

dh tð Þ dt

R L

h tð Þ ¼

R L

δ tð Þ

Then, the complimentary solution hc(t) satisﬁes

dhcðtÞ dt

R L

hcðtÞ ¼ 0

To obtain complementary solution, let us assume

Substituting this into

yields

yc tð Þ ¼ Bekt

dhcðtÞ dt

R L

hcðtÞ ¼ 0

Bekt ¼ 0

R (cid:14) L Bekt ¼ 0

Bkekt þ (cid:13) R L

k þ

Thus, k ¼ (cid:2)R

L and

hc tð Þ ¼ Be(cid:2)R

Ltu tð Þ

The particular solution hp(t) is zero since hp(t)) cannot contain δ(t). Thus,

h tð Þ ¼ Be(cid:2)R

Ltu tð Þ

To ﬁnd the constant B, substituting h tð Þ ¼ Be(cid:2)R

Ltu tð Þinto

dh tð Þ dt

R L

h tð Þ ¼

R L

δ tð Þ

2.5 Systems Described by Differential Equations

yields

(cid:2)B

R L

e(cid:2)

LtuðtÞ þ Be(cid:2) Lt duðtÞ dt

Be(cid:2)

Lt duðtÞ dt ¼ Be(cid:2)

e(cid:2)

R L δðtÞ ¼

þ B LtR L

LtuðtÞ ¼ R L

δðtÞ

R L

δðtÞ

such that B ¼ R

L and hence

h tð Þ ¼

R L

e(cid:2)R

Ltu tð Þ:

(ii) The step response s(t) is given by

sðtÞ ¼

Ð 0 hðτÞdτ ¼ (cid:13) τ j t 0 ¼ 1 (cid:2) e(cid:2)

R L

e(cid:2)

t 0

R L

¼ (cid:2)e(cid:2)

dτ (cid:14)

R Lt

uðtÞ

2.5.7 Solution of Differential Equations Using MATLAB

The response y(t) of a system described by Eq. (2.54) for an input x(t) can be determined by using the MATLAB command dsolve(‘eqn1’,‘eqn2’, …) which accepts symbolic equations representing ordinary differential equations and initial conditions. Several equations or initial conditions may be grouped together, sepa- rated by commas, in a single input argument.

Example 2.30 Verify the result of Example 2.28 using MATLAB.

Solution The relation between the output y(t) and the input x(t) is related by

dy tð Þ dt

þ y tð Þ ¼ e(cid:2)2tu tð Þ

y 0ð Þ ¼ 0

The output response y(t)

is determined and displayed by the MATLAB

commands

y = dsolve(‘Dy+y=exp(-2*t)’, ,‘y(0)=0’, ‘t’); disp ([‘y(t) = (‘,char(y), ‘)u(t) ’ ]);

The displayed output is

(cid:10) y tð Þ ¼ e(cid:2)t (cid:2) e(cid:2)2t

(cid:11)

u tð Þ

2 Continuous-Time Signals and Systems

Example 2.31 Consider the RLC circuit of Example 2.26 and determine the current y(t) for the input voltage x(t) ¼ 10e(cid:2)3t u(t) where the initial inductor current is zero and the initial capacitor voltage is equal to 5 volts.

Solution The relation between the output y(t) and the input x(t) is related by

d2y dt2 þ 3 y 0ð Þ ¼ 0,

dy þ 2y ¼ dt (cid:9) (cid:9) dy tð Þ (cid:9) (cid:9) dt

t¼0

dx dt ¼ 5:

The current y(t) is determined and displayed by the MATLAB commands

y = dsolve(‘D2y+3Dy+2y=-30exp(-3t)’, ‘y(0)=0’, ‘Dy(0)=5’, ‘t’); disp ([‘y(t) = (’, char(y), ‘)u(t) ’ ]);

The displayed y(t) is

(cid:10)

y tð Þ ¼ 25e(cid:2)2t (cid:2) 10e(cid:2)t (cid:2) 15e(cid:2)3t

(cid:11)

u tð Þ

2.5.8 Determining Impulse Response and Step Response

for a Linear System Described by a Differential Equation Using MATLAB

In general, for a system described by Eq. (2.54), the impulse response can be determined by using the following MATLAB command:

h ¼ impluse b; a; t

For example, for the differential equation given by

d2y dt2 þ 3

dy dt

þ 2y ¼ x tð Þ

we use the following MATLAB program 2.4 to determine impulse response and step response.

2.6 Block-Diagram Representations of LTI Systems Described by Differential Equations

Program 2.4

clc; clear all; close all; th=0:.01:4; b = [1]; a = [1 3 2]; h=impulse(b,a,th); tx=[0:0.01:4]; x=[ones(1,length(tx))]; [y ty]=convint(x,tx,h,th); figure,plot(th,h) xlabel(‘Time’) ylabel(‘Amplitude’); figure, plot(ty,y); xlabel(‘Time’) ylabel(‘Amplitude’);

The impulse response and step response obtained from the above program are

shown in Figure 2.29(a) and (b), respectively.

2.6 Block-Diagram Representations of LTI Systems

Described by Differential Equations

The block diagram of a continuous system describes how the internal operations are ordered, whereas the differential equation description gives only the input and output relation. Hence, the block diagram is more detailed representation of continuous- time systems than the differential equation description. Integrators are preferred to differentiators in the block-diagram representation of continuous-time systems as the integrators can be easily built from analog components and noise in a system will be smoothed out.

Let us deﬁne the following three basic elements adder, scalar multiplier, and integrator used in the block-diagram representation of continuous-time systems (Figure 2.30).

Consider the system described by Eq. (2.54), which is repeated here for conve-

nience assuming M ¼ N:

n¼0

dkyðtÞ dtk ¼

n¼0

dnxðtÞ dtn

ð2:71Þ

2 Continuous-Time Signals and Systems

0.25

0.2

0.15

0.1

0.05

0.5

0.45

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

e d u t i l

p m A

e d u t i l

p m A

0.5

1.5

2.5

3.5

Time (a)

4 Time

(b)

Figure 2.29 (a) impulse response, (b) step response

If it is assumed that the system is at rest, then the Nth integral of dny tð Þ dtn is x N(cid:2)n ð

and the Nth integral of is dnx tð Þ dtn Eq. (2.71), we obtain the integral description of the system as

Þ tð Þ, Þ tð Þ. Hence, taking the Nth integral of

is y N(cid:2)n ð

2.7 Singularity Functions

Figure 2.30 Block- diagram representation of basic elements (a) adder, (b) multiplier, (c) integrator

(a)

(b)

n¼0

any N(cid:2)k ð

Þ tð Þ ¼

n¼0

bnx N(cid:2)n ð

Þ tð Þ

Since y(0)(t) ¼ y(t), Eq. (2.72) can be rewritten as

y tð Þ ¼

h X

n¼0

1 aN

bnx N(cid:2)n ð

Þ tð Þ (cid:2)

N(cid:2)1

n¼0

any N(cid:2)n ð

Þ tð Þ

ð2:72Þ

ð2:73Þ

The direct form I and the direct form II implementations of Eq. (2.73) are shown

in Figure 2.31(a) and (b), respectively.

Example 2.32 Obtain the block-diagram representation of a system described by

d2y dt2 þ 2

dy dt

þ y ¼

d2x dt2 (cid:2)

dx dt

(cid:2) 6x tð Þ

Solution

2.7 Singularity Functions

The unit impulse δ(t) is one of a class of signals known as singularity functions. Consider a LTI system for which the input and the output are related by

y tð Þ ¼

dx tð Þ dt

ð2:74Þ

The unit impulse response of the considered system is the derivative of the unit

impulse which is referred as the unit doublet u1(t).

2 Continuous-Time Signals and Systems

(a)

(b)

Figure 2.31 Block-diagram representation for continuous-time system described by integral Eq. (2.73) (a) direct form I, (b) direct form II

2.7 Singularity Functions

Figure 2.32 Block diagram representation in direct form II for a second order continuous time system described by the above differential equation is shown in Figure 2.32

By the convolution representation of LTI systems, we represent

dx tð Þ dt

¼ x tð Þ∗u1 tð Þ

for any input signal x(t). Similarly, for an LTI system described by

y tð Þ ¼

d2x tð Þ dt2

we obtain

d2x tð Þ dt2 ¼

d dt

(cid:13)

(cid:14)

dx tð Þ dt

¼ x tð Þ∗u1 tð Þ∗u1 tð Þ

¼ x tð Þ∗u2 tð Þ

ð2:75Þ

ð2:76Þ

ð2:77Þ

where u2(t) ¼ u1(t) ∗ u1(t) is the second derivative of unit impulse.

Thus, uk(t) for k > 0 is the kth derivative of the unit impulse and is the impulse

response of a LTI system that takes the kth derivative of the input:

uk tð Þ ¼ u1 tð Þ∗u1 tð Þ … … k times

ð2:78Þ

If we consider a system described by Eq. (2.75) with input x(t) ¼ 1,then

dx tð Þ dt

¼ x tð Þ∗u1 tð Þ ¼

(cid:2)1

u1 τð Þx t (cid:2) τ ð

Þdτ ¼

(cid:2)1

u1 τð Þdτ ¼ 0

ð2:79Þ

Hence, the unit doublet has zero area. In addition to the singularity functions, the successive integrals of the unit

impulse function, it is known that

u tð Þ ¼

(cid:2)1

δ τð Þdτ,

ð2:80Þ

2 Continuous-Time Signals and Systems

and hence the unit step function is the impulse response of an integrator, and we have

x tð Þ∗u tð Þ ¼

(cid:2)1

x τð Þdτ

ð2:81Þ

Similarly,

the impulse response of a system consisting of two integrators in cascade can be denoted by u(cid:2)1(t) which can be expressed as the convolution of u(t) with itself:

ð2:82Þ

ð2:83Þ

u(cid:2)1 tð Þ ¼ u tð Þ∗u tð Þ ¼

u τð Þdτ

(cid:2)1

Since u(t) ¼ 0 for t < 0 and u(t) ¼ l, u(cid:2)1(t) can be expressed as

u(cid:2)1 tð Þ ¼ tu tð Þ

Example 2.33 For a system if y(t) ¼ x(t) * h(t), show that

(cid:13) ð

(cid:2)1

x τð Þdτ

(cid:14) ∗ dh tð Þ dt

dx tð Þ dt

∗

(cid:2)1

h τð Þdτ

y tð Þ ¼ x tð Þ∗h tð Þ,

(i) y tð Þ ¼

(ii) y tð Þ ¼

Solution (i)

¼ x tð Þ∗u(cid:2)1 tð Þ∗u1 tð Þ∗h tð Þ (cid:2)1 x τð Þdτ∗ dh tð Þ dt

y tð Þ ¼ x tð Þ∗h tð Þ,

(ii)

¼ x tð Þ∗u1 tð Þ∗h tð Þ∗u(cid:2)1 tð Þ t

dx tð Þ dt

∗

h τð Þdτ

(cid:2)1

2.8 State-Space Representation of Continuous-Time LTI

Systems

2.8.1 State and State Variables

The state of a system at time t0 is the minimal information required that is sufﬁcient to determine the state and the output of the system for all times t (cid:4) t0 for the known system input at all times t (cid:4) t0.The variables that contain this information are called state variables.

2.8 State-Space Representation of Continuous-Time LTI Systems

2.8.2 State-Space Representation of Single-Input Single-

Output Continuous-Time LTI Systems

Consider a single-input single-output continuous-time LTI system described by the following Nth-order differential equation:

dNy tð Þ dtN þ aN(cid:2)1

dN(cid:2)1y tð Þ dtN(cid:2)1 þ … þ a1

dy tð Þ dt

þ a0y tð Þ ¼ ℧ tð Þ

ð2:84Þ

where y(t)) is the system output and ʊ(t) is the system input.

Deﬁne the following useful set of state variables

x1 tð Þ ¼ y tð Þ, x2 tð Þ ¼

dy tð Þ dt

, x3 tð Þ ¼

d2y tð Þ dt2

, … , xN tð Þ ¼

dN(cid:2)1y tð Þ dtN(cid:2)1

ð2:85Þ

Taking derivatives of the ﬁrst N (cid:2) 1 state variables of the above, we get

dx1 tð Þ dt

¼ x2 tð Þ,

dx2 tð Þ dt

¼ x3 tð Þ, … ,

dxN(cid:2)1 tð Þ dt

¼ xN:

Rearranging Eq. (2.84) and using Eq. (2.86), we obtain

dxN tð Þ dt

¼ (cid:2)a0x1 tð Þ (cid:2) a1x2 tð Þ (cid:2) (cid:6) (cid:6) (cid:6) (cid:2) aN(cid:2)1 xN(cid:2)1 tð Þ þ ℧ tð Þ

y tð Þ ¼ x1 tð Þ

dt by _x,

Denoting dxðtÞ Eqs. (2.86), (2.87), and (2.88) can be written in matrix form as 3 2

6 6 6 6 6 6 4

_x1ðtÞ _x2ðtÞ ⋮ _xN(cid:2)1ðtÞ _xNðtÞ

7 7 7 7 7 7 5

6 6 6 6 6 6 4

…

0 0 1 ⋮ ⋮ ⋮ ⋱ ⋮

…

0 0 (cid:2)a0 (cid:2)a1 (cid:2)a2

… … (cid:2)aN(cid:2)1

7 7 7 7 7 7 5

6 6 6 6 6 6 4

x1ðtÞ x2ðtÞ ⋮

xN(cid:2)1ðtÞ xNðtÞ

7 7 7 7 7 7 5

y tð Þ ¼ 1

0 0

…

(cid:3)

7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 4

x1 tð Þ

x2 tð Þ ⋮

xN(cid:2)1 tð Þ

xN tð Þ

ð2:86Þ

ð2:87Þ

ð2:88Þ

7 7 7 7 7 7 5

℧ðtÞ

6 6 6 6 6 6 4

0 0 ⋮

ð2:89aÞ

ð2:89bÞ

100

2 Continuous-Time Signals and Systems

Deﬁne a Nx1 dimensional vector called state vector as

X tð Þ ¼

6 6 6 6 6 6 6 4

7 7 7 7 7 7 7 5

x1 tð Þ

x2 tð Þ ⋮

xN(cid:2)1 tð Þ

xN tð Þ

The derivative of X(t) becomes

dXðtÞ dt

¼ _X ðtÞ ¼

6 6 6 6 6 6 6 6 4

7 7 7 7 7 7 7 7 5

_x1ðtÞ _x2ðtÞ ⋮

_xN(cid:2)1ðtÞ _xNðtÞ

More compactly Eqs. (2.89a) and (2.89b) can be written as

_X tð Þ ¼ AX tð Þ þ b℧ tð Þ

y tð Þ ¼ cX tð Þ

ð2:90Þ

ð2:91Þ

ð2:92Þ ð2:93Þ

where

6 6 6 6 6 6 6 4

A ¼

…

0 1 0 ⋮ ⋮ ⋮ ⋱ ⋮

…

(cid:2)a0 (cid:2)a1 (cid:2)a2

… (cid:2)aN(cid:2)1

7 7 7 7 7 7 7 5

; b ¼

6 6 6 6 6 6 6 4

7 7 7 7 7 7 7 5

0 ⋮

; c ¼ 1

0 …

(cid:3)

Eqs. (2.92) and (2.93) are called N-dimensional state-space representation or state

equations of the system.

In general state equations of a system are described by

_X tð Þ ¼ AX tð Þ þ b℧ tð Þ y tð Þ ¼ cX tð Þ þ d℧ tð Þ

ð2:94Þ ð2:95Þ

2.8 State-Space Representation of Continuous-Time LTI Systems

101

Example 2.34 Obtain the state-space representation of a system described by the following differential equation:

d3y tð Þ dt3 þ 2

d2y tð Þ dt2 þ 3

dy tð Þ dt

þ 4y tð Þ ¼ ℧ tð Þ

Solution: The order of the differential equation is three. Hence, the three-state

variables are

x1 tð Þ ¼ y tð Þ, x2 tð Þ ¼

dy tð Þ dt

, x3 tð Þ ¼

d2y tð Þ dt2

The ﬁrst derivatives of the state variables are

_x 1 tð Þ ¼ x2 tð Þ _x 2 tð Þ ¼ x3 tð Þ _x 3 tð Þ ¼ (cid:2)4x1 tð Þ (cid:2) 3x2 tð Þ (cid:2) 2x3 tð Þ þ ℧ tð Þ

The state-space representation in matrix form is given by

6 4

_x 1 tð Þ _x 2 tð Þ _x 3 tð Þ

7 5 ¼

6 4

7 5

6 4

7 5 þ

6 4

x1 tð Þ

x2 tð Þ

-4

-3

-2

7 5℧ tð Þ

y tð Þ ¼ 1

0 0

6 4 (cid:3)

x3 tð Þ 2

7 5

x1 tð Þ

x2 tð Þ

x3 tð Þ

Example 2.35 Obtain the state-space representation for the electrical circuit shown in Figure 2.33 considering Vc1, i1, and Vc2 as state variables and vc1 as output y(t).

vc1 –

vs(t)

–

Vc2

Figure 2.33 Third-order electrical circuit

102

2 Continuous-Time Signals and Systems

Solution The state variables for the circuit are

x1 ¼ Vc1

x2 ¼ i1

x3 ¼ Vc2

From the relationship between the voltage Vc1 and current i1, we obtain

dx1 tð Þ dt

¼ x2

Kirchhoff’s voltage equation around the closed loop gives

(cid:2)Vs þ x1 þ

dx2 tð Þ dt

þ x3 ¼ 0

This equation can be rewritten as

dx2 tð Þ dt

¼ (cid:2)x1 (cid:2) x3 þ Vs

The current i3 ¼ x3 and the current i2 ¼ dx3 tð Þ dt By Kirchhoff’s current law

implying that

Hence,

i1 ¼ i2 þ i3

x2 ¼

dx3 tð Þ dt

þ x3

dx3 tð Þ dt

¼ x2 (cid:2) x3

The voltage Vc1 is taken as the output y(t):

y tð Þ ¼ x1

The state-space representation of the circuit is given by

2.8 State-Space Representation of Continuous-Time LTI Systems

103

1Ω

i1(t )

iL(t )

vs(t )

1Ω

−

vc(t )

Figure 2.34 Electrical circuit of Example 2.36

6 6 4

_x 1 tð Þ

_x 2 tð Þ

_x 3 tð Þ

7 7 5 ¼

6 6 4

(cid:2)1

0 (cid:2)1

7 7 5

6 6 4

x1 tð Þ

x2 tð Þ

7 7 5 þ

6 6 4

7 7 5Vs

1 (cid:2)1 2

y tð Þ ¼ 1

(cid:3)

6 6 4

x3 tð Þ 3

7 7 5

x1 tð Þ

x2 tð Þ

x3 tð Þ

Example 2.36 Obtain state-space representation of the circuit shown in Figure 2.34 considering the current through the inductor and voltage across the capacitor as state variables and voltage across the capacitor as the output y(t).

Solution The state variables for the circuit are

x1 tð Þ ¼ iL tð Þ

x2 tð Þ ¼ Vc tð Þ

Since the voltage across the capacitor is equal to the voltage across the series inductor and resistor branch,

we obtain

dx1 tð Þ dt

þ x1 tð Þ ¼ x2 tð Þ

This equation can be rewritten as

104

2 Continuous-Time Signals and Systems

dx1 tð Þ dt Kirchhoff’s voltage equation around the closed loop gives

¼ (cid:2)x1 tð Þ þ x2 tð Þ

Hence

(cid:2)Vs tð Þ þ i1 tð Þ þ x2 tð Þ ¼ 0

i1 tð Þ ¼ (cid:2)x2 tð Þ þ Vs tð Þ

By Kirchhoff’s current law

Therefore

i1 tð Þ ¼ x1 tð Þ þ

dx2 tð Þ dt

(cid:2)x2 tð Þ þ Vs tð Þ ¼ x1 tð Þ þ

dx2 tð Þ dt

This equation can be rewritten as

dx2 tð Þ dt

¼ (cid:2)x1 tð Þ (cid:2) x2 tð Þ þ Vs tð Þ

The voltage Vc tð Þ is taken as the output y(t):

y tð Þ ¼ x2 tð Þ

The state-space representation of the circuit is given by ”

”

_x 1 tð Þ _x 2 tð Þ

(cid:2)1

(cid:2)1 (cid:2)1

”

x1 tð Þ

x2 tð Þ

” # 0

vs tð Þ

y tð Þ ¼ 0

(cid:3)

x1 tð Þ

x2 tð Þ

2.8.3 State-Space Representation of Multi-input Multi-output

Continuous-Time LTI Systems

The state-space representation of continuous-time system with m inputs and l output and N state variables can be expressed as

2.9 Problems

where

6 6 6 6 4

A ¼

105

ð2:96Þ ð2:97Þ

7 7 7 7 5

N(cid:7)m

7 7 7 7 5

_X tð Þ ¼ AX tð Þ þ B℧ tð Þ y tð Þ ¼ CX tð Þ þ D℧ tð Þ

a11

a12

…

a1N

…

a22

a2N a21 ⋮ ⋮ ⋱ ⋮

aN1 2

c11

aN2

c12

…

aNN

c1N

6 6 6 6 4

C ¼

…

c22

c21 c2N ⋮ ⋮ ⋱ ⋮

7 7 7 7 5

N(cid:7)N

6 6 6 6 4

B ¼

b11

b12

…

b1m

…

b22

b2m b21 ⋮ ⋮ ⋱ ⋮

bN1 2

d11

d12

bN2

… bNm … d1m … d2m d21 ⋮ ⋮ ⋱ ⋮

d22

6 6 6 6 4

D ¼

cl1

cl2

…

clN

l(cid:7)N

dl1

dl2

…

dlm

l(cid:7)m

2.9 Problems

Check the following for linearity and time-invariance:

(i) y tð Þ ¼ dx tð Þ dt (ii) y(t) ¼ tx(t) (iii) y(t) ¼ t2x2(t) (iv) dy tð Þ (v) y(t) ¼ ln (x(t)) (vi) y(t) ¼ x(t) þ cons tan t

dt þ ty tð Þ ¼ x tð Þ

Determine which of the following systems are linear and which are nonlinear:

(i) dy tð Þ dt þ 3y tð Þ ¼ x2 tð Þ (ii) dy tð Þ dt þ y2 tð Þ ¼ x tð Þ (cid:15) dy tð Þ (iii) dt (iv) dy tð Þ

þ 3y tð Þ ¼ x tð Þ dt þ sin tð Þy tð Þ ¼ dx tð Þ

(cid:16)

dt þ 3x tð Þ

An ampliﬁer has an output y(t) ¼ cos (ωt). If y(t) is limited by clipping resulting

in the clipped output yc(t), check for linearity and time-invariance of yc(t).

An input signal x(t) and two possible outputs of a linear time-invariant system are shown in Figure P2.1. Which outputs are possible given that the system is linear and time invariant?

106

2 Continuous-Time Signals and Systems

x (t )

y (t )

(i)

(ii)

Figure P2.1 Input signal and two possible outputs of problem 4

If x(t) ¼ u(t + 1) (cid:2) u(t (cid:2) 1), compute (x *x *x)(t) and sketch.
Determine graphically the convolution y(t) ¼ x(t) * h(t) for the following:

(i)

(ii)

(iii)

(iv)

(v)

x tð Þ ¼ u tð Þ-u t-4ð

Þ-u t-6ð

h tð Þ ¼ u t-4ð x tð Þ ¼ e(cid:2)tu tð Þ h tð Þ ¼ e(cid:2)2tu tð Þ xðtÞ ¼ 2uðt (cid:2) 1Þ (cid:2) 2uðt (cid:2) 2Þ hðtÞ ¼ uðt þ 1Þ (cid:2) 2uðt (cid:2) 1Þ þ uðt (cid:2) 2Þ x tð Þ ¼ u tð Þ

(

h tð Þ ¼

e(cid:2)t, (cid:2)et, x tð Þ ¼ u t þ 1 ð 1 3

h tð Þ ¼

t (cid:4) 0,

t (cid:5) 0 Þ (cid:2) u t (cid:2) 1 ð

t u tð Þ (cid:2) u t (cid:2) 3 ð

Consider a continuous-time LTI system with the step response

s tð Þ ¼ e(cid:2)tu tð Þ

Determine and sketch the output of this system to the input

x tð Þ ¼ u t (cid:2) 1

Þ (cid:2) u t (cid:2) 3 ð

Consider h(t) be the triangular pulse and x(t) be the unit impulse train as shown in

Figure 2.16 of Example 2.17. Determine y(t) ¼ x(t) * h(t) and sketch it.

(i) for T ¼ 3 (ii) for T ¼ 3 2

An LTI system consists of two subsystems in cascade. The impulse responses of

the subsystems are, respectively, given by

h1 tð Þ ¼ δ tð Þ (cid:2) 2e(cid:2)2tu tð Þ;

h2 tð Þ ¼ etu tð Þ

Find the impulse response of the overall system.

2.9 Problems

107

If y(t) ¼ x(t) * h(t) shows that the area of the convolution y(t) is the product of the

areas of the signals that are being convolved x(t) and h(t), that is,

(cid:2)1

y τð Þdτ ¼

(cid:13) ð

(cid:13) (cid:14) ð

(cid:14)

h τð Þdτ

x τð Þdτ

(cid:2)1

Compute and sketch the periodic convolution of the square-wave signal x(t)

shown in Fig. P2.2 with itself.

Figure P2.2 Square wave signal of problem 11

x(t)

-2

-1

Consider an LTI system with impulse response h(t):

(i) Check for its stability, if h(t) is periodic and nonzero. (ii) Check for causality of the inverse of the LTI system if h(t) is causal. (iii) Check for its stability if h(t) is causal.

Consider the system described by

dy tð Þ dt

þ 3y tð Þ ¼ x tð Þ þ

dx tð Þ dt

Determine the impulse response h(t) of the system.

Consider the system described by

dy tð Þ dt

þ y tð Þ ¼ x tð Þ

y 0ð Þ ¼ 0

(i) Determine the step response of the system. (ii) Determine the impulse response from the step response.

Determine the response y(t) of the OP-Amp circuit shown in Figure P2.3 for an

input x(t) ¼ u(t)

108

2 Continuous-Time Signals and Systems

Figure P2.3 OP-Amp circuit of problem 15

Determine the impulse response y(t) of

the OP-Amp circuit shown in

Figure P2.4.

Figure P2.4 OP-Amp circuit of problem 16

Determine the impulse response h(t) for a system described by

d2y dt2 þ 3

dy dt

þ 2y ¼

dx dt

Draw block diagrams for direct form II implementation of the corresponding

systems

(i) d2y

dt2 þ 5dy

dt þ 4y ¼ dx

dt þ x

2.10 MATLAB Exercises

109

dt þ 3y ¼ 2dx

(ii) dy dt þ x tð Þ (iii) d2y/dt2 (cid:2) ady/dt ¼ a dx/dt + abx(t)

For a given signal x(t)

(i) Show that x tð Þu1 tð Þ ¼ x 0ð Þu1 tð Þ (cid:2) dx tð Þ dt (ii) Determine the value of

(cid:9) (cid:9) (cid:9) t¼0

δ tð Þ:

(cid:2)1

x τð Þu2 τð Þdτ:

(iii) Find an expression for x(t) u2(t) similar to (i).

Obtain the state-space representation for the electrical circuit shown

In Figure P2.5 considering i1, V1, and V2 as state variables and current through the

inductor as the output y(t)

Figure P2.5 Electrical circuit of prblem 20

−

1 +

1Ω

1 H

i 1

1Ω

v 2 −

2.10 MATLAB Exercises

Verify the solution of problem 2 using MATLAB.
Write a MATLAB program to compute the convolution of the input x(t) and the

impulse response h(t) shown in Figure M2.1.

0.5

1.5

Figure M2.1 Input and Impulse response of MaTLAB exercise 2

110

2 Continuous-Time Signals and Systems

If x(t) ¼ u(t (cid:2) 1) (cid:2) u(t (cid:2) 2), write a MATLAB program to compute the result y10(t)

convolving ten x(t) functions together, that is

y10 tð Þ ¼ x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ∗x tð Þ

and comment on the result.

Write a MATLAB program to ﬁnd the impulse response h(t) and step response for

a system described by

d2y tð Þ dt2 þ 5

dy tð Þ dt

þ 6y tð Þ ¼

dx tð Þ dt

þ x tð Þ:

Further Reading

Oppenheim, A.V., Willsky, A.S.: Signals and Systems. Prentice-Hall, Englewood Cliffs (1983)
Hsu, H.: Signals and Systems Schaum’s Outlines, 2nd edn. McGraw-Hill, New York (2011)
Kailath, T.: Linear Systems. Prentice-Hall, Englewood Cliffs (1980)
Zadeh, L., Desoer, C.: Linear System Theory. McGraw-Hill, New York (1963)

Chapter 3 Frequency Domain Analysis of Continuous- Time Signals and Systems

The continuous-time signals and systems are often characterized conveniently in a transform domain. This chapter describes the transformations known as Fourier series and Fourier transform which convert time-domain signals into frequency- domain (or spectral) representations. The frequency domain representation of continuous-time signals are described along with the conditions for the existence of Fourier series for periodic signals and Fourier transform for nonperiodic signals and their properties. Finally, the frequency response of continuous-time systems is discussed.

3.1 Complex Exponential Fourier Series Representation

of the Continuous-Time Periodic Signals

It is recalled from Chapter 1 that a signal x(t) is periodic if

ð x tð Þ ¼ x t þ T

Þ for all t

ð3:1Þ

The fundamental period T0 is small minimum, positive nonzero value of T for is referred to as the fundamental angular

which Eq. (3.1) is satisﬁed, and Ω0 ¼ 2π T 0 frequency.

A sinusoidal signal x(t) ¼ cos (ω0t) and the complex exponential signal x tð Þ

¼ ejΩ0t are the two examples of periodic signals.

The complex exponentials related harmonically are expressed by

xn tð Þ ¼ ejnΩ0t,

n ¼ 0, (cid:2) 1, (cid:2) 2, (cid:3) (cid:3) (cid:3)

ð3:2Þ

The fundamental frequency of each of these signals is a multiple of Ω0, and hence

each is periodic with period T0.

111

ð3:3Þ

ð3:4Þ

ð3:5Þ

ð3:6Þ

112

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Thus, a linear combination of complex exponentials related harmonically can be

written as

x tð Þ ¼

n¼(cid:4)1 anejnΩ0t

Eq. (3.3) is the Fourier series representation of a periodic signal x(t). The Fourier coefﬁcients ak can be determined as follows: Multiplying Eq. (3.3) both sides by e(cid:4)jmΩ0t, we obtain

x tð Þe(cid:4)jmΩ0t ¼

n¼(cid:4)1 an ejnΩ0te(cid:4)jmΩ0t

Integrating Eq. (3.4) both sides from 0 to T0, we have

T 0

x tð Þe(cid:4)jmΩ0t ¼

T 0

n¼(cid:4)1 an ejnΩ0te(cid:4)jmΩ0tdt

Interchanging the integration and summation, Eq. (3.5) can be rewritten as

T 0

x tð Þe(cid:4)jmΩ0tdt ¼

n¼(cid:4)1 ak

(cid:2)

T 0

(cid:3)

ej n(cid:4)mð

ÞΩ0t

T 0

ej n(cid:4)mð

ÞΩ0tdt ¼

T 0

Hence,

cos

n (cid:4) m

ÞΩ0t

Þ dt þ j

0 ð

T 0

sin

ð ð

n (cid:4) m

ÞΩ0t

Þ dt

ð3:7Þ

T 0

ej n(cid:4)mð

ÞΩ0t dt ¼

(cid:4)

T 0 m ¼ n 0 m 6¼ n

Thus, Eq. (3.6) becomes

T 0

x tð Þe(cid:4)jmΩ0t dt ¼ anT 0

Therefore, the Fourier coefﬁcients an are given by

an ¼ 1 T 0

T 0

x tð Þe(cid:4)jnΩ0t dt

ð3:8Þ

ð3:9Þ

ð3:10Þ

Thus, the Fourier series of a periodic signal is deﬁned by Eq. (3.3) referred to as

the synthesis equation and Eq. (3.10) as the analysis equation.

3.1 Complex Exponential Fourier Series Representation of the Continuous…

113

3.1.1 Convergence of Fourier Series

The sufﬁcient conditions for guaranteed convergence of Fourier series are the following Dirichlet conditions:

x(t) must be absolutely integral over any period, that is,

T 0

j x tð Þ j dt < 1:

ð3:11Þ

which guarantees that each Fourier coefﬁcient ak has ﬁnite value. 2. x(t) must have ﬁnite number of maxima and minima during any single period of it. 3. x(t) must have ﬁnite number of discontinuities in any ﬁnite interval of time and

each of these discontinuities being ﬁnite.

3.1.2 Properties of Fourier Series

Linearity Property If x1(t) and x2(t) are two continuous-time signals with Fourier series coefﬁcients an and bn, then the Fourier series coefﬁcients of a linear combi- nation of x1(t) and x2(t), that is, c1x1(t) þ c2x2(t), are given by

where c1 and c2 are arbitrary constants.

c1 an þ c2bn

Time Shifting Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the x(t (cid:4) t0) are given by e

(cid:4)jn 2π T0

, the Fourier coefﬁcients an of a periodic signal x(t) are given

t0 an. Since Ω0 ¼ 2π T 0

Proof by

an ¼ 1 T 0

T 0

x tð Þe

(cid:4)jn 2π T0

t dt

Let the Fourier series coefﬁcients of x(t (cid:4) t0) be ban

ban ¼ 1 T 0

T 0

ð x t (cid:4) t0

Þe

(cid:4)jn 2π T0

tdt

Letting τ ¼ t (cid:4) t0

114

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

T 0

2π T 0

ban ¼ 1 T 0

(cid:4)jn

ban ¼ e

(cid:4)jn

2π T 0

ban ¼ e

x τð Þe ð

t0 1 T 0 t0an

(cid:4)jn

2π T 0

(cid:4)jn

2π T 0

t0dτ

T 0

(cid:4)jn

x τð Þe

2π T 0

dτ

Conjugate Property If x(t) is a continuous-time periodic signal with the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the x*(t) are given by a∗ (cid:4)n. Proof an ¼ 1 T 0

tdt, then replacing n by (cid:4)n, we get

x tð Þe

(cid:4)jn 2π T0

T 0

a(cid:4)n ¼ 1 T 0

T 0

x tð Þejn2π

tdt

Taking both sides conjugate of this equation, we have

a∗ (cid:4)n

¼ 1 T 0

T 0

x∗ tð Þe

(cid:4)jn2π T0

tdt

Symmetry for Real Valued Signal If x(t) is a continuous-time real valued signal with the Fourier coefﬁcients an, then

an ¼ a∗ (cid:4)n

where * stands for the complex conjugate.

Proof From conjugate property, we know that

a∗ (cid:4)n

¼ 1 T 0

T 0

x∗ tð Þe

(cid:4)jn 2π T0

t dt

Since x tð Þ is real x∗ tð Þ ¼ x tð Þ, we get ð

a∗ (cid:4)n

¼ 1 T 0 ¼ an

T 0

(cid:4)jn

x tð Þe

2π T 0

implies that Re(an) ¼ Re(a–n), i.e., the real part of an is even, andIm(an) ¼ (cid:4)(Im (a(cid:4)n)), i.e., the imaginary part of an is odd.

If x(t) is a continuous-time real and even signal, i.e., x*(t) ¼ x(t) x(t) ¼ x((cid:4)t), it

can be easily shown that an¼ a(cid:4)n and an ¼ a∗ n .

Similarly, if x(t) is a continuous-time real and odd signal, i.e., x*(t) ¼ x(t) x(t) ¼

(cid:4)x((cid:4)t), it can be easily shown that an ¼ (cid:4)a(cid:4)n and an ¼ (cid:4) a∗

(cid:5)

(cid:6)

3.1 Complex Exponential Fourier Series Representation of the Continuous…

115

Frequency Shifting Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the ejK 2π Proof The Fourier coefﬁcients an of a periodic signal x(t) are given by

tx tð Þ are given by an(cid:4)K.

an ¼ 1 T 0

T 0

x tð Þe

(cid:4)jn2π T0

tdt

Let dn be the Fourier coefﬁcients of ejK 2π

tx tð Þ, then

T 0

0 ð T 0

dn ¼ 1 T 0 ¼ 1 T 0 0 ¼ an(cid:4)K

x tð ÞejK

2π T 0

(cid:4)jn

2π T 0

x tð Þej n(cid:4)Kð

Þ 2π T 0

Thus, it is proved.

Time Reversal Property If x(t) is a continuous-time periodic signal with the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the x((cid:4)t) are given by a(cid:4)n.

Time Scaling Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the x(αt) α > 0 are given by an with period T 0 α . Since Ω0 ¼ 2π T 0

, the Fourier coefﬁcients an of a periodic signal x(t) are

Proof given by

an ¼ 1 T 0

T 0

x tð Þe

(cid:4)jn2π T0

tdt

Let the Fourier series coefﬁcients of x(αt) be ban, then

ban ¼

T 0

x αtð

Þe

(cid:4)jn2παt

T0 dt

T 0

ban ¼ 1 T 0 ¼ an

(cid:4)jn

x τð Þe

2π T 0

dτ

Letting τ ¼ αt

Hence, bT 0 ¼ T0 α .

116

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Differentiation in Time If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the d dt x tð Þ are given by jn 2π T 0

Proof The Fourier series representation of of x(t) is

x tð Þ ¼

n¼(cid:4)1 an ejnΩ0t

Differentiating this equation both sides with respect to t, we obtain

Since Ω0 ¼ 2π T 0

d dt

x tð Þ ¼

n¼(cid:4)1

jnΩ0anejnΩ0t

d dt

x tð Þ ¼

n¼(cid:4)1

2π T 0

anejnΩ0t

dt x tð Þ. Comparing this with the Fourier giving the Fourier series representation of d series representation of coefﬁcients an, it is clear that the Fourier coefﬁcients of dt x tð Þ are jn 2π d Integration Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefﬁcients an, then the Fourier series coefﬁcients of the x(t) are given by T 0

T 0

jn2π an

Proof The Fourier series representation of x(t) is

x tð Þ ¼

n¼(cid:4)1 anejnΩ0t

Integrating this equation both sides with respect to t, we obtain

x tð Þdt ¼

ð X1

anejnΩ0t

n¼(cid:4)1

an jnΩ0

ejnΩ0t

x tð Þdt ¼

n¼(cid:4)1

T 0 jn2π anejnΩ0t

Since Ω0 ¼ 2π T 0

3.1 Complex Exponential Fourier Series Representation of the Continuous…

117

giving the Fourier series representation of series representation of coefﬁcients an, it is clear that the Fourier coefﬁcients of dt are T 0

x(t)dt. Comparing this with the Fourier x(t)

jn2π an

Periodic Convolution If x1(t) and x2(t) are two continuous-time signals with common period T0 and Fourier coefﬁcients an and bn, respectively, then the Fourier series coefﬁcients of the convolution integral of x1(t) and x2(t) are given by T0anbn.

Proof The periodic convolution integral of two signals with common period T0 is deﬁned by

y tð Þ ¼

ð x1 τð Þx2 t (cid:4) τ

Þdτ

T 0

ð3:12Þ

Let cn be the Fourier series coefﬁcients of y(t), then

cn ¼ 1 T 0

T 0

y tð Þe(cid:4)jnΩ0tdt ¼ 1 T 0

(cid:2)

T 0

Letting t (cid:4) τ ¼ t1, Eq. (3.13) becomes

(cid:3)

ð x1 τð Þx2 t (cid:4) τ

Þdτ

e(cid:4)jnΩ0tdt

ð3:13Þ

cn ¼ 1 T 0

(cid:2) ð

T 0

x1 τð Þx2 t1ð Þdτ

(cid:3) ð e(cid:4)jnΩ0 τþt1

Þdt1

Interchanging the order of integration, the above equation can be rewritten as

cn ¼ 1 T 0

(cid:2) ð

T 0

(cid:3) ð (cid:2)

(cid:3)

x1 τð Þe(cid:4)jnΩ0τdτ

x2 t1ð Þe(cid:4)jnΩ0t1dt1

ð3:14Þ

By deﬁnition of Fourier series (cid:2) ð

T 0

(cid:3)

1 T 0 h Ð

T 0

x1 τð Þe(cid:4)jnΩ0τdτ i

T 0

¼ an

x2 t1ð Þe(cid:4)jnΩ0t1dt1

¼ T 0bn

Hence cn¼ T0anbn.. Thus, it is proved.

Multiplication Property If x1(t) and x2(t) are two continuous-time signals with Fourier coefﬁcients an and bn, respectively, then the Fourier series coefﬁcients of a new signal x1(t)x2(t) are given l¼(cid:4)1 al bn(cid:4)l implying that signal multiplication in the time domain is equiv- by alent to discrete-time convolution in the frequency domain. Proof Let dn be the Fourier coefﬁcients of the new signal x1(t)x2 (t). By deﬁnition of Fourier series representation of a signal, we have

118

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

T 0

x1 τð Þx2 tð Þe(cid:4)jnΩ0tdt

dn ¼ 1 T 0

T 0

¼ 1 T 0 0 P1

l¼(cid:4)1 al ejlΩ0tx2 tð Þe(cid:4)jnΩ0tdt

l¼(cid:4)1 al

T 0

1 T 0

x2 tð Þe(cid:4)j n(cid:4)l

ÞΩ0tdt

l¼(cid:4)1 al bn(cid:4)l

Parseval’s Theorem If x(t) is a continuous-time signal with period T0 and Fourier coefﬁcients an, then the average power P of x(t) is given by

P ¼ 1 T 0

T 0

x tð Þ j

j2dt ¼

n¼(cid:4)1 anj

Proof The average power P of a periodic signal x(t) is deﬁned as

P ¼ 1 T 0

T 0

x tð Þ

j2dt

ð3:15Þ

ð3:16Þ

Assuming that x(t) is complex valued x(t)x*(t) ¼ |x(t)|2 and x*(t) can be expressed

in terms of its Fourier series as

x∗ tð Þ ¼

Eq. (3.16) can be rewritten as

n¼(cid:4)1 a∗

(cid:4)nejnΩ0t

P ¼ 1 T 0

T 0

x tð Þx∗ tð Þ dt

Substituting Eq. (3.17) in Eq. (3.18), we obtain

P ¼ 1 T 0

x tð Þ

T 0

n¼(cid:4)1 a∗

(cid:4)nejnΩ0t

Interchanging the order of integration, Eq. (3.19) can be rewritten as

P ¼

n¼(cid:4)1 a∗

(cid:4)n

T 0

1 T 0

x tð Þe

(cid:4)jn 2π T0

tdt

ð3:17Þ

ð3:18Þ

ð3:19Þ

ð3:20Þ

By deﬁnition of the Fourier series

an ¼ 1 T 0

T 0

x tð Þe

(cid:4)jn2π T0

tdt

Thus,

3.1 Complex Exponential Fourier Series Representation of the Continuous…

119

Table 3.1 Some properties of continuous-time Fourier series

Property Linearity Time shifting

Time reversal Conjugate Symmetry

Frequency shifting Time scaling

Differentiation in time

Integration

Periodic convolution property

Multiplication property

Periodic signal c1x1(t) þ c2x2(t) x(t (cid:4) t0) x((cid:4)t) x*(t) x(t) real xe(t) (x(t) real) xo(t) (x(t) real)

ejK 2π

tx tð Þ

x(αt), α > 0 (periodic with period T 0 α ) dt x tð Þ d Ð x(t)dt

ð x1 τð Þx2 t (cid:4) τ

Þdτ

T 0 x1(t)x2(t)

Fourier series coefﬁcients c1an + c2bn (cid:4)jn 2π t0 an e T0 a(cid:4)n a∗ (cid:4)n 8 an ¼ a∗

< (cid:4)n Re½an(cid:5) ¼ Re½a(cid:4)n(cid:5) Im½an(cid:5) ¼ (cid:4)Im½a(cid:4)n(cid:5) : j an j¼j a(cid:4)n j arg½an(cid:5) ¼ (cid:4)arg½a(cid:4)n(cid:5) Re[an] jIm[an] an(cid:4)K

jn 2π an T 0 T 0 jn2π an T0anbn. X1

l¼(cid:4)1

al bn(cid:4)l

P ¼ 1 T 0

T 0

x tð Þ

j2dt ¼

n¼(cid:4)1

anj

The properties of continuous-time Fourier series are summarized in Table 3.1. ð1

Parseval’s Theorem

1 T 0

x tð Þ

j2dt ¼

(cid:4)1

n¼(cid:4)1 anj

Half-Wave Symmetry If the two halves of one period of a periodic signal are of identical shape, except that one is the negation of the other, the periodic signal is said to have a half-wave symmetry. Formally, if x(t) is a periodic signal with period T0, then x(t) has half- wave symmetry if x t (cid:4) T 0 2

¼ (cid:4)x tð Þ

(cid:5)

(cid:6)

Example 3.1 Prove that Fourier series representation of a periodic signal with half- wave symmetry has no even-numbered harmonics.

Proof A periodic signal x(t) with half-wave symmetry is given by

120

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

(

x tð Þ (cid:4)x tð Þ ð

0 (cid:6) t < T 0=2 T 0=2 (cid:6) t < T 0

x tð Þe(cid:4)jnΩ0tdt

x tð Þ ¼

an ¼ 1 T 0 ¼ 1 T 0

T 0 ð T 0=2

x tð Þe(cid:4)jnΩ0tdt þ 1 T 0

T 0

T 0=2

x tð Þe(cid:4)jnΩ0tdt

For

T 0=2 (cid:6) t < T 0 (cid:5)

Substituting x tð Þ ¼ x t (cid:4) T 0 2

(cid:8)

(cid:7) x tð Þ ¼ x t (cid:4) T 0 2

(cid:6)

in the above equation, we get

an ¼ 1 T 0

T 0=2

x tð Þe(cid:4)jnΩ0tdt (cid:4) 1 T 0

T 0

T 0=2

(cid:8)

(cid:7) x t (cid:4) T 0 2

e(cid:4)jnΩ0tdt

By using time shifting property, we obtain

an ¼ 1 T 0

T 0=2

x tð Þe(cid:4)jnΩ0tdt (cid:4) e(cid:4)jnΩ0

T 0=2

T0 2

1 T 0

x tð Þe(cid:4)jnΩ0tdt

Since Ω0 ¼ 2π T 0

T 0=2

x tð Þe(cid:4)jnΩ0tdt

T 0=2

an ¼ 1 T 0 ð

¼ 1 (cid:4) e(cid:4)jnπ T 0 ¼ 1 (cid:4) (cid:4)1ð T 0

x tð Þe(cid:4)jnΩ0tdt (cid:4) e(cid:4)jnπ 1 T 0

T 0=2

x tð Þe(cid:4)jnΩ0tdt

Þn

0 ð

T 0=2

x tð Þe(cid:4)jnΩ0tdt

8 <

0 2 T 0

T 0=2 0

for even n

x tð Þe(cid:4)jnΩ0tdt

for odd n

Example 3.2 Find Fourier series of the following periodic signal with half-wave symmetry as shown in Figure 3.1.

Figure 3.1 Periodic signal with half-wave symmetry

3.1 Complex Exponential Fourier Series Representation of the Continuous…

121

Solution The period T0 ¼ 6 and Ω0 ¼ 2π 6 Fourier coefﬁcients are given by

¼ π 3

an ¼

<

:

2 T 0

T 0=2

for even n

x tð Þe(cid:4)jnΩ0tdt

for odd n

For odd n

T 0=2

x tð Þe(cid:4)jnΩ0tdt

an ¼ 2 T 0 ð ¼ 2 6

(cid:4)jn

(cid:4)e

3tdt

0 ð

1 (cid:9)

¼ 1 3 ¼ 1

(cid:10)

jnπ e(cid:4)j2nπ=3 (cid:4) e(cid:4)jnπ=3

Example 3.3 Find Fourier series of the following periodic signal with half-wave symmetry (Figure 3.2)

Solution The period T0 ¼ 8 and Ω0 ¼ 2π 8 Fourier coefﬁcients are given by

¼ π 4

an ¼

8 <

2 T 0

T 0=2 0

for even n

x tð Þe(cid:4)jnΩ0tdt

for odd n

For odd n

x(t)

-1

Figure 3.2 Periodic signal with half-wave symmetry

122

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

T 0=2

x tð Þe(cid:4)jnΩ0tdt

an ¼ 2 T 0 ¼ 2 8

¼ 1 4

0 ð

¼ 1 8 ”

(cid:4)jn

4tdt

t 2

(cid:4)jn

4tdt

π 4t

(cid:4)jn

(cid:11) (cid:11) (cid:11) (cid:11)

þ 1

(cid:4)jn

4tdt

¼ 1 8

(cid:4)t jnπ=4 ”

jnπ=4

(cid:11) (cid:11) (cid:11) (cid:11)

π 4t

(cid:10)

Þ (cid:4) 1

(cid:4)jn

¼ 1 8 ¼ j (cid:4)kþ1 nπ

nπ j(cid:4)k þ 16 n2π2e (cid:9) þ 2 n2π2 j (cid:4)kð

Example 3.4 Consider the periodic signal x(t) given by

ð x tð Þ ¼ 2 þ j2

Þe(cid:4)j3t (cid:4) j3e(cid:4)j2t þ 6 þ j3ej2t þ 2 (cid:4) j2

Þej3t

(i) Determine the fundamental period and frequency of x(t) (ii) Show that x(t) is a real signal (iii) Find energy of the signal Solution (i) x(t) is the sum of two periodic signals with periods T 1 ¼ 2π

3 and T 2

¼ 2π

2 The ratio T 1 T 2

¼ 2

3 is a rational number.

The fundamental period of the signal x(t) is 3T1 ¼ 2T2 ¼ 2π. The fundamental

frequency Ω0 ¼ 2π

2π ¼ 1.

(ii) x(t) is exponential Fourier series representation of the form

x tð Þ ¼

n¼(cid:4)1

an ejnΩ0t

where

a(cid:4)3 ¼ 2 þ j2; a(cid:4)2 ¼ (cid:4)j3; a0 ¼ 6; a3 ¼ 2 (cid:4) j2; a2 ¼ j3

and an¼ 0 for all other. It is noticed that an ¼ a∗

(cid:4)n for all n. Hence, x(t) is a real signal.

3.1 Complex Exponential Fourier Series Representation of the Continuous…

123

(iii) The average power of the signal x(t) is

E ¼ 1 T 0

T 0

x tð Þ

j2dt ¼

n¼(cid:4)1

anj

By Parseval’s theorem,

anj

E ¼

n¼(cid:4)1 p(cid:9)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 22 þ 22

(cid:10)

þ 32 þ 62 þ 32 þ

p(cid:9)

(cid:10) 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 22 þ 22

¼ 8 þ 9 þ 36 þ 9 þ 8 ¼ 70

Example 3.5 Find the Fourier series coefﬁcients for each of the following signals: (i) x(t) ¼ cos(Ω0t) þ sin (2Ω0t) (ii) x(t) ¼ 2 cos(Ω0t) þ sin2 (2Ω0t)

Solution (i) x tð Þ ¼

n¼(cid:4)1 an ejnΩ0t

x tð Þ ¼ 1 2 a1 ¼ 1 2

ejΩ0t þ 1 2 ; a(cid:4)1 ¼ 1 2

e(cid:4)jΩ0t þ 1 2j ; a2 ¼ 1 2j

ej2Ω0t (cid:4) 1 2j ; a(cid:4)2 ¼ (cid:4) 1 2j

e(cid:4)j2Ω0t

an ¼ 0 for all other n.

(ii) x tð Þ ¼ 2 cos Ω0t

Þ þ sin 2 2Ω0t

Þ ¼ 2 cos Ω0t

x tð Þ ¼

anejnΩ0t

Þ þ 1

½ 2 1 (cid:4) cos 4Ω0t

(cid:5)

n¼(cid:4)1 (cid:2)

x tð Þ ¼ 2

(cid:3)

(cid:2)

(cid:3)

e(cid:4)j4Ω0t (cid:3)

ej4Ω0t þ 1 2

1 2

e(cid:4)jΩ0t

ejΩ0t þ 1 2

þ 1 2 (cid:2) (cid:4) 1 2

(cid:4) 1 1 2 2 ¼ ejΩ0t þ e(cid:4)jΩ0t þ 1 1 ej4Ω0t þ 1 2 2 2 ; a1 ¼ 1; a(cid:4)4 ¼ (cid:4)1 4

a(cid:4)1 ¼ 1; a0 ¼ 1 2

e(cid:4)j4Ω0t

; a4 ¼ (cid:4)1 4

an ¼ 0 for all other n.

124

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Example 3.6 Find Fourier series coefﬁcients of the following continuous-time periodic signal and plot the magnitude and phase spectrum of it:

x tð Þ ¼ 2s in 2πt (cid:4) 3

Þ þ s in 6πt

Solution

ð ej 2πt(cid:4)3

ð e(cid:4)j 2πt(cid:4)3

ð ej 6πt

ð e(cid:4)j 6πt

x tð Þ ¼ 2 2j ¼ 1 j ¼ (cid:4) 1 2j

Þ (cid:4) 2 2j Þ (cid:4) 1 j Þ (cid:4) 1 j

Þ þ 1 2j Þ þ 1 2j Þ þ 1 j

Þ (cid:4) 1 2j Þ (cid:4) 1 2j Þ þ 1 2j

ð e(cid:4)j3ej 2πt

ð ej3e(cid:4)j 2πt

ð ej 6πt

ð e(cid:4)j 6πt

ð ej3e(cid:4)j 2πt

ð e(cid:4)j3ej 2πt

ð ej 6πt

Since Ω0¼ 2π

x tð Þ ¼ (cid:4) 1 2j

ð e(cid:4)j 3Ω0t

Þ (cid:4) 1 j

ð ej3e(cid:4)j Ω0t X1

Þ þ 1 j anejnΩ0t

x tð Þ ¼

ð e(cid:4)j3ej Ω0t

Þ þ 1 2j

ð ej 3Ω0t

n¼(cid:4)1 ¼ j 2

ð Þ ej π

¼ 1 a(cid:4)3 ¼ (cid:4) 1 2j 2 a(cid:4)1 ¼ (cid:4)ej3 ¼ jej3 ¼ e(cid:4)j1:7124 j a1 ¼ e(cid:4)j3 ¼ (cid:4)je(cid:4)j3 ¼ ej1:7124 j ¼ (cid:4) j ¼ 1 a3 ¼ 1 2j 2 2

ð ej (cid:4)π

an¼ 0 for all other n. The magnitudes of Fourier coefﬁcients are

j j

a(cid:4)3 a(cid:4)1

j ¼ a3j j ¼ a1j

j ¼ 0:5 j ¼ 1:0

The magnitude spectrum and phase spectrum are shown in Figures 3.3 and 3.4,

respectively.

Since x(t) is a real valued, its magnitude spectrum is even and the phase spectrum

is odd.

Example 3.7 (i) Obtain x(t) for the following non-zero Fourier series coefﬁcients of a continuous-

time real valued periodic signal x(t) with fundamental period of 8.

a1 ¼ a∗ (cid:4)1

¼ j, a5 ¼ a(cid:4)5 ¼ 1

3.1 Complex Exponential Fourier Series Representation of the Continuous…

125

Figure 3.3 Magnitude spectrum of x(t)

Figure 3.4 Phase spectrum of x(t)

Figure 3.5 Magnitude spectrum of a signal

(ii) Consider a continuous periodic signal with the following magnitude spectra shown in Figure 3.5. Find the DC component and average power of the signal.

Solution (i)

x tð Þ ¼

n¼(cid:4)1

anejnΩ0t

x tð Þ ¼ e(cid:4)j5Ω0t þ ej5Ω0t þ j ejΩ0t (cid:4) e(cid:4)jΩ0t

¼ 2 cos 5Ω0t ð ð ¼ 2 cos 5Ω0t

ð Þ (cid:4) 2 s in Ω0t Þ þ 2 cos Ω0t þ

ð (cid:9)

(cid:10)

126

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

(ii) The DC component is given by

a0 ¼ 1:

By using Parseval’s relation, the average power is computed as X1 j2 ¼ 12 þ 22 þ 12 þ 22 þ 12 ¼ 11

n¼(cid:4)1 anj

Example 3.8 Which of the following signals cannot be represented by the Fourier series?

(i) x(t) ¼ 4 cos(t) þ 6 cos(t) (ii) x(t) ¼ 3 cos(πt) þ 6 cos(t) (iii) x(t) ¼ cos(t) þ 0.75 (iv) x(t) ¼ 2 cos(3πt) þ 3 cos(7πt) (v) x(t) ¼ e(cid:4)|t| sin (5πt)

Solution (i) x(t) ¼ 4 cos(t) þ 6 cos(t) is periodic with period 2π. (ii) x(t) ¼ 3 cos(πt) þ 6 cos(t) The ﬁrst term has period

The second term has period

T 1 ¼ 2π π

¼ 2

T 2 ¼ 2π 1

¼ 2π

The ratio T 1 T 2

¼ 2

π is not a rational number. Hence, x(t) is not a periodic signal.

(iii) x(t) ¼ cos(t) þ 0.75 is periodic with period 2π. (iv) x(t) ¼ 2 cos(3πt) þ 3 cos(7πt)

The ﬁrst term has period

The second term has period

T 1 ¼ 2π 3π

¼ 2 3

T 2 ¼ 2π 7π

¼ 2 7

The ratio T 1 T 2

¼ 7

3 is a rational number. Hence, x(t) is a periodic signal.

(v) Due to decaying exponential function, it is not periodic. So Fourier series cannot

be deﬁned for it.

Hence, (ii) and (v) cannot be represented by Fourier series. Since the remaining

three are periodic; they can be represented by Fourier series.

3.1 Complex Exponential Fourier Series Representation of the Continuous…

127

Example 3.9 Find the Fourier series of a periodic square wave with period T0 deﬁned over one period by

Solution For n¼0,

For n 6¼ o,

an ¼ 1 T 0

(cid:4) x tð Þ ¼ 1

j t j< T 0=4

0 T 0=4 <j t j(cid:6) T 0=2

a0 ¼ 1 T 0

T 0=4

(cid:4)T 0=4

dt ¼ 2T 0=4 T 0

¼ 1 2

T 0=4

(cid:4)T 0=4 (cid:2)

e(cid:4)jnΩ0tdt ¼ (cid:4) 1

e(cid:4)jnΩ0t

jnΩ0T 0 (cid:3)

(cid:11) T 0=4 (cid:11) (cid:11) (cid:11) (cid:11) (cid:4)T 0=4

ejnΩ0T 0=4 (cid:4) e(cid:4)jnΩ0T 0=4 2j

¼ 2

nΩ0T 0 ð ¼ 2 sin nΩ0T 0=4 nΩ0T 0

Since Ω0 ¼ 2π T 0

an ¼

sin

(cid:7)

2π T 0 nπ 2π T 0 4

B B @

(cid:8)

T 0=4

n 6¼ 0

C C A

T 0

¼ 1 2

sin

nπ (cid:9) (cid:10) nπ 2 nπ=2

Example 3.10 Find the Fourier series of the periodic signal shown in Figure 3.6. Solution The period T0 ¼ 2. Ω0 ¼ 2π T 0

¼ π:

For n¼0,

a0 ¼ 1 2

(cid:4)1

tdt ¼ 0

For n 6¼ o,

128

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure 3.6 Periodic signal

an ¼ 1 2

(cid:4)1

”

te(cid:4)jnπtdt

¼ 1 2

(cid:4)jnπ e(cid:4)jnπt

(cid:11) (cid:11) (cid:11) (cid:11)

(cid:4)1

(cid:11) (cid:11) (cid:11) (cid:11)

þ 1 jnπ

(cid:4)1

(cid:4) e(cid:4)jnπt Þ2 ð (cid:4)jnπ

e(cid:4)jnπtdt

(cid:4)1

þ 0

”

¼ 1 2

¼ (cid:4) 1 2

(cid:4)jnπ e(cid:4)jnπt

(cid:4)jnπ e(cid:4)jnπt (cid:4)1 e(cid:4)jnπ þ ejnπ jnπ

Since e(cid:4)jnπ + e jnπ ¼ 2((cid:4)1)n

an ¼

(cid:4)1ð

Þnþ1

jnπ

3.2 Trigonometric Fourier Series Representation

The trigonometric Fourier series representation of a periodic signal x(t) is expressed by

x tð Þ ¼ a0 2

Xþ1 n¼1

ð an cos nΩ0t

Þ þ bn sin nΩ0t

ð3:21Þ

The Fourier coefﬁcients an and bn are given by

an ¼ 2 T 0 bn ¼ 2 T 0

T 0

0 ð

T 0

ð x tð Þ cos nΩ0t

Þdt

ð x tð Þ sin nΩ0t

Þdt

ð3:22aÞ

ð3:22bÞ

3.2 Trigonometric Fourier Series Representation

129

3.2.1 Symmetry Conditions in Trigonometric Fourier Series

If x(t) is an even periodic signal, then

a0 ¼ 2 T 0 ð

T 0=2

an ¼ 4 T 0

T 0=2

x tð Þdt

ð x tð Þ cos nΩ0t

Þdt

ð3:23aÞ

ð3:23bÞ

bn¼ 0 for all n. If x(t) is an odd periodic signal, thena0¼ 0;an ¼ 0 for all n

bn ¼ 4 T 0

T 0=2

ð x tð Þ sin nΩ0t

Þdt

ð3:24Þ

Therefore, for every even signal bn ¼ 0. Hence, Fourier series of an even signal contains DC term and cosine terms only. Fourier series of an odd signal contains sine terms only.

Example 3.11 Find the trigonometric Fourier series representation of the periodic signal shown in Figure 3.7 with A ¼ 3 and period T0¼ 2π. Solution The period T0¼2π. Ω0 ¼ 2π T 0

¼ 1.

The periodic signal x(t) deﬁned over one period is

(cid:4)

x tð Þ ¼

(cid:4)3 (cid:4)π (cid:6) t < 0 3

0 (cid:6) t < π

Since x(t) has odd symmetry, a0 ¼ 0 and an ¼ 0

bn ¼ 4 2π (cid:2) ¼ 6 π ¼ 6 nπ ( 0 12 nπ

Þ dt

(cid:4)3 sin ntð (cid:3) (cid:11) (cid:11)

0 (cid:4)π

(cid:4)π cos ntð n

(cid:5) 1 (cid:4) cos nπð for n even

for n odd

Figure 3.7 Periodic signal

130

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure 3.8 Periodic triangle wave

The trigonometric Fourier series representation of x(t) is given by

x tð Þ ¼

n¼1

bn sin ntð

Þ ¼ 12 π

n¼1

ð ð sin 2n (cid:4) 1 Þ ð 2n (cid:4) 1

Þt

Example 3.12 Find the trigonometric Fourier series representation of the periodic triangle wave shown in Figure 3.8 with A ¼ 2 period T0 ¼ 2. Solution The period T0¼2. Ω0 ¼ 2π T 0 The periodic triangle wave x(t) deﬁned over one period with A¼2 is

¼ π.

(cid:4)

x tð Þ ¼

4t 4 1 (cid:4) t

j t j< 1=2 Þ 1=2 < t < 3=2

Since x(t) has odd symmetry, a0 ¼ 0 and an ¼ 0

ð cos nπt

Þ dt

1=2

(cid:4)1=2

bn ¼ 4 2

1=2

(cid:4)1=2

”

ð 4t sin nπt

Þ dt

þ 1 nπ i

(cid:11) (cid:11)1=2 (cid:4)1=2 (cid:11) (cid:11)1=2 (cid:4)1=2 (cid:3)

ð nπ cos nπt

¼ 8 (cid:4)t h ¼ 8 0 þ 1 (cid:2) ¼ 8 0 þ 2

Þ n2π2 sin nπt (cid:9) (cid:10) nπ 2

n2π2 sin (cid:9) (cid:10) nπ 2

¼ 16

n2π2 sin

The trigonometric Fourier series representation of x(t) is

(cid:2)

x tð Þ ¼ 16 π2

sin πtð

Þ (cid:4) 1 9

ð sin 3πt

Þ þ 1 25

ð sin 5πt

Þ (cid:4) 1 49

ð sin 7πt

Þ þ (cid:3) (cid:3) (cid:3)

(cid:3)

Example 3.13 Find the trigonometric Fourier series representation of the following periodic signal with period T0 ¼ 2.

3.2 Trigonometric Fourier Series Representation

131

x tð Þ ¼

< :

ð cos 3πt 0

(cid:4)1 (cid:6) t (cid:6) (cid:4)1 2 (cid:6) t < 1 2 1=2 (cid:6) t < 1

Þ (cid:4)1 2

Solution The period T0¼2. Ω0 ¼ 2π T 0

¼ π.

Since x(t) has even symmetry, bn ¼ 0 a0 ¼ 2 2 ð

0 dt

1=2

þ2 2

ð(cid:4)1=2

(cid:4)1

0 dt þ 2 2

1=2

(cid:4)1=2

ð cos 3πt

Þ dt

(cid:11) (cid:11) (cid:11)

1=2

(cid:4)1 2

ð ¼ sin 3πt 3π ¼ (cid:4) 2 3π

an ¼ 2 2

1=2

(cid:4)1=2

ð cos 3πt

ð Þ cos nπt

Þ dt

For n ¼ 1,

For n ¼ 2,

For n ¼ 3

a1 ¼

1=2

(cid:4)1=2

ð cos 3πt

Þ cos πtð

Þ dt ¼ 0

a2 ¼

1=2

(cid:4)1=2

ð cos 3πt

ð Þ cos 2πt

Þ dt ¼ 6 5π

a3 ¼

1=2

(cid:4)1=2

ð cos 3πt

ð Þ cos 3πt

Þ dt ¼ 1 2

For n ¼ 4,5,6,… .

1=2

an ¼

cos 3πt ð (cid:9) (cid:10) (cid:4)1=2 nπ 6 cos 2 n2π (cid:4) 9π

ð Þ cos nπt

Þ dt

The trigonometric Fourier series representation of x(t) is

132

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

x tð Þ ¼ a0 2

Xþ1

n¼1

an cos nπt

(cid:4)1 3π

þ 6

ð 5π cos 2πt

Þ þ 1 2

ð cos 3πt

(cid:9) (cid:10) nπ 6 cos 2 n2π (cid:4) 9π cos nπt

n¼4

Example 3.14 If the input to the half-wave rectiﬁer is an AC signal x(t) ¼ cos(2πt), ﬁnd trigonometric Fourier series representation of output signal of the half-wave rectiﬁer. Solution The output y(t) of half-wave rectiﬁer is

(cid:4) y tð Þ ¼ x tð Þ 0

for x tð Þ (cid:7) 0 for x tð Þ < 0

The sinusoidal input and output of half-wave rectiﬁer are shown in Figure 3.9. Since y(t) is a real and even function, its Fourier coefﬁcients are real and even.

The period T0¼1. Ω0 ¼ 2π T 0

¼ 2π

Input x(t)=cos(2p t)

0.5

−0.5

−1 −1.5

0.5

−0.5

−1 −1.5

−1.25

−1

−0.75

−0.5

−0.25

0.25

0.5

0.75

1.25

1.5

Output y (t )

−1.25

−1

−0.75

−0.5

−0.25

0 t (sec)

0.25

0.5

0.75

1.25

1.5

Figure 3.9 Input and output of half-wave rectiﬁer

3.3 The Continuous Fourier Transform for Nonperiodic Signals

133

For n ¼ 0,

a0 ¼ 1 1

1=4

(cid:4)1=4

ð cos 2πt

Þdt ¼ 1 π

Hence, the DC component is 1 π For n6¼0

an ¼ 2 1

1=4

”

(cid:4)1=4 ð

1=4

¼ 2 1 ”

ð cos 2πt

ð Þ cos nΩ0t

Þdt

1=4

Þt

Þdt

ð cos 2π n þ 1

Þt

Þdt þ

ð cos 2π n (cid:4) 1

(cid:4)1=4 ¼ sin 2π n þ 1 ð ð ð 2π n þ 1 Þ (cid:9) π

Þt

(cid:11) (cid:11) Þ (cid:11) (cid:11) (cid:10)

1=4

(cid:4)1=4

(cid:4)1=4 þ sin 2π n (cid:4) 1 ð ð Þ 2π n þ 1 ð (cid:10) 3 π

(cid:9)

Þt

(cid:11) (cid:11) Þ (cid:11) (cid:11)

1=4

(cid:4)1=4

2 sin

n þ 1 Þ

2 sin

2 2π n þ 1 ð (cid:10) (cid:6)(cid:9) π n 2 n þ 1

cos

¼ 1 π

cos

(cid:6)(cid:9) π n 2 n (cid:4) 1

(cid:4)

Þt

n þ 1 2 2π n þ 1 Þ ð (cid:10) 3

(cid:10)

(cid:6)(cid:9) π n 2 cos π 1 (cid:4) n2 ð

5 ¼

The trigonometric Fourier series representation of y(t) is

x tð Þ ¼ a0 2

¼ 1 2π

Xþ1

n¼1 X1

n¼4

Þ an cos nπt ð (cid:10) (cid:6)(cid:9) π n 2 cos π 1 (cid:4) n2 ð

(cid:6)

(cid:5)

nπt

Þ cos

3.3 The Continuous Fourier Transform

for Nonperiodic Signals

Consider a nonperiodic signal x(t) as shown in Figure 3.10 (a) with ﬁnite duration, i.e., x(t) ¼ 0 for |t| >T1. From this nonperiodic signal, a periodic signal ~x tð Þ can be constructed as shown in Figure 3.10(b).

The Fourier series representation of ~x tð Þ is

~x tð Þ ¼

n¼(cid:4)1 anejnΩ0t

ð3:25aÞ

134

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

x(t)

− 1

(a)

~ (t)

− 2 0

− 0

− 1

2 0

(b)

Figure 3.10 (a) Nonperiodic signal (b) periodic signal obtained from (a)

an ¼ 1 T 0

T 0=2

(cid:4)T 0=2

~x tð Þe(cid:4)jnΩ0tdt

ð3:25bÞ

Since ~x tð Þ ¼ x tð Þ for j t j< T 0

2 and also since x(t) ¼ 0 outside this interval, so

an ¼ 1 T 0

T 0=2

(cid:4)T 0=2

x tð Þe(cid:4)jnΩ0tdt ¼ 1 T 0

ð1

(cid:4)1

x tð Þe(cid:4)jnΩ0tdt

ð3:26Þ

we have

Deﬁne

Then

X jΩð

Þ ¼

ð1

(cid:4)1

x tð Þe(cid:4)jΩtdt

an ¼ 1 T 0

ð X jnΩ0

and ~x tð Þ can be expressed in terms of X( jΩ), that is,

~x tð Þ ¼

ÞejnΩ0t

n¼(cid:4)1 X1

ð X jnΩ0

1 T 0 ð n¼(cid:4)1 X jnΩ0

ÞejnΩ0tΩ0

¼ 1 2π

ð3:27Þ

ð3:28Þ

ð3:29Þ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

135

As T0 tends to inﬁnity, ~x tð Þ ¼ x tð Þ and summation becomes integration,

Eq. (3.29) becomes

x tð Þ ¼ 1 2π

ð1

(cid:4)1

X jΩð

ÞejΩtdΩ

ð3:30Þ

Eq. (3.27) is referred to as the Fourier transform of x(t), and Eq. (3.30) is called

the inverse Fourier transform.

3.3.1 Convergence of Fourier Transforms

The sufﬁcient conditions referred to as the Dirichlet conditions for the convergence of Fourier transform are:

x(t) must be absolutely integrable, that is,

ð1

(cid:4)1

j x tð Þ j dt < 1

ð3:31Þ

x(t) must have a ﬁnite number of maxima and minima within any ﬁnite interval.
x(t) must have a ﬁnite number of discontinuities within any ﬁnite interval, and

each of these discontinuities is ﬁnite.

Although the above Dirichlet conditions guarantee the existence of the Fourier transform for a signal, if impulse functions are permitted in the transform, signals which do not satisfy these conditions can have Fourier transforms. Example 3.15 Determine x(0) and X(0) using the deﬁnitions of the Fourier trans- form and the inverse Fourier transform Solution By the deﬁnition of Fourier transform, we have

X jΩð

½ Þ ¼ F x tð Þ

(cid:5) ¼

ð1

(cid:4)1

x tð Þe(cid:4)jΩtdt

Substituting Ω ¼ 0 in this equation, we obtain

X 0ð Þ ¼

ð1

(cid:4)1

x tð Þ dt

By the deﬁnition of the inverse Fourier transform, we have

x tð Þ ¼ 1 2π

ð1

(cid:4)1

X jΩð

Þ ejΩtdΩ

136

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Substituting t¼0, it follows that

x 0ð Þ ¼ 1 2π

ð1

(cid:4)1

X jΩð

Þ dΩ

3.3.2 Fourier Transforms of Some Commonly Used

Continuous-Time Signals

The unit impulse The Fourier transform of the unit impulse function is given by

½ F δ tð Þ

(cid:5) ¼

ð1

(cid:4)1

δ tð Þ e(cid:4)jΩtdt ¼ 1

ð3:32Þ

implying that the Fourier transform of the unit impulse contribute equally at all frequencies Example 3.16 Find the Fourier transform of x(t) ¼ e(cid:4)btu(t) for b > 0.

Solution

ð1

½ F x tð Þ

(cid:5) ¼

ð1

e(cid:4)btu tð Þe(cid:4)jΩtdt ¼

(cid:4)1 ð1

ð e(cid:4) jΩþb

Þtdt ¼

e(cid:4)bte(cid:4)jΩtdt (cid:11) (cid:11) (cid:11) (cid:11)

0 ð e(cid:4) jΩþb

Þt

(cid:4)1 jΩ þ b

¼ 1

b þ jΩ b > 0

Example 3.17 Find the Fourier transform of x(t) ¼ 1. Solution By deﬁnition of the inverse Fourier transform and sampling property of the impulse function, we have

F(cid:4)1 δ Ωð Þ ½

(cid:5) ¼ 1 2π

ð1

(cid:4)1

δ Ωð ÞejΩtdΩ ¼ 1 2π

¼ δ Ωð Þ and thus F 1½ (cid:5) ¼ 2πδ Ωð Þ

(cid:13) (cid:14) Hence, F 1 2π

Example 3.18 Find the Fourier transforms of the following (i) sin (Ω0t) (ii) cos (Ω0t)

Solution

(i)

sin Ω0t ð

Þ ¼ ejΩ

0t0(cid:4)e(cid:4)jΩ 2j

0 t

By the sampling property of the impulse function, we have

3.3 The Continuous Fourier Transform for Nonperiodic Signals

137

F(cid:4)1½δðΩ (cid:4) Ω0Þ(cid:5) ¼ 1 2π

δðΩ (cid:4) Ω0ÞejΩtdΩ ¼ 1

2π ejΩ0t

ð1

(cid:4)1

(cid:14)

(cid:13) 2πejΩ0t Þ ð Hence, F 1 ¼ δ Ω (cid:4) Ω0 (cid:13) (cid:14) Þ, F e(cid:4)jΩ0t Thus F ejΩ0t ̀ ¼ 2πδ Ω (cid:4) Ω0 ð ½ ð δ Ω (cid:4) Ω0 Þ Therefore, F sin Ω0t

(cid:13)

(cid:5) ¼ π j

(cid:14) ̀ ¼ 2πδ Ω þ Ω0 ð

ð Þ (cid:4) δ Ω þ Ω0

(cid:5) Þ

(ii) cos ðΩ0tÞ ¼ ejΩ

0 tþe(cid:4)jΩ 2

ð F cos Ω0t

(cid:2)

(cid:5) ¼ F ejΩ0t þ e(cid:4)jΩ0t

(cid:3)

ð ¼ π δ Ω (cid:4) Ω0

ð Þ þ δ Ω þ Ω0

(cid:5) Þ

Example 3.19 Find the Fourier transform of the rectangular pulse signal shown in Figure 3.11

Solution

(cid:4) x tð Þ ¼ 1 0

XðjΩÞ ¼ F½xðtÞ(cid:5) ¼

ð (cid:4)1 T 1

xðtÞe(cid:4)jΩtdt

tj j (cid:6) T 1 tj j > T 1 ð1

e(cid:4)jΩtdt

(cid:4)T 1 ¼ (cid:4) 1

(cid:4)T 1

jΩe(cid:4)jΩtjT 1 ejΩT1 (cid:4) e(cid:4)jΩT1 2jΩ sin ðΩT 1Þ Ω

¼ 2

Since sinc tð Þ ¼ sin πtð πt

function as

sin ΩT 1 ð Ω

can be written in terms of the sinc

sin ΩT 1 ð Ω

¼ 2T 1 sin c

(cid:8)

(cid:7)

ΩT 1 π

Figure 3.11 Rectangular pulse signal

x(t)

− 1

138

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure 3.12 Fourier transform of a signal

( Ω)

-Ω

Hence,

X jΩð

½ Þ ¼ F x tð Þ

(cid:5) ¼ 2T 1sinc

(cid:8)

(cid:7)

ΩT 1 π

Example 3.20 Consider the Fourier transform X( jΩ) of a signal shown in Fig- ure 3.12. Find the inverse Fourier transform of it.

Solution

X jΩð

Þ ¼

(cid:4)

Ωj Ωj

j (cid:6) Ω j > Ω

By the inverse Fourier transform deﬁnition, we have

X jΩð

ÞejΩtdΩ

ð1

(cid:4)1 ð Ω

x tð Þ ¼ 1 2π

¼ 1 2π ¼ 1 2π

¼ 1 2π ¼ sin Ωtð πt

(cid:2)

(cid:4)Ω

ejΩtdΩ (cid:3)

(cid:11) (cid:11)

(cid:4)Ω

jΩejΩt Ω ejΩt þ e(cid:4)jΩt jt

(cid:3)

Ω π sinc

(cid:7) (cid:8) Ωt π

Example 3.21 Determine the Fourier transform of Gaussian signal x tð Þ ¼ (cid:4)t2 e2σ2

Solution

Letting b ¼ 1 2σ2

X jΩð

½ Þ ¼ F x tð Þ

(cid:5) ¼

ð1

(cid:4)1

(cid:4)t2 e2σ2 e(cid:4)jΩtdt

3.3 The Continuous Fourier Transform for Nonperiodic Signals

139

X jΩð

Þ ¼

Ð 1 (cid:4)1 e(cid:4)bt2 e(cid:4)jΩtdt Ð 1 ð (cid:4)1 e(cid:4)b t2þ jΩ=b Þ dt Ð 1 Þ2(cid:4)Ω2=4bdt ð Þ (cid:4)1 e(cid:4)c tþ jΩ=2b ð Ð 1 ð (cid:4)1 e(cid:4)b t(cid:4) jΩ=2b

¼ e(cid:4)Ω2=4b ﬃﬃﬃ b

dτ ¼

Þ2(cid:4)Ω2=4bdt Þ

Letting τ ¼ t

ﬃﬃﬃﬃﬃ b,

(cid:9)

(cid:10)

t(cid:4)ðjΩ=2bÞ

(cid:4)Ω2=4b

XðjΩÞ ¼ e(cid:4)Ω2=4b ¼ e(cid:4)Ω2=4b p ﬃﬃﬃ b

Ð 1 (cid:4)1 e

(cid:4)b (cid:9)

ð1

τ(cid:4)ðjΩ=2

(cid:4)

(cid:4)1

(cid:10)

ﬃﬃ Þ b

dτ

ð e(cid:4) τ(cid:4) jΩ=2

ﬃﬃ p b

Þ2

ﬃﬃﬃ π

dτ ¼

X jΩð

Þ ¼ e(cid:4)Ω2=4b p ﬃﬃﬃ b

ﬃﬃﬃ π

ð1

(cid:4)1

Since

Substituting b ¼ 1 2σ2

XðjΩÞ ¼ e(cid:4)Ω2=4b ﬃﬃﬃ b p ﬃﬃﬃﬃﬃ e(cid:4)σ2Ω2=2 2π ¼ σ

p ﬃﬃﬃ π

3.3.3 Properties of the Continuous-Time Fourier Transform

Linearity If x1(t) and x2(t) are two continuous-time signals with Fourier transforms X1( jΩ) and X2( jΩ), then the Fourier transform of a linear combination of x1(t) and x2(t) is given by

F a1x1 tð Þ þ a2x2 tð Þ

(cid:5) ¼ a1X1 jΩð

Þ þ a2X2 jΩð

ð3:33Þ

where a1 and a2 are arbitrary constants.

Example 3.22 Find the Fourier transform of an impulse train with period T as given by

x tð Þ ¼

k¼(cid:4)1

ð δ t (cid:4) kT

140

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Solution A periodic signal x(t) with period T is expressed by

x tð Þ ¼

P1 k¼(cid:4)1 akejkΩ0t Ω0 ¼ 2π

Taking Fourier transform both sides, we obtain

½ F x tð Þ

(cid:5) ¼ F

”

k¼(cid:4)1

akejkΩ0t

(cid:13)

Using F ejΩ0t

(cid:14) ð ̀ ¼ 2πδ Ω (cid:4) Ω0

Þand the linearity property, we have

”

k¼(cid:4)1

akejkΩ0t

¼ 2π

k¼(cid:4)1

akδ Ω (cid:4) kΩ0

If x(t) is an impulse train with period T as given by

x tð Þ ¼

k¼(cid:4)1

ð δ t (cid:4) kT

Since

k¼(cid:4)1

ð δ t (cid:4) T

Þ ¼ 1 T

k¼(cid:4)1 ejkΩ0t

”

k¼(cid:4)1

ð δ t (cid:4) kT

¼ 2π T

k¼(cid:4)1

ð δ Ω (cid:4) kΩ0

Symmetry for Real Valued Signal If x(t) is a continuous-time real valued signal with Fourier transform X( jΩ), then

X (cid:4)jΩð

Þ ¼ X∗ jΩð

ð3:34Þ

where * stands for the complex conjugate.

Proof

X∗ jΩð

Þ ¼ ¼

(cid:13) Ð 1 (cid:4)1 x tð Þe(cid:4)jΩtdt Ð 1 (cid:4)1 x∗ tð ÞejΩtdt

(cid:14)∗

Since x(t) is real x*(t) ¼ x(t), we get

X∗ jΩð

Þ ¼

ð1

(cid:4)1

x tð ÞejΩtdt ¼ X (cid:4)jΩð

The X( jΩ) can be expressed in rectangular form as

3.3 The Continuous Fourier Transform for Nonperiodic Signals

141

X jΩð

½ Þ ¼ Re X jΩð

(cid:5) þ jIm

(cid:13)

X jΩð

If x(t) is real, then

Re X jΩð ½ Þ ½ Im X jΩð

(cid:5) ¼ Re X (cid:4)jΩð ½ (cid:5) ½ (cid:5) ¼ (cid:4)Im X (cid:4)jΩð Þ

(cid:5)

implying that the real part is an even function of Ω and the imaginary part is an odd function of Ω.

For real x(t) in polar form

X jΩð Þ j ½ arg X jΩð

j ¼ X (cid:4)jΩð ½ (cid:5) ¼ (cid:4)arg X (cid:4)jΩð Þ

(cid:5)

indicating that the magnitude is an even function of Ω and the phase is an odd function of Ω.

Symmetry for Imaginary Valued Signal If x(t) is a continuous-time imaginary valued signal with Fourier transform X( jΩ), then

X∗ jΩð

Þ ¼ (cid:4)X (cid:4)jΩð

ð3:35Þ

Proof

X∗ jΩð

Þ ¼

(cid:14)∗

(cid:13) Ð 1 (cid:4)1 x tð Þe(cid:4)jΩtdt Ð 1 (cid:4)1 x∗ tð ÞejΩtdt

Since x(t) is purely imaginary, we get x(t) ¼ (cid:4)x*(t), and we get ð1

X∗ jΩð

Þ ¼ (cid:4)

(cid:4)1

x tð ÞejΩtdt ¼ (cid:4)X (cid:4)jΩð

Re X jΩð ½ ½ Im X jΩð Þ

(cid:5) ¼ (cid:4)Re X (cid:4)jΩð Þ ½ (cid:5) ½ (cid:5) ¼ Im X (cid:4)jΩð Þ

(cid:5)

Symmetry for Even and Odd Signals (i) If x(t) is a continuous-time real valued and has even symmetry, then

X∗ jΩð

Þ ¼ X jΩð

(ii) If x(t) is a continuous-time real valued and has odd symmetry, then

X∗ jΩð

Þ ¼ (cid:4)X jΩð

ð3:36aÞ

ð3:36bÞ

Proof Since x(t) is real x(t) ¼ x*(t) and x(t) has even symmetry x(t) ¼ x((cid:4)t), we get

142

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

X∗ jΩð

Þ ¼

(cid:3)∗

x tð Þe(cid:4)jΩtdt

(cid:2)

ð1

(cid:4)1

ð1

(cid:4)1 ð1

¼ (cid:4)

(cid:4)1

x∗ tð Þ ejΩtdt

x tð Þ ejΩtdt

ð x (cid:4)t

Þ e(cid:4)jΩ (cid:4)tð

Þdt

Letting τ ¼(cid:4)t, we obtain

X∗ jΩð

Ð 1 (cid:4)1 x τð Þe(cid:4)jΩτdτ

Þ ¼ ¼ X jΩð

The condition X*( jΩ) ¼ X( jΩ) holds for the imaginary part of X( jΩ) to be zero.

Therefore, if x(t) is real valued and has even symmetry, then X( jΩ) is real.

Similarly, for real valued x(t) having odd symmetry, it can be shown that X*

( jΩ)¼(cid:4)X( jΩ) and X( jΩ) is imaginary. Time Shifting If x(t) is a continuous-time signal with Fourier transform X( jΩ), then the Fourier transform of x(t-t0) the delayed version of x(t) is given by

ð F x t (cid:4) t0

(cid:5) ¼ e(cid:4)jΩt0X jΩð

ð3:37Þ

Proof

ð F x t (cid:4) t0

(cid:5) ¼ Þ

ð1

(cid:4)1

ð x t (cid:4) t0

Þe(cid:4)jΩtdt

Letting τ ¼ t(cid:4)t0, we obtain

ð F x t (cid:4) t0

Ð 1 (cid:4)1 x τð Þe(cid:4)jΩ τ(cid:4)t0 Ð 1 (cid:4)1 x τð Þe(cid:4)jΩτdτ

(cid:5) ¼ ¼ e(cid:4)jΩt0 ¼ e(cid:4)jΩt0X jΩð

Þdτ

Therefore, time shifting results in unchanged magnitude spectrum but introduces

a phase shift in its transform, which is a linear function of Ω. Example 3.23 Find the Fourier transform of δ(t(cid:4)t0)

Solution

δ jΩð

½ Þ ¼ F δ tð Þ

(cid:5) ¼ 1

Hence, F½δðt (cid:4) t0Þ(cid:5) ¼ e(cid:4)jΩt0δðjΩÞ ¼ e(cid:4)jΩt0

3.3 The Continuous Fourier Transform for Nonperiodic Signals

143

Frequency Shifting If x(t) is a continuous-time signal with Fourier transform X ( jΩ), then the Fourier transform of the signal ejΩ0tx tð Þ is given by

(cid:13)

F ejΩ0tx tð Þ

(cid:14)

ð ¼ X j Ω (cid:4) Ω0

ð3:38Þ

Proof

ð1

(cid:4)1 ð1

F ejΩ0tx tð Þ

(cid:5) ¼

ejΩ0tx tð Þe(cid:4)jΩtdt

ð x tð Þe(cid:4)j Ω(cid:4)Ω0

Þtdt

(cid:4)1 ¼ X j Ω (cid:4) Ω0 ð ð

Thus, multiplying a sequence x(t) by a complex exponential ejΩ0t

in the time

domain corresponds to a shift in the frequency domain.

Time and Frequency Scaling If x(t) is a continuous-time signal with Fourier transform X( jΩ), then the Fourier transform of the signal x(at) is given by

where a is a real constant.

Proof

½ F x atð

(cid:5) ¼ 1

(cid:7) aj j X j

(cid:7) (cid:8) Ω

(cid:8)

½ F x atð

(cid:5) ¼

ð1

(cid:4)1

x atð

Þ e(cid:4)jΩtdt

Letting τ ¼ at, we obtain

ð3:39Þ

ð1

F½xðatÞ(cid:5) ¼ 1 a ð1 (cid:4)1 ¼ (cid:4)1 (cid:4)j xðτÞe a

(cid:4)1

xðτÞe (cid:5) (cid:6) Ω a

dτ

(cid:5) (cid:6) Ω a

(cid:4)j

dτ

for a > 0

f or a < 0

Thus,

½ F x atð

(cid:5) ¼ 1

(cid:7) aj j X j

(cid:8)

(cid:7) (cid:8) Ω

Differentiation in Time If x(t) is a continuous-time signal with Fourier transform X ( jΩ), then the Fourier transform of the d dt x tð Þ is given by (cid:3)

x tð Þ

¼ jΩX jΩð

ð3:40Þ

(cid:2) F d dt

144

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Proof By the deﬁnition of the inverse Fourier transform, it is known that

x tð Þ ¼ 1 2π

ð1

(cid:4)1

X jΩð

Þ ejΩtdΩ

Differentiating this equation both sides with respect to t, we obtain

x tð Þ ¼ 1 2π

ð1

(cid:4)1

x tð Þ ¼ jΩx tð Þ

d dt d dt

X jΩð

ÞjΩejΩtdΩ

Taking the Fourier transform of this equation both sides, we get

(cid:3)

x tð Þ

¼ jΩX jΩð

(cid:2) F d dt

Thus, differentiation in the time domain corresponds to multiplication by jΩ in

the frequency domain.

By repeated application of this property, we obtain

(cid:3)

(cid:2) F dn dtnx tð Þ

¼ jΩð

ÞnX jΩð

Example 3.24 Determine the Fourier transform of x(t) ¼ u(t)

Solution Decomposing the unit step function into even and odd components, it is written as

u tð Þ ¼ xe tð Þ þ xo tð Þ

where the even component xe tð Þ ¼ 1 (cid:5) þ F xe tð Þ ½ Þ Þ þ Xo jΩð

(cid:5) ¼ F xe tð Þ ½ ¼ Xe jΩð

Hence, F u tð Þ

(cid:5)

2 and the odd component xo tð Þ ¼ u tð Þ (cid:4) 1

2πδ Ωð Þ ¼ πδ Ωð Þ

Xe jΩð

(cid:5) ¼ 1 F 1½ (cid:5) ¼ 1 ½ Þ ¼ F xe tð Þ 2 2 xo tð Þ ¼ d u tð Þ ¼ δ tð Þ dt

d dt ¼ jΩXo jΩð

(cid:14) (cid:13) Thus, F d dtxo tð Þ

Therefore, U jΩð

jΩXo jΩð

Xo jΩð

(cid:5) ¼ 1

Þ ¼ F δ tð Þ ½ Þ ¼ 1 jΩ Þ þ Xo jΩð

(cid:5) ¼ Xe jΩð

Þ ¼ F u tð Þ ½ ¼ πδ Ωð Þ þ 1 jΩ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

145

Example 3.25 Determine the Fourier transform of (i) x(t) ¼ sin(Ω0t)u(t) (ii) x(t) ¼ cos(Ω0t)u(t)

Solution (i)

sin Ω0t ð

ð F sin Ω0t

Þu tð Þ

(cid:2)

(cid:3)

Þu tð Þ ¼ ejΩ0t (cid:4) e(cid:4)jΩ0t (cid:4) 1 (cid:5) ¼ 1 2j 2j

2j F ejΩ0tu tð Þ

(cid:14)

(cid:13)

u tð Þ (cid:13)

(cid:14) F ejΩ0tu tð Þ

since F u tð Þ

(cid:5) ¼ πδ Ωð Þ þ 1 jΩ

By frequency shifting property, we get

½ 2j π

½ 2j

ð F sin Ω0t

Þu tð Þ

(cid:5) ¼

(ii)

δ Ω (cid:4) Ω0 ð

ð Þ (cid:4) δ Ω þ Ω0

(cid:2)

(cid:5) þ 1 2j

δ Ω (cid:4) Ω0 ð

ð Þ (cid:4) δ Ω þ Ω0

(cid:5) þ

(cid:5)

1 j Ω (cid:4) Ω0 ð Ω0 2 (cid:4) Ω2

(cid:6)

Ω0

(cid:3)

(cid:4)

1 j Ω þ Ω0 ð

cos Ω0t ð

ð F cos Ω0t

Þu tð Þ

(cid:3)

(cid:2) Þu tð Þ ¼ ejΩ0t þ e(cid:4)jΩ0t þ 1 (cid:5) ¼ 1 2 2

2 (cid:14) F ejΩ0tu tð Þ

(cid:13)

u tð Þ (cid:13)

F e(cid:4)jΩ0tu tð Þ

Since F u tð Þ By frequency shifting property, we get

(cid:5) ¼ πδ Ωð Þ þ 1 jΩ

ð F cos Ω0t

Þu tð Þ

(cid:5) ¼

2 π

δ Ω (cid:4) Ω0 ½ ð

ð Þ þ δ Ω þ Ω0

δ Ω (cid:4) Ω0 ½ ð

ð Þ þ δ Ω þ Ω0

(cid:2)

(cid:5) þ 1 Þ 2 (cid:5)

(cid:5) þ Þ

1 j Ω (cid:4) Ω0 ð jΩ (cid:6) 2 (cid:4) Ω2

Ω0

Example 3.26 Determine the Fourier transform of (i) x tð Þ ¼ e(cid:4)bt sin Ω0t (ii) x tð Þ ¼ e(cid:4)bt cos Ω0t

Þu tð Þ b > 0 Þu tð Þ b > 0

ð ð

(cid:14)

(cid:3)

1 j Ω þ Ω0 ð

Solution (i)

(cid:13)

F e(cid:4)bt sin Ω0t

Þu tð Þ

(cid:2) ¼ F e(cid:4)btejΩ0t 2j

Let x1(t) ¼ e(cid:4)btu(t)

u tð Þ

(cid:14)

(cid:3)

(cid:2)

(cid:2) ¼ F e(cid:4)bt ejΩ0t (cid:4) ejΩ0t

(cid:3)

(cid:3) u tð Þ

(cid:2) (cid:4) F e(cid:4)btejΩ0t 2j

(cid:3)

u tð Þ

146

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

½ F x1 tð Þ

(cid:5) ¼ ¼ 1

0 e(cid:4)bte(cid:4)jΩtdt ¼ b þ jΩ b > 0:

0 e(cid:4) bþjΩ

Þtdt

By frequency shifting property

(cid:14)

(cid:13)

F ejΩ0tx1 tð Þ (cid:3)

(cid:2) F e(cid:4)btejΩ0t 2j

u tð Þ

ð ¼ X1 j Ω (cid:4) Ω0

(cid:7)

1 b þ j Ω (cid:4) Ω0 ð

¼ 1 2j

Similarly,

Hence,

(cid:2) F e(cid:4)btejΩ0t 2j

(cid:3)

u tð Þ

¼ 1 2j

(cid:7)

1 b þ j Ω þ Ω0 ð

(cid:8)

(cid:13)

F e(cid:4)bt sin Ω0t

(cid:7)

(cid:14)

Þu tð Þ

¼ 1 2j

(ii)

(cid:13)

F e(cid:4)bt cos Ω0t

Þu tð Þ

(cid:14)

1 b þ j Ω (cid:4) Ω0 ð Ω0 Þ2 þ Ω0

b þ jΩ

(cid:8)

1 b þ j Ω þ Ω0 ð

(cid:8)

(cid:7)

(cid:4) 1 2j

b > 0

(cid:2)

(cid:2) ¼ F e(cid:4)bt ejΩ0tþe(cid:4)jΩ0t

(cid:3)

u tð Þ

(cid:2) ¼ F e(cid:4)btejΩ0t 2 (cid:7)

¼ 1 2

1 b þ j Ω (cid:4) Ω0 ð b þ jΩ

b þ jΩ

Þ2 þ Ω0

(cid:3)

(cid:3) u tð Þ (cid:2)

(cid:3)

u tð Þ

þ F e(cid:4)bte(cid:4)jΩ0t 2 (cid:7) (cid:8)

þ 1 2

1 b þ j Ω þ Ω0 ð

(cid:8)

b > 0

Differentiation in Frequency If x(t) is a continuous-time signal with Fourier transform X( jΩ), then the Fourier transform of the -jtx(t) is given by

F (cid:4)jtx tð Þ

(cid:5) ¼ d

dΩ X jΩð

ð3:41Þ

Proof By the deﬁnition of the Fourier transform, it is known that

3.3 The Continuous Fourier Transform for Nonperiodic Signals

147

X jΩð

Þ ¼

ð1

(cid:4)1

x tð Þe(cid:4)jΩtdt

Differentiating this equation with respect to Ω, we have

ð1

(cid:4)1

(cid:4)jtx tð Þe(cid:4)jΩtdt

Þ ¼

d dΩ X jΩð (cid:5) ¼ d dΩ X jΩð

implying that F (cid:4)jtx tð Þ

Thus, differentiation in the frequency domain corresponds to multiplication by

(cid:4)jt in the time domain. dΩ X jΩð

F (cid:4)jtx tð Þ

(cid:5) ¼ d

Þ can also be expressed as

½ F tx tð Þ

(cid:5) ¼ j

d dΩ X jΩð

Example 3.27 Find the Fourier transform of the following continuous-time signal:

x tð Þ ¼ tn(cid:4)1 ð n (cid:4) 1

Þ! e(cid:4)btu tð Þ

Solution For n ¼ 1, x tð Þ ¼ e(cid:4)btu tð Þ,

b > 0

X jΩð

Þ ¼ 1

b þ jΩ

For n ¼ 2, x(t) ¼ te(cid:4)btu(t) By differentiation in frequency property,

(cid:13) Þ ¼ F te(cid:4)btu tð Þ

(cid:14)

X jΩð

(cid:3)

(cid:2)

¼ j

d dΩ

1 b þ jΩ

Þ(cid:4)1

d ð dΩ b þ jΩ ð Þ b þ jΩ (cid:4)1ð

Þ(cid:4)2

¼ j

¼ j 1

1 b þ jΩ

Þ2

For n ¼ 3, x tð Þ ¼ t2

2! e(cid:4)btu tð Þ

148

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

(cid:3)

(cid:2) Þ ¼ F t2

2!e(cid:4)btu tð Þ

X jΩð

”

¼ j 2

d dΩ

1 b þ jΩ

Þ2

Þ(cid:4)2

d ð dΩ b þ jΩ ð Þ b þ jΩ (cid:4)2ð

Þ(cid:4)3j

¼ j 2 ¼ j 2

1 b þ jΩ

Þ3

For n ¼ 4, x tð Þ ¼ t3

3! e(cid:4)btu tð Þ (cid:2) Þ ¼ F t3

X jΩð

(cid:3)

”

¼ j 3

d dΩ

1 b þ jΩ

Þ3

3!e(cid:4)btu tð Þ

Þ(cid:4)3

d ð dΩ b þ jΩ ð Þ b þ jΩ (cid:4)3ð

Þ(cid:4)4j

¼ j 3 ¼ j 3

1 b þ jΩ

Þ4

Thus for n, x tð Þ ¼ tn(cid:4)1 n(cid:4)1

Þ! e(cid:4)btu tð Þ

(cid:2) Þ ¼ F tn(cid:4)1 ð n (cid:4) 1

X jΩð

Þ!e(cid:4)btu tð Þ

(cid:3)

”

j n (cid:4) 1

d dΩ

1 b þ jΩ

Þn(cid:4)1

Þ(cid:4)nþ1

d ð dΩ b þ jΩ (cid:5) ð (cid:4) n (cid:4) 1

ð Þ b þ jΩ

Þ(cid:4)nþ1(cid:4)1j

j n (cid:4) 1 j n (cid:4) 1 1 b þ jΩ

Þn

Integration If x(t) is a continuous-time signal with Fourier transform X( jΩ), then x τð Þ dτ is given by

the Fourier transform of the

(cid:4)1 (cid:2) ð

(cid:3)

x τð Þ dτ

¼ 1

jΩ X jΩð

ð3:42Þ

(cid:4)1

3.3 The Continuous Fourier Transform for Nonperiodic Signals

149

Proof Letting y tð Þ ¼

(cid:4)1

x τð Þ dτ and differentiating both sides, we obtain

d dt

y tð Þ ¼ x tð Þ

Now taking the Fourier transform of both sides, it yields

(cid:2) F d dt

(cid:3)

y tð Þ

½ ¼ F x tð Þ

(cid:5) ¼ X jΩð

jΩY jΩð

Þ ¼ X jΩð

Hence

½ F y tð Þ

(cid:14)

(cid:13) Ð

(cid:5) ¼ F ¼ 1

t (cid:4)1 x τð Þ dτ Þ

jΩX jΩð

Parseval’s theorem If x(t) is a continuous-time signal with Fourier transform X ( jΩ), then the energy E of x(t) is given by

ð1

(cid:4)1

E ¼

x tð Þ

j2dt ¼ 1 2π

ð1

(cid:4)1

X jΩð

j2dΩ

ð3:43Þ

where |X( jΩ)|2 is called the energy density spectrum. Proof The energy E of x(t) is deﬁned as ð1

E ¼

(cid:4)1

x tð Þ

j2dt

ð3:44Þ

Assuming that x(t) is complex value x(t)x∗(t) ¼ |x(t)|2 and x*(t) can be expressed

in terms of its Fourier transform as

x∗ tð Þ ¼ 1 2π

ð1

(cid:4)1

X∗ jΩð

Þe(cid:4)jΩtd Ω

ð3:45Þ

Eq. (3.44) can be rewritten as

ð1

(cid:4)1

E ¼

x tð Þx∗ tð Þ dt

ð3:46Þ

Substituting Eq (3.45) in Eq. (3.46), we obtain

150

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

ð1

(cid:4)1

(cid:2) x tð Þ 1 2π

ð1

(cid:4)1

E ¼

(cid:3)

X∗ jΩð

Þe(cid:4)jΩtd Ω

ð3:47Þ

Interchanging the order of integration, Eq. (3.47) can be rewritten as

E ¼ 1 2π

ð1

(cid:4)1

X∗ jΩð

(cid:2) ð1

(cid:4)1

(cid:3)

x tð Þe(cid:4)jΩtd t

dΩ

ð3:48Þ

By deﬁnition of the Fourier transform

ð X jΩð

Þ ¼

ð1

(cid:4)1

x tð Þe(cid:4)jΩtdt

Thus,

ð1

(cid:4)1

E ¼

x tð Þ

j2dt ¼ 1 2π

ð1

(cid:4)1

X jΩð

j2dΩ

Example 3.28 Consider a signal x(t) with its Fourier transform given by 8 <

X jΩð

Þ ¼

2 Ωj 1 0

j (cid:6) 1 1 < Ωj otherwise

j (cid:6) 2

(i) Determine the energy of the signal x(t) (ii) Find x(t) Solution (i) By Parseval’s theorem, the energy E of x(t) is given by

Ð 1 (cid:4)1 x tð Þ j

E ¼

j2dt ¼ 1 2π ¼ 1 2π ¼ 8 2π ¼ 5 π

ð1

X jΩð

(cid:4)1

j2dΩ ð

4dΩ þ 1 2π

(cid:4)1 þ 1 2π

þ 1 2π

1dΩ þ 1 2π

ð(cid:4)1

(cid:4)2

1dΩ

(ii)

can be written as

8 <

X jΩð

Þ ¼

2 Ωj 1 0

j (cid:6) 1 1 < Ωj otherwise

j (cid:6) 2

3.3 The Continuous Fourier Transform for Nonperiodic Signals

151

X jΩð

Þ ¼ X1 jΩð

Þ þ X2 jΩð

(cid:4) Þ ¼ 1 Ωj 0 Ωj

j (cid:6) 1 j > 1

j (cid:6) 2 j > 2

X1 jΩð

X2 jΩð

where

which are depicted as

1 ( W)

-2

h Since F sin Ω0t

ð πt

(cid:4)

¼ 1 Ωj 0 Ωj

j (cid:6) Ω0 j > Ω0

By linearity property,

2 ( W)

Þ þ F(cid:4)1 X2 jΩð ð

x tð Þ ¼ F(cid:4)1 X1 jΩð ð ¼ sin tð Þ πt

Þ þ sin 2tð πt

The Convolution Property If x1(t) and x2(t) are two continuous-time signals with Fourier transforms X1( jΩ) and X2( jΩ), then the Fourier transform of the convolution integral of x1(t) and x2(t) is given by

F x1 tð Þ∗x2 tð Þ

(cid:5) ¼ X1 jΩð

ÞX2 jΩð

ð3:49Þ

Hence, convolution of two sequences x1(t) and x2(t) in the time domain is equal to

the product of their frequency spectra. Proof By the deﬁnition of convolution integral,

y tð Þ ¼ x1 tð Þ∗x2 tð Þ ¼

ð1

(cid:4)1

ð x1 τð Þx2 t (cid:4) τ

Þ dτ

ð3:50Þ

Taking the Fourier transform of Eq. (3.50), we obtain

152

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Y jΩð

½ Þ ¼ F y tð Þ

(cid:5) ¼

ð1

(cid:4)1

½ x1 τð Þx2 t (cid:4) τ ð

Þ dτ

(cid:5)e(cid:4)jΩt dt

ð3:51Þ

Interchanging the order of integration, Eq. (3.51) can be rewritten as

Y jΩð

Þ ¼

By the shifting property,

ð1

(cid:4)1 ð1

(cid:4)1

ð1

(cid:13)

(cid:4)1

x1 τð Þ

ð x2 t (cid:4) τ

Þe(cid:4)jΩtdt

(cid:14)

dτ

(cid:13)

ð x2 t (cid:4) τ

Þe(cid:4)jΩtdt

(cid:14)

¼ e(cid:4)jΩτX2 jΩð

Þ.

Hence, Eq. (3.52) becomes

ð3:52Þ

YðjΩÞ ¼

ð1

(cid:4)1

x1ðτÞe(cid:4)jΩtX2ðjΩÞdτ ¼ X2ðjΩÞ

ð1

(cid:4)1

x1ðτÞe(cid:4)jΩτdτ

ð3:53Þ

By X1 jΩð

Þ ¼

the ð1

(cid:4)1

Thus

deﬁnition x1 τð Þe(cid:4)jΩτdτ

continuous-time

Fourier

transform,

Y jΩð

Þ ¼ X1 jΩð

ÞX2 jΩð

Example 3.29 Determine the Fourier transform of the triangular output signal y(t) of an LTI system as shown in Figure 3.13.

Solution A triangular signal can be represented as the convolution of two rectan- gular pulse signals x1(t) and x2(t) deﬁned by (cid:4) x1 tð Þ ¼ x2 tð Þ ¼ 1 0

tj j < 1 tj j > 1

y tð Þ ¼ x1 tð Þ∗x2 tð Þ

By deﬁnition,

Figure 3.13 Triangular output signal

y(t)

-2

3.3 The Continuous Fourier Transform for Nonperiodic Signals

153

X1 jΩð

Þ ¼

Ð 1 (cid:4)1 x1 tð Þe(cid:4)jΩtdt ¼ (cid:5) jΩ ejΩ (cid:4) e(cid:4)jΩ

(cid:6)

¼ 2

¼ 1

(cid:4)1 e(cid:4)jΩtdt sin Ω Ω

By the convolution property, Y( jΩ) the Fourier transform of y(t) is given by

Y jΩð

Þ ¼ X1 jΩð

ÞX2 jΩð

sin Ωð Þ Ω

¼ 2

¼ 4

(cid:8) 2

sin Ωð Þ Ω sin 2 Ωð Þ Ω2

Example 3.30 Consider an LTI continuous-time system with the impulse response h tð Þ ¼ sin Ω0t

. Find the output y(t) of the system for an input

ð t

Solution Since F sin ðΩ0tÞ

πt

x tð Þ ¼ sin 2Ω0t

(cid:4)

1 jΩj (cid:6) Ω0 jΩj > Ω0 0

(cid:2) (cid:5) ¼ F sin Ω0t

(cid:3)

H jΩð

Þ ¼ F h tð Þ ½ (cid:4)

Þ ¼ F x tð Þ ½ (cid:4)

π Ωj 0 Ωj

j (cid:6) Ω0 j > Ω0 (cid:2) (cid:5) ¼ F sin 2Ω0t

π Ωj 0 Ωj

j (cid:6) 2Ω0 j > 2Ω0

(cid:3) Þ

W )

(

shown as

X jΩð

(

) W

-2W0

2W 0

154

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

The output y(t) of the system is given by

y tð Þ ¼ x tð Þ∗h tð Þ

By convolution property, we have

Y jΩð

Þ ¼ F y tð Þ ½ (

ÞH jΩð

(cid:5) ¼ X jΩð j (cid:6) Ω0 j > Ω0

π2 Ωj 0 Ωj

which is depicted as

( W)

-W0

Therefore,

y tð Þ ¼ F(cid:4)1 Y jΩð Þ

Duality property For a given Fourier transform pair

Þ ¼ π sin Ω0t

F x tð Þ $

X jΩð

By interchanging the roles of time and frequency, a new Fourier transform pair is

obtained as

F X jtð Þ $

2πx (cid:4)Ωð

For example, the duality exists between the Fourier transform pairs of Examples

3.19 and 3.20 as given by

(cid:4)

x tð Þ ¼ 1 0

tj j (cid:6) T 1 tj j > T 1

F $

X jΩð

Þ ¼ 2T 1 sin c

(cid:8)

(cid:7)

ΩT 1 π

3.3 The Continuous Fourier Transform for Nonperiodic Signals

155

x tð Þ ¼

Ω π sin c

(cid:7) (cid:8) Ωt π

F $

X jΩð

(cid:4) Þ ¼ 1 0

tj j (cid:6) Ω tj j > Ω

The Modulation Property Due to duality between the time domain and frequency domain, the multiplication in the time domain corresponds to convolution in the frequency domain.

If x1(t) and x2(t) are two continuous-time signals with Fourier transforms X1( jΩ) and X2( jΩ), then the Fourier transform of the product of x1(t) and x2(t) is given by

(cid:5) ¼ 1

Þ∗X2 jΩð

F x1 tð Þx2 tð Þ

½ 2π X1 jΩð This can be easily proved by dual property. Eq. (3.54) is called the modulation property since the multiplication of two signals often implies amplitude modulation. Example 3.31 Find the Fourier transform of ejΩ0tx tð Þ Solution Let x1 tð Þ ¼ ejΩ0t and x2(t) ¼ x(t)

ð3:54Þ

(cid:5)

X1ðjΩÞ ¼ F½ejΩ0t(cid:5) ¼ 2πδðΩ (cid:4) Ω0Þ X2ðjΩÞ ¼ F½xðtÞ(cid:5) ¼ XðjΩÞ F½x1ðtÞx2ðtÞ(cid:5) ¼ F½ejΩ0txðtÞ(cid:5) ¼ 1 2π

½2πδðΩ (cid:4) Ω0Þ∗XðjΩÞ(cid:5)

¼ XðjðΩ (cid:4) Ω0ÞÞ

Example 3.32 Let y(t) be the convolution of two signals x1(t) and x2(t) deﬁned by

x1 tð Þ ¼ sin c 2tð x2 tð Þ ¼ sin c tð Þ cos 3πt

Determine the Fourier transform of y(t).

Solution By dual property, the Fourier transform of sinc(t) is given by

F sinc tð Þ

(cid:5) ¼ rect

(cid:7) (cid:8) Ω 2π

The Fourier transform of x1(t) is given by

F x1 tð Þ ½

(cid:5) ¼ X1 jΩð

Þ (cid:5) (cid:8)

(cid:7)

rect

Þ ¼ F sin c 2tð ½ Ω=2 ¼ 1 2π 2 (cid:7) (cid:8) Ω ¼ 1 4π 2

rect

156

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

The Fourier transform of x2(t) is given by

F x2 tð Þ ½

(cid:5) ¼ X2 jΩð

By modulation property

½ (cid:2)

Þ ¼ F sin c tð Þ cos 3πt (cid:7) ¼ F sin c tð Þ ej3πt þ e(cid:4)j3πt

(cid:5)

(cid:8)

(cid:3)

X2 jΩð

(cid:2)

(cid:7) Þ ¼ F sin c tð Þ ej3πt þ e(cid:4)j3πt

(cid:8)

(cid:3)

(cid:7) ð

(cid:2)

rect

(cid:8)

Ω (cid:4) 3π 2π

¼ 1 2

þ rect

(cid:3)

(cid:8) Þ

(cid:7) ð

Ω þ 3π 2π

By convolution property (cid:13) y tð Þ ¼ F x1 tð Þ∗x2 tð Þ ½

¼ 1 4

rect

(cid:2)

(cid:7) (cid:8) Ω 4π

rect

Ω (cid:4) 3π 2π

(cid:5) ¼ X1 jΩð (cid:7) ð

ÞX2 jΩð (cid:8) Þ

(cid:5) (cid:7)

þ rect

(cid:8)

(cid:3)

Ω þ 3π 2π

1 ( W)=1

rect W 4

-2p

2( W

) =

1 2

rect

(W − 3 ) 2

rect

(W + 3 ) 2

-2p

There is no overlap between the two transforms X1( jΩ) and X2( jΩ), and hence

X1 jΩð

ÞX2 jΩð

Þ ¼ 0

Therefore,

YðjΩÞ ¼ F½x1ðtÞ∗x2ðtÞ(cid:5) ¼ X1ðjΩÞX2ðjΩÞ

¼ 0

3.3 The Continuous Fourier Transform for Nonperiodic Signals

157

Example 3.33 Determine the value of

ð1

(cid:4)1

sin c2 2tð

Þ dt

Solution Since the Fourier transform of sinc (2t) is 1 theorem,

(cid:5) (cid:6) 2 rect Ω 4π

, using Parseval’s

(cid:7) (cid:8) Ω 4π

dΩ

rect2

(cid:7) (cid:8) 2 1 2

1dΩ

ð1

(cid:4)1 ð

2π

(cid:4)2π

ð1

(cid:4)1

sin c2 2tð

Þdt ¼ 1 2π

¼ 1 8π ¼ 4π 8π ¼ 1 2

Example 3.34 Consider a signal x(t) with its Fourier transform given by

(cid:4)

X jΩð

Þ ¼

π Ωj 0 Ωj

j (cid:6) Ω0 j > Ω0

Find the Fourier transform the system output y(t) given by

y tð Þ ¼ x tð Þ cos Ωct

Þ where Ωc > Ω0

Solution By the deﬁnition of the inverse Fourier transform, we have

πejΩtdΩ

ejΩ0t (cid:4) e(cid:4)jΩ0t

(cid:6)

ðΩ0

¼ 1 2

x tð Þ ¼ 1 2π (cid:4)Ω0 (cid:7) (cid:8) (cid:5) 1 jt ð ¼ sin Ω0t t Þ ¼ sin Ω0t

ð Hence, y tð Þ ¼ x tð Þ cos Ωct t By modulation property, we obtain

ð cos Ωct

YðjΩÞ ¼ XðjΩÞ∗F½cos ðΩctÞ(cid:5)

¼ XðjΩÞ∗½πδðΩ (cid:4) ΩcÞ þ πδðΩ þ ΩcÞ(cid:5)

158

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

−W − W0 −W −W + W0

W − W 0

W + W0

The Fourier transform properties of continuous-time signals are summarized in

Table 3.2 and 3.3.

Duality : for given Fourier transform pair x tð Þ $F X jΩð

Parseval’s theorem

X jtð Þ $F 2π x (cid:4)Ωð Ð 1 (cid:4)1 x tð Þ j

Þ j2dt ¼ 1 2π

ð1

(cid:4)1

X jΩð j

j2dΩ

Table 3.2 Some properties of continuous-time Fourier transforms

Property Linearity Time shifting Symmetry

Aperiodic signal a1x1(t)+a2x2(t) x(t (cid:4) t0) x∗ðtÞ xðtÞ real xe(t) (x(t) real) xo(t) (x(t) real)

Time reversal Frequency shifting Time and frequency scaling

Differentiation in time

Differentiation in frequency

Integration

Convolution property Modulation property

x((cid:4)n) ejΩ0tx tð Þ x(at) dt x tð Þ (cid:4)jtx(t) ð t

x τð Þ dτ

(cid:4)1 x1(t) ∗ x2(t)

Fourier transform a1X1( jΩ)+a2X2( jΩ) e(cid:4)jΩt0 X jΩð Þ X∗ (cid:4)jΩð 8

Þ X∗ jΩð Þ ¼ X (cid:4)jΩð (cid:5) (cid:5) ¼ Re X (cid:4)jΩð Þ Re X jΩð ½ Þ ½ (cid:13) X (cid:4)jΩð (cid:5) ¼ (cid:4)Im Im X jΩð Þ Þ ½ Þ X jΩð Þ j ¼ X (cid:4)jΩð j j j ½ (cid:5) ¼ (cid:4)arg X (cid:4)jΩð Þ ½ arg X jΩð Þ

(cid:5)

< : Re[X( jΩ) jIm[X( jΩ)] X(e(cid:4)jω) X( j(Ω(cid:4)Ω0)) (cid:5) (cid:5) (cid:6) aj j X j Ω 1 a jΩX( jΩ) dΩ X jΩð Þ d Þ jΩ X jΩð

(cid:6)

Xl( jΩ)X2( jΩ)

x1(t) x2(t)

2π ½X1ðjΩÞ∗X2ðjΩÞ(cid:5)

3.4 The Frequency Response of Continuous-Time Systems

159

Table 3.3 Basic Fourier transform pairs

b > 0

Signal δ(t) e(cid:4)btu tð Þ 1 ejΩ0t X1

k¼(cid:4)1 akejkΩ0t

ð δ t (cid:4) kT

k¼(cid:4)1 sin(Ω0t) cos(Ω0t) (cid:4) x tð Þ ¼ 1 0

and x(t) ¼ x(t+T0) (cid:5) (cid:6) π sin c Ωt Ω π

δ(t(cid:4)t0) u(t)

Sgn(t)

te(cid:4)btu tð Þ

b > 0

Þ! e(cid:4)btu tð Þ

tn(cid:4)1 n(cid:4)1 ð (cid:4)t2 e2σ2

tj j < T 1 tj j > T 1 Periodic square wave with period T0 x tð Þ ¼

tj j < T 1

(cid:4)

0 T 1 tj j (cid:6) T 0=2

Fourier transform 1

1 bþjΩ 2πδ(Ω) 2πδ(Ω(cid:4)Ω0) X1

2π

akδ Ω (cid:4) kΩ0

k¼(cid:4)1 X1

k¼(cid:4)1

ð δ Ω (cid:4) kΩ0

2π T π[δ(Ω(cid:4)Ω0)(cid:4)δ(Ω(cid:4)Ω0)] π[δ(Ω(cid:4)Ω0)+δ(Ω(cid:4)Ω0)]/j (cid:6) 2T 1 sin c ΩT 1

(cid:5)

k¼(cid:4)1

2 sin kΩ0T 1 ð k

δ Ω (cid:4) kΩ0 ð

(cid:4) Þ ¼ 1 Ωj 0 Ωj

j (cid:6) Ω j > Ω

X jΩð

e(cid:4)jΩt0 πδ Ωð Þ þ 1 jΩ

2 jΩ 1 bþjΩ 1 Þn bþjΩ p ﬃﬃﬃﬃﬃ 2π

Þ2

Ω 6¼ 0

e (cid:4)σ2Ω2 2

3.4 The Frequency Response of Continuous-Time Systems

As in chapter 2, the input output relation of a useful class of continuous-time LTI systems satisﬁes the linear constant coefﬁcient differential equation

N n¼0

dnyðtÞ dtn

M n¼0

dnxðtÞ dtn

ð3:55Þ

where coefﬁcients an and bn are real constants. From convolution property, it is known that

Y jΩð

Þ ¼ H jΩð

ÞX jΩð

ð3:56Þ

which can be rewritten as

160

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

H jΩð

Þ Þ ¼ Y jΩð Þ X jΩð

Applying Fourier transform to both sides of Eq. (3.55), we obtain

(cid:2)

N n¼0

dny tð Þ dtn

(cid:3)

(cid:2)

¼ F

(cid:3)

M n¼0

dnx tð Þ dtn

By linearity property, Eq. (3.58) becomes

N n¼0

(cid:3)

(cid:2) anF dny tð Þ dtn

M n¼0

(cid:3)

(cid:2) bnF dnx tð Þ dtn

From the differentiation property, Eq. (3.59) can be rewritten as

ð3:57Þ

ð3:58Þ

ð3:59Þ

N n¼0

an jΩð

ÞnY jΩð

Þ ¼

M n¼0

bn jΩð

ÞnX jΩð

ð3:60Þ

which can be rewritten as h

Y jΩð

an jΩð

Þn

¼ X jΩð

M n¼0

bn jΩð

Þn

ð3:61Þ

N n¼0

Thus, the frequency response of a continuous-time LTI system is given by

H jΩð

Þ ¼ Y jΩð X jΩð

Þ Þ

n¼0 bn jΩð k¼0 an jΩð

Þn Þn

ð3:62Þ

The function H(jΩ) is a rational function being a ratio of polynomials in (jΩ).

3.4.1 Distortion During Transmission

Eq. (3.56) implies that the transmission of an input signal x(t) through the system is changed into an output signal y(t). The X( jΩ) and Y( jΩ) are the spectra of the input and output signals, and H( jΩ) is the frequency response of the system.

During the transmission, the input signal amplitude spectrum |X( jΩ)| is changed to|X( jΩ)||H(jΩ)|. Similarly, the input signal phase spectrum ∠X(jΩ) is changed to ∠X( jΩ) þ ∠ H(jΩ). An input signal spectral component of frequency Ω is modiﬁed in amplitude by a |H(jΩ)| factor and is shifted in phase by an angle ∠H(jΩ).

Thus, the output waveform will be different from the input waveform during

transmission through the system introducing distortion.

3.4 The Frequency Response of Continuous-Time Systems

161

Example 3.35 Consider an LTI system described by the following differential equation

d2y tð Þ dt2

þ 3

dy tð Þ dt

þ 2y tð Þ ¼ 4

dx tð Þ dt

(cid:4) x tð Þ

Find the Fourier transform of the impulse response of the system.

Solution Apply the Fourier transform on both sides of the differential equation, then we obtain

(cid:3)

þ 2y tð Þ

þ 3

(cid:2) F d2y tð Þ dy tð Þ dt2 dt (cid:2) (cid:3) þ 3F dy tð Þ dt

(cid:3)

(cid:2) F d2y tð Þ dt2

(cid:3) (cid:4) x tð Þ

(cid:2) ¼ F 4

dx tð Þ dt (cid:3) (cid:2) (cid:5) ¼ 4F dx tð Þ dt

½ þ 2F y tð Þ

½ (cid:4) F x tð Þ

(cid:5)

jΩð

Þ þ 3jΩY jΩð

Þ2Y jΩð (cid:13) (cid:4)Ω2 þ j3Ω þ 2

Þ þ 2Y jΩð (cid:14) Y jΩð

Þ ¼ j4Ω (cid:4) 1

(cid:5)X jΩð

Þ ¼ 4jΩX jΩð

Þ (cid:4) X jΩð

The Fourier transform of the impulse response H(jΩ) is given by

H jΩð

Þ ¼ Y jΩð X jΩð

Þ Þ

¼ 1 (cid:4) j4Ω

Ω2 (cid:4) j3Ω (cid:4) 2

Example 3.36 Find the frequency response H( jΩ) of the following circuit

Solution

i tð Þ ¼ C

(cid:4)vi tð Þ þ L

dv0 tð Þ dt di tð Þ dt

þ v0 tð Þ R

þ v0 tð Þ ¼ 0

¼ vi tð Þ (cid:4) v0 tð Þ

d2v0 tð Þ dt2

dv0 tð Þ dt

þ v0 tð Þ ¼ vi tð Þ

di tð Þ dt þ L R

162

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Taking Fourier transform both sides of this equation, we obtain

Þ2v0 jΩð

Þ þ jΩL v0 jΩð LC (cid:4)jΩð R (cid:8) (cid:7) 1 (cid:4) LCΩ2 þ jΩL R

Þ þ v0 jΩð

Þ ¼ vi jΩð

v0 jΩð

Þ ¼ vi jΩð

H jΩð

Þ Þ ¼ v0 jΩð Þ vi jΩð

1 1 (cid:4) LCΩ2 þ jΩL R

3.5 Some Communication Application Examples

3.5.1 Amplitude Modulation (AM) and Demodulation

Amplitude Modulation

In amplitude modulation, the amplitude of the carrier signal c(t) is varied in some manner with the baseband signal (message signal) m(t) also known as the modulat- ing signal.

The AM signal is given by

m(t)

s(t)

c(t)

s tð Þ ¼ m tð Þ:c tð Þ c tð Þ ¼ cos Ωct þ θc ð

For convenience, if it is assumed that θc ¼ 0

c tð Þ ¼ cos Ωct CðjΩÞ ¼ F½cðtÞ(cid:5) ¼ π½δðΩ (cid:4) ΩcÞ þ δðΩ þ ΩcÞ(cid:5) (cid:5) ¼ 1

½ Þ ¼ F m tð Þc tð Þ

Þ∗C jΩð

S jΩð

(cid:5)

M jΩð

Þ∗δ Ω (cid:4) Ωc

2π M jΩð ½ ð ð Þ ¼ M j Ω (cid:4) Ωc

S jΩð

Þ ¼ 1 2

M j Ω (cid:4) Ωc ½

ð ð Þ þ M j Ω þ Ωc Þ

(cid:5)

ð3:63Þ

ð3:63aÞ

ð3:64Þ

ð3:65Þ

ð3:66Þ

ð3:67Þ

ð3:68Þ

Eq. (3.61) implies that the AM shifts the message signal so that it is centered at (cid:2)Ωc. The message signal m(t) can be recovered if Ωc >Ωm so that the replica spectra

3.5 Some Communication Application Examples

163

− Ω

0.5

− Ω − Ω

− Ω

− Ω Ω

Ω − Ω

Ω Ω

Figure 3.14 Amplitude modulation

s(t)

w(t)

Low pass filter

( Ω)

(t)ˆ

c(t)

Figure 3.15 Amplitude demodulation

do not overlap. The AM modulation in frequency domain is illustrated in Figure 3.14.

Amplitude Demodulation The message signal m(t) can be extracted by multiplying the AM signal s(t) by the same carrier c(t) and passing the resulting signal through a low-pass ﬁlter as shown in Figure 3.15.

164

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

w tð Þ ¼ s tð Þc tð Þ

¼ m tð Þc2 tð Þ ¼ m tð Þ cos 2 Ωct ð (cid:2) ¼ m tð Þ 1 þ cos 2Ωct

Þ ð

(cid:3)

ð3:69Þ

The amplitude demodulation in frequency domain is illustrated in Figure 3.16.

3.5.2 Single-Sideband (SSB) AM

The double-sideband modulation is used in Section 3.5.1. By removing the upper sideband by using a low-pass ﬁlter with cutoff frequency Ωc or a lower sideband by high-pass ﬁlter with cutoff frequency Ωc, single-sideband modulation that requires half the bandwidth can be used. The frequency domain representation of single- sideband modulation is shown in Figure 3.17. However, the single-sideband mod- ulation requires nearly ideal ﬁlters and increases the transmitter cost.

3.5.3 Frequency Division Multiplexing (FDM)

In frequency division multiplexing, multiple signals are transmitted over a single wideband channel using a single transmitting antenna. Different carriers with ade- quate separation are used to modulate for each of these signals with no overlap between the spectra of the modulated signals. The different modulated signals are summed before sending to the antenna. At the receiver, to recover a speciﬁc signal, the corresponding frequency is extracted through a band-pass ﬁlter. The FDM spectra for three modulated signals are shown in Figure 3.18.

3.6 Problems

Find the exponential Fourier series representation for each of the following

signals: (i) x(t) ¼ cos(Ω0t) (ii) x(t) ¼ sin(Ω0t) (iii) x tð Þ ¼ cos 2t þ π 6 (iv) x(t) ¼ sin2(t) (v) x(t) ¼ cos(6t) þ sin (4t) ½ (vi) x tð Þ ¼ 1 þ cos 2πt

(cid:5)

(cid:6)

(cid:5) (cid:5) sin 5πt þ π 4

(cid:6)

3.6 Problems

165

0.5

− Ω − Ω

− Ω

− Ω Ω

Ω − Ω

Ω Ω

− Ω

0.5

− Ω

Figure 3.16 Illustration of amplitude demodulation in frequency domain

166

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

0.5

− Ω

− Ω Ω

Ω Ω

Figure 3.17 Single-sideband AM

0.5

Figure 3.18 Illustration of FDM for three signals

What signal will have the following Fourier series coefﬁcients

an ¼ 1 4

sin 2 nπ=2 ð Þ2 ð nπ=2

If x1 (t) and x2(t) are periodic signals with fundamental period T0, ﬁnd the Fourier

series representation of x(t) ¼ x1(t)x2(t). 4. Consider the periodic signal x(t) given by

ð x tð Þ ¼ 2 þ j2

Þe(cid:4)j3t (cid:4) j3e(cid:4)j2t þ 6 þ j3ej2t þ 2 (cid:4) j2

Þej3t

Determine the trigonometric Fourier series representation of the signal x(t). 5. Find the Fourier series of a periodic signal x(t) with period 3 deﬁned over one

period by

(cid:4)

x tð Þ ¼ t þ 2 (cid:4)2 (cid:6) t (cid:6) 0 0 (cid:6) t (cid:6) 1 2 (cid:4) 2t

Find the Fourier series of a periodic signal x(t) with period 6 deﬁned over one

period by

3.6 Problems

167

< :

x tð Þ ¼

(cid:4)3 (cid:6) t (cid:6) (cid:4)2 t þ 2 (cid:4)2 (cid:6) t (cid:6) (cid:4)1 (cid:4)1 (cid:6) t (cid:6) 1 1 (cid:6) t (cid:6) 2 2 (cid:6) t (cid:6) 3

1 (cid:4)t þ 2 0

Find Fourier series of the periodic signal shown in Figure P3.1.

Figure P3.1 Periodic signal of problem 7

Determine the exponential Fourier series representation of the periodic signal

depicted in Figure P3.2

Figure P3.2 Periodic signal of problem 8

Determine trigonometric Fourier series representation of the signal shown in

Figure P3.3

Figure P3.3 Periodic signal of problem 9

Plot

the magnitude and phase spectrum of the periodic signal shown in

Figure P3.4

168

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure P3.4 Periodic signal of problem 10

Find the Fourier transform of x tð Þ ¼ eatu (cid:4)t

a > 0

Find the Fourier transform of xðtÞ ¼

(cid:4)

ð Þ t jtj (cid:6) 1 jtj > 1 0

Find the Fourier transform of rectangular pulse given by

(cid:4)

x tð Þ ¼ 1 0

tj j (cid:6) T tj j > T

Find the Fourier transform of xðtÞ ¼ 4
Find the Fourier transform of the complex sinusoidal pulse given by

π2t2 sin 2ð2tÞ

(cid:4) x tð Þ ¼ ej5t 0

tj j (cid:6) π otherwise

Find Fourier transform of the following signal shown in Figure P3.5

Figure P3.5 Signal x(t)

Consider the following two signals x(t) and y(t) as shown in Figure P3.6. Determine Fourier transform y(t) using the Fourier transform of x(t), time shifting property, and differentiation property

3.6 Problems

169

Figure P3.6 Signals x(t) and y(t)

Find the inverse Fourier transform of

(cid:4)

XðjΩÞ ¼ 2cos ðΩÞ jΩj (cid:6) π 0

jΩj > π

Find the inverse Fourier transform of

X jΩð

Þ ¼

(cid:4)jΩ

jΩð

Þ2 þ 3jΩ þ 2

Consider the following communication system shown in Figure P3.7 to transmit

two signals simultaneously over the same channel.

Figure P3.7 Communication system

Plot the spectra of x(t), y(t), and z(t) for given the following spectra of the two

input signal shown in Figure P3.8.

170

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure P3.8 Spectra of x1(t) and x2(t)

Determine y(t) of an LTI system with input

x tð Þ ¼ anejnΩ0t and the following H( jΩ) depicted in Figure P3.9 where H( jΩ) is H

( jnΩ0) evaluated at frequency nΩ0.

Figure P3.9 H( jΩ) of LTI system of problem 21

Sketch amplitude single-sideband modulation and demodulation if the message

signal m(t) ¼ cos(Ωmt).

Further Reading

Lanczos, C.: Discourse on Fourier Series. Oliver Boyd, London (1966)
Körner, T.W.: Fourier Analysis. Cambridge University Press, Cambridge (1989)
Walker, P.L.: The Theory of Fourier Series and Integrals. Wiley, New York (1986)
Churchill, R.V., Brown, J.W.: Fourier Series and Boundary Value Problems, 3rd edn. McGraw-

Hill, New York (1978)

Papoulis, A.: The Fourier Integral and Its Applications. McGraw-Hill, New York (1962)
Bracewell, R.N.: Fourier Transform and Its Applications, rev, 2nd edn. McGraw-Hill, New York

(1986)

Morrison, N.: Introduction to Fourier Analysis. Wiley, New York (1994)
Lathi, B.P.: Linear Systems and Signals, 2nd edn. Oxford University Press, New York (2005)
Oppenheim, A.V., Willsky, A.S.: Signals and Systems. Englewood Cliffs, NJ, Prentice- Hall

(1983)

Chapter 4 Laplace Transforms

The Laplace transform is a generalization of the Fourier transform of a continuous time signal. The Laplace transform converges for signals for which the Fourier transform does not. Hence, the Laplace transform is a useful tool in the analysis and design of continuous time systems. This chapter introduces the bilateral Laplace transform, the unilateral Laplace transform, the inverse Laplace transform, and properties of the Laplace transform. Also, in this chapter, the LTI systems, including the systems represented by the linear constant coefﬁcient differential equations, are characterized and analyzed using the Laplace transform. Further, the solution of state-space equations of continuous time LTI systems using Laplace transform is discussed.

4.1 The Laplace Transform

4.1.1 Deﬁnition of Laplace Transform

The Laplace transform of a signal x(t) is deﬁned as

X sð Þ ¼ L x tð Þ

g ¼

(cid:2)1

x tð Þe(cid:2)stdt

ð4:1Þ

The complex variable s is of the forms ¼ σ + jΩ, with a real part σ and an imaginary part Ω. The Laplace transform deﬁned by Eq. (4.1) is called as the bilateral Laplace transform.

171

172

4 Laplace Transforms

4.1.2 The Unilateral Laplace Transform

The unilateral Laplace transform plays an important role in the analysis of causal systems described by constant coefﬁcient linear differential equations with initial conditions.

The unilateral Laplace transform is mathematically deﬁned as

X sð Þ ¼ L x tð Þ

g ¼

0þ

x tð Þe(cid:2)stdt

ð4:2Þ

The difference between Eqs. (4.1) and (4.2) is on the lower limit of the integra- tion. It indicates that the bilateral Laplace transform depends on the entire signal, whereas the unilateral Laplace transform depends on the right-sided signal, i.e., x(t) ¼ 0 for t < 0.

4.1.3 Existence of Laplace Transforms

The Laplace transform is said to exist if the magnitude of the transform is ﬁnite, that is, |X(s)| < 1. Piecewise continuous A function x(t) is piecewise continuous on a ﬁnite interval a (cid:3) t (cid:3) b, if x is continuous on [a,b], except possibly at ﬁnitely many points at each of which x has a ﬁnite left and right limit. Sufﬁcient Condition The sufﬁcient condition for existence of Laplace transforms is that if x(t) is piecewise continuous on (0, 1) and there exist some constants k and M such that |x(t)| (cid:3) Mekt, then X(s) exists for s > k.

Proof As x(t) is piecewise continuous on (0, 1), x(t)e(cid:2)st is integrable on (0, 1).

L x tð Þ g f

j ¼

(cid:2) (cid:2) (cid:2) (cid:2)

(cid:2) (cid:2) (cid:2) (cid:2) (cid:3) (cid:4)

(cid:2) (cid:2) Þt 1 0

x tð Þe(cid:2)stdt (cid:3)

e(cid:2) s(cid:2)k ð

0 M k (cid:2) s

x tð Þ j

je(cid:2)stdt (cid:3)

Mekte(cid:2)stdt

M k (cid:2) s

0 (cid:2) 1

Þ ¼

M s (cid:2) k

ð4:3Þ

For s > k, |ℒ{x(t)}| < 1 .

4.1.4 Relationship Between Laplace Transform and Fourier

Transform

When the complex variable s is purely imaginary, i.e., s ¼ jΩ, Eq. (4.1) becomes

4.1 The Laplace Transform

X jΩð

Þ ¼

(cid:2)1

x tð Þe(cid:2)jΩtdt

Eq. (4.4) is the Fourier transform of x(t), that is,

(cid:2) (cid:2) X sð Þ s¼jΩ

¼ F x tð Þ f

If s is not purely imaginary, Eq. (4.1) can be written as

X σ þ jΩ

Þ ¼

Eq. (4.6) can be rewritten as

X σ þ jΩ

Þ ¼

(cid:2)1

x tð Þ e(cid:2) σþjΩ ð

Þtdt

x tð Þe(cid:2)σte(cid:2)jΩtdt

173

ð4:4Þ

ð4:5Þ

ð4:6Þ

ð4:7Þ

(cid:2)1 The right hand side of Eq. (4.7) is the Fourier transform of x(t)e(cid:2)σt. Thus, the Laplace transform can be interpreted as the Fourier transform of x(t) after multipli- cation by a real exponential signal.

4.1.5 Representation of Laplace Transform in the S-Plane

The Laplace transform is a ratio of polynomials in the complex variable, which can be represented by

X sð Þ ¼

N sð Þ D sð Þ

ð4:8Þ

where N(s) is the numerator polynomial and D(s) represents the denominator polynomial. The Eq. (4.8) is referred to as rational. The roots of the numerator polynomial are referred to as zeros of X(s) ¼ 0 because for those values of s, X(s) becomes zero. The roots of the denominator polynomial are called the poles of X(s), as for those values of s, X(s) ¼ 1. A rational Laplace transform can be speciﬁed by marking the locations of poles and zeros by x and o in the s-plane, which is called as pole-zero plot of the Laplace transform. For a signal, the Laplace transform con- verges for a range of values of s. This range is referred to as the region of convergence (ROC), which is indicated as shaded region in the pole-zero plot.

174

4 Laplace Transforms

4.2 Properties of the Region of Convergence

Property 1 ROC of X(s) consists of strips parallel to the jΩ axis. The ROC of X(s) contains the values of s ¼ σ + jΩ for which the Fourier transform of x(t)e(cid:2)σt converges. Thus, the ROC of X(s) is on the real part of s not on the frequency Ω. Hence, ROC of X(s) contains strips parallel to the jΩ axis

Property 2 ROC of a rational Laplace transform should not contain poles.

In the ROC, X(s) should be ﬁnite for all s since X(s) is inﬁnite at a pole and

Eq. (4.1) does not converge at a pole. Hence, the ROC should not contain poles Property 3 ROC is the entire s-plane for a ﬁnite duration x(t), if there is at

least one value of s for which the Laplace transform converges.

Proof A ﬁnite duration signal is zero outside a ﬁnite interval as shown in Figure 4.1. Let us assume that x(t)e(cid:2)σt is absolutely integrable for some value of σ ¼ σ1 such that

x tð Þ

je(cid:2)σ1t < 1

ð4:9Þ

Then, the line ℜe(s) ¼ σ1 is in the ROC. For ℜe(s) ¼ σ2 also to be in the ROC, it

is required that

x tð Þ

je(cid:2)σ2t ¼

x tð Þ

je(cid:2)σ1te(cid:2) σ2(cid:2)σ1 ð

Þt < 1

ð4:10Þ

If σ2 > σ1 such that e(cid:2) σ2(cid:2)σ1

Þt is decaying, then the maximum value of e(cid:2) σ2(cid:2)σ1

Þt

becomes e(cid:2) σ2(cid:2)σ1

Þt1 for nonzero x(t) over the interval.

Hence,

x tð Þ

je(cid:2)σ2t < e(cid:2) σ2(cid:2)σ1

Þt1

x tð Þ

je(cid:2)σ1t

ð4:11Þ

The RHS of Eq. (4.11) is bounded and hence the LHS. Thus, the ℜe(s) > σ1 must also be in the ROC. Similarly, if σ2 < σ1, it can be shown that x tð Þe(cid:2)σ2t is absolutely integrable. Hence, the ROC is the entire s-plane.

Figure 4.1 Finite duration signal

4.2 Properties of the Region of Convergence

175

Property 4:

If ROC of a right-sided signal contains the line ℜe(s) ¼ σ1, then ℜe(s) > σ1will also be in the ROC for all values of s. Proof For a right-sided signal, x(t) ¼ 0 prior to some ﬁnite time t1 as shown in Figure 4.2

If the Laplace transform converges for some value of σ ¼ σ1, then

If x(t) is right sided, then

(cid:2)1

x tð Þ

je(cid:2)σ1t < 1

x tð Þ

je(cid:2)σ1t < 1

ð4:12Þ

ð4:13Þ

For σ2 > σ1, x tð Þ e(cid:2)σ2t is absolutely integrable as e(cid:2)σ2t decays faster than e(cid:2)σ1t as

t ! 1. Thus, ℜe(s) > σ1 will also be in the ROC for all values of s.

Property 5:

If ROC of left-sided signal contains the line ℜe(s) ¼ σ1, then ℜe(s) < σ1 will also be in the ROC for all values of s.

Proof For left-sided signal, x(t) ¼ 0 after some ﬁnite time t2 as shown in Figure 4.3. This can be proved easily with the same argument and intuition for the property 4.

Figure 4.2 Right-sided signal

x(t)

Figure 4.3 Left-sided signal

x(t)

176

4 Laplace Transforms

Property 6:

If ROC of a two-sided signal contains the line ℜe(s) ¼ σ0, then ROC will contain a strip, which includes the line. Proof A two-sided signal is of inﬁnite duration for both t > 0 and t < 0 as shown in Figure 4.4(a)

Let us choose an arbitrary time t0 that divides the signal into as sum of right-sided signal and left-sided signal as shown Figure 4.4(b) and (c). The Laplace transform of x(t) converges for the values of s for which both the right-handed signal and left- handed signal converge. It is known from property 4 that the ROC of Laplace transform of right-handed signal Xr(s) consists of a half plane ℜe(s) > σr for some value σr; and from property 5, it is known that Xr(s) consists of a half plane ℜe (s) > σl for some value σl. Then the overlap of these two half planes is the ROC of the two-sided signal x(t) as shown in Figure 4.4(d) with the assumption that σr < σl. If σr is not less than σl, then there is no overlap. In this case, X(s) does not exist even Xr(s) and Xl(s) individually exist.

4.3 The Inverse Laplace Transform

From Eq. (4.6), it is known that the Laplace transform X(σ þ jΩ) of a signal x(t) is given by

X σ þ jΩ

Þ ¼

(cid:2)1

x tð Þ e(cid:2)σte(cid:2)jΩtdt

ð4:14Þ

Applying the inverse Fourier transform on the above relationship, we obtain

xðtÞe(cid:2)σt ¼ F (cid:2)1fXðσ þ jΩÞg ¼

(cid:2)1

1 2π

ðXðσ þ jΩÞÞ ejΩtdΩ

ð4:15Þ

Multiplying both sides of Eq. (4.15) by eσt, it follows that ð

xðtÞ ¼

1 2π

(cid:2)1

ðXðσ þ jΩÞÞ eðσþjΩÞtdΩ

ð4:16Þ

As s ¼ σ + jΩ and σ is a constant, ds ¼ j dΩ. Substituting s ¼ σ + jΩ ds ¼ j dΩ in Eq. (4.16) changing the variable of

integration from s to Ω, we arrive at the following inverse Laplace transform

x tð Þ ¼

ðσþj1

σ(cid:2)j1

1 2πj

X sð Þ estds

ð4:17Þ

4.3 The Inverse Laplace Transform

177

e R

e n a l p

) d (

) c (

) t ( x

) b (

) a (

l a n g i s

d e d i s

o w

e h t

f o C O R

) d (

. l a n g i s

d e d i s

t f e L ) c (

. l a n g i s

d e d i s

t h g i R

) b (

l a n g i s

d e d i s

o w T ) a (

4 4

e r u g i F

178

4 Laplace Transforms

4.4 Properties of the Laplace Transform

Linearity If x1(t) and x2(t) are two signals with Laplace transforms X1(s) and X2(s) and ROCs R1 and R2, respectively, then the Laplace transform of a linear combina- tion of x1(t) and x2(t) is given by

Lfa1x1ðtÞ þ a2x2ðtÞg ¼ a1X1ðsÞ þ a2X2ðsÞ

ð4:18Þ

whose ROC is at least (R1 \ R2), a1 and a2 being arbitrary constants.

Proof

Lfa1x1ðtÞ þ a2x2ðtÞg ¼

fa1x1ðtÞ þ a2x2ðtÞg e(cid:2)stdt

(cid:2)1 ð

¼ a1

(cid:2)1

x1ðtÞe(cid:2)stdt þ a2

(cid:2)1

x2ðtÞe(cid:2)stdt

¼ a1X1 sð Þ þ a2X2 sð Þ

ð4:19Þ

ð4:20Þ

The result concerning the ROC follows directly from the theory of complex

variables concerning the convergence of a sum of two convergent series.

Time Shifting If x(t) is a signal with Laplace transform X(s) and ROC R, then for any constant t0 (cid:4) 0, the Laplace transform of x(t – t0) is given by

L x t (cid:2) t0 ð

g ¼ e(cid:2)st0X sð Þ

whose ROC is the same as that of X(s).

Proof

L x t (cid:2) t0 ð

g ¼

(cid:2)1

Substituting τ ¼ t – t0,

x t (cid:2) t0 ð

Þ e(cid:2)stdt

ð4:22Þ

L x t (cid:2) t0 ð

g ¼

(cid:2)1

¼ e(cid:2)st0

Þdτ

x τð Þ e(cid:2)s τþt0 ð

(cid:2)1 ¼ e(cid:2)st0 X sð Þ

x τð Þ e(cid:2)sτdτ

ð4:23Þ

Shifting in the s-domain. If x(t) is a signal with Laplace transform X(s) and ROC R, then the Laplace transform of the signal es0tx tð Þ is given by

L es0tx tð Þ

g ¼ X s (cid:2) s0 ð

ð4:24Þ

whose ROC is the R þ ℜe(s)

4.4 Properties of the Laplace Transform

Proof

L es0tx tð Þ

g ¼

(cid:2)1 1

es0tx tð Þe(cid:2)stdt

x tð Þe(cid:2) s(cid:2)s0 ð

Þtdt

(cid:2)1 ¼ X s (cid:2) s0 ð

179

ð4:25Þ

Time Scaling. If x(t) is a signal with Laplace transform X(s) and ROC R, then the Laplace transform of the x(at) for any constant a, real or complex, is given by

whose ROC is the R a

Proof

L x atð f

g ¼

(cid:3) (cid:4) s a

1 a

L x atð f

g ¼

(cid:2)1

x atð

Þ e(cid:2)stdt

Letting τ ¼ at; dt ¼ dτ/a Then,

L x atð f

g ¼

τ a

dτ

1 a sτ a dτ

(cid:2)1 x τð Þ e(cid:2)s ð 1 a 1 a

(cid:2)1 (cid:3) (cid:4) s a

x τð Þe(cid:2)

ð4:26Þ

ð4:27Þ

ð4:28Þ

Differentiation in the Time Domain. If x(t) is a signal with the Laplace transform X (s) and ROC R, then

(cid:5) (cid:6) dx dt

¼ sX sð Þ

ð4:29Þ

with ROC containing R.

Proof This property can be proved by differentiating both sides of the inverse Laplace transform expression

Then,

x tð Þ ¼

ðσþj1

σ(cid:2)j1

1 2πj

X sð Þ estds

dx dt

1 2πj

ðσþj1

σ(cid:2)j1

sX sð Þestds

ð4:30Þ

180

4 Laplace Transforms

From the above expression, it can be stated that the inverse Laplace transform of

sX(s) is dx dt.

Differentiation in the s-Domain. If x(t) is a signal with the Laplace transform X(s), then

dX sð Þ ds

¼ L (cid:2)tx tð Þ f

ð4:31Þ

Proof From the deﬁnition of the Laplace transform,

X sð Þ ¼ L x tð Þ

g ¼

(cid:2)1

x tð Þ e(cid:2)stdt

Differentiating both sides of the above equation, we get

dX sð Þ ds

(cid:2)tx tð Þ e(cid:2)stdt

(cid:2)1 ¼ L (cid:2)tx tð Þ g f

with ROC ¼ R

Division by t If x(t) is a signal with Laplace transform X(s), then

(cid:6)

(cid:5)

x tð Þ t

X uð Þ du

provided that lim t!0

h x tð Þ t

exists.

Proof Let x1 tð Þ ¼ x tð Þ

t , then (t) ¼ t x1(t). By using the differentiation in the s-domain property,

which can be rewritten as

X sð Þ ¼ (cid:2)

d ds

L x1 tð Þ

dL x1 tð Þ f

g ¼ (cid:2)X sð Þds

Integrating both sides of Eq. (4.35) yields ð

dL x1 tð Þ f

g ¼ (cid:2)

X sð Þ ds

L x1 tð Þ

g ¼ (cid:2) ð

X uð Þ du

L x1 tð Þ

g ¼

X uð Þ du

ð4:32Þ

ð4:33Þ

ð4:34Þ

ð4:35Þ

ð4:36Þ

4.4 Properties of the Laplace Transform

181

Integration. If x(t) is a signal with Laplace transform X(s) and ROC R, then

(cid:5)

(cid:6)

x tð Þ dt

(cid:2)1

1 s

X sð Þ

ð4:37Þ

with ROC containing R \ ℜe(s) > 0.

Proof This property can be proved by integrating both sides of the inverse Laplace transform expression

x tð Þ ¼

ðσþj1

σ(cid:2)j1

1 2πj

X sð Þestds

Then,

ð τ

(cid:2)1

x τð Þ dτ ¼

ðσþj1

σ(cid:2)j1

1 s

1 2πj

X sð Þ estds

Consequently, the inverse Laplace transform of 1

s X sð Þ is

Ð τ (cid:2)1 x τð Þ dτ:

Convolution in the Time Domain. If x1(t) and x2(t) are two signals with Laplace transforms X1(s) and X2(s) with ROCs R1 and R2, respectively, then

L x1 tð Þ ∗ x2 tð Þ

g ¼ X1 sð ÞX2 sð Þ

ð4:38Þ

with ROC containing R1 \ R2

Proof

L x1 tð Þ ∗ x2 tð Þ

g ¼

(cid:7)

(cid:2)1

(cid:8)

x1 τð Þx2 t (cid:2) τ ð

Þ dτ

e(cid:2)stdt

ð4:39Þ

Changing the order of integration, Eq. (4.39) can be rewritten as (cid:8) ð

(cid:7)

L x1 tð Þ ∗ x2 tð Þ

g ¼

x1 τð Þ

x2 t (cid:2) τ ð

Þe(cid:2)stdt

dτ

ð4:40Þ

(cid:2)1

Let t1 ¼ t – τ; dt1 ¼ dt;

L x1 tð Þ ∗ x2 tð Þ

g ¼

(cid:2)1 ð 1

(cid:2)1

x1 τð Þ e(cid:2)sτ

(cid:7) ð

(cid:2)1

(cid:8)

dτ

(cid:8)

x2 t1ð Þe(cid:2)st1 dt1 (cid:7) ð

(cid:2)1

e(cid:2)sτx1 τð Þ X2 sð Þdτ ¼

e(cid:2)sτx1 τð Þdτ

X2 sð Þ

ð4:41Þ

¼ X1 sð ÞX2 sð Þ

The ROC includes R1 \ R2 and is large if pole-zero cancellation occurs.

Convolution in the Frequency Domain If x1(t) and x2(t) are two signals with Laplace transforms X1(s) and X2(s), then

182

4 Laplace Transforms

L x1 tð Þx2 tð Þ

g ¼

cþj1

c(cid:2)j1

1 2πj

X1 pð ÞX2 s (cid:2) p

Þdp

ð4:42Þ

Proof Let x(t) ¼ x1(t)x2(t)

ℜe(s) > σ2, respectively, then

with Laplace transforms X1(s) and X2(s) and areas of convergence ℜe(s) > σ1 and

L x tð Þ f

g ¼

x1 tð Þ x2 tð Þ e(cid:2)stdt

ð4:43Þ

According to the inverse integral, ð

x1 tð Þ ¼

1 2πj

cþj1

c(cid:2)j1

X1 pð Þ eptdp, c > σ1

ð4:44Þ

Substituting this relationship in Eq. (4.43), it follows that

L x tð Þ f

g ¼

x2 tð Þe(cid:2)st

(cid:9)

1 2πj

cþj1

c(cid:2)j1

(cid:10)

X1 pð Þeptdp

ð4:45Þ

Permuting the sequence of integration, we obtain

X sð Þ ¼

cþj1

c(cid:2)j1

1 2πj

X1 pð Þdp

x2 tð Þe(cid:2) s(cid:2)p

Þtdt

ð4:46Þ

where

X2 s (cid:2) p ð

Þ ¼

x2 tð Þe(cid:2) s(cid:2)p

Þtdt

ð4:47Þ

This integral converges for ℜe(s (cid:2) p) > σ2. By substituting Eq. (4.47) in

Eq. (4.46), yields the following proving the property

X sð Þ ¼

cþj1

c(cid:2)j1

1 2πj

X1 pð ÞX2 s (cid:2) p

Þdp

ð4:48Þ

The above properties of the Laplace transform are summarized in Table 4.1.

4.4.1 Laplace Transform Properties of Even and Odd

Functions

Even Property If x(t) is an even function such that x(t) ¼ x(–t), then X(s) ¼ X((cid:2)s).

4.4 Properties of the Laplace Transform

Proof

Consider

Let t1 ¼ (cid:2)t; then,

(cid:2)1

X sð Þ ¼

X1 sð Þ ¼

x tð Þ e(cid:2)stdt

x (cid:2)t ð

Þ e(cid:2)stdt

x t1ð Þ est1dt1

(cid:2)1 ¼ X (cid:2)s ð

Since x(t) ¼ x((cid:2)t) then L{x(t)} ¼ L{x((cid:2)t)} Thus, X(s) ¼ X((cid:2)s).

Odd Property If x(t) is an odd function such that x(t) ¼ (cid:2)x((cid:2)t), then X(s) ¼ (cid:2)X((cid:2)s).

Proof

Consider

X sð Þ ¼

(cid:2)1

x tð Þ e(cid:2)stdt

183

ð4:49Þ

ð4:50Þ

ð4:51Þ

ð4:52Þ

Let t1 ¼ (cid:2)t; then

X1 sð Þ ¼

(cid:2)1

(cid:2)x (cid:2)t ð

Þ e(cid:2)stdt

ð4:53Þ

(cid:2)1

(cid:2)x t1ð Þ est1dt1 ¼ (cid:2)

x t1ð Þ est1 dt1

(cid:2)1 ¼ (cid:2)X (cid:2)s ð

ð4:54Þ

Since x(t) ¼ (cid:2)x((cid:2)t) then L{x(t)} ¼ (cid:2)L{x((cid:2)t)} Thus, X(s) ¼ (cid:2)X((cid:2)s).

4.4.2 Differentiation Property of the Unilateral Laplace

Transform

Most of the properties of the bilateral transform tabulated in Table 4.1 are the same for the unilateral transform. In particular, the differential property of unilateral Laplace transform is different as it requires that x(t) ¼ 0 for t < 0 and contains no impulses and higher-order singularities.

If x(t) is a signal with unilateral Laplace transform X (s), then the unilateral

Laplace transform of dx

dt can be found by using integrating by parts as

184

4 Laplace Transforms

Table 4.1 Some properties of the Laplace transform

Property Linearity Time shifting Shifting in the s-domain Time scaling

Differentiation in the time domain

Differentiation in the s-domain

Integration

Convolution

Signal a1x1(t) + a2x2(t) x(t – t0) es0tx tð Þ x(at)

dx dt (cid:2)tx(t) ð

x tð Þdt

(cid:2)1 x1(t) * x2(t)

Laplace transform ROC a1X1(s) + a2X2(s) At least R1 \ R2 e(cid:2)st0 X sð Þ X(s – s0) (cid:11) (cid:12) a X s 1 a sX(s)

Same as R Shifted version of R R a At least R

dX sð Þ ds 1 s X sð Þ

R R \ ℜe(s) > 0.

X1(s)X2(s)

R1 \ R2

dx dt

e(cid:2)stdt ¼ x tð Þe(cid:2)st 1 0þ

(cid:2) (cid:2)

þ s

x tð Þe(cid:2)stdt

¼ sX sð Þ (cid:2) x 0þð

ð4:55Þ

Applying this second time yields the unilateral Laplace transform of d2x

dt2 as given

d2x dt2 e(cid:2)stdt ¼ s2X sð Þ (cid:2) sx 0þð

Þ (cid:2) _x 0þð

ð4:56Þ

where _x 0þð

Þ is the dx

dt evaluated at t ¼ 0+.

Similarly, applying this for third time yields the unilateral Laplace transform

ð4:57Þ

d3x dt3 e(cid:2)stdt ¼ s3XðsÞ (cid:2) s2xð0þÞ (cid:2) s_xð0þÞ (cid:2) €xð0þÞ dt2 evaluated at t ¼ 0+.

where €x 0þð

Þ is the d2x

Continuing this procedure for the nth time, the unilateral Laplace transform of dnx dtn

is given by ð

dnx dtn e(cid:2)stdt ¼ snX sð Þ (cid:2) sn(cid:2)1x 0þð

Þ (cid:2) sn(cid:2)2 _x 0þð

Þ (cid:2) sn(cid:2)3€x 0þð

Þ: …

ð4:58Þ

Example 4.1 Determine whether the following Laplace transforms correspond to the even time function or odd time function and comment on the ROCs.

Ks (a) X sð Þ ¼ Þ s(cid:2)2 sþ2 Þ ð ð (b) X sð Þ ¼ K sþj2 Þ s(cid:2)j2 ð ð Þ s(cid:2)2 sþ2 Þ ð ð (c) Comment on the ROCs

4.4 Properties of the Laplace Transform

185

Solution (a) X (cid:2)s ð

(b) X (cid:2)s

Þ ¼

(cid:2)sþ2 ð

(cid:2)Ks Þ (cid:2)s(cid:2)2 Þ ð

Þ ¼ X sð Þ; ; (cid:2)X (cid:2)s ð Hence, the corresponding x(t) is an odd function. Þ Þ ¼ X sð Þ

Þ ¼ K (cid:2)sþj2 ð (cid:2)sþ2 ð

Þ (cid:2)s(cid:2)j2 ð Þ (cid:2)s(cid:2)2 ð

Ks Þ s(cid:2)2 ð

Þ ¼

sþ2 ð

Hence, the corresponding x(t) is an even function.

(c) The ROCs for (a) and (b) are shown in Figure 4.5(a) and (b), respectively. From Figure 4.5(a) and (b), it can be stated that for the time function to be even or odd, the ROC must be two sided.

Example 4.2 A real and even signal x(t) with its Laplace transform X(s) has four 2 ejπ=4, with no zeros in the ﬁnite s-plane and X poles with one pole located at 1 (0) ¼ 16. Find X(s). Solution Since X(s) has four poles with no zeros in the ﬁnite s-plane, it is of the form

X sð Þ ¼

s (cid:2) p1

Þ s (cid:2) p2 ð

K Þ s (cid:2) p3 ð

Þ s (cid:2) p4 ð

As x(t) is real, the poles of X(s) must occur as conjugate reciprocal pairs. Hence, p2 ¼ p1 ∗ ; p4 ¼ p3 ∗ and X(s) becomes

X sð Þ ¼

s (cid:2) p1

Þ s (cid:2) p1 ∗ ð

K Þ s (cid:2) p3 ð

Þ s (cid:2) p3 ∗ ð

(a)

(b)

s plane

-2

-j2

s plane

Figure 4.5 (a) ROC of X sð Þ ¼

Ks Þ s(cid:2)1 ð

sþ2

Þ. (b) ROC of X sð Þ ¼ K sþj2

ð sþ2 ð

Þ s(cid:2)j2 Þ ð Þ s(cid:2)2 Þ ð

186

4 Laplace Transforms

Since x(t) is even, the X(s) also must be even, and hence the poles must be

symmetric about the jΩ axis. Therefore, p3 ¼ (cid:2)p1 ∗ .

Thus,

XðsÞ ¼

ðs (cid:2) p1Þðs (cid:2) p∗

K 1 Þðs þ p∗

1 Þðs þ p1Þ

Assuming that the given pole location is that of p1, that is, p1 ¼ 1 We obtain

2 ejπ=4,

XðsÞ ¼

(cid:7)

s (cid:2)

(cid:8)

(cid:7)

s (cid:2)

ejπ=4

1 2

ejπ=4 ¼

1 2

(cid:3)

cos

1 2

þ jsin

1 2 (cid:4)

e(cid:2)jπ=4 (cid:7)

1 2

(cid:8)

(cid:7)

s þ

(cid:8)

ejπ=4

1 2

1 ﬃﬃﬃ p þ j 2

1 p

ﬃﬃﬃ 2

K (cid:7) (cid:8)

s þ

1 2

e(cid:2)jπ=4 (cid:8)

1ﬃﬃﬃ p 2

1ﬃﬃﬃ p þ j 2 K (cid:7) (cid:8)

S2 þ

XðsÞ ¼

(cid:7)

S2 (cid:2)

1ﬃﬃﬃ p s þ 2

1 4

(cid:8)

1ﬃﬃﬃ p s þ 2

1 4

when s ¼ 0, X 0ð Þ ¼ K

1=16 ¼ 16, and therefore K ¼ 1.

Hence,

X sð Þ ¼

1 (cid:3) (cid:4)

(cid:3) s2 (cid:2) 1ﬃﬃ 2

p s þ 1 4

(cid:4)

s2 þ 1ﬃﬃ p s þ 1 4 2

4.4.3

Initial Value Theorem

For a signal x(t) with Laplace transform X(s) and x(t) ¼ 0 for t < 0, then

x 0þð

Þ ¼ lims!1sX sð Þ

ð4:59Þ

Proof To prove the theorem, let the following integral ﬁrst be evaluated by using integration by parts

1 0þ

dx dt

e(cid:2)stdt ¼ x tð Þ e(cid:2)st 1 0þ

(cid:2) (cid:2)

x tð Þs e(cid:2)stdt

0þ ¼ (cid:2)x 0þð

Þ þ sX sð Þ

ð4:60Þ

As s tends to 1, the Eq. (4.60) can be expressed as

0þ

dx dt

lim s!1

e(cid:2)stdt ¼ lim s!1

½sXðsÞ (cid:2) xð0þÞ(cid:5)

ð4:61Þ

4.5 Laplace Transforms of Elementary Functions

187

As the integration is independent of s,

the calculation of the limit and the integration can be permuted provided that the integral converges uniformly. If L{x(t)} exists, then

is valid. Hence, we get

lim s!1

dx dt

e(cid:2)st ¼ 0

x 0þð

Þ ¼ lims!1sX sð Þ

ð4:62Þ

ð4:63Þ

4.4.4 Final Value Theorem

For a signal x(t) with Laplace transform X(s) and x(t) ¼ 0 for t < 0, then

x 1ð

Þ ¼ lims!0sX sð Þ

ð4:64Þ

Proof To prove this, the following integration is to be evaluated

0þ

dx dt

lim s!0

e(cid:2)stdt ¼ sX sð Þ (cid:2) x 0þð

ð4:65Þ

Again one can permute the sequence of determining the limit and the integration

provided the integral converges. The result is

0þ

dx dt

dt ¼ lim s!0

sX sð Þ (cid:2) x 0þð ½

(cid:5), Þ

and after integration it follows that

x 1ð

Þ (cid:2) x 0þð

Þ ¼ lim s!0 ¼ lim s!0

sX sð Þ (cid:2) x 0þð

(cid:5)

sX sð Þ (cid:2) x 0þð

(cid:5)

Therefore,

x 1ð

Þ ¼ lim s!0

sX sð Þ

4.5 Laplace Transforms of Elementary Functions

Unit Impulse Function The unit impulse function is deﬁned by

(cid:5)

δ tð Þ ¼

1 0

for t ¼ 0 elsewhere

ð4:66Þ

ð4:67Þ

ð4:68Þ

ð4:69Þ

188

4 Laplace Transforms

By deﬁnition, the Laplace transform of δ (t) can be written as

ð4:70Þ

ð4:71Þ

ð4:72Þ

X sð Þ ¼ L δ tð Þ

g ¼

δ tð Þe(cid:2)stdt

0 ¼ 1

The ROC is the entire s-plane.

Unit Step Function The unit step function is deﬁned by

(cid:5)

u tð Þ ¼

1 0

for t (cid:4) 0 elsewhere

The Laplace transform of u(t) by deﬁnition can be written as

X sð Þ ¼ L u tð Þ

g ¼

0 1

0 1 s Hence, the ROC for X(s) is ℜe(s) > 0.

u tð Þ e(cid:2)stdt

1e(cid:2)stdt ¼ (cid:2)

(cid:2) (cid:2) e(cid:2)st 1 0

1 s

Example 4.3 Find the Laplace transform of x(t) ¼ δ(t – t0).

Solution By using the time shifting property, we get

L δ t (cid:2) t0 ð

g ¼ e(cid:2)st0L δ tð Þ Þ

g ¼ e(cid:2)st0

The ROC is the entire s-plane.

Example 4.4 Find the Laplace transforms of the following: (i) x(t) ¼ (cid:2)e(cid:2)αtu((cid:2)t) (iv) x(t) ¼ eαtu(t) Solution (i) X sð Þ ¼ L (cid:2)e(cid:2)αtu (cid:2)t

(ii) x(t) ¼ eαtu((cid:2)t)

e(cid:2)αtu (cid:2)t ð

g ¼ (cid:2)

(iii) x(t) ¼ e(cid:2)αtu(t)

Þe(cid:2)stdt

Because u((cid:2)t) ¼ 1 for t < 0 and u((cid:2)t) ¼ 0 for t > 0,

X sð Þ ¼ (cid:2)

0(cid:2)

(cid:2)1

e(cid:2) sþα ð

Þtdt

1 s þ α

4.5 Laplace Transforms of Elementary Functions

189

The ROC for X(s) is ℜe(s) < (cid:2) α

(ii) X sð Þ ¼ L eαtu (cid:2)t

g ¼

eαtu (cid:2)t ð

Þe(cid:2)stdt

Because u((cid:2)t) ¼ 1 for t < 0 and u((cid:2)t) ¼ 0 for t > 0,

0(cid:2)

e(cid:2) s(cid:2)α ð

Þtdt

X sð Þ ¼ (cid:2)

¼ (cid:2)

(cid:2)1 1 s (cid:2) α

The ROC for X (s) is ℜe(s) < α

(iii) Let x1(t) ¼ u(t), then

X1 sð Þ ¼ L u tð Þ

g ¼

1 s

By using the shifting in the s-domain property, we get

X sð Þ ¼ L e(cid:2)αtu tð Þ f

g ¼ X1 s þ α ð

Þ ¼

1 s þ α

The ROC for X(s) is ℜe(s) > (cid:2) α

(iv) Let x1(t) ¼ u(t), then

X1 sð Þ ¼ L u tð Þ

g ¼

1 s

By using the shifting in the s-domain property, we get

X sð Þ ¼ L eαtu tð Þ f

g ¼ X1 s (cid:2) α ð

Þ ¼

1 s (cid:2) α

The ROC for X(s) is ℜe(s) > α.

Example 4.5 Find the Laplace transform of

x tð Þ ¼

Solution Let x1(t) ¼ u(t), then

t n(cid:2)1 ð n (cid:2) 1

Þ! u tð Þ

X1 sð Þ ¼ L u tð Þ

g ¼

1 s

and for n ¼ 2, x2(t) ¼ tu(t).

190

4 Laplace Transforms

By using the differentiation in the s-Domain property, we get

X2 sð Þ ¼ L tu tð Þ

g ¼ (cid:2)

dX1 ds

1 s2

Similarly, for n ¼ 3, x3 tð Þ ¼ t2 3(cid:2)1 ð Again, by using the differentiation in the s-domain property, we get

Þ! u tð Þ.

(cid:5)

t2 n (cid:2) 1 ð o

X3 sð Þ ¼ L

Þ!u tð Þ 1 sn. Example 4.6 Find the Laplace transform of

In general, X sð Þ ¼ L t n(cid:2)1 ð Þ n(cid:2)1 ð

Þ!u tð Þ

(cid:6)

¼ (cid:2)

dX2 ds

1 s3

x tð Þ ¼

t n(cid:2)1 Þ ð n (cid:2) 1 ð

Þ! e(cid:2)αtu tð Þ

Solution Let x1 tð Þ ¼ t n(cid:2)1 ð n(cid:2)1

Þ! u tð Þ, then (cid:5)

X1 sð Þ ¼ L

t n(cid:2)1 ð n (cid:2) 1

Þ!u tð Þ

(cid:6)

1 Sn

By using the shifting in the s-domain property, we get

X sð Þ ¼ L

(cid:5)

(cid:6)

t n(cid:2)1 ð n (cid:2) 1

Þ!e(cid:2)αtu tð Þ

¼ X1 s þ α ð

Þ ¼

1 s þ α

Þn

for ℜe sð Þ > (cid:2)α

Example 4.7 Find the Laplace transform of x(t) ¼ sin ωt u(t)

Solution

(cid:5)

L sin ωt u tð Þ (cid:14)

(cid:16)

(cid:5) ¼ L

(cid:15) L ejωtu tð Þ

1 2j

(cid:6)

u tð Þ

(cid:15)(cid:17)

ejωt (cid:2) e(cid:2)jωt 2j (cid:14)

(cid:2) L e(cid:2)jωtu tð Þ

Using the shifting in the s-domain property, we get

(cid:14)

L ejωtu tð Þ

(cid:15)

(cid:16)

1 2j

(cid:14)

(cid:2) L e(cid:2)jωtu tð Þ

(cid:15)

(cid:17)

(cid:9)

(cid:10)

1 s þ jω

1 s (cid:2) jω (cid:2) ω

1 2j s2 þ ω2 for ℜe sð Þ > 0

Therefore,

L sin ωt u tð Þ

(cid:5) ¼

s2 þ ω2 for ℜe sð Þ > 0

4.5 Laplace Transforms of Elementary Functions

191

Example 4.8 Find the Laplace transform of x(t) ¼ cos ωt u(t).

Solution

(cid:5)

(cid:6)

L cos ωt u tð Þ (cid:16) L ejωtu tð Þ

(cid:5) ¼ L (cid:15)

(cid:14)

1 2

ejωt þ e(cid:2)jωt 2 L e(cid:2)jωtu tð Þ

(cid:14)

u tð Þ (cid:15)(cid:17)

Using the shifting in the s-domain property, we get

(cid:14)

(cid:16) L ejωtu tð Þ

(cid:15)

(cid:14)

þ L e(cid:2)jωtu tð Þ

(cid:15)

(cid:17)

1 2

(cid:9)

(cid:10)

1 2

1 s (cid:2) jω þ s s2 þ ω2

1 s þ jω for ℜe sð Þ > 0

Therefore,

L cos ωt u tð Þ

(cid:5) ¼

s2 þ ω2 for ℜe sð Þ > 0

Example 4.9 Find the Laplace transform of x(t) ¼ e(cid:2)αtsin ωt u(t). Solution Let x1(t) ¼ sin ωt u(t), then

X1 sð Þ ¼ L sin ωt u tð Þ f

g ¼

ω s2 þ ω2

By using the shifting in the s-domain property, we get

X sð Þ ¼ L e(cid:2)αt sin ωt u tð Þ

g ¼ X1 s þ α ð

Þ ¼

ω Þ2 þ ω2 s þ α ð

for ℜe sð Þ > (cid:2)α

Example 4.10 Find the Laplace transform of x(t) ¼ e(cid:2)αtcos ωt u(t). Solution Let x1(t) ¼ cos ωt u(t), then

X1 sð Þ ¼ L cos ωt u tð Þ f

g ¼

s s2 þ ω2

By using the shifting in the s-domain property, we get

X sð Þ ¼ L e(cid:2)αt cos ωt u tð Þ

g ¼ X1 s þ α ð

Þ ¼

s þ α

Þ2 þ ω2

for ℜe sð Þ > (cid:2)α

Example 4.11 Find the Laplace transform of sin t t

Solution

L sin t

g ¼

1 s2 þ 1

192

4 Laplace Transforms

(cid:6)

(cid:5)

sin t t

1 u2 þ 1 (cid:2) (cid:2) ¼ tan (cid:2)1u 1 s ¼

(cid:2) tan (cid:2)1s

Example 4.12 Find the Laplace transform of e4t(cid:2)e(cid:2)3t

Solution

(cid:14) (cid:15)

L e4t

(cid:5)

e4t (cid:2) e3t t

¼ (cid:6)

1 s (cid:2) 4 ð

¼ ln u (cid:2) 4 ð

(cid:14) ; L e(cid:2)3t

(cid:15)

1 s þ 3

1 u (cid:2) 4

du (cid:2)

s (cid:2) (cid:2) Þ 1 s

s (cid:2) (cid:2) Þ 1 (cid:2) ln u þ 3 ð s

1 u þ 3

¼ (cid:2)ln

¼ ln

ð ð

Þ Þ

s (cid:2) 4 ð s þ 3 ð s þ 3 Þ s (cid:2) 4 Þ

Example 4.13 Consider the signal x(t) ¼ etu(t) + 2e2tu(t).

(a) Does the Fourier transform of this signal converge? (b) For which of the following values of a does the Fourier transform of x(t) e(cid:2)αt

converge? (i) α ¼ 1 (ii) α ¼ 2.5

Sketch the location of the poles and zeros of X(s) and the ROC.

Solution

(a) The Fourier transform of the signal does not converge as x(t) is not absolutely

integrable due to the rising exponentials. (b) (i) For α ¼ 1, x(t) e(cid:2)αt ¼ u(t) þ 2etu(t).

Although the growth rate has been slowed, the Fourier transform still does not

converge.

(ii) For α ¼ 2.5, x(t) e(cid:2)αt ¼ e(cid:2)1.5tu(t) + 2e(cid:2)0.5tu(t), the Fourier transform

converges

X sð Þ ¼

1 s (cid:2) 1

2 s (cid:2) 2

2s (cid:2) 3 Þ s (cid:2) 2 ð

s (cid:2) 1

(cid:12) (cid:11) 2 s (cid:2) 3 2 Þ s (cid:2) 2 ð

s (cid:2) 1

and its pole-zero plot and ROC are as shown in Figure 4.6.

4.5 Laplace Transforms of Elementary Functions

Figure 4.6 Pole-zero plot and ROC

193

1.5

It is noted that if α > 2, s ¼ α þ jΩ is in the region of convergence, as it is shown

in part (b) (ii), the Fourier transform converges.

Example 4.14 The Laplace transform H(s) of the impulse response h(t) for an LTI system is given by

H sð Þ ¼

1 s þ 2 ð

ℜe sð Þ > (cid:2)2

Determine the system output y(t) for all t if the input x(t) is given by x(t) ¼ e(cid:2)3t/2

2e(cid:2)t for all t.

Solution From the convolution integral,

y tð Þ ¼

(cid:2)1

h τð Þx t (cid:2) τ ð

Þdτ

Let x(t) ¼ eαt, then

y tð Þ ¼

(cid:2)1

h τð Þeα t(cid:2)τ

Þdτ ¼ eαt

(cid:2)1

h τð Þe(cid:2)ατdτ

h τð Þ e(cid:2)ατdτ can be recognized as H(s)|s ¼ α.

(cid:2)1 Hence, if x(t) ¼ eαt, then

y tð Þ ¼ eαt H sð Þ s¼α ½

(cid:5)

Using linearity and superposition, it can be recognized that if x(t) ¼ e(cid:2)3t/2 þ 2e(cid:2)t,

then

So that

y tð Þ ¼ e(cid:2)3t=2H sð Þ s¼(cid:2)3=2

(cid:2) (cid:2)

þ 2e(cid:2)tH sð Þ s¼(cid:2)1

y tð Þ ¼ 2e(cid:2)3t=2 þ 2e(cid:2)t

for all t

194

4 Laplace Transforms

Example 4.15 The output y(t) of a LTI system is

(cid:11)

y tð Þ ¼ 2 (cid:2) 3e(cid:2)t þ e(cid:2)3t

(cid:12)

u tð Þ

for an input

(cid:11) x tð Þ ¼ 2 þ 4e(cid:2)3t

(cid:12)

u tð Þ

Determine the corresponding input for an output

y1 tð Þ ¼ 1 (cid:2) e(cid:2)t (cid:2) te(cid:2)t

Þu tð Þ

Solution For the input

x(t) ¼ (2 þ 4e(cid:2)3t)u(t), the Laplace transform is

X sð Þ ¼

2 s

4 s þ 3

6 s þ 1 Þ ð s s þ 3 Þ ð

The corresponding output has the Laplace transform

Y sð Þ ¼

2 s

(cid:2)

3 s þ 1

1 s þ 3

6 Þ s þ 3 s s þ 1 ð ð

Hence, H sð Þ ¼

Y sð Þ X sð Þ

1 s þ 1 ð

Þ2

ℜe sð Þ > 0

Now, the output y1(t) ¼ (1 – e(cid:2)t– te(cid:2)t)u(t) has the Laplace transform

Y 1 sð Þ ¼

1 s

(cid:2)

1 s þ 1

Þ2 ¼

1 s s þ 1 ð

Þ2

ℜe sð Þ > 0

Hence, the Laplace transform of the corresponding input is

X1 sð Þ ¼

Y 1 sð Þ H sð Þ

1 s

ℜe sð Þ > 0

The inverse Laplace transform of X1 (s) gives

x1 tð Þ ¼ u tð Þ:

4.6 Computation of Inverse Laplace Transform Using

Partial Fraction Expansion

The inverse Laplace transform of a rational function X(s) can be easily computed by using the partial fraction expansion.

4.6 Computation of Inverse Laplace Transform Using Partial Fraction Expansion

195

4.6.1 Partial Fraction Expansion of X(s) with Simple Poles

Consider a rational Laplace transform X(s) of the form

X sð Þ ¼

N sð Þ Þ: … … s (cid:2) pn ð

Þ s (cid:2) p2 ð

s (cid:2) p1

ð4:73Þ

with the order of N(s) is less than the order of the denominator polynomial.

The poles p1, p2, … .., pn are distinct. The rational Laplace transform X(s) can be expanded using partial fraction

expansion as

X sð Þ ¼

k1 s (cid:2) p1

k2 s (cid:2) p2 ð

þ (cid:6) (cid:6) (cid:6)

kn s (cid:2) pn

ð4:74Þ

The coefﬁcients k1, k2, … ., kn are called the residues of the partial fraction

expansion. The residues are computed as

ki ¼ s-pi ð

(cid:2) (cid:2) Þ X sð Þ s¼pi

i ¼ 1, 2, … , n

ð4:75Þ

With the known values of the coefﬁcients k1, k1, … ., kn inverse transform of each term can be determined depending on the location of each pole relative to the ROC.

4.6.2 Partial Fraction Expansion of X(s) with Multiple Poles

Consider a rational Laplace transform X(s) with repeated poles of the form

X sð Þ ¼

N sð Þ

s (cid:2) p1

Þr s (cid:2) p2

Þ: … … s (cid:2) pn ð

ð4:76Þ

with multiplicity r poles at s ¼ p1.

The X(s) with multiple poles can be expanded as

Y sð Þ ¼

k11 s (cid:2) p1 ð

k12 s (cid:2) p1

Þ2 þ (cid:6) (cid:6) (cid:6)

k1r s (cid:2) p1

Þr þ

k2 s (cid:2) p2

þ (cid:6) (cid:6) (cid:6)

kn s (cid:2) pn ð

ð4:77Þ

The coefﬁcients k2, … ., kn can be computed using the residue formula used in

Section 4.6.1. The residues k11, k12, …, klr are computed as

k1r ¼ s (cid:2) p1 ð

(cid:2) (cid:2) ÞrX sð Þ s¼pi

k1 r(cid:2)1 ð

Þ ¼

½ ð

s (cid:2) p1

ÞrX sð Þ

(cid:2) (cid:2) (cid:5) s¼pi

1 1!

d ds

ð4:78Þ

ð4:79Þ

196

4 Laplace Transforms

k1 r(cid:2)2 ð

Þ ¼

1 2!

d2 ds2

½ ð

s (cid:2) p1

ÞrX sð Þ

(cid:2) (cid:2) (cid:5) s¼pi

ð4:80Þ

and so on.

Example 4.16 Find the time function x(t) for each of the following Laplace trans- form X(s)

(a) X sð Þ ¼

(b) X sð Þ ¼

(d) X sð Þ ¼

s þ 2 s2 þ 7s þ 12 s2 þ s þ 1 s2 s (cid:2) 1 Þ s2 (cid:2) s þ 1 s þ 1 ð s þ 1 s2 þ 5s þ 6

ℜe sð Þ > (cid:2)3

0 < ℜe sð Þ < 1

Þ2 (cid:2)1 < ℜe sð Þ ℜe sð Þ < (cid:2)3

Solution (a) X sð Þ ¼ sþ2

s2þ7sþ12

Using partial fraction expansion,

s þ 2 s2 þ 7s þ 12

k1 s þ 3

k2 s þ 4

k1 þ k2 ¼ 1 4k1 þ 3k2 ¼ 2

Solving for k1 and k2, k1 ¼ (cid:2) 1; k2¼2. Thus, X sð Þ ¼ sþ2 s2þ7sþ12 ¼ (cid:2) 1 Using Table 4.2, we obtain

sþ3 þ 2 sþ4

x(t) ¼ (cid:2)e(cid:2)3tu(t) þ 2e(cid:2)4tu(t)

(b) X sð Þ ¼

s2 (cid:2) s þ 1 s2 s (cid:2) 1 Þ

1 s (cid:2) 1

(cid:2)

1 s s (cid:2) 1 ð

1 s2 s (cid:2) 1

Using partial fraction expansion,

1 s s (cid:2) 1 ð

k1 s

k2 s (cid:2) 1

Solving for k1 and k2, k1 ¼ (cid:2)1; k2 ¼ 1.

1 s s (cid:2) 1 ð

(cid:2)1 s

1 s (cid:2) 1

Using partial fraction expansion,

1 s2 s (cid:2) 1

k1 s (cid:2) 1

k11 s

k12 s2

4.6 Computation of Inverse Laplace Transform Using Partial Fraction Expansion

197

Table 4.2 Elementary functions and their Laplace transforms

Signal δ(t) u(t)

(cid:2)u((cid:2)t) δ(t (cid:2) t0) e(cid:2)αtu(t) (cid:2)e(cid:2)atu((cid:2)t) t n(cid:2)1 Þ ð n (cid:2) 1 t n(cid:2)1 Þ ð n (cid:2) 1 ð sin ωt u(t) cos ωt u(t) e(cid:2)αt sin ωt u(t) e(cid:2)αt cos ωt u(t)

Þ! u tð Þ Þ! e(cid:2)αtu tð Þ

Solving for k1, k11, and k12

Laplace transform 1 1 s 1 s e(cid:2)st0 1 sþα 1 sþα 1 sn

1 sþα ð

Þn

ω s2þω2 s s2þω2 ω Þ2þω2 sþα Þ2þω2

sþα ð

ROC All s ℛe(s) > 0 ℛe(s) > 0 All s ℛe(s) > (cid:2) α ℛe(s) < (cid:2) α ℛe(s) > 0

ℛe(s) > (cid:2) α

ℜe(s) > 0 ℜe(s) > 0 ℜe(s) > (cid:2) α ℜe(s) > (cid:2) α

k1 ¼ 1; k11 ¼ (cid:2)1; k12 ¼ (cid:2)1: 1 s (cid:2) 1

1 s2ðs (cid:2) 1Þ

1 s2

1 s

(cid:2)

Thus,

X sð Þ ¼

s2 (cid:2) s þ 1 s2 s (cid:2) 1 Þ

ð 1 s (cid:2) 1

(cid:2)

1 s s (cid:2) 1 ð

1 s2 s (cid:2) 1

1 s (cid:2) 1

(cid:2)

1 s

(cid:2)

1 s2

1 s (cid:2) 1 1 s (cid:2) 1

(cid:2)

1 s 1 s2

Using Table 4.2, we obtain

x tð Þ ¼ (cid:2)etu (cid:2)t

Þ (cid:2) tu tð Þ

Þ2 ¼ 1 (cid:2) 3s

sþ1 ð

Þ2

sþ1 ð

Using partial fraction expansion,

3s s þ 1

Þ2 ¼

k11 s þ 1

k12 s þ 1

Þ2

198

4 Laplace Transforms

Solving for k11 and k12, we get

k11 ¼ 3; k12 ¼ (cid:2)3

Hence,

X sð Þ ¼

s2 (cid:2) s þ 1 s þ 1 ð

Þ2 ¼ 1 (cid:2)

3 s þ 1

Þ2

Using Table 4.2, we obtain

x tð Þ ¼ δ tð Þ (cid:2) 3e(cid:2)tu tð Þ þ 3te(cid:2)tu tð Þ

(d) X sð Þ ¼ sþ1

s2þ5sþ6

Using partial fraction expansion,

s þ 1 s2 þ 5s þ 6

k1 s þ 3 k1 þ k2 ¼ 1

k2 s þ 2

2k1 þ 3k2 ¼ 1

Thus, X sð Þ ¼ sþ1

Solving for k1 and k2, k1 ¼ 2; k2 ¼ (cid:2)1. s2þ5sþ6 ¼ 2 Using Table 4.2, we obtain

sþ3 (cid:2) 1 sþ2

x tð Þ ¼ (cid:2)2e(cid:2)3tu (cid:2)t

Þ þ e(cid:2)2tu (cid:2)t

Example 4.17 Find the inverse Laplace transform of the following

X sð Þ ¼

s2 þ s þ 1 s2 (cid:2) s þ 1

Solution

X sð Þ ¼ 1 þ

¼ 1 þ

2s s2 (cid:2) s þ 1 2s

(cid:11)

(cid:12)

(cid:3) (cid:4) ﬃﬃ p 2 3 2

s (cid:2) 1 2

¼ 1 þ 2

(cid:11)

s (cid:2) (cid:12) s (cid:2) 1 2

1 2 (cid:3) (cid:4) ﬃﬃ p 3 2

2 þ

(cid:11)

(cid:12)

(cid:3) (cid:4) ﬃﬃ p 2 3 2

s (cid:2) 1 2

4.6 Computation of Inverse Laplace Transform Using Partial Fraction Expansion

199

Using Table 4.2, we obtain

x tð Þ ¼ δ tð Þ þ 2e(cid:2)t=2 cos

(cid:7)

ﬃﬃﬃ 3 2

(cid:8)

u tð Þ þ

2ﬃﬃﬃ p e(cid:2)t=2 sin 3

(cid:7)

ﬃﬃﬃ 3 2

(cid:8)

u tð Þ

Example 4.18 Determine x(t) for the following conditions if X(s) is given by

X sð Þ ¼

1 Þ s þ 3 ð

s þ 2

(a) x(t) is right sided (b) x(t) is left sided (c) x(t) is both sided

Solution Using partial fraction, X(s) can be written as

(a)

X sð Þ ¼

1 Þ s þ 3 ð

1 s þ 2

(cid:2)

1 s þ 3

s þ 2

If x(t) is right sided, (cid:5)

x tð Þ ¼ L(cid:2)1

(cid:6)

1 s þ 2

(cid:2) L(cid:2)1

(cid:5)

(cid:6)

1 s þ 3

¼ e(cid:2)2tu tð Þ (cid:2) e(cid:2)3tu tð Þ

The ROC is to the right of the rightmost pole as shown in Figure 4.7(a).

(b) If x(t) is left sided,

x tð Þ ¼ L(cid:2)1

(cid:5)

1 s þ 2

(cid:6)

(cid:5)

(cid:2) L(cid:2)1

1 s þ 3

(cid:6)

¼ (cid:2)e(cid:2)2tu (cid:2)t

(cid:11)

Þ (cid:2) (cid:2)e(cid:2)3tu (cid:2)t

(cid:12)

The ROC is to the left of the leftmost pole as shown in Figure 4.7(b)

(a)

(b)

s plane

-3

-2

(c)

s plane

-3

-2

-3

-2

Figure 4.7 (a) ROC of right-sided x(t) (b) ROC of left-sided x(t) (c) ROC of both-sided x(t)

200

4 Laplace Transforms

(cid:5)

(cid:6)

Þ (cid:6)

1 s þ 2

1 s þ 3

L(cid:2)1

(

e(cid:2)2tu tð Þ for right sided

(

(cid:2)e(cid:2)2tu (cid:2)t Þ for left sided e(cid:2)3tu tð Þ for right sided

(cid:2)e(cid:2)3tu (cid:2)t

Þ for left sided

Hence, if x(t) is chosen as,

x tð Þ ¼ (cid:2)e(cid:2)2tu (cid:2)t

Þ (cid:2) e(cid:2)3tu tð Þ

The ROC is as shown in Figure 4.7(c)

Example 4.19 Determine x(t) ﬁrst for the following and verify the initial and ﬁnal value theorems.

(i) X sð Þ ¼ 1 sþ4 (ii) X sð Þ ¼ sþ5 Þ sþ4 Þ ð

sþ3 ð

Solution (i) x tð Þ ¼ L(cid:2)1

n o 1 sþ4

¼ e(cid:2)4tu tð Þ

x 0þð

Þ ¼ lim t!0

e(cid:2)4t ¼ 1:

x 0þð

Þ ¼ lim s!1

sX sð Þ ¼ lim s!1

s s þ 4

¼ lim s!1

1 þ

4 s e(cid:2)4t ¼ 0:

x 1ð

Þ ¼ lim t!1

1 þ

4 1

1 1 þ 0

¼ 1:

x 1ð

Þ ¼ lim s!0

sX sð Þ ¼ lim s!0

s s þ 4

¼ lim s!0

(ii) x tð Þ ¼ L(cid:2)1

sþ5 Þ sþ4 ð

sþ3 ð

¼ L(cid:2)1

n o 2 sþ3

(cid:2) L(cid:2)1

1 þ

4 0 n o 1 sþ4

1 1 þ 1

¼ 0:

1 þ

4 0

¼ 2e(cid:2)3tu tð Þ (cid:2) e(cid:2)4tu tð Þ

4.7 Inverse Laplace Transform by Partial Fraction Expansion Using MATLAB

201

(cid:11)

(cid:12)

x 0þð

Þ ¼ lim t!0 (cid:5)

2e(cid:2)3t (cid:2) e(cid:2)4t (cid:6)

¼ 1:: (cid:5)

(cid:6)

(cid:5)

(cid:2) lim s!1

(cid:6)

s s þ 4

2s s þ 3

¼ lim s!1

x 0þð

Þ ¼ lim s!1

sX sð Þ ¼ lim s!1

¼ lim s!1

s þ 5 Þ s þ 4 Þ ð 1

s þ 3 ð 2

¼ 2 (cid:2) 1 ¼ 1:

(cid:2) lim s!1

1 þ

3 s Þ ¼ lim t!1 (cid:5)

(cid:11)

1 þ

4 s 2e(cid:2)3t (cid:2) e(cid:2)4t (cid:6)

(cid:12)

x 1ð

¼ 0: (cid:5)

x 1ð

Þ ¼ lim s!0

sX sð Þ ¼ lim s!0

¼ lim s!0

s þ 5 Þ s þ 4 ð 1

s þ 3

ð 2

(cid:2) lim s!0

1 þ

3 s

4 s

(cid:6)

(cid:5)

(cid:2) lim s!0

(cid:6)

s s þ 4

2s s þ 3

¼ lim s!0

¼ 0 (cid:2) 0 ¼ 0:

4.7

Inverse Laplace Transform by Partial Fraction Expansion Using MATLAB

The MATLAB command residue can be used to ﬁnd the inverse transform using the power series expansion. To ﬁnd the partial fraction decomposition of Y(s), we must ﬁrst enter the numerator polynomial coefﬁcients and the denominator polynomial coefﬁcients as vectors.

The following MATLAB statement determines the residue (r), poles (p), and

direct terms (k) of the partial fraction expansion of H(s).

k; p; const ½

(cid:5) ¼ residue N; Dð

Þ;

where N is the vector of the numerator polynomial coefﬁcients in decreasing order and the vector D contains the denominator polynomial coefﬁcients in decreasing order.

Example 4.20 Find the inverse Laplace transform of the following using MATLAB

X sð Þ ¼

s þ 1 s2 þ 5s þ 6

Solution The following MATLAB statements are used to ﬁnd the Laplace trans- form of given X(s) ½ ½

(cid:5); % coefﬁcients of the numerator polynomial in decreasing order (cid:5); % coefﬁcients of the denominator polynomial in decreasing

N ¼ D ¼ 1

1 5 6

order

[k, p, const] ¼ residue (N and D); % computes residues, poles, and constants Execution of the above statements gives the following output

4 Laplace Transforms

202

k ¼

2.0000 (cid:2)1.0000

p ¼

(cid:2)3.0000 (cid:2)2.0000

Thus, the partial fraction decomposition of X(s) is

X sð Þ ¼

2 s þ 3

(cid:2)

1 s þ 2

After getting the partial fraction expansion of X(s), the following MATLAB

statements are used to obtain x(t), that is, the inverse Laplace transform of X(s).

syms s t X=2/(s+3)-1/(s+2); ilaplace(X)

Execution of the above three MATLAB statements gives

x tð Þ ¼ 2e(cid:2)3t (cid:2) e(cid:2)2t

4.8 Analysis of Continuous-Time LTI Systems Using

the Laplace Transform

4.8.1 Transfer Function

It was stated in Chapter 2 that a continuous time LTI system can be completely characterized by its impulse response h(t). The output signal y(t) of a LTI system and the input signal x(t) are related by convolution as

y tð Þ ¼ h tð Þ ∗ x tð Þ

By using the convolution property, we get

Y sð Þ ¼ H sð ÞX sð Þ

ð4:81Þ

ð4:82Þ

indicating the Laplace transform of the output signal y(t) is the product of the Laplace transforms of the impulse response h(t) and the input signal x(t). The transform H(s) is called the transfer function or the system function and expressed as

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

203

Figure 4.8 Sallen-Key low-pass ﬁlter circuit

H sð Þ ¼

Y sð Þ X sð Þ

ð4:83Þ

The roots of the denominator of the transfer function are called poles. The roots of the numerator are called zeros. The places where the transfer function is inﬁnite (the poles) determine the region of convergence.

Example 4.21 Obtain the system function of the following Sallen-Key low-pass ﬁlter circuit.

Solution For the circuit shown in Figure 4.8, the following relations can be established:

Vi (cid:2) V1 ¼ r1Ir1 Ic2 ¼ sc2V2; Ic1 ¼ sc1 V1 (cid:2) V0 Ir1 ¼ Ir2 þ Ic1

; V1 (cid:2) V2 ¼ r2Ir2 Þ; ; V2 ¼ V0

ð ; Ir2 ¼ Ic2

;

Rewriting the current node equations, we get

Vi (cid:2) V1 r1

V1 (cid:2) V0 r2

þ sc1 V1 (cid:2) V0 ð

Which can be rewritten as

Vir2 ¼ r1 þ r2 þ r1r2sc1

ÞV1 (cid:2) r1 þ r1r2sc1

ÞV0

¼ sc2V0;

V1 (cid:2) V0 r2 ð

V1 ¼ 1 þ r2sc2

ÞV0

Substituting the above Eq. for V1, the input-output relation can be written as

Vir2 ¼ r1 þ r2 þ r1r2sc1 (cid:8)

ð ½ (cid:9) (cid:7)

Þ 1 þ r2sc2 ð

Þ (cid:2) r1 (cid:2) r1r2sc1 (cid:10)

(cid:5)V0

Vi ¼ 1 þ

þ r1sc1

1 þ r2sc2

Þ (cid:2)

(cid:2) r1sc1

r1 r2

204

Thus,

4 Laplace Transforms

V0 sð Þ Vi sð Þ

(cid:9)

(cid:7)

1 þ

r1 r2

(cid:8)

þ r1sc1

1 þ r2sc2 ð

Þ (cid:2)

(cid:10)

r1 r2

(cid:2) r1sc1

1 Þs þ s2c2c1r2r1 1 þ c2 r1 þ r2 ð

Hence, the system function is given by

H sð Þ ¼

1 r1r2c1c2

s2 þ

r1 þ r2 r1r2c1

s þ

1 r1r2c1c2

4.8.2 Stability and Causality

Stabile LTI System A continuous-time LTI system is stable if and only if the impulse response is absolutely integrable, that is,

(cid:2)1

j h tð Þ j dt < 1:

ð4:84Þ

The Laplace transform of the impulse response is known as the system function,

which can be written as

H sð Þ ¼

(cid:2)1

h tð Þe(cid:2)stdt

ð4:85Þ

A continuous-time LTI system is stable if and only if the transfer function has ROC that includes the imaginary axis (the line in complex where the real part is zero).

Causal LTI System A continuous time LTI system is causal if its output y(t) depends only on the current and past input x(t) but not the future input. Hence, h(t) ¼ 0 for t < 0.

For causal system, the system function can be written as

HðsÞ ¼

hðtÞe(cid:2)stdt

ð4:86Þ

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

205

s plane

Figure 4.9 ROC of a causal LTI system

If s ¼ σ + jΩ and h(t)e(cid:2)σt being absolutely integrable for convergence of

H(s) leads to the following convergence condition,

j h tð Þe(cid:2)σt j dt < 1

ð4:87Þ

Any large value of σ satisﬁes the above equation. Thus, the ROC is the region to the right of a vertical line that passes through the point ℜe(s) ¼ σ as shown in Figure 4.9.

In particular, if H(s) is rational, H sð Þ ¼ N sð Þ

D sð Þ, then the system is causal if and only if its ROC is the right-sided half plane to the right of the rightmost pole and the order of numerator N(s) is no greater than that of the denominator D(s), so that the ROC is a right-sided plane without any poles (even at s ! 1).

Stable and Causal LTI System As the ROC of a causal system is to the right of the rightmost pole and for a stable system, the rightmost pole should be in the left half of the s-plane and should include the jΩ axis; all the poles of a system should lie in the left half of the s-plane (the real parts of all poles are negative, ℜe(sp) < 0 for all sp) for a system to be causal and stable as shown in Figure 4.10.

Stable and Causal Inverse LTI System For a causal stable system, the poles must lie in the left half of the s-plane. But it is known that the poles of the inverse system are zeros of the original system. Therefore, the zeros of the original system should be in the left half of the s-plane.

206

4 Laplace Transforms

s plane

Figure 4.10 ROC of a stable and causal LTI system

Example 4.22 Consider a LTI system with the system function

H sð Þ ¼

s (cid:2) 1 s2 (cid:2) s (cid:2) 6

Show the pole-zero locations of the system and ROCs for the following:

(a) the system is causal (b) the system is stable, noncausal (c) the system is neither causal nor stable

Solution

(a) The system function can be rewritten as

H sð Þ ¼

s (cid:2) 1 Þ s (cid:2) 3 ð

s þ 2

Since the ROC of a causal system is to the right of the rightmost pole, the

pole-zero plot and ROC of the system are shown in Figure 4.11.

(b) For a stable system, the ROC should include the imaginary axis. The pole-zero

plot and the ROC are shown in Figure 4.12.

(c) Since the system is neither causal nor stable, the ROC should not include the imaginary axis and not to the right of the rightmost pole. Hence, the pole-zero plot and the ROC are shown in Figure 4.13.

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

207

-2

s-plane

Figure 4.11 Pole-zero plot and ROC of the causal system

s-plane

-2

Figure 4.12 Pole-zero plot and ROC of the noncausal, stable system

4.8.3 LTI Systems Characterized by Linear Constant

Coefﬁcient Differential Equations

The system function for a system characterized by a linear constant coefﬁcient equation can be obtained by exploiting the properties of the Laplace transforms. The Laplace transform transforms a differential equation into an algebraic equation in the s-domain making it easy to ﬁnd the time-domain solution of the differential equation.

208

4 Laplace Transforms

s-plane

-2

Figure 4.13 Pole-zero plot and ROC of the system neither causal nor stable

Consider a general linear constant coefﬁcient differential equation of the form

dny dtn þ an(cid:2)1

¼ bm

d n(cid:2)1 Þy ð Þ þ (cid:6) (cid:6) (cid:6): þ a2 dt n(cid:2)1 ð dmx dtm þ bm(cid:2)1

d2y dt2 þ a1

dy dt

þ a0y

d m(cid:2)1 Þx ð Þ þ (cid:6) (cid:6) (cid:6): þ b2 dt m(cid:2)1 ð

d2x dt2 þ b1

dx dt

þ b0x

ð4:88Þ

Taking the Laplace transform of both sides of equation repeated use of the

differentiation property and linearity property, we obtain

ðansn þ an(cid:2)1sðn(cid:2)1Þ þ (cid:6) (cid:6) (cid:6): þ a2s2 þ a1s þ a0ÞYðsÞ

¼ ðbmsm þ bm(cid:2)1sðm(cid:2)1Þ þ (cid:6) (cid:6) (cid:6): þ b2s2 þ b1s þ b0ÞXðsÞ

Thus, the system function is given by

HðsÞ ¼

YðsÞ XðsÞ

bmsm þ bm(cid:2)1sðm(cid:2)1Þ þ (cid:6) (cid:6) (cid:6): þ b2s2 þ b1s þ b0 ansn þ an(cid:2)1sðn(cid:2)1Þ þ (cid:6) (cid:6) (cid:6): þ a2s2 þ a1s þ a0

ð4:89Þ

ð4:90Þ

The system function is rational for a system characterized by a differential equation with the roots of the numerator polynomial as zeros and the roots of the denominator polynomial as poles.

Eq. (4.90) does not specify any ROC since a differential equation by itself does not constrain any region of convergence. However, with the additional knowledge of stability and causality, the ROC can be speciﬁed.

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

209

Example 4.23 Consider a continuous LTI system described by the following dif- ferential equation d2y

dt2 (cid:2) dy

dt (cid:2) 6y ¼ x.

(a) Determine the system function (b) Determine h(t) for each of the following:

(i) the system is stable, noncausal (ii) the system is causal (iii) the system is neither causal nor stable

Solution Taking the Laplace transform of both sides of the given differential equation, we obtain

s2Y sð Þ (cid:2) sY sð Þ (cid:2) 6Y sð Þ ¼ X sð Þ

The above relation can be rewritten as (cid:12)

(cid:11)

s2 (cid:2) s (cid:2) 6

Y sð Þ ¼ X sð Þ

Now, the system function is given by

H sð Þ ¼

Y sð Þ X sð Þ

1 s2 (cid:2) s (cid:2) 6

The pole-zero plot for the system function is shown in Figure 4.14.

(b) The partial fraction expansion of H(s) yields

H sð Þ ¼

1 5 s (cid:2) 3 ð

(cid:2)

1 5 s þ 2 ð

-2

Figure 4.14 Pole-zero plot of the system function

210

4 Laplace Transforms

(i) For H(s) to be stable, noncausal, the ROC is (cid:2)2 < ℜe(s) < 3.

Hence, h tð Þ ¼ (cid:2)1

5 e3tu (cid:2)t ð

Þ (cid:2) 1

5 e(cid:2)2tu tð Þ

(ii) For the system to be causal, the ROC is ℜe(s) > 3.

Therefore, h tð Þ ¼ 1

5 e(cid:2)2tu tð Þ. (iii) For the system to be neither causal nor stable, the ROC is ℜe(s) < (cid:2) 2.

5 e3tu tð Þ (cid:2) 1

Hence, h tð Þ ¼ (cid:2)1

5 e3tu (cid:2)t ð

Þ þ 1

5 e(cid:2)2tu (cid:2)t

4.8.4 Solution of linear Differential Equations Using Laplace

Transform

The stepwise procedure to solve a linear differential equation is as follows:

Step 1: Take Laplace transform both sides of the equation. Step 2: Simplify the algebraic equation obtained for Y(s) in the s-domain. Step 3: Find the inverse transform of Y(s) to obtain y(t), the solution of the

differential equation.

Example 4.24 Find the solution of the following differential equation using Laplace transform:

d2y dt2 (cid:2) 5

dy dt

þ 6y ¼ 0

y 0ð Þ ¼ 2, _y 0ð Þ ¼ 1:

Solution

Step 1: Laplace transform both sides of given differential equation is

s2 (cid:2) 2s (cid:2) 1 (cid:2) 5 sY sð Þ (cid:2) 2

Þ þ 6Y sð Þ ¼ 0

Step 2: Simplifying the expression for Y(s),

s2 (cid:2) 5s þ 6

ÞY sð Þ (cid:2) 2s (cid:2) 1 þ 10 ¼ 0

Y sð Þ ¼

2s (cid:2) 9 s2 (cid:2) 5s þ 6

2s (cid:2) 9 Þ s (cid:2) 2 ð

s (cid:2) 3

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

211

Step 3: Expanding Y(S) using partial fraction expansion,

Y sð Þ ¼

k1 s (cid:2) 3

k2 s (cid:2) 2

k1 s (cid:2) 2 ð

s (cid:2) 3

Þ þ k2 s (cid:2) 3 ð Þ s (cid:2) 2 ð

Comparing the numerator polynomial with the numerator polynomial of Y(s) of

step 2, we get

Solving for k1 and k2,

Thus,

k1 þ k2 ¼ 2

(cid:2)2k1 (cid:2) 3k2 ¼ (cid:2)9

k1 ¼ (cid:2)3, k2 ¼ 5:

Y sð Þ ¼

(cid:2)3 s (cid:2) 3

5 s (cid:2) 2

Inverse transform of Y(s) gives the solution of the differential equation as

y tð Þ ¼ (cid:2)3e3t þ 5e2t

Example 4.25 Find the solution of the following differential equation using Laplace transform

dy dt

þ y ¼ 2te(cid:2)t

y 0ð Þ ¼ (cid:2)2

Solution

Step 1: Laplace transform both sides of given differential equation is

sY sð Þ (cid:2) (cid:2)2ð

Þ þ Y sð Þ ¼

2 s þ 1

Þ2

Step 2: Simplifying the expression for Y(s),

212

4 Laplace Transforms

s þ 1

ÞY sð Þ þ 2 ¼

s þ 1

Þ2

s þ 1

ÞY sð Þ ¼

2 s þ 1

Y sð Þ ¼

s þ 1

Þ2 (cid:2) 2 2 Þ s þ 1 ð

Y sð Þ ¼

2 s þ 1

Þ3 (cid:2) (cid:2)2s2 (cid:2) 4s Þ3 s þ 1 ð

2 s þ 1

Þ2 (cid:2) 2 s þ 1

Step 3: Expanding Y(S) using partial fraction expansion,

Y sð Þ ¼

k11 s þ 1

k12 s þ 1

k11 s2 þ 2s þ 1 ð

Þ3

k13 s þ 1

Þ2 þ Þ þ k12 s þ 1 ð Þ3 s þ 1

Þ þ k13

Comparing the numerator polynomial with the numerator polynomial of Y(s) of

step 2, we get

k11 þ k12 þ k13 ¼ 0

2k11 þ k12 ¼ (cid:2)4

k11 ¼ (cid:2)2

Solving for k11, k12, and k13, k11 ¼ (cid:2)2, k12 ¼ 0, and k13 ¼ 2. Thus,

Y sð Þ ¼

(cid:2)2 s þ 1

s þ 1

Þ3

Inverse transform of Y(s) gives the solution of the differential equation as

y tð Þ ¼ (cid:2)2e(cid:2)t þ t2e(cid:2)t

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

213

Figure 4.15 Series RLC circuit

Example 4.26 Consider the following RLC circuit with R ¼ 5 ohms, L ¼ 1h, and C ¼ 0.25 f.

(a) Determine the differential equation relating Vin and Vc. (b) Obtain Vc tð Þ using Laplace transform for Vin tð Þ ¼ e(cid:2)tu tð Þ with Vc 0ð Þ ¼ 1,

_V c

0ð Þ ¼ 2 (Figure 4.15).

Solution (a) By applying Kirchhoff’s voltage law, we can arrive at the following

differential equation:

di dt

þ Ri þ

1 C

idt ¼ vin

It is known that i ¼ C dV c

dt , hence the above differential equation can be rewritten

d2vc dt2 þ RC For given values of R, L, and C, the differential equation becomes

þ vc ¼ vin

dvc dt

d2vc dt2 þ 5

dvc dt

þ 4vc ¼ 4vin

(b) For given vin, d2vc

dt2 þ 5dvc

dt þ 4vc ¼ 4e(cid:2)tu tð Þ

Laplace transform both sides of given differential equation is

(cid:11)

s2Vc sð Þ (cid:2) s (cid:2) 2 þ 5 sVc sð Þ (cid:2) 1

Þ þ 4Vc sð Þ ¼

4 s þ 1

Simplifying the expression for vc(s)

214

4 Laplace Transforms

(cid:11) s2Vc sð Þ (cid:2) s (cid:2) 2 þ 5 sVc sð Þ (cid:2) 1

Þ þ 4vc sð Þ ¼

4 s þ 1

s2 þ 5s þ 4 ð

ÞVc sð Þ (cid:2) s (cid:2) 2 (cid:2) 5 ¼

4 s þ 1

s2 þ 5s þ 4 ð

ÞVc sð Þ ¼

þ s þ 7

4 s þ 1 s2 þ 8s þ 11

Vc sð Þ ¼

s2 þ 5s þ 4

Þ s þ 1 ð

Vc sð Þ ¼

s2 þ 8s þ 11 Þ2 s þ 4 s þ 1 ð

Expanding vc(s) using partial fraction expansion

Vc sð Þ ¼

k11 s þ 1

k12 s þ 1

Þ2 þ

k2 s þ 4

Solving for k11, k12, and k2, we obtain k11 ¼ 14 Thus,

3, k2 ¼ (cid:2)5 9.

9 , k12 ¼ 4

Vc sð Þ ¼

14 9 s þ 1 ð

4 3 s þ 1 ð

Þ2 (cid:2)

5 9 s þ 4 ð

Inverse transform of Vc(s) gives V c tð Þ as

vc tð Þ ¼

14 9

e(cid:2)t þ

4 3

te(cid:2)t (cid:2)

5 9

e(cid:2)4t

Example 4.27 Consider the following parallel RLC circuit with R¼1 ohm, L¼1 h.

(a) Determine the differential equation relating Is and IL. (b) Obtain zero-state response for IL(t) using Laplace transform for Is(t) ¼ e(cid:2)3tu(t). (c) Obtain zero-input response for IL(t) using Laplace transform with IL(0) ¼ 1

(Figure 4.16).

Figure 4.16 Parallel RLC circuit

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

215

Solution (a) The Is and IL are related by the following differential equation

(b) For the given Is,

dIL dt

þ IL ¼ IS

dIL dt

þ IL ¼ e(cid:2)3tu tð Þ

Applying the unilateral Laplace transform to the above differential equation, we

obtain

sIL sð Þ (cid:2) IL 0ð Þ þ IL sð Þ ¼

1 s þ 3

Since IL(0) ¼ 0 for zero state, s þ 1

ÞIL sð Þ ¼ 1 sþ3

IL sð Þ ¼

1 Þ s þ 3 ð

s þ 1

By using partial fraction expansion, IL(s) can be expanded as

IL sð Þ ¼

1 2 s þ 1 ð

(cid:2)

1 2 s þ 3 ð

The inverse unilateral Laplace transform gives

IL tð Þ ¼

1 2

e(cid:2)tu tð Þ (cid:2)

1 2

e(cid:2)3tu tð Þ

solution of the following differential equation for zero-input response

dIL dt

þ IL ¼ 0

IL 0ð Þ ¼ 1

The unilateral Laplace transform of the above differential equation is

Thus,

sIL sð Þ (cid:2) 1 þ IL sð Þ ¼ 0

IL sð Þ ¼

1 s þ 1

216

4 Laplace Transforms

The inverse transform of IL(s) is zero-input response given by

IL tð Þ ¼ e(cid:2)tu tð Þ

4.8.5 Solution of Linear Differential Equations Using

Laplace Transform and MATLAB

Example 4.28 Find the solution of linear differential equation considered in Exam- ple 4.24 using MATLAB Solution The following MATLAB statements are used to ﬁnd the solution of the differential equation considered in Example 4.24

syms s t Y Y1 = s*Y - 2;%Laplace transform of

with y(0)=2

Y2 = sY1 - 1; %Laplace transform of Sol = solve(Y2 - 5Y1 + 6*Y, Y);%Y(s) y = ilaplace(Sol,s,t);%inverse Laplace transform of Y(s)

with

Execution of the above MATLAB statements gives the solution of the differential

equation as

y tð Þ ¼ (cid:2)3e3t þ 5e2t

Example 4.29 Find the solution of linear differential equation considered in Exam- ple 4.25 using MATLAB Solution The following MATLAB statements are used to ﬁnd the solution of the differential equation considered in Example 4.25

syms s t Y f = 2texp(-t);%input signal F = laplace(f,t,s);% finds Laplace transform of input with y(0)=-2 Y1 = s*Y + 2;% Laplace transform of Sol = solve(Y1 + Y-F, Y);%Y(s) y = ilaplace(Sol,s,t);% inverse of Y(s)

Execution of the above MATLAB statements gives the solution of the differential

equation as

y tð Þ ¼ (cid:2)2e(cid:2)t þ t2e(cid:2)t

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

217

4.8.6 System Function for Interconnections of LTI Systems

Series Combination of Two LTI Systems Impulse response of the series combination of two LTI systems is

h tð Þ ¼ h1 tð Þ ∗ h2 tð Þ

ð4:91Þ

and from convolution property of the Laplace transform, the associated system function is (Figure 4.17)

H sð Þ ¼ H1 sð ÞH2 sð Þ

Parallel Combination of Two LTI Systems Impulse response of the parallel combination of two LTI systems is

h tð Þ ¼ h1 tð Þ þ h2 tð Þ

ð4:92Þ

ð4:93Þ

and from linearity property of the Laplace transform, the associated system function is (Figure 4.18)

H sð Þ ¼ H1 sð Þ þ H2 sð Þ

ð4:94Þ

Figure 4.17 Series combination of two LTI systems

Figure 4.18 Parallel combination of two LTI systems

218

4 Laplace Transforms

4.9 Block-Diagram Representation of System Functions

in the S-Domain

Consider the system function

H sð Þ ¼

Y sð Þ X sð Þ

bmsn þ bm(cid:2)1s n(cid:2)1 ansn þ an(cid:2)1s n(cid:2)1

Þ þ (cid:6) (cid:6) (cid:6): þ b2s2 þ b1s þ b0 Þ þ (cid:6) (cid:6) (cid:6): þ a2s2 þ a1s þ a0

ð4:95Þ

Let us deﬁne the following basic elements for addition, multiplication, differen-

tiation, and integration in the s-domain (Figure 4.19)

The block-diagram representation of the above system function can be obtained as the interconnection of these basic elements similar to the block-diagram repre- sentation of differential equations in the time domain carried out in Section 2.6 of Chapter 2. The block-diagram representation of the system function is shown in Figure 4.20.

Example 4.30 Is the system represented by the following block-diagram stable? (Figure 4.21)

Solution Let the signal at the bottom node of the block diagram be denoted by E(s). Then we have the following relations

Figure 4.19 Block- diagram representation basic elements (a) adder, (b) multiplier, (c) differentiator, (d) integrator

(a)

(b)

Differentiation

(c)

Integration

(d)

4.9 Block-Diagram Representation of System Functions in the S-Domain

219

X(s)

Y(s)

bm-1

1/an

-an-1

-a1

1/s

-a0

Figure 4.20 Block-diagram representation of the system function

Figure 4.21 Block-diagram representation of a second-order system function

X sð Þ (cid:2) 2sE sð Þ (cid:2) E sð Þ ¼ s2E sð Þ s2E sð Þ (cid:2) sE sð Þ (cid:2) 6E sð Þ ¼ Y sð Þ

Eliminating the auxiliary signal E(s), we get

H sð Þ ¼

Y sð Þ X sð Þ

s2 (cid:2) s (cid:2) 6 s2 þ 2s þ 1

From the system function H(s) found in the previous part, we see that the poles of the system are at s ¼ (cid:2)1. Since the system is given to be causal, and the rightmost pole of the system is left of the imaginary axis, the system is stable.

220

4 Laplace Transforms

4.10 Solution of State-Space Equations Using Laplace

Transform

For convenience, the state-space equations from Chapter 2 are repeated here:

_X tð Þ ¼ A X tð Þ þ b℧ tð Þ

y tð Þ ¼ cX tð Þ

Taking Laplace transform both sides of Eq. (4.96), we obtain

SX Sð Þ (cid:2) X 0ð Þ ¼ AX sð Þ þ b℧ Sð Þ

(cid:12)

Eq. (4.98) can be rewritten as

SI (cid:2) A

(cid:5)X sð Þ ¼ X 0ð Þ þ b℧ Sð Þ

where I is the identity matrix. From Eq. (4.99), we get

X sð Þ ¼ SI (cid:2) A ½

(cid:5)(cid:2)1 X 0ð Þ þ b℧ Sð Þ

(cid:5)

X sð Þ ¼ SI (cid:2) A ½

(cid:5)(cid:2)1X 0ð Þ þ SI (cid:2) A

(cid:5)(cid:2)1b℧ Sð Þ

Taking inverse Laplace transform both sides of Eq. (4.100b) yields

L(cid:2)1 X sð Þ ½

h (cid:5) ¼ L(cid:2)1 SI (cid:2) A ½

(cid:5)(cid:2)1X 0ð Þ

h þ L(cid:2)1 SI (cid:2) A ½

(cid:5)(cid:2)1b℧ Sð Þ

h L(cid:2)1 SI (cid:2) A ½

(cid:5)(cid:2)1X 0ð Þ

¼ eAtX 0ð Þ

By using convolution theorem, we obtain

ð4:96Þ ð4:97Þ

ð4:98Þ

ð4:99Þ

ð4:100aÞ

ð4:100bÞ

ð4:101Þ

ð4:102Þ

L(cid:2)1 SI (cid:2) A ½

(cid:5)(cid:2)1b℧ Sð Þ

eA t(cid:2)τ ð

Þb℧ τð Þ dτ

ð4:103Þ

Thus,

L(cid:2)1 X sð Þ ½

(cid:5) ¼ X tð Þ ¼ eAt X 0ð Þ þ

eA t(cid:2)τ ð

Þb℧ τð Þ dτ

ð4:104Þ

Example 4.31 Consider the electrical circuit given in Example 2.36. Find Vc tð Þ if Vs tð Þ ¼ u tð Þ under an initially relaxed condition.

4.10 Solution of State-Space Equations Using Laplace Transform

221

Solution

”

½sI (cid:2) A(cid:5) ¼

(cid:2)

(cid:2)1 1

(cid:2)1 (cid:2)1 ”

½sI (cid:2) A(cid:5)(cid:2)1 ¼

1 ðs þ 1Þ2 þ 1

eAt ¼ L(cid:2)1½½sI (cid:2) A(cid:5)(cid:2)1(cid:5) ¼ e(cid:2)t

s þ 1

(cid:2)1

”

cost

s þ 1 (cid:2)1

s þ 1

sint

cost

XðtÞ ¼ eAtXð0Þ þ

0 eAðt(cid:2)τÞ

VsðτÞdτ

(cid:2)sint ” # 0

Since the circuit is initially relaxed, eAtX(0) ¼ 0. Therefore,

X tð Þ ¼

eA t(cid:2)τ ð

(cid:9) (cid:10) Þ 0 1

Vs τð Þdτ

Since Vs tð Þ ¼ u tð Þ

X tð Þ ¼

”

0 ð

”

Þ cos t (cid:2) τ ð Þ Þ sin t (cid:2) τ ð

Þ sin t (cid:2) τ ð Þ Þ cos t (cid:2) τ ð

e(cid:2) t(cid:2)τ ð (cid:2)e(cid:2) t(cid:2)τ ð e(cid:2) t(cid:2)τ ð e(cid:2) t(cid:2)τ ð Ð t

Vc tð Þ ¼ x2 tð Þ ¼ ð

0 e(cid:2) t(cid:2)τ

Þ cos t (cid:2) τ ð ð

Þ dτ

e(cid:2) t(cid:2)τ ð e(cid:2) t(cid:2)τ ð

Þ sin t (cid:2) τ ð Þ Þ cos t (cid:2) τ ð

” # 0

dτ

e(cid:2) t(cid:2)τ ð

Þ cos t (cid:2) τ ð

Þdτ ¼

e(cid:2) t(cid:2)τ ð

Þ cost cos τ þ sint sin τ ð

Þ dτ

e(cid:2) t(cid:2)τ ð

Þcost cos τ dτ þ ð

e(cid:2)tcost eτ cos τ dτ þ

e(cid:2) t(cid:2)τ ð

Þsint sin τ dτ

e(cid:2)tsint eτ sin τ dτ

integration by parts gives ð

(cid:9)

e(cid:2)tcost eτ cos τ dτ ¼ e(cid:2)tcost eτ cos τj t 0 þ Ð

(cid:16)

¼ e(cid:2)tcost etcost (cid:2) 1 þ eτ sin τj t

0 (cid:2)

(cid:10)

eτ sin τ dτ (cid:17)

0 eτ cos τ dτ

This equation can be written as

222

e(cid:2)tcost eτ cos τ dτ ¼ cos 2t þ sint cost (cid:2) e(cid:2)t cost

4 Laplace Transforms

e(cid:2)tsint eτ sin τ dτ

0 e(cid:2)tcost eτ cos τ dτ ¼ (cid:9) ¼ e(cid:2)tsint eτ sin τj t (cid:9)

0 (cid:2)

cos 2t þ sint cost (cid:2) e(cid:2)tcost 2 (cid:10)

eτ cos τ dτ

(cid:10)

eτ sin τ dτ

¼ e(cid:2)tsint etsint (cid:2) eτ cos τj t

0 (cid:2)

which can be written as

2 Ð

0 e(cid:2)tsint eτ sin τ dτ ¼ sin 2t (cid:2) sint cost þ e(cid:2)tsint sin 2t (cid:2) sint cost þ e(cid:2)tsint Þ 2

0 e(cid:2)tsint eτ sin τ dτ ¼

Hence,

Vc tð Þ ¼ x2 tð Þ ¼

cos 2t þ sint cost (cid:2) e(cid:2)tcost 2

sin 2t (cid:2) sint cost þ e(cid:2)tsint 2

1 2

1 þ e(cid:2)tsint (cid:2) e(cid:2)tcost ð

Þ, t > 0

4.11 Problems

Find the Laplace transforms of the following:

(i) x(t) ¼ e(cid:2)2tu(t) þ e3tu((cid:2)t) (ii) x(t) ¼ etu(t) þ e(cid:2)3tu((cid:2)t)

(cid:3) hint : sin 2t

(cid:4) Þ

Find the Laplace transform of sin 2t
Find the Laplace transform of cos 4t(cid:2) cos 5t
Show that the ROC for the Laplace transform of a noncausal signal is the region

t ¼

1 2 1(cid:2) cos 2t ð t

to the left of a vertical line in the s-plane.

Find Laplace transform of a periodic signal with period T, that is, x(t+T) ¼ x(t).
By ﬁrst determining x(t), verify the ﬁnal value theorem for the following with

comment

X sð Þ ¼

1 s2 þ 1

4.11 Problems

223

The transfer function of an LTI system is

1 s þ α Find the impulse response and region of convergence and the value of α for

H sð Þ ¼

the system to be causal and stable.

Consider a continuous LTI system described by the following differential

equation

d3y dt3 þ 6

d2y dt2 þ 11

dy dt

þ 6y ¼ x

(a) Determine the system function (b) Determine h(t) for each of the following:

(i) The system is causal (ii) The system is stable (iii) The system is neither causal nor stable

Find the solution of the following differential equation using Laplace transform

d2y dt2 (cid:2) 2

dy dt

þ 2y ¼ cos t

y 0ð Þ ¼ 1, _y 0ð Þ ¼ 0:

Consider the following RC circuit with R¼2 ohms, C¼0.5 f.

(a) Determine the differential equation relating Vi and Vc. (b) Obtain Vc tð Þ using Laplace transform for Vi tð Þ ¼ e(cid:2)3tu tð Þ with Vc 0ð Þ ¼ 1.

Consider the following RLC circuit.

Obtain y(t) using Laplace transform for x(t)¼u(t).

224

4 Laplace Transforms

Determine the differential equation characterizing the system represented by the

following block diagram

X(s)

1/s

Y(s)

-5

-7

1/s

-12

Consider the following state-space representation of a system. Determine the

system output y(t) with the initial state condition X 0½ (cid:5) ¼

3 (cid:2)3 (cid:2)47 3

7 7 5e(cid:2)t

6 6 4

_x 1 tð Þ _x 2 tð Þ _x 3 tð Þ

7 7 5 ¼

6 6 4

7 7 5

6 6 4

x1 tð Þ

x2 tð Þ

7 7 5 þ

6 6 4

(cid:2)1 (cid:2)3 (cid:2)3

y tð Þ ¼ 1

0 0

6 6 (cid:5) 4

x3 tð Þ 3

7 7 5

x1 tð Þ

x2 tð Þ

x3 tð Þ

Further Reading

225

4.12 MATLAB Exercises

Write a MATLAB program for magnitude response of Sallen-Key low-pass ﬁlter.
Verify the solution of problem 8 using MATLAB.
Verify the solution of problem 9 using MATLAB.

Further Reading

Doetsch, G.: Introduction to the theory and applications of the Laplace transformation with a

table of Laplace transformations. Springer, New York (1974)

LePage, W.R.: Complex variables and the Laplace transforms for engineers. McGraw-Hill,

New York (1961)

Oppenheim, A.V., Willsky, A.S.: Signals and systems. Prentice-Hall, Englewood Cliffs (1983)
Hsu, H.: Signals and systems, 2nd edn. Schaum’s Outlines, Mc Graw Hill (2011)
Kailath, T.: Linear systems. Prentice-Hall, Englewood Cliffs (1980)
Zadeh, L., Desoer, C.: Linear system theory. McGraw-Hill, New York (1963)

Chapter 5 Analog Filters

Filtering is an important aspect of signal processing. It allows desired frequency components of a signal to pass through the system without distortion and suppresses the undesired frequency components. One of the most important steps in the design of a ﬁlter is to obtain a realizable transfer function H(s), satisfying the given frequency response speciﬁcations. In this chapter, the design of analog low-pass ﬁlters is ﬁrst described. Second, frequency transformations for transforming analog low-pass ﬁlter into band-pass, band-stop, or high-pass analog ﬁlters are considered. The design of analog ﬁlters is illustrated with numerical examples. Further, the design of analog ﬁlters using MATLAB is demonstrated with a number of examples. Also, the design of special ﬁlters by pole and zero placement is illustrated with examples.

5.1

Ideal Analog Filters

An ideal ﬁlter passes a signal for one set of frequencies and completely rejects for the rest of the frequencies.

Low-Pass Filter The frequency response of an ideal analog low-pass ﬁlter HLP (Ω) that passes a signal for Ω in the range –Ωc (cid:1) Ωc can be expressed by

HLP Ωð Þ ¼

(cid:1)

1, Ωj 0, Ωj

j (cid:1) Ωc j > Ωc

ð5:1Þ

The frequency Ωc is called the cutoff frequency.

The impulse response of the ideal low-pass ﬁlter corresponds to the inverse

Fourier transform of the frequency response shown in Figure 5.1.

227

228

5 Analog Filters

Stop band Pass band Stop band

Figure 5.1 Frequency response of ideal low-pass ﬁlter

Hence,

h tð Þ ¼

ðΩc

(cid:3)Ωc

1 2π

ejΩctdΩ ¼

sin Ωct πt

sinc function can be deﬁned as

sinc xð Þ ¼

sin πx πx

therefore from sinc function we can express Eq. (5.2) as

Thus,

sin Ωct πt

Ωc π sinc

(cid:3) (cid:4) Ωct π

Ωc π sinc

(cid:3) (cid:4) Ωct π

hlp tð Þ ¼

ð5:2Þ

ð5:3Þ

ð5:4Þ

ð5:5Þ

The impulse response for Ωc ¼ 200 Hz is shown in Figure 5.2.

The ﬁlter bandwidth is proportional to Ωc and the width of the main lobe is . The impulse response becomes narrow with increase in the

proportional to 1 Ωc bandwidth.

High-Pass Filter The following system (Figure 5.3) is generally used to obtain high-pass ﬁlter from a low-pass ﬁlter. The frequency response of an ideal analog high-pass ﬁlter HHP (Ω) that passes a signal for |Ω| > Ωc can be expressed by

HHP Ωð Þ ¼

(cid:1)

0, Ωj 1, Ωj

j (cid:1) Ωc j > Ωc

ð5:6Þ

and is shown in Figure 5.4. The frequency Ωc is called the cutoff frequency

5.1 Ideal Analog Filters

Figure 5.2 Impulse response for Ωc ¼ 200 Hz

229

Figure 5.3 System to obtain a high-pass ﬁlter from low-pass ﬁlter

Figure 5.4 Frequency response of ideal high-pass ﬁlter

230

5 Analog Filters

From Figure 5.3, the frequency response of the ideal high-pass ﬁlter can also be

expressed as

HHP Ωð Þ ¼ 1 (cid:3) HLP Ωð Þ

ð5:7Þ

Therefore, the impulse response of an ideal high-pass ﬁlter is given by the inverse

Fourier transform of Eq. (5.7).

Hence, the impulse response of the ideal high pass ﬁlter is given by

hhp tð Þ ¼ δ tð Þ (cid:3)

Ωc π sinc

(cid:3) (cid:4) Ωct π

Band-Pass Filter The frequency response of band-pass ﬁlter can be expressed by

(cid:1)

HBP Ωð Þ ¼

which is shown in Figure 5.5.

1, Ωc1 (cid:1) Ωj 0, Ωj

j < Ωc1 and Ωj

j (cid:1) Ωc2

j > Ωc2

ð5:8Þ

ð5:9Þ

From Figure 5.5, the frequency response of the ideal band-pass ﬁlter can be

expressed as

hBP tð Þ ¼

(cid:3)Ωc1

(cid:3)Ωc2

1 2π

ejΩtdΩ þ

ðΩc2

Ωc1

1 2π

ejΩtdΩ

ð5:10Þ

hBPðtÞ ¼

1 2π

ejΩt jt

(cid:5) (cid:5) (cid:5) (cid:5) (cid:5)

(cid:3)Ωc1 (cid:3)Ωc2

(cid:5) (cid:5) (cid:5) (cid:5) (cid:5)

1 2π

ejΩt jt

(cid:7)

(cid:6) e(cid:3)jΩc1t (cid:3) e(cid:3)jΩc2t

1 j2πt ½sin Ωc2 t (cid:3) sin Ωc1 t(cid:4) tπ

Ωc2 Ωc1 1 j2πt

(cid:6) ejΩc2t (cid:3) ejΩc1t

(cid:7)

Thus, the impulse response of an ideal band-pass ﬁlter is

Figure 5.5 Frequency response of ideal band-pass ﬁlter

5.1 Ideal Analog Filters

hBP tð Þ ¼

Ωc2 π sinc

(cid:4)

(cid:3)

Ωc2 t π

Ωc1 π sinc

(cid:3)

(cid:4)

(cid:3)

Ωc1 t π

231

ð5:11Þ

Band-Stop Filter

The band-stop ﬁlter can be realized as a parallel combination of low-pass ﬁlter with cutoff frequency Ωc1 and high-pass ﬁlter cutoff frequency Ωc2. The frequency response of band-stop ﬁlter can be expressed by

(

HBS Ωð Þ ¼

1, Ωj 0, Ωc1 < Ωj

j (cid:1) Ωc1 and Ωj j < Ωc2

j (cid:5) Ωc2

ð5:12Þ

which is shown in Figure 5.7.

From Figure 5.6, the frequency response of the ideal band-stop ﬁlter can also be

expressed as

HBS Ωð Þ ¼ HLP Ωð Þ þ HHP Ωð Þ

ð5:13Þ

Therefore, the impulse response of an ideal band-stop ﬁlter is given by the inverse

Fourier transform of Eq. (5.13).

Hence, the impulse response of the ideal band-stop ﬁlter is given by

hBS tð Þ ¼ δ tð Þ þ

Ωc1 π sinc

(cid:4)

(cid:3)

Ωc1t π

Ωc2 π sinc

(cid:3)

Ωc2 t π

(cid:4) :

ð5:14Þ

Figure 5.6 System with summation of high-pass ﬁlter and low-pass ﬁlter

Figure 5.7 Frequency response of ideal band

232

5 Analog Filters

5.2 Practical Analog Low-Pass Filter Design

A number of approximation techniques for the design of analog low-pass ﬁlters are well established in the literature. The design of analog low-pass ﬁlter using Butterworth, Chebyshev I, Chebyshev II (inverse Chebyshev), and elliptic approx- imations is discussed in this section.

5.2.1 Filter Speciﬁcations

The speciﬁcations for an analog low-pass ﬁlter with tolerances are depicted in

Figure 5.8, where Ωp - Passband edge frequency Ωs - Stopband edge frequency δp- Peak ripple value in the passband δs - Peak ripple value in the stopband Peak passband ripple in dB ¼ αp ¼ (cid:3)20 log10(1 – δp) dB Minimum stopband ripple in dB ¼ αs ¼ (cid:3)20 log10 (δs) dB Peak ripple value in passband δp ¼ 1 (cid:3) 10(cid:3)αp=20 Peak ripple value in stopband δs ¼ 10(cid:3)αs=20

|H (jW)|

1+ 1-

Transition band

Pass band

Stop band

Figure 5.8 Speciﬁcations of a low-pass analog ﬁlter

5.2 Practical Analog Low-Pass Filter Design

233

5.2.2 Butterworth Analog Low-Pass Filter

The magnitude-square response of an Nth-order analog low-pass Butterworth ﬁlter is given by

Ha jΩð

j2 ¼ Þ

1 þ Ω=Ωc ð

Þ2N

ð5:15Þ

Two parameters completely characterizing a Butterworth low-pass ﬁlter are Ωs and N. These are determined from the speciﬁed band edges Ωp and Ωc, peak passband ripple αp, and minimum stopband attenuation αs. The ﬁrst (2N (cid:3) 1) derivatives of |Ha( jΩ|2 at Ω ¼ 0 are equal to zero. Thus, the Butterworth low-pass ﬁlter is said to have a maximally ﬂat magnitude at Ω ¼ 0. The gain in dB is given by 10log10|Ha( jΩ|2. At Ω ¼ Ωc, the gain is 10log10(0.5) ¼ (cid:3) 3 dB; therefore, Ωc is called the 3 dB cutoff frequency. The loss in dB in a Butterworth ﬁlter is given by

(cid:8) α ¼ 10log 1 þ Ω=Ωc

Þ2N

(cid:9)

For Ω ¼ Ωp, the passband attenuation is given by (cid:11)

(cid:8) αp ¼ 10log 1 þ Ωp=Ωc

(cid:10)

(cid:9)

For Ω ¼ Ωs, the stopband attenuation is

(cid:8) αs ¼ 10log 1 þ Ωs=Ωc

Þ2N

(cid:9)

Eqs.(5.17) and (5.18) can be rewritten as

(cid:10)

Ωp=Ωc

(cid:11)

¼ 100:1αp (cid:3) 1

Ωs=Ωc

Þ2N ¼ 100:1αs (cid:3) 1

From Eqs.(5.19) and (5.20), we obtain

(cid:10)

Ωs=Ωp

(cid:11)

(cid:4)1=2N

(cid:3)

100:1αs (cid:3) 1 100:1αp (cid:3) 1

Eq. (5.21) can be rewritten as

(cid:10)

log Ωs=Ωp

(cid:11)

1 2N

log

(cid:4)

(cid:3)

100:1αs (cid:3) 1 100:1αp (cid:3) 1

From Eq. (5.22), solving for N we get

ð5:16Þ

ð5:17Þ

ð5:18Þ

ð5:19Þ

ð5:20Þ

ð5:21Þ

ð5:22Þ

234

5 Analog Filters

(cid:8)

(cid:9)

log 100:1αs (cid:3)1 100:1αp (cid:3)1 2log Ωs=Ωp ð

ð5:23Þ

N (cid:5)

Since the order N must be an integer, the value obtained is rounded to the next higher integer. This value of N is used in either Eq. (5.19) or Eq. (5.20) to determine the 3 dB cutoff frequency Ωc. In practice, Ωc is determined by Eq. (5.20) that exactly satisﬁes stopband speciﬁcation at Ωc, while the passband speciﬁcation is exceeded with a safe margin at Ωp. We know that |H( jΩ)|2 may be evaluated by letting s ¼ jΩ in H(s)H((cid:3)s), which may be expressed as

H sð ÞH (cid:3)s

Þ ¼

(cid:10)

1 1 þ (cid:3)s2=Ω2 c

(cid:11) N

ð5:24Þ

If Ωc ¼ 1, the magnitude response |HN( jΩ)| is called the normalized magnitude

response. Now, we have

where

(cid:10) 1 þ (cid:3)s2

(cid:11)N

Y2N

k¼1

s (cid:3) sk

(

sk ¼

ej 2k(cid:3)1 ð

Þπ=2N

for n even

ej k(cid:3)1 ð

Þπ=N

for n odd

ð5:25Þ

ð5:26Þ

Since |sk| ¼ 1, we can conclude that there are 2N poles placed on the unit circle in

the s-plane. The normalized transfer function can be formed as

HN sð Þ ¼

l¼1

s (cid:3) pl

ð5:27Þ

where pl for l ¼ 1, 2,.., N are the left half s-plane poles. The complex poles occur in conjugate pairs.

For example, in the case of N ¼ 2, from Eq. (5.26), we have (cid:4)

(cid:4)

sk ¼ cos

(cid:3) ð

Þπ

2k (cid:3) 1 4

(cid:3) ð

Þπ

2k (cid:3) 1 4

þ j sin

k ¼ 1, … ::, 2N

The poles in the left half of the s-plane are

s2 ¼ (cid:3)

1ﬃﬃﬃ p þ 2

jﬃﬃﬃ p ; s3 ¼ (cid:3) 2

1ﬃﬃﬃ p (cid:3) 2

jﬃﬃﬃ p 2

Hence,

5.2 Practical Analog Low-Pass Filter Design

235

and

p1 ¼ (cid:3)

1ﬃﬃﬃ p þ 2

jﬃﬃﬃ p ; p2 ¼ (cid:3) 2

1ﬃﬃﬃ p (cid:3) 2

jﬃﬃﬃ p 2

HN sð Þ ¼

1 ﬃﬃﬃ p 2

s þ 1

s2 þ

In the case of N ¼ 3, (cid:3) ð

sk ¼ cos

(cid:4)

Þπ

k (cid:3) 1 3

(cid:3)

Þπ

k (cid:3) 1 3

þ j sin

(cid:4)

k ¼ 1, … ::, 2N

The left half of s-plane poles are

ﬃﬃﬃ 3

1 2

s3 ¼ (cid:3)

;

s4 ¼ (cid:3)1;

1 s5 ¼ (cid:3) 2

(cid:3)

ﬃﬃﬃ 3

Hence

and

p j

ﬃﬃﬃ 3

;

1 2

p1 ¼ (cid:3)

p2 ¼ (cid:3)1;

p3 ¼ (cid:3)

ﬃﬃﬃ 3

;

1 2

(cid:3)

HN sð Þ ¼

s þ 1

Þ s2 þ s þ 1 ð

The following MATLAB Program 5.1 can be used to obtain the Butterworth

normalized transfer function for various values of N.

Program 5.1 Analog Butterworth Low-Pass Filter Normalized Transfer Function

N=input(‘enter order of the ﬁlter’); [z,p,k] = buttap(N)% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den]=zp2tf(z,p,k); %Print coefﬁcients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den); sos=zp2sos(z,p,k);%determines coefﬁcients of second order sections

The normalized Butterworth polynomials generated from the above program for

typical values of N are tabulated in Table 5.1.

The magnitude response of the normalized Butterworth low-pass ﬁlter for some typical values of N is shown in Figure 5.9. From this ﬁgure, it can be seen that the response monotonically decreases both in the passband and the stopband as Ω

236

5 Analog Filters

Table 5.1 List of normalized Butterworth polynomials

N 1 2

3 4 5 6

Denominator of HN(s) S þ 1 ﬃﬃﬃ p s2 þ s þ 1 2 (s þ 1)(s2 + s þ 1) (s2 þ 0.76537s þ 1)(s2 þ 1.8477s þ 1) (s þ 1) (s2 þ 0.61803s þ 1)(s2 þ 1.61803s þ 1) ﬃﬃﬃ (cid:10) p s2 þ 1:931855s þ 1 Þ s2 þ s þ 1 2 ð (s þ 1) (s2 þ 1.80194s þ 1)(s2 þ 1.247s þ 1)(s2 þ 0.445s þ 1)

s2 þ 0:51764s þ 1 ð

(cid:11)

Figure 5.9 Magnitude response of typical Butterworth low-pass ﬁlter

increases. As the ﬁlter order N increases, the magnitude responses both in the passband and the stopband are improved with a corresponding decrease in the transition width. Since the normalized transfer function corresponds to Ωc ¼ 1, the transfer function of the low-pass ﬁlter corresponding to the actual Ωc can be obtained by replacing s by (s/Ωc) in the normalized transfer function. Example 5.1 Design a Butterworth analog low-pass ﬁlter with 1 dB passband ripple, passband edge frequency Ωp ¼ 2000π rad/sec, stopband edge frequency Ωs ¼ 10,000π rad/sec, and a minimum stopband ripple of 40 dB. Solution Since αs ¼ 40 dB, αp ¼ 1 dB, Ωp ¼ 2000π, and Ωs ¼ 10,000π,

5.2 Practical Analog Low-Pass Filter Design

237

(cid:4)

(cid:3)

100:1αs (cid:3) 1 100:1αp (cid:3) 1

log

(cid:4)

(cid:3)

104 (cid:3) 1 100:1 (cid:3) 1

¼ log

¼ 4:5868:

Hence from (5.23),

(cid:9)

(cid:8) log 104(cid:3)1 100:1(cid:3)1 2log 5=1ð Þ

N (cid:5)

4:5868 1:3979

¼ 3:2811

Since the order must be an integer, we choose N ¼ 4. The normalized low-pass Butterworth ﬁlter for N ¼ 4 can be formulated as

HN sð Þ ¼

1 Þ s2 þ 1:8477s þ 1 s2 þ 0:76537s þ 1 ð

From Eq. (5.20), we have

Ωc ¼

(cid:10)

Ωs 104 (cid:3) 1

(cid:11)

1=2N ¼

(cid:10)

10000π (cid:11) 104 (cid:3) 1

1=8 ¼ 9935

The transfer function for Ωc ¼ 9935 can be obtained by replacing s by (s/Ωc) ¼

(s/9935)in HN (s).

Ha sð Þ ¼

(cid:10)

(cid:11) 2

s 9935

(cid:8)

þ 0:76537

(cid:9)

s 9935

(cid:6)

(cid:10)

(cid:11) 2

s 9935

þ 1

(cid:8)

þ 1:8477

(cid:9)

s 9935

þ 1

(cid:10)

s2 þ 7:604 (cid:6) 103s þ 9:8704225 (cid:6) 107

s2 þ 1:8357 (cid:6) 104s þ 9:8704225 (cid:6) 107

9:7425 (cid:6) 1015 (cid:10) (cid:11)

(cid:11)

5.2.3 Chebyshev Analog Low-Pass Filter

Type 1 Chebyshev Low-Pass Filter The magnitude-square response of an Nth-order analog low-pass Type 1 Chebyshev ﬁlter is given by

H Ωð Þ j

j2 ¼

1 (cid:10) Ω=Ωp 1 þ ε2T 2 N

(cid:11)

ð5:28Þ

where TN(Ω) is the Chebyshev polynomial of order N

238

5 Analog Filters

(

T N Ωð Þ ¼

ð (cid:10)

cos N cos (cid:3)1Ω Ωj Þ, (cid:11) , Ωj cosh Ncosh(cid:3)1Ω

j (cid:1) 1 j > 1

The loss in dB in a Type 1 Chebyshev ﬁlter is given by

(cid:10) α ¼ 10log 1 þ ε2T 2 N

(cid:10)

Ω=Ωp

(cid:11)

For Ω ¼ Ωp, TN(Ω) ¼ 1, and the passband attenuation is given by

(cid:10)

αp ¼ 10log 1 þ ε2

(cid:11)

From Eq. (5.31), ε can be obtained as p

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 100:1αp (cid:3) 1

ε ¼

For Ω ¼ Ωs the stopband attenuation is

αs ¼ 10logð1 þ22T 2

nðΩs=ΩpÞÞ

Since (Ωs/Ωp) > 1, the above equation can be written as (cid:6)

(cid:10)

αs ¼ 10log 1 þ22cosh2 Ncosh(cid:3)1 Ωs=Ωp

(cid:11) (cid:11)

(cid:7)

ð5:29Þ

ð5:30Þ

ð5:31Þ

ð5:32Þ

ð5:33Þ

ð5:34Þ

Substituting Eq. (5.32) for ε in the above equation and solving for N, we get

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 100:1αs (cid:3)1 cosh(cid:3)1 100:1αp (cid:3)1 (cid:11) (cid:10) cosh(cid:3)1 Ωs=Ωp

N (cid:5)

ð5:35Þ

We choose N to be the lowest integer satisfying (5.35). In determining N using the above equation, it is convenient to evaluate cosh(cid:3)1(x) by applying the identity cosh(cid:3)1 xð Þ ¼ ln x þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x2 (cid:3) 1

(cid:8)

(cid:9)

The poles of the normalized Type 1 Chebyshev ﬁlter transfer function lie on an

ellipse in the s-plane and are given by (cid:1)

(cid:13)

xk ¼ (cid:3)sinh (cid:1)

yk ¼ cosh

(cid:3) (cid:4) sinh(cid:3)1 1 2 (cid:3) (cid:4) sinh(cid:3)1 1 2

1 N

(cid:13)

sin

ð2k (cid:3) 1Þπ 2N

for k ¼ 1, 2, ::::, N

ð5:36Þ

cos

ð2k (cid:3) 1Þπ 2N

for k ¼ 1, 2, :::, N

ð5:37Þ

Also, the normalized transfer function is given by

HN sð Þ ¼

H0 Πk s (cid:3) pk ð

ð5:38Þ

where

5.2 Practical Analog Low-Pass Filter Design

pk ¼ (cid:3)sinh

(cid:1)

(cid:13)

(cid:3) (cid:4) sinh(cid:3)1 1 2

1 N

sin

Þπ

2k (cid:3) 1 2N

(cid:1)

þ j cosh

(cid:13)

(cid:3) (cid:4) sinh(cid:3)1 1 2

1 N

cos

and

H0 ¼

1 2N(cid:3)1

1 ε

239

Þπ

2k (cid:3) 1 2N ð5:39aÞ

ð5:39bÞ

As an illustration, consider the case of N ¼ 2 with a passband ripple of 1 dB. From

Eq. (5.32), we have

Hence

1 ε ¼

1 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 100:1αp (cid:3) 1

¼ 1:965227

(cid:3) (cid:4) sinh(cid:3)1 1 ε

¼ sinh(cid:3)1 1:965227 ð

Þ ¼ 1:428

Therefore, from (5.39a), the poles of the normalized Chebyshev transfer function

are given by

pk ¼ (cid:3)sinh 0:714

Þ sin

Þπ

2k (cid:3) 1 ð 4

þ j cosh 0:714 ð

Þ cos

Þπ

2k (cid:3) 1 4

k ¼ 1, 2

Hence

p1 ¼ (cid:3)0:54887 þ j0:89513, p2 ¼ (cid:3)0:54887 (cid:3) j0:89513

Also, from (5.39b), we have

H0 ¼

1 2

1:965227

Þ ¼ 0:98261

Thus for N ¼ 2, with a passband ripple of 1 dB, the normalized Chebyshev

transfer function is

HN sð Þ ¼

0:98261 Þ s (cid:3) p2 ð

s (cid:3) p1

0:98261 s2 þ 1:098s þ 1:103

Similarly for N ¼ 3, for a passband ripple of 1 dB, we have

pk ¼ (cid:3)sinh 1:428=3 ð

Þ sin

Þπ

2k (cid:3) 1 ð 6

þ j cosh 1:428=3 ð

Þ cos

Þπ

2k (cid:3) 1 6

k ¼ 1, 2, 3

Thus,

240

5 Analog Filters

p1 ¼ (cid:3)0:24709 þ j0:96600; p2 ¼ (cid:3)0:49417; p3 ¼ (cid:3)0:24709 (cid:3) j0:966:

Also, from (5.39b),

H0 ¼

1 4

1:965227

Þ ¼ 0:49131

Hence, the normalized transfer function of Type 1 Chebyshev low-pass ﬁlter for

N¼3 is given by

HN sð Þ ¼

0:49131 Þ s (cid:3) p2 ð

Þ s (cid:3) p3 ð

0:49131 s3 þ 0:988s2 þ 1:238s þ 0:49131

s (cid:3) p1

The following MATLAB Program 5.2 can be used to form the Type1 Chebyshev

normalized transfer function for a given order and passband ripple.

Program 5.2 Analog Type 1 Chebyshev Low-Pass Filter Normalized Transfer Function

N=input(‘enter order of the ﬁlter’); Rp=input(‘enter passband ripple in dB’); [z,p,k] = cheb1ap(N,Rp)% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den]=zp2tf(z,p,k); %Print coefﬁcients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den);

The normalized Type 1 Chebyshev polynomials generated from the above pro- gram for typical values of N and passband ripple of 1 dB are tabulated in Table 5.2. The typical magnitude responses of a Type 1 Chebyshev low-pass ﬁlter for N ¼ 3, 5, and 8 with 1 dB passband ripple are shown in Figure 5.10. From this ﬁgure, it is seen that Type 1 Chebyshev low-pass ﬁlter exhibits equiripple in the passband with a monotonic decrease in the stopband. Example 5.2 Design a Type 1 Chebyshev analog low-pass ﬁlter for the speciﬁca- tions given in Example 5.1.

Table 5.2 List of normalized Type 1 Chebyshev transfer functions for passband ripple ¼ 1 dB

N 1 2 3 4 5

Denominator of HN(s) S þ 1.9652 s2 þ 1.0977s þ 1.1025 s3 þ 0.98834s2 þ 1.2384s þ 0.49131 s4 þ 0.95281s3 þ 1.4539s2 þ 0.74262s þ 0.27563 s5 þ 0.93682s4 þ 1.6888s3 þ 0.9744s2 þ 0.58053s þ 0.12283

H0 1.9652 0.98261 0.49131 0.24565 0.12283

5.2 Practical Analog Low-Pass Filter Design

241

Figure 5.10 Magnitude response of typical Type 1 Chebyshev low-pass ﬁlter with 1 dB passband ripple

Solution Since αs ¼ 40 dB, αp ¼ 1 dB, Ωp ¼ 2000π, and Ωs ¼ 10,000π,

¼ cosh(cid:3)1 196:52 ð

cosh(cid:3)1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 100:1αs (cid:3) 1 100:1αp (cid:3) 1 (cid:10) cosh(cid:3)1 Ωs=Ωp

¼ cosh(cid:3)1 (cid:11)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 104 (cid:3) 1 100:1 (cid:3) 1 ¼ cosh(cid:3)1 5ð Þ ¼ 2:2924 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 104 (cid:3) 1 100:1 (cid:3) 1

¼ 2:6059

cosh(cid:3)1

N (cid:5)

cosh(cid:3)1 5ð Þ

Since the order of the ﬁlter must be an integer, we choose the next higher integer value 3 for N. The normalized Type 1 Chebyshev low-pass ﬁlter for N ¼ 3 with a passband ripple of 1 dB is given from Table 5.2 as

HN sð Þ ¼

0:49131 s3 þ 0:988s2 þ 1:238s þ 0:49131

The transfer function for Ωp ¼ 2000π is obtained by substituting s ¼ (s/Ωp) ¼

(s/2000π) in HN(s):

Ha sð Þ ¼

(cid:8)

(cid:9) 3

(cid:8)

0:49131 2

(cid:9)

(cid:8)

(cid:9)

þ 0:988

s 2000π

s 2000π 1:2187 (cid:6) 1011 s3 þ 6:2099 (cid:6) 103s2 þ 4:889 (cid:6) 107s þ 1:2187 (cid:6) 1011

s 2000π

þ 1:238

þ 0:49131

242

5 Analog Filters

Type 2 Chebyshev Filter The squared-magnitude response of Type 2 Chebyshev low-pass ﬁlter, which is also known as the inverse Chebyshev ﬁlter, is given by

H Ωð Þ j

j2 ¼

(cid:3)

(cid:4)

1 þ ε2

Ωs=Ωp Þ ð T 2 N Ωs=Ω T 2 Þ ð N

ð5:40Þ

The order N can be determined using Eq. (5.35). The Type 2 Chebyshev ﬁlter has both poles and zeros, and the zeros are on the jΩ axis. The normalized Type 2 Chebyshev low-pass ﬁlter, or the normalized inverse Chebyshev ﬁlter (normalized to Ωs ¼ 1), may be formed as

HNðsÞ ¼ H0

Πkðs (cid:3) zkÞ Πkðs (cid:3) pkÞ

k ¼ 1, 2, ::, N

ð5:41Þ

where

zk ¼ j

1 cos 2k(cid:3)1 ð N

Þπ

for

k ¼ 1, 2, ::, N

ð5:42aÞ

pk ¼

σk k þ Ω2 σ2 k (cid:3) (cid:4)

þ j

(cid:13)

σk ¼ (cid:3)sinh

(cid:1)

1 N

Ωk ¼ cosh

(cid:1)

1 N

sinh(cid:3)1 1 δs (cid:3) (cid:4)

(cid:13)

sinh(cid:3)1 1 δs

ωk k þ Ω2 σ2

for

k ¼ 1, 2, ::, N

ð5:42bÞ

sin

Þπ

2k (cid:3) 1 2N

for

k ¼ 1, 2, ::, N

ð5:42cÞ

cos

Þπ

2k (cid:3) 1 2N

for

k ¼ 1, 2, ::, N

ð5:42dÞ

δs ¼

H0 ¼

1 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 100:1αs (cid:3) 1 Πkð(cid:3)pkÞ Πkð(cid:3)zkÞ

ð5:42eÞ

ð5:42fÞ

For example, if we consider N ¼ 3 with a stopband ripple of 40 dB, then from

(5.42e),

Hence,

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 100:1αs (cid:3) 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 104 (cid:3) 1

¼ 99:995

1 δs

(cid:3) (cid:4)

sinh(cid:3)1 1 δs

¼ 5:28829

Using (5.42c) and (5.42d), we have

5.2 Practical Analog Low-Pass Filter Design

243

σk ¼ (cid:3)sinh 5:28829=3 ð

Þ sin

Ωk ¼ cosh 5:28829=3 ð

Þ cos

Hence

2k (cid:3) 1 6 2k (cid:3) 1 6

Þπ

for

k ¼ 1, 2, 3

for

k ¼ 1, 2, 3

σ1 ¼ (cid:3)1:41927, σ2 ¼ (cid:3)2:83854, σ3 ¼ (cid:3)1:41927 Ω1 ¼ (cid:3)2:60387, Ω2 ¼ (cid:3)2:83854, Ω3 ¼ 2:60387

Thus, from (5.42b), the poles are

p1 ¼ (cid:3)0:16115 þ j0:29593, p2 ¼ (cid:3)0:3523, p3 ¼ (cid:3)0:16115 þ j0:29593

Also, using (5.42a), the zeros are given by

p(cid:8)

z1 ¼ (cid:3)j 2=

ﬃﬃﬃ 3

(cid:9) , z2 ¼ j 2=

p(cid:8)

ﬃﬃﬃ 3

(cid:9)

Finally, from (5.42f),

H0 ¼ 0:03

Therefore, the normalized Type 2 Chebyshev low-pass ﬁlter for N ¼ 3 with a

stopband ripple of 40 dB is given by

HN sð Þ ¼

0:03 s (cid:3) z1 Þ s (cid:3) z2 Þ ð ð s (cid:3) p1 Þ s (cid:3) p3 Þ s (cid:3) p2 ð ð

0:03 s2 þ 1:3333 s3 þ 0:6746s2 þ 0:22709s þ 0:04

The following MATLAB Program 5.3 can be used to form the Type2 Chebyshev

normalized transfer function for a given order and stopband ripple.

Program 5.3 Analog Type 2 Chebyshev Low-Pass Filter Normalized Transfer Function

N=input(‘enter order of the ﬁlter’); Rs=input(‘enter stopband attenuation in dB’); [z,p,k] = cheb2ap(N,Rs);% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den]=zp2tf(z,p,k); %Print coefﬁcients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den);

The normalized Type 2 Chebyshev transfer functions generated from the above program for typical values of N with a stopband ripple of 40 dB are tabulated in Table 5.3.

244

5 Analog Filters

Table 5.3 List of normalized Type 2 Chebyshev transfer functions for stopband ripple ¼ 40 dB

Order N 1

HN (s) 0:01 s þ 0:01

0:01s2 þ 0:02 s2 þ 0:199s þ 0:02

0:03s2 þ 0:04 s3 þ 0:6746s2 þ 0:2271s þ 0:04 0:01s4 þ 0:08s2 þ 0:08 s4 þ 1:35s3 þ 0:9139s2 þ 0:3653s þ 0:08

0:05s4 þ 0:2s2 þ 0:16 s5 þ 2:1492s4 þ 2:3083s3 þ 1:5501s2 þ 0:6573s þ 0:16

0:01s6 þ 0:18s4 þ 0:48s2 þ 0:32 s6 þ 3:0166s5 þ 4:5519s4 þ 4:3819s3 þ 2:8798s2 þ 1:2393s þ 0:32

Figure 5.11 Magnitude response of typical Type 2 Chebyshev low-pass ﬁlter with 20 dB stopband ripple

The typical magnitude response of a Type 2 Chebyshev low-pass ﬁlter for N ¼ 4 and 7 with 20 dB stopband ripple is shown in Figure 5.11. From this ﬁgure, it is seen that Type 2 Chebyshev low-pass ﬁlter exhibits monotonicity in the passband and equiripple in the stopband. Example 5.3 Design a Type 2 Chebyshev low-pass ﬁlter for the speciﬁcations given in Example 5.1.

Solution The order N is chosen as 3, as in Example 5.2, since the equation for order ﬁnding is the same for both Type 1 and Type 2 Chebyshev ﬁlters. The normalized

5.2 Practical Analog Low-Pass Filter Design

245

Type 2 Chebyshev low-pass ﬁlter for N ¼ 3 with a stopband ripple of 40 dB has already been found earlier and is given by

HN sð Þ ¼

0:03 s2 þ 1:3333 s3 þ 0:6746s2 þ 0:2271s þ 0:04

For Ωs ¼ 10,000π, the corresponding transfer function can be obtained by substituting s ¼ (s/Ωs) ¼ (s/10000π) in the above expression for HN(s). Thus, the required ﬁlter transfer function is

Ha sð Þ ¼

(cid:10)

(cid:11) 3

s 10000π

(cid:10)

(cid:11)

0:03 (cid:10) s 10000π

s 10000π (cid:11) 2

þ 0:04 (cid:8) þ 0:22709

þ 0:6746

(cid:9)

s 10000π

þ 0:04

9:4252 (cid:6) 102s2 þ 1:2403 (cid:6) 1012 s3 þ 2:1193 (cid:6) 104s2 þ 2:2413 (cid:6) 108s þ 1:2403 (cid:6) 1012

5.2.4 Elliptic Analog Low-Pass Filter

The square-magnitude response of an elliptic low-pass ﬁlter is given by

Ha jΩð

j2 ¼

1 1 þ ε2UN Ω=Ωp

(cid:10)

(cid:11)

ð5:43Þ

where UN(x) is the Jacobian elliptic function of order N and ε is a parameter related to the passband ripple. In an elliptic ﬁlter, a constant k, called the selectivity factor, representing the sharpness of the transition region is deﬁned as

k ¼

Ωp Ωs

ð5:44Þ

A large value of k represents a wide transition band, while a small value indicates

a narrow transition band.

For a given set of Ωp, Ωs, αp, and αs, the ﬁlter order can be estimated using the

formula

(cid:9)

(cid:8)

log 16 (cid:6) 100:1αs (cid:3)1 100:1αp (cid:3)1 log10 1=ρ Þ ð

N ﬃ

where ρ can be computed using

0 ¼

ﬃﬃﬃﬃ k0 ﬃﬃﬃﬃ p k0

1 (cid:3) (cid:8) 2 1 þ

(cid:9)

ð5:45Þ

ð5:46Þ

246

5 Analog Filters

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 (cid:3) k2

k0 ¼

ρ ¼ ρ

0 þ 2 ρ 0ð

Þ5 þ 15 ρ 0ð

Þ9 þ 150 ρ 0ð

Þ13

ð5:47Þ

ð5:48Þ

The following MATLAB Program 5.4 can be used to form the elliptic normalized transfer function for given ﬁlter order and passband ripple and stopband attenuation. The normalized passband edge frequency is set to 1.

Program 5.4 Analog Elliptic Low-Pass Filter Normalized Transfer Function

N=input(‘enter order of the ﬁlter’); Rp=input(‘enter passband ripple in dB’); Rs=input(‘enter stopband attenuation in dB’); [z,p,k] = ellipap(N,Rp,Rs)% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den] =zp2tf(z,p,k); %Print coefﬁcients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den);

The normalized elliptic transfer functions generated from the above program for

typical values of N and stopband ripple of 40 dB are tabulated in Table 5.4.

The magnitude response of a typical elliptic low-pass ﬁlter is shown in Figure 5.12, from which it can be seen that it exhibits equiripple in both the passband and the stopband. Example 5.4 Design an elliptic analog low-pass ﬁlter for the speciﬁcations given in the Example 5.1.

Table 5.4 List of normalized elliptic transfer functions for passband ripple ¼ 1 dB and stopband ripple ¼ 40 dB

Order N 1

HN(s)

1:9652 s þ 1:9652

0:01s2 þ 0:9876 s2 þ 1:0915s þ 1:1081

0:0692s2 þ 0:5265 s3 þ 0:9782s2 þ 1:2434s þ 0:5265

0:01s4 þ 0:1502s2 þ 0:3220 s4 þ 0:9391s3 þ 1:5137s2 þ 0:8037s þ 0:3612

0:0470s4 þ 0:2201s2 þ 0:2299 s5 þ 0:9234s4 þ 1:8471s3 þ 1:1292s2 þ 0:7881s þ 0:2299

0:01s6 þ 0:1172s4 þ 0:28s2 þ 0:186 s6 þ 0:9154s5 þ 2:2378s4 þ 1:4799s3 þ 1:4316s2 þ 0:5652s þ 0:2087

5.2 Practical Analog Low-Pass Filter Design

247

Figure 5.12 Magnitude response of typical elliptic low-pass ﬁlter with 1 dB passband ripple and 30 dB stopband ripple

Solution

and

k ¼

Ωp Ωs

2000π 10000π ¼ 0:2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 (cid:3) k2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 (cid:3) 0:04

k0 ¼

¼ 0:979796:

Substituting these values in Eq. (5.46) and Eq. (5.47), we get

and hence

N ¼

ρ 0 ¼ 0:00255135, ρ ¼ 0:0025513525

(cid:8)

(cid:9)

log 16 (cid:6) 104(cid:3)1 100:1(cid:3)1 0:0025513525Þ ¼ 2:2331:

log10ð

Choose N ¼ 3. Then, for N ¼ 3, a passband ripple of 1 dB, and a stopband ripple of 40dB, the normalized elliptic transfer function is as given in Table 5.4. For Ωp ¼ 2000π, the corresponding transfer function can be obtained by substituting s ¼ (s/Ωp) ¼ (s/2000π) in the expression for HN(s). Thus, the required ﬁlter transfer function is

248

5 Analog Filters

(cid:10)

(cid:11)

Ha sð Þ ¼

(cid:8)

(cid:9) 3

0:0692 (cid:8)

þ 0:97825

þ 0:5265 (cid:8)

s 2000π

s 2000π (cid:9) s 2000π 4:348 (cid:6) 102s2 þ 1:306 (cid:6) 1011 s3 þ 6:1465 (cid:6) 103s2 þ 4:9087 (cid:6) 107s þ 1:306 (cid:6) 1011

s 2000π

þ 1:2434

(cid:9)

þ 0:5265

5.2.5 Bessel Filter

Bessel ﬁlter is a class of all-pole ﬁlters that provide linear phase response in the passband and characterized by the transfer function

HaðsÞ ¼

1 a0 þ a1s þ a2s2 þ (cid:8) (cid:8) (cid:8) þ aN(cid:3)1sN(cid:3)1 þ aNsN

where the coefﬁcients an are given by

an ¼

Þ! 2N (cid:3) n 2N(cid:3)nn! N (cid:3) n ð

Þ!

ð5:49Þ

ð5:50Þ

The magnitude responses of a third-order Bessel ﬁlter and Butterworth ﬁlter are shown in Figure 5.13 and the phase responses of the same ﬁlters with the same order are shown in Figure 5.14. From these ﬁgures, it is seen that the magnitude response

Figure 5.13 Magnitude response of a third-order Bessel ﬁlter and Butterworth ﬁlter

5.2 Practical Analog Low-Pass Filter Design

249

Figure 5.14 Phase response of a third-order Bessel ﬁlter and Butterworth ﬁlter

of the Bessel ﬁlter is poorer than that of the Butterworth ﬁlter, whereas the phase response of the Bessel ﬁlter is more linear in the passband than that of the Butterworth ﬁlter.

5.2.6 Comparison of Various Types of Analog Filters

The magnitude response and phase response of the normalized Butterworth, Chebyshev Type 1, Chebyshev Type 2, and elliptic ﬁlters of the same order are compared with the following speciﬁcations:

filter order ¼ 8, maximum passband ripple ¼ 1 dB and minimum stopband ripple

¼ 35 dB:

The following MATLAB program is used to generate the magnitude and phase responses for these speciﬁcations.

Program 5.5 Magnitude and Phase Responses of Analog Filters of Order 8 with a Passband Ripple of 1 dB and a Stopband Ripple of 35 dB

clear all;clc; [z,p,k]=buttap(8); [num1,den1]=zp2tf(z,p,k);[z,p,k]=cheb1ap(8,1); [num2,den2]=zp2tf(z,p,k);[z,p,k]=cheb2ap(8,35); [num3,den3]=zp2tf(z,p,k); [z,p,k]=ellipap(8,1,35);

250

5 Analog Filters

[num4,den4]=zp2tf(z,p,k); omega=[0:0.01:5]; h1=freqs(num1,den1,omega);h2=freqs(num2,den2,omega); h3=freqs(num3,den3,omega);h4=freqs(num4,den4,omega); ph1=angle(h1);ph1=unwrap(ph1); ph2=angle(h2);ph2=unwrap(ph2); ph3=angle(h3);ph3=unwrap(ph3); ph4=angle(h4);ph4=unwrap(ph4); ﬁgure(1),plot(omega,20log10(abs(h1)),‘-’);hold on plot(omega,20log10(abs(h2)),‘—’);hold on plot(omega,20log10(abs(h3)),‘: ’);hold on plot(omega,20log10(abs(h4)),‘-.’); xlabel(‘Normalized frequency’);ylabel(‘Gain,dB’);axis([0 5 -80 5]); legend(‘Butterworth’,‘Chebyshev Type 1’,‘Chebyshev Type 2’,‘Ellip- tic’);hold off ﬁgure(2),plot(omega,ph1,‘-’);hold on plot(omega,ph2,‘—’);hold on plot(omega,ph3,‘: ’);hold on plot(omega,ph4,‘-.’) xlabel(‘Normalized frequency’);ylabel(‘Phase,radians’);axis([0 5 -8 0]); legend(‘Butterworth’,‘Chebyshev Type 1’,‘Chebyshev Type 2’,‘Elliptic’);

The magnitude and phase responses for the above speciﬁcations are shown in Figure 5.15. The magnitude response of Butterworth ﬁlter decreases monotonically both in passband and stopband with wider transition band. The magnitude response of the Chebyshev Type 1 exhibits ripples in the passband, whereas the Chebyshev Type 2 has approximately the same magnitude response to that of the Butterworth ﬁlter. The transition band of both the Type 1 and Type 2 Chebyshev ﬁlters is the same, but less than that of the Butterworth ﬁlter. The elliptic ﬁlter exhibits an equiripple magnitude response both in the passband and the stopband with a transition width smaller than that of the Chebyshev Type 1 and Type 2 ﬁlters. But the phase response of the elliptic ﬁlter is more nonlinear in the passband than that of the phase response of the Butterworth and Chebyshev ﬁlters. If linear phase in the passband is the stringent requirement, then the Bessel ﬁlter is preferred, but with a poor magnitude response.

Another way of comparing the various ﬁlters is in terms of the order of the ﬁlter required to satisfy the same speciﬁcations. Consider a low-pass ﬁlter that meets the passband edge frequency of 450 Hz, stopband edge frequency of 550 Hz, passband ripple of 1 dB, and stopband ripple of 35 dB. The orders of the Butterworth, Chebyshev Type 1, Chebyshev Type2, and elliptic ﬁlters are computed for the above speciﬁcations and listed in Table 5.5. From this table, we can see that elliptic ﬁlter can meet the speciﬁcations with the lowest ﬁlter order.

5.2 Practical Analog Low-Pass Filter Design

251

Figure 5.15 A comparison of various types of analog low-pass ﬁlters: (a) magnitude response and (b) phase response

252

5 Analog Filters

Table 5.5 Comparison of orders of various types of ﬁlters

Filter Butterworth Chebyshev Type 1 Chebyshev Type 2 Elliptic

Order 24 9 9 5

5.2.7 Design of Analog High-Pass, Band-Pass,

and Band-Stop Filters

The analog high-pass, band-pass, and band-stop ﬁlters can be designed using analog frequency transformations. In this design process, ﬁrst, the analog prototype low- pass ﬁlter speciﬁcations are derived from the desired speciﬁcations of the analog ﬁlter using suitable analog-to-analog transformation. Next, by using the speciﬁca- tions so obtained, a prototype low-pass ﬁlter is designed. Finally, the transfer function of the desired analog ﬁlter is determined from the transfer function of the prototype analog low-pass transfer function using the appropriate analog-to-analog frequency transformation. The low-pass to low-pass, low-pass to high-pass, low-pass to band-pass, and low-pass to band-stop analog transformations are considered next.

Low Pass to Low Pass

Let Ωp ¼ 1 and bΩp be the passband edge frequencies of the normalized prototype low-pass ﬁlter and the desired low-pass ﬁlter, as shown in Figure 5.16. The trans- formation from the prototype low pass to the required low pass must convert bΩ ¼ 0 to Ω ¼ 0 and bΩ ¼ (cid:9)1 to Ω ¼ (cid:9)1. The transformation such as s ¼ kbs or Ω ¼ k bΩ achieves the above transformation for any positive value of k. If k is chosen to to Ωs ¼ bΩs= bΩp. Since be Ωp ¼ 1 is the passband edge frequency for the normalized Type I Chebyshev and elliptic low-pass ﬁlters, we have the design equations for these ﬁlters as

, then bΩp gets transformed to Ωp ¼ 1, and bΩs

1= bΩp

(cid:11)

(cid:10)

Ωp ¼ 1, Ωs ¼ bΩs= bΩp:

ð5:51aÞ

(cid:11) (cid:10) bs

Also, the transfer function HLP

for these ﬁlters is related to the corresponding

normalized low-pass transfer function HN (s) by

(cid:11) (cid:10) bs

HLP

¼ HN sð Þc

s¼bs=bΩ p

ð5:51bÞ

However, in the case of a Butterworth ﬁlter, since Ω ¼ 1 corresponds to the cutoff for the Butterworth ﬁlter is

frequency of the ﬁlter, the transfer function HLP related to the normalized low-pass Butterworth transfer function HN (s) by

(cid:11) (cid:10) bs

5.2 Practical Analog Low-Pass Filter Design

253

Figure 5.16 Low-pass to low-pass frequency transformation. (a) Prototype Low-pass ﬁlter frequency response. (b) Low-pass ﬁlter frequency response

(a)

(b)

(cid:10) (cid:11) bs

HLP

¼ HN sð Þc

s¼bs=bΩ c

ð5:51cÞ

where bΩc is the cutoff frequency of the desired Butterworth ﬁlter and is given by Eq. (5.19). For similar reasons, for the Type 2 Chebyshev ﬁlter is related to the normalized transfer function HN (s) by (cid:10) (cid:11) bs

the transfer function HLP

(cid:10) (cid:11) bs

ð5:51dÞ

¼ HN sð Þc

HLP

s¼bs=bΩ s

Low Pass to High Pass (Figure 5.17)

Let the passband edge frequencies of the prototype low-pass and the desired high- pass ﬁlters be Ωp ¼ 1 and bΩp, as shown in Figure 5.17. The transformation from prototype low pass to the desired high pass must transform bΩ ¼ 0 to Ω ¼ 1 and bΩ ¼ 1 to Ω ¼ 0. The transformation such as s ¼ k=bs or Ω ¼ k= bΩ achieves the

254

5 Analog Filters

(a)

(b)

Figure 5.17 Low-pass to high-pass frequency transformation. (a) Prototype low-pass ﬁlter fre- quency response. (b) High-pass ﬁlter frequency response

above transformation for any positive value of k. By transforming bΩp to Ωp ¼ 1, the constant k can be determined as k ¼ bΩp.Thus, design equations are

and the desired transfer function HHP HN (s) by

Ωp ¼ 1, Ωs ¼ bΩp= bΩs, (cid:11) (cid:10) bs

is related to the low-pass transfer function

ð5:52aÞ

(cid:10) (cid:11) bs

HHP

¼ HN sð Þj

s¼bΩp=bs

ð5:52bÞ

5.2 Practical Analog Low-Pass Filter Design

255

(a)

(b)

Figure 5.18 Low-pass to band-pass frequency transformation. (a) Prototype low-pass ﬁlter fre- quency response. (b) Band-pass ﬁlter frequency response

The above equations (5.52a) and (5.52b) hold for all ﬁlters except for Butterworth

and Type 2 Chebyshev ﬁlter. For Butterworth

(cid:11) (cid:10) bs HLP (cid:11) (cid:10) bs

HHP

¼ HN sð Þc (cid:5) (cid:5) (cid:5) ¼ HN sð Þ (cid:5)

s¼bs=bΩ c

s¼bΩp=bs

ð5:53aÞ

ð5:53bÞ

For Type 2 Chebyshev ﬁlter, the design equations are

256

and

Ωp ¼ bΩs= bΩp, Ωs ¼ 1

(cid:10) (cid:11) bs

HHP

¼ HN sð Þc

s¼bΩ s=bs

5 Analog Filters

ð5:53cÞ

ð5:53dÞ

Example 5.5 Design a Butterworth analog high-pass ﬁlter for the following speciﬁcations:

Passband edge frequency: 30.777 Hz Stopband edge frequency: 10 Hz

Passband ripple: 1 dB Stopband ripple: 20 dB

Solution For the prototype analog low-pass ﬁlter, we have

Ωp ¼ 1, Ωs ¼ bΩp= bΩs ¼ 3:0777,

αp ¼ 1 dB, αs ¼ 20 dB

Substituting these values in Eq. (5.23), the order of the ﬁlter is given by

(cid:9)

(cid:8) log 102(cid:3)1 100:1(cid:3)1 (cid:10) 2 log 3:077 1

(cid:11) ¼ 2:6447

N (cid:5)

Hence, we choose N ¼ 3. From Table 5.1, the third-order normalized Butterworth

low-pass ﬁlter transfer function is given by

HN sð Þ ¼

s þ 1

Þ s2 þ s þ 1 ð

Substituting the values of Ωs and N in Eq. (5.20), we obtain

(cid:4) 6

(cid:3)

3:0777 Ωc

¼ 102 (cid:3) 1

Solving for Ωc, we get Ωc ¼ 1.4309. The analog transfer function of the low-pass ﬁlter is obtained from the above ¼ s

transfer function by substituting s ¼ s Ωc

1:4309; hence,

HLP sð Þ ¼

2:93 s3 þ 2:8619s2 þ 4:0952s þ 2:93

From the above transfer function, the analog transfer function of the high-pass _ Ωp s

ﬁlter can be obtained by substituting s ¼

3:0777 s

5.2 Practical Analog Low-Pass Filter Design

257

HHP sð Þ ¼

s3 s3 þ 4:3017s2 þ 9:2521s þ 9:9499

Example 5.6 Design a Butterworth analog high-pass ﬁlter for the speciﬁcations of Example 5.5 using MATLAB

Solution The following MATLAB code fragments can be used to design HHP (s)

[N,Wn]=buttord(1,3.0777,1,20,‘s’); [B,A]=butter(N,Wn,‘s’); [num, den]=1p2hp(B,A,3.0777);

The transfer function HLP(s) of the analog low-pass ﬁlter can be obtained by displaying numerator and denominator coefﬁcient vectors B and A and is given by

HLP sð Þ ¼

2:93 s3 þ 2:8619s2 þ 4:0952s þ 2:93

The transfer function HHP (s) of the analog high-pass ﬁlter can be obtained by displaying numerator and denominator coefﬁcient vectors num and den and is given by

HHP sð Þ ¼

s3 s3 þ 4:3017s2 þ 9:2521s þ 9:499

Low Pass to Band Pass The prototype low-pass and the desired band-pass ﬁlters are shown in Figure 5.18. In this ﬁgure, bΩp1 is the lower passband edge frequency, bΩp2 the upper passband edge frequency, bΩs1 the lower stopband edge frequency, and bΩs2 the upper stopband edge frequency of the desired band-pass ﬁlter. Let us denote by Bp the bandwidth of the passband and by bΩmp the geometric mean between the passband edge frequencies of the band-pass ﬁlter, i.e.,

Bp ¼ bΩp2 (cid:3) bΩp1 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ q bΩp1 bΩp2

bΩmp ¼

ð5:54aÞ

ð5:54bÞ

Now, consider the transformation

(cid:10) bs2 þ bΩ2 Bpbs As a consequence of this transformation, it is seen that bΩ ¼ 0, bΩp1, bΩmp, bΩp2, and 1 transform to the frequencies Ω ¼ (cid:3)1, (cid:3)1, 0, þ1, and 1, respectively, for

ð5:55Þ

S ¼

(cid:11)

258

5 Analog Filters

the normalized low-pass ﬁlter. Also, the transformation (5.55) transforms the fre- quencies bΩs1 and bΩs2 to Ω0

s , respectively, where

s and Ω00

and

Ω0

s ¼

bΩ2 s1 (cid:3) bΩp1 (cid:10) bΩp2 (cid:3) bΩp1

bΩp2 (cid:11) bΩs1

¼ A1

sayð

Ω00

s ¼

bΩ2 s2 (cid:3) bΩp1 (cid:10) bΩp2 (cid:3) bΩp1

bΩp2 (cid:11) bΩs2

¼ A2

sayð

ð5:56Þ

ð5:57Þ

In order to satisfy the stopband requirements and to have symmetry of the stopband edges in the low-pass ﬁlter, we choose Ωs to be themin{|A1|, |A2|}. Thus, the spectral transformation (5.55) leads to the following design equations for the normalized low-pass ﬁlter (except in the case of the Type 2 Chebyshev ﬁlter)

Ωp ¼ 1, Ωs ¼ min A1j

j; A2j

ð5:58aÞ

where A1 and A2 are given by (5.56) and (5.57), respectively, and the desired high- pass transfer function HBP can be obtained from the normalized low-pass transfer function HN (s) using (5.55). In the case of the Type 2 Chebyshev ﬁlter, the equation corresponding to (5.58a) is

(cid:11) (cid:10) bs

Ωp ¼ max 1= A1j

j; 1= A2j

g, Ωs ¼ 1

ð5:58bÞ

Example 5.7 Design a Butterworth IIR digital band-pass ﬁlter for the following speciﬁcations:

Lower passband edge frequency: 41.4 Hz Upper passband edge frequency: 50.95 Hz Lower stopband edge frequency: 7.87 Hz Upper stopband edge frequency: 100 Hz Passband ripple: 2 dB Stopband ripple: 10 dB

Solution We have

A1 ¼

A2 ¼

(cid:3)ð0:0787Þ2 þ ð0, 414Þð0:5095Þ 0:0787ð0:5095 (cid:3) 0, 414Þ

¼ 27:25

(cid:3)1 þ ð0, 414Þð0:5095Þ 1ð0:5095 (cid:3) 0, 414Þ

¼ (cid:3)8:26

For the prototype analog low-pass ﬁlter, Ωp ¼ 1, Ωs ¼ min {|A1|, |A2|} ¼ 8.26;

αp ¼ 2 dB αs ¼ 10 dB

5.2 Practical Analog Low-Pass Filter Design

259

Substituting these values in Eq. (5.23), the order of the ﬁlter is given by h

N ¼

101(cid:3)1 log10 100:2:(cid:3)1 2log10 8:26 Þ ð

¼ 0:5203

Let us choose N ¼ 1 The transfer function of the ﬁrst-order normalized Butterworth low-pass ﬁlter is

given by

HN sð Þ ¼

1 s þ 1

Substituting the values of Ωs and N in Eq. (5.20), we obtain (cid:3)

(cid:4) 2

8:26 Ωc

¼ 101 (cid:3) 1

Solving for Ωc, we get Ωc ¼ 2.7533 The analog transfer function of the low-pass ﬁlter can be obtained from the above ¼ s

transfer function by substituting s ¼ s Ωc

2:7533

HLP sð Þ ¼

2:7533 s þ 2:7533

To arrive at the analog transfer function of the band-pass ﬁlter, variable s in the

above normalized transfer function is to be replaced by (cid:10)

(cid:11)

(cid:3)

(cid:4)

S ¼

bΩp2 s2 þ bΩp1 (cid:11) (cid:10) bΩp2 (cid:3) bΩp1 s

s2 þ 2109:3 9:55s

HBP sð Þ ¼

26:2943 s s2 þ 26:2943s þ 2109:3

Example 5.8 Design a band-pass Butterworth ﬁlter for the speciﬁcations of Exam- ple 5.7 using MATLAB

Solution The following MATLAB code fragments can be used to design HBS (s): Bandwidth ¼ bw ¼ 50.95–41.4 ¼ 9.55; Ωo ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Þ 41:4 50:95 Þ ð ð

¼ 45:9271.

[N,Wn]=buttord(1,8.26,2,10,‘s’); [B,A]=butter(N,Wn,‘s’); [num, den]=1p2bp(B,A, 45.9271,9.55);% [num, den]=1p2bp(B,A, Ωo, bw);

The transfer function HLP(s) of the analog low-pass ﬁlter can be obtained by

displaying numerator and denominator coefﬁcient vectors B and A. It is given by

260

5 Analog Filters

HLP sð Þ ¼

2:7533 s þ 2:7533

The transfer function HBP (s) of the analog band-pass ﬁlter can be obtained by displaying numerator and denominator coefﬁcient vectors num and den. It is given by

HBP sð Þ ¼

26:2943 s s2 þ 26:2943s þ 2109:3

Low Pass to Band Stop The prototype low-pass and the desired band-stop ﬁlters are shown in Figure 5.19. In this ﬁgure, bΩp1 is the lower passband edge frequency, bΩp2 the upper passband edge frequency, bΩs1 the lower stopband edge frequency, and bΩs2 the upper stopband edge frequency of the transformation

the desired band-stop ﬁlter. Let us now consider

kbs (cid:10) bs2 þ bΩ2 where bΩms is the geometric mean between the stopband edge frequencies of the band-stop ﬁlter, i.e.,

ð5:59Þ

S ¼

(cid:11)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ bΩs1 bΩs2

bΩms ¼

ð5:60Þ

As a consequence of this transformation,

is seen that bΩ ¼ 0 and 1 transformed to the frequency bΩ ¼ 0 for the normalized low-pass ﬁlter. Now, we transform the lower stopband edge frequency bΩs1 to the stopband edge frequency Ωs of the normalized low-pass ﬁlter; hence,

Ωs ¼

k bΩs2 (cid:3) bΩs1

k Bs

ð5:61aÞ

is the bandwidth of the stopband. Also, the upper stopband

(cid:10) bΩs2 (cid:3) bΩs1 where Bs ¼ edge frequency bΩs2 is transformed to

(cid:11)

k (cid:3) bΩs2 (cid:3) bΩs1

¼ (cid:3)

k Bs

¼ (cid:3)Ωs

ð5:61bÞ

Hence, the constant k is given by

k ¼ BsΩs ¼

(cid:10)

bΩs2 (cid:3) bΩs1

(cid:11)

Ωs

As a consequence,

the passband edge frequencies

transformed to

ð5:61cÞ

bΩp1 and bΩp2

are

5.2 Practical Analog Low-Pass Filter Design

and

Ω0

p ¼

Ω00

p ¼

(cid:10)

(cid:11)

bΩs2 (cid:3) bΩs1 bΩs1

bΩp1 bΩs2 (cid:3) bΩ2

bΩs2 (cid:3) bΩs1 bΩs1

bΩp2 bΩs2 (cid:3) bΩ2

Ωs ¼

1 A1

Ωs

Ωs ¼

1 A2

Ωs

261

ð5:62aÞ

ð5:62bÞ

(cid:5) (cid:5) (cid:5)

In order to satisfy the passband requirement as well as to satisfy the symmetry requirement of the passband edge of the normalized low-pass ﬁlter, we have to (cid:5) (cid:5) choose the higher of Ω0 (cid:5) as Ωp. Since for the normalized ﬁlter (except for p the case of Type 2 Chebyshev ﬁlter), Ωp ¼ 1, we have to choose Ωs to be the lower of {|A1|, |A2|}. Hence, the design equations for the normalized low-pass ﬁlter (except for the Type 2 Chebyshev) (Figure 5.19) are

(cid:5) (cid:5) (cid:5) (cid:5) (cid:5) and Ω00 (cid:5) p

Ωp ¼ 1, Ωs ¼ min A1j

j; A2j

ð5:63aÞ

where

A1 ¼

bΩs1 (cid:10) bΩs2 (cid:3) bΩs1

bΩs2 (cid:3) bΩ2 (cid:11) bΩp1

, A2 ¼

bΩs1 (cid:10) bΩs2 (cid:3) bΩs1

bΩs2 (cid:3) bΩ2 (cid:11) bΩp2

and the transfer function of the required band-stop ﬁlter is

(cid:10) (cid:11) bs

HBS

¼ HN sð Þc

(cid:10)

(cid:11)

s¼

^Ω s2 (cid:3) ^Ω ^s 2 þ ^Ω s1

s1 ^Ω

Ωs

For the Type 2 Chebyshev ﬁlter, Eq. (5.63a) would be replaced by

ð5:63bÞ

ð5:63cÞ

Ωp ¼ max 1= A1j

j; 1= A2j

g, Ωs ¼ 1 j

ð5:63dÞ

Example 5.9 Design an analog band-stop Butterworth ﬁlter with the following speciﬁcations:

Lower passband edge frequency: 22.35 Hz Upper passband edge frequency: 447.37 Hz Lower stopband edge frequency: 72.65 Hz Upper stopband edge frequency: 137.64 Hz

Passband ripple: 3 dB Stopband ripple: 15 dB

262

5 Analog Filters

Figure 5.19 Low-pass to band-stop frequency transformation. (a) Prototype low-pass ﬁlter frequency response. (b) Band-stop ﬁlter frequency response

(a)

(b)

Solution From Eq. (5.63b), we have

A1 ¼

bΩs1 (cid:10) bΩs2 (cid:3) bΩs1

bΩs2 (cid:3) bΩ2 (cid:11) bΩp1

¼ 6:5403, A2 ¼

bΩs1 (cid:10) bΩs2 (cid:3) bΩs1

bΩs2 (cid:3) bΩ2 (cid:11) bΩp2

¼ (cid:3)6:5397

Now using (5.63a), we get the speciﬁcations for the normalized analog low-pass

ﬁlter to be

Ωp ¼ 1, Ωs ¼ min A1j

j; A2j

αp ¼ 3 dB, αs ¼ 15 dB

5.2 Practical Analog Low-Pass Filter Design

263

Substituting these values in Eq. (5.23), the order of the ﬁlter is given by

(cid:9) (cid:8) log 101:5(cid:3)1 100:3(cid:3)1 2 log 6:5397 ð

N (cid:5)

¼ 0:9125

We choose N ¼ 1. The transfer function of the ﬁrst-order normalized Butterworth

low-pass ﬁlter is

HN sð Þ ¼

1 s þ 1

(cid:8)

(cid:9) 2

¼ 101:5 (cid:3) 1: Substituting the values of Ωs and N in Eq. (5.20), we obtain 6:5397 Ωc Solving for Ωc, we get Ωc ¼ 1.1818. The analog transfer function of the low-pass

ﬁlter is obtained from HN(s) by substituting s ¼ s Ωc

¼ s

1:1818

HLP sð Þ ¼

1:1818 s þ 1:1818

To arrive at the analog transfer function of the band-stop ﬁlter, we use, in the the low-pass to band-stop transformation given by (5.63c),

above expression, namely,

(cid:10)

S ¼

to obtain

(cid:11)

bΩs2 (cid:3) bΩs1 s2 þ bΩs1

Ωss bΩs2

64:99

Þ 6:5397 ð s2 þ 10000

Þs

425s s2 þ 10000

HBS sð Þ ¼

s2 þ 10000 s2 þ 360s þ 10000

Example 5.10 Design a band-stop Butterworth ﬁlter for the speciﬁcations of Exam- ple 5.9 using MATLAB

Solution The following MATLAB code fragments can be used to design HBS(s): Bandwidth ¼ bw ¼ 447.37 – 22.35; Ωo ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Þ 72:65 137:64 Þ ð ð

¼ 100.

[N,Wn]=buttord(1,6.5397,3,15,‘s’); [B,A]=butter(N,Wn,‘s’); [num, den]=1p2bs(B,A,100,425.02);% [num, den]=1p2bs(B,A, Ωo, bw); The transfer function HLP (s) of the analog low-pass ﬁlter can be obtained by displaying numerator and denominator coefﬁcient vectors B and A and is given by

HLP sð Þ ¼

1:1818 s þ 1:1818

264

5 Analog Filters

The transfer function HBS(s) of the analog band-stop ﬁlter can be obtained by displaying numerator and denominator coefﬁcient vectors num and den and is given by

HBS sð Þ ¼

s2 þ 10000 s2 þ 360s þ 10000

5.3 Effect of Poles and Zeros on Frequency Response

Frequency response of a system can be obtained by evaluating H(s) for all values of s ¼ jΩ.

5.3.1 Effect of Two Complex System Poles on the Frequency

Response

Consider the following system function with complex poles

H sð Þ ¼

(cid:10)

s (cid:3) α þ jΩ ð

1 (cid:10) Þ

s (cid:3) α (cid:3) jΩ ð

ð5:64Þ

with placement of poles as shown in Figure 5.20(a). The magnitude response of

the system with pole locations shown in Figure 5.20(a) is given by

H jΩð Þ

j ¼

1 dd0

ð5:65Þ

and is shown in Figure 5.20(b), and its phase response is shown in Figure 5.20(c)

5.3.2 Effect of Two Complex System Zeros on the Frequency

Response

Consider the following system function with complex zeros

(cid:8) s (cid:3) ðα þ jΩÞÞðs (cid:3) ðα (cid:3) jΩÞÞ

HðsÞ ¼

ð5:66Þ

with placement of zeros as shown in Figure 5.21(a). The magnitude response of the system with zeros locations shown in Figure 5.21(a) is given by

and is shown in Figure 5.21(b), and its phase response is shown in Figure 5.21(c)

H jΩð

j ¼ rr0 Þ

ð5:67Þ

5.4 Design of Specialized Analog Filters by Pole-Zero Placement

265

(a)

(b)

(c)

Figure 5.20 (a) Pole locations of H(s). (b) Magnitude response. (c) Pole locations of H(s)

5.4 Design of Specialized Analog Filters by Pole-Zero

Placement

There are certain specialized ﬁlters often used in signal processing applications in addition to the ﬁlters designed in the previous sections. These specialized ﬁlters can be directly designed based on placement of poles and zeros.

266

5 Analog Filters

(a)

(b)

(c)

Figure 5.21 (a) Zero locations of H(s). (b) Magnitude response. (c) Phase response

5.4.1 Notch Filter

The notch ﬁlter removes a single frequency f0, called the notch frequency. The notch ﬁlter can be realized with two zeros placed at (cid:9)jΩ0,

where

Ω0 ¼ 2πf 0 ð As such a ﬁlter does not have unity gain at zero frequency. The notch will not be sharp. By placing two poles close to the two zeros on the semicircle as shown in Figure 5.22(a), the notch can be made sharp with unity gain at zero frequency as shown in Figure 5.22(b).

5.5 Problems

267

Figure 5.22 (a) Placing two poles close the two zeros on the semicircle. (b) Magnitude response of (a)

Example 5.11 Design a second-order notch ﬁlter to suppress 50 Hz hum in an audio signal Solution Choose Ω0 ¼ 100π. Place zeros at s ¼ (cid:9)jΩ0 and poles at –Ω0 cos θ (cid:9) jΩ0 sin θ.

Then, the transfer function of the second-order notch ﬁlter is given by

H sð Þ ¼

s (cid:3) jΩ0 s þ Ω0 cos θ þ jΩ0 sin θ s2 þ Ω0 s2 þ 2Ω0 cos θ

Þs þ Ω0

Þ s þ jΩ0 ð Þ s þ Ω0 cos θ (cid:3) jΩ0 sin θ ð

2 ¼

s2 þ 98775:5102

s2 þ 628:57 cos θ

Þs þ 98775:5102

5.5 Problems

Test the impulse response of an ideal low-pass ﬁlter for the following properties:

(i) Real valued (ii) Even (iii) Causal

268

5 Analog Filters

Consider the ﬁrst-order RC circuit shown in the ﬁgure below

(i) Determine H(Ω), the transfer function from vs to vc. Sketch the magnitude

and phase of H(Ω).

(ii) What is the cutoff frequency for H(Ω)? (iii) Consider the following system:

(a) Draw the corresponding RC circuit and determine H(Ω), the transfer

function from v to vs. Sketch the magnitude and phase of H(Ω).

(b) What is the corresponding cutoff frequency?

Design a continuous time low-pass ﬁlter with the following transfer function

H Ωð Þ ¼

α α þ jΩ

with the following speciﬁcations

Find the range of values of α that meets the speciﬁcations.

Consider the following system

Further Reading

269

If H(Ω) is an ideal band-pass ﬁlter, determine for what values of α, it will act as

an ideal band-stop ﬁlter.

Design an elliptic analog high-pass ﬁlter for the speciﬁcations of Example 5.5

Further Reading

Raut, R., Swamy, M.N.S.: Modern Analog Filter Analysis and Design: a Practical Approach.

Springer, WILEY- VCH Verlag & Co. KGaA, Weinheim, Germany (2010)

Antoniou, A.: Digital Filters: Analysis and Design. McGraw Hill Book Co., New York (1979)
Parks, T.W., Burrus, C.S.: Digital Filter Design. Wiley, New York (1987)
Temes, G.C., Mitra, S.K. (eds.): Modern Filter Theory and Design. Wiley, New York (1973)
Vlach, J.: Computerized Approximation and Synthesis of Linear Networks. Wiley, New York

(1969)

Mitra, S.K.: Digital Signal Processing. McGraw-Hill, New York (2006)
Chen, C.T.: Digital Signal Processing, Spectral Computation and Filter Design. Oxford Univer-

sity Press, NewYork/Oxford, UK (2001)

Chapter 6 Discrete-Time Signals and Systems

Discrete-time signals are obtained by the sampling of continuous-time signals. Digital signal processing deals basically with discrete-time signals, which are processed by discrete-time systems. The characterization of discrete-time signals as well as discrete-time systems in time domain is required to understand the theory of digital signal processing. In this chapter, time-domain sampling and the funda- mental concepts of discrete-time signals as well as discrete-time systems are con- sidered. First, the sampling process of analog signals is described. Next, the basic sequences of discrete-time systems and their classiﬁcation are emphasized. The input-output characterization of linear time-invariant (LTI) systems by means of convolution sum is described. Further, sampling of discrete-time signals is intro- the state-space representation of discrete-time LTI systems is duced. Finally, described.

6.1 The Sampling Process of Analog Signals

6.1.1

Impulse-Train Sampling

The acquisition of an analog signal at discrete-time intervals is called sampling. The sampling process mathematically can be treated as a multiplication of a continuous- time signal x(t) by a periodic impulse train p(t) of unit amplitude with period T. For example, consider an analog signal xa(t) as shown in Figure 6.1(a), and a periodic pulse train p(t) of unit amplitude with period T as in Figure 6.1 (b) is referred to as the sampling function, the period T as the sampling period, and the fundamental frequency ωT ¼ (2π/T ) as the sampling frequency in radians. Then, the sampled version xp(t) is shown in Figure 6.1 (c).

271

272

ax t ( )

6 Discrete-Time Signals and Systems

p(t)

…

T 2T

(b)

(a)

xp(t)

… t

(c)

Figure 6.1 (a) Continuous-time signal, (b) pulse train, (c) sampled version of (b)

In the time domain, we have

where

xp tð Þ ¼ xa tð Þp tð Þ

p tð Þ ¼

n¼(cid:2)1

δ t (cid:2) nT ð

ð6:1Þ

ð6:1aÞ

xp(t) is the impulse train with the amplitudes of the impulses equal to the samples of xa(t) at intervals T, 2T, 3T, … .

Therefore, the sampled version of signal xp(t) mathematically can be represented as

xp tð Þ ¼

n¼(cid:2)1

xa nTð

Þδ t (cid:2) nT

ð6:2Þ

6.1.2 Sampling with a Zero-Order Hold

In Section 6.1.1, the sampling process establishes a fact that the band-limited signal can be uniquely represented by its samples. In a practical setting, it is difﬁcult to generate and transmit narrow large amplitude pulses that approximate impulses.

6.1 The Sampling Process of Analog Signals

273

Lowpass filter

Sample and hold

Quantizer

Encoder

x (t)

Analog input

• (cid:129) (cid:129)

x (n)

Logic circuit

Digital output code

Figure 6.2 A block diagram representation of an analog-to-digital conversion process.

Hence, it is more convenient to implement the sampling process using a zero-order hold. It samples analog signal at a given sampling instant and holds the sample value until the succeeding sampling instant. A block diagram representation of the analog- to-digital conversion (ADC) process is shown in Figure 6.2. The amplitude of each signal sample is quantized into one of the 2b levels, where b is the number of bits used to represent a sample in the ADC. The discrete amplitude levels are encoded into distinct binary word of length b bits.

A sequence of samples x(n) is obtained from an analog signal xa(t) according to

the relation

x nð Þ ¼ xa nTð

Þ (cid:2)1 < n < 1:

ð6:3Þ

In Eq. (6.2), T is the sampling period, and its reciprocal, FT ¼ 1/T is called the sampling frequency, in samples per second. The sampling frequency FT is also referred to as the Nyquist frequency.

Sampling Theorem The sampling theorem states that an analog signal must be sampled at a rate at least twice as large as highest frequency of the analog signal to be sampled. This means that

FT (cid:3) 2f max

ð6:4Þ

where fmax is maximum frequency component of the analog signal. The frequency 2fmax is called the Nyquist rate.

For example, to sample a speech signal containing up to 3 kHz frequencies, the required minimum sampling rate is 6 kHz, that is, 6000 sample per second. To sample an audio signal having frequencies up to 22 kHz, the required minimum sampling rate is 44 kHz, that is, 44000 samples per second.

A signal whose energy is concentrated in a frequency band range fL < |f| < fH is often referred to as a band-pass signal. The sampling process of such signals is generally referred to as band-pass sampling. In the band-pass sampling process, to prevent aliasing effect, the band-pass continuous-time signal can be sampled at sampling rate greater than twice the highest frequency ( fH):

FT (cid:3) 2f H

ð6:5Þ

274

6 Discrete-Time Signals and Systems

The bandwidth of the band-pass signal is deﬁned as

Δf ¼ f H (cid:2) f L

ð6:6Þ

Consider that the highest frequency contained in the signal is an integer multiple

of the bandwidth that is given as

The sampling frequency is to be selected to satisfy the condition as

f H ¼ c Δfð

FT ¼ 2 Δfð

Þ ¼

f H c

ð6:7Þ

ð6:8Þ

6.1.3 Quantization and Coding

Quantization and coding are two primary steps involve in the process of A/D conversion. Quantization is a nonlinear and non-invertible process that rounds the given amplitude x(n) ¼ x(nT) to an amplitude xk that is taken from the ﬁnite set of values at time t ¼ nT. Mathematically, the output of the quantizer is deﬁned as

xq nð Þ ¼ Q x nð Þ

(cid:4) ¼ bxk

ð6:9Þ

The procedure of the quantization process is depicted as

x x 3

………………

The possible outputs of the quantizer (i.e., the quantization levels) are indicated (cid:5) (cid:5) (cid:5) bxL where L stands for number of intervals into which the

by bx1 bx2 bx3 bx4 signal amplitude is divided. For uniform quantization,

bxkþ1 (cid:2) bxk ¼ Δ xkþ1 (cid:2) xk ¼ Δ

k ¼ 1, 2, (cid:5) (cid:5) (cid:5), L: for finite xk, xkþ1:

ð6:10Þ

where Δ is the quantizer step size.

The coding process in an ADC assigns a unique binary number to each quanti- zation level. For L levels, at least L different binary numbers are needed. With word length of n bits, 2n distinct binary numbers can be represented. Then, the step size or the resolution of the A/D converter is given by

Δ ¼

A 2n

ð6:11Þ

where A is the range of the quantizer.

6.1 The Sampling Process of Analog Signals

275

(a)

(b)

(c)

Figure 6.3 (a) Quantizer, (b) mathematical model, (c) power spectral density of quantization noise

Quantization Error Consider an n bit ADC sampling analog signal x(t) at sampling frequency of FTas shown in Figure 6.3(a). The mathematical model of the quantizer is shown in Figure 6.3(b). The power spectral density of the quantization noise with an assump- tion of uniform probability distribution is shown in Figure 6.3(c).

If the quantization error is uniformly distributed in the range (‐Δ/2, Δ/2) as shown in Figure 6.3(b), the mean value of the error is zero, and the variance (the quantization noise power) σ2

e is given by ðΔ=2

Pqn ¼ σ2

e ¼

2 nð ÞP eð Þde ¼

Δ2

(cid:2)Δ=2

The quantization noise power can be expressed by

σ2 e ¼

quantization step2 12

A2 12

(cid:6)

1 22n ¼

A2 12

2(cid:2)2n

ð6:12Þ

ð6:13Þ

276

6 Discrete-Time Signals and Systems

The effect of the additive quantization noise on the desired signal can be quantiﬁed by evaluating the signal-to-quantization noise (power) ratio (SQNR) that is deﬁned as

SQNR ¼ 10log10

Px Pqn

is the signal power and Pqn ¼ σ2

e ¼ E e2

q nð Þ

ð6:14Þ

is the

(cid:2) where Px ¼ σ2 x ¼ E x2 nð Þ quantization noise power.

(cid:3)

6.2 Classiﬁcation of Discrete-Time Signals

6.2.1 Symmetric and Anti-symmetric Signals

A real valued signal x(n) is said to be symmetric if it satisﬁes the condition

x (cid:2)nð

Þ ¼ x nð Þ

ð6:15aÞ

Example of a symmetric sequence is shown in Figure 6.4 On the other hand, a signal x(n) is called anti-symmetric if it follows the condition

x (cid:2)nð

Þ ¼ (cid:2)x nð Þ

ð6:15bÞ

An example of anti-symmetric sequence is shown in Figure 6.5.

6.2.2 Finite and Inﬁnite Length Sequences

A signal is said to be of ﬁnite length or duration if it is deﬁned only for a ﬁnite time interval:

· · ·

-1

· · ·

Figure 6.4 An example of symmetric sequence

6.2 Classiﬁcation of Discrete-Time Signals

277

· · ·

-1

· · ·

Figure 6.5 An example of anti-symmetric sequence

(cid:2)1 < N1 (cid:7) n (cid:7) N2 < 1

ð6:16Þ

The length of the sequence is N ¼ N2 (cid:2) N1 þ 1. Thus, a ﬁnite sequence of length N has N samples. A discrete-time sequence consisting of N samples is called a N- point sequence. Any ﬁnite sequence can be viewed as an inﬁnite length sequence by adding zero-valued samples outside the range (N1, N2). Also, an inﬁnite length sequence can be truncated to produce a ﬁnite length sequence.

6.2.3 Right-Sided and Left-Sided Sequences

A right-sided sequence is an inﬁnite sequence x(n) for which x(n) ¼ 0 for n < N1, where N1 is a positive or negative integer. If N1 (cid:3) 0, the right-sided sequence is said to be causal. Similarly, if x(n) ¼ 0 for n > N2, where N2 is a positive or negative integer, then the sequence is called a left-sided sequence. Also, if N2 (cid:7) 0, then the sequence is said to be anti-causal.

6.2.4 Periodic and Aperiodic Signals

A sequence x(n) ¼ x(n + N ) for all n is periodic with a period N, where N is a positive integer. The smallest value of N for which x(n) ¼ x(n + N ) is referred as the fundamental period. A sequence is called aperiodic, if it is not periodic. An example of a periodic sequence is shown in Figure 6.6. Proposition 6.1 A discrete-time sinusoidal sequence x(n) ¼ A sin (ω0n + θ) is periodic if and only if ω0

2π is a rational number.

The rational number is deﬁned as the ratio of two integers. For the given periodic signal x(n) ¼ A sin (ω0n þ θ), its fundamental period N is obtained from the following relationship

278

· · ·

6 Discrete-Time Signals and Systems

· · ·

-5

-4

-3

· -2

· -1

· 0

-3

· 4

· 5

· 6

-9

· 10

· 11

· 12

Figure 6.6 An example of a periodic sequence

ω0 m 2π ¼ N 2π ω0

N ¼

The fundamental period of a discrete-time sinusoidal sequence satisfying the proposition 6.1 is calculated by setting m equal to a small integer that results in an integer value for N.

The fundamental period of a discrete-time complex exponential sequence can

also be calculated satisfying the proposition 6.1.

Example 6.1 Determine if the discrete-time sequences are periodic: (cid:4) (cid:5) (i) x nð Þ ¼ cos πn 4 (ii) x(n) ¼ sin2n (cid:4) (cid:5) (iii) x nð Þ ¼ sin πn 4 (iv) x nð Þ ¼ ej 5πn 8 þθ

þ cos 2n:

Solution (i) The value of ω0 in x(n) is π

Since ω0

2π ¼ 1

discrete-time sequence. The fundamental period of x(n) is given by N ¼ 2π ω0 For m ¼ 1, N ¼ 2π4 period N ¼ 8,

(cid:4) (cid:5) π ¼ 8. Hence, x nð Þ ¼ cos πn 4

8 is a rational number, it is periodic m: is periodic with fundamental

(ii) x(n) ¼ sin 2n is aperiodic because ω0N ¼ 2N ¼ 2πm is not satisﬁed for any

(iii)

(iv) The value of ω0 in x(n) is 5π 8

integer value of m in making N to be an integer. (cid:4) (cid:5) sin πn (cid:4) (cid:5) 4 aperiodic signals is aperiodic, the signal x nð Þ ¼ sin πn 4

is periodic and cos 2n is aperiodic. Since the sum of periodic and þ cos 2n is aperiodic. 16 is a rational number, it is periodic m: Þ is periodic with funda-

discrete-time sequence. The fundamental period of x(n) is given by N ¼ 2π ω0 For m ¼ 5, N ¼ 8 2π mental period N ¼ 16.

5π 5 ¼ 16: Hence, x nð Þ ¼ ej 5πn ð

: Since ω0

2π ¼ 5

8 þθ

6.2 Classiﬁcation of Discrete-Time Signals

279

6.2.5 Energy and Power Signals

The total energy of a signal x(n), real or complex, is deﬁned as

E ¼

n¼(cid:2)1

x nð Þ

By deﬁnition, the average power of an aperiodic signal x(n) is given by

P ¼ Lt

N!1

1 2N þ 1

n¼(cid:2)N

x nð Þ

ð6:17Þ

ð6:18aÞ

The signal is referred to as an energy signal if the total energy of the signal satisﬁes the condition 0 < E < 1. It is clear that for a ﬁnite energy signal, the average power P is zero. Hence, an energy signal has zero average power. On the other hand, if E is inﬁnite, then P may be ﬁnite or inﬁnite. If P is ﬁnite and nonzero, then the signal is called a power signal. Thus, a power signal is an inﬁnite energy signal with ﬁnite average power.

The average power of a periodic sequence x(n) with a period I is given by

P ¼

1 I

XI(cid:2)1

n¼0

x nð Þ j

ð6:18bÞ

Hence, periodic signals are power signals.

Example 6.2 Determine whether the sequence x(n) ¼ anu(n) is an energy signal or a power signal or neither for the following cases:

ðaÞjaj < 1, ðbÞjaj ¼ 1, ðcÞjaj > 1:

Solution For x(n) ¼ a(cid:2)nu(n), E is given by (cid:6) (cid:6)2

E ¼

(cid:6) (cid:6) x nð Þ

(cid:6) (cid:6)2

(cid:6) (cid:6)

(cid:2)1

P ¼ limN!1

1 2N þ 1

(cid:6) (cid:6)

(cid:2)1

(cid:6) (cid:6)2

x nð Þ

¼ limN!1

1 2N þ 1

(cid:6) (cid:6)

a2n

(cid:6) (cid:6)

(a) For |a| < 1,

(cid:6) (cid:6)

(cid:2)1

(cid:6) (cid:6)2

x nð Þ

E ¼

anj

j2 ¼

1 (cid:2) aj j2 is finite

P ¼ limN!1

1 2N þ 1

n¼0

(cid:6) (cid:6) a2n

(cid:6) (cid:6)

¼ limN!1

1 2N þ 1

1 (cid:2) aj j2 Nþ1

1 (cid:2) aj j2 ¼ 0

280

6 Discrete-Time Signals and Systems

The energy E is ﬁnite and the average power P is zero. Hence, the signal x(n) ¼ an

u(n) is an energy signal for |a| < 1.

(b) For |a| ¼ 1,

E ¼

P ¼ limN!1

1 2N þ 1

1 0 anj (cid:6) (cid:6) N

j2 ! 1 (cid:6) (cid:6)

a2n

n¼0

¼ limn!1

N þ 1 2N þ 1

1 2

The energy E is inﬁnite, and the average power P is ﬁnite. Hence, the signal x

(n) ¼ anu(n) is a power signal for |a| ¼ 1.

E ¼

anj

j2 ! 1

P ¼ limN!1

1 2N þ 1

n¼0

(cid:6) (cid:6)

a2n

(cid:6) (cid:6)

¼ limN!1

1 2N þ 1

aj j2 Nþ1 ð

Þ (cid:2) 1

aj j2 (cid:2) 1

! 1

The energy E is inﬁnite and also the average power P is inﬁnite. Hence, the signal

x(n) ¼ anu(n) is neither an energy signal nor a power signal for |a| > 1.

Example 6.3 Determine whether the following sequences (i) x(n) ¼ e–nu(n), (ii) x(n) ¼ enu(n), (iii) x(n) ¼ nu(n), and (iv) x(n) ¼ cosπn u(n)

are energy or power signals or neither energy nor power signals.

Solution

(i) x(n) ¼ e(cid:2)nu(n). Hence, E and P are given by

(cid:2)1

E ¼

P ¼ limN!1

¼ limN!1

n¼0

x nð Þ

j2 ¼ 1 2N þ 1 1 2N þ 1

1 (cid:2) e(cid:2)2 is finite

e(cid:2)2n ¼

x nð Þ

e(cid:2)2n

¼ limN!1

1 2N þ 1

1 (cid:2) e(cid:2)2 Nþ1 Þ

1 (cid:2) e(cid:2)2 ¼ 0

The energy E is ﬁnite and the average power P is zero. Hence, the signal x(n) ¼

e(cid:2)nu(n) is an energy signal.

6.3 Discrete-Time Systems

281

(ii) x(n) ¼ e+nu(n). Therefore, E and P are given by

E ¼

1 (cid:2)1 x nð Þ j

P ¼ lim N!1

1 2N þ 1

1 0 e2n ! 1

x nð Þ

j2 ¼ lim N!1

1 2N þ 1

n¼0

e2n ¼ lim n!1

1 2N þ 1

e2 Nþ1 ð

Þ (cid:2) 1

e2 (cid:2) 1

! 1

j2 ¼ XN

n¼0

The energy E is inﬁnite and also the average power P is inﬁnite. Hence, the signal

x(n) ¼ enu(n) is neither an energy signal nor a power signal.

(iii) x(n) ¼ nu(n). Hence, E and P are given by

1 0 n2 ! 1

E ¼

(cid:2)1 jxðnÞj2 ¼

P ¼ limN!1

¼ limN!1

1 2N þ 1 1 2N þ 1

(cid:2)1

n¼0

jxðnÞj2

n2 ¼ limN!1

NðN þ 1Þð2N þ 1Þ 6ð2N þ 1Þ

! 1

The energy E is inﬁnite and also the average power P is inﬁnite. Hence, the signal

x(n) ¼ nu(n) is neither an energy signal nor a power signal. (iv) x(n) ¼ cos πn u(n). Sincecosπn ¼ ((cid:2)1)n, E and P are given by

E ¼

1 (cid:2)1 x nð Þ j

cos πn

j2 ¼

1 0 (cid:2)1ð

Þ2n ! 1

j2 ¼ 1 2N þ 1 1 2N þ 1

1 0 X

(cid:2)1

n¼0

P ¼ limN!1

¼ limN!1

x nð Þ

(cid:2)1ð

Þ2n ¼ limN!1

N þ 1 2N þ 1

1 2

The energy E is not ﬁnite and the average power P is ﬁnite. Hence, the signal

x(n) ¼ cos πnu(n) is a power signal.

6.3 Discrete-Time Systems

A discrete-time system is deﬁned mathematically as a transformation that maps an input sequence x(n) into an output sequence y(n). This can be denoted as

y nð Þ ¼ ℜ x nð Þ

(cid:4)

ð6:19Þ

where ℜ is an operator.

282

6 Discrete-Time Signals and Systems

6.3.1 Classiﬁcation of Discrete-Time Systems

Linear Systems A system is said to be linear if and only if it satisﬁes the following conditions:

ℜ x1 nð Þ þ x2 nð Þ

ℜ ax nð Þ ½ (cid:4) ¼ ℜ x1 nð Þ ½

(cid:4) ¼ aℜ x nð Þ (cid:4) ½ (cid:4) þ ℜ x2 nð Þ ½

(cid:4) ¼ y1 nð Þ þ y2 nð Þ

ð6:20Þ ð6:21Þ

where a is an arbitrary constant and y1(n) and y2(n) are the responses of the system when x1(n) and x2(n) are the respective inputs. Equations (6.20) and (6.21) represent the homogeneity and additivity properties, respectively.

The above two conditions can be combined into one representing the principle of

superposition as

ℜ ax1 nð Þ þ bx2 nð Þ

(cid:4) ¼ aℜ x1 nð Þ ½

(cid:4) þ bℜ x2 nð Þ ½

(cid:4)

ð6:22Þ

where a and b are arbitrary constants.

Example 6.4 Check for linearity of the following systems described by the follow- ing input-output relationships: X

(i) y nð Þ ¼ (ii) y(n) ¼ x2(n) (iii) y(n) ¼ x(n (cid:2) n0), where n0 is an integer constant

x kð Þ

k¼(cid:2)1

Solution (i) The outputs y1(n) and y2(n) for inputs x1(n) and x2(n) are, respectively,

given by

y1ðnÞ ¼

x1ðkÞ

k¼(cid:2)1 Xn

y2ðnÞ ¼

x2ðkÞ

k¼(cid:2)1

The output y(n) due to an input x(n) ¼ ax1(n) þ bx2(n) is then given by

y nð Þ ¼

ax1 kð Þ þ bx2 kð Þ ¼ a

x1 kð Þ þ b

x2 kð Þ

k¼(cid:2)1

Hence the system described by y nð Þ ¼

x kð Þ is a linear system.

¼ ay1 nð Þ þ by2 nð Þ X

k¼(cid:2)1

6.3 Discrete-Time Systems

283

(ii) The outputs y1(n) and y2(n) for inputs x1(n) and x2(n) are given by

y1 nð Þ ¼ x2 y2 nð Þ ¼ x2

1 nð Þ 2 nð Þ

The output y(n) due to an input x(n) ¼ ax1(n) þ bx2(n) is then given by

y nð Þ ¼ ax1 nð Þ þ bx2 nð Þ ay1 nð Þ þ by2 nð Þ ¼ ax2

1 nð Þ þ bx2

2 nð Þ 6¼ y nð Þ

Þ2 ¼ a2x2

1 nð Þ þ 2abx1 nð Þx2 nð Þ þ b2x2

2 nð Þ

Therefore, the system y(n) ¼ x2(n) is not linear.

(iii) The outputs y1(n) and y2(n) for inputs x1(n) and x2(n), respectively, are given by

y1 nð Þ ¼ x1 n (cid:2) n0 y2 nð Þ ¼ x2 n (cid:2) n0

The output y(n) due to an input x(n) ¼ ax1(n) þ bx2(n) is then given by

y nð Þ ¼ ax1 n (cid:2) n0

Þ þ bx2 n (cid:2) n0 ð

Þ ¼ ay1 nð Þ þ by2 nð Þ

Hence, the system y(n) ¼ x(n (cid:2) n0) is linear.

Time-Invariant Systems A time-invariant system (shift invariant system) is one in which the internal param- eters do not vary with time. If y1(n) is output to an input x1(n), then the system is said to be time invariant if, for all n0, the input sequence x1(n) ¼ x(n (cid:2) n0) produces the output sequence y1(n) ¼ y(n (cid:2) n0), i.e.,

ℜ x n (cid:2) n0 ð

(cid:4) ¼ y n (cid:2) n0 ð Þ

where n0 is a positive or negative integer.

Example 6.5 Check for P

1 k¼(cid:2)1 x kð Þ

time-invariance of

the system deﬁned by

y nð Þ ¼

Solution From given Eq., the output y(n) of the system delayed by n0 can be written as

Xn(cid:2)n0

y n (cid:2) n0 ð

Þ ¼

x kð Þ

k¼(cid:2)1

For example, for an input x1(n) ¼ x(n (cid:2) n0), the output y1(n) can be written as

y1 nð Þ ¼

x k (cid:2) n0 ð

k¼(cid:2)1

284

6 Discrete-Time Signals and Systems

Substitution of the change of variables k1 ¼ k (cid:2) n0 in the above summation yields

Xn(cid:2)n0

y1 nð Þ ¼

x k1ð

Þ ¼ y n (cid:2) n0 ð

k1¼(cid:2)1

Hence, it is a time-invariant system.

Example 6.6 Check for time-invariance of the down-sampling system with a factor of 2, deﬁned by the relation

y nð Þ ¼ x 2nð

Þ (cid:2)1 < n < 1

Solution For an inputx1(n) ¼ x(n (cid:2) n0), the output y1(n) of the compressor system can be written as

From given equation,

y1 nð Þ ¼ x 2n (cid:2) n0 ð

y n (cid:2) n0 ð

Þ ¼ x 2 n (cid:2) n0 ð

Comparing the above two equations, it can be observed that y1(n) 6¼ y(n (cid:2) n0).

Thus, the down-sampling system is not time invariant.

Causal System A system is said to be causal if its output at time instant n depends only on the present and past input values, but not on the future input values.

For example, a system deﬁned by

y nð Þ ¼ x n þ 2

Þ (cid:2) x n þ 1 ð

is not causal, as the output at time instant n depends on future values of the input. But, the system deﬁned by

y nð Þ ¼ x nð Þ (cid:2) x n (cid:2) 1

is causal, since its output at time instant n depends only on the present and past values of the input.

Stable System A system is said to be stable if and only if every bounded-input sequence produces a bounded-output sequence. The input x(n) is bounded if there exists a ﬁxed positive ﬁnite value βx such that

x nð Þ

j (cid:7) β

< 1 for all n

ð6:23Þ

Similarly, the output y(n) is bounded if there exists a ﬁxed positive ﬁnite value βy

such that

y nð Þ

j (cid:7) β

< 1 for all n

ð6:24Þ

and this type of stability is called bounded-input bounded-output (BIBO) stability.

6.3 Discrete-Time Systems

285

Example 6.7 Check for stability of the system described by the following input- output relation

y nð Þ ¼ x2 nð Þ

Solution Assume that the input x(n) is bounded such that |x(n)| (cid:7) βx < 1 for all n

Then,

y nð Þ

j ¼ x nð Þ j

j2 (cid:7) β2 x

< 1

Hence, y(n) is bounded and the system is stable.

Example 6.8 Check for stability, causality, linearity, and time-invariance of the system described by ℜ[x(n)] ¼ ((cid:2)1)nx(n)

This transformation outputs the current value of x(n) multiplied by either (cid:8)1. It is stable, since it does not change the magnitude of x(n) and hence satisﬁes the

conditions for bounded-input bounded-output stability.

It is causal, because each output depends only on the current value of x(n).

Let

y1ðnÞ ¼ ℜ½x1ðnÞ(cid:4) ¼ ð(cid:2)1Þnx1ðnÞ

y2ðnÞ ¼ ℜ½x2ðnÞ(cid:4) ¼ ð(cid:2)1Þnx2ðnÞ

Then, ℜ ax1 nð Þ þ bx2 nð Þ

(cid:4) ¼ (cid:2)1ð

Þnax1 nð Þ þ (cid:2)1ð

Þnbx2 nð Þ ¼ ay1 nð Þ þ by2 nð Þ

Hence, it is linear.

y nð Þ ¼ ℜ x nð Þ

(cid:4) ¼ (cid:2)1ð

Þnx nð Þ

ℜ x n (cid:2) 1 ð

(cid:4) ¼ (cid:2)1ð

Þnx n (cid:2) 1 ð

ℜ x n (cid:2) 1 ð

(cid:4) 6¼ y n (cid:2) 1 ð

Therefore, it is not time invariant.

Example 6.9 Check for stability, causality, linearity, and time-invariance of the system described by ℜ[x(n)] ¼ x(n2) Solution Stable, since if x(n) is bounded, x(n2) is also bounded.

It is not causal, since, for example, if n ¼ 4, then the output y(n) depends upon the

future input because y(4) ¼ ℜ[x(4)] ¼ x(16)

y1 ¼ ℜ x1 nð Þ ½

(cid:4) ¼ x1 n2ð ℜ ax1 nð Þ þ bx2 nð Þ

Þ; y2 nð Þ ¼ ℜ x2 nð Þ ½ (cid:4) ¼ ax1 n2ð

Þ þ bx2 n2ð

(cid:4) ¼ x2 n2ð

Þ;

Therefore, it is linear.

¼ ay1 nð Þ þ by2 nð Þ

y nð Þ ¼ ℜ x nð Þ ½ ℜ x n (cid:2) 1 ð

(cid:4) ¼ x n2ð

(cid:4) 6¼ y n (cid:2) 1 ð

Hence, it is not time invariant.

286

6 Discrete-Time Signals and Systems

6.3.2

Impulse and Step Responses

Let the input signal x(n) be transformed by the system to generate the output signal y(n). This transformation operation is given by

y nð Þ ¼ ℜ x nð Þ

(cid:4)

ð6:25Þ

If the input to the system is a unit sample sequence (i.e., impulse input δ(n)), then the system output is called as impulse response and denoted by h(n). If the input to the system is a unit step sequence u(n), then the system output is called its step response. In the next section, we show that a linear time-invariant discrete-time system is characterized by its impulse response or step response.

6.4 Linear Time-Invariant Discrete-Time Systems

Linear time-invariant systems have signiﬁcant signal processing applications, and hence it is of interest to study the properties of such systems.

6.4.1

Input-Output Relationship

An arbitrary sequence x(n) can be expressed as a weighted linear combination of unit sample sequences given by

x nð Þ ¼

k¼(cid:2)1

x kð Þδ n (cid:2) k ð

Now, the discrete-time system response y(n) is given by

y nð Þ ¼ ℜ x nð Þ

(cid:4) ¼ ℜ

h X

k¼(cid:2)1

x kð Þδ n (cid:2) k ð

From the principle of superposition, the above equation can be written as

y nð Þ ¼

x kð Þℜ δ n (cid:2) k ½ ð

(cid:4)

k¼(cid:2)1 Let the response of the system due to input δ(n (cid:2) k) be hk(n), that is,

ð6:26Þ

ð6:27Þ

ð6:28Þ

hk nð Þ ¼ ℜ δ n (cid:2) k

(cid:4)

Then, the system response y(n) for an arbitrary input x(n) is given by X

y nð Þ ¼

k¼(cid:2)1

x kð Þhk nð Þ

6.4 Linear Time-Invariant Discrete-Time Systems

287

Since δ(n (cid:2) k) is a time-shifted version of δ(n), the response hk(n) is the time- shifted version of the impulse response h(n), since the operator is time invariant. Hence,hk(n) ¼ h(n ‐ k). Thus,

y nð Þ ¼

k¼(cid:2)1

x kð Þh n (cid:2) k ð

ð6:29Þ

The above equation for y(n) is commonly called the convolution sum and

represented by

y nð Þ ¼ x nð Þ∗h nð Þ

ð6:29aÞ

where the symbol * stands for convolution. The discrete-time convolution operates on the two sequences x(n) and h(n) to produce the third sequence y(n).

Example 6.10 Determine discrete convolution of the following sequences for large value of n:

(cid:4) (cid:5) h nð Þ ¼ 1 5

x nð Þ ¼ (cid:2)1ð

nu nð Þ Þnu nð Þ

Þn(cid:2)ku n (cid:2) k ð

Þ ¼ (cid:2)1ð

Þn

(cid:7) (cid:8) 1 5

(cid:2)1ð

Þ(cid:2)k

k¼0 (cid:10)

(cid:9)

(cid:5)

(cid:4) nþ1 1 (cid:2) (cid:2) 1 5 (cid:7) (cid:8) 1 5

1 (cid:2) (cid:2)

Solution

y nð Þ ¼ x nð Þ∗h nð Þ P

1 k¼(cid:2)1 x kð Þh n (cid:2) k (cid:7) (cid:8) k 1 5

u kð Þ (cid:2)1ð

k¼(cid:2)1

¼ (cid:2)1ð

Þn

¼ (cid:2)1ð

Þn

¼ (cid:2)1ð

Þn

(cid:10)

nþ1

(cid:7) (cid:8) k 1 5

(cid:2)

k¼0 (cid:9)

(cid:5)

(cid:4) 1 (cid:2) (cid:2)1 5 1 5

1 þ

nþ1 tends to zero and hence,

y nð Þ ¼ (cid:2)1ð

Þn 1 1:2

(cid:5) For large n, (cid:2)1 5

(cid:4)

Example 6.11 Determine discrete convolution of the following two ﬁnite duration sequences:

h nð Þ ¼

x nð Þ ¼

(cid:7) (cid:8) 1 3 (cid:7) (cid:8) n 1 5

u nð Þ

288

6 Discrete-Time Signals and Systems

Solution The impulse response h(n) ¼ 0 for n < 0; hence the given system is causal; and x(n) ¼ 0 for n < 0, therefore the sequence x(n) is causal sequence: P

y nð Þ ¼ x nð Þ∗h nð Þ ¼

n P

(cid:4) (cid:5) ¼ 1 3

(cid:4) (cid:5) 3 5

n k¼0

n(cid:2)k

(cid:4) (cid:5) (cid:4) (cid:5) k 1 n 1 k¼0 5 3 n 1 (cid:2) 3=5 ð 1 (cid:2) 3=5 ð

Þnþ1 Þ

(cid:4) (cid:5) ¼ 1 3

6.4.2 Computation of Linear Convolution

Matrix Method If the input x(n) is of length N1 and the impulse sequence h(n) is of length N2, then the convolution sequence is of length N1 + N2 (cid:2) 1. Thus, the linear convolution given by Eq. (6.29) can be written in matrix form as

xð0Þ

xð1Þ

xð2Þ

⋮

xðN1 (cid:2) 1Þ ⋮

(cid:5) (cid:5) (cid:5)

0 (cid:5) (cid:5) (cid:5) (cid:5) (cid:5) (cid:5) ⋮ (cid:5) (cid:5) (cid:5) ⋮ (cid:5) (cid:5) (cid:5) ⋮ (cid:5) (cid:5) (cid:5) ⋮ ⋮ ⋮

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

(cid:5) (cid:5) (cid:5)

xð0Þ

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

yð0Þ

yð1Þ

yð2Þ

yð3Þ

⋮

yðN1 (cid:2) 1Þ

yðN1Þ ⋮

yðN1 þ N2 (cid:2) 2Þ

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

(cid:6)

xð0Þ

xð1Þ

xð2Þ

⋮

xðN1 (cid:2) 1Þ

xð0Þ

xð1Þ

xð2Þ

⋮

xðN1 (cid:2) 1Þ

⋮

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

⋮

hð0Þ

hð1Þ

hð2Þ

hð3Þ

⋮

hðN2 (cid:2) 1Þ

⋮

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

The following example illustrates the above procedure for computation of linear

convolution.

ð6:30Þ

6.4 Linear Time-Invariant Discrete-Time Systems

289

Example 6.12 Find the convolution of h(n) ¼ {(cid:2)3, 6, 3}.

the sequences x(n) ¼ {6, (cid:2)3} and

Solution Using Eq. (6.19), the linear convolution of x(n) and h(n) is given by

6 6 6 6 6 4

7 7 7 7 7 5

yð0Þ

yð1Þ

yð2Þ

yð3Þ

6 6 6 6 6 4

(cid:2)3

0 (cid:2)3

7 7 7 7 7 5

6 6 6 6 6 4

0 (cid:2)3 (cid:2)6

7 7 7 7 7 5

(cid:2)3

6 6 6 6 6 4

7 7 7 7 7 5

(cid:2)18

(cid:2)9

Thus,

y nð Þ ¼ x nð Þ∗h nð Þ ¼ (cid:2)18; 45; 0; (cid:2)9

Graphical Method for Computation of Linear Convolution Evaluation of sum at any sample n consists of the following four important operations: (i) Time reversing or reﬂecting of the sequence h(k) about k ¼ 0 sample to give

h(–k).

(ii) Shifting the sequence h(–k) to the right by n samples to obtain h(n – k). (iii) Forming the product x(k)h(n – k) sample by sample for the desired value of n. (iv) Summing the product over the index k in y(n) for the desired value of n.

The length of the convolution sum sequence y(n) is given by n ¼ N1 + N2 (cid:2) 1,

where N1 is length of the sequence x(n) and N2 is length of the sequence h(n).

Example 6.13 Compute the convolution of the sequences of Example 6.12 using the graphical method.

Solution The sequences x(n) and h(n) are shown in Figures 6.7.

( )x n

-3

( )h n

-3

Figure 6.7 Sequences x(n) and h(n)

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6 Discrete-Time Signals and Systems

Figure 6.8 Convolution of sequences x(n) and h(n)

6.4 Linear Time-Invariant Discrete-Time Systems

291

Figure 6.9 Sequence generated by the convolution

6.4.3 Computation of Convolution Sum Using MATLAB

The MATLAB function conv(a,b) can be used to compute convolution sum of two sequences a and b as illustrated in the following example.

Example 6.14 Compute convolution sum of the sequences x(n) ¼ {2,(cid:2)1,0,0} and h(n) ¼ {(cid:2)1,2,1}, using MATLAB.

Program 6.1. Illustration of convolution

a=[ 2 -1 0 0 ];% ﬁrst sequence b=[-1 2 1];% second sequence c=conv(a,b);% convolution of ﬁrst sequence and second sequence len=length(c)-1; n=0:1:len; stem(n,c) xlabel(‘Time index n’); ylabel(‘Amplitude’); axis([0 5 -3 5])

6.4.4 Some Properties of the Convolution Sum

Starting with the convolution sum given by (6.30), namely, y(n) ¼ x(n) * h(n), we can establish the following properties:

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6 Discrete-Time Signals and Systems

The convolution sum obeys the commutative law

x nð Þ∗h nð Þ ¼ h nð Þ∗x nð Þ

ð6:31aÞ

The convolution sum obeys the associative law

x nð Þ∗h1 nð Þ

Þ∗h2 nð Þ ¼ x nð Þ∗ h1 nð Þ∗h2 nð Þ Þ

ð6:31bÞ

The convolution sum obeys the distributive law

x nð Þ∗ h1 nð Þ þ h2 nð Þ

Þ ¼ x nð Þ∗h1 nð Þ þ x nð Þ∗h2 nð Þ

ð6:31cÞ

Let us now interpret the above relations physically.

The commutative law shows that the output is the same if we interchange the roles of the input and the impulse response. This is illustrated in Figure 6.10.
To interpret the associative law, we consider a cascade of two systems whose impulse responses are h1(n) and h2(n). Then y1(n) ¼ x(n) * h(n) if x(n) is the input to the system with the impulse response h1(n). If y1(n) is now fed as the input to the system with impulse response h2(n), then the overall system output is given by

y nð Þ ¼ y1 nð Þ∗h2 nð Þ ¼ x nð Þ∗h1 nð Þ

(cid:4)∗h2 nð Þ

¼ x nð Þ∗ h1 nð Þ∗h2 nð Þ

(cid:4),

by associative law

¼ x nð Þ∗h nð Þ

ð6:32Þ

This equivalence is shown in Figure 6.11. Hence, if two systems with impulse responses h1(n) and h2(n) are cascaded, then the overall system response is given by

Figure 6.10 Interpretation of the commutative law

( )x n

nh ( ) 1

y n 1( )

y n ( )

nh ( ) 2

( )x n

≡

h n ( )

h n h n ( )* ( ) 1

( )y n

Figure 6.11 Interpretation of the associative law

6.4 Linear Time-Invariant Discrete-Time Systems

293

( )x n

y n 1( )

h n 1( )

h n 2 ( )

y n 2 ( )

Figure 6.12 Interpretation of distributive law

( )y n

( )x n

≡

h n ( ) 1

h n ( ) 2

( )y n

Figure 6.13 Input-output relations for Example 6.15

( )x n

)-y n (

( ) x n

y n ( )

y n 1( )

y n 2( )

( ) h n

( ) h n 1

h nð Þ ¼ h1 nð Þ∗h2 nð Þ

ð6:33Þ

This can be generalized to a number of LTI systems in cascade.

We now consider the distributive law given by (6.31c). This can be easily interpreted as two LTI systems in parallel and that the overall system impulse response h(n) of the two systems in parallel is given by

h nð Þ ¼ h1 nð Þ þ h2 nð Þ

ð6:34Þ

This is illustrated in Figure 6.12.

Example 6.15 Consider the system shown in Figure 6.13 with h(n) being real. If y2(n) ¼ y1(–n), ﬁnd the overall impulse response h1(n) that relates y2(n) to x(n).

Solution

y nð Þ ¼ x nð Þ∗h nð Þ

From Figure 6.13, we have the following relations:

y1(n) ¼ y((cid:2)n) ∗ h(n)

y2 nð Þ ¼ y1 (cid:2)nð

Þ ¼ y nð Þ∗h (cid:2)nð

Þ∗h (cid:2)nð ¼ x nð Þ∗h nð Þ ¼ x nð Þ∗ h nð Þ∗h (cid:2)nð ð

Þ Þ ¼ x nð Þ∗h1 nð Þ

Hence, the overall impulse response ¼ h1(n) ¼ h(n) ∗ h(–n)

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6 Discrete-Time Signals and Systems

x n ( )

( )nh

y n ( )

Figure 6.14 Interconnection of three causal LTI systems

Example 6.16 Consider the cascade interconnection of three causal LTI systems as shown in Figure 6.14. The impulse response h2(n) is given by

h2 nð Þ ¼ u nð Þ (cid:2) u n (cid:2) 2

and the overall impulse response h(n) ¼ {1,5,10,11,8,4,1}. Determine the impulse response h1(n).

Solution Let the overall impulse response of the cascaded system be h(n). Hence,

h nð Þ ¼ h1 nð Þ∗h2 nð Þ∗h2 nð Þ

Since the convolution is associative in nature,

h nð Þ ¼ h1 nð Þ∗ h2 nð Þ∗h2 nð Þ ð

Let h3(n) ¼ h2(n) * h2(n). Since h2(n) is nonzero for n ¼ 0 and 1 only, h3(n) can be written as

h3 nð Þ ¼

h2 kð Þh2 n (cid:2) k ð

k¼0

Therefore, h3 0ð Þ ¼

X 1

k¼0

h2 kð Þh2 (cid:2)kð

Þ ¼ 1:1 þ 1:0 ¼ 1

h3 1ð Þ ¼

h3 2ð Þ ¼

k¼0 X1

k¼0

h2 kð Þh2 1 (cid:2) k ð

Þ ¼ 1:1 þ 1:1 ¼ 2

h2 kð Þh2 2 (cid:2) k ð

Þ ¼ 1:0 þ 1:1 ¼ 1

Thus, we obtain

h3 nð Þ ¼ 1; 2; 1 f

Now, h(n) is nonzero in the interval 0 to 6 and h3(n) is nonzero in the interval

0 to 2:

h nð Þ ¼ h1 nð Þ∗h3 nð Þ

Hence, h1(n) will be nonzero in the interval 0 to 4. Then, we have

6.4 Linear Time-Invariant Discrete-Time Systems

295

h nð Þ ¼ h1 nð Þ∗h3 nð Þ ¼

k¼0

h1 kð Þh3 n (cid:2) k ð

Let h1(n) ¼ {a1, a2, a3, a4, a5}. Therefore, we have

h 0ð Þ ¼

h1 kð Þh3 (cid:2)kð

Þ ¼ a1 (cid:5) 1 ¼ 1

k¼0

) a1 ¼ 1:

h 1ð Þ ¼

h1 kð Þh3 1 (cid:2) k ð

Þ ¼ a1 (cid:5) 1 þ a2 (cid:5) 2 ¼ 5

k¼0

) a2 ¼ 3

h 2ð Þ ¼

h1 kð Þh3 2 (cid:2) k ð

Þ ¼ a1 (cid:5) 1 þ a2 (cid:5) 2 þ a3 (cid:5) 1 ¼ 101

k¼0

) a3 ¼ 3:

h 3ð Þ ¼

h1 kð Þh3 3 (cid:2) k ð

Þ ¼ a2 (cid:5) 1 þ a3 (cid:5) 2 þ a4 (cid:5) 1 ¼ 11

k¼0

) a4 ¼ 2:

h 4ð Þ ¼

h1 kð Þh3 4 (cid:2) k ð

Þ ¼ a3 (cid:5) 1 þ a4 (cid:5) 2 þ a5 (cid:5) 1 ¼ 8

k¼0

) a5 ¼ 1:

Thus,

h1 nð Þ ¼ 1; 3; 3; 2; 1 f

6.4.5 Stability and Causality of LTI Systems in Terms

of the Impulse Response

The output of a LTI system can be expressed as (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) k¼(cid:2)1

h kð Þx n (cid:2) k ð

(cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:7)

y nð Þ

j ¼

h kð Þ

j x n (cid:2) k j

k¼(cid:2)1

For bounded input x(n)

we have

x nð Þ

j (cid:7) β

< 1

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6 Discrete-Time Signals and Systems

y nð Þ j

j (cid:7) β

k¼(cid:2)1

h kð Þ

ð6:35Þ

h kð Þ j j

ð6:36Þ

It is seen from (6.35) that y(n) is bounded if and only if

k¼(cid:2)1 bounded. Hence, the necessary and sufﬁcient condition for stability is that

k¼(cid:2)1

S ¼

h kð Þ

j < 1:

The output y(n0) of a LTI causal system can be expressed as

y n0ð

Þ ¼

h kð Þx n0 (cid:2) k ð

k¼(cid:2)1

¼ h (cid:2)1ð

Þx n0 þ 1

ð þh 0ð Þx n0ð

Þ þ … … : þ h (cid:2)2ð Þx n0 þ 2 Þ þ h 2ð Þx n0 (cid:2) 2 ð For a causal system, the output at n ¼ n0 should not depend on the future inputs.

Þ þ h 1ð Þx n0 (cid:2) 1

Þ þ … ::

Þx n0 þ 1

Þ þ h (cid:2)1ð

Hence, in the above equation, h(k) ¼ 0 for k < 0.

Thus, it is clear that for causality of a LTI system, its impulse response sequence

h nð Þ ¼ 0

for n < 0:

ð6:37Þ

Example 6.17 Check for the stability of the systems with the following impulse responses:

(i) Ideal delay, h(n) ¼ δ(n (cid:2) nd); (ii) forward difference, h(n) ¼ δ(n þ 1) (cid:2) δ(n). (iii) Backward difference, h(n) ¼ δ(n) (cid:2) δ(n (cid:2) 1); (iv) h(n) ¼ u(n). (v) h(n) ¼ anu(n), where |a| < 1, and (vi) h(n) ¼ anu(n), where |a| (cid:3) 1.

Solution Given impulse responses of the systems, stability of each system can be tested by computing the sum

k¼(cid:2)1

S ¼

h kð Þ j j

In case of (i), (ii), and (iii), it is clear that S < 1. As such, the systems

corresponding to (i), (ii), and (iii) are stable.

For the impulse response given in (iv), the system is unstable since

S ¼

n¼0

u nð Þ ¼ 1:

This is an example of an inﬁnite-duration impulse response (IIR) system. In case of (v), S ¼

aj jn. For |a| < 1, S < 1, and hence the system is stable.

n¼0

This is an example of a stable IIR system.

Finally, in case of (vi), |a| (cid:3) 1, and the sum is inﬁnite, making the system

unstable.

6.5 Characterization of Discrete-Time Systems

297

Example 6.18 Check the following systems for causality:

(cid:4) (cid:5) (i) h nð Þ ¼ 3 (cid:4) (cid:5) 4 (iii) h nð Þ ¼ 1 2

(cid:4) (cid:5) nu n þ 2 nu nð Þ, (ii) h nð Þ ¼ 1 ð (cid:4) (cid:5) 2 nu (cid:2)n (cid:2) 1 Þ, (iv) h nð Þ ¼ 3 4

(cid:4) (cid:5) Þ þ 3 4 nj j, and (v) h(n) ¼ u(n þ 1) (cid:2) u(n)

nu nð Þ,

Solution (i) h(n) ¼ 0 for n < 0; hence the system is causal. (ii) h(n) 6¼ 0 for n < 0; hence the system is not causal. (iii) h(n) 6¼ 0 for n < 0; thus, the system is not causal. (cid:4) (cid:5) (iv) h nð Þ ¼ 3 4 (v) h(n) ¼ u(n þ 1) – u(n), and h(n) 6¼ 0 for n < 0; so, the system is not causal.

nj j; hence h(n) 6¼ 0 for n < 0; so, the system is not causal.

Example 6.19 Check the following systems for stability: (cid:4) (cid:5) nu n (cid:2) 1 (i) h nð Þ ¼ 1 ð (cid:4) (cid:5) 3 (iv) h nð Þ ¼ sin nπ 4

Þ, (ii) h(n) ¼ u(n þ 2) (cid:2) u(n (cid:2) 5), (iii) h(n) ¼ 5nu((cid:2)n (cid:2) 3), (cid:4) (cid:5) nj j cos πn 4

(cid:4) (cid:5) u nð Þ, and (v) h nð Þ ¼ 1 2

Solution

(i) The system is stable, since S ¼ (ii) h(n) ¼ u(n þ 2) – u(n – 5). The system is stable, since S is ﬁnite.

h kð Þ j

k¼(cid:2)1

j < 1:

X(cid:2)3

h nð Þ j

j ¼

5n ¼

n¼(cid:2)1

n¼3

(cid:7) (cid:8) 1 5

< 1: Therefore,

(iii) h(n) ¼ 5nu(–n – 3). Hence,

(cid:4) (cid:5) (iv) h nð Þ ¼ sin nπ 4

the system is stable. u nð Þ

Summing |h(n)| over all positive n, we see that S tends to inﬁnity. Hence, the

system is not stable.

(cid:4) (cid:5) (v) h nð Þ ¼ 1 2

(cid:4) (cid:5) nj j cos πn 4

(cid:4) (cid:5) |h(n)| is upper bounded by 1 2

system is stable.

nj j. Thus, S ¼

k¼(cid:2)1

h kð Þ j

j < 1: Hence the

6.5 Characterization of Discrete-Time Systems

Discrete-time systems are characterized in terms of difference equations. An impor- tant class of LTI discrete-time systems is one that is characterized by a linear difference equation with constant coefﬁcients. Such a difference equation may be of two types, namely, non-recursive and recursive.

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6 Discrete-Time Signals and Systems

6.5.1 Non-Recursive Difference Equation

A non-recursive LTI discrete-time system is one that can be characterized by a linear constant coefﬁcient difference equation of the form

y nð Þ ¼

bmx n (cid:2) m ð

m¼(cid:2)1

ð6:38Þ

where bm’s represent constants. By assuming causality, the above equation can be written as

y nð Þ ¼

m¼0

bmx n (cid:2) m Þ ð

ð6:39Þ

In addition, if x(n) ¼ 0 for n < 0 and bm ¼ 0 for m > N, then Eq. (6.39) becomes

y nð Þ ¼

m¼0

bmx n (cid:2) m Þ ð

ð6:40Þ

Thus an LTI, causal, non-recursive system can be characterized by an Nth-order linear non-recursive difference equation. The Nth-order non-recursive difference equation has a ﬁnite impulse response (FIR). Therefore, an FIR ﬁlter is characterized by a non-recursive difference equation.

6.5.2 Recursive Difference Equation

The response of a discrete-time system depends on the present and previous values of the input as well as the previous values of the output. Hence a linear time-invariant causal, recursive discrete-time system can be represented by the following Nth-order linear recursive difference equation:

y nð Þ ¼

m¼0

bmx n (cid:2) m ð

Þ (cid:2)

amy n (cid:2) m Þ ð

m¼1

ð6:41Þ

where am and bm are constants. An Nth-order recursive difference equation has an inﬁnite impulse response. Hence, an inﬁnite impulse response (IIR) ﬁlter is charac- terized by a recursive difference equation.

Example 6.20 An initially relaxed LTI system was tested with an input signal x(n) ¼ u(n) and found to have a response as shown in Table 6.1.

(i) Obtain the impulse response of the system. (ii) Deduce the difference equation of the system.

6.5 Characterization of Discrete-Time Systems

299

Table 6.1 Response of an LTI system for an input x(n) ¼ u(n)

n y(n)

1 1

2 2

3 4

4 6

5 10

… … … …

100 10

… … … …

Solution

(i) From Table 6.1, it can be observed that the response y(n) for an input x(n) ¼ u(n)

is given by

y nð Þ ¼ 1; 2; 4; 6; 10; 10; 10; … … ::

Similarly, for an input x(n) ¼ u(n(cid:2)1), the response y(n-1) is given by

y n (cid:2) 1 ð

Þ ¼ 0; 1; 2; 4; 6; 10; 10; 10; … … ::

For an input x(n) ¼ u(n)(cid:2)u(n(cid:2)1), the response of an LTI system is the impulse

response h(n) given by

h nð Þ ¼ y nð Þ (cid:2) y n (cid:2) 1

Þ ¼ 1; 1; 2; 2; 4

(ii) The difference equation is given by

y nð Þ ¼

h mð Þx n (cid:2) m ð

m¼0

Hence, the difference equation of the system can be written as

yðnÞ ¼ xðnÞ þ 1xðn (cid:2) 1Þ þ 2xðn (cid:2) 2Þ þ 2xðn (cid:2) 3Þ þ 4xðn (cid:2) 4Þ

6.5.3 Solution of Difference Equations

A general linear constant coefﬁcient difference equation can be expressed as

y nð Þ ¼ (cid:2)

k¼1

aky n (cid:2) k ð

Þ þ

k¼0

bkx n (cid:2) k ð

ð6:42Þ

The solution of the difference equation is the output response y(n). It is the sum of

two components which can be computed independently as

y nð Þ ¼ yc nð Þ þ yp nð Þ

ð6:43aÞ

where yc(n) is called the complementary solution and yp(n) is called the particular solution.

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6 Discrete-Time Signals and Systems

The complementary solution yc(n) is obtained by setting x(n) ¼ 0 in Eq. (6.42).

Thus yc(n) is the solution of the following homogeneous difference equation:

k¼0

aky n (cid:2) k ð

Þ ¼ 0

ð6:43bÞ

where a0 ¼ 1. To solve the above homogeneous difference equation, let us assume that

yc nð Þ ¼ λn

ð6:43cÞ

where the subscript c indicates the solution to the homogeneous difference equation. Substituting yc(n) in Eq. (6.43b), the following equation can be obtained:

k¼0 akλn(cid:2)k ¼ 0

(cid:2)

¼ λn(cid:2)N λN þ a1λN(cid:2)1 þ … :: þ aN(cid:2)1λ þ aN

(cid:3)

¼ 0

ð6:44Þ

which takes the form:

λN þ a1λN(cid:2)1 þ … :: þ aN(cid:2)1λ þ aN ¼ 0

ð6:45Þ

The above equation is called the characteristic equation, which consists of N roots represented by λ1, λ2, … … , λN. If the N roots are distinct, then the complementary solution can be expressed as

yc nð Þ ¼ α1λ n

1 þ α2λ n

2 þ … :: þ αNλ n

ð6:46Þ

where α1, α2, … … , αN are constants which can be obtained from the speciﬁed initial conditions of the discrete-time system. For multiple roots, the complementary solution yc(n) assumes a different form. In the case when the root λ1 of the characteristic equation is repeated m times, but λ2, … … , λN are distinct, then the complementary solution yc(n) assumes the form

(cid:4) α1 þ α2n þ … :: þ αmnm(cid:2)1

(cid:5)

λ n 1

þ β

2 þ … þ β λ n

N(cid:2)M

λ n N(cid:2)M

ð6:47Þ

In case the characteristic equation consists of complex roots λ1, λ2 ¼ a (cid:8) jb, then the complementary solution results in yc(n) ¼ (a2 + b2)n/2 (C1 cos nq + C2 sin nq), where q ¼ tan–1b/a and C1 and C2 are constants.

We now look at the particular solution yp (n) of Eq. (6.42) The particular solution yp(n) is any solution that satisﬁes the difference equation for the speciﬁc input signal x(n), for n (cid:3) 0, i.e.,

y nð Þ þ

k¼1

aky n (cid:2) k ð

Þ ¼

k¼0

bkx n (cid:2) k ð

ð6:48Þ

6.5 Characterization of Discrete-Time Systems

301

The procedure to ﬁnd the particular solution yp(n) assumes that yp(n) depends on the form of x(n). Thus, if x(n) is a constant, then yp(n) is implicitly a constant. Similarly, if x(n) is a sinusoidal sequence, then yp(n) is implicitly a sinusoidal sequence and so on.

In order to ﬁnd out the overall solution, the complementary and particular

solutions must be added. Hence,

y nð Þ ¼ yc nð Þ þ yp nð Þ

ð6:49Þ

Example 6.21 Determine impulse response for the case of x(n) ¼ δ(n) of a discrete- time system characterized by the following difference equation:

y nð Þ þ 2y n (cid:2) 1

Þ (cid:2) 3y n (cid:2) 2 ð

Þ ¼ x nð Þ

ð6:50Þ

Solution First, we determine the complementary solution by setting x(n) ¼ 0 and y(n) ¼ λn in Eq. (6.50), which gives us

λn þ 2λn(cid:2)1 (cid:2) 3λn(cid:2)2 ¼ λn(cid:2)2 λ2 þ 2λ (cid:2) 3

(cid:4)

(cid:5)

¼ λn(cid:2)2 λ (cid:2) 1 ð

Þ λ þ 3 ð

Þ ¼ 0

Hence, the zeros of the characteristic polynomial λ2 þ 2λ (cid:2) 3 are λ1 ¼ (cid:2)3 and

λ2 ¼ 1.

Therefore, the complementary solution is of the form

yc nð Þ ¼ α1 (cid:2)3ð

Þn þ α2 1ð Þn

ð6:51Þ

For impulse x(n) ¼ δ(n), x(n) ¼ 0 for n > 0 and x(0) ¼ 1. Substituting these

relations in Eq. (6.50) and assuming that y((cid:2)1) ¼ 0 and y((cid:2)2) ¼ 0, we get

y 0ð Þ þ 2y (cid:2)1ð

Þ (cid:2) 3y (cid:2)2ð

Þ ¼ x 0ð Þ ¼ 1,

i.e., y(0) ¼ 1. Similarly y(1) þ 2y(0) – 3y(–1) ¼ x(1) ¼ 0 yields y(1) ¼ –2. Thus, from Eq. (6.51), we get

α1 þ α2 ¼ 1 and (cid:2)3α1 þ α2 ¼ (cid:2)2

Solving these two equations, we obtain α1 ¼ 3/4 and α2 ¼ 1/4. Since x(n) ¼ 0 for n > 0, there is no particular solution. Hence, the impulse

response is given by

h nð Þ ¼ yc nð Þ ¼ 0:75 (cid:2)3ð

Þn þ 0:25 1ð Þn

ð6:52Þ

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6 Discrete-Time Signals and Systems

Example 6.22 A discrete-time system is characterized by the following difference equation:

y nð Þ þ 5y n (cid:2) 1

Þ þ 6y n (cid:2) 2 ð

Þ ¼ x nð Þ

ð6:53Þ

Determine the step response of the system, i.e., x(n) ¼ u(n).

Solution For the given difference equation, total solution is given by

y nð Þ ¼ yc nð Þ þ yp nð Þ

First, we determine the complementary solution by setting x(n) ¼ 0 and y(n) ¼ λn

in Eq. (6.53), which give us

λn þ 5λn(cid:2)1 þ 6λn(cid:2)2 ¼ λn(cid:2)2 λ2 þ 5λ þ 6

(cid:4)

(cid:5)

¼ 0

Hence, the zeros of the characteristic polynomial λ2 + 5λ þ 6 are λ1 ¼ –3 and

λ2 ¼ –2.

Therefore, the complementary solution is of the form yc(n) ¼ α1((cid:2)3)n + α2((cid:2)2)n The particular solution for the step input is of the form yp(n) ¼ K For n > 2, substituting yp(n) ¼ K and x(n) ¼ 1 in Eq. (6.53), we get 12, and yp nð Þ ¼ 1 12.

K þ 5K þ 6K ¼ 1; K ¼ 1

Therefore, the solution for given difference equation is

y nð Þ ¼ α1 (cid:2)3ð

Þn þ α2 (cid:2)2ð

Þn þ

1 12

ð6:54Þ

For n ¼ 0, Eq. (6.53) becomes y(0) þ 5y(–1) þ 6y(–2) ¼ x(0) Assuming y(–1) ¼ y(–2) ¼ 0, from the above equation, we get y(0) ¼ x(0) ¼ 1

and for n ¼ 1, y(1) þ 5y(0) þ 6y(–1) ¼ x(1) ¼ 1, i.e., y(1) ¼ –4.

Then, we get from Eq. (6.54) α1 þ α2 þ 1 12 ¼ 1

1 12 Solving these equations, we arrive at α1 ¼ 27 12 and α2 ¼ (cid:2)16 12 . Then, the step response is given by

(cid:2)3α1 (cid:2) 2α2 þ

¼ (cid:2)4

y nð Þ ¼

27 12

(cid:2)3ð

Þn (cid:2)

16 12

(cid:2)2ð

Þn þ

1 12

ð6:55Þ

6.5 Characterization of Discrete-Time Systems

303

Example 6.23 A discrete-time system is characterized by the following difference equation:

y nð Þ (cid:2) 2y n (cid:2) 1

Þ þ y n (cid:2) 2 ð

Þ ¼ x nð Þ (cid:2) x n (cid:2) 1

ð6:56Þ

Determine the response y(n), n (cid:3) 0 when the system input is x(n) ¼ (–1)nu(n) and

the initial conditions are y((cid:2)1) ¼ 1 and y((cid:2)2) ¼ (cid:2)1.

Solution For the given difference equation, the total solution is given by

y nð Þ ¼ yc nð Þ þ yp nð Þ

First, determine the complementary solution by setting x(n) ¼ 0 and y(n) ¼ λn in

Eq. (6.56); this gives

λn (cid:2) 2λn(cid:2)1 þ λn(cid:2)2 ¼ λn(cid:2)2 λ2 (cid:2) 2λ þ 1

(cid:4)

(cid:5)

¼ 0

Hence, the zeros of the characteristic polynomial λ2 (cid:2) 2λ þ 1 are λ1 ¼ λ2 ¼ 1. It has repeated roots; thus, the complementary solution is of the form yc(n) ¼ 1n(α1 + nα2). The particular solution for the step input is of the form yp(n) ¼ K((cid:2)1)nu(n). Substituting x(n) ¼ (–1)nu(n) and yp(n) ¼ K(-1)nu(n) in Eq. (6.56), we get K(–1)nu(n) – 2K(–1)n–1u(n – 1) þ K(–1)n–2u(n – 2) ¼ (–1)nu(n) – (–1)n–1u(n – 1) For n ¼ 2, the above equation becomes K + 2K + K ¼ 2; K ¼ 1 2. Therefore, the particular solution is given by

yp nð Þ ¼

1 2

(cid:2)1ð

Þnu nð Þ

Then, the total solution for given difference equation is

y nð Þ ¼ 1n α1 þ nα2 ð

Þ þ

1 2

(cid:2)1ð

Þnu nð Þ:

ð6:57Þ

For n ¼ 0, Eq. (6.56) becomes

y 0ð Þ (cid:2) 2y (cid:2)1ð

Þ þ y (cid:2)2ð

Þ ¼ 1

Using the initial conditions y(–1) ¼ 1 and y(–2) ¼ –1, we get y(0) ¼ 4. Then, for n ¼ 1, from Eq. (6.56), we get y(1) ¼ 5. Thus, we get from Eq. (6.57)

α1 þ 1=2 Þ ¼ 4 ð α1 þ α2 (cid:2) 1=2

Þ ¼ 5

Solving these two equations, we arrive at α1 ¼ (7/2) and α2 ¼ 2. Thus, the

response of the system for the given input is

304

6 Discrete-Time Signals and Systems

(cid:7) y nð Þ ¼ 1n 7 2

(cid:8)

þ 2n

(cid:2)1ð

Þnu nð Þ

1 2

ð6:58Þ

6.5.4 Computation of Impulse and Step Responses Using

MATLAB

The impulse and step responses of LTI discrete-time systems can be computed using MATLAB function:

y ¼ filter b; a; x

where b and a are the coefﬁcient vectors of difference equation describing the system, x is the input data vector, and y is the vector generated assuming zero initial conditions. The following example illustrates the computation of the impulse and step responses of an LTI system.

Example 6.24 Determine the impulse and step responses of a discrete-time system described by the following difference equation:

y nð Þ (cid:2) 2y n (cid:2) 1

Þ ¼ x nð Þ þ 0:1x n (cid:2) 1

Þ (cid:2) 0:06x n (cid:2) 2

Solution Program 6.2 is used to compute and plot the impulse and step responses, which are shown in Figure 6.15a and b, respectively.

Program 6.2: Illustration of Impulse and Step Response Computation

clear;clc; ﬂag=input(‘enter 1 for impulse response, and 2 for step response’); len=input(‘enter desired response length=‘); b=[l -2];%b coefﬁcients of the difference equation a=[l 0.l -0.06]; %a coefﬁcients of the difference equation if ﬂag==l; x=[l,zeros(l,len-l)]; end if ﬂag==2; x=[ones(1,len)]; end y=ﬁlter(b,a,x); n=0:1:len-1; stem(n,y) xlabel(‘Time index n’); ylabel(‘Amplitude’);

6.6 Sampling of Discrete-Time Signals

305

Figure 6.15 (a) Impulse response and (b) step response for Example 6.24

6.6 Sampling of Discrete-Time Signals

It is often necessary to change the sampling rate of a discrete-time signal, i.e., to obtain a new discrete-time representation of the underlying continuous-time signal of the form x’(n) ¼ xa(nT’). One approach to obtain the sequence x’(n) from x(n) is to reconstruct xa(t) from x(n) and then resample xa(t) with period T’ to obtain x’(n).

306

6 Discrete-Time Signals and Systems

x n ( )

dx n ( )

x nM= (

)

Sampling period T

Sampling period

T =’

Figure 6.16 Block diagram representation of a down sampler

x n ( )

ex n ( )

x n L ( /

)

Sampling period T

Sampling period

T =’

Figure 6.17 Block diagram representation of an up-sampler

Often, however, this is not a desirable approach, because of the non-ideal analog reconstruction ﬁlter, DAC, and ADC that would be used in a practical implementation. Thus, it is of interest to consider methods that involve only discrete-time operation.

6.6.1 Discrete-Time Down Sampler

The block diagram representation of a down sampler, also known as a sampling rate compressor, is depicted in Figure 6.16.

The down-sampling operation is implemented by deﬁning a new sequence xd(n) in which every Mth sample of the input sequence is kept and (M(cid:2)1) in-between samples are removed to obtain the output sequence, i.e., xd(n) is identical to the sequence obtained from xa(t) with a sampling period T’ ¼ MT

xd nð Þ ¼ x nMð

ð6:59Þ

For example, if x(n) ¼ {2,6,3,0,1,2,–5,2,4,7,–1,1,–2,…}, then xd(n) ¼ {2, 1, 4, –2, …} for M ¼ 4, i.e., M(cid:2)1 ¼ 3 samples are left in

between the samples of x(n) to get xd(n).

6.6.2 Discrete-Time Up-Sampler

The block diagram representation of an up-sampler, also called a sampling rate expander or simply an interpolator, is shown in Figure 6.17.

The output of an up-sampler is given by

xe nð Þ ¼

k¼(cid:2)1

x kð Þδ n (cid:2) kL ð

Þ ¼ x

(cid:9) (cid:10) n L

n ¼ 0, (cid:8) L, (cid:8) 2L, … … ::

ð6:60Þ

¼ 0

otherwise

6.7 State-Space Representation of Discrete-Time LTI Systems

307

Eq. (6.60) implies that the output of an up-sampler can be obtained by inserting (L – 1) equidistant zero-valued samples between two consecutive samples of the input sequence x(n), i.e., xe(n) is identical to the sequence obtained from xa(t) with a sampling period T’ ¼ T/L. For example, then xe(n) ¼ {2,0,0,0,1,0,0,0,4,0,0,0,–2,0,0,0, …} for L ¼ 4, i.e., L(cid:2)1 ¼ 3 zero- valued samples are inserted in between the samples of x(n) to get xe(n).

if xe(n) ¼ {2,1,4,–2,

… .},

6.7 State-Space Representation of Discrete-Time

LTI Systems

6.7.1 State-Space Representation of Single-Input

Single-Output Discrete-Time LTI Systems

Consider a single-input single-output discrete-time LTI system described by the following Nth-order difference equation:

y nð Þ þ a1y n (cid:2) 1

Þ þ a2y n (cid:2) 2 ð

Þ þ … þ aNy n (cid:2) N

Þ ¼ ℧ nð Þ

ð6:61Þ

where y(n) is the system output and ℧(n) is the system input. Deﬁne the following useful set of state variables:

x1 nð Þ ¼ y n (cid:2) N Þ, x2 nð Þ ¼ y n (cid:2) N þ 1 ð xN nð Þ ¼ y n (cid:2) 1 Þ

ð ð

Þ, x3 nð Þ ¼ y n (cid:2) N þ 2 ð

Þ, … ,

ð6:62Þ

Then from Eqs. (6.61) and (6.62), we get

x1 n þ 1

Þ ¼ x2 nð Þ

x2 n þ 1

Þ ¼ x3 nð Þ

⋮

xN(cid:2)1 n þ 1 ð

Þ ¼ xN nð Þ

Þ ¼ (cid:2)aNx1 nð Þ (cid:2) aN(cid:2)1 x2 nð Þ (cid:2) (cid:5) (cid:5) (cid:5) (cid:2) a1xN nð Þ þ ℧ nð Þ

xN n þ 1

and

y nð Þ ¼ xN n þ 1

Eqs. (6.63a) and (6.63b) can be written in matrix form as

ð6:63aÞ

ð6:63bÞ

308

6 Discrete-Time Signals and Systems

7 7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 6 4

⋮

℧ nð Þ

ð6:64aÞ

℧ nð Þ

ð6:64bÞ

ð6:65Þ

ð6:66aÞ

ð6:66bÞ

6 6 6 6 6 6 6 6 4

x1 n þ 1

x2 n þ 1 ⋮

xN(cid:2)1 n þ 1 Þ ð

7 7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 6 4

⋮

(cid:5) (cid:5) (cid:5)

⋱ ⋮

(cid:5) (cid:5) (cid:5)

7 7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 6 4

x1 nð Þ

x2 nð Þ ⋮

xN(cid:2)1 nð Þ

xN n þ 1

(cid:2)aN (cid:2)aN(cid:2)1 (cid:2)aN(cid:2)2

(cid:5) (cid:5) (cid:5) (cid:2)a1

xN nð Þ

y nð Þ ¼ (cid:2)aN (cid:2)aN(cid:2)1 (cid:2)aN(cid:2)2 (cid:5) (cid:5) (cid:5) (cid:2)a1

(cid:4)

6 6 6 6 6 6 6 4

x1 nð Þ

x2 nð Þ ⋮

xN(cid:2)1 nð Þ

xN nð Þ

7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 4

7 7 7 7 7 7 7 5

0 ⋮

Deﬁne a Nx1 dimensional vector called state vector as

X nð Þ ¼

6 6 6 6 6 6 6 4

7 7 7 7 7 7 7 5

x1 nð Þ

x2 nð Þ ⋮

xN(cid:2)1 nð Þ

xN nð Þ

More compactly Eqs. (6.64a) and (6.64b) can be written as

X n þ 1

Þ ¼ AX nð Þ þ b℧ nð Þ

y nð Þ ¼ cX nð Þ þ d℧ nð Þ

where

A ¼

6 6 6 6 6 6 6 4

0 ⋮

(cid:5) (cid:5) (cid:5)

0 1 (cid:5) (cid:5) (cid:5) ⋮ ⋱ ⋮

(cid:5) (cid:5) (cid:5)

(cid:2)aN (cid:2)aN(cid:2)1 (cid:2)aN(cid:2)2

(cid:5) (cid:5) (cid:5) (cid:2)a1

7 7 7 7 7 7 7 5

; b ¼

7 7 7 7 7 7 7 5

;

6 6 6 6 6 6 6 4

0 ⋮

c ¼ (cid:2)aN (cid:2) aN(cid:2)1 (cid:2) aN(cid:2)2 (cid:5) (cid:5) (cid:5) (cid:2) a1

(cid:4); d ¼ 1:

Equation (6.66a) and (6.66b) is called N-dimensional state-space representation

or state equations of the discrete-time system.

6.7 State-Space Representation of Discrete-Time LTI Systems

309

Example 6.25 Obtain the state-space representation of a discrete-time system described by the following differential equation:

y n (cid:2) 3 ð

Þ þ 2y n (cid:2) 2 ð

Þ þ 3y n (cid:2) 1 ð

Þ þ 4y nð Þ ¼ ℧ nð Þ

Solution The order of the differential equation is three. We have to choose three state variables:

x1(n) ¼ y(n (cid:2) 3), x2(n) ¼ y(n (cid:2) 2), x3(n) ¼ y(n (cid:2) 1)

Let Then

x1 n þ 1

Þ ¼ x2 nð Þ

x3 n þ 1

Þ ¼ (cid:2)

x2 n þ 1 1 4

Þ ¼ x3 nð Þ 1 2 y nð Þ ¼ x3 n þ 1

x2 nð Þ (cid:2)

x1 nð Þ (cid:2)

3 4

x3 nð Þ þ

℧ nð Þ

1 4

The state-space representation in matrix form is given by

7 7 7 5 ¼

6 6 6 6 4

6 6 6 4

x1 n þ 1

x2 n þ 1

x3 n þ 1

(cid:11)

y nð Þ ¼ (cid:2)

1 4

(cid:2)

1 2

(cid:2)

3 4

(cid:2) 3

7 7 7 5 þ

(cid:2)

1 4 2

(cid:12)

6 6 6 4

3 4

(cid:2)

1 2 x1 nð Þ

x2 nð Þ

x3 nð Þ

7 7 7 5 þ

6 6 6 6 4

7 7 7 7 5

6 6 6 4

x1 nð Þ

x2 nð Þ

x3 nð Þ

7 7 7 7 5

1 4

℧ nð Þ

(cid:11) (cid:12) 1 4

℧ nð Þ

6.7.2 State-Space Representation of Multi-input Multi-output

Discrete-Time LTI Systems

The state-space representation of discrete-time system with m inputs and 1 output and N state variables can be expressed as

X n þ 1

Þ ¼ AX nð Þ þ B℧ nð Þ

ð y nð Þ ¼ CX nð Þ þ D℧ nð Þ

ð6:67aÞ ð6:67bÞ

where

310

6 Discrete-Time Signals and Systems

6 6 6 6 4

A ¼

C ¼

a11

a12

(cid:5) (cid:5) (cid:5)

a1N

a22

a21 a2N ⋮ ⋮ ⋱ ⋮

(cid:5) (cid:5) (cid:5)

7 7 7 7 5

6 6 6 6 4

B ¼

b11

b12

(cid:5) (cid:5) (cid:5)

b1m

b22

b21 b2m ⋮ ⋮ ⋱ ⋮

(cid:5) (cid:5) (cid:5)

aN1

c11

aN2

c12

(cid:5) (cid:5) (cid:5) aNN 3

(cid:5) (cid:5) (cid:5)

c1N

N(cid:6)N

bN1

bN2

(cid:5) (cid:5) (cid:5)

d11

d12

(cid:5) (cid:5) (cid:5)

d1m

bNm 3

c22

c2N c21 ⋮ ⋮ ⋱ ⋮

(cid:5) (cid:5) (cid:5)

7 7 7 7 5

D ¼

d22

d2m d21 ⋮ ⋮ ⋱ ⋮

(cid:5) (cid:5) (cid:5)

7 7 7 7 5

6 6 6 6 4

7 7 7 7 5

N(cid:6)m

cl1

cl2

(cid:5) (cid:5) (cid:5)

clN

l(cid:6)N

dl1

dl2

(cid:5) (cid:5) (cid:5)

dlm

l(cid:6)m

6.8 Problems

Determine if the following discrete-time signals are periodic:

(cid:5)

(cid:4)

(cid:5) (cid:4) 6 þ π (i) x nð Þ ¼ sin πn (cid:4) 3 (ii) x nð Þ ¼ cos 3πn 10 þ =0 (cid:5) (cid:4) 2 þ θ (iii) x nð Þ ¼ cos n (cid:5) (iv) x nð Þ ¼ ej πn 4 þ =0 (v) x(n) ¼ 6((cid:2)1)n (cid:4) (cid:5) (vi) x nð Þ ¼ sin 3πn (cid:5) (cid:4) 8 (vii) x nð Þ ¼ sin 3πn (cid:5) 8 (viii) x nð Þ ¼ ej 7πn 4

(cid:4)

(cid:5)

(cid:4) cos 63πn (cid:4) 64 þ cos 63πn (cid:5) 64

(cid:5)

(cid:4) þ ej 3πn 4

Determine if the following discrete-time signals are energy or power signals or

neither. Calculate the energy and power of the signals in each case: (cid:5) (cid:4) (cid:5) (ix) x nð Þ ¼ cos πn 2 (x) x(n) ¼ ((cid:2)1)n 3ð Þn

þ sin 3πn 4

0 (cid:7) n (cid:7) 10

(cid:4)

<

(xi) x nð Þ ¼

11 (cid:7) n (cid:7) 15

: 8 <

cos

otherwise (cid:9) (cid:10) πn 15

(xii) x nð Þ ¼

: 0 (cid:5) (cid:4) 2 þ π (xiii) x nð Þ ¼ ej πn

(cid:2) 10 (cid:7) n (cid:7) 0

otherwise

Determine if the following discrete-time signals are even, odd, or neither even

nor odd:

(i) x nð Þ ¼ sin 4nð

(cid:4)

(cid:5) Þ þ cos 2πn 3

6.8 Problems

311

(cid:4) (cid:5) (cid:5) þ cos 2πn (ii) x nð Þ ¼ sin πn (cid:5) (cid:5) (cid:4) 3 30 þ cos 3πn (iii) x nð Þ ¼ sin 3πn 4 8 Þn n (cid:3) 0 (cid:2)1ð n < 0 0

(iv) x nð Þ ¼

(cid:4) (cid:4)

(cid:13)

Check the following for linearity, time-invariance, and causality: (i) y(n) ¼ 5nx2(n). (ii) y(n) ¼ x(n)sin2n. (iii) y(n) ¼ e–nx(n + 3)

(cid:4) (cid:5) 5. Given the input x(n) ¼ u(n) and the output y nð Þ ¼ 1 2

n(cid:2)1u n (cid:2) 1 ð

Þ of a system,

(i) Determine the impulse response h(n) (ii) Is the system stable? (iii) Is the system causal?

Check for stability and causality of a system for the following impulse

responses: (cid:4) (cid:5) (i) h nð Þ ¼ e2n sin πn 2

u n (cid:2) 1 ð

(cid:4) (cid:5) Þ (ii) h nð Þ ¼ sin πn 2

u nð Þ

Determine if the following signals are periodic, and if periodic, ﬁnd its period: (cid:5) (cid:4) þ sin 3πn 4

(cid:4) (cid:5) (ii) sin n (b) e jπn/3 (c) sin πn 4

Determine the convolution of the sum of the two sequences:

x1(n) ¼ (3,2,1,2) and x2(n) ¼ (1,2,1,2).

Determine the convolution of the sum of the two sequences x1(n) and x2(n), if

x1(n) ¼ x2(n) ¼ cnu(n) for all n, where c is a constant.

Determine the impulse response (i.e., when x(n) ¼ δ(n) of a discrete-time system

characterized by the following difference equation:

y nð Þ þ y n (cid:2) 1

Þ (cid:2) 6y n (cid:2) 2 ð

Þ ¼ x nð Þ

A discrete-time system is characterized by the following difference equation:

6y nð Þ (cid:2) y n (cid:2) 1

Þ (cid:2) y n (cid:2) 2 ð

Þ ¼ 6x nð Þ

Determine the step response of the system, i.e., x(n) ¼ u(n), given the initial conditions y(-1) ¼ 1 and y(-2) ¼ -1.

A discrete-time system is characterized by the following difference equation:

y nð Þ (cid:2) 5y n (cid:2) 1

Þ þ 6y n (cid:2) 2 ð

Þ ¼ x nð Þ

Determine the response of the system for x(n) ¼ nu(n) and initial conditions y(-1) ¼ 1 and y(-2) ¼ 0.

312

6 Discrete-Time Signals and Systems

Determine the response of the system described by the following difference

equation:

y nð Þ þ y n (cid:2) 1

Þ ¼ sin 3n u nð Þ

Obtain the state-space representation of a discrete-time system described by the

following differential equation:

2y nð Þ þ 3y n (cid:2) 1

Þ þ y n (cid:2) 2 ð

Þ ¼ ℧ nð Þ

6.9 MATLAB Exercises

Using the function impz, write a MATLAB program to determine the impulse

response of a discrete-time system represented by

y nð Þ (cid:2) 5y n (cid:2) 1

Þ þ 6y n (cid:2) 2 ð

Þ ¼ x nð Þ (cid:2) 2x n (cid:2) 1

Write a MATLAB program to illustrate down-sampling by an integer factor of 4 of a sum of two sinusoidal sequences, each of length 50, with normalized frequencies of 0.2 Hz and 0.35 Hz.
Write a MATLAB program to illustrate up-sampling by an integer factor of 4 of a sum of two sinusoidal sequences, each of length 50, with normalized frequencies of 0.2 Hz and 0.35 Hz.

Further Reading

Linden, D.A.A.: Discussion of sampling theorem. Proceedings of the IRE. 47, 1219–1226 (1959)
Proakis, J.G., Manolakis, D.G.: Digital Signal Processing Principles, Algorithms and Applica-

tions, 3rd edn. Prentice-Hall, India (2004)

Crochiere, R.E., Rabiner, L.R.: Multirate Digital Signal Processing. Prentice-Hall, Englewood

Cliffs (1983)

Hsu, H.: Signals and Systems, Schaum’s Outlines, 2nd edn, Mc Graw Hill, New York (2011)
Mandal, M., Asif, A.: Continuous and Discrete Time Signals and Systems. Cambridge, UK;

New York: Cambridge University Press, (2007)

Chapter 7 Frequency Domain Analysis of Discrete- Time Signals and Systems

This chapter discusses the transform domain representation of discrete-time sequences by discrete-time Fourier series (DTFS) and discrete-time Fourier trans- form (DTFT) in which a discrete-time sequence is mapped into a continuous function of frequency. We ﬁrst obtain the discrete-time Fourier series (DTFS) expansion of a periodic sequence. The periodic convolution and the properties of DTFS are discussed. The Fourier transform domain representation of discrete-time sequences are described along with the conditions for the existence of DTFT and its properties. Later, the frequency response of discrete-time systems, frequency domain representation of sampling process, and reconstruction of band-limited signals from its samples are discussed.

7.1 The Discrete-Time Fourier Series

If a sequence x(n) is periodic with period N, then x(n) ¼ x(n + N ) for all n. In analogy with the Fourier series representation of a continuous periodic signal, we can look for a representation of x(n) in terms of the harmonics corresponding to the funda- mental frequency of (2π/N ). Hence, we may write x(n) in the form

x nð Þ ¼

bkej2πkn=N

ð7:1aÞ

It can easily be veriﬁed from Eq. (7.1a) that x(n) ¼ x(n + N ). Also, we know that there are only N distinct values for e j2πkn/N, corresponding to k ¼ 0, 1, … . N (cid:2) 1, these being 1, e j2πn/N, …, e j2πn(N (cid:2) 1)/N. Hence, we may rewrite (7.1a) as

x nð Þ ¼

N(cid:2)1

k¼0

akej2πkn=N

ð7:1bÞ

It should be noted that the summation can be taken over any N consecutive values of k. Eq. (7.1b) is called the discrete-time Fourier series (DTFS) of the periodic

313

314

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

sequence x(n) and ak as the Fourier coefﬁcients. We will now obtain the expression for the Fourier coefﬁcients ak. It can easily be shown that {e j2πkn/N} is an orthogonal sequence satisfying the relation

N(cid:2)1

n¼0

ej2πkn=Ne(cid:2)j2πl n=N ¼

(cid:2)

k 6¼ l 0 N k ¼ l

0 (cid:3) k; l (cid:3) N (cid:2) 1

ð7:2Þ

Now, multiplying both sides of (7.1b) by e(cid:2)j2πln/N and summing over n between

0 and (N (cid:2) 1), we get

N(cid:2)1

N(cid:2)1 n¼0

n¼0 x nð Þ e(cid:2)j2πln=N ¼ ¼ ¼ alN, using Eq: 7:2ð

k¼0 akej2πkn=Ne(cid:2)j2πln=N N(cid:2)1 P n¼0 ej2πkn=Ne(cid:2)j2πln=N

N(cid:2)1 k¼0 ak

N(cid:2)1

Þ:

Hence,

ak ¼

N(cid:2)1

n¼0

1 N

x nð Þe(cid:2)j2πkn=N,

k ¼ 0, 1, 2, … , N (cid:2) 1

ð7:3Þ

It is common to associate the factor (1/N ) with x(n) rather than ak. This can be

done by denoting Nak by X(k); in such a case, we have

x nð Þ ¼

N(cid:2)1

k¼0

1 N

X kð Þ ej2πkn=N

where the Fourier coefﬁcients X(k) are given by X

X kð Þ ¼

N(cid:2)1

n¼0

x nð Þ e(cid:2)j2πkn N ,

k ¼ 0, 1, 2, … , N (cid:2) 1

ð7:4Þ

ð7:5Þ

It is easily seen that X(k + N ) ¼ X(k), that is, the Fourier coefﬁcient sequence X(k) is also periodic of period N. Hence, the spectrum of a signal x(n) that is periodic with period N is also a periodic sequence with the same period. It is also noted that since the Fourier series of a discrete periodic signal is a ﬁnite sequence, the series always converges, and the Fourier series gives an exact alternate representation of the discrete sequence x(n).

7.1.1 Periodic Convolution

In the case of two periodic sequences x1(n) and x2(n) having the same period N, linear convolution as deﬁned by Eq. (6.29) does not converge. Hence, we deﬁne a different form of convolution for periodic signals by the relation

7.1 The Discrete-Time Fourier Series

315

Table 7.1 Some important properties of DTFS

Property Linearity Time shifting

Periodic sequence ax1(n) þ bx2(n) a and b are constants x(n (cid:2) m) ej 2π Frequency shifting Nð Þln x nð Þ Periodic convolution XN(cid:2)1

x1 mð Þx2 n (cid:2) m Þ ð

DTFS coefﬁcients aX1(k) þ bX2(k) e(cid:2)j 2π Nð ÞkmX kð Þ X(k (cid:2) l )

X1(k)X2(k)

Multiplication

m¼0 x1(n)x2(n)

x*(n) x*((cid:2)n) Re x nð Þ g f j lm x nð Þ g f

Symmetry properties

xe nð Þ

XN(cid:2)1

X1 lð ÞX2 k (cid:2) l

1 N

l¼0 X*((cid:2)k) X*(k)

X kð Þ þ X∗

X kð Þ (cid:2) X∗

Xe kð Þ ¼

Xo kð Þ ¼

1 ð 2 1 ð 2 Re X kð Þ g f j Im X kð Þ g f

(cid:2)kð

ð7:6Þ

x nð Þ þ x∗

(cid:2)nð

(cid:4) Þ

1 2

¼ xo nð Þ

x nð Þ (cid:2) x∗

1 2 If x nð Þis real

(cid:2)nð

(cid:4) Þ

xe nð Þ ¼

xo nð Þ ¼

1 x nð Þ þ x (cid:2)nð ½ 2 1 ½ 2

x nð Þ (cid:2) x (cid:2)nð Þ

(cid:4)

(cid:4) Þ

Re X kð Þ g f j Im X kð Þ f

y nð Þ ¼

N(cid:2)1

m¼0

x1 mð Þx2 n (cid:2) m ð

Þ ¼

N(cid:2)1

m¼0

x1 n (cid:2) m

Þx2 mð Þ

The above convolution is called periodic convolution. It may be observed that y (n) ¼ y(n + N ), that is, the periodic convolution is itself periodic of period N. Some important properties of the DTFS are given in Table 7.1. In this table, it is assumed that x1(n) and x2(n) are periodic sequences having the same period N. The proofs are omitted here, since they are similar to the ones that will be given in Section 7.2 for the corresponding properties of the DTFT.

Example 7.1 Determine the Fourier series representation for the following discrete- time signal:

x nð Þ ¼ 3 sin

(cid:3) (cid:4) πn 4

sin

(cid:5)

(cid:6)

2πn 5

316

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Solution

x nð Þ ¼ 3 sin

(cid:3) (cid:4) πn 4 (cid:5)

sin (cid:6)

(cid:5)

(cid:6)

2πn 5

cos

(cid:2) cos

3n π 20

(cid:6)

(cid:5)

13nπ 20

(cid:6)

(cid:5)

3nπ 20 þ e(cid:2)j

3nπ 20 (cid:2) ej

13nπ 20 (cid:2) e(cid:2)j

13nπ 20

3nπ 20 þ ej

17nπ 20 (cid:2) ej

13nπ 20 (cid:2) ej

7nπ 20

(cid:6)

(cid:5)

3 2

3 4

X(0) ¼ X(1) ¼ X(2) ¼ X(4) ¼ X(5) ¼ X(6) ¼ X(8) ¼ X(9) ¼ X(10) ¼ X (11) ¼ X(12) ¼ X(14) ¼ X(15) ¼ X(16) ¼ X(18) ¼ X(19) ¼ 0, X(3) ¼ Xð17Þ ¼ 3 2, X(7) ¼ Xð13Þ ¼ (cid:2)3 2

Example 7.2 Discrete-time signal x(n) is periodic of period 8, and x(n) ¼ n for 0 (cid:3) n (cid:3) 7.

Solution The sequence is periodic with period N ¼ 8.

X kð Þ ¼

N(cid:2)1

n¼0

x nð Þe(cid:2)j2πkn=N, k ¼ 0, 1, 2, … , N (cid:2) 1

Using above equation, the DTFS coefﬁcients are computed as

X(4) ¼ (cid:2)4

X(0) ¼ 28., X(1) ¼ (cid:2)4þ 9.6569 j, X(5) ¼ (cid:2)4 (cid:2) 1.6569 j X(2) ¼ (cid:2)4 þ 4 j X(3) ¼ (cid:2)4 þ 1.6569 j, X(7) ¼ (cid:2)4 (cid:2) 9.6569 j

X(6) ¼ (cid:2)4 (cid:2) 4 j

X(k) ¼ {28, (cid:2)4þ 9.6569j, (cid:2)4 þ 4 j, (cid:2)4 þ 1.6569 j, (cid:2)4, (cid:2)4 (cid:2) 1.6569 j,

(cid:2)4(cid:2)4i,(cid:2)4 (cid:2) 49.6569 j}

7.2 Representation of Discrete-Time Signals and Systems

in Frequency Domain

7.2.1 Fourier Transform of Discrete-Time Signals

The discrete-time Fourier transform (DTFT) of a ﬁnite energy sequence x(n) is deﬁned as

(cid:8)

F x nð Þ ½

(cid:7) (cid:4) ¼ X ejω

n¼(cid:2)1

x nð Þe (cid:2)jωn Þ ð

ð7:7Þ

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

317

From X(e jω), x(n) can be computed as ð π

x nð Þ ¼

1 2π

(cid:2)π

(cid:8)

(cid:7) X ejω

e jωnð

Þdω

ð7:8Þ

Eq. (7.8) is called the inverse Fourier transform.

Convergence of the DTFT The existence of DTFT of x(n) depends on the convergence of the series in Eq. (7.7). Now, we look at the condition for convergence. 1

(cid:7)

(cid:8)

Þ denote the partial sum of the weighted

Let Xk ejω

x nð Þe (cid:2)jωn ð

complex exponentials in Eq. (7.7). Then for uniform convergence of X(e jω), (cid:7) Xk ejω

(cid:7) ¼ X ejω

(cid:8)

ð7:9Þ

k¼(cid:2)1

lim k!1

Hence, for uniform convergence of X(e jω), x(n) must be absolutely summable,

i.e.,

Then

(cid:9) (cid:9)

(cid:7) X ejω

(cid:8)

(cid:9) (cid:9)

n¼(cid:2)1

x nð Þ j

j < 1,

ð7:10Þ

(cid:9) (cid:9) (cid:9) (cid:9) (cid:9)

n¼(cid:2)1

x nð Þe(cid:2)jωn

(cid:9) (cid:9) (cid:9) (cid:9) (cid:9) (cid:3)

(cid:9) (cid:9) j e(cid:2)jωn

(cid:9) (cid:9)

(cid:3)

x nð Þ

n¼(cid:2)1

x nð Þ

j < 1 ð7:11Þ

guaranteeing the existence of X(e jω), for all values of ω. Consequently, Eq. (7.10) is only a sufﬁcient condition for the existence of the DTFT, but is not a necessary condition.

7.2.2 Theorems on DTFT

We will now consider some important theorems concerning DTFT that can be used in digital signal processing. All these properties can be proved using the deﬁnition of DTFT. The following notation is adopted for convenience:

(cid:8)

(cid:7) X ejω

¼ F x nð Þ ½ (cid:10) (cid:7)

(cid:4) (cid:8)

(cid:11)

ð7:12aÞ

ð7:12bÞ

x nð Þ ¼ F(cid:2)1 X ejω

Linearity If x1(n) and x2(n) are two sequences with Fourier transforms X1(e jω) and X2(e jω), then the Fourier transform of a linear combination of x1(n) and x2(n) is given by

318

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

F a1x1 nð Þ þ a2x2 nð Þ

(cid:7)

(cid:4) ¼ a1X1 ejω

(cid:8)

(cid:7)

þ a2X2 ejω

(cid:8)

ð7:13Þ

where a1and a2 are arbitrary constants. Time Reversal If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of time reversed sequence x((cid:2)n) is given by

F x (cid:2)nð

(cid:7) (cid:4) ¼ X e(cid:2)jω

(cid:8)

ð7:14Þ

Time Shifting If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of the delayed sequence x(n (cid:2) k), where k an integer, is given by

F x n (cid:2) k ð

(cid:7)

(cid:4) ¼ e(cid:2)jωkX ejω

(cid:8)

ð7:15Þ

Therefore, time shifting results in a phase shift in the frequency domain.

Frequency Shifting If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of the sequence ejω0n x(n) is given by (cid:4)i (cid:10)

(cid:11)

F ejω0nx nð Þ

(cid:3) ¼ X ej ω(cid:2)ω0 ð

ð7:16Þ

Thus, multiplying a sequence x(n) by a complex exponential ejω0n in the time

domain corresponds to a shift in the frequency domain. Differentiation in Frequency If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of the sequence nx(n) is given by

F nx nð Þ

(cid:4) ¼ j

(cid:7)

dω X ejω

(cid:8)

ð7:17Þ

Convolution Theorem If x1(n) and x2(n) are two sequences with Fourier trans- forms X1(e jω ), then the Fourier transform of the convolution of x1(n) and x2(n) is given by

)and X2(e jω

F x1 nð Þ∗x2 nð Þ

(cid:4) ¼ X1 ejω

(cid:7)

(cid:8)

(cid:7) X2 ejω

ð7:18Þ

Hence, convolution of two sequences x1(n) and x2(n) in the time domain is equal to the product of their frequency spectra. In the above equation, since X1(e jω) and X2(e jω) are periodic in ω with period 2π, the convolution is a periodic convolution. Windowing Theorem If x(n) and w(n) are two sequences with Fourier transforms X(e jω) and W(e jω), then the Fourier transform of the product of x(n) and w(n) is given by

F x nð Þw nð Þ

(cid:7) (cid:4) ¼ X ejω

(cid:8)

(cid:7) ∗W ejω

(cid:8)

ð π

(cid:2)π

1 2π

(cid:8)

(cid:7) X ejθ

(cid:3)

W ej ω(cid:2)θ ð

(cid:4) dθ

ð7:19Þ

The above result is called the windowing theorem.

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

319

Correlation Theorem If x1(n) and x2(n) are two sequences with Fourier transforms X1(e jω) and X2(e jω), then the Fourier transform of the correlation rx1x2 lð Þ of x1(n) and x2(n) deﬁned by

rx1x2 lð Þ ¼

n¼(cid:2)1

x1 nð Þx2 n (cid:2) l ð

is given by

F rx1x2 lð Þ

h (cid:4) ¼ F

n¼(cid:2)1

x1 nð Þx2 n (cid:2) l ð

(cid:7) ¼ X1 ejω

(cid:8)

(cid:7) X2 e(cid:2)jω

(cid:8)

ð7:20aÞ

ð7:20bÞ

which is called the cross energy density spectrum of the signals x1(n) and x2(n). Parseval’s Theorem If x(n) is a sequence with Fourier transform X(e jω), then the energy E of x(n) is given by

E ¼

(cid:2)1

x nð Þ

j2 ¼

ð π

(cid:2)π

1 2π

(cid:9) (cid:7) (cid:9) X ejω

(cid:8)

(cid:9) (cid:9)2

dω

where |X(e jω)|2 is called the energy density spectrum. Proof The energy E of x(n) is deﬁned as

1 (cid:2)1 x nð Þ j

1 (cid:2)1 x nð Þ

E ¼

j2 ¼ ð π

1 2π

(cid:2)π

(cid:2)1 x nð Þx∗ nð Þ (cid:8)

(cid:7) X∗ ejω

e(cid:2)jωndω

ð7:21Þ

ð7:22Þ

using Eq. (7.8).

Interchanging the integration and summation signs, the above equation can be

rewritten as

ð π

(cid:2)π ð π

(cid:2)π

1 2π

E ¼

(cid:7) X∗ ejω

(cid:8)X1

x nð Þ e(cid:2)jωndω

(cid:8)

(cid:7) X∗ ejω

(cid:2)1 (cid:7) X ejω

(cid:8)

dω

(cid:9) (cid:9)

(cid:7) X ejω

(cid:8)

(cid:9) (cid:9)2dω

Thus,

E ¼

(cid:2)1

x nð Þ

j2 ¼

ð π

(cid:2)π

1 2π

(cid:9) (cid:7) (cid:9) X ejω

(cid:8)

(cid:9) (cid:9)2dω

ð7:23Þ

320

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Table 7.2 Some properties of discrete-time Fourier transforms

Property Linearity Time shifting Time reversal Frequency shifting

Differentiation in the frequency domain

Sequence a1x1(n) þ a2x2(n) x(n (cid:2) k) x((cid:2)n) ejω0nx nð Þ nx(n)

x1(n) * x2(n) x1(n)x2(n) X 1

x1 nð Þx2 n (cid:2) l Þ ð

Convolution theorem Windowing theorem Correlation theorem

Table 7.3 Some useful DTFT pairs

(cid:2)1

x(n) δ(n) 1 ((cid:2)1 < n < 1)

|a| < 1

anu(n), sin ωcn Þ ð πn

(cid:2)

1 0 (cid:3) n (cid:3) L otherwise 0 e(cid:2)jω0n

(cid:8)

DTFT a1X1(e jω) þ a2X2(e jω) e(cid:2)jωkX(e jω) X(e(cid:2)jω) (cid:7) X ej ω(cid:2)ω0 ð dω X ejωð j d Þ X1(e jω) X2(e jω) X1(e jω) ∗ X2(e jω) X1(e jω)X2(e(cid:2)jω)

DTFT 1 X

k¼(cid:2)1

2πδ ω þ 2πk

1 1(cid:2)ae(cid:2)jω (cid:2) ωj 1 0 ωc < ωj sin ω Lþ1 Þ=2 ð sin ω=2

j < ωc

j < π

e(cid:2)jωL=2

k¼(cid:2)1

2πδ ω (cid:2) ω0 þ 2πk

The above theorems concerning DTFT are summarized in Table 7.2

Parseval’s theorem

(cid:2)1

x nð Þ

j2 ¼

ð π

(cid:2)π

1 2π

(cid:9) (cid:9)

(cid:7) X ejω

(cid:9) (cid:8) (cid:9)2

dω

Using the deﬁnitions of DTFT pair given by (7.7) and (7.8), we may establish the

DTFT pairs for some useful functions. These are given in Table 7.3

7.2.3 Some Properties of the DTFT of a Complex Sequence x

(n)

From Eq. (7.7), the DTFT of a time reversed sequence x((cid:2)n) can be written as

F x (cid:2)nð

(cid:4) ¼

n¼(cid:2)1

x (cid:2)nð

Þe(cid:2)jωn ¼

(cid:2)1

l¼1

(cid:7)

x lð Þejωl ¼ X e(cid:2)jω

(cid:8)

ð7:24aÞ

Similarly, expressed as

the DTFT of

the complex conjugate sequence x*(n) can be

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

321

F x∗ nð Þ

(cid:4) ¼

n¼(cid:2)1

x∗ nð Þe(cid:2)jωn ¼

(cid:3)

n¼(cid:2)1

x nð Þejωn

(cid:4)∗

(cid:7) ¼ X∗ e(cid:2)jω

(cid:8)

ð7:24bÞ

From the above two equations, it can be easily shown that

F x∗ ½

(cid:2)nð

(cid:7)

(cid:4) ¼ X∗ ejω

(cid:8)

ð7:25Þ

The sequence x(n) can be represented as a sum of conjugate symmetric sequence

xe (n) and a conjugate antisymmetric sequence xo(n) as

where

and

The DTFT X(e jω

x nð Þ ¼ xe nð Þ þ xo nð Þ

xe nð Þ ¼

xo nð Þ ¼

1 2

x nð Þ þ x∗ ½

(cid:2)nð

(cid:4)

x nð Þ (cid:2) x∗ ½

(cid:2)nð

(cid:4)

) can be split into (cid:7) X ejω

(cid:8)

(cid:7) ¼ Xe ejω

(cid:8)

(cid:7) þ Xo ejω

(cid:8)

ð7:26Þ

ð7:27Þ

ð7:28Þ

ð7:29Þ

where Xe(e jω) and Xo(e jω) are the DTFTs of xe(n) and xo(n), respectively. Using Eqs. (7.7), (7.25), and (7.27), Xe(e jω) can be expressed as

(cid:8)

(cid:7) Xe ejω

¼ F xe nð Þ ½

(cid:4)

1 2

F x nð Þþ

F x∗ ½

(cid:2)nð

(cid:4)

Þ ¼

1 2

(cid:8)

(cid:10)

(cid:7) X ejω

(cid:7) þ X∗ ejω

(cid:8)

(cid:11)

(cid:10) ¼ Re X ejω

(cid:7)

(cid:8)

(cid:11)

ð7:30Þ

In a similar way, using Eqs. (7.7), (7.25), and (7.28), Xo(e jω) can be written as

Xoðejω

Þ ¼ F½xoðnÞ(cid:4) 1 2

ðF½xðnÞ(cid:4) (cid:2) F½x∗

(cid:4)

ð(cid:2)nÞ(cid:4)

1 2

½Xðejω

Þ (cid:2) X∗

ðejω

Þ(cid:4) ¼ jIm½Xðejω

Þ(cid:4)

ð7:31Þ

A complex sequence x(n)can be decomposed into a sum of its real and imaginary

parts as

x nð Þ ¼ xR nð Þ þ jxI nð Þ

ð7:32Þ

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

322

where

and

xR nð Þ ¼

jxI nð Þ ¼

1 2

x nð Þ þ x∗ nð Þ ½

(cid:4)

x nð Þ (cid:2) x∗ nð Þ

(cid:4)

The DTFT of xR (n) can be written as (cid:12) (cid:4) ¼ F 1 ð 2 (cid:10) (cid:7) 1 X ejω 2

F Re x nð Þ ð

(cid:13)

x nð Þ þ x∗ nð Þ Þ

(cid:8)

(cid:7) þ X∗ e(cid:2)jω

(cid:8)

(cid:11)

Similarly, the DTFT of jxI (n) can be expressed as

(cid:12) F½jImðxðnÞÞ(cid:4) ¼ F 1 2 1 Þ (cid:2) X∗ 2

½Xðejω

ðxðnÞ (cid:2) x∗

(cid:13)

ðnÞÞ

ðe(cid:2)jω

Þ(cid:4)

ð7:33Þ

ð7:34Þ

ð7:35Þ

ð7:36Þ

The above properties of the DTFT of a complex sequence are summarized in

Table 7.4.

7.2.4 Some Properties of the DTFT of a Real Sequence x(n)

Since e–jωn ¼ cosωn – jsinωn, the DTFT X(e jω) given by Eq. (7.7) can be expressed as

(cid:8)

(cid:7) X ejω

n¼(cid:2)1

x nð Þ cos ωn (cid:2) j

n¼(cid:2)1

x nð Þ sin ωn

ð7:37Þ

The Fourier transform X(e jω) is a complex function of ω and can be written as the

sum of the real and imaginary parts as

Table 7.4 Some properties of DTFT of a complex sequence

Sequence x∗(n) x∗((cid:2)n) xR(n) ¼ Re [x(n)] jxI(n) ¼ j Im [x(n)] xe nð Þ ¼ 1 x0 nð Þ ¼ 1

2 x nð Þ þ x∗ (cid:2)nð ½ Þ (cid:4) 2 x nð Þ þ x∗ (cid:2)nð ½ Þ

(cid:4)

DTFT X∗(e(cid:2)jω) X∗(e jω) 2 X ejωð 1 ½ 2 X ejωð 1 ½ Re[X(e jω)] j Im [X(e jω)]

Þ þ X∗ e(cid:2)jω ð Þ (cid:2) X∗ e(cid:2)jω ð

(cid:4)

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

323

(cid:8)

(cid:7) X ejω

(cid:7) ¼ XR ejω

(cid:8)

(cid:7) þ j XI ejω

(cid:8)

From Eq. (7.37), the real and imaginary parts of X(e jω) are given by X

(cid:8)

(cid:7) XR ejω

n¼(cid:2)1

x nð Þ cos ωn

and

(cid:8)

(cid:7) XI ejω

¼ (cid:2)

n¼(cid:2)1

x nð Þ sin ωn

ð7:38Þ

ð7:39Þ

ð7:40Þ

Since cos((cid:2)ωn) ¼ cosωn and sin((cid:2)ωn) ¼ (cid:2)sinωn, we can obtain the following

relations from Eqs. (7.39) and (7.40):

(cid:8)

(cid:7) XR e(cid:2)jω (cid:8) (cid:7) XI e(cid:2)jω

n¼(cid:2)1

x nð Þ cos ωn ¼ XR ejω

(cid:7)

(cid:8)

x nð Þ sin ωn ¼ (cid:2)XI ejω

(cid:7)

(cid:8)

ð7:41aÞ

ð7:41bÞ

indicating that the real part of DTFT is an even function of ω, while the imaginary part is an odd function of ω. Thus, (cid:7) X ejω

(cid:7) ¼ X∗ e(cid:2)jω

ð7:42Þ

(cid:8)

In polar form, X(e jω) can be written as (cid:9) (cid:7) (cid:9) ¼ X ejω

(cid:7) X ejω

(cid:8)

(cid:9) (cid:9)ejθω

where

and

(cid:9) (cid:7) (cid:9) X ejω

(cid:8)

(cid:9) (cid:9)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ (cid:4)2 XR ejωð

(cid:4)2 þ XI ejωð ½

(cid:7)

θ ωð Þ ¼ ∠X ejω

(cid:8)

¼ phase of X ejω

(cid:7)

(cid:8)

¼ tan (cid:2)1 XI ejωð Þ XR ejωð Þ

ð7:43Þ

ð7:44Þ

ð7:45Þ

Using the above relations, it can easily be seen that |X (e jω)| is an even function of

ω, whereas the function θ(ω)is an odd function of ω.

Now, the DTFT of xe (n), the even part of the real sequence x(n) is given by

F xe nð Þ

(cid:4) ¼

1 2

F x nð Þ ½

(cid:4) þ F x (cid:2)nð ½

(cid:4)

Þ ¼

(cid:8)

(cid:10)

(cid:7) X ejω

(cid:7) þ X e(cid:2)jω

(cid:8)

(cid:11)

(cid:7) ¼ XR ejω

(cid:8)

ð7:46Þ

1 2

Thus, the DTFT of even part of a real sequence is the real part of X (e jω). Similarly, the DTFT of xo (n), the odd part of the real sequence x(n), is given by

324

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Table 7.5 Some properties of DTFT of a real sequence

Þ þ jXI ejωð

Þ ¼ XR ejωð

(cid:4) ¼ X ejωð F x nð Þ ½ (cid:4) ¼ XR ejωð F xe nð Þ Þ ½ (cid:4) ¼ jXI ejωð F xo nð Þ Þ ½ XR(e jω) ¼ XR(e(cid:2)jω) XI(e jω) ¼ (cid:2) XI(e(cid:2)jω) X(e jω) ¼ X∗(e(cid:2)jω) |X(e jω)| ¼ |X(e(cid:2)jω)| ∠X(e jω) ¼ (cid:2) ∠ X(e(cid:2)jω)

F xo nð Þ

(cid:4) ¼

1 2

(cid:8)

(cid:10) (cid:7) X ejω

(cid:7) (cid:2) X e(cid:2)jω

(cid:8)

(cid:11)

(cid:7) ¼ jXI ejω

(cid:8)

ð7:47Þ

Hence, the DTFT of the odd part of a real sequence is jXI (ejω). The above properties of the DTFT of a real sequence are summarized in

Table 7.5.

Example 7.3 A causal LTI system is represented by the following difference equation:

y nð Þ (cid:2) ay n (cid:2) 1

Þ ¼ x n (cid:2) 1 ð

(i) Find the impulse response of the system h(n), as a function of parameter a. (ii) For what range of values would the system be stable?

Solutions (i) Given

y nð Þ (cid:2) ay n (cid:2) 1

Þ ¼ x n (cid:2) 1 ð

Taking Fourier transform on both sides of above equation, we get

(cid:8)

(cid:7) Y ejω

(cid:7)

(cid:2) ae(cid:2)jωY ejω

(cid:8)

(cid:7)

¼ e(cid:2)jωX ejω

(cid:8)

From the above relation, we arrive at

e(cid:2)jω 1 (cid:2) ae(cid:2)jω P

(cid:7) H ejω

(cid:8)

Þ Þ

Y ejωð X ejωð n¼(cid:2)1 ane(cid:2)jωn ¼ 1 1 (cid:2) ae(cid:2)jω

F anu nð Þ

(cid:4) ¼

n¼(cid:2)1 ae(cid:2)jω ð

Þn

From the above equation and time shifting property, the impulse response is

given by

(cid:7)

h nð Þ ¼ F(cid:2)1 H ejω

(cid:8)

¼ F(cid:2)1

(cid:6)

(cid:5)

e(cid:2)jω 1 (cid:2) ae(cid:2)jω

¼ an(cid:2)1u n (cid:2) 1 ð

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

325

(ii) Now,

n¼1

h nð Þ

j ¼

n¼1

aj jn(cid:2)1 < 1

for aj j < 1:

Thus, the system is stable for |a| < 1.

Example 7.4 Find the impulse response of a system described by the following difference equation:

y nð Þ (cid:2)

5 6

y n (cid:2) 1 ð

Þ þ

1 6

y n (cid:2) 2 ð

Þ ¼

1 3

x n (cid:2) 1 ð

Solution Taking Fourier transform on both sides of given difference equation, we get

Yðejω

Þ (cid:2)

5 6

e(cid:2)jωYðejω

Þ þ

1 6

e(cid:2)2jωYðejω

Þ ¼

e(cid:2)jωXðejω

1 3

From the above relation, we arrive at

H ejωð

Þ ¼

Y ejωð X ejωð

Þ Þ

1 (cid:2) 5=6 ð

2 1 (cid:2) 1=2 ð

Þe(cid:2)jω (cid:2)

The impulse response h(n) is given by

Þe(cid:2)jω 1=3 ð Þe(cid:2)jω þ 1=6 ð 2 1 (cid:2) 1=3 ð

Þe(cid:2)jω

Þe(cid:2)2jω

(cid:5)

h nð Þ ¼ F(cid:2)1 (cid:10) (cid:7) (cid:8) ¼ 2 1 2

2 1 (cid:2) 1=2 ð (cid:11) (cid:7) (cid:8) n (cid:2) 1 u nð Þ 3

Þe(cid:2)jω

(cid:6)

(cid:2) F(cid:2)1

(cid:6)

(cid:5)

2 1 (cid:2) 1=3 ð

Þe(cid:2)jω

Example 7.5 Find the DTFT of x nð Þ ¼ nþm(cid:2)1 ð n! m(cid:2)1 ð

Þ! Þ! anu nð Þ,

aj j < 1

Solution Let x1(n) ¼ anu(n)

The Fourier transform of x1(n) is given by

(cid:8)

(cid:7) X1 ejω

n¼0

að Þne(cid:2)jωn ¼

n¼0

(cid:7) ae(cid:2)jω

(cid:8)

1 1 (cid:2) ae(cid:2)jω

For m ¼ 2,

x nð Þ ¼ n þ 1 ð

Þanu nð Þ

Using the differentiation property of DTFT, the Fourier transform of nanu(n) is

given by

326

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

dX1 ejωð

Þ dω ¼ j

d dω

(cid:6)

(cid:5)

1 1 (cid:2) ae(cid:2)jω

ae(cid:2)jω 1 (cid:2) ae(cid:2)jω

Þ2

Using linearity property of the DTFT, the Fourier transform of x(n) is denoted by

(cid:8)

(cid:7) X ejω

ae(cid:2)jω 1 (cid:2) ae(cid:2)jω

1 1 (cid:2) ae(cid:2)jω

Þ2

Þ2 þ

For m ¼ 3,

x nð Þ ¼

(cid:5) ð

n þ 2

(cid:6)

Þ n þ 1 ð 2

anu nð Þ ¼

(cid:10) n2anu nð Þ þ 3nanu nð Þ þ 2anu nð Þ

1 2

n2 þ 3n þ 2 2 (cid:11)

anu nð Þ

Using the differentiation and linearity properties of DTFT, the Fourier transform

of x(n) is given by

X ejωð

Þ ¼

”

1 2

d dω

ae(cid:2)jω 1 (cid:2) ae(cid:2)jω

Þ2

3ae(cid:2)jω 1 (cid:2) ae(cid:2)jω

Þ2 þ

ae(cid:2)jω 1 þ ae(cid:2)jω Þ Þ3 þ 1 (cid:2) ae(cid:2)jω ð

3ae(cid:2)jω 1 (cid:2) ae(cid:2)jω

Þ2 þ

2 1 (cid:2) ae(cid:2)jω

Þ3

1 1 (cid:2) ae(cid:2)jω

Þ3

2 1 (cid:2) ae(cid:2)jω

In general, for m ¼ k, the Fourier transform of x(n) is given by

(cid:8)

(cid:7) X ejω

1 1 (cid:2) ae(cid:2)jω

Þk , where k is any integer value:

Example 7.6 Let G1(e jω) denote the DTFT of the sequence g1(n) shown in Figure 7.1 (a). Express the DTFT of the sequence g2(n) in Figure 7.1b in terms of G1(e jω). Do not evaluate G1(e jω ). Solution From Figure 7.1(b), g2(n) can be expressed in terms of g1(n) as

g2 nð Þ ¼ g1 nð Þ þ g1 n (cid:2) 4

Applying DTFT on both sides, we obtain (cid:7)

(cid:8)

(cid:7) G2 ejω

(cid:7) ¼ G1 ejω

þ e(cid:2)j4ωG1 ejω

(cid:8)

(cid:7)

¼ 1 þ e(cid:2)j4ω

(cid:8)

(cid:7) G1 ejω

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

327

1( ) g n

2 ( ) g n

(a)

(b)

Figure 7.1 (a) Sequence g1(n). (b) Sequence g2(n)

Example 7.7 Evaluate the inverse DTFT of each of the following DTFTs:

(a) X1ðejωÞ ¼

k¼(cid:2)1 (cid:7) Solution (a) X1 ejω

δðω þ 2πkÞ X1

(cid:8)

k¼1

δ ω þ 2πk ð

(b) X2 ejωð

Þ ¼ (cid:2)αe(cid:2)jω 1(cid:2)αe(cid:2)jω ð

Þ2 ,

αj

j < 1

From Table 7.3,

F 1ð Þ (cid:2)1 < n < 1

Þ ¼

k¼(cid:2)1

2πδ ω þ 2πk

Hence,

F(cid:2)1 δ ω þ 2πk ð

(cid:4) ¼

1 2π , (cid:2)1 < n < 1

(b) X2 ejωð

Þ ¼ (cid:2)αe(cid:2)jω 1(cid:2)αe(cid:2)jω ð

Þ2 , αj

j < 1

From Example 7.5,

1 1 (cid:2) αe(cid:2)jω

Þm $

Þ! n þ m (cid:2) 1 ð Þ! n! m (cid:2) 1

αnu nð Þ

For m ¼ 2,

1 1 (cid:2) αe(cid:2)jω ð 1 1 (cid:2) αe(cid:2)jω

Þ! n þ 1 n! 1ð Þ!

Þ2 $Þ 2$ n þ 1 ð

αnu nð Þ

Þαnu nð Þ

328

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

-1

· -2

-3

-2

-3

· 4

-1

Figure 7.2 A length-9 sequence x(n)

Then

(cid:2)α 1 (cid:2) αe(cid:2)jω (cid:2)αe(cid:2)jω 1 (cid:2) αe(cid:2)jω

Þ2 $ (cid:2) n þ 1 ð

Þαnþ1u nð Þ

Þ2 $ (cid:2)nαnu n (cid:2) 1

Example 7.8 A length-9 sequence x(n) is shown in Figure 7.2

If the DTFT of x(n) is X(e jω), calculate the following functions without comput-

ing X(e jω ).

(a) X(ej0)

(b) X(ejπ)

(c)

(cid:7) X ejω

(cid:8) dω (d)

ðπ

(cid:2)π

ðπ

(cid:2)π

(cid:9) (cid:9)

(cid:7) X ejω

(cid:8)

(cid:9) (cid:9)2dω (e)

(cid:9) (cid:9) (cid:9) (cid:9)

dX ejωð d

(cid:9) (cid:9) 2 (cid:9) (cid:9)

dω

ðπ

(cid:2)π

Solution From the given data,

x((cid:2)3) ¼3, x((cid:2)2) ¼0, x((cid:2)1) ¼ 1, x(0) ¼ (cid:2)2, x(1) ¼ (cid:2)3, x (2) ¼ 4, x (3) ¼ 1, x

(4) ¼ 0, x(5) ¼ (cid:2)1 (a) X(e j0)

From the deﬁnition of Fourier transform,

X ejωð

Þ ¼

x nð Þe(cid:2)jωn

n¼(cid:2)1

X ej0ð

Þ ¼

x nð Þ

n¼(cid:2)1

¼ 3 þ 0 þ 1 (cid:2) 2 (cid:2) 3 þ 4 þ 1 þ 0 (cid:2) 1

(cid:4) ¼ 3

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

329

(b) X(e jπ)

From the deﬁnition of Fourier transform,

X ejπð

Þ ¼

x nð Þe(cid:2)jπ

n¼(cid:2)1 X1

Þ ¼ (cid:2)

x nð Þ ¼ (cid:2)3

n¼(cid:2)1

X ejπð

ðπ

(cid:8)

(cid:7) X ejω

dω

(c)

(cid:2)π From the deﬁnition of inverse Fourier transform,

Hence,

x nð Þ ¼

ð π

(cid:2)π

1 2π

(cid:7) X ejω

(cid:8) ejωndω

ð π

(cid:2)π

(cid:8)

(cid:7) X ejω

ejωndω ¼ 2πx 0ð Þ ¼ (cid:2)4π

ðπ

(d)

(cid:9) (cid:9)

(cid:7) X ejω

(cid:8)

(cid:9) (cid:9)2

dω

(cid:2)π From the deﬁnition of Parseval’s theorem,

n¼(cid:2)1

x nð Þ

j2 ¼

1 2π

ðπ

(cid:2)π

(cid:9) (cid:9)

(cid:7) X ejω

(cid:9) (cid:8) (cid:9)2

dω

Hence, Ð π (cid:2)π X ejωð j

j2dω ¼ 2π

1 n¼(cid:2)1 x nð Þ j

¼ 2π 9 þ 0 þ 1 þ 4 þ 9 þ 16 þ 1 þ 0 þ 1

Þ ¼ 82π

ðπ

(cid:9) (cid:9) (cid:9) (cid:9)

(e)

(cid:9) (cid:9) (cid:9) (cid:9)

dω

dX ejωð dω

(cid:2)π From differentiation property and Parseval’s theorem of DTFT,

330

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

( ) H e w

) ( H e w

/ 2p

(a)

/ 3p (b)

Figure 7.3 (a) Fourier transform of h1(n) (b) Fourier transform of h2 (n)

( ) H e w

) ( H e w

/ 3p

(a)

/ 2p

(b)

Figure 7.4 (a) Fourier transform of h1(n) (b) Fourier transform of h2(n)

(cid:9) (cid:9) 2 (cid:9) (cid:9)

dX ejωð dω

ðπ

(cid:9) (cid:9) (cid:9) (cid:9)

(cid:2)π

dω ¼ 2π

nx nð Þ

n¼1 ½

¼ 2π 81 þ 0 þ 1 þ 0 þ 9 þ 64 þ 9 þ 0 þ 25

(cid:4) ¼ 189π

Example 7.9 (a) The Fourier transforms of the impulse responses, h1(n) and h2 (n), of two LTI systems are as shown in Figure 7.3. Find the Fourier transform of the impulse response of the overall system, when they are connected in cascade.

(b) The Fourier transforms of the impulse responses h1(n) and h2(n) of two LTI systems are as shown in Figure 7.4. Find the Fourier transform of the overall system, when they are connected in parallel.

Solution (a) The impulse response h(n) of the overall system is given by

h nð Þ ¼ h1 nð Þ∗h2 nð Þ

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

331

Figure 7.5 (a) Fourier transform of the impulse response of the cascade system (b) Fourier transform of the impulse response of the parallel system

)wJeH (

p 3

(a)

)wJeH (

1.0

p 2

p 3

p 2

(b)

Then, by the convolution property of the Fourier transform, the Fourier transform

of the impulse response of the cascade system is given by

(cid:7) H1 ejω

(cid:8)

(cid:7) H2 ejω

(cid:8)

The Fourier transform of impulse response of the cascade system is shown in

Figure 7.5(a).

(b) The impulse response h(n) of the overall system is given by

h nð Þ ¼ h1 nð Þ þ h2 nð Þ

Hence, the Fourier transform of impulse response of the cascade system is

given by

(cid:8)

(cid:7) H1 ejω

(cid:7) þ H2 ejω

(cid:8)

The Fourier transform of the impulse response of the parallel system is shown in

Figure 7.5(b).

332

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

7.3 Frequency Response of Discrete-Time Systems

For an LTI discrete-time system with impulse response h(n) and input sequence x(n), the output y(n) is the convolution sum of x(n) and h(n) given by

y nð Þ ¼

h kð Þx n (cid:2) k ð

k¼(cid:2)1

ð7:48Þ

To demonstrate the eigenfunction property of complex exponential for discrete-

time systems, consider the input x(n) of the form

x nð Þ ¼ ejωn, (cid:2)1 < n < 1

ð7:49Þ

ð7:51aÞ

ð7:51bÞ

Then from Eq. (7.48), the output is given by

y nð Þ ¼

k¼(cid:2)1

h kð Þejω n(cid:2)k

Þ ¼

The above equation can be rewritten as (cid:7)

y nð Þ ¼ H ejω

k¼(cid:2)1

(cid:8)

ejωn,

where

h kð Þe(cid:2)jωk

ejωn

ð7:50Þ

(cid:8)

(cid:7) H ejω

h nð Þe(cid:2)jωn:

n¼(cid:2)1 H(e jω) is called the frequency response of the LTI system whose impulse response is h(n), e jωn is an eigenfunction of the system, and the associated eigen- value is H(e jω). In general H(e jω) is complex and is expressed in terms of real and imaginary parts as

(cid:8)

(cid:7) H ejω

(cid:7) ¼ HR ejω

(cid:8)

(cid:7) þ jHI ejω

(cid:8)

ð7:52Þ

where HR(e jω) and HI(e jω) are the real and imaginary parts of H(e jω), respectively. Furthermore, due to convolution, the Fourier transforms of the system input and

output are related by

(cid:8)

(cid:7) Y ejω

(cid:7) ¼ H ejω

(cid:8)

(cid:7) X ejω

(cid:8)

ð7:53Þ

where X(e jω) and Y(e jω) are the Fourier transforms of the system input and output, respectively. Thus,

(cid:8)

(cid:7) H ejω

Y ejωð X ejωð

Þ Þ

ð7:54Þ

7.3 Frequency Response of Discrete-Time Systems

333

The frequency response function H(e jω) is also known as the transfer function of the system. The frequency response function provides valuable information on the behavior of LTI systems in the frequency domain. However, it is very difﬁcult to realize a digital system since it is a complex function of the frequency variable ω. In polar form, the frequency response can be written as

(cid:8)

(cid:7) H ejω

(cid:9) (cid:7) (cid:9) ¼ H ejω

(cid:8)

(cid:9) (cid:9)ejθ ωð Þ

ð7:55aÞ

where |H(e jω)|, the amplitude response term, and θ(ω), the phase-response term, are given by

jHðejω

Þj2 ¼ jHRðejω Þj2 þ jHIðejω (cid:5) θ ωð Þ ¼ tan (cid:2)1 HI ejωð Þ HR ejωð Þ

(cid:6)

Þj2

ð7:55bÞ

ð7:55cÞ

Phase and Group Delays If the input is a sinusoidal signal given by

x nð Þ ¼ cos ωnð

Þ,

for (cid:2) 1 < n < 1,

ð7:56aÞ

then from Eq. (7.55a), the output is (cid:7)

(cid:9) (cid:9) y n½ (cid:4) ¼ H ejω0

(cid:8)

(cid:9) (cid:9) cos ωn þ θ ωð Þ

The above equation can be rewritten as

(cid:8)

(cid:7)

(cid:9) (cid:9) y n½ (cid:4) ¼ H ejω0 (cid:9) (cid:9)

(cid:9) (cid:7) (cid:9) ¼ H ejω0

(cid:5)

(cid:5) (cid:9) θ ωð Þ (cid:9) cos ω n þ ω (cid:9) (cid:8) (cid:7) (cid:7) (cid:9) cos ω n (cid:2) τp ωð Þ

(cid:6)

(cid:8)

ð7:56bÞ

ð7:57aÞ

It can be clearly seen that the above equation expresses the phase response as a

time delay in seconds which is called as phase delay and is deﬁned by

τP ωð Þ ¼ (cid:2)

θ ωð Þ ω

ð7:57bÞ

An input signal consisting of a group of sinusoidal components with frequencies within a narrow interval about ω experiences different phase delays when processed by an LTI discrete-time system. As such, the signal delay is represented by another parameter called group delay deﬁned as

τg ωð Þ ¼ (cid:2)

dθ ωð Þ dω

ð7:57cÞ

334

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Example 7.10 Determine the magnitude and phase response of a system whose (cid:7) (cid:8) nu nð Þ impulse response is given by h nð Þ ¼ 1 2

(cid:7) (cid:8) Solution For h nð Þ ¼ 1 2

nu nð Þ, the frequency response is given by

H ejωð

Þ ¼

(cid:5) (cid:6) 1 2

e(cid:2)jωn ¼

(cid:5)

(cid:6) n

e(cid:2)jω

1 2 1 1 (cid:2) 0:5 cos ω þ j0:5 sin ω

n¼(cid:2)1

n¼(cid:2)1 1 1 (cid:2) 0:5e(cid:2)jω ¼

The magnitude response is given by

(cid:9) (cid:9)

(cid:7) H ejω

(cid:9) (cid:8) (cid:9)

1 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Þ2 sin 2ω 1 (cid:2) 0:5 cos ω ð

Þ2 þ 0:5ð

The phase response is

(cid:3)r

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Þ2 (cid:2) 2 0:5ð 1 þ 0:5ð Þ cos ω

θ ωð Þ ¼ (cid:2) tan (cid:2)1

0:5 sin ω 1 (cid:2) 0:5 cos ω

The magnitude and phase values are tabulated in Table 7.6 for various values of ω

and plotted in Figure 7.6(a) and (b), respectively.

Table 7.6 Magnitude and phase

ω |H(e jω)| θ(ω)

π 4

0 2 1.3572 00 (cid:2)28.675

π 2

3π 4

0.8944 (cid:2)26.565

0.7148

(cid:2)14.640

0.67 00

5π 4 0.715 14.640

3π 2 0.894 26.565

7π 4 1.3572 28.6750

2π 2 00

1.8

1.6

1.4

1.2

0.8

e d u t i n g a M

0.2 0.4 0.6 0.8

1.2 1.4

1.6 1.8

1 ω/π (a)

-10

s e e r g e d

, e s a h P

-20

-30 0

0.2 0.4 0.6 0.8

1 ω/π (b)

1.2 1.4 1.6 1.8

Figure 7.6 (a) Magnitude and (b) phase responses of h(n) of Example 7.10

7.3 Frequency Response of Discrete-Time Systems

Figure 7.7 (a) Impulse response of h1(n) (b) impulse response of h2(n)

2 ( ) h n

1( ) h n

(a)

-2

(b)

335

Example 7.11 Compute the magnitude and phase responses of the impulse responses given in Figure 7.7, and comment on the results.

Solution Since h1(n) is an even function of time, it has a real DTFT indicating that the phase is zero, that is, the phase is a horizontal line; h2(n) is the right-shifted version of h1(n). Hence, from time shifting property of DTFT, the transform of h2(n) is obtained by multiplying the transform of h1(n) by e–j2ω. This changes the slope of the phase linearly and can be veriﬁed as follows:

The frequency response of h1(n) is

H1 ejωð

Þ ¼ e2jω þ 2ejω þ 3 þ 2e(cid:2)jω þ e(cid:2)2jω

¼ e2jω þ e(cid:2)2jω The magnitude response of H1(e jω

) is

Þ þ 2 ejω þ e(cid:2)jω

Þ þ 3 ¼ 2 cos 2ω þ 4 cos ω þ 3

(cid:9) (cid:9)

(cid:7) H1 ejω

(cid:8)

(cid:9) (cid:9)

¼ 2 cos 2ω þ 4 cos ω þ 3

The phase response of H1(e jω) is zero. The frequency response of h2(n) is

H2 ejωð

Þ ¼ e(cid:2)2jωH1 ejωð Þ ¼ e(cid:2)2jω 2 cos 2ω þ 4 cos ω þ 3

ð The magnitude response of H2(e jω

) is

(cid:9) (cid:9)

(cid:7) H2 ejω

(cid:8)

(cid:9) (cid:9)

¼ 2 cos 2ω þ 4 cos ω þ 3

The phase response of H2(e jω) is given by

(cid:7) ∠H2 ejω

(cid:8)

¼ ∠e(cid:2)2jω

¼ (cid:2)2ω:

The magnitude and phase responses of h1(n) and h2(n) are shown in Figure 7.8(a), (b), (c), and (d). From the magnitude and phase responses of h1(n) and h2(n), it is observed that h1(n) has zero phase and h2(n) has a linear phase response, whereas both h1(n) and h2(n) have the same magnitude responses.

336

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Figure 7.8 (a) Magnitude response of h1(n). (b) Phase response of h1(n). (c) Magnitude response of h2(n). (d) Phase response of h2(n)

Example 7.12 The trapezoidal integration formula is represented by a recursive difference equation as y(n) – y(n – 1) ¼ 0.5x(n) þ 0.5x(n – 1). Determine H(e jω) of the trapezoidal integration formula.

7.3 Frequency Response of Discrete-Time Systems

337

Figure 7.8 (continued)

Solution Given

y nð Þ (cid:2) y n (cid:2) 1

Þ ¼ 0:5x nð Þ þ 0:5x n (cid:2) 1

338

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Taking Fourier transform on both sides of the above equation, we get

Y ejωð

Þ (cid:2) e(cid:2)jωY ejωð Þ 1 (cid:2) e(cid:2)jω Y ejωð ð (cid:8) (cid:7) H ejω

Þ ¼ 0:5X ejωð Þ ¼ 0:5X ejωð ð ð

Y ejωð X ejωð

¼ 0:5

Þ þ 0:5e(cid:2)jωX ejωð Þ 1 þ e(cid:2)jω ð 1 þ e(cid:2)jω 1 (cid:2) e(cid:2)jω (cid:12)

Þ Þ

(cid:8)

(cid:13)

(cid:7)

”

¼ 0:5

e(cid:2)jω=2 ejω=2 þ e(cid:2)jω=2 e(cid:2)jω=2 ejω=2 (cid:2) e(cid:2)jω=2

cos ω=2 Þ ð sin ω=2 Þ ð

The magnitude response is given by

(cid:9) (cid:7) (cid:9) H ejω

(cid:8)

(cid:9) (cid:9)

¼ 0:5

¼ (cid:2)j0:5

(cid:9) (cid:9) (cid:9) (cid:9)

cos ω=2 Þ ð sin ω=2 Þ ð

(cid:9) (cid:9) (cid:9) (cid:9)

(cid:7)

The phase response is given as follows: If 0 < ω < π, then both cos ω/2 and sin ω/2 are positive, and hence the phase is (cid:8)

(cid:2)π (cid:7) (cid:8) 2 If π < ω < 2π, then cos ω/2 is negative, but sin ω/2 is positive; hence the phase is π 2

7.3.1 Frequency Response Computation Using MATLAB

The M-ﬁle function freqz(h, w) in MATLAB can be used to determine the values of the frequency response of an impulse response vector h at a set of given frequency points ω. Similarly, the M-ﬁle function freqz(b, a, ω) can also be used to ﬁnd the frequency response of a system described by the recursive difference equation with the coefﬁcients in vectors b and a. From frequency response values, the real and imaginary parts can be computed using MATLAB functions real and imag, respec- tively. The magnitude and phase of the frequency response can be determined using the functions abs and angle as illustrated in the following examples:

Example 7.13 Determine the magnitude and phase response of a system described by the difference equation, y(n) ¼ 0.5x(n) þ 0.5x(n – 2). Solution If x(n) ¼ δ(n), then the impulse response h(n) is given by

h nð Þ ¼ 0:5δ nð Þ þ 0:5δ n (cid:2) 2

Hence, h(n) sequence is [0.5 0 0.5]. When this sequence is used in Program 7.1 given below, the resulting magnitude and phase responses are as shown in Figure 7.9 (a) and (b), respectively.

7.3 Frequency Response of Discrete-Time Systems

339

Figure 7.9 (a) Magnitude response of h(n) sequence. (b) Phase response of h(n) sequence

340

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Program 7.1

clear;clc; w=0:0.05:pi; h=exp(jw); %set h=exp(jw) num=0.5+0h.^-1+0.5*h.^-2; den=1; %Compute the frequency responses H=num/den; %Compute and plot the magnitude response mag=abs(H); ﬁgure(1),plot(w/pi,mag); ylabel(‘Magnitude’);xlabel(‘\omega/\pi’); %Compute and plot the phase responses ph=angle(H)*180/pi; ﬁgure(2),plot(w/pi,ph); ylabel(‘Phase, degrees’); xlabel(‘\omega/\pi’)

Example 7.14 Determine the magnitude and phase responses of a system described by the following difference equation:

y nð Þ (cid:2) 2:1291y n (cid:2) 1 ð ¼ 0:0534x nð Þ (cid:2) 0:0009x n (cid:2) 1

Þ þ 1:7834y n (cid:2) 2

Þ (cid:2) 0:0009x n (cid:2) 2

Þ (cid:2) 0:5435y n (cid:2) 3

Þ Þ þ 0:0534x n (cid:2) 3

Comment on the frequency response of the system.

Solution The following MATLAB program 7.2 is used and the resultant magnitude response and phase response are shown in Figure 10(b) and (b), respectively.

Program 7.2

clear;close all; num=[0.0534 -0.0009 -0.0009 0.0534];% numerator coefﬁcients den=[1 -2.1291 1.7834 -0.5435];% denominator coefﬁcients w=0:pi/255:pi; %Compute the frequency responses H=freqz(num,den,w); %Compute and plot the magnitude response mag=abs(H); ﬁgure(1),plot(w/pi,mag); ylabel(‘Magnitude’);xlabel(‘\omega/\pi’); %Compute and plot the phase responses ph=angle(H)*180/pi; ﬁgure(2),plot(w/pi,ph); ylabel(‘Phase, degrees’);xlabel(‘\omega/\pi’);

7.3 Frequency Response of Discrete-Time Systems

341

Figure 7.10 (a) Magnitude response (b) phase response

The frequency response shown in Figure 7.10 characterizes a low-pass ﬁlter with

nonlinear phase.

Example 7.15 Determine the magnitude and phase responses of a system described by the following difference equation:

342

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

y nð Þ (cid:2) 3:0538y n (cid:2) 1 ¼ x nð Þ (cid:2) 4x n (cid:2) 1

Þ þ 3:8281y n (cid:2) 2 Þ (cid:2) 4x n (cid:2) 3 ð

Þ (cid:2) 2:2921y n (cid:2) 3 Þ: Þ þ x n (cid:2) 4 ð

Þ þ 6x n (cid:2) 2 ð

Þ þ 0:5507y n (cid:2) 4

Comment on the frequency response of the system.

Solution Program 7.2 with variables num ¼ [ 1 (cid:2)4 6 (cid:2)4 1] and den ¼ [1 (cid:2)3.0538 3.8281 (cid:2)2.2921 0.5507] is used, and the resultant magnitude and phase responses are shown in Figure 7.11(a) and (b), respectively. It is observed from this ﬁgure that the frequency response characterizes a narrowband band-pass ﬁlter.

Example 7.16 An LTI system is described by the following difference equation:

y nð Þ ¼ x nð Þ þ 2x n (cid:2) 1

Þ þ x n (cid:2) 2 ð

(a) Find the frequency response H(e jω) and group delay grd [H(e jω)] of the system. (b) Determine the difference equation of a new system such that the frequency response H1(e jω) of the new system is related to H(e jω) as H1(e jω) ¼ H(e j(ω þ π)).

Solution (a)

y nð Þ ¼ x nð Þ þ 2x n (cid:2) 1 h nð Þ ¼ δ nð Þ þ 2δ n (cid:2) 1

Þ þ x n (cid:2) 2 ð Þ Þ þ δ n (cid:2) 2 ð

¼ 2e(cid:2)jω

H ejωð (cid:12)

Þ ¼ 1 þ 2e(cid:2)jω þ e(cid:2)2jω

ejωð

(cid:5) (cid:6) 1 2

(cid:5) (cid:6) 1 2 ¼ 2e(cid:2)jω cos ω þ 1 ð

Þ þ 1 þ

(cid:13)

e(cid:2)jω

Hence,

Therefore,

H ejωð Þ j ∠H ejωð

j ¼ 2 cos ω þ 1 ð Þ ¼ (cid:2)ω

group delay ¼ grad H ejω

(cid:10)

(cid:7)

(cid:8)

(cid:11)

¼ (cid:2)

d∠H ejωð dω

¼ 1

(b) By frequency shifting property, e(cid:2)jπnh(n) $ H(e j(ω+π)). Therefore,

h1 nð Þ ¼ e(cid:2)jπnh nð Þ ¼ (cid:2)1ð Þnh nð Þ Þ þ δ n (cid:2) 2 ¼ δ nð Þ (cid:2) 2δ n (cid:2) 1 Þ ð ð

Hence, the difference equation of the new system is

y nð Þ ¼ x nð Þ (cid:2) 2x n (cid:2) 1

Þ þ x n (cid:2) 2 ð

Þ:

7.3 Frequency Response of Discrete-Time Systems

343

Figure 7.11 (a) Magnitude response (b) phase response

344

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

7.4 Representation of Sampling in Frequency Domain

As mentioned in Section 6.1, mathematically, the sampling process involves multi- plying a continuous-time signal xa(t) by a periodic impulse train p(t)

p tð Þ ¼

n¼(cid:2)1

δ t (cid:2) nT ð

ð7:58Þ

As a consequence, the multiplication process gives an impulse train xp(t), which

can be expressed as

xp tð Þ ¼ xa tð Þp tð Þ P

(cid:2)1 xa tð Þδ t (cid:2) nT ð

Since xa(t) δ(t – nT) ¼ xa(nT) δ(t – nT), the above reduces to

xp tð Þ ¼

(cid:2)1

xa nTð

Þδ t (cid:2) nT

ð7:59Þ

ð7:60Þ

If we now take the Fourier transform of (7.59), and use the multiplication

property of the Fourier transform, we get

Xp jΩð

1 2π Xa jΩð ½ where * denotes the convolution in the continuous-time domain and Xp(jΩ), Xa(jΩ), and P(jΩ) are the Fourier transforms of xp(t), xa(t), and p(t), respectively. Since p(t) is periodic with a period T, it can be expressed as a Fourier series

Þ∗P jΩð

ð7:61Þ

Þ ¼

(cid:4)

p tð Þ ¼

1 T

(cid:2)1

ej 2π Tð Þkt

Since the Fourier transform of f

tð Þ ¼ ejΩT t is given by F(jΩ) ¼ 2πδ(Ω – ΩT), we

see that the Fourier transform of p(t) is given by

P jΩð

Þ ¼

k¼(cid:2)1

2π T

δ Ω (cid:2) kΩT ð

where ΩT ¼ 2π T

Substitution of (7.62) in (7.61) yields

Xp jΩð

Þ ¼

h Xa jΩð

Þ∗

1 T

k¼(cid:2)1

δ Ω (cid:2) kΩT ð

ð7:62Þ

ð7:63Þ

Since the convolution of Xa(jΩ) with a shifted impulse δ(Ω – kΩT) is the shifted

function Xa(j(Ω – kΩT)), the above reduces to

Xp jΩð

Þ ¼

k¼1

1 T

Xa jΩ (cid:2) jkΩT

ð7:64Þ

7.4 Representation of Sampling in Frequency Domain

345

Eq. (7.64) shows that the spectrum of xp(t) consists of an inﬁnite number of shifted copies of the spectrum of xa(t), and the shifts in frequency are multiples of ΩT; that is, Xp(jΩ) is a periodic function with a period of ΩT ¼ 2π/T. Since the continuous Fourier transform of δ(t – nT) is given by

F δ t (cid:2) nT ð

(cid:4) ¼ e(cid:2)jΩTn,

we have from Eq.(7.60) that

Since

Xp jΩð

Þ ¼

n¼(cid:2)1

xa nTð

Þe(cid:2)jΩTn

x nð Þ ¼ xa nTð

Þ, (cid:2)1 < n < 1

and the fact that the DTFT of the sequence x(n) is given by (cid:7) X ejω

x nð Þe(cid:2)jωn,

(cid:8)

n¼(cid:2)1

we obtain

or equivalently

(cid:8)

(cid:7) X ejω

(cid:9) (cid:9)

¼ Xp jΩð

Ω¼ω=T

(cid:7) Þ ¼ X ejω

(cid:8)(cid:9) (cid:9)

Xp jΩð

ω¼ΩT

ð7:65Þ

ð7:66Þ

ð7:67Þ

ð7:68aÞ

ð7:68bÞ

Hence, we have from (7.68a) and (7.64) that

(cid:8)

(cid:7) X ejω

1 T

k(cid:2)1

Xa jΩ (cid:2) jkΩT

(cid:9) (cid:9) (cid:9)

Ω¼ω=T

k(cid:2)1

1 T

(cid:6)

(cid:5)

(cid:2) j

2πk T

ð7:69Þ

On the other hand, the above equation can also be expressed as

(cid:8)

(cid:7) X ejΩT

k(cid:2)1

1 T

Xa jΩ (cid:2) jkΩT

ð7:70Þ

From Eq.(7.69) or (7.70), it can be observed that X(e jω) is obtained by frequency

scaling Xp ( jΩ) using Ω ¼ ω/T.

As mentioned earlier, the continuous-time Fourier transform Xp(jΩ) is periodic with respect to Ω having a period of ΩT ¼ (2π/T). In view of the frequency scaling, the DTFT X(e jω) is also periodic with respect to ω with a period of 2π.

346

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

(a)

(b)

Figure 7.12 (a) Spectrum of an analog signal (b) spectrum of the pulse train

Figure 7.13 Spectrum of an undersampled signal, showing aliasing (fold-over region). Signals in the fold-over region are not recoverable

7.4.1 Sampling of Low-Pass Signals

Sampling Theorem If the highest component of frequency in analog signal xa(t) is Ωm, then xa(t) is uniquely determined by its samples xa(nT), provided that

ΩT (cid:5) 2Ωm

ð7:71Þ

where ΩT is called the sampling frequency in radians. Eq. (7.71) is often referred as the Nyquist condition. The spectra of the analog signal xa(t) and the impulse train p(t) with a sampling period T ¼ 2π/ΩT are shown in Figure 7.12(a) and (b), respectively.

Undersampling If ΩT < 2Ωm, then the signal is undersampled, and the corresponding spectrum Xp(jΩ) is as shown in Figure 7.13. In this ﬁgure, the image frequencies centered at ΩT will alias into the baseband frequencies, and the information of the desired signal is indistinguishable from its image in the fold-over region.

7.5 Reconstruction of a Band-Limited Signal from Its Samples

347

Figure 7.14 Spectrum of an oversampled signal

Oversampling If ΩT > 2Ωm, then the signal is oversampled, and its spectrum is shown in Fig- ure 7.14. Its spectrum is the same as that of the original analog signal, but repeats itself at every multiple of ΩT. The higher-order components centered at multiples of ΩT are called image frequencies.

7.5 Reconstruction of a Band-Limited Signal

from Its Samples

According to the sampling theorem, samples of a continuous-time band-limited signal (i.e., its Fourier transform Xa(jΩ) ¼ 0 for |Ω| > |Ωm|) taken frequently enough are sufﬁcient to represent the signal exactly. The original continuous-time signal xa(t) can be fully recovered by passing the modulated impulse train xp(t) through an ideal low-pass ﬁlter, HLP(jΩ), whose cutoff frequency satisﬁes Ωm (cid:3) Ωc (cid:3) ΩT/2. Consider a low-pass ﬁlter with a frequency response:

HLPðjΩÞ ¼

(cid:2)

T jΩj (cid:3) Ωc jΩj > Ωc 0

ð7:72Þ

Applying the inverse continuous-time Fourier transform to HLP(jΩ), we obtain the

impulse response hLP(t) of the ideal low-pass ﬁlter given by

hLPðtÞ ¼

(cid:2)1

1 2π

HLPðjΩÞejΩtdΩ ¼

ðΩc

(cid:2)Ωc

T 2π

ejΩtdΩ ¼

sin ðΩctÞ ðπt=TÞ

, (cid:2) 1 < t < 1

ð7:73Þ

For a given sequence of samples x(n), we can form an impulse train xp(t) in which successive impulses are assigned an area equal to the successive sequence values, i.e.,

xp tð Þ ¼

n¼(cid:2)1

x nð Þδ t (cid:2) nT ð

ð7:74Þ

The nth sample is associated with the impulse at t ¼ nT, where T is the sampling period associated with the sequence x(n). Therefore, the output xa(t) of the ideal low-pass ﬁlter is given by the convolution of xp(t) with the impulse response hLP(t) of the analog low-pass ﬁlter:

348

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

( ( t

ADC

( ( nx

( JeH

(

( ( ny

(( tyr

DAC

1 =T

0001

sec

(a)

2 =T

( ) aX jW

0001

sec

10000p

(b)

Figure 7.15 (a) Discrete time system (b) spectrum of input xa(t)

xa tð Þ ¼

n¼(cid:2)1

x nð ÞhLP t (cid:2) nT ð

ð7:75Þ

Substituting hLP(t) from Eq.(7.73) in Eq. (7.75) and assuming for simplicity that

Ωc ¼ ΩT/2 ¼ π/T, we get

xa tð Þ ¼

n¼(cid:2)1

x nð Þ

sin π t (cid:2) nT ½ ð π t (cid:2) nT ð

Þ=T

(cid:4)

ð7:76Þ

The above expression indicates that the reconstructed continuous-time signal xa(t) is obtained by shifting in time the impulse response hLP(t) of the low-pass ﬁlter by an amount nT and scaling it in amplitude by the factor x(n) for all integer values of n in the range –1 < n < 1 and then summing up all the shifted versions. Example 7.17 Consider the system shown in Figure 7.15(a), where H(e jω) is an ideal LTI low-pass ﬁlter with cutoff of π/8 rad/sec, and the spectrum of xa(t) is shown in Figure 7.15(b).

(i) What is the maximum value of T to avoid aliasing in the ADC? (ii) If 1/T ¼ 10 kHz, then what will be the spectrum of yr(t).

Solution (i) From Figure 7.15(b), Ωm ¼ 10 k π. The given T1 ¼ 0.0001 sec. Then ΩT ¼ 2π T 1 The condition to avoid aliasing in the ADC is ΩT ¼ 2Ωm (Figure 7.16)

¼ 20 kπ.

(ii) T ¼ 1

10K ¼ 0:0001 sec

7.6 Problems

Figure 7.16

349

-10000

10000

(a)

X e W (

j T

)

1 T

-20000

-10000

10000

20000

(b)

)jX e w (

w )jH e (

· 2p-

p- 8

p 8 (c)

w= WT

)jY e w (

2- p

p 8

p 8 (d)

rY e W (

j T

)

rH jW (

)

1 T

2 p T 8

10000p

2 p T 8

(e)

p 8

p 8 (f )

7.6 Problems

Obtain the DTFS representation of the periodic sequence shown in Figure P7.1

350

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

…….

0 1 2 3

4 5 6 7 8 9

Figure P7.1 Periodic sequence with period N ¼ 5

Find the Fourier coefﬁcients (cid:7) (cid:8) x nð Þ ¼ sin 5π n 4

in DTFS representation of

the sequence

Find the DTFT for the following sequences:

(a) x1(n) ¼ u(n) – u(n – 5) (cid:7) (cid:8) (c) x3 nð Þ ¼ n 1 2

nj j

(d) x4(n) ¼ |a|nsin ωn, |α| < 1

(b) x2(n) ¼ αn(u(n) – u(n – 8)), |α| < 1

Let G1(e jω) denote the DTFT of the sequence g1(n) shown in Figure P7.2(a). Express the DTFTs of the remaining sequences in Figure P7.2 in terms of G1(e jω). Do not evaluate G1(e jω).

1( ) g n

(a)

2 ( ) g n

3 ( ) g n

Figure P7.2 Sequences g1(n), g2(n), and g3(n)

Further Reading

351

Determine the inverse DTFT of each of the following DTFTs:

(a) H1(e jω) ¼ 1 þ 4 cos ω þ 3 cos 2ω (b) H2(e jω) ¼ (3 þ 2 cos ω þ 4 cos (2ω)) cos (ω/2)e(cid:2)jω/2 (c) H3(e jω) ¼ e(cid:2)jω/4 (d) H4(e jω) ¼ e(cid:2)jω[1 þ 4 cos ω]

A continuous-time signal xa(t) has its spectrum Xa(jΩ) as shown in Figure P7.3(a). The signal xa(t) is input to the system shown in Figure P7.3(b). H(e jω) in Figure P7.3(b) is an ideal LTI low-pass ﬁlter with a cutoff frequency of (π/2). Sketch the spectrums of x(n), y(n), and yr(t).

p5000

(a)

xa (t)

( )nx

)wJeH (

ADC

( )ny

yr (t)

DAC

1 =T

0001

sec

2 =T

(b)

Figure P7.3 (a) Spectrum of signal. (b) Signal reconstruction

0001

sec

Further Reading

Morrison, N.: Introduction to Fourier Analysis. Wiley, New York (1994)
Mitra, S.K.: Digital Signal Processing. McGraw-Hill, New York (2006)
Oppenheim, A.V., Schafer, W.: Discrete-Time Signal Processing, 2nd edn. Prentice-Hall, Upper

Saddle River (1999)

Chapter 8 The z-Transform and Analysis of Discrete Time LTI Systems

The DTFT may not exist for all sequences due to the convergence condition, whereas the z-transform exists for many sequences for which the DTFT does not exist. Also, the z-transform allows simple algebraic manipulations. As such, the z-transform has become a powerful tool in the analysis and design of digital systems. This chapter introduces the z-transform, its properties, the inverse z-transform, and methods for ﬁnding it. Also, in this chapter, the importance of the z-transform in the analysis of LTI systems is established. Further, one-sided z-transform and the solution of state- space equations of discrete-time LTI systems are presented. Finally, transformations between continuous-time systems and discrete-time systems are discussed.

8.1 Deﬁnition of the z-Transform

The z-transform of an arbitrary discrete-time signal x(n) is deﬁned as

X zð Þ ¼ Z x nð Þ

(cid:2) ¼

n¼(cid:3)1

x nð Þz(cid:3)n

ð8:1Þ

where z is a complex variable. For the existence of the z-transform, Eq. (8.1) should x nð Þz(cid:3)n is absolutely converge. It is known from complex variables that if convergent, then Eq. (8.1) is convergent. Eq. (8.1) can be rewritten as

n¼(cid:3)1

X zð Þ ¼

n¼0

x nð Þz(cid:3)n þ

(cid:3)1

n¼(cid:3)1

x nð Þz(cid:3)n

ð8:2Þ

By ratio test, the ﬁrst series is absolutely convergent if

limn !1

(cid:2) (cid:2) (cid:2) (cid:2)

x n þ 1 ð x nð Þ

z(cid:3) nþ1 ð z(cid:3)n

(cid:2) (cid:2) (cid:2) (cid:2) ¼ limn!1

(cid:2) (cid:2) (cid:2) (cid:2)

x n þ 1 ð x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) z(cid:3)1

(cid:2) (cid:2) < 1

353

354

8 The z-Transform and Analysis of Discrete Time LTI Systems

zj j > limn!1

(cid:2) (cid:2) (cid:2) (cid:2)

x n þ 1 ð x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ r1 sayð

Similarly, the second series in Eq. (8.2) is absolutely convergent if (cid:2) (cid:2) x n þ 1 ð (cid:2) (cid:2) x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) z(cid:3)1

(cid:2) (cid:2) < 1

limn!1

zj j < limn!1

(cid:2) (cid:2) (cid:2) (cid:2)

x n þ 1 ð x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ r2 sayð

Thus, in general, Eq. (8.1) is convergent in some annulus

r1 < zj j < r2

ð8:3aÞ

ð8:3bÞ

ð8:4Þ

The set of values of z satisfying the above condition is called the region of convergence (ROC). It is noted that for some sequences r1 ¼ 0 or r2 ¼ 1. In such cases, the ROC may not include z ¼ 0 or z ¼ 1, respectively. Also, it is seen that no z-transform exists if r1 > r2.

The complex variable z in polar form may be written as

z ¼ rejω

ð8:5Þ

where r and ω are the magnitude and the angle of z, respectively. Then, Eq. (8.1) can be rewritten as

(cid:4)

(cid:3) X rejω

n¼(cid:3)1

x nð Þ reð

Þ(cid:3)jωn ¼

n¼(cid:3)1

x nð Þe(cid:3)jωnr(cid:3)n

ð8:6Þ

When r ¼ 1, that is, when the contour |z| ¼ 1, a unit circle in the z-plane, then

Eq. (8.5) becomes the DTFT of x(n).

Rational z-Transform In LTI discrete-time systems, we often encounter with a z-transform which is a ratio of two polynomials in z:

X zð Þ ¼

N zð Þ D zð Þ

b0 þ b1z(cid:3)1 þ b2z(cid:3)2 þ (cid:4) (cid:4) (cid:4) þ bMz(cid:3)M 1 þ a1z(cid:3)1 þ a2z(cid:3)2 þ (cid:4) (cid:4) (cid:4) þ aNz(cid:3)N

ð8:7Þ

The zeros of the numerator polynomial N(z) are called the zeros of X(z) and those of the denominator polynomial D(z) as the poles of X(z). The numbers of ﬁnite zeros and poles in Eq. (8.7) are M and N, respectively. For example, the function X zð Þ ¼

Þ has a zero at z ¼ 0 and two poles at z ¼ 1 and z ¼ 2.

z Þ z(cid:3)2 ð

z(cid:3)1 ð

8.2 Properties of the Region of Convergence for the z-Transform

355

8.2 Properties of the Region of Convergence

for the z-Transform

The properties of the ROC are related to the characteristics of the sequence x(n). In this section, some of the basic properties of ROC are considered.

Property 1: ROC should not contain poles.

In the ROC, X(z) should be ﬁnite for all z. If there is a pole p in the ROC, then X(z) is not ﬁnite at this point, and the z-transform does not converge at z ¼ p. Hence, ROC cannot contain any poles. Property 2: The ROC for a ﬁnite duration causal sequence is the entire z-plane

except for z ¼ 0.

A causal ﬁnite duration sequence of length N is such that x(n) ¼ 0 for n < 0 and for

n > N (cid:3) 1. Hence X(z) is of the form

XðzÞ ¼

N(cid:3)1 n¼0 xðnÞz(cid:3)n

¼ xð0Þ þ xð1Þz(cid:3)1 þ (cid:4) (cid:4) (cid:4) þ xðN (cid:3) 1Þz(cid:3)Nþ1

ð8:8Þ

It is clear from the above expression that X(z) is convergent for all values of z except for z ¼ 0, assuming that x(n) is ﬁnite. Hence, the ROC is the entire z-plane except for z ¼ 0 and is shown as shaded region in Figure 8.1. Property 3: The ROC for a noncausal ﬁnite duration sequence is the entire

z-plane except for z ¼ 1.

A noncausal ﬁnite duration sequence of length N is such that x(n) ¼ 0 for n (cid:5) 0

and for n (cid:6) (cid:3)N. Hence, X(z) is of the form

X zð Þ ¼

(cid:3)1 n¼(cid:3)N x nð Þz(cid:3)n

¼ x (cid:3)Nð

ÞzN þ (cid:4) (cid:4) (cid:4) þ x (cid:3)2ð

Þz2 þ x (cid:3)1ð

Þz

Figure 8.1 ROC of a ﬁnite duration causal sequence

Im(z)

ð8:9Þ

Re(z)

356

8 The z-Transform and Analysis of Discrete Time LTI Systems

Figure 8.2 ROC of a ﬁnite duration noncausal sequence

Im(z)

Re(z)

It is clear from the above expression that X(z) is convergent for all values of except for z ¼ 1, assuming that x(n) is ﬁnite. Hence, the ROC is the entire z-plane except for z ¼ 1 and is shown as shaded region in Figure 8.2. Property 4: The ROC for a ﬁnite duration two-sided sequence is the entire

z-plane except for z ¼ 0 and z ¼ 1.

A ﬁnite duration of length (N2 + N1 þ 1) is such that x(n) ¼ 0 for n < (cid:3)N1 and for

n > N2, where N1 and N2 are positive. Hence, x(z) is of the form

X zð Þ ¼

N2 n¼(cid:3)N1

x nð Þz(cid:3)n

¼ x (cid:3)N1 ð

ÞzN1 þ (cid:4) (cid:4) (cid:4) þ x (cid:3)1ð

Þz þ x 0ð Þ þ x 1ð Þz(cid:3)1 þ (cid:4) (cid:4) (cid:4) þ x N2ð

ÞzN2

ð8:10Þ

It is seen that the above series is convergent for all values of z except for z ¼ 0 and

z ¼ 1. Property 5: The ROC for an inﬁnite duration right-sided sequence is the

exterior of a circle which may or may not include z ¼ 1.

For such a sequence, x(n) ¼ 0 for n < N. Hence, X(z) is of the form

X zð Þ ¼

n¼N

x nð Þz(cid:3)n

ð8:11Þ

If N (cid:5) 0, then the right-sided sequence corresponds to a causal sequence and the

above series converges if Eq (8.3a) is satisﬁed, that is,

zj j > limn!1

(cid:2) (cid:2) x n þ 1 ð (cid:2) (cid:2) x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ r1

ð8:12Þ

Hence, in this case the ROC is the region exterior to the circle |z| ¼ r1 or the region |z| > r1 including the point at z ¼ 1. However, if N is a negative integer, say, N ¼ (cid:3)N1, then the series (8.12) will contain a ﬁnite number of terms involving positive powers of z. In this case, the series is not convergent for z ¼ 1, and hence the ROC is the exterior of the circle |z| ¼ r1 but will not include the point at z ¼ 1.

8.2 Properties of the Region of Convergence for the z-Transform

357

Figure 8.3 ROC of an inﬁnite duration causal sequence

Region of Convergence

As an example of an inﬁnite duration causal sequence, consider

(

r n 1 0

x nð Þ ¼

n¼0

1 z(cid:3)n ¼ r n

Then X zð Þ ¼

n (cid:5) 0, n < 0: (cid:3)

r1z(cid:3)1

n¼0

(cid:4)n

1 1 (cid:3) r1z(cid:3)1

ð8:13Þ

Eq. (8.13) holds only if |r1z(cid:3)1| < 1. Hence, the ROC is |z| > r1. The ROC is indicated by the shaded region shown in Fig. 8.3 and includes the region |z| > r1. It can be seen that X(z) has a zero at z ¼ 0 and pole at z ¼ r1. The zero is denoted by O and the pole by X. Property 6: The ROC for an inﬁnite duration left-sided sequence is the

interior of a circle which may or may not include z ¼ 0.

For such a sequence, x(n) ¼ 0 for n > N. Hence, X(z) is of the form

X zð Þ ¼

n¼(cid:3)1

x nð Þz(cid:3)n

ð8:14Þ

If N < 0, then the left-sided sequence corresponds to a noncausal sequence and the

above series converges if Eq. (8.3b) is satisﬁed, that is,

zj j < limn!(cid:3)1

(cid:2) (cid:2) (cid:2) (cid:2)

x n þ 1 ð x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ r2

ð8:15Þ

Hence, in this case, the ROC is the region interior to the circle |z| ¼ r2 or the

region |z| < r2 including the point at z ¼ 0.

However, if N is a positive integer, then the series (8.14) will contain a ﬁnite number of terms involving negative powers of z. In this case, the series is not convergent for z ¼ 0, and hence the ROC is the interior of the circle |z| ¼ r2 but will not include the point at z ¼ 0.

358

8 The z-Transform and Analysis of Discrete Time LTI Systems

Figure 8.4 ROC of an inﬁnite duration noncausal sequence

Region of Convergence

As an example of an inﬁnite duration noncausal sequence, consider (

x nð Þ ¼

0 (cid:3)r n

n (cid:5) 0, 2 n (cid:6) (cid:3)1:

Then,

X zð Þ ¼

(cid:3)1 n¼(cid:3)1 (cid:3)r(cid:3)n 1 1 (cid:3) r2z(cid:3)1 ¼

z z (cid:3) r2

2 z(cid:3)n ¼ (cid:3)r(cid:3)1 2 z

1 m¼0 r(cid:3)m

2 zm

for

zj j < r2

ð8:16Þ

ð8:17Þ

Hence, the ROC is |z| < r2, that is, the interior of the circle |z| ¼ r2. The ROC and

the pole and zero of X(z) are shown in Fig. 8.4. Property 7: The ROC of an inﬁnite duration two-sided sequence is a ring in

the z-plane.

In this case, the z-transform X(z) is of the form

X zð Þ ¼

n¼(cid:3)1

x nð Þz(cid:3)n

ð8:18Þ

and converges in the region r1 < |z| < r2, where r1 and r2 are given by (8.3a) and (8.3b), respectively. As mentioned before, the z-transform does not exist if r1 > r2.

As an example, consider the sequence

(

x nð Þ ¼

r n 1 (cid:3)r n 2

n (cid:5) 0, n < (cid:3)1:

Then,

X zð Þ ¼

z z (cid:3) r1

z z (cid:3) r2

z 2z (cid:3) r1 (cid:3) r2 Þ ð z (cid:3) r1 Þ z (cid:3) r2 Þ ð ð

ð8:19Þ

ð8:20Þ

8.2 Properties of the Region of Convergence for the z-Transform

359

Figure 8.5 ROC of an inﬁnite duration two-sided sequence

Region of Convergence

where the region of convergence is r1< |z| < r2. Thus, the ROC is a ring with a pole on the interior boundary and a pole on the exterior boundary of the ring, without any pole in the ROC. There are two zeros, one being located at the origin and the other in the ROC. The poles and zeros as well as the ROC are shown in Figure 8.5.

Example 8.1 Determine the z-transform and the ROC for the following sequence:

x nð Þ ¼ 2n

for n (cid:5) 0

Solution From the deﬁnition of the z-transform,

X zð Þ ¼

x nð Þz(cid:3)n ¼

2nz(cid:3)n ¼

n¼(cid:3)1 1 1 (cid:3) 2z(cid:3)1,

n¼0 (cid:2) (cid:2) < 1

2z(cid:3)1

(cid:2) (cid:2)

(cid:3)

(cid:4)n

2 z(cid:3)1

n¼0

Thus, the ROC is |z| > 2.

Example 8.2 Determine the z-transform and the ROC for the following sequence:

x nð Þ ¼

< :

(cid:5) (cid:6) n 1 (cid:3) 5 (cid:5) (cid:6) n 1 3

(cid:3)

for n (cid:5) 0

for n < 0

Solution

X zð Þ ¼

n¼(cid:3)1

x nð Þz(cid:3)n ¼

(cid:5) (cid:6) n 1 5

(cid:3)

n¼0

respectively.

(cid:2) (cid:2) Thus, the ROC is 1 5

Þz(cid:3)1 þ 1 þ 1=5 ð (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) < zj j < 1 3

1 (cid:3) 1=3 ð

Þz(cid:3)1,

z(cid:3)n þ (cid:2) (cid:2) (cid:2) (cid:2)

for

(cid:3)

X(cid:3)1

z(cid:3)n

(cid:5) (cid:6) n 1 3 n¼(cid:3)1 (cid:2) (cid:2) (cid:2) (cid:2) (cid:2) < zj j and zj j < 1 (cid:2) (cid:2) (cid:2) 3

1 5

(cid:2) (cid:2) (cid:2) (cid:2)

360

8 The z-Transform and Analysis of Discrete Time LTI Systems

8.3 Properties of the z-Transform

Properties of the z-transform are very useful in digital signal processing. Some important properties of the z-transform are stated and proved in this section. We will denote in the following the ROC of X(z) by R(r1 < |z| < r2) and those of X1(z) and X2(z) by R1 and R2, respectively. Also, the region (1/(r2) < |z| < 1/(r1) is denoted by (1/R).

Linearity If x1(n) and x2(n) are two sequences with z-transforms X1(z) and X2(z) and ROCs R1 and R2, respectively, then the z-transform of a linear combination of x1(n) and x2(n) is given by

Zfa1x1ðnÞ þ a2x2ðnÞg ¼ a1X1ðzÞ þ a2X2ðzÞ

ð8:21Þ

whose ROC is at least (R1 \ R1) and a1 and a2 being arbitrary constants.

Proof

Zfa1x1ðnÞ þ a2x2ðnÞg ¼

¼ a1

1 n¼(cid:3)1 fa1x1ðnÞ þ a2x2ðnÞgz(cid:3)n P

1 n¼(cid:3)1 x1ðnÞz(cid:3)n þ a2

1 n¼(cid:3)1 x2ðnÞz(cid:3)n

¼ a1X1 zð Þ þ a2X2 zð Þ

ð8:22Þ

ð8:23Þ

The result concerning the ROC follows directly from the theory of complex

variables concerning the convergence of a sum of two convergent series.

Time Reversal If x(n) is a sequence with z-transform X(z) and ROC R, then the z-transform of the time reversed sequence x((cid:3)n) is given by

Z x (cid:3)nð

(cid:3) g ¼ X z(cid:3)1 Þ

(cid:4)

ð8:24Þ

whose ROC is 1/R. Proof From the deﬁnition of the z-transform, we have P

Z x (cid:3)nð

(cid:2) ¼

1 n¼(cid:3)1 x (cid:3)nð

Þz(cid:3)n ¼

Hence,

1 m¼(cid:3)1 x mð Þ zm 1 m¼(cid:3)1 x mð Þ z(cid:3)1

ð8:25Þ

Þ(cid:3)m

(cid:3) (cid:2) ¼ X z(cid:3)1 Þ

(cid:4)

Z x (cid:3)nð

ð8:26Þ

Since (r1 < |z| < r2), we have (1/(r2) < |z(cid:3)1| < 1/(r1)). Thus, the ROC of Z [x((cid:3)n)]

is 1/R.

8.3 Properties of the z-Transform

361

Time Shifting If x(n) is a sequence with z-transform X(z) and ROC R, then the z-transform of the delayed sequence x(n (cid:3) k), k being an integer, is given by

Z x n (cid:3) k ð

(cid:2) ¼ z(cid:3)kX zð Þ

ð8:27Þ

whose ROC is the same as that of X(z) except for z ¼ 0 if k > 0 and z ¼ 1 if k < 0

Proof

Z x n (cid:3) k ð

g ¼

n¼(cid:3)1

x n (cid:3) k ð

Þz(cid:3)n

Substituting m ¼ n (cid:3) k,

Z x n (cid:3) k ð

1 m¼(cid:3)1 x mð Þz(cid:3) m þ k P 1 m¼(cid:3)1 x mð Þz(cid:3)m

(cid:2) ¼ ¼ z(cid:3)k ¼ z(cid:3)kX zð Þ

Þ ¼ z(cid:3)k

1 m¼(cid:3)1 x mð Þz(cid:3)m

ð8:28Þ

ð8:29Þ

ð8:30Þ

It is seen from Eq. (8.30) that in view of the factor z(cid:3)k, the ROC of Z [x(n (cid:3) k)] is the same as that of X(z) except for z ¼ 0 if k > 0 and z ¼ 1 if k < 0. It is also observed that in particular, a unit delay in time translates into the multiplication of the z-transform by z(cid:3)1. Scaling in the z-Domain If x(n) is a sequence with z-transform X(z), then Z{anx(n)} ¼ X(a(cid:3)1z) for any constant a, real or complex. Also, the ROC of Z{anx(n)} is |a|R, i.e., |a|r1 < |z| < |a|r2.

Proof

Z anx nð Þ

g ¼

x nð Þ

n¼(cid:3)1

anx nð Þz(cid:3)n

n¼(cid:3)1 (cid:7) (cid:8) z a

(cid:3)n

(cid:7) (cid:8) z a

¼ X

ð8:31Þ

ð8:32Þ

Since the ROC of X(z) is r1 < |z| < r2, the ROC of X(a(cid:3)1z) is given by r1 < |a–1z| < r2,

that is,

aj jr1 < zj j < aj jr2

Differentiation in the z-Domain If x(n) is a sequence with z-transform X(z), then

Z nx nð Þ

g ¼ (cid:3)z

dX zð Þ dz

ð8:33Þ

whose ROC is the same as that of X(z). Proof From the deﬁnition,

Z x nð Þ

(cid:2) ¼

n¼(cid:3)1

x nð Þz(cid:3)n

362

8 The z-Transform and Analysis of Discrete Time LTI Systems

Differentiating the above equation with respect to z, we get

dX zð Þ dz

n¼(cid:3)1

(cid:3)nð

Þx nð Þ z(cid:3)n(cid:3)1

ð8:34Þ

Multiplying the above equation both sides by (cid:3)z, we obtain

(cid:3)z

dX zð Þ dz

¼ (cid:3)z

n¼(cid:3)1

(cid:3)nð

Þx nð Þz(cid:3)n(cid:3)1

ð8:35Þ

which can be rewritten as

(cid:3)z

dX zð Þ dz

n¼(cid:3)1

nx nð Þz(cid:3)n ¼ Z nx nð Þ

ð8:36aÞ

Now, the region of convergence ra < |z| < rb of the sequence nx(n) can be found

using Eqs. (8.3a) and (8.3b).

ra ¼ limn!1

(cid:2) (cid:2) (cid:2) (cid:2)

n þ 1

Þx n þ 1 ð nx nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ limn!1

(cid:2) (cid:2) (cid:2) (cid:2)

(cid:2) (cid:2) (cid:2) (cid:2) ¼ r1

x n þ 1 ð x nð Þ

and

rb ¼ limn!(cid:3)1

(cid:2) (cid:2) (cid:2) (cid:2)

n þ 1

Þx n þ 1 ð nx nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ n

(cid:2) (cid:2) (cid:2) (cid:2)

x n þ 1 ð x nð Þ

(cid:2) (cid:2) (cid:2) (cid:2) ¼ r2

lim n!(cid:3)1

Hence, the ROC of Z[nx(n)] is the same as that of X(z). By repeated differentiation of Eq. (8.36a), we get the result

(cid:9)

(cid:10) Z nkx nð Þ

(cid:11)

¼ (cid:3)z

(cid:12)

d X zð Þ dz

ð8:36bÞ

It is to be noted that the ROC of Z[nkx(n)] is also the same as that of X(z).

Convolution of Two Sequences If x1(n) and x2(n) are two sequences with z-trans- forms X1(z) and X2(z), and ROCs R1 and R2, respectively, then

Z x1 nð Þ∗x2 nð Þ ½

(cid:2) ¼ X1 zð ÞX2 zð Þ

whose ROC is at least R1 \ R2.

Proof

X zð Þ ¼

n¼(cid:3)1

x nð Þz(cid:3)n

ð8:37Þ

ð8:38Þ

8.3 Properties of the z-Transform

363

The discrete convolution of x1(n) and x2(n) is given by

x1 nð Þ∗x2 nð Þ ¼

k¼(cid:3)1

x1 kð Þx2 n (cid:3) k ð

Þ ¼

k¼(cid:3)1

x2 kð Þx1 n (cid:3) k ð

ð8:39Þ

Hence, the z-transform of the convolution is

Z x1 nð Þ∗x2 nð Þ ½

(cid:2) ¼

h X

n¼(cid:3)1

k¼(cid:3)1

x2 kð Þx1 n (cid:3) k ð

i z(cid:3)n

ð8:40Þ

Interchanging the order of summation, the above equation can be rewritten

Z x1 nð Þ∗x2 nð Þ ½

(cid:2) ¼

Hence,

1 k¼(cid:3)1 x1 kð Þ 1 k¼(cid:3)1 x1 kð Þ 1 k¼(cid:3)1 x1 kð Þz(cid:3)k

1 Þz(cid:3)n n¼(cid:3)1 x2 n (cid:3) k ð 1 m¼(cid:3)1 x2 mð Þz(cid:3) mþk P 1 m¼(cid:3)1 x2 mð Þz(cid:3)m

ð8:41Þ

Z x1 nð Þ∗x2 nð Þ ½

(cid:2) ¼ X1 zð ÞX2 zð Þ

ð8:42Þ

Since the right side of Eq. (8.42) is a product of the two convergent sequences X1(z) and X2(z) with ROCs R1 and R1, it follows from the theory of complex variables that the product sequence is convergent at least in the region R1 \ R2. Hence, the ROC of Z[x1(n) ∗ x2(n)] is at least R1 \ R2.

Correlation of Two Sequences If x1(n) and x2(n) are two sequences with z-trans- forms X1(z) and X1(z), and ROCs R1 and R2, respectively, then

Z rx1x2 lð Þ

(cid:3)

(cid:2) ¼ X1 zð ÞX2 z(cid:3)1

(cid:4)

whose ROC is at least R1 \ (1/R2)

Proof Since

rx1x2 lð Þ ¼ x1 lð Þ∗x2 (cid:3)l ½ ð

Þ, (cid:2) ¼ Z x1 lð Þ∗x2 (cid:3)l Þ

Z rx1x2 lð Þ

(cid:2),

¼ Z x1 lð Þ (cid:2)Z x2 (cid:3)l ð ½ ¼ X1 zð ÞX2 z(cid:3)1 ð

Þ,

using Equation 8:37

(cid:2), ð using Equation 8:24

ð8:43Þ

ð8:44Þ

ð8:45Þ

Since the ROC of X2(z) is R2, the ROC of X2(z(cid:3)1) is 1/R2 from the property concerning time reversal. Also, since the ROC of X1(z) is R1, it follows from Eq. (8.45) that the ROC of Z rx1x2 lð Þ

(cid:2) is at least R1 \ (1/R2).

364

8 The z-Transform and Analysis of Discrete Time LTI Systems

Conjugate of a Complex Sequence If x(n) is a complex sequence with the z-transform X(z), then

Z x∗ nð Þ ½

(cid:2) ¼ X z∗ ½

(cid:2)

∗

with the ROCs of both X(z) and Z[x*(n)] being the same

Proof The z-transform of x*(n) is given by

Z x∗ nð Þ ½

(cid:2) ¼

n¼(cid:3)1

h X

x∗ nð Þz(cid:3)n

i∗

n¼(cid:3)1

x nð Þ z∗ ð

Þ(cid:3)n

ð8:46Þ

ð8:47Þ

ð8:48Þ

In the R.H.S. of the above equation, the term in the brackets is equal to x (z*).

Therefore, Eq. (8.48) can be written as

Z½x∗

ðnÞ(cid:2) ¼ ½Xðz∗

Þ(cid:2)

∗

¼ X∗

ðz∗

ð8:49Þ

It is seen from Eq. (8.49) that the ROC of the z-transform of conjugate sequence is

identical to that of X(z).

Real Part of a Sequence If x(n) is a complex sequence with the z-transform X(z), then

Z Re x nð Þ g f

(cid:2) ¼

whose ROC is the same as that of X(z).

Proof

Z Re x nð Þ f

(cid:2) ¼ Z

1 2

(cid:11)

X zð Þ þ X∗ z∗ ½

(cid:2)

ð8:50Þ

(cid:12)

x nð Þ þ x∗ nð Þ

1 2

ð8:51Þ

Since the z-transform satisﬁes the linearity property, we can write Eq. (8.51) as

Z Re x nð Þ f

(cid:2) ¼

1 2

Z x nð Þ

(cid:2) þ

Z x∗ nð Þ ½

(cid:2)

1 2

X zð Þ þ X∗ z∗

(cid:2), using 8:49

ð8:52Þ

ð8:53Þ

It is clear that the ROC of Z[Re{x(n)}] is the same as that of X(z).

8.4 z-Transforms of Some Commonly Used Sequences

365

Imaginary Part of a Sequence If x(n) is a complex sequence with the z-transform X(z), then

Z Im x nð Þ f

(cid:2) ¼

1 2j

X zð Þ (cid:3) X∗ z∗ ½

(cid:2)

whose ROC is the same as that of X(z).

Proof Now

x nð Þ (cid:3) x∗ nð Þ ¼ 2jIm x nð Þ

Im x nð Þ f

g ¼

1 2j

x nð Þ (cid:3) x∗ nð Þ

Thus,

Hence,

ð8:54Þ

ð8:55Þ

ð8:56Þ

Z½ImfxðnÞg(cid:2) ¼ Z

(cid:11)

1 2j

fxðnÞ (cid:3) x∗

ðnÞg(cid:2)

ð8:57Þ

Again, since the z-transform satisﬁes the linearity property, we can write

Eq. (8.57) as

Z Im x nð Þ f

(cid:2) ¼

1 2j 1 2j

Z x nð Þ

(cid:2) (cid:3)

Z x∗ nð Þ ½

(cid:2)

1 2j

X zð Þ (cid:3) X∗ z∗

(cid:2), using 8:49

ð8:58Þ

Again, it is evident that the ROC of the above is the same as that of X(z). The

above properties of the z-transform are all summarized in Table 8.1.

8.4

z-Transforms of Some Commonly Used Sequences

Unit Sample Sequence The unit sample sequence is deﬁned by

(cid:13)

δ nð Þ ¼

1 0

for n ¼ 0 elsewhere

ð8:59Þ

By deﬁnition, the z-transform of δ(n) can be written as

X zð Þ ¼

n¼(cid:3)1

x nð Þz(cid:3)n ¼ 1z0 ¼ 1

ð8:60Þ

It is obvious from (8.60) that the ROC is the entire z-plane.

366

8 The z-Transform and Analysis of Discrete Time LTI Systems

Table 8.1 Some properties of the z-transform

Property Linearity Time shifting

Sequence a1x1(n) þ a2x2(n) x(n (cid:3) k)

ROC

z-Transform a1X1(z) þ a2X2(z) At least R1 \ R2 z(cid:3)kX(z).

Same as R except for z ¼ 0 if k > 0 and for z ¼ 1 if k < 0 1 R |a|R

x((cid:3)n) anx(n)

nx(n)

X(z(cid:3)1) X(a(cid:3)1z)

(cid:3)z dX zð Þ dz

x1(n) ∗ x2(n)

X1(z)X2(z)

At least R1 \ R2

n¼(cid:3)1

x1ðnÞx2ðn (cid:3) lÞ

rx1x2 ðlÞ ¼ x∗(n)

Re[x(n)]

Im[x(n)]

x∗((cid:3)n)

X1(z)X2(z(cid:3)1)

At least R1 \ 1/R2

[X(z∗)]∗

2 X zð Þ þ X∗ z∗ð ½

(cid:2) At least R

2j X zð Þ (cid:3) X∗ z∗ð ½

(cid:2) At least R

X∗(1/z∗)

1 R

Time reversal

Scaling in the z-domain Differentiation in the z-domain Convolution theorem Correlation theorem

Conjugate com- plex sequence Real part of a complex sequence Imaginary part of a complex sequence Time reversal of a complex conju- gate sequence

Unit Step Sequence The unit step sequence is deﬁned by

(cid:13)

u nð Þ ¼

1 0

for n (cid:5) 0 elsewhere

The z-transform of x(n) by deﬁnition can be written as

X zð Þ ¼

1 n¼(cid:3)1 x nð Þz(cid:3)n ¼ 1 þ z(cid:3)1 þ z(cid:3)2 þ (cid:4) (cid:4) (cid:4) 1 1 (cid:3) z(cid:3)1 ¼ Hence, the ROC for X(z) is |z| > 1

z z (cid:3) 1

(cid:2) (cid:2) < 1

z(cid:3)1

for

(cid:2) (cid:2)

ð8:61Þ

ð8:62Þ

Example 8.3 Find the z-transform of x(n) ¼ δ(n (cid:3) k)

Solution By using the time shifting property, we get

Z δ n (cid:3) k ½ ð

(cid:2) ¼ z(cid:3)kZ δ nð Þ

(cid:2) ¼ z(cid:3)k

ð8:63Þ

The ROC is the entire z-plane except for z ¼ 0 if k is positive and for z ¼ 1 if k is

negative

8.4 z-Transforms of Some Commonly Used Sequences

367

Example 8.4 Find the z-transform of x(n) ¼ (cid:3)u((cid:3)n (cid:3) 1)

Solution We know that Z u nð Þ

z(cid:3)1 for Hence, using the time shifting property

(cid:2) ¼ z

zj j > 1 from Eq. (8.62)

Z u n (cid:3) 1 ð

(cid:2) ¼ z(cid:3)1

z z (cid:3) 1

1 z (cid:3) 1

for

zj j > 1

ð8:64Þ

Now, using the time reversal property (Table 8.1), we get

Z½uð(cid:3)n (cid:3) 1Þ(cid:2) ¼

1 z(cid:3)1 (cid:3) 1

z 1 (cid:3) z

for jzj < 1

Hence,

Z (cid:3)u (cid:3)n (cid:3) 1 ð

(cid:2) ¼

z z (cid:3) 1

for

zj j < 1

ð8:65Þ

Example 8.5 Find the z-transform of the sequence x(n) ¼ {bnu(n)}

Solution Let x1(n) ¼ u(n). From Eq. (8.62), Z u nð Þ

(cid:2) ¼ X1 zð Þ ¼ z

z(cid:3)1 for

zj j > 1

Using the scaling property, we get

Z bnu nð Þ

(cid:3) (cid:2) ¼ X1 b(cid:3)1z

(cid:4)

z z (cid:3) b

for zj j > bj j

Example 8.6 Find the z-transform of x(n) ¼ nu(n)

x1(n) ¼ u(n). Again,

using

Eq.

(8.62), we

have

Solution Let Z u nð Þ

zj j > 1 (cid:2) ¼ X1 zð Þ ¼ z z(cid:3)1 Using the differentiation property,

for

Z nx nð Þ

(cid:2) ¼ (cid:3)z

dX zð Þ dz

we get

Z nu nð Þ

(cid:2) ¼ (cid:3)z

dX1 zð Þ dz

¼ (cid:3)z

d dz

(cid:5)

(cid:6)

z z (cid:3) 1

Þ2

for

zj j > 1

Example 8.7 Obtain the z-transform of the following sequence:

x nð Þ ¼

(

n2u nð Þ

elsewhere

368

Solution

8 The z-Transform and Analysis of Discrete Time LTI Systems

X zð Þ ¼

n¼(cid:3)1

x nð Þz(cid:3)n ¼

n¼0

n2u nð Þz(cid:3)n

Let x(n) ¼ n2x1(n), where x1(n) ¼ u(n). Then

X1 zð Þ ¼

z z (cid:3) 1

for zj j > 1

Using the differentiation property that

(cid:5)

ZT if x nð Þ $

X zð Þ,

ZT then n2x nð Þ $

X (cid:3)z

(cid:6)

d dz

X zð Þ

we get

X zð Þ ¼ (cid:3)z

d dz

(cid:5)

(cid:3)z

d ½ dz

(cid:6)

X1 zð Þ (cid:2)

”

¼ (cid:3)z

d dz

z z (cid:3) 1

Þ2

z z þ 1 Þ ð Þ3 z (cid:3) 1 ð

The ROC of X(z) is the same as that of u(n), namely, |z| > 1

Example 8.8 Find the z-transform of x(n) ¼ sin ωn u(n)

Solution

Zfsin ωn uðnÞg ¼ Z

(cid:13)

ejωn (cid:3) e(cid:3)jωn 2j

(cid:14)

uðnÞ

1 2j

½ZfejωnuðnÞg (cid:3) Zfe(cid:3)jωnuðnÞg(cid:2)

Using the scaling property, we get

(cid:9) (cid:15) Z ejωnu nð Þ

(cid:16)

(cid:15)

(cid:3) Z e(cid:3)jωnu nð Þ

(cid:16)

(cid:10)

1 2j

Therefore,

z z (cid:3) e(cid:3)jω

1 2j

(cid:11) z z (cid:3) ejω (cid:3) z sin ω z2 (cid:3) 2z cos ω þ 1

(cid:12)

for

zj j > 1

Z sin ωn u nð Þ f

(cid:2) ¼

z sin ω z2 (cid:3) 2z cos ω þ 1

for

zj j > 1

Example 8.9 Find the z-transform of x(n) ¼ cos ωn u(n).

Solution Z cos ωn u nð Þ f

2 Using the scaling property, we get

n (cid:2) ¼ Z ejωnþe(cid:3)jωn

u nð Þ

¼ 1

2 Z ejωnu nð Þ ½

g þ Z e(cid:3)jωnu nð Þ

(cid:2)

½ZfejωnuðnÞg þ Zfe(cid:3)jωnuðnÞg(cid:2) ¼

1 2

z z (cid:3) e(cid:3)jω

(cid:11) z z (cid:3) ejω þ zðz (cid:3) cos ωÞ z2 (cid:3) 2zcos ω þ 1

(cid:12)

for jzj > 1

8.4 z-Transforms of Some Commonly Used Sequences

369

Therefore,

Z cos ωn u nð Þ f

(cid:2) ¼

z z (cid:3) cos ω ð z2 (cid:3) 2z cos ω þ 1

for

zj j > 1

Example 8.10 Find the z-transform of the sequence x(n) ¼ [u (n) (cid:3) u (n (cid:3) 5)]

Solution

XðzÞ ¼

n¼ 0

z(cid:3)n ¼ 1 þ z(cid:3)1 þ z(cid:3)2 þ z(cid:3)3 þ z(cid:3)4 ¼

z ðz (cid:3) 1Þ

ð1 (cid:3) z(cid:3)5Þ ¼

1 z4

z5 (cid:3) 1 z (cid:3) 1

The ROC is the entire z-plane except for z ¼ 0

(cid:9) (cid:10) Example 8.11 Determine X(z) for the function x nð Þ ¼ (cid:3) 1 2

nu (cid:3)n (cid:3) 1

Solution From Eq. (8.65), we have

Z (cid:3)u (cid:3)n (cid:3) 1 ð

(cid:2) ¼

z z (cid:3) 1

for

zj j < 1

Now using the scaling property (Table 8.1), (cid:11) (cid:12) 1 2

u (cid:3)n (cid:3) 1 ð

Z (cid:3)

(cid:13)

(cid:14)

2z 2z (cid:3) 1

for

zj j < 1 2

Thus the ROC is zj j < 1 2

Example 8.12 Consider a system with input x(n) and output y(n). If its impulse response h(n) ¼ Ax(L (cid:3) n), where L is an integer constant, and A is a known constant, ﬁnd Y(z) in terms of X(z).

Solution

h nð Þ ¼ Ax L (cid:3) n Þ y nð Þ ¼ x nð Þ∗h nð Þ

By the convolution property of the z-transform, we have

Y zð Þ ¼ H zð ÞX zð Þ

where

H zð Þ ¼ Z Ax L (cid:3) n f

g ¼ A

n¼(cid:3)1

x (cid:3) n (cid:3) L ð ð

Þz(cid:3)n

Letting n (cid:3) L ¼ m in the above, we have

H zð Þ ¼ A

1 m ¼ (cid:3)1 x (cid:3)mð

Þ ¼ Az(cid:3)L Þz(cid:3) mþL ð P 1 m¼(cid:3)1 x (cid:3)mð

1 m¼(cid:3)1 x (cid:3)mð Þz(cid:3)m

Þz(cid:3)m

¼ Az(cid:3)L ¼ Az(cid:3)LX z(cid:3)1

Þ ¼ Az(cid:3)LX 1=z

370

8 The z-Transform and Analysis of Discrete Time LTI Systems

Hence,

Y zð Þ ¼ Az(cid:3)LX 1=z

ÞX zð Þ

A list of some commonly used z-transform pairs are given in Table 8.2

Initial Value Theorem If a sequence x(n) is causal, i.e., x(n) ¼ 0 for n < 0, then

x 0ð Þ ¼ Lt z!1

X z½ (cid:2)

ð8:66Þ

Proof Since x(n) is causal, its z-transform X[z] can be written as

X z½ (cid:2) ¼

n¼0

x nð Þ:z(cid:3)n ¼ x 0ð Þ þ x 1ð Þz(cid:3)1 þ x 2ð Þz(cid:3)2 þ (cid:4) (cid:4) (cid:4)

ð8:67Þ

Now, taking the limits on both sides

(cid:15)

Lt z!1

X zð Þ ¼ Lt z!1

x 0ð Þ þ x 1ð Þz(cid:3)1 þ x 2ð Þz(cid:3)2 þ (cid:4) (cid:4) (cid:4)

(cid:16)

¼ x 0ð Þ

ð8:68Þ

Hence, the theorem is proved.

Example 8.13 Find the initial value of a causal sequence x(n) if its z-transform X(z) is given by

X zð Þ ¼

0:5z2

z (cid:3) 1

Þ z2 (cid:3) 0:85z þ 0:35 ð

Table 8.2 Some commonly used z-transform pairs

x(n) δ(n) u(n)

nu(n)

(cid:3)anu((cid:3)n (cid:3) 1)

(cid:3)nan{u((cid:3)n (cid:3) 1)}

{cosωn} u(n)

{sinωn} u(n)

X(z) 1

Þ2

1 1 (cid:3) z(cid:3)1 z(cid:3)1 1 (cid:3) z(cid:3)1 1 1 (cid:3) az(cid:3)1 az(cid:3)1 ð1 (cid:3) az(cid:3)1Þ2

1 (cid:3) z(cid:3)1cos ω 1 (cid:3) 2z(cid:3)1cos ω þ z(cid:3)2 z(cid:3)1sin ω 1 (cid:3) 2z(cid:3)1cos ω þ z(cid:3)2

ROC Entire z-plane |z| > 1

|z| > 1

|z| < |a|

|z| > 1

8.5 The Inverse z-Transform

371

Solution The initial value x(0) is given by

x 0ð Þ ¼ lim z!1

X zð Þ ¼ lim n!1

0:5z2

z (cid:3) 1

Þ z2 (cid:3) 0:85z þ 0:35 ð

¼ lim z!1

0:5z2 z z2ð Þ

¼ 0

8.5 The Inverse z-Transform

The z-transform of a sequence x(n), Z[x(n)], deﬁned by Eq. (8.1), is

X zð Þ ¼

m¼(cid:3)1

x mð Þz(cid:3)m

ð8:69Þ

Multiplying the above equation both sides by zn (cid:3) 1 and integrating both sides on a closed contour C in the ROC of the z-transform X(z) enclosing the origin, we get P

Þ CX zð Þzn(cid:3)1dz ¼ ¼

1 m¼(cid:3)1 x mð Þz(cid:3)mzn(cid:3)1dz 1 m¼(cid:3)1 x mð Þz(cid:3)mþn(cid:3)1dz

Multiplying both sides of Eq. (8.70) by 1

1 2πj

X zð Þzn(cid:3)1dz ¼

1 2πj

2πj, we arrive at X

x nð Þz(cid:3)mþn(cid:3)1dz

By Cauchy integral theorem, we have þ

1 2πj

m¼(cid:3)1 þ 1 2πj

z(cid:3)mþn(cid:3)1dz ¼

1 0

for m ¼ n for m 6¼ n

X zð Þzn(cid:3)1dz ¼ x nð Þ:

m¼(cid:3)1

(cid:13)

Thus, the inverse z-transform of X(z), denoted by Z(cid:3)1[X(z)], is given by

Z(cid:3)1 X zð Þ ½

(cid:2) ¼ x nð Þ ¼

1 2πj

X zð Þzn(cid:3)1dz

ð8:73Þ

It should be noted that given the ROC and the z-transform X(z), the sequence x(n) is unique. Table 8.2 can be used in most of the cases for obtaining the inverse transform. We will consider in Section 8.6 different methods of ﬁnding the inverse transform.

ð8:70Þ

ð8:71Þ

ð8:72Þ

372

8 The z-Transform and Analysis of Discrete Time LTI Systems

8.5.1 Modulation Theorem in the z-Domain

The z-transform of the product of two sequences (real or complex) x1(n) and x2(n) is given by

Z x1 nð Þx2 nð Þ

(cid:2) ¼

1 2πj

X1 vð ÞX2

(cid:7) (cid:8) z v

v(cid:3)1dv

ð8:74Þ

where C is a closed contour which encloses the origin and lies in the ROC that is common to both X1(v) and X2

(cid:3) (cid:4) z v

Proof Let x(n) ¼ x1(n)x2(n)

The inverse z-transform of x1(n) is given by

x1 nð Þ ¼

1 2πj

X1 vð Þ vn(cid:3)1dv

ð8:75Þ

Using Eq. (8.75), we get

x nð Þ ¼ x1 nð Þx2 nð Þ ¼

1 2πj

X1 vð Þ vn(cid:3)1x2 nð Þdv

ð8:76Þ

Taking the z-transform of Eq. (8.76), we obtain

XðzÞ ¼

1 n¼(cid:3)1 xðnÞz(cid:3)n ¼ þ

1 2πj

X1ðvÞ½

1 n¼(cid:3)1

n¼(cid:3)1

(cid:11)

X1ðvÞ vn(cid:3)1x2ðnÞdv

1 2πj v(cid:3)nx2ðnÞz(cid:3)n(cid:2)v(cid:3)1dv

(cid:12)

z(cid:3)n

ð8:77Þ

Using the scaling property, we have that

n¼(cid:3)1

v(cid:3)nx2 nð Þ z(cid:3)n ¼ X2

(cid:7) (cid:8) z v

Hence, Eq. (8.77) becomes

X zð Þ ¼

1 2πj

X1 vð ÞX2

(cid:7) (cid:8) z v

v(cid:3)1dv

which is the required result.

8.5.2 Parseval’s Relation in the z-Domain

If x1(n) and x2(n) are complex valued sequences, then

8.5 The Inverse z-Transform

373

n¼(cid:3)1

x1 nð Þx2

∗ nð Þ

(cid:2) ¼

1 2πj

X1 vð Þ X2

(cid:5) (cid:6)

∗ 1 v∗

v(cid:3)1dv

ð8:78Þ

where C is a contour contained in the ROC common to the ROCs of X1(v) and X∗ (cid:5) (cid:6) 1 v∗

Proof From Eq. (8.77), we have

Z x1 nð Þx2 nð Þ

(cid:2) ¼

1 2πj

X1 vð ÞX2

(cid:7) (cid:8) z v

v(cid:3)1dv

Hence,

Z x1 nð Þx2

∗ nð Þ

(cid:2) ¼

1 2πj

X1 vð ÞX2

(cid:5) (cid:6) ∗ z∗ v∗

v(cid:3)1dv

ð8:79Þ

where we have used the result concerning the z-transform of a complex conjugate (see Table 8.1). That is,

n¼(cid:3)1

x1 nð Þx2

∗ nð Þ

(cid:2) z(cid:3)n ¼

1 2πj

Letting z ¼ 1 in Eq. (8.80), we get

n¼(cid:3)1

x1 nð Þx2

∗ nð Þ

(cid:2) ¼

1 2πj

X1 vð ÞX2

(cid:5) (cid:6)

∗ 1 v∗

v(cid:3)1dv

(cid:5) (cid:6) ∗ z∗ v∗

v(cid:3)1dv

ð8:80Þ

Hence, the theorem. If x1(n) ¼ x2(n) ¼ x(n) and the unit circle is included by the ROC of X(z), then by

letting v ¼ e jω in (8.78), we get the energy of sequence in the z-domain to be

n¼(cid:3)1

x nð Þ

j2 ¼

1 2πj

(cid:5) (cid:6)

X zð ÞX∗ 1 z∗

z(cid:3)1dz

ð8:81Þ

For the energy of real sequences in the z-domain, the above expression becomes þ

n¼(cid:3)1

x nð Þ j

j2 ¼

1 2πj

(cid:3) X zð ÞX z(cid:3)1

(cid:4)

z(cid:3)1dz

ð8:82Þ

The Parseval’s relation in the frequency domain is given by

n¼(cid:3)1

jxðnÞj2 ¼

ðπ

(cid:3)π

1 2π

jXðejω

Þj2dω

374

8 The z-Transform and Analysis of Discrete Time LTI Systems

Thus,

n¼(cid:3)1

jxðnÞj2 ¼

1 2πj

XðzÞXðz(cid:3)1Þz(cid:3)1dz ¼

ðπ

(cid:3)π

1 2π

jXðejω

Þj2dω

ð8:83Þ

8.6 Methods for Computation of the Inverse z-Transform

8.6.1 Cauchy’s Residue Theorem for Computation

of the Inverse z-Transform

By Cauchy’s residue theorem, the integral in Eq. (8.73) for rational z-transforms yields Z(cid:3)1[X(z)] ¼ x(n) ¼ sum of the residues of the function [X(z)zn(cid:3)1] at all the poles pi enclosed by a contour C that lies in the ROC of X(z) and encloses the origin. The residue at a simple pole pi is given by (cid:9)

(cid:10)

(cid:9) X zð Þzn(cid:3)1

res z¼p

¼ lim z!pi

z (cid:3) pi

Þ X zð Þzn(cid:3)1

ð8:84Þ

while for a pole pi of multiplicity m, the residue is given by

(cid:9)

res z¼p

X zð Þzn(cid:3)1

(cid:10)

1 m (cid:3) 1

Þ! lim

z!pi

dm(cid:3)1 dzm(cid:3)1

(cid:9)

z (cid:3) pi

ÞmX zð Þzn(cid:3)1

(cid:10)

ð8:85Þ

We will now consider a few examples of ﬁnding the inverse z-transform using the

residue method. Example 8.14 Assuming the sequence x(n) to be causal, ﬁnd the inverse z-transform of

X zð Þ ¼

z z þ 1 Þ ð Þ3 z (cid:3) 1 ð

Solution Since the sequence is causal, we have to consider the poles of X(z)zn(cid:3)1 for only n (cid:5) 0. For n (cid:5) 0, the function X(z)zn(cid:3)1 has only one pole at z ¼ 1 of multiplicity 3. Thus, the inverse z-transform is given by

x nð Þ ¼

1 3 (cid:3) 1

Þ! lim

z!1

d2 dz2

”

z (cid:3) 1

”

Þ3 z z þ 1 ð z (cid:3) 1 ð

Þ3 zn(cid:3)1

x nð Þ ¼

1 3 (cid:3) 1

1 2! lim

z!1

¼ n2

d2 dz2

z!1

Þ! lim d2 dz2 z þ 1 ½ ð

z (cid:3) 1

Þzn

(cid:2) ¼

Þ3 zn(cid:3)1

Þ3 z z þ 1 ð z (cid:3) 1 ð (cid:9)

lim z!1

n n þ 1 ð

1 2

Þzn(cid:3)1 þ n n (cid:3) 1

Þzn(cid:3)2

(cid:10)

8.6 Methods for Computation of the Inverse z-Transform

375

It should be mentioned that if x(n) were not causal, then X(z)zn(cid:3)1 would have had a multiple pole of order n at the origin, and we would have to ﬁnd the residue of X(z)zn(cid:3)1 at the origin to evaluate x(n) for n < 0. Example 8.15 If x(n) is causal, ﬁnd the inverse z-transform of

X zð Þ ¼

1 Þ z þ 0:4 2 z (cid:3) 0:8 ð ð

Solution Since the sequence is causal, we have to consider the poles of X(z)zn(cid:3)1 for Þ zn(cid:3)1, we see that for n (cid:5) 1, X(z)zn(cid:3)1 has only n (cid:5) 0. Hence X zð Þzn(cid:3)1 ¼ two simple poles at 0.8 and (cid:3)0.4. However for n ¼ 0, we have an additional pole at the origin. Hence, we evaluate x(0) separately by evaluating the residues of X zð Þz(cid:3)1 ¼

1 Þ zþ0:4 2 z(cid:3)0:8 ð ð

Þ. Thus,

x 0ð Þ ¼

2 z (cid:3) 0:8 ð

1 (cid:3) Þ z þ 0:4

(cid:4)

j þ z¼0

1 2z z (cid:3) 0:8

1 2 (cid:3)0:8 ð

Þ 0:4ð

1 Þ (cid:3)1:2 2 (cid:3)0:4 ð ð

j Þ z¼(cid:3)0:4 1 Þ 1:2ð 2 0:8ð

1 2z z þ 0:4

j Þ z¼0:8

¼ 0

For n > 0,

x nð Þ ¼

zn(cid:3)1 2 z (cid:3) 0:8 ð Þn(cid:3)1 (cid:3)0:4 2 (cid:3)1:2 Þ ð

j Þ z¼0:4

0:8n(cid:3)1 2 1:2ð Þ

zn(cid:3)1 j Þ z¼0:8 2 z þ 0:4 ð (cid:7) : 0:8n(cid:3)1 (cid:3) (cid:3)0:4

1 2:4

(cid:8)

Þn(cid:3)1

Hence for any n (cid:5) 0,

x nð Þ ¼

1 2:4

(cid:7) : 0:8n(cid:3)1 (cid:3) (cid:3)0:4

(cid:8)

Þn(cid:3)1

u n (cid:3) 1

8.6.2 Computation of the Inverse z-Transform Using

the Partial Fraction Expansion

Partial fraction expansion is another technique that is useful for evaluating the inverse z-transform of a rational function and is a widely used method. To apply the partial fraction expansion method to obtain the inverse z-transform, we may consider the z-transform to be a ratio of two polynomials in either z or in z(cid:3)1. We now consider a rational function X(z) as given in Eq. (8.7). It is called a proper rational function if M > N; otherwise, it is called an improper rational function. An improper rational function can be expressed as a proper rational function by dividing

376

8 The z-Transform and Analysis of Discrete Time LTI Systems

the numerator polynomial N(z) by its denominator polynomial D(z) and expressing X (z) in the form

X zð Þ ¼

XM(cid:3)N

k¼0

f kz(cid:3)k þ

N1 zð Þ D zð Þ

ð8:86Þ

where the order of the polynomial N1(z) is less than that of the denominator polynomial. The partial fraction expansion can be now made on N1(z)/D(z). The z inverse z-transform of the terms in the sum is obtained from the pair δ n½ (cid:2) $ 1 (see Table 8.1) and the time-shift property (see Table 8.2). Let X(z) be a proper rational function expressed as

X zð Þ ¼

N zð Þ D zð Þ

b0 þ b1z(cid:3)1 þ b2z(cid:3)2 þ (cid:4) (cid:4) (cid:4) þ bMz(cid:3)M 1 þ a1z(cid:3)1 þ a2z(cid:3)2 þ (cid:4) (cid:4) (cid:4) þ aNz(cid:3)N

ð8:87Þ

For simpliﬁcation, eliminating negative powers, Eq. (8.87) can be rewritten as

XðzÞ ¼

NðzÞ DðzÞ

b0zN þ b1zN(cid:3)1 þ b2zN(cid:3)2 þ (cid:4) (cid:4) (cid:4) þ bMzN(cid:3)M zN þ a1zN(cid:3)1 þ a2z(cid:3)2 þ (cid:4) (cid:4) (cid:4) þ aN

ð8:88Þ

Since X(z) is a proper fraction, so will be [X(z)/z]. If all the poles pi are simple,

then, [X(z)/z] can be expanded in terms of partial fractions as

where

X zð Þ z

i¼1

ci z (cid:3) pi

ci ¼ z (cid:3) pi ð

(cid:2) (cid:2) X zð Þ (cid:2) (cid:2) z

z¼v

ð8:89Þ

ð8:90Þ

If [X(z)/z] has a multiple pole, say at pj, with a multiplicity of k, in addition to (N(cid:3)k) simple poles at pi, then the partial fraction expansion given in Eq. (8.89) has to be modiﬁed as follows.

X zð Þ z

cj1 z (cid:3) pj

(cid:3)

cj2 z (cid:3) pj

(cid:4)

2 þ (cid:4) (cid:4) (cid:4) þ

cjk (cid:3) z (cid:3) pj

(cid:4)

k þ

N(cid:3)k

i¼1

ci z (cid:3) pi

where ci is still given by (8.90) and cjk by (cid:13)

1 k (cid:3) j

Þ!

d k(cid:3)j ð dzk(cid:3)j

(cid:3)

(cid:4)

z (cid:3) pj

k X zð Þ z

(cid:14)(cid:2) (cid:2) (cid:2) (cid:2) z¼pj

cjk ¼

Hence,

ð8:91Þ

ð8:92Þ

8.6 Methods for Computation of the Inverse z-Transform

X zð Þ ¼

cj1z z (cid:3) pj

cj2z (cid:3) z (cid:3) pj

(cid:4)

2 þ (cid:4) (cid:4) (cid:4) þ

(cid:3)

cjkz z (cid:3) pj

(cid:4) k þ

N(cid:3)k

i¼1

ciz z (cid:3) pi

377

ð8:93Þ

Then inverse z-transform is obtained for each of the terms on the right-hand side of (8.91) by the use of Tables 8.1 and 8.2. We will now illustrate the method by a few examples. Example 8.16 Assuming the sequence x(n) to be right-sided, ﬁnd the inverse z-transform of the following:

X zð Þ ¼

z Þ z (cid:3) b ð

z (cid:3) a

Solution The given function has poles at z ¼ a and z ¼ b. Since X(z) is a right-sided sequence, the ROC of X(z) is the exterior of a circle around the origin that includes both the poles. Now X(z)/z can be expressed in partial fraction expansion as

X zð Þ z

a a (cid:3) b

1 z (cid:3) a

(cid:3)

b a (cid:3) b

1 z (cid:3) b

Hence,

X zð Þ ¼

a a (cid:3) b

1 1 (cid:3) az(cid:3)1 (cid:3)

b a (cid:3) b

1 1 (cid:3) bz(cid:3)1

We can now ﬁnd the inverse transform of each term using Table 8.2 as

x nð Þ ¼

a a (cid:3) b

anu nð Þ (cid:3)

b a (cid:3) b

bnu nð Þ

Example 8.17 Assuming the sequence x(n) to be causal, ﬁnd the inverse z-transform of the following:

X zð Þ ¼

10z2 (cid:3) 3z 10z2 (cid:3) 9z þ 2

Solution Dividing the numerator and denominator by z2, we can rewrite X(z) as

10 (cid:3) 3z(cid:3)1 10 (cid:3) 9z(cid:3)1 þ 2z(cid:3)2 4 2 (cid:3) z(cid:3)1 (cid:3) 2 1 (cid:3) 0:5z(cid:3)1 (cid:3)

5 5 (cid:3) 2z(cid:3)1 1 1 (cid:3) 0:4z(cid:3)1

378

8 The z-Transform and Analysis of Discrete Time LTI Systems

Each term in the above expansion is a ﬁrst-order z-transform and can be recog-

nized easily to evaluate the inverse transform as

Z(cid:3)1 X zð Þ f

g ¼ x nð Þ ¼ 2 0:5ð

Þnu nð Þ (cid:3) 0:4ð

Þnu nð Þ:

Example 8.18 Assuming the sequence x(n) to be causal, determine the inverse z-transform of the following:

X zð Þ ¼

z z þ 1 Þ ð Þ3 z (cid:3) 1 ð

Solution Since X(z)/z can be written in partial fraction expansion as

X zð Þ z

z þ 1 z (cid:3) 1

Þ3 ¼

A z (cid:3) 1

B z (cid:3) 1

Þ2 þ

C z (cid:3) 1

Þ3

Solving for A, B, and C, we get A ¼ 0, B ¼ 1, C ¼ 2. Hence, X(z) can be expanded

X zð Þ ¼

z z (cid:3) 1

Þ2 þ

2z z (cid:3) 1

Þ3

Making use of Table 8.2, the inverse z-transform of X(z) can be written as

Z(cid:3)1 X zð Þ f

g ¼ x nð Þ ¼ nu nð Þ þ n n (cid:3) 1

Þu nð Þ ¼ n2u nð Þ

Example 8.19 If x(n) is a right-handed sequence, determine the inverse z-transform for the function:

X zð Þ ¼

1 þ 2z(cid:3)1 þ z(cid:3)3 Þ 1 (cid:3) 0:5z(cid:3)1 ð

1 (cid:3) z(cid:3)1

Solution

X zð Þ ¼

1 þ 2z(cid:3)1 þ z(cid:3)3 Þ 1 (cid:3) 0:5z(cid:3)1 ð

1 (cid:3) z(cid:3)1

z3 þ 2z2 þ 1 Þ z (cid:3) 0:5 ð

z z (cid:3) 1 ð

Now, X(z)/z can be written in partial fraction expansion form as

X zð Þ z

z3 þ 2z2 þ 1 Þ z (cid:3) 0:5 ð ð

z2 z (cid:3) 1

A z

B z2 þ

C z (cid:3) 1

D z (cid:3) 0:5

Solving for A, B, C, and D, we get A ¼ 6, B ¼ 2, C ¼ 8, D ¼ (cid:3)13. Hence,

X zð Þ ¼

z3 þ 2z2 þ 1 Þ z (cid:3) 0:5 ð

z z (cid:3) 1 ð

¼ 6 þ

2 z

8z z (cid:3) 1

(cid:3)

13z z (cid:3) 0:5

8.6 Methods for Computation of the Inverse z-Transform

379

Since the sequence is right-handed and the poles of X(z) are located z ¼ 0, 0.5,

and 1, the ROC of X(z) is |z| > 1. Thus, from Table 8.2, we have

Z(cid:3)1 X zð Þ f

g ¼ x nð Þ ¼ 6δ nð Þ þ 2δ n (cid:3) 1

Þ þ 8u nð Þ (cid:3) 13 0:5ð

Þnu nð Þ

Example 8.20 Assuming h(n) to be causal, ﬁnd the inverse z-transform of

H zð Þ ¼

Solution Expanding H(z)/z as

Þ2

z (cid:3) 1 z2 (cid:3) 0:1z (cid:3) 0:56

H zð Þ z

ð z z (cid:3) 0:8 ð

Þ2 z (cid:3) 1 Þ z þ 0:7 ð

A z

B z (cid:3) 0:8

C z þ 0:7

Solving for A, B, and C, we get A ¼ (cid:3)1.78, B ¼ 0.033, and C ¼ 2.75 Therefore, H(z) can be expanded as

H zð Þ ¼ (cid:3)1:7857 þ

0:0333z z (cid:3) 0:8 Þ ð

2:7524z z þ 0:7 Þ ð

Hence,

Z(cid:3)1 H zð Þ f

g ¼ h nð Þ ¼ (cid:3)1:7857δ nð Þ þ 0:0333 0:8ð

Þnu nð Þ þ 2:7524 (cid:3)0:7

Þnu nð Þ

8.6.3

Inverse z-Transform by Partial Fraction Expansion Using MATLAB

The M-ﬁle residue z can be used to ﬁnd the inverse z-transform using the power Series expansion.

The coefﬁcients of the numerator and denominator polynomial written in

descending powers of z for Example 8.20 can be

num= [1 -2 1]; den= [1 -0.1 -0.56];

The following MATLAB statement determines the residue (r), poles (p), and

direct terms (k) of the partial fraction expansion of H(z).

[r,p,k]= residuez(num,den);

380

8 The z-Transform and Analysis of Discrete Time LTI Systems

After execution of the above statements, the residues, poles, and constants

obtained are

Residues: 0.0333 Poles: 0.8000 Constants: (cid:3)1.7857

–0.7000

2.7524

The desired expansion is

H zð Þ ¼ (cid:3)1:7857 þ

0:0333z z (cid:3) 0:8 Þ ð

2:7524z z þ 0:7 Þ ð

ð8:94Þ

8.6.4 Computation of the Inverse z-Transform Using

the Power Series Expansion

The z-transform of an arbitrary sequence deﬁned by Eq. (8.1) implies that X(z) can be expressed as power series in z(cid:3)1 or z. In this expansion, the coefﬁcient of the term indicates z(cid:3)n the value of the sequence x(n). Long division is one way to express X(z) in power series. Example 8.21 Assuming h(n) to be causal, ﬁnd the inverse z-transform of the following:

H zð Þ ¼

z2 þ 2z þ 1 z2 þ 0:4z (cid:3) 0:12

Solution We obtain the inverse z-transform by long division of the numerator by the denominator as follows:

1 þ 1:6z(cid:3)1 þ 0:48z(cid:3)2 þ 0z(cid:3)3 þ 0:0576z(cid:3)4 þ (cid:4) (cid:4) (cid:4)

z2 þ 0:4z (cid:3) 0:12j

z2 þ 2z þ 1 z2 þ 0:4z (cid:3) 0:12 1:6z þ 1:12 1:6z þ 0:64 (cid:3) 0:192z(cid:3)1 (cid:3)0:48 þ 0:192z(cid:3)1 0:48 þ 0:19z(cid:3)1 (cid:3) 0:0576z(cid:3)2 0:0576z(cid:3)2

0:0576z(cid:3)2 þ 0:02304z(cid:3)3 (cid:3) 0:006912z(cid:3)4 (cid:3) 0:02304z(cid:3)3 þ 0:006912z(cid:3)4

… … … …

8.6 Methods for Computation of the Inverse z-Transform

381

Hence, H(z) can be written as

H zð Þ ¼ 1:0 þ 1:6z(cid:3)1 þ 0:48z(cid:3)2 þ 0z(cid:3)3 þ 0:0576z(cid:3)4 þ (cid:4) (cid:4) (cid:4)

implying that

h n½ (cid:2)

g ¼ 1:0; f

1:6;

0:48;

0:0576; …

for n (cid:5) 0

Example 8.22 Find the inverse z-transform of the following:

(cid:3) X zð Þ ¼ log 1 þ bz(cid:3)1

(cid:4)

bj j < zj j

Solution We know that power series expansion for log(1 þ u) is

log 1 þ u ð

u2 2 (cid:3)1ð

u3 þ (cid:3) 3 Þnþ1un n

Þ ¼ u (cid:3) X1

n¼1

(cid:3) (cid:4) (cid:4) (cid:4)

u5 5

u4 4 , uj j < 1

Letting u ¼ bz(cid:3)1, X(z) can be written as

(cid:3) X zð Þ ¼ log 1 þ bz(cid:3)1

(cid:4)

n¼1

(cid:3)1ð

Þnþ1bnz(cid:3)n

bj j < zj j

From the deﬁnition of z-transform of x(n), we have

X zð Þ ¼

n¼1

x nð Þz(cid:3)n

Comparing the above two expressions, we get x(n), i.e., the inverse z-transform of

X(z) ¼ log (1 þ bz(cid:3)1) to be

(

x nð Þ ¼

bn Þnþ1 n

(cid:3)1ð 0

n > 0 n (cid:6) 0

ð8:95Þ

Example 8.23 Find the inverse z-transform of

X zð Þ ¼

z z (cid:3) b

for

zj j > bj j

Solution The sequence is a right-sided causal sequence as the region of conver- gence is |z| > |b|. We can use the long division as we did in Example 8.21 to express z/ (z(cid:3)b) as a series in powers of z(cid:3)1. Instead, we will use binomial expansion.

382

8 The z-Transform and Analysis of Discrete Time LTI Systems

X zð Þ ¼

z z (cid:3) b

1 1 (cid:3) bz(cid:3)1

¼ 1 þ bz(cid:3)1 þ b2z(cid:3)2 þ (cid:4) (cid:4) (cid:4) ¼

n¼0 bnz(cid:3)1

for

zj j > bj j

(cid:2) (cid:2) for bz(cid:3)1

(cid:2) (cid:2) < 1

Hence,

Z(cid:3)1 X zð Þ f

g ¼ x nð Þ ¼ Z(cid:3)1

(cid:13)

(cid:14)

z z (cid:3) b

¼ bnu nð Þ:

Example 8.24 Find the inverse z-transform of

X zð Þ ¼

z z (cid:3) b

for zj j < bj j

Solution Since the region of convergence is |z| < |b|, the sequence is a left-sided sequence. We can use the long division to obtain z/(z(cid:3)b) as a power series in z. However, we will use the binomial expansion.

X zð Þ ¼

z z (cid:3) b (cid:5) z b P

z b

¼ (cid:3)

1 1 (cid:3) z=bð (cid:7) (cid:8) z z b b n¼(cid:3)1 bnz(cid:3)n

1 þ

for

¼ (cid:3)

þ (cid:4) (cid:4) (cid:4)

zj j < bj j

(cid:2) (cid:2) (cid:2) < 1

(cid:2) (cid:2) z (cid:2) b

for

Hence,

Z(cid:3)1 X zð Þ f

g ¼ x nð Þ ¼ Z(cid:3)1

(cid:13)

(cid:14)

z z (cid:3) b

¼ (cid:3)bnu (cid:3)n (cid:3) 1 ð

Example 8.25 Using the z-transform, ﬁnd the convolution of the sequences:

x1 nð Þ ¼ 1; (cid:3)3; 2 f

g and x2 nð Þ ¼ 1; 2; 1

Solution

Step 1: Determine z-transform of individual signal sequences

X1 zð Þ ¼ Z x1 nð Þ

(cid:2) ¼

2 n¼0 x1 nð Þz(cid:3)1 ¼ x1 0ð Þ þ x1 1ð Þz(cid:3)1 þ x1 2ð Þz(cid:3)2

¼ 1 (cid:3) 3z(cid:3)1 þ 2z(cid:3)2

and

X2 zð Þ ¼ Z x2 nð Þ

(cid:2) ¼

2 n¼ 0 x2 nð Þz(cid:3)1 ¼ x2 0ð Þ þ x2 1ð Þz(cid:3)1 þ x2 2ð Þz(cid:3)2

¼ 1 þ 2z(cid:3)1 þ z(cid:3)2

8.6 Methods for Computation of the Inverse z-Transform

383

Step 2: Obtain X(z) ¼ X1(z)X2(z)

X zð Þ ¼ 1 (cid:3) 3z(cid:3)1 þ 2z(cid:3)2

Þ 1 þ 2z(cid:3)1 þ z(cid:3)2 ð

¼ 1 (cid:3) z(cid:3)1 (cid:3) 3z(cid:3)2 þ z(cid:3)3 þ 2z(cid:3)4

Step 3: Obtain the inverse z-transform of X(z)

(cid:9)

x nð Þ ¼ Z(cid:3)1 1 (cid:3) z(cid:3)1 (cid:3) 3z(cid:3)2 þ z(cid:3)3 þ 2z(cid:3)4

(cid:10)

¼ 1; (cid:3)1; (cid:3)3; 1; 2

8.6.5

Inverse z-Transform via Power Series Expansion Using MATLAB

The M-ﬁle impz can be used to ﬁnd the inverse z-transform using the power series expansion.

The coefﬁcients of the numerator and denominator polynomial for Example 8.21

can be written as

num = [1 2 1]; den = [1 0.4 -0.12];

The following statement can be run to obtain the coefﬁcients of the inverse z-

transform:

h = impz(num,den);

where h is the vector containing the coefﬁcients of the inverse z-transform. The ﬁrst 11 coefﬁcients of the inverse z-transform of Example 8.21 obtained after execution of the above MATLAB statements are

Columns 1 through 9 1.0000 1.6000 0.4800 0 0.0576 -0.0230 0.0161 -0.0092 0.0056 Columns 10 through 11 -0.0034 0.0020

8.6.6 Solution of Difference Equations

Using the z-Transform

Example 8.26 Determine the impulse response of the system described by the difference equation:

y nð Þ (cid:3) 3y n (cid:3) 1

Þ (cid:3) 4y n (cid:3) 2 ð

Þ ¼ x nð Þ þ 2x n (cid:3) 1

Þ:

Assume that the system is relaxed initially.

384

8 The z-Transform and Analysis of Discrete Time LTI Systems

Solution Let X(z) ¼ Z[x(n)] and Y(z) ¼ Z[y(n)]. Taking z-transform on both sides and using the time shifting property, we get (cid:3)

(cid:4)

1 (cid:3) 3z(cid:3)1 (cid:3) 4z(cid:3)2

(cid:3) Y zð Þ ¼ 1 þ 2z(cid:3)1

X zð Þ

Since X(z) ¼ 1, we have

YðzÞ ¼

YðzÞ z

YðzÞ ¼

1 þ 2z(cid:3)1 1 (cid:3) 3z(cid:3)1 þ 4z(cid:3)2 z þ 2 ðz (cid:3) 4Þðz þ 1Þ ð6=5Þ 1 (cid:3) 4z(cid:3)1 (cid:3)

¼ ð1=5Þ 1 þ z(cid:3)1

ð6=5Þ z (cid:3) 4

(cid:3)

ð1=5Þ z þ 1

We now take inverse transform of the above and use Table 8.2 to obtain y(n),

which is the impulse response of the system as

h nð Þ ¼ y nð Þ ¼ 6=5

Þ4nu nð Þ (cid:3) 1=5

Þ (cid:3)1ð

Þnu nð Þ

Example 8.27 Determine the response y(n), n (cid:5) 0 of the system described by the second-order difference equation

y nð Þ (cid:3) 3y n (cid:3) 1

Þ (cid:3) 4y n (cid:3) 2 ð

Þ ¼ x nð Þ þ 2x n (cid:3) 1

for the input x(n) ¼ 4n u(n)

Solution Applying z-transform to both sides of the equation, we have

(cid:9)

Y zð Þ 1 (cid:3) 3z(cid:3)1 (cid:3) 4z(cid:3)2

(cid:10)

(cid:9) ¼ X zð Þ 1 þ 2z(cid:3)1

(cid:10)

Given that x(n) ¼ 4n u(n), we have

X zð Þ ¼

1 1 (cid:3) 4z(cid:3)1

Substituting for X(z) in the expression for Y(z) and simplifying, we get

Y zð Þ z

ð z (cid:3) 4 ð

z2 þ 2 ð

Þ Þ2 z þ 1

Y zð Þ z

(cid:3)1 25 z þ 1 ð

26 25 z (cid:3) 4

24 5 z (cid:3) 4 ð

Þ2

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

385

Hence,

Y zð Þ ¼

(cid:3)z 25 z þ 1 ð

26z 25 z (cid:3) 4 ð

24z 5 z (cid:3) 4 ð

Þ2

By applying inverse z-transforms, we get

y nð Þ ¼

(cid:3)1 25

(cid:3)1ð

Þnu nð Þ þ

n 4ð Þnu nð Þ þ

6 5

26 25

4ð Þnu nð Þ

Example 8.28 Find the impulse response of the system

y nð Þ ¼ 3y n (cid:3) 1

Þ þ 2y n (cid:3) 2 ð

Þ þ x nð Þ

Solution Taking z-transforms on both sides of the above equation, and using the factZ[δ(n)] ¼ 1, we get

Y zð Þ ¼

1 1 (cid:3) 3z(cid:3)1 (cid:3) 2z(cid:3)2 ¼

z2 z2 (cid:3) 3z (cid:3) 2 0:135 1 þ 0:56z(cid:3)1

0:86 1 (cid:3) 3:56z(cid:3)1 þ Hence, the impulse response is given by

Y zð Þ ¼

h nð Þ ¼ y nð Þ ¼ 0:86 3:56

Þnu nð Þ þ 0:135 (cid:3)0:561

Þnu nð Þ

8.7 Analysis of Discrete-Time LTI Systems

in the z-Transform Domain

8.7.1 Transfer Function

It was stated in Chapter 6 that an LTI system can be completely characterized by its impulse response h(n). The output signal y(n) of a LTI system and the input signal x (n) are related by convolution as

y nð Þ ¼ h nð Þ∗x nð Þ

ð8:96Þ

Taking z-transform on both sides of the above equation and using the convolution

property, we get

Y zð Þ ¼ H zð ÞX zð Þ

ð8:97Þ

indicating the z-transform of the output sequence y(n) is the product of the z-trans- forms of the impulse response h(n) and the input sequence x(n). The quantities h(n) and H(z) are two equivalent descriptions of a system in the time domain and

386

8 The z-Transform and Analysis of Discrete Time LTI Systems

z-domain, respectively. The transform H(z) is called the transfer function or the system function and expressed as

H zð Þ ¼

Y zð Þ X zð Þ

ð8:98aÞ

Or equivalently,

H zð Þ ¼

(cid:9)

1 þ

M k¼0 bkz(cid:3)k P N k¼1 akz(cid:3)k

(cid:10)

ð8:98bÞ

where the constants ak and bk are real.

The above transfer function is a ratio of polynomials in z(cid:3)1 and, hence, is a

rational transfer function or system function.

Example 8.29 The following are known about a LTI discrete-time system: (i) y(n) ¼ δ(n) þ a(0.25)nu(n) (ii) y(n) ¼ 0 for all n if x(n) ¼ ((cid:3)2)n for all n

for x(n) ¼ (0.5)nu(n)

Find the value of the constant a.

Solution It is given that for the input x(n) ¼ (0.5)nu(n), the output of the LTI system is y(n) ¼ δ(n) þ a(0.25)nu(n). From this fact, the transfer function H(z) is given by

H zð Þ ¼

Y zð Þ X zð Þ

1 þ a (cid:3) 0:25z(cid:3)1 Þ 1 (cid:3) 0:5z(cid:3)1 ð

1 (cid:3) 0:25z(cid:3)1

It is also given that the output y(n) ¼ 0 for the input x(n) ¼ ((cid:3)2)n for all n. Since the function z n 0 is an eigenfunction for a discrete-time LTI system, the output to this Þz n 0 . From this, it can be inferred that H((cid:3)2) ¼ 0. Using this in the above input is H z0ð transfer function, the value of a is calculated to be (cid:3)1.125

8.7.2 Poles and Zeros of a Transfer Function

As mentioned earlier, the zeros of a system function H(z) are the values of z for which H(z) ¼ 0, while the poles are the values of z for which H(z) ¼ 1. Since H(z) is a rational transfer function, the number of ﬁnite zeros and the number of ﬁnite poles are equal the numerator and denominator polynomials, respectively.

to the degrees of

In MATLAB, tf2zp command can be used to ﬁnd the zeros, poles, and gains of a rational transfer function. z plane command can be used for plotting pole-zero plot of a rational transfer function.

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

387

Example 8.30 Determine the pole-zero plot using MATLAB for the system described by the system function

H zð Þ ¼

Y zð Þ X zð Þ

z (cid:3) 1 8z2 (cid:3) 6z þ 1

Solution The coefﬁcients of the numerator and denominator polynomial can be written as

numerator = [0 1 -1]; denominator = [8 -6 1];

The following MATLAB statement yields the poles and zeros and gain of the

system:

[z,p,gain] = tf2zp (numerator, denominator) zeros, z = 1 poles, p = [0.500 0.250] and gain = 0.1250

The MATLAB command z-plane (z, p) plots the poles and zeros as shown in

Figure 8.6.

Figure 8.6 Pole-zero plot of Example 8.30

388

8 The z-Transform and Analysis of Discrete Time LTI Systems

8.7.3 Frequency Response from Poles and Zeros

By factorizing the numerator and denominator polynomials of Eq. (8.98b), the transfer function can be written in pole-zero form as

HðzÞ ¼ b0zðN(cid:3)MÞ

i¼1 QN

i¼1

ðz (cid:3) ziÞ

ðz (cid:3) piÞ

ð8:99Þ

where zi and pi are the zeros and poles of H(z). It should be noted that the zeros are either real or occur in conjugate pairs. The frequency response of the system can be obtained by letting z ¼ e jω

in the transfer function H(z), that is,

(cid:4)

(cid:3) H ejω

¼ H zð Þ

(cid:2) (cid:2)

z¼ejω

Hence,

Hðejω

Þ ¼ b0ejωðN(cid:3)MÞ

i¼1 QN

i¼1

ðejω (cid:3) ziÞ

ðejω (cid:3) piÞ

ð8:100Þ

The contribution of the zeros and poles to the system frequency response can be

visualized from the above expression.

The magnitude of the frequency response can be expressed by

jHðejω

Þj ¼ jb0jjejω

jðN(cid:3)MÞ

i¼1 QN

i¼1

jðejω (cid:3) ziÞj

jðejω (cid:3) piÞj

ð8:101Þ

The zeros contribute to pulling down the magnitude of the frequency response, whereas the poles contribute to pushing up the magnitude of the frequency response. The size of decrease or increase in the magnitude response depends on how far the zero or the pole is from the unit circle. A peak in |H(e jω)| appears at the frequency of a pole very close to the unit circle.

To illustrate this, consider the following example.

Example 8.31 Consider a system with the transfer function

H zð Þ ¼

0:1 z2 þ 2z þ 1 ð 1:2z2 þ 1

ð8:102Þ

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

389

The numerator and denominator polynomials coefﬁcients in descending powers

of z can be written as

num=[1 2 1]; den=[1.2 0 1];

Then, as used in Example 8.30, using the MATLAB commands tf2zp and z-plane, the pole zero plot can be obtained as shown in Figure 8.7(a). The magnitude and phase responses of the above system transfer function are obtained using the above num and den vectors using the MATLAB command freqz. The magnitude and phase responses are shown in Figure 8.7(b) and (c), respectively.

Figure 8.7(a) indicates that the system has zeros of order 2 at z ¼ (cid:3)1 and two poles on the imaginary axis close to the unit circle. In the magnitude response of Figure 8.7(b), a peak occurs at ω ¼ π/2. This can be attributed to the fact that the frequency of the poles is π/2. The magnitude response is small at high frequencies due to the zeros.

8.7.4 Stability and Causality

The stability of a LTI system can be expressed in terms of the transfer function or the impulse response of the system. It is known from Section 6.4.5 that a necessary and sufﬁcient condition for a LTI system to be BIBO (bounded-input bounded-output) stable is that its impulse response be absolutely summable, i.e.,

n¼(cid:3)1

h nð Þ j

j < 1

H zð Þ ¼

n¼(cid:3)1

h nð Þ z(cid:3)n

H zð Þ

j (cid:6)

n¼1

h nð Þz(cid:3)n

j ¼

n¼1

h nð Þ

j z(cid:3)n j

On the unit circle (i.e., |z| ¼ 1), the above expression becomes

H zð Þ

j (cid:6)

h nð Þ

n¼(cid:3)1

ð8:103Þ

ð8:104Þ

ð8:105Þ

ð8:106Þ

Therefore, for a stable system, the ROC of its transfer function H(z) must include

the unit circle. Thus we have the following theorem.

BIBO Stability Theorem A discrete LTI system is BIBO stable if and only if the ROC of its system function includes the unit circle, |z| ¼ 1.

390

8 The z-Transform and Analysis of Discrete Time LTI Systems

Figure 8.7 (a) Pole-zero plot, (b) magnitude response, (c) phase response

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

391

We know from Section 6.4.5 that for a discrete LTI system to be causal h(n) ¼ 0 for n < 0. Thus, the sequence should be right-sided. We also know from Section 8.2 that the ROC of a right-sided sequence is the exterior of a circle whose radius is equal to the magnitude of the pole that is farthest from the origin. At the same time, we also know that for a right-sided sequence, the ROC may or may not include the point z ¼ 1. But we know from Section 8.2 that a causal system cannot have a pole at inﬁnity. Thus, in a causal system, the ROC should include the point z ¼ 1. Thus, we may summarize the result for causality by the following theorem:

Causality Theorem A discrete LTI system is causal if and only if the ROC of its system function is the exterior of a circle including z ¼ 1. An alternate way of stating this result is that a system is causal if and only if its ROC contains no poles, ﬁnite or inﬁnite.

Thus the conditions for stability and causality are quite different. A causal system could be stable or unstable, just as a noncausal system could be stable or unstable. Also, a stable system could be causal or noncausal just as an unstable system could be causal or noncausal. However, we can conclude from the above two theorems that a causal stable system must have a system function whose ROC is |z| ¼ r, where r < 1. Hence, we can summarize this result as follows.

Condition for a System to Be Both Causal and Stable A causal LTI system is BIBO stable if and only if all its poles are within the unit circle.

As a consequence, for a LTI system with a system function H(z) to be stable and causal, it is necessary that the degree of the numerator polynomial in z not exceed that of the denominator polynomial. As such, an FIR system is always stable, whereas if an IIR system is not designed properly, it may be unstable.

Example 8.32 Given the system function

H zð Þ ¼

(cid:3)

z 4z (cid:3) 3 ð (cid:4) z (cid:3) 1 ð 3

Þ z (cid:3) 4

Find the various regions of convergence for H(z), and state whether the system is stable and/or causal in each of these regions. Also, ﬁnd the impulse response h(n) in each case.

Solution The system function can be expressed in partial fraction in the form

H zð Þ ¼

z (cid:3) z (cid:3) 1 3

(cid:4) þ

3z z (cid:3) 4

(cid:3)

1 1 (cid:3) 1 3z(cid:3)1

(cid:4) þ 3

1 1 (cid:3) 4z(cid:3)1

The system function has two zeros, viz., z ¼ 0, 3

4, and two poles at z ¼ 1 zj j < 1 Hence, there are three regions of convergence: (i) (iii) |z| > 4. Let us consider each of these regions separately.

3 , 4: < zj j < 4, and

3, (ii) 1

392

8 The z-Transform and Analysis of Discrete Time LTI Systems

(i)

zj j < 1 3

In this region, there are no poles including the origin, but has poles exterior to it. Hence, the system is noncausal. Also, it is an unstable system, since the ROC does not include the unit circle. By using Table 8.2, we get

(cid:11)

h nð Þ ¼ (cid:3)

(cid:5) (cid:6) n 1 3

(cid:12)

þ 3 4ð Þn

u (cid:3)n (cid:3) 1 ð

(ii) 1 3

< zj j < 4

This region includes the unit circle and hence the system is stable. However, since the pole |z| ¼ 4 is exterior to this region, it is noncausal, and the corresponding sequence is two-sided. Again by using Table 8.2, we have

h nð Þ ¼

(cid:5) (cid:6) n 1 3

u nð Þ (cid:3) 3 4ð Þnu (cid:3)n (cid:3) 1

(iii) |z| > 4

This region does not include the unit circle, and hence the system is unstable. However, in this region, there are no poles, ﬁnite or inﬁnite, and hence, the system is causal. The impulse response of the system is obtained from H(z) using Table 8.2 as

h nð Þ ¼

(cid:5) (cid:6) n 1 3

u nð Þ þ 3 4ð Þnu nð Þ

Example 8.33 The rotational motion of a satellite was described by the difference equation

y nð Þ ¼ y n (cid:3) 1

Þ (cid:3) 0:5 y n (cid:3) 2

Þ þ 0:5 x nð Þ þ 0:5 x n (cid:3) 1

Is the system stable? Is the system causal? Justify your answer.

Solution Taking the z-transform on both sides of the given difference equation, we get

Y zð Þ ¼ z(cid:3)1Y zð Þ (cid:3) 0:5z(cid:3)2Y zð Þ þ 0:5X zð Þ þ 0:5z(cid:3)1X zð Þ

H zð Þ ¼

Y zð Þ X zð Þ

0:5 1 þ z(cid:3)1 1 (cid:3) z(cid:3)1 þ 0:5z(cid:3)2 ¼

0:5 z þ 1 Þz ð z2 (cid:3) z þ 0:5

The poles of the system are at z ¼ 0.5 (cid:7) 0.5j as shown in Figure 8.8 All poles of the system are inside the unit circle. Hence, the system is stable. It is

causal since the output only depends on the present and past inputs.

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

393

Figure 8.8 Poles of Example 8.33

0.5

-0.5

0.5 ´

Example 8.34 Consider the difference equation

y nð Þ (cid:3)

7 3

y n (cid:3) 1 ð

Þ þ

2 3

y n (cid:3) 2 ð

Þ ¼ x nð Þ

(a) Determine the possible choices for the impulse response of the system. Each choice should satisfy the difference equation. Speciﬁcally indicate which choice corresponds to a stable system and which choice corresponds to a causal system. (b) Can you ﬁnd a choice which implies that the system is both stable and causal? If

not, justify your answer.

Solution (a) Taking the z-transform on both sides and using the shifting theorem,

we get

(cid:5)

1 (cid:3)

z(cid:3)1 þ

7 3 Y zð Þ X zð Þ

(cid:6)

z(cid:3)2

2 3

Y zð Þ ¼ X zð Þ

1 (cid:3)

7 3

z(cid:3)1 þ

z(cid:3)2

2 3

H zð Þ ¼

z2 (cid:5)

z (cid:3) 2

Þ z (cid:3)

(cid:6)

1 3

The system function H(z) has a zero of order 2 at z ¼ 0 and two poles at z ¼ 1/3, 2. Hence, there are three regions of convergence, and thus, there are three possible choices for the impulse response of the system. The regions are

(i) R1: zj j < 1 3,

(ii) R2: 1 3

< zj j < 2, and

(iii) R3: |z| > 2.

The region R1 is devoid of any poles including the origin and hence corresponds to an anti-causal system, which is not stable since it does not include the unit circle. Region R2 does include the unit circle and hence corresponds to a stable system; however, it is not causal in view of the presence of the pole z ¼ 2. Finally, the region R3 does not have any poles including at inﬁnity and hence corresponds to a causal system; however, since R3 does not include the unit circle, the system is not stable.

394

8 The z-Transform and Analysis of Discrete Time LTI Systems

(b) There is no ROC that would imply that the system is both stable and causal. Therefore, there is no choice for h(n) which make the system both stable and causal.

Example 8.35 A system is described by the difference equation

y nð Þ þ y n (cid:3) 1

Þ ¼ x nð Þ, y nð Þ ¼ 0,

for n < 0:

(i) Determine the transfer function and discuss the stability of the system. (ii) Determine the impulse response h(n) and show that it behaves according to the

conclusion drawn from (i).

(iii) Determine the response when x(n) ¼ 10 for n (cid:5) 0. Assume that the system is

initially relaxed.

Solution (i) Taking the z-transforms on both sides of the given equation, we get

Hence,

Y zð Þ þ Y zð Þz(cid:3)1 ¼ X zð Þ

H zð Þ ¼

Y zð Þ X zð Þ

z z þ 1

The pole is at z ¼ (cid:3)1, that is, on the unit circle. So the system is marginally stable

or oscillatory.

(ii) Since h(n) ¼ 0 for n < 0,

h nð Þ ¼ Z(cid:3)1

(cid:11)

(cid:12)

z z þ 1

¼ (cid:3)1ð

Þnu nð Þ

This impulse response conﬁrms that the impulse response is oscillatory.

(iii) Since

Thus,

x nð Þ ¼ 10

for n (cid:5) 0,

X zð Þ ¼

10z z (cid:3) 1

Y zð Þ ¼ H zð ÞX zð Þ ¼

z z þ 1

10 z (cid:3) 1

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

395

Y zð Þ z

5 z þ 1

5 z (cid:3) 1

Therefore,

y nð Þ ¼ Z(cid:3)1 Y zð Þ

(cid:2) ¼ 5 (cid:3)1ð ½

Þn þ 5

(cid:2)u nð Þ

8.7.5 Minimum-Phase, Maximum-Phase, and Mixed-Phase

Systems

A causal stable transfer function with all its poles and zeros inside the unit circle is called a minimum-phase transfer function. A causal stable transfer function with all its poles inside the unit circle and all the zeros outside the unit circle is called a maximum-phase transfer function. A causal stable transfer function with all its poles inside the unit circle and with zeros inside and outside the unit circle is called a mixed-phase transfer function. For example, consider the systems with the following transfer functions:

H1 zð Þ ¼

H2 zð Þ ¼

Y zð Þ X zð Þ

z þ 0:4 z þ 0:3 0:4z þ 1 z þ 0:5

H3 zð Þ ¼

Y zð Þ X zð Þ

0:4z þ 1 Þ ð z þ 0:5 Þ ð

z þ 0:4 z þ 0:3

Þ Þ

ð ð

ð8:107Þ

ð8:108Þ

ð8:109Þ

The pole-zero plot of the above transfer functions are shown in Figure 8.9 (a), (b), and (c), respectively. The transfer function H1(z) has a zero at z ¼ (cid:3)0.4 and a pole at z ¼ (cid:3)0.3, and they are both inside the unit circle. Hence, H1(z) is a minimum-phase function. The transfer function H2(z) has a pole inside the unit circle, at z ¼ (cid:3)0.5, and a zero at z ¼ (cid:3)2.5, outside the unit circle. Thus, H2(z) is a maximum-phase function. The transfer function H3(z) has two poles, one at z ¼ (cid:3)0.3 and the other at z ¼ (cid:3)0.5, and two zeros one at z ¼ (cid:3)0.4, inside the unit circle, and the other at z ¼ (cid:3)2.5, outside the unit circle. Hence, H3(z) is a mixed-phase function.

8.7.6

Inverse System

Let H(z) be the system function of a linear time-invariant system. Then its inverse system function HI(z) is deﬁned, if and only if the overall system function is unity when H(z) and HI(z) are connected in cascade, that is, H(z) HI(z) ¼ 1, implying

396

8 The z-Transform and Analysis of Discrete Time LTI Systems

Figure 8.9 Pole-zero plot of (a) a minimum-phase function, (b) a maximum-phase function, and (c) a mixed-phase function

HI zð Þ ¼

1 H zð Þ

In the time domain, this is equivalently expressed as

hI nð Þ∗h nð Þ ¼ δ nð Þ

ð8:110Þ

ð8:111Þ

Example 8.36 A system is described by the following difference equation:

y nð Þ ¼ x nð Þ (cid:3) e(cid:3)8αx n (cid:3) 8

where the constant α > 0. Find the corresponding inverse system function to recover x(n) from y(n). Check for the stability and causality of the resulting recovery system, justifying your answer.

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

397

Solution

Y zð Þ ¼ X zð Þ (cid:3) e(cid:3)8αz(cid:3)8X zð Þ; Y zð Þ X zð Þ

(cid:3)

¼ 1 (cid:3) e(cid:3)8αz(cid:3)8

(cid:4)

The corresponding inverse system

H1 zð Þ ¼

1 1 (cid:3) e(cid:3)8αz(cid:3)8

X zð Þ Y zð Þ

The recovery system is both stable and causal, since all the poles of the system

HI(z) are inside the unit circle.

8.7.7 All-Pass System

Consider a causal stable Nth-order transfer function of the form

H zð Þ ¼ (cid:7)

aN þ aN(cid:3)1z(cid:3)1 þ (cid:4) (cid:4) (cid:4) þ z(cid:3)N 1 þ a1z(cid:3)1 þ (cid:4) (cid:4) (cid:4) þ aNz(cid:3)N ¼ (cid:7) Dðz(cid:3)1Þ ¼ 1 þ a1z þ a2z2 þ (cid:4) (cid:4) (cid:4) þ aNzN

M zð Þ D zð Þ

¼ zN½aN þ aN(cid:3)1z(cid:3)1 þ (cid:4) (cid:4) (cid:4) þ z(cid:3)N(cid:2) ¼ zNMðzÞ

Now

Hence,

and

Therefore,

Thus,

for all values of ω.

(cid:3)

M zð Þ ¼ z(cid:3)ND z(cid:3)1

(cid:4)

H zð Þ ¼ (cid:7)z(cid:3)N D z(cid:3)1 ð D zð Þ

(cid:4)

(cid:3) H z(cid:3)1

¼ (cid:7)zN D zð Þ D z(cid:3)1 ð

(cid:3)

H zð Þ H z(cid:3)1

(cid:4)

¼ 1

H ωð Þ

(cid:3)

j2 ¼ H ejω

(cid:4)

(cid:3) H e(cid:3)jω

(cid:4)

¼ 1

ð8:112Þ

ð8:113Þ

ð8:114Þ

ð8:115Þ

ð8:116Þ

ð8:117Þ

ð8:118Þ

398

8 The z-Transform and Analysis of Discrete Time LTI Systems

In other words, H(z) given by (8.112) passes all the frequencies contained in the input signal to the system, and hence such a transfer function is an all-pass transfer function, and the corresponding system is an all-pass system. It is also seen from (8.112) that if z ¼ pi is a zero of D(z), then z ¼ (1/pi) is a zero of M(z). That is, the poles and zeros of an all-pass function are reciprocal of one another. Since all the poles of H(z) are located within the unit circle, all the zeros are located outside the unit circle.

If x(n) is the input sequence and y(n) the output sequence for an all-pass system,

then

Thus,

Since |H(e jω)| ¼ 1, we get

Y zð Þ ¼ H zð ÞX zð Þ:

(cid:4)

(cid:3) Y ejω

(cid:3) ¼ H ejω

(cid:4)

(cid:3) X ejω

(cid:4)

(cid:2) (cid:2)

(cid:3) Y ejω

(cid:4)

(cid:2) (cid:2)

(cid:2) (cid:3) (cid:2) ¼ X ejω

(cid:4)

(cid:2) (cid:2)

ð8:119Þ

ð8:120Þ

ð8:121Þ

We know from Parseval’s relation that the output energy of a LTI system is given

Hence,

n¼(cid:3)1

y nð Þ j

1 2π

j2 ¼ ð π

(cid:2) (cid:2)

(cid:3)π

(cid:2) (cid:2)

(cid:3) Y ejω

(cid:4)

(cid:2) (cid:2)2

dω

ð π

1 2π (cid:3) X ejω

(cid:3)π (cid:2) (cid:4) (cid:2)2 dω

n¼(cid:3)1

y nð Þ j

j2 ¼

n¼(cid:3)1

x nð Þ

ð8:122Þ

ð8:123Þ

ð8:124Þ

Thus, the output energy is equal to the input energy for an all-pass system. Hence,

an all-pass system is that it is a lossless system.

Example 8.37 A discrete-time system with poles at z ¼ (cid:3)0.6 and z ¼ (cid:3)0.7 and zeros at z ¼ (cid:3)1/0.6 and z ¼ (cid:3)1/0.7 is shown in Figure 8.10. Demonstrate algebra- ically that magnitude response is constant.

Solution For given pole-zero pattern, the system function is given by

Hap zð Þ ¼

0:42 þ 1:3z(cid:3)1 þ z(cid:3)2 1 þ 1:3z(cid:3)1 þ 0:42z(cid:3)2

Substituting z ¼ e jω in the above transfer function, we get

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

399

Figure 8.10 Pole-zero plot of a second-order all-pass system

(cid:4)

(cid:3) Hap ejω

Hap e(cid:3)jω Þ ¼ ð (cid:2) (cid:2) (cid:2)2 (cid:2)

Hap ωð Þ

0:42 þ 1:3e(cid:3)jω þ e(cid:3)2jω 1 þ 1:3e(cid:3)jω þ 0:42e(cid:3)2jω 0:42 þ 1:3ejω þ e2jω 1 þ 1:3ejω þ 0:42e2jω ÞH e(cid:3)jω ð

¼ H ejωð

Þ ¼ 1

8.7.8 All-Pass and Minimum-Phase Decomposition

Consider an Nth-order mixed-phase system function H(z) with m zeros outside the unit circle and (n(cid:3)m) zeros inside the unit circle. Then H(z) can be expressed as (cid:3) (cid:4)

(cid:3)

(cid:4)

(cid:3)

(cid:4)

H zð Þ ¼ H1 zð Þ z(cid:3)1 (cid:3) a∗

z(cid:3)1 (cid:3) a∗ 2

(cid:4) (cid:4) (cid:4) z(cid:3)1 (cid:3) a∗ m

ð8:125Þ

where H1(z) is a minimum-phase function as its N poles and (n(cid:3)m) zeros are inside the unit circle. Eq. (8.125) can be equivalently expressed as

H zð Þ ¼ H1 zð Þ 1 (cid:3) z(cid:3)1a1 (cid:3) (cid:3)

z(cid:3)1 (cid:3) a∗ 1 1 (cid:3) z(cid:3)1a1

(cid:8)

(cid:4)

Þ 1 (cid:3) z(cid:3)1a2 Þ(cid:4) (cid:4) (cid:4) 1 (cid:3) z(cid:3)1am ð (cid:3) z(cid:3)1 (cid:3) a∗ 1 Þ 1 (cid:3) z(cid:3)1a2 ð

ð (cid:4) (cid:4) (cid:4) z(cid:3)1 (cid:3) a∗ m Þ(cid:4) (cid:4) (cid:4) 1 (cid:3) z(cid:3)1am

(cid:4)

ð8:126Þ

In the above equation, the factor H1(z)(1(cid:3)z(cid:3)1a1)(1(cid:3)z(cid:3)1a2)(cid:4) (cid:4) (cid:4)(1(cid:3)z(cid:3)1am) is also a minimum-phase function, since |a1|, |a2|, …, |am| are less than 1, the zeros are inside

400

8 The z-Transform and Analysis of Discrete Time LTI Systems

the unit circle, and the factor

(cid:3)

(cid:3) (cid:4)

(cid:4)

(cid:3)

z(cid:3)1 (cid:3) a∗ 1 1 (cid:3) z(cid:3)1a1 ð

z(cid:3)1 (cid:3) a∗ 2 Þ 1 (cid:3) z(cid:3)1a2 ð

(cid:4) (cid:4) (cid:4) z(cid:3)1 (cid:3) a∗ m Þ(cid:4) (cid:4) (cid:4) 1 (cid:3) z(cid:3)1am

(cid:4)

is all-pass.

Thus, any transfer function H(z) can be written as

H zð Þ ¼ Hmin zð ÞHap zð Þ

ð8:127Þ

Hmin(z) has all the poles and zeros of H(z) that are inside the unit circle in addition to the zeros that are conjugate reciprocals of the zeros of H(z) that are outside the unit circle, while Hap(z) is an all-pass function that has all the zeros of H(z) that lie outside the unit circle along with poles to cancel the conjugate reciprocals of the zeros of H(z) that lie outside the unit circle, which are now contained as zeros in Hmin(z).

Example 8.38 A signal x(n) is transmitted across a distorting digital channel char- acterized by the following system function:

Hd zð Þ ¼

1 (cid:3) 0:5z(cid:3)1 ð

Þ 1 (cid:3) 1:25ej0:8πz(cid:3)1 Þ 1 (cid:3) 1:25e(cid:3)j0:8πz(cid:3)1 ð ð 1 (cid:3) 0:81z(cid:3)2 Þ

Consider the compensating system shown in Figure 8.11. Find H1C(z) such that the overall system function G1(z) is an all-pass system.

Solution

(cid:3)

Hdmin1 zð Þ ¼

Hd zð Þ ¼ Hdmin1 zð ÞHap zð Þ 1 (cid:3) 0:5z(cid:3)1 Þ ð 1:25 ð 1 (cid:3) 0:8z(cid:3)2 Þ ð Þ z(cid:3)1 (cid:3) 0:8ej0:8π z(cid:3)1 (cid:3) 0:8e(cid:3)j0:8π Þ ð ð (cid:4) 1 (cid:3) 0:8e(cid:3)j0:8π 1 (cid:3) 0:8ej0:8πz(cid:3)1 z(cid:3)1 ð ð

Hap zð Þ ¼

(cid:3)

Þ2 1 (cid:3) 0:8ej0:8πz(cid:3)1

H1C zð Þ ¼

1 Hdmin1 zð Þ

1:25

Þ2 1 (cid:3) 0:5z(cid:3)1

Þ 1 (cid:3) 0:81z(cid:3)2 Þ 1 (cid:3) 0:8e(cid:3)J0:8πz(cid:3)1 ð

(cid:3) (cid:4)

1 (cid:3) 0:8e(cid:3)j0:8πz(cid:3)1

(cid:4)

Þ 1 (cid:3) 0:8eJ0:8πz(cid:3)1 ð

Then,

is an all-pass system.

Figure 8.11 Compensating system

G1 zð Þ ¼ Hd zð ÞH1C zð Þ ¼ Hap zð Þ

( )zG

( )nxd

( )nx

( )zH d

( )zH C1

( )nxca

8.8 One-Sided z-Transform

401

8.8 One-Sided z-Transform

The unilateral or one-sided z-transform, which is appropriate for problems involving causal signals and systems, is evaluated using the portion of a signal associated with nonnegative values of time index (n (cid:5) 0). It gives considerable meaning to assume causality in many applications of the z-transforms. Deﬁnition The one-sided z-transform of a signal x[n] is deﬁned as

Zþ x nð Þ ½

(cid:2) ¼ Xþ zð Þ ¼

n¼0

x nð Þz(cid:3)n

ð8:128Þ

which depends only on x(n) for (n (cid:5) 0). It should be mentioned that the two-sided z- transform is not useful in the evaluation of the output of a non-relaxed system. The one-sided transform can be used to solve for systems with nonzero initial conditions or for solving difference equations with nonzero initial conditions. The following special properties of X+ (z) should be noted.

The one-sided transform X+ (z) of x(n) is identical to the two-sided transform X(z) of the sequence x(n)u(n). Also, since x(n)u(n) is always causal, its ROC and hence that of X+ (z) are always the exterior of a circle. Hence, it is not necessary to indicate the ROC of a one-sided z-transform.
X+ (z) is unique for a causal signal, since such a signal is zero for n < 0.
Almost all

the properties of the two-sided transform are applicable to the

one-sided transform, one major exception being the shifting property.

Shifting Theorem for X+ (z) When the Sequence is Delayed by k If

Zþ x nð Þ ½

(cid:2) ¼ Xþ zð Þ,

then

Zþ x n (cid:3) k ð

(cid:2) ¼ z(cid:3)k

(cid:9) Xþ zð Þ þ

n¼1

x (cid:3)nð

Þzn, k > 0

ð8:129Þ

However, if x(n) is a causal sequence, then the result is the same as in the case of

the two-sided transform and

Zþ x n (cid:3) k ð

(cid:2) ¼ z(cid:3)kXþ zð Þ

ð8:130Þ

Proof By deﬁnition,

Zþ x n (cid:3) k ð

(cid:2) ¼

n¼0

x n (cid:3) k ð

Þ z(cid:3)n

Letting (n(cid:3)k) ¼ m, the above equation may be written as

402

8 The z-Transform and Analysis of Discrete Time LTI Systems

Zþ x n (cid:3) k ð

(cid:2) ¼ z(cid:3)k

m¼0

h ¼ z(cid:3)k Xþ zð Þ þ

(cid:3)1

x mð Þ z(cid:3)m þ X

n¼1

x (cid:3)nð

m¼(cid:3)k i

Þzn

x mð Þ z(cid:3)m

which proves (8.129). If the sequence x(n) is causal, then the second term on the right side of the above equation is zero, and hence we get the result (8.130).

Shifting Theorem for X+ (z) When the Sequence is Advanced by k If

Zþ x nð Þ ½

(cid:2) ¼ Xþ zð Þ,

then

Zþ x n þ k ð

h (cid:2) ¼ zk Xþ zð Þ (cid:3)

k(cid:3)1

n¼0

i ,

x nð Þz(cid:3)n

k > 0

ð8:131Þ

Proof By deﬁnition

Zþ x n þ k ð

(cid:2) ¼

n¼0

x n þ k ð

Þz(cid:3)n

Letting (n + k) ¼ m, the above equation may be written as (cid:9) P

(cid:10)

Zþ x n þ k ð

(cid:2) ¼ zk

h P

¼ zk

1 m¼k x mð Þz(cid:3)m 1 m¼0 x mð Þz(cid:3)m (cid:3) P

¼ zk Xþ zð Þ (cid:3)

k(cid:3)1 n¼0 x nð Þz(cid:3)n

k(cid:3)1 m¼0 x mð Þz(cid:3)m i

thus establishing the result (8.131).

Final Value Theorem If a sequence x(n) is causal, i.e., x(n) ¼ 0 for n < 0, then

limn!1x nð Þ ¼ limz!1 z (cid:3) 1

ÞX zð Þ

ð8:132Þ

The above limit exists only if the ROC of (z(cid:3)1) X(z) exists.

Proof Since the sequence x(n) is causal, we can write its z-transform as follow

Z x nð Þ

(cid:2) ¼

Also

Z x n þ 1 ð

(cid:2) ¼

n¼0

x nð Þ z(cid:3)n ¼ x 0ð Þ þ x 1ð Þz(cid:3)1 þ x 2ð Þz(cid:3)2 þ (cid:4) (cid:4) (cid:4):

ð8:133Þ

x n þ 1 ð

Þ z(cid:3)n ¼ x 1ð Þ þ x 2ð Þz(cid:3)1 þ x 3ð Þz(cid:3)2 þ (cid:4) (cid:4) (cid:4):

ð8:134Þ

8.8 One-Sided z-Transform

Hence, we see that

Thus,

403

Z x n þ 1 ð

(cid:2) ¼ z Z x nð Þ ½

(cid:2) (cid:3) x 0ð Þ

(cid:2)

ð8:135Þ

Z x n þ 1 ð

(cid:2) (cid:3) Z x nð Þ ½

(cid:2) ¼ z (cid:3) 1 ð

ÞZ x nð Þ ½

(cid:2) (cid:3) zx 0ð Þ

Substituting (8.133) and (8.135) for the L.H.S., we have

x 1ð Þ (cid:3) x 0ð Þ

(cid:2) þ x 2ð Þ (cid:3) x 1ð Þ

(cid:2)z þ x 3ð Þ (cid:3) x 2ð Þ

(cid:2)z2 þ (cid:4) (cid:4) (cid:4) ¼ z (cid:3) 1

ÞX zð Þ (cid:3) zx 0ð Þ

Taking the limit as z ! 1, we get

x 1ð Þ (cid:3) x 0ð Þ

(cid:2) þ x 2ð Þ (cid:3) x 1ð Þ

(cid:2) þ x 3ð Þ (cid:3) x 2ð Þ

(cid:2) þ (cid:4) (cid:4) (cid:4) ¼ limz!1 z (cid:3) 1

ÞX zð Þ (cid:3) x 0ð Þ

Thus,

Hence,

(cid:3)x 0ð Þ þ x 1ð

Þ ¼ limz!1 z (cid:3) 1

ÞX zð Þ (cid:3) x 0ð Þ

x 1ð

Þ ¼ limz!1 z (cid:3) 1

ÞX zð Þ

limn!1x nð Þ ¼ limz!1 z (cid:3) 1

ÞX zð Þ

It should be noted that the limit exists only if the function (z(cid:3)1) X(z) has an ROC that includes the unit circle; otherwise, system would not be stable and the limn!1x (n) would not be ﬁnite. Example 8.39 Find the ﬁnal value of x(n) if its z-transform X(z) is given by

X zð Þ ¼

0:5z2

z (cid:3) 1

Þ z2 (cid:3) 0:85z þ 0:35 ð

Solution The ﬁnal value or steady value of x(n) is given by

x nð Þ ¼ limz!1 z (cid:3) 1

ÞX zð Þ ¼

0:5 1 (cid:3) 0:85 þ 0:35

¼ 1

The result can be directly veriﬁed by taking the inverse transform of the

given X(z)

Example 8.40 The following facts are given about a discrete-time signal x(n) with X (z) as its z-transform:

(i) x(n) is real and right-sided. (ii) X(z) has exactly two poles.

404

8 The z-Transform and Analysis of Discrete Time LTI Systems

(iii) X(z) has two zeros at the origin. (iv) X(z) has a pole at z ¼ 0:5ejπ 3. (v) X 1ð Þ ¼ 8 3.

Determine X(z) and specify its region of convergence.

Solution It is given that x(n) is real and that X(z) has exactly two poles with one of jπ the pole at z ¼ 1 3 . Since, x [n] is real, the two poles must be complex conjugates of 2 e each other. Thus, the other pole of the system is at Z ¼ 1 2 e (z) has two zeros at the origin. Therefore, X(z) will have the following form:

(cid:3)jπ 3 . Also, it is given that X

X zð Þ ¼

(cid:5)

z (cid:3)

Kz2 (cid:5) (cid:6)

z (cid:3)

(cid:6)

(cid:3)jπ 3

1 2

jπ 3

1 2

Kz2 1 2

z2 (cid:3)

z þ

1 4

for some constant K to be determined. Finally, it is given that Xð1Þ ¼ 8 this in the above equation, we get K ¼ 2. Thus,

Substituting

X zð Þ ¼

2z2 2 z þ 1 z2 (cid:3) 1 4 zj j > 1 2

its ROC is

(note that both poles have

Since x[n] is right-sided,

magnitude 1 2).

8.8.1 Solution of Difference Equations with Initial

Conditions

The one-sided z-transform is very useful in obtaining solutions for difference equations which have initial conditions. The procedure is illustrated with an example.

Example 8.41 Find the step response of the system

y nð Þ (cid:3)

(cid:5) (cid:6) 1 2

y n (cid:3) 1 ð

Þ ¼ x nð Þ

with the initial condition y((cid:3)1) ¼ 1

Solution Taking one-sided z-transforms on both sides of the given equation and using (8.129), we have

8.9 Solution of State-Space Equations Using z-Transform

405

Y þðzÞ (cid:3)

(cid:5) (cid:6) 1 2

½z(cid:3)1Y þðzÞ þ yð(cid:3)1Þ(cid:2) ¼ XþðzÞ

Substituting for X+ (z) and y((cid:3)1), we have (cid:11)

(cid:12)

(cid:5) (cid:6) 1 2

1 (cid:3)

z(cid:3)1

Y þðzÞ ¼

1 2

1 1 (cid:3) z(cid:3)1

Hence,

Y þðzÞ ¼

1 2

(cid:6) þ

(cid:5)

1 (cid:3)

z(cid:3)1

1 1 2 2 1 (cid:3) z(cid:3)1 (cid:3)

1 (cid:6)

z(cid:3)1

ð1 (cid:3) z(cid:3)1Þ

(cid:5)

1 (cid:3)

1 2

(cid:5)

1 (cid:3)

(cid:6)

z(cid:3)1

1 1 2

Taking the inverse transform, we get

”

(cid:11)

(cid:5)

Z(cid:3)1 þ

YðzÞ

¼ yðnÞ ¼ 2 (cid:3)

(cid:12)

Þnþ1

uðnÞ

1 2

8.9 Solution of State-Space Equations Using z-Transform

For convenience, the state-space equations of a discrete-time LTI system from Chapter 6 are repeated here:

ð8:136aÞ ð8:136bÞ

ð8:137aÞ ð8:137bÞ

X n þ 1

Þ ¼ A X nð Þ þ b℧ nð Þ

y nð Þ ¼ cX nð Þ þ d℧ nð Þ

Taking unilateral z-transform Eqs. (8.136a) and (8.136b), we obtain

where

zX zð Þ (cid:3) zX 0ð Þ ¼ AX zð Þ þ b℧ zð Þ Y zð Þ ¼ cX zð Þ þ d℧ zð Þ

X zð Þ ¼

6 6 6 6 4

7 7 7 7 5

X1 zð Þ X2 zð Þ ⋮ XN(cid:3)1 zð Þ XN zð Þ

Eq. (8.137a) can be rewritten as

406

8 The z-Transform and Analysis of Discrete Time LTI Systems

zI (cid:3) A ½

(cid:2)X zð Þ ¼ zX 0ð Þ þ b℧ zð Þ

ð8:138Þ

where I is the identity matrix. From Eq. (8.138), we get

X zð Þ ¼ zI (cid:3) A ½

(cid:2)(cid:3)1zX 0ð Þ þ zI (cid:3) A

(cid:2)(cid:3)1b℧ zð Þ

ð8:139Þ

The inverse one-sided z-transform of Eq. (8.139) yields

z(cid:3)1 X zð Þ ½

(cid:2) ¼ z(cid:3)1

zI (cid:3) A ½

(cid:2)(cid:3)1zX 0ð Þ

þ z(cid:3)1

h ½

zI (cid:3) A

(cid:2)(cid:3)1b℧ zð Þ

ð8:140Þ

h ½

z(cid:3)1

zI (cid:3) A

(cid:2)(cid:3)1zX 0ð Þ

¼ An X 0ð Þ

ð8:141Þ

By using convolution theorem, we obtain

h ½

z(cid:3)1

zI (cid:3) A

(cid:2)(cid:3)1 b℧ zð Þ

n(cid:3)1

k¼0

An(cid:3)1(cid:3)k b℧ kð Þ

ð8:142Þ

Thus,

z(cid:3)1 X zð Þ ½

(cid:2) ¼ X nð Þ ¼ An X 0ð Þ þ

n(cid:3)1

k¼0

An(cid:3)1(cid:3)k b℧ kð Þ

n > 0:

ð8:143Þ

Substituting Eq. (8.143) into Eq. (8.136b), we get

y nð Þ ¼ cAn X 0ð Þ þ

n(cid:3)1

k¼0

An(cid:3)1(cid:3)k b℧ kð Þ þ d℧ nð Þ

n > 0:

ð8:144Þ

Example 8.42 Consider an initially relaxed discrete time system with the following state-space representation. Find y(n).

”

x1 n þ 1 x2 n þ 1

0 1 3

(cid:3)

(cid:11)

1 yðnÞ ¼ (cid:3) 3

4 3

1 4 3 (cid:11) (cid:12)

3 ” 5 x1 nð Þ x2 nð Þ

” # 0

℧ nð Þ

(cid:12)

x1ðnÞ x2ðnÞ

þ ℧ðnÞ

8.9 Solution of State-Space Equations Using z-Transform

407

Solution

A ¼

zI (cid:3) A

(cid:2) ¼

”

0 1 3

(cid:3) (cid:11)

1 4 3 (cid:12)

0 z

zI (cid:3) A

(cid:2)(cid:3)1 ¼

z (cid:3)1 z (cid:3) 4 1 3 3

; b ¼

(cid:11) (cid:12) 0

(cid:11)

; c ¼ (cid:3)

(cid:12)

4 3

1 3

”

(cid:3)

(cid:3)1

0 1 (cid:3) 3

1 4 3

1 (cid:5)

z (cid:3) 1

Þ z (cid:3)

z (cid:3)1 4 1 3 3

z (cid:3) 2

z (cid:3)

6 4

(cid:6)

1 3

4 3 1 3

; d ¼ 1: 3

7 5

(cid:3) 3

7 7 7 7 7 7 7 7 7 5

6 6 6 6 6 6 6 6 6 4

6 6 6 6 6 6 4

4 3

z (cid:3) (cid:5)

z (cid:3) 1

Þ z (cid:3)

(cid:6)

1 3

1 (cid:5)

z (cid:3) 1 ð

Þ z (cid:3)

(cid:6)

1 3

(cid:3)1=3 (cid:5)

z (cid:3) 1

Þ z (cid:3)

(cid:6)

z (cid:3)

1 3 3=2 1 3 1=2 1 3 i (cid:2)(cid:3)1 z

z (cid:3)

(cid:3)1=2 z (cid:3) 1

(cid:3)

1=2 z (cid:3) 1 h

zI (cid:3) A

z (cid:5)

z (cid:3) 1 ð

Þ z (cid:3)

(cid:6)

1 3 3

3=2 z (cid:3) 1

(cid:3)

3=2 z (cid:3) 1

(cid:3)

7 7 7 7 7 7 5

3=2 1 z (cid:3) 3 1=2 1 3

z (cid:3)

7 7 7 7 7 7 7 7 7 5

3 z 2

z (cid:3)

1 z 2

z (cid:3) 3

1 3

7 7 7 5

An ¼ z(cid:3)1

¼ z(cid:3)1

½ 2

6 6 6 6 6 6 6 6 6 4

(cid:3)z

1 2 z (cid:3) 1

(cid:3)z

1 2 z (cid:3) 1

3 z 2

z (cid:3)

1 z 2

z (cid:3)

(cid:5) (cid:6) n 1 3 (cid:5) (cid:6) n 1 3

6 6 6 4

6 4

(cid:3)

1 2 1 2 1 2 1 2

3 2 1 2 3 3 2 3 2

(cid:3) (cid:4) 7 n 5 þ 1 3

6 4

1 3

1 3 3 2 3 2 2

3 2 z (cid:3) 1

(cid:3)

3 2 z (cid:3) 1

(cid:3)

3 2 1 2

(cid:5) (cid:6) n 1 3 (cid:5) (cid:6) n 1 3 (cid:3)3 2 (cid:3)1 2

7 5

(cid:3)

3 2 1 2

408

8 The z-Transform and Analysis of Discrete Time LTI Systems

Since the system is initially relaxed, cAnX(0) ¼ 0 2 2

7 (cid:3) (cid:4) 7 5 þ 1 3

n(cid:3)1(cid:3)k

6 6 4

7 7 5

= ;

(cid:3)

3 2 1 2

” # 0

(cid:3) (cid:4) þ 1 3

n(cid:3)1(cid:3)k

(cid:11)

(cid:3)

1 3

(cid:12)

4 3

(cid:12)

(cid:3)

6 6 4

< : 2 (cid:12) (cid:3) 6 6 4

1 2 1 2 1 2 1 2 n(cid:3)1(cid:3)k

(cid:3)

3 2 3 2 3 3 2 3 2

7 7 5

6 6 4

3 2 1 2

7 7 5

3 (cid:3) 2 1 (cid:3) 2

” # 0

(cid:11)

cAn(cid:3)1(cid:3)k b ¼ (cid:3)

(cid:11)

¼ (cid:3)

1 3

4 3

(cid:5) (cid:6) 1 3

1 6

3 2

(cid:3)

Hence,

1ð Þk þ 1

þ 1

y nð Þ ¼

Xn(cid:3)1 3 2 ”

k¼0 Xn(cid:3)1

(cid:3)

3 2

(cid:3)

n(cid:3)k

1 6

1 2

Xn(cid:3)1

n(cid:3)1(cid:3)k

(cid:5) (cid:6) 1 3 (cid:5) (cid:6) 1 3 (cid:5) (cid:6) 1 3 k¼0 (cid:5) (cid:6) 1 3 1 (cid:3) 3 2 2 (cid:5) (cid:6) (cid:5) n 1 (cid:3) 3n 1 3 1 (cid:3) 3 (cid:5) (cid:6) 1 3

1 (cid:3) 3n

k¼0 Xn(cid:3)1 3 2

k¼0 P

n(cid:3)1 k¼0

3 2

n (cid:3)

n þ

1 2

1 4

n X

(cid:6)

n(cid:3)1

k¼0

3k þ 1

þ 1

n > 0:

Þ þ 1

n > 0:

8.10 Transformations Between Continuous-Time Systems

and Discrete-Time Systems

The transformation of continuous-time system to discrete-time system arises in various situations. In this section, two techniques, namely,

(i) Impulse invariance technique (ii) Bilinear transformation technique

are discussed for transforming continuous-time system to discrete-time system.

8.10 Transformations Between Continuous-Time Systems and Discrete-Time Systems

409

8.10.1 Impulse Invariance Method

In this method, the impulse response of an analog ﬁlter is uniformly sampled to obtain the impulse response of the digital ﬁlter, and hence this method is called the impulse invariance method. The process of designing an IIR ﬁlter using this method is as follows: Step 1: Design an analog ﬁlter to meet the given frequency speciﬁcations. Let Ha(s) be the transfer function of the designed analog ﬁlter. We assume for simplicity that Ha(s) has only simple poles. In such a case, the transfer function of the analog ﬁlter can be expressed in partial fraction form as

Ha sð Þ ¼

k¼1

Ak s (cid:3) pk

ð8:145Þ

where Ak is the residue of H(s) at the pole pk. Step 2: Calculate the impulse response h(t) of this analog ﬁlter by applying the

inverse Laplace transformation on H(s). Hence,

ha tð Þ ¼

k¼1

Akepkt ua tð Þ

ð8:146Þ

Step 3: Sample the impulse response of the analog ﬁlter with a sampling period T.

Then, the sampled impulse response h(n) can be expressed as

h nð Þ ¼ ha tð Þjt¼nT P

N k¼1 AkepkT ð

n u nð Þ

ð8:147Þ

Step 4: Apply the z-transform on the sampled impulse response obtained in Step 3, to form the transfer function of the digital ﬁlter, i.e., H(z) ¼ Z[h(n)]. Thus, the transfer function H(z) for the impulse invariance method is given by

H zð Þ ¼

k¼1

Ak 1 (cid:3) epkT z(cid:3)1

ð8:148Þ

This impulse invariant method can be extended for the case when the poles are

not simple.

Example 8.43 Consider a continuous system with the transfer function:

H sð Þ ¼

s þ b

s þ b ð

Þ2 þ c2

410

8 The z-Transform and Analysis of Discrete Time LTI Systems

Solution The inverse Laplace transform of H(s) yields

(cid:13)

h tð Þ ¼

e(cid:3)bt cos ctð Þ 0

for t (cid:5) 0 otherwise

Sampling h(t) with sampling period T, we get

(

e(cid:3)bnT cos cnTð 0 X1

for n (cid:5) 0

otherwise

Þz(cid:3)n

e(cid:3)bnT cos cnTð (cid:11) e(cid:3)bnT z(cid:3)n1 2

(cid:3)

ejcnT þ e(cid:3)jcnT

(cid:12)

(cid:4)

h nTð

Þ ¼

H zð Þ ¼

n¼0 X1

1 2

n¼0 (cid:11)

h (cid:7)

e(cid:3) b(cid:3)jc ð

ÞT Z(cid:3)1

(cid:8)

(cid:7)

þ e(cid:3) bþjc ð

(cid:8)

ÞT Z(cid:3)1 (cid:12)

1 2

1 ð

ÞTz(cid:3)1 (cid:3)

1 ð

1 (cid:3) e(cid:3) bþjc

ÞTz(cid:3)1

1 (cid:3) e(cid:3) b(cid:3)jc

1 (cid:3) e(cid:3)bT cos cTð

Þz(cid:3)1

1 (cid:3) 2e(cid:3)bT cos cTð

Þz(cid:3)1þe(cid:3)2bTz(cid:3)2

Disadvantage of Impulse Invariance Method The frequency responses of the digital and analog ﬁlters are related by

(cid:4)

(cid:3) H ejω

1 T

k¼(cid:3)1

(cid:6)

(cid:5)

ω þ 2πk T

ð8:149Þ

From Eq. (8.149), it is evident that the frequency response of the digital ﬁlter is not identical to that of the analog ﬁlter due to aliasing in the sampling process. If the analog ﬁlter is band-limited with (cid:7) (cid:8) ω

Ha j

¼ 0

(cid:2) (cid:2) (cid:2) ¼ Ωj

(cid:2) (cid:2) (cid:2)

j (cid:5) π=T

then the digital ﬁlter frequency response is of the form

(cid:4)

(cid:3) H ejω

Ha j

1 T

(cid:7) (cid:8)(cid:8) ω

ωj

j (cid:6) π

T In the above expression, if T is small, the gain of the ﬁlter becomes very large. This can be avoided by introducing a multiplication factor T in the impulse invariant transformation. In such a case, the transformation would be

ð8:150Þ

ð8:151Þ

and H(z) would be

h nð Þ ¼ Tha nTð

ð8:152Þ

8.10 Transformations Between Continuous-Time Systems and Discrete-Time Systems

411

H zð Þ ¼ T

k¼1

Ak 1 (cid:3) epkT z(cid:3)1

Also, the frequency response is

(cid:4)

(cid:3) H ejω

1 T

(cid:7) (cid:8) ω j

ωj

j (cid:6) π

ð8:153Þ

ð8:154Þ

Hence, the impulse invariance method is appropriate only for band-limited ﬁlters, i.e., low-pass and band-pass ﬁlters, but not suitable for high-pass or band-stop ﬁlters where additional band limiting is required to avoid aliasing. Thus, there is a need for another mapping method such as bilinear transformation technique which avoids aliasing.

8.10.2 Bilinear Transformation

In order to avoid the aliasing problem mentioned in the case of the impulse invariant method, we use the bilinear transformation, which is a one-to-one mapping from the s-plane to the z-plane; that is, it maps a point in the s-plane to a unique point in the z-plane and vice versa. This is the method that is mostly used in designing an IIR digital ﬁlter from an analog ﬁlter. This approach is based on the trapezoidal rule. Consider the bilinear transformation given by

S ¼

2 T

ð ð

z (cid:3) 1 z þ 1

Þ Þ

ð8:155Þ

Then a transfer function Ha (s) in the analog domain is transformed in the digital

domain as

(cid:2) (cid:2) H zð Þ ¼ Ha sð Þ

S¼

2 T

ð ð

z(cid:3)1 zþ1

Þ Þ

Also, from Eq. (8.155), we have

Z ¼

2 T

ð ð

1 þ s 1 (cid:3) s

Þ Þ

ð8:156Þ

ð8:157Þ

We now study the mapping properties of the bilinear transformation. Consider a

point s ¼ (cid:3)σ + jΩ in the left half of the s-plane. Then, from Eq. (8.157), (cid:2) (cid:2) (cid:2) (cid:2)

(cid:2) (cid:2) (cid:2) (cid:2) > 1

zj j ¼

1 (cid:3) σ þ jΩ 1 þ σ (cid:3) jΩ

Þ Þ

ð ð

ð8:158Þ

Hence, the left half of the s-plane maps into the interior of the unit circle in the z-plane (see Figure 8.12). Similarly, it can be shown that the right-half of the s-plane

412

8 The z-Transform and Analysis of Discrete Time LTI Systems

Left half s-plane

Im z

z-plane

-1

jΩ

(cid:2)

Figure 8.12 Mapping of the s-plane into the z-plane by the bilinear transformation

maps into the exterior of the unit circle in the z-plane. For a point z on the unit circle, z ¼ e jω, we have from Eq. (8.155)

Thus,

S ¼

2 T

ejω (cid:3) 1 ejω þ 1

Þ Þ

ð ð

¼ j

2 T

tan

Ω ¼

2 T

tan

ω ¼ 2 tan (cid:3)1

(cid:5) (cid:6) ΩT 2

ð8:159Þ

ð8:160Þ

ð8:161Þ

showing that the positive and negative imaginary axes of the s-plane are mapped, respectively, into the upper and lower halves of the unit circle in the z-plane. We thus see that the bilinear transformation avoids the problem of aliasing encountered in the impulse invariant method, since it maps the entire imaginary axis in the s-plane onto the unit circle in the in the z-plane. Further, in view of the mapping, this transfor- mation converts a stable analog ﬁlter into a stable digital ﬁlter. Example 8.44 Design a low-pass digital ﬁlter with 3 dB cutoff frequency at 50 Hz and attenuation of at least 10 dB for frequency larger than 100 Hz. Assume a suitable sampling frequency.

Solution Assume the sampling frequency as 500 Hz. Then,

ωc ¼

2π f c FT 2π f s FT

2π (cid:8) 50 500 2π (cid:8) 100 500 T ¼ 1=500 ¼ 0:002

ωs ¼

¼ 0:2π

¼ 0:4π

Prewarping of the above normalized frequencies yields

8.11 Problems

413

(cid:5)

(cid:6)

ΩC ¼ tan

(cid:5)

Ωs ¼ tan

0:2π 2

0:4π 2

(cid:6)

2 tan 0:1π ð T

2 tan 0:2π ð T

¼ 325

¼ 727

Substituting these values in Ωs=Ωc

ð (cid:3)

Þ2N ¼ 100:1αs (cid:3) 1 and solving for N, we get (cid:4)

N ¼

log 101 (cid:3) 1 2log 0:727=0:325

0:9542 0:6993

¼ 1:3643:

Hence,

the order of the Butterworth ﬁlter is 2. The normalized low-pass

Butterworth ﬁlter for N ¼ 2 is given by

HN sð Þ ¼

1 ﬃﬃﬃ p 2

s þ 1

s2 þ

The transfer function Hc(s) corresponding to Ωc ¼ 0.325 is obtained by substitut-

ing s ¼ (s/Ωc) ¼ (s/0.325) in the expression for HN(s); hence,

Ha sð Þ ¼

0:1056 s2 þ 0:4595s þ 0:1056

The digital transfer function H(z) of the desired ﬁlter is now obtained by using the

bilinear transformation (8.157) in the above expression:

H zð Þ ¼ Ha sð Þj

s¼ 0:1056z2 þ 0:2112z þ 0:1056 1000459:6056z2 (cid:3) 1999999:7888z þ 999540:6056

2 T

z(cid:3)1 Þ ð zþ1 Þ ð

H zð Þ ¼

8.11 Problems

Find the z-transform of the sequence x(n) ¼ {1,2,3,4,5,6,7}.
Find the z-transform and ROC of the sequence x(n) tabulated below.

n x(n)

(cid:3)2 1

(cid:3)1 2

0 3

1 4

2 5

3 6

4 7

Find the z-transform of the signal x(n) ¼ [3(3)n(cid:3)4(2)n].
Find the z-transform of the sequence x(n) ¼ (1/3)n(cid:3)1(cid:3)u(n(cid:3)1).

414

8 The z-Transform and Analysis of Discrete Time LTI Systems

Find the z-transform of the sequence

(cid:13)

x nð Þ ¼

1, 0,

0 (cid:6) n (cid:6) N (cid:3) 1 otherwise

Find the z-transform of the following discrete-time signals, and ﬁnd the ROC for

each.

(cid:3)

(cid:4) (i) x nð Þ ¼ (cid:3)1 2 (cid:3) (cid:4) (ii) x nð Þ ¼ 1 4 (iii) x nð Þ ¼ n þ 0:5

(cid:3) (cid:4) nu nð Þ þ 3 1 4 δ nð Þ þ δ n (cid:3) 2 ð (cid:3) (cid:4) nu n (cid:3) 1 Þ 1 ð 2

(cid:3)nu (cid:3)n (cid:3) 1 (cid:3) (cid:4) Þ (cid:3) 1 3 (cid:3) (cid:4) Þ (cid:3) 1 3

Þ δ n (cid:3) 3 ð Þ δ n (cid:3) 3 ð

Find the z-transform of the sequence x(n) ¼ nan(cid:3)1u(n(cid:3)1).
Find the z-transform of the sequence x(n) ¼ (1/4)n+1u(n).
Find the z-transform of the signal x(n) ¼ [(4)n+1(cid:3)3(2)n(cid:3)1].
Determine the z-transform and the ROC for the following time signals. Sketch

(cid:3)

the ROC, poles, and zeros in the z-plane. 4 n (cid:3) π (i) x nð Þ ¼ sin 3π 8 (cid:3) Þ sin 3π

(ii) x nð Þ ¼ n þ 1

u n (cid:3) 1 ½ (cid:4) 4 n þ π

u n þ 2 ½

(cid:4)

(cid:2)

(cid:2):

Find the inverse z-transform of the following, using partial fraction expansions:

(i) X zð Þ ¼ zþ0:5 Þ z(cid:3)2 ð (ii) X zð Þ ¼ 1þz(cid:3)1

zþ0:2

Þ ,

1þ3z(cid:3)1þ2z(cid:3)2 , Þ ,

z(cid:3)3

(iii) X zð Þ ¼ z2þz Þ z(cid:3)2 ð (iv) X zð Þ ¼ z zþ1 Þ ð Þ z(cid:3)1

ð z(cid:3)1 2

zj j > 2

zj j > 3

zj j > 1 2

Find the inverse z-transform of the following using the partial fraction

expansion.

z Þ z(cid:3)4 ð

(i) X zð Þ ¼

Þ , (ii) X zð Þ ¼ z2þ2z(cid:3)3 Þ z(cid:3)4 ð

z(cid:3)1

(iii) X zð Þ ¼

z(cid:3)1 Þ z(cid:3)3 ð ð z 3z2(cid:3)4zþ1 ,

zj j < 1

Þ , zj j < 1 3

for að Þ

zj j > 4 and bð Þ

zj j < 1

Determine all the possible signals that can have the following z-transform:

X zð Þ ¼

z2 z2 (cid:3) 0:8z þ 0:15

8.11 Problems

415

Find the stability of the system with the following transfer function:

H zð Þ ¼

z z3 (cid:3) 1:4z2 þ 0:65z (cid:3) 0:1

The transfer function of a system is given as

H zð Þ ¼

z þ 0:5 Þ z (cid:3) 2 ð

z þ 0:4

Specify the ROC of H(z) and determine h(n) for the following conditions:

(i) The system is causal. (ii) The system is stable. (iii) Can the given system be both causal and stable?

A causal LTI system is described by the following difference equation:

y nð Þ (cid:3)

1 4

y n (cid:3) 2 ð

Þ ¼ x n (cid:3) 2 ð

Þ (cid:3)

1 4

x nð Þ

Determine whether the system is an all-pass system.

In the system shown in Figure P8.1, if S1 is a causal LTI system with system

function,

(cid:5)

H zð Þ ¼ 1 (cid:3)

(cid:6)

(cid:5)

1 (cid:3)

z(cid:3)1

1 2

z(cid:3)1

3 4

(cid:6)

(cid:3)

(cid:4)

1 (cid:3) 3z(cid:3)1

Determine the system function for a system S2 so that the overall system is an

all-pass system.

( )nx

( )ny

Figure P8.1 Cascade connection of two systems S1 and S2

The transfer function of a system is given by

H zð Þ ¼

1 z2 þ 5z þ 6

Determine the response when x(n) ¼ u(n). Assume that the system is initially

relaxed.

416

8 The z-Transform and Analysis of Discrete Time LTI Systems

A causal LTI system is described by the following difference equation:

y nð Þ (cid:3) y n (cid:3) 1

Þ (cid:3) y n (cid:3) 2 ð

Þ ¼ x n (cid:3) 1 ð

Is it a stable system? If not, ﬁnd a noncausal stable impulse response that

satisﬁes the difference equation.

Using the one-sided z-transform, solve the following difference equation:

y nð Þ (cid:3)

(cid:5) (cid:6) 1 9

y n (cid:3) 2 ð

Þ ¼ u nð Þ,

y (cid:3)1ð

Þ ¼ 0, y (cid:3)2ð

Þ ¼ 2

Consider a continuous system with the transfer function

H sð Þ ¼

s þ 1

Þ s2 þ s þ 1 ð

Determine the transfer function and pole-zero pattern for the discrete-time

system by using the impulse invariance technique.

Design a low-pass Butterworth ﬁlter using Bilinear transformation for the

following speciﬁcations:

Passband edge frequency: 1000 Hz Stopband edge frequency: 3000 Hz Passband ripple: 2 dB Stopband ripple: 20 dB Sampling frequency: 8000 Hz

Consider an initially relaxed discrete time with the following state-space repre-

sentation. Find y(n).

”

x1 n þ 1

x2 n þ 1

(cid:11)

0 1 8 (cid:12)

1 3 4 ”

(cid:3)

3 ” 5 x1 nð Þ x2 nð Þ

” # 0

℧ nð Þ

y nð Þ (cid:3)

1 8

3 4

x1 nð Þ

x2 nð Þ

þ ℧ nð Þ

8.12 MATLAB Exercises

Write a MATLAB program using the command residuez to ﬁnd the inverse of the

following by partial fraction expansion:

Further Reading

417

X zð Þ ¼

16 (cid:3) 4z(cid:3)1 þ z(cid:3)2 8 þ 2z(cid:3)1 (cid:3) 2z(cid:3)2

Write a MATLAB program using the command impz to ﬁnd the inverse of the

following by power series expansion:

X zð Þ ¼

15z3 15z3 þ 5z2 (cid:3) 3z (cid:3) 1

Write a MATLAB program using the command z-plane to obtain a pole-zero plot

for the following system:

H zð Þ ¼

1 þ 1 1 þ 5

3 z(cid:3)1 þ 5 2 z(cid:3)1 (cid:3) 1

7 z(cid:3)2 (cid:3) 3 3 z(cid:3)2 (cid:3) 3

2 z(cid:3)3 5 z(cid:3)3

Write a MATLAB program using the command freqz to obtain magnitude and

phase responses of the following system:

H zð Þ ¼

1 (cid:3) 3:0538z(cid:3)1 þ 3:8281z(cid:3)2 (cid:3) 2:2921z(cid:3)3 þ 0:5507z(cid:3)4 1 (cid:3) 4z(cid:3)1 þ 6z(cid:3)2 (cid:3) 4z(cid:3)3 þ z(cid:3)4

Further Reading

Lyons, R.G.: Understanding Digital Signal Processing. Addison-Wesley, Reading (1997)
Oppenheim, A.V., Schafer, R.W.: Discrete-Time Signal Processing, 2nd edn. Prentice-Hall,

Upper Saddle River (1999)

Mitra, S.K.: Digital Signal Processing. McGraw-Hill, New York (2006)
Hsu, H.: Signals and Systems, Schaum’s Outlines, 2nd edn. McGraw-Hill, New York (2011)
Kailath, T.: Linear Systems. Prentice-Hall, Englewood Cliffs (1980)
Zadeh, L., Desoer, C.: Linear System Theory. McGraw-Hill, New York (1963)

Index

A Aliasing, 273, 347, 410–412 All-pass decomposition, 399–400 All-pass system, 397–400, 415 Amplitude, 5 Amplitude demodulation, 163–165 Amplitude modulation (AM), 155, 162–164 Analog ﬁlter design, 237, 238, 240, 242, 244,

250, 252–255, 258, 260–263 band-pass, 227, 230, 252–264, 269, 411 band-stop, 31, 227, 231, 252–264, 411 Butterworth low-pass ﬁlter, 62, 163, 228,

232–237, 249, 263, 413

Chebyshev analog low-pass ﬁlter

type 1 Chebyshev low-pass ﬁlter, 237,

238, 240, 255, 258, 261 type 2 Chebyshev ﬁlter, 242, 244,

elliptic analog low-pass ﬁlter, 227,

250, 253, 261

245–247, 251

high-pass, 31, 227–229, 231, 252–264,

269, 411

low-pass, 227, 229, 231–264 notch, 31, 266–267 speciﬁcations of low-pass ﬁlter, 232, 259,

261, 263 transformations

low-pass to band-pass, 252 low-pass to band-stop, 252, 260,

262, 263

low-pass to high-pass, 252, 254 low-pass to low-pass, 252, 253

Analog ﬁlter types comparison, 249–252 Application examples, vii, 10, 30, 162–164 Associative property, 51–52, 292

B Band-pass ﬁlter, 230, 257–260, 269, 411 Band-stop ﬁlter, 231, 260, 261, 263, 264 Basic continuous-time signals

complex exponential function, 24 ramp function, 22 real exponential function, 23–24 rectangular pulse function, 22–23 signum function, 23 sinc function, 24–27, 137 unit impulse function, 21–22, 136, 187 unit step function, 20, 22, 29, 98, 188

Basic sequences

arbitrary, 21, 50, 77, 113, 139, 176, 178, 282, 286, 318, 353, 360, 380 exponential and sinusoidal, 277, 278, 301 unit sample, 286, 365 unit step, 20, 49, 68, 286

Bessel ﬁlter, 248–250 BIBO stability theorem, 284, 285, 389, 391 Bilateral Laplace transform, 171, 172 Bilinear transformation, 411, 413, 416 Block diagram representation

described by differential equations, 82–93 Butterworth analog low-pass ﬁlter, 233–237

C Cauchy’s residue theorem, 374–375 Causal and stable conditions, 391 Causality, 41, 77, 82, 86–87, 107, 204–206, 208, 285, 297, 298, 311, 391, 401

Causality for LTI systems, 77, 294–297 Causality theorem, 391 Characteristic equation, 300

419

420

Index

Chebyshev analog low-pass ﬁlter, 237–245 Classiﬁcation of signals

analog and digital signals, 5, 271–276, 346 causal, non-causal and anti-causal signals, 12 continuous time and discrete time signals, 5 deterministic and random signals, 20 energy and power signals, 13–20 even and odd signals, 9–12, 141 periodic and aperiodic signals, 6–9

Commutative property, 46 Complex Exponential Fourier Series, 111–128 Computation of convolution integral using

MATLAB, 70–74 Computation of convolution sum using MATLAB, 291 Computation of linear convolution

graphical method, 289 matrix method, 288

Conjugate of complex sequence, 364 Continuous Fourier Transform

convergence of Fourier transform, 135–136

Continuous Fourier transform properties

convolution property, 151 differentiation in frequency, 146, 147 differentiation in time, 143 duality, 154 frequency shifting, 143 integration, 148 linearity, 139, 151, 158, 160 modulation, 155, 158 Parseval’s theorem, 149, 150, 157 symmetry properties, 119, 158 time and frequency scaling, 143, 158 time reversal, 158 time shifting, 142, 158, 168 Continuous time signal, 113–133

complex exponential Fourier series, 111–128 convergence of Fourier series, 113 properties of Fourier series, 113–128 trigonometric Fourier series, 128–133, 166

symmetry conditions, 129–133

Continuous-time systems causal system, 48, 77 invertible system, 49, 79 linear systems, 42–48 memory and memoryless system, 49 stable system, 49, 78 time–invariant system, 43–48, 105

Convergence of the DTFT, 317 Convolution integral

associative property, 51–52 commutative property, 50 distributive property, 50–51, 75 graphical convolution, 58–70

Convolution of two sequences, 151, 318, 362 Convolution sum, 271, 287, 289, 332 Convolution theorem, 220, 318, 320,

366, 406

Correlation, 30, 31, 33, 319, 363, 366 Correlation of discrete-time signals, vii Correlation of two sequences, 363 Correlation theorem, 319

D Direct form I, 96 Direct form II, 95–97 Discrete-time Fourier series (DTFS)

Fourier coefﬁcients, 350 multiplication, 315 periodic convolution, 313–316, 318 symmetry properties, 315

Discrete-time Fourier transform (DTFT)

linearity, 315

Discrete time LTI systems in z-domain,

385–400

Discrete-time signal, 271, 315–331, 414 Discrete-time signals classiﬁcation

energy and power signals, 13, 279–281 ﬁnite and inﬁnite length, 276 periodic and aperiodic, 6–9, 278 right-sided and left-sided, 277 symmetric and anti-symmetric, 276

Discrete-time system characterization

non-recursive difference equation, 298 recursive difference equation, 298, 336

Discrete-time systems classiﬁcation

causal, 284, 298 linear, 282, 286–289, 291–297 stable, 284 time-invariant, 283–284 Discrete transformation, 281 Distributive property, 50–51, 75 Down-sampler, 306

E Elementary operations on signals, 1–5 Elliptic analog low-pass ﬁlter, 246 Energy and power signals, 13–20, 279 Examples of real world signals and systems

audio recording system, 32 global positioning system, 33 heart monitoring system, 34–36 human visual system, 36 location-based mobile emergency

services system, 33–34 magnetic resonance imaging, 36–37

Index

421

F Filtering, 31, 227, 233, 235, 236 Final value theorem, 187, 200, 222 Fourier transform, 318 Fourier transform of discrete-time signals,

158, 315, 317–331, 335, 342

convergence of the DTFT, 317 properties of DTFT

for a complex sequence, 320–322 for a real sequence, 322–331

theorems on DTFT

convolution theorem, 318, 320 correlation theorem, 319, 320 differentiation in frequency, 158, 318 frequency shifting, 315, 318, 320, 342 linearity, 317, 320, 326 Parseval’s theorem, 319, 320, 328, 329 time reversal, 318, 320 time shifting, 320, 324, 335 windowing theorem, 318

Frequency division multiplexing (FDM), 32,

164, 166

Frequency response computation using

MATLAB, 338, 341, 342, 346 Frequency response from poles and zeros, 264–265, 388–389 Frequency response of continuous time

systems, 111, 159–162 Frequency response of discrete-time systems computation using MATLAB, 338, 341,

342, 346

Frequency shifting, 115, 143, 145, 158, 315,

318, 342

G Generation of continuous-time signals using

MATLAB, 28–30 Graphical convolution, 58–70

H Half-wave symmetry, 119–124, 126,

127, 132

High-pass ﬁlter, 231, 257

I Imaginary part of a sequence, 365 Impulse and step responses, 286 Impulse and step responses computation using MATLAB, vii, 92, 225, 304, 305, 312

Impulse response, 49, 53–57, 62, 66–68, 73–75, 77–79, 88, 92–95, 106, 109, 153, 161, 193, 202, 204, 217, 227, 229, 267, 286–288, 294–297, 324, 330, 332

Impulse step responses, 304, 305 Initial Value Theorem, 186–187, 370–371 Input-output relationship, 271, 282, 286–288,

293

Interconnected systems, 74–76 Inverse discrete Fourier transform, 111, 135,

136, 138 Inverse Fourier transform, 317 Inverse Laplace transform

partial fraction expansion, 194–202, 209,

211, 375–379, 416

partial fraction expansion using MATLAB,

201–202

partial fraction expansion with multiple

poles, 195–201

partial fraction expansion with simple

poles, 195, 376 Inverse system, 79–81, 205, 395, 396 Inverse z-transform

Cauchy’s residue theorem, 374–375 modulation theorem, 372 Parseval’s relation, 126, 372–374, 398 partial fraction expansion, 375–379, 414 partial fraction expansion using MATLAB,

379–380

power series expansion, 379–383 power series expansion using MATLAB, 383

L Laplace transform, 117, 144, 151, 155, 172, 174–191, 200, 208, 215, 222, 402, 403

block diagram representation, 218–219 deﬁnition of, 171–225 existence of, 172 inter connection of systems, 218 properties, 171, 178–187

convolution in the frequency domain,

117, 155, 181, 182

convolution in the time domain,

151, 181

differentiation in the s-domain, 180,

184, 190

differentiation in the time domain,

144, 179, 184 division by t, 180 ﬁnal value theorem, 187, 200, 222,

402, 403

422

Index

Laplace transform (cont.)

initial value theorem, 186–187 integration, 184, 187 linearity, 178, 184, 208

properties of even and odd functions,

182–183

shifting in the s-domain, 178, 184,

189–191

time scaling, 179, 184 time shifting, 178, 184, 188

region of convergence, 174–176, 203,

208, 223

ﬁnite duration signal, 174 left sided signal, 175–177 right sided signal, 172, 175–177 strips parallel to the jΩ axis, 174 two sided signal, 176, 177 region of convergence (ROC), 173 relationship to Fourier transform, 172–173 representation of Laplace transform in the

s-plane, 173, 188

system function, 202, 204

block diagram representation, 218–219 interconnection of systems, 218

table of properties, 184 transfer Function, 202–204, 223, 235 unilateral Laplace transform, 171, 172,

183–186, 215

differentiation property, 183–186, 215

Linear constant coefﬁcient difference

Modulation property, 155–158 Modulation theorem, 372 Multiplexing and demultiplexing, 32

N Nonperiodic signals, 111, 133–158 Non-recursive difference equation, 298 Notch ﬁlter, 266–267

O One-sided z-transform, 384, 393, 401, 404–405

properties

shifting theorem, 384, 393, 401 ﬁnal value theorem, 402–404 solution of linear difference equations

with initial conditions, 401, 404–405

P Parseval’s relation, 126, 372–374, 398 Parseval’s theorem, 118, 149, 150, 157, 319,

320, 328 Partial fraction expansion, 378 Partial fraction expansion using MATLAB,

201–202, 379–380 Particular solution, 85, 89, 90, 300–303 Periodic convolution, 76, 107, 117, 119,

313–316, 318

equations, 298, 299 Linear constant-coefﬁcient differential

equations, 82–84, 159, 171, 207–210

Linearity, 41, 43, 44, 50, 86, 105, 113, 119,

Phase and group delays, 333 Poles and zeros, 173, 192, 227, 235, 240,

242, 243, 246, 265, 386–388, 395, 398, 400, 414

139, 140, 151, 158, 160, 178, 184, 193, 208, 217, 282, 285, 311, 315, 317, 320, 326, 360, 364–366

Linearity property of the Laplace transform,

217

Low-pass to band-pass, 257 Low-pass ﬁlter, 164, 203, 225, 227–229, 231–264, 267, 268, 341, 351

LTI discrete-time systems, vii, 271, 286–289, 291–297, 304, 332, 333, 354, 386

LTI systems with and without memory, 77

Pole-zero pairing, 388 Pole-zero placement, 227, 264–267 Power series expansion using MATLAB, 417 Properties of the convolution integral, 50–57, 151 Properties of the convolution sum, 291–295 Properties of the impulse function

sampling property, 21, 52, 67, 136 scaling property, 22 shifting property, 21, 324, 335

Proposition, 7, 18, 278

M Matrix method, 288 Maximum-phase systems, 395 Minimum-phase decomposition, 399–400 Minimum-phase systems, 395, 399 Mixed-phase systems, 395 Modulation, 158 Modulation and demodulation, 31, 170

Q Quantization and coding, 274–276 Quantization error, 275

R Rational transfer function, 386 Rational z-transform, 354, 374 Real part of a sequence, 364

Index

423

Reconstruction of a band-limited signal from

its samples, 350

Recursive difference equation, 298, 336, 338 Region of convergence (ROC), 173–181,

184, 185, 188, 192, 195, 197, 199, 203–208, 210, 222, 354–364, 366, 369–371, 373, 377, 379, 389, 391, 392, 401–404, 413–415

Representation of signals in terms of impulses, 41–42

S Sampling

continuous time signals, vii, 5, 271, 273,

305, 344

discrete time signals, 271–276, 305–307 frequency domain, vii, 306, 344–347 Nyquist rate, 273

Sampling frequency, 271, 273–275, 346,

412, 416 Sampling in frequency-domain aliasing, 273, 347, 410 over-sampling, 347 sampling theorem, 273, 346 under sampling, 346

Sampling period, 271, 273, 306, 346, 409 Sampling theorem, 273–274, 346, 347 Scalar multiplication, 93 S-domain, 178, 180, 184, 189–191 Single-side-band (SSB) AM, 164 Singularity functions, 41, 95 Solution of difference equations characteristic equation, 300 complementary solution, 85, 90, 300 using MATLAB, 91–92 particular solution, 300–303

Solution of linear differential equations

using MATLAB, 216 Stability and causality, 208, 295–297, 311,

389, 391, 396

Stability and causality of LTI systems in terms

of the impulse response, 295–297

Stability for LTI systems, 77–79 Step and impulse responses, 49 Step response, 49, 68, 89, 91–94, 106,

inverse, 79–82, 205, 395, 396 maximum-phase, 395 minimum-phase, 395 mixed-phase, 395, 399 stable, 49, 78, 204–210, 218, 219, 223,

285, 296, 297, 311, 324, 325, 389, 391–394, 397, 415, 416

T Theorems on DTFT, 317–320, 329 Time-invariant, 41, 43–48, 50, 77–82, 88,

283, 286–289, 291–297, 395

Time reversal, 3–5, 115, 119, 158, 318, 320,

360

Time scaling, 2–3, 115, 119, 179, 184 Time shifting, 2, 17, 18, 113, 119, 120, 142, 158, 168, 178, 184, 188, 315, 318, 324, 335 Transfer function, 202–204, 223, 227,

234–241, 243–246, 248, 252, 254, 256–259, 261, 263, 267, 385–386, 388, 389, 394, 395, 397–399, 409, 411, 413, 415

Transformation, 31, 42, 111, 227, 252–255,

257, 260, 263, 281, 285, 286, 325, 408–413, 416

U Under sampling, 346 Unilateral Laplace transform

differentiation property, 183–186, 208

Unit doublet, 95, 97 Unit impulse response, 95 Unit ramp, 22 Unit sample sequence, 286, 365 Unit step response, 68 Unit step sequence, 286 Unstable system, 78, 391, 392 Up-sampling, 312

W Windowing theorem, 318, 320

107, 110, 286, 302, 304, 305, 311, 404

Systems

all-pass, 397–400, 415 causal, 48, 77, 78, 86, 107, 172, 204–210, 219, 223, 284, 288, 294–298, 324, 391–394, 397, 415, 416

Z Zero locations, 206, 264, 266 Zero-order hold, 272–274 Z-transform

deﬁnition of, 353, 359, 360, 381 properties of, 360–366

424

Index

Z-transform (cont.)

region of convergence (ROC), 354–359, 361–366, 368–371, 374, 377, 379, 413, 414

Z-transform properties

conjugate of a complex sequence, 364 convolution of two sequences, 362 correlation of two sequences, 363 differentiation in the z-domain, 361

imaginary part of a sequence, 365 linearity, 360, 364–366 real part of a sequence, 364 scaling in the z-domain, 361 time reversal, 360, 363, 366 time shifting, 361, 366

Z-transforms of commonly-used sequences

unit step sequence, 366

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