Signals and Systems 9788177583809, 9788131799758, 8177583808

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Signals and Systems

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Signals and Systems

Smarajit Ghosh Professor Department of Electrical and Electronics Sikkim Manipal Institute of Technology Majitar, Rangpo, East Sikkim

PEARSON

Copyright © 2006 Dorling Kindersley (India) Pvt. Ltd. Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material present in this eBook at any time. ISBN 9788177583809 eISBN 9788131799758 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

rro 11tlj aaugliter Saraaa

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Preface

Signals and Systems has been written for undergraduate engineering courses, in India as well as abroad. This course is common to electrical engineering, electrical and electronics engineering, electronics and communication engineering, and computer science. Recently, this course has gained tremendous importance due to the increasing practical use of signals analysis in digital signal processing (DSP), where the study of signals is a prerequisite. Even where analog methods are still used, like in telecommunication, medical and defence applications, the knowledge of the frequency-domain description of analog signals is mandatory. In addition, the study of systems is required for computer simulation, filter implementation, and the assessment of system responses for analysis and synthesis. The chapters in this book have been organized to present the subject matter in a structured and simple manner. An introductory chapter on mathematical preliminaries has been included for beginners. An appendix on the basics of MATLAB in signals has also been provided: Pedagogical features, such as numerous solved problems, short questions and answers, and multiple-choice questions, are included for the benefit of students. All feedback and suggestions regarding this book are welcome. SMARAJIT GHOSH

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Acknovvledgements

It

is my great pleasure to thank Prof. A. K. Sarkar (Dean 10) and Prof. Ved Prakash of BITS, Pilani, who have encouraged me to write this book. I am deeply indebted to Prof. H.O. Gupta, Head of Electrical Engineering of lIT, Roorkee, for his valuable comments and suggestions. I am thankful to Prof. P. P. Sengupta and Prof. A. K. Dey of NIT, Durgapur; Prof. A. K. Mukhopadhaya, Vice Chancellor of Tripura University; Prof. A. K. Deb, Prof. G. Sarkar and S. Sengupta from the Department of Applied Physics, University College of Science and Technology, Calcutta University, for their support. I also wish to record my gratitude to the editorial team of Pearson Education for their valuable contribution in bringing out this book. Finally, I wish to express my sincere gratitude to my mother, Smt. Beena Ghosh, and my wife, Dr. (Mrs.) Kuntal Ghosh, for their patience and constant motivation during the preparation of this book. SMARAJIT GHOSH

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Contents

Preface

vii

Acknowledgements

ix

MATHEMATICAL PRELIMINARIES ...................................................................................... 1 Matrices Equality of Two Matrices Vector Square Matrix Diagonal Matrix Identity Matrix or Unity Matrix Zero Matrix Determinant of a Matrix Singular Matrix Transpose Symmetric Matrix Conjugate Matrix Conjugate Transpose Hermitian Matrix Addition of Matrices Multiplication of a Matrix by a Scalar Multiplication of a Matrix by a Matrix Power of a Matrix Rank of a Matrix Minor of a Matrix Co-factor Adjoint Matrix Inverse of a Matrix More Properties of Matrices Trigonometry Formulae Calculus

I

2 2 2

2 2 2

3 3 3 3 3

3 3 4 4 4 4 4 4 5 5

6

xii

Contents

1. FUNDAMENTALS OF SIGNALS AND SYSTEMS ............................................................... 7 1.1 1.2 1.3

Signals and Systems Classification of Signals Continuous Time Signals 1.3.1 Classification of Continuous Time Signals 1.4 Basic Continuous Time Signals 1.4.1 Unit Step Function 1.4.2 Ramp Function 1.4.3 Unit Impulse Function 1.4.4 Complex Exponential Signals 1.4.5 General Complex Exponential Function 1.4.6 Real Exponential Function 1.4.7 Sinusoidal Signal 1.5 Classification of Continuous Time Systems 1.5.1 Static and Dynamic Systems 1.5.2 Causal and Non-Causal Systems 1.5.3 TIme Invariant and Time Variant Systems 1.5.4 Linear and Non-Linear Systems 1.5.5 Linear Time Invariant System 1.5.6 Stable and Unstable Systems 1.5.7 Invertible and Non-Invertible Systems 1.5.8 Feedback System 1.6 Discrete Time Signals 1.7 Concept of Frequency in Discrete Time Signals 1.8 Standard Discrete Time Signals 1.8. I Unit Sample Sequence 1.8.2 Unit Step Sequence 1.8.3 Unit Ramp Sequence 1.8.4 Exponential Sequence 1.9 Classification of Discrete Time Signals 1.9. I Even and Odd Signals 1.9.2 Periodic Signals and Non-Periodic Signals 1.9.3 Deterministic and Random Signals 1.9.4 Energy Signals and Power Signals 1.9.5 Multichannel and Multidimensional Signals 1.10 Discrete Time Systems 1.1 I Representation of Discrete Time Systems l.l1.1 Adder 1.11.2 Constant Multiplier 1.l1.3 Signal Multiplier 1. 11.4 Unit Delay Block 1.11.5 Unit Advance Block 1.12 Classifications of Discrete Time Systems I. 12. I Static and Dynamic Systems 1.12.2 Causal and Non-Causal Systems 1.12.3 Time Invariant and Time Variant Systems

7 7 7 9 13 13 14 14 15 15 16 16 17 17 17 18 18 18 19 19 19 19 24 25 25 25 26 26 28 28 30 31 31 32 32 33 33 33 . 33 33 33 34 34 34 35

Contents

1.13 1.14 1.15 1.16 1.17

1.11.4 Linear and Non-Linear Systems 1.12.5 Stable and Unstable Systems Nyquist Rate Sampling Theorem Aliasing Convolution Correlation 1.17.1 Cross-correlation and Auto-correlation Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

xiii 35 37 40 40 40 41 41 41

43 48 48 51 52 54

2. FOURIER SERIES ............................................................................................................... 55 2.1 2.2 2.3 2.4

2.5

Fourier Series Dirichlet Conditions Determination of Fourier Co-efficients Wave Symmetry 2.4.1 Even or Mirror Symmetry 2.4.2 Odd or Rotation Symmetry 2.4.3 Half Wave Symmetry 2.4.4 Quarter Wave Symmetry Exponential Form of Fourier Series Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

55 56 56 61 62 63 64 64 66

69 76 77 77

78 79

3. FOURIER TRANSFORM ...................................................................................................... 80 3.1 3.2 3.3

Fourier Transform Condition for the Existence of Fourier Integral Fourier Transform of Some Functions 3.3 .1 Fourier Transform of Gate Function 3.3.2 Fourier Transform of Impulse Function 3.3.3 Fourier Transform of Shifted Impulse Function 3.3.4 Fourier Transform of One-Sided Exponential Function 3.4.5 Fourier Transform of Two-Sided Exponential Function 3.3.6 Fourier Transform of sgn(t) e-a(t) 3.3.7 Fourier Transform of Signum Function 3.3.8 Fourier Transform off (t) = 1 3.3.9 Fourier Transform of u(t)

80 82 82 82 84 85 85 86 87 88 88 89

xiv 3.4

3.5 3.6

3.7 3.8 3.9

Contents Fourier Transfonnation Theorem 3.4.1 Linearity 3.4.2 Time Scaling 3.4.3 Time Differentiation 3.4.4 Time Shifting Property 3.4.5 Translation in the Frequency Domain 3.4.6 Modulation Theorem 3.4.7 Symmetry or Duality Property 3.4.8 Time Convolution Property 3.4.9 Frequency Convolution 3.4.10 Frequency Differentiation 3.4.11 Time Integration 3.4.12 Fourier Transfonn of f (-t) 3.4.13 Symmetry Properties of Fourier Transfonn Fourier Transfonn of Periodic Signals 3.5.1 Fourier Transform in Tenns of Fourier Series Co-efficients Energy Density and Power Spectral Density 3.6.1 Energy in a Signal 3.6.2 Power in a Signal 3.6.3 Energy Spectrum and Energy Spectral Density 3.6.4 Parseval's Relation for Energy Signals 3.6.5 Power Spectral Density 3.6.7 Power Spectrum for Periodic Signal 3.6.8 Parseval's Relation for Periodic Signals Nyquist Theorem 3.7.1 Proof of Nyquist Theorem 3.7.2 Practical Implementation System Analysis Using Fourier Transfonn Relation between Differential Equation and System Function Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

89 89 90 90 91 91 92 92 93 94 94 95 96 96 98 98 99 99 99 100 101 102 102 103 104 104 105 105 106 106 109 110

111 111 114

4. LAPLACE TRANSFORM ................................................................................................... 115

4.1 4.2 4.3 4.4

Laplace Transfonn Region of Convergence (ROC) Inverse Laplace Transformation Basic Properties of Laplace Transforms

115 116 116 116

4.5

Laplace Transfonn of a Derivative [ d:~t) ]

119

4.6

Laplace Transfonn of an Integral

J

f(t)dt

120

Contents 4.7

4.8

4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23

Laplace Transform of Some Common Time Function 4.7.1 Unit Step Function 4.7.2 Impulse Function 4.7.3 Ramp Function 4.7.4 Parabolic Function 4.7.5 f{t) =eatu(t) with a > 0 4.7.6 f(t) = e-atu(t) 4.7.7 Sinusoidal Function 4.7.8 Cosine Function 4.7.9 Hyperbolic Sine and Cosine Functions 4.7.10 Damped Sine and Cosine Functions 4.7.11 Damped Hyperbolic Sine and Cosine Function 4.7.12 Laplace Transform of (' Laplace Transform of Two-Sided Functions (BLT) 4.8.1 ROC forf(t) =ebtu(_t) 4.8.2 ROC forf(t) =eatu(t) 4.8.3 ROC for Two-Sided Function Initial Value Theorem Final Value Theorem Partial Fraction Expansions Relation between Step Response and Impulse Response Application of Laplace Transforms in Circuit Pure Resistive Element Pure Inductive Element Pure Capacitive Element Step Response of Series R-L Circuit Step Response of Series R-C Circuit Step Response of Series R-L-C Circuit Impulse Response of Series R-L Circuit Impulse Response of Series R-C Circuit Pulse Response of Series R-L Circuit Pulse Response of Series R-C Circuit Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

xv 122 122 122 122 123 123 124 124 125 125 126 126 127 131 131 132 133 134 134 135 138 140 140 140 141 142 142 143 145 146 146 147 150 170

171 172 173 177

5. SYSTEM MODELLING ...................................................................................................... 178 5.1 5.2 5.3 5.4 5.5 5.6

Transfer Function Impulse Response and Transfer Function Properties of Transfer Function (TF) Definition of Basic Elements of Block Diagram Basic Definition of Signal Flow Graph (SFG) Mason's Gain Formula

178 179 180 181 182 185

xvi

Contents

5.7

Modelling of Mechanical Systems 5.7.1 Translational Motion 5.7.2 Rotational Motion 5.8 Modelling of Electrical Systems 5.8.1 Resistor 5.8.2 Inductor 5.8.3 Capacitor 5.9 Analogous Systems 5.9.1 Force-Voltage (f- v) Analogy 5.9.2 Force-Current (f-i) Analogy 5.10 Representation by Nodal Method Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

187 187 188 190 190 190 190 190 191 191 192 195

200 201 202 205 209

6. z-TRANSFORM ..................................................................................................................210

6.1 6.2 6.3

z-Transform Region of Convergence (ROC) Properties of z-transform 6.3.1 Linearity 6.3.2 Time Shifting 6.3.3 Scaling in z-domain 6.3.4 Time Reversal 6.3.5 Differentiation in z-domain 6.3.6 Convolution in Time Domain 6.3 .7 Correlation of Two Sequences 6.3.8 Multiplication of Two Sequences 6.3 .9 Conjugate of a Complex Sequence 6.3.10 Real Part of a Sequence 6.3.11 Imaginary Part of a Sequence 6.3.12 Initial Value Theorem 6.3.13 Final Value Theorem 6.3 .14 Partial Sum 6.3.15 ParsevaJ's Theorem 6.4 z-Transform of Right-Sided Exponential Sequences 6.5 z-Transform of Left-Sided Exponential Sequences 6.6 Finite Length Sequence 6.7 z-Transform of Unit Sample Sequence 6.8 z-Transform of Delayed Unit Sample Sequence 6.9 z-Transform of Unit Step Sequence 6.10 z-Transform of Folded Unit Step Sequence 6.11 z-Transform of the Signal x(n) = Nanu(n)

210 210 211 211 212 212 212 213 213 215 215 216 217 217 218 218 219 219 220 222 222 223 223 224 225 225

Contents 6.12 6.13

z- Transfonn of Unit Ramp Sequence z- Transfonn of Causal cosine Sequence

6.14 6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22 6.23

z-Transfonn of Causal sine Sequence z-Transfonn of an cos(nco)u(n) z-Transfonn of an sin (nco) u(n) Inverse z-transfonn Inverse z-transfonn Using Partial Fraction Expansion Inverse z-transfonn Using Power Series Expansion System Function and Pole-Zero Plots from z-transfonn Pole-Zero Plot System Function of the LTI System Causality and Stability in Tenns of z-transfonn Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

xvii 225 226 227 227 228 231 232 234 235 235 236 237 240

248 249 250

253 254

7. CONVOLUTION ................................................................................................................. 255 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Convolution Theorem for Continuous System Properties of Linear Convolution Graphical Convolution Discrete Convolution 7.4.1 Properties of Discrete Convolution Important Properties of Systems Convolution Theorem in z-transfonn Circular or Periodic Convolution Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

255 256 256 261 262 262 264 265

267 270

271 272 273 273

8. STABILITY ......................................................................................................................... 274 8.1 8.2

Effect of Location of Poles on Stability Routh-Hurwitz Criterion 8.2.1 Hurwitz's Criterion 8.2.2 Routh's Stability Criterion Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

274 275 276 277

287 288 288 289 292

xviii

Contents

9. STATE VARIABLE APPROACH (CONTINUOUS SYSTEMS) ........................................... 293

9. 1 9.2 9.3 9.4 9.5

Advantages and Disadvantages of Modem Control Theory Concepts of State, State Variables and State Model State Model Non-Uniqueness of the State Model Different Representations of a State Model 9.5.1 State Space Representation Using Phase Variables in Controllable Conical Form (CCF) 9.5.2 Phase Variable CCF Form for Numerator Terms 9.5.3 Phase Variable OCF Form 9.5.4 Cascade Decomposition 9.5.5 Parallel Decomposition 9.5.6 Jordan's Canonical Form 9.6 Eigen Value 9.7 Transfer Function Derivation from the State Model 9.8 Solution of the State Equation 9.8.1 Solution of Homogeneous State Equation 9.8.2 State Transition Matrix 9.8.3 Properties of STM, I/> (t) 9.8.4 Solution of Non-homogeneous State Equation 9.9 Controllability 9.10 Observability Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

293 293 294 297 298 298 300 302 304 306 309 311 311

312 312 313 313 315 316 317 318

325 326 327 328 330

10. STATE VARIABLE METHODS (DISCRETE CASE) .......................................................... 331

10.1 10.2 10.3 10.4 10.5 10.6 10.7

Delay Elements State Model of a First-Order System State Model of a Second-Order System Non-Uniqueness State Model Canonical Form of State Model Transfer Function from State Model Solution of State Equation 10.7.1 z-Transform Method 10.7.2 Series Expansion Method Additional Solved Examples Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

331 331 332 339 339 341 342 342 343

346 354 355 355 357 357

Contents

xix

11. DISCRETE FOURIER TRANSFORM AND FAST FOURIERTRANSFORM ..................... 358 11 .1 11.2 11.3 11.4 11.5 11.6

11 .7 11.8 11 .9 11.10 11.11 11.12 11.13 11.14

Fourier Transform of Discrete Time Signals Properties of Fourier Transform Inverse Fourier Transform Magnitude/Phase Transfer Functions Using Fourier Transform Discrete Fourier Transform Properties of Discrete Fourier Transform 11.6.1 Periodicity 11.6.2 Linearity 11 .6.3 Circular Symmetries of a Sequence 11 .6.4 Symmetry Properties 11.6.5 Circular Convolution 11 .6.6 Time Reversal of a Sequence 11.6.7 Circular Time Shift Sequence 11.6.8 Circular Frequency Shift of a Sequence 11.6.9 Circular Correlation of Two Sequences 11.6.10 Multiplication of Two Sequences 11.6.11 ParsevaJ's Theorem Relationship Between DFT and z-Transform Fast Fourier Transform (FFT) 11.8.1 Properties of FFT Radix-2 FFT Algorithm Radix-2 DIT-FFT Algorithm Computational Complexity Compared to Direct Computation Memory Requirement and In Place Computations Bit Reversal Radix -2 DIF FFT Algorithm Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

358 359 360 360 361 365 365 366 366 370 372 377 377 377 377 378 378 378 379 380 381 381 388 390 390 391

395 396 397 397 398

12. STRUCTURES AND DESIGN OF DIGITAL FILTERS ....................................................... 399 12.1

12.2 12.3

12.4 12.5 12.6

Classification of Filters 12.1.1 Analog Filter 12.1.2 Digital Filter Review of Analog Filters Specification of the Frequency Response Characteristics of the Filters 12.3.1 Low Pass Filter 12.3.2 High Pass Filter 12.3.3 Band Pass Filter 12.3.4 Band Stop Filter Specifications of Phase Response Structure of Digital Filter Describing Equation

399 399 399 400 404 405 405 406 407 407 408 408

xx

Contents 12.7 Structure of FIR Filter 12.7.1 Direct Form Structure of FIR Filter 12.7.2 Cascade Form Structure of FIR Filter 12.7.3 Frequency Sampling Structure for FIR Filter 12.7.4 Lattice Structure for FIR Filter 12.8 Structure for IIR Filter 12.8.1 Direct Form Structure for IIR Filter 12.8.2 Cascade Structure for IIR Filter 12.8.3 Parallel Form Structure for IIR Filter 12.8.4 Lattice Structure of IIR Filter 12.9 Realization Procedure for Digital Filter 12.9.1 Recursive Realization 12.9.2 Non-Recursive Realization 12.9.3 FFT Realization 12.10 Notch Filter 12.11 Comb Filter 12.12 All-pass Filter 12.13 Design of an IIR Filter 12.13.1 Impulse Response Invariance Method 12.13.2 Bilinear Transformation Method 12.13.3 Digital Butterworth Filter 12. 13.4 Digital Chebyshev Filter 12. 14 Design of FIR Filters 12.14.1 Fourier Series Method for Design of FIR Filters Significant Points Short Questions and Answers Exercises Multiple Choice Questions Answers

409 410 411 412 414 415 416 421 423 424 426 426 427 427 427 429 432 432 433 433 439 443 447 447

453 453 455 455 456

APPENDIX ................................................................................................................................. 457 INDEX ........................................................................................................................................ 461

Preliminaries

Mathematical

MATRICES

I

A rectangular array of elements that may be real numbers, complex numbers, functions or operators is called a matrix. Let us consider a matrix A as follows:

a l2 a'n a22 ···· · ···· a2n .......... .. ... .. ... .......... a m 2 ······· ··

~mm

mxn

The order of the matrix is m x n since it has n rows and n columns. The matrix A can also be written as A

= [a;.] ~ mxn

Equality of Two Matrices Two matrices are called equal if and only if their corresponding elements are equal. Therefore, equal matrices should have equal number of rows and columns. Vector A matrix having only one column is known as a column vector. If the column vector has known as n vector or n dimensional vector. The column vector is given below:

II

elements, it

a, a2

B= an nx'

If the matrix has only one row, it is known as row vector such as a, a2 ..... an 1 x n Square Matrix If the number of rows is equal to the number of columns of any matrix, the matrix is called a square matrix.

2

Signals and Systems

Diagonal Matrix If the other elements except the main diagonal elements in a square matrix is zero, the matrix is called a diagonal matrix.

Oij = 1 if i = j

where

=0 if i 7= j and Oij is known as Kroneker delta.

Identity Matrix or Unity Matrix If the main diagonal elements of matrix are unity and other elements are zero, the matrix is known as an identity or unity matrix .

Zero Matrix A matrix whose elements are all zero is known as zero matrix.

Determinant of a Matrix There exists a determinant for every matrix if the matrix is square. The properties of determinant are given below:

I. The determinant changes its sign if any two constructive rows or columns are interchanged. 2. The value of the determinant is zero if any row or column consists of only zero elements. 3. The value of the determinant is also zero if the elements of any row or any column are exactly k times of another row or column. 4 . If any determinant is multiplied by a constant, then only one row or only one column is multiplied by that constant. 5. If A and B are two square matrices and both are n x n, the following relation holds good.

IA x BI

=IAIBI

Singular Matrix A square matrix is known as singular if the associated determinant is zero.

Transpose If the rows and columns of an m x n matrix are interchanged, the resulting matrix is known as the transpose of the matrix A. The transpose of A is denoted by AT. al2

If

A

n

= fa" afl

a22

al a2n

ami

a m2

a"l11

1 nxm

then

Mathematical Preliminaries all

AT

= [ a: 2

alii 2I alii

a lll

amn

3

1 nx m

Again

Symmetric Matrix If a square matrix A is equal to its transpose, then the matrix A is known as a symmetric matrix. For symmetric matrix A = AT. Again , if A = _AT, the matrix A is termed as a skew matrix.

Conjugate Matrix If the elements of a matrix (A) are complex numbers, the elements of the conjugate of matrix (A) are the conjugate of the elements of the matrix (A).

Conjugate Transpose The conjugate of the transpose of a matrix A is known as the conjugate transpose.

Hermitian Matrix If the elements of a matrix (A) are complex numbers and A of aij' A is called a Hermitian Matrix. If A

=A * or a ii = i(j where iiij are complex conjugate

=-A *, A is known as skew Hermitian Matrix.

Addition of Matrices

Multiplication of a Matrix by a Scalar If a matrix A = [aij]m x 11 is multiplied by a scalar k, we can write kA

=k [aij]1I/ x

11

Multiplication of a Matrix by a Matrix If

A

= [aij],n x ll and B = [bij]

k

111 X 1/'

then AB

=C = 'Lai/blj 1=1

where i

= 1,2, .. .. ., m and} = 1,2, .. ..... , n.

For multiplication, the following properties hold good: I. (AB)C = A(Be) 2. (A+B)C 3. C(A+B)

= AC + BC = CA + CB

If AB = BA, then A and B are known as commutative. Generally, AB may not be equal to BA.

4

Signals and Systems

Power of a Matrix The n th power of a square matrix A is defined by An

= A.A-A.A-A .. ... .... ..

Rank of a Matrix A matrix (A) has a rank m if there exists a m x m submatrix B of A such that its determinant is non zero and the determinant of every k x k submatrix where k ;0: m + 1 of A is zero.

Minor of a Matrix If the ith row and /h column are deleted from an n x n matrix, then there results a matrix of order (n - 1) x (n - I) and the determinant of this matrix is called the minor (Mu) of the matrix A.

Co-factor The co-factor Au of the element aU of the matrix is defined by

Adjoint Matrix Adjoint of matrix A is defined as a matrix where the element of the i 1h row and the/ h column equals to Ai;'

:::::. :,~~. 1 ...... ...... ... ...

.

Ann n X n

Inverse of a Matrix For a square matrix A, a matrix B exists if and only if AB = BA inverse of A.

=I and B is denoted by A- I and it is called

Therefore A A- I = A-I A = I where I is the identity matrix. If A is non-singular and AB

=C, then we can write B =A- Ie.

Again, (AB)- I = B- 1 A-I and (AB) (A- I B- 1) = I. The following relations hold good: (A- Ir l =A (A-1)T

=(ATrl

(A-I)'

= (A' rl

By definition, the inverse of the matrix [A]mxn where m A-I =provided IAI :;:. 0

=n is given by

Mathematical Preliminaries More Properties of Matrices (A + B)T = AT + BT (AB)T =BT AT (A + B)* = A* + B* (AB) = B*A*

TRIGONOMETRY FORMULAE

=sin (A) cos (B) +sin (B) cos (A) cos (A+B) =cos (A) cos (B) -sin (A) sin (B)

• sin (A+B) •

=[tan (A) + tan (B)]/[I-tan (A) tan (B)] sin (A-B) =sin (A) cos (B) -sin (B) cos (A)

• tan (A+B) •

• cos (A-B) = cos (A) cos (B) +sin (A) sin (B) • tan (A-B)

=[tan (A) -tan (B)]/[I+tan (A) tan (B)]

• sin (2A) = 2 sin (A) cos (A) • cos (2A)

=cos2 (A) -sin2 (A)

• tan (2A) =2 tan (A)I[I-tan 2 (A)]

• I. cos (2A)

= [cos2 (A) -sin2(A)] + [cos2 (A) + sin2(A)] =2cos 2(A)

• l--cos (2A) = [cos2 (A) -sin 2 (A)] - [cos 2 (A) + sin 2 (A)] = 2sin2 (A) • cos (A) = ± -.,/[(1 +cos (2A»I2]

=± -.,/[(l-cos (2A»/2] cos (AI2) =± v'[(cos (A) +1)/2]

• sin (A) •

• sin (AI2)

= ± -"/[(cos (A) -1)12]

=± v'[(cos (A) -1)/(cos (A) +1)] ·tan (AI2) =± Isin (A)I/(cos (A) +1) =± sin (A)/(cos (A) +1)

• tan (AI2) •

• tan (AI2) = ± (cos (A) -1 )/lsin (A)I = ± (cos (A )-1)/sin (A)

=2sin (A) cos (B) sin (Y) =2sin (B) cos (A)

• sin (X) + sin (Y) • sin (X) -

• cos (X) + cos (Y)

=2cos (A) cos (B)

=2sin (A) sin (B) tan (X) + tan (Y) =2(tan (A) [1+ tan 2 (B)])/[I-tan 2 (A) tan 2(B)]

• cos (X) -cos (Y) •

• tan (X) -tan (Y)

= 2(tan (B) [1+ tan 2 (A)])/[I-tan 2 (A) tan 2 (B)]

5

6

Signals and Systems

CALCULUS R.3.2 Integration

R3.1 Derivative

-.!L dx

(C)

=0 where C is a contant.

feUxdx =:..-a ax

d (e ax) =ae ax -dx

--d (x) a =a x 1na dx d{sin (ax)} () -----'----'---'--'- = a cos ax dx

d{cos(ax)} dx

---'---'-~

=- a SIn. (ax )

~!!~J~~2J = a sec 2 (ax) dx

d {sin- 1(ax)}

dx d{cos-1(ax)} - - - -dx d{tan - 1(ax)} dx

a

=

a --:01+(ax)2

=- -

=

du dy dy dx

=-.

~

SIn

f ~ 1- (ax)2 1 -dx=-lsin-'(ax) a f

~

-1

1 .I dx= -- cos- (ax) 1-(ax)2 a

1 + (ax)2

d(~) v du _u dv v _ dx dx Tx- v 2 ~

cos(ax) -f . (ax)dx =- a sin (ax) fcos ()d ax x = - - a ftan (ax) dx =- a Inlcos (ax)1

a

d(uv) dv du - - =u - +v dx dx dx

du dx

Jxndx=~ n+l f Cdx= Cx n+l

!L(xn)=nx n- 1 dx

f udv = uv - Jvdu

1 Fundamentals of Signals and Systems

In almost all of the electrical engineering fields, the concept and theory of signals and systems has tremendous

usage. The scientific disciplines and many other engineering fields also use signals and systems. Research work is going on in this area. The aim of this chapter is to introduce the basics of signals and systems.

1.1

SIGNALS AND SYSTEMS

If a function represents a physical quantity or variable containing the information about the behaviour and nature of the phenomenon, it is known as a signal. Since a function is mathematically represented as a function of an independent variable, it is denoted by x(t). In a R-L circuit, the current through the resistor or voltage across the inductor is called a signal. On the contrary, a device or a set of rules defining the functional relation between the input and output is known as a system.

1.2 CLASSIFICATION OF SIGNALS There are two main broad classifications of signals. One is continuous time signal and the other is discrete time signal.

1.3 CONTINUOUS TIME SIGNALS If a signal is defined at all values of t where t is a continuous variable, this is known as a continuous time signal. Fig. 1.1 is known as continuous time signal. fit)

fit)

1.0

,

(a) Exponential signal

(b) Sinusoidal Signal

Fig. 1.1

Continuous signal

,

8

Signals and Systems

The signal shown in Fig. 1.1 is continuous in time as well as in amplitude. Fig. 1.2 given below shows a signal which is continuous in time but discrete in amplitude. Discrete amplitude steps

3.0 - - - - - - - - - - - - - - - r-----"'-,

2.0 -------.,...,_ - - . J

1.0

f----i

---r----~--------------------.----------~.

t

Time

Fig. 1.2 Continuous in time but discrete in amplitude

Example 1.1 Fig. E 1.1 shows a continuoustime signal. Sketch and level of the following signals: (a) x(t-3)

(b) x( ~)

Solution

(c) x(3t) and x(-t)

2

/1

4

2

(±)

4

x(t)

(a) x(t-3)

(b) x

A(t)

3

5

4

3 Fig. E1.1

• t

6

Fig. E1.1 (a) x(t)

4

2

3

4

5

6

7

Fig. E1.1(b)

8

9

10

11

12

4

5

6

Fundamentals of Signals and Systems (c).\.(3t)

(d)

X

(-t)

x(t)

x(-t)

-4

4

-+---'---'---- - ~

2

3

3

Fig. E1.1(c)

1.3.1

9

2

2

3

Fig. E1.1(d)

Classification of Continuous Time Signals

The continuous time signals are classified as follows: (i) Analog signal, (ii) One-sided signal, (iii) Real and complex signal, (iv) Even signal and odd signal, (v) Periodic and non-periodic signal, (vi) Piecewise continuous signal, (vii) Deterministic and random signal, (viii) Multichannel signals, (ix) Multidimensional signals, (x) Energy and power signals. These signals are discussed below. 1.3.1.1 Analog signals A continuous time signal x (t) is said to be an analog signal if it takes any value in the interval (a ,b) where a and b may be - 0 0 and 00 respectively. 1.3.1.2 One-sided signals If a continuous time signal X(/) has zero value for t < 0, the signal is said to be one-sided (positive sided) signal. On the other hand, if the signal x(t) has zero value for t > 0, it is also known as one-sided (negative sided) signal. A continuous time signal x(t) is said to be perpetual if it is represented by the same equal value for all time. The signal x: (t) =A sin %t u (I) is one sided whereas x 2 (t) =A sin %t is perpetual shown in Fig. 1.3 (a) and Fig. 1.3 (b) respectively.

X1(t)

,, ,, ,

, , '-

(a) ,11

,- ' '

,, ,, ,

=A sin %tu(t)

(b) }.2 = A sin wat Fig. 1.3

One·sided and perpetual signal

1.3.1.3 Real and complex signals A continuous time signal x(t) is said to be real, if its vaiue is a real number. On the other hand, it is said to be complex if its value is a complex number. A complex signal x(t) is represented by ( I. I ) x(t) =Xl (1) + jx 2 (t) where j = H In Equatiu,. ( !.1) , x l (t) and x2(t) are continuous real signals.

10

Signals and Systems

1.3.1.4 Even and odd signals

A signal x (t) is said to be even if x(t)

=x(-t) is satisfied.

(1.2)

On the other hand, a signal x(t) is said to be odd if x(t) = -x(-t) is satisfied.

( 1.3)

Figure 1.4(a) and Fig. 1.4(b) show even and odd signals respectively. x(t)

x(t)

(a) Even signal

(b) Odd signal

Fig. 1.4

Even and odd signals

1.3.1-.5 Periodic and non-periodic signals A continuous time signal is said to be periodic with period T" for which the signal is advanced in time and hence it remains unchanged. The signal x(t) must satisfy the following condition to be periodic in nature. x(t)

=x(t + Tp) for -

00

to

(1.4)

00

The examples of periodic signals are sine and cosine signals having period 2nl%. If a continuous time signal x (t) does not satisfy Eq. (1.4), the signal is termed as non-periodic or aperiodic signal.

Example 1.2 If Yl(t) and Y2(t) are two periodic signals with two periods T] and Tz respectively, find the suitable condition so that y(t) y](t) + Yz(t) will be periodic. Determine its fundamental period if so.

=

olution

Here the given signal Y] (t) and Y2 (t) are periodic with periods T] and Tz respectively. Therefore, we can write

=Y] (t + 1) =Y] (t + pT1) yz(t) =h(t + 1) =yz(t + qTz) yet) =y] (t) + yz (t) yet) =y] (t + pT]) + Y2(t + qT2 )

YI (t)

and Since,

where p is a positive integer where q is a positive integer we can write yet) as (l)

Now yet) will be periodic if and only if yet

+ 1) = y] (t + pT]) + Y2(t +qT2)

(2)

Equation (I) and (2) is valid if and only if pT]

i.e.,

=qT2 =T

-T] = -q = rational number Tz

p

(3)

Therefore, we can conclude that the sum of two periodic signals will be periodic if the ratio of their respective period is a rational number only. The fundamental period is the least common mUltiple 0f T] lind T2 .

Fundamentals of Signals and Systems

11

1.3.1.6 Piecewise continuous signals If a signal x(t) does not have any definite values at certain points for certain values of t but has definite value at remaining points, the signal is said to be a piecewise continuous signal. Figure 1.5 shows piecewise continuous signals.

r

(a)

Fig. 1.5

,, ,, ,

(b)

Piecewise continuous signals

1.3.1.7 Deterministic and random signals The signals which can be described uniquely by a mathematical expression, table, graph or a well defined rule are known as deterministic signals. Therefore, it is possible to model a deterministic signal by a known function of time t. On the other hand, if the signals can 'lot be described by a formula or graph, they are known as random signals. These signals take random values at a given time. They should be characterized statistically. The sound signal in a radio, data signals in a computer and picture signal in TV are treated as random signals. Noise signals are also examples of random signals. 1.3.1.8 Multichannel Signals If the different signals are recorded from the same source, they are known as multichannel signals. ECG signals recorded from 3 leads or 12 leads for the same person results in 3 channel or 12 channel signal. 1.3.1.9 Multidimensional signals Generally, most of the signals are function of time only, i.e., they are function of single variable. The brightness of a picture during scanning is a function of x and y co-ordinates. Therefore, the picture can be described by a function f(x,y) depending on two variables. The intensity of a TV signal also varies from frame to frame. Therefore, it becomes a function of f(x ,y,t). There may be many signals at input and output of a real system. Each of these is termed as a channel. Three signals in a colour TV picture tube comes out from RED, BLUE and GREEN channels. Therefore, total signal of colour TV can be written as IR (X, y 't)]

I(x,y,t)

= [ IB (x,y,t)

( 1.5)

10 (x,y,t)

Therefore, Eq. (1.5) represents a three dimensional signal. 1.3.1.10 Energy and power signals If a voltage vet) is applied across a resistor R and produces a current i(t) through it. the instantaneous power per ohm is given by

p (t)

= vet )i(t) =i\t) R

(1.6)

We can write the total energy E and average power P on a per ohm basis as follows: ~

E=

fi 2 (t)dt joules

( 1.7)

12

Signals and Systems T

1 2

P = lim - fi 2 (t)dt watts

and

T-7 ~

T

(1.8)

T

2

The normalized energy content E of an arbitrary continuous-time signal x(t) is defined as ( 1.9) We can define the normalized average power P of x(t) as T

f

1 2

P = lim T-7 ~

T

( 1.10)

IX (t)1 2 dt

T

2

=

A signal x(t) is said to be an energy signal if 0 < E < 00 and P O. The signal x (t) is said to be power signal if 0 < P < 00 and E = 00. If the signal Ix(t) I does not satisfy any of the above conditions, the signal is neither a energy signal nor a power signal. All short time or transient signals are energy signals whereas periodic signals cos mot and sin mot etc. are power signals. I

I

The signal s t -"2, (t2 +a 2 ) -"2 are neither energy nor power signals.

Example 1.3 Find whether the following signals are energy or power signals or not.

=A sin(mot + 1/» =e-btu(t) and y(t) = tnu(t) , n > O.

(a) y(t) (b) y(t) (c)

olution (a) The signal yet) = Asin(mot + 1/» is a periodic signal with period To = 2n . The average power of yet)

mo is given by

I

T,

T,

0

0

Pay = ~. f[y(t)f dt=(J)Jl· fA

To

2n

2 sin 2 {m

A2

T, 0

I

A2

ot + I/»dt= ~ f - [I-cos2{mot + 1/»] dt=- 0)

(b) Decreasing (0"< 0)

Fig.1.11

Real exponential function

1.4.7 Sinusoidal Signal The expression for a continuous time sinusoidal function is given by x(t)

=A COS(OJl + 4»

0.26)

where A is the amplitude having real values, OJ is the angular frequency in radians per second and 4> is the phase angle. The signal x(t) = A cos(OJt+ 4» is periodic with period T" = 2n where T" is known as the OJ

fundamental period. The reciprocal of fundamental period (T,,) is known as fundamental frequency (f) having unit Hz. Therefore, Again,

f=

-~ T

OJ= 2nf

(1.27)

( 1.28)

Fundamentals of Signals and Systems where OJ is known as the fundamental angular frequency. Figure 1.12 shows the signal xU). Again , x(t) =A cos(OJt + 1jJ) = Re {Aej(WI+--.---.--.--,---.--.-~~-L--~--L-~--~--L-~~~--o-~~~r-~n

-1 1

2

3

-------- -2 ------------- -3

------------------ -4 )----------------------- -5 ---------------- -- ----- -- --- -6

Fig. 1.23

Odd signal

4

5

6

7

8

9

Sample

30 1.9.2

Signals and Systems Periodic Signals and Non-Periodic Signals

A discrete signal x(n) is said to be periodic if x(n + N)

=x(n) for all n

( 1.57)

were' N' is the period of the signal. The signal x(n) is said to be non-periodic if it does not satisfy the relation of Eq. (1.57). A discrete time sinusoid is periodic for all values of 'n' only if its frequency is rational i.e., f= m N where N is the fundamental period and 'm' is some integer.

Example 1.7

If nof2n is a rational number then the sequence x(n)

( 1.58)

=e jQoll will be periodic (no ;eO).

x(n) will be periodic if the following relation holds good:

=e jQon i.e., e jQon e jQoN =e jQon i.e., e jQoN = I Equation (1) holds good for no =O. Here no ;e O. Therefore, Eq. (1) will hold good if ejQo(I1+N)

(1)

noN = m2n where m is a positive number.

no = m =rational number

i.e.,

2n

Hence we can conclude that if

Example 1.8

no

2n

N

= is a rational number then x(n) = e jQon will be periodic.

Determine whether the following signals are periodic or not: (i) cos( 0.1 nn), (ii) cos( 7nn), (iii) sin(5n) and (iv)

cosC~ )cos ( ~~).

We know that a discrete time signal will be periodic if its frequency is represented by the ratio of two integers i.e.,

f

= m where m and N both are integers. N

(i) cos( 0.1 nn) Here

OJ

= 0.1 n

= OJ =O.ln =~ = ~

cycles per sample. Here m = 1 and N = 20. Therefore,! is expressed 2n 2n 20 N as the ratio of two integers. Here period is 20 samples. The given sequence is periodic. We have f

(ii) cos( 7nn): Here

OJ

=7n

Fundamentals of Signals and Systems

31

w 7n 7 m . We have! = - =- =- = - cycles per sample. Here m = 7 and N = 2. Therefore,f IS expressed 2n 2n 2 N as the ratio of two integers. Here period is 2 samples. The given sequence is periodic. (iii) cos( 5n): w=5

Here

~=~= m

cycles per sample. Here m = 5 and N = 2n. Therefore,fis not the ratio of 2n 2n N two integers. The given sequence is non-periodic. We have!=

(iv)

cosCno)cosC~) Let the sequence be as cos w(cos llh.

n

I

Here

w( = -

We have

! _ WI

and llh = 10 10

1-

_ 10 _ I _ m( 2n - 2n - 20n - N(

Here N( is not an integer. Therefore, cos( I~) is not periodic.

n We have

!2

= w2 =10 =J.-= m2 2n

2n

20

N2

Here!2 is the ratio of two integers and hence cos( ~~) is periodic. Therefore, the given sequence cos (I~ )cos ( ~~ ) is non-periodic because it is the product of a periodic and a non-periodic sequence.

1.9.3 Deterministic and Random Signals The signals which can be described uniquely by a mathematical expression, table, graph or a well defined rule are known as deterministic signals. On the other hand, if the signals cannot be described by a formula or graph, they are known as random signals.

1.9.4 Energy signals and Power signals The energy of discrete time signal x(n) can be represented by (1.59) n=-oo

If the energy of the signal x(n) is finite i.e., O 2fm' The required conditions for recovery of the analog signal f(t) from the sampled signal/(r) are given below: IF(w)1

• The analog signal should be band limited to some maximum frequency f m. • The sampling rate (fs) must be greater than 2fm' • We will be able to get hold of an ideal low pass filter.

f---

Fig. 3.26 shows a typical band limited signal.

3.7.1

Fig.3.26

Typical band limited signal

Proof of Nyquist Theorem

The Fourier transform of impulse sampled signal is given by

l

00

(t) = f(t)· 0PT(t) = f(t) L 0 (t - nT) n=-oo 00

let) = Lf(nt)o(t-nT) n=-oo

Fourier transform of I (t) can be obtained using frequency convolution theorem F[f*U)] = F[f(t)]* F[Op,(t)] where F[f(t)] = F(w) 27r We can prove that 00

F[OPT(t)]

= LWoO(W -

nwo)

n=-oo 00

F(w)* LWoO(W - nw o ) Therefore,

F [f*(t)]

=

n;-oo

27r

=~

f

F(w - nw o )

(3.64)

n=-oo

= -27r

. I' f . IS samp 109 requency 10 rad/sec. T From Eq. (3.64), we conclude that the Fourier transform of the sampled signal consists of several versions of F(w)which are displaced from each other by COo. h were Wo

F[f*(t)]

I = -[ .. . + F(w+ 2wo) + F(w+ COo) + F(w) + F(w- wo) + F(w- 2COo) + ... ] T

(3.65)

Fig. 3.27 shows the Fourier transformfsampled signal. From Fig. 1.27 we can recover F(w) if and only if

Fourier Transform

105

FT of b and limited signal f(t)

+ aim

-aim

-aim

0

'"""'
·>----mo--->~I

Fig.3.27

Fourier transform of sampled signal f'(t)

% - rom > rom % > 2rom fo> 2:fm

i.e., i.e.,

Therefore, for recovery of signal,

Is> 2fm

where Is is the sampling frequency.

(3.66)

3.7.2 Practical Implementation Figure 3.28 shows that the signal before reaching the sampling switch comes to a low pass filter (LPF) which makes the signal band limited. This filter is also known as anti-aliasing filter. After sampling of the signal another filter for recoveringf(t) from/(t) which is known as a reconstruction filter. Sampler Analog signal

LPF (ANTI-ALIASING) FILTER

Fig. 3.28

fit)

fit) LPF '---f*-(t-)-l (RECONSTRUCTION) f - - - - FILTER

Reconstruction of signal f(t)

3.8 SYSTEM ANALYSIS USING FOURIER TRANSFORM Figure 3.29 shows a system having impulse response. The system function H( ro) of the system is the Fourier transform of h(t). In Fig. 3.29, the input is x(t) and its output y(t) can be obtained by convolving x(t) with h(t).

R(w)

I

- - - >, Fig. 3.29

H(m)

•

Y(m)

Impulse response

We can write Y( ro) = x( ro) * H( ro) i.e., yet) = rl[x(ro) * H(ro)] where X(ro) is the Fourier transform of x(t), Y(ro) is the Fourier transform of y(t) and H(ro) is the system function .

106 3.9

Signals and Systems RELATION BETWEEN DIFFERENTIAL EQUATION AND SYSTEM FUNCTION

Let us consider the following dIfferential equation relating the input x(t) and y(t) d 2 y(t) dy(t) () b d 3x(t) b d 2x(t) b dx(t) b ( d 3y(t) a3--3-+a2--2-+al --+aoy t = 3--3-+ 2--2-+ 1 - - + OX t) ~

~

~

~

~

~

Taking Fourier transform of both sides, we get (/3(jW)3 y (W) + (/2(jW)2 y (W) + (/1 (jw)Y(w) + aoY(w) = b 3(jw)3 X (W) + bijw)2X(W) + bl(jw)X(w) + boX(w) = b 3(jW)3X(W) + b 2(jw)2X(W) + bl(jw)X(w) + boX(w) y(w) _ b 3(jw)3 +b2(jw)2 +bl(jw)+bo H (w) _- - 3 2 X(w) a 3(jw) +a 2(jw) +al(jw)+ao

i.e.,

Example 3.1

(3.67)

If the input of the differential equation

d 3v(t) d 2y(t) dy(t) -'-3-+4--2-+3--+ y(t) dt dt dt find the steady state response.

dx(t) = x(t)+3-dt

(1r)

. IS x(t) = 3 cos 4t+- , 4

Solution Let the input be ejW,,1 so that the output is H(w)e(jw ol+8). [(jwo)3H(W) + 4(jWO)2H(W) + 3jrooH(w) + H(w)] e jWo1 = e jWo1 (1 + 3jwo) i.e., [(jroo)3 + 4(jroo)2 + 3jroo + 1] H(w) = e jWo1(1 + 3jroo) H(w) =

i.e.,

1+3j w o = 1+3j(4) = 1+ jl2 (jWo)3+4(jWo)2+3jwo+1 (j4)3+ 4 (j4)2+ 3j (4)+1 -52-63

=

12.04 < 85.24° 81.69 < -129.54°

=0.147 < 214.780

yet) = 3 x 0.147 cos (4t + 45° + 214.78°) = 0.441 cos (4t +259.78°) Ans.

ADDITIONAL SOLVED EXAMPLES

Example 3.2 Determine the Fourier transform of a triangular function shown in Fig. E3 . 1.

f(t)

A

Mathematically the function is defined as for for

- T 5, t 5,

o}

05, t 5, T

T

Fig. E3.1

Fourier Transform f (t)

The function f(t) and its first and second derivative has been shown in Fig. E3.2. d 2 f(t)

A

----:iT = 1..

F[d 2 f(t)]

dt 2

2A 0 (t + T) - T O(t)

A

A + y:0(t - T)

= A ejroT _ 2A + A e- jroT T

T

-T

T

T

f'(t)

= A (ejroT _2+e- jroT ) = A (e jroT +e- jroT -2) T

107

T

AfT

A 2A = - (2cos roT - 2) = -(cos roT -1)

T

F[d

T

f(t)] =(jro)2F(ro) dt 2

2

(jro)2F(ro) F(ro)

= 2A T

= =

F(w)

T

-T

(cos roT - 1)

2A

T(jro)

-AfT

.

2

f"(t)

(cos.roT-I)

2A

4A

.

(l-cosroT) = - 2 - SIn roT roT

-2-

= (i)'

'in' w;

2

roT 2

AfT

AfT

-T

T

-

AT(~:lJ

=

- 2AIT

=AT Sa 2 (ro;)

Fig. E3.2

Example 3.3

Determine the Fourier transform of the trapezoidal function shown in Fig. E3.3. f(t)

/1

-b

-a

1

~

a

Fig. E3.3

f

(t),f'(t) and !"(t) have been shown in Fig. E3 .3(a).

(i) :.

A f"(t) = - - [O(t + b) - O(t + a) - O(t - a) +O(t - b)]

b-a

b

•

t

108

Signals and Systems f(l)

/1

-a

-b

~

1

a

b

a

b

•

1

f'(I)

-a

-b

f"1 ()

-a

a

-b

b

Fig. E3.3 (a)

Example 3.4 Determine system function of a system having impulse response of h(t)= e- 3tu(t) . If the input to this system is x(t) e-2tu(t), find its output using Fourier transform.

=

Fourier transfonn of e- 3tu(t) is _1_ . jw+3

Again input is

X(I)

=e-2tu(t).

H(w) = Yew) = _1_ X(w) jw+ 3

(I)

I X(w) = - jw+2

(2)

Using Eq. (2), Eg. (1) becomes Yew)

yet)

Example 3.5

A system has H( w)

=

(i) its impulse response and (ii) its response to the input e-2tu(t).

=H(w)X(w) = =e-2tu(t) -

1 (jw + 2) (jw + 3)

--- -jw + 2

jw + 3

e-3t u(t)

jw + 1 . Find the following using Fourier transform: (jw + 2) (jw + 3)

Fourier Transform

109

olution (i) Here h(t) is the inverse Fourier transform of H(m) . H(m) =

Here

..

h(t)

}m + I

I

1

=---+-(jw+2) (jm+3) }w+2 }(0+2

=_e- 2tu(t) + e-3tu(t)

(ii) Here input is x(t) = e-2tu(t) . I X(m) = - }w+2

H(m)

Since

= Y(m) = X(w)

Y(w)

=[

}w +

II

I

1],X(w)=---,,--jrLl + 1

(jw+2)(jw+3) I

= yet) =

}w+l (jw+2)(jw+3)

(jm+2)2 -te- 2t u(t)

+

}w+2

}w + 3

2e- 2tu(t)

sgn (t)

-1

--=,

~,

ill, +T/2

u(t)

- 2e- 3t u(t)

II

POINTS

-

I

I(w) plot

I(w)

1ft)

$'

-T/2

-2

+--+--

SIGNIFICANT 1ft) plot

(jw+2)" (jm+3)

2

/I~

2

jw

1

no(w) +-.JW

"

/j~.

(I)

(V

1

e-

O.

f{t) = e atu{t) with a> 0

This function is shown in Fig. 4.2. Mathematically this function is expressed by 0 { f(t) = e al LT[f(t)]

=

t

~ 00,

where Re(s) > a.

1

=

eal e - SI dt

1

=

0

1= 00

e -(s - a)1 dt

= [e ~(~-:)I l~

0

e - (s-a)1 =e -(a+jw-a)11

/.' -(a-a)1 - jWII Th erelOre,e e Hence

(4.29)

t> _0

f(t)e - SI dt

o

At

t+", e

dt -

e

dt

1

126

Signals and Systems

(4.35)

Similarly,

[1

LT[coshbtl =LT - (bt e +e -bt )] 2 LT[cosh btl

4.7.10

= -1(1 - - + -1) 2 s-b

1 = s2 -b 2

s+b

(4.35a)

Damped Sine and Cosine Functions

(a) The damped sine function is represented

J(t)

=e-at sin bt =e

-at [

e jbt -e - jbt 2j

1

-_ - 1 [ e -(a- jb) -e -(a+ jb)t] 2j LT[e - "' It

. Sin

1[1 1[

2jb

= 2j (s+a)2+b 2

LT[e -atcos btl

4.7.11

1]

btl = - - - - 2j s+(a- jb) s+(a+ jb)

1

s+a = ------..,.. 2 (s+a)2 +b

Damped Hyperbolic Sine and Cosine Function

(a) The damped hyperbolic sine function is represented as

J(t) = e-at sinh bt

=e-at

LT [e-at sinh btl

= fo

oo

-1 [bt e -e -ht ] 2

e- at sinhbte- st dt

(4.36)

Laplace Transform

=

+[r

=

+[ s+~ s+~

127

e -[(s+aH]t dt-r e -[(s+a)+b]t dtJ

-b

+b ]

LT [e-at sinh btl

b = ---:--~ 2

(4.37)

LT [e- at cosh btl

s+a = ---:--~ 2

(4.38)

(s+a)2 _b

Similarly, (s+a)2 _b

4.7.12 Laplace Transform of f Let

f(t)

=('

F(s) = LT[f(t)]

= fo~ tne -

SI

dt

x = st,

Let

dt

= -I dx and t = -k s

F(s) =

r(n+l)

Since

s

r~(~)n e- x Jo s

dx =_1 S

S"+I

r~ xn e- x dx Jo

= fo~ xn e- x dx

LT[tn] = f(n ~ 1)

(4.39)

sn

Table 4.1 ((I)

u(t) e±at

Laplace transforms of some useful functions F (s) 1 8

1 8+a W

sin wt

8 2 +w2

cos wt

8 2 +W2

8

(Contd . . .)

128

Signals and Systems

Table 4.1

(Contd .. .)

Nt) t

f2 n!

e-{1/

sin wt 8 +a (8 +a)2 +0)2

1

8(t)

sinh bt

82

8

cosh bt e-{1i

e-{1/

Example 4.3

b _b 2

82

sinh bt

_b 2 b

8+a

cosh bt

Detennine the Laplace transfonnation off(t)

olution

ff

=3 -

2e- 41

~

LT[f(t)]

=F (s) =

f

(t)e- sl dt

0

~

= (3 - 2e -41 ) e -

sl

dt

o

f ~

=

o

3

2e-(s+4)1 dt

0

2

= ---- = s s+4

Example 4.4

f ~

3e- lsl dt-

3s+12-2s s(s+4)

Detennine the Laplace transfonnation of f(t) = 5 cos

LT[f (t)]

=F (s)

s+12

= - -s(s+4) OJt

+ 4 sin

OJt.

Laplace Transform

=5 LT [cos aN] + 4 LT[sin (Of] =

Example 4.5

i

5s 4(0 5s+4(O + =--+(02 s2 +(02 i +(02

Obtain the Laplace transfonn of the gate function shown in Fig. E4.1 f(t)

f(t)

------ --- -.---------.-------1 .0

Fig. E4.1(a)

Fig. E4.1

Using delayed unit step function, the Fig. E4.1 be constituted from Fig. E4.1 (a). ..

f(t-D=u(t-T I )-u(t-T2)

LT [t(t - D]

..

=LT [u (t s

TI )]

-

s

Example 4.6 Determine the Laplace transfonn of the wave shown in Fig. E4.2.

olution The Figure shown in Fig. E4.2 can be constructed from Fig. E4.2(a) and Fig. E4.2(b). ..

f(t)

f(t)

1.0 -------

= fl(t) + h(t)

2n ( T) . T2n t + u (T). =u (t) SIn t - 2 Sin T t - 2

Using Laplace transfonn methods, we get F(s)

LT [ u (t - T2 )]

=LT [J(t)]

Fig. E4.2

129

130

Signals and Systems f(t)

u(t) sin 2n

-----

1.0 -------

~

T

o

Fig. E4.2(a)

f(/)

( "2T) Sin. T2n

u t-

(t -

T) "2

1.0

o

TI2

Fig. E4.2(b)

Example 4.7

Determine the Laplace transform of the wave shown in Fig. E4.3. f(t)

1.0

3T I I

,, ,, -1 .0 ,

T

2T

I

~-------

Fig. E4.3

4T

Laplace Transform

131

Figure E4.3 can be constructed from Fig. E4.3(a) f(t)

-------- - [2u(t- T)]

1.0

~--,

-1 .0

-2.0 . . ______

,,

i----i--------- u(t)

,, ,, ,, ,

.L..:_ _I__________

__ ___ +_

[-

2u(t- T)]

I I I

t Fig. E4.3(a)

f(t) = u(t) - 2u (t- T) + 2u (t-

2D + ~ .... .

1 2e - Ts 2e Ts F(s) = LT [f(t)] = - - - - + + ... . . .

s

s s F(s) = 1.[1- 2e- Ts (1- e- 7S + e-2Ts + e-3Ts _ e-4Ts + ...... .)] s

4.8

LAPLACE TRANSFORM OF TWO-SIDED FUNCTIONS (BLT)

Let us discuss about the Laplace transform of two-sided function. Let us first consider a negative function and then a positive function . Then we will consider the two-sided function.

4.8.1

f(t)

ROC for f(t) = ~tu(-t)

Let us consider a functionf(t) =ebtu(_t) which has been shown in Fig. 4.6. This a negative function of time and b is a positive quantity. The Laplace transform of this function is given by LT[e bt u(-t)] =

J

=

f

~

f(t)e- st dt

o

~

Fig. 4.6

f(t)e-stdt+ Jf(t)e - stdt

((I) = ebt u(-I)

o

o bt e- si dt = J 0e-(s -b)1dt = [_e_ _(S_b)I]O = [e(-S(S_-bb)1 ]O- ~ = Je __ _~

_~

-(s-b) _~

132

Signals and Systems t~

At

00,

=e-(s -b)1 = e-(a+jW-b)'1 1= 00

=e - (a-b)1 e-jW'I

jw (= 00

= e -(a-b)1 e - j·WI I

Therefore,

/=- 00

will be zero if and only if cr < b. Hence

=

e -ocl

-

-(s-b)

where Re(s) < b. Figure 4.7 shows the ROC of e

4.8.2

e-{)

=

s-b

(4.40) Flg. 4.7 ROCofebtu(-t)

bIU(_t).

ROC for f(t) = eBtu(t)

f(t)

Let us consider a functionf(t) = ealu(t) which has been shown in Fig. 4.8. Here a is a positive quantity. The Laplace transform of this function is given by

a,

LT[e u(-t)]

= =

f ""

f(t)e- S' dt

o

f

f(t)e- S' dt

!

= At Therefore,

t ~ e- (a-a)1 e-

Fig. 4.8 f (t) =e8tu(t)

f 00

,-(' -0" dl

00,

jwt;oo

e -(s - a)1

+ f(t)e- S' dt o

= [,-:~-;" =

I

e -(a+ jw-a )t

It;oo

=

e -(a-all e -

jwt;oo jw

will be zero if and only if cr > a.

Hence

=

e -001 -e - 0 -(s-a)

where Re(s) > a. Figure 4.9 shows the ROC of e

= s-a

(4.41)

I

"I

Roe

--------~---;~~~~a

alu(t).

Fig. 4.9

ROC of e at u(t)

Laplace Transform 4.8.3

ROC for Two Sided Function

133

f(t)

Let us consider the bilateral transfer function. Mathematically, we can write this function as f(t) = e hl for t < 0 and f(t) =

eal

for t > O.

Figure 4.10 shows the function.

f ~

LT[f(t)]

=

=

f (t) e- sl dt

o

f 0

f

-----1-------.

f (t) e- s l dt

+

f (t) e - sl dt

Fig. 4.10

o

= f eh1e- S1dt+ f eate-S1dt ~

jw

o

1 1 1 1 =--+-= - s-b s-a s-a s -b

ROC

I

(4.42)

'I '

The first LT exits only if (J < b and the second one exist only if (J> a. Hence the both conditions will be satisfied if and only if a < (J < b. Therefore, Re (s) must lie between a and b. For b < a, the Laplace transform does not exist. Fig. 4.11 shows the ROC of the two-sided function.

Example 4.8

-----+-~~~~-~a

Fig. 4.11

ROC of Two-sided Function

Find the Laplace transform of f(t)

=e 31 u( -t) + / u(t)

olution We know that

LT[e hl u(-t)] = - _1_ with ROC Re(s) < band s-b s-a

Therefore,

LT[e 31 u(-t)] LT [e l u(

t

~

= __1s -3

jw

with ROC Re (s) > a . with ROC Re(s) < 3 and

-t)] = _1_ with ROC Re(s) > l. s-a

Fig. E4.4 shows the combined ROC for which the LT exists. The region of convergence is I < Re (s) < 3.

1

1 s-3 s-l

F(s) = - - + - =

-(s-I)+(s-3)

--------- ~ - -

(s-I)(s-3)

Fig. E4.4

-2

= --------

(s-\)(s-3)

134

Signals and Systems

Example 4.9

Detennine the Laplace transfonn of

f(t)

= e-{)I u(-t) + e 51 u(t)

Let f(t) = e bl u( -t)e al u(t). Laplace transform of f(t) will exist if and only if a < b. Here b = -6 and a = 5. Hence a > b. Therefore, Laplace transform off(t) does not exist.

4.9

INITIAL VALUE THEOREM LT[df(t)] dt

Since

= f~ df(t) e- sl dt =sF(s) - f(O) Jo

dt

Taking Lt [LT{df(t)}] dt

s~

Lt f~ df (t) e -SI dt S-+~ Jo dt s ~ 00, e-SI ~ 0

or,

At

Lt [sF(s) - f(O)]

= s~ Lt [sF(s)- f(O)] =

Lt [sF(s) - f (0)] s-+~

=0

Since f (0) is a constant,

f(O) = Lt sF(s) s-+~

Lt f(t)

1-lO

=

Lt sF(s)

(4.43)

s~

Equation (4.43) gives the initial value of the time domain solutionf(t)

4.10

FINAL VALUE THEOREM Lt [df(t)] = f~ df(t)e-Sldt=sF(s)_f(O) dt Jo dt

Since

HO

Taking Lt [LT{df(t)}] 5-+0 dt

or, Since at s or, or, or,

~

Lt s-+o 0, e-51 ~ 0

f

=

Lt [sF(s) - f(O)] s-+O

df(t) -sl - - - e dt = Lt [sF(s)-f(O)] dt s-+o f(oo) - f(O)

=-f(O) + s-+o Lt sF(s)

Lt sF(s) =f(oo) s-+o Lt [f(t)] t~ oo

= s...-tO Lt sF(s)

Equation (4.44) gives the initial value of the time domain solution f(t)

(4.44)

Laplace Transform Example 4.10

135

Determine the initial and final value of the current where /(s)

=

032 s (i +0.45s+0.818)

Solution (i) Applying the initial value of theorem, it can be written that, /(0)

=

Lt sF(s) = Lt sl(s)

s~

L

s ~oo

[

1

032

L

= H~ s S(S2 +0.455+0.818) = H~ i Lt 032

L[1

o

s~ .. s2

=---

= -~-=-"'-------"-----~

t s~..

0.45

1+0+0

0.188]

+-- + -~

S

032

+0.45s+0.818

s2

=0. (ii) Using final value of theorem, it can be written as /(00) = Lt sFs(s) = Lt sICs ) s

-

~O

s~

Lt s [

HO

= Lt HO s2

032

s(s2 +0.45s+0.818)

032 + 0.45s + 0.818

=

1

032 0.818

=0.6357 A

Therefore the initial current is 0 A and final current is 0.6357 A.

4.11

PARTIAL FRACTION EXPANSIONS

The Laplace transformation of any function may contain the ratio of two polynomials in s-domain i.e. F(s)

=

pe s ) Q(s)

(4.45)

where Q(s ) is of a higher degree than pes) . Therefore, the expansion of quotient into the sum of several fraction is required to obtain inverse transforms. The application of partial fraction expansion method and another method popularly known as Heaviside expansion theorem is discussed here. Equation (4.45) can be written as F(s)

pes) = ----:......::---(s - s) )(s - s2 ) .. .... .. (s - s,,)

136

Signals and Systems (4.46)

(a) General Case Let us consider an example F(s)= P(s)= s+2 Q(s) S2 +s-6

(4.47)

Factorizing the denominator F(s)

= - s+2 - -

a) a2 = --+ -

(s+3)(s-2)

s+3

s-2

(4.48)

s + 2 = a) (s - 2 ) + a2 ( s + 3 )

=(a) + a2 ) + ( 3a2 a) + a 2 = I 3a2 - 2a) =2 s+2

and

2a) )

(4.49) (4.50)

Solving equations (4.49) and (4.50), the following is obtained, I 4 a) = - and a2 = -

5

5

Putting the value of a) and a 2 in Eq. (4.47),we get

4 F(s)

= _5_ + _5_ s+3

s-2

(4.51)

The inverse Laplace transform ofEq. (4.51) gives the result as follows: f(t)

= -I e_31 + _4 e 21

(b) Heaviside Expansion Theorem (i) Roots are real and unequal Consider the example of Eq. (4.48), F(s)

5

5

= ~+~ s+3

s-2

In general,

Using this formula, we have a) = [(S+3)

a)

(S+2)] = s+21 (s2 +s-6) s=-3 s-2 s=-3

-3+2 1 =-3-2 5

= -

(4.52)

Laplace Transform

and

a 2 =(S-2)( [

1 ::~1'=2 5=2

(s+2) ) S2 +s-6

137

2+2 4 = 5 = 2+3

=

4

F(s)

= _5_+_5_

f(t)

= ~e-s, +i e"

s+3

s-2

5

5

(ii) The roots of Q(s) are real and equal To determine the co-efficients of fractions where roots are not equal is same as above. For equal and repeated roots, the general formula is

where p denotes number of times the roots are repeated and} F(s)

= P(s) = Q(s)

F(s)

= 1,2, .. . p.

(s+l) (s+3)2(s+2)

(4.52a)

A22 + ---A2I = -Al- + ----(s+2)

(s+3)2

(s+3)

S+ I I = _ -_2_+_1_ =-1 (S+3)2 5 =_2 (-2+3)2

-3+ 1 -2 - - - - - -- -2 - -3+2 - -1 - . A22

= [ -d

ds

{ (s+3) 2 . - -s+ -1- -}] (s + 2)(s + 3)2

= [~(~)] ds s+2

=

(-3+2)2 -1+2

=~-=I

1

= [1(S+2)-I(S+

5=-3

(-3+2)-(-3+ I)

for} 5=-3

(s+2)2

I)]

=2

,1 =-3

138

Signals and Systems F(s)

= --1- + - -2 - + -I s+2

(s+3)2

(4.53)

s+3

The inverse Laplace transform will be ! (1) = -le- 21 + Ie -3r + 2e -3r

(4.54)

(iii) The roots ofQ(s) are complex Let

I

F(s) =

= -----;:(s+2)z +

s2 +4s+6

(J2f

1 = -,------=,...-;-------==(s+2+ jJi)(s+2- jJi)

F(s)

=

a, + _ _a=-z----==_ s+2+jJ2 s+2-jJ2

a,

= [(s+2+ jJi)F(S)L _z_j J2

a,

= [

(4.55)

I ] s+2- jJi s=-z- j/i

1

-I

j

= -2- jJi +2- jJi = 4jJi = 4Ji

*(

j a2 = a, = 4jJ2

)* = 4J2 -j

j F(s)

= 4J2

s+2+Ji

[since the roots are complex conjugates]

j

+

412

s+2- jJi

Taking Laplace inverse transformation, we have !(t)

= _j_e-(z+j J2 )r _ _j_e-(z-j J2 )t 4Ji

(4.56)

4Ji

the above methods except (a) were originally discovered by Oliver Heaviside and hence they are known as Heaviside expansion theorem.

4.12

RELATION BETWEEN STEP RESPONSE AND IMPULSE RESPONSE

Let input to the network shown in Fig. 4.12 be Vi (t), let the output or response of the network be va(t). The Laplace transform of vi (I) and Va (I) be Vi (s) and Va(s) respectively. The ratio of Laplace transform of output to that of input putting initial conditions tozero is known as transfer function.

--II

: =

L

.

S

Again , /(s)

=

2s+3 QI Q2 =-+(s +I)(s +3) s+1 s+3

Q1

= [(S+I)

(2S+3)] =[2S+3] =_-_2_+_3 =~ (s+1)(s+3) s=-I s+3 s=-I -1+3 2

Q2

=[(S+3)

2s+3 ] = [2S+3] = -6+3 =_-3 =2 (s+I)(s+3) s=-3 s+1 s=-3 -3+1 -2 2

3

/(s)

=--L+_2_ s+1

s+3

Taking inverse Laplace transform, we have

.( ) =-e 1 _I +-e 3 -31

It

2

2

. 1 0 3 0 13 = - e + - e =- + - =2 2 2 2 2

1(0)

Hence the result is verified.

Example 4.19 function and

Find by convolution integral, the Laplace inverse of

~-}- as the second function . (s+2)

Solution Here

F(s)

Let

F, (s)

= - --1 -

(s+l)(s+2)

= - 1-

(s+l)

and F2 (s)

Nt) = e-I andf2 (t) = e-21 Lr' [

1

(s+I)(s+2)

]

=Lrl [FI (s) F2(s)]

= - 1-

(s+2)

1 1 ,taking - - as one (s + l)(s+2) (s+ I)

Laplace Transform 1

=} Nt-r)Nr) dr o

f 1

f 1

= e-(I-r) e- 2t dr= e- I

o

e- r dt

0

-_ e-I [ -e - r]1 -_ - e-I (-I e - I) -_ e-I (1 -e- I)

o

Example 4.20 Solve the following differential equations, using Laplace transform technique d2x - 2 + xa

dt

= cos t; x (0)

=0 and !!~' dt I = I 1=0

Solution The Laplace transform of the given differential equation can be written as

[ S2 X (S)-SX(0)- dX(O)] + X(s) dt Since x (0)

=0 and x'(O) = I,

s2X(s) -1 + X(s) or,

(s2 +1) X(s)

X (s)

x(t)

Let

=-/s +1 =-/s +1 S s2 + s + 1 =--+ I =- --s2 + 1 s2 + 1 s2 + s+ l I s

=

(s2+1)

2 = -"2 - -+ - - - 2 s +1 (s2+1)

=Lr\ ( - /-) s +1

+ LT-\ [

1)2]

= 'int+

Lr' [(" :

F\(s)

= _s_

and F2 (s)

f\(t)

=cos t andf2 (t) = sin t

s2 + 1

= _ _I _

LT-{ (,2 : I)] = LT-' [F, (s) F, (s)]

f 1

=

o

fl (t-r) f2 (r) dr

S 2]

(s2 +1)

s2 + 1

155

156

Signals and Systems

f

=

t

cos (t-r) sinr dr

o

±f t

=

2cos (t-r) sinrdr

o

f

= ± [sin (r + t t

r) + sin (r - t + r)] d r

o

=

±[[

1

-,intdr+ ['in(2r-t)dr

I. = -tsmt +--1{[COS(2r-t)]~}

2

2

= -It .sm t 2

-2

-1 [cos t - cos t ]

4

1 . =-t sm t 2

(I

x(t) = sin t + ±t sin t = Example 4.21

Tf}in t.

Solve the differential equation given below using Laplace transform. d i 4 di S· + -+ I 2

= S u(t)

i(O)

= 1 and

2

dt

dt

Assume

di (0) dt

=2

olution The Laplace transform of the given differential equation can be written as [ s2!(S)-Si(0)- di(O)] + 4[s!(s)-i(0)] + S!(s) = dt

2. s

Putting initial conditions,

,

[s-/(s) - s- 2] + 4[s! (s) -I] + S/(s) 7

or,

!(s) (s- + 4s + S)

or,

!(s)

or

S

=-

s

= -S + s + 6 s

=

S2 + 6s + S

s2 + 6s + S s(s2+4s+S)

= !:L + s

a2

s+2+j

s(s+2 + j)(s+ 2 - j) +

a3

s+2-j

Laplace Transform

0, =

Q

[

157

(s2 +6S+S)] [s2 +6S+S] S S. s(s2 +4s+S) 5=0 = S2 +4s+S 5=0 = 5" = 1

6s 2 = [(S+2+ j)_-,--(s2_+__+---,S)_] s(s+2+ j)(s+2- j) 5=-2-) = (_2_j)2 +6(-2-j)+S (-2- j)(-2- j+2- j)

=

4+4j -1-12-6j + S (-2 - j)( -2j)

-4 - j2

= (-2 - j)( -2j)

j

Since, /(s)

=_1 +

.

]

s

s+2+ j

-j

+ --"-s+2- j

Taking inverse Laplace transform it can be written as ..

= l+je-2I[e-jl-ejl]=1+2e-2Isint

i(t)=I+j[e-(2+ j )l- e -(2- j )l]

Example 4.22 Plot a function which is given by fCtV 31 for t > 0 andfit)e 31 for t < O. Find the Laplace transform

f(t)

if possible and also show the ROC.

Solution Fig. E4.7 shows the plot of f(t) f(t) Here

>/ \ 0 and for t < O.

=e-31 u(t) + e 31 u(-t). = -3 and b = 3. Therefore, b >

We know that

Therefore,

Q.

=- -

LT [eat u( -t)]

= _ 1_

LT

s-b

s-o

[e 3I u(-t)] = -

LT [e-3I u(t)] =

/

\

e-3t u(t)

Fig. E4.7

1

LT [e u(-t)]

hI

t)

with ROC Re(s) < band

with ROC Re(s) > Q .

_1_ with ROC Re(s) < 3 and s-3

1 =_l-with ROC Re(s) > -3. s-(-3) s+3

F(s) = _ 1_ _ _ 1_ = 2-D -- with ROC: -3 < Re(s) < 3. s+3 s-3 s-9 & S2 +4s+3 F' d h . I f ·. Show all the ROC condltIons lor yes) = 2 . In t e Inverse Lap ace trans orm (s+2) (s+4) for each of the ROC conditions.

4 23 ExampIe.

158

Signals and Systems

Given that yes)

=

s2+4s+3 (S+2)2 (s+3)

B = (s + 2)2

C =(s + 4)

and

I

Case 1: Re(s) > -2

,

I

s+4

2

s2 +4s+3 ] = (s+2)2(s+4) ;=-2 4

-\

= _4_ +

+--

(s+2)2

s2 + 4s + 3 I =3 (S+2)2 (s+4) 5=-4 4

ds

s+2

s+2

= (S+2)2 (s+4) 5= -2

d [ (s+ "2)2 A =-

yes)

=-- +

+ 4s + 3

52

jw

ABC

Fig.E4.8

3

2 + _4_ (s+2)2 s+4

1_. I ,

In this case both the poles lie on the left of ROC and hence both of them contribute to positive time function. y(t) =..!..e- 21 u(t)-..!..te- 2r

4 Case 2: Re(s) < -4

2

u(t)+~e-41 u(t)

--7';","~---t-----""(j

4

In this case both the poles lie on the right of ROC and hence both of them contribute to negative time function. y(t)

=-..!..

I

Flg.E4.9

e-21 u(-t) - [- ~te-21 U(-t)] -l..e-41 u(-t)

4 2 4

jw

1 -'1 1 -21 u(-t)--e 3 -41 u(-r) = -e - u(-t)+-te

4 Case 3: -4 < Re(s) < -2

2

jw

4

In this case the pole corresponding to s = - 2 lies to the right of ROC and hence it contributes negative time function. On the other hand, the pole corresponding to s =-4 lies to the left of ROC and hence it contributes to the positive time function . y(t )

'-21 21 3 -41 u(t) = --e u(-t)+ -' -te u(-t)+-e

Example 4.24

4

2

Fig. E4.10

4

Find the inverse Laplace transform when F(s)

(i) ROC Re(s) > -5 and (ii) ROC Re(s) < -5.

= - 's+5

with

Solution (i) We know that when Re(s) > a, the function is a positive time function.

LT[e at u(t)]

1

=-s-a

Laplace Transform I

F(s) = - =

Here

s+5

1 (5) Re(s) > - 5.

s--

f(t) = e-51 u(t) because a = - 5.

Figures E4.11(a) and E4.1I(b) show the ROC and plot off(t) vs t. jro

f(t)

-----.:5~~44_ 3 Both poles will lie to the left of ROC and hence both of them will contribute to positive time function. f(t)

1

_I

3

31] u(t)

= [ 4 e + 4e-

162

Signals and Systems jw

f(t)

,

I

3,

--+---)('-'-T-T;-r-r--a

~ ~

,

Fig. E4.16

Case 1

Case 2: Re(s) < -1 Both poles will lie to the right of ROC and hence both of them will contribute to negative time function. 1 f(t) = [ -"4e

-1 -"4e 3 31] u(-t) f(t)

jw

,

~ ~-1

3

I

~77"77'7.if--+-------*-+-a

-1

Fig E4.17

Case 2

Case 3: -1 < Re(s) < 3 jw

f(t)

1 - e-tu(t)

4

-3/4

Fig.E4.18

Case3

Laplace Transform

163

The pole corresponding to s =3 lies to the right of ROC and hence it contributes to a negative time function. On the other hand, the pole corresponding to s = -1 lies to the left of ROC and hence it contributes to a positive time function . I_I 3 31 fi(1) = -e u(t) - -e u(-t)

4

4

Example 4.27 s +3

(i) If F(s) = (ii) If F(s)

(s + pes + 2)

=

, find the inverse Laplace transform for -2 < Re(s) 0 for t < 0

eft

Also find the ROC of the above cases. [Ans.: (i)

2~ + 20 ROC:

(J>

s + 25 Q2. Find the Laplace transform of f(t)

0 (ii) _ 1_ , ROC: (J> -2 (iii) _ I_ _ _ 1_ , ROC: a < (J < fJ] s +2 s- a s - f3 =e-5tu(-t) + te-7tu(t) if it exists. Find the ROC also if it exists. [Ans.: LT Exists F(s)

=

-2 ROC: -7 < (J< -5] s2 + 12s + 3s

Q3. A funtionf(t) is given by f(t)

={

e- ar

for t > 0

eClt

fort 0

Plot the funtion and find its Laplace transform if it exists. Also sketch the ROC. f(t)

jw

[Ans. = - - - - - - - - - 1 . - - - - - = + - t

Q4. Findf(t) if F(s)

=

(s

Ans. F(s)

=

-2a

2

s -a

s+I for the following ROCs: + 2)(s + 3)

2

ROC:

+a

G]

Laplace Transform

173

(i) Re(s) < (-3) (ii) Re(s) > (-2) (iii) Re (s) lying between -2 and -3. [Ans.: (i) (e- 2t -2e- 3t )u(t), (ii) (_e- 2t + 2e- 3t )u(t), (iii) e- 2t u(-t) + 2e- 3t u(t)] Q5. Findf(t) if F(s)

=

(s

s+4 for -3 < Re (s) < -2. + 2) (s + 3) [Ans.: f(t)

Q6. Findf(t) if F(s)

=

(s

=_2e- 2tu(_t) + (_I)e- 3tu(t)]

s.~:_4_ _

+ 2)

(s

+ 3)

[Ans.: f(t)

= e-2tu(_t) -2te- 2tu(-t) + e-3tu(t)]

Q7. Find the response of the system having equation d 2 y(t)

dy(t)

dt 2

dt

- ---- + - - - - 2y(t)

= -duCt)- + 2u(t) dt

where u(t) is a unit step function. The initial conditions are yeO)

= I and c!.Y. .(t)1 = I . dt

[A ns.: y ( t ) = ( --I

4

Q8.

t=O

+ -3 e 2t 8

- I e -?t) - u(t) ]

8

The output off system is 2(l_e-O· St ) for t;;:: 0 for an input 2u(t). Find its output if input is (e- t + e- 3t ) u(t) where u(t) is a unit step funtion. [Ans.: y(t) 1.6e-O-St_e- t -O.6e- 3t ]

=

MULTIPLE CHOICE QUESTIONS I. The Laplace transform of a functionf (t) (t > 0) is given by ~

(a) ff(t)e st dt

(c)

f f(t)e- st dt

(d)

o 0 2. The Laplace transform of a unit step function is (a)

.!. s

(b) I

(c) s

f o

0

~

(b) f f(t)e-st dt

f(t)e- st dt

(d) 2 s

3. The Laplace transform of shifted unit step function u (t - a) is (a) _1_

(b)

S2

.leat s

(c)

'!'e-at

(d) None of these

.!.

(d)

s

4. The Laplace transform of a unit impulse function is (a) 1

(b)

i

(c)

s

5. The advantage(s) of Laplace transform is (are): (a) (b) (c) (d)

it gives total solution more systematically it gives solution in frequency domain only initial conditions are incorporated in the very first step all of these

l-

174

Signals and Systems

6. The Laplace transform of df(t) is (a) sF(s) - f(O)

dt (b) sF(s)

(c) sF(s) + f(O)

(d) F(s) - f(O)

(c) A + a

(d)

7. The Laplace transform off (t) = Aeal is (a) -

A

A

(b) s

s-a

s

A

s-a

8. The Laplace transform of f(t) = 5 sin 3 t is equal to (a)

~

(b)

s- - 4

~~

(c)

s- + 1

-~ s2 + 9

(d) None of these

9. Laplace transform and Fourier integrals are related through (a) time domain (b) frequency domain (c) both frequency and time domains (d) none of these 10. The Laplace transform of eat is (a) _1_

(b) _I_

s-a

(d) None of these

s+a

11. The Laplace transform of (t) eat is I 1 at (a) - -(b) - e (s - a)2

(c)

s2

(s + a)2

(d) None of these

12. The Laplace transform off(t) e-at is (a) F(s)e at

(b) F(s) e----o-u-tP-- Mass (M) Spring CK)

Current (i) Capacitance (C) Reciprocal of

Torque (1') Inertia (J). Damper (D)

Inductance Damper (D) Displacement Velocity

5.10

UJ

Conductance

(~ )

Spring (10

Flux Linkage (lfI)

(x)

(u=:)

Voltage

(v)

Displacement (8)

= -dlfl

Velocity

dt

(w =~~)

REPRESENTATION BY NODAL METHOD

This method is very simple to get mathematical model of both types of mechanical system. Let us consider a mechanical system as shown in Fig. 5.24. The following steps are followed:

=

Step 1

Total number of nodes Total number of displacements = 3 One reference node is taken in addition i.e., total number of nodes becomes 4.

Step 2

MI

is connected between XI and reference is connected between x2 and reference is connects between x3 and reference

M2 M3

Step 3

is connected between XI and reference is connected between XI and x2 K2 is connected between x2 and reference K23 is connected between X2 and x3 K3 is connected between x3 and reference. KI

KI2

Step 4

is connected between XI and X2 is connected between x2 and reference is connected between x3 and reference.

DI2 D2 D3

T f

Fig. 5.24

Step 5

Let us connect Ms , Ks and Ds as discussed in step 2, 3 and 4 respectively.

Step 6

The force F is connected between xI and reference because F is connected to M in Fig. 5.25.

System Modelling

193

Fig. 5.25

Step 7

Applying force equation at node x I' (S.3S)

(S.36)

d2X3

K23 (X2 -X 3)

=M3 7

+

dX3

D3

.

dr + K3 x 3

(S .37)

Note: The same method can also be applied to rotational systems.

Example 5.7 Find the system equation of the mechanical system shown in Fig. ES.7. Also find the force-voltage (i-v) and force-current (i-i) analogy. Step I

There are two nodes Xl and x2 and an additional node known as reference node .

Step 2

Ml is connected between xl and reference M 2 is connected between X2 and reference.

Step 3

Kl is connected between Xl and reference K12 is connected between Xl and X2 K2 is connected between x2 and reference.

Step 4

D I is connected between X I and reference D12 is connected between xl and x 2 D2 is connected between x2 and reference.

Step 5

Figure ES.7(a) shows the schematic diagram.

/

L.L.LL. lal 0)

+ j sin

0)

lal

= Jcos 2

Z[anu(n)]

= l-e}WI.

1

ROC: Izl > 1

(6.62)

Z[{3nu(n)]

1 = 1- {3z 1 = -----,,--~ 1- e- jw Z-I

ROC: Izl > I

(6.63)

0)

z

+ sin 2 0) = 1

Substituting the individual z-transform values from above Eqs. (6.62) and (6.63) to Eq. (6.61), the expressions of X(z) , we get ROC: Izl > I

1- Z -I cosO) cos (nO)u(n) ~ - - - ; - - - - ' " 1- 2z 1 cos 0) + Z 2

ROC: Izl > 1

(6.64)

z-Transform 6.14

227

z-TRANSFORM OF CAUSAL SINE SEQUENCE

The causal sine sequence is given by x(n)

=sin (nco)u(n)

Using Euler's identity, we can write cos (nco) as follow : sin (nco) x(n)

e jnw -e -jnw

= --...:..2j

= J... [e jnw _e-jnw]u(n) =J...ejnwu(n)_ J...e-jnwu(n) 2j

2j

2j

The z-transfonn of above equation is given below. X(z)

=Z

~e - jnwu(n)] = Z [~ejnwu(n)] -

[J...ejnwu(n)_

2j

2j

2j

Z

[~e - jnwu(n)] 2j

·

Let us put a = e jW and {3 = e-jW in the above expression, we get X(z)

We know that

Z[anu(n)]

= Z [21j anu(n)] =

1

I-az

(6.65)

Z [21j {3n U(n)]

ROC: Izl > lal

_I

= ejW =cos co + sin co lal = Jcos 2 co+sin 2 co =1

where

a

Z[~u(n)]

= 1- e J1'w Z _I

Similarly,

ROC: Izl > 1

(6.66)

ROC: Izl > 1

(6.67)

Substituting the individual z-transfonn values from above Eqs. (6.66) and (6.67) to Eq. (6.65), we get X(z)

= -21_.} [1 -e ~w z -I

-

-1--_1"""jW-_-:-1 ] -e Z

ROC: Izl > 1

ROC: Izl > 1 .

sm (nco)u(n)

6.15

-I

~

z-TRANSFORM OF

Let xl(n)

z

.

z sm co 1 2 1- 2z - cos CO + z-

ROC: Izl > 1

a" cos (nco) u(n)

=cos (nco)u(n). Now we can write X1(z) =

1

-I

-z cosco

1- 2z- 1cosO) + Z-2

ROC:lzl>1

(6.68)

228

Signals and Systems x(n) = ancos (nw)u(n) = anxl(n)

The given expression is

X(z)

=Z[a"xl(n)]

Using scaling property, we can write ROC: lal r l < Izl < lalrz

ROC: Izl > l.1al

i.e.,

Hence, the obtained z-transformation pair is

1_(~)-1 cosw a"cos (nw)u(n) ~

1- 2

(~)

~ cos W +

(~ )

ROC: Izl > lal

z

(6.69)

6.16 z-TRANSFORM OF a" sin (nw) u(n) Letxl(n)

=sin (nw)u(n). Now we can write -I

Xl () z

.

___z~-sl-n-w---, =_1-2z I cosw+z 2

ROC: Izl > 1

The given expression is x(n)= a"cos (nw)u(n) = anxI(n)

Using scaling property, we can write ROC: lalrl < Izl < lalrz

i.e.,

ROC: Izl > l.lal

(6.70)

Hence, the obtained z-transformation pair is

(~rlsinw

r

ansin (nw)u(n) ~ -----'--""-0-----------_::_ ROC: Izl > 1.1 a I

1- 2( Table 6.1 gives the standard z-transform pairs.

~

cosw + (

~ rz

(6.71)

z-Transform Table 6.1

Sr. No.

Standard z-transform pairs

$(ra)

1

8(n)

2

u(n)

3

anu(n)

4

-a nu(-n-l)

5

nanu(n)

6

-na nu(-n-l)

7

cos (nw)u(n)

8

sin (nw)u(n)

9

an cos (nw)u(n)

10

an sin (nw)u(n)

11

An for 0 ::; n ::; N-l

Example 6.4

229

x(z)

ROC

1 1

Complete z-plane Iz I > 1

l-z- 1

1 1 -az -1 1 1 -az -1 az- 1 (l_az- 1 )2 az- 1

(1- az- 1 )2 1- z -1 cosw 1- 2z -1 cosw +z-2 z -1 sin w

1- 2z- 1 cosw +z-2

1-(~rl cosw

1-2(~rl cosw+(~r2

(~rl sinw 1-2(~rl cosw+(~r2 I-aN z - N

Izl>

lal

Izl


lal

Izl
1 Izl > 1

Izl>

lal

Izi >

lal

Izl > 0

1-az- 1

Find the z-transfonn of the following signal with the help of linearity and shifting properties. x(n)

I

={o

for 0 ~ N-I elsewhere

The given sequence is a pulse of N samples where each sample has amplitude of I which can be obtained from two unit step sequence as follows: x(n)

=u(n) -

u(n-N)

Taking z-transfonn of the above expression is given by X(z)

We know that

Z [u(n)]

=Z {u(n)} =- - -1 - I l-z

u(n - N)

=Z [u(n)] -

ROC : Izl > 1

Z [u(n - N)]

(1)

(2)

230 c:::::::J

Signals and Systems

Using shifting property of z-transforrn we get

Z

[u(n - N)]

=z-NZ[u(n)] =Z-N [1- ~

-I ]=1~-:-I

ROC: Izl > 1

(3)

Therefore, substituting Eqs. (2) and (3) in Eq. (1), we get X(z)

Example 6.5

1

I-z- N

Z-N

=I -I - - 1 -I =- I -I -z -z -z

ROC: Izi > 1

Find the z-transforrn and ROC of the signal x(n)

Solution We know that Therefore,

= [4.(Sn) 1

Z[anu(n)]

=

ZWu(n)]

= __1--;-

l-az I-Sz

3(4n)] u(n)

ROC: Izl > lal

1

ROC: Izl > S

1

ROC: Izl > 4

and The required z-transforrn of the signal x(n) is given by X(z)

=Z [x(n)] =4Z Wu(n)] =

Example 6.6

4

----;-

I-Sz

3

3Z Wu(n)]

ROC: Izl > S

1

Find the z-transforrn as well as ROC for the following sequences:

(i) GYu(-n) and (ii) GJ[u(-n)-U(n-8)]

(i) The given sequence is

r

x(n) =GJu(-n) :. X(z) =z[Gu(-n)] =I_I! z ROC: 1,1 < t 3

(ii) The given sequence is x(n)

=GY[u(n)-U(n-8)]

X(z)

= Z [GY u(n)] -

Z [GY U(n-8)]

(I)

We know that

z[G)"u(n)] = I-(±k We can write Z [

(~r u(n - 8)]aS follows:

ROC IZI>i

(2)

z-Transform

(t z-'

=1

1

r -I

-3 z

ROC:IZI>l 3

231

(3)

Substituting Eqs. (2) and (3) in Eq. (1), we get x(z)

Example 6.7

=

1

1

ROC:lzl>3

1-(~)z-1

Find the z-transformation of x(n)

= ~ for n ~ O. n!

olution Form the definition of z-transformation of x(n) we can write x(z)

= ~ x(n)z-n = ~ ~z-n = ~ (bz-I)n £...

£"'n'

n=O'

n=-~ ~

£...

n=O

n'.

n

eX=L~

Again,

(I)

(2)

n=O n!

Using Eq. (2), we can write Eq. (1) as follows : X(z) = ebz- I The required transform of x(n) =

6.17

INVERSE

~ for n ~ 0 is i n!

z- I .

~TRANSFORM

We have already studied z-transform as well as its properties. It is true that we need to transfer the signal to z-domain and after that we should get it back in time domain. Using inverse z-transform we can convert the signal from z-domain to time domain. To get inverse z-transform, the following three methods are used: • Contour integration • Partial fraction expansion • Power series expansion The contour integration is the basic method to get the inverse z-transform. The inversion formula based on contour integration is x(n)

= _1_. 1, X(z)z"-'dz which is based on Cauchy residue theorem. 2lrJ j

c

The z-transforms that we use are almost rational. The partial fraction expansion is the more simple to get inverse z-transform.

232 6.18

Signals and Systems INVERSE z-TRANSFORM USING PARTIAL FRACTION EXPANSION

In partial fraction expansion method, the function X(z) is expressed as follows:

X(z) = a,X,(z) + ~X2(Z) + 0.4 , therefore, the second term of Eq. (1) will represent the causal part of hen).

=

:. IZT {

II} =

1-0.4z-

(O.4r u(n) for ROC: Izl > 0.4.

Since Iz I < 2, the first term of this equation will represent anti-causal part. :. IZT {

1 1-2z

-I}=

-(2ru(- n - 1) for ROC: Izl < 2.

Therefore, the unit sample response is H(z) is given by hen) = -4(2tu(- n - 1) + 3(0.4tu(n) This gives the required sample response for a stable system. (ii) Causal System: For a causal system, the ROC will not contain any poles. Therefore, the ROC for the system to be causal will be Iz I > 2. The unit sample response is given by hen) = 4(2tu(n) + 3(0.4)"u(n) The ROC of H(z) does not include the unit circle because ROC: Iz I > 2. Therefore, the system is unstable. (iii) Anti-causal System: We know that for an anti-causal system the ROC will be interior to the circle. Hence ROC: Iz 1< 0.4. Therefore, ROC of Iz I < I Pole nearest to origin I is considered. The unit sample response is given by hen)

=-4(2)"u(- n -

1) - 3(0.4)"u(- n - 1)

Since the ROC of H(z) is Iz I < 0.4, it does not include unit circle at Izl = 1. Therefore, the system is unstable.

Example 6.11 yen)

=0.8y(n -

Find the complete response of the system 1) - 0.157y(n - 2) + x(n - 1) + x(n - 2) to the input x(n)

=nu(n). Is the system stable?

240

Signals and Systems

Solution The difference equation of the system is yen)

=O.By(n -

1) - 0 .157y(n - 2) + x(n - I) + x(n - 2)

(I)

Taking z-transform of Eg. (I), we get

0.8z- 1Y( z) - 0.15z-2y(z) + z- IX(z) + Z-2X (Z) rI - 0.8z- 1 + 0.15z-2]y(z) =(Z-1 + Z-2)X(Z)

Y(z) =

i.e.,

Y(d

i.e.,

X(.::)

=

Z- I+ Z- 2

1-0.8z- 1 +O.l5z

30z I-O.5z

(2)

=------

X(z)

Y(_)_

Z2 -0.8z+C.15

2

z+I

Y(z)

i.e.,

z -

z+1

(z-0.3)(z-0.5)

-I +

-13.26z 1-0.3z

= 30(0.5)" u(n) -

-1 +

5.71z- 1

-26.53z

I-z

-1 +

(l-z-1)

2

13.26(0.3)"u(n) - 26.53(1)"u(n)

+ 5.71n(l)nu(n).

From Eq. (2) the poles of H(z) are at 0.3 and 0.5 and one zero is at -1. The both poles at 0.3 and 0.5 of the system are inside the unit circle I z I = 1. We know that for a stable system both poles should lie inside the unit circle. Therefore, the given system is stable. Since the input to the system is x(n) =nu(n), its z-transfoml is given by -1

i.e., i.e.,

Y(z)

X(z)

z =Z{nu(n)} = (1-z1 2 )

X(z)

= -~-?

=H(z)X(z) =

7

(3)

(z-It

z +1 (z - O.3)(z -

x~= z(z + 1) 0.5) (z - I) (z - O.3)(z - 0 .5)(z _1)2 -13.26 -26.53 5.71

Y(z) 30 -= ---+-- - - + - - + -

I.e.,

z

z-0.5

z-O.3

z-1

--

(z-I)2

ADDITIONAL SOLVED EXAMPLES

Example 6.12 Find the z-transform and its ROC of the following sequence: x(n)

={4, 2, - t, 0, 3, -

4}

Solution The z-transform X(z) of x(n) is given by X(z)

= L,x(n)z-n = n=- ~

3

L,x(n)z-n n=-2

=x(-2)Z2 + x( -1 )Zl + x(O) + x( l)z-2 + x(3)Z-3

=4i + 2z =

4z 2

I + OZ-I + 3z-2 - 4z- 3

+ 2z - 1 + 3z-2 - 4z-3

Therefore, X(z) will be finite if and only if z is not equal to 0 or

00 •

Its ROC is given by 0 < Iz I < 00.

(4)

z-Transform

241

(a) We know that

{3" u(n) ~ _ z _ z-a

(I)

Differentiating both sides of Eq. (I) with respect to {3, we get

() n {3 "-' un (b) x(n) X(z)

=u(n) -

d ( -z-)_ __z---" d{3 z -a (z-al

c ~-

u(n - 4)

=Z[u(n)] -

Z[u(n - 4)]

= - - -1- I I-z

Z

-4

(c) x(n) = n[u(n) - u(n - 4)] = nx,(n)

1 I- z

- ---I

Z3 Z- Z3 = - 1 = -=ROC: Iz I> I.

z -I

z -I

z -I

(Z- Z3)_

4 d Z3) . I} X( z-) - zX ,()_ z--z - - - - - z {(Z-1)(l+3Z- )-(ZJ .

z-I

dz

we know that

-,

az (1- az -')2

•

n

na u(n) ~

V( )

i.e.,

I'Z

i.e.,

I'Z

V( )

Example 6.14 (a) x(n)

(b) x(n)

(c) x(n)

ROC: Izl > lal

-13.26z -26.53z 5.71z =-30z -+ + +--2 z-0.5

z-O.3

z-1

(z-I)

30 -13.26 -26.53 5.7Iz-' =-+ +--+---:-::1 1 1-0.5z 1-0.3zl-zO-Z-')2 =30(O.5)"u(n) - 13.26(O.3)"u(n) - 26.53(l)"u(n) + 5.7In(l)"u(n).

Find the z-transform of the following:

(I)" = ('5I)" = (4'I)" = 4'

(z-It

u(n) +

u(n) +

u(n)+

olution (a) The z-transform of

(II" '5 j (I)" 4' (I)" '5 (!

u(n)

u(- n - I)

u(-

I! -

I)

r

u(n) is given by

( -I )" u(n) ~ 4

1 11 ' ROC : Izl > -4 1__ 7-1 4~

(I)

242

Signals and Systems

)n u(n)~

1 1 1 ,ROC:lzl>5 l- - z-1 5 5 Therefore, ROC of Eqs. (1) and (2) overlap and we can write 1 (-

Similarly,

z

X(z)

(b) The z-transform of

Z(2Z-~) 20

z

=1-~Z-1 + 1-~Z-1 = -Z--~ + -z--~ =(Z-~)(Z-~)

1 ROC: Izl >4

(~r u(n) is given by

(-l)n u(n) ~ 5

( -1

Similarly,

(2)

4

)n u(-n-I)~

1 1 1 , ROC: Izl >l--z-1 5 5 -1

1

1

-I

-4 z

(3)

1 ,ROC:lzl2

(ii)

Here X(z)

=

2

2

i _~.

Therefore, given X(z) is in proper form.

l- - z 9

X(z) =

1

z - -"" 3'v _

x(n)

Example 6.19

=

z(z-D

I

z- -1

(z+

9

X(z)

2

l_'!'Z-1

2

i.e.,

3

1 1

1+ -z -I 3

= 1-

D(z-~)

(-31) z

Z

= --1 z+ -3

t

-I

= IZT\X(z)} = tzT IX(z)} =

fIT {I-(_Jz_, }=(-±)"

u(n) with

ROC Izl > -

~

Find the pole-zero diagram of the following system:

(i) yen) = 0 .6y(n - 1) + x(n) + x(n - 1) (ii) y en) =x(n) + 6x(1I - 1) + 12x(11 - 2) + 8.x(n - 3) Assume yen) = 0 for n < O. Plot the pole-zero diagram and also find the values of yen) for n = 0, 1, 2, 3, 4, 5 for x(n) = D(n).

The gi yen difference equation is yen)

=0 .6y(n -

1)

+ x(n) + x(n

Taking l-transform of the above equation we get

Y(z)

- 1)

=0.6z- 1Y(z) + X(z) + z-IX(z)Y

z-Transform i.e.,

(I - 0.6z- l )y(z) Y(z) = (

i.e.,

Y(z)

i.e.,

X(z)

I+Z-I 1-0.6z- 1

= (1

+ Z-I)X(Z)

Im(z)

)X(Z)=(~)X(Z)

+j

z-0.6

z+1

= z - 0.6

(I)

Therefore, H(z) has one pole at z = 0.6 and one zero at z = -I. Fig. E6.2 shows the pole-zero diagram of the given system. Since the input to the system is x(n) =ben)

----iJ----t--*--If------ Re(z) -1

-j

(2)

X(Z)=Z{b(n)} = I

..

Using Eq. (2), we can write the Eq. (1) as Fig. ES.2

Y(z)=~ z-0.6

Y(z)

i.e.,

z

z+1 A B ---=-+--

z

z(z-0.6) A

z-0.6

= _~ and B =[(Z-0.6)X Y(Z)]

= [ZX Y(Z)] Z

5

Y(z)

3

2=0

Z

=~ 2=0.6

3

8

-3

3

-- = -- + --Z z z-0.6

8 i.e.,

Y(z)

= -~+

3

3 1- 0.6z- 1 Taking inverse z-transform of Eq. (3), we get

y(n)=-~b(n)+~(0.6tu(n) forn~O. We know thatb(n) = { yeO) = y(l)

=

3 I for

o

for

3 n=0

n:;t

0

-~b(0)+~(0.6)o = -~+~ = I 3

3

3 3

-~b(l)+~(0.6)1 = 0+~xO.6 = 1.6 333

y(2) =

-~b(2)+~(0.6)2 =0+~XO.62

y(3) =

-~b(3)+~(0.6)3 = 0+~XO.63 = 0.576

y(4) =

-~b(4)+~(0.6)4 =0+~XO.64

333 3

3

247

=0.96

3

333

=0.3456

v(5) = - ~ b(5) + ~ (0.6)5 = 0 + ~ X0.6 5 = 0.20736 . 3 3 3 (ii) The given difference equation is yen) = x(n) + 6x(n - 1) + 12x(n - 2) + 8x(n - 3)

248

Signals and Systems Taking z-transform of the above equation we get Y(z)

= X( z) + 6z- I X(z) + 12z-2X(z) + Z-3X(Z)

=(l + 6z- 1 + 12z- 2 + Z-3)X(Z) i.e.,

H(z)

= Y(z) = 1+6z-1 + 12z-2 +Z-3 = Z-3 +6z 2 +12z+8 = (Z+2)3 Z3

X(z)

(3)

Z3

Therefore, H(z) has three zeros at z =-1 and three poles at z = O. The pole-zero plot has shown in Fig. E6.3. Im(z) Since the input to the system is x(n) =8 (n) ..

(4)

X(z)=Z{8(n)}=1

+2j

Using Eq. (4), we can write Eq. (3) as Y(z) = 1 + 6z- 1 + 12z-2 + Z-3 :. yen)

=8(n) + 68(n -

We know that 8(n)

1) + 128(n - 2) + 88(n - 3)

I

={o

for

n =0

for

n:;t: 0

-~>-----lIi-----If----

-2

:. y(2)

Re(z)

-2j

:. yeO) = 8(0) + 68(-1) + 128(-2) + 88(-3) = 1 :. y(l) = 8(1) + 60(0) + 128(-1) + 88(-2)

2

=6

Fig. E6.3

=8(2) + 60(1) + 128(0) + 88(-1) = 12

: . y(3) = 8(3) + 60(2) + 128(1) + 88(0) = 8 :. y(4) :. y(5)

=8(4) + 60(3) + 128(2) + 88(1) =0

=8(5) + 60(4) + 128(3) + 88(2) =0

II

SIGNIFICANT

Sr. No.

,,(n)

1

D(n)

2

u(n)

3

anu(n)

4

-a nu(_ n - 1)

5

nanu(n)

6

-nanu(- n - 1)

7

cos (nw)u(n)

II

POINTS ROC

X(Ii)

1

Complete z-plane 1

1-z-1 1 1 -az -1 1 1 -az -1 az- 1 (1- az- 1 )2

az - 1 (1-az-1 )2 l-z- 1 cosw

1- 2z- 1 cosw +z-2

Iz I > 1 Izl > lal Izl


lal Iz I < lal Iz I > 1 (Contd ... )

z-Transform Sr. No.

x 1

1_2z-1 cosw+z- 2

~

1:-

(~r1 cosw

1-2(~r1 cosw+(~J2

(~J1 sinw 1-2(~J1 cosw+(~r2 l_aNz - N

N-l

249

1-az -1

Izl

>

lal

Izl

>

lal

Izl

>0

SHORT QUESTIONS AND ANSWERS Q1. What do you mean by z-transform of the discrete signal x(n) ? ADS.: The z-transform of discrete time signal x(n) and it is given by X(z)

= Lx(n) z-n 11=- 00

where z is the complex variable and the z-transform of x(n) is also represented by the operator Z i.e., X(z) = Z[x(n)]

Here X(Z) and x(n) are known as z-transform pair which is represented as x(n) ~ X(z)

Q2. What do you mean by Region of Convergence? ADs.: The region of convergence are the values of z for which the z-transform converges. Eq. (6.1) shows that z-transform is an infinite power series which is not always convergent for all values of z. Therefore, the region of convergence should be mentioned along with the z-transformation. Q3. State the following properties of z-transform: (i) Linearity (ii) Time shifting (iii) Time reversal and (iv) Differentiation in z-domain.

ADs.: (i) The linearity property states that if x\(n) ~ X\(z) and x 2(n) ~ X 2(z) then a\x\(n) + a2x 2(n) ~ a1X\(z) + a2XzCz) where a\ and a 2 are constants.

(ii) The time shifting property of z-transform states that if x(n)

~ X(z) ,

then we can write

x(n - k) ~ Z-kX(Z) where k is an integer which is the shift in time in x(n) in samples.

(iii) The time reversal property of z-transform states that if x(n)

~

X(Z-l)

ROC: ~ I (ii) x(n) = [-2 + (0.5)"]U(- n - J) , ROC: Iz 1 .l 2 (ii) x(n) =

u(n) -

(I4)"

3 -

u(n)

(- 2I)" u(n), ROC: Izl >-2I ]

252

Signals and Systems

Q9. Find the expression for yen), the linear convolution of x(n) and hen) for the following combinations of x(n) and hen) where a, band m are constants with 0 < a, b < 1 and yen) is a unit step sequence: (i) x(n) = anu(n) and hen) = o(n + m) (ii) x(n) = anu(-n) and hen) = bmu(-n) [Ans.: (i) yen)

=all +lnu(n + m), (ii) yen) = ab ll -

ball u(-n)] a-b

QI0. Find inverse z-transform of the following: X(z)

!)

= (z _~) (z_

17 ( 4" 1 )11 u(n) ] 3 5 ( 1 )11 u(n)-16 [ Ans.:x(n)=O(n+2)+4"0(n+l)+"2"2

QU. If the difference equation of the system is yen)

=0.5 yen -

1) + x(n), find the following:

(i) System functions (ii) Pole-zero plot of the system function and (iii) Unit sample response of the given system Im(z)

[ Ans. : (i) H(z) =

1- 0.5z- 1

(ii) --+1+---*-.... Re(z) (iii) hen) = (0.5)2 u(n)] 0.5

Q12. For the following transfer function of the LTI System H(z) =

3 __

1- 3.5 z

;Z-I+

-2' find the ROC of H(z) and determine unit sample response of hen) for the

1.5z •

following conditions: (i) stable system (ii) causal system and (iii) anticausal system 2

[Ans.: H(z)

= 1- 3z-1

(i) hen)

=-2(3)llu(-

+

1

1- 0.5z-1

n .- I)

+ (0.5)"u(n)

ROC: 0.5 < Izl < 3, (system stable) (ii) hen)

=2 (3)"u(n) + (0.5)" u(n)

ROC : Izl > 3 (system unstable) (iii) hen)

=-2(3)"u(- n -

1) - (0.5)"u(- n - 1)

ROC: Izl < 0.5 (system unstable)]

z-Transform MULTIPLE CHOICE QUESTIONS 1. z-transfonn X(z) of the sequence x(n) is given by (a) X(z)

(b) X(z)

= Lx(n)z-n

= Llx(n)lz- n

(d) X(z)

= Lx(n)z-n

o

(c) X(z)

o

= Lx(n)z-n

2. z-transfonn of the signal {... , 0, ~, 3, 2, 0, 0, 0, ...} is (a)

1+1.+~ z Z2

(b)

1+1 z

(c)

3. The ROC ofQ2 is (a) whole of z-plane (c) whole of z-plane except of z =0

1+1+2. Z

Z2

(b) whole of z-plane except z =00 (d) none of these

4. z-transfonn of 1

(b)lzl-4

=

(a) 0,0

(d) X(-ll) H

_z_, then Z[u(-n)] is

(a) - z I-z

11. If X(z)

X~Z)

2

2°.5z (z -1)(z - 0.95z

+ 0.45)

(b) 0,0.5

z-1

(d)

Z2

z-1

(d) Izl , 0, then the Z transform of sequence x(n) is

z

1 (a) - z-3 13. If x(n)

(b) - z-3

(c)

1 (z-3)

(d)

1

1 2 (z -3)

= 3'" for n < 0 and x(n) = for n > 0, then the z transform of sequence x(n) is

(a) _3_

(b)

3- z

_z_

(c)

z- 3

1

(d)

z- 3

Z2

(z - 3)2

14. Z transform of [ax(n) + by(n)] is (b) ax(n) + y(n) (d) none of these

(a) ax(n) - by(n) (c) ax(n) + by(n)

15. Z-transform of x(k ± ko) is (a) z± kOx(Z) (b) Z±kOx(Z)

16. If X(z)

= (l-az3az-\ 3 )

Ml

=

7

1 + Z- I .

l--z 12

_I

1

+-z

-2'

~O

Answers

1. (e) 7. (d) 13. (a)

-~ 3

~=

the poles of H(z) are at

12

1 1 (a) z = - and z = -3 5 (c) z =

(d) none of these

and lal < Izl, then the initial value of x(n) is

W2 17. If H(z)

(c) Z-l X(z)

(b)

and z =L

z= 1 3

and z = _

-.L 4

(d) z= ·landz=± -l

4

3

4

J

2. (e)

3. (e)

8. (d)

9. (e)

14. (e)

15. (a)

4. (a) 10. (b) 16. (e)

5. (a) 11. (e) 17. (e)

6. (e) 12. (b)

7 CONVOLUTION

In

Fourier transform, Laplace transform and z-transform we have already introduced convolution theorems as a part of their properties. The actual significance of convolution theorem as well as its use in digital signal processing have not been introduced. The aim of this chapter is to introduce convolution in detail.

7.1

CONVOLUTION THEOREM FOR CONTINUOUS SYSTEM

If the impulse response of an analog or continuous time system be h(t), it means that for an input 0(1) its output is h (I), Hence for linear system its output for a shifted impulse 0 (t -7) is h (I -7). The following are the mathematical representation of these operations: o(t) ~ h(t)

o(t - 7) ~ h (t-7)

x (t)o(t - 7) ~ x(7)h(t - 7)

f 00

Obviously,

f 00

x(T)o(t-T)dT

~

x(T)h(t-T)dT

From last two statements, we can say that for an input x (7)0(1 -T) the output is x (7)h(t -7) and for an input of continuous summation or integration of x(T) 0 (t -T), the output is also continuous summation of x (T) h(t - T). Therefore, we can write by the linearity of the system

f 00

y(I) =

x(7)h(t -7)OT

(7.1 )

The right-hand side of Eq. (7.1) is defined as convolution of x(t) with h(t) and hence we can say that the output y(t) is the convolution of h(t) and x(1) .

,,

yet)

= Convolution of h (t) and x (I) = x (t)*h(t)

f

(7.2)

00

=

x(7)h(t -DdT

The symbol .*, in Eq. (7.2) stands for convolution.

(7.3)

256

Signals and Systems

7.2 PROPERTIES OF LINEAR CONVOLUTION The linear convolution is the convolution process for the continuous time systems. The convolution of continuous time (CT) signals possesses the following properties: Commutative Property: xl(t)*x2(t) = x2(t)*xl(t) Associative Property: XI(t)*[X2(t) *x3(t)] = [x l (t)*X2(t)] *x3(t) Distributive Property: xl(t)*[x2(t) + x3(t)] = XI(t)*x2(t) + xI(t)*x3(t) If any function is convolved with impulse function 8(t), it results the function itself. Again, convolution of any function with 8 (t - to), it results the shifted function. If the original function is impulse, this is also true. Mathematically, we can write x(t)*o(t) = x(t)

= x(t -fo) 8(t)*8(t) = 8(t)

x(t)*8(t - to)

8(t)*8(t - to) = 8(t - to) 8(t - t l )*8(t - t 2)

7.3

= 8(t -

tl - t 2)

GRAPHICAL CONVOLUTION

In this article graphical convolution is explained here. Let us consider two sided signal. Let us consider a signal shown in Fig. 7.1. let XI(t) and x2(t) are identical to x(t). Therefore,

f ~

X I(t) * x1(t) =

XI (A)X1(t

- A)dA for a given t.

X1(A)

x(t)

IA I

-T Fig. 7.1

+T Signal x(t)

I

-T Fig. 7.2

+T

• A

Plot of x dA)

The meaning of the words 'graphically process' is to find XI(A), x2(t - A) and to multiply these two. The multiplication of these two gives the area under the func'tion for a given 't'. From Fig. 7.2, XI(A) is nothing but x(t) with the name of x-axis changed from t to A. In Fig. 7.3, x2(A) is the function x(t) and X1(-A) is the mirror image of Xz{A) shown in Fig. 7.4. In Fig. 7 .5(b) xl( -A) is advanced to the right by an amount '(' giving x2(t-A) for positive '!' and Fig. 7.5(c) shows the plot of multiplication of XI(A) and Xit-A). The area under XI(A) and x2(t - A) is given by =

f

XI (A)X2 (t -Il)dll

= Area of rectangle in Fig. 7.5

Convolution

257

X2(A)

_

_

-----1_ _ _ _ _ _ _ ---.--J _ _ _ __

-T

Fig.7.4

Pial of XiA)

Fig. 7.3

----'_ _ -------.-

+T

A

Pial of x2 (-A)

This will continue upto t =2T. For t > 2T: The area under XI(A) and xit-A) will be zero because x2(t-A) will go out of the range of XI(A). Therefore, for positive 't', the convolution is given by

yet) = XI (t)*xi t ) = A2 (2T -

(7.4)

t)

For negative t. this process gives the same result except that xi -A) in Fig. 7.4 will be shifted to the left only. y(t) =A 2 (2T

+ t)

(7.5)

where t has been changed by -t in Eq. (7.4). X1(A)

A (a) ;

,

-T

,,,

---

(b)

-T -T+ I

y(l)

T

(e)

-T+ I 0 Fig.7.S

o

-2T

T

Produclofx t (A)andx2(I-A)

Fig. 7.6

+2T

Plot of y(l) = x,(l) • x2 (I)

Therefore from Eqs. (7.4) and (7.5) we can write that y(t) = {

A 2 (2T-t)

for t > 0 but t < 2 T

A2(2T+t)

for t < 0 but t > -2T

(7.6)

258

Signals and Systems

Compactly, we can write that yet) = {

Fig. 7.6 gives the plot of yet)

Example 7.1

A 2 (2T-ltl)

o

for

O0 £

Therefore, sign is negative

Again

8£ - 5 2.5x---£ lim _---",---'£"-=-__ 8£ - 5

=

lim 2.5(8£ - 5) - £2 -> 0 8£ - 5

E

= lim £ -> 0

-12 .5 -5

= 2.5

£

Therefore, the array is given by S5 S4

4 2

8

s3

£

2.5

S2

-00

sl

2.5

sO

3

0 0

0

0

0 0 There are two changes of sign in first column and hence the system is unstable. Method 2: I Put s =Z The polynomial becomes s5 + 2s4 + 4s 3 + 8s2 + 3s +1 =0

279

280

Signals and Systems I 2 4 8 3 +-+-+-+-+1 =0 Z5 Z4 Z3 Z2 Z

or;

-

or,

Z5 + 3z4 + 8Z3 + 4Z2 + 2Z + I

=0

: . Routh's array becomes 8

2

3

4

I

6.67

1.67 0

3.25

I

0

- 0.382

0

0

0

0

There are two changes of sign. Hence the system is .unstable.

Example 8.7

Examine stability of s5 + 2s4 + 2s 3 + 4i + 4s + 8 = 0 using Routh's method.

Solution

S5 05 4

s3

I 2 4 2 4 8 10 0 0 Routh's array failed

To eliminate the above difficulty, form on equation by using the co-efficients of arrow which is just above the row of zeros. The equation is known as auxiliary equation. ..

A(s) dA(s) ds

=2s4 + 4s2+ 8 =8s 3 + 8s

The Routh 's array becomes S5

I

S4

2

s3

8

3 4 8 8 0

s2

2

8 0

Sl

-6 0

2

sO

8 There are two change of sign. The system is unstable. Example 8.8 Examine the stability of 054 + 5s 3 + 2i + 3s + 2 =0 using Routh's method. Find the number of roots lying in the right half of the s-plane.

Solution 2 2 3 0

2 -4.14 0 SO 2 There are two changes of sign. Hence the system is unstable. There are two roots in the right half of s-plane. Sl

Stability

281

Example 8.9 Examine the stability of the system having characteristic equation s5 + S4 + 3s3 + 3i + 4s + 8

=0

Hence Routh's array fails. Method 1:

S5

I

3

.1'4

2

3 8

s3

£ 3£-4 £

S2

lim

£

_ 4( 3£ : 4 ) - 8£ _---'----;;-=-c----;_ _

£ -> 0

Routh's array becomes

3£ + 4

0 0

3£+4 £

sO

4 · 3£ +I1m

0

-4C££-4 )

sl

£->0

-4

4

8

= 3 + I'1m -4 = £->0 £

00

2

= lim

-4(3£ + 4) - 8£ = _ 4 £ -> 0 3£ + 4

£

3

.1'5 .1'4 .1'3

£

i

+00

.1'1

-4

4

3 8 -4 0 8 0 0

8 There are two changes of sign. Hence the system is unstable. There are two roots lying in the right half of s-plane Method 2: .1'0

1 Put s =Z or, or,

1 1 3 3 3 -- + -- + - - + .~ + - + 8 = 8 Z5 Z4 Z3 Z2 Z 8Z s +3Z 4 + 3Z 3 + 3Z 2 + Z + 1 =0

282

Signals and Systems

Routh's array becomes

Z5 Z4 Z3 Z2 Zl ZO

1 3 3 1 -5 1.67 0 2 1 0.83 0 8 3

There are two changes of sign. Hence the system is unstable. Those are two roots lying in the right half of splane.

Example 8.10 stable.

Determine range of vales of 'K' so that system having characteristic equations will be S

(s2

+ 2s + 3 ) (s + 2 ) + K =0

The given characteristic equation is

=0 S(S3 + 4i + 7s + 6) + K =0

s(s2

or, or,

+ 2s + 3 ) (s + 2) + K

S4

+ 4s 3 + 7 S2 + 6s + K = 0

The Routh's array becomes S4

s3 s2

1 4

sl

5.5 33-4K 5.5

sO

K

7 6 K

K 0 0

0

The system will be stable if no changes of sign occurs in the first column

.. and

K>O 33-4K >0

5.5

or,

33 -4K > 0 4K < 33

or,

..

33 4

K 0 or, T < 2

Ans.

Stability

S6

s3

I 7 14 8 2 10 8 0 2 10 8 0 0 0 0 0

A(s)

=2S4 + lOi + 8 =0

s5 S4

f-

287

Row of Zero

dA(s) 3 - - =8s +20s ds Routh's array becomes s6 s5 S4 s3 s2 Sl sO

I 7 14 8 2 10 8 0 2 10 8 0 8 20 0 0 5 8 7.2 0 8

There are no changes of sign and hence the system is stable. Since there is a row of zero, system may be marginally stable or unstable. To verify this, let us solve A(s) =0 2S4 + lOi + 8

..

=0

i=

-1O±JI0 2 -64 4

=

-1O±J36 4

=

-1O±6 4

i=-I,-4

s

-

=±j, ±j2

Since the roots on imaginary axis are not repeating, the system is stable.

Stability

SIGNIFICANT

POINTS

If a linear time invariant system satisfies the following conditions:

-

• even after excitation by a bounded input, output must be bounded, and • in the absence of input, output must be zero despite the initial conditions, the system is said to be stable.

Absolute Stability The stability of the system with respect to variation of parameters is known as absolute stability and it said to be absolutely stable when it is stable for all values of the parameters. Relative Stability It measures the degree of stability and it is a quantitive measure of how fast the transients die out in a system.

288

Signals and Systems

Conditional Stability conditional stability.

The stability of the system with respect to its parameter variation is known as

Hurwitz's Criterion Hurwitz had given the sufficient conditions for having all roots of the characteristic equation in left half of s-plane. The disadvantages of the Hurwitz's criterion are: (i) It is very complicated and time consuming for solving higher order systems. (ii) This method is unable to judge the exact number of poles located in right half of s-plane. (iii) It is very tough to predict marginal stability.

Routh-Hurwitz Criterion It determines the poles of a characteristic equation with respect to the left and right half of the s-plane without solving the equation. Any change of sign in the first column of Routh's array indicates (i) the system is unstable, and (ii) the number of changes of sign gives the number of roots lying in the right half of s-plane.

SHORT QUESTIONS AND ANSWERS Q1. What do you mean by stability? If a linear time invariant system satisfies the following conditions: • even after excitation by a bounded input, output must be bounded, and • in the absence of input, output must be zero despite the initial conditions, the system is said to be stable. Q2. State Hurwitz's criterion and give its disadvantages. Hurwitz had given the sufficient conditions for having all roots of the characteristic equation in left half of s-plane. The disadvantages of the Hurwitz's criterion are: (i) It is very complicated and time consuming for solving higher order systems. (ii) This method is unable to judge the exact number of poles located in right half of s-plane. (iii) It is very tough to predict marginal stability. Q3. State Routh-Hurwitz Criterion. It determines the poles of a characteristic equation with respect to the left and right half of the s-plane without solving the equation. Any change of sign in the first column of Routh's array indicates (i) the system is unstable and (ii) the number of changes of sign gives the number of roots lying in the right half of s-plane.

EXERCISES I. Find the stability of the system by Routh's criterion. The characteristic equation of the system is s4 + 6s 3 + 26i + 56s + 82

=0

[Ans.: Stable)

Stability

289

2. If the characteristic equation of the system is s3- - 4s 2+ s + 10 =0, find the stability of the system by Routh's criterion. [Ans. : Unstable] 3. If the characteristic equation of the system is s6 + 4s 5 + 3s4 -16i - 64s - 48 =0, find the stability of the system by Routh's criterion. Also find the number of roots of this equation with positive real part, zero real part and negative real part. [Ans.: Unstable, Roots with positive Real part is one, Roots with negative Real part are three, Roots with zero Real part are two] 4. Determine range of values of K, marginal value of K and frequency of sustained oscillations for unity feedback system having G(s)

K =- - - ---

s(1 + O.4s)(1 + 0.25s) [Ans.: 0 < K < 6.5, Kmar = 6.5, frequency of oscillations is 3.162 rad/sec.] 5. It a unity feedback system is marginally stable and oscillates with frequency 4 rad/sec., determine Kmar and q where G(s)

4 = ----::2,-------(s +qs+2k)s

[Ans. : Kmar = 8, q = 0.25]

MULTIPLE CHOICE QUESTIONS 1. A positive feedback signal improves the performance of automatic control system. (a) false (b) true

2. For given system Routh 's array is given below: 3 2 7

S4 s3

i Sl

-5

sO

7

The system is (a) marginally stable

7 0

(b) unstable

(c) stable

(d) conditionally stable

3. The number of sign changes in the entries in the first column of Routh 's array denotes (a) the number of roots of characteristics polynomial in RHP (b) the number of open loop poles in RHP (c) the number of zeros the system in RHP (d) the number of open-loop zeros in RHP 4. The transfer function of a system is the valid condition is (a) a3' a2' ai ' ao > 0 and a2 a l (c) a 3, a 2, ai' ao > 0 and a2 a l

-

-

3

K 2

a3s +a2s +als+ao

a3 ao < 0 a 3 ao =0

.

For the system to be absolutely stable,

(b) a 3, a 2, ai' ao > 0 and a2 a l (d) a2' ao > 0 and a3' a l < 0

-

a 3 ao > 0

290

Signals and Systems

5. A closed loop system is shown below. Find the valid condition for which the system will be very much underdamped. + R(s) ----1-(

Fig. 1

(a) T)

= T2

(b) T) < T2

(c) T)

> T2

(d) T]« T2

6. Routh's array of characteristic equation is given below. The number of roots lying on the right-hand-side of s-plane is:

S4

9

s3 S2

-4

8

)

2

s SO (a) 0

(b) 1

(c) 3

(d) 2

7. Routh's array is given below

s6

1

s5

3 18

15

S4

3 18

15

s3

0

The auxiliary equation of this array is (a) 3s4 + 15i + 9 = 0 (c)

3s4

+ 18i + 15

=0

9 23

15

0 (b) 2s4 + 18s2 + 15

=0

(d) s4+9i+15=O

8. If some pole of a system lies on the imaginary axis, the system is (a) absolutely stable (b) conditionally stable (c) marginally stable (d) unstable 9. The main cause of absolute instability is (a) error detector where the two signals are compared (b) parameters of the controlled system (c) parameters of the controlling system (d) none of these 10. The characteristic equation of a feed back control system is

s3 + Ki + 5s + 10 =0

For the system to be critically stable, the value of K should be WI M2 M3

~4

Stability

291

II. The open loop transfer function of a unity feedback control system is G) (s

K(s + 1O)(s + 20) = -----,:-----s2(s

+ 2)

The closed loop system will be stable if the value of K is (a)

2

(b) 3

(c)

4

(d) 5

12. A feedback control system shown below is stable for all values of K, if +

K(1 +Ts)

R(s) ----i~1

s2(1 + s)

I---_-C(s)

Fig. 2

(a) T

=0

(b) T < 0

(d) 0 < t < I

(c) T> I

13. The characteristic equation of a unity feedback system is given by S3

(a) (b) (c) (d)

The The The The

system system system system

+ i + 4s + 4

=0

has one pole in the RH s-plane. has no poles in the RH s-plane. exhibits oscillatory nature is asymptotically stable

14. The control system shown in the figure below has an internal rate feedback loop. The closed loop system for open and closed conditions of switch will be respectively +

K

R(s)----i~1

s2(s + 10)

I--_-+-- C(s)

Switch

Fig. 3

(a) unstable and stable (c) stable and unstable

(b) unstable and unstable (d) stable and stable

15. For the block diagram shown below, the limiting values of K for stability of inner loop is found to be X < K < Y. The overall system will be stable if and only if (a) 4X' < K < 4Y (b) 2X < K < 2Y (c) X < K < Y

X

(d) -

2

Y

< K ---()-----o-.......--o-----o-----o

----~-an-1

-a n -2

yes)

-an -3

Fig.9.3

SFG

With help of the above representations, the Eq. (9.16) can be written as

xI

'\-2

x n_1 XII

0

1

0

0

=

...... 0

..... .I

-an_I

= Ax + Bu

-a n- 2 . .. -a l

0 0

xI

x2

+

.. .... 0 0

X

...... 0 ...... 0 ...... 0

-an

..

0

XII _ I XII

[u]

(9.17)

0 b (9.18)

State Variable Approach (Continuous Systems)

299

The matrix for output equation of the system is given below: xI

X2 y= [1 00 ........ 0]

y

+ [0] [u]

(9.19)

=ex + Du

(9.20)

The matrix A has all ones in the upper off-diagonal and other elements except the last row. The co-efficient of the last row has negative co-efficient. This form is known as the Bush or Comparison form. The other name is phase variable Controllable Comparison form (CCF).

Example 9.3

Express the following transfer function in CCF form. Draw the signal flow graph. Y(s) U(s)

=

3 s4+2s 3+3s+2

From given transfer function we can write (s4 +2s 3 +3s+2) Y(s) = 3U(s)

or, Let

d 4y dt 4

d 3y

dy

=3u XI = Y · dy XI = - = x 2 dt

(I)

· d 2y x2 = dt 2 = x3

(3)

· d 3y 4 X3 = dt 3 = X

(4)

· d 4y d 3y dy X4 =-4 = 3u-23 -3--2y = 3u - 2x1 -3X2 -2x4 dt dt dt

(5)

+ 2 dt 3 + 3 dt + 2y

(2)

uo~-------_--l(n)x2(n)

2n

n=-oo

11.3

z

X l (}.,)X (m)dm

-Tr

INVERSE FOURIER TRANSFORM

The inverse Fourier transform is given by TC

x(n)

= J _ JX(OJ)e-jwndOJ

(11.9)

2n -TC

Since X (ro) is the continuous function of ro, the integration of the above equation is possible. Since the range of ro is from -n to + n, the limits of integrations are from -n to + n.

11.4

MAGNITUDE I PHASE TRANSFER FUNCTIONS USING FOURIER TRANSFORM

The output of a linear time invariant system is given by y(n)

= Lh(k) x(n -

k)

(11.1 0)

k=-~

If the system is excited by sinusoid phasor x(n) = e jwn for - 0 0 < n < y(n)

i.e.,

=

~

Lh(k)e-jW(n-k)

00.

The output becomes

= Lh(k)e-jwne - jWk ~

(1l.l1)

Discrete Fourier Transform and Fast Fourier Transform

= Lh(k)e- JWk

361

~

where

H(w)

(I1.12)

k=- oo

H(w) is called Fourier transform of h(k) where h(k) is the unit sample response. It is also called transfer function of the system which is a complex valued function of w in the range -1! ~ W ~ 1!. We can also express the transfer function H(w) in polar form as H(w)

= IH(w)1e-JLH(W)

(11.13)

where IH(w)1 and LH(w) is the magnitude and angle of H(w) respectively. Using Euler's identity we can write Eq. (11 .12) as ~

H(w)

= Lh(k) (cos wk- jsin wk) k=-lXl

= Lh(k) cos wk- j Lh(k) sin wk k=-~

(11.14)

k= -~

The real and imaginary part of H(w) are given by ~

HR(w)

= Lh(k) cos wk

(11.15)

k= - oo ~

and

H/(w) = Lh(k) sin wk

(11.16)

k=-~

where

(11.17)

and

(11.18)

11.5

DISCRETE FOURIER TRANSFORM

We know that the Fourier transform of discrete sequence x(n) is given by X(w)

=

Lx(n) e- Jwn n=-oo

where X(w) is the continuous function of wand the range of w is from -1! to +1! or 0 to 21!. Since X( w) is the continuous function of w, it is not possible to calculate it on a digital computer or a digital signal processor. Hence it is necessary to calculate it only for discrete values of w.When Fourier transform is calculated at discrete points, it is called discrete Fomier transform (OFT) which is denoted by X(k) and is expressed as

= L x(n)e N-J

X(k)

n=O

_J2Trkn N

where k = 0, 1, 2, ....... , N-1.

(11.19)

In Eq. (I1.19) X(k) is called OFT which is computed at k = 0, 1,2, .... .. ., N-I i.e., N discrete points. Therefore, X(k) is the sequence of N samples. The sequence x(n) can be obtained back from X(k) which is known as inverse discrete Fourier transform (10FT) which is given by

362

Signals and Systems N-J

x(n)

j2trkll

= IX(k)e-N -

where n

11=0

=0, 1,2, ....... , N-1.

(11.20)

where x(n) is a sequence of N samples. Therefore, both X(k) and x(n) contain N samples. Let us put

WN

=e

- j 2tr

(11.21)

N

which is known as the twiddle factor. Therefore, we can write Eq. (11 . 19) and (11.20) as

=I

N-J

X(k)

x(n)W~"

where k = 0, 1,2, .... . .. , N-1.

( 11.22)

X (k)WNkn where n = 0, 1,2, .. . .... , N-1.

(11.23)

11=0

I

N-J

x(n) =

and

11=0

We can represent sequence x(n) as a vector x N of N samples i.e.,

XN

n=O n= I n=2

=

X(O) X(l) X(2)

(11.24 )

n=N-I X(N)

NxJ

We can also represent X(k) as a vector XN of N samples i.e.,

XN

k=O

X(O)

k=1

X(l)

k=2

X(2)

= .

k = N-I X(N)

Therefore, it is possible to represent

W~ n

as a matrix

(11.25)

Nxl

[wN ] of size N x N as follows:

n=O n=l n=2

WN =

k=O

W~

k =1

w~

k=2

W~

k= N-I w~

W~

w~

wk wh wh wt

wff- 1 W~(N-l) .... . ....

n=N-l

W~

wff-1 W2(N-1J

(11.26)

N

W(N-l)(N - l) N

NxN

363

Discrete Fourier Transform and Fast Fourier Transform

The individual elements in Eq. (11.26) are represented by W kn with ' k' rows and 'n ' columns. Therefore, N we can represent the N - DFT as follows:

X N = [WN ]X N

(11.27)

The IDFT of equation (11.27) is represented by (11.2g)

= WNkn

W~

where

(11.29)

To check the periodicity property of WN let us put N

= g in Eg. (11.22). We get

-j2rr

W8=

e

8

- jrr

=e

4

(11.30)

I 2 15 Tabie 11.3 shows values of WgO , Wg , Wg , ...... , Wg

kn

o

kn -

8

-

e

-j.!!xkn 4

o _ eO

Comments Phasor magnitude 1 and angle 0

8 -

1

Phasor magnitude 1 and angle _ Ii 4

2

Phasor magnitude 1 and angle _ Ii 2 3 _

3

8 -

4

8 =

5

4

5 _

8 -

e e

e

3 4

- j .-

- jf(

Phasor magnitude 1 and angle _ 3n 4 Phasor magnitude 1 and angle

--It

.51r

-j-

4

Phasor magnitude 1 and angle _ 5n 4

6

Phasor magnitude 1 and angle _ 3n 2

7

Phasor magnitude 1 and angle _ 7n 4

8

8 8 =

e

-j21r

Phasor magnitude 1 and angle -2n (Contd ... )

364

Signals and Systems

(Table Contd .. .) -j.!!..xlen

len

kn _ e 8 -

9

i =e 10 _

8

11

811=e

12

12 8

13

13 _ 8 -

-j( 211+!:) 4

Phasor magnitude 1 and angle - (211" + ~ ) Phasor magnitude 1 and angle -(211" + %-)

_j( 2.11+ 311" ) 4

=e e

Phasor magnitude 1 and angle -(211" + 3: )

-j(2" +1r)

Phasor magnitude 1 and angle -(211" + 11")

_j( 211+ 511") 4

Phasor magnitude 1 and angle -(211" + 5:_)

_j( 2Jr+ 3; )

14

14 8

15

13 _ 8 -

Example 11.1

Comments

e-j( 211+!:.2 )

10

-

..

Phasor magnitude 1 and angle -(211" + 3; )

=e e

_j(211"+711") 4

Phasor magnitude 1 and angle -(211" + 7: )

Find DFT of the following sequence: X(n)

={I , 2, 3, 4}

The matrix [w 4 ] will be of 4 x 4 size and its individual element is given below: n=O n=} n=2 n=3 k =0

W4 =

k

=1

k=2 k=3

W40

WO4

WO4

W40

WO4 WO4

Wi4

W42

W43

W44

W46

WO4

W43 W46 W49

-[: -

1

-j -1 1 -1 1 1 j -1

Here we have

x,=[i]

Since

x4

x,

wl

)1] -}

= [W4] x4 we have

=[I

}

-j -1 -1

j

1 -1

lJm

Discrete Fourier Transform and Fast Fourier Transform

1+2+3+41 [

=

11.6

365

10

1-2j -3+4j

-2-2j

1+2)-3-4) [-2-2J =

1-2+3-4

-2

PROPERTIES OF DISCRETE FOURIER TRANSFORM

There is similarity among most of the properties of OFT and z-transform due to existence of some relationship of each other. OFT also differs in some properties like circular convolution property. The following properties play an important role in practical techniques for processing a signal: • • • • • • • • • •

11.6.1

Periodicity Linearity Circular symmetry of a sequence Symmetry properties Circular convolution of two sequences Time reversal of a sequence Circular frequency-shift of a sequence Circular correlation of two sequences Multiplication of two sequences Parseval's theorem

Periodicity

If a discrete time signal x(n) is periodic, its OFT will also be periodic in nature. Let x (n) and X(k) be the OFT . . OFT pair I.e., x(n) H X(k) N If

x (n + N) == x(n) for all n, then X(k

Proof: The OFT of x(n) is given by

+ N)

=X(k) for all k.

X(k)

= L x(n)W~n N- J

(11.31)

n=O

Therefore,

X(N

+ k)

N-J

N- J

n=O

n=O

= Lx(n)W~N+k)n = Lx(n)w~nw:n -j2tr

Nn _

Now,

N

-

e

-nN N

.

= e- J2Tr11

= cos 27m -

j sin2n n = I

(11.32)

(11.33)

Using Eq. (11.33), we have from Eq. (11.32) as X(N

+ k)

=

Lx(n)W~n = N-J

n=O

X(k)

( 11.34,)

366

Signals and Systems

11.6.2

linearity

property we can write DFT

(/jX j (n)+a 2X 2(n) H

where aj and

(/2

a j X j (k)+a2 X 2(k)

N

are constants.

Proof: The OFT of x(n) is given by N-J

X(k)

k

= L,x(n)WN n n=O

Here Therefore.

X(k)

N- J

N-j

N-J

n=O

n=O

n=O

= L,[aJxJ(n)+a2x2(n)]W~n = L,aJxJ(n)W~n + L,a2x2(n)W~n N-J

N-J

n=O

n=O

=al~>J(n)W~n +a2Lx2(n)w~n Hence, we can write that DFT

a Jx J(n)+a2 x 2(n) H

N

a JX J(k)+a 2 X 2(k)

(11.35)

11.6.3 Circular Symmetries of a Sequence We know that x(n) is the sequence containing N simples and X(k) is the N point OFT. If we take 10FT, it gives a periodic sequence xp (n) which is related to x(n) as, ~

xp (n)

= Lx(n-IN)

(11.36)

I=-~

Therefore, x p (n) is a periodic repetition of x(n) which contains such infinite periodic repetitions. Wf; can conclude that the OFT of xp(n) gives the same OFT X(k) which is obtained due to x(n) i.e. x(n)

DFT H

X(k)

(11.37)

X (k)

(11.38)

N xp (n)

= x(n)

DFT H

N

where x(n) and 9n) are related by Eq. (11.36) Also, we can write for 0

~

n $. N - 1

otherwise

(11.39)

Discrete Fourier Transform and Fast Fourier Transform

367

Let xI' (n) be shifted by k units to the right giving the new sequence x;,(Il) i.e.

( IIAO)

x;(n) = xI' (n - k)

(IIAI)

= Lx(n-k-IN) 1= -=

The corresponding sequence x'(n) can be obtained from Eq. (11.39) as X'(n) =

X;, {

for O::S; n ::s; N -I

(n)

o

( IIA2)

otherwise

The sequence x'(n) is related to x(n) by the circular shift which has been illustrated in Fig. 11.1; Fig. 11.1 (a) shows sequence x(n) having N = 4 samples. Figure 11.1 (b) shows the periodic repetition " IX(/, i' x/n). Due to shifting of x/n) to the right by two samples, it results the sequence x;(n) The sequence x'(Il) is shown in Fig. ll.l (d) as per Eq. (11.42). The sequence x'(n) is basically one period of ';'(n) from 0 S II S 3. I_ 05n53 _I

x(n)

t

4

3 2

______________-+__

~~

J________ n

__

023 (a) The sequence x(n) between 0 5 n 5 N - 1 Here N = 4

4

4

4

3

3 2

3

2

-4 -3 -2 -1 0

2

2

3

4

5

6

7

(b) xp(n) is periodic repetition of x(n) after N samples. x(n) and xp(n) results in same DFT X(k)

x'p(n) = xp(n-2)

4

4

3

4

3 2

-2 -1 0

3 2

2

3

4

5

6

7

(c) The sequence xp(n) is shifted to right by two samples. Observe new sequence xp(n) formed in interval 0 5 n 5 N - 1

368

Signals and Systems I,O$n$3' 'In)

1

4

1

3

2

--------------~~--~~----~

o

2

3

n

(d) Sequence x(n). It is related to x(n) by circular shift of two samples

Fig. 11.1

The sequence x'(n) is related to x(n) by a circular shift which is given below. x'(n)

=x(n -

k, modulo N)

(11.43)

The shorthand notation for (n - k, modulo N) is (n - k)N i.e. x'(n)

=x(n -

k )N

Using Eg. (11.44), let us evaluate samples of x'(n) for k

(11.44)

=2 and N =4. the equation (11.44) becomes,

=x(n - 2)4 x' (0.) =X(-2)4 =x(2)

(11.45)

x'(n)

Hence,

}

(11.46)

x'(I)=x(-1)4=x(3) x' (2)

=x(0)4 =x (0)

x'(3)=X(1)4=x(1) wherex'(n) is obtained by shifting x(n) circularly by two samples. To understand the relationship represented by Eq. (11.46), let us plot x(n) anti-clockwise along the circular as shown in Fig il .2(a). Figure 11 .2 (b) shows the sequence x(n) delayed circularly by one sample, i.e., x(n-I)4 where this shift is anti-clockwise. Figure 11.2(c) shows the sequence x(n) delayed circularly by two samples, i.e., .t(n - 2 )4 which is the sequence x'(n) as per Eq. (11.46). Therefore, all the relations of Eq. (11.46) are satisfied. x(1)

=3

x(O) = 4 __ - -0- - _______ ,

___ - -0- - ______ ,,'.

, ,, ,,, ,

x(2)

=2

,

,, ,

¢, x(O)=4

Q

,, ,

,,

I

,,

,

---- -- - 0 - - - - --,' x(3) = 1 (a) The samples of sequence x(n) plotted circularly anticlockwise i.e., x((n»4

I

,,

,

,, ,

, I

x(1)=3

Q,

,, ,

,, ,

¢

x(3)

I

,,

,, ,,

,

------0---x(2)

I

I

,-

=2

(b) The sequence x(n) shifted circularly by one sample anti-clockwise i.e., x((n - 1 »4

=1

Discrete Fourier Transform and Fast Fourier Transform x(3)

=3

x(2) = 2

, __ ---0----

x(O) = 4

,, ,, , , ,, \

___ - -0- -_ -"~"~ll,

x'(1)

, ,,

, \ \

x'(O) ¢ x(2)

,, ,,

, "

,,

,, ,,

\

¢ x'(2) \ \

=2

x(3)

,, ,,

,

\ \

\

=1 Q \ \ \

,, ,

¢x(1)=3

,,

,

,

x'(3)

"---0----

x(O) = 4 (d) The sequence x(n) shifted circularly clockwise by one sample i.e., x«n + 1))4

(c) x«n - 2))4

Circular shift of sequence x(n). Here x'(n) is obtained by shifting x(n) Circularly by two samples

Figure 11.2(d) indicates the sequence x(n) advanced by one sample, i.e., x(n+ I )4' The shift is clockwise.

x(1)

Consider the sequence x(n) x(4-1)=x(l)

=x(2) 3) =x(3)

x(4 - 2) x(4 -

1 $. n $. N - 1

,,

, ,,

, ,,

Circularly even sequence A sequence is said to be circularly even if it is symmetric about the point zero on the circle i.e., x(N - n) = x(n)

=7

_--0--_

11.6.3.1

Here

, ,,

'---0---'

x(1) = 3

Fig. 11.2

369

x(2) = 9

, \

\

¢ \

¢x(O) = 5

,

\

(11.47)

= (5, 7, 9, 7 )

\

,, ,

,

=x(2) i.e., x(l) =x(3)

,, ,, , ,,

"--0---'

i.e .,x(3)=x(1)

i.e., x(2)

,, ,

x(3) = 7

Fig.11.3

Circularly even sequence

This sequence is circularly even which is symmetric about the point zero on the circle shown in Fig. 11.3.

11.6.3.2 Circularly odd sequence

x(1)=-7

A

,_---0---,

sequence is circularly odd if it is not symmetric about x(O) on the circle i.e., x(N - n)

=-x(n)

Consider the sequence x(n) Here

=-x(l) 2) =-x(2) 3) =-x(3)

l$.n$.N-l

x(4 -

(11.48)

= (5, -7, 9, 7)

x(4 - 1) x(4 -

,,

Thus the given sequence is circularly odd.

=-x(l) i.e. x(2) =-x(2)

i.e., x(3)

i.e. x(l) = -x(3)

x(2) = 9

Q

¢x(O)

,, ,,

,, ,, ,

=5

, ,, "

'---0---' x(3) = 7

Fig. 11.4 Circularly odd sequence

11.6.3.3 Circularly folded sequence A circularly folded sequence is represented as x (-n)N which is obtained by plotting x(n) in clockwise direction along the circle. It is represented as,

370

Signals and Systems x(-n)N = x(N - n)

(11.49)

O~n~N-l

The 4 point sequence x(n) has been shown in Fig. 11.5 (a). x(3) _--0-- _

x(1 ) _--0--_

,, ,,

,, ,, , ,,

,, ,,

, ,,

\

\ \

x(2)Q

¢x(O)

,, ,, \

,

,,

, ,,

,, ,,

, \

\

x(2)Q

¢x(O)

\ \ \ \

,,

,

,,

,

,,

,,

,,

,,

,, ,,

'------0-----

----0----

x(1 )

x(3) (a) Th6 sequence x(n) plotted across the circle

(b) The sequence x(n) folded circularly to get x( - n)4

Fig. 11.5

The samples of x(n) are plotted anti-clockwise in Fig. 11.5 (a) whereas Fig. 11.5(b) shows circularly folded sequence x(-n)4' Here x(n) is circularly folded and the samples of x(n) are plotted clockwise in Fig. 11.5(b). The table given below summarizes these operations. N-point sequence plotted across the circle anticlockwise which means positive direction. Sequence x(n) shifted anti-clockwise (positive direction) by 'k' samples. Anti-clockwise shift indicates delay. Circular folding. Sequencex(n) plotted across circle in clockwise direction, i.e., negative direction.

11.6.4

Symmetry Properties

Let x(n) be complex valued which can be expressed as x(n) = xR(n) + j Xl (n) Let its DFT be also complex valued and it is expressed as

o ~ n ~ N-l

(11.50)

X(k) = XR(k) + j Xl(k)

O~k~N-I

(11.51)

Proof:

The DFT of x(n) is given by X(k)

N-J

N-J

n=O

1/=0

- j2lC kn

= Lx(n)W~n = Lx(n)e-NN-J

-j2lCkn

N= L[xR(n)+ jXI (n)] e 1/=0

Discrete Fourier Transform and Fast Fourier Transform

371

. . ( 2,. kn . , 2,. k n) =~I 1~[XR(n)+ jXj(n)] cos N - - j SIn-N -

N-l[

kn +xjsm. 2n kn] -J£... .~[xR(n)sm . 2n = I xR(n)cos -2n ----- - kn-- - x ,cos -2n -- kn] --11=0

N

N

N

11=0

=X R (k) + JXj (k)

N

Comparing the real and imaginary parts, we get XR(k)

2,. kn . 2,. kn] = ~I[ £... xR(n)cos N -- +xl SIn -- j,/

(11.52)

11=0

and

N-I[

X/(k) = L

. 2,. kn xR(n)SIn - f j - x l

2,. kn] COS -ij-

(11.53 )

11=0

Similarly, we will get the real and imaginary parts of x(n).

and

xR(n)

= ~ [~[ XR(k) cos '?:.~k1l. -

X/(k) sin '?:."Nkl1 ]]

(11.54)

xln)

kn + Xj(k) cos 2,. = -I [~I[ £... XR(k) SIn -2,. -- --knJ] --

(11.55)

.

N 11=0

11.6.4.1 Symmetry property for real valued x(n) relation holds good: X(N-k)

N

N

This property states that for x(n), the following

=X *(k) =X(-k)

(11.56 )

Proof: The DFf of x(n) is given by N -I

X(k )

=L

x(n)

W~/1

( 11.57)

11=0

Putting N-k in place k in Eg. (11.57), we get The DFf of x(n) is given by N -I

X(N-k)

N- I

= Lx(n) W~N -k)/1 = L 11=0

Since,

W: /1

=e

_pn Nil . N

x (n)W: /IWNkll

II = ()

= e- j2nll = cos (2Jrn) -

J sin (2Jrn)

=I

(11.58 )

Using (11.58), we can write Eg. (11 .57 ) as N- I

X(N-k)

= Lx(n) WNkll =

X(-k)

( 11.59)

11=0

The complex conjugate of X(k) is given by

= L x(n) WNkl1 N-J

X*(k)

11=0

( 11.60)

372

Signals and Systems

Therefore, from Eq. (11.59) and (11.60), we can write X(N-k)

=X(-k) =X* (k)

(11.61)

11.6.4.2 Real and even sequence If x(n) is real and even i.e., x(n)

=x(N-n)

X(k)

N-1 kn) = Lx(n) cos ~

O:S;n:S;N-l

then this property states that

(2

(11.62)

n=O

11.6.4.3 Real and odd sequence If x(n) is real and odd i.e., x(n)

=-x(N-n)

X(k)

=- j

then this property states that N-1 ( ) Lx(n) sin 21C kn N

(11.63)

n=O

11.6.4.4 Purely imaginary sequence If x(n) is purely imaginary sequence i.e., x(n)=jx1(N-n)

O:S;n:S;N-l

then this property states that XR(k)

= N-1 L x(n) SIn. (21C-Nkn) n=O

and

X[(k)

N-1 =~ x(n) cos

(21C

k ) N n

(11.64)

(11.65)

11.6.5 Circular Convolution If x 1(n)

DFf H

N

X 1(k) and x2(n)

DFT H

N

X 2(k), then this property states that

A1(n)@A2(n)

DFf H

N

X 1(k)X 2(k)

where x1(n) (N) x2(n) means circular convolution of x1(n) and x2(n).

Proof: The DFf of x 1(n) is given by N-1

X 1(k)

- j2/r kn

= LX1(n)e-N-k =0, 1,2, ...... , N-l n=O

(1l.66)

Discrete Fourier Transform and Fast Fourier Transform

373

Similarly, the DFf of xl(n) is given by - j211 km

N-I X 2(k)

=

LX2(m)e-N- k

= 0, 1,2, ... ... , N-I

(11.67)

m=O

Let the sequence x 3(x) has DFf X 3(x) where X 3(k)

=X1(k) X 2(k)

(11.68)

Again, from the definition of IDFT we get IN-I

X3(l)

L X (k)e

=-

N I=~

3

j21111t

(11.69)

N

Using Eq. (11.68), we can write Eq. (11.69) as ' x3(n)

IN-I j211kl LX1(k)X 2 (k)e N

=N

(11.70)

k=O

Substituting X1(k) and X 2(k) in Eq. (11.70) we get 1 N-I[N-I -j211kll][N _1 -j211km] j211kl x3(l)=N LLX1 (n)e N Lx 2 (m)e N eN k=O n=O

m =0

N-I N-J N-I [ j211 k(l-n_m)] =.1 Lxl(n) Lx2 (m)L e N

N

If l-n- m

n=O

m=O

=N, 2N, 3N, ....... i.e., multiple of N, then e

(11.71)

k=O

j211 kU-n-m) N

= I. Hence,

N-J[ j211kU-n-m)] LeN =N

01.72)

k=O

j211 k(l-n-m)

When

e N-J [

"#

N

j211 k (I-n-m)

k=O

"e

N-J[ j211 k (I - n-m) £..,

N

1- e 1'2 11 k(1 -It-m )

--

{N

0

if if

I N-J

N-I

N-I[ j211 k(l-n-m)

= N ~XI(n) n~X2(m) ~

e

N

N-I N-I = LXJ(n) L X2(m)

n=O

j211 k(/-n-m) =

I-e

N

m=O

1

N-I

0(11.73)

N

l-n-m is a multiple of N l-m-n is not a multiole of N

Using Eq. (11.74), we can write Eq. (11.71) as X3 (l)

I _I

j211 k(l-n-m) =

I-e

1

k=O

1 =

LeN

i.e., l-n-m is not a multiple of N.' Therefore, we can write that

I, we have

(11.74)

N-I

= ~ ~XI(n),?;X2(m)N (11.75)

374

Signals and Systems

when I-II-m is a multiple of N. This integer multiple may be positive or negative. Let us put /-n-m = -qN Therefore. m =I-n +qN Using Eq. (11.77). we get from Eq. (11.75)

(11.76)

(11.77)

N-I

= L, x l(n)x2(l-I1+qN)

x3(1)

(11. 78)

n=O

Due to change of index the summation of x 2( ) has been dropped because it is redundant. )c2(i-n+qN) represents a periodic sequence with period N having n samples. We can write xil-n+qN) as x2(1-I1+qN) x2(1-I1, modulo N) (11.79)

=

(11.80)

= x 2(I-n)N

Using Eq. (11.80), we have from Eq. (11.78) N-I

x3(1) = Lxl(n) x2(l-n)N =xl(n)

® x2 (n)

(11.81)

n=O

where

IV) denotes the circular convolution of xl(n) and xin)

Example 11.2

For the following two sequences xl(n)

= {3, 1,3,1 land xin) = {I, 2, 3, 4} find out the i

i

,_-- -0-- __ ,,','\

/ / " ___ --0-- __ ''','\

,, ,,

,, ,,

,

x1(2) = 3

I

,,

I

x2(2) = 3

,

I I

,, ,

,

I

,, ,

, I

¢ X1(O) = 3

Q

I

, ,,

I I

¢, X2(O) = 1

Q I I

,, ,,

, ,, ,,

----0------

, ,,

I

,

"----0----

x1(3) = 4

x2(3) = 4

(a) Sequence x1(n)

(b) Sequence x2(n) ,-_---0---_,,',,'\

,

,

,, ,

,,

,,

I

I

I

x2(2)

=3 Q

,,

I

I

I

I

,

I

,, '---0---' x2(1) = 2

(c) x2((-n»4 : Sequence x2((n» is folded circularly which is obtained by plotting x2(n) circularly

Fig. E11.1

I

,

Discrete Fourier Transform and Fast Fourier Transform 3

Again,

x3(m) =x)(n) (N)Xin)

= ~»(n) x2(m-4)4 n=O

3

(i) x3(0)

= Lx)(n) x 2(-n»)4 n=O

x2(3) = 4

_----0----_ 4xl=4

,,

x1(1) = 1

,/

//

,,

,,

""

/""" ...... --0- .............. . . ''

''

" I , '

X2(2)=3~3x3=9QX1(2)=3 I I

\

\

,, ,,

'\

I I I

? X2(O) = 1

" I I

\

'

,

~

'\\

,

I I

, I

x1(3) = 1

,

,, ,, , ,

I

I

............... --0- ........ ",;"

,

,, ,,

x1(O)=3?3xl=3

I '

I

,,

,,

,

2 x 1 =2

'"

............ ---4(r---- ............

x2(1) = 2

Fig. E11.2

x3 (0)

=3+ 2 + 9 + 4 =18

3

(ii) x 3(1)

= L,x)(n) X2(O-n»)4" n=O

/"

", ,,--

------:~,~ ~-: -- ---

,,'

, ,,

,,

I

" " I

x2(3) = 4~4X3= ~

\ ,

,

,. ........... --0-- ..........

''

'

X1(O) = 3

\

, x1(3) = 1

"'"

I

~3 x2 = 6 ~ x2(1) = 2

I ' ,, ' ,, , ,, I

"""--0---/,' \

\

,':

\

\,

,, ,, ,

,,

''

I

12~ x2(2) = 3

,

' '

\

" I

,,

3x I =3

'-'-,---~-------x2(2) = 3

Fig. E11.3

I

I

,, /-

,,

,

375

376

Signals and Systems x 3(1)

= I + 6 + 3 + 12 = 22

:1

(iii) x3(2)

=L

XI (n)x 2 (2-n)

11=0

x2(1) = 2

_- ---0 ---- _ 2x 1=2

,,

x1(1) = 1

, ,,

, ,, ,,, , ,,

,

X2(O) = 1 91 x 3 = 3

,, ,

,,

,,

/,-- --0-- -- __ ,

," ,,

,,

,, ,,

,

,

",

Q, x1(2) = 3 ,,

,,

,,

,, X1(O) = 363 x 3= 9 ,, ,, ,

,

,, ,,

,

'---- --0--- /"

,, ,

6, x2(2) = 3

,/

x2(3) = 1

,

,

,/

,

4 x 1=4 /'

-----0----x2(3) = 4

Fig. E11.4

3

(iv) x3(3)

= LXl(n) x 2 (3-n) 11=0

x2(2) = 3

_----0--- __ 3 x 1=3 ,

, /, / " / /

x1(1) = 1

,, ,,

//_ - - -0-- - __ ",

,

,,

x2( 1) = 292 x 3 = 6

,, ,, ,, ,

,,

"

,, ,,

Qx2(2) = 3 \,

,, ,, ,

,

,

""

,, ,

,,

,, ,, ,,

,, ,, ,

,

X1(O) = 3 64 , x 3 = 126, x2(3) = 4

,,

,,

:

x1(3) = 1 , / /

"

:'

- -- --0- -/

:' ,

, 1 x I =1

--------- ---0-------X2(O) =1 Fig. E11.S

:

/

Discrete Fourier Transform and Fast Fourier Transform x3(3)

377

= 1 +6+3+ 12=22

x 3(n) = {18, 22, 17,22}

i

11.6.6 Time Reversal of a Sequence This property states that if we reverse the N point discrete-time sequence, it is equivalent to reversing the Discrete Fourier Transform (OFT) values. If x(n)

OFT H

X(k), then

N

x(N -n)

OFT

X(N -k)

H

(11.82)

N

Circular Time Shift Sequence

11.6.7

This property states that if OFT

x(n) H

X(k)

N

OFf

then

_/km

x[(n-m)]N H e N X(k)

(11.83)

N

where x[(n-m)]N is a circularly time shifted version of x(n).

11.6.8

Circular Frequency Shift of a Sequence

This property states that if OFT

x(n) H

X(k)

N ./7rklll

then

e

N

OFf

_j27r kill

X[(k - m)] N

x(n) H e N

(11.84 )

N

where X[(k -m)] N is a circularly frequency shifted version of X(k).

11.6.9

Circular Correlation of Two Sequences

Let xl(n) and x2(n) be two complex valued sequences. This property states that if xl(n) OFT

x 2(n) H

N

OFT H

XI(k) and

N X 2(k) then n- I

R tI X2 (m)

= L,xl(n)x;[(n-m)]N 11=0

OFT H

N

XI(k)X ; (k)

(11.85)

378

Signals and Systems

where x;[(n - m)]N is the complex conjugate of x2[(n -m)]N and X; (k) is the complex conjugate of X 2(k) and R XlX2 (m) is the circular correlation sequence.

11.6.10 Multiplication of Two Sequences If x](n)

OFT H

N

OFT

X 2(k) then

X](k) andx 2(n) H

N

x](n)

®

DFT

x2(n) H

N

X](k) X 2(k)

01.86)

where (11) denotes circular convolution. This property is dual convolution property which is defined as x](n)

11.6.11

(f.i;

OFT

xin) H

N

X](k)X 2(k)

01.87)

Parseval's Theorem

Let x] (n) and x 2(n) be two complex valued sequences. OFT OFT If x](n) H X](k) and x2(n) H X 2(k) then

N

N

(11.88) This is the general expression of Parseval's theorem. When x](n) x2(n) =x(n) and X](k) =X2(k) X(k), we get the special form of Parseval's theorem.

=

11.7

=

RELATIONSHIP BETWEEN OFT AND z-TRANSFORM

Let us develop the relationship between the DFT and z-transform. The z-transform of a discrete time sequence of finite duration is given by ~

X(z)

= L, x(n) z-n

(11.89)

n=-oo

Let us consider a finite duration sequence x(n), 0:::; n :::; N-1. The above equation reduces to N-]

X(z)

= L,x(n)z-n

(11.90)

n=O

OFT of x(n) is given by N-]

X(k) = L,x(n)e n=O

Let us put z = e

.21C k

-j-

N

in Eq. (11.90), we get

_j21Ckn N

(11.91)

Discrete Fourier Transform and Fast Fourier Transform .2TCk

X(z)lz

N-I

N-I

11=0

11=0

379

.2TCkn

= e- IN = L.,x(n) z-n = L.,x(n)e - l--':;-

(11.92)

If we compare Eq. (11.92) with Eq. (11.91), we get (11.93)

X(k) = X(z)l z=e- /: k

It is possible to determine uniquely z-transform by its N-point DFT. It is possible to express X(z) as a function of the X(k). 10FT of x(n) is given by . 2TCkll

N- I

x(n)

~ l-= ~1 4.J X(k) e N

N k=O Substituting the value of x(n) in its expression of z-transform, we get

1- ( e

1

N-I

N

k=O

=-

jN-I+I

/TCk N

L X(k)

Z-I .2TC k

l-e

.1N Z-I

I [I - lIN [I -

e j2TC N Z - -N z- N = ~ ~-I X(k) ~.~---

N -I =~ ~ X(k) N k.. k =O

Therefore,

.2TCk

I-e

X(k)

=

J N

z-I

N k..

k=O

.2TCk

I-e

[I-~-~][~.~~;:r-- l k-O

I-e

1 N

1

z- I (11.94 )

N Z-I

The above Eq. (11.94) gives the relationship between z-transform and DFT of a finite-duration discrete-time sequence.

11.8

FAST FOURIER TRANSFORM (FFT)

Fast Fourier Transform (FFT) has an important role in DSP. Large number of application such as filtering, spectrum analysis, etc. require calculation of DFT where largl'! number of computations are required. Using special algorithms, it is possible to reduce the number of computations. These special algorithms altogether are known as FFT.

380

Signals and Systems

11.8.1

Properties of FFT

The N point DFf sequence is given by N- I

X(k) =

k =0, 1,2, ... .. 00, N-l

Lx(n)W~n

(11.95)

n=O

where

.2n

-j-

WN = e

(11.96)

N

which is known as the twiddle factor and it exhibits symmetry and periodicity properties.

11.8.1.1 Periodicity property of WN This property states that W~+N = W~ Proof: WN is represented by WN

=e

.21r

-j-

N

W~+N = e

..

-/Ir(k+Nl N

Since e-j21r = 1, we have from Eq. (11.97)

=e

_/lrk N e

_/Ir N 1V

=e

_/lrk e- j2tr N

W~+N. /~k .(e-I~r .W~ 11.8.1.2 Symmetry property Proof:

k

This property states that WN

+!::!.. 2

(11.97)

(11.98)

= - W~

WN is represented by k+!::!..

..

WN 2=e

Since e-

jlr

_/Ir ( k+!::!.. ) _/lrk _/IrN _/lrk . N 2 =e Ne 2N=e N e-jlr

=-I, we have from Eq. (11.99)

(11.99)

(11.1 00)

We can also prove that -

2

We know that

WN

=e

.21r

-j -

N

Putting N by N/2, we get

W~ e- ?.(e•

The FFf algorithm uses the above properties.

I

12 ,:]'

.W~

(11.101)

Discrete Fourier Transform and Fast Fourier Transform

381

11.8.1.3 Calssification of FFT Algorithms The FFT algorithms depends on two methods. The first method is divide and conquer approach whereas the second one is based on linear filtering. In the first method, the N-point DFT is successively divided to 2 point DFTs so that the calculation is reduced. There Radix -2, Radix,-4, decimation in time, decimation in frequency etc. type of FFT algorithms are developed. In second method, there are Goertzel algorithms and the chirp- z algorithms.

11.9

RADIX-2 FFT ALGORITHM

This algorithm is based on the divide and conquer approach and the N point DFT is decomposed into smaller DFTs and hence the number of computations is reduced. The value of N is selected such that N = 2 '. This algorithm is known as Radix-2 or radix of these algorithms is '2' because the N point DFT is decomposed successively such that smallest DFT size will be N =2.

11.10

RADIX-2 DIT-FFT ALGORITHM

The meaning of DIT is Decimation in Time and let the N point data sequence x(n) be splitted into two N

2

point data sequences!l(n) and!2(n) such that!l(n) contain even number of sequence and!2(n) contain odd number of sequence. Therefore, we can write that n=O ,

1" 2········· '2Ii_I)

N n =0, I, 2,.········, 2-- 1

(11.1 02)

Equation. (Il.I 02) shows that the time domain sequence is splitted into two sequences. The above splitting operation is called decimation. It is called decimation in time domain because it is done on time domain sequence. The N point DFT of x(n) is given by N-l

X(k)

= Lx(n)W:", k = 0,

1,2,3, .... ... , N-I

(11.103)

n=O

Splitting the sequence x(n) into even and odd number of sequence, we get X(k)

=

L,x(n) 11

even

w: + L,x(n) n

W:"

n odd

~ -l

~ -I

2

2

= LX (2m) W~mk

+ LX (2m + I) WN (2m + I)k

m=O

(11.104 )

111=0

In Eq. (11.104), let us put !l(m) =x(2m) and!2(m) =x(2m + 1). We have from Eq. (11.104) ~-I 2

X(k)

= Lfl(m)(W~'k) 111=0

~- I 2

2

2

+ .2J2(m)(W~lk) W: m=O

(11.105)

382

Signals and Systems

Since W~ = WN , we have from Eq. (11.105) -

2

~-1

~-1

2

X(k)

2

= L/l (m) W.t + W~ L/2 (m) W.t 2

m=O

111=0

2

k =0, 1,2,3, .. ......, N-I

i.e.,

(11.1 06)

where F 1(k) is the N -point OFT offl(m) and F 2(k) is the N -point Off ofh(m).

2

2

F 1(k) and F 2(k) are periodic with period N because F 1(k) and F 2(k) are N -point OFfs. Hence, we can write

2

Substituting k by k +

2

Fl( k

+N~) = Fl(k)}

F2( k

+"2 ) = F2(k)

(11.107)

~ ~n Eq. (11.1 06) we get (11.108)

k+!::!.. We know that WN 2

=- W~' we have from the Eq. (11.108)

X( k + ~) = F,(k) - W~F2(k)

k =0, 1,2,3, . ....... , N-I

(11.109)

From Eq. (11 .109), we can say that X(k) is N-point Off. Therefore, we can determine F,(k) and F 2(k) taking k =0 to N -1 because they are N -point OFfs. Therefore, N-point Off we can get from Eqs. (11.106) and 2 2 (11.1 09)

where k =0, 1, .... ,

N "2 -1'

First Stage of Decimation

Figure 11.6 shows that 8-point Off can be determined directly. x(O) x(1 ) x(2)

a-point OFT

xC?)

X(O) X(1) X(2)

XC?)

Fig. 11.6

Discrete Fourier Transform and Fast Fourier Transform

383

As per Eqs. (11.106) and (11.107), we can say that X(k) can be obtained from F,(k) and Fz(k) where F, (k) and F 2(k) are two 4-point DFfs. The symbolic diagram operation has been shown in Fig. 11.7. The sequences offI(m) and/2(m) are given below: II(m)=X(2n)={X(0)'X(2)'X(4)'X(6)}}

N

.

(11.110)

- - POInt sequences 12 (m) = x(2n + 1) = {x(1) , x(3), x(5), x(7)} 2

I'1(m) = x(2n) I '1(0) = x(O) '1(1) = x(2) '1(2) = x(4) '1(3) = x(6)

F1(0)

N . t - - POIn 2 OFT

F1(1)

i.e. 4 - point OFT

F1(2)

,r-------------------

X(O) X(1) X(2)

F1(3)

These values are obtained by putting k = 0, 1,2, 3 in Eq. 5.5.7

X(3)

I'2(m) = x(2n + 1) I '2(0) = x(1) '2(1) = x(3) '2(2) = x(5) '2(3) = x(7)

F2(0) N . t - - POIn 2 OFT i.e. 4- point OFT

F2(1)

X(1 + 4) = X(5)

F2(2) F2(3)

X(2 + 4) = X(6)

,

I1- _ __________________

X(3 + 4) = X(7)

These values are obtained by putting k = 0, 1.2,3 in Eq. 5.5.8

This block combines two 4-point OFTs according to Eq. 3.5.7 and Eq. 3.5.8

Fig. 11.7

Second Stage of Decimation Let us split/l(n) into the following even and odd numbered samples:

N}

n=O, I , ........ · '4 -

11=01 ......... !i " '4

(11.111)

vlI(n) and v,z{n) are the even and odd numbered sequences of II(n) containing N samples because II(n) 4

. -N sampIes. contaInS 2 Now let us split/2 (n) into the following even and odd numbered samples:

384

Signals and Systems

N}

n =0, 1......... , , 4 n =0 1......... "

vZ)(n) and v22 (n) are the even and odd numbered sequences of f2(n) containing

~

(11.112)

N , 4

samples because f2(n)

. 2 N sampIes. contams

Earlier we have obtained X(k) and X ( k +

~.

and the length of DFf was

~) from F1(k) and F 2(k) withiJ(n) andf2(n) as decimated sequences

As before. we can obtain F)(k) and Fl (k +

~)

from v)) (n) and v 12 (n) .

Therefore. we can write that F1(k)

= V))(k) + WNk V)2(k)

k

N =0.1 ...... 4-1

(1l.l13)

2

and

+ ~) =

Fl (k

V l1 (k) -

W~VI2(k)

k =0. I ......

~ -1

(11.114)

2

In Eqs. (11.113) and (11.114). the N point DFf are obtained from N point DFf. V ll (k) and Vdk) are the 2 4

N point DFf of v ll (n) and v 12 (n) respectively. Similarly. we can write the following for F 2(k) also: F 2(k)

k =V21 (k) + W~V22(k)

k

=0.1 ... ... 4N -1

(11.115)

k

=0.1 ...... 4N -1

01.116)

2

and

In Eqs. (11.115) and (I I.l 16). the

!i 4

+ ~)

Fz( k

!i 2

=V2)(k) - W~V22(k) z

point DFf are obtained from N point DFf. V2)(k) and V22 (k) are the

4

point DFf of v 21 (n) and vzz(n) respectively.

Figure 11.8 shows F)(k) and Fz(k) are two 4 points DFf which are shown by two separate blocks.

'1(0) '1(1 ) '1(2)

4 point

OFT

'1(3)

(a) F1(x)

F 1(0)

'2(0)

F2 (0)

F 1(1)

'2(1 )

F2 (1)

F1(2)

'2(2)

F2(2)

F1(3)

'2(3)

F2(3)

(b) F2(x)

Fig.11.8 A Symbolic Representation 0'4 point OFT using direct computation

Discrete Fourier Transform and Fast Fourier Transform For N

385

=8, Vll (k) and V I2 (k) are Ii =2 point OFfs. 4

The symbolic diagram of operation has been shown in Fig. 11.9(a). From Fig. 11.9(a) the sequences nll(n) and v 12 (n) for 4 point!l(n) are given by }

v ll (n)=!1(2n)=x(4n)={x(0),x(4)},n=0,1

N

.

- -- pomt sequences

v 12 (n)=!1(2n+I)=x(4n+2)={x(2),x(6)},n=0,1

(ll.l17)

4

Therefore, the two N point OFf can be computed separately and combining them we will get

4

fi 2

point OFf.

The symbolic diagram of an operation has been shown in Fig. 1l.9(b). From Fig. 1l.9(b) the sequences v21 (n) and v22 (n) for 4 pointHn) are given by v 21 (n)=!1(2n)=x(4n+I)={x(l),x(5)},n=0,]

}

N

.

- -- pomt sequences v I2 (n) = !1(211 + I) = x(4n + 3) = {x(3) , x(7)},n = 0, 1 4

I

v,,(n)

=f,(2n) =x(4n)

1

'i._ point 4 OFT i.e. 2-point OFT

v,,(O) = f,(O) = x(O) v,,(1) = f,(2) = x(4)

V11 (0)

i-------- ---------:

F (0) } These values

,

1

f-V _ 1_1(_1)--+-:--r-'(n+~)w~n !!..-I

=

~[x(n)+ (_1)k x(n+ ~)]w~n

(11.139)

Now let us split X(k) into even and odd numbered samples i.e .•

!i.- I X(2k) =

~[x(n) + (_1)2k x(n+ ~)] w~kn

k=O,I ... .. . N_ I

2

(I 1.140)

392

Signals and Systems !!.-l

and Since (_1)2k

X(2k+1)

= ~[x(n)+(_1)(2k+l)x(n+~)]W~n(2k+l)

(11.141)

= 1 always and (_1)2k+1 =-1. Again we know that WN2kn _- Wkn N

(11.142)

2

We have from Eqs. (I1.140) and (l1.l41) ~-l

X(2k) =

~[x(n) + x (n + ~)] w~n

(I1.143)

N

k=O, 1, .. ... 2-1 !!.-l

X(2k+ 1)=

~{[x(n)+(-I)kx(n+~)]w~}wf

(11.144)

N

k=0,1 , ...... ·'2 - 1 Eqs. (I1.143) and (l1.l44) together give DFf X(k). Let us define the two N point sequences gl(n) and g2(n) 2 as

glen) and

=x(n)+x(x+~}n=o,

g2(n)=[X(n)-x(n+~)]w~"

1, ...... ,

~ -1

n=O, 1, ...... ,

(11.145)

~ -1

(11.146)

Using Eqs. (11.14S) and (11.146), we have from Eqs. (11.143) and (11 .144) as can be written as

!!.-l 2

X(2k)

= L,gl (n)W~?2'

N

k=O, 1 ....... '2-1

(11.147)

n=O

and

N

k=O,1 ....... '2-1

(11.148)

n=O

The N-point DFf is split into two N points DFfs. This is called Decimation in Frequency (DIF) FFf 2 because X(k) is decimated with k even and k odd where k represents frequency components in the range of 0 to 2n. For an example of eight-point DFf, the values of glen) of Eqs. (11.147) and (11.148) becomes

gl (n) gl (0) gl (1) g) (2) gl (3)

=x(n) + x(n + 4) from Eq. (I1.14S) with N =8 =x(O) + x(4)

=x(l) + xeS) =x(2) + x(6)

=x(3) + x(7)

(11.149)

Discrete Fourier Transform and Fast Fourier Transform

393

and gin) = [x(n) - x(n + 4) ] W8n from Eq. (11.146) with N = 8 o g2 (0) = [ x(O) - x(4 ) ] Ws (11.150)

g2 (I) = [x(l) - x(5) ] W81 g2 (2) = [ x(2) - x(6) ] W82 g2 (3) = [ x(3) - x(7) ] W83

The partial signal flow graph for computation of gl(n) and g2(n) according to Eqs. (11.149) and (11.150) is shown in Fig. 11.15. which also shows that from gl(n), the 4 point OFT X (2k) is obtained and similarly, from g2(n), the four-point OFT X(2k + 1) is obtained. X(k) is splitted into even numbered, values and odd numbered values shown in Fig. 11.14 and these two sets of OFTs are computed separately by four-point OFTs. The sequences gl(n) and g2(n) are required for these four-point DFTs. The computation of Fig. 11.15 is opposite to that shown in Fig. 11.12 for DIT FFT.

I X(2k)) I r- - - - - - - - - - - - - - - - - - - - - - - - - :

91 (0)

x(O) x( 1)

X(O) g1(1)

X(2)

----'~_\_--__j'-1>----+--______l

4-point

OFT

X(4) X(6)

I X(2k + 1) I X(1) X(3) x(6)

~_-+-----I___>--_+,-=g:..:::c2 -,---(2-,---)- l

4-point

OFT

X(5)

W3

xC?)

: 8 : g2(3) ----i-:-4-----+_..,.1>----;........:=-~

,

XC?)

,

1 _________________________ ,

This block splits x(n) into g1 (n) g2(n) according to Eq. (11.149) and Eq. (11.150)

Fig.11.15

a=X=Aoa+b

First stage decimation of OIF FFT algorithm for N = 8

Figure 11.16 shows the basic butterfly operation in DIF FFT algorithm which is similar to the butterflies shown in Fig. 11.11. Here the sequences X(2k) and X(2k + 1) are further decimated into their even and odd numbered values. There will be v = log2N decimation stage or N-point OFT but opposite in direction. The signal flow graph for N = 8 point DIF FFT has been shown in Fig. 11 .17.

bW~

B=(a-b)

-1

Fig. 11.16

Butterfly operation in OfF FFT

W~

0

woo

o o o o

:o

o

o

o

8: .b .:

p

7

\"

w~

o

.

92(2)

92(1)

92(0)

91(3)

91(2)

91(1)

91(0)

Input sequence in natural order

I

v=1 First stage of decimation

Fig.11.17

':P

wg

• "'0 •

v=2 Second stage of decimation

'_~ __ mmmm~'I_~m _i

Q:

Q:

i

0

~X(6)

X(3)

OFT sequence shuffled in bit reversed order

v=3 Third stage of decimation

:d'mm:tr X)(7)

:---------------------i

: ________________~~ ___ :

';;0----:- X(1)

•

i'L~ 8: X(5)

0::::

if

~X(2)

0::

."'0 .•

• ~X(4)

if

':;0

0

.~X(O)

Two-point OFT

Q

':;0

\ ( ---y

)(

6

6

)(

Q

Q

Split input sequence into two sequences

Signalflow graph for N = 8 point DIF FFT

TT_ mmm_m m____:~t: __:.,(3)

x(6)

x(5)

X

X'

/ Y \ \wJi

X

X

x(4) -+-.-!...._D_ig_ita_1~I--_ _ _ >

_

Fig. 12.2

12.2

Filter

Output Discrete-time Signals

Block Diagram Digital Filter

REVIEW OF ANALOG FILTERS

Analog filters are the following tive kinds: • Low pass analog filters • High pass analog filters • Band pass analog filters • Band stop analog filters • All pass analog filters Table 12.1 gives the pole-zero configuration as well as the corresponding frequency response of different kind analog filters.

Table 12.1

Pole-zero configuration and corresponding frequency responses of various kinds of analog filters ,,.

Name oftlae analog filter

fimctiora

Frequ.eney rnpoMe

H(S)

H(JfD)

traJN~

Pole-zero lDcaIIoM of IH(J.) I j(O

IH(j(O) I Low pass

X---

r.02

filter (LPF) 82

/~

+(~ )s +r6

\

,

\

0

'( ,

(0

,,

8

2

X---

+( ~) +r5

I

----- ----1

0

/~

/ j

(ro IQ)8

filter (BPF) 82

+( ~ ) +1'6 0

s-plane Here two zeros p

\ \

, '~, ,

,,

(0

j(O

IH(j(O)1

Band Pass

P

j(O

High pass filter (HPF)

0

-

IH(j(O)1 s2

s-plane

\

ro

X---

/~ ,,

\

'~, (0

,,

s-plane Here one zeros p

'-

Contd...

Structures and Design of Digital Filters Na.meoflhe analog filter

lis tral'lll(er function

Frequency response

H(8)

HUm)

Pole·~ero

locations of IHUm) I jm

IH(jm)1

Band stop

s2 +r6

filter (BSF) s2

+( ~) +r6 0

T fO

x'

,

-

:' ~

filter (APF)

s2 - (ro/Q)s s2

'( ,

p ,

m

-jm

+r~

/

+(~ )s +r~

1

s-plane

,, ,

IH(jm) I

All pass

401

,,'kZ-k k=' From Eq. (12.66) we can conclude that the overall IIR system can be realized as a cascade of two functions H,(z) and Hiz) . H,(z) represents the zeros of H(z) and it is all the zero system. H 2(z) represents the poles of H(z) and it is all the pole system.

12.8.1.1

Direct form-I structure for IIR filter

We will at first prepare the direct form structures of

H,(z) and H 2(z) . Let us write H,(z) of Eq. (12.65) as H,(z) bo + b,z-' + b 2z- 2 + b 3z-3 + ... + bMz-M

=

We know that H, (z)

}) (z) =-

X, ( z) Using Eq. (12.68), we get from Eq. (12.67) as

(12.68)

(12 .67) x1(n)

bo

-----.---------1

t---+-Y1(n)

-}) (--z) =b0 + bI Z- , + b2Z-2 + b3Z-3 + ... + bMZ-M XI (z)

i.e., YI(z)

=boXI(z) + b, Z-IX,(Z) + b2z- 2 XI(z) + ... + bMz-MX,(z) 02 .69)

The inverse z-transform of Eq. (12.69) gives Y,(n)

=boXI(n) + blx,(n -

1) + b2x l (n - 2) + ...

+ bMx,(n - m) (12.70) Fig. 12.18 gives the direct form realization of Eq. (12.70). From Eq. (12.66) we can write that Hiz)

=-

1 -r;;-- - ---

I+L

I----------~

(12.71)

+

Gk Z - k

k =1

The following relation is also valid: H 2(z) = Yz ( z ) X z (z)

(12.72)

Fig. 12.18

Direct form of realization of H dz)

Structures and Design of Digital Filters

417

From Eq. (12.71) and (12.72), we get

=

Yz (z)

~

Xz(z)

i.e.,

Y2(z)

[I

I

1+ £.J ak Z-

k

k=l

+ ~ ak Z-k ]

=Xz(z) N

L ak Z-k Yz (z) + X z(z)

i.e.,

Yz(z) = -

i.e.,

Y2(z) = -a1z- 1Y2(z) - a 2 Z- 2Yz(z) - a 3z- 3y 2(z) - .. . - aNz-NYiz)

k=l

+ Xz(z)

(12.73)

The inverse z-transform of the Eq. 02.73) gives Y2(n)

=-alYz(n -1) --azyzen -

2) --a3Yz(n - 3) - .. . -aNyin - N) + x2(n)

Figure 12.19 gives the direct form realization of Hiz). } - - - - - - - - - , - - - - - - - . . Y2(n)

Fig.12.19

Since, H(z)

Direct form realization of H2 (z)

=HI (z)Hz(z), gives the cascading of H1(z) and Hiz) to get H(z). H ()_ Y1(Z) 1 Z --X 1(z)

Fig.12.20

H(z)

Y1(n) = x2(n)

Y2(z) H 2(z) = - X2(Z)

=H,(z)Hz{z) i.e., cascading of two systems

Y2(n) = yen)

(12.74)

418

Signals and Systems

Figure 12.21 shows the connections of HJ(z) and Hiz). r----------------------------------

bo

x(n) = x1(n)

-a1

+ /-----------1

+

,

I

,

,-----------------------------------, All zero system

Fig. 12.21

All pole system

Direct form-' realization of fiR system

From above we can conclude that this realization requires (M + N + I) multiplication, (M + N) number of additions and (M + N + I) number of memory locations.

12.8.1.2 given by

Direct form-II structure for IIR filter

The overall system function of H(z) of IIR system is

(12.75)

We can write H(z) as

where (12.77)

Structures and Design of Digital Filters M

H 2(Z) --

and

419

Y(z) - " b Z - k - £... k W(z) k;1

(12.78)

From Eq. (12.77), we get W(z)

[1 + ~akz-k] =

X(z) N

i.e.,

W(z) =

i.e.,

W(z)

L akz-k W(z) + X(z)

k;1

=-aIZ- 1W(z) -

a 2z-2W(Z) - .. . - aNz-NW(Z) + X(z)

(12.79)

The inverse z-transform of Eq. (12.79) gives wen)

=x(n) -

Qlw(n - I) - a2w(n - 2) - .. . - aNw(n - N)

02.80)

The implementation of direct form of Eq. (12.80) is shown in Fig. 12.22. x(n)

Fig.12.22

--~

r----------.---

wen)

Direct form implementation of Eq. (12.80). All pole system.

From Eq. (12.78), we get M

Y(z)

= L bkz-kW(Z)

Y(z)

=boW(z) + blz- 1W(Z) + b2z- 2W(Z) + ... + bMz-MW(Z)

k;1

i.e., The inverse z-transfonn of this equation gives yen) = bowen) + b l wen - I) + b 2 wen - 2) + ... + b M wen - M) The direct form of realization of Eq. (12.81) is shown in Fig. 12.23.

02.81)

420

Signals and Systems bo

w(n)

}---.-y(n)

+

Fig.12.23

Direct form implementation ofEq. (12.81)

Figure 12.24 shows the direct form of realization of IIR system. y(n)

x(n)

+

-aN-1

+

,

,, ,

1______ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

-----------------------------------

All pole system

All zero system

Fig. 12.24

Direct form realization of IIR system

Structures and Design of Digital Filters

421

In Fig. 12.24, w(n) is delayed and used in an all pole system. It is also delayed and used in a all zero system. Therefore, we can merge the two delay elements of an all pole and zero system into a single delay e I e men t Fig. 12.25 shows the resulting realization. bo

w(n)

\ - - - - - Yin)

x(n)--~

Fig. 12.25

Direct form of realization of IIR system (M = N is considered)

The realization shown in Fig. 12.25 is called the direct form-II realization of IIR system. This structure needs {M, N} memory locations which is less than that of direct form-I. This form is also called canonical form because it reduces memory locations.

12.8.2

Cascade Structure for IIR Filter

From Eq. (12.64), we get M

"b H(z)

= Y(z) = X(z)

-k

£.., k Z k=I_ N-_-

1"

+ £.., ak Z

-k

=

b

h

0

-1

+ l~j +

l+a j z

b

-2 2 Z_2

+a2 z

b-M

+" . + M~N

(12.82)

+ " '+aNz

k=1

We can represent the numerator and denominator of Eq. (12.82) as multiplication of second-order polynomials. Therefore, we have (12.83)

422

Signals and Systems

where Hk(z) =

bkO

-I

+ bkl _ZI +

1+aklZ

b -2 k2 Z

-2'

+ak2 Z

k = 1,2, ..... .. , K

(12 .84)

Figure 12.26 shows the realization ofEq. (12.83) where HI(z), H 2(z), . .. ., HK(z) are connected in cascade. x(n) = x1(n)

Yk(n) = yen)

Fig. 12.26 Cascade form realization of IIR filter

HI (z), H 2(z) , . . ..., H K(Z) can be realized in direct form-I and direct form-II structures. HK(z) can be written

as

( 12.85) where Hkl(z)

and

Wk(Z)

=- - = Xk(z)

l+aklZ

-I

-2

+ak2 Z

Yk (z) b b -I b -2 H k2 () Z = - - = kO + kl Z + k2Z Wk(z)

(12.86) (12.87)

The direct form-II second-order subsystems Hkl(z) and H k2 (z) has been shown in Fig. 12.27 in cascade form. bkO

bk2

Fig. 12.27 Direct form-II realization of second order subsystems use in cascade connection

The following equations describe the cascade structure: wk(n) = -akl wk(n - 1) - ak2win - 2) + xin)

and

Yk(n) = bkOwk(n) + bk1wk(n - I)

+ bk2wk(n - 2)

(12.88) (12.99)

Structures and Design of Digital Filters

423

The following equations represent second-order subsystems: Yk-I (n) = xk (n)} ydn) = Xk+1 (n)

(12.100)

yo(n)=x(n) yen) = xk (k)

12.8.3

Parallel Form Structure for IIR Filter

From Eq. (12.64), we get M

Y( ) H(z)=~= X(z)

"" bkz- k £..J k=I

N

-

b

b

0+ IZ

-I

+

b

2Z

-2

+ ... +

b-M MZ

(12.101)

l+Lakz-k - l+alz-l+a2z-2+" ' +aNz-N k=1

Let us expand the above system function in partial fractions: H(z)

=A + H1(z) + H 2(z) + H 3(z) + ... + Hk (z)

(12.102)

where A is the constant and H1(z), H 2(z), .....etc. are the second-order subsystem which can be represented as

=

bkO + bk1z- 1 (12.103) 1 -I -2 +aklZ +ak2 Z The co-efficients are real and these second-order subsystems are formed by combining complex conjugate poles. In parallel connection, the additions of these system functions gives the result. Figure 12.28 shows the realization of H(z) represented in Eq. (12.102). H (z) k

c

,,

x(n)

--..1.-------1., Fig. 12.28

HiJ.z)

8}---'"

I---~.

y(n)

Parallel form realization structure of IIR filter

It is possible to realize each of H1(z), H 2(z), .... etc. by direct form-lor direct form-II . Figure 12.29 shows the direct form-II realization of Hk(z) .

424

Signals and Systems +

Fig. 12.29

Direct form-II realization of second order subsystem

The following equations represent the above parallel fonn structure: Wk (n) = -aklwk (n -1) - ak2 wk (n -1) + Yk-l (n)

I

Yk(n) = bkOWk(n~ + bklwk(n -1) + bk2wk (n - 2)

(12.104)

y(n)=Ax(n)+ LYk(n) k=1

12.8.4 Lattice Structure of IIR Filter The lattice realization of FIR filters have been discussed already. Now let us find the lattice representation of IIR filters. For simplicity, let us consider all the poles of IIR filters . From Eq. (12.64), we get H(z)

= --N;-;-l=------

(12.105)

1+ L,akZ-k k=1

From Eq. (12.64), we get Y(z) =

- ----~- -

X(z)

N

i.e.,

Y(z)

+ L akz-ky(Z) = X(z)

(12.106)

k=1

The inverse z-transfonn of Eq. (12.106) gives N

yen)

=- L

aky(n - k)

+ x(n)

(12. 107)

+ yen)

(12.108)

k=1

From Eq. (12.107), we have by interchanging x(n) and yen) N

x(n)

=- L

ak x(n - k)

k=1

Equation (12.108) can also be written in the following fonn: N

yen)

=x(n) + L

k=1

ak x(n - k)

(12.109)

Structures and Design of Digital Filters

425

We have already seen the realizations of FIR systems for first-and second-order equations. In the present case all the poles of IIR system can be obtained by interchanging the roles of x(n) and y(n). Fig. 12.30 shows the lattice structure for first order IIR system by interchanging the inputs and outputs. fo(n) = y(n)

----,-----~-:-:-___j

Fig. 12.30 A single'stage lattice realization of IIR system

Let us compare the single-stage lattice realization of IIR system given in Fig. 12.30 with that of an FIR system. Only the positions of input and output are interchanged. Let us redraw the above structure shown in Fig. 12.31 properly. ....--:-:---,.-------r--~

fo(n)

= y(n)

g1(n) ...- - - (

Fig. 12.31

A single'stage lattice realization of al/ pole IIR Filter

Feedback is present because it is an IIR system. Putting N = 1 in Eq. (12.107) we get y(n)

=x(n) -

(12.110)

G1(k)y(n - 1)

Comparing Fig. 12.31 and Eg. (12.110) we get (12.111)

K\ = G\(1) Figure 12.32 shows the two-stage lattice structure. fo(n) = y(n) ----.------..,..--:-;---(

\-------..,.-,-:--___j +

- - - f2(n)

= x(n)

g2(n)

Fig.12.32

Two·stage lattice realization of al/ pole IIR system

The following equations describes the IIR system: !,(n)=x(l1) J;(n) = 12 (n)- K 2 g,(n-l)

g2 (n) =

KJ, (n) + gl (11-1)

lo(n) = J; (11) - K,go(n-I)

(12.112)

gl (11) = Kdo (n) + go(n-I) y(n) = lo(n) = go(n)

From Eq. (12.l12). we get

y(n) = -K\(l + K 2 )y(11 - 1) - K2Y(n - 2) + x(n)

(12.113)

426

Signals and Systems

With N = 2, we get from Eqs. (12.107) 2

y(n) = LaN(k)y(n-k)+x(n) k=)

= -a)(1)y(n - 1) - az(2)y(n - 2)

+ x(n)

02.114)

Comparing Eqs. (12.113) and (12.114), we get a) (1)

= K) (I + K 2 )}

(12.115)

a2(2)=K2

From Eq. (12.115) we get K)

=

K2

= a 2(2)

a)(1)

1+ a 2 (1)

}

02.116)

Similarly, for higher values of N we can do the realization for IIR systems.

12.9

REALIZATION PROCEDURE FOR DIGITAL FILTER

Although digital filters can be decided in different ways but the relationship between the input and output decides the realization of digital filters . There are three broad categories for realization procedures: • recursive realization • non-recursive realization • FFT realization

12.9.1

Recursive Realization

If the present output of the digital filter is dependent on both the past outputs as well as present and past inputs, this realization procedure of digital filter is known as recursive realization and it needs the feedback of output. Figure 12.33 shows the recursive realization of digital filters. The system function H(z) is given by Y(z)

H(z) = - - = X(z)

b o +b)z-) +b2z- 2 +b3z- 3 l+a)z

_)

+a2 Z

-2

_

+a3 Z 3

bo

Input

h(n)

Fig. 12.33

Block diagram of a recursively realized digital filters

(12.117)

427

Structures and Design of Digital Filters

12.9.2 Non-Recursive Realization In this realization the present output yen) depends on present and past values of inputs but it does not depend on the past values of outputs. In non-recursive realization, there is no feedback of outputs. Figure 12.34 shows non-recursive realization. The system function H( z) in non-recursive realization is given by H(z)

=bo + b]z- ] + b2z- 2 + b3z- 3

(12 .118)

where bo, b) , b2 and b3 are the filter coefficients. bo

Input s(n)

Fig. 12.34

Block diagram of a non-recursively realized digital filter

12.9.3 FFT Realization The input signal in this realization is transformed by an FFT algorithms and hence the computational speed also increases. An inverse FFT transformation is carried out after filtering the spectrum of the signal. Figure. 12.35 shows the block diagram of FFT realization. Modified spectrum of input signal

Spectrum of input signal

nput s(n)

Modification of Spectium according to Gain Requirement

FFT Transformation

Fig. 12.35

Inverse FFT Transformation

Output y(n)

Block diagram of FFT realization

IH(w)1

12.10 NOTCH FILTER A filter which contains one or more notches or ideally perfect nulls in its frequency response characteristics is known as a notch filter. The frequency response characteristics of a notch filter is shown in Fig. 12.36 having two null frequency roo and w) Amplitudes are zero at null frequency.

w

Fig_ 12.36

Frequency response characteristic of a notch filter

428

Signals and Systems

These filters are useful for eliminating some specific frequency components. These filters are used in instrumentation and recording system to eliminate the harmonics. A pair of complex-conjugate zeros on unit circle is introduced for inserting a null in the frequency response of a filter at frequency Wo. ZI =e jWO andz 2 = (ZI)* =e-jWo

Let

The system function of an FIR filter is given by H(z)

i.e. ,

H(z)

=bo (1 -

zlz-I)(1 - Z2Z-I)

(12.119)

=bo(l- ej(J)o Z-I)(1 -

=bo[l -

(ej(J)o + e-j(J)O)z-1 + Z-2]

=bo[ I -

2z- 1 cos Wo + Z-2]

e-j(J)o Z- I)

(12.120)

where bo is a constant. The frequency response of a notch filter without poles is shown in Fig. 12.37 having a null at W = !!... . 4

1.2 1.0

Jr

~ 0.8

Jr/2

:s' Q)

0.6

0

'cCl 0.4

Jr/2

"0

2

I1l

::2'

0.2 0

-Jr -Jr

- Jr/2

0

Jr/2

Jr

-Jr/2

-Jr

(w)

(a) Magnitude response Fig. 12.37

Frequency

0

Jr/2

(w) Jr

(b) Phase response characteristics

Frequency response characteristics of a notch filter without poles ( Null at w =

~)

The other frequency components at the desired null are severely attenuated because the notch has a relatively high bandwidth. We need to resort to a more sophisticated longer FIR filter to reduce the bandwidth of the null. To improve the frequency response characteristics we insert poles in the system function H(z). Let us place a pair of complex conjugate poles at PI = rej(J)o and P2 = (P,)* = re-j(J)o

which introduce resonance in the vicinity at the null. Therefore, the bandwidth of the notch is reduced. Therefore, the resulting system function is given by H(z)

i.e. ,

=b

(1- ZIZ-1 )(1- Z2 Z-') o (1 - PI Z-1)(1 . - P2 Z- I)

H(Z)=bo [

=b

(l-ej(J)°z-1 )(1_. e-jilloz - I ) 0 (1 - re jillo Z-1)(1 - re -jillo Z-I)

1-2 z-lcosWO+z- 2 ] 1-2rz- Icosw o +r 2z- 2

(12.121)

Figure 12.38 shows the frequency response characteristics of a notch filter with pole at 0.85 and null at W

= !!.... The bandwidth of the null is reduced after introduction of the poles. 4

Structures and Design of Digital Filters

429

1.2 1.0 Q)

0.8 -g= :0: :3

Ol::r:

c~

ro -

:2

0.6 0.4 0.2 0 -1C

-1C12

1C12

0

Frequency (w)

(a) Magnitude characteristic

1

1C r1C12 rQ)-

~a

.r::~

0

a.~

----

1C12 r-1C f-

I

r= 0.85

""

"' I

-

I

o

-1C12

-1C

"1C12

Frequency

(w)

(b) Phase characteristics Fig. 12.38

12.11

Frequency response characteristics of a notch filter with pole at r = 0.85 and null at w = ~ . 4

COMB FILTER

It is the simplest fonn of a notch filter in which the nulls occur periodically across the frequency band. These filters are used for reduction of power line hannonics, suppression of clutter from fixed objects in Moving Target Indication (MTI) Radar. It has also a wide range of practical applications. Let us consider a moving average FIR filter which is a simple fonn of comb filter having the difference equation yen)

The z-transfonn ofEq. (12.122) gives

I

M

= --L,x(n-k)

(12.122)

M + 1 k=O M

Y(z) = _1_ L, Z- k x(z)

M

i.e., i.e.,

+ 1 k=O

M

H(z) = Y(z) = _1_ L, Z-k

X(z) H(z) = Y(z) =

X(z)

M

+ 1 k=O

_1_[ 1M+I

(z -I )M+I ]

l-z- 1

(12.123)

430

Signals and Systems

To get frequency response characteristic of FIR filter, let us put z = e jro in Eq. (12.123). We can write H(w)

Y( z) 1 [I_(e-JIO)M+I] =-=. X(z) M+I l-e jW

I e = - -[ M+I

jW - I) . (M+

e

2

e

-

.

~ 2

-

jlO ( M+I

)

= M+l

2

jlO e

= _ I [Sinm(

e e

-jlO . (M+I)

. 1 jlO-

-~

e 2j

-

e

M

-

2

2

- jlO ( M+l) 2

e

_jlO

e 2j

2 -

T

2

.

-jlO( ~ )

2

)1e-jW~

(12.124)

sin~ 2 Equation (12.124) suggests that the filter has zeros on the unit circle at

M+I

j21Cm

Z

= e M+l

,

m =0, 1, ..... , M

(12.125)

Figure 12.39 shows the plot of magnitude characteristics of frequency response of Eq. (12.124). It also shows that the periodically spaced zeros in frequency at wm

= 2;rm

M+l

for m =1,2,3, ... , M

1.2 B

1.0

~ 0.8 Q)

"0

2

0.6

'cOl 0.4