the discrete fourier transform: theory, algorithms and applications

The Discrete Fourier Transform

The Discrete Fourier Transform Theory, Algorithms and Applications

D. Sundararajan

,@ World Scientific I t Singapore • New Jersey 'London • Hong Kong

Published by

World Scientific Publishing Co. Pte. Ltd.

P O Box 128, Farrer Road, Singapore 912805

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THE DISCRETE FOURIER TRANSFORM Theory, Algorithms and Applications

Copyright © 2001 by World Scientific Publishing Co. Pte. Ltd.

All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 981-02-4521.-1

Printed in Singapore by UtoPrint

To my mother Dhanabagyam and my late father Duraisamy

Preface

Fourier transform is one of the most widely used transforms for the analysis and design of signals and systems in several fields of science and engineering. The primary objective of writing this book is to present the discrete Fourier transform theory, practically efficient algorithms, and basic applications using a down-to-earth approach. The computation of discrete cosine transform and discrete Walsh-Hadamard transforms are also described.

The book is addressed to senior undergraduate and graduate students in engineering, computer science, mathematics, physics, and other areas who study the discrete transforms in their course work or research. This book can be used as a textbook for courses on Fourier analysis and as a supplementary textbook for courses such as digital signal processing, digital image processing, digital communications engineering, and vibration analysis. The second group to whom this book is addressed is the professionals in industry and research laboratories involved in the design of general- and special-purpose signal processors, and in the hardware and software applications of the discrete transforms in various areas of engineering and science. For these professionals, this book will be useful for self study and as a reference book.

As the discrete transforms are used in several fields by users with different mathematical backgrounds, I have put considerable effort to make things simpler by providing physical explanations in terms of real signals, and through examples, figures, signal-flow graphs, and flow charts so that the reader can understand the theory and algorithms fully with minimum effort. Along with other forms of description, the reader can easily understand that the mathematical version presents the same information in a

vii

Vlll Preface

more abstract and compact form. In addition, I have deliberately made an attempt to present the material quite explicitly and describing only practically more useful methods and algorithms in very simple terms.

With the arrival of more and more new computers, the user needs a deep understanding of the algorithms and the architecture of the computer used to achieve an efficient implementation of algorithms for a given application. By going through the mathematical derivations, signal-flow graphs, flow charts, and the numerical examples presented in this book, the reader can get the necessary understanding of the algorithms. Large number of exercises are given, analytical and programming, that will further consolidate the readers' confidence. Answers to selected analytical exercises marked * are given at the end of the book. Answers are given to all the programming exercises on the Internet at www.wspc.com/others/software/4610/. Important terms and expressions are defined in the glossary. A list of abbreviations is also given. For readers with little or no prior knowledge of discrete Fourier analysis, it is recommended that they read the chapters in the given order.

I assume the responsibility for all the errors in this book and would very much appreciate receiving readers' suggestions and pointing of any errors (email address: [email protected]). I thank my friend Dr. A. Pedar for his help and encouragement during the preparation of this book. I thank my family for their support during this endeavor.

D. Sundararajan

http://www.wspc.com/others/software/4610/

mailto:[email protected]

Contents

Preface vii

Abbreviations xiii

Chapter 1 Introduction 1 1.1 The Transform Method 1 1.2 The Organization of this Book 3

Chapter 2 The Discrete Sinusoid 7 2.1 Signal Representation 7 2.2 The Discrete Sinusoid 11 2.3 Summary and Discussion 27

Chapter 3 The Discrete Fourier Transform 31 3.1 The Fourier Analysis and Synthesis of Waveforms 32 3.2 The DFT and the IDFT 37 3.3 DFT Representation of Some Signals 44 3.4 Direct Computation of the DFT 51 3.5 Advantages of Sinusoidal Representation of Signals 54 3.6 Summary 58

Chapter 4 Properties of the D F T 61 4.1 Linearity 61 4.2 Periodicity 62 4.3 Circular Shift of a Time Sequence 62 4.4 Circular Shift of a Spectrum 66

ix

x Contents

4.5 Time-Reversal Property 69 4.6 Symmetry Properties 71 4.7 Transform of Complex Conjugates 81 4.8 Circular Convolution and Correlation 82 4.9 Sum and Difference of Sequences 85 4.10 Padding the Data with Zeros 86 4.11 Parseval's Theorem 90 4.12 Summary 91

Chapter 5 Fundamentals of the P M DFT Algorithms 95 5.1 Vector Format of the DFT 96 5.2 Direct Computation of the DFT with Vectors 101 5.3 Vector Format of the IDFT 104 5.4 The Computation of the IDFT 104

5.5 Fundamentals of the PM DIT DFT Algorithms 106 5.6 Fundamentals of the PM DIF DFT Algorithms 112 5.7 The Classification of the PM DFT Algorithms 114 5.8 Summary 117

Chapter 6 The u X 1 P M D F T Algorithms 121 6.1 The u x 1 PM DIT DFT Algorithms 122 6.2 The 2 x 1 PM DIT DFT Algorithm 125 6.3 Reordering of the Input Data 128 6.4 Computation of a Single DFT Coefficient 130 6.5 The u x 1 PM DIF DFT Algorithms 132 6.6 The 2 x 1 PM DIF DFT Algorithm 134 6.7 Computational Complexity of the 2 x 1 PM DFT Algorithms . 135

6.8 The 6 x 1 PM DIT DFT Algorithm 138 6.9 Flow Chart Description of the 2 x 1 PM DIT DFT Algorithm . 141 6.10 Summary 149

Chapter 7 The 2 x 2 P M D F T Algorithms 151 7.1 The 2 x 2 PM DIT DFT Algorithm 151

7.2 The 2 x 2 PM DIF DFT Algorithm 154 7.3 Computational Complexity of the 2 x 2 PM DFT Algorithms . 158 7.4 Summary 161

Contents x '

Chapter 8 D F T Algorithms for Real Data - I 163 8.1 The Direct Use of an Algorithm for Complex Data 163 8.2 Computation of the DFTs of Two Real Data Sets at a Time . . 166 8.3 Computation of the DFT of a Single Real Data Set 169 8.4 Summary 173

Chapter 9 D F T Algorithms for Real Data - II 175 9.1 The Storage of Data in PM RDFT and RIDFT Algorithms . . 175 9.2 The 2 x 1 PM DIT RDFT Algorithm 176 9.3 The 2 x 1 PM DIF RIDFT Algorithm 180 9.4 The 2 x 2 PM DIT RDFT Algorithm 187 9.5 The 2 x 2 PM DIF RIDFT Algorithm 190 9.6 Summary and Discussion 193

Chapter 10 Two-Dimensional Discrete Fourier Transform 195 10.1 The 2-D DFT and IDFT 195 10.2 DFT Representation of Some 2-D Signals 196 10.3 Computation of the 2-D DFT 200 10.4 Properties of the 2-D DFT 205 10.5 The 2-D PM DFT Algorithms 212 10.6 Summary 220

Chapter 11 Aliasing and Other Effects 225 11.1 Aliasing Effect 226 11.2 Leakage Effect 231 11.3 Picket-Fence Effect 244 11.4 Summary and Discussion 246

Chapter 12 The Continuous-Time Fourier Series 249 12.1 The 1-D Continuous-Time Fourier Series 249 12.2 The 2-D Continuous-Time Fourier Series 262 12.3 Summary 268

Chapter 13 The Continuous-Time Fourier Transform 273 13.1 The 1-D Continuous-Time Fourier Transform 273 13.2 The 2-D Continuous-Time Fourier Transform 282 13.3 Summary 284

xii Contents

Chapter 14 Convolution and Correlation 287 14.1 The Direct Convolution 287 14.2 The Indirect Convolution 289 14.3 Overlap-Save Method 292 14.4 Two-Dimensional Convolution 295 14.5 Computation of Correlation 298 14.6 Summary 301

Chapter 15 Discrete Cosine Transform 303 15.1 Orthogonality Property Revisited 303 15.2 The 1-D Discrete Cosine Transform 305 15.3 The 2-D Discrete Cosine Transform 309 15.4 Summary 310

Chapter 16 Discrete Walsh-Hadamard Transform 313 16.1 The Discrete Walsh Transform 313 16.2 The Naturally Ordered Discrete Hadamard Transform 320 16.3 The Sequency Ordered Discrete Hadamard Transform 325 16.4 Summary 329

Appendix A The Complex Numbers 333

Appendix B The Measure of Computational Complexity 341

Appendix C The Bit-Reversal Algorithm 343

Appendix D Prime-Factor D F T Algorithm 347

Appendix E Testing of Programs 349

Appendix F Useful Mathematical Formulas 353

Answers to Selected Exercises 357

Glossary 365

Index 369

Abbreviations

dc Constant D C T Discrete cosine transform D F T Discrete Fourier transform DIF Decimation-in-frequency DIT Decimation-in-time D W T Discrete Walsh transform FT Fourier transform FS Fourier Series IDFT Inverse discrete Fourier transform I m Imaginary part of a complex number lsb Least significant bit LTI Linear time-invariant msb Most significant bit N D H T Naturally ordered discrete Hadamard transform P M Plus-minus R D F T Discrete Fourier transform of real data R e Real part of a complex number RIDFT Inverse discrete Fourier transform of the transform of real data SDHT Sequency ordered discrete Hadamard transform SFG Signal-flow graph 1-D One-Dimensional 2-D Two-Dimensional

Xlll

Chapter 1

Introduction

Fourier analysis is the representation of signals in terms of sinusoidal waveforms. This representation provides efficiency in the manipulation of signals in a large number of practical applications in science and engineering. Although the Fourier transform has been a valuable mathematical tool in the linear time-invariant (LTI) system analysis for a long time, it is the advent of digital computers and fast numerical algorithms that has made the Fourier transform the single most important practical tool in many areas of science and engineering. Fourier representation of signals is extremely useful in spectral analysis as well as a frequency-domain tool.

In this book, we will be dealing mostly with the discrete Fourier transform (DFT), which is the discrete version of the Fourier transform. The main purpose of this book is to present: (i) the DFT theory and some basic applications using a down-to-earth approach and (ii) practically efficient DFT algorithms and their software implementations. In the rest of this chapter, we explain the transform concept and describe the organization of this book.

1.1 T h e Transform M e t h o d

Transform methods are used to reduce the complexity of an operation by changing the domain of the operands. The transform method gives the solution of a problem in an indirect way more efficiently than direct methods. For example, multiplication operation is more complex than addition operation. In using logarithms, we find the logarithm of the two operands to be multiplied, add them, and find the antilogarithm to get the product.

l

2 Introduction

By computing the common logarithm of a number, for example, we find the exponent to which 10 must be raised to produce that number. When numbers are represented in this form, by a law of exponents, the multiplication of numbers reduces to the addition of their exponents. In addition to providing faster implementation of operations, the transformed values give us a better understanding of the characteristics of a signal. The reader might have used the log-magnitude plot for better representation of certain functions.

The output of an LTI system can be found by using the convolution operation, which is more complex than the multiplication operation. When a given signal is represented in terms of complex exponentials (a functionally equivalent mathematical representation of sinusoidal waveforms), the response of a system is found by multiplying the complex coefEcients of the complex exponentials representing the input signal by the corresponding complex coefficients representing the system impulse response. This is because a complex exponential input signal is a scaled version of itself at the output of an LTI system. Note that this procedure is very similar to the use of logarithms just described: in using common logarithms, we represent numbers as powers of ten to get advantages in number manipulation whereas, in using the Fourier transform, we represent signals in terms of complex exponentials to get advantages in signal manipulation and understanding. We use transform methods quite often in system analysis. Apart from logarithms, we usually prefer to use the Laplace transform to solve a differential equation rather than using a direct approach. Similarly, we prefer to use the z-transform to solve a difference equation.

The time- and frequency-domain approaches are two different ways of presenting the interaction between signals and systems. An arbitrary signal can be considered as a linear combination of frequency components. The time-domain representation is the superposition sum of the frequency components. The DFT is the tool that separates the frequency components. Viewing the signal in terms of its frequency components gives us a better understanding of its characteristics. In addition, it is easier to manipulate the signal. After manipulation, the inverse DFT (IDFT) operation can be used to sum all the frequency components to get the processed time-domain signal. Obviously, this procedure of manipulating signals is efficient only if the effort required in all the steps is less than that of the direct signal manipulation. The manipulation of signals, using DFT, is efficient because of the availability of fast algorithms.

The. Organization of this Book 3

1.2 The Organization of this Book

In Fourier analysis, the principal object is the sinusoidal waveform. Therefore, it is imperative to have a good understanding of its representation and properties. In Chapter 2, The Discrete Sinusoid, we describe the discrete sinusoidal waveform and, its representation and properties. The two principal operations, in Fourier analysis, are the decomposition of an arbitrary waveform into its constituent sinusoids and the building of an arbitrary waveform by summing a set of sinusoids. The first operation is called signal analysis and the second operation is called signal synthesis. The discrete mathematical formulation of these two operations are, respectively, called DFT and IDFT operations. In Chapter 3, The Discrete Fourier Transform, we derive the DFT and the IDFT expressions and provide examples of finding the DFT of some simple signals analytically. The advantages of sinusoidal representation of signals are also listed. The existence of fast algorithms and the usefulness of the DFT in applications is due to its advantageous properties. In Chapter 4, Properties of the DFT, we present the various properties and theorems of the DFT.

In Chapter 5, Fundamentals of the P M D F T Algorithms, we present the fundamentals of the practically efficient PM family of DFT algorithms. The classification of the PM DFT algorithms is also presented. In Chapter 6, The u X 1 P M DFT Algorithms, the subset of u x 1 PM DFT algorithms for complex data are derived and the software implementation of an algorithm is presented. In Chapter 7, The 2 x 2 P M DFT Algorithms, the 2 x 2 PM DFT algorithms for complex data are derived. When the data is real, usually it is, there are more efficient ways of computing the DFT and IDFT rather than using the algorithms for complex data directly. In Chapter 8, D F T Algorithms for Real Data - I, the efficient use of DFT algorithms for complex data for the computation of the DFT of real data (RDFT) and for the computation of the IDFT of the transform of real data (RIDFT) is described. In Chapter 9, D F T Algorithms for Real Data - II, the PM DFT and IDFT algorithms, specifically suited for real data, are deduced from the corresponding algorithms for complex data.

In the analysis of a 1-D signal, the signal, which is an arbitrary curve, is decomposed into a set of sinusoidal waveforms. In the analysis of a 2-D signal, typically an image, the signal, which is an arbitrary surface, is decomposed into a set of sinusoidal surfaces. In Chapter 10, Two-Dimensional

4 Introduction

Discrete Fourier Transform, the theory and properties of the 2-D DFT is presented. The practically efficient way of computing the 2-D DFT is to compute the row DFTs followed by the computation of the column DFTs and vice versa. Using this approach, the 2-D PM DFT algorithms are derived.

In practice, most of the naturally occurring signals are continuous-time signals. It is by representing this signal by a set of finite samples, we are able to use the DFT. This creation of a set of samples to represent a continuous-time signal necessitates sampling and truncation operations. These operations introduce some errors in the signal representation but, fortunately, these errors can be reduced to a desired level by using an appropriate number of samples of the signal taken over proper record length. Therefore, the level of truncation and the number of samples used are a trade-off between accuracy and computational effort. A good understanding of the effects of truncation and sampling is essential in order to analyze a signal with minimum computational effort while meeting the required accuracy level. In Chapter 11, Aliasing and Other Effects, the problems created by sampling and truncation operations, namely aliasing, leakage, and picket-fence effects, are discussed.

The continuous-time Fourier series (FS) is the frequency-domain representation of a periodic continuous-time signal by an infinite set of harmonically related sinusoids. In Chapter 12, The Continuous-Time Fourier Series, the approximation of the continuous-time Fourier Series, 1-D and 2-D, by the DFT coefficients is described. The inability of the Fourier representation to provide uniform convergence in the vicinity of a discontinuity of a signal is also discussed. The continuous-time Fourier transform (FT) is the frequency-domain representation of an aperiodic continuous-time signal by an infinite set of sinusoids with continuum of frequencies. In Chapter 13, The Continuous-Time Fourier Transform, the approximation of the samples of the continuous-time Fourier transform, 1-D and 2-D, by the DFT coefficients is described.

A major application of the DFT is the fast implementation of fundamentally important operations such as convolution and correlation. In Chapter 14, Convolution and Correlation, the fast implementation of the convolution and correlation operations, 1-D and 2-D, using the DFT is presented.

The even extension of a signal eliminates discontinuity at the edges, if present, thereby enabling the signal to be represented by a smaller set

The Organization of this Book 5

of DFT coefficients. This special case of the DFT is called the discrete cosine transform and it is widely used in practice for signal compression. In Chapter 15, Discrete Cosine Transform, the computation of the discrete cosine transform, 1-D and 2-D, is presented.

While the sinusoids are the basis waveforms in the DFT representation of signals, a set of orthogonal rectangular waveforms is used to represent signals in the discrete Walsh-Hadamard transforms. These transforms, often used in image processing, are computationally efficient since only addition operations are required for their implementation. Algorithms for their computations are very similar to those of the DFT algorithms. The study of these transforms provides a contrast in representing an arbitrary waveform using a different set of orthogonal waveforms. In Chapter 16, Discrete Walsh-Hadamard Transform, the computation of the discrete Walsh-Hadamard transforms, 1-D and 2-D, is described.

In the Appendices, the complex numbers, the measure of computational complexity, the bit-reversal algorithm, the prime-factor DFT algorithm for a data size of six, and the testing of programs are briefly described. A list of useful mathematical formulas is also given.

The theory of the Fourier analysis is that any periodic signal satisfying certain conditions, which are met by most signals of practical interest, can be represented uniquely as the sum of a constant value and an infinite number of sinusoids with frequencies those are integral multiples of the frequency of the signal under analysis. In short, almost everything that is said in this book is concerned with this one line.

Chapter 2

The Discrete Sinusoid

A signal represents some information. Manipulation of signals, such as removing noise from a signal, is a major activity in applications in science and engineering. An arbitrary signal can be easily manipulated only by representing it as a linear combination of simple and mathematically well-defined signals. There are many ways a signal can be represented. The proper representation of a signal is crucial for the efficient manipulation of it. For the analysis and design of LTI systems, most often, signals are represented as a linear combination of the impulse signal in the time-domain and the sinusoidal signal in the frequency-domain.

In Sec. 2.1, we briefly describe the time- and frequency-domain representations of signals. The sinusoidal waveform is the principal object in Fourier analysis. In Sec. 2.2, we study the characteristics of the discrete sinusoidal waveform and its representation by complex exponentials. The orthogonality property of the sinusoids is also presented.

2.1 Signal Representation

Time-domain signal representation

Signals occur, mostly, in a form that is called the time-domain representation. In this form, the signal amplitude, x(t), is represented against time, t. If the signal is denned at all instants of time, it is referred as a continuous-time or analog signal. If the signal is defined only at discrete instants of time, then the signal is referred as a discrete signal. It is assumed that the interval between instants of time is uniform. Figures 2.1(a)

7

The Discrete Sinusoid

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0 4 8 12 16 20 24 28 n

(c)

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(d)

Fig. 2.1 (a) The continuous-time cosine signal, x{t) = cos(f t) . (b) The continuous time sine signal, x(t) = s i n ( | t ) . (c) The discrete cosine signal, x(n) = cos( f^n). (d) Thi discrete sine signal, x(n) = sin(y^n).

The

and (b) show, respectively, one cycle of the continuous-time cosine and sine signals x(t) = cos(ft) and x{t) = sin(ft). Figures 2.1(c) and (d) show, respectively, the discrete cosine and sine signals x(n) = cos(f^n) and x(n) — s in (^n) , obtained by sampling the continuous signals shown in Figs. 2.1(a) and (b) with a sampling interval of 0.25 seconds. For the most part, we deal with discrete signals in this book. However, the relationship between the continuous-time and discrete signal representations will be presented. The time and amplitude variables of a digital signal take on only discrete values and this form is suitable for processing using digital devices. We use the term time-domain although the independent variable is

Signal Representation 9

not time for all the signals. For example, in a speech signal, the amplitude of the signal varies with time whereas the intensity values of an image vary with two spatial coordinates. Signals such as speech signal, which vary with respect to a single independent variable, is called a one-dimensional (1-D) signal. An image is a two-dimensional (2-D) signal since it varies with respect to two independent variables.

A discrete signal is represented, mathematically, as a sequence of numbers {x(n), —oo < n < oo}, where the independent variable n is an integer and x(n) denotes the nth element of the sequence. Although it is not strictly correct, x{n) is also used to refer a sequence as the sequence x(n). The element x(n) of a sequence is often referred as the nth sample of the sequence regardless of the way the sequence is obtained. Usually, a discrete sequence is obtained by sampling an analog signal. However, discrete signals can also be generated directly. Even if the signal is obtained by sampling a continuous-time signal, the sampling instant is shown explicitly only when it is required as x(nTs), where Ts is the sampling interval.

The unit-impulse signal, shown in Fig. 2.2(a), is defined as

r , . ( 1 for n = 0 ' ( B ) = \ 0 f o r n ^ O

In practice, the input signal to a system, most often, is quite arbitrary and it is difficult to represent and manipulate it analytically. To circumvent this problem, it is a necessity to represent the signal as a linear combination of elementary signals. An arbitrary signal can be represented as the sum of delayed and scaled unit-impulses. An arbitrary discrete signal, {x(—1) = - l , x ( 0 ) = l,ar(l) = -3 ,z(2) = 2}, is shown in Fig. 2.2(b). This signal can be expressed as

2 x(n) = ] C x(m)ti(n-™>) o r x(n) = -6(n+l)+6(n)-36(n-l)+26(n-2)

m=—1

and the constituent impulses are shown in Figs. 2.2(c) to (f). With this type of representation, if the unit-impulse response of an LTI system is known, the response of the system to an arbitrary input sequence can be obtained by summing the responses to all the individual impulses.

10 The Discrete Sinusoid

1

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Fig. 2.2 (a) Unit-impulse signal, S(n), —5 < n < 5. (b) An arbitrary discrete signal. (c), (d), (e), and (f): The representation of the signal shown in (b) in terms of delayed and scaled impulses, (c) -S(n + 1), (d) 8(n), (e) -3<J(n - 1), and (f) 2S(n - 2).

Frequency-domain signal representation

An alternate representation of signals is called the frequency-domain representation. In this representation, the variation of a signal in terms of frequency is used to characterize the signal. At each frequency, the amplitude and phase or, equivalently, the amplitudes of the cosine and sine components of the sinusoid are required for representing a signal. Figure 2.3(a) shows two sinusoidal waveforms that are the components of a periodic signal. The sum of the projection of the two waveforms on the time axis is the time-domain representation of the signal, shown in Fig. 2.3(b). Figure 2.3(c) shows the representation of the signal on the frequency axis. The amplitude and the phase shift of the first sinusoid are, respectively, 1 and -60 degrees and those of the second sinusoid are, respectively, 1 and 90 degrees. It is evident that either representation completely specifies the signal. The independent variable is time in the time-domain representation. In the frequency-domain, the independent variable is frequency thereby explicitly specifying the frequency components of a signal. This book is about the frequency-domain representation of signals. Therefore, we have to explore the characteristics of the sinusoidal waveform.

The Discrete Sinusoid 11

frequency, Hz 0 0 time

• (1,-60) •(1,90)

1 2 frequency, Hz

(c)

Fig. 2.3 (a) Two sinusoidal components of a periodic signal, (b) Time-domain representation of the signal, (c) Frequency-domain representation of the signal.

2.2 The Discrete Sinusoid

The two waveforms we usually remember are the cosine and sine waveforms. We have already seen one cycle of the discrete versions of the cosine, cos(j^n), and sine, sin(j^n), waveforms, respectively, in Figs. 2.1(c) and (d). The magnitude of a peak value from the horizontal axis is called the amplitude of the waveform. The wave oscillates with equal amplitudes about the horizontal axis. There are two zero crossings in a cycle. In order to compare the positions of two or more waveforms of the same frequency along the horizontal axis, we have to specify a reference position. Let the occurrence of the positive peak of the waveform at Oth instant (n = 0) be the reference point and we define the phase shift of the waveform zero. Therefore, the phase shift of the cosine waveform is zero and it is used as the reference waveform in this book (The sine waveform can also be considered as the reference waveform.). The phase shift of a waveform is defined as the amount of the shift of the cosine waveform to the right or left to

(a)

(b)


obtain that waveform. If the shifting is to the right we define the phase shift to be negative and a shift to the left is positive. For example, the sine waveform has a -90 degrees or — | radians phase shift since we have to shift the cosine wave to the right by that amount to get the sine wave. What is called a sinusoid is a cosine or sine wave with arbitrary phase shift. The cosine and sine waveforms are important special cases of the sinusoid with phase shifts of zero and -90 degrees, respectively.

The polar form

A discrete sinusoidal waveform is mathematically characterized as

x(n) = Acos(u>n + 0), n = - c o , . . . , — 1 , 0 , 1 , . . .,oo (2.1)

where A is the amplitude (half the peak-to-peak length), cu is the angular frequency of oscillation in radians per sample, and 9 is the phase shift in radians. The cyclic frequency of oscillation / is ^ cycles per sample. The period N is 4 samples (The period of a discrete sinusoidal waveform is j only when \ is an integer. We will consider the more general case later.). For the waveform shown in Fig. 2.1(c), the amplitude is 1, the phase shift is 0 (that is the positive peak of the waveform occurs at the point n = 0), the angular frequency, u, is f radians per sample, and the cyclic frequency / is ^ cycles per sample. The period is 32 samples, that is, the waveform repeats any 32-point sequence of its sample values, at intervals of 32 samples, indefinitely, x(n) = x(n ± 32) for any n. The interval between two samples is ^ = 11.25 degrees. Therefore, the values of the cosine and sine functions at intervals of 11.25 degrees can be read from this figure. For the waveform shown in Fig. 2.1(d), the amplitude is 1 and the phase shift is — ^ radians (shift of the cosine waveform by six samples (6/u — —6) to the right), that is, the positive peak of the waveform occurs after f radians from the point n = 0. The angular frequency, ui, is y^, and the cyclic frequency / is ^ cycles per sample. The period is 24 samples, that is, the waveform repeats any 24-point sequence of its sample values, at intervals of 24 samples, indefinitely. The values of the sine and cosine functions at intervals of 15 degrees can be read from this figure. A shift by an integral number of periods does not change a sinusoid. If a sinusoid is given in terms of a phase-shifted sine wave, then it can be, equivalently, expressed in terms of a phase-shifted cosine wave as x (n) = Asm(un+9) = A cos(am+(#-§)) . Conversely, x(n) = Acos(um + 9) = Asm(uin + (9 + f ) ) .


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Fig. 2.4 The sinusoid, x(n) = 5cos(f n - ^ ) .

Example 2.1 Determine the amplitude, angular and cyclic frequencies, the period, and the phase shift of the following sinusoid.

x{n) = —5 cos(—n + —)

Solution By adding a phase shift of —7r (as the amplitude is always a positive quantity, —ACOS(LJTI + 8) = Acos(am + 8 ± 7r)), we get

x(n) = 5cos(— n ——) T: O

Now, A = 5,w = j radians per sample, / = 2V = g cyc^es P e r sample, N = 8 samples, and the phase shift is — radians. One cycle of the sinusoid is shown in Fig. 2.4. I

In Fig. 2.4 (and in most of the figures in this book), we have shown the corresponding continuous waveform for clarity. However, it should be remembered that, in discrete signal analysis, a signal is represented only by its samples.

Even if the peak value does not occur at a sample point, for a given LJ, the amplitude and phase of a sinusoid can be obtained by solving the equations

x(n) = A cos(um + 8) and x(n + 1) = Acos(u(n + 1) + 8)

Values x{n) and x(n+l) are, respectively, the nth and the (n+l ) th samples, assuming that the number of samples in a period is, at the least, one more than twice the number of cycles. Solving these equations for 6 and A, we get

_x x(n) cos(w(n + 1)) - x(n + 1) cos(wn) 8 — tan

x(ri) sin(w(n + 1)) — x(n + 1) sin(wn) (2.2)


A = *(n) (2 3) cos(am + 0) v ' '

Since the tangent function has period TT, the signs of the numerator and denominator must be taken into account in determining the angle 9.

Example 2.2 Let a;(l) = 1 and x(2) = —1 be the two samples of a sinusoid with frequency / = \ cycles/sample. Find the polar form of the sinusoid. Solution u) = 2nf = f radians/sample. Substituting the values in Eqs. (2.2) and (2.3), we get

0 _ t a n - l l c O 8 ( 7 r ) - ( - 1 ) C O S ( ? ) _ * A - 1 - , / n

* - t a n l s i n ( 7 r ) - ( - l ) s i n ( f ) - 4 ' A ~ cos(f - f) " VZ

Therefore, the sinusoid is x(n) = \ /2cos ( |n — f ) . I

The rectangular form

In the polar form, a sinusoid is represented by its amplitude and phase. In the rectangular form, a sinusoid is represented in terms of the amplitudes of its cosine and sine components. By expanding Eq. (2.1), the rectangular form of representing a sinusoid is obtained as

x(n) = Ccos(am) + -Dsin(wn),

where C = A cos 6 and D = —A sin 9. The inverse relation is A = VC2 + D2

and 9 = c o s " 1 ^ ) = s i i T 1 ^ ) .

Example 2.3 Express the following sinusoid in rectangular form.

x(n) = 5cos(— n —)

Solution

C = 5 c o s ( - y ) = -2 .5 , J? = - 5 s i n ( - y ) = 5 ^

Therefore, the sinusoid, in the rectangular form, is given by

x(n) = -2 .5cos( -n) + 5 — s i n ( - n )


1.7678

0

-2 5 D 2 4

n

(a)

6

3.0619

- # ° -4.3301

3 2 4

(b)

6

Fig. 2.5 (a) The cosine component, x(n) = — 2.5cos( j n ) , and (b) the sine component,

x(n) = \ / 3 (2 .5 ) s in (^n) , of the sinusoid shown in Fig. 2.4.

The sinusoid, and its cosine and sine components are shown, respectively, in Figs. 2.4, 2.5(a), and (b). We can easily verify that each sample value of the sinusoid is the sum of the corresponding samples of its cosine and sine components. I

Example 2.4 Express the following sinusoid in polar form.

i n = cos(—n) + sm(—n) 6 6

Solution

A = \ / l 2 + l 2 = y/2, 9 = cos _1 I —= I = sin x I —=. ) = — — radians

\V2j \V2j 4

Hence, the sinusoid is given by x(n) = y/2cos(fn — j) in the polar form. I

The rectangular form of a sinusoid shows clearly that a sinusoid is a linear combination of sine and cosine waveforms of the same frequency. This point is so important that we provide an alternate viewpoint. Any function x(n) can be expressed as the sum of an even function *W+2*(-") a n d

an odd function x\n)-*\-n) _ N0te that, for a periodic function with period N, x(N — n) can also be used instead of x(—n). Therefore, an arbitrary sinusoid, which is neither odd nor even, can be expressed as the sum of an odd function and an even function. For a sinusoid, the odd function is a sine function and the even function is a cosine function of the same frequency. Acosjun + 6) + Acos(w(N -n) + 6)

2 vlcos(um + 0) - Acos{u(N - n) +6)

= Acos(6)cos(u:n)

= -Asm(0)sm(un)


Example 2.5 Use even and odd split to find the sample values of the cosine and sine components of the sinusoid, x(n) = \ /2cos(fn — J ) . Solution The sample values of the sinusoid for n = 0,1,2,3 are {1,1, —1, —1}. Using the even and odd split, we get the sample values of the cosine and sine components, respectively, as {1,0, - 1 , 0 } and {0 ,1 ,0 , -1} . I

The sum of sinusoids of the same frequency

The sum of discrete sinusoids of the same frequency but arbitrary amplitudes and phases is a sinusoid of the same frequency. Let

#i (n) = A\ cos(um + 9\) and #2(71) = A2 cos(om + 92)

Then, £3 (n) = x\ (n) + x2(n) = A3 cos(un + #3). Expressing the waveforms in rectangular form, we get

x\(n) = a\ cos(um) + 61 sin(um), a\ = A\ cos(#i), 61 = —Ai sin(#i)

£2(71) = a2 cos(wn) + 62 sin(um), 02 = A2 cos(92), b2 = —A2 sin(#2)

x3(n) = a3 cos(wn) + 63 sin(um), a3 = A3 cos(03), b3 = -A3 sin(03)

It is obvious that a3 = ai + 02 and 63 = &i +62- Converting from rectangular form to the polar form, we get

A3 = y/(ai + a2)2 + (&i + b2)

2 = yJA\ +A\+ 2AXA2 cos(91 - 92)

_, Ai cos(^i) + A2 cos(92) . _, Ai sin(0i) + A2 sin(02) 63 — cos -——. - — = sin -—'—

A3 A3

By repeatedly adding, any number of sinusoids of the same frequency can be combined into a single sinusoid.

Example 2.6 Determine the sinusoid that is the sum of the two sinusoids

Xl(n) = - 4 . 3 c o s ( | n - f$) and x2(n) = 3 . 2 c o s ( f n - f ) . Solution The first sinusoid can also be expressed as xi (n) = 4.3 cos( |n + ^jf )• Now,

A 1 = 4 . 3 , A 2 = 3 . 2 , <?i = ^ , 02 = - |


3.7239

1.1129

-2.15

-4.1535 0 2 4 6

n

2.9782

1.6583

^T-0.6329

-2.5534

(b)

0 2 4 6 n

(c)

Fig. 2.6 (a) The sinusoid x i (n) = 4 . 3 c o s ( | n + ^ f 1 ) . (b) The sinusoid X2(n) 3 . 2 c o s ( f n - | ) . (c) The sum of n ( n ) and x2(n), i 3 ( n ) = 3.0447 cos ( fn - 2.5656).

Substituting the numerical values in appropriate equations, we get

A3 = W4.3 2+3.22+ 2(4 .3) (3 .2)cos(^ + | ) = 3.0447

_ 1 4 . 3 c o s ( ^ ) + 3 . 2 c o s ( - f ) _, 4 .3s in (^ f )+3 .2s in ( - f ) 63 = cos Hr^rTT^ — =sin — —

3.0447 = —2.5656 radians

3.0447

The waveforms of the two sinusoids and their sum, X3(n) = 3.0447 cos(fn-2.5656), are shown, respectively, in Figs. 2.6(a), (b), and (c). I

Periodicity

The condition for a discrete sinusoid to be periodic is that the cyclic frequency / is a rational number (a ratio of two integers). For a discrete sinusoid to be periodic with period N,

Acos(un + 6) = Acos(uj(n + N) + 9)

Since a sinusoid is periodic only with an integer multiple of 27r, this implies LJN = 2nfN = 2irl, where / and N are integers. That is, / = jj.

Example 2.7 Is the waveform periodic? If periodic, what is the period? (a) x(n) = cos(^Ln) (b) x(n) = 4cos( |n) Solution (a) From inspection, the cyclic frequency, / = ^ , is a rational number. Therefore, the waveform, shown in Fig. 2.7(a), is periodic with a period


(a)

•S- o

(b)

Fig. 2.7 (a) The sinusoid x(n) = cos( -n) is periodic with a period of AT = 16 samples. (b) The sinusoid x(n) = 4cos(|n) is not periodic.

of 16 samples. The waveform repeats any 16-point sequence of its sample values, at intervals of 16 samples, indefinitely. (b) Prom inspection, the cyclic frequency, / = j ^ , is an irrational number. Therefore, the waveform, shown in Fig. 2.7(b), is not periodic. I

Highest frequency for unique representation

A sinusoid which completes k cycles in its period has a distinct set of 2k +1 sample values. Due to the representation of a waveform by a finite number of samples, in practice, sinusoids with a finite number of frequencies only can be uniquely identified. Consider the following identities with positive integers N, m, and k.

, 2 T T . ,2TJ\ cos(—(k + mN)n + 0) = cos(—fcn + 0)

•N . 2 T T 2TT

cos(—(mN — k)n + 6) = cos(—kn — 6) -N N

(2.4)

(2.5)

With increasing frequencies, oscillations increase only up to k — y (with N even), decrease afterwards, and cease at k = N. This pattern continues forever. Therefore, sinusoids with k up to ^ ~ 1 on^y (f°r cosine waves k up to Y) c a n D e uniquely identified with N samples.

Example 2.8 Specify two higher frequency sinusoids with the same set of sample values as that of

x[n) = 4 s m ( - n + - )

Solution Two of the innumerable number of sinusoids with the same set of sample


0 1 2 3 n

Fig. 2.8 The sinusoids x(n) = 4s in( | ra + *•), x(n) = 4s in(5 | ra + | ) , and x(n) ~ —4sin(3^n — ^ ) . All the three sinusoids have the same set of sample values.

values are

/ \ J . /57T 71". . . .37T 7T.

x(n) = 4sm(—n + - ) , x(n) = - 4 s i n ( y n - - )

The three waveforms are shown in Fig. 2.8. I

Harmonically related sinusoids

Harmonically related sinusoids are a set of sinusoids, called harmonics, comprising a fundamental harmonic with a frequency / and other harmonics having frequencies nf, where n is a positive integer. The frequency of the second harmonic is 2 / , that of the third harmonic is 3 / , and so on. The nth harmonic completes n cycles during the period of the fundamental. For example, the fundamental 3cos(| |-n + f ) , the second harmonic -2cos(ff2n + f ) , and the third harmonic cos(|f3n + f ) are shown in Fig. 2.9. The sum of discrete sinusoids with harmonically related frequencies is not sinusoidal, but it is periodic.

Example 2.9 Find the period of the combination of the sinusoids. Plot one period of the waveform of the first, the sum of the first and second, and the sum of the three of the frequency components.

. , 8 . . .2TT , 1 . .2TT . 1 . ,27rr „ x(n) = ^ ( s m ( - n ) - - s i n ( - 3 n ) + - s i n ( - 5 n ) )

Solution This is an approximation of the triangular waveform with amplitude one. The period is 16. Only the fundamental harmonic is shown in Fig. 2.10(a) with dotted line. The error in this representation is shown in Fig. 2.10(b). Figures 2.10(c) and (d) show, respectively, an approximation and the error with the sum of the first and third harmonics. Figures 2.10(e) and (f) show,


0 5 10 15 n

Fig. 2.9 The fundamental x(n) = 3cos(y|-7i + | - ) , the second harmonic x(n) = -2cos ( | ^2 ra + j), and the third harmonic x(n) = c o s ( ^ 3 n + j). All the three sinusoids are periodic with a period of N = 16 samples.

respectively, the sum of the first, third, and fifth harmonics, and the resulting error in the approximation. Note that the error in the approximation of the triangular waveform reduces as more and more harmonics are used.

I

To find the period of the combination of sinusoids of various rational cyclic frequencies: (i) cancel out any common factors of the numerators and denominators of each of the frequencies and (ii) divide the greatest common divisor of the numerators by the least common multiple of the denominators of the frequencies. This yields the fundamental frequency. The denominator of the fundamental frequency is the fundamental period.

Example 2.10 Find the fundamental cyclic frequency of the sum and the harmonic numbers of the two sinusoids.

xi(n) = c o s ( y n - — ) , x2{n) = s i n ( y n - - )

Solution The cyclic frequency of the waveforms are / i = | and fi = \- There are no common factors of the numerators and denominators. The least common multiple of the denominators (5,3) is 15. The greatest common divisor of the numerators (2,1) is one. Therefore, the fundamental cyclic frequency is TE-. T n e fundamental period is 15 samples. Frequency / i is the 6th harmonic and /•! is the 5th harmonic. The first sinusoid completes 6 cycles


t ° 10 15

0.2

0

-0.2 10 15

(b)

(e)

0.1

0

-0.1

0.05

I o -0.05

10 15

(d)

5 10 15 n

(f)

Fig. 2.10 The approximation of a triangular wave, (a) The approximation with the fundamental and (b) the resulting error, (c) The approximation with the fundamental and third harmonics, and (d) the resulting error, (e) The approximation with the fundamental, third, and fifth harmonics, and (f) the resulting error.

(Fig. 2.11(a)) and the second sinusoid completes 5 cycles (Fig. 2.11(b)) in the period. The combined waveform, shown in Fig. 2.11(c), completes one cycle. |

The complex sinusoid

It is a well-known fact that physical quantities such as force, velocity, etc., which require more than one parameter to describe, are compactly represented using vectors. Therefore, a sinusoid, which, at a given frequency, is characterized by its amplitude and phase shift, is also compactly represented and efficiently manipulated using vectors. A complex number, which is a two-element vector, is the ideal entity for the representation of a sinu-


*M *M *i* 0 5 10 15 0 5 10 15 0 5 10 15

n n n (a) (b) (c)

Fig. 2.11 (a) The sinusoid xi(n) = c o s ( ^ n — f^) completes 6 cycles during 15 samples. (b) The sinusoid X2{n) = s i n ( ^ n - | ) completes 5 cycles during 15 samples, (c) The sum of the two sinusoids completes one cycle during 15 samples. All the three waveforms are periodic with a period of N = 15 samples.

soid. With this representation, we get the advantage of manipulating both the amplitude and phase of a sinusoid at the same time.

Although physical systems always interact with real signals, it is often mathematically convenient to represent real signals in terms of complex signals. We are looking for a single entity that represents both cosine and sine functions in a compact form and is much easier to manipulate. This function is the complex exponential or complex sinusoid representing a rotating vector as described in Appendix A. The complex exponential function is a functionally equivalent mathematical representation of the sinusoidal waveform, but is more convenient for manipulation than the cosine and sine functions.

The complex exponential function with an imaginary argument is given

by

x(n) = Ae>tun+V = Aei6eiwn, n = - o o , . . . , - 1 , 0 , 1 , . . . ,oo

The term eJW" is the complex sinusoid with unit amplitude and zero phase shift. This form of the sinusoid is more commonly used in theoretical and practical Fourier analysis due to its compact form and ease and efficiency of manipulation. By multiplying with the complex (amplitude) coefficient Ae^, we can generate a complex sinusoid with arbitrary amplitude and phase shift. The complex coefficient Aeje is a single complex number containing both the amplitude and phase of a sinusoid.

The complex conjugate of the complex exponential is Ae~j(-un+0\ By adding the complex exponential with its conjugate and dividing by two,


due to Euler's identity, we get

x ( n ) = ^(ei(^n+fi) + e-j(«n+«)j = ^ c o s ( W n + 9)

The right side of this equation is a real sinusoid described earlier, which is of interest in practical applications. Note that the terms appearing on the left-hand side are complex conjugates which combine to represent the real function A cos (am + 6). The plot of the amplitude and phase (or the real and imaginary parts) of the complex coefficients of the complex sinusoids of a signal against frequency is its complex spec t rum.

Example 2.11 Find the spectrum of the signal.

/ \ /I* 27T. x(n) = 8 c o s ( - n - —)

Solution The waveform is shown in Fig. 2.12(a). Since u = 2nf = | , / = i cycles/sample. Expressing the waveform in terms of complex sinusoids, we get

a;(n) = ^ ( e ^ t « - ¥ ) + e - i ( f « - ¥ ) )

The complex frequency coefficients are 4e~J~3L and 4e-7"3L. Therefore, the amplitude and phase of the spectrum at / = ^ are 4 and — p radians and those at / = — are 4 and ^-, respectively. The spectrum, in terms of amplitude and phase, is shown in Fig. 2.12(b). The real and imaginary parts of the spectrum and the corresponding cosine and sine components of the sinusoid are shown, respectively, in Figs. 2.12(c) and (d).

It should be observed that the real part (as well as the amplitude) of the spectrum is even-symmetric and the imaginary part (as well as the phase) is odd-symmetric. We use four real values to represent a real sinusoid instead of two. This redundancy, in terms of storage and operations, can be eliminated as described in a later chapter. I

Both the dc and the frequency component with frequency index y (with N even) have only real spectral values, A or —A, for real signals. That means the phase shift of these frequency components is 0 or ±n. It should be noted that real sinusoids are easy to visualize, while complex sinusoids are easy to manipulate.


8

0

-8 10 15

(a)

. 3.4641 00

a 0 1 -2 ^ -3.4641

o

•

• real o imaginary

• . , 9

-1/16 0 1/16 / cycles/sample

(c)

4

2.0944

0

-2.0944

•amplitude ophase

-1/16 0 1/16 / cycles/sample

(b)

• cpsine osine

(d)

Fig. 2.12 (a) The sinusoid x(n) = 8 c o s ( | n - ^p) . (b) The spectrum of the sinusoid showing the amplitude and phase, (c) The spectrum showing the real and imaginary parts, (d) The cosine and sine components of the sinusoid.

Orthogonality

If the sum of pointwise products of two real discrete signals (for complex signals, the pointwise products of a signal with the complex conjugate of the other signal) is zero over a specified interval, the signals are said to be orthogonal in that interval. This property allows us to represent and manipulate a single frequency component of a signal as though other frequency components do not exist.

The sum of the samples of a discrete cosine or sine function is zero over an integral number of periods with any staring point. This is obvious due to the symmetry of these functions about the horizontal axis. The product of two harmonically related discrete cosine and sine signals, over an integral number of periods, will be odd. Therefore, they are orthogonal. Figures 2.13(a) and (b) show, respectively, signals cos(f n) and sin(f n). Figure 2.13(c) shows the product of these signals, cos(f n) sin(fn) = 0.5(sin(^n) - sin(f n)). From the odd symmetry of the waveform, it is obvious that the sum of the samples is zero.

Consider the products of the form cos(Zam) cos(mwn), I, m = 0 ,1 , and sin(Jwn) sin(mum), /, m = 1,2,.. . , y

2

1, where u> = ^ and N is even.


* °

S °

f °

• « • B • B

S B B S

0 5 10 15 n

(a)

B B S B S B

S B S B

0 5 10 15 n

(d)

B B B B B

B • B B

0 5 10 15 n

(g)

1

1

+ + + +

+ + + +

+4-+

0 5 10 15 n

(b)

1

t ° +++

+ + + +

+ + + +

+4.+

0 5 10 15 n

(e)

1

S ° 1

+ + + +

+ + + +

+4-+

0 5 10 15 n

(h)

1

1

o sum = 0 o o o o

o o o o

0 5 10 15 n

(c)

1

5 0

1

o o o o

oo o°

0 5 10 15 n

(f)

1

1 ° 1

3 ° sum = 0 °

o o o o

o o

0 5 10 15 n

(i)

Fig. 2.13 (a) The sinusoid, x(n) = cos(^n) . (b) The sinusoid, x(n) = s in (^n) . (c) The product of the waveforms of (a) and (b). (d) The sinusoid, x(n) = sin( J n ) . (e) The sinusoid, x(n) = s i n ( Ju ) . (f) The product of the waveforms of (d) and (e). (g) The sinusoid, x(n) = cos(^n) . (h) The sinusoid, x(n) = cos(^n). (i) The product of the waveforms of (g) and (h). The sum of the samples of each of the waveforms shown in (c), (f), and (i) is zero.

These products can be written as sums using trigonometric identities as

cos(Zom) cos(mam) = 0.5(cos((Z — m)um) + cos((Z + m)um))

sin(Zo;n) sin(mwn) = 0.5(cos((Z - m)um) — cos((Z + m)um))

In both cases, if I =£ m, the functions are cosines and, as noted above, the sum of the sample values over an integral number of periods is zero. Figures 2.13(d) and (e) show, respectively, signals sin(Jn) and s in( |n ) . Figure 2.13(f) shows the product of these signals. Figures 2.13(g) and (h) show, respectively, signals cos( |n) and cos( |n) . Figure 2.13(i) shows the product of these signals. The sum of the sample values of all the rightmost


0.1464 0

(a)

Fig. 2.14 (a) The product cos(£ n)cos( Jn) , with N = 16, and the sum of the samples is 8. (b) The product sin(f n)sm(^n), with N = 8, and the sum of the samples is 4.

waveforms in Fig. 2.13 is zero. If I = m ^ 0 or I = m ^ ^> *^e s u m °f *^e

second term evaluates to zero (since it is a cosine) while the first term is cos (0) and the sum of N samples multiplied by 0.5 equals y . Figures 2.14(a) and (b) demonstrate this property through specific examples. The first one shows the product cos(fn) cos( |n) = cos2(f-n) = 0.5(1 + cos( | n ) ) (a cosine wave of amplitude 0.5 with a positive bias of 0.5 superimposed on it) and the second one shows the product s i n ( | n ) s i n ( | n ) = s in 2 ( |n ) = 0.5(1 — cos( |n)) (an inverted cosine wave of amplitude 0.5 with a positive bias of 0.5 superimposed on it). It can be seen that the sum of the sample values is 8, with N = 16, in Fig. 2.14(a) and the sum is 4, with JV = 8, in Fig. 2.14(b). Figures 2.13(f), 2.13(i), 2.14(a), and 2.14(b) illustrate the property that the sum of products of two even or two odd functions (as the product function is even) can be computed using just over half the samples of the product function. If/ = m = 0 o r / = m = y , with N even, the sum of the samples of the product of two cosines is N.

For the two complex exponential signals eJ~^mn and ej~&ln over a period of N samples, the orthogonality condition is given by

J V - l

n = 0

'-(m-l) Mo N for m = I for m / I

where m, / = 0 , 1 , . . . , N -1. The expression in the summation can be written as cos (^mn) cos(%ln) + sin(%-mn) sin(^-in) -jcos(^mn) sin(%ln) +j sin(^-mn) cos(^Zn). The sum of this expression over the interval n = 0 to N - 1 is equal to N when m = I and zero otherwise. This result is also established by using the closed-form expression for the sum of a geometric

5ummorj/ and Discussion 27

progression, when m / / (when m — l, the summation is equal to N).

£ e i^(m-0n = e

2 i r ( _ t ) = 0, for m * l

n=0 l-eJ—w—

2.3 Summary and Discussion

• In addition to appreciating the necessity of representing arbitrary signals as a linear combination of simple signals, in this chapter, we learned the properties of a discrete sinusoid and the efficient way to represent it mathematically.

• Most signals that occur in practical applications are continuous-time signals having arbitrary amplitude profile. These signals are very difficult to manipulate analytically and numerical methods are resorted to using the discrete signal obtained by sampling the signal at periodic intervals.

• While numerical processing enables us to manipulate arbitrary signals, it requires considerable computational effort. When the discrete signals are represented in terms of their frequency components, it turns out that the computational effort required for signal processing operations is reduced significantly.

• In the frequency-domain representation of signals, a signal is represented as a linear combination of a set of sinusoids. A sinusoid itself is a linear combination of sine and cosine waveforms of the same frequency. The sine and cosine waveforms are defined as the vertical and horizontal projection, respectively, of a point moving around a circle, with center at origin, at uniform angular velocity. Therefore, the most efficient way of representing a sinusoid is using the function eiuin which represents a point moving around the unit circle at angular velocity u.

• The representation of signals in terms of sinusoids is very efficient because of the advantageous properties of the sinusoids. The orthogonality property implies that the representation and manipulation of a sinusoid at a specific frequency can be carried out with the assumption that no other sinusoids are present in a signal.

• A time-domain signal can be decomposed into a set of scaled and delayed impulses or sinusoids with various frequencies, amplitudes,


and phases. The representation using sinusoids is more efficient for the analysis and design of signals and systems. However, signals do not occur in this form naturally. Therefore, we have to derive the frequency-domain representation of a signal from its time-domain representation. The Fourier transform is the tool to do this job. In the next chapter, we shall find out the relationship between the time-domain and the frequency-domain representation of signals.

Reference

(1) Cadzow, J. A. and Van Landingham, H. F. (1985) Signals, Systems, and Transforms, Prentice-Hall, New Jersey.

Exercises

2.1 Determine the amplitude, angular and cyclic frequencies, the period, and phase of the sinusoid. 2.1.1. x(n) = 7 s i n ( f n + f ) * 2.1.2 x(n) = - 7 . 1 s i n ( f n - 2=) 2.1.3 i(r») = -2.2cos^fn) 2.1.4 i (n) = 3 c o s ( | n + f ) 2.1.5 x(n) = -1 .1 sin(ffn)

2.2 Given two adjacent samples and the frequency of a sinusoid, find the polar form of the sinusoid. 2.2.1. x(0) = 2>/2,a:(l) = (-%/2)(V3 - l),w = f

2.2.2 x(0) = ^ , x ( l ) = =^,u= 2f

* 2.2.3 ar(l) = ( ^ ) ( V 5 - l) ,z(2) = (=^)(V3 - l),a, = f

2.2.4 x(2) = ( = ^ ) , x ( 3 ) = ( = ^ ) ( V 3 + l),u = f 2.2.5 a:(0) = >/3,a:(l) - ( ^ ) ( \ / 3 - l),w = |

2.3 Express the sinusoid in rectangular form. 2.3.1 x(n) = 5 c o s ( | n + f ) 2.3.2 x(n) = cos(f n - \) 2.3.3z(n) = 4sin(fn + f ) 2.3.4 x{n) = 2cos(f |n) 2.3.5 x(n) = 7 c o s ( f f n - f ) .

Exercises 29

* 2.3.6z(n) = 3cos(fn + f ) 2.3.7 x(n) = s i n (^n )

2.4 Express the sinusoid in polar form. 2.4.1 x(n) = -4cos ( fn ) +4s in ( fn ) 2.4.2 x(n) = cos (^n) + v^Jsin(^n) 2.4.3 x(n) = \/3cos(ff n) + sin(ff n) 2.4.4 x(n) = cos(f n) - V ^ i n ^ n ) * 2.4.5 x(n) = - c o s ( f n ) - s i n ( f n ) 2.4.6 i (n) = -3s in ( fn ) . 2.4.7 x(n) = -2cos ( fn) .

2.5 Determine the sinusoid that is the sum of the pair of sinusoids. 2.5.1 2 . 3 c o s ( f n + f ) , - 7 . 1 s i n ( | n - ^ ) 2.5.2 2 .7cos(fn) , -s in(fn) * 2.5.3 4 . 7 c o s ( f n - j | ) , c o s ( f n - f) 2.5.4 2.2cos(fn), l . lcos(fn) 2.5.5 1.2sin(|n),3.1sin(fn)

2.6 Is the waveform periodic? If periodic, what is the period? 2.6.1 x(n) = cos(0.37rn) 2.6.2 x(n) = cos(\/37m) 2.6.3 x(n) = cos(^ |n)

* 2.6.4 x(n) = s i n ( i ^ n ) 2.6.5 x(n) = 3 + cos(7rn) + cos(2n)

2.7 Find the polar form of three higher frequency sinusoids with the same set of sample values as that of x(n). 2.7.1 x(n) = 5cos ( fn+ f) * 2.7.2 s(n) = s i n ( ^ n - f )

2.8 Find the fundamental cyclic frequency of the sum and the harmonic numbers of the two sinusoids. * 2.8.1 c o s ( ^ n ) , c o s ( ^ n ) 2.8.2 cos(^n) ,cos(fn) 2.8.3 5,s infn

2.9 Find the spectrum of the signal in terms of: (i) amplitude and phase and (ii) real and imaginary parts. 2.9.1 x(n) = 6 c o s ( f n + f )


* 2.9.2 a;(n) = 4 c o s ( f n - f ) 2.9.3 x(n) = cos (^n) + \ / 3 s i n ( ^ n ) 2.9.4 x(n) = lOsin(fn) 2.9.5 x(n) = -2cos( fn) 2.9.6 x\n) = - 2 2.9.7 x(n) = 3cos(7rn)

Programming Exercises

2.1 Write a program to generate the sample values of a real discrete sinusoid, Acos(jj-n + 6), with period N, phase 6, and amplitude A.

2.2 Write a program to generate the sample values of a complex discrete sinusoid, Ae^^n+8\ with period N, phase 6, and amplitude A.

Chapter 3


In this chapter, the tools that enable the conversion between the time-and frequency-domain representations of discrete signals, the DFT and the IDFT, are derived. These are the mathematical formulations of the problem of waveform analysis and synthesis, respectively, with harmonically related discrete sinusoids as the basis functions. An infinite number of sinusoids with various frequencies, in general, are required for the accurate analysis and synthesis of an arbitrary waveform. But, out of necessity, only a finite number of sinusoids are used in practice. The consequences of this limitation will be considered in Chapter 11. For the time being, we assume that, unless otherwise stated, the constituent frequency components of a real periodic signal, with N (N even) samples in a cycle, are sinusoids with frequency indices 0 , 1 , . . . , y only. Note that the sinusoid with zero frequency index is a dc signal and the sinusoid that can be represented with the frequency index y is a cosine waveform. The DFT is defined for any length. But, due to the availability of fast algorithms with high regularity, the lengths those are integral powers of two are most often used in practice. Therefore, we put emphasis on these lengths in the development of the DFT theory and algorithms.

In Sec. 3.1, the DFT and the IDFT expressions are obtained through a simple example. In Sec. 3.2, the DFT and the IDFT expressions are formally derived and the symmetry property of the DFT kernel matrix is analyzed. In Sec. 3.3, the closed-form expressions of the DFT of some simple signals are derived. In Sec. 3.4, the direct computation of the DFT is described. In Sec. 3.5, the advantages of sinusoidal representation of signals are listed.

31

32 The Discrete Fourier Transform

3.1 The Fourier Analysis and Synthesis of Waveforms

Consider the waveform shown in Fig. 3.1(d) with sample values 2 + ^ , | , 2— ^ , and — | . The waveform is periodic with a period of 4 samples and one cycle of which is shown. The problem of Fourier analysis is to find the sinusoids, the superposition summation of which yields this waveform. The solution is shown in Fig. 3.1: (a) a dc component xa(n) = 1, (the spectral value X(0) = 1), (b) a sinusoid xb(n) = cos(f n - f ) , (X(l) = ^ - j \ and X(3) = ^ + j \ ) , and (c) a sinusoid xc(n) = cos(7rn), (X(2) = 1). The sums of the corresponding sample values of these three waveforms are equal to the sample values of the waveform shown in Fig. 3.1(d), x(n) — xa(n) + xb(n) + xc(n).

The spectrum of x(n) is displayed in Fig. 3.1(e) with a scale factor of 4. The real part of the spectrum X(k) is marked with the symbol '• ' and the imaginary part is indicated by the symbol V . The two problems in Fourier analysis are: (i) given the time-domain waveform of a signal, how do we get the coefficients of the individual sinusoids (analysis) and (ii) given the coefficients of the individual sinusoids, how do we get the combined time-domain waveform (synthesis). We have started with the problem and the solution in order to easily understand the problem formulation and the method of obtaining the solution. The periodic sequence shown in Fig. 3.1(d) can be considered as the output of a single complex signal generator or, equivalent^, the output of three simple signal generators connected in series as shown, respectively, in Figs. 3.2(a) and (b).

The Fourier analysis of waveforms

Each frequency component contributes to the value of the time-domain waveform at each sample point. By multiplying each of the complex frequency coefficient by the sample value of the corresponding complex sinusoid and summing the products must be equal to the time-domain sample value. Therefore, we set up four simultaneous equations with four time-domain samples.

X(0)e^(«)(o) + x ( l ) e ^ ( O W + X(2)eJT(°)(2) +X(3)e^(o)(3) = x ( 0 )

l ( 0 ) e ^ « ( o ) + X ( l ) e ^ ( i ) W + I ( 2 ) e * T P ) P ) + X ( 3 ) e ^ W ( 3 ) = x ( l ) X ( 0 ) e ^ ( 2 ) W + X ( l ) e ^ ( 2 ) ( 1 ) + X ( 2 ) e ^ ( 2 ) W +X(3)e^(2)(3) = X(2) X(0)e^(3)(o) + X(i)e»*(3)(D + X(2)<W><2> +X(3)e^(3)(3) = x ( 3 )

The Fourier Analysis and Synthesis of Waveforms 33

(b)

(c)

(d)

4

«. 1.73? X 1 0

-1

•

o • .

•

9

•

o


• o

(e)

Fig. 3.1 (a) A dc signal, (b) The sinusoid cos(f n - | ) . (c) The sinusoid COS(TTO). (d) The sum of the signals shown in (a), (b), and (c). (e) The spectrum of the signal shown in (d).


x(n) = 1 + cos(f n - f ) + cos(7rn)

x(n) = cos(nn)

x(n) = cos( |n - | )

• • • •

(a)

x(n) = 1

(b)

Fig. 3.2 The signal shown in Fig. 3.1(d) can be considered as the output of a single complex signal generator, (a), or, equivalently, it can be considered as the output of three simple signal generators connected in series, (b).

The contributions of the frequency components to the sample x{0) are each coefficient multiplied by unity, because the zeroth sample values of the complex sinusoids are unity. Therefore, the sum of these products must be equal to x(0). For the waveforms shown in Fig. 3.1, we get

(i)(i) + ( ^ - j i)( i) + (i)(i) + ( ^ +; l ) ( i ) = 2 + ^

In general, the computational complexity of solving a set of N simultaneous equations with N unknowns is 0(N3). Fortunately, the orthogonality property of the sinusoids can be used to reduce the complexity to 0(N2). Let us evaluate the coefficient X(l). If we multiply the terms of each column of the four simultaneous equations by the sample values of the conjugate of the complex sinusoid with frequency index one, the sum of all columns, except the second (with frequency coefficient -X"(l)), on the left-hand side vanish due to the orthogonality property and the problem is

The Fourier Analysis and Synthesis of Waveforms 35

reduced to summing the following two columns.

j r ( l ) e W > ) ( i ) e - ^ ( ° ) W = z(0)e-''T(°)(i) X( l )e iT(DW e - iT( i ) ( i ) = xtfe-*3?™1) X(l)e^(2)( i ) e - i¥(2) ( i ) = o;(2)e-^(2)(i)

X( l )e^(3)(D e - i¥(3)( i ) = a:(3)c-^(3)(i)

The result of summing these equations is given by

a ( l ) = i ( 0 ) e - ^ ( » ) W + i ( l ) e - ^ ( 1 ) W + x ( 2 ) e - ^ ( J ) ( 1 ) + i ( 3 ) e - ^ ( 3 ) ( 1 )

Substituting specific values, we see that the equation is satisfied.

(2 + ^ ) ( 1 ) + ( l ) ( - j ) + (2 - ^ ) ( - l ) + (-I)O-) = V3-jl = 4X(1)

Now, we can deduce a formula that can be used to compute any of the four frequency coefficients.

1 3

X(k) = -Y,<n)e-^nk, fc = 0,l,2,3 (3.1) n=0

The Fourier synthesis of waveforms

The four equations set up for the analysis problem constitute the synthesis of the waveform from the coefficients. However, it is instructive to develop the synthesis problem formulation independently. Due to the orthogonality property, multiplying the sample values of an arbitrary waveform by the corresponding sample values of the conjugate of the complex sinusoid at a frequency and summing the products must be equal to the coefficient of that frequency component with a scale factor N, the number of samples (for the present example N = 4). Therefore, we set up four simultaneous equations with four frequency coefficients.

a:(0)e-^(°)(°) +ar(l)e-^<°X1) + z ( 2 ) e - ^ ( ° K 2 ) +ar(3)e-^(°)(3> = 4X(0) a ; ( 0 ) e - ^ W W + x ( l ) e - ^ ( 1 ) ( 1 ) + x ( 2 ) e - ^ W ( 2 ) + a ; ( 3 ) e - ^ ( i ) ( 3 ) = 4X(l) a ; ( 0 ) e - ^ T ( 2 ) ( o ) + x ( i ) e - ^ ( 2 ) ( i ) + a ; ( 2 ) e - ^ ( 2 ) ( 2 ) + a . ( 3 ) e - j ^ (2)(3) = 4 X ( 2 )

x(0 )e^ ' " ( 3 ) ( ° )+a ; ( l ) e - ^ (3 ) ( i ) + a ; (2 ) e - i ¥ (3 ) (2 ) + a ; (3 ) e - ^ (3 ) (3 ) = 4 X ( 3 )

The contributions of the data samples to the coefficient X(0) is each data sample multiplied by unity, because the sample values of the conjugate of the complex sinusoid with frequency index zero are unity. Therefore, the


sum of these products must be equal to AX (0). For the waveform shown in Fig. 3.1, we get

(2 + ^ ) ( 1 ) + (\)(1) + (2 - ^ ) ( 1 ) + ( - ! ) ( ! ) = 4(1)

Let us evaluate the time-domain sample a;(l). If we multiply the terms of each column of the four simultaneous equations by the sample values with index 1 of the complex sinusoids, the sum of all columns, except the second (with data sample x(l)) , on the left-hand side vanish due to orthogonality property and the problem is reduced to summing the following two columns.

a r ( l )e -^(o) ( iV¥(°) ( i ) = 4X(0)e^(o)(i) a ; ( l ) e -^ ( i ) ( i ) e^ ( i ) ( i ) = 4 X ( l ) e ^ ( i ) W

^ l J e - ^ e W V T P X i ) = 4X(2)e^(2)(1) i ( l )e - 'T(3)W e iT(3)W = 4Y(3)e*T(3)(i)

The result of summing these equations is given by

4ar(l) = 4(I(0)e^T(»)(i) +X(l)e^W + X ( 2 ) e ^ ( 2 ) W + X ( 3 ) e ^ ( 3 ) M )

Substituting specific values for the example, we see that the equation is satisfied.

x(i) = (i)(i) + ( i ( V 5 - ji))U) + ( i ) ( - i ) + (\(V3 + ji))(-j) = \

Now, we can deduce a formula that can be used to compute any of the four time-domain samples.

3

x(n) = ^X(ifc)e J 'T"*, n = 0,1,2,3 (3.2) Jfe=0

The DFT and IDFT expressions

Equations 3.1 and 3.2, respectively, are the mathematical formulations of the Fourier analysis and synthesis problems using harmonically related discrete sinusoids as the basis functions. All that is to be said is about some conventions. The division operation in Eq. (3.1) can be shifted to Eq. (3.2). The complex exponential can be written in a more compact form using the abbreviation Wi = e - - 7 ^ . With these observations, the analysis equation

The DFT and the IDFT 37

becomes

3

X{k) = ^x{n)WZk, ft = 0,1,2,3 (3.3) n=0

and the synthesis equation becomes

1 3

ar(n) = i 5 ] j r ( * ) W 7 n * , n = 0 , l ,2 ,3 (3.4) k=o

These two equations, called, respectively, DFT and IDFT, are a transform pair. What one does the other undoes. In the literature, one can find some variations in the definitions and a particular variation must be consistently used.

The frequency composition of real signals

Although we use complex sinusoids in the problem formulation, for real signals, the complex sinusoids combine to yield a real signal. For the present example, expanding Eq. (3.4), we get

-, i

x(n) = ±{X(0) + ^2(X(k)Wrn" + X(4 - fc)W4-"(4-fc)) + X(2)W4-2"},

fc=i

= J{X(0) + 2 |X(l) | co s ( | n + /.X(1)) + X{2) cos(Trn)}, n = 0,1,2,3

In general, we get

N _ 1

x(n) = ±{X(0) + 2 J2 {\X(k)\ coS(^nk + ZX(fe))) + X(j) cos(Trn)},

where n = 0 , 1 , . . . , N — 1. Therefore, an iV-point real signal, where iV is even, is composed of ^ + 1 sinusoidal waveforms.

3.2 The D F T and the IDFT

In this section, we examine the DFT and IDFT expressions in detail. A bandlimited periodic time-domain sequence x(n), with period N, can be represented in terms of a summation of a set of N harmonically related


complex sinusoids with coefficients X(k) as

J V - l

x(n) = J2 X(k)ej^nk, n = 0 , 1 , . . .,N - 1 (3.5) fc=o

In order to evaluate a specific coefficient X(k) in terms of x(n), we replace the index k by m on the right side of Eq. (3.5) (We use a dummy variable to avoid confusion between two independent frequency-domain variables. Note that similar substitutions will be made in other derivations.), multiply both sides of the nth equation by e~i~R~nk, and sum them over the interval n = 0 to n = N — 1. This process yields, for a specific k,

N-l N-l N-l

n—0 m = 0

N-l N-l

= J2 X(TO) Yl e^n(m- f c ) = NX(k)

71=0 71=0 771=0

7 V - 1 J V - l

771=0 71=0

with the fact that

N-l

for m / k Y ej%-n(m-k) = f N for m = k

7J=0

(Summations of complex exponential terms similar to the inner summation and the evaluation of the outer summation will occur several times in the proofs of DFT properties that the reader is well advised to get thoroughly familiar with it.) Therefore, we get

X(k) = ^ x ( n ) e - ^ n k

71=0

The constant factor ^ is usually suppressed and the iV-point DFT of the sequence x(n) is defined as

N-l

X{k)=J2x(n)WNk> fc = 0 , l , . . . , i V - l (3.6) 71=0


x(n) = 1+ cos(f n - f ) + cos(7rn) ra)

( ] 1 2

*

0 1 2

0 1 2

a:(n) = l . . . • • • •

a : ( n ) = c o s ( f n - f ) . . . - ^ .

x(n) = cos(7rn)

(b)

x{n) = cos(7rn)

x(n) = cos(f n - f )

x(n) = 1

(a)

fc

*

summing unit x(n) = 1 + cosff n — | )

+ cos(7rn)

Fig. 3.3 (a) The operation of DFT is similar to that of a set of bandpass filters with a narrow passband. (b) The operation of IDFT is similar to that of a summing unit. It should be noted that the output of the DFT are the coefficients of the corresponding sinusoids. Similarly, the input to the IDFT operation is a set of coefficients.

where WN = e J '« . Inserting the constant ^ in Eq. (3.5), we get the iV-point IDFT of the frequency coefficients X(k) as

1 7V_1

i (n ) = - ^ X ( i ) V , n = 0 , l , . . . , A T - l (3.7) N

The IDFT operation is summing over the responses at all frequency samples for determining each time-domain sample. The DFT operation is summing over the responses at all time samples for determining each frequency coefficient. The separation of the various frequency components by the DFT is similar to the operation of a set of bandpass filters with a narrow passband as shown in Fig. 3.3(a), for the waveform in Fig. 3.1. The combining of the various frequency components by the IDFT is similar to the operation of a summing unit as shown in Fig. 3.3(b).


It can be verified that Eqs. (3.6) and (3.7) form a transform pair. Substituting for X(k) in the ID FT definition, we get

fc=0 ro=0

Changing the order of summation, we get

m=0 Jfe=0

Since the inner summation evaluates to N for m — n and zero otherwise, the expression on the right-hand side is equal to x{n) and the transform relationship is established.

For N = 8, we have shown the cosine and sine components of the DFT basis functions

Xk(n) = ej^~nk = cos(-nk) + j s i n ( — nk), n,k = 0 , 1 , . . . , N — 1

in Fig. 3.4. The sample values of sines and cosines used in DFT for evaluating frequency coefficients are represented compactly by the complex exponential and the notation used is Wfik = cos(^nfc) — j sin(^nA;), called the twiddle factor. The sine function of the twiddle factor is the negative of that shown in Fig. 3.4 because of the conjugation of the basis function. Note that, from the sample values of the fundamental shown in Fig. 3.4(b), the sample values of the other waveforms in Fig. 3.4 can be easily deduced as the waveforms are harmonically related.

The discrete complex exponential is a periodic function of nk with a period N. Therefore, the twiddle factors are periodic, as shown in Fig. 3.5 for N = 8. They are iVth roots of unity. The roots of unity are called twiddle factors, because multiplying a number by a twiddle factor is changing the phase of that number or rotating a vector without changing its magnitude. Frequencies, called DFT bins, at which the spectral values are computed are also shown in Fig. 3.5. For example, the point marked 0 corresponds to dc, that is / = 0. The point marked | corresponds to the fundamental frequency f = h cycles per sample. This is also the frequency increment of the DFT spectrum.


I e—e—e——e—e—o o — e -* -1

(a)

r^z^^

\ o H 1 p"

0

- . 1 ^ o K 1

p^? * S

•cc^r^

"X 2

\/7 ^

^ ^ i ^ . ^ ^ " * - «_-^*^-_o -®^^

4 6 n

(b)

* *P7 /7C \ /

\5G/ V w

(c) (d)

, j f y / V W » ;Ry?^*^ (e)

~1 i ^QxC^ >0*C-^

(f)

(g) (h)

Fig. 3.4 The cosine and sine components of the DFT basis functions, with N = 8 and frequency indices varying from 0 to 4, are shown, respectively, in (a), (b), (c), (d), and (e). Note that (f), (g), and (h) are, respectively, the same for cosine and the negative for sine as that of (d), (c), and (b). This duplication is due to the use of complex exponentials in the DFT problem formulation. The sine function with frequency indices 0 and 4 have all sample values equal to zero.

Half-wave symmetry of the DFT kernel matrix

The DFT equation, with N = 8, is denned as 7

*(fc) = 5>(n)W8"*, fc = 0 , l , . . . ,7 n=0

Writing the eight individual equations, we get

X(0) = x(0)Wg + x(l)Wg + x(2)W° + z(3)W° + a;(4)W80 + x(5)Wg

+ x(6)W8° +x(7)Wg

X(l) = x(0)W° + x{l)Wl + x(2)W* + x(3)W$ + x(4)W84 + x(5)W8

5

+ x(6)Wg +z(7)W87


2

8 8 " " 8

w84 = w8

12 = --- , i w?k Q9w§ = wi = ---

8

o g o 8

Fig. 3.5 The periodicity of the twiddle factors, with N = 8, and the discrete frequencies at which the spectral values are computed.

X(2) = x(0)Wg +x(l)W£ + x(2)W£ + x(3)W* + ar(4)W| + x(5)W810

+ x(6)W^+x(7)W^

X(3) = x{0)Wg + x(l)W$ + x(2)W£ + x(S)Wg + z(4)W812 + x(5)W8

15

+ x{6)W^8 + x{7)W£1

JT(4) = s(0)W8° + x(l)W£ + x{2)Wi + x(3)W812 + z(4)W8

16 + x(5)W820

+ x(6)W|4 + x(7)W£8

X(5) = x(0)W° + x(l)Wi + x{2)W™ + x(3)W£5 + x{4)Wg° + x(5)W£5

+ x(Q)Wi° + x(7)Wi5

X(6) = z(0)W8° + x(l)W86 + x(2)W8

12 + x(3)W818 + x(4)W8

24 + x(5)W|°

+ x(6)W836 + x(7)W8

42

X(7) = x(0)W° 4- x(l)W87 + x(2)W8

14 + z(3)W821 + x(4)W28 + x(5)W8

35

+ x(6)W842 + x(7)W$9

The even-indexed sines and cosines of the twiddle factors complete an even number of cycles. Therefore, they are even half-wave symmetric, x{n) = x(n ± y ) . At n = y , we start getting the first-half sample values since they start a new cycle.

cos(^(2fc)(n±y)) = cos(~2kn) and sin(-^(2ft)(n±y)) = sin(-^2fcn)


Since the odd-indexed sines and cosines complete an odd number of cycles, they are odd half-wave symmetric, x(n) = —x{n ± T£). At n = %, we start getting the first-half sample values with sign reversed since they start the second half of a cycle.

cos ( -^ (A) (n±y) ) = -cos(—kn) and sin(—(k)(n±—)) •sin —«n) v N

The waveforms shown in Fig. 3.4 exhibit this property. The samples of all the waveforms at a sample point also exhibit this same half-wave symmetry property. Therefore, the twiddle factor matrix is symmetric. The DFT equations can be rewritten showing the half-wave symmetry explicitly, using matrices, as

X(0)1 X(l) X(2) X(3) X(4) X(5) X(6) X(7) .

' W8° W8° W8° W8° Wi W8° W8° W8° ' 8° Ws1 W£ Wi -Wg -Wg1 -Wi -W$

w$ wi wi w§ w§ wi wi wi wi wi wi wi -wi -wi -wi -wi wi -wi wi -wi wi -wi wi -wi wi -w^ wi -wi -wi w£ -wi wi wi -wi wi -wi wi -wi wi -wi wi -wi wi -wi -wi wi -wi wi

• x(0)

x(l) x{2) x(3) x{4) x(5) x(6)

. *(7)

The twiddle factor matrix is called the kernel matrix of the DFT. Exploiting the symmetry, as will be seen in Chapter 5, this equation can be written more compactly. Note that, in the DFT definition, the time-domain index starts from zero. It does not necessarily mean that the signal always start from zeroth instant. Since periodicity is implied in computing the DFT, we can always get the time-domain samples starting from the index zero. Of course, we can also change the limits of the summation as orthogonality of complex exponentials holds for any interval of N samples. A particular case of interest is described below.

Center-zero format of the DFT and IDFT

With N even, the DFT and IDFT expressions are sometimes written as

"r--i N N IV

(3.8) X(k)= Y, *(«)W *. * = - ( T ) ' - ( f - 1 ) ' • • • ' ? - X

n=(-f)


f -1 x(n) = ± J2 ^WW^k,n = -(j),-(^-l),...,~ -1(3.9)

The values of these forms of DFT and IDFT results in a better display with the value with index zero in the middle and these forms are also convenient to derive certain derivations. The spectrum of the waveform shown in Fig. 3.1(d) in the usual format is X(0) = 4, X(l) = V3 - jl, X{2) = 4, X(3) — y/3+jl. The same spectrum in center-zero format is X(—2) — 4. X(-l) = y/3 + jl, X(0) = 4, X{1) = V3 - jl. Getting one format of the spectrum or the signal from the other involves a circular shift by ^ positions (swapping of the positive and negative halves).

3.3 D F T Representation of Some Signals

In this section, we derive the DFT of some simple signals analytically. Although the primary use of the DFT, in practice, is to analyze arbitrary waveforms through a numerical procedure, finding the DFT of some simple signals analytically improves our understanding and the resulting closed-form solutions serve as test cases for the algorithms.

The impulse, x(n) — 6(n)

o X{k) = ^ 1 = 1 and S(n) & 1

n=0

(The double-headed arrow indicates that the two quantities are a DFT pair, that is the frequency-domain function is the DFT of the time-domain function.) Since the impulse signal is zero except at n = 0, for all k, the DFT coefficient is unity. All the frequency components exist with equal amplitude and zero phase.

Example 3.1 Figures 3.6(a) and (b) show, respectively, the unit-impulse signal and its spectrum, with N = 16. The representation of the impulse signal, in terms of complex exponentials, is given by

1 1 5

k=0

DFT Representation of Some Signals 45

_ 1 c

•

0

c 'Sf

0 0

5

5

10 n

(a)

10 n

15

15

_ 1

0

* 0 , f 0

f f

5 10 k

(b)

5 10 k

15

15

(c) (d)

Fig. 3.6 (a) The unit-impulse signal, with N = 16, and (b) its spectrum, (c) A dc signal, with N = 16, and (d) its spectrum.

To find the real sinusoids that constitute the impulse signal, we add the corresponding positive and negative frequency components. For example, the sinusoid with frequency index one is obtained as

The representation of the impulse signal, in terms of real sinusoids, is given by

1 7 2?r S(n) = —(1 + 2 5T(cos(— nk)) + cosfrn)), n = 0 , 1 , . . . , 15

A dc component with amplitude ^ , a cosine waveform with amplitude ^ and frequency index 8, and cosine waveforms with amplitude | and frequency indices from 1 to 7 make up the impulse. We can find out the real sinusoids using the first half of the frequency coefficients due to the conjugate symmetry of the spectrum of real-valued data. For example, at frequency index one, the value of the complex coefficient is 1, with magnitude 1 and phase zero. A sinusoid with these characteristics is a cosine wave with amplitude | (taking into account the scale factor).

Due to the duality (explained in the next chapter) of the time- and frequency-domains, the DFT of the dc signal, shown in Fig. 3.6(c), is an impulse, shown in Fig. 3.6(d). The impulse at frequency index zero corresponds to a complex exponential with its exponent zero, yielding a dc signal in the time-domain. I


The complex exponential

The complex exponential signal, as we have already mentioned, is the standard unit in Fourier analysis, although we are interested in real sinusoids. Consider the signal x(n) = ej~&~mn,n = 0 , 1 , . . . , JV—1, where m is a positive integer. From the DFT definition, we get

N - l N-l

X{k)= J2ej2^mne-^kn= Y,e~j¥{k~m)n> k = 0,l,...,N-1 n=0 n=0

Due to the orthogonality property, we get

e ^ r a n <S> N5(k - m) and e~^mn o NS(k - (TV - m))

Example 3.2 Figures 3.7(a) and (b) show, respectively, the waveform eJ if"" and its spectrum, with N = 16. This is a complex exponential with frequency index one and amplitude one and, therefore, its spectrum consists of an impulse at frequency index k — 1 with amplitude sixteen.

Figures 3.7(c) and (d) show, respectively, the waveform 3e J ( i t 1 4 n + e ) = 3 e if eii?-i4n a n d its spectrum. This is a complex exponential with frequency index 14 and complex amplitude 3e J t . Therefore, its spectrum consists of an impulse at frequency index A; = 14 with amplitude (3)(16)(cos | + j s i n | ) = 24(V3 + j l ) .

Figures 3.7(e) and (f) show, respectively, the waveform e _ ^ i t 1 4 " + ^^ = e_Jte^ie"14™ and its spectrum. This is a complex exponential with frequency index —14 and complex amplitude e --7^. Therefore, its spectrum consists of an impulse at frequency index k = 16 — 14 = 2 with amplitude 1 6 ( c o s ( - f ) + j s i n ( - f ) ) = 8 ( l - j V 3 ) . I

The real sinusoid

The DFT of a real sinusoid, x(n) = cos (^mn + 9), n = 0 , 1 , . . . , TV - 1, is obtained by expressing it as a combination of complex sinusoids as

x(n) = hejeej^mn + e ^ e " ^ " 1 " )

Li

From previous results for the complex exponentials, we get

cos(-jj-mn + 6)& -^{ejeS(k - m) + e-jeS(k - (TV - m)))


16 • real o imaginary

10 15

3

I ° -3

(a)

o

° • » o

g o •» 10 15

24

(b)

41.569 '

o

10 15

(c)

8

(d)

o -13.856

10 15

(e) (0

Fig. 3.7 (a) The complex exponential e'TB"™ and (b) its spectrum, (c) The complex exponential 3e J ' ' i6 '1 4 n+6') and (d) its spectrum, (e) The complex exponential

e-j 'Cif-i4n+f) a n d ^ its spectrum.

With 0 = 0 and 0 = - | , we get, respectively,

27T N cos(—mn) •£> —(6(k -m) + 6(k - (N - m)))

2TT N sin(—mn) ^ — (-jS(k - m) + jd{k - (N - m)))

Example 3.3 Figures 3.8(a) and (b) show, respectively, the cosine waveform cos(ffn) and its spectrum, with N = 16. The magnitude and the phase shift of the frequency coefficient (8) at frequency index A; = 1 are 8 and 0 degrees, respectively. A sinusoid with these characteristics is a cosine wave with amplitude one. Remember that the spectrum of a cosine wave is pure real, that of a sine wave is pure imaginary, and that of a sinusoid


1

0

-1

10 15


10 15

(a)

X

8

0

- 8 L

(b)

10 15

1 . •

0

-1

(c)

10 15

X

•6.928

(d)

10 15

(e) (f)

Fig. 3.8 (a) The sinusoid c o s ( ^ n ) and (b) its spectrum, (c) The sinusoid s i n ( ^ 2 n )

and (d) its spectrum, (e) The sinusoid c o s ( | | 4 4 n 4- | ) and (f) its spectrum.

other than a cosine or sine consists of both real and imaginary parts. Figures 3.8(c) and (d) show, respectively, the sine waveform sin(f|-2n)

and its spectrum. The magnitude and the phase shift of the frequency coefficient (-J8) at frequency index k — 2 are 8 and - f radians, respectively. A sinusoid with these characteristics is a sine wave with amplitude one.

Figures 3.8(e) and (f) show, respectively, the sinusoid cos(^f 14n + f ) and its spectrum. The magnitude and the phase of the frequency coefficient (6.9282 - j4) at frequency index k = 2 are 8 and - f radians, respectively. This is a sinusoid with amplitude one and phase — f radians. There are two points to observe: (i) a real sinusoid with frequency index in the second half of the frequency range cannot be distinguished from a sinusoid with a corresponding frequency in the first half, as mentioned in Section 2.2, and (ii) the phase shift is negated. The sinusoid yields a spectrum that is the same as that of cos(f |2n — | ) . I


0.9619 0.8536

0.6913

0.5

0.3087

0.1464 0.0381

10 15 10 15

(a) (b)

Fig. 3.9 (a) The Hann function, 0.5 - 0.5cos(^n), and (b) its spectrum.

The Hann function

2n x(n) = 0 .5-0 .5cos(—n) , n = 0 , 1 , . . .,N - 1

From previous results, we get

0.5 - 0.5 cos(—n) & N(0.56(k) - 0.25(<5(fc - 1) + 6(k - (N - 1))))

Example 3.4 Figures 3.9(a) and (b) show, respectively, the waveform and its spectrum, with N = 16. A value of 8 at A; = 0 indicates a dc signal with amplitude 0.5. A value of -4 at A; = 1 and A; = 15 indicates a sinusoid with frequency index one, amplitude 0.5, and phase 180 degrees, that is -0.5cos(ffn). I

Rectangular waveform

x(n) _ f 1 forn = 0 , l , . . . , L - l ~ { 0 for n = L, L + 1 , . . . , N - 1

L-\

X(k) = £ \ -J¥nk = l-e~^Lk = _^(-lifcsin(ffcf)

n=0 o J N K s i n ( ^ | )

sin(ffc)


1

*? "sr

ok . . •••• 0 10 20 30

n

(a)

1

oL , «« 0 10 20 30

n

(c)

Fig. 3.10 (a) The rectangular function with four zeros and N = 32 and (b) its spectrum. (c) The rectangular function with eight zeros and (d) its spectrum.

(The rewriting of the numerator and denominator terms of the sum in terms of sine functions, by decomposing each of the complex exponential terms into two with the same argument but of opposite sign, occurs in other derivations also.) It is instructive to study the pattern of the spectrum for various values of L. With L = N, X(0) = N and X(k) — 0 otherwise as shown earlier for the dc signal.

Example 3.5 Figure 3.10(a) shows the signal with L = 28 and N = 32, and its spectrum is shown in Fig. 3.10(b). Only the first half of the spectrum is shown as the spectrum is symmetric. The signal is almost dc and, therefore, the spectral value at k = 0 is very large compared with the rest of the values. Therefore, we have plotted the log-magnitude of the spectrum on the y-axis. Figure 3.10(c) shows the signal with L = 24 and its spectrum is shown in Fig. 3.10(d). The spectral value at A; = 0 is reduced and other spectral values are increased. With L = %, x(n) is a square wave and the DFT is given by

r - J . ^ ( JV-2) fe S l n ( | f c ) s i n ( ^ )

For even values of k, the spectral value is zero. Eventually, with L = 1, X(k) = 1 as shown earlier for the impulse signal. I

=? 1

g1 0 5 10

k

(b)

15

5 10 k

(d)

15

Direct Computation of the DFT 51

3.4 Direct Compu ta t i on of the D F T

In this section, we analyze the computational burden in computing the DFT from its definition and study the details of the direct implementation.

Example 3.6 Compute the DFT of the 4-point sequence {2,3,1,4}. Compute the ID FT of the transform to get back the time-domain data. Solution

3

X(k) = Y^x{n)Wf, fc = 0 , l ,2 ,3 n=0

X(0) = 2x1 + 3x1 + 1x1 + 4x1 = 10 X(l) = 2x 1 + 3 x (-j) + lx (-l) + 4x j = l + jl X(2) = ' 2x 1 + 3 x (-l) + lx 1 + 4 x (-l) = -4 X(3) = 2xl + 3xj + lx(-l) + 4x(-j) = l-jl

The IDFT gives the original input samples.

1 3

x(n) = -Y^X(k)W^k, n = 0 , l ,2 ,3 fc=0

x(0) = J ( 1 0 x l + (l + j l ) x l + ( - 4 ) x l + ( l - j l ) x 1) = 2

x(l) = i ( 1 0 x l + (l + j l ) x j + ( - 4 ) x ( - l ) + ( l - j l ) x ( - j ) ) = 3

x{2) = i ( i 0 x l + (l + j l ) x ( - l ) + ( - 4 ) x l + ( l - j l ) x ( - l ) ) = l

ar(3) = i ( 1 0 x l + (l + j l ) x ( - j ) + ( - 4 ) x ( - l ) + ( l - j l ) x j ) = 4 l

For the computation of N frequency coefficients, N2 complex multiplications and N(N — 1) complex additions are required. Therefore, the computational complexity of the direct DFT computation is 0(N2).

Direct implementation of the DFT

Although, most of the times, we would be using fast algorithms to compute the DFT, we may have to compute the DFT directly from the definition at times. We can verify the output of the fast algorithms when they are


M 3 t a r t J

call twid_fac

I call in_put

I call dir.dft

I call out.put

I f Stop ")

3.11(a)

( s t a r t )

read value xr(i), i = 0,l,...,N-l

X read value xi(i),

i = 0,l,...,N-l

I ( Return)

3.11(c)

f Start *)

arp = 0.0 2TT

* = 0 JV

•(Return J

tfc(i) — cos(arg) tfs(i) = sin(arg) arg = arg + inc

i = i + l

3.11(b)

("start J

print value XR(i),XI{i),

i = 0,l,...,N-l

I (Return J

3.11(d)

developed or implemented. In addition, we can compute the DFT for any length whereas fast algorithms are available only for certain data lengths.

A set of flow charts for the direct implementation of the DFT is shown in Fig. 3.11. The main module, shown in Fig. 3.11(a), invokes four modules to get the computation done. The next module, shown in Fig. 3.11(b), sets up the twiddle factor arrays, that is the computation and storage of the sample values of cosines and sines over one cycle, with argument starting from zero with increment of ^ . This module is called twid-fac. Variable i is used as a loop counter and the loop is terminated when i equals N, the data size. Arrays tjc and tjs, each of size N, are used to store the sample values of

Direct Computation of the DFT 53

( S t a r t )

k = 0

• N n o / \ Return)* (k < JV>

Iyes

n = l,nk = k, tr = xr(0),ti = xi(Q)

XR(k) = tr,XI(k) = ti k = k + l

ind — nk mod N c = tfc(ind), s = tfs(ind)

tr = tr + xr(n) * c + xi(n) * s ti = ti + xi(n) * c — xr(n) * s

nk = nk + k,n = n + 1 i

3.11(e) Fig. 3.11 (a) The main module for direct implementation of the DFT. (b) The twiddle factor module, (c) The input module, (d) The output module, (e) The DFT module.

cosines and sines, respectively. The setting up of the twiddle factor arrays can be made faster by using identities such as cos(^(iV — n)) = cos (^n ) and s in(^( iV-7j)) = - s i n ( ^ n ) . Note that this module can be eliminated by computing the twiddle factors each time they are required. However, it is costly in terms of run-time.

The next module, shown in Fig. 3.11(c), reads the real and imaginary parts of the complex input data, respectively, into the arrays xr and xi, each of size N. It is assumed that the real parts of the input data are stored before the imaginary parts in the input file. If the data is real, initialize the values of the array xi to zero and read the data into the array xr. This module is called in-put.

The next module, shown in Fig. 3.11(e), computes the DFT coefficients. This module is called dir-dft. The computation is carried out in two nested


loops, the outer loop controlling the frequency index and the inner loop controlling the access of the data values. In each iteration of the outer loop, one coefficient is computed. The real and imaginary parts of the coefficients are stored, respectively, in arrays XR and XI, each of size N. The access of correct twiddle factor values is carried out using the mod function. Inside the inner loop, each coefficient is computed according to the DFT definition. The next module, shown in Fig. 3.11(d), prints the real and imaginary parts of the coefficients, respectively, from the arrays XR and XI, one coefficient in each iteration. This module is called out-put.

3.5 Advantages of Sinusoidal Representation of Signals

The DFT is a tool to obtain the representation of a signal in terms of a set of harmonically related discrete sinusoids. In general, a signal is represented in other than its naturally occurring form to gain some advantages in signal processing and understanding. The sinusoidal representation of signals is the predominant one when it comes to the analysis and design of LTI systems because of the following advantages this representation provides.

(1) Efficient signal manipulation: When excited with a sinusoidal signal, the output of a stable LTI system is a sinusoid of the same frequency as that of the input. Therefore, the input and output are related only by a complex constant, representing the amount of scaling of the amplitude and change in the phase shift of the input signal. This is due to the fact that the derivative and integral of a sinusoid is another sinusoid of the same frequency. This characteristic along with the linearity and time-invariant properties of LTI systems makes it efficient to implement fundamental operations such as convolution. An arbitrary signal is represented as a sum of sinusoids and the sum of the responses of a system to all the individual sinusoids is the response to the arbitrary signal.

The frequency-domain representation provides a better understanding of the signal characteristics.

In addition, a signal can be stored in a highly compressed form in the frequency-domain representation because of the tendency of most practical signals to have most of the energy concentrated in the lower part of the spectrum.

(2) Availability of fast algorithms for the computation of the DFT: The availability of fast algorithms for computing the DFT makes its use to

Advantages of Sinusoidal Representation of Signals 55

2.866

•£ 1.134*1 0.5

-0.5

7

| 6.64

« 6.36 6.16 6.04

•

•

0.5

• * R * . § .

1 x(n)

(b)

•

•

•

1.5

(a)

Fig. 3.12 (a) An arbitrary waveform and its dc component with amplitude one. (b) The least squares error in approximating the waveform with only its dc component, the amplitude varying from 0.5 to 1.5.

process signals more efficient compared with alternate methods. The IDFT can be computed by a DFT algorithm with trivial modifications.

(3) The accuracy of representation: Let

N N N x{n) and xa{n), n = - ( — ) , - ( — - 1 ) , . . . , — - 1

be the given real signal and an approximation to it, respectively, with N even. Then, the least squares error between x(n) and xa(n) is defined as

o 1

error = ] P (x(n) - xa(n))2

For a given number of coefficients, there is no better approximation for the signal than that provided by the DFT coefficients when the least squares error criterion is applied.

Example 3.7 Consider the signal we analyzed in Section 3.1, which is shown again in Fig. 3.12(a). We found that three frequency components are required to construct the signal accurately. Assume that we are constrained to use only the dc component, with a value of 1, to approximate the signal. This optimum value is found by Fourier analysis and there is no other value that will reduce the least squares error. The least squares errors, for the value of the dc component varying from 0.5 to 1.5, are plotted in Fig. 3.12(b). The optimum value of 1 yields the minimum least squares error. I


Let the signal x{n) be represented by N DFT coefficients exactly. Then,

- l

N <W = 7J E *(Wv"*, n = - ( ^ ) , - ( ^ - l ) , . . . , £ - l *=-(f)

Let us approximate the signal x(ri) by xa(n) with M < N frequency coefficients Xa(k), where M is odd (For convenience, we have put some constraints on the number of coefficients N and M. However, it should be noted that this property is valid for any N and M.). Then,

M - l

X°(n) = jt E *«(*W**, n = - ( y ) , - ( y - l ) , . . . , y - l

In order to prove that the DFT provides coefficients with optimum values with respect to the least squares error criterion, we express the error equation in a form so that the result is evident. Substituting for xa(n), we get

SS-—1 M - l

errors £ (x(n) - 1 £ Xa{k)W^nkf „=-(f) k=-(^i)

Expanding and rearranging, we get

= E *»-f E *»(*) E *(»w* n=-(f) ^ - ( ^ i ) «=-(f)

M - l M - l ^ — 1

+]J» E E *.(*>*.« E ^n(fc+,)

The second term is simplified by the fact that the summation involving n is X(—k). The third term is simplified because the summation involving the complex exponential is equal to N when I = —k and zero otherwise. Therefore, we get

IV i M - l M - l

= E * » - ; ! E Xa(k)X(-k) + ± J2 Xa(k)Xa(-k) „=-(f) *=-(^) *=-(^)

Advantages of Sinusoidal Representation of Signals 57

M - l

Adding and subtracting the term ^ YJ^III M-I ) X(k)X(-k) and using the M - l M - l

fact that E ^ M ^ I ) Xa(fc)X(-fc) = E f c =2_ (M^ } * « ( - * ) * ( * ) , we get

JV _ j M - l

= E * » - ^ E *(*)*(-*) M - l

"4 E (Xa(-k)X(k) + Xa(k)X(-k)-Xa(k)Xa(-k)-X(k)X(-k))

Factoring the last term, we get

= E a;2(") + ^ E (Xa(k)-X(k))(Xa(-k)-X(-k)) n=-(f) * = - ( ^ )

Af —1

- 1 J2 X(k)X(-k)

Using the fact that the two expressions forming the product in the last two terms are complex conjugates, we get

< ¥ . _ ! M - l M - l

error = £ x2(n)+± £ \(Xa(k)-X(k))f-± £ \X(k)f

The error is minimum only when Xa(k) = X(k), since x{n) and X(fc) are fixed. This implies that an optimal representation of a signal is provided by the DFT coefficients with respect to the least squares error criterion. When constrained to use only M < N coefficients to represent a signal, the minimum error is obtained by using the M largest coefficients (not necessarily the first M coefficients).

(4) The finality of coefficients: Because of the orthogonality property, the coefficient value of a specific frequency component is independent of the number of coefficients used to approximate a signal. This property allows the computation of additional coefficients to improve the approximation of a signal without the need to compute already found coefficients again.

(5) Representation of signal power: The signal power can be expressed in terms of the frequency coefficients.


(6) Complete representation: The least squares error in the approximation of a signal with the frequency coefficients can be made arbitrarily small by increasing the number of coefficients.

(7) Conditions for existence: The sufficient conditions under which the sinusoidal representation is possible are such that almost all signals of practical interest satisfy them.

3.6 S u m m a r y

• The DFT transforms an JV—point arbitrary time-domain sequence into a set of JV frequency coefficients. The JV coefficients are the representation of the given time-domain sequence in the frequency-domain. These coefficients represent the magnitudes and phases (or the amplitudes of cosine and sine components) of a set of harmonically related sinusoidal sequences whose superposition summation yields the time-domain sequence they represent.

• The representation of a finite length sequence in terms of infinite length sinusoidal sequences is obtained by assuming that the finite length sequence can be extended periodically.

• The IDFT transforms the JV frequency coefficients back into the original set of JV time-domain samples by the process of superposition summation of the sinusoids represented by the JV frequency coefficients.

• A signal is completely characterized either by the JV time-domain samples or by the corresponding JV frequency coefficients.

• The representation of a signal by the DFT coefficients provides minimum least squares error. In the next chapter, we study the properties of the DFT.

References

(1) Guillemin, E. A. (1952) The Mathematics of Circuit Analysis, John Wiley, New York.


Exercises 59

Exercises

3.1 Formulate the Fourier analysis problem explicitly in terms of cosines and sines. Find the coefBcients of sines and cosines for the waveform in Fig. 3.1(d).

3.2 For the waveform shown in Fig. 3.1(d), verify the value of the coefficients X(0), X(2), and X(3) using Eq. (3.1).

3.3 For the waveform shown in Fig. 3.1(d), verify the value of the data samples a;(0), x{2), and x(3) using Eq. (3.2).

3.4 Given the coefficients with N = 4, find the individual sinusoids. Note that the coefficients are as defined by Eq. (3.3). 3.4.1 X(0) = -2 ,X(1) = 2v/2(l - jl),X(2) = 1,X(3) = 2^2(1 + j l ) *3.4.2 X(0) = 1,X(1) = 3(V3 + j l ) ,X(2) = -1 ,X(3) = 3(V5 - j l ) 3.4.3 X(0) = 3,X(1) = | ( 1 + jy/3),X{2) = -2 ,X(3) = | ( 1 - jy/i)

3.5 Write the DFT equation in matrix form with N = 8 showing the numerical values of the twiddle factors.

3.6 Prove that the sum of the cosines, the frequency components of an impulse signal, yields the impulse signal.

3.7 Derive the DFT of the dc signal, x{n) = 1, n = 0 , 1 , . . . , N - 1, from the definition.

3.8 Find the DFT of 3.8.1 x(n) = 5{n - 2) with TV = 8 *3.8.2 x(n) = 5{n + 2) with N = 8

3.9 Find the DFT of x(n) = ( -1 )" with N even.

3.10 Find the DFT of cos(fn) with N = 10 and N = 20.

*3.11 Find the DFT of x{n) = 3 e - ^ " + f ) with N = 16.

3.12 Find the DFT of 2 s i n ( ^ n + | ) with AT = 16.

*3.13 Find the DFT of x{n) = e~Sn, n = 0 , 1 , . . . , N - 1, where (J is a constant.

3.14 Find the DFT of x(n) = %,n = 0 , 1 , . . .,N - 1.


3.15 Find the IDFT of 3.15.1 X(k) = 3, 0 < k < 4. 3.15.2 X(0) = 8 and X(k) = 0 otherwise, 0 < k < 4. 3.15.3 X(3) = 26 and X{k) = 0 otherwise, 0 < k < 16. 3.15.4 X(2) = ( | + j&), X(U) = {\ - j&), and X(k) = 0 otherwise, 0 < A;<16. * 3.15.5 X(5) = ( - ^ - j ^ ) , X ( l l ) = ( - ^ + j ^ ) , and X(fc) = 0 otherwise, 0 < k < 16. 3.15.6 X{5) = 4, X ( l l ) = 4, and X(fc) = 0 otherwise, 0 < A; < 16. 3.15.7 X(10) = 8, X(22) = 8, and X(k) = 0 otherwise, 0 < k < 32. 3.15.8 X(3) = - j 8 , X(13) = j8, and X(k) = 0 otherwise, 0 < k < 16. 3.15.9 X(2) = - ; 5 and X(k) = 0 otherwise, 0 < Jb < 16.

3.16 Compute the DFT. Verify the transform pair by computing the IDFT of the transform to get back x(n). 3.16.1 x(n) = {3 ,2 ,1 , -4} 3.16.2 x{n) = {2 + j3,10 + j2 , - 4 + jl, 6 - j4}

3.17 Find the 4-point DFT of x(-2) = l , a : ( - l ) = -4,a;(0) = 3,a;(l) = 2 using the center-zero format of the DFT. Using the center-zero format of the IDFT, verify the transform pair by computing the IDFT of the transform to get back x(n).


3.1 Write a program for the direct implementation of the DFT.

3.2 Write a program for the direct implementation of the IDFT.

Chapter 4

Properties of the DFT

In this chapter, we study the properties of the DFT. The existence of advantageous properties is the major reason for the widespread use of the DFT. In applications of the DFT or in deriving DFT algorithms, the properties are repeatedly used.

4.1 Linearity

The linearity property implies that the DFT of a linear combination of a number of signals is the same linear combination of the DFTs of the individual signals. Let X(k) and Y(k), respectively, be the DFT of x(n) and y(n). It is assumed that the lengths of the sequences x(n) and y(n) are equal. If the sequences are not of equal length, they are assumed to be appended by sufficient number of zeros so that their lengths become equal. The DFT of ax(n) + by{ri), where a and b are real or complex constants, is given by

N-l

Y,(ax(n) + by(n))WZk

Due to the linearity property of the summation operation, we can rewrite this equation as

N-l N-l

a ^ x(n)WHk + b Y^ y(n)W$k = aX(k) + bY(k) n=0 n=0

That is, ax(n) + by{n) <s> aX(k) + bY(k).

61

62 Properties of the DFT

Example 4.1 Let a = 1, 6 = j , x(n) = {1,2,1,3}, and y(n) = {2,1,1,4}. X(fc) = {7, j , - 3 , -j} and Y(fc) = {8,1 + j3, -2,1 - j3}. Then, the DFT <rfar(n)+jj/(n) = {1 + j2,2 + jl,l + jl,3 + j4} is X(k) + jY{k) = {7 + j8,-3+j2,-3-j2,3 + j0}. I

Linearity applies in both the time- and frequency-domains.

4.2 Periodicity

A sequence x(n) is denned periodic with period N if x(n) = x(n + aN), where a is any integer. In DFT analysis, both the input data and its DFT are assumed periodic of period N, the length of the sequence. This property follows from the fact that the discrete complex exponential, W£k, is periodic with a period N. By substituting k + aN for k in Eq. (3.6), we get

J V - l N-l

X(k + aN)=Yl x(n)W$k+aN) = £ x[n)W%k = X(k), n=0 rc=0

where k = 0 , 1 , . . . N — 1. Similarly, by substituting n+aN for n in Eq. (3.7), we get

x(n + aN) = 1 £ X(k)W^n+aN)k = 1 £ X ( f e ) ^ " f c = *(«), fc=0 fc=0

where n = 0 , 1 , . . . N — 1. The periodicity of x(n) and X(fc) with a period of 8 is shown in Fig. 4.1.

Example 4.2 Consider the DFT pair

x(n) = {2+jl,3+j2,l-jl,2-j3}&X(k) = {8-jl,6+jl,-2+jl,-4+j3}

Now, x(4) = x(0) = 2 + jl and X ( - l ) = X(3) = - 4 + j 3 . I

4.3 Circular Shift of a Time Sequence

Consider the sequence, x(n) = {x(0),x(l),x(2),x(3),x(4),x(5),x(6),x(7)}, shown in Fig. 4.1. The circular shift of the sequence to get x(n - 1) = {x(7), x(0), x(l), x(2), x(3), x(4), x(5), x(6)} means the rotation of the circle by one position in the counterclockwise direction as shown in Fig. 4.2. The

Circular Shift of a Time Sequence 63

x(3) = x ( l l ) = • • • X(3) = X(11) = --- '

x(4) = se(12) = • • • X(4) = X(12) = --- *

x(2) = ar(10) = • X{2) = X{10) =

•

t

x{n) X(k)

x{5) = ar(13) = • • • X(S) = X{13) = ••• *

x(6) =x(U) = • X(6) = X(U) =

x(l) = x(9) = ••• *X(1) = X(9) = -

x(0) = ar(8) = *X(Q)=X(8)

x(7) = x(15) = • • *X(7) = X(15) =

Fig. 4.1 The periodicity of the time- and frequency-domain sequences x(n) and X(k) with a period of 8.

x{2) X(3)W$ *

*(3) X(4)W$ *

ar(4) X(5)W8

5 #

x(l)

X(2)Wi •

a;(n - 1)

X{k)Wg

•

X{G)W$

x(0)

x(7) *X(0)W$

1(6)

*X(7)W87

Fig. 4.2 The delayed time-domain sequence x(n - 1) and its DFT, X{k)w£.


sequence x{n + m) can be obtained by rotating the circle in Fig. 4.1 in the clockwise direction by m positions. If there are N values in a sequence, then only N — 1 unique shifts are possible. x(n — m) is the signal delayed by m sample intervals and x(n + m) is the signal advanced by m sample intervals. Substituting n — m for n in the IDFT definition, we get

x(n -m) = 1 £ X(k)W^n-m)k = 1 X)(W3?*A:(*))W^, * = 0 fc=0

as both x(n) and Wj^"* are periodic functions, where n = 0 , 1 , . . . , N — 1.

x(n =F m) & W£mkX(k)

In shifting a waveform, obviously, its form and amplitude remain unchanged. Therefore, the shift results only in a change of the phase shift in the frequency-domain representation. The multiplication by the complex exponential in the frequency-domain is just changing the phase. The delay (advance) of the signal x{n) by m samples produces a phase shift of—j^mk (jij-mk) radians. Figure 4.2 shows the DFT of x(n — 1) in terms of X(k) multiplied by appropriate complex exponential.


z(n) = {0 ,1 ,1 ,0}^X(fc) = { 2 , - 1 - j , 0 , - 1 + j }

Now,

s ( n - l ) = {0,0,l,l} & W*X(k) = (-j)kX(k) = {2,-l + j,0,-l-j}

ar(n + l) = {l,l,0,0} <* W^kX(k) = (j)kX(k) = {2,1-j,0,l+j} 1

Example 4.4 Figure 4.3(a) shows one cycle of the cosine waveform cos(^-n). Its frequency coefficients 8 and 8 are shown in Fig. 4.3(b). Figure 4.3(c) shows the signal delayed by one sample, cos(f | (n — 1)), and its DFT coefficients, shown in Fig. 4.3(d), are related to the coefficients of the original signal by the constants, respectively, Wl& and W^ (a delay of one sample interval and, k = 1 and k = 15). These constants represent phase delays of - 1 ^ = -22.5 degrees and - 1 5 ^ = -337.5 degrees, respectively. The DFT coefficients, shown in Fig. 4.3(d), of the shifted signal are 8 ^ = 8cos(-22.5) + j8sin(-22.5) = 7.391 - j'3.0615 and 8W$ = 8cos(-337.5) + j8sin(-337.5) = 7.391 + J3.0615.

Circular Shift of a Time Sequence 65

S ° • real . o imaginary

10 15

(a)

5 10 *; (b)

15

1

£ ° -1

0

1

* ° -1

r\ I ^

0

1

S ° - 1

\ \ ^

0

5

4 v / \ / V

5

f\ I \ 1 \

1

5

10 n

(c)

A /* \ * \ \7

10 n

(e)

A t\ J \ \S \

10 n

(g)

15

\ \ \ l \J

15

n /

/ J

15

I

1

7.391

0 -3.0615

•

0

0

8

1 ° -8

0

5.6569

0

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0

0

•

5

5

5

10 k

(d)

10 k

(f)

10 k

(h)

•

o

15

o

15

•

o

15

Fig. 4.3 (a) The cosine waveform cos(^|-n) and (b) its DFT. (c) The delayed cosine waveform cos(|^(n - 1)) and (d) its DFT. (e) The sine waveform sin( | |3n) and (f) its DFT. (g) The sine waveform, advanced by 2 samples, sin(||-3(n + 2)) and (h) its DFT.

Figure 4.3(e) shows the sine waveform s in(^3n) . Its frequency coefficients —j8 and j8 are shown in Fig. 4.3(f). Figure 4.3(g) shows the signal advanced by two samples, s in( | |3(n + 2)), and its DFT coefficients, shown in Fig. 4.3(h), are related to the coefficients of the original signal by the constants, respectively, W{~e

6 and W^?G (an advance of two sample intervals and, k = 3 and k — 13). These constants represent phase shifts of 6 ^ = 135 degrees and 2 6 ^ = 10^f = 225 degrees. The DFT coefficients,


Table 4.1 Computing the 4-point DFTs of overlapping segments of a sequence, x(n).

x(n) 2 + j \ X(k) lO + j'13

X(k)

1 + J 4 - 3 + j O 13 + jl8

X(k)

4 + ^3 2 - j 5

- 5 + j O I 8 + 7 I 8

X(k)

3 + j 5 - 1 - J 4

- 5 + JO 0+jO

18 + j l 8

5 + j 6

1 - J 2 0+jO 0+jO

6 + J4

- 2 - j6 0 + jO

4 + J3

- 6 + j2

shown in Fig. 4.3(h), of the shifted signal are -j8W166 = -j'8cos(135) +

j(-j8)sin(135) = 5.657+j5.657andj8W^26 = j8cos(225)+j(j'8)sin(225) = 5.657 - J5.657. I

DFT of overlapping segments of a sequence

If the iV-point DFTs of overlapping segments of a sequence are required, the first iV-point DFT can be computed using a fast DFT algorithm and the rest can be computed at much less computational cost. With an overlap of N - 1 samples, the next set of data is different from the current set in that the first value is lost and a new value is added at the end. Let the DFT of x(n),n = 0 , 1 , . . .,N - 1 be X(k). The DFT of x(n),n = N, 1,2,. . . , N - 1 is X(k) + x(N) - x(0). If we circularly left shift the data by one sample interval, we get x(n),n = 1,2,. . . , iV. The DFT of this sequence is (X(k) + x(N) - ^ ( O ) ) ^ * , k = 0 , 1 , . . . , JV - 1. Consider the sequence x(n) shown in the first row of Table 4.1. The DFT of the first four values is computed directly. The DFT of adjacent set of 4 input values can be computed using the formula given above. Table 4.1 shows the DFT of four adjacent sets. Note that the computational complexity of computing the DFT of each set except the first is only 0(N).

4.4 Circular Shift of a Spectrum

The converse of the previous theorem is that if a time-domain signal is multiplied by a complex exponential to get a new time-domain signal W^mnx(n), then the spectrum of the new signal is that of the original signal circularly shifted by m positions.

W±mnx{n)&X(k±m)

Circular Shift of a Spectrum 67

X(l)

X(2) x(3)Wi3 *

*(3) x(4)W8"

4 *

X(4) x(5)W8~

5 *

x(2)Ws~2

•

X(fc - 1) x(n)W8-

n

X(5) x(6)Wf6

X(0)

X{7) *x(0)Wi

X(6) 'x(7)W8~

7

Fig. 4.4 The delayed DFT sequence X(k — 1) and the corresponding time-domain sequence, x(ra)Vl^~n.

Figure 4.4 shows the spectrum shifted by one position in the counterclockwise direction and the corresponding time-domain sequence. Substituting k — m for k in the DFT definition, we get

J V - l N-l

^ ( f c - m ) ^ ^ ^ ) ^ - " 1 ^ ^ ^ ^ " ^ ) ) ^ , k = 0,l,...,N-l n=0 n=0

For a given k, W^ ~m' corresponds to complex sinusoid with frequency index {k — m). Therefore, the index of the frequency coefficient computed by the summation for the value of k is (k — m) and the frequency scale is shifted by m positions.

Example 4.5 The cosine waveform, cos(y|n), shown in Fig. 4.5(a) has DFT coefficient 8 at k = 1 and k = 15 shown in Fig. 4.5(b). Let us assume that we want to shift the spectrum circularly by one position to the right, that is we want 8 and 8 at k = 2 and k = 0. To achieve this, we have to multiply the cosine waveform by the complex exponential with frequency index 1, e-7w". The complex exponential and its spectrum are shown, respectively, in Figs. 4.5(c) and (d). The product of the two signals shown in Figs. 4.5(a) and (c) and the resulting shifted spectrum are


10 15 Ow o '

0

• real . o imaginary

10 L2 15

1

0

-1

(a)

(c)

5 10 15 n

(g)

16

0HL2J 0

8 •

X

0 b t o

X

(b)

10 L i 15

(d)

10 15

(f)

0 w » * » » » » » o » o » » » » « - 8 - 6 - 4 - 2 0 2 4 6

(h)

Fig. 4.5 (a) The cosine waveform cos(||-n) and (b) its DFT. (c) The complex exponential e*'ff"' and (d) its DFT. (e) The product of the waveforms shown in (a) and (c), and (f) its DFT, which is a shifted version of that shown in (b). (g) The waveform shown in (e) multiplied by ( — 1)" and (h) its DFT, which is the same as that shown in (f) but approximately centered.

Time-Reversal Property 69

shown, respectively, in Figs. 4.5(e) and (f). We can specify the spectrum from looking at Fig. 4.5(e). The waveform corresponding to the real part consists of a dc and a cosine wave with frequency index two, each with an amplitude of 0.5. The imaginary part is a sine waveform with frequency index two and amplitude 0.5. These waveforms produce a spectral value of 8 at frequency indices 0 and two, which is a shifted version of the spectrum shown in Fig. 4.5(b).

A particular case, which is often used in practice, is the multiplication N n

of the input signal with W^ = ( -1)" , with N even. Figure 4.5(g) shows the signal that is the product of the signal in Fig. 4.5(e) and (-1)"- Figure 4.5(h) shows its spectrum which is the same as that of Fig. 4.5(f) with a shift of ~ = 8 sample intervals. Since we are multiplying the given signal by a complex sinusoid with frequency index ^» the spectrum is shifted by half the number of sample intervals ( ^ ) so that the origin is at about the center (center-zero format). Note that neither the magnitude nor the phase of the DFT coefficients changes. Only the position of the coefficients on the frequency scale changes. I

4.5 Time-Reversa l P r o p e r t y

The time reversal of the sequence x(n) = {ar(0),x(l),x(2),a;(3),a;(4),a;(5), x(6),x(7)} is x(&-n) = {x(0),x(7),x(6),x(5),x(4),x(3),x{2),x(l)} shown in Fig. 4.6, which is the placing of the values of the sequence in a clockwise direction starting from x(0). The DFT of a reversed sequence is the reversal of the DFT coefficients of the original sequence as shown in Fig. 4.6. If x(n) & X(k), then x(N - n) <=> X(N — k). By changing the frequency index k to N — k in Eq. (3.6), we get

X(N - k) = £ x(n)W$N-V = £ * ( n ) W - » \ k = 0 , 1 , . . . , N - 1 n=0 n=0

The sum of the products a;(n)W_nfc is, obviously, equal to the sum of the products x(N - n)Wnk. That is,

J V - l

X{N-h)=^x{N- n)W£k

n=0


x(5) X(5) *

x(4) X(4) *

x(3) X{3) *

x(6) X(6)

•

x(8 - n) X(8 - k)

x(2) X(2)

x(7) *X(7)

x(0) •X(0)

x(l) *X(1)

Fig. 4.6 The time-reversal, x(8 - n), of the sequence x(n) and its DFT, X(8 - k).


x{n)= {2-jl, 2 + j 3 , 2 - j 2 , 4 - j 2 } & X ( k ) = {10- j2 ,5+J3 , - 2 - j 4 , - 5 - j l }

Now,

x ( 4 - n ) = { 2 - j l , 4 - j 2 , 2 - j 2 , 2 + j 3 } ^

X(4- fc ) = { 1 0 - j 2 , - 5 - j l , - 2 - j 4 , 5 + j3} I

Computing the DFT twice in succession

Another interesting property is obtained by computing the DFT twice in succession.

x(N-n) = i (DFT(DFT(ar(n)))) = -i(DFT(X(fc)))

x{n) = l (DFT(X(AT-fc)))

If x{n) is even, then

x(n) = i (DFT(DFT(x(n)) ) ) = i ( D F T ( X ( * ) ) )

Symmetry Properties 71

By substituting N — n for n in the IDFT equation, we get

x(N-n) = i E *(W ( J V _ n )* = ^ E * ( * « * , n = 0,1,...N-l k=0 fc=0

Example 4.7 The DFT of x(n) = {2 - j l , 3 - j 2 , l + j l , 2 + j3} is X(k) = {8 + j l , - 4 - j 3 , - 2 - j l , 6 - j l } . If we compute the DFT of X(k) and divide by N = 4, we get z(4 - n) = {2 - j l , 2 + j 3 ,1 + j l , 3 - j2} . If we compute the DFT of X(4 - A;) = {8 + j l , 6 - j l , - 2 - j l , - 4 - j3} and divide by TV = 4, we get x(n). I

This property brings out the almost duality of the time- and frequency-domain sequences. If x(n) •& X(fc),then jjX(N=fn) <S> x(N±k). If x(n) is even, then ^X{n) & x(k).

4.6 Symmetry Properties

The symmetry properties can be used to reduce the computational effort and storage requirements in signal representation and manipulation. Before we describe these properties, some definitions of symmetry are given for a periodic sequence of period N. A sequence is even-symmetric if x(n) = x(N — n). For even N, an example of an even-symmetric sequence is {9,1,5,3,7,3,5,1}. The values at the 0th and the y t h positions can be arbitrary. The other values are even-symmetric with respect to these positions. A sequence is odd-symmetric if x(n) = —x(N — n). For even N, an example of an odd-symmetric sequence is {0,1,4,3,0, —3, —4, —1}. The values at the 0th and the ^ t h positions must be zero to satisfy the definition. The other values are odd-symmetric with respect to these positions. A periodic sequence is even half-wave symmetric if x(n) = x(n ± y ) . A periodic sequence is odd half-wave symmetric if x(n) = —x(n ± y ) . Conjugate or hermitian symmetry, x{n) = x*(N — n), implies that the real part is even-symmetric and the imaginary part is odd-symmetric. Antihermitian symmetry, x{n) = —x*(N — n), implies that the real part is odd-symmetric and the imaginary part is even-symmetric.


Real signal

The definition of DFT can be rewritten expressing x{n), X(k), and W#* in terms of their real and imaginary parts as

2n . . . 27T Xr(k) + jXi(k) = 2 J (%r(n) + jxi(n))(cos jrnk — j sin -rj-nk) (4.1)

n=0

If x(n) is real, its imaginary parts are zero. Then, Eq. (4.1) reduces to

JV-i 2TT 2TT Xr(A;) + jXi{k) = y ^ a;r(n)(cos -Tr A; — j'sin -r^nk)

n=0

While we are going to establish mathematically, it is obvious that, since the cosine is an even function of k, the real part of the spectrum is an even function. Similarly, since the sine is an odd function of k, the imaginary part of the spectrum is an odd function. Substituting k = N — k, we get

JV-i „ „

Xr{N - k) + jXi(N -k)=Y^ xr(n)(cos jfn(N -k)-j sin - n ( J V - *)) n=0

Conjugating both sides and simplifying, we get

Jv-i 2TT 2TT

Xr(N — k) — jX{(N — k) = y ^ ay(n)(cos — nk — j sin —nfc) n=0

= Xr(k)+jXi(k) (4.2)

Therefore, Xr(k) = Xr{N - k) and X{{k) = -Xj(iV - fc). By substituting k = f - Jb, we get X r ( f -A) = Xr(f+k) and X f ( f - * ) = -Xt(% + k).

x(n) r ea l^X(fc ) hermitian

Example 4.8

x(n) = {2,1,4,3} & X{k) = {10,-2 + j2,2,-2 - j2}

Figures 4.7(a) and (b) show, respectively, an arbitrary real signal and its hermitian-symmetric spectrum. I

This symmetry implies that the DFT coefficients X ( 0 ) , X ( y ) , and X(k), k = 1,2,. . . , y — 1 are sufficient to uniquely specify the spectrum of a real signal. Since the coefficients X(0) and X ( y ) are real, for even N, and the rest of the required coefficients are generally complex, a total of N


5 4

* o k> •

• • peal 0 o imaginary

10

* ° 8 . o

10

(d)

15

8

5 4

0

• •

•

(b)

•

•

•

15

10 15

I °

(e)

10 15

4

0 o

(f)

10 15

S °

(g)

10 15

(h)

0 * • • • • o » o » »

10 15

(i) G)

Fig. 4.7 (a) A real signal and (b) its hermitian-symmetric spectrum, (c) An even-symmetric real signal and (d) its real and even-symmetric spectrum, (e) An odd-symmetric real signal and (f) its imaginary and odd-symmetric spectrum, (g) A real signal with even half-wave symmetry and (h) its spectrum with even-indexed harmonics only, (i) A real signal with odd half-wave symmetry and (j) its spectrum with odd-indexed harmonics only.


storage locations are adequate to store an TV-point real-valued signal or its spectrum.

Real signal with even symmetry

Since x(n)sin(^nfc) is an odd function of n, the imaginary part of the DFT is zero. Therefore, as x(n) cos(^-nfc) is an even function of n, we get, for JV even,

N-i 2 ? r

Xr(k) = Xr(N — k)='^2 x(n) cos -rj-nk

= z(0) + (-l) f cx(y) + 2 ^ x ( n ) c o s ^ n f c n=l

x(n) real and even •& X(k) real and even

Example 4.9

x(n) = {2,1,4,1} «- X(k) = {8, - 2 , 4 , - 2 }

Figures 4.7(c) and (d) show, respectively, an even-symmetric real signal and its real and even-symmetric spectrum. I

Real signal with odd symmetry

Since x(n) cos(^nfc) is an odd function of n, the real part of the DFT is zero. Therefore, as x(n)sm(^nk) is an even function of n, we get, for N even,

Jv-i 2 n f-i 2 ? r

Xi(k) = -Xi(N - k) = - ^ x(n) sin -rrnA; = - 2 ^ x(n) sin -rjnk n=0 n=l

x(n) real and odd <S> -XX&) imaginary and odd

Example 4.10

x{n) = {0,3,0, - 3 } <£• X(k) = {0, - j 6 , 0 , j6}

Figures 4.7(e) and (f) show, respectively, an odd-symmetric real signal and its imaginary and odd-symmetric spectrum. I


Real signal with even half-wave symmetry

The computation of the spectrum is equivalent to computing the DFT over two periods of a periodic signal with period y . The DFT equation can be written as

*(*) = £ ^ n ) + (-!)**(» + J)WN (4-3) n=0

Obviously, an even half-wave symmetric signal has odd-indexed DFT coefficients with zero value.

x(n) even half-wave symmetric •& X(k) even-indexed only and hermitian

Example 4.11

x(n) = {1,2,1,2} & X(k) = {6,0, - 2 , 0 }

Figures 4.7(g) and (h) show, respectively, an even half-wave symmetric real signal and its hermitian-symmetric spectrum consisting of even-indexed harmonics only. I

Real signal with odd half-wave symmetry

An odd half-wave symmetric signal has even-indexed DFT coefficients with zero value, which is evident from Eq. (4.3).

x(n) odd half-wave symmetric <$ X(k) odd-indexed only and hermitian

Example 4.12

x(n) - {1,2, - 1 , - 2 } &X(k) = {0,2 - j'4,0,2 + j4}

Figures 4.7(i) and (j) show, respectively, an odd half-wave symmetric real signal and its hermitian-symmetric spectrum consisting of odd-indexed harmonics only. I

Imaginary signal

This case is similar to that of the real-valued signal except that the data values are multiplied by the complex constant j . Therefore,

Xr(k) = -Xr(N - k) and X^k) = Xt(N - k)


x(n) imaginary & X(k) antihermitian

Example 4.13

x(n) = {J2,jl,j4,j3} &X(k) = {jlO, - 2 - j2, j2,2 - j2}

Figures 4.8(a) and (b) show, respectively, an imaginary signal and its antihermitian spectrum. I

Imaginary signal with even symmetry

x(n) imaginary and even O X(k) imaginary and even

Example 4.14

x(n) = {J2,jl,j4,jl} & X(k) = {j8, -J2J4, -j2}

Figures 4.8(c) and (d) show, respectively, an imaginary signal with even symmetry and its imaginary and even-symmetric spectrum. I

Imaginary signal with odd symmetry

x(n) imaginary and odd •£> X(k) real and odd

Example 4.15

x{n) = {0, j3 ,0 , - j 3 } & X(k) = {0,6,0, - 6 }

Figures 4.8(e) and (f) show, respectively, an imaginary signal with odd symmetry and its real and odd-symmetric spectrum. I

Imaginary signal with even half-wave symmetry

x(n)even half-wave symmetric<^-X(fe)even-indexed only and antihermitian

Example 4.16

z(n) = {jl,j2,jl,j2}&X{k) = {j6,0,-j2,0}


o ^ X ^ ^ 10

(a)

(i)

15

10 15

3? 4

-4

•real o imaginary

5 10 k

(b)

15

8

5 ° $< o •

-4 , 5 10 15

k

(d) 8f

jX. op» o • o <

5 10 k

(0

15

4 0

-4(< £ Oh. •

5 10 15 k

4 0

-4

• :

(h)

o o

0 o

. • 5 10

k 15

G)

Fig. 4.8 (a) An imaginary signal and (b) its antihermitian-symmetric spectrum, (c) An even-symmetric imaginary signal and (d) its imaginary and even-symmetric spectrum. (e) An odd-symmetric imaginary signal and (f) its real and odd-symmetric spectrum. (g) An imaginary signal with even half-wave symmetry and (h) its spectrum with even-indexed harmonics only, (i) An imaginary signal with odd half-wave symmetry and (j) its spectrum with odd-indexed harmonics only.


Figures 4.8(g) and (h) show, respectively, an imaginary signal with even half-wave symmetry and its antihermitian spectrum consisting of even-indexed harmonics only. I

Imaginary signal with odd half-wave symmetry

x(n)odd half-wave symmetric <4- X(fc)odd-indexed only and antihermitian

Example 4.17

x(n) = {jl,j2, -jl, -j2} & X(k) = {0,4 + j2,0, - 4 + j2}

Figures 4.8(i) and (j) show, respectively, an imaginary signal with odd half-wave symmetry and its antihermitian spectrum consisting of odd-indexed harmonics only. I

Complex signal

Example 4.18

x{n) = {2+J4,0+j2,4+j3,1+j2} &X(k) = {7+ j l l , -2+J2,5+J3, -2+jO}

Figures 4.9(a) and (b) show, respectively, an arbitrary complex signal with no symmetry and its spectrum, which also has no symmetry. I

Complex signal with even symmetry

Using the results for real and imaginary signals, we get

x(n) even &X (k) even

Example 4.19

x{n) = {2+jA, 1+J2,4+J3,1+J2} &X{k) = {8+ j l l , - 2 + j l , 4+J3, -2+jl}

Figures 4.9(c) and (d) show, respectively, a complex signal with even symmetry and its even-symmetric spectrum. I


_ 1

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^ V

0

2

I o V^ vlj V 0

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1.5

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a/30? / W ty

^ / \ &

5

'a^> 1\ ^ V

5

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H u 5

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yw ^ %

10

(a)

a f*

A./ / ^\^y

10 n

(c)

, ^ l

r \ 10

n

!)dt y>Af

10

n

(g)

\j£^m\ V w\j

J* (

15

^/V n V

15

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15

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8

J? 0

o

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0

8

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4

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a

o

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5

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(b)

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k

(d)

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10 k

(0

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10 k

(h)

•

15

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15

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a

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0) (J) Fig. 4.9 (a) A complex signal and (b) its spectrum, (c) An even-symmetric complex signal and (d) its even-symmetric spectrum, (e) An odd-symmetric complex signal and (f) its odd-symmetric spectrum, (g) A complex signal with even half-wave symmetry and (h) its spectrum with even-indexed harmonics only, (i) A complex signal with odd half-wave symmetry and (j) its spectrum with odd-indexed harmonics only.


Complex signal with odd symmetry

x(n) odd O X(fc) odd

Example 4.20

x(n) = {0,1 + j2 ,0 , - 1 - j2} &X{k) = {0,4 - j2 ,0 , - 4 + j2}

Figures 4.9(e) and (f) show, respectively, a complex signal with odd symmetry and its odd-symmetric spectrum. I

Complex signal with even half-wave symmetry

x(n) even half-wave symmetric O X(k) even-indexed only

Example 4.21

x(n) = {2 + j3 ,1 - j l , 2 + j 3 , 1 - j l } <S> X(k) = {6 + j4,0,2 + j8,0}

Figures 4.9(g) and (h) show, respectively, a complex signal with even half-wave symmetry and its spectrum consisting of even-indexed harmonics only.

I

Complex signal with odd half-wave symmetry

x(n) odd half-wave symmetric O- X (k) odd-indexed only

Example 4.22

x(n) = {1 + j l , 2 + j2, - 1 - j l , - 2 - j2} & X(k) = {0,6 - j2,0, - 2 + j6}

Figures 4.9(i) and (j) show, respectively, a complex signal with odd half-wave symmetry and its spectrum consisting of odd-indexed harmonics only.l

Representation of a signal defined over a finite range

In DFT analysis, it is assumed that the signal is periodic. If the objective is to represent the signal only over the defined finite range, then the function can be arbitrarily extended and the denned range along with the extension can be considered as the fundamental period of the signal. The extension

Transform of Complex Conjugates 81

* 0

0

0

5

10

10 n

(a)

20 «

15

_ ^ 30

Si X OP o (

0

- . 16

• o

real . imaginary

5 10

(b)

•0 0 10 20

k

15

o

30

(c) (d)

Fig. 4.10 (a) A real signal and (b) its spectrum with several nonzero coefficients, (c) The extension of the signal shown in (a), and (d) its spectrum with only a pair of nonzero coefficients.

of the signal can be made in several ways and the DFT representation of all of them will represent the signal correctly in the defined range.

Given the freedom to arbitrarily extend the signal, it is desirable to extend so that the signal is adequately represented by as few coefncients as possible. The basic consideration is to extend the signal so that the extension does not create any discontinuities. The more smoother the extension the smaller is the set of coefficients required to represent the signal.

Example 4.23 Figures 4.10 (a) and (b) show, respectively, half period of a sine waveform and its spectrum. In this representation, the half period waveform is considered as the fundamental period. If we extend the signal as shown in Fig. 4.10(c) and compute the DFT, we get the spectrum shown in Fig. 4.10(d). The difference between the spectra is that the first spectrum has several nonzero frequency coefficients whereas the second spectrum has only two and is, obviously, a more efficient representation. I

4.7 Transform of Complex Conjugates

The conjugation operation reflects the vector represented by a complex number about the real axis, that is the imaginary part is negated. Let x(n) &X(k). Then,

x*(n)&X*(N -k) and x*(N - n) <* X*(k)


Conjugating both sides of Eq. (3.6), we get

J V - l J V - l

X*(k) = Y, x*{n)W^nk = J2 **(N ~ n)W^k

n=0 n=0

Conjugating both sides of Eq. (3.6) and substituting k = N — k, we get

J V - l

X*(N-k)=J2x*(n)W%k, k = 0,l,...N-l 71=0

Example 4.24

x{n) = {2 + jl,3 + j2,l-jl,2-j3}&

X(k) = {8 - j l , 6 + i l , - 2 + j l , - 4 + j3}

x*(n) = {2-jl,Z-j2,l + jl,2+j3}&

X*(4-k) = {8 + j l , - 4 - A - 2 - j l , 6 - j l }

Note that for a real signal x*(n) = x{n) and X{k) = X*(N — k).

ar*(4-n) = {2 - j l , 2 + j 3 , l + j l , 3 - j2} &

X*(k) = {8 + jl,6-jl,-2-jl,-A-j3}

Figures 4.11(a) and (b) show, respectively, a signal and its spectrum. Figures 4.11(c) and (d) show, respectively, the signal x*(16 - n) and its spectrum which is the same as that shown in Fig. 4.11(b) with the spectral values conjugated. Figures 4.11(e) and (f) show, respectively, the signal x*(n) and its spectrum which is the same as that shown in Fig. 4.11(b) with the spectral values conjugated and frequency-reversed. I

4.8 Circular Convolution and Correlation

As the DFT of a data set and the IDFT of a transform are periodic quantities, the convolution operation carried out using the transform as a tool results in a periodic output sequence. This convolution is referred to as circular, cyclic, or periodic convolution. The linear convolution, which is of interest in the analysis of LTI systems, can be simulated by the circular convolution. The method to do that is discussed in Chapter 14. In this section, we just present the theorems.

Circular Convolution and Correlation 83

n * _ * • A Q O

S u B ' 5 " " a ° »

T 2 0

-2C T * o

5 10 n

(a)

s " ' » • ; ' "

15

4

X 0 - 2

5 10 15 n

(c)

w S ' S ' 9 ' " ' 5 ° a

5 10 n

15

(•)

16

8 • real . o imaginary

» « » » » • " • t t t • • 0 5 10

k

(b)

16

5 8

*~ Qp • • • • • • o • • • • -8

15

5 10 k

(d)

15

? i1

16 8 0

- 8

• ) • • • • • • O B • • • • •

O -_ . . a

5 10 k

(0

15

Fig. 4.11 (a) A complex signal x(n) and (b) its spectrum, X(k). (c) The signal i*(16 —n) and (d) its spectrum, X"(k). (e) The signal x*(n) and (d) its spectrum, X*(16-ifc).

Circular convolution in the time-domain

Let x{n) 4> X{k) and h(n) 4> H(k), n,k = 0 , 1 , . . . , JV - 1. Then, the circular convolution of x(n) and h(n) is given by

J V - l J V - 1

y(n) = Y^ x(m)h(n — rn) = Y^ /i(m)a;(n - m), n = 0 , 1 , . . . , N — 1 m = 0 m = 0

This convolution can be implemented using the transform as a tool as given

by

7 V - 1

»(») = E *(*)# (*w* fc=0


That is the circular convolution of two time-domain sequences is obtained by taking the IDFT of the product of the DFTs of the individual sequences. Substituting the corresponding DFT expressions for X(k) and H(k), we get

1 jv-i jv-i tf-i

y^ = ]v £ < £ x(m)W^}{^2 h(l)WlN

k}W^k

k=0 m=Q 1=0

Rearranging the summation, we get

m=0 1=0 ifc=0

The rightmost summation is equal to N for / = n — m and zero otherwise. Therefore,

7 V - 1

y(ri) = 22 x(m)h(n — m) m=0

Example 4.25 Convolve x(n) = {1,4,2,0} and h{n) = {2,3,0,1}.

X ( * ) = { 7 , - 1 - J 4 > - 1 , - 1 + J 4 } and tf(fc) = {6,2 - j2 , - 2 , 2 + j2}

X(k)H(k) = { 4 2 , - 1 0 - J 6 , 2 , - 1 0 + J"6}

The product is obtained by multiplying the corresponding terms in the two sequences. The IDFT of X(k)H(k) is the convolution sum, y(n) = {6,13,16,7} I

Circular convolution in the frequency-domain

The circular convolution of two frequency-domain sequences, X(k) and H(jfe), k = 0 , 1 , . . . , N - 1, divided by N is obtained by taking the DFT of the product of the IDFTs of the individual sequences.

1 J V _ 1 1 N~l Y(k) x(n)h(n) & - J2 X(m)H(k - " » ) = - £ H(m)X(k - m) = -j^,

m=0 m=0

where Y(k) is the convolution of the sequences X(k) and H(k) and k = 0 , l , . . . , i V - l .

Example 4.26 The product of x(n) and h(n) given in Example 4.25 is {2,12,0,0}. The DFT of this sequence is {14,2-j 12, -10 ,2 +jl2}= ^ . 1

Sum and Difference of Sequences 85

Circular correlation of time-domain sequences

The circular cross-correlation of two time-domain sequences x(n) and h(n), n = 0 , 1 , . . . , N — 1 is given by

7 V - 1

yXh(n) = ^2x*(m)h(n + m), n-0,l,...,N-l, 771=0

where x*(m) is the complex conjugate of x(m). Note that, for real signals, x*(m) = x(m). This equation can also be written, in terms of the convolution operation, as

J V - l

yxh(n) = ^x*(N -m)h(n-m), n = 0,1, . . . ,JV - 1, 771=0

where x*(N — m) = ID FT of X*(k). Therefore, the circular correlation of two time-domain sequences can be obtained by taking the ID FT of the product of the complex-conjugate of the DFT of the first sequence and the DFT of the second sequence.

yxh(n) = IDFToi (X*(k)H(k))

Vhx(n) = y*xh(N-n)= IDFT of (H*(k)X(k))

Example 4.27 For the sequences in Example 4.25

yxh(n) = {14,5,8,15} and yhx(n) = {14,15,8,5} I

The autocorrelation operation is the same as the cross-correlation operation with x(ri) — h(n).

yxx(n)= IDFTof(|X(fc)|2)

Example 4.28 For the sequence {2,3,0,1}, we get yxx{n) ={14,8,6,8}.l

4.9 Sum and Difference of Sequences

Since A; = 0, the values of all the Wfik terms in the DFT definition is just one and the value of X(0) is the sum of the input sequence x(n).

N-l

X(0) = £ x(n) 71=0


With an even N and k = y-, the input sequence values are alternately multiplied by 1 and - 1 . Therefore,

N - 2 JV-1

*(y> = £ *(») - £ *(») n=0,2 n=l ,3

Example 4.29

x(n) = {2,1,3,4} & Jf(fc) = {10 , -1 + ^ 3 , 0 , - 1 - ^ 3 }

X(0) = 2 + l + 3 + 4 = 10 and X(2) = (2 + 3) - (1 + 4) = 0 I

In the case of the IDFT,

*(°) = Jf E XW

With an even N,

^y 2 TV —1

*(y) = ^( £*(*)- £*(*)) fc=0,2 fc=l,3

Example 4.30 For the x(n) and X(k) given in the earlier example,

a:(0) = i ( 1 0 - l + j 3 + 0 - l - j 3 ) = ^ = 2

a:(2) = | (10 + l - j 3 + 0 + l + i 3 ) = j = 3 '

These values can be used as a preliminary check of the output of an algorithm to compute the DFT or the IDFT.

4.10 Padding the Data with Zeros

Padding the data with zeros at the end

Let x(n) O- X(k), n, k = 0 , 1 , . . . , N — 1. If we pad x(n) with zeros to get y(n), n = 0 , 1 , . . . , mN — 1 defined as

, . _ f x(n) for n = 0 , l , . . . , i V - l VW - | Q otherwise

Padding the Data with Zeros 87

where m is any positive integer, then,

Y(mk)=X{k), fc = 0 , l , . . . , iV

The DFT of the signal y(n) is given by

mN-l

Y(k)= Y, y(n)W£kN, h = 0,1,-••,mN-l

n=0

Since y(n) is zero for n > N — 1, we get

J V - l

Y(k)=Y,y(n)w™N, k = 0,l,...,mN-l

n=0

Substituting mk for A; and simplifying, we get

J V - l

Y(mk)= Y^y{n)Wtf =X(k), k = 0 , 1 , . . . ,JV - 1 n=0

Example 4.31 Let m = 2 and x(n) = {2,1,4,3}. X(k) = {10, -2 + j 2 , 2 , - 2 - j 2 } . Then,

y(n) = { 2 , 1 , 4 , 3 , 0 , 0 , 0 , 0 } ^

Y{k) = {10,*,-2 + j 2 , * , 2 , * , - 2 - j 2 , * }

By zero padding, the frequency increment of the spectrum is halved. Therefore, the spectral values with indices 0,1,2,3 in X(k) become spectral values with indices 0,2,4,6 in Y{k). I

Figures 4.12(a) and (b) show, respectively, a signal with eight samples and its spectrum. Figures 4.12(c) and (d) show, respectively, the same signal padded up with eight zeros at the end and the corresponding spectrum. The even-indexed spectral values are the same as those shown in Fig. 4.12(b). The odd-indexed spectral values are not specified by this theorem. By zero padding at the end, we get interpolation of the spectral values.

A similar effect is observed in zero padding a spectrum. Figures 4.12(e) and (f) show, respectively, a spectrum with eight samples and the corresponding time-domain signal. Figures 4.12(g) and (h) show, respectively, the same spectrum padded up with eight zeros in the middle of the spectrum (at the end in the center-zero format) and the corresponding time-domain signal. The even-indexed signal values are one-half of those shown

Properties of the DFT

~2W ? n X: £ 0 K

-2

5 f

^ ~

2 4 6 n

(a)

£ °

5 10 15 n

(c)

-4 o 2 4 6

k

(e)

5 10 15 k

(9)

£ ° • real o imaginary

5

Sf 0

-5

(b)

8 " oS « .

o ° 8 o

5 10 15

(d)

(h)

Fig. 4.12 (a) A real signal and (b) its spectrum, (c) The signal shown in (a) with zero padding at the end and (d) its spectrum with even-indexed values same as in (b). (e) A spectrum and (f) the corresponding time-domain signal, (g) The spectrum shown in (e) with zero padding and (h) the corresponding time-domain signal with even-indexed values one-half of those shown in (f).

Padding the Data with Zeros 89

in Fig. 4.12(f). By zero padding at the end in the frequency-domain, we get interpolation of the time-domain samples.

Padding the data with zeros in between the samples

Let x(n) O X(k), n, k = 0 , 1 , . . . , N — 1. If we pad x(n) with zeros to get y{n), n = 0 , 1 , . . . , mN — 1 defined as

{? , . _ . x{n) for n = 0 , 1 , . . . , N

' n otherwise

where m is any positive integer, then,

Y{k) = X(k modN), k = 0 , 1 , . ..,mN - 1

The DFT of the sequence y(n) is given by

m J V - l

Y{k)= J2 y(n)W^kN, k = 0,1,...,mN-1

n=0

Since we have nonzero input values only at intervals of m, we can substitute n = ran. Then, we get

J V - l i V - l

Y(k) = £ y(mn)W™Zk = £ y(mn)WRk = X(k mod N), n=0 n=0

where k = 0 , 1 , . . . , mN — 1. Y(k) can be obtained by repeating X(k) m times. This is due to the periodicity of Wfik.

Example 4.32 Let m = 2 and x(n) = {2,1,3,4}. X{k) = {10 , -1 + j 3 , 0 , - l - j 3 } . Then,

y(n) = { 2 , 0 , 1 , 0 , 3 , 0 , 4 , 0 } ^

Y{k) = { 1 0 , - l + j 3 , 0 , - l - A 1 0 , - l + A 0 , - l - j 3 } I

Figures 4.13(a) and (b) show, respectively, the signal in Fig. 4.12(a) padded up with eight zeros placed in between the samples and the corresponding spectrum. The spectrum shown in Fig. 4.12(b) repeats itself, in Fig. 4.13(b).

A similar effect is observed in padding with zeros in the frequency-domain. Figures 4.13(c) and (d) show, respectively, the spectrum in Fig. 4.12(e) padded up with eight zeros placed in between the samples and


2

S o H

-2

•

0

4 3~ X

- 4 3

•

o

•

5

5

• • • •

•

10 n

(a)

10 k

(c)

•

15

o

15

5

sr ° -5

• • •

o° o

0 5

0.5

| 0

-0.5

/ \ / \ \

V 3 5

o . . o

• • • • •

0 ° o

10 15 k

(b)

/ \ / \ / \ / / \S

10 15 n

(d)

Fig. 4.13 (a) The signal shown in Fig. 4.12(a) with zero padding in between the samples and (b) its spectrum which is the same as that shown in Fig. 4.12(b), but repeats, (c) The spectrum shown in Fig. 4.12(e) with zero padding in between the samples and (d) the corresponding time-domain signal which is the same as that shown in Fig. 4.12(f) with one-half amplitude, but repeats.

the corresponding time-domain signal. The signal shown in Fig. 4.12(f) repeats itself, in Fig. 4.13(d) with one-half amplitude. Note that the indices of the frequency coefficients in Fig. 4.13(c) are 2 and 14 whereas they are 1 and 7 in Fig. 4.12(e). With the same frequency coefficients but with a frequency index of 2 and double the number of samples, we get two cycles of the same waveform with one-half amplitude as shown in Fig. 4.13(d).

4.11 Parseval ' s T h e o r e m

This theorem implies that the sums of the squared magnitudes of the input and DFT sequences are related by the constant N, the number of samples. That is the signal power can also be computed from the DFT coefficients of the sequence. Let x(n) & X(k), n, k = 0 , 1 , . . . , N — 1. Then,

J V - l i V - l

E wi2 = i E i*(*)i 71=0 fc=0

Summary 91

Since the squared magnitude can be computed by multiplying a complex number by its conjugate, we can write the left summation as

5>(n)|2=5>(n)z>) n=0 n=0

Substituting the corresponding IDFT expressions for x(n) and x*(n) , we get

J V - l J V - l J V - l

n = 0 fc=0 m = 0

N-lN-l JV-l

= ^ E E ^ * H E ^ n ( M

fc=0 m=0 n=0

If fe = m, this expression becomes

Otherwise, it evaluates to zero due to the orthogonal property.


{2,1,4,3} O {10, - 2 + j2, 2, - 2 - j2}

The sum of the squared magnitude of the data sequence is 30 and that of the DFT coefficients divided by 4 is also 30. I

The generalized form of this theorem applies for two different signals x(n) and y(n) as given by

J V - l 1 J V - l

n = 0 fc=0

4.12 Summary

• In this chapter, we have studied several properties of the DFT. Either in applications of the DFT or in developing fast DFT algorithms, these properties are repeatedly used. In the next chapter, we shall find how the properties are used to develop fast DFT algorithms.


Reference

(1) Brigham, E. O. (1988) The Fast Fourier Transform and Its Applications, Prentice-Hall, New Jersey.

Exercises

4.1 Verify the linearity property ax{n) + by(n) & aX(k) + bY(k). 4.1.1 x(n) = {2+j3,l-j4,2-j2,l+jl},y(n) = {2-j3,l-jl,2+j4,l-j2}, a = j3, and 6 = 2. 4.1.2 X(k) = {l-j4,2+j3,l-jl,2-j2}, Y(k) = {3 - j l , 4 + jl,2 + j2,1 + j3}, a = -j3, and b = 4.

4.2 Verify the periodicity of the DFT and the IDFT. * 4.2.1 Compute the DFT of x(n) = {1 - j4 ,2 + j3,1 - jl, 2 + j 4 } . Find X(23), X( -47) . 4.2.2 Compute the IDFT of X(k) = {2 - jA, 2 + j2,1 + j5,1 - jl}. Find 3(27), x(-35) .

4.3 Verify the time-domain shift property. 4.3.1 Find the DFT of x(n) = {2+jl, 1+J2, l-jl,4-j2}. Compute the DFT of x(n + 3) using the shift property. * 4 . 3 . 2 T h f i D F T o f x ( n - 2 ) i s { 2 + j 3 , l - i l , 3 - j l , 2 + j 2 } . Findx(n + 1) using the shift property. 4.3.3 Compute the DFT of Ae~^e^n. Using the shift property, compute t h e D F T o f 4 e ^ ' f e ^ ( " - 2 ) . 4.3.4 Compute the DFT of — 2sin(|7i). Using the shift property, compute the DFT of - 2 sin(f (n + 1)). 4.3.5 Let x(n) = {2,1,3,5,7,2,3,2,1,2,3}. Compute 4-point DFT of the overlapping segments of x{n) with an overlap of 3 data values. 4.4 Verify the frequency-domain shift property. 4.4.1 Let x{n) = {1 - jA, 2 - j2,1 - j3,2 + j5}. Find X(k). What is the input that would generate the spectrum, (a) X(k + 1). (b) X(k — 1). (c) X(k-2). 4.4.2 Find the DFT of x(n) = {1 - jl, 2 + jl, 1 - j3,2 + j4}. Deduce the spectrum of: (a) x(n) cos(^Ln) and (b) x(n)sm(^n).

4.5 Compute the DFT of x(n) = {1 - jl, 2 - j2,1 + j3,2 - jl}. Deduce the DFT of x (4 -n).

Exercises 93

4.6 Find the symmetry of x(n). Compute the DFT and verify that the transform exhibits the anticipated property. 4.6.1 x(n) = { 2 , - 4 , 2 , - 4 } . 4.6.2 s(n) = {0 , -3 ,0 ,3} . * 4.6.3 x{n) = { -1 ,2 ,1 , -2} . 4.6.4z(n) = {2,3 , l ,4} . 4.6.5z(n) = {l ,2 ,3 ,2} .

4.7 Find the symmetry of x(n). Compute the DFT and verify that the transform exhibits the anticipated property. 4.7.1 x(n) = {-j4,-j2,j4,j2}. * 4.7.2 x(n) = {0 , - j 4 , 0 , j 4} . 4.7.3 x(n) = {J2,j3,j2, j3} . 4.7.4 z(n) = { j l , - j 2 , - j 3 , j 4 } . 4.7.5 x(n) = { - j 2 , j l , - j 4 , j l } .

4.8 Find the symmetry of x(n). Compute the DFT and verify that the transform exhibits the anticipated property. * 4.8.1 x(n) = {0,1 - j2,0, - 1 + j2} . 4.8.2 x(n) = {2 + j 3 ,1 - j l , - 2 - j 3 , - 1 + j l } . 4.8.3 x(n) = {1 + j3 ,2 + j2,4 - j3 ,2 - j l } . 4.8.4 x{n) = {2 + j3 ,1 + j2, - 2 - j4 ,1 + j2}. 4.8.5 x{n) = {1 - j2 ,2 - j 3 ,1 - j2 ,2 - j3} .

4.9 Find the DFT of x{n) = {0,1,1,0} and x{n) = { 0 , 1 , 1 , 0 , - 1 , - 1 } . What are the number of nonzero frequency coefficients in each case.

4.10 Find the DFT of x{n) = {1 - j4 ,2 + j 3 ,1 - j l , 3 + j3} . Deduce the DFT of x*{n) and a;* (4 - n)

4.11 Find, X(fc), the DFT of x(n) = {3+j3 ,2+ j2 , l - j 4 , 2 - j l } . Find the DFT of: (a) X(k) and (b) X(4 - k). What is the relation of the resulting sequences to x(n).

* 4.12 Find the circular convolution of the time-domain sequences, x(n) — {3,2, - 1 , 2 } and h(n) = {3, - 1 ,2 ,4} using the DFT.

4.13 Prove that the circular convolution of two frequency-domain sequences, X(k) and H(k), k = 0 , 1 , . . . , N-l is N times the DFT of the product of the corresponding time-domain sequences, x(n) and h(n), n = 0 , 1 , . . . , N — 1.


4.14 Find the circular convolution of the frequency-domain sequences, X(k) = { 2 + j l , l - j 4 , 2 + j 5 , 3 + j 2 } a n d # ( f c ) = { 2 + j 3 , l - j 2 , l - j l , 2 + j 4 } using the DFT.

4.15 Find the circular cross-correlation of the time-domain sequences, x{n) and h(n), using the DFT. Deduce the correlation of h(n) and x(n). * 4.15.1 x(n) = {3,4, - 1 , 2 } and h(n) = {3, - 1 , 3 , 4 } . 4.15.2z(n) = {2 + j l , l - j 4 , 2 + j5 ,3+j2}and/ i (7 i ) = {2 + j 3 , l - j 2 , l -j l , 2 + j4} .

4.16 Find the circular autocorrelation of the time-domain sequence using the DFT. 4.16.1 x(n) = {2 ,4 , -1 ,2} . 4.16.2 x(n) = {j2, j 4 , - j l , j 2 } . 4.16.3 x{n) = {2 + j l , 1 - j4 ,2 + j5 ,3 + j2}.

4.17 Find X(0) and X(4) with x(n) = {2 + j l , 1 - j4 ,2 + j5 ,3 + j2 ,3 -j 2 , 2 - j l , l - j 3 , 2 + j 2 } .

4.18 Find x(0) and x(4) with X(k) = {2 - j l , 4 + j4 ,2 - j 5 ,3 - j2,3 -jl,2 + jl,l-j3,2-j2}.

4.19 Let x(n) = {2 + j l , 4 - j 4 ,1 - j '5,3 - j2}. Compute X(k). Deduce y(0), Y(2), F(4), and F(6) from X{k) if j/(n) = {2 + j l , 4 - j 4 , l - j 5 , 3 -j2,0,0,0,0}.

4.20LetX(ifc) = { 2 - j l , 3 - j 4 , l - ; ' 5 , 3 - j 2 } . Compute x{n). Deduce y(0), y(2), 2/(4), and 2/(6) from x(n) if Y{k) = {2- j ' 1 , 3 - j 4 , l - j 5 , 0 , 0 , 0 , 0 , 3 -J2}.

4.21 Let x{n) = {2 + j l , 4 - j l , 1 - j 2 ,3 - j2} . Compute X(k). Deduce y(fc)fromX(fc)if2/(n) = {2 + j l , 0 , 4 - j l , 0 , l - j 2 , 0 , 3 - j 2 , 0 } .

4.22 Let X(k) = {1 - j l , 2 - j4 ,1 - j 5 ,3 + j2} . Compute x{n). Deduce y(n) from x(n) if Y(k) = {1 - j l , 0,2 - j4 ,0 ,1 - j5 ,0 ,3 + j2 ,0} .

4.23 Verify ParsevaFs relation. * 4.23.1 x{n) = {1 - j3 ,1 + j4 ,1 - j 2 ,3 - j 2} .

4.23.2 x(n) = {1 - j 3 ,1 + j4 ,1 - j2 ,3 - j2} and h(n) = {1 + j 3 ,1 + j l , 3 -

A i - J ' 2 } -

Chapter 5

Fundamentals of the P M DFT Algorithms

If we compute the DFT or IDFT from definition, the computational complexity is 0(N2) and they will not be efficient for signal analysis. Therefore, the task now is to develop fast algorithms for these operations. Due to the similarity of the DFT and the IDFT computation, a DFT algorithm can be used to compute the IDFT with trivial modifications. Therefore, we concentrate on the development of DFT algorithms.

The computation required for an TV—point DFT is the evaluation of N sums, each of N products. The data values are the same and a set of TV twiddle factors is repeatedly used in forming the products for computing each of the N sums. Therefore, there is a large amount of redundancy in the computation. For example, with N = 8, the data sample x(l) is multiplied by Wg and — W$ in order to compute the coefficients X(l) and X(5), respectively. We can use any method to find the sum of products, but the object is to reduce the number of multiplication and addition operations. Typically, a multiplication operation is split and is carried out in stages. For example, a multiplication by W^ is split into multiplications by W^ and Wtf. This approach helps the usage of partial results for the computation of more than one coefficient. A multiplication is carried out after adding all possible terms. Now, the way to split the multiplications and share the partial results is found using the DFT properties.

An alternate view of the problem of the computation of the DFT is that a larger DFT is decomposed into a large number of smaller DFTs, in particular into two-point DFTs, and the coefficients of the smaller DFTs are combined to form the coefficients of the larger DFT. Again, the DFT properties are used to find the procedure of decomposition and combina-

95

96 Fundamentals of the PM DFT Algorithms

tion. In short, the classical divide-and-conquer strategy of developing fast algorithms is used.

In Sec. 5.1, first, we reformulate the DFT definition using vector input and output quantities with each vector having two complex values. Then, the reformulation is extended to any vector length. In Sec. 5.2, the direct implementation of the reformulated DFT definition is described. In Sec. 5.3, the reformulation of the IDFT definition in terms of vector quantities is presented. The computation of the IDFT using DFT is described in Sec. 5.4. In Sees. 5.5 and 5.6, the PM DIT DFT and the PM DIF DFT type of algorithms, respectively, are developed for N = 8. In Sec. 5.7, the set of PM DFT algorithms is classified.

5.1 Vector Format of the DFT

In order to derive more efficient DFT algorithms, we have to reformulate the DFT definition using vector input and output quantities. Before we derive the general form, let us consider the computation of a 4-point DFT using input and output vector quantities, each with two elements. The definition of a 4-point DFT to compute X(k) from x(n) is shown in matrix form below, explicitly showing the half-wave symmetry of the twiddle factors.

* ( 0 ) 1 X(l) X{2) X{2)\

• w 4 ° W4° wl

. w2

w% wl

-w% -wl

w% -w$

w2 -w%

w2 ' -wl -w%

wl .

r x(o) x(l) x(2)

1.3(3)

The half-wave symmetry of the twiddle factor matrix along the rows can be used to rewrite the equation as

X(0)1 X(l) X(2) X(3) J

r W2 w2 w2 w2

w° •

wl -w2 -wl .

• a,(0) " . a,(l) .

where a0(0) = x(0)+x{2), ai(0) = x(0)-x(2), a0( l) = a?(l) + a;(3), a i ( l ) = a;(l) — x(3), and q = k mod 2. Let us say we want to compute the DFT of x(n) = {2+ j l , 3—j l ,4+ j ' 2 , —1+J2}. Then, the input vectors are given as ag(n) = {(a0(0),ai(0)),(ao(l),ax(l))} = { ( 6 + A - 2 - j l ) , ( 2 + i M - j 3 ) } . The half-wave symmetry of the twiddle factor matrix along the columns can

Vector Format of the DFT 97

o(0) = {o0(0),o1(0)} =

{s(0) + x(2),a;(0)-x(2)}

o( l ) = {o0( l) ,o1( l)} =

{x(l) + x(3),x(l)-x(3)} (a)

A(0) = {Ao{0)MO)} = {X(0),X(2)} Ao(0) = a o ( 0 ) + a o ( l ) A1(0) = a o ( 0 ) - a o ( l )

i l ( l ) = {^,( l ) ,A 1 ( l )} = {X(l) ,X(3)} i40(l) = ai(0) + W4

1ai(l) Ai(l) = a i (0 ) -W 4

1 a i ( l )

(2 + j l ) ± ( 4 + j2) =

{6 + j 3 , - 2 - j l }

( 3 - j l ) ± ( - l + j 2 ) =

{ 2 + j l , 4 - j 3 }

{8 + j4 ,4 + j2}

^ b { - 5 - j 5 , l + j3}

(b) Fig. 5.1 (a) The SFG for the computation of 4-point DFT using vectors of length two. (b) A specific example of (a).

be used to rewrite the equation as

P(0) AP(1)

w2 ( - l ) 0 p a , (0)

L (-l)lpag(l) J

where Ao(0) = X(0), A^O) = X{2), A0{l)=X(l), Ai(l) = X(3), p = 0,1 and q — k mod 2. The matrix equation can be written in algebraic form as

MQ = £ ( - i ) p n a»wT*, * = o,i (5.1) n=0

where p = 0,1 and 9 = k mod 2. Equation (5.1) seems more complex than the direct definition, Eq. (3.3). But it is actually easy and more efficient to implement. For JV = 4, the definition of the DFT, Eq. (5.1), using vectors itself is the optimum algorithm. The computation is shown in Figs. 5.1(a) and (b) using a signal-flow graph (SFG).

The signal-flow graph

Although mathematical equations describe the DFT algorithms completely, the SFG is the most convenient way to describe them as it gives all the details at a glance. The SFG depicts the process of combining the input values to generate the output values using nodes and arrows. We describe here the


SFG for the vector length of two. It consists of unfilled circles indicating nodes, and arrows indicating the signal flow path. Each node, except at the beginning, represents an add-subtract operation. The nodes at the beginning of a SFG just receive the input vector values. In Figs. 5.1(a) and (b), the input vectors are placed close to the input nodes. The arrows terminating at an upper node originate from the nodes whose first element of their vectors contributes to form the elements of the vector at the upper node by add-subtract operation. The lower node receives the second elements of its source vectors and the elements of the vector at a lower node are also formed by add-subtract operation. At both type of nodes, the result of the add operation forms the first element and the result of the subtract operation forms the second element of the vector. Each data value along an arrow is multiplied by a twiddle factor, W^, where N is the data length. The exponent of the twiddle factor, s, is indicated close to the arrowhead. No value or a zero near an arrowhead indicates that the value of the twiddle factor is unity. Operations other than add-subtract and multiplication, and any other objects will be indicated by special symbols.

The output vectors and the equations relating them to the input vectors are shown at the output nodes in Fig. 5.1(a). Figure 5.1(b) shows the operation of Eq. (5.1) for a specific example. Before we proceed to generalize the DFT definition for any vector length, we try to answer the question why do we usually use complex signals in DFT algorithms although we are almost always interested in real signals.

Why do we use a complex signal

A complex signal is an ordered pair of real signals. For most part, we are interested in processing real signals. But there is nothing wrong in processing two real signals at a time if that procedure is advantageous. The DFT of two real signals is combined in the DFT of a complex signal. The individual DFTs can be split. If a signal is complex, its DFT is complex whereas if it is real, its DFT is mostly complex. By keeping the input and output values of the same (complex) type, we get algorithms that are very regular and simpler. So much is the difference in the regularity and simplicity, that although it may look odd, it is assumed that, in designing DFT algorithms, the signal is complex as though the complex signal is naturally occurring, but not the real signal. We can compute the DFT with both the input and output quantities real. However, the algorithm

Vector Format of the DFT 99

will be irregular and practically inefficient. The reason is that representing sinusoids by complex exponentials is natural and most efficient.

Vector format of the DFT

We present an equivalent vector format of Eq. (3.6) for any value of u, where u is the vector length and it is a factor of N, the data length. For following the derivation easily, apply each of the steps to the specific case of TV = 8 and u = 2. The individual DFT equations with N = 8 are given in Chapter 3. In addition, refer to the computation of the 4-point DFT presented at the beginning of this section. Each sum of N products in Eq. (3.6) can be rewritten as the sum of ^ products each of which is formed using a sum of u products.

U — 1 , r U — 1 T.J

X(k) = W°Nk Y, x(0 + s-)W°N

k" +WkkJ2x(l + s-)WN - + • " + 8=0 8=0

8=0

n=0 s=0 n=0 8=0

For each value of n, the inner summation is a u-point DFT and when evaluated will give rise to u distinct values for k = 0 , 1 , . . . , u — 1. Let a(n),n = 0 , 1 , . . . , ^ — 1 denote the nth input vector (Vectors will be shown in boldface.) consisting of the u-point DFT values aq (n), q = 0 , 1 , . . . , u — 1 as defined below.

a(n) = { a 0 ( n ) , a i ( n ) , . . . A - i W } = V i ( n + 8-)B'u,«,

z—' u 8=0

0 = 0 , 1 , . . . , u - l , n = 0 , l , . . . , 1 (5.2) u

Now, the last but one equation can be rewritten as

X(k) = J2 aq(n)W£k, fc = 0 , l , . . . , i V - l n=0


Since the values of the u-point DFT aq(n) are periodic with a period of u, for values of k equal to or greater than u, the values aq (n) repeat. Therefore,

q = k mod u

Replacing the variable k by k + p— in the last two equations, we get

£ - 1

X(k+p*) = ^ r a , W ^ , i = 0 , l , . . , - - l , n=0 U

N p = 0 , 1 , . . . , u — 1, q = (k + p—) mod u

u

The N frequency coefficients X{k) can also be represented by ^ vectors each with u elements. Let

A(k) = {A0(k),A1(k),...,Au_1(k)} = {X(k+p-)}, u

N k = 0 , 1 , . . . , 1, p = 0 , 1 , . . . ,u - 1 (5.3)

u

With vector input quantities aq (n) as defined by Eq. (5.2) and vector output quantities Ap(k) as defined by Eq. (5.3), the DFT as defined by Eq. (3.6) can be equivalently written as

Ap(k) = J2 WZnaq(n)Wtf, k = 0 ,1 , . . . , ^ - 1 (5.4) n=0

where q= (k+ p^) mod u and p = 0 , 1 , . . . , u — 1. Thus, the use of vector input and output quantities changes the form of the DFT definition from Eq. (3.6) to Eq. (5.4).

Vector format of the DFT with u = 2

The use of vectors, each with two elements, provides the highest efficiency since the frequency coefficients of a two-point DFT are the most closely related. For the specific value of u = 2, Eqs. (5.2), (5.3), and (5.4) become, respectively, as

N N a(n) = {o0(n),ai(n)} = {x(n) + x(n+ -^),x(n) -x(n+ —)},

n = 0 , l , . . . , y - l (5.5)

Direct Computation of the DFT with Vectors 101

A(k) = {MQMk)} = {X(k),X(k+ j)}, fc = 0 , l , . . . , ^ - l (5.6)

N 4(*) = £ (-l)pna„(n)WZk, k = 0 ,1 , . . . , - - 1

n=0

(5.7)

where p — 0,1 and q = (k + p y ) mod 2. The vector format definition of the 8-point DFT with u = 2 is shown below in matrix form.

Ap(0)

4,(1) 4(2) 4(3)

- wi wi wi wi' wi wi wi wi wi wi wi wi

. wi wi wi wi .

r (-l)<*a,(0)

( - l W i ) (-1)2%(2)

L (-l)3"a9(3)

(5.8)

5.2 Direct Computation of the DFT with Vectors

In this section, we present an implementation of the DFT expression, Eq. (5.7), for N that is an integral multiple of 4. It can be easily modified for an even N. This implementation will be useful for testing programs. For these lengths, this implementation is much more efficient than that given in Chapter 3.

The main module, shown in Fig. 5.2(a), invokes four modules to get the computation done. The second module, shown in Fig. 5.2(b), sets up the twiddle factor array tf of size JV, that is the computation and storage of the sample values of the cosine function over one cycle, with argument starting from zero with increment of ^ . This module is called twid-fac. Variable i is used as a loop counter and the loop is terminated when i equals N, the data size. The setting up of the twiddle factor array can be made faster by using identities such as cos(^(iV — n)) = cos (^n) . Note that this module can be eliminated by computing the values of the twiddle factors each time they are required. However, it is costly in terms of run-time.

The next module, shown in Fig. 5.2(c), reads the real and imaginary parts of the complex input data, respectively, into the arrays xr and xi, each of size N, the data size. It is assumed that the real parts of the input data are stored before the imaginary parts in the input file. If the data is real, initialize the values of the array xi to zero and read the data into the array xr. This module is called in-put.


( Start")

call twid_fac I

call in.put

1 call vec_dft

1 call out_put

f Stop ^

5.2(a)

( s ta r t *)

read value ar(t), i = 0 , l , . . . , i V - l

1 read value xi{i),

i = 0,l,...,N-l

(Return J

('start ")

arp = 0.0

mc — 2ff

z = 0

{^ReturnJ

tf(i) = cos(arg) arg — arg + inc

i = i + l

5.2(b)

Tstart J

print value xr(i),xi(i),

i = 0,l,...,N-l

X

5.2(c)

(Return)

5.2(e)

The next module, shown in Fig. 5.2(d), computes the DFT coefficients. This module is called vec-dft. First, input vectors are formed by computing the 2-point DFT of pairs of values taken from the first and second half of the input data. The vectors are stored in arrays ar and ai, each of size N. The coefficient computation is carried out in two nested loops, the outer loop controlling the frequency index and the inner loop controlling the data index. In each iteration of the outer loop, the DFT values with indices A; and k + y are computed. The real and imaginary parts of the DFT coefficients are stored, respectively, in arrays xr and xi. The sum of products for the even and odd indexed terms are computed separately.

Direct Computation of the DFT with Vectors

('start')

n l = 0,n2 = N

n o

ar(nl) — xr(nl) + xr[n2),ar(n2) = xr(nl) — xr{n2) ai(nl) = xi(n 1) + xi(n2),ai{n2)

i,l = n l + l ,n2 = r = xi(nl) — xi(n2)

h2 + l

( Return)

n l = 0, n2 = nl + 1, n3 = y,nfc = 0, sumer = 0, sumei = 0, sumor = 0, sumoi = 0

i/(fc mod 2 ^ 0), {nl = f, n2 = n l + 1, n3 = iV}

c = t/(n& mod iV), a = */((nfc + f ) mod JV)) sumer = sumer + ar(nl) * c - oi(nl) * s sumei = sumei + ai(nl) * c + ar(nl) * s

c = i/(nfc mod N), s = tf((nk + f ) mod N)) sumor = sumor + ar(n2) * c — ai{n2) * s sumoi = sumoi + ai(n2) *c + or(n2) * s nk = nk + k, nl = n l + 2, n2 = n2 + 2

_L xr(k + ~) = sumer — sumor,xi(k + y ) = sumei — sumoi Tl^r

xr(k) = sumer + sumor,xi(k) — sumei + sumoi, k = k + 1

5.2(d)

Fig. 5.2 The flow chart description of the computation of the DFT with vector length two. (a) The main module, (b) The twiddle factor module, (c) The input module. (d) The DFT module, (e) The output module.


The sum of these yields a;(fe) and difference yields x(k + —•). The access of correct twiddle factor values is carried out using the mod function. Inside the inner loop, the coefficient computation is carried out according to the DFT definition. The next module, shown in Fig. 5.2(e) prints the real and imaginary parts of the DFT values, respectively, from the arrays xr and xi, one complex coefficient in each iteration. This module is called out-put.

5.3 Vector Format of the IDFT

A set of expressions similar to those given by Eqs. (5.5), (5.6), and (5.7) can be obtained for computing the IDFT.

B(k) = {B0(k),B1(k)} = {X(k) + X(k+j),X{k)-X(k+^)},

* = 0 , l , . . . , y - l (5.9)

b(n) = {b0(n),b1(n)} = {x(n),x(n+j)}, n = 0 , 1 , . . . , y - 1 (5.10)

M«) = ^ £ ( - l ) P * ^ ( * W n * , n = 0 , l , . . . , y - l (5.11) fc=0

where p = 0,1 and q = (n + p y ) mod 2. The SFG for the computation of IDFT with TV = 4 is shown in Fig. 5.3(a) and a specific example, which computes the IDFT of the DFT values shown in Fig. 5.1(b), is shown in Fig. 5.3(b). By dividing the output values by N = 4, we get the time-domain values.

5.4 The Computation of the IDFT

Any DFT algorithm can be used to compute the IDFT with very little additional effort. Let x{n) — xr{n) + jxi(n) and X(k) = Xr(k) + jXi(k). The IDFT is defined as

N-l

(n)+jxi(n) = ^Y^Xr(k) + 3Uk))W^nk, =o n = 0 , l , . . . , J V - l (5.12)

k=0

The Computation of the IDFT 105

B(0)={Bo (0), B1 (0) }= cv » MWMbo (0), h (0)}={ar(0), x(2)}) {X{0)+X(2),X(0)-X(2)} X / * " 460(0) = Bo(0) + B0(l)

461(0) = flo(0)-B0(l) B(1)={B0(1), Bj.(1)}= </ > ^b4(ft(l)={b0(l), 61(l)}={ar(l), x(3)})

{ X ( 1 ) + X ( 3 ) , X ( 1 ) - X ( 3 ) } 4fe0(l) = B1(0) + W4-1B1(l)

( a ) 461(l) = 5 1 ( 0 ) - W 7 1J B i ( l )

(8 + j 4 ) ± ( 4 + j 2 ) = o ^ — ^ ^ { 8 + j4,16 + j8}

{12+ j6 ,4 + j2}

( - 5 - j5) ± (1 + j3) = c^ ^ {12 - j4 , - 4 + j8}

{ - 4 - A - 6 - J 8 } (b)

Fig. 5.3 (a) The SFG for the computation of 4-point IDFT. (b) A specific example of (a) for computing the IDFT of the DFT values in Fig. 5.1(b).

By conjugating both sides, we obtain

1 J V _ 1

xr(n) - jXi(n) = T7 E ^(k) - 3Xi(k))Wp (5.13) N

k=0

Equation (5.13) represents a DFT with the input and output data conjugated, in addition to a constant divisor. This implies that a DFT algorithm can be used to compute the IDFT by conjugating the input, computing the DFT, and conjugating the output. An alternate algorithm is obtained by multiplying both sides of Eq. (5.13) by j .

N-l

Xi{n) + jxr(n) = - J2 (Xi(k) +jXr(k))Wtf (5.14) fc=0

Equation (5.14) represents a DFT with the real and imaginary parts of the input and output data interchanged. This implies that a DFT algorithm can be used to compute the IDFT by suitably reading the input and writing the output. The output values must be divided by N in order to get the actual values of the IDFT. Figure 5.4 shows the computation of the IDFT using the DFT for the same input used in Fig. 5.3(b).


{6 + j l 2 , 2 + j 4 } <v » p{4 + j 8 ; 8 + jl6}

{-2-j4,-8-j6}o/ >\{-4 + jl2,8-j4}

Fig. 5.4 A specific example of computing 4-point IDFT of the DFT values in Fig. 5.1(b) using the DFT.

5.5 Fundamentals of the P M DIT DFT Algorithms

Efficient DFT algorithms are obtained by decomposing Eq. (5.4) into several stages of computation. In this section, we derive the DIT type of algorithms. In order to do that, we have to study two properties on which this type of algorithms is based. For simplicity, we will develop the algorithm for the specific case with N = 8 and u = 2.

Shift of data vectors

Assume that the input vector values aq(0),aq(l), and o,(2) are arbitrary and those of aq(3) are zero. Let the DFT of aq(n) shown in Fig. 5.5(a) be Ap(k) as shown in Fig. 5.5(b). If we circularly shift the input vectors by one position in the counterclockwise direction, then the input vectors are placed as shown in Fig. 5.5(c). The DFT of the shifted vectors can be expressed in terms of Ap(k) as shown in Fig. 5.5(d). This result is justified as follows. The expression, for example, for Ap(l) of the vectors shown in Fig. 5.5(a) is given, from Eq. (5.7), as

a i ( 0 ) ( - l ) ° W 8 ° 4 < l l ( l ) ( - l ) l p W 81 ^ 1 ( 2 ) ( - l ) 2 W 8

2 ^ 1 ( 3 ) ( - l ) 3 W 83 (5.15)

The expression for AP{1) of the vectors shown in Fig. 5.5(c) is given by

O i ( 3 ) ( - l ) 0 p W 8 ° + a 1 ( 0 ) ( - l ) 1 ^ 81 ^ 1 ( l ) ( - l ) 2 p W 8

2 ^ i ( 2 ) ( - l ) 3 W 83 (5.16)

If we multiply Eq. (5.15) by (-1)PW81, we get

^ ^ ( - ^ ^ ^ ^ ( ^ ( - ^ ^ ^ ( ^ ( - ^ ^ ^ ^ ( ^ ( - l ) ^ ^ 4 (5.17)

The first three terms in this expression are exactly same as the corresponding terms in Eq. (5.16). As all the elements of the input vector aq(3) are

Fundamentals of the PM DIT DFT Algorithms 107

a,(l) Ap(l) • •

a„(2)« aq(n) • aq(0) Ap(2)» Ap(k) • Ap{0)

ag(3) = 0 Ap{3)

(a) (b)

aq(0) W%W£AP(1)

a g ( l )« aq(n-l) • aq(3) W2PW8

2AP(2). WiW8kAp(k) •WfW§Av(p)

= 0

o,(2) W%WiAp(3)

(c) (d)

Fig. 5.5 (a) A set of 4 input vectors with the last vector having all zero elements and (b) its DFT. (c) The shifted vectors of (a) by one position in the counterclockwise direction and (d) its DFT in terms of the DFT vectors shown in (b).

zero, the term involving this vector does not contribute to the summation. Therefore, Eqs. (5.16) and (5.17) yield the same value.

Zero padding of data vectors

Assume that we zero pad a set of ^ input vectors by inserting vectors with all zero elements in between two vectors to get a set of ^ input vectors. Then, the DFT of the new set of input vectors is defined as

AiZ)(k) = E Wraq(n)W?*, k = 0,1,..., 1 n=0 U


Since the odd indexed input vectors are zero, we get

4 2 ) ( f c ) = E Wfnaq(2n)WZk, k = 0 , 1 , . . . , ^ - 1 n=0

Now, for k = 0 ,1 , . . . , — - 1, we get

u

T 7 - 1

nk 4z)(k) = J2w?p)na<>(2n)wN n=0

R , r * = £ , £ + l , . . . , 2 £ - l , w e g e t

A^{k) = Y, Wgv+Vnaq(2n)W%k rnk

n=0

Since Ap(k) is periodic of period u with respect to the variable p, we get

A^{k) = A{2p) m o d u(k), A^(k + —) = Apjp+i) mod «(*),

JV p = 0 , 1 , . . . , u — 1, fc = 0 , 1 , . . . , 1

For an even vector length, when the number of input vectors is doubled by inserting vectors with all zero elements in between the vectors, the even subscripted vector elements of the smaller DFT repeat for the first-half of the DFT of the zero padded input vectors and the odd subscripted DFT vector elements repeat for the second half. For example,

{(a,b),(e,f)}&{(A,B),(E,F)}

implies {(a,b),{0,0),(e,f),(0,0)}^{(A,A),(E,E),(B,B),(F,F)}

The 2 x 1 PM DIT DFT algorithm

The naming of the algorithm will be explained at the end of this chapter. The basis of this type of algorithms is to divide the computation into the computation of the DFTs of the even and the odd indexed input vectors. These DFTs are then combined to form the required DFT. Because the time-domain vectors are divided into two smaller groups, this approach is called decimation-in-time (DIT). For example, in order to compute the DFT of four input vectors, the input vectors are split into two groups of


a b

e

f &

A B

E F

a b

0 0

e /

0 0

(a)

(b)

c d

9 h

C D

G H

A A

E E

B B

F F

c d

0 0

9 h

0 0

& C C

G G

D D

H H

(c)

0 0

c d

0 0

9 h

• & w°c -w$c

W^G -WIG

WiD -WiD

WiH -WiH

(d)

a b

c d

e /

9 h

a b

0 0

e /

0 0 +

0 0

c d

0 0

9 h o

A A

E E

B B

F F + wic

-wic W£G

-WIG WiD

-WiD WiH

-WiH

A + WiC A - WiC

E + W£G E - W£G

(e)

B + WiD B - WiD

F + WiH F - WiH

Fig. 5.6 Fundamentals of the 2 x 1 PM DIT DFT algorithm, (a) The D F T of the even and odd indexed input vectors in (e). (b) The DFT of the zero padded even indexed input vectors, (c) The DFT of the zero padded odd indexed input vectors, (d) The DFT of zero padded and shifted odd indexed input vectors, (e) The D F T of 4 input vectors is obtained as the sum of the DFTs of zero padded even and odd indexed input vectors.

zero padded even and odd indexed vectors as shown in Fig. 5.6(e). The sum of the two groups of the input vectors is equal to the given input vectors. If we compute the DFT of the two groups of input vectors and add, due to the linearity property of the DFT, we get the DFT of the given input vectors. Note that, due to zero padding, the computation of the DFT of each group of vectors can be reduced to the computation of a 2-vector DFT.

Consider the DFT vector pairs shown in Fig. 5.6(a). Due to zero padding


Fig. 5.7 The SFG of the 2 x 1 PM DIT DFT algorithm, with N = 8. Twiddle factor of the form Wg is represented only by its exponent s near the arrowheads. For example, the number 1 represents Wg.

property described above, we can deduce the DFT of the vectors shown on the left side of Figs. 5.6(b) and (c) as shown on the right side. The DFT of the shifted input vectors shown in Fig. 5.6(d) can be obtained by multiplying the DFT vectors shown in Fig. 5.6(c) with appropriate twiddle factors. Due to the linearity property of the DFT, the sum of the DFTs shown in Figs. 5.6(b) and (d) is the DFT of the sum of the input vectors shown in those figures. The point is that we are able to compute a longer DFT (a 4-vector DFT) by combining the coefficients of two 2-vector DFTs as shown in Fig. 5.6(e).

The SFG of the 2 x 1 PM DIT DFT algorithm is shown in Fig. 5.7. Note that the computation module shown, with N = 4, in Fig. 5.1 is used four times, with N = 8. We will say more about the structure of the SFG in the next few chapters. While Figs. 5.6 and 5.7 depict the same algorithm, the SFG shown in Fig. 5.7 is a more compact description whereas Fig. 5.6 shows explicitly the use of properties in deriving the algorithm.

In order to understand the algorithm, let us use an algorithm trace. A trace of an algorithm is a table of values of the variables at various stages of the algorithm from input to output. Figure 5.8 shows the trace of the algorithm shown in the SFG of Fig. 5.7. The input data is {x(n),n = 0,1,2,3,4,5,6,7} = {3 + j l , l + j 2 , 2 - j l , - 2 + j3,4 + j2,l + j4,2-j2, —1 + j3} . With N = 8 and u = 2, four vectors are required and the input values are read into the vector locations as shown in column 1. For example, a;(0) = 3 + j l and x(4) = 4 + j 2 are stored in the storage locations of the first vector. Vector formation consists of computing 2-point DFT of the values stored in each vector locations. For example, the first vector


Input values stored in vector

locations

x(0)=3+j l z(4)=4+j2

x{l)=l+j2 x(5)=l+j4

x ( 2 ) = 2 - j l x (6)=2- j2

x(3)=-2+j3 ar(7)=-l+j3

Vector formation

and swapping

a0(0)=7+;3 o i ( 0 ) = - l - j l

a 0 (2 )=4- ;3 Oi(2)=0+jl

a 0 ( l )=2+;6 o i ( l ) = 0 - ; 2

a 0 (3)=-3+j6 a i ( 3 ) = - l + j 0

Stage 1 output

11+JO 3+J6

0 - j l - 2 - j l

-1+J12 5+jO

0 - j l 0 - J 3

Stage 2 output

X(0)=A0(0)=W+jl2 X(4)=A1(0)=12-jl2

X(l)=A0(l)=-^-j(l + ^)

X(5)=Ml)=±-j(l-±)

X(2)=A0(2)=3+jl X(6)=Ai(2)=3+j l l

X ( 3 ) = A 0 ( 3 ) = ( - 2 - ^ H ( l - ^ )

X(7)=A1(3M-2+^)-j(l + ^)

Fig. 5.8 The trace of the algorithm shown in Fig. 5.7.

is formed by adding and subtracting x(0) and x(4). In addition, vectors are swapped to store them in a special order. More about this will be said in the next chapter but, for the moment, remember that in Fig. 5.7 we have split the even and odd indexed input vectors into two groups. After vector formation and swapping, we get the values shown in column 2. Columns 3 and 4, respectively, show the values after the first and second stage operations shown in SFG of Fig. 5.7 are carried out. Column 4 shows the DFT values.

We have already shown that any DFT algorithm can be used to compute the IDFT with trivial modifications. Figure 5.9 shows the trace of the algorithm of the SFG shown in Fig. 5.7 in computing the IDFT. The DFT values are the output values shown in Fig. 5.8. At the end of the algorithm, we should get back the input values shown in Fig. 5.8. The trivial modification required is the interchange of the real and imaginary parts at the input and at the output of the algorithm. With these changes and carrying out the operations of the SFG in Fig. 5.7, we get the values shown in various columns in Fig. 5.9. Note that there is an additional division operation by N = 8 in computing the IDFT.


Interchange of real and imaginary parts, vector formation and swapping

a 0 ( 0 ) = 0 + j22 ai(0) = 2 4 - j 2

a0(2) = 1 2 + j 6 ai(2) = - 1 0 + j0

a0( l) = - 2 + j0

Oi(l) = - \ / 2 - j V 2

a0(3) = - 2 - j 4

Oi(3) = 3%/2-j3\/2

Stage 1 output

12 + j28 -12 + jW

24 + J8 24 - j l 2

- 4 - j4 0 + j4

-4v/2(l +jl)

2y/2(l + jl)

Stage 2 output

8 + j24 16 + j32

16 + J8 32 + j 8

- 8 + jl6 - 1 6 + J16

24 - j l 6 24 - j 8

Interchange of real and

imaginary parts and division by 8

x(0)=3 + jl x(4) = 4 + j2

x(l) = l+j2 x(5) = 1 + j4

x(2) = 2-jl x(6) = 2 - j2

x(3) = -2 + j3 x(7) = -l + j3

Fig. 5.9 The trace of the computation of IDFT using the DFT algorithm shown in Fig. 5.7.

5.6 Fundamentals of the P M DIF DFT Algorithms

In this section, we derive the DIF type of algorithms. This type of algorithms are also based on certain properties of the DFT. Assume that N = 8 and u = 2.

Shift of transform vectors

Let the vector elements AP(Q) = 0 and Ap{2) = 0 and the vector elements Ap{\) and Ap(3) have arbitrary values as shown in Fig. 5.10(a). The corresponding vector elements aq(n) are as shown in Fig. 5.10(b). With vector length 2, the even indexed DFT vectors are computed using the even subscripted input vector elements. The even subscripted input vector elements with zero values ensure that the even indexed DFT vectors are zero. Figure 5.10(c) shows the placement of the output vectors advanced by one sample. The corresponding input vectors, the input vectors of Fig. 5.10(b) multiplied by appropriate twiddle factors with the swapping of the even and odd subscripted input elements, are shown in Fig. 5.10(d).

Fundamentals of the PM DIF DFT Algorithms 113

Ap(l) (0,ai(l))

Ap(2)* Ap(k) •ApiO) (0 ,a i (2)) . aq(n) • (0,a1(0)) = 0 = 0

Ap(3)

(a)

Ap(2) = 0

(0,a! (3))

(b)

W£(ai(l),0)

4>(3)« Ap(k +1) • ^ ( 1 ) W82(oi(2),0)« W£aq(n) • W8°(ai(0),0)

Ap(0) = 0

(c)

W 83 M3) ,0)

(d)

Fig. 5.10 (a) A set of 4 DFT vectors with the elements of the even indexed vectors zero and (b) the corresponding input vector elements, (c) The shifted vectors of (a), and (d) the corresponding input vector elements in terms of the input vector elements in (b).

Compression of data vectors

If we are interested only in computing the even indexed DFT vectors, for example, then the number of output vectors becomes one-half and the input vectors can be compressed to reduce their number also to one-half so that a one-half length DFT can be used. Consider the matrix form of the DFT expression Eq. (5.8). For example, Ap(0) and Ap(2) can be computed using the equation

4e)(o) A(e) .rip (l)

w2 w? w? wl 4 J

(-1)°P4C)(0) (-i)^4c)(i)

where Ap{0) = Ape)(0) and Ap(2) = Ap

e\l), and aqc\n) = {(o0(0) +

o0(2), a0(0) - a0(2)), (a0(l) + a0(3), a0(l) - a0(3))}, and q = k mod 2. For


the computation of odd indexed DFT vectors, odd subscripted input vector elements combine to produce the new set of input vectors of one-half the number.

The 2 x 1 PM DIF DFT algorithm

The basis of this type of algorithms is to divide the DFT vectors into two groups and compute the values of these groups with reduced computation. Since we divide the problem by partitioning the DFT vectors, this type of algorithms is called decimation-in-frequency (DIF) algorithms.

Figure 5.11(a) shows a 4-vector DFT pair. Let us split the output vectors into two groups as shown. Now, the corresponding input vectors are shown, respectively, in Figs. 5.11(b) and (c). This is because the computation of even indexed DFT vectors requires only the first element of the input vectors and the computation of the odd indexed DFT vectors requires only the second element of the input vectors. By multiplying the input vectors with appropriate twiddle factors, according to the shift property described above, and swapping the two elements of each input vector, we can get a shifted version of the odd indexed DFT vectors as shown in Fig. 5.11(d). Using the compression property described above, the computation of the zero padded 4-vector DFTs shown in Figs. 5.11(b) and (d) can be reduced to the computation of two-vector DFTs as shown in Figs. 5.11(e) and (f), respectively. As in the case of the DIT algorithm, Figure 5.11 shows explicitly the use of properties in deriving the DIF algorithm. The SFG shown in Fig. 5.12 is a more compact description.

Figure 5.13 shows the trace of the algorithm for computing the DFT of the same input values shown in Fig. 5.8. The difference in the first column of Fig. 5.13 is that the input vectors are stored in natural order. We carry out the operations of the first and second stages of the SFG to get the values shown in columns 2 and 3. The output vectors in column 3 have to swapped to put them in natural order.

5.7 The Classification of the P M D F T Algorithms

Each member of the set of algorithms described in this book is designated, hereafter, as radix-z uxv PM algorithm. The computation of 2-point DFT of data values a and b is just a plus b and a minus b. This is the fundamental

The Classification of the PM DFT Algorithms 115

a b

c d

e

f 9 h

A B

0 0

E F

0 0

a 0

c 0

e 0

9 0

<$

+ (a)

<£•

(b)

A B

C D

E F

G H

0 0

C D

0 0

G H

A B

0 0

E F

0 0

0 b

0 d

0

/

0 h

(c) D

G H

W$b 0

Wid 0

wif 0

Wih 0

<& c D

0 0

G H

0 0

(d)

a + e a — e

c + g

c-9 <&

A B

E F

(e)

Wib + Wif W§b- Wif

WU + Wih Wid-Wih

4> c D

G H

(f)

Fig. 5.11 Fundamentals of the 2 x 1 PM DIF DFT algorithm, (a) A set of input vectors and its DFT vectors, and the splitting of the even and odd indexed D F T vectors, (b) The zero padded even indexed DFT vectors and the corresponding input vectors, (c) The zero padded odd indexed DFT vectors and the corresponding input vectors, (d) The zero padded and shifted odd indexed DFT vectors and the corresponding input vectors. (e) The compression of the input vectors to generate the even indexed DFT vectors. (f) The compression of the input vectors to generate the odd indexed DFT vectors.


Fig. 5.12 The SFG of the 2 x 1 PM DIF DFT algorithm, with N = 8. Twiddle factor of the form Wg is represented only by its exponent s near the arrowheads. For example, the number 1 represents Wg.

Input vectors

a0(0) = 7 + j 3 ai(0) = - l - j l

a0(l) = 2 + j6

O l ( l ) = 0 - j 2

o 0 (2 )=4- j3 a1(2)=0+jl

a0(3) = - 3 + j 6 ai(3) = - l + j 0

Stage 1 output

11+JO 3 + J 6

- 1 + J 1 2 5 + j O

0 - j l -2-jl

_ 3 _ -• 3 V2 J V2

Stage 2 output

X(0) = Ao(0) = 10 + jl2 X(4) = Ai{0) = 12 - jl2

X(2) = A0(2) = 3 + jl X(6)=Ai(2) = 3 + jll

X(l) = A0(l) = - ^ - j ( l + ^)

X(3) = A0(3) = (-2- ^) - j(l- ^)

X(7)=A1(3) = (-2+ ±) - j(l + ^)

Fig. 5.13 The trace of the algorithm shown in Fig. 5.12 for the same input values in Fig. 5.8.

Summary 117

operation in the algorithm. Hence, the name plus-minus (PM). The term radix-z specifies the way a problem is decomposed into smaller problems. While we can design algorithms with any radix, the practically most useful value for z is two and, therefore, the algorithms described in this book are exclusively of the type radix-2. In a radix-2 algorithm, a problem is split into two smaller independent problems of the same type, each of half the size, recursively from the input or output end. The solutions of two smaller problems are combined to form the solution of a larger problem. A radix-2 algorithm requires that the length of the data N divided by the vector length u is equal to an integral power of 2, that is ^ = 2 m where m is an integer.

The letter u indicates the number of elements in a vector. Although general expressions are developed in the following chapter, with one exception for vector length six, all the algorithms described, in detail, in this book use a vector length of two since that length is practically the most useful. It is possible to combine fully or partially the multiplication operations of two or more consecutive stages of an algorithm in order to reduce the total number of arithmetic operations. However, the partial merging of multiplication operations of adjacent stages leads to practically inefficient and irregular algorithms. Therefore, the letter v, in the range 1 to m = log2 —, specifies the number of adjacent stages whose multiplications are combined. Although the set of PM algorithms is quite large, the practically more useful algorithms are those with z = 2, u — 2, and v = 1 or v = 2. A good understanding of these algorithms will enable the reader to derive algorithms with other parameters, if necessary.

5.8 S u m m a r y

• The problem of computing the DFT of N values is the problem of evaluating N sums, each of N products. The challenge in designing fast algorithms is to find the way to develop partial results and share each partial result to compute as many coefficients as possible.

• The problem of computing the DFT can also be considered as the decomposition of a larger DFT into smaller DFTs and combining the coefficients of the smaller DFTs to produce the coefficients of the larger DFT.

• In this chapter, the DFT and IDFT definitions using vector inputs


and outputs were derived. The computation of the IDFT using the DFT was presented. The fundamental principles and the relevant theorems on which the fast algorithms are based were explained. The classification of PM DFT algorithms was presented. In the next few chapters, the PM DFT algorithms will be described in detail.

References

(1) Sundararajan, D., Ahmad, M. O. and Swamy, M. N. S. (1994) "Computational Structures for Fast Fourier Transform Analyzers", U.S. Patent, No. 5,371,696.

(2) Sundararajan, D., Ahmad, M. O. and Swamy, M. N. S. (1998) "Vector Computation of the discrete Fourier Transform", IEEE Trans. Cir. and Sys. II, vol. CAS-45, No.4, pp. 449-461.

Exercises

5.1 Using the SFG shown in Fig. 5.1, compute the DFT of: 5.1.1. x(n) = {2,3,5,4}. * 5.1.2. x(n) = {2+jl,l-j2,2 + j2,l-j3}.

5.2 Starting from the DFT definition, derive the expression Eq. (5.7) with N = 8.

5.3 Starting from the IDFT definition, derive the expression Eq. (5.11) with N = 8.

5.4 Using the SFG shown in Fig. 5.3, compute the IDFT of X(k) = {2-jl,l-j2,3 + j2,l-j3}.

* 5.5 Using the SFG shown in Fig. 5.4, compute the IDFT of X(k) = {2 - jl, 1 - j2,3 + j2 ,1 - j3} using DFT.

5.6 Verify that the transform vectors shown in Fig. 5.5(d) corresponds to the input vectors shown in Fig. 5.5(c).

5.7 Compute the input vectors, {aq(0), aq(l)}, with vector length 2 if x(n) = {1 - j4 ,2 + j3 ,1 - j l , 2 - j l } . Find the DFT using the vectors. Deduce the DFT if the input vectors are {aq(0),0,aq(l),0}.

Exercises 119

5.8 Using the SFG shown in Fig. 5.7, compute the DFT of x(n) = {1 -j4,2-j3,l-jl,2-j4,2-j2,l+j3,l-jl,2 + j2}.

* 5.9 Using the SFG shown in Fig. 5.7, compute the IDFT of X(k) = { l - j 3 , 2 - j 3 , 3 - j l , 2 - j 4 , 3 + j 3 , 2 - j 2 , l + j 3 , 2 + ; 4 } .

* 5.10 Using the SFG shown in Fig. 5.12, compute the DFT of x(n) = {1 - jA, 2 - j3,1 - j l , 2 - j4 ,2 - j 2 ,1 + j 3 ,1 - j l , 2 + j2}.

5.11 Using the SFG shown in Fig. 5.12, compute the IDFT of X(k) = { l - j 3 , 2 - j 3 , 3 - j l , 2 - j 4 , 3 + j 3 , 2 - j 2 , l + j3 ,2 + j4} .

Programming Exercise

5.1 Write a program to directly implement Eq. (5.7) to compute the DFT.

Chapter 6

The u X 1 P M DFT Algorithms

The problem of computing the DFT is multiplication of the data with samples of complex exponentials of various frequencies and summing it up. In the direct computation of the defining equation, the multiplication and addition operations required to determine each frequency coefficient are carried out separately. In fast computation of the DFT, the multiplication and addition operations are carried out in parts over several stages. This spreading of the computation makes it possible to compute partial results and use them for the computation of several frequency coefficients. In the last chapter, we derived the vector form of the DFT definition. In addition, fast DIT and DIF algorithms, for small values of N, were developed using the basic principles. In this chapter, we will study more formally the class of algorithms in which the computation is split over several stages without merging the multiplication operations of adjacent stages.

The general version of the u x 1 PM DIT DFT algorithms is derived in Sec. 6.1. Then, we deduce the DIT algorithm for the most efficient vector length u = 2 and give a detailed description in Sec. 6.2. The necessity of reordering the input vectors for in-place computation is explained in Sec. 6.3. In Sec. 6.4, the computation of a single DFT coefficient is traced in the SFG of the algorithm. The general version of the u x l PM DIF DFT algorithms is derived in Sec. 6.5. The 2 x 1 PM DIF DFT algorithm is deduced in Sec. 6.6. The computational complexity of the 2 x 1 PM DFT algorithms is presented in Sec. 6.7. The 6 x 1 PM DIT DFT algorithm, which provides the computation of DFT for lengths with a factor of three, is described in Sec. 6.8. A flow chart description of the 2 x 1 PM DIT DFT algorithm is given in Sec. 6.9.

121

122 The uxl PM DFT Algorithms

6.1 T h e u x l P M D I T D F T Algor i thms

Decomposing the summation in Eq. (5.4) into those corresponding to the even and odd indexed input values aq(n) and simplifying, we get

M. _ , • 3 , 1 x

/•nk Mk)= Y, W^naq(2n)Wf n=0

+ W*W% J2 W^naq(2n + l)Wf, (6.1) n=0

where q = ((k + p^-) mod u), p = 0 , 1 , . . . , u — 1, and A; = 0 , 1 , . . . , — — 1. The DFT values Ap(k + ^ ) , k = 0 , 1 , . . . , ^ - 1, can be obtained by replacing k with A; + ^ in Eq. (6.1). The resulting equation can be written as

) = * 2u

Ap(k+^) = £ W^+l>ag(2n)WZk

71=0

- N i l

+ WZWluWk Y, W£*>+l>aq{2n + l)W£, (6.2) n=0

where q = ((k+ £ j + p f ) mod u), p = 0 , 1 , . . . , u - 1 , and k = 0 , 1 , . . . , ^ - 1 . Let the DFT of the even indexed input vector elements, aq(n),n =

0 , 2 , . . . , ^ - 2 , be represented by Ay (k), and that of the odd indexed input vector elements, aq(n), n — 1 ,3 , . . . , ^ — 1, be represented by Ay(k). For even indexed input vectors,

4*)(fc)= Y Wra9{2n)W£

where q = ((A; + p ^ ) mod u), p — 0 , 1 , . . . , u — 1, and k = 0 , 1 , . . . , ^ — 1. For even values of p, as Ay (k) is periodic of period u with respect to the variable p,

4?mod„(*) = £ W^aq{2n)Wf,

Theuxl PM DIT DFT Algorithms 123

where q = ((fc + p%) mod u), p = 0 , 1 , . . . , u - 1, and fc = 0 , 1 , . . . , ^~ - 1. For odd values of p,

< + i ) „.„«(*> = E ^ ^ " a ^ W f , n=0

where 9 = ((fc+P^ + ^ ) mod u), p = 0 , 1 , . . . , « - l , and fc = 0 , 1 , . . . , ^ - 1 -Similarly, for odd indexed input vectors, for even values of p,

4?mod«(*)= E ^i2p)"^(2n + l)Wf,

where g = ((fc + p ^ ) mod u), p = 0 , 1 , . . . , u - 1, and fc = 0 , 1 , . . . , ^ - 1. For odd values of p,

ra=0

where g = ((fc+P^ + ^j) modu) ,p = 0 , 1 , . . . , u - l , and fc = 0 , 1 , . . . , ^ - 1 . Equations (6.1) and (6.2), using the smaller size DFTs defined above,

can be written as

Mk) = AeJmo<iuW + WSW^A{0Jmodu(k) (6.3)

Ap(k+^) = < + 1 ) m o d u ( f c ) + ^ ^ ^ ^ + 1 ) m o d u ( f c ) , ( 6 . 4 )

where p = 0 , 1 , . . . , u - 1 and fc = 0 , 1 , . . . , | £ - 1. The problem of computing an ——vector DFT has been decomposed into a problem of computing two ^—vector DFTs. This process of decomposition can be continued recursively, until the problem is reduced to a set of 1-vector DFTs.

The u X 1 PM DIT DFT butterfly

As the DFT of a single vector is itself, we never compute the DFT of a set of vectors using the definition. Therefore, the process, which is repeated several times over a number of stages, is only the decomposition of a larger DFT into two smaller DFTs. The basic computation that is repeatedly used in the decomposition process is called the butterfly computation. In general, the butterfly computation at the rth stage, for an even vector length (For an odd vector length, the first and third equations of Eq. (6.5)

124 The u X 1 PM DFT Algorithms

comprise the input-output relation of a butterfly with p — 0 , 1 , . . . , u — 1. However, the use of odd vector lengths is not preferred since it is less advantageous.), can be deduced from Eqs. (6.3) and (6.4) as

4r+1)W = A^modu(h) + wsw^modu(i) 4+}\h) = A%modu(h) - WTO4?mod„(0 4r+1)(D = ^+1)modJh) + WZWluW^Afi odu(l) A^(l) = A$p+1)modu(h)-W£WWA$p+1)modu(l),

(6.5)

where s is an integer whose value depends on the stage of computation r and the index h, and p = 0 , 1 , . . . , | — 1. Equation (6.5) characterizes the u x 1 PM DIT DFT butterfly and represents u 2-point DFTs after the input values indexed I are multiplied by appropriate twiddle factors. At a time, in the butterfly computation, we use only two vectors and the output quantities can be overwritten in the locations of the input quantities as they are no longer required. Therefore, only two indices are required which are in general represented as h and I.

The computational stages

We have just split the problem into two (Eqs. (6.3) and (6.4)) in order to form the output of the last stage. We assumed that the output vectors of the previous stage are available. Next, we break each of the two problems of the previous stage into two. Keeping on doing this, we get more and more independent problems but each with a smaller size. We stop the process of breaking down when each problem becomes a 1-vector DFT. To reach that level of decomposition, we need m stages of computation specified as r = 1,2,... ,m, where m = log2 ^ . As the number of problems increases, the number of butterflies used to decompose each problem reduces in the same proportion. Therefore, the number of butterflies remains the same in each stage. Remember that two vectors are processed in a butterfly and we have ^ vectors in any stage. Therefore, the number of butterflies that constitutes a stage is ^ . Only the grouping of the butterflies changes from stage to stage. In the last stage, the twiddle factor exponent s is the same as the index h. And it is true for any stage provided the base N in WN is changed according to the stage. As we want to use one set of twiddle factors, we generate them only with the base TV. Therefore, instead of varying the base N, we multiply the exponent by 2 for each stage. As the

The 2 x 1 PM DIT DFT Algorithm 125

size of the problem is reduced, the difference between the indices h and / is also reduced by a factor of two. The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given as follows.

h = i m o d 2 r " 1 , z = 0 , 1 , . . . , ^ - 1 / = h + r-1 (6.6) a = h(2m~r)

Using the expressions given above, algorithms can be derived for any value of N and u for which ^ is an integral power of 2. As a consequence of splitting the even and odd indexed input vectors over m stages, the input vectors to the first stage are placed in the bit-reversed order.

6.2 The 2 x 1 P M DIT DFT Algorithm

The specific algorithm with u = 2, that is each vector with two complex elements, is practically very useful. Therefore, we give a detailed description of this algorithm.

The 2 x 1 PM DIT DFT butterfly

The butterfly input-output relations at the rth stage can be obtained, by substituting u = 2 in Eq. (6.5), as

4r+1)W = A$\h) + WNA%\l) Af+1\h) = 4\h)-W^\l) 4r+1)(0 = 4 r , w + < * 4 r ) ( D (6-7)

A?+1\l) = AP(h)-W8N

+*AP(l), where s is an integer whose value depends on the stage of computation r and the index h. These equations characterize the 2 x 1 PM DIT DFT butterfly shown in Fig. 6.1. This butterfly computes two 2-point DFTs after the input values indexed I are multiplied by appropriate twiddle factors. The butterfly structure comprises two input nodes, at each of which it receives the two complex numbers of an input vector. The input vector at the upper node is A^r'(h) and its elements are A£'(h) and AY'(h). The input vector at the lower node is A^r'(l) and its elements are A£'(l) and A{ >(l). The first and second elements of A^r' (I) are multiplied, respectively, by twiddle

126 The u x 1 PM DFT Algorithms

= {A^(h),A[r\h)} o ^ - *^o = {A^+1\h),A[r+1\h)}

= {4r)(0,4r)(0} o ^ > > ={k+1)d),Ar+1\i)}

Fig. 6.1 The SFG of the butterfly of the 2 x 1 PM DIT DFT algorithm, where 0 < s < ^-. Twiddle factors are represented only by their exponents. The symbol s represents vvN.

factors W^ and {-j)W8N to produce W8

NA<£\l) and ( - j ) W ^ r ) ( Z ) . Only one of the two twiddle factors needs to be generated as they are trivially related. The two-point DFT of (A^ (h), W^A^ (I)) constitute the first and the second elements, respectively, of the butterfly output vector A^r+1'(h) at the upper output node. The two-point DFT of (A[r)(h), (-j)Wfr A{{] (/)) constitute the first and the second elements, respectively, of the butterfly output vector A^r+1'(l) at the lower output node.

The computation involved in a butterfly includes 2 complex multiplications, each requiring two real additions and four real multiplications, and two 2-point DFTs, each requiring four real additions. Four complex numbers or eight real numbers are loaded and stored. Therefore, a butterfly requires 8 real multiplications, 12 real additions, and, assuming that sufficient number of registers is available in the processor, 16 data transfer operations between the memory and the processor (The loading of the twiddle factors is ignored since it occurs once for a set of butterflies and amounts to less than N data transfers, whereas the number of load and store operations for the data is a function of N log2 N.). While the computation of two 2-point DFTs is required in all the butterflies, the complex multiplication operation requires fewer number of real operations for some butterflies. These special butterflies requiring reduced computation are with s = 0 and s = ^-. The first special butterfly requires 8 real additions while the second special butterfly requires 4 real multiplications and 12 real additions.


The number of 2 x 1 PM DIT butterflies in each of the m stages is ^ . For example, with N — 16, there are 3 stages specified as, r = 1,2, and 3. Four butterflies make up a stage. Indices h and I, and the twiddle


Fig. 6.2 The SFG of the 2 x 1 PM DIT DFT algorithm, with N = 16. Twiddle factors are represented only by their exponents. For example, the number 4 represents Wj6.

factor exponent s for each group of butterflies are given, respectively, by (h = i mod 2 r - \ i = 0 , 1 , . . . ,3), I = h + 2r-\ and s = h(23-r). The SFG of the 2 x 1 PM DIT DFT algorithm, with N = 16, is shown in Fig. 6.2. The three stages are demarcated by unfilled circles. Except for the last stage, the output vectors of a butterfly are the input vectors for butterflies of the succeeding stage.

Example 6.1 Figure 6.3 shows the trace of the algorithm shown in Fig. 6.2. The values, given to a precision of two decimal places, were obtained from running a program with a much higher precision. The first column values are 16 complex input values to the DFT. The input values are stored sequentially in the storage locations assigned for 8 vectors each with 2 complex elements, the first half values in the first element of the vectors and the second half values in the second element. Vectors are formed by adding and subtracting the elements stored in the individual vector locations. For example, the elements of the vector in the first row of the second column are (2 + jO) + (1 + JO) and (2 + jO) - (1 + JO). In addition


x(0) x(8)

x(l) x(9)

x{2) x(10)

x(3) ar(ll)

s(4) x(12)

s(5) x(13)

*(6) x(14)

x(7) z(15)

2 + jO 1 + j O

3 + jO 3 + jO

4 + jO 3 + jO

4 + jO 1 + j O

2 + j l 1 + j l

1 + J 3 2 + J2

4 + J3 2 + J2

3 + j l 1 + j l

3 + jO 1 + jO

3 + J2 1 + jO

7 + jO 1 + jO

6 + j5 2 + j l

6 + jO 0 + jO

3 + J5 - 1 + j l

5 + jO 3 + jO

4 + J2 2 + jO

O

OS

1 +

Vs.

Vs.

to

to

1 - j l 1 + j l

13+J5 1 - J 5

2 - J 2 0 + J 2

9 + j 5 3 - J 5

1 + j l - 1 - j l

>_i

to

1 +

Vs.

Vs.

to

to

CO

C

O

+ 1

Vs.

Vs.

to

to

19.00 + J7.00 -7.00 - J3.00

1.00 - J3.83 l.OO + j l .83

-5.00 - J3.00 5.00 - j l .00

2.41 - J0.41 -0.41 + J2.41

18.00 + J7.00 0.00 + J3.00

1.71 - J2.54 0.29 + J4.54

1.00 - J6.00 5.00 - J4.00

- I . 7 1 - j 4 . 5 4 -0.29 + J2.54

37.00 + J14.00 1.00 + jO.OO

1.61-J6.82 0.39 - jO.83

-8.54 - J7.95 -1.46 + J1.95

-2.43 - jO.57 7.26 - jO.26

-4.00 - J3.00 -10.00-J3.00

5.08 - jO.18 -3.08 + J3.83

- 1 . 3 6 - j l . 7 1 11.36 - jO.29

0.83 + J0.18 -1.66 + J4.64

X(0) X(8)

X(l) X{9)

X{2) XilO)

X{2) X(ll)

X(4) X(12)

X(5) X(13)

X(6) X(14)

X(7) X(15)

Fig. 6.3 The trace of the 2 x 1 PM DIT DFT algorithm, with N = 16.

to forming the vectors, the vectors are swapped at the same time so that they are placed in the bit-reversed order. The result of vector formation and swapping is shown in column two. The third, fourth, and fifth column vectors are obtained after the first, second, and third stage operations of the algorithm are carried out. The 16 DFT coefficients are stored sequentially, the first half in the first element of the vectors and the second half in the second element as shown. I

6.3 Reordering of the Input Data

The input vectors, in Fig. 6.2, are placed in bit-reversed order (see Appendix C) while the output vectors are placed in natural order. The bit-reversed order is necessary for in-place computation, that is with storage locations just enough for y vectors. This advantage is quite significant and the overhead of placing the input vectors in bit-reversed order is very small.

Figure 6.4 shows the binary representation of the indices of the vectors

http://-I.71-j4.54

Reordering of the Input Data 129

Fig. 6.4 Reordering of the input vectors in the 2 x 1 PM DIT DFT algorithm, with N = 16.

in all the stages. In the last stage, the output vectors are in natural order. These vectors are formed by butterflies, which require the DFT of even indexed input vectors at the upper node and the DFT of the odd indexed input vectors at the lower node. The butterflies are placed in natural order. Therefore, the DFT of the even and the odd indexed vectors are placed in natural order as indicated by the two boldface bits on the left side of each index. To compute the DFT of the even and odd indexed input vectors we require the DFT of the even of even, even of odd, odd of even, and odd of odd indexed input vectors in natural order. This is a recursive process that continues until the input data is split into groups of one, as at the input of stage 1. The repeated splitting of the input data into even and odd indexed sets is required. The question is what is the order of the indices we get at the input due to this process and what is the relation between the input and the output orders.

If we circularly left shift the naturally ordered binary numbers in the last stage by one position, the msbs come to the position of the lsbs and we get the required order for our algorithm at the end of stage 2. Therefore,


the lsbs of the new order are fixed and again we circularly left shift the two sets of numbers with two boldface bits to get the proper order at the end of the stage 1. Now the position of the second bit gets fixed. Only one bit, shown in boldface, remains to be considered and, as shifting one bit makes no difference, we get the input order. Circularly left shifting by one bit repeatedly, the first time we made the msbs of the natural order as the lsbs of the new order, the second time we moved the bits next to the right of msbs in the natural order to the left of lsbs in the new order, and finally we left the lsbs of the natural order in the msb positions of the new order. The net result is that we have reversed the position of the bits in the natural order.

6.4 Computation of a Single D F T Coefficient

To get further insight into the algorithm, let us consider how a specific coefficient is computed. By highlighting the relevant signal flow paths, we have shown the grouping of partial results and multiplication by common factors in the computation of X(13) in Fig. 6.5. By definition,

7

X(13) = ^ ( 5 ) = 5^(-l)na1(n)W18

an

n=Q

= a, (0) - oi (1) Wft + oi (2) Wft0 - oi (3) W?e5

+ ai{4)W™ - ai(5)W?i + oi(6)W-§» - 0l(7)W?fl6

This computation is split over three stages. At the end of the first stage, we get

= MO) + ^ ( 4 ) 0 - M2) + ai(6)W?6)W?fl

- M l ) + «i(5)^i46)^i5

6 + (oi(3) + ai(7)W?e)W7e

This grouping and finding common factors is continued. Eventually, the number of groups is reduced from y to one. The four groups of partial

results in this equation, designated ^ ( l ) . ^ " ^ 1 ) . ^ ^ 1 ) . ^ 0 ^ 1 ) . r e" spectively, are the coefficients indexed 1 of the DFT of the input values indexed even of even, even of odd, odd of even, and odd of odd. These

Computation of a Single DFT Coefficient 131

stage 1 stage 2 stage 3 A(0)

A{1)

A(2)

A{3)

A(4)

A(5)

A!(5) = X(13)

A(6)

A(7) Fig. 6.5 The computation of the coefficient X(13) by the 2 x 1 PM DIT DFT algorithm, with N = 16.

quantities are grouped as

= (4ee)(i) - 4eo)(W26) - (4oe)(i) - 400)(i)w?«)n

The two groups of partial results in this equation, designated A\e'(l), A[0' (1), respectively, are the coefficients indexed 5 of the DFT of the input values indexed even and odd. To get the desired coefficient, these quantities are grouped as

= (A|e )(l) - 4°Hl)W*e) = A1(5) = X(13)

Essentially, instead of multiplying a data value by a single twiddle factor, the multiplication operation is spread over several stages by splitting the twiddle factor. For example, the term -a i (7 )W^ | = ai(7)W™ = aMW&W&W&W&W?,.

The figure also shows clearly why this algorithm is more efficient than the direct method. The partial results computed in the first stage for the computation of X(13) are used for the computation of one-quarter of the


DFT coefficients. Second stage partial results are used to compute one-eighth of the DFT coefficients and so on. This sharing of the partial results contributes to the reduction of the computation compared with the direct computation since, in that case, no sharing is done.

Seven complex multiplications and fifteen complex additions are required to compute a single coefficient. The multiplications in the first stage are trivial. The number of multiplications in the other stages is ^ — 1. In the second stage, the complex multiplication requires less computational effort. Compare this with N complex multiplications required in the direct computation of the DFT. The reduction of the number of multiplications reduces the round-off errors. The spreading of the computation over several stages also reduces the errors and makes the scaling problem less severe in fixed-point implementations.

The DFT algorithms are so efficient that, unless we want to compute very few coefficients, it is better to use an algorithm to compute all the coefficients even if all of them are not required. To compute a single DFT coefficient, we have to use the appropriate computational path in the SFG of an algorithm. This method is more efficient than using the DFT definition because the computational effort is reduced by more than half in terms of arithmetic operations. In addition, we have to compute very few twiddle factors.

6.5 The u X 1 P M DIF DFT Algorithms

The DIT algorithm, just described, is based on splitting a set of input vectors into two groups, computing their DFTs separately, and combining the two DFTs to get the DFT of the input data. The DIF algorithms , on the other hand, are based on combining the first- and second-half of the input vectors to make two independent sets of input vectors and computing their DFTs separately to get the even and odd indexed DFT vectors of the input data. The DIT and DIF algorithms are dual and one can easily be deduced from the other.

Decomposing the summation in Eq. (5.4) corresponding to those of the first- and the second-half of the input vectors aq(n) yields

Ap(k)= £ WZnaq(n)WZk+ ]T W ^ a , ( n ) ^ , n=0 „ = £

The n x l PM DIF DFT Algorithms 133

where q = ((k + p~) modu), k = 0 , 1 , . . . , ^ - 1, and p = 0, l , . . . , u - 1. Replacing n by n + |jj in the second summation on the right side gives

Ap{k)= £ Wraq(n)WZk+ £ Wpu

{n+^aq{n + ^ ) W ^ ) k

n=0 n=0 2 u ' ' , J V

The pair of summations can be combined into a single one, giving

JV__1 2 u *•

Ap(k) = Y, Wr(aq(n) + W^Wpag(n + ^ ) « * (6.8) n=0

The even and odd indexed DFT values Ap(k) are readily obtained from Eq. (6.8) by replacing A; with 2k and k with 2k + 1, respectively.

N--1

Ap(2k) = £ Wr(aq(n) + wik+p&)aq(n + ^))Wf, n=0

N q = (2(k + p—)) mod u (6.9)

2u

r - l

Ap(2k + 1 ) = £ W T ( o , ( n ) + W ^ ^ W ^ i n + — ))W%Wf, 71=0

q = (2(A; + p — ) + 1) mod u, (6.10) 2u

where k = 0 , 1 , . . . , ~ — 1 and p = 0 , 1 , . . . , u — 1. The problem of computing an ^—vector DFT, therefore, has been decomposed into a problem of computing two ^ —vector DFTs.

The u X 1 PM DIF DFT butterfly

In general, the input-output relations of the butterfly computation at the rth stage, for an even vector length, can be deduced from Eqs. (6.9) and (6.10) as

aqr+1\h) = a$modu(h)+WZa%modu(l)

a™(h) = a%modu(h)-WZa$modu(l)

4 " + 1 ) ( 0 = WhaVmodJh) + WtW}uW^Z\+1)modu(l) {- }

aqr^(1) = W^q+l) m o d u(h) - W<W}uW^q+1) m o d „(/),


where s is an integer whose value depends on the stage of computation r and the index h, and q = 0 , 1 , . . . , ^ — 1. These equations characterize the u x 1 PM DIF DFT butterfly and represent u 2-point DFTs and the multiplication of the input values by appropriate twiddle factors.


There are m stages of computation specified as r = 1,2, . . . ,m, where m = log2 ^ . The number of butterflies that constitute a stage is £.. The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given as follows.

h = z m o d 2 m - r , i = 0 , 1 , . . . , £. - 1 I = h + 2m~r (6.12) s = h(2r~1)

As a consequence of splitting the even and the odd indexed output vectors over the m stages, the output vectors from the last stage are placed in the bit-reversed order.

6.6 The 2 x 1 P M DIF DFT Algorithm

The 2 x 1 PM DIF DFT butterfly

The 2 x 1 PM DIF butterfly relations are deduced by substituting a value 2 for u in Eq. (6.11).

a£+1\h) = aP(h) + air\l)

a[r+1\h) = 4 r ) ( / i ) - a « ( Z )

4 r + 1 ) ( 0 = ^ « M ( f c ) + < * a M ( / ) ( 6 J 3 )

where s is an integer whose value depends on the stage of computation r and the index h. These equations characterize the 2 x 1 PM DIF DFT butterfly shown in Fig. 6.6. This butterfly computes two 2-point DFTs after the input values are multiplied by appropriate twiddle factors. The operation of this butterfly is similar to that shown in Fig. 6.1. The difference is that the twiddle factors corresponding to the second element of the two input vectors are W^ and (—j)W^, respectively. The input and output vectors

Computational Complexity of the 2 X I PM DFT Algorithms 135

= {aW(/ l),ai r )(ft)} o , >^o ={4 r + 1 ) ( / i ) , a ( r + 1 ) ( / i ) }

a ( r ) ( 2 { a « ( / ) , a W ( 0 } o - ^ ^ o ° ( r f= (?4 r + 1 , (0 ,«i r + 1 ) (0}

Fig. 6.6 The SFG of the butterfly of the 2 x 1 PM DIF DFT algorithm, where 0 < s < x- Twiddle factors are represented only by their exponents. The symbol s represents

are represented by the symbol a just to indicate that the decomposition process of an ^—vector DFT starts from the input side.


For example, with N = 16, there are 3 stages specified as r = 1,2, and 3. Four butterflies make up a stage. Indices h and /, and the twiddle factor exponent s of each group of butterflies are given, respectively, by (ft = (i mod 2 3 " r ) , (i = 0 , 1 , . . . , 3)), I = h + 23~r, and s = h(2r-1). The SFG of the 2 x 1 PM DIF DFT algorithm, with N = 16, is shown in Fig. 6.7. The grouping of the butterflies is just the reverse of that shown in Fig. 6.2, since the decomposition of the problem starts from the input side. The input and output vectors have the same interpretations as in Fig. 6.2, except that the input vectors are placed in the natural order and the output vectors are placed in the bit-reversed order.

Example 6.2 Figure 6.8 shows the trace of the algorithm shown in Fig. 6.7. Vector formation is similar to that in Example 6.1 except that the vectors, shown in the second column, are placed in natural order. The third, fourth, and fifth column vectors are obtained after the first, second, and third stage operations of the algorithm are carried out. The output vectors have to be swapped to put them in sequential order. I

6.7 Computa t iona l Complexi ty of t he 2 x 1 P M D F T Algor i thms

In this section, the computational complexity of the 2 x 1 PM DIT DFT algorithms is described. The computational complexity of the DIF algorithms is the same as that of the DIT algorithms.


Fig. 6.7 The SFG of the 2 x 1 PM DIF DFT algorithm, with N = 16. Twiddle factors are represented only by their exponents. For example, the number 4 represents W±6.

One-butterfly implementation

There are m = log2 y stages each with —• butterflies. Each butterfly requires 8 real multiplications and 12 real additions. Further, 2N real additions are required in the vector formation stage to compute y 2-point DFTs. Therefore, the expressions for the number of real multiplications and real additions are 2Nrn and TV (3m + 2), respectively.

Two-butterfly implementation

In the first butterfly of each group, the twiddle factors are 1 and — j . These butterflies can be implemented separately. This type of implementation is called a two-butterfly implementation since one general and one special butterfly are used. There is one butterfly with twiddle factors of 1 and -j in the last stage, two in the last but one stage, four in the preceding stage, and so on. Thus, the saving in real multiplications is 8(2° + 21 + 22 + ... + 2 m _ 1 ) = 8(2m - 1) = (47V - 8). Therefore, the total number

Computational Complexity of the 2 x 1 PM DFT Algorithms 137

x(0) x(8)

ar(l) x(9)

x(2) x{W)

ar(3) x( l l )

*(4) s(12)

*(5) x(13)

x(6) a;(14)

*(7) z(15)

2 + jO 1 + j O

3 + jO 3 + jO

4 + jO 3 + jO

4 + jO 1 + jO

2 + j l 1 + j l

1 + J3 2 + j2

4 + J3 2 + J2

3 + j l 1 + j l

3 + jO 1+jO

6 + j O 0 + j O

7 + jO 1+jO

5 + jO 3 + jO

3 + J2 1+ jO

3 + J5 - 1 + j l

6 + j 5 2+jl

4 + J2 2 + j O

6.00 + j'2.00 0.00 - J2.00

9.00 + J5.00 3.00 - j'5.00

13.00 + J5.00 1.00 - J5.00

9.00 + j'2.00 1.00 - J2.00

l.OO-jl.OO 1.00 + jl.OO

1.31 + J0.54 - I .31- j0 .54

0.00 - j'2.83 1.41+J1.41

-0 .70- j3 .54 3.00 - J2.01

19.00 + J7.00 -7.00 - J3.00

18.00 + j"7.00 0.00 + J3.00

- 5 . 0 0 - j 3 . 0 0 5.00 - j l .00

-3.54 - j'4.95 0.71 - J6.36

1.00 - J3.83 l.OO + j l .83

0.61 - j3.00 2.01 + J4.08

2.41 - j'0.41 -0.41 + J2.41

- 4 . 8 4 - j 0 . 1 6 2.23 + J1.24

37.00 + j'14.00 1.00 + jO.OO

-4.00 - j/3.00 -10.00-J3.00

-8.54 - J7.95 -1.46 + J1.95

- l . 3 6 - j l . 7 1 11.36 - jO.29

1.61 0.39

J6.82 J0.83

5.08 -jO. 18 -3.08 + J3.83

-2.43 - J0.57 7.26 - jO.26

0.83 + J0.18 -1.66 + J4.64

X(0) X(8)

X(4) X(12)

X{2) X(10)

X(6) X(U)

X(l) X(9)

X(5) X(13)

X(3) X ( l l )

X(7) X(15)

Fig. 6.8 The trace of the 2 x 1 PM DIF DFT algorithm, with N = 16.

of real multiplications required for a 2 x 1 algorithm with two-butterfly implementation is 2Nm -AN + 8 = 2N(m - 2) + 8. Since N - 2 complex multiplications are saved in this type of implementation, the savings in real additions is 2N — 4. Therefore, the total number of real additions required is 3Nm + 4.

Three-butterfly implementation

If we set up another special butterfly (a three-butterfly implementation) to process multiplications by the twiddle factors ± 4 s — J 775, we can further save some real multiplications. There is one such butterfly in the last stage, two in the last but one stage, four in the preceding stage, and so on. Therefore, by separately implementing such butterflies, the saving in real multiplications is given by

4(2° + 21 + 22 + ... + 2m~2) = ( 2 m + 1 - 4) = N - 4

http://-I.31-j0.54

http://-0.70-j3.54

http://-5.00-j3.00

http://-l.36-jl.71


Table 6.1 Figures of Computational Complexity for the 2 x 1 PM DFT Algorithm with Various Number of Butterflies.

Number of multiplications | Number of additions N 1-B_fly 2-B_fly 3-B_fly 1-Bjfly 2- fc 3-B_fly 16 32 64 128 256 512 1024 2048 4096

96 256 640 1536 3584 8192 18342 40960 90112

40 136 392 1032 2568 6152 14344 32776 73736

28 108 332 908 2316 5644 13324 30732 69644

176 448 1088 2560 5888 13312 29696 65536 143360

148 388 964 2308 5380 12292 27652 61444 135172

There is no further reduction in the number of real additions. Therefore, for the 3-butterfly implementation, the expressions for the number of real multiplications and real additions are N(2m — 5) + 12 and 3Nm + 4, respectively. The number of each of the two operations of real multiplication and real addition required by the 2 x 1 PM algorithm for complex data for various values of N and for various number of butterflies is given in Table 6.1.

Twiddle factor generation

Twiddle factors for the general butterflies indexed h and 2 m ^ + 2 — h for the DIT algorithms (h and ^qr - h f° r the DIF algorithms) are trivially related and, therefore, they need to be referenced or computed only once if the two butterflies are processed at the same time. Twiddle factors for the first butterfly are of the form Wfc and W^ and those of the second butterfly

— - 8 — - 8

are of the form WJ and W£ . The size of the look-up table to store the twiddle factors is ^ which is one-eighth of that of the data. Twiddle factors can also be generated as they are required but the computation time will increase.

6.8 The 6 x 1 P M DIT D F T Algorithm

The DFT lengths that are most often used in practice are integral powers of two. However, for a specific requirement, it is relatively straightforward


to develop efficient algorithms for DFT lengths with a factor such as 3,5,7, etc. For example, using a vector with six elements, that is u = 6, we can design algorithms with DFT lengths 12, 24, 48, 96,192, 384, etc. The input vectors are formed by computing 6-point DFTs of the input data as defined in Eq. (5.2). The prime-factor algorithm described in Appendix D can be used to compute the 6-point DFTs efficiently.


The input-output relations of a butterfly of the 6 x 1 PM DIT DFT algorithm can be easily deduced from Eq. (6.5), by substituting u = 6, as

4 r + 1 )

4 r + 1 )

A(r+1)

4 r + 1 )

A ^

4 r + 1 )

A{r+1

A{r+l

A3 A ( r+1

4 r + 1

4 r + 1

4 r + 1

w (h)

(h)

(h)

(h)

w Hi) Hi) Hi) Hi) Hi) Hi)

A(r

A{r

4 ' 4 r

4 r

4 r

4 r

A[r

A(r

A3 A3

A{r

A5

\h)

\h)

{h)

(h)

(h)

(h)

(h)

(h)

(h)

(h)

(h)

(h)

+ w^4p)(0 -WS,A£HI)

„.«+£ .(VI + <+^4r)(o -< + -4 r ) (0 • "JV " 4 ' ( 0

-W8N

+^A^(l)

+ WaN

+^A^Hl)

+ w^+ f +"4 r )(o -w#"«+ i a4 r )(o + wJV

+6+i=4r)(o - ^ + - + - 4 r ) ( o ,

where s is an integer whose value depends on the stage of computation r and the index h. These equations characterize the 6 x 1 PM DIT DFT butterfly. This butterfly computes six 2-point DFTs after the input values indexed I are multiplied by appropriate twiddle factors. The differences between this butterfly and that shown in Fig. 6.1 are: (i) three 2-point DFTs are computed at each output node rather than one and (ii) the twiddle factors are different. This butterfly requires 24 real multiplications, 36 real additions, and assuming that sufficient number of registers is available in the processor, 48 data transfer operations between the memory and the processor. There are ^ log2 y butterflies. Two special butterflies with reduced computation can be derived when s = 0 and s = ^ . The first but-


terfiy requires 16 real multiplications and 32 real additions while the second butterfly requires 20 real multiplications and 36 real additions. There are Y — 1 butterflies of the first type and ^ — 1 of the second type. The six twiddle factors required for the butterfly operation can be obtained by gen-erating only three of them. Specifically, if the twiddle factors Wfc, WN

12, 8+ &- 8+ ^-+ ^-

and WN 6 are generated, then the remaining twiddle factors WN

6 12,

WN e , and WN

B 12 are obtained simply by multiplying, respectively,

the first three twiddle factors with W£ = —j.


For example, with N = 48, there are 3 stages specified as r = 1,2, and 3. Four butterflies make up a stage. Indices h and /, and the twiddle factor exponent s of each group of butterflies are given, respectively, by (h = i mod 2 r - \ (i = 0 , 1 , . . . , 3)), l = h + 2r~\ and s = h(23~r). The SFG of the 6 x 1 PM DIT DFT algorithm, with N = 48, is the same as that shown in Fig. 6.2 with u = 2 and N = 16, but the number of 2-point DFTs computed by each butterfly is six rather than two.

The computational complexity

For a 1-butterfly implementation, the number of real multiplications and real additions required are given, respectively, by 2Nm + ^ and 3Nm + 6N, where m = log2 j - - For a 2-butterfly implementation, the number of real multiplications and real additions required are given, respectively, by 2iVm+8 and 3JVm+ ^§^+4. For a 3-butterfly implementation, the number of real multiplications and real additions required are given, respectively, by 2Nm - f + 12 and 3Nm + ^ - + 4. The number of each of the two operations of real multiplication and real addition required by the 6 x 1 PM DIT algorithm for complex data for various values of N and various number of butterflies is given in Table 6.2.

Example 6.3 A trace table of the 6 x 1 PM DIT DFT algorithm, with N = 24, is shown in Fig. 6.9. The result of vector formation and swapping is shown in column two. The third and fourth column vectors are obtained after the first and second stage operations of the algorithm are carried out.

Flow Chart Description of the 2 x 1 PM DIT DFT Algorithm 141


N 24 48 96 192 384 768 1536 3072 6144

Number of multiplications 1-B_fly

128 352 896 2176 5120 11776 26624 59392 131072

2-B_fly 104 296 776 1928 4616 10760 24584 55304 122888

3-B_fly 100 284 748 1868 4492 10508 24076 54284 120844

Number of additions 1-BJy

288 720 1728 4032 9216 20736 46080 101376 221184

2- & 3-B_By

276 692 1668 3908 8964 20228 45060 99332 217092

6.9 Flow Chart Description of the 2 x 1 P M DIT D F T Algorithm

In this section, we present a flow chart description of the 3-butterfly implementation of the 2 x 1 PM DIT DFT algorithm. This will further aid the understanding of the algorithm and will enable the reader to implement the algorithm in a preferred programming language. The algorithm consists of a main module, which invokes five modules as shown in Fig. 6.10(a). The modules are (i) twid-fac, (ii) iti-put, (iii) make.vec, (iv) dit21Sl, and (v) out-put. The data length N must be an integral power of two and greater than 4. The data structure for storing the data consists of four arrays each of size y . Two arrays, apr and amr, store the first half and the second half of the real part of the input data, respectively. Two arrays, api and ami, store the first half and the second half of the imaginary part of the input data, respectively. Two twiddle factor arrays tfc and tfs, each of size y , store, respectively, the cosine and sine values of arguments from ^ to j .

The twidjac module is shown in Fig. 6.10(b). We have to create a table consisting of cosine and sine values of ^ to \ with increments of ^ . The value cos J is stored in the element tfc(0). We initialize variables arg and inc to ^F, and a counter i to 1, which controls the number of iterations. In each iteration, the array elements with index i are assigned with cosine and sine values of the current argument. Variable arg and the iteration counter i are updated and the iteration continues until i < %-. It will be more efficient to fill the first half of the arrays by invoking cosine and sine functions and using the relations s i n ( ^ ( y - n ) ) = - ^ ( c o s ( ^ n ) - s i n ( ^ n ) )


x(0) x(4) x(8) x(12) z(16) z(20)

*(1) x{h) x(9) x(13) *(17) a:(21)

x(2) x(6) x(10) z(14) ar(18) z(22)

x(3) x{7) x(U) z(15) x(19) x(23)

2 + j l 1 - J 3 3 + J 2 1 + J 2 3 + j l 1+J2

1 - J 2 O - j l 1 - J l 3 4-J3

- 2 - j2 3 + J 2

3 - j l 1+J2 2 + jO

2-n - 1 + J3

2 + j O

2 + J 2 2 + J 3 3 - j l

- 2 + J2 - 2 - j 3 - 1 - J 2

11.00+ J5.00 -4.46 - J3.00 -6.20 + J2.00

5.00 + J3.00 4.20 + J2.00 2.46 - J3.00

9.00 + J3.00 1.13 - j'2.23 7.33 - j l .04

-1.00 + jl.OO - l . 3 3 - j 7 . 9 6

2.87 + j l .23

6.00 - j l .00 -1.73 - j'3.00 -0.46 4-j7.20 -6.00 - J9.00

6.46 - j'3.20 1.73 - j'3.00

2.00 + jl.OO 10.06 - j'4.43 1.60 + j'7.23 4.00 - J5.00

-3.60 4-j3.77 -2.06 + J9.43

20.00 4-j8.00 -3.43 - J4.87 -2.04 + J7.13

2.00 + J2.00 -8.96 4-j8.87 10.43 -j '3.13

-4.60 - J5.50 6.00+J4.00 0.60-J5.50

-4.33 - jO.50 4.00 + j2.00 4.33-J0.50

8.00 + jO.OO 6.60+J9.43

11 .53- j l .96 4.00 - j2.00

-7.53 4-j4.96 1.40-J4.43

4.77-J11.87 - 1 1 . 0 - J 1 3 . 0 8.23 - jlO.13 -8.23 4-j5.87 - l . 0 0 - j 5 . 0 0 -4.77 4-j4.13

28.00 4- J8.00 2.03 - j'3.83

-0.90 + J12.90 12.00 4- j'8.00

-6.10 4-jl8.10 -17.03 4-j4.83

-3.06 -j 18.20 -7.06 -j 16.07

1.17 4-J6.24 -6.13 4-j7.20

8.26 4- J5.07 6.83 - J2.24

7.00 4- jO.OO 0.00 - J2.00 7.00 4- jO.OO

-13.86 - J9.73 4.00 4- j'6.00

13.86 - j'6.27

-10.974-j2.59 3.47 4-J5.93

10.01 - J3.26 22.97 4- J5.41

-12.13-J6.93 - l . 3 5 4-j2.26

X(0) X(4) X(8) X(12) X(16) X(20)

X(l) X(5) X(9) X(13) X(17) X(21)

X{2) X(6) X(10) X(U) X(18) X(22)

X(3) X(7) X ( l l ) X(15) X(19) X(23)

Fig. 6.9 The trace of the 6 x 1 PM DIT DFT algorithm, with N = 24.

and c o s ( ^ ( y - n)) = - ^ ( c o s ^ n ) 4- s in (^n) ) to fill the second half at the same time. Note that this module can be eliminated by computing the cosine and sine values as required. However, it will be costly in terms of run-time.

The in.put module for reading the data is shown in Fig. 6.10(c). It is assumed that the input data is available with N real values followed by N imaginary values. There are y elements in each of the four arrays, apr,amr,api, and ami. Each quarter of the input values are read into, respectively, into the four arrays. If the input data is real-valued, arrays api and ami must be initialized to zero.

The make-vec module for making and swapping vectors is shown in

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( s t a r t )

lra.ll twiri facl

|call in-put|

leal I m a k e veri

hall HiUl 311

|call out-put|

( Stop*)

6.10(a)

f Start")

read value apr(i), ^ = 0 , 1 , . . . , £ - 1

T read value amr(i), t = 0 , l , . . . , g - l

T read value api(i), |i = 0, ! , . . . , £ - !

T read value amz(i), i = 0 , l , . . . , * _ - l

I (Return)

6.10(c)

( S t a r t )

T ar-^ — i*r\n — -ZL mc N '

t/c(0) = cos( | )

no •T Return)

£/c(i) = cosfarp tfs(i) = sin(ar</ ar<7 = org + inc

i = i + 1 T

6.10(b)

("start)

print value apr(i), » = 0,1,...,£

print value i = 0,1

1 v amr(i), •• • ' T |

aj»(*)> - 1

amz(i), - 1

( Return)

6.10(f)

Fig. 6.10 The flow chart description of the 2 x 1 PM DIT DFT algorithm, (a) The main module (b) The twiddle factor module, (c) The input module, (d) The module for making and swapping vectors, (dl) The swap module, (e) The dit21.31 DFT module. (el) The first special butterfly module. (e2) The second special butterfly module. (e3) The general butterflies module, (f) The output module.

http://lra.ll

144 The u x l PM DFT Algorithms

(Return J

brev = brev + bit

call swap

hbs -

f Start")

'' = 0, bs = 1,5s = 2, inc = N

' 4

( Return j

call b-flyl(bs,gs) call b_fly2(hbs, bs, gs)

call b-fly3(hbs, bs, gs, inc)

hbs = bs,bs — gs,gs = 2 * gs, inc = ^

6.10(d) 6.10(e)

Fig. 6.10(d). The algorithm actually finds the bit-reversal of the first y even numbers of a list of y • Therefore, the outer loop is limited to j - as the sequence counter i is incremented twice in each loop. The bit-reversal algorithm is described in Appendix C. In the inner loop, the first zero bit of the previous bit-reversal, brev, is found. Until that time each nonzero bit of brev is set to zero. The variable bit is used to check each bit and it is divided by two in each iteration. When bit > brev, the iteration ends and the bit-reversal of i is found with the operation brev = brev + bit. Now, the swap module, shown in Fig. 6.10(dl), is called. If the bit-reversal brev is equal to i, then another vector index in the upper half of the list is found and 2-point DFTs of each vector is computed. If brev is greater than i, then the pair of vectors with indices i and brev are swapped and the 2-point DFTs of each vector is computed. If brev is less than i, then swapping and the 2-point DFT of a pair of vectors in the upper half of the


f Start J

no

£1 = amrii), t2 = amriq), £3 = ami(i), £4 = amiiq) £5 = aprii), amrii) = £5 — tl,aprii) = £5 + t l £5 = apriq), amriq) = £5 - £2,apr(g) = £5 + £2 £5 = apt(t) ,amm) = £5 - £3,api(t) = £5 + £3

' " " • ' - £5+ £4 t5 = api(i) ,omim = £5 — £3,api(t) = £5 = api(q),ami(q) = £5 — £4,api(g)

u; = i, q — brev ifiw > q){w = w+% + l,q = q+% + l}

t l = apr(tu),t2 = api(u;),£3 = amr(tu £4 = ami(w),£5 = apr(g),£6 = amriq

amriw) = £5 — £6,apr(w) = £5 + £6 £5 = apiiq), *6 = amiiq)

amiiw) = £5 — £6,api(u;) = £5 + £6 amrfg) = £1 — t3,apriq) = £1 + £3 omi(g) = £2 - £4, apiiq) = £2 + £4

3. i = i + 1, gr = 6rew 4- ^

£1 = ap7-(i),t2 = api(i),t3 = amr(j) £4 = amz(i),£5 = apr(</),£6 = amriq)

amrii) = £5 — £6,ajw(z) = £5 + £6 £5 = api(g),£6 = amiiq)

amiii) = £5 — £6,api(i) = £5 + £6 amriq) — £1 — £3,apr(g) = t l + t3 ami\q) = t2 — tA, apiiq) = £2 + £4

i = i + l

I ( Return)

6.10(dl)


list is carried out. In all the iterations, the vector whose index is the next higher odd number and the vector whose index is the bit-reversal of the odd number are swapped and 2-point DFTs computed.

Figure 6.10(e) shows the dit21Sl module. The term dit21 indicates that the algorithm is the 2 x 1 PM DIT DFT algorithm. This is a three-butterfly implementation, which is indicated by the number 3. The last digit 1 indicates that each stage is implemented separately. Two general and two special types of butterflies are used as presented in Section 6.2. This flow chart should be followed with the SFG shown in Fig. 6.2. The variable bs represents the span of a butterfly. The variable gs represents the span of a group of butterflies. The variable hbs is equal to ^ - Variable inc represents the difference between two indices of the twiddle factors in the look-up table for a particular stage. These variables are initialized for first stage values. The loop steps through various stages. The butterfly span in the last stage of the algorithm is maximum at ^ , and therefore, the value ^- 1S u s e d to end the iteration. As the first butterfly in each group is a special one, module b-flyl is invoked to process this type. This module processes all the butterfly of this type in a loop as shown in Fig. 6.10(el). The second special butterfly is required from Stage 2 onwards. The module, b-fly2, is invoked and it processes all the butterflies of this type in a loop as shown in Fig. 6.10(e2).

Two general type butterflies are used in the module called b-fly3, shown in Fig. 6.10(e3). Unlike in the previous cases, the number of butterflies in each group is more than one. Therefore, two loops are required for implementation. Two butterflies of the general type are processed at the same time. This reduces the fetching of the twiddle factors as the twiddle factors are related in a trivial manner. For each pair of butterflies in the first group, the indices of the nodes and the twiddle factors are set. The inner loop steps through all the butterflies with the same twiddle factors in all groups. The outer loop steps through each butterfly of the first group. The values are updated and the iteration continues until all the butterflies in a stage are processed. The ouLput module is shown in Fig. 6.10(f). In each iteration of the first loop, a pair of values apr(i) and api(i) are printed. In the second loop, a pair of values amr(i) and ami(i) are printed.

Flow Chart Description of t/ie 2 x 1 PM DIT DFT Algorithm

C Start J

(Return)

£1 = apr(h),t2 = apr(l),t3 = amr(h) apr(h) = £1 + t2,amr(h) =tl-t2

£1 — api(h),t2 = api(l),H = ami(h) api(h) = £1 + £2, ami(h) = tl-t2

£1 = amr(l),t2 = ami (I) apr(l) = £3 + t2,amr(l) = £3 - £2 api(l) = £4 - tl,ami(l) = £4 + £1

ft = ft + gs, I = I + gs

6.10(el)

f Start")

I ft = hbs, I = h + bs,c = £/c(0)

(Return)

£5 = apr(Z),£4 = apiVt) £1 = c * (£5 + £4), £3 = c * (£4 - 45)

£5 = amr(l),H = ami(l) t2 = c * (£4 - £5), £4 = c * (£4 + £5)

£5 + £4,api(J) = £ 5 - £ 4 £5 = amr(h),amr(l) = £5 — £2, apr(Z) = £5 + £2 £5 = api(h), api(h) = £5 + £3, ami(h) = £5 — £3 £5 = apr(h),apr(h) = £5 + £l,amr(ft) = £5 - £1

ft = ft + <7S, / = I + gs

6.10(e2)


(Return)

h = i,l = h + bs,rh = bs — i, rl — rh + bs c = tfc(J), s = tfs(j)

, . \ U U • • , -i • . -

£5 = apr(rl),t4 = api(rl),tl = s * £5 + c* £4 £3 = s * £4 — c* £5, £5 = amr(rZ),£2 = ami(rl)

tA = s * £5 + c * £2, £2 = s * £2 — c * £5 £5 = ami(rh),ami(rl) = £5 + £4,api(r/) = £5 — £4 £5 = amr^r/i^amr^rZ) = £5 — t2,apr(rl) = £5 + £2 £5 = api(rh), apiOrh) = £5 + £3, ami(rh) = £5 — £3 £5 = apr(rh), apr{rh) = £5 + £1, amr(rh) = £5 — £1

£5 = apr(Z), £4 = api(l), £1 = c * £5 + s * £4 £3 = c * £4 - s * £5, £5 = amr{l), £2 = ami(l)

£4 = c * £5 + s * £2, £2 = c * £2 — s * £5 £5 = ami(h), ami (I) — £5 + £4, api(Z) = £5 - £4 £5 = omr(/i),omr(I) = £5 — £2,opr(Z) = £5 + £2 £5 = api(h), api(h) = £5 + £3, ami(h) = £5 — £3 £5 = opr(ft), apr(/i) = £5 + £1, amr{h) = £5 — £1 h = h + gs,l = I + gs,rh = rh + gs, rl = rl + gs

6.10(e3)

Implementation issues

Factors affecting the execution time of any DFT algorithm are (i) arithmetic operations, (ii) data movement operations, and (iii) overhead operations such as bit-reversals, data swapping, and array-index updating. With the fast execution of arithmetic operations in modern computers, all the operations contribute significantly to the execution time of an algorithm. For example, the 2 x 1 PM algorithm requires 4.17iVlog2 N, 4JVlog2./V, and y log2 N operations, respectively, for the arithmetic, data movement,

Summary 149

and the array-index updating for large values of N. The last two operation counts can be reduced by a factor of two by implementing the operations of two consecutive stages together. This involves executing four 2 x 1 PM butterflies, two from each stage of a pair of adjacent stages, at the same time. Modern arithmetic units typically have 16 or more registers. Single-stage implementation of a 2 x 1 PM algorithm requires 7 registers whereas two-stage implementation requires about twice as many registers. Therefore, implementing pairs of consecutive stages concurrently results in efficient register utilization and thereby reduces the execution time significantly. The process of implementing the operations of two consecutive stages at a time is almost a mandatory requirement in order to achieve a fast execution on modern processors, since the execution is very fast once the operands are in the registers.

In conclusion, for good performance of the algorithms, the requirements are the availability of: (i) sufficient number of registers, (ii) a plus-minus instruction, (iii) a multiply-add instruction, and (iv) efficient generator for twiddle factors. Further speedup can be achieved by using one or more hardware butterflies, or parallel processors.

6.10 S u m m a r y

• Fast DFT algorithms essentially spread the computation over several stages, develop partial results in each stage, and use them in an optimum way to compute all the N frequency coefficients.

• In this chapter, the u x 1 PM DFT algorithms were presented. The algorithms are an interconnected network of butterflies, each butterfly representing a set of operations. The particular case with vector length two was described in detail. DFT algorithms with vector length two are highly recommended for practical use as they are simple, regular, and very efficient.

References

(1) Sundararajan, D., Ahmad, M. 0 . and Swamy, M. N. S. (1994) "Computational Structures for Fast Fourier Transform Analyzers", U.S. Patent, No. 5,371,696.


(2) Sundararajan, D., Ahmad, M. O. and Swamy, M. N. S. (1998) "Vector Computation of the discrete Fourier Transform", IEEE Trans. Cir. and Sys. II, vol. CAS-45, No.4, pp. 449-461.

Exercises

6.1 Derive the 2 x 1 PM DIT DFT algorithm directly from the definition, with N = 16.

6.2 Derive the 2 x 1 PM DIF DFT algorithm directly from the definition, with TV = 16.

6.3 Derive the 6 x 1 PM DIT DFT algorithm directly from the definition, with N = 48.


6.1 Write a 3-butterfly program to implement the 2 x 1 PM DIT DFT algorithm.

6.2 Write a two stages at a time 3-butterfly program to implement the 2 x 1 PM DIT DFT algorithm.

6.3 Write a program to compute a single DFT coefficient using the 2 x 1 PM DIT DFT algorithm.


Chapter 7

The 2 x 2 P M DFT Algorithms

In this chapter, we describe the 2 x 2 PM DFT algorithms, which reduce the number of arithmetic operations further compared with the 2 x 1 PM DFT algorithms. The reduction is brought about by merging the multiplication operations of pairs of adjacent stages. The 2 x 2 PM DIT DFT algorithm is derived in Sec. 7.1. The 2 x 2 PM DIF DFT algorithm is described in Sec. 7.2. In Sec. 7.3, the computational complexity of the algorithms is presented.

7.1 The 2 x 2 P M DIT DFT Algorithm

Before we derive the 2 x 2 PM DIT algorithm, however, it is instructive to find how the number of multiplications are reduced through a careful examination of the SFG of the 2 x 1 PM DIT DFT algorithm (Fig. 6.2). It can be observed that, for any stage (except the first one), half of the vector output values from the previous stage are multiplied in the current stage by twiddle factors of the form Wfr and W^+ T . Instead of carrying out the two multiplications in the current stage, if we premultiply the two data values that produce an output vector in the previous stage by Wjy, then no multiplications would be required in the current stage. A multiplication by Wj§ is still required, but it is trivial. The twiddle factor W^ shifted from the current stage is combined with the twiddle factor of the previous stage for one of the data values, whereas Wjy itself becomes the twiddle factor for the other data value, since there is no twiddle factor associated with this data value. Therefore, for every two multiplications eliminated in the current stage one multiplication is added in the previous stage.

151

152 The 2 x 2 PM DFT Algorithms

The combining of the DFTs of two sets of vectors, each of ^ vectors, to produce the DFT of a set of % vectors in the last stage of the 2 x 1 PM DIT DFT algorithm is given by Eqs. (6.3) and (6.4), with u = 2, as

Ap(k) = Aie){k) + WZWk4°\k) (7.1)

Ap(k+j) = A[e\k) + WZWlWkA[°\k), (7.2)

where k = 0 , 1 , . . . , ^ — 1 and p = 0,1. To obtain the first and the third quarter values of Ap(k), Eqs. (7.1) and (7.2) can be used with k = 0 ,1 , . . . , Y — 1. The second and the fourth quarter values of Ap(k) can be obtained by replacing k with k + y in these equations. The resulting equations are given by

Ap(k + ^) = 4°\k+j) + WZWkN

+*Ai°\k + j) (7.3)

Ap(k+*£) = A[e\k+j) + W>W}WkN

+*A{°\k + j), (7.4)

where A; = 0 ,1 , . . . , j - — 1. The quantities on the right side of Eqs. (7.1) through (7.4) represent the output values of the previous stage. These values can be obtained recursively as

4*)(fc) = AM(k) + W$W™4e°\k) (7.5)

4eHk+j) = A\ee\k) + W$W}W*NkA[e°\k) (7.6)

Ap°\k) = 4°e\k) + WiW™4°0)(k) (7.7)

Ap°Hk+j) = A[°e\k) + WZWlW™A[°°\k), (7.8)

where k = 0 , 1 , . . . , f - 1 and A^(k), A^(k), Aêo\k), and A^00\k) represent the DFT of the vectors a(n) such that (n mod 4) is 0, 1 ,2 , and 3, respectively. The twiddle factors associated with the terms A^'^k), A[°\k), 40)(fc + f ) , and A[°\k + f ) in Eqs. (7.1) through (7.4) can be merged with those in Eqs. (7.7) and (7.8) as

WkNAp°\k) = Wk

NA^\k) + W*W$A(°0\k) (7.9)

W^+%Ap°Hk + j) = WÂ[oe\k) + WiWiW3N

k+Â{oo)(k) (7.10)



The input-output relation of a butterfly of the 2 x 2 PM DIT algorithm can be deduced from Eqs. (7.1) through (7.6), (7.9), and (7.10) as

A[r+1

Ar 4 r + 1

A<[+1

4 r + 2

A^2

^ (r+2

\hl)

\hl) 15 (11)

"(ll)

\h2)

\K2)

"{12)

"(12)

\hl)

\hl)

\h2)

\h2)

4 r ) («) + WftM<p)(/l) A^(hl)-WlfA^{ll)

A^(hl) + WÂ^{11)

A[r)(hl)-W2J+*A[r)(ll) A^(hl)-W2N

8+Â[r\ll)

W^m + WlfA^m = W°NA%\h2)-W%°A%\l2)

-__»-l-«- .M „.3s+W

4 r + 2 )ai) = 4r+2)(zi) = 4r+2)«2) = A{{+2\12) =

W?*AP{h2) w°+a

+ W%+ 8 •A«(B) NrA(r)m_wÂ(;){l2)

1 \ . A(r+1) / l n \ ^ ^ ( w j + ^ ^ C W )

A£+1\hl)-A£+1\h2)

A^+1)(hl) + WÂ^+1)(h2)

4r + 1 )(/il)-^|4r + 1 )(^2) 4r+1)(Zl) + 4r+1)(Z2) 4r + 1)(n)-4r + 1)(/2) 4r+1)(/l) + wJ"4''+1)(/2) A(r+1)(n)-^|4r+1)(Z2),

(7.11)

where s is an integer whose value depends on the stage of computation r and the index hi. The 2 x 2 PM DIT DFT butterfly is shown in Fig. 7.1. The butterfly computes eight 2-point DFTs along with the multiplication of the input values by appropriate twiddle factors. This butterfly requires 24 real multiplications, 44 real additions, and, assuming that sufficient number of registers is available in the processor, 32 data transfer operations between the memory and the processor. Two special butterflies requiring reduced computation can be derived from Eq. (7.11) when s = 0 or s = ^ . The first special butterfly requires 4 real multiplications and 36 real additions while the second requires 20 real multiplications and 44 real additions.


Fig. 7.1 The SFG of the butterfly of the 2 x 2 PM DIT DFT algorithm, where 0 < s < 4r. Twiddle factors are represented only by their exponents. The symbol s represents


There are m stages specified a s r = l , 2 , . . . , m , where m = log2 y . Starting from the output side, pairs of adjacent stages are formed each using ^ 2 x 2 PM DIT butterflies. If m is odd, f 2 x 1 PM DIT butterflies are used for the first stage. The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given as follows: hi = i mod 2 r - 2 , i = 0 , 1 , . . . , f - 1 , 1 1 = hl + l ( 2 r - 2 ) , hi = hl + 2(2r~2), 12 = hl + 3(2 r _ 2 ) , and s = / i l (2 m _ r ) , where r is the stage number of the second of the two adjacent stages combined. The SFG of the 2 x 2 PM DIT DFT algorithm with N = 16 (m = 3) is shown in Fig. 7.2.

7.2 The 2 x 2 P M DIF D F T Algorithm

To derive the 2 x 2 PM DIF DFT algorithm, consider the decomposition of an Y _ v e c t o r DFT into two ^-vector DFTs in the first stage of the 2 x 1 PM DIF DFT algorithm. Equations (6.9) and (6.10) of the u x l P M DIF algorithm, with u — 2, is given as

Ap(2k) = J2 W r { a o N + w f + p f a 0 ( n + j)}Wf (7.12) n=0

f-1 N

Ap(2k + 1) = Y^ Win{a1(n) + W^+p^Wla1(n + -)}W^Wf ,(7.13) n=0

The 2 x 2 PM DIF DFT Algorithm

stage 1 stage 2 stage 3

155

A(0)

A(l)

A(2)

A(3)

A(i)

A(5)

A(6)

%*>M7)

Fig. 7.2 The SFG of the 2 x 2 PM DIT DFT algorithm, with N = 16. Twiddle factors are represented only by their exponents. For example, the number 4 represents W*6.

where fc = 0 , l , . . . , ^ — 1 andp = 0 ,1. Combining the input values a,o(n), ao(rc + ;f-), ao(n + y ) , and ao(n + ^ ) such that the summation is only over Y terms, Eq. (7.12) becomes

8 -1

Ap(2k) = £ W 2p n { M n ) + W*+p*a0(n + ^ ) )

n=0

+ W p * W4*(oo(n + £ ) + W * + p f a0(n + ^ ) ) } W £ * , (7.14) o o 2

where k = 0 , 1 , . . . , ^ - 1. Changing A; to 2k in Eq. (7.14) gives 4

a A

Ap(4fc) = £ WriMn) + W^a0(n+ ^ ) ) n=0

+ W 2f e + p f ( a o ( n + ^ ) + < ^ a 0 ( n + ^ ) ) } W | f c , (7.15)


where A; = 0 ,1 , . . . , y - 1. Replacing k by 2A; + 1 in Eq. (7.14) yields

AP(4A; + 2)= 5 3 Win{(a0(n) - W^ao(n + ^ ) ) n=0

+ W*^*^(ooCn + y ) - < f a0(n + ™))}Wf?Wf, (7.16)

where A; = 0 ,1 , . . . , y - 1. Rewriting Eq. (7.13) such that the summation is only over j - terms yields

-l

Ap(2k + 1 ) = £ Wr ,{(ai(n) + W2*+PTW4

1oi(n + £ ) ) n=0

+ < • WtfWftaiC" + ^ ) + W*+p * W 4 V(r*+^))TOW^,(7 .17) O 0 2

N. jfe+p«. , 37V -j) + W2

+PiWlai(n+—

where fc = 0 ,1 , . . . , ^ - 1. Changing k to 2fc in Eq. (7.17) gives

f—I AP(4A: + 1 ) = £ W2

p"{(ai(n) + W ^ W ^ n + £ ) ) n=0

+ W*+p • W2(ai(n +j) + wf* Wlai{n + ™))}WZWf, (7.18)

where A; — 0 , 1 , . . . , T - 1. Replacing A; by 2A: + 1 in Eq. (7.17) yields

AP(4A: + 3 ) = £ Win{(ai(n)-W^W\a^n +j)) n=0

+ W^WlWHa^n +*L)-WZ*Wlal(n +^-))}W%nWt ,(7.19) O O 4

where k = 0 ,1 , . . . , y - 1.

The 2 x 2 PM DIF DFT Algorithm 157

The 2 x 2 PM DIF DFT butterfly

The input-output relation of a butterfly of the 2 x 2 PM DIF algorithm can be deduced from Eqs. (7.15), (7.16), (7.18), and (7.19) as

4r+1)(w) a[p+1)(W)

a{;+1\h2) <4r+1)(fc2)

a{;+1\n) alr+1)(/l) 4r+1)(<2) a^+1)(Z2)

4r+2)(»D a^ 2 )(hl)

«ra)ci) 4r+2)(Zl)

4r+2)(M) a< r + 2 )( / i2)

ar2 )( /2) 4r+2)(Z2)

= 4 r )(w)+4 r )(M)

= 4p)(w)-4r)(w) = <4r)(/il) + w|<4 r ) (Zi2) = ajr)(M)-W$ajr)(/»2) = 4 r )(U)+a«(/2) = 4r)(n)-aW(/2) = ajp)(n) + w|a[p)(J2) = 4r)(/l)-wlotr)(Z2)

= air+1\hl) + a£+1\ll) = a£+1\hl) - a£+1\ll) = Wf?a[r+1\hl) + wZ+*a<r1\ll) = Wfta[r+1\hl)-W2

N8+*a<f+1\ll)

= wUr+1)m+w?*a£+lHi2) = W^+1\h2)-W8

N+^a^+1\l2)

= W*N°a<f+l\h2) + W3N

8+3-?a?+1\l2)

= Wfra^m-W^a^W,

where s is an integer whose value depends on the stage of computation r and the index hi. The 2 x 2 PM DIF DFT butterfly is shown in Fig. 7.3. The butterfly computes eight 2-point DFTs along with the multiplication of the input values by appropriate twiddle factors. The number of operations in the DIF butterfly is the same as that in the DIT butterfly.


There are m stages specified as r = 1,2,. . . , m, where m = log2 y . Starting from the input side, pairs of adjacent stages are formed using | 2 x 2 PM DIF DFT butterflies. If m is odd, f- 2 x 1 PM DIF DFT butterflies are used for the last stage. The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given as follows: (hi = i mod 2 m - r - \ i = 0 , 1 , . . . , f - 1), Zl = hi + l ( 2 " J - r - 1 ) , h2 =


a W ( M ) O r ^ h U ^o a(r+2)(hl)

4oa(r+V(l2) 8

Fig. 7.3 The SFG of the butterfly of the 2 x 2 PM DIF DFT algorithm, where 0 < s < y . Twiddle factors are represented only by their exponents. The symbol s represents W

hi + 2{2m-r~1), I2 = hl+ 3 ( 2 m - r - 1 ) , s = hl(2r~1), where r is the stage number of the first of the two adjacent stages combined. The SFG of the 2 x 2 PM DIF DFT algorithm with N = 16 (m = 3) is shown in Fig. 7.4.

7.3 Computa t iona l Complexi ty of t he 2 x 2 P M D F T Algor i thms

One-butterfly implementation

M = log2 N, odd: There are j - butterflies in each of the ^-^- combined stages. Each butterfly requires 24 real multiplications and 44 real additions. The vector formation stage requires 2N additions. The number of real additions and multiplications required is, respectively, given as

N(M-l)lt „Ar A T ( l l M - 3 ) , N(M-l)nA 3 ^ , , , , , — ^ '-M + 2N = N± '- and — '-2A= -N(M -I) 8 2 4 8 2 2 v '

M even: There are y butterflies in each of the ^f^- combined stages. The vector formation and first stages require 4N additions. The number of real additions and multiplications required is, respectively, given as

N{M-2).. . . . . . ( 1 1 M - 6 ) , N{M-2)nA 3 . . . . . _. - v

2 ; 44 + 4iV = A^ '- and — 2

y24 = - i V ( M - 2)

Computational Complexity of the 2 x 2 PM DFT Algorithms 159

Fig. 7.4 The SFG of the 2 x 2 PM DIF DFT algorithm, with N = 16. Twiddle factors are represented only by their exponents. For example, the number 4 represents Wf6.

Two-butterfly implementation

M odd: Each of the first special butterfly saves 20 real multiplications and 8 real additions. The total number of such butterflies in each pair of adjacent stages is 4° + 4X+. . . +4 a" _ 1 = |(iV - 2). By subtracting the number of operations saved from the corresponding value for the one-butterfly implementation, the number of real additions and multiplications required is, respectively, given as

N 8 N 20 _ ( 3 3 M - 2 5 ) + - and - ( 9 M - 2 9 ) + -

M even: The total number of special butterflies in each pair of adjacent stagesis4° + 41+. . . + 4 2 l — ^(N-4). The number of real additions and multiplications required is, respectively, given as

— ( 3 3 M - 2 6 ) + - and - ( 9 M - 2 8 ) + y

160 The2x2 PM DFT Algorithms


Number of multiplications | Number of additions N 16 32 64 128 256 512 1024 2048 4096

1-B_fly

48 192 384 1152 2304 6144 12288 30720 61440

2-B_fly

28 92 284 732 1884 4444 10588 23900 54620

3-B_fly

24 88 264 712 1800 4360 10248 23560 53256

1-B_fly

152 416 960 2368 5248 12288 26624 60416 129024

2- & 3-BJ

144 376 920 2200 5080 11608 25944 57688 126296

Three-butterfly implementation

M odd: Each of the second special butterfly saves 4 real multiplications. The total number of such butterflies in each pair of adjacent stages is 4° + 4}+... +4 a 2 = ^j(iV—8). The number of real multiplications required is given as iV( |M — 5) + 8. The number of real additions remains the same as in the case of the 2-butterfly implementation. M even: The total number of special butterflies is ^(N — 4). The number of real multiplications is the same as in the case of M odd.

The number of each of the two operations of real multiplication and real addition required by the 2 x 2 PM algorithms for various values of N and for various number of butterflies is given in Table 7.1. The implementation of the 2 x 2 PM DFT algorithms consists of a three-loop construct that is similar to the implementation of the 2 x 1 PM DFT algorithms.

Twiddle factor generation

The twiddle factors for the general butterflies with indices h and 2 m ^ + 3 - h

for the DIT algorithms (h and ^h - h for the DIF algorithms) are triv

ially related and, therefore, they need to be computed or referenced only

once if the butterflies are processed at the same time. The twiddle fac

tors of the first butterfly are of the form Wft, Wjf, W%8, W£ +s, and

WN* + , whereas they are of the form W$ ",W^ , WNB , WJ ",

and WN* for the second butterfly. If a look-up table is used to store the

twiddle factors, a table to store y real constants is required for straightfor-

Summary 161

ward access of all the twiddle factors. The table size can be further reduced using trigonometric identities to compute some of the twiddle factors. The access of almost double the number of twiddle factors is a drawback of this algorithm compared with the 2 x 1 PM DFT algorithms. Therefore, this algorithm is not recommended when all the twiddle factors are computed. Even when twiddle factors are stored in a table, the reduction in execution time is not much and it comes only with doubling the size of the twiddle factor table. Therefore, for software applications, this algorithm is not of much advantage. The butterfly of this algorithm requires fewer number of components and this is an advantage when the whole butterfly is implemented in hardware, in high speed applications.

7.4 S u m m a r y

• In this chapter, the 2 x 2 PM DIT DFT and DIF DFT algorithms were derived. These algorithms reduce the number of arithmetic operations by merging the multiplication operations of adjacent stages.

• The 2 x 2 PM butterfly is highly suitable for hardware implementation.

• In the next chapter, we present efficient procedures for the use of the algorithms for complex data in processing real-valued signals.

References


(2) Sundararajan, D., Ahmad, M. 0 . and Swamy, M. N. S. (1998) "Vector Computation of the discrete Fourier Transform", IEEE Trans. Cir. and Sys. II, vol. CAS-45, No.4, pp. 449-461.

P r o g r a m m i n g Exercise


Chapter 8

DFT Algorithms for Real Data - I

We assumed, thus far, that the input data is complex-valued in developing DFT algorithms. However, signals are real-valued in most practical applications. There are two methods used to compute the DFT of real data using the algorithms for complex data. The first method is to use the algorithms directly. The relatively more efficient second method is to use the algorithms by packing the real data to form complex data. In Sec. 8.1, the direct use of an algorithm for complex data is presented. In Sec. 8.2, the computation of DFTs of two real data sets at a time is presented. In Sec. 8.3, the computation of the DFT of a single real data set is described.

8.1 The Direct Use of an Algorithm for Complex Data

The algorithms for complex data can be used for real data by converting the real numbers into complex numbers with the value of the imaginary parts zero. Although this approach requires about double the memory and the processing compared with more efficient approaches, we recommend this if the speed and storage requirements are not critical as it is the simplest approach.

In order to understand the more efficient algorithms for real data, we have to look at the trace tables of algorithms. Figure 8.1 shows the trace of the 2 x 1 PM DIT DFT algorithm for real input data with N = 32. Figure 8.2 shows the trace of the 2 x 1 PM DIF IDFT algorithm for the transform of real data, shown in Fig. 8.1. The first column values, in Fig. 8.1, are 32 real input values. The result of vector formation and swapping is shown in column two. The third, fourth, fifth, and sixth column

163

164 DFT Algorithms for Real Data - I

x(0) x(16) a:(l) *(17) x{2) x(18) x(3) x(19) x(4) z(20)

ar(5) x(21)

ar(6) x(22)

x(7) ar(23)

*(8) z(24)

x(9) ar(25) ar(10) a: (26) x ( l l ) z(27)

x(12) a: (28)

x(13) x (29)

ar(14) ar(30)

ar(15) x(31)

1C 9h-8

1C 9H

8-6

13 21-5

13 8|-3

0

17

1 4-31

13 6-3

20 + j'O 0 + jO

-8 + J4 -8-j4 19 + jO 7 + jO 5 + 5 - j 6

J6-15

15+jO -11+jO

2 + j 5 2 - j 5

15 + jO-11.00-jll.OC l l+ j0 -11 .00+j l l .0c | -15 . -3 + jO -3 + jO 22+jO 12+jO

-1+jS - 1 - J 3 12 + jO 0 + j O 2 + j 6 2 - j 6

18 + jO 8 + j O

- 3 -- 3 + J3I 18 + jO 4 + jO

3 - j 7 3 + j7

39.00 + jO.OO 1.00 + jO.OO

-0.22 + J4.71 -15.78+ J3.29

0 .00-J7 .00-15 0.00 + J7.00

.78--0.22 •

•J3. • j4.71

2S-16

30.00 4-jO.OO 0.00 + jO.OO

-0.12 + J7.12 4.12 + j2.

-14.70 -.71.62 .86 + J8.20 88-16

4.12-J2.88 - 0 . 1 2 - j 7 . 1 2 34.00 + jO.OO 10.00 + jO.OO 4.66 + J5.83

-6.66 + J0.17 12.00 + jO.OO 12.00+ JO.O0

- 6 . 6 6 - j O . 17 4.66 - J5.83

36.00 + JO.O0 0.00 + jO.OO

j"3-5 83 - j 10.07 0.17 + j4.07 8.00 - j'4.00 8.00 + J4.00

- 0 . 1 7 - j 4 . 0 7 -5.83 + J10.07

69.00 + jO.OC 9.00 + jO.OO

2.39 + J11.33 -2.83 - j'1.92

.56 - J7.0C 15.56 - J7.00

.86 - J8.2C 14.70 + J1.62

1.00 + jO.OO 1.00 + jO.OO

15.56+ J7.0C 56 + j7.00

-2.83 + J1.92 2.39-J11.33 70.00 + jO.OO -2.00 + jO.OO - 4 . 5 8 - j l . 2 5 13.90 + J12.9C 14.83 9.17

-J8.4S + j"8.49

1-10.48 -2.83

- jl .57 + J1.23

10.00 10.00

+ jO.OC + jO.OO

-2.83 h 10.48

- j'1.23 + J1.57

9.17 14.83

- J8.49 + j'8.49

13.90 --4.58

-J12.90 + J1.25

139+jO.OO -1.00+jO.OO -2.35+j'l l .Ol 7.13+J11.66

- 5 . 1 0 - j 2 0 . 5 1 -26.01+j"6.51 -26.45 - j'3.68 -7 .27- j l2 .72

8.07 - j"7.07 -6.07 + j'7.07

-17 .29+ j'0.05 -12.10-j '3 .29

11.23-j'4.72 19.89+ j 18.72

-12 .78 - j 14.23 7.11+J18.06

9.00 + J2.00 9.00 - j'2.00

7 . 1 1 - j 18.06 -12.78+j"14.23 1 9 . 8 9 - j 18.72

11.23+J4.72 -12.10+J3.29 -17.29-jO.05 -6.07 - J7.07

8.07 + j7.07 -7.27 + J12.72 -26.45+ J3.68 -26.01-j'6.51 -5.10 + J20.51 7.13-J11.66

- 2 . 3 5 - j l l . 0 1

X(0) X(16) X( l ) X(17) X(2) X(18) X(3) X(19) X(4) X(20) X(5) X(21)

X(6) X(22) X(7) X(23) X(8) X(24) X(9) X(25) X(10) X(26) X ( l l ) X(27) X(12) X(28)

X(13) X(29) X(14) X(30) X(15) X(31)

Fig. 8.1 The trace of the 2 x 1 PM DIT DFT algorithm for real input data, with TV = 32, clearly showing the redundancies.

http://-0.12-j7.12

http://-0.17-j4.07

http://-5.10-j20.51

http://-7.27-jl2.72

http://-2.35-jll.01

The Direct Use of an Algorithm for Complex Data 165

138.00 + J0.00 140.00 + J0.00 4.78 + J22.67 -9.48 - jO.65

-31.11 - j'14.0 20.91 - J27.02

-33.72 -J16.40 -19.18+ J9.04

2.00 + jO.OO 14.14 -jU.14 -29.39 -J3.24 -5.19 + J3.34

31.12+ J14.00 -8.66 -j'23.44 -5.67 + J3.83

-19.89 -J32.2S 18.00 + J0.00 0.00 + J4.00

- 5 . 6 7 - j 3 . 8 3 19.89 -J32.29 31.12 - j 14.00

8.66 -J23.4A -29.39 +J3.24

5.19 + J3.34 2.00 + jO.OO

-14.14-j l4 . l4 -33.72 +J16.4C

19.18+J9.04 -31.11 + j 14.00 -20.91 -;27.02

4.78 - j'22.67 9.48 - jO.65

156.00 + jO.OO 120.00+ J0.00

-0.89 + j 18.84 10.45 + j'26.50 0.01 - J28.00

-62.23 + jO.OO -63.11-J13.16 -^.33 -j/19.64

4.00 + jO.OO 0.00 + jO.OO

-63.11+J13.16 4.33 -j 19.64 0.01 + j 28.00 62.23 + jO.OO

-0.89 - J18.84 -10.45+ J26.5 136.00 + jO.OO 144.00 + jO.OO

18.62 + J23.32 -36.96 -j'28.29

48.00 + jO.Ol 11.32 -J33.93 -26.63 -J0.68 -15.31 - J5.60 39.99+ JO.O0

0.00 + jO.OO -26.63 +J0.68

15.31 -J5.60 48.00 - jO.Ol

-11.32 -J33.93 18.62-J23.32 36.96 - J28.29

160.00 + J0.00 152.00+ J0.00 -64.0 + j'32.00

62.22+ J5.68 0.02+jO.OO

0.00 - j'56.00 -64.0 - J32.00 -£2.22 + j5.68 120.00+ JO.O0 120.00+ JO.O0 16.00 + J40.00

-16.97+j 16.97 -88.01+ JO.O0

0.00 -j'88.01 16.00 - J40.00 16.97+ J16.97 175.99+ JO.O0 96.01 + JO.O0

-8.01 + J24.0C 45.25 + J22.64

96.00 + JO.O0 O.OO + jO.02

-8.01 - j'24.00 -45.25 + J22.64

144.00+ JO.O0 144.00 + JO.O0

-24.00 -J24.00 -22.63 -j'56.57

63.99+ JO.O0 0.00 - J31.98

-24.00+ J24.00 22.63 - j'56.57

160.02 + jO.OO 159.98 + jO.OO

-128.0 + JO.O0 0.00 + j'64.00

208.00+ J0.00 96.00 + jO.OO 79.96 + jO.OO 0.00 + J96.03 31.99 + J0.00

208.01 + jO.OO 32.00 + jO.OO 0.00 + J80.00

208.01 + jO.OO 31.99 + JO.O0

-48.00 + jO.OO 0.00 - jO.Ol

271.99+ J0.00 79.99 + jO.OO

-16 .02+ J0.00 0.00+j'48.00 95.99 + ;0.00 96.03 + jO.OO 31.97 + J0.00 0.00 + J96.01

207.99 + jO.OO 80.01 + jO.OO

-48.01 + jO.OC 0.00 - J48.0C

175.98 + jO.OO 112.02 + jO.OO 48.00 + JO.O0

0.00-J112.00

32.02 288.02

255.97 288.02

64.00 -0.01

159.98 256.00

287.96 128.04

127.96 64.01

160.01 256.01

223.98 127.99

95.98 223.98

31.99 127.99

128.01 288.00

128.00 32.01

-0.03 192.03

0.02 192.04

32.01 31.98

224.03 0.00

Fig. 8.2 The trace of the 2 x 1 PM DIF IDFT algorithm for the transform of real data, with N = 32, clearly showing the redundancies.

http://-5.67-j3.83

http://-14.14-jl4.l4


vectors are obtained after the first, second, third, and fourth stage operations of the algorithm are carried out. The redundancies are evident. The imaginary parts of the input are all zero. Duplication of values exists in various stages of the algorithm. For example, the second half of the output is redundant and, therefore, we need not compute and store these values.

In Fig. 8.2, the result of vector formation is shown in column one. The second, third, fourth, and fifth column vectors are obtained after the first, second, third, and fourth stage (including swapping of output vectors) operations of the algorithm are carried out. The output values have to be divided by the length, N = 32, to get back the input values shown in Fig. 8.1. Note that the output values are not precise. For example, the first value is 32.02 instead of 32 because we input the DFT values with only a precision of 2 digits after the decimal point. The second half of the input vectors is redundant. The output of the various stages exhibits symmetry. The imaginary parts of the output values are zero.

The redundancies can be eliminated by: (i) packing the real data to form complex data or (ii) cutting out the redundancies in each stage of the algorithm. The first approach is described in this chapter and the second approach is presented in the next chapter.

8.2 Computation of the DFTs of Two Real D a t a Sets at a Time

In this method of computing the DFT of real data (RDFT), we compute the DFTs of two real data sets using a DFT algorithm for complex data for the same data length, thereby removing the factor of two redundancy in computation and storage. One set of real input data is stored in the real part of the complex input data and the other set is stored in the imaginary part. The DFT is computed and then the individual DFTs of the two sets are separated by using the linearity property of the DFT and the fact that the DFT of a real data set is hermitian-symmetric.

Let x(n) and y(n) be the data sets, each with N real values. Let X(k) and Y(k) be their respective DFTs. Due to the linearity theorem,

c(n) = x(n) + jy(n) & X{k) + jY{k) = C{k)

Since x(n) and y(n) are real sequences, due to the hermitian-symmetric

Computation of the DFTs of Two Real Data Sets at a Time 167

(start}

Read x(n),y(n), n = 0 ,1 , . . . , N — 1 i

Form c(n) = x(n) + jy{n), n = 0 , 1 , . . . , N — 1

X Compute C(k) = DFT(c(n))

X X(0) = Re(C(0)) ,X(£) = Re(C( f ) )

Y(0) = Im(C(0)) ,Y(f) = Im(C(f ) )

X(k) =

* Compute

C(k)+C*(N-k) y ^ _ C(k)-C'(N-k) h _ = 1,2,.. N • , 2 - 1

X Write JST(fc),y(fc), A; = 0 , 1 , . . . , TV

± ( Return)

Fig. 8.3 The flow chart of the algorithm to compute two RDFTs, each of length N, at the same time using a DFT algorithm for the same data length.

property, X{N-k) = X*{k) and Y{N - k) = Y*(k). Therefore,

C(N-k) = X*(k)+jY*(k)

Conjugating both sides, we get

C*(N-k)=X(k)-jY(k)

Solving for X(k) and Y{k), with C{k) = X(k)+jY(k), we get

X(k) = C{k)+C^N-k)

Yft) = g(fc)-C"CAr-fc) (8.1)

The flow chart of the algorithm to compute RDFT, for an even N, is shown in Fig. 8.3. Only half of the DFT values are computed since each DFT is hermitian-symmetric.


( s t a r t )

Read X(k), Y(k), k = 0 , 1 , . . . , f

± C(0) = X(0)+jY(0),C(f) = X ( f ) + j Y ( £ )

X Form C(k) = X{k) + jY(k),

C(N -k) = X*(k) + jY* (k), k = 1,2,. . . , f - 1

X Compute c(n) = IDFT(C(fc))

± Form x(n) = Re(c(n)), j/(n) = Im(c(n)), n = 0 , 1 , . . . , iV - 1

X Write x(n),y(n), n = 0,l,...,N -1

TReturnJ

Fig. 8.4 The flow chart of the algorithm to compute two RIDFTs, each of length N, at the same time using an IDFT algorithm for the same data length.

The flow chart of the algorithm to compute RIDFT, for an even N, is shown in Fig. 8.4. Only half of the DFT values are input since the DFTs are hermitian-symmetric. Flow charts in Figs. 8.3 and 8.4 can be modified for an odd N.

Example 8.1 Find the DFT of the sequences x{n) — {2,3,4,2} and y{n) = {3,1,1,2}. Compute x(n) and y(n) back from the DFT values. Solution Form the complex values c(n) = x(n)+jy(n) — {2+jS,3+jl,4+jl,2+j2}. The DFT of c{n) is C{k) = {11 + P, - 3 + j l , 1 + j l , - 1 + ; 3 } . To separate X(k) and Y(k), we use Eq. (8.1).

X(0) X(l) X(2) X(3)

Re(C(0)) (-3+jlH(-l-J3)

Re(C(2)) X*(l)

11

-2-jl 1

- 2 + j l

Computation of the DFT of a Single Real Data Set 169

Y(0) = Im(C(0)) = 7 y ( i ) = (-3+ii)-(-i-J3) = 2 + j l

y(2) = Im(C(2)) = 1 y(3) = y*( i ) = 2 - j i

To compute the RIDFT, X(fc)+.7 Y(fc) is formed and the IDFT is computed. The result gives x(n) as the real part and y(n) as the imaginary part. For this example, X{k)+jY(k) = {11+j7,-3+jl,l+jl,-l+j3}. The IDFT of this sequence yields x(n) + jy(n) = {2 + j3,3 + jl, 4 + j l , 2 + j2} . I

The number of real additions required to separate the two DFTs is 2N — 4. Therefore, for computing an JV-point RDFT, the number of operations required is N — 2 more than one-half of that of the algorithm for complex data used.

8.3 Computation of the DFT of a Single Real Data Set

In this method of computing the RDFT, the redundancy is removed by: (i) computing the DFTs of even and odd indexed input values using a single DFT algorithm for half data length rather than two, as in the case of the algorithm for complex data with zero imaginary parts and (ii) computing only half of the DFT values in combining the two smaller DFTs.

Let x(n) be the data set with N real values whose DFT, X(k), is to be computed. Leta;(e)(n) = x(2n) anda;(°)(n) = x(2n + l),n = 0 , 1 , . . . , f - 1 . Let X<e)(fc) and X^ik) be their respective DFTs. Then,

x (e ) (n) + jx^ (n) o X ^ (k) + jXM (k)

As described in the last section, X^(k) and X^(k) can be computed using a single DFT algorithm for data length —. The DFTs can be combined to get X(k) = X^(k) + W#X(°)(fc). The flow chart of the algorithm to compute RDFT, for N an integral multiple of 8, is shown in Fig. 8.5. It can be modified for any even N. This flow chart shows how we can implement this algorithm with minimum arithmetic operations and twiddle factor access. Note that, in merging the DFTs of the even and odd indexed data, only half the number of multiplications shown in Fig. 8.5 are actually required.

In computing the RIDFT, the redundancy is removed by: (i) computing only half of the input values for the two independent IDFTs computing the


( s t a r t )

^r^ Read real x(n), n = 0,1,. . . ,N — 1

• Form c(n) = x(2n) + jx(2n + 1), n = 0 ,1 , . . . , ^ - 1

I Compute C{k) = DFT (c(n))

I XW(0) = Re(C(0)),XW(f) = Re(C(f))

XW(Q) = Im(C(0)),X(°)(g) = Im(C(f))

Compute xM(fc) = C ^ + C ; ^ - * > ,

X(0) ( f c ) = g(fc)-^(f-fc)! fc==i)2,...,f - 1

X(0) = lW(0) + X(°)(0),X(f) = X«(0) - X(°)(0),

X(f) = X(°)(f)-jX(°)(f) V

X(f) = xW(f) + w£x(°)(%),x(3-£) = x*^(f) - Wf 1 *^*) I

X(fc)

X(£

= XW(ifc)

* ( £ + fc) = X '

+ w#x(°) - * ) =

{ 4

x<e

- * )

Compute

(fc),X(f->(?-*)-- jW#X*

-*) =

-m v 4

x*w kx{o

-k),

(*)"

Hf fc =

- W ^

-fc) ,

1,2,.

fe^*(o)

N • • > 8

(*),

- 1

I Write X(k), k = 0,l,...,f

(Return)

Fig. 8.5 The flow chart of the algorithm to compute a RDFT of length N using a DFT algorithm for half data length.

Computation of the DFT of a Single Real Data Set 171

even and odd indexed data values and (ii) computing the two IDFTs in a single IDFT algorithm for half data length, as described in the last section. Given X(k), we have

X{k) = X(e\k) + WJt,X(°\k)

X{j + k) = XM(k)-Wkx(°\k)

Since the DFT of real data is hermitian-symmetric, the last equation can be written as

x*(^ -k) = xW(fc) - wkx(°\k)

Solving for xM(fc) and X(°)(fc), we get

*•>(*, = *""+*•<?-*> (8.2)

xMm = ( i w - r f f - w ^ t _ 0 j l N (g3)

The IDFT of C(k) = (XW(fe) + jX^(k)) is c{n) = {x^{n) + jx^(n)). The flow chart of the algorithm to compute RIDFT, for N an integral multiple of 8, is shown in Fig. 8.6. It can be modified for any even Ar.

Example 8.2 Find the DFT of the sequence x(n) = {2,3,3,1,4,1,2,2}. Compute x{n) back from the DFT values. Solution Form the complex data c(n) = x^ (n) + jy^ (n) = {2 + j3,3 + jl, 4 + jl,2 + j2}. The DFT of c{n) is C(fc) = {11 + j 7 , - 3 + j l , l + j l , - l +j3}. To separate X{k) and Y(k), we use Eq. (8.1). XW(0) = l l ,XW(l) = - 2 - j l , X W ( 2 ) = 1,XW(3) = - 2 + j l and Jf(°)(0) = 7,X<°>(1) = 2 + jl, Jf (°)(2) = 1, X(°)(3) = 2 - j l . Combining the two DFTs, we get

X{0) = 11 + 7 = 18 X(l) = ( - 2 - j l ) + - L ( l - j l ) ( 2 + j l ) = 0.12-J1.71 X(2) = l-ji X(3) = (-2 + j l ) - i ( l + j l ) ( 2 - j l ) = -4.12 + J0.29 X(4) = 1 1 - 7 = 4 X{5) = X*(3) = -4.12-j0.29 X(6) = X*(2) = 1 + j l X(7) = X*(l) = 0.12 + J1.71

http://-4.12-j0.29


( s t a r t )

ReadX(fc), fc = 0 , l , . . . , f

I X ( e ) ( 0 ) = X ( 0 ) + X ( f ) ) X ( o ) ( 0 ) = X ( 0 ) - X ( f ) ;

XW(f) = Re(X(f )),X(°)(f) = -Im(X(f))

XW(f) = +

*(*>+**(¥>, *(.)(*) _ ( * ( * ) -**(^ jwr- i

3 Compute X«(re) = *(*)+**(*-*)

x(«)(fc) = *(*)-**(*-*) jy-*

X( e ) (^ - fc) = ^(f- fe)+x ,(f+fe)

XW(f -'fc) = [x^-k\x*^+k))jW*N, re = 1 , 2 , . . . , -g 1

C*(0) = XU (0) + jX(°) (0), C( f ) =JrW(f)+ ^ W ( f )

Form C(k) = X^(k) +jX<-°\k), re = 1,2,

<?(f - Jfe) = XW*(re) + jX(°)*(re),

Compute c(n) = IDFT(C(ifc))

Form x(2n) = Re(c(n)),x{2n + 1) = Im(c(n)), n = 0 , l , . . . , f - 1

Write x(n), n = 0 , 1 , . . . , AT - 1

( Return)

Fig. 8.6 The flow chart of the algorithm to compute a RIDFT of length AT using an IDFT algorithm for half data length.

Summary 173

Note that the products ^ - ( 1 - j l ) (2 + jl) and ^ ( 1 + j l ) (2 - jl) are complex conjugates.

To compute the RIDFT, we use Eqs. (8.2) and (8.3), to get X^(k) and XM(k)fromX(k).

X(e)(0) = i M i = 1 1 X(e)(l) = (0.12-jl.71)+(-4.12-j0.29) _ _2 - j l

XW(2) = R e ( i ( f ) ) = 1 l W ( 3 ) = X<e\l) = -2 + jl X(°)(0) = ± t ! = 7 X(°)(l) = (0-H-Jl-71?-(-4.12-j0.29)^(1+jl) = g + ^

XW(2) = - I m ( X ( f ) ) = 1 XW(3) = X*(°)(l) = 2 - j l

X^(k) +jX^(k) is formed and the IDFT is computed. The result gives x^ (n) as the real part and x^ (n) as the imaginary part. For this example,

X^ (k)+jX(°\k) = {11+ J7,-3+jl,l+jl,-l+j3}

The IDFT of this sequence yields c(n) = x^ (n) + jx^ (n) = {2 + j3 ,3 + jl,4 + jl,2+j2). I

The additional operations required to compute an TV-point RDFT, apart from that of a ^--point DFT algorithm for complex data, are as follows. To separate the DFTs X^(k) and X^(k), TV — 4 real additions are required. The number of real multiplications needed is TV-6. The number of addition operations due to complex multiplications is y — 2. The number of addition operations required to combine X(e\k) and X(°)(fc) is TV — 2. Therefore, the additional operations required is TV — 6 multiplications and ^ — 8 additions.

8.4 S u m m a r y

• In this chapter, two methods were presented for the computation of the RDFT and the RIDFT.

• If the speed and storage requirements are not critical, the real data can be assumed to be complex data with zero imaginary parts and algorithms for complex data can be used directly.


• The second method involves the indirect use of algorithms for complex data. The first type of algorithm is to compute two RDFTs or RIDFTs at a time using a single algorithm for complex data for the same data length. The second type of algorithm is to compute a single RDFT or RIDFT using a single algorithm for complex data for half the data length.

• By using appropriate PM algorithms for complex data, efficient algorithms for real data when the number of samples is not an integral power of 2 can be realized.

• An alternative, which is described in the next chapter, is to deduce algorithms specifically suited for real data from the algorithms for complex data by cutting out the redundancies in each stage.

Reference

(1) Brigham, E. 0 . (1988) The Fast Fourier Transform and Its Applications, Prentice-Hall, New Jersey.

P r o g r a m m i n g Exercises

8.1 Write a program to compute two iV-point RDFTs at a time using the Appoint 2 x 1 PM DIT DFT algorithm for complex data.

8.2 Write a program to compute two AT-point RIDFTs at a time using the Appoint 2 x 1 PM DIT DFT algorithm for complex data.

8.3 Write a program to compute an Af-point RDFT using the ^-point 2 x 1 PM DIT DFT algorithm for complex data.

8.4 Write a program to compute an AT-point RIDFT using the ^-point 2 x 1 PM DIT DFT algorithm for complex data.

Chapter 9

DFT Algorithms for Real Data - II

In this chapter, we deduce DFT and IDFT algorithms, specifically intended for real-valued data, from the algorithms for complex-valued data by removing the redundant processing in each stage. These algorithms provide an alternative in computing the RDFT and RIDFT to the indirect use of algorithms for complex data described in the last chapter. In Sec. 9.1, the storage scheme for real data and its DFT is presented. In Sec. 9.2, the 2 x 1 PM DIT RDFT algorithm is described. In Sec. 9.3, the 2 x 1 PM DIF RIDFT algorithm is derived. In Sec. 9.4, the 2 x 2 PM DIT RDFT algorithm is described. In Sec. 9.5, the 2 x 2 PM DIF RIDFT algorithm is derived.

9.1 The Storage of D a t a in P M R D F T and R I D F T Algori thms

Due to the hermitian-symmetric property, the storage of the first half of the spectrum of real data is sufficient. The first half and the second half of the real input data can be stored, respectively, in the locations assigned for the real and imaginary parts of the complex vectors. (For following the derivations presented in this chapter, the trace tables shown in Figs. 8.1 and 8.2 will be helpful.) Therefore, the storage of data to compute the RDFT is given by

N N N a (n) = {a(n),a{n + —)} = {(x(n) + x(n + —),x{n) -x(n + —)),

(x(n + —) + x(n + -r-),x(n + —) - x(n + -j-))}, N

n = 0 , ! , . . . , - - ! (9.1)

175

176 DFT Algorithms for Real Data - II

A'(0) = {A(OMo(£)} = { ( X ( 0 ) , X ( f ) ) , X ( f ) } r q _-A'(k) = A(k) = {X(k),X(k+%)}, fc = l , 2 , . . . , £ - l {y-Z)

The storage of data to compute the RIDFT is given by

B'(0) = { B ( 0 ) , B ( f ) } = { ( X ( 0 ) + X ( f ),X(0)-X(f ) ) , (2Re(X(f )) ,2Im(X(£)))} (9.3)

B'(fc) = B(fc) = {(X(k) + X(k + f ) ) , (X(k) - X(k + f ) ) } ,

where k = 1,2, . . . , f - 1.

b'(n) = {b(n),b(n + ^ ) }

iV AT 37V = {(x(n),x(n + -)), (x(n + -),x{n + ~^))}, (9.4)

where n = 0 , 1 , . . . , ^ — 1.

9.2 The 2 x 1 P M DIT R D F T Algorithm

Algorithm that is suited to real data is obtained by just eliminating half the number of butterflies in each group of butterflies of every stage of an algorithm for complex data along with the necessary changes in the storage of the data values. The combining of the DFTs of two sets of vectors, each of ^ vectors, to produce the DFT of a set of y vectors in the last stage of the 2 x 1 PM DIT DFT algorithm is given as

Ap(k) = 4e)(fc) + (-ir<40 )(fe) (9.5)

Ap(k + ^) = A[e\k) + (-iyW^A[°\k) (9.6)

The index k runs only from 1 to y — 1, since we need only half of the DFT values and p = 0,1 (The computation of the output vectors with indices 0, Y, and ^ is a special case that will be explained later.).

In order to obtain the first half output vectors as defined in Eq. (9.2), the second quarter vectors are derived from the third quarter vectors using the conjugate symmetry property as

N N 3N N N N N

-£) = {X(k+j),x(k+^)} = {X(^-C--k)),x(-H--

The 2 x 1 PM DIT RDFT Algorithm 177

A'W(l) = J ^^C A'(r+1)(/1) = {A'Wi),4r\i)} o ^ > > *o{4r+1\ii),4r+1\ii)}

Fig. 9.1 The SFG of the butterfly of the 2 x 1 PM DIT RDFT algorithm, where 1 < s < •^. Twiddle factors are represented only by their exponents. The symbol s represents

TV N SN N N N N Ap(^-k) = {X(^-k),X(^-k)} = {XQ-Q+k)),X(-+(--k))}

Therefore, Eqs. (9.5) and (9.6), for the vectors defined in Eq. (9.2), can be expressed as

A'p(k) = A'(e\k) + (-iyW*4°Hk) (9.7)

A'p(^-k) = {4e\k)-(-l)*W?*40Hk)}*, (9.8)

where k = 1 , . . . , j - — 1 and p = 0,1.

The 2 x 1 PM DIT RDFT butterfly

In general, the relation governing the basic computation at the r th stage can be obtained from Eqs. (9.7) and (9.8) as

4r+1\h) = 4r\h) + w^4r\i) 4r+l\h) = 4r\h)-w^4r\i) 4r+1\n) = {4r\h)-w^4r\Dr 4r+l\n) = (4r\h) + ws

N+*4r\i))*,

where s is an integer whose value depends on the stage of the computation r and the index h. These equations characterize the input-output relation of the 2 x 1 PM DIT RDFT butterfly, shown in Fig. 9.1. There are three differences at the lower node between this butterfly and the butterfly of the corresponding algorithm for complex data: (i) the result of the subtract operation is stored as the first element in the output vector, (ii) the elements of the output vector are conjugated, and (iii) the storage locations of the input and output vectors are different.



There are m stages specified as r = 1,2,. . . , m, where m = log2 y-. A stage consists of Y butterflies. The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given, for the last m — 2 stages, as follows (The butterflies of the first two stages and the first butterfly of each group of butterflies of the other stages are special cases and they will be explained later.).

h = imod2r'2, i = 0 ,1 , . . . , f - 1 s = 2m~rh

11 = 2r~1 - h I = h + T-1

(9.10)

The computation and storing of A' (^ — k) in the last stage will erase some of the input values for the computation of the butterfly indexed ^- — k. Therefore, two butterflies with indices A; and y — k must be computed at the same time in the algorithm implementation. The input values for the computation of the butterfly indexed y — k can be temporarily stored in the processor registers. A similar procedure is followed in implementing the other stages.

For example, with N = 32, there are 4 stages specified as r = 1,2,3, and 4. Four butterflies make up a stage. Indices h, I and 11, and the twiddle factor exponent s for each group of butterflies are given, respectively, by (h = i mod 2 r - 2 , {i = 0,1,2,3)), I = h+2r~2,11 = 2r~1-h, and s = 2l'rh. Fig. 9.2 shows the SFG of the 2 x 1PM DIT RDFT algorithm, with N = 32.

Special butterflies

Unlike in the algorithms for complex data, the data values at one end of an algorithm for real data is real and it is mostly complex at the other end. This lack of symmetry of the data forces the use of special butterflies, the / - and g-butterflies, those are specific for the algorithms of real data. These butterflies process data values that may be pure real, pure imaginary, or complex. The input/output relationship of an /-butterfly is as follows. Let


stage 1 s tage 2 s tage 3 s tage 4

Fig. 9.2 The SFG of the 2 x 1 PM DIT RDFT algorithm, with N = 32. Twiddle factors are represented only by their exponents. For example, the number 4 represents W£2.

A'W{n) = {(Ar$0){n), Atd(0)(n)), (Ar'^in), A< (0 )(n))}. Then,

Ar'0{1\n) = Ar; (0)(n) + Ari (0)(n)

Ai'^in) = Ar'(0)(n)-Ar'{0)(n)

Ari(1) (n) + j A»i(1) (n) = AZQ(0) (n) - j Aif] (n)

This butterfly requires 2 operations of real addition. In the implementation, the vector formation, its scrambling, and the computation of the first stage are all carried out at the same time to reduce data transfers between the memory and the processor. An /-butterfly carries out the same processing, but with reduced computation, that is carried out by a butterfly in the first stage of the corresponding algorithm for complex data.

The input/output relationship of a 3-butterny is as follows. Let the first element of the vectors A,{-T\h) and A'W(Z) be {Ar'0

(r) (h), Ai'0(r) (h))


and (Ar'0ir)(l),A4r\l)), respectively. Then,

A4rHh)+ArSrHl) Ar$r){h)-Ar$r){l) Atf\h)-jAi$r\l) 4r\h) + Wi4r\l) A'±r\h)-WZA'{r\l)

This butterfly requires 2 operations of real multiplication and 8 operations of real addition. The g-butterfly carries out the same processing, but with reduced computation, that is carried out in the first and the middle butterflies of each group of butterflies, from the second stage onwards, in the corresponding algorithm for complex data.

The software implementation of the PM RDFT algorithms is similar to that of the PM DIT DFT algorithms for complex data with the differences as explained above. A faster version can be obtained by implementing two adjacent stages at a time. The number of real multiplications required by an RDFT algorithm is one-half and that of additions is N — 2 fewer than one-half of that of the corresponding algorithm for complex data.

Example 9.1 A trace table of the algorithm, shown in Fig. 9.2, is shown in Fig. 9.3. The first column shows how the 32 real data values are read into the storage locations of 8 vectors each consisting of two complex elements. When a value is pure real, it is shown separately with the use of a vertical line. The second column shows the values after vectors have been formed and swapped. The third, fourth, fifth, and sixth columns show the values, respectively, after the first, second, third, and fourth stage operations of the algorithm are carried out. Compare this table with that given in Fig. 8.1 to find out how redundancy at each stage is eliminated. I

9.3 The 2 x 1 P M DIF RIDFT Algorithm

While the DFT of real data is hermitian-symmetric, the data itself is usually arbitrary. Therefore, it is relatively easier to eliminate the redundancies starting from the DFT side of the SFG of the algorithm of complex data. That is the reason we used the DIT approach to derive the RDFT algorithm and we are going to use the DIF approach for deriving the RIDFT algorithm.

Ar^+1\h) =

Ar'{r+1)(h)+jAi'}r+1)(h) =

4r+1\D = 4r+1\i) =

The 2 x 1 PM DIF RIDFT Algorithm

Input

x(2)=2.00 g(10)=4.00

a:(0)=1.00 x(8)=3.00

z(16)=9.00 s(24)=7.00

x(l)=8.00 g(9)=1.00

ar(17)=9.00 a;(25)=4.00 *(18)=0.00 s(26)=9.00

z(3)=5.00 z(ll)=4.00 x(4)=9.00 x(12)=0.00 z(5)=4.00 E ( 1 3 ) = 0 . 0 0

a;(6)=5.00 g(14)=1.00 x(7)=7.00 j(15)=7.00

z(19)=8.00 s(27)=1.00 a;(20)=4.00 z(28)=6.00 z(21)=2.00 z(29)=6.00 z(22)=8.00 ar(30)=1.00 x(23)=4.00 g(31)=0.00

Vectors Swapped

Stage 1 Output

10.00 10.00

13.00 6.00

2.00 13.00

13.00 2.00

17.00 5.00

6.00 6.00

13.00 5.00

11.00 7.00

-8 .00 -4 .00

5.00 -6 .00

2.00 -5 .00

-3 .00 0.00

-1 .00 -3 .00

2.00 -6 .00

-3 .00 3.00

3.00 7.00

20.00 0.00 -8.00 + J4.00

19.00 7.00 5.00 + j'6.00

15 .00-11 .0 2.00 + J5.00

15.00 11.00 -3.00 - jO.OO

22.00 12.00 -l.OO+j'3.00

12.00 0.00 2.00+J6.00

18.00 8.00 -3.00 - J3.00

18.00 3.00-

4.00 J7.00

Stage 2 Output

39.00 0.00-

1.00 j'7.00

-0.22 + J4.71 -15.78+ J3.29

30.00 0.00 - 1 1 . 0 - j l l . O

-0.12 + J7.12 4.12 + J2.88 34.00

12.00 -10.00 jO.OO

4.66 + J5.83 -6.66 + J0.17

36.00 0.00 8.00 - J4.00

-5 .83- j l0 .07 -0.17 + J4.07

-15.56 - J7 .00 15.56 - J7 .00

-16 .86-J8 .20 -14.70 + j 1.62

Stage 3 Output

69.00 9.00 1.00-jO.OO

2.39 + J11.33 - 2 . 8 3 - j l . 9 2

70.00 10.00 -

-2.00 jO.OO

- 4 . 5 8 - j l . 2 5 13.90+ J12.90

14.83 - J8.49 9.17 + J8.49

-10.48 -j 1.57 -2.83 + j 1.23

Stage 4 Output

X(0)= X(8)=

139| X(16)= 9.00 + j2.00

X(l)= *(17)

-2.35 + j l l .Ol = 7.13 + j l l . 66

X{2)=-

*(18)=

-5 .10- ;20 .51 -26 .01+ J6.51

X(3)=

-y(i9)=

-26.45 - J3.68 = -7 .27 - J12 .72

•Y(20) 8.07 - J7.07

6.07 + J7.07 X(5)=

*(21)=

-17 .29+ J0.05 = -12.10 - J 3 . 2 9

X{6)= Jf(22)=

11.23 - J4.72 = 19.89 + j 18.72

X(7)=-A" (23)=

-12.78 - j'14.23 7.11 + J18.06

Fig. 9.3 The trace of the 2 x 1 PM DIT RDFT algorithm, with N =

http://-5.83-jl0.07

http://-2.83-jl.92

http://-4.58-jl.25

182 DFT Algorithms for Real Data - IT

To derive the 2 x 1 PM DIF RIDFT algorithm, consider the decomposition of an y-vector IDFT into two ^-vector IDFTs in the first stage of the 2 x 1 PM DIF IDFT algorithm.

M2n) = i E (-Vpk{Bo(k) + (-l)n+p%B0(k + ^)}W2nk (9.H) k=0

bp(2n + l) = ^^{-l^W^iBiW fc=0

+ (-l)"+^{j)Bi{k+^)}Wsnk, (9-12) 4 2

where n = 0 , 1 , . . . , ^ — 1 andp = 0,1. The input vectors, B'(k), as defined by Eq. (9.3) are the first half vectors. The expressions inside the pair of braces in Eqs. (9.11) and (9.12) require the first and the third quarter input vectors to generate the first half input vectors (since the other half is redundant) of the two independent IDFTs, each of size ^ vectors. Thus, we have to derive the third quarter input vectors from the second quarter input vectors. The third quarter input vectors can be written as

N TV 1/V N 3/V [j + k) = {X(k+I-) + X(k+^),X(k+1)-X(k + —

N N N N

= {x(^-C-i-k)) + x(- + (- + k)), 2 v 4 " v 2 v 4 N N N N

* ( y - ( j - *)) " X{- + (-+ k))}, (9.13)

where q = 0,1. The second quarter input vectors can be expressed as

N N 37V N 3N B9Q-k) = {X(j-k) + X(^--k),X(J-k)-X(—-k)}

N N N N = {X(j-(j + k)) + X(- + (--k)),

N N N N X(j - (j + *)) - X{- + (-- k))} (9.14)

Thus, from Eqs. (9.13) and (9.14), the relation between the third quarter and the second quarter input vectors is given by

N N N * 'Jl+k) = {B0C--k),-B1(j-^* Bq{~ +k) = {B0(- - k), - B i ( T - *)} • (9-15)

The 2 x 1 PM DIF RIDFT Algorithm 183

{B'0^(h),B'lr\h)} o . ^ > ° { f l ^ W . - B ^ W }

- * - T Fig. 9.4 The SFG of the butterfly of the 2 x 1 PM DIF RIDFT algorithm, where

1 < s < 4r. Twiddle factors are represented only by their exponents. The symbol - s

represents W^'.

Therefore, Eqs. (9.11) and (9.12) can be expressed as

bpi2n) = jj E ("1)Pfc{^o(fc) + ( - l ) " + ^ ( 5 0 ( f - fc))*}WV**(9.16) *=o

&p(2n + l) = i £ ( - l ) * * W t f { i ? i ( * ) fc=0

- ( - l ) " + ^ ( j ) ( B i ( ^ - k))*}W£nk, (9.17) 4 2

where n = 0 , l , . . . , ^ — 1 and p = 0,1. The combination of the inputs inside the pair of braces with indices k = 0, k = y , and k = ^ is a special case that will be explained later.

The 2 x 1 PM DIF RIDFT butterfly

In general, the relation governing the basic computation, for the vectors defined in Eq. (9.3), at the rth stage can be obtained from Eqs. (9.16) and (9.17) as

B$r+1\h) = B'U{h) + {B$r\ll))* B'}r+1\h) = B$'Hh) - (B$r\ll))* B$r+1\t) = W^B'}r\h)-W-8-%(B'Sr\ll))*

B'lr+1)(l) = WJB'lr\h) + W--*{B$r)(ll))*,

(9.18)

where s is an integer whose value depends on the stage of the computation r and the index h. These equations characterize the input-output relation of the 2 x 1 PM DIF RIDFT butterfly, shown in Fig. 9.4.



There are m stages specified as r = 1,2,. . . , m, where m = log2 Y . A stage consists of Y butterflies. The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given, for the first m — 2 stages, (The butterflies of the last two stages and the first butterfly of each group of butterflies of the other stages are special cases and they will be explained later.) as follows.

h = i m o d 2 r o - r - 1 , i = 0 , 1 , . . . , f - 1

S = T~lh (9 19)

I = h + 2m-r-1

For example, with N = 32, there are 4 stages specified as r = 1,2,3, and 4. Four butterflies make up a stage. Indices h, I, and 11, and the twiddle factor exponent s of each group of butterflies are given, respectively, by h = i mod 2 3 - r (i = 0,1,2,3), / = h + 23~r, 11 = 24~r - h, and s = 2r~1h. The SFG of the 2 x 1 PM DIF RIDFT algorithm, with N = 32, is shown in Fig. 9.5.

Special butterflies

The input/output relationship of a p-butterfly is as follows. LetB'W(/i) = {(Br$rHh),B4r\h)),(Br'{r)(h),Bi'ir\h))} and B'W(/) = {Br$r\l) + jB4r)(1),Br[{r)(I) + jBifr)(I)}. Then,

Br'0{r+1\h) = Br'0

ir)(h) + Br'!r\h)

Bi'0ir+1\h) = B4r)(h)-Br'lr\h)

Br'ir+l)(h) = B4r)(l)+Br'0{r\l)

Bi'}r+1\h) = Bi$r\l) + Bi'tr\l)

Br'0{r+1)(l) = Bi'Q

ir\h)-Bi'lr\h)

Bi$r+1\l) = B4r)(h)+Bi'ir\h)

Br'}r+1Hl) = V2(Br'}r)(l)-Bi'}r\l))

Bi'}r+1)(l) = V2(Br'lr)(l) + Bi'lr\l))

Note that Bi'£r+1\h) and Bi'{r+1'(l) are pure imaginary and the rest of the values are pure real. The p-butterfly requires 2 operations of real multiplication and 8 operations of real addition. It should be noted that all


stage 1 stage 2 s tage 3 s tage 4

Fig. 9.5 The SFG of the 2 x 1 PM DIF RIDFT algorithm, with N = 32. Twiddle factors are represented only by their exponents. For example, the number —4 represents

the operations of this butterfly, except 4 additions, can be merged in the twiddle factor computation of the preceding stage for about half the number of g—butterflies. The p-butterfly carries out the same processing, but with reduced computation, that is carried out in the first and the middle butterflies of each group of butterflies, up to the last but one stage, in the corresponding DIF IDFT algorithm.

The input/output relationship of an /-butterfly is as follows. Let B ' ^ - 1 ) ^ ) = {(Br(, (m_1)(n) lBi|fTO-1)(n)), (Bri ( m-1 )(n),B* ,

1( r B _ 1 )(n))}.

Then,

Br'0(m)(n) = flr(,(,n-1)(n) + Br'1

( m-1 )(n)

B # m ) ( n ) = B r ( , ( m - 1 ) ( n ) - B r i ( m - 1 ) ( n ) Br'im\n) = B»2 m - 1 >(n) -£ t ' 1

( m - 1 ) (n)

Bi'}m\n) = S ^ m - 1 ) ( n ) + B < ( m - 1 ) ( n )

The /-butterfly requires 4 operations of real addition and the computation


Vectors Stage 1 Stage 2 Stage 3 Stage 4 Output Output Output Swapped

Output

138.00 18.00

140.00 J4.00

4.78 + J22.67 -9.48 - jO.65

-31.11- j '14 .0 20.91 -J27.02 - 3 3 . 7 2 - J 1 6 . 4 -19.18 +J9.04

2 + j O 14.14-j'14.14 -29.39 - j'3.24 -5 .19+J3 .34

31.12+J14.00 -8.66 - j"23.44 -5.67 + J3.83

-19.89 -J32.29

156.00 120.00 4.00 jO.OO

-0.89 + J18.84 10.45 + j'26.50 0.01 - j/28.00

-62.23 - jO.OO -63.11 -j'13.16 -4.33 - j/19.64

136.00 39.99

144.00 jO.OO

18.62+J23.32 -36.96 -J28.29

48.00 +J0.01 11.32-J33.93 -26.63 - jO.68 -15.31-j"5.60

160.0C 0.02

152.00 -J56.00

-64.00 +J32.00 62.22+J5.68

120.00 -88.01

120.00 -j'88.01

16.00+J40.00 -16.97 + J16.97

175.99 96.0C

96.01 J0.02

-8.01+^24.00 45.25+J22.64 144.0C 63.99

144.00 -J31.98

-24.00 - j'24.00 -22 .63-J56 .57

160.02 -128.0

208.00 79.96

31.99 32.00

208.0: -48.0C

271.99 -16.05

95.9£ 31.97

207.99 -48.0:

159.98 j'64.00

96.00 J96.03

208.01 J80.00

31.99 -jO.01

79.99 J48.0

96.03 j96.01

80.01 -J48.0

175.98 112.02 48.0C-jll2.01

32.02 95.98

255.97 31.99

64.00 128.01

159.98 128.00

287.96 -0.03

127.96 0.02

160.01 32.01

223.98 224.03

288.02 223.98

288.02 127.99

- 0 . 0 1 288.00

256.00 32.01

128.04 192.03

64.01 192.04

256.01 31.98

127.99 0.00

Fig. 9.6 The trace of the 2 x 1 PM DIF RIDFT algorithm, with N = 32.

is confined to the values of a single vector. The processing of this stage, unscrambling of the output vectors, and dividing the output values by N can be carried out at the same time to reduce the data transfer operations. An /-butterfly carries out the same processing, but with reduced computation, that is carried out by a butterfly of the last stage in the corresponding DIF IDFT algorithm.

Example 9.2 A trace table of the algorithm, shown in Fig. 9.5, is shown in Fig. 9.6. Half of the DFT values are read into the storage locations of 8 vectors each consisting of two complex elements as shown in the last column of Fig. 9.3. The first column shows the values after vectors have been formed. When a value is pure real or pure imaginary, it is shown separately with a vertical line. The second, third, fourth, and fifth, columns show the values, respectively, after the first, second, third, and fourth stage operations of the algorithm are carried out. Note that the swapping operation is carried out along with the processing of stage 4. The output values have to be divided by N = 32. Since the input data has been given only to a precision of two digits, the output is not exact as expected. For example, a;(0) = 32.02 instead of 32 in the last stage. Compare this table with that given in Fig. 8.2 to find out how the redundancy at each stage is eliminated.

http://48.0C-jll2.01


9.4 The 2 x 2 P M DIT R D F T Algorithm

By merging the multiplication operations of adjacent stages of the 2 x 1 PM DIT RDFT algorithm, we obtain the 2 x 2 PM DIT RDFT algorithm.

The 2 x 2 PM DIT RDFT butterfly

In general, the relation governing the basic computation at the r th stage is given as

A'<r+1\hO) -

A[{r+1HhO) --

4r+1\n) -. 4r+1\ll) = <{r+1)(h2) -

A'}r+1\h2) -.

4r+1)(/3) = A'}r+l\l3) --

4r+2\m) --A'}r+2\hO) --

4r+2\l3) --4r+2)(l3) --4r+2\ii) = A'}r+2\ll) =

4r+2\h2) = A[(r+2\h2) =

= 4r\hO) + WfrA$rHhl) -- 4r\ho)-w^4r\hi) = (4r\ho)-w2

N8+^4r\hi))*

= (A'1{r)(hO) + W2

Na+^A'}r\hl))*

= Wfr4r\h2) + W3N°4r\h3)

-- Wfr4r\h2)-W*f4r){h3) = (W8

N-%A'}r\h2)-W3N

s+%A'tr\h3))*

-- {W8N-*A'±r\h2) + W3

Na+^A[{r)(h3))*

-- 4r+1)(h0)+4r+1)(h2) = 4r+1)(h0)-4r+1\h2) = (4r+1\hO)-Wj4r+1\h2))* -- (4r+1) (hO) + W$ 4r+1) (h2))* = 4r+1\ii)+4r+1\i3) = 4r+1\n)-4r+l\i3) -- (4r+1\ii)-wl$4r+1)(i3))* = {4r+1\n) + w$4r+1\i3))*,

where s is an integer whose value depends on the stage of the computation r and the index hO. These equations characterize the input-output relation of the 2 x 2 PM DIT RDFT butterfly, shown in Fig. 9.7.


There are m stages specified as r = 1,2,. . . , m, where m = log2 y . Starting from the output side, pairs of adjacent stages are formed using ^ 2 x 2 PM DIT RDFT butterflies. If m is even, f 2 x 1 PM DIT RDFT g-


© A'(r+V(l3)

A'(r+2\h2)

Fig. 9.7 The SFG of the butterfly of the 2 x 2 PM DIT RDFT algorithm, where 1 < s < j ^ . Twiddle factors are represented only by their exponents. The symbol s represents W*.

butterflies are used to form the second stage (the first stage always consists of /-butterflies). The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given, for the last m — 3 stages for an odd m and for the last m — 2 stages for an even m, as follows (The butterflies of the first three stages for an odd m and the first two stages for an even m, and the first butterfly of each group of butterflies of the other stages are special cases and they will be explained later.).

ftO s

hi hi h3 13 11

= = = = = = =

i m o d 2 r - 3 , » = 0 , 1 , . . . , ^ 2m~rh0 hQ+l(2r'3) hO + 2(2 r-3) hO + 3(2 r-3) 2 ' - 1 - hO 13 - 2(2 r-3) ,

(9.21)

where r is the stage number of the second of the two adjacent stages combined.

To illustrate the SFG of the algorithm, consider a specific case with N = 32. There are 4 stages specified as r = 1,2,3, and 4. Two butterflies make up two adjacent stages. The indices of the nodes of the butterfly and the twiddle factor exponent s can be readily computed from Eq. (9.21) for the specific value m = 4. Fig. 9.8 shows the SFG of the 2 x 2 PM DIT RDFT algorithm, with N = 32.


stage 1 stage 2

o'(0) o

a'(4)o

a'(2) o - ^ >

a'(6) o - - U

a ' ( l ) o •* » o

a'(5)c

o'(3)o

a'(7)o

stage 4

g > * *oA'(0)

•-oA'(l)

Fig. 9.8 The SFG of the 2 x 2 PM DIT RDFT algorithm, with N = 32. Twiddle factors are represented only by their exponents. For example, the number 4 represents W£2.

Special butterflies

The /-butterfly used in processing the first stage is identical to that in the 2 x 1 PM DIT RDFT algorithm. However, one 5-butterfly, out of a group of three ^-butterflies and one regular butterfly, is different from that of the 2 x 1 PM DIT RDFT algorithm. The difference in this 5-butterfly is due to the merging of the twiddle factors of the regular butterfly with its twiddle factor W£ . This results in the multiplication of the inputs to the lower

output node of this butterfly by Wtf and W^6, respectively. The output vector at the lower node of this type of ^-butterfly is computed as follows.

A'(r+1) ^ 0

A'lr+1\l) = Wl,A'{r\h)-W*%A^\l)


This special 2 X 2 PM DIT RDFT butterfly requires 12 operations of real multiplication and 34 operations of real addition.

9.5 The 2 x 2 P M DIF RIDFT Algorithm

By merging the multiplication operations of the adjacent stages of the 2 x 1 PM DIF RIDFT algorithm, we obtain the 2 x 2 PM DIF RIDFT algorithm.

The 2 x 2 PM DIF RIDFT butterfly

In general, the relation governing the basic computation at the r th stage is given as

j ' (r+l)

?'(r+l) B'^'ihO)

B'^'ihO) =

B$r+1\h2) =

B'}r+1\h2) =

B'Q{r+1)(ll) =

B?r+1\ll) =

B'0(r+1)(l3) =

B?r+1)(l3) = j'(r+2) (

J3°(p+2)(W))

<+2)(w) j'(H-2)

B'^'ihO)

B'o B'o B[

Bi

B'0

B'o

B[ B[ B'„

(hO) + (B'^(l3))*

(W))- ( i# p ) ( J3) )*

(hO)-W-^(B'1ir\l3))*

(hO) + W~f(B'1{r\l3))*

(ll) + (B$r\h2))*

(U)-(itfp)(/i2))* (ll)-W-*(B?r\h2))*

(Zl) + ^ ^ " ( B ; W ( / i 2 ) ) *

r+l) (hO) + (B'0(r+1\ll))*

= B'Jr+1)(hO)-(B'0{r+1\ll)y

- 2 8 - * V(r+1) , W^'B[(r+1){hO) - W-Z8~(B[(r+l,(ll)) -2s D'(r+1) B[(r+i>(hl) = W^'B'i (ho) + wN

2s-* (B[(r+1,(ll)Y

B'0{r+2\h2)

B '(r+2) (h2) =

W^B$r+1)(h2) + W^+%(B'0{r+1)(l3))

W^B^>(h2) w~8+^(BTL,m) B'0

{r+2\h3) = W^sB'}r+1)(h2)-W-38-^(B'}r+1)(l3))* -3s-*

B[{r+2>(h3) = Wj3'B?r+1\h2) + W-''--*{B'l(r+1>(l3))*, D'(r+1),

(9.22)

where s is an integer whose value depends on the stage of the computation r and the index hO. These equations characterize the input-output relation of the 2 x 2 PM DIF RIDFT butterfly, shown in Fig. 9.9.


B'W(ft2)o

^ M _ o B ^ ) ( / l 0 )

J * B'(r+2)(W)

B'^+2)(/i2)

Fig. 9.9 The SFG of the butterfly of the 2 x 2 PM DIF RIDFT algorithm, where 1 < s < jg. Twiddle factors are represented only by their exponents. The symbol —s represents W^3 •


There are m stages specified as r = 1,2,. . . , m, where m = log2 y . Starting from the input side, two adjacent stages are formed using ^ 2 x 2 PM DIF RIDFT butterflies. If m is even, f 2 x 1 PM DIF RIDFT ^-butterflies are used to form the last but one stage (the last stage always consists of f-butterflies). The expressions for the twiddle factor exponent and the indices of the nodes of each group of butterflies are given, for the first m — 3 stages for an odd m and for the first m — 2 stages for an even m, as follows (The butterflies of the last three stages for an odd m and for the last two stages for an even m, and the first butterfly of each group of butterflies of the other stages are special cases and they will be explained later.).

= 0 1 2- - 1 ftO s

h\ Kl hZ IZ 11

= i mod 2 m - r - 2 , = 2r-1M) = M + l ( 2 m - r - 2 ) = h0 + 2(2m" r-2) = hO + 3 (2 m - r - 2 ) = 2 m ~ r - hO = 13- 2 (2 m - ' - 2 ) ,

(9.23)

where r is the stage number of the first of the two adjacent stages combined. To illustrate the SFG of the algorithm, consider a specific case with

N = 32. There are 4 stages specified as r = 1,2,3, and 4. Two butterflies make up two adjacent stages. The indices of the nodes of the butterfly and

192 DFT Algorithms for Real Data - / /

stage 1 stage 2 stage 3 stage 4

Fig. 9.10 The SFG of the 2 x 2 PM DIF RIDFT algorithm, with JV = 32. Twiddle factors are represented only by their exponents. For example, the number —4 represents

the twiddle factor exponent s can be computed readily from Eq. (9.23) for the specific value m = 4. Figure 9.10 shows the SFG of the 2 x 2 PM DIF RIDFT algorithm, with JV = 32.

Special butterflies

The /-butterfly used in processing the last stage is identical to that of the 2 x 1 PM DIF RIDFT algorithm. However, one ^-butterfly, out of a group of three (/-butterflies and one regular butterfly, is different from that of the 2 x 1 PM DIF RIDFT algorithm. The difference in this (/-butterfly is due to the merging of the twiddle factor of the regular butterfly with its twiddle factor WN

8 . This results in the multiplication of the inputs to the lower — M. 31V

input node of this butterfly by WN16 and WN

ie , respectively. The output

Summary and Discussion 193

vectors of this type of y-butterfly is computed as follows. Let

tr + jti = 2Wrt1{B4r)(l)+jB4r\l))

sr+jsi = > /2Wtf (Br ; ( r ) ( / )+ j£» i ( r ) (0 )

Then,

Br't+l)(h) Bi'^W

Br[ir+1){h) Bi[{r+1\h) Br'f+1\l)

B4r+1)d) Br[ir+1\l) Bi'{r+1\l)

=

=

=

=

=

=

=

=

Br$r){h) + Br'fr\h) Br$r){h)-Br'tr)(h) tr ti Bif\h)-Bi'lr\h) Bi'(T)(h) + Bi'}r\h) sr — si sr + si

This special 2 x 2 PM DIF RIDFT butterfly requires 12 operations of real multiplication and 34 operations of real addition. This operation count can further be reduced by merging some of the operations of this butterfly with that of the previous stage.


• In this chapter, algorithms specifically meant for real data were deduced from the algorithms for complex data. As in the case of the PM algorithms for complex data, efficient PM algorithms for real data when the number of samples is not an integral power of 2 can be realized.

• To make a choice between the indirect use of complex algorithms described in the last chapter and the algorithms specifically designed for real data, we have to compare the characteristics of the algorithms. The algorithms specifically designed for real data are not as regular as that of the algorithms for complex data, but they are slightly more efficient in terms of the number of arithmetic operations. The algorithms for complex data are very regular but their indirect use requires additional processing. The additional processing complexity is 0(N) compared with the total complexity 0(N log2 N) of the algorithm. As N becomes large, the additional


processing tends to become a smaller proportion of the total complexity.

• Between algorithms of the same order of complexity, in general, the algorithm with better regularity should be preferred even if it requires slightly more number of arithmetic operations as irregularity also increases execution time in terms of overhead operations. Therefore, for large N, the indirect use of algorithms for complex data is preferred. For small N, PM RDFT and PM RIDFT algorithms are preferred, since the arithmetic advantage is significant and the irregularity may not be a problem as the algorithm can be implemented as one block in hardware or software. While the user has to make the final choice for a specific application, we recommend the PM RDFT and PM RIDFT algorithms for N less than or equal to 16 and the indirect use of algorithms for complex data for N greater than or equal to 32.

References


(2) Sundararajan, D., Ahmad, M. O. and Swamy, M. N. S. (1997) "Fast Computation of the discrete Fourier Transform of Real Data", IEEE Trans. Signal Processing, vol. 45, No. 8, pp. 2010-2022.


9.1 Write a 3-butterfly program to implement the 2 x 1 PM DIT RDFT algorithm.

9.2 Write a 3-butterfly program to implement the 2 x 1 PM DIF RIDFT algorithm.

9.3 Write a 3-butterfly program to implement the 2 x 2 PM DIT RDFT algorithm.

9.4 Write a 3-butterfly program to implement the 2 x 2 PM DIF RIDFT algorithm.

Chapter 10

Two-Dimensional Discrete Fourier Transform

The theory of 2-D signals, for the most part, is a straightforward extension of the theory of 1-D signals. In this chapter, we refer to the spatial time-domain 2-D discrete signal as image. In Sec. 10.1, the definitions of 2-D DFT and IDFT are given. In Sec. 10.2, the physical interpretation of 2-D DFT is presented. The DFT of some simple 2-D signals are derived analytically. In Sec. 10.3, the row-column approach of computing the 2-D DFT is described. The properties and theorems of 2-D DFT are presented in Sec. 10.4. The 2-D PM DFT algorithms are developed in Sec. 10.5.

10.1 The 2-D D F T and IDFT

The 2-D DFT of an iV x JV image (for simplicity, we assume that, unless otherwise stated, the dimensions are the same in the two directions) {x(ni, n2) , n i , n2 = 0 , 1 , . . . , N — 1} is defined as

J V - l N-l

X(kuk2)=Y, E x ( n i ' r i 2 W * 1 ^ 2 ^ i ^ 2 = 0 , l , - , i V - l (10.1) n i=0 ri2=0

The 2-D IDFT is defined as

.. N-l N-l

*(m,na) = j^Y, J2X^k^WNnikxWNn2k2^ ki =0ifc2=0

n i , n 2 = 0 , 1 , . . . , 7 V - 1 (10.2)

195

196 Two-Dimensional Discrete Fourier Transform

Center-zero format of the 2-D DFT and IDFT

The 2-D DFT, in the center-zero format with N even, is defined as

X(kuk2) = £ E <riun2)W^W^k\ « ! = - « - 712

N N -, N ,

The 2-D IDFT, in the center-zero format with N even, is defined as

*(m,n2) = 4 E E ^i-M^"1*1^"2*2, «1——"2" «2 ——"2"

N N , N n

n 1 ,n 2 = - T , - y + l , . . . , y - l

Getting one format from the other involves the swapping of the quadrants of the image or the spectrum.

10.2 D F T Representation of Some 2-D Signals

The physical interpretation of the 1-D DFT representation of a signal is that the signal, which is a curve, is a linear combination of a set of sinusoidal curves of various frequencies, phase shifts, and amplitudes. The physical interpretation of the 2-D DFT representation of a signal is that the signal, which is a surface, is a linear combination of a set of sinusoidal surfaces of various frequencies, phase shifts, amplitudes, and directions. Given an image, therefore, the problem of Fourier analysis is the determination of the coefficients of its constituent sinusoidal surfaces. The problem of Fourier synthesis is, given the coefficients of a set of sinusoidal surfaces, the building of the corresponding image. A sinusoidal surface is a stack of shifted sinusoidal waveforms of the same amplitude and frequency. Assuming N is even, the constituent sinusoidal surfaces of a real image is obtained from Eq. (10.2).

x(nun2) = j^{X(0,0) + X{-, 0) cos{-Kni) + X(0, - ) cos(7m2)

+ X{-^, y ) cos(7r(ni + n2))

DFT Representation of Some 2-D Signals 197

* _ 1 27T + 2 J2 (l*(*i.O)| cos(— mfci + Z(X(fc!,0)))

2 27T

+ 2 ^ (|X(0,fc2)| cos( — n2fc2 + Z(X(0, fe))) fe2=i N

y N 2ir N N + 2 ^ ( | X ( Y , f c 2 ) | c o s ( - ( n 1 - + n2fc2) + Z(X(-,fe2)))

fc2=i

+ 2 ^ ^( |X(fc1 , fc2) |cos(-^(n1fc1+n2fe2) + Z(X(fc1,A;2)))}, fc1=ifc2=i

m , n 2 = 0 , l , . . . , i V - l (10.3)

Note that X ( 0 , 0 ) , X ( f ,0),X(0, f ) , X ( f , f ) are real for real images. Therefore, an N x N real image, where TV is even, consists of ^ - + 2 sinusoidal surfaces. To avoid the aliasing effect, the indices, fci and fc2, of the constituent frequency components of a 2-D signal must be less than y . Note that in Eq. (10.3), we have used the coefficients of the upper half of the spectrum but the left half also will do.

The DFT of x(ni ,n 2 ) = <5(rii,n2) is given by

o o X(k1,k2)= ^2 53 X = 1 a n d S(ni,n2)<$l

m=0n2=0

Since the impulse signal is zero except with n\ = 0 and n2 = 0, for all &i,fc2, the DFT coefficient is unity.

Example 10.1 Identify the sinusoidal surfaces that constitute the 4 x 4 impulse signal. Solution Figures 10.1(a) and (b) show, respectively, the 2-D impulse signal and its spectrum. (In the representation of the image matrix, we assume, unless otherwise stated, that the origin is at the upper left-hand corner. The index ni increases downward and the index n2 increases to the right.) All the frequency components exist and have equal amplitude and zero phase. In


X(h 1 0 0 0 0 0 0 0

7*2)

0 0 0 0 0 0 0 0

(c)

(g)

(k)

0 0 0 0

0 0 0 1

0 0 0 0

0 1 0 0

(o)

0 0 0 0 0 0 0 0

0

rH

0

rH

0 0 0 0

(s)

*>TS

0 0 1 0

0 0 0 0

0 0 0 0

0 0 0 0

<$

0 1 0 1

0 0 0 0

0 0 0 0

0 0 0 0

W 8

<S>

<$

x{ 1 0 0 0

rc, 0 0 0 0

,n2) 0 0 0 0

0 0 0 0

(a)

x(ni,n2)

1-H

1 1 1

rH

1 1 1

rH

1-H

1

h-i

1 1 1 1

(d)

1 16

rH

-1 1 -1

1 -1 1 -1

1 -1 1 -1

--1 1 -1

(h) 1-H

0 -1 0

1 0

-1 0

1 1 0 0

-1 -1 0 0

(1)

1 8

1 0 -1 0

0 1 0

-1

-1 0 1 0

0 -1 0 1

(p)

1 0

-1 0

-1 0 1 0

rH

0 -1 0

T

0 1 0

(t)

t>

X(kufa) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (b)

X 0 0 0 0

(* 0 0 0 0

,fc2) 1 0 0 0 0 0 0 0

(e)

0 0 0 0

1 0 0 0

0 0 0 0

1 0 0 0

(i)

0 0 0 0

0 0 0 0

0 0 1 0

0 0 0 0

(m)

0 0 0 0

0 1 0 0

0 0 0 0

0 0 0 1

(q)

0 0 0 0

0 0 1 0

0 0 0 0

0 0 1 0

(u)

w 16

<S>

w 16

<£>

<£>

s(ni,n2) 1 1 1 1

-1 1 -1 1 -1 1 -1 1

-1 -1 -1 -1

(f)

1 1 1 1

0 0 0 0

-1 -1 -1 -1

0 0 0 0

(j)

1 -1 1

-1

-1 1

-1 1

1-H

-1 1 -1

-1 1 -1 1

(n)

1 8

1 0

-1 0

0 -1 -1 0 0 1 1 0

0

1-H

0 -1

(r)

1 -1 1

-1

0 0 0 0

-1 1

-1 1

0 0 0 0

(v)

Fig. 10.1 (a) The 4 x 4 2-D impulse signal, (b) The DFT of the impulse, (c), (e), (g), (i), (k), (m), (o), (q), (s) and (u) are DFT coefficients of the impulse and (d), (f), (h), 0)i (')> (n)> (P)> (r)> M a n ^ (v)> respectively, are the corresponding sinusoidal surfaces.

DFT Representation of Some 2-D Signals 199

terms of complex exponentials, the impulse signal is represented as 3 3

*(ni,«2) = i £ ^ e i» f (m* 1 +n 9 fa ) > n i , n 2 = 0,1,2,3 kr =0*2=0

We can find the real sinusoidal surfaces that constitute the impulse signal using this equation or Eq. (10.3) with N = 4. Each separate frequency coefficient or a pair and the corresponding sinusoidal surface are shown in Figs. 10.1(c) to (v). Figure 10.1(c) shows X(0,0) = 1 and the corresponding surface is the dc component having sample value i - at all points, shown in Fig. 10.1(d). Figure 10.1(e) shows X(0,2) = 1 and the corresponding surface is ^ cos(7rn2), shown in Fig. 10.1(f). This is a stack, in the vertical direction, of four cosine waveforms with frequency index two. Figure 10.1(k) shows X(1,0) = 1,X(3,0) = 1 and the corresponding surface is | c o s ( f n i ) , shown in Fig. 10.1(1). This is a stack, in the horizontal direction, of four cosine waveforms with frequency index one. Figure 10.1(o) shows X( l ,3 ) = 1,X(3,1) = 1 and the corresponding surface is | c o s ( | ( n i + 3n2)), shown in Fig. 10.1(p). This is a stack, in the horizontal direction, of four cosine waveforms with frequency index one with a shift of §3n2 . Figure 10.1(s) shows X(l, 2) = 1, X(3,2) = 1 and the corresponding surface is | c o s ( | ( n i + 2n2)), shown in Fig. 10.1(t). This is a stack, in the horizontal direction, of four cosine waveforms with frequency index one with a shift of |2n2- The other sinusoidal surfaces, shown in Fig. 10.1, can be identified similarly. I

The DFT of z (m ,n 2 ) = e ^ i n i + m " 2 \ n i , n2 = 0 , 1 , . . . , TV - 1, where I and m are positive integers, is similar to that of the 1-D complex sinusoids.

ej^(ln1+mn,) ^ ^ 2 ^ _ j ^ _ m )

With — I and —m, we get

e i ^ ( - ( n i - m n 2 ) ^ ^ 2 ^ _ ^ _ ^ ^ _ ( j y _ m ) )

The DFT of x(nun2) = cos (^ ( /m +mn2) + 6),nl,n2 = 0 , 1 , . . .,N - 1, where I and m are positive integers, is deduced from the previous result.

27T

cos( —(Zni + mn2) + 6) O-TV2

—(e^(J(fci - I, fc2 - m) + e-j9<5(fci - (N - I), h - (JV - m)))


With 0 = 0, cos(^(ln1 + mn2))<^^-(d(ki-l,k2-m) + 6{k1-(N-l),k2-(N-m))). With0 = -f, sin(^(Zm+mn2)) <S> ^-(-jS(k1-l,k2-m)+j6{k1-{N-l),k2-{N-m))).

Example 10.2 Consider the 64x64 sinusoidal surface shown in Fig. 10.2(a) and its DFT shown in Fig. 10.2(b). The coefficients X(0,1) = O-j'2048 and X(0,63) = 0+J2048 represent a stack, along the ni direction, of 64 shifted (with zero shift) sine waveforms with frequency index one and amplitude one, resulting in the sinusoidal surface, x{n\,n2) = sin( |yn2) . An alternate view of a sinusoidal surface is of a plane wave with frequency \f{u\ + w2) in the direction 6 = t a n - 1 ^ to the n\ axis. The sinusoidal surface shown in Fig. 10.2(a) is a plane wave with frequency | f radians per sample in the direction 90 degrees to the n\ axis.

Consider the sinusoidal surface shown in Fig. 10.2(c) and its DFT shown in Fig. 10.2(d). The coefficients X ( l , l ) = 0 - j'2048 and X(63,63) = 0 + j'2048 represent a stack, in the n2 direction, of 64 phase shifted sine waves, s in ( | jn i ) , (with | f n 2 radians shift) with frequency index one and amplitude one, resulting in the sinusoidal surface, x(ni,n2) = s in ( | j n i + |j7i2). This sinusoidal surface is a plane wave with frequency 2 ^ 2 radians per sample in the direction 45 degrees to the n\ axis.

Consider the sinusoidal surface shown in Fig. 10.2(e) and its DFT shown in Fig. 10.2(f). The coefficients X(2,1) = 1448.2-J1448.2 and X(62,63) = 1448.2 + j'1448.2 represent a stack, in the n2 direction, of 64 phase shifted cosine waves, cos( | |2n i ) , (with (§fn2 — j) radians shift) with frequency index two and amplitude one, resulting in the sinusoidal surface, x(ni,n2) = cos(|^2ni + | f n2 — \). This sinusoidal surface is a plane wave with frequency 2sxl radians per sample in the direction t a n - 1 \ = 26.565 degrees to the ni axis. I

10.3 Computation of the 2-D DFT

The direct computation of Eq. (10.1), obviously, has the complexity of 0(AT4). The basis functions, arfcllfc2(ni,n2) = e ^ ( * i n i + * 2 n 2 ) , n i , n 2 , h, k2

= 0 , 1 , . . . , N — 1, are separable. Therefore, the 2-DFT can be obtained by computing the 1-D DFT of each row of the image followed by the computation of the 1-D DFT of each column of the resulting data and vice

Computation of the 2-D DFT 201

« 0

-1

32

", ° 0 32

63 [ »X(0,63) = 0+/2048

1 f»X(0,l) = 0-;2048

60

(a) (b)

(c)

63 X(63,63) = 0+/2048 <

1 \ •X(l, l) = 0-;2048

1 63

(d)

63 | X(62,63) = 1448.2+/1448.2

1 f •X(2,l)= 1448.2-;1448.2

62

(•) (f)

Fig. 10.2 (a) The 2-D sinusoid 1(111,712) = s i n ( | | r i 2 ) . (b) The DFT of the signal shown in (a), (c) The 2-D sinusoid 1(711,712) = s i n ( | | n i + §Jri2). (d) The DFT of the signal shown in (c). (e) The 2-D sinusoid x ( n i , n 2 ) = cos ( | f 2ni + | j n 2 - \). (f) The D F T of the signal shown in (e).


versa. This method is called the row-column method. Equation (10.1) can be rewritten in two different forms as

J V - l J V - l

X(kuk2) = X ^ E ^ ' ^ T O 1 * 1 } ^ * ' (10.4) n 2 =0 m = 0

JV-l JV-l

X(h,k2) = J2{T,x^n^WN2k2}WNkl (10-5) n i = 0 712=0

The expression inside the braces is 1-D DFT of each column in Eq. (10.4) and of each row in Eq. (10.5). The DFT of the N xN matrix x(rii, n2) can be computed in two stages. One way is to compute the 1-D DFT of each column of the matrix to get

J V - l

X(h,n2)= ^2 x(ni,n2)W^k\ kltTi2 = 0,l,...,N-l ni=0

Then, compute the 1-D DFT of each row of the resulting matrix to get

J V - l

X(h,k2)= J2x(kuri2)W^k\ k1,k2 = 0,l,-..,N-l 712=0

Just by decomposing the problem into 2JV 1-D DFTs, the computational complexity of 2-D DFT has been reduced to 0(N3).

Example 10.3 Compute the 2-D DFT of the following matrix of data using the row-column method.

n2 ->

2 1 4 3 " 1 0 2 2 3 1 0 1 2 0 1 3 .

Compute the IDFT of the transform to get back the original image. Solution As the kernel is separable and symmetric, Eq. (10.1) can be written in

Computation of the 2-D DFT 203

ix form as

" 1 1

1 -3 1 - 1

. 1 3

1 - 1

1 - 1

1 "

3 - 1

-3 .

" 2 1 4 3 " 1 0 2 2 3 1 0 1

. 2 0 1 3 .

" 1 1 1

. 1

1

-3 - 1

3

1 - 1

1 - 1

1

3 - 1

-3

Post-multiplying the image matrix by the right kernel matrix is computing 1-D DFT of the rows and we get

1 1 1 1

1

-3 - 1

3

1 - 1

1 - 1

1 "

3 - 1

-3 .

' 10 5 5 6

-2+J2 -1+J2

3 1+J"3

2 1 1 0

-2-j2 -1-32

3 1 - J 3

Pre-multiplying the partially transformed image matrix by the left kernel matrix is computing 1-D DFT of the columns. The 2-D DFT of the image is given by

26 5 + j l

4 5 - j l

l + j ' 7 - 6 + j 4

1 - J 3 - 4 - j O

4

1 - J ' l 2

1 + j l

1-J '7 - 4 + jO

1 + J 3 - 6 - j4

The original image can be obtained by computing row and column 1-D IDFTs of the transform matrix and vice versa. Note that the kernel matrices of the IDFT are the same as those of the DFT with the difference that each matrix is to be conjugated. However, similar to the case of computing the 1-D IDFT, the 2-D IDFT can be obtained using the DFT simply by swapping the real and imaginary parts in reading the input and writing the output values. The DFT after swapping the real and imaginary parts is given by

0 + J26 1 + J 5 O+j/4

- 1 + J 5

7 + j l 4 - j 6

- 3 + j l 0 - J 4

0 + j 4 - 1 + j l

0 + j 2 1 + j l

- 7 + j l 0 - j 4 3 + j l

- 4 - j6


The result of computing the column DFTs yields

0 + j'40 0 + J 2 0 0 + J 2 0 0+J24

8-j8 8 - j 4

0+jU 12 + J4

O+j/8 0 + j ' 4 0+;?4 0 + j0

- 8 - j8 - 8 - j 4 0 + J12

- 1 2 + J4

The result of computing the row DFTs gives

0+J32 0 + J 1 6 0+J16 0 + j0 0+j '48 0 + J 1 6 O + j'32 0 + j0

0 + j64 0 + j48 0 + j32 0 + j32 0 + ;0 0 + jl6

0 + j"16 0 + j48

These values divided by iV2 = parts yield the input image.

16 and swapping the real and imaginary

At high frequencies, the magnitude of the spectrum decreases rapidly for practical images and, therefore, the high-frequency components look vague when displayed in image form. To get a better contrast, we display the function log10(l+|X(A;i, /t2)|) instead of \X(ki,k2)\. For the example image, the magnitude of the spectrum, |X(fci,A;2)|, in the center-zero format, is

2.00 3.16 4.00 3.16 1.41 7.21 5.10 4.00 4.00 7.07 26.00 7.07 1.41 4.00 5.10 7.21

whereas log10(l + |X(fci,fc2)|) is

0.48 0.62 0.70 0.62 0.38 0.91 0.79 0.70 0.70 0.91 1.43 0.91 0.38 0.70 0.79 0.91

It can be easily seen that the ratio between the smallest and largest coefficient has been reduced.

Properties of the 2-D DFT 205

10.4 Properties of the 2-D D F T

Linearity

The DFT of a linear combination of a set of discrete images is equal to the same linear combination of the individual DFTs of the images. Let xi{ni,n2) <$ Xi(ki,k2) and £ 2(ni ,n 2) <$ X2(ki,k2). Then,

axi(ni,ri2) + bx2(ni,n,2) O a-X"i(fci,fc2) + 6X2(^1, fc2)

where a and 6 are real or complex constants. It is assumed that the images have same dimensions. If it is not so, sufficient zero padding must be done. Linearity holds in both the spatial time- and frequency-domains.

Example 10.4 Compute the DFT of the first two images. Using the DFTs obtained, deduce the DFT of the third image by applying the linearity property.

X\

The indivic

(ni ,n2) 1 2 ' 3 4

lual DF

X i ( * i J

= x^

I s are

c2) = 1

(ni,n2)

" 4 1 ' 2 3

3 - 2 " 4 0

1

,X2(ku

x3(ni,n2) =

' 1 + J 4 2 + jl'

. 3 + J

k2) =

2 4 + j'i

10 2 ' 0 4

1

The DFT of X3(ni,n2) = xi(ni,n2) + jx2(ni,n2) is

* 3 ( * i , fo)=Ji :i(h,) c2)-rjX2(k l.fa) = " io+. -4 +

;'10 - 2 JO 0

7'2" 7 4 .

Periodicity

The 2-D DFT of an N x N image and the IDFT of its spectrum are assumed periodic in both the horizontal and vertical directions with period N. The right and left edges, and the top and bottom edges of an image are considered adjacent. This follows from the definition of DFT and IDFT since the discrete complex exponential is periodic.

-X"(*i,*a) = X(k1+aN,k2) = X(ki,k2 + bN) = X{k1+aN,k2+bN) x(ni,n2) = x(ni+aN,n2) = x(ni,n2 + bN) = x(n!+aN,n2+bN)

where a and b are arbitrary integers.


Spatial circular shift of an image

The shift of an image can be carried out in two steps: (i) shift each row to the right or left as required and (ii) shift each column upwards or downwards as required. Let 2(711,712) •£> X(ki,k2). Then, x{n\ — I,n2 — m) 4^ X(kuk2)W^ll+k2m). As in the case of 1-D DFT, the spatial shift of #(711,712) does not affect the magnitude of the spectrum but only changes the phase.

Example 10.5 If we circularly shift a 4 x 4 image matrix to the right by one position, it corresponds to adding a phase shift of 0, — f, — n, and — ^L

to the zeroth, first, second, and third columns, respectively, of the DFT matrix. The new DFT matrix can be simply obtained by multiplying the zeroth, first, second, and third columns, respectively, by 1,— j , — 1, and j . The right shift of the example image and the corresponding spectrum are shown below.

2j (7 i i ,n 2 - 1) =

3 2 1 4 2 1 0 2 1 3 1 0 3 2 0 1

<$

(-j)k*X(kuk2) =

26 5 + j l

4 5 - j l

7 - j l 4 + j6

- 3 - j l 0 + j 4

- 4 - 1 + j 'l

- 2 - 1 - j l

7 + j l 0 - J 4

-3 + j l 4 - j 6

Spatial circular shift of a spectrum

Let x(n1,n2)&X(kuk2). Then, x(nun2)ej^ln^+mn^ &X{h -l,k2 -

m). With N even and I = m = f , x (n i ,n 2 ) ( - l ) ( " 1 + " 2 ) <S> X(h - f , k2 -y ) . The spectrum is put in the center-zero format.

Example 10.6 If we multiply the example image by (-l)("i+"=), we get

{-l)(ni+n^x(nun2) =

2 1 3 2

- 1 0

- 1 0

4 - 2

0 - 1

- 3 2

- 1 3

4>


X ( f c i - 2 , * 2 - 2 ) =

2

+n 4

-jl

1 + J3 - 6 - jl

1-J7 - 4

4 5 - j l

26 5 + j l

1 - J 3 - 4

1+J7 - 6 + > 4

Reversal property

Let ar(ni,n2) <3> X(fci, fc2). Then, a r ( i V - n u N - n 2 ) <£> X(N-kuN-k2).

Example 10.7 This property is illustrated for the example image as

•Y(4 - fc i , 4 -*2 )

n2) -

26 5 - j l

4 5 + j l

' 2 3 4 2 3 1 3 1 0

. 1 2 2

i - i 7 - 6 - j 4

l + j ' 3 - 4 - JO

1 " 0 1 0 .

>£>

4

l+jl 2

1 - j l

1 + J 7 - 4 + jO

1 - J 3 - 6 + j 4

Symmetry

Similar to the 1-D case, the DFT of a real 2-D data is conjugate-symmetric. The DFT values at diametrically opposite points form complex conjugate pairs.

X*{N-k1,N-k2) = X{kuk2)

This symmetry can also be expressed as

N N * N X(-±k1,-±k2) = X*(-Tk1,- + k2) ~2

Since the spectrum of a real image is conjugate-symmetric, half of the spectral values are redundant.

Example 10.8 Underline the left half of the non-redundant DFT values of the example image.


Solution

* 1

1 26

5 + j l 4

L 5 - j l

1 + J7 - 6 + j4

1 - J 3 - 4 - j O

4

i - i i 2

1 + j l

1 - J7 - 4 + jO

1+J3 - 6 - J 4

Therefore, the storage of the DFT values of columns (rows) 1 to y - 1 and the first y + 1 values of the zeroth and the y t h columns (rows) is sufficient requiring TV real storage locations as for the image matrix. The computation of the DFT requires the computation of N + 2 1-D DFT of real-valued data and the computation of ^ — 1 1-D DFT of complex-valued data. Other symmetry properties are similar to those of the 1-D DFT.

Complex conjugates

Let x(nun2) •£> X(k1,k2). Then, x*(n1,n2) <& X*(N -ki,N - k2) and x*(N-m,N- n2) O X*(kuk2).

Circular convolution in the spatial time-domain

Let x(ni,n2) & X(ki,k2) and h(ni,n2) <S> H(ki,k2), ni,n2,ki,k2 = 0 , 1 , . . . , N — 1. Then, the circular convolution of x(ni,n2) and h(ni,n2) is given by

7V-1 N - l

y(ni,n2) = E ^2 x(mi,m2)h{ni-mi,n2-m2) mi = 0 7712=0

N - l N-l

= E E h(mi,m2)x(n\ —m\,n2 — m2), 7711=0 7712 = 0

where n\, n2 = 0 , 1 , . . . , N - 1. This convolution can be implemented using the transform as a tool as given by

1 N-I N-i

y{nun2) = T ^ E E X{kuk2)H{kuk2)W^W^k* kl = 0 * 2 = 0


Substituting the corresponding DFT expressions for X(ki, k2) and H(ki, k2), we get

y(nun2) = ^ £ £ ^ £ E ^ ^ W * 1 ^ * 2 } fci=0fc2=0 m i = 0 r o j = 0

JV-l JV-l

x{j2 E M'i,/2)^fcit<2*2}^nifciw-"2*2

l1=0h=0

-. JV-l JV-l N-1N-1

= w E E E Eaf(mi'maWi'Z2) m i = 0 m 2 =0 (i=0 '2=0

j v - i J v - i V~* V~~* TT^(ii + mi-ni)*iTi^((2+m2-n2)fc2

kl =0*2=0

The rightmost two summations evaluate to N2 for li = ni — mi and Z2 = ri2 — m2, and to zero otherwise. Therefore,

J V - l J V - l

y(ni,ri2) = £ E ximi>m2:)h{ni—mi,n2 — m2) 7711=0 7712=0

Circular convolution in the spatial frequency-domain

J V - l J V - l

x(ni,n2)h(ni,n2) <£> j= E ^2 x{™1,m2)H{ki - mi,k2 - m2) 7711=0 7712=0

Circular cross-correlation in the spatial time-domain

The circular cross-correlation of x(ni,n2) and h(ni,n2) is given by

J V - l J V - l

Vxh{ni,n2) = £ y ^ a r * ( m i , m 2 ) / i ( n i + m i , n 2 + m 2 ) 7711=0 7712=0

•»x*(*i,*2)ir(*i,fc2)

2/hx("i,n2) = y*xh{N - nuN -n2) & H*(k1,k2)X(kl,k2)


The autocorrelation operation is the same as the cross-correlation operation with x{n\,ri2) = / i(ni,n2) .

ifc*(ni,n2) = IDFT of (\X(kuk2)\2)

Sum and difference of sequences

N-l J V - 1

x(o,o)= J2 E*-'12) m = 0 n2—0

With N even,

x(f'Y)=EE^'^)(-1)("1+"2)

n i=0 ri2=0

1 JV-1 JV-1

fc, = 0 * 2 = 0

With JV even,

fc1=0fc2=0

T/ie difference

x{ni,n2)-x{n1 - l , n 2 ) <3> X ( f c i , A ; 2 ) ( l - e - ^ * 1 ) x ( n i , n 2 ) - x ( n 1 , n 2 - 1 ) <S> X(fci,fc2)(l - e - ^ * 2 )

Image rotation

When an image is rotated by an angle 9 about the origin, its spectrum is also rotated by the same angle. A rotation of a multiple of 90 degrees of the image corresponds to exactly the same amount of rotation of the spectrum but rotation by other angles requires interpolation. Consider the following image and its spectrum.


x(m,n2) =

10 11 8 9 14 15 12 13

2 3 0 1 6 7 4 5

<3>

9 8

11 10

13 12 15 14

1 0 3 2

5 4 7 6

&

8 - j 8 0 0 0

- 8 0 0 0

8 + J8 0 0 0

angle of 90 degrees in t

X(ki, k2)

0 0 0

- J"32 -

= 0 0 0

32

0 0 0

32 + j32

X(kuk2) = 120

32 - j32 - 3 2

L 32+J32 The image and its spectrum rotated by an terclockwise direction are

x{n[,n'2) =

8 + j8 - 8

8 - J 8 120 32

This rotation is obtained through the transformation a^n'j, n2) = x(ni, n2), where n\ = —n2 — 1 and n'2 = ni. Remember that the image is periodic. When an index is negative we add the length N to make it positive.

Separable signals

As the kernel of the 2-D DFT is separable, the DFT of a separable signal x(ni,n2) — x(ni)x(n2) is also separable. If x(rii,n2) = x(ni)x(n2), x{ni,n2) & X(ki,k2), x{n,\) <£> X(ki), anda;(n2) <=> X(k2), thenX(fci,fc2) = X(k!)X{h2).

JV-1 N-l

X{kuk2) = J2 E <nun2)W^W^k> 711=0 712=0

= ( Yl *(m W*11 { £ x(n2)Wtfk> \ = X(h)X(k, l?l l=0 J (.712=0 J

Example 10.9 Find the DFT of x(m) = {2,2,1,3} and x(n2) = {1,4,2,3}. Then, deduce the DFT of x(ni,n2) = x(ni)x(n2) using the separability theorem. Solution X(kx) = {8,1 + j l , - 2 , 1 - j l } and X{k2) = {10, - 1 - j l , - 4 , - 1 + j l } .

X(k1,k2) = X(k1)X(k2)

80 - 8 - j 8 -32 - 8 + j8 10 + jlO - j 2 - 4 - j 4 - 2

-20 2 + j2 8 2 - J 2 L 1 0 - j l O - 2 - 4 + j 4 j2 J


Parseval's theorem

This theorem implies that the sums of the squared magnitudes of the input and DFT sequences are related by the constant factor TV2, the number of samples. That is the signal power can also be computed from the DFT coefficients of the input sequence.

N-l N-l 1 J V - 1 N-l

£ £ \x{nun,)\* = ± £ £ \X(kl,k2)f ni=0n2=0 ki =0*2=0

Since the squared magnitude can be computed by multiplying a complex number by its conjugate, we can write the left summation as

J V - l N-l

^ ^ a;(ni,n2)x*(ni,n2) r»i=0n2=0

Substituting the corresponding IDFT expressions for 2(7x1,712) and x* (711,712), and rearranging the summations, we get

1 N-l N-l N-l N-l N-l N-l W£ £ £ £*(*!, W(*3,fc4) £ ^W-^r-^W-n^-k.)

fc1=0 *2=0 fc3=0 fc4=0 ni=On2=0

If fci = £3 and A2 = ki, this expression becomes

fc1=0fe2=0 fci=0fe2=0

Otherwise, the expression evaluates to zero. The generalized form of this theorem applies for two different signals as given by

J V - l N-l 1 N-l N-l

E E a;(ni'n2)2/*(ni,n2) = jyj E E X(fcl'fc2)y*(fcl'fc2) ni=0n2=0 *M=0fc2=0

10.5 The 2-D P M D F T Algorithms

The DFT of 2-D data is usually obtained by computing the 1-D DFT of each row of data followed by the 1-D DFT of each column of the resulting data and vice versa. The SFG of a 2-D DFT algorithm with the same

The 2-D PM DFT Algorithms 213

R_vector R_stage 1 R_stage 2 Rjs tage 3 Col_by_Col

o(0) = a;(0,0) ± x(0,8)

o(4) = x(0,4) ± x(0,12)

a(2) = ar(0,2) ± ar(0,10)

o ( 6 ) = o x(0,6)±x(0,14)

o( l ) = x(0, l )±x(0,9)

o(5) = x(0,5) ± x(0,13)

o(3) = x(0,3) ± x(0,ll)

o ( 7 ) = o x(0,7) ± x(0,15)

f A(0) = ,OX(0,0),I(0,8)

, Mi) = 1X(0,1),X(0,9)

, A{?) = 2X(0,2),X(0,10)

t A(3) = '3X(0,3),X(0,11)

A(4) = X(0,4),X(0,12)

A(5) = X(0,5),X(0,13)

A(6) = X(0,6),X(0,14)

A{7) = X(0,7),X(0,15)

Fig. 10.3 The SFG of the 2 x 1 PM DIT DFT algorithm for a 1 X 16 2-D signal. Twiddle factors are represented only by their exponents. For example, the number 1 represents

number of elements has the same structure as that of the SFG of the 1-D DFT algorithm. Consider the 1-D data

x{n) = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]

The DFT can be computed using an algorithm such as the 2 x 1 PM DIT DFT algorithm shown in Fig. 6.2. This data can also be considered as 2-D data with 1 row and 16 columns as

x(0, n) = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]

According to row-column method, we can compute the DFT by computing 16 column DFTs each of size one and one row DFT of size 16. As the DFT of a single data element is itself, there is no computation required for the 16 column DFTs. Therefore, even if we consider the data as 2-D, the algorithm, shown in Fig. 10.3, is the same as that shown in Fig. 6.2. As 2-D data, we compute the vectors for the row DFT of size 16, followed by


C-vector R_vector R_stage 1 R-stage 2 Col_by_Col a(0) =

z(0,0) ± x(l,0)

o(4) = x(0,4) ± z(l,4)

o(2) = x(0,2)±a;(l,2)

a ( 6 ) = o x(0,6)±x(l ,6)

o(l) = z ( 0 , l ) ± x ( l , l )

a(5) = x(0,5)±x(l ,5)

o(3) = z(0,3) ± x(l,3)

o ( 7 ) = o x(0,7)±x(l ,7)

f A(0) = U ( 0 , 0 ) , I ( 0 , 4 )

A(l) = X(1,0),X(1,4)

r A(2) = 1 1 ( 0 , 1 ) ^ ( 0 , 5 )

A(3) = X(1,1),X(1,5)

Af4) = X(0,2),X(0,6)

A(5) = X(1,2),X(1,6)

Af6) = X(0,3),X(0,7)

*±>A(7) = 6 X(l,3),X(l,7)

Fig. 10.4 The SFG of the 2 x 1 PM DIT DFT algorithm for a 2 x 8 2-D signal. Twiddle factors are represented only by their exponents. For example, the number 1 represents

^ s ' -

the stages 1,2, and 3 of row DFT computation. The DFT values are read column-by-column. Consider the same data with 2 rows and 8 columns as shown below.

' 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

The process of 2-D DFT computation is shown in Fig. 10.4. If we read the data row-by-row and then put it in bit-reversed order, we get the appropriate data in each vector locations for computing the column DFTs. In addition to that, the individual columns of data are placed in bit-reversed order that is required for computing the row DFTs. There are eight column DFTs each of size two and are carried out when vectors are formed in the stage marked C.vector. After computing the column DFTs, the problem reduces to the computation of two 1-D row DFTs each of size 8. Including the vector formation stage, we need three stages of computation. After


computing the row vectors in the stage marked R_vector, the upper nodes of the butterflies contain the vectors in the required order (0, 2, 1, 3) for the computation of the first row DFT carried out in the next two stages. The second row DFT is carried out using the vectors at the lower nodes of the butterflies. It is clearly seen that each set of alternate butterflies in the last two stages constitutes a row DFT and we get the 2-D DFT column-by-column in natural order. Consider the same data with 4 rows and 4 columns as shown below.

" 0 1 2 3 " 4 5 6 7 8 9 10 11

. 12 13 14 15 .

The SFG, shown in Fig. 10.5, shows the computation of the 4 independent column DFTs each of size 4 followed by the computation of the 4 independent row DFTs each of size 4. While the butterflies of each of the column DFTs appear together, every fourth butterfly in the last stage computes each of the 4 independent row DFTs. Note that there are only four butterflies in a stage which means one butterfly to each row DFT. Consider the same data with 8 rows and 2 columns as shown below.

0 2 4 6 8

10 12 14

I—>

3 5 7 9

11 13 15

The SFG, shown in Fig. 10.6, shows the computation of the 2 independent column DFTs each of size 8 followed by the computation of the 8 independent row DFTs each of size 2. The last stage is the vector formation stage for the row DFTs and no more stages are required since each row has only two elements.

In summary, we read the data row-by-row (column-by-column), compute the required number of column (row) DFTs, followed by the required number of row (column) DFTs, and read the DFT values column-by-column (row-by-row). The vectors for the individual column (row) DFTs are placed

216 Two-Dimensional Discrei le Fourier Transform

C - v e c t o r

o(0) = a;(0,0)±x(2,0)

a(4) = z(l,0) ± z(3,0)

o(2) = ar(0,2) ± ar(2,2)

0(6) = » x(l,2) ± ar(3,2)

«(!) = z(0,l) ± z(2,l)

o(5) = a;(l,l)±ar(3,l)

o(3) = z(0,3) ± x(2,3)

o ( 7 ) = c x(l ,3)±x(3,3)

C-stage 1 R_vector Rjs tage 1 Col_by_Col

r AW = '0X(0,0),X(0,2)

A(l) = X(1,0),X(1,2)

t)X(2,0),X(2,2)

Af3) = X(3,0),X(3,2)

AU) = X(0,1),X(0,3)

ii(5) = X(1,1),X(1,3)

X(2,1),X(2,3)

1 Z(3,1),X(3,3)

Fig. 10.5 The SFG of the 2 x 1 PM DIT DFT algorithm for a 4 x 4 2-D signal. Twiddle factors are represented only by their exponents. For example, the number 1 represents

consecutively in bit-reversed order whereas the vectors for the individual row (column) DFTs are placed at intervals, the interval being the number of rows (columns), in the bit-reversed order. The reader is urged to compute the 2-D DFTs using the SFGs shown in Figs. 10.4, 10.5, and 10.6 and verify that DFT values are the same as those obtained by direct computation. With the sizes of the row and column DFTs being powers of two, the similarities and differences between 1-D and 2-D DFT computation are: (i) the bit-reversal operation required is the same and independent data swapping is eliminated in the first vector formation stage, (ii) the structure of the SFGs is the same for the same number of data elements, and (iii) the twiddle factors are different. One large 1-D DFT is computed in the 1-D case whereas a number of smaller 1-D DFTs are computed in the 2-D case. We can choose any algorithm for computing the 1-D DFTs required in computing the 2-D DFT. With a good knowledge of 1-D algorithms, it is simple to deduce the SFG for the computation of the 2-D DFT. In


C_vector Cs tage 1 C_stage 2 R_vector Col_by_Col o(0) = <*- ^ - ^-Pr +-pA(0) =

x(0,0)±x(4,0)^y<^0 \ A \ / X(0,0),X(0,1)

a(4) = o ^ — - ^ o v \ S > A \ M-pA{\) = x(2,0)±x(6,0) Z y V f \ \ / / *(1,0),X(1,1)

a(2) = <*- £?cX N ( ijNv \ Y / pA{2) = x{lfi)±x{5,0)^^>^0 / \ ~ \ y y / X(2,0),X(2,1)

a(6) = 0^— ^ c A 'T^\ A A )T y°A(3) = x(3,0)±x(7,0) Z 6 V Y Y V A-(3,0),A-(3,1)

a(l) = 0^- £ * v >-*/ A A R vAU) = x ( 0 , l ) ± x ( 4 , l ) \ > < ^ , 0 X *6 / y y \ X(4,0),X(4,1)

a(5) =0^— ^ r 0 ^ y<C * /> / A T "-^(5) = x(2,l)±z(6,l) 2 V / V f / / \ \ A(5,0),^(5,l)

o(3) = o^- y / >< V 0 ^ - / V ^ o ^ ( 6 ) = a;(l,l) ± x(5,l) ^ ^ " * 0 / \ ~ / \ X(6,0),X(6,1)

a(7) =0^— -5^°^ n ^ - » A M ( 7 ) = z(3,l)±:r(7,l) 6 X(7,0),X(7,1)

Fig. 10.6 The SFG of the 2 x 1 PM DIT DFT algorithm for a 8 x 2 2-D signal. Twiddle factors are represented only by their exponents. For example, the number 1 represents Wl.

the implementation of the algorithm, the modification required in a 1-D algorithm is the setting up of an additional loop with appropriate index increments and an intermediate vector formation stage.

Computation of the DFTs of two real 2-D signals at a time

To compute the 2-D DFT of real signals, the algorithm for complex data can be used directly if the speed and storage requirements are not critical. The two DFTs at a time method, described in Chapter 8, can be used to compute the DFTs more efficiently. Let a:(ni,n2) and y(ni,n2) be the real data sets, each with N x N values, whose DFTs are to be computed. Let X(ki,k2) and Y(fci,fc2) be their respective DFTs. Let C(fci,fc2) be the DFT of s(ni,n2)+jy(ni,n2). Then,

x { k u h ) = (C(kuk2) + C*(N-h,N-k2))

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Table 10.3 The DFT of the real data corresponding to the imaginary part of 8 x 8 2-D data shown in Table 10.1. Real part in the upper half of the table and imaginary part in the lower half of the table. Origin at upper left-hand corner. Note the conjuagte-symmetry of the DFT.

240.00 -15.66 -22.00

-4.34 28.00 -4.34

-22.00 -15.66

0.00 6.00 2.00

-6.00 0.00 6.00

-2.00 -6.00

-5.76 8.56

-4.76 -26.49

2.00 0.51

-17.14 16.73

-35.46 17.49

-20.90 13.66 19.56 32.90 -0.17

-32.73

4.00 -9.14 10.00

-14.90 -10.00 19.14 12.00 4.90

18.00 1.93 2.00 1.83 0.00

16.07 -12.00

-3.83

-14.24 -9.51 11.14

-22.56 2.00

-8.73 -13.24 17.49

-15.46 -2.34 5.83

-0.51 11.56 7.27 1.10

-13.10

-16.00 -9.07

-16.00 5.07

32.00 5.07

-16.00 -9.07 0.00

-19.07 8.00 4.93 0.00

-4.93 -8.00 19.07

-14.24 17.49

-13.24 -8.73 2.00

-22.56 11.14 -9.51 15.46 13.10 -1.10 -7.27

-11.56 0.51

-5.83 2.34

4.00 4.90

12.00 19.14

-10.00 -14.90 10.00 -9.14

-18.00 3.83

12.00 -16.07

0.00 -1.83 -2.00 -1.93

-5.76 16.73

-17.14 0.51 2.00

-26.49 -4.76 8.56

35.46 32.73

0.17 -32.90 -19.56 -13.66 20.90

-17.49

the DFT of the complex data in Table 10.2 as follows for two specific cases.

y ( o , o ) =( 2 1 0 + ^ - ( 2 1 0 - J ' 2 4 0 ^ 2 4 o

j 2

= (-4.93 + J 6 . 4 9 ) - (6.73 -J15.80) = u ^

It is left as an exercise to verify the values in Table 10.3 and derive the DFT of the real part of the data shown in Table 10.1.

Computation of the DFT of a single real 2-D signal

Let us assume that we compute the column DFTs first in computing the DFT of a single real 2-D signal. Each pair of adjacent columns are combined so that a 1-D algorithm for complex data can be used to compute the column DFTs. The individual DFTs are split as described in Chapter 8. Then, a complex algorithm is used to compute the row DFTs as the result of column DFTs is, in general, complex-valued.


Computational complexity

The computational complexity of an TV x N 2-D DFT of complex data is 2N times that of the 1-D DFT algorithm used. The twiddle factors can be generated once for all the rows and once for all the columns. This overhead is, therefore, becomes negligible. Consequently, no table is necessary for the storage of twiddle factors. Further, the use of one-dimensional indexing and the 1-D PM DFT algorithms makes the row-column approach of computing the 2-D DFT practically very efficient. A single algorithm for complex data is enough for both the DFT and IDFT computation. As in the case of the 1-D IDFT, for the 2-D IDFT computation using DFT, the data is to be read and written with the imaginary and real parts interchanged.

10.6 Summary

• In this chapter, the theory, properties, and algorithms of the DFT of 2-D signals were presented. In the case of 2-D signals, Fourier analysis decomposes an arbitrary signal into a set of sinusoidal surfaces of various frequencies, amplitudes, phases, and directions. Some examples of simple 2-D signals and their DFTs were given.

• The 2-D DFT is a straightforward extension of the 1-D DFT. Therefore, practically efficient 2-D DFT algorithms are obtained by the repeated use of the 1-D DFT algorithms. The computation of the 2-D DFT of real-valued signals is also similar to that of the 1-D DFT.

References

(1) Gonzalez, R. C. and Woods, P. (1987) Digital Image Processing, Addison Wesley, Reading, Mass.

(2) Jain, A. K. (1989) Fundamentals of Digital Image Processing, Prentice-Hall, New Jersey.

(3) Sundararajan, D., Ahmad, M. 0 . and Swamy, M. N. S. (1994) "Computational Structures for Fast Fourier Transform Analyzers", U.S. Patent, No. 5,371,696.

Exercises 221

Exercises

10.1 Verify that Eqs. (10.1) and (10.2) form a transform pair.

10.2 Prove Eq. (10.3) from Eq. (10.2).

10.3 Describe the sinusoidal surfaces shown in Figs. 10.1(h), (j), (n), (r), and (v).

10.4 Find the DFT of the N x N 2-D dc signal, x(nun2) = - 3 , from definition.

10.5 Find the DFT of N x N delayed impulse signal, S(ni - l,n2 - m), from definition.

10.6 From definition, derive the DFT of a;(ni,n2) = e ^ ' m + m n 2 ) , n i , n 2 = 0 , 1 , . . . , N — 1, where I and m are positive integers.

10.7 Compute the 2-D DFT of (a) x(m,n2) = 2e-'w(4m+3n2)) 0 < m < 64, 0 < n2 < 64. (b) x(ni,n2) = 3e^(" 1 +" 2 ) , 0 < nx < 32, 0 < n2 < 32. (c) x{m,n2) = 4e^'f (3"2), 0 < ni < 16, 0 < n2 < 16. *(d) x(nun2) = - 5 e - ^ f 2 n i + e ) , 0 < m < 16, 0 < n2 < 16. (e) x(rn,n2) = 2sin(ff(4m +3n 2 ) + f ) , 0 < m < 16, 0 < n2 < 16. *(f) x(nun2) = 2sin(ff (4m +3n 2 ) + f ) , 0 < ni < 32, 0 < n2 < 32. (g) x(nun2) = 5cos(f |(2ni +5n 2 ) - | ) , 0 < nx < 32, 0 < n2 < 32. (h) x(m,n2) = s in (^3n i ) , 0 < nx < 16, 0 < n2 < 16. (i) x(m,n2) — cos(fni), 0 < n\ < 16, 0 < n2 < 16. (j) x(ni,n2) = cos(7rn2), 0 < n\ < 16, 0 < n2 < 16. (k) x(ra,n2) = 4, 0 < ri! < 4, 0 < n2 < 4. (1) a;(0,0) = 3 and x(ni,n2) = 0 otherwise, 0 < n\ < 4, 0 < n2 < 4.

10.8 Compute the 2-D IDFT of (a) X{ki,h») = 7, 0 < h < 5, 0 < fc2 < 5. (b) X(0,0) = 8 and X{kuk2) = 0 otherwise, 0 < h < 3, 0 < k2 < 3. (c) X(2,1) = 26 and X(k!,k2) = 0 otherwise, 0 < fci < 16, 0 < k2 < 16. (d) X(3,0) = - 4 6 and X(kltk2) = 0 otherwise, 0 < A;i < 16, 0 < k2 < 16. (e) X(13,15) = 2 3 ( ^ - j^~) and X(kuk2) = 0 otherwise, 0 < h < 16, 0 < k2 < 16. *(f) X(2,3) = ( i +j&), X(14,13) = ( | - j&), and X{kuk2) = 0 otherwise, 0 < ki < 16, 0 < k2 < 16.


(g) X( l ,5 ) = ( - ^ -jj-), X(15 , l l ) = ( - ^ +j±), and X(kuk2) = 0 otherwise, 0 < ki < 16, 0 < k2 < 16. (h) X(5,4) = 4, X( l l ,12) = 4, and X(jfci,*2) = 0 otherwise, 0 < fei < 16, 0 < k2 < 16. (i) X(10,8) = 16, X(22,24) = 16, and X(kuk2) = 0 otherwise, 0 < h < 32, 0 < k2 < 32. (j) X(2,3) = -j8, AT(14,13) = j&, and X(h,k2) = 0 otherwise, 0 < h < 16, 0 < k2 < 16. (k) X(2,6) = - 3 , X(14,10) = - 3 , and X(kuk2) = 0 otherwise, 0 < h < 16, 0 < k2 < 16. *(1) X(2,2) = -j5 and X(ku k2) = 0 otherwise, 0 < fci < 16, 0 < k2 < 16.

10.9 Compute the DFT of the following image using the row-column method. Then, compute the IDFT of the transform using the DFT and verify that the original image is obtained.

ni n2

2 1 4 2

1 0 1 0

3 1 0 1

3 2 1 2

10.10 Compute the DFT of the first two images. Using the resulting transforms, deduce the transform of the third image by applying the linearity theorem.

x — - 1 - 3 - 4 - 2

2 3 1 4 z =

-1-J2 - 3 - J 3 - 4 - j l - 2 - J 4

*10.11 For the image of Exercise 10.9, deduce the transform values X(7,21), X(—4, —3), and X(l, —3), using the periodicity property.

10.12 For the image of Exercise 10.9, deduce the image values x(7,42), x{—9, —4), and x(2, —7), using the periodicity property.

10.13 For the image of Exercise 10.9, deduce the spectrum of x(ni - 2, n2 -!-3), using the shift property.

10.14 For the image of Exercise 10.9, deduce the spectrum if the image is multiplied by 2e~j2^(-3ni~2n^, using the shift property.

Exercises 223

10.15 For the image of Exercise 10.9, deduce the spectrum of x (ni,n2)

*10.16 Underline the upper half of the nonredundant transform values of the image of Exercise 10.9.

10.17 Find the DFT of x(m) = | c o s ( ^ n i ) , 0 < ni < 4 and x(n2) = i s i n ^ r ^ ) , 0 < n2 < 4. Then, deduce the DFT of ar(ni,n2) = x(ni)x(n2) using the separability theorem.

*10.18 For the image of Exercise 10.9, verify the Parseval's theorem.


10.1 Write a program to directly implement the 2-D DFT using the row-column method.

10.2 Write a program to directly implement the 2-D ID FT using the row-column method.

10.3 Write a program for the computation of the 2-D DFT of a complex-valued image using the 2 x 1 PM DIT DFT algorithm.

Chapter 11

Aliasing and Other Effects

Thus far, we have assumed that signals are periodic and composed of ~ harmonically related sinusoidal components for N time-domain samples taken over a period. We made these assumptions in order to develop the discrete version of the Fourier analysis that is suitable for numerical computation. However, in practice, signals are mostly continuous and aperiodic. The accurate representation of these signals may require an infinite number of samples in the frequency- and time-domains. Therefore, in trying to approximate arbitrary signals in terms of finite discrete signals, we may end up with signals having frequency components with frequencies that exceeds the allowable bandwidth for a given N, the number of samples. In addition, due to the finite nature of the DFT, the signal may have to be truncated. In this chapter, we are going to study the effects created by these problems and the remedies.

The problems are concerned with the selection of appropriate (i) sampling rate, (ii) record length, and (iii) frequency spacing or increment, in the representation of a signal. These parameters are specified by the theory of the Fourier analysis. But, due to the discrete and finite nature of the DFT, we are forced to violate the theory. Therefore, in practice, the problem usually reduces to the selection of these parameters so that the accuracy of the frequency-domain representation of a signal is adequate. Consequently, a good understanding of these problems along with the knowledge of the characteristics of the signals encountered in a given application will enable the use of the DFT in an efficient manner. In addition, errors arise in DFT processing because of the digital representation of the data and coefficients, and the round off of numbers in arithmetic operations, as in any numer-

225

226 Aliasing and Other Effects

te-VCV;

infinity

^ * '

(a)

band-limiled spectrum

infinity (b)

/ V V w * infinity

Fig. 11.1 (a) Significant magnitude of frequency components in an infinite range. Aliasing is unavoidable, (b) Nonzero value of frequency components only in a finite range. Aliasing is avoidable by providing a sufficient frequency range, (c) Magnitude of frequency components falls off as frequency increases. Aliasing can be made negligible by an appropriate choice of the frequency range.

ical computation using digital devices. In Sec. 11.1, the aliasing effect is described. In Sec. 11.2, the leakage effect is analyzed. In Sec. 11.3, the picket-fence effect is explained.

11.1 Aliasing Effect

The theory of the Fourier analysis is that a periodic signal can be represented uniquely by a spectrum with an infinite number of harmonically related sinusoids as shown in Fig. 11.1(a). Since we can deal with only a finite number of frequency components in the DFT, we are faced with the following two alternatives: (i) we can represent a signal uniquely with a finite number of frequency components if the signal is band-limited as shown in Fig. 11.1(b) or (ii) we give up unique and unambiguous representation of the signal. The DFT spectral representation of a signal will be in gross error due to aliasing if its spectrum is as shown in Fig. 11.1(a) (spectrum has significant values up to infinity). Fortunately, in practice, the spectrum of

Aliasing Effect 227

signals tends to fall off to insignificant levels at high frequencies as shown in Fig. 11.1(c) so that the error is within tolerable limits in the representation of the signal with sinusoids whose frequencies span only a finite range.

In order to understand the problem of fixing the frequency range, we want to get answers to two questions: (i) for a given N, what is the highest frequency component present in a signal so that it can be unambiguously represented by the DFT coefficients, and (ii) if a signal is composed of frequency components of higher frequencies than that can be allowed for its proper representation, what problem it creates and what are the possible remedies to that problem.

The highest frequency component for unambiguous signal representation

Let us assume that we use N = 4 samples to represent a signal. Figures 11.2(a) and (b) show, respectively, the representation of sine and cosine signals with zero frequency. Figures 11.2(c) and (d) show, respectively, the representation of sine and cosine signals with frequency index one. The set of samples, in each of these cases, is distinct and unambiguously represent the corresponding signal. Figures 11.2(e) and (f) show, respectively, the representation of sine and cosine signals with frequency index two. A cosine wave with frequency index two has a distinct set of samples and can be unambiguously represented with four samples. The set of samples in Figs. 11.2(a) and (e) are the same. Therefore, we cannot represent a sine wave of frequency index two with four samples. If we use five samples we could represent the sine wave with frequency index two unambiguously. With two samples per cycle, the representation is not possible while just more than an average of two samples per cycle is adequate. This implies that the sampling rate or frequency must be greater than two times of the frequency of the highest frequency component of a signal. That is, the minimum number of samples required is two times the frequency index plus one. For example, with frequency index one, we need three samples. With frequency index ~ , we need JV + 1 samples which exceeds our assumption of N samples. Our conclusion is that the index of the highest frequency component a signal is composed of must be less than y with N time-domain samples, in order to represent the signal unambiguously with N DFT coefficients. This is called the sampling theorem and the frequency with index y is called the folding frequency.


£ 0

(a)

(c)

(e)

* ° -1

4 0

(f)

Fig. 11.2 (a), (c), and (e): Sine waves with frequency indices 0, 1, and 2, respectively, with 4 samples, (b), (d), and (f): Cosine waves with frequency indices 0, 1, and 2, respectively, with 4 samples. Waveforms shown in (a), (b), (c), (d), and (f) only have distinct set of sample values.

The conclusion we arrived is a theoretical answer to our first question. In practice, due to the inability of physical devices to produce ideal behavior as represented mathematically, typically the index of the highest frequency component present in a signal is limited to ^ in order to represent the signal in the frequency-domain with reasonable accuracy. This implies that at the least four samples per cycle of a sinusoid are available. If the number of samples per cycle is decreased, it results in more inaccuracy in the signal representation. On the other hand, if we do have more than the necessary number of samples per cycle then the computational effort for processing the signal is increased unnecessarily. The number of samples per cycle, for a practical problem, has to be determined considering both these aspects.

We have concluded that aliasing is avoided in the frequency-domain if the sampling interval, Ts, in the time-domain is less than ^ where fh is the highest frequency of the component of a signal. Aliasing is avoided in the time-domain if the sampling interval in the frequency-domain, fs, is less than ^, where T is the record length of an aperiodic signal.

Aliasing Effect 229

We present an analogy to explain the aliasing problem. With a word length of one bit, we can represent two binary numbers (0,1). With a word length of two bits, we can represent four binary numbers (00,01,10,11). In general, with a word length of N bits, we can represent (2^) binary numbers. We select an appropriate word length for a specific problem. If we select an insufficient word length, overflow will occur and the problem cannot be solved correctly. On the other hand, if we select a longer word length than necessary the processing cost is increased unnecessarily. Similarly, with a set of N samples, we can distinctly represent sinusoids with frequency index up to and including ^ ~~ !• F° r example, with 256 samples, sinusoids with frequency index up to and including 127 can be distinctly represented. Therefore, we are able to represent a larger set of sinusoids distinctly with a larger set of samples. Ultimately, with continuous signal representation, we can represent an infinite number of sinusoids distinctly as the number of samples becomes unlimited. The point is that we do not need, for practical applications, an infinite number of distinct sinusoids since the signal representation with some error is acceptable.

The folding of frequencies

The second question is what happens if the signal contains frequency components with index greater than %• Simply, the frequency coefficients in the valid range are corrupted and we cannot recover the original time-domain signal from these corrupted coefficients. This happens because of the periodicity of the discrete complex exponentials or discrete cosines and sines. Eqs. (2.4) and (2.5) imply that a set of N discrete time-domain samples can represent an infinite number of sinusoids. Therefore, in the frequency-domain, an infinite number of sinusoids contribute to each DFT coefficient making it impossible to discriminate the individual sinusoids. The impersonating of high frequency sinusoids as low frequency sinusoids, due to sampling a signal with a sampling interval that is not small enough, is called the aliasing effect.

Figure 11.3(a) shows a sine waveform with N = 4 and frequency index one. The DFT of this signal {0 , - j 2 ,0 , j 2} , shown in Fig. 11.3(b), correctly indicates a sine wave with frequency index 1. Figure 11.3(c) shows a sine waveform with N = 4 and frequency index three. The samples we obtain in Fig. 11.3(c) are exactly the negative values of that shown in Fig. 11.3(a) and the DFT yields the coefficients of the signal shown in Fig. 11.3(a)


2 r s pal it o Ima ginary

(a)

1 2

(b)

it 2 n

(c) (d)

Fig. 11.3 (a) and (c): Sine waves with frequency indices 1 and 3, respectively, with 4 samples, (b) and (d): The DFT coefficients of the signals shown in (a) and (c), all located at the same set of frequency samples.

with signs reversed as shown in Fig. 11.3(d). Looking at the set of time-domain samples or spectra shown in Fig. 11.3, we will assume the presence of a sine waveform with frequency index 1 whether it is true or not. The samples or spectra could represent sine waveforms with frequency indices 1,3,5,7, . . . , or the sum of a set of these waveforms. Figure 11.4 shows folding of frequencies due to the aliasing effect, for real signals, with N — 16 samples in a period. If we draw an horizontal line starting from a frequency index in the vertical axis, then we get the indices of frequencies those are all aliases to it. For example, frequencies with indices 15,17,31,33, . . . , have alias representation at index 1. For a complex signal with TV samples, aliasing occurs when the index of a frequency component exceeds N — 1.

•*. 8 •" 7 ^ 6 •2,5 S-4 5 3 • a p . » f r 6 1 ". 0

o o o

o o A / = 1 6 o o

o o o o o

o o o 15 a 17 —e—e—

31 n 33 16 24

Input frequency, k 32

Fig. 11.4 The folding of frequencies due to aliasing effect, with N = 16. Frequency components with index 0 to 8 have true representations. Frequency components with an higher index, shown by unfilled circles, have alias representations.

Leakage Effect 231

Reducing the aliasing effect

In practice, aliasing will be present because if a signal is time-limited then it cannot be band-limited and vice versa. Therefore, the aim is to reduce it to allowable limits. One solution to reduce aliasing is to ensure that the signal is composed of frequency components with index less than y by prefiltering it with a low-pass filter. The second solution is to see that the number of samples, N, is more than twice the index of the highest frequency component present in the signal, if prior knowledge of the signal is available. For example, high quality audio signals are band-limited to about 20 kHz. For practical use, a sampling rate of 80 kHz is adequate. If the highest frequency component of a signal is unknown, compute the DFT of the signal with some arbitrary number of samples. Then, repeat the process with double the number of samples, keeping the record length the same. If the magnitudes of the DFT coefficients become negligible towards the end of the spectrum (for real signals, close to the frequency with index y ) , then it is an indication that the value of N is sufficient. It may be necessary to reduce the sampling interval further in order to reduce the error due to aliasing created by the leakage effect described in the next section.

11.2 Leakage Effect

In the last section, we considered the restriction on the range of frequencies for the unique representation of a signal with a finite number of samples. In this section, we consider the appropriate time-domain range over which samples must be taken for the accurate representation of a signal. Again, due to the finite nature of the DFT, we are forced to chop off some part of the signal and hence we tend to violate the theory of Fourier analysis in order to make the problem suitable for numerical analysis.

In the case of an aperiodic signal, since the period is infinity, we may have to take samples over the entire time-domain range for the proper representation of the signal. Figure 11.5(a) shows an aperiodic signal. In this case, in practice, we are forced to cutoff some part of the signal since only a finite number of samples can be processed numerically. This is called truncation and we represent only part of the signal correctly. Similar to the case of the aliasing problem, the frequency-domain representation will be in gross error due to truncation if the signal is as shown in Fig. 11.5(a)


? "tM^Hm^lM^ (a)

infinity

# M ^ V x(t) time-limited

T NT ' R (b)

infinity

infinity

Fig. 11.5 (a) A signal with significant time-domain values in an infinite range, (b) A signal with nonzero time-domain values only in a finite range, (c) A signal with time-domain values falling off rapidly with time.

(time-domain signal has significant values up to infinity). The signal, shown in Fig. 11.5(b), is time-limited and hence is nonzero only over some finite range. If we limit the range over which samples are taken up to the point marked T, then we have truncated the signal. If we limit the range over which samples are taken up to the point marked NT, then there is no truncation. However, even though there is no truncation with the range NT, sometimes samples are taken over a longer interval such as marked R in order to have a denser spectrum. The duration of signals with nonzero values may be infinity. Nevertheless, in practice, they tend to fall off to insignificant levels after some duration, such as marked NEGT in Fig. 11.5(c). This characteristic enables the use of samples taken over a finite range, such as marked NEGT, to approximate the frequency-domain representation with a desired accuracy. To represent a periodic signal correctly, we must take samples over an integral number of periods. Otherwise, we are making an analysis over a false period producing an incorrect spectrum.

Leakage Effect 233

Multiplication of the signal with a rectangular window

In order to limit the number of samples, we may have to chop off some part of the signal. The untruncated signal has some spectrum whereas the DFT produces a spectrum that corresponds to the truncated signal since that was the signal used in the DFT computation. We have to model the process of truncation so that we know the relation between the spectra of the untruncated and truncated signals. The input signal to the DFT can be considered as the product of the actual signal multiplied by a rectangular window that has a value of one for a limited interval and zero otherwise. Therefore, the DFT produces a spectrum that is the circular convolution of the spectrum of the actual input signal and that of the rectangular window signal. Figures 11.6(a) and (b) show, respectively, a sine wave and its spectrum. Figures 11.6(c) and (d) show, respectively, a rectangular window and its spectrum. Now, extend periodically both the sine wave and the window signal. We get a true sine wave and a signal with all its sample values unity. The multiplication of these two signals does not alter the sine wave. Therefore, the convolution of the spectra of these signals in the frequency-domain does not alter the spectrum of the sine wave, except for a scale factor.

Figures 11.6(e) and (f) show, respectively, a part of the sine wave chopped off and its spectrum {1, —jl,—l,jl}. The spectrum is different from that shown in Fig. 11.6(b) as the time-domain signal is different because of truncation. This signal can be considered as the product of the sine wave shown in Fig. 11.6(a) with the window shown in Fig. 11.6(g). The circular convolution of the spectrum of the window shown in Fig. 11.6(h), {3)-J , I j j } , with the spectrum of the untruncated signal, {0,- j '2,0, j2} , results in {4, ~j4, - 4 , j 4} , which is the same, except for a scale factor, as the spectrum of the truncated signal shown in Fig. 11.6(f).

The spectrum of the actual signal is altered in two ways due to truncation. The first difference is that the amplitude of the spectrum of the sine signal has been reduced by a factor of two, (-J2J2) to (-jl,jl). That is smoothing of the original spectrum resulting in the loss of detail of the spectrum. This is because the truncated window shown in Fig. 11.6(g), which is a filter, has a less sharp response (a wider passband) compared with that of the window shown in Fig. 11.6(c). The second difference is that the spectrum has spread out, that is the energy has leaked into nearby frequencies. Hence, this effect is called the leakage effect. This is because


= 1

2 n

(a)

_ 2

-* X

1 • real . o imaginary

(b)

(c)

(e)

1 0

-1

(d)

(0

(g) (h)

Fig. 11.6 (a) A sine wave and (b) its spectrum with nonzero coefficients at k = 1 and k = 3. (c) An ideal rectangular window with all its sample values unity and (d) its spectrum with just one nonzero coefficient at k = 0. (e) A truncated sine wave and (f) its spectrum with nonzero coefficients at all k. (g) A rectangular window with a zero sample at the end and (h) its spectrum with nonzero coefficients at all k.

the window shown in Fig. 11.6(c) is an ideal narrowband filter passing only one frequency whereas the filter shown in Fig. 11.6(g) has a nonzero stopband response passing frequencies in the neighborhood. Due to the aliasing effect, a set of frequencies fold back on to a single frequency in the spectrum whereas, due to the leakage effect, a single frequency component produces response at a set of frequencies in the spectrum. This creation of new frequencies has to be taken into account in considering the aliasing effect. Similar to the reduction of the aliasing effect by reducing the sampling interval, the leakage effect can be reduced by sampling over a more appropriate record length. Once truncation occurs, the only choice is whether it is desirable to reduce the second error, which is to reduce the magnitude

Leakage Effect 235

of the new frequency components produced at the cost of increasing the first error, the smearing of the spectrum. To study this choice, we have to understand the properties of various windows.

Before we do that, let us look at another truncation model. It is to reduce the record length we truncate a signal. Therefore, if the spectrum is sufficiently dense, we can compute the spectrum of the nonzero part of a signal after truncation. In the truncation model presented so far, we assumed that the signal is sampled correctly and the window is imperfect. In computing the DFT of the nonzero part of a truncated signal, we assume that the window is perfect and the signal is not sampled over an integral number of cycles.

The frequency response of the DFT

The DFT of the complex sinusoid, x{n) — e-7"^"', is given by

X{k) = jS^'e-^"* n = 0

J V - 1

n = 0

I - e-J^-(k-l)N

l-e-i^-(k-i)

= sirnr(fc-Q fa(1_j,K>_n

sin £(*;-/)

If I is an integer, the sinusoid completes an integral number of cycles in the period N and we get an impulse at the corresponding frequency index properly representing the sinusoid. If I is not an integer, the sinusoid does not complete an integral number of cycles in the period N and, as a result, it is represented not as a sinusoid at a single frequency but as a combination of a number of sinusoids those have bins. The energy of a sinusoid, in this case, is leaked to neighboring frequencies. The set of narrowband filters, the DFT, does not have a sharp response for frequency components at other than bin frequencies and has a worst response for frequency components with frequencies at the midpoint between bins.


1

0.08 0

o 8 $

M ° + • o 9 r +

. . + -*•

« » 9 " « x 9 >

9 " 8 X

8 " • rectangular + o hamming

+ Hann x triangular

X ® 9 9 X 9

X 9 L--

8 * 0

+

= 32 = 32

5n + " ? ,

0 4 8 12 16 20 24 28 32 n

Fig. 11.7 Time-domain representation of different windows.

Windows

Rectangular window

The rectangular window is denned as

. . . f 1 for n — 0 , 1 , . . .,L — 1 rectw(n) = \

[ 0 for n = L,L + 1,...,N - 1

The time-domain representation of this window is shown in Fig. 11.7 with L = 32 and TV = 32. The DFT of this window, with u = ^ , is given by

The magnitude of the DFT of this window is shown in Fig. 11.8(a) with L = 16 and N = 64. The magnitude of the largest side lobe is -13.339 dB (slightly greater than one-fifth of the main lobe magnitude). The response to complex sinusoids is the same as given in the last subsection, which can be obtained by replacing k = k — I and N = L in Eq. (11-1). The function in Eq. (11.1) has a value of L at k = 0. X(k) = 0 for A; = £ , ^ , . . . . Main lobe width is ^ . Side lobe width is ^ . Apart from k = 0, peaks occur at k = 2T' I f ' • • •• Magnitude at peaks is given by s i n ^( 2 f c + 1 ^ , fc = 1,2,.. .. For large values of L, the side lobe peak magnitudes are approximately | ^ , 2 i a + U — 3N_ 5N_ 5 7 r v • • a t ft — 2L > 2L '

There are other windows, some of which we present in this section, apart from the rectangular window which is inherent whenever we truncate a signal. These windows are smoother and, therefore, their spectra have

Leakage Effect 237

0

pa - 2 0

£ - 4 0

| - 6 0

-80

rectangular N = 64 L= 16

(a)

Harm

L= 16

-32.192

16

(c)

24

0

« - l u T3

- -20 o ^ - 3 0 J <

* _40

-50

"•» triangular > N = 64 \ L=16

\ 1 \ ^ -25.913

lA ^ \ 1 \ /A *•* \f\ A II \t \l \

12 18 24 30

(b)

0

pa -10

§ - 2 0

$ - 3 0 >3

-40

' •• hamming * . N = 64

. L=16

•

• -40.160

• • • • • • « « « • • • • • • • • .

0 4 10 16 k

(d)

22 28

Fig. 11.8 Harm, and

(a), (b), (c), and (d): The magnitude spectrum of the rectangular, triangular, hamming windows, respectively.

wider main lobe and smaller side lobes. The truncated data is modified by multiplying it with the window samples so that the sample values of the modified data is smoothly reduced to zero or near zero at both the beginning and end of the record. The major advantage of these other windows is the possibility of detecting a weak frequency component that may be masked by the large side lobes of the rectangular window. Therefore, the rectangular window is preferred: (i) if there is no leakage, (ii) if the leakage is within tolerable limits, or (iii) if leakage can be reduced to tolerable limits by using a more appropriate record length. The other windows when used with no leakage or very little leakage leads to the undesirable effect of smearing the spectrum due to a wider main lobe with no advantage. There are several windows with different characteristics. A choice has to be made depending on the requirements.


Triangular window

One approach to have smaller side lobes is to multiply the DFT of the rectangular window by itself. This increases the ratio between the main and side lobe amplitudes at the cost of doubling the main lobe width. The multiplication of the DFTs results in the convolution of rectangular windows in the time-domain. The triangular window is defined as

{ \n for n = 0 , 1 , . . . , §

triw(L -n) for n = \ + 1, \ + 2 , . . . , L - 1 0 forn = L,L + l,...,N-l

The time-domain representation of this window is shown in Fig. 11.7 with L — 32 and N = 32. The circular convolution of a rectangular window

red (n)-i l ^v n = 0,1,. ..,^ - 1

with itself followed by a circular right shift of one position yields the triangular window with a scale factor. For example,

xrectw(n) = {1,1,0,0} e> Xrect„(k) = {2,l-jl,0,l+jl} Lxtriw(n + 1) = {1,2,1,0} ^ X?ectJk) = {4,-j2,0,j2} 2

%xM„(n) = {0,1,2,1} & X r2

e r f J fc )e -^ f c = { 4 , - 2 , 0 , - 2 }

Therefore, the DFT of the triangular window is given in terms of that of the rectangular window as

2

That is, with u = ^ ,

2 /sinfrjfcf) X^ ( f c )-Z^ s i n (f) ) 6 (1L2)

The magnitude of the DFT of this window is shown in Fig. 11.8(b) with L = 16 and N = 64. The magnitude of the largest side lobe is -25.913 dB. The response to complex sinusoids is obtained, by replacing k by k — I and N = L in Eq. (11.2), as

2 / _ s i n - V ^ \ Mk_l}

L\smUk-l)

Leakage Effect 239

Let us consider the reason for the differences in the spectral behavior of the rectangular and triangular windows. Because of discontinuity, according to the theory of Fourier analysis, the amplitude of the spectrum of the rectangular window decreases at the rate which is a function of j , where k is the frequency index. This is evident from Eq. (11.1) and from Fig. 11.8(a). With no discontinuity, the amplitude of the spectrum of the triangular window decreases at the rate which is a function of -^ • This is evident from Eq. (11.2) and from Fig. 11.8(b). In addition, from Eqs. (11.1) and (11.2), we find that the main lobe width is ^ for the rectangular window and ^ for the triangular window. Therefore, the main lobe width is 8 in Fig. 11.8(a) and it is 16 in Fig. 11.8(b).

Hann window

Another approach to reduce the size of the side lobes of the rectangular window is to make the frequency response to be the sum of the scaled and shifted responses of three rectangular windows so that the side lobes tend to cancel out. This implies the multiplication of the rectangular window by another function (a linear combination of complex exponentials) in the time-domain. The multiplication in the time-domain results in the convolution of the DFTs of the two functions in the frequency-domain. The Hann window is defined as

hnn rr7-)- / 0.5 - 0.5 cos (^n) forn = 0 , l , . . . , L - l

The time-domain representation of this window is shown in Fig. 11.7 with L = 32 and N = 32. Since

cos(—n) = ,

the response to complex sinusoids of this window is given in terms of that of the rectangular window as

Xhan„ (*) = 0.5Xrect„ (k) - 0.25Xrectw (* + 1) - 0.2bXrectw {k - 1)

The magnitude of the DFT of this window is shown in Fig. 11.8(c) with L = 16 and N = 64. The magnitude of the largest side lobe is -32.192 dB. The main lobe width is 16 samples.


Hamming window

The hamming window is defined as

'.54 - 0.46cos(^n) for n = 0 , 1 , . . . , L - 1 hamw(n) = <

v ' { 0 for n = £ , £ + ! , . . . , JV - 1

The time-domain representation of this window is shown in Fig. 11.7 with L = 32 and N = 32. The response to complex sinusoids of this window is given in terms of that of the rectangular window as

Xhamw (k) = 0.54Xrect„ (*) - 0.23Xrect„ (k + 1) - 0.23Xrectw (fc - 1)

The magnitude of the DFT of this window is shown in Fig. 11.8(d) with L = 16 and N = 64. The magnitude of the largest side lobe is -40.160 dB. The main lobe width is 16 samples.

The frequency response of the rectangular and Hann windows

The frequency response of the rectangular and Hann windows to complex exponential, x{n) = e-7^"', with frequency indices 1.0, 1.3 and 1.5 and N = 64 are given in Fig. 11.9. With frequency index 1.0, there is no leakage and the response with the rectangular window is ideal as shown in Fig. 11.9(a) whereas that with the Hann window gives responses at two adjacent frequencies also as shown in (b). As can be seen from (c), with frequency index 1.3, the response with the rectangular window is relatively sharp but with a large leakage effect. There is equal response, in (e), at frequency indices 1 and 2 as the frequency of the complex sinusoid is 1.5, that is in between these frequencies. As can be seen from (d) and (f), the Hann window provides a much reduced leakage but the response is less sharp close to the frequency of the complex sinusoid.

The frequency response can be explained in terms of the sine function. The DFT detects frequency components with integer frequency indices exactly. That is the zero crossings of the sine function, the frequency response of the DFT shown by a continuous curve, occurs at all the bin frequencies except at the frequency of interest, where the peak of the main lobe is the sample value. With a non-integer frequency index, the sine function is centered at that frequency giving nonzero response at all bin frequencies. The difference between the continuous response curves is that the response

Leakage Effect 241

pa "

£ - 2 0

-30

rectangular W=64 /=1.0

(a)

AAV 2 3

k

(e)

-6 -20

-40

-60

-6 -20

-40

-60

-6 -20

-40

-60 (

3 1

3 1

3 1

2

2

2

Hann \ W=64 \ /=1.0

Y\n\ 3 4 5

k

(b)

^. /=1.3

v \ r > 3 4 5

k

(d)

\ . /=1 .5

nr 3 4 5

k

(f)

Fig. 11.9 (a), (c), and (e): The frequency response of the rectangular window with the frequency index of the complex exponential signal 1, 1.3, and 1.5, respectively. The response is relatively sharp with large sidelobes. (b), (d), and (f): The frequency response of the Hann window is relatively broad with small sidelobes.

of the Hann window, which is a combination of three sine functions, has a broader main lobe and smaller side lobes.

Truncation and spectral resolution

The clear identification of the individual components of a spectrum is called the spectral resolution. The best spectral resolution is obtained with no truncation. This implies that the record length must be, at the least, one complete cycle of a periodic signal and the whole nonzero part of an aperiodic signal. With truncation, the resolution is affected in two ways. Zero padding of a signal makes its spectrum denser, but it can not improve the resolution.


"fir

£ "><"

"? X

X

;rr

• 0 20

4 0

-4

• A

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0 16

•'• •"• - . A • • • • • < ^ * L * • •

40 n

(a)

40 n

(c)

0 20

4 0 f" -\ - .„

- . ^ . — — 0 16

11 i n fo i T I , O

40 n

(•>

40 n

(g)

60

60

60

60

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0

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. • • '

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0

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0

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7 9

• •

7 9

• •

7 10

•

7 10

^ CM ;+c

20 k

(b)

20

(d)

20 k

(f)

20

(h)

30

30

30

.". •• 30

f,-\ T\

truncated signal of (a), and (d) its spectrum, (e) The signal, 2cos( | j7n)+2cos( | j l0n) , and (f) its spectrum, (g) The truncated signal of (e), and (h) its spectrum.

Figures 11.10(a) and (b) show, respectively, the signal, 2cos(g|7n) + 2cos( |f9n), and its spectrum indicating the presence of frequency components with frequency indices 7 and 9. Figures 11.10(c) and (d) show, respectively, a truncated version of the signal and its spectrum indicating just one peak. This is due to the convolution of the spectrum of the untrun-cated signal with that of the rectangular window with L = 16 and N = 64. We cannot clearly identify the two frequency components.

Figures 11.10(e) and (f) show, respectively, the signal, 2cos(|f-7n) + 2cos(|^10n), and its spectrum indicating the presence of frequency components with frequency indices 7 and 10. Figures 11.10(g) and (h) show, respectively, a truncated version of the signal and its spectrum with two peaks. Therefore, with truncation of the signal, we are able to distinguish neighboring sinusoids only when the separation between them is relatively

Leakage Effect 243

- 2 r « •"• £ 0 • • '

_2L»— 16

2r A

°tv>: 16

20 40 n

(a)

40

(c)

40

60

60

60

60

5 I t210° [

* 1 0 "

18

(b)

18

30

• •

0 7 • • . " • • • • . - " • • " • • •

18 30 k

(d)

30

(e) (0

Fig. 11.11 (a) The signal, 2 c o s ( | | 7 n ) + 0 . 1 c o s ( | j l 8 n ) , and (b) its spectrum, (c) The truncated signal of (a), and (d) its spectrum, (e) The signal shown in (c) after the application of the Hann window and (f) its spectrum.

large. As the rectangular window has the smallest main lobe width, it gives the best response in this respect.

Figures 11.11(a) and (b) show, respectively, the signal,

2 cos(-^7n) + 0.1 cos(-^18n), 64 64

and its spectrum indicating the presence of frequency components with frequency indices 7 and 18. Figures 11.11(c) and (d) show, respectively, a truncated version of the signal and its spectrum indicating the presence of the frequency component with index 7 clearly. It is almost impossible to infer the presence of the second frequency component. As its magnitude is small, it is masked by the interaction of the large side lobes of the component with a large magnitude. Figures 11.11(e) and (f) show, respectively, a truncated version of the signal using the Hann window and its spectrum, although blurred, indicating the presence of the two frequency components. As the magnitude of the side lobes of all the windows is smaller than that of the rectangular window, the other windows provide a better resolution in this respect. We conclude that a careful selection of the window has to be made to suit the specific application requirements.


Reduction of leakage

To reduce leakage, the natural solution is to use an appropriate record length, if possible. Similar to the case of the aliasing problem, prior knowledge of a signal will enable to fix the record length or a trial and error procedure can be used, keeping the sampling interval fixed. Otherwise, leakage can be reduced at the expense of increased smearing of the spectrum by the application of windows. As mentioned earlier, if there is no leakage or negligible leakage then choose the rectangular window. Otherwise choose any of the other windows that reduces the side lobes to the desired level. Windows can be applied in the time-domain by multiplying the data, x(n), with window samples, w(n). That is, instead of computing the DFT of x(n), we compute the DFT of x(n)w(n). Note that the window w(n) must be centered on the truncated signal x{n) in the defined interval of x(n), wherever it may be on the time scale. Windows can also be applied, in the frequency-domain, in terms of the frequency coefficients of the rectangular window. Although our discussion concentrated on the truncation problem in the time-domain, it should be noted that windows can be used in both the time- and frequency-domains.

11.3 Picket-Fence Effect

In this section, we consider the appropriate frequency spacing or increment for the unique representation of a signal. The theory of Fourier analysis is that the frequency increment between the harmonic components must be the reciprocal of the period of the waveform under analysis. Here again, we run into a problem because we tend to violate the theory by not providing proper frequency spacing (because of not using a long enough record length), in order to make the problem suitable for numerical analysis.

The DFT coefficients specify the spectrum only at certain frequencies. The computation of the coefficients at these frequencies is carried out exactly the same way whether a signal is periodic or not. For periodic signals, with input samples taken over an integral number of periods, the frequency increment is correct. If we assume that the signal is periodic, then the spectrum of the signal shown in Fig. 11.12(a) is just the dc component as shown in Fig. 11.12(b). If we mean that the same set of samples represents an aperiodic signal, then its spectrum is continuous. The DFT coefficients are only a finite set of samples of the continuous spectrum, even if it is band-

Picket-Fence Effect 245

16 i

= 1

* 0 . 0

_ 1 e K 0

0

_ 1 e

0 16

5

N=16

n

(a)

W = 32

16 n

(c)

N = 64

32 n

<•)

10

48

15

31

63

3f

5 *

}

0 . 0

16 *

0 •

•

0

16 * •

0

• • 2

5

10

• • 4 k

(b)

10 k

(d)

20 k

(0

• 6

_ j

8

. • . 15

30

Fig. 11.12 (a) A signal and (b) its spectrum, (c) The signal shown in (a) with 16 zeros appended and (d) its spectrum, (e) The signal shown in (a) with 48 zeros appended and (f) its spectrum. With more and more zero padding, a denser spectrum is obtained.

limited. Therefore, the spectrum shown in Fig. 11.12(b) is also the correct frequency-domain representation of the aperiodic signal, but is very coarse. It seems as though we look at the spectrum through a picket-fence. With a coarse spectrum, we may miss some of its features. For example, if there is a peak of the spectrum in between two frequencies it will be missed.

How to make the spectrum denser

As the frequency spacing is inversely proportional to the record length of the time-domain signal, we get a coarse spectrum with a short record length. Only by increasing the input record length, we can have a denser spectrum. The best thing to do is to reduce the level of truncation, if employed, and get samples over a longer record length. If the nonzero part of the record of a signal is finite, then pad the signal with zeros at the end to make the record length longer. If required, a window should first be employed and then zero padding carried out. Figure 11.12(c) shows the signal in Fig. 11.12(a) padded with 16 zeros. The corresponding spectrum, shown in Fig. 11.12(d), presents a denser spectrum (only half of the spectrum is


shown since it is symmetric). With more and more zero padding, we get a denser spectrum as shown in Figs. 11.12(e) and (f). The signal shown in Fig. 11.12(e) is a more accurate representation of the given aperiodic signal than that shown in Fig. 11.12(c).

A periodic signal is composed of components with frequencies which are integral multiples of its frequency and of no other. Therefore, we should set the record length equal to an integral number of cycles and there should be no zero padding. If we truncate a periodic signal, we are making an analysis with a false period. An aperiodic signal is composed of components of all frequencies. Whether we truncate an aperiodic signal or not, the frequency coefficients computed by DFT are part of the spectrum. The point is that the signal processed by the DFT should be sufficiently close to the actual signal, with the appropriate selection of the record length and the number of samples so that the aliasing and leakage errors are small enough and the spectrum is sufficiently denser.


• In this chapter, we studied the aliasing, leakage, and picket-fence effects. These effects occur, in the analysis of a signal, due to the setting of parameters such as record length, sampling interval, and frequency spacing not in accordance with the theory of the Fourier analysis, in order to use the DFT. It is normal that the DFT solution is not exact as the solution given by analytical method. But, the analytical method is almost impossible to use with practical signals. What is important is to ensure that the error in DFT representation is within the required limits. Fortunately, we can reduce the errors to any desired level by increasing the number of samples and/or the record length and using other techniques presented.

• Noting that DFT means the same computation at any time, the accuracy control lies in the data modeling. Although sampling and truncation are unavoidable to analyze a signal using the DFT, the user has to ensure that the digital signal is sufficiently close to the actual signal to keep the errors within tolerable limits. Therefore, if the DFT of a signal seems to be in gross error it is because of improper data modeling, such as insufficient sampling rate, too much of truncation, insufficient number of bits to represent data values,

Exercises 247

etc. The situation is like using a computer to solve a problem. A computer executes some instructions in a predefined way. If the execution of a program yields improper output, the reason is either the programmer not putting up the proper algorithm, or proper codes according to the language used, or proper input data.

• In the next two chapters, we describe the approximation of the continuous-time Fourier series and Fourier transform of signals and provide examples illustrating the effects described in this chapter.

Reference


Exercises

11.1 What is the minimum number of samples in a period required to avoid aliasing. What are the DFT coefficients. Find the DFT coefficients with 4, 8, 16, and 32 samples in a period. (a) x(t) = -4sin(107ri) (b) x(t) = cos(107ri) * (c)x(t) = 2cos(147rt) (d) x(t) = 5sin(167rt) (e) x(t) =3cos(327it) (f) x(t) =4sin(127rt) (g) x(t) =2cos(24?ri) (h) x(t) = - 3

11.2 With JV = 8 and L = 8, list the DFT coefficients of the signal using the rectangular, triangular, Hann, and hamming windows. (a) x(n) = cos(^2n) * (b) x(n) = cos(^2.5n)

* 11.3 The frequency of the significant highest possible frequency component of an aperiodic signal of record length 0.1 seconds is 5 kHz. A minimum frequency increment of 20 Hz is required. Determine the number of samples required to approximate the spectrum using the DFT. Is zero


padding required? If so, what is the minimum record length required?

11.4 The frequency of the significant highest possible frequency component of an aperiodic signal of record length 0.1 seconds is 12.5 kHz. A minimum frequency increment of 5 Hz is required. Determine the number of samples required to approximate the spectrum using the DFT. Is zero padding required? If so, what is the minimum record length required?

P r o g r a m m i n g Exercises

11.1 Write a program to apply the Hann window to a signal, with L = JV'.

11.2 Write a program to apply the hamming window to a signal, with L = N.

11.3 Write a program to apply the triangular window to a signal, with L = N.

Chapter 12

The Continuous-Time Fourier Series

The FS is the frequency-domain representation of a continuous-time periodic signal in terms of an infinite set of harmonically related sinusoids in addition to a dc value. Both the DFT and the FS do the same function, that is, providing the sinusoidal representation of signals. However, the DFT analyzes a discrete signal whereas the FS analyzes a continuous-time signal. Therefore, the essential differences to be considered in approximating the FS coefficients by those of the DFT are: (i) the integral evaluated in the case of the FS is approximated by a numerical integration procedure and (ii) a finite set of coefficients is used in DFT and an infinite set of coefficients is used in FS.

In Sec. 12.1, we start with the two forms of the definitions of the 1-D FS, describe the Gibbs phenomenon, present the relation between the DFT and the FS, and conclude with an example of approximating the FS coefficients by those of the DFT. In Sec. 12.2, we describe the approximation of the 2-D FS coefficients by those of the DFT.

12.1 The 1-D Continuous-Time Fourier Series

The FS represents a continuous-time periodic signal, x(t), with period T as a sum of an infinite set of harmonically related sinusoids in addition to a dc value. The frequency of the fundamental harmonic is u>o = ^ . The frequencies of the other harmonics are integral multiples of w0. The sufficient conditions, called Dirichlet conditions, a signal should satisfy so that it can be represented by a Fourier series are: (i) the signal x(t) has an absolutely convergent integral, that is / 0 |x(t)|di < oo, (ii) the signal has

249

250 The Continuous- Time Fourier Series

a finite number of maxima and minima in one period, and (iii) the signal has a finite number of finite discontinuities in one period. These conditions are met by most signals of practical interest.

The trigonometric form

A real periodic signal, x(t), satisfying Dirichlet conditions, can be equiva-lently expressed in terms of cosine and sine waveforms as

oo

x(t) = Xc(0) + Y^(Xc(k) cos(k(j0t) + Xa(k) sm(kuj0t)), (12.1)

where

1 / T l - t - i

Xc(0) = TJ x(f)dt, o rti+T

Xc(k) = — I x(t) cos(fcw0*) dt, k = 1,2,. . . , T Jti

ti+T

{'] '" . ~ " , ' ~.~. , o o tl

2 ftl+T

X„(k) = — x(t) sm(kcj0t) dt, k=l,2,...,oo 1 Jh

and ti is arbitrary.

The exponential form

As in the case of the DFT, it is more efficient to represent a sinusoid in terms of complex exponentials. Substituting the fact that

ejk<jj0t i e—jkwot eJkuiot _ e—jku>ot cos(ku>ot) — and sin(kuiot) =

into Eq. (12.1), we get

0 0 pjkuot _i_ p—jkuiot pjkuot _ p—jkuot

x(t) = Xc(0) + ]T(Xe(fc)- \ + * . (* ) a ) fc=i J

Rearranging the terms, we get

*(*) = XM + f v ^ W - ^ W + xc(k)+jx8(k)e_jkuot)

k=l

The 1-D Continuous-Time Fourier Series 251

Let XC8(0) = Xe(0), XC8(k) = *°W-/M*>, and Xcs(-k) = *°W+J'*-W. Note that Xcg(fc) and Xcg(—fc) are complex conjugates, and Xc{k) = 2Re(Xc8(k)) and X8(k) = -21m(Xcs(k)). Now, x(t) can be expressed as

oo

x(t) = XC8(0) + £ ( X C 8 ( A ; ) e w + X „ (-*)e-'fa*>«)

By allowing the summation index, k, to run from — oo to oo, we can rewrite the equation as

oo

x(t) = J2 Xcs(k)ej^ot (12.2) fc= — oo

We use the denning equations for Xc(k) and X8(k) to get XC8(k) as

1 / - t l + T xcs(k) = = / 2;(i)(cos(fcw0i) - jsin(fcw0i))^,

1 /"'1+T

= - / x{t)e-jkuot dt, k = - o o , . . . , - 1 , 0 , 1 , . . . , oo (12.3)

Gibbs phenomenon

The failure of the Fourier representation to provide uniform convergence in the vicinity of a discontinuity of a signal is called the Gibbs phenomenon. In the vicinity of a discontinuity, the Fourier representation deviates at the least about 9% from the original signal irrespective of the number of coefficients used. It is not surprising since it is impossible to provide a pointwise match in the vicinity of a discontinuity with a set of basis functions each of those is continuous. Where there is no discontinuity, by increasing the number of frequency coefficients to represent a signal, the Fourier representation can be made to correspond to the original signal to any tolerance.

Let us assume that we are approximating a signal using only part of its spectrum (2JV+1 complex frequency coefficients). The part of the spectrum used can be considered as obtained by multiplying the original spectrum by a rectangular window whose value is unity during the part of the spectrum used and zero otherwise. This multiplication operation in the frequency-domain is equivalent to convolution of the corresponding signals in the time-domain. Now, we are going to derive the convolution expression that

252 The Continuous-Time Fourier Series

will help us to understand the characteristics of the reconstructed signal. The FS synthesis expression with 2iV + 1 coefficients is given by

N

k=-N

Replacing Xcs(k) by its definition and interchanging the order of summation and integration, we get

N

*N(t) = i f x(l) J2 ejk»°Wdl 1 Jo k=-N

Let us evaluate the summation operation separately.

N y ^ ejkuj0{t-i)

k=-N

2N y^ ej(k-N)w0(t-i)

k=0

l_ e i (2iV+l)wo(t-0 _ e-jNwQ{t-l)_ sin( (2N+l)u0{t-l)

) I — eJuo(t-l) s i n ( ^ i )

Now, the convolution expression characterizing the reconstructed signal using only 2N + 1 coefficients is given by

XN (t) = ±j\(l) s i n ( (2Af+ iHf t - i ? )

8 i n ( i - 2 ^ 1 )

i rT

= - / x(l)xs(t - l)dl 1 Jo

The reconstructed signal using 27V + 1 coefficients is the result of the convolution of the original signal x(l) and the sine function xs(l). The sine function, for the specific value of t = 3.14, is shown, with T = 6.28 seconds, N = 10, and I ranging from 0 to T, in Fig. 12.1(a) and with N = 20 in Fig. 12.1(b). The sine function has one main lobe (the region between the

(a) (b)

Fig. 12.1 (a) and (b): The sine function is(3.14 — I), respectively, with N = 10 and AT = 20.


(a) (b)

Fig. 12.2 (a) and (b): The reconstructed square wave with JV = 10 and JV = 20, respectively.

two adjacent zero crossings about / = 3.14) and large number of side lobes (the other regions between adjacent zero crossings), which are alternately positive and negative. The total area of the sine function during a period is T for any value of JV since

f Y" gifc-oC*-*) a = f (l + 2cos(wo(t-0) J° k=-N J0

+ 2 cos(2cj0(t - I)) + • • • + 2 cos(Noj0(t - I))) dl = / dl = T Jo

An inspection of the figures shows that the total area of the side lobes pointing downwards is more than that of the side lobes pointing upwards. To compensate for this, the area of the main lobe is more than T. The result of this is that the synthesized signal exhibits at the least about 9% deviation at a discontinuity of the signal. Figures 12.2(a) and (b) show, respectively, the Fourier synthesis of a square wave with N = 10 and JV = 20. As the number of coefficients is increased, the amplitude of the deviation reaches a finite limit since the ratio between the amplitudes of the main and side lobes reaches a finite limit, but the frequency of the oscillations increase. This behavior is different from that of the other parts of the signal, where there is no discontinuity, in that the match between the signal and its Fourier synthesis improves as the number of coefficients is increased.

The relation between the DFT and the FS

Let us approximate the integral in Eq. (12.3) using the rectangular rule of numerical integration. We divide the period T into JV intervals, each of those is of width Ts = ^ , and represent the signal at JV points as


x(0),x(£),x(2^),...,x((N-l)$). Then, Eq. (12.3) is approximated as

T *-> v " N N n=0 n=0

Since x(nTs) represents the nth sample, the equation can be rewritten as

1 J V _ 1

Xca(k) = -J£x(n)e-i¥nk, k = 0,l,...,N-l N

71=0

This equation analyzes a waveform in terms of sinusoids. The inverse operation, that is the synthesis of a waveform from the coefficients, Eq. (12.2), is approximated as

J V - l

x(n) = Y, Xcs{k)e^nk, n = 0,l,...,N-l.

One difference between the DFT equation and the numerical approximation to the FS is the factor jj. Therefore, by dividing the DFT coefficients by N, we get an approximation to the complex FS coefficients. Comparing the approximation of the synthesis equation with the IDFT equation, we find that there is no constant divisor ^ in the synthesis equation. For N even, comparing the coefficients of the DFT with that of the FS, we get, for real signals,

Xc(k) = |pRe(X(fc)), Xs(k) = -^Im(X(k)), k = 1,2, . . . , y - 1

Xcs(k) = ^ - , * = 0 , l , . . . , y - l and R e ( X c s ( ^ ) ) = ^

Errors arise in DFT processing because of the digital representation of the data and coefficients, and the round off of numbers in arithmetic operations, as in any numerical computation using digital devices.

Aliasing effect

Once we represent the signal by N samples, we are able to compute only N distinct frequency coefficients. If the input signal is bandlimited to components with frequency indices less than ^ i then there is no problem.


Otherwise, the aliasing effect, described in terms of time-domain signals in the last chapter, corrupts the spectrum. Consider the synthesis of a signal,

oo

x(t) = Y, Xcs(iwlaot

l=—oo

where u>o = %jr. Let Ts — ]j. To generate only samples of the signal at intervals of Ts, we get

oo oo

x(nTs) = J2 Xca{l)eiluonT' = £ Xe.(I)e>*,n

l=—oo l=—oo

Let I = k + rN, where r = —oo,... , —1 ,0 ,1 , . . . , oo. Then,

JV-1 oo N-l oo

x(nTs) = Y, J2 Xcs(k+rN)ej^k+rN)n = Y, H XC8(k + rN)ej^kn

k=0 r=—oo fc=0 r=—oo

We use X)^l-oo Xcs{k + rN) to get the samples of continuous-time signal x(t). Comparing this equation with that of the IDFT, we find that

oo

X(k) = N ] T XC8(k + rN), fc = 0 , 1 , . . . , J V - l r = —oo

This equation shows how the DFT coefficients are corrupted due to aliasing.

Example 12.1 The aliasing effect Figure 12.3(a) shows the FS magnitude spectrum, which is aperiodic, of the signal x(t) = 4 + 6 sin(27r£) + 4 sin 2(27ri) + 2 sin 3(27ri). The signal has frequency components, apart from dc, with frequencies 1 Hz, 2 Hz, and 3 Hz. Therefore, we need a minimum of seven samples in a period to represent it accurately. Figure 12.3(b) shows the scaled DFT periodic spectrum which is a superposition sum of the aperiodic spectrum placed at intervals of nine samples. As the interval of nine samples is more than adequate to prevent overlap of nonzero values of the spectrum, there is no aliasing effect. Figure 12.3(c) shows the scaled DFT periodic spectrum with a period of seven samples and there is no aliasing effect. Figure 12.3(d) shows the scaled DFT periodic spectrum which is a corrupted periodic version of the aperiodic spectrum. This is because the period of five samples is inadequate to prevent the overlap of nonzero values of the aperiodic spectrum. The magnitude of the spectral value with index 2 is one. We can recover the original signal from its samples with N = 7 and N = 9 using a lowpass


4

"B2 X

FS aperiodij spgctrum

4

S 2 * 1

• • • • • • DFT, pegodic^spejtrurr^ /V= 9

U al ias ing . . -10 0

k

(a)

10 -10 0 k

(b)

10

^ 3

& 2

1

DFT periodic spectrum, N = 7

No aliasinq

4

§.3

1

DFT periodic spectrum, N= 5 Aliasing

-10 0 k

(c)

10 -10 0 k

(d)

10

Fig. 12.3 (a) The FS aperiodic spectrum of the signal x(t) = 4 + 6sin(27rt)-f 4 sin(4n-t)+ 2sin(67rt). (b) The scaled DFT periodic spectrum, with N = 9, of i ( t ) with no overlapping of nonzero values and hence no aliasing, (c) The scaled DFT periodic spectrum, with N = 7, of x(t) with no aliasing, (d) The scaled DFT periodic spectrum, with N = 5, of x(t) with overlapping of nonzero values resulting in a corrupted spectrum.

filter (because the sampling process has not altered the original spectrum, except for repeating), whereas it is not possible with N = 5. I

The aliasing effect can also be explained in the following way. The creation of N samples for numerical approximation can be considered as multiplying one period of the continuous-time signal x(t) with the sampling signal, a set of unit-impulses. Then, due to the convolution theorem, we get a periodic spectrum obtained by superposition sum of an infinite number of shifted aperiodic FS spectrum. Remember that the FS of the sampling signal is also a set of impulses and the convolution of a spectrum with an impulse is just translating the origin of the spectrum to the location of the impulse. The amount of shifting of the spectrum is inversely proportional to the sampling interval. The sampling interval, in order to avoid aliasing, must be sufficiently short so that there is no overlap of nonzero values of the spectra due to the superposition operation.

If the data samples are not taken over an integral number of periods, then leakage effect also must be accounted for.


Continuous-time frequency

The DFT produces N output values corresponding to N input values and the DFT computation is independent of the sampling interval Ts. Therefore, the actual frequency can be found only if Ts or T = NTS is known. The fundamental frequency /o = ^ is also the frequency increment in Hz. The DFT frequency index k represents a frequency of A;/0 Hz. The DFT time index n represents nTs seconds. The folding frequency is y / 0 and the highest usable frequency is ( y - l) /o.

Sample value at a discontinuity

At points of discontinuity of the function x(t), the average value

should be assigned to a sample, where x(t+) and x{t~) are the limiting values of x(t) as t tends to tn from the right and left, respectively. In order that the value x(tn) provides minimum least squares error,

(x(tn)-x(tt))2 + (x(tn)-x(t-))2

must be minimum. Differentiating this expression with respect to x(tn) and equating to zero, we get the result. In practice, it may be difficult to set the sample value at each discontinuity to the average of the two immediately adjacent values. This will introduce some error, which can be reduced by increasing the number of samples.

Example 12.2 The approximation of the FS coefficients of a square wave.

x(t) = | A for 0 < t < £ 0 for y < t < T

Solution The waveform is odd-symmetric and odd half-wave symmetric with a dc bias. Therefore, we expect only the odd-indexed sine waves as the frequency components, apart from the dc component.

1 /"= A Xcs(0) = -Jo Adt=-


•5- _ K , Period, IT =1 •g. 0.5 '/v=4 > "? n K Period T=1

0 2 4 6

(b)

io.; • I M I M I

„ Period, r= 1 'W=16"

0 4 8 12

(c)

Fig. 12.4 (a), (b), and (c): The continuous-time square wave with period T = 1 second and its representation by 4, 8, and 16 samples, respectively.

X (k) - - r Ae-f^dt-i ~j£ hrkodd Xcs{k) - r ^ Ae d t - \ Q i o l k e v e r even and k ^0

37T 7T 2 2 4 , . 2?r 1 . 2TT 1 . 2TT

+ — ( s i n — i + - s m 3 — i + - s i n 5 — < + •••)

x(t) = h j — e

~ 2 + 7T l S m r ' ' 3 " ~ T ' ' 5 " " " T

Assuming that A = 1 and T = 1 second, we get

J' V ' -|- " j — g^ V * j g ^ T * A „-2* — < 7T

1 2 1 1 a:(t) = - + -(sin(27rt) + - sin3(27rf) + - sin5(27ri) + • • •) (12.4)

2 7r 3 5

Figure 12.4 shows one cycle of the square wave with 4, 8, and 16 samples. Note that the sample values at discontinuities are the average values of the right and left limits. Figure 12.5 shows the coefficients of the constituent sine waveforms of the signal along with that of the dc component, the exact coefficients marked 'o' and the coefficients computed through the DFT marked ' x ' . The even harmonic coefficients are zero. The DFT coefficients corresponding to the samples shown in Fig. 12.4(a) are {2, — j'1,0,,7'1}. Re-

0.5 > 0 0.637

oFS »DFT

A/=4

0.5 ^ *"

00.604 W=8

00.212 »0.104

0 1 2 k

(b)

3 4

0.5 „ A/=16 ° 9 o

• • • " •

1 3 5 7 k

(c)

Fig. 12.5 (a), (b), and (c): The exact and the scaled DFT values of the dc and the sine components, which constitute the square wave shown in Figs. 12.4(a), (b), and (c), respectively.

The 1-D Continuous-Time. Fourier Series 259

ferring to Fig. 12.5(a), we see that the dc value is 0.5 for both the FS and the DFT. The dc value is not affected due to aliasing since the waveform consists of no other cosine component. However, the magnitude of the harmonic with k = 1 is \ for the FS and 0.5 for the DFT. This discrepancy is due to the aliasing effect. With four time-domain samples, we can only represent frequency components with indices one and two, in addition to dc value. Therefore, all the odd frequency components, in the case of the DFT, fold back on to that of the fundamental harmonic and we get

The value of the summation can be obtained from Eq. (12.4) by substituting t = 0.25.

-, 1 2 „ 1 1 1

^ S + ^ - a + s - ^ - ) Therefore, we get Xs(l) = 0.5. In Fig. 12.5(b), we have doubled the number of samples. Therefore, the aliasing effect is reduced. For example, the frequency component with k = 3 and magnitude ^ folds back on the frequency component with k = 1 in Fig. 12.5(a). In Fig. 12.5(b), that frequency component has its independent representation. Frequency component with ifc = 1 is mixed up with frequency components starting with k = 7, whose magnitudes are ^ or smaller, and, therefore, the value computed by the DFT with k = 1 becomes more accurate. The approximation becomes much better in Fig. 12.5(c) with more number of samples. In a practical application, sufficient number of samples must be used to satisfy the accuracy requirements.

Figure 12.6 shows parts of the reconstructed waveforms, using the DFT coefficients shown in Fig. 12.5. Remember that the magnitudes of the even

f N=* \ 0,5r N=a \ 0 ,5f " - i s 0 0.5 0 0.5 0 0.5

i t t

(a) (b) (c)

Fig. 12.6 (a), (b), and (c): Parts of the reconstructed square waveforms using frequency coefficients shown in Figs. 12.5(a), (b), and (c), respectively.


1

S 0 . 5

0 0 0.5 1

t

Fig. 12.7 The reconstructed square waveform, using frequency coefficients shown in Fig. 12.5(c) modified by the Hann window.

harmonics are zero. The frequency components of the waveform consist of only sine waves with odd frequencies and a dc value. Therefore, with TV = 4 samples, we see a sine wave and a dc value representing the waveform. It can be seen that, as the number of coefficients is increased, the reconstructed waveform gets more closer to the original waveform except at the discontinuities due to Gibbs phenomenon.

In order to reduce the overshoot, the coefficients can be multiplied by window samples before used for reconstructing the signal. Figure 12.7 shows the reconstructed signal using the Hann window. Note that the reduction of the overshoot is achieved at the cost of reducing the rise time of the waveform. The function, used in the reconstruction, corresponding to this window has a wider main lobe (therefore, a slower rime time) and smaller side lobes (therefore, a smaller overshoot) compared with the sine function of the rectangular window.

Let us consider how the DFT representation is related to that of the FS in terms of actual frequency. The DFT representation of the signal, for example, with N = 8 is given as

1 2TT 2TT

x(n) = -+ 0.6036 s i n ( - n ) + 0.1036 sin(3—n) (12.5) 2 8 8

With T = 1 second and N = 8, the sampling interval is | seconds. In the DFT representation, we express the frequency as a multiple of ^ radians per sample. To get the actual frequency, we have to divide this value by the sampling interval and we get 2TT radians per second. This corresponds to a fundamental cyclic frequency of 1 Hz. The actual cyclic frequency of the third harmonic is 3 Hz. I


1.1002 1

0.5

LH

1.0901 1

0.5

LH = 13

0 0.0753 0.1907 0 0.0382 0.0969 t

0 0.0192 0.0486 t

(b)

Fig. 12.8 The initial part of the square wave, its FS reconstruction, and the last harmonic used (with some offset), (a) Using up to the 3rd harmonic (LH = 3). (b) Using up to the 7th harmonic (LH = 7). (c) Using up to the 15th harmonic (LH = 15).

Gibbs phenomenon again

The FS reconstructed waveform of the square wave, using up to the third harmonic, is given by

1 2 1 x(t) = - + -(sin(27rf) + - sin3(27ri))

2 7T 3 (12.6)

To find out the occurrence of the peak values of this waveform, we have to differentiate the expression with respect to t and equate it to zero and we get

cos(27ri)+cos(27r(3)i) = ^ g l M = 0 (12.7)

It is obvious that the first time this expression is equal to zero is when t = 0.125. For this value of t, we get x(t) = 1.1002. Figure 12.8(a) shows the initial part of the square wave, the FS reconstructed waveform, and the last harmonic used (third) in the reconstruction. Note that the third harmonic is shown with some offset. Remember that the coefficient of the harmonics are determined with respect to the least squares error criterion. Therefore, all the reconstructed waveforms are constrained to start from a value of 0.5 at t = 0, as the FS assumes the average value at a discontinuity. The waveform still keeps raising even after it reaches the value of 1 as the sinusoids are continuous signals and cannot suddenly stop. The last harmonic is the first one to have a negative slope and it starts correcting the overshoot. However, its amplitude is small and it takes some time to reach a deep negative slope. After some time, the last but one harmonic comes into play and so on. Eventually, the overshoot is corrected and an undershoot occurs.


Figures 12.8(b) and (c) show the initial part of the square wave, the FS reconstructed waveform, and the last harmonic used in the reconstruction, the last harmonic being 7 and 15, respectively. We can easily calculate, as shown earlier, that the first peak value occurs for these cases at t = 0.0625 and t = 0.03125, respectively. That is, the value of t for the occurrence of the first peak value is reduced by a factor of 2 with the doubling of the number of harmonics used for the reconstruction. The overshoot is also corrected faster by a factor of two. With an infinite number of harmonics used, the overshoot occurs almost at t = 0 and it exists only for a moment. The largest overshoot converges to the value of 1.0895 with a relatively small number of harmonics used. While the overshoot converges to a limit, the area under the overshoot is reduced and eventually becomes zero confirming that the FS provides a complete representation for any waveform with respect to the least squares error criterion. It should be noted that the FS representation provides uniform convergence when the original waveform is continuous. It is obvious because when the waveform is continuous the size of discontinuity between adjacent points is zero and the overshoot=size of discontinuity x 0.0895=0.

12.2 The 2-D Continuous-Time Fourier Series

The 2-D FS is a straightforward extension of the 1-D FS. A periodic signal, x(ti, t2), satisfying the Dirichlet conditions, with periods T\ and T2, can be equivalently expressed as

00 00

x(h,t2) = Y, E XC8(kx,k2)ejmkltlejuJ2k2t\

where wi = §£ and W2 = §£, and

Xc.(*i,fe) = = V f ' f ^ x{tut2)e-^k^e-^k^dtldt2 J1J2 Jo Jo

Example 12.3 Find the 2-D FS of the continuous-time signal, a period of which is defined as

2ir x{h,t2) = s in(— t i ) , 0 < h < 3, 0 < t2 < 2

o

Compute the 2-D DFT of the signal with 4 samples in each direction.


Solution

. ,, 27r 2TT . X{tl,t2) = B i n ( l y t i + O y t a )

j2^ ;

^e . ( l ,0 ) = - | , X c g ( - l , 0 ) = |

It is instructive to derive the FS coefficients directly from the definition. The frequency components in the h direction is analyzed as

xc8(kut2) = lj s in(y*i)e_j (¥fc l i l )d ii

Let wi = ?f. Then,

\ g— j k iu i t i

XC8{k±,t2) = 3 /-. _ k2\ 2 (~J fci"i sin(o;iti) - ui cos(aJiti))|jj

= 0, fa ^ ±1

The numerator and the denominator evaluate to zero for fa = ± 1 . Therefore, we apply l'hopital's rule by differentiating the numerator and the denominator separately with respect to fa and evaluate the limit to get

Xca(±l,t2) = T^

The frequency components in the t2 direction is analyzed as

(^7> J\^_ |2

i' -J7Tk2 '° = 0, k2 # 0

The numerator and the denominator evaluate to zero for k2 = 0. Therefore, we apply l'hopital's rule and evaluate the limit to get

X„(±1,0) = ^


The DFT coefficients with 4 samples in each direction are given below.

k2 ->•

0 0 0 0 -38 0 0 0

0 0 0 0 j8 0 0 0

The coefficients are the same as the analytically driven FS coefficients with a scale factor of 16. In this case, there is no aliasing and we are able to get the exact FS coefficients using the DFT. The DFT representation of the signal is given by

x(ni,n2) = sin(—ni)

The corresponding continuous-time frequency of the signal is obtained as ui = ^ | = Tp radians per second. I

Example 12.4 Find the 2-D FS of the continuous-time signal, a period of which is defined as

x(tut2) = ^ - s i n ( - ^ 2 ) , 0 < h < 3, 0 < t2 < 5 3 5

Compare the scaled DFT coefficients with the corresponding FS coefficients. Solution This signal is separable and, hence, the FS is found by multiplying the individual FS of the two 1-D signals ^ and sin(^-t2).

1 .Y c . (±ki ,±l ) = ( I )(^-) = ±-

27r(±fci)A 2 ' 47r (±* i ) ' fei^O

*c.(0,±l) = (i)(^) = TJ The 4 x 4 samples of the signal are

"1 r

I

n2 ->

0 0 0 0 0 0.25 0 -0.25 0 0.5 0 -0 .5 0 0.75 0 -0.75

The 2-D Continuous- Time Fourier Series 265

Note that the sampling interval in the t\ direction is § seconds whereas it is | seconds in the ti direction. That is, there is no necessity to set the sampling interval the same in the two directions. In each direction, the sampling interval has to be determined considering the frequency content of the signal in that direction. The DFT of the 4 x 4 samples is found to be

k2 ->•

*i

4-0 0 0 0

- j 3 0 l + jl 0

jl 0 -l+jl 0

J'3 - 1 - j l

-n i - j i

The DFT coefficients are very inaccurate because we have not put the average values at discontinuities. The proper sample values of the signal are

r i 2 ->•

Til

1 " 0

0 0

. 0

0.5 0.25

0.5 0.75

0 0 0 0

-0 .5 -0.25 -0 .5

-0.75

The average values at discontinuities must be computed based on the continuous-time signal. When the signal has a discontinuity in two directions, we have to set the sample value to the average of 4 values as

x(ni,m)tllta = (x(tt,4) + x(tt,1£)+x(ti,4) + x{ti,t2))/4:

The DFT coefficients of the signal with proper values at discontinuities are

k2 -t fci I

0 0 0 0

- j 4 0 1 0 0 0

- 1 0

jA -1

0 1

The coefficients in the first row are exact as there is no aliasing. However, the other coefficients are not exact due to the aliasing effect. Note that


Table 12.1 Comparison of the exact 2-D FS coefficients (second row) of Example 12.4 with those obtained from the DFT coefficients with 4 x 4 (third row), 8 x 8 (fourth row), and 16 X 16 (fifth row) samples.

0,1 1,1 2,1 3,1 4,1 5,1 6,1 7,1 8,1 -jO.25 0.0796 -jO.25 0.0625 -jO.25 0.0754 -jO.25 0.0786

0.0398 0

0.0312 0.0377

0.0265 0.0199 0.0159 0.0133 0.0114 0.0099

0.0129 0.0234

0 0.0156 0.0104 0.0065 0.0031

X(2,1) = 0 whereas tha t of the analytical value is nonzero. This is because

the spectrum changes sign a t the folding frequency and we get the average

value which is zero. As mentioned earlier, it may not be convenient, in

practice, t o set the sample values at discontinuities t o the average values.

Therefore, t o reduce the error due to this problem, the number of samples

must be increased. The reconstructed function using the D F T coefficients

is given by

2w 2TT 2W 2TT 2TT a ; (n i ,n 2 )=0 .5s in( -—n2)+0.125cos(—-ni+—-n2)-0 .125cos(—-niH—r^ n 2) v ' 4 4 4 4 4

It can be verified t ha t the samples corresponding to this function are the

same as given earlier. Table 12.1 shows the analytically obtained FS coef

ficients and those obtained by the D F T with various number of samples.

We have just shown the first half of the frequency coefficients of the fre

quency components with the frequency index fc2 = 1. Wi th more number

of samples, the coefficients become more accurate.

Figures 12.9(a) and (b) show, respectively, the signal of size 32 x 32

without and with proper sample values a t discontinuities. Figure 12.10(a)

shows the spectrum of the signal shown in Fig. 12.9(b). Figure 12.10(b)

(a) (b)

Fig. 12.9 (a) The discrete representation of the 2-D signal x{t\,t2) = ^-sin(^Lt2), with 32 x 32 samples, (b) The same as in (a) with average values at discontinuities.


(a) (b)

Fig. 12.10 (a) The scaled DFT spectrum of signal in Fig. 12.9(b). Note that X(0,1) and X(0,31) are not plotted to make the figure more clear, (b) The spectrum in the center-zero format.

shows the spectrum in the center-zero format. From these figures, we see that the value of the coefficients is decreasing towards the folding frequency. The problem of fixing the sampling frequency is the same as that in the 1-D case. In 2-D, we have to keep increasing the sampling frequency in two directions, rather than one, until the spectral values are sufficiently small close to the folding frequency.

Figure 12.11(a) shows the reconstructed signal of Fig. 12.9(a) with 32 x 32 DFT coefficients. The signal has ripples in the neighborhood of the discontinuity due to Gibbs phenomenon. Figure 12.11(b) shows the reconstructed signal after applying the Hann window to the DFT coefficients.

(a) (b)

Fig. 12.11 The reconstruction of signal in 12.9(a) with 32 x 32 DFT coefficients, (b) The reconstruction after applying the Hann window.


As expected, the ripples are reduced at the cost of reducing the rise time. Note that, for 2-D signals, we apply the 1-D window in each direction. For this example, we applied the window only in the ki direction as there is no truncation of the spectrum in the other direction. I

12.3 S u m m a r y

• In this chapter, we studied the trigonometric and complex exponential forms of the FS. The failure of the FS to provide uniform convergence in the vicinity of a discontinuity was discussed. It was shown how the FS coefficients are approximated by the DFT coefficients. The errors arising in the resulting procedure were analyzed.

• Fourier analysis is the representation of a signal in terms of sinusoids. As the function being the same, the DFT and the FS are closely related and the latter can be approximated to a desired accuracy by the DFT coefficients with proper choice of the number of samples taken over an integral number of periods of the time-domain continuous-time signal.

References



Exercises

12.1 Find the FS representation, x(t) = cos31.

12.2 Find the FS representation, x(t) = sin4t.

12.3 Square wave with even symmetry Deduce the FS representation from the result of Example 12.2.

A f <t<T

Exercises 269

* 12.4 Sawtooth wave Find the FS representation analytically.

x{t) = { ±t 0<t<T

Compute the DFT with 4, 8, and, 16 samples and compare the scaled DFT and the FS coefficients with A = 1 and T = 1.

12.5 Triangular wave Find the FS representation analytically.

M - J 7T* 0 < i < f XW ~ \ 2A(1 - f ) \ < t < T

Compute the DFT with 4, 8, and, 16 samples and compare the scaled DFT and the FS coefficients with A — 1 and T = 1.

12.6 Half-wave rectified sine wave Find the FS representation analytically.

*«> = { o A s i n ( ^ i ) 0 < i < f

J<t<T


12.7 Using the results of Exercise 12.6, deduce the FS representation of half-wave rectified cosine wave.

x(t) = { Acos (^ t ) 0 < i < ^ a n d ^ < t < T

0 ? < * < f

12.8 Using the results of Exercise 12.6, deduce the FS representation of full-wave rectified sine wave

x(t) = { A\ s i n ( ^ i ) | 0 < t < T

12.9 Using the results of Exercise 12.7, deduce the FS representation of full-wave rectified cosine wave

x(t) = { A\cos{^-t)\ 0<t<T


12.10 Square curve Find the FS representation analytically.

x(t) = { t2 0<t<T

Compute the DFT with 4, 8, and, 16 samples and compare the scaled DFT and the FS coefficients with T = 1.

* 12.11 Even symmetric pulse train Find the FS representation analytically.

A 0<t<w x{t) = { 0 w <t<T-w

A T-w<t<T

Compute the DFT with 4, 8, and, 16 samples and compare the scaled DFT and the FS coefficients with A — 1, T = 1, and w — | .

12.12 Pulse train Find the FS representation analytically.

x(t) = | A 0<t<2w 0 2w < t < T

Compute the DFT with 4, 8, and, 16 samples and compare the scaled DFT and the FS coefficients with A = 1,T — 1, and w = | .

12.13 Half inverted cosine wave Find the FS representation analytically.

/jt, f -Acosi^t) 0<t < f X(i) = l 0 f<t<T


12.14 Using the results of Exercise 12.13, deduce the FS representation of two inverted half cosine waves

<t) = { -AcosC^-t) 0 < < < £ Acos(^f) %<t<T

Exercises 271

12.15 Find the complex FS coefficients.

27T 7T 27T /(*i,*2) = cos(—h + - + —2*2), 0 < h < 7, 0 < h < 5

7 3 5

12.16 Find the FS representation analytically.

f(tut2) = 2tit2, 0 < *i < 3, 0 < *2 < 4

Compare the first 4 x 4 scaled DFT coefficients computed using 32 x 32 samples with the FS coefficients.

12.17 Find the FS representation analytically.

f(h,t2) = *i*2, 0 < *i < 2, 0 < *2 < 3

Compare the first 4 x 4 scaled DFT coefficients computed using 32 x 32 samples with the FS coefficients.

* 12.18 Find the FS representation analytically.

/(*i,*2) = * i + * 2 , 0 < * i < 3 , 0 < * 2 < 2

Compare the scaled DFT coefficients using 4 x 4, 8 x 8, 16 x 16 samples with the FS coefficients.

Chapter 13

The Continuous-Time Fourier Transform

The FT is the frequency-domain representation of a continuous-time aperiodic signal in terms of sinusoids with continuum of frequencies. Both the DFT and the FT do the same function, that is providing the sinusoidal representation of signals. However, the DFT analyzes a periodic version of a finite discrete signal whereas the FT analyzes an aperiodic continuous-time signal. Therefore, the essential differences to be considered in approximating the samples of the FT by those of the DFT are: (i) the integral evaluated in the case of the FT is approximated by a numerical integration procedure, (ii) the limit of the integration in DFT is finite whereas it is infinity in the case of FT, and (iii) a finite set of coefficients is used in the DFT and coefficients at continuum of frequencies are used in the case of the FT.

In Sec. 13.1, we start with the fact that FT is a limiting case of the FS, present the relation between the DFT and the FT, and conclude with an example of approximating the samples of the FT by those of the DFT. In Sec. 13.2, the approximation of the 2-D FT by the DFT is described.

13.1 The 1-D Continuous-Time Fourier Transform

The FT as a limiting case of the FS

The FT is the same as the FS with the period of the waveform approaching infinity. Consider the FS of the pulse train shown in Fig. 13.1 with increasing periods. Keeping the pulse width the same but doubling the period, the magnitudes are reduced by a factor of two and there are double

273

274 The Continuous-Time Fourier Transform

KMM1 ? M d Kft iriod, T= 4

(a) (b) (c)

Fig. 13.1 (a) Pulse train with period T = 1 second, (b) Pulse train with period T = 2 seconds, (c) Pulse train with period T — 4 seconds.

the number of frequency components as shown in Figs. 13.2(a) and (c). Note that the spectrum of a pulse train is of infinite duration and only a part of the spectrum is shown in Fig. 13.2. It is easily visualized that magnitudes reduce with increasing number of frequency components, as we build the same pulse using a larger number of frequency components. If we keep doubling the period, a similar phenomenon repeats as shown in Fig. 13.2(e). The important observation is that the relative variations of the magnitudes (the shape of the envelope of the frequency spectrum, shown by a continuous line) of the frequency components remain the same.

(a)

(c)

(e)

(b)

(d)

(0

Fig. 13.2 (a), (c), and (e): The FS magnitude spectrum of the waveforms shown in Figs. 13.1(a), (b), and (c), respectively, (b), (d), and (f): The FS magnitude spectrum, multiplied by the period T.

The 1-D Continuous-Time Fourier Transform 275

As the period approaches infinity, the individual magnitudes approach zero still maintaining the same relative variations of the magnitudes. The frequency increment between adjacent frequency components tends to zero and, hence, the spectrum becomes continuous. The envelopes of the spectra shown in Figs. 13.2(a), (c), and (e) are the magnitude of the FT of the signal shown in Fig. 13.1 with an infinite period, with different scale factors. The envelopes of the spectra shown in Figs. 13.2(b), (d), and (f) are the magnitude of the FT of the pulse shown in Fig. 13.1 with an infinite period (These spectra are the same as those on the left side of Fig. 13.2 except that the magnitudes are multiplied by the period.).

One difference between the FS and the FT is that the FS gives the amplitudes of the various sinusoidal components. The amplitudes of the sinusoidal components of an aperiodic signal tend to zero. However, the limit of the product of the period of the signal (as the period approaches infinity) and the infinitesimal amplitudes of the spectral components yields a finite continuous curve, representing the spectral density. That is, the FT yields a relative amplitude spectrum. Although the FT yields the spectral density of a signal, it is still called the spectrum. Essentially, there is no significant difference between the FS and FT and the continuous spectrum of the FT is used in the same way as the discrete spectrum of the FS in applications. Remember that the relative variations of the amplitudes of the various frequency components determine the essential characteristics of a spectrum. The FS coefficients of a periodic signal can be easily deduced from the FT of the corresponding aperiodic signal as shown in Example 13.5, since the FS coefficients are samples of the FT at discrete intervals of ^ , apart from the constant multiplier y . This argument can be mathematically put as follows. Substituting for Xcs(k) with ^ replaced by | £ in (12.2), we get

°° u rti+T

;(t) = Y eJ^ot{ o / x(l)e-jkuoldl} t™

2 n Jti

As T tends to oo, kuiQ becomes a continuous variable LJ, ^ = wo tends to a differential dw, and the summation becomes an integral. Therefore, we get

^{ir- / x{l)e-iu,dl] = £- / { / x(l)e-juldl}ejutcLj -oo i 7 r J-oo to J-oo J-oo

i r°°


Therefore, the FT of the signal x(t) is defined as

/

oo

x(t)e-jutdt (13.1)

-oo

The inverse FT of the transform Xft(u) is defined as

1 r°° x(t) = — / Xft(u)e?utdw (13.2)

The sufficient conditions for the existence of FT are essentially the same as those for the FS. Gibbs phenomenon is also common to both the FS and the FT.

Example 13.1 Find the FT of the unit impulse signal x(i) — 5{i).

/

oo 8{t)e-jultdt = 1 and 5{t) & 1

•oo

That is, the unit impulse signal is composed of complex sinusoids of all frequencies from u — — oo to a; = oo in equal proportion. I

Example 13.2 Find the inverse FT of the transform Xft(uj) = 8(UJ).

1 f°° 1 x(t) = — 6(oj)ejutdui = — and 1 O 2ir6(u)

2-K J_O0 2n

That is, the dc signal has nonzero spectral component only at u = 0. I

Example 13.3 Find the FT of the signal x{t) = ePUot. This signal can be considered as the product of signals x{t) = 1 and x(t) =

ejwot Therefore, using the shift theorem e^mtx(t) <S> Xft(u)—w0) and from the result of Example 13.2, we get

eiuot « • 2-K6(UJ - u>0)

That is, the spectrum of the complex sinusoid with u = u>o is a single impulse at u — UIQ. I

Example 13.4 Find the FT of the signal x(t) = COS(CJO^)-

/

OO I /-OO

cos{u0t)e-jutdt =- (e^ot + e-juJot)e-iutdt -OO 2 ./-oo

— Tr(6(u! - UIQ) + 6(iJ + W0))


Hence, cos(ui0t) O ir(5(u - ui0) + S(u + u>0))-Similarly, sin(u;o*) 4* {-j)ir(6(u ~ wo) - &(w + wo))-

Example 13.5 Find the FT of the signal x(t) =

From the Xft(u) obtained, deduce the complex FS coefficients Xcs(k) if one period of a periodic signal is defined as

{ 1 0<t<a 0 a<t<T-a 1 T-a<t<T

where T = 5 seconds and a = | seconds. Solution As the signal is even-symmetric,

Xft(u) = 2 / cos(wi) dt = 2 — ^ — -Jo u

Since Xcg(fc) = ^Xft(kuJo), with a = | , T = 5, and w get

^ . « = I S l ^ = , M 0 and 5

The relation between the DFT and the FT

Comparing with the approximation of the FS by the DFT, the difference in the approximation of the samples of the FT is that we truncate the signal, which could be of infinite length, to a finite length T. Now, the signal is assumed periodic of period T. The consequence of truncation is that the spectrum is distorted, since the resulting spectrum is the convolution of the spectra of the given signal and the rectangular window. But for truncation, the approximation of the samples of the FT by the DFT is the same as that of the FS.

Let us approximate the integral in Eq. (13.1) using the rectangular rule of numerical integration. The summation interval can start from zero, since we assume periodicity, although the input signal can be nonzero in any interval. We divide the period T into N intervals of width Ts = jj and represent the signal at N points as x(0),x(^-),x(2^),.. .,x((N-l)^). Ts

in seconds represents the sampling interval in the time-domain and -r£r =

J 1 for — a <t < a 1 0 elsewhere

koj0 = k~, we

* » ( 0 ) = Yo


Y~ in radians per second represents the sampling interval in the frequency-domain. Now, Eq. (13.1) is approximated as

2 *• N~1

Xft(-£Tr) = T.Yi<nTB)e-i3B-nh1 k = 0 , 1 , . . .,N - 1 (13.3)

8 n=0

Equation (13.2) is approximated as

<nTs) = WF ^ XMj£fy¥nk> n = 0,l,...,N-l (13.4) 8 fc=o s

Therefore, for a direct comparison with the samples of the FT, the DFT coefficients must be multiplied by the sampling interval Ts. For a direct comparison with the samples of the inverse FT, the IDFT values must be multiplied by £-.

Example 13.6 Find the FT analytically.

. , f e " ' for t > 0 x(t) = J 0 for * < 0

Let the record length T — 8 seconds and the number of samples N = 1024. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 8 scaled DFT coefficients and the corresponding FT coefficients. Solution

/•OO y-OO -I

Xft(w)= e-*e->utdt= e^1+ju)tdt= -

Jo Jo 1 + JUJ

The magnitude of the transform is

1 N/TT w

Figure 13.3 shows the exponential signal e - t with 16 samples taken in a duration of T = 1 second. This signal has nonzero sample values up to infinity, but we have truncated it to one second duration. Note that, at the discontinuity, we have used the average value (0.5). In order to approximate the spectrum of this signal using the DFT, we have to select appropriate sampling interval and record length. This signal is neither time-limited nor band-limited. First, we try to find the appropriate sampling interval by increasing the number of samples over the one second duration. For this


Fig. 13.3 The exponential waveform x(t) = e _ t , 0 < t < oo with 16 samples over the range 0 < t < 1.

signal, we compute the DFT with N = 16, 32, 64, and 128. The scaled magnitude of the spectra are shown, respectively, in Figs. 13.4(a), (b), (c), and (d) (focussing on the spectrum close to the folding frequency in Figs. 13.4(a), (b), and (c)). As the sampling interval is decreased, the frequency range is increased and the spectrum has reduced aliasing effect as the spectral values close to the folding frequency tend to become very small. Remember that if we make the frequency range infinity, then there is no aliasing effect. From looking at the figures, it is obvious that a sampling interval of j^g seconds is sufficient. In this case, we have the analytical result also plotted for comparison. In the case of an arbitrary signal, we

*2 0.03 5 0.02 &0 .01

- F T •DFT

T=\ W=16

3 0.009 5- 0.006 & 0.003

0 10 20

k

(c)

30

2 0.015 5; 0.01 S" 0.005

0

.0.5

X

/V = 32

5 10 15 k

(b)

W=128

20 40 60 k

(d)

Fig. 13.4 The magnitude of the F T and the scaled DFT spectrum of the waveform in Fig. 13.3. (a) With T = 1 and N = 16. (b) With T = 1 and N = 32. (c) With T = 1 and N = 64. (d) With T = 1 and N = 128. In (a), (b), and (c), the spectrum close to the folding frequency is focussed.


have to keep reducing the sampling interval until the spectral values close to the folding frequency tend to become negligible.

To fix the record length, keeping the sampling interval the same, we keep increasing the number of samples until two consecutive spectra are essentially the same. The reason for this procedure is as follows. By truncating a signal, we get the convolution of the true spectrum and the spectrum of the rectangular window. As we reduce truncation, by sampling the signal over a longer interval, the spectrum of the rectangular window distorts the true spectrum of the signal to a lesser extent. Ideally, with no truncation, the spectrum of the rectangular window is an impulse and it does not distort the true spectrum of a signal. For the present example, Figs. 13.5(a), (b), and (c) show, respectively, the spectra with N = 256, 512, and 1024 samples (focussing on the low frequency part of the spectrum in Figs. 13.5(a) and (b)). We find that increasing the number of samples beyond 1024 results in little change in the spectrum from that for N = 1024, as the distortion of the spectrum due to truncation becomes negligible. Note that the dc value computed through the DFT is almost 1, the correct value, only for N = 1024. Therefore, for this signal, we conclude that the sampling interval of j-|g seconds and the record length of eight seconds results in a DFT spectrum that is very close to the analytically derived spectrum. Table 13.1

0.5

-FT •DFT T=2 N=256

10

0.5 T = 4 N = 512

.0.5 r=8 N = 1024

(b)

50 100 150 200 250 k

300 350 400 450 500

(c)

Fig. 13.5 The magnitude of the F T and the scaled DFT spectrum of the waveform in Fig. 13.3. (a) With T = 2 and N - 256. (b) With T = 4 and TV = 512. (c) With T = 8 and N = 1024. In (a) and (b), the low frequency part of the spectrum is focussed.


Table 13.1 Comparison of the first 8 exact values (the first two rows) of the spectrum in Fig. 13.5(c) with the corresponding scaled DFT values (the second two rows).

0.0920 - jO.2890 0.6185-J0.4858 0.0609 - jO.2391

0.2884 - jO.4530 0.0431 - jO.2031

0.1526-jO.3596" 0.0320 -jO.1761

0.9997 0.6183 - jO.4856 0.2883-jO.4529 0.1526 - jO.3595 0.0920 - jO.2889 0.0609 - jO.2390 0.0431 - jO.2030 0.0320 - j'0.1760

shows a comparison of the first eight exact and the scaled DFT values of the spectrum shown in Fig. 13.5(c). This example demonstrates the fact that even if a signal is neither time-limited nor band-limited, for all practical purposes, it can be considered as both time-limited and band-limited by using appropriate sampling interval and record length. The frequency increment is 2- radians per second in the DFT spectrum shown in Fig. 13.5(c). In this case, the frequency increment is also small enough to represent the continuous spectrum adequately as the spectrum is very smooth. To obtain a denser spectrum, we have to increase the record length either by reducing the level of truncation or zero padding.

Figure 13.6(a) shows the reconstructed signal using the DFT coefficients computed with T = 1 and N = 16. In this case, the representation is poor. Note that the reconstructed signal passes through all the sampling points, as it should be, despite significant aliasing. If the parameters are fixed, then we can get a better reconstructed signal by applying a window to the frequency coefficients. Figure 13.6(b) shows the reconstructed signal using the Hann window. The price we paid for reducing the ripples is a slow rate of rise. Of course, the best thing to get a good reconstructed signal is to use the appropriate parameters. Figure 13.6(c) shows the reconstructed

0.5

T=\ N=16

0.5 t

(a) (b)

0 2 4 6 8

(c)

Fig. 13.6 The reconstructed signal using D F T coefficients computed with T = 1 and N = 16. The reconstructed signal, using DFT coefficients computed with T = 1 and N = 16 after applying the Hann window. The reconstructed signal using DFT coefficients computed with T = 8 and N = 1024.


signal using DFT coefficients computed with T = 8 and N = 1024, which is quite close to the original signal (Notice the overshoot and undershoot at the discontinuity due to Gibbs phenomenon.). I

In summary, a trial and error procedure is used to approximate the spectrum of an arbitrary signal by the DFT. First, select the appropriate sampling interval. This is done by selecting a reasonable record length and trying with different sampling intervals so that spectral values close to the folding frequency are sufficiently small. The second step is to keep the sampling interval the same and try different record lengths. The record length is chosen so that changes in the spectrum with a longer record length is negligible.

13.2 The 2-D Continuous-Time Fourier Transform

The 2-D FT is a straightforward extension of the 1-D FT.

/

oo roo

/ x(t1,t2)e-ju'ltle-j"*t*dt1dt2 -oo J—oo

-I rOO /"OO

x(ti,t3) = T 1 / ^ ( w i . w a J e ^ V ^ d w i d w a ^™ J—oo J—oo

Example 13.7 Find the 2-D FT analytically.

{I u ^ _ J - for 0 < ii < 3, 0 < i2 < 2

*(*i . ' 2 ) -<{ n eiSeWhere

Let the record length Ti = 12 and T2 = 8 seconds and the number of samples Ni,N2 = 32. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 4 x 4 scaled DFT coefficients and the corresponding FT coefficients. Reconstruct the signal using the DFT coefficients with and without applying the Hann window. Solution As the signal is separable, we use the 1-D FT results to get

v i \ sin(1.5wi)sin(w2) _,-awl _,-W2 , n

Xft(u!i,u2) = 4—i '-—^—-e J^le JW2, wi, ui2 ^ 0 U)\ Ul2


(a) (b)

,-0.5 Z-0.5

(c) (d)

Fig. 13.7 (a) A 32 x 32 2-D signal in the center-zero format, (b) Its magnitude spectrum in the center-zero format, (c) The reconstructed signal using the DFT coefficients. (d) The reconstructed signal using the 1-D Hann window in the two directions.

Xft(0,uj2) = 6 ^ ) e - ' - , a* # 0, W2

Xf t(0,0) = 6

Figure 13.7(a) shows the signal with 32 x 32 samples in the center-zero format. Note that the sample values along the border are set 0.5 and those at the corners are set 0.25, the average values at discontinuities. The given signal is defined in a 3 x 2 area. However, the signal appears square because we used a sampling interval of | seconds in the t\ direction and \ seconds in the £2 direction. The scaled DFT magnitude spectrum is shown in Fig. 13.7(b) in the center-zero format. The frequency increment in the fci direction is ff radians per second and it is ^ in the fc2 direction. Table 13.2 gives a comparison of the first 4 x 4 exact and the scaled 2-D DFT values of the spectrum shown in Fig. 13.7(b). The reconstructed signal using 64 x 64 samples is shown in Fig. 13.7(c). We can observe considerable amount of


Table 13.2 Comparison of the first 4 x 4 exact FT values (the first four rows) and the corresponding scaled D F T values (the second four rows) of Example 13.7.

3.8197-

-1.2732-

3.8074-

-1.2362-

6 -J3.8197 -J3.8197 -jl .2732

6 -J3.8074 -J3.7705 -jl .2362

3.8197-j3.8197 -jf4.8634

-2.4317-j2.4317 -1.6211

3.8074-J3.8074 -j4.8322

-2.3927-j2.3927 -1.5689

^/3.8197 -2.4317-j2.4317

-2.4317 -O.8106+J0.8106

^3.7705 -2.3927-j2.3927

-2.3695 -0.7769-f-j0.7769

- l .2732- j l .2732 -1.6211

-0.8106+j0.8106 jO.5404

- l .2362- j l .2362 -1.5689

-0.7769+J0.7769 jO.5094

ripples. Figure 13.7(d) shows the reconstructed signal after applying the 1-D Hann window to each row and column of the spectrum. I

In approximating the FT of a 2-D signal with the DFT coefficients, we use the same procedure as that in the case of 1-D signals. The difference is that we have to use appropriate parameters in two directions rather than one. Therefore, for 2-D signals, we have to ensure that the sampling interval is small enough in both the directions so that the errors due to the aliasing effect is negligible. The record length must be long enough in both the directions so that the leakage effect due to truncation is insignificant and the spectrum is sufficiently dense in both the directions. On the other hand, we cannot be too conservative and set the parameters much more than sufficient because the memory and execution time requirements are very high for a 2-D signal even if we use a fast in-place algorithm.

13.3 S u m m a r y

• In this chapter, we learned that the continuous-time Fourier transform is a limiting case of the continuous-time Fourier series. The approximation of the samples of the spectrum of a continuous-time aperiodic signal, both 1-D and 2-D, by the DFT coefficients was presented.

• Fourier analysis is the representation of the signal in terms of sinusoids. As the function being the same, the DFT and the FT are closely related and the samples of the latter can be approximated to a desired accuracy by the DFT coefficients with proper choice of the record length and the number of samples.

http://-2.4317-j2.4317

http://-2.3927-j2.3927

http://-2.4317-j2.4317

http://-2.3927-j2.3927

http://-0.7769-f-j0.7769

http://-l.2732-jl.2732

http://-l.2362-jl.2362

Exercises 285

References



Exercises

13.1 Find the FT analytically.

J e~°-2t cos(2i) + e-°'3* sin(3i) for 0 < t < oo XW ~ { 0 for t < 0

Let the record length T — 40 seconds and the number of samples N = 512. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 8 scaled DFT coefficients and the corresponding FT coefficients.


"U x,t)=i —1*1 for |i | < 1 ' n elsewhere

Let the record length T — 32 seconds and the number of samples N = 512. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 8 scaled DFT coefficients and the corresponding FT coefficients.


x(t) = e~2W

Let the record length T = 8 seconds and the number of samples N = 256. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 8 scaled DFT coefficients and the corresponding FT coefficients.


x(t) = e"4t2



* 13.5 Find the FT analytically.

(f\ _ / cos(lOi) for - 1 < t < 1 \ 0 elsewhere



u\ _ J fe~* for 0 < t < oo X[t) ~ { 0 for t < 0



"{ . , e _ t l e - t a for 0 <ti < oo, 0 < t2 < oo

*&'**) " I o elsewhere

Let the record length Ti,T2 = 8 seconds and the number of samples Ni,N2 — 128. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 4 x 4 scaled DFT coefficients and the corresponding FT coefficients.

* 13.8 Find the FT analytically.

( l - | * i | ) ( l - M ) for|ii|<l,|i2|<l ' ' 0 elsewhere

Let the record length 7i ,T 2 = 4 seconds and the number of samples Ni,N% = 64. Compute the samples of the spectrum of the signal using the DFT. Tabulate the first 4 x 4 scaled DFT coefficients and the corresponding FT coefficients.

Chapter 14

Convolution and Correlation

The linear convolution operation, relating the input, output, and the impulse response of an LTI system, is of fundamental importance. The operation of linear correlation is very similar to convolution and is used as a similarity measure between signals. Use of DFT makes the computation of these very important operations more efficient than implementing directly. In Sec. 14.1, the linear convolution operation is presented. In Sec. 14.2, the computation of convolution using the DFT is described. In Sec. 14.3, the overlap-save method of convolution of long sequences is explained. In Sec. 14.4, the convolution of 2-D signals is presented. In Sec. 14.5, the computation of correlation is described.

14.1 The Direct Convolution

In practice, the input signal to a system is usually arbitrary. The system response is obtained by representing the input signal in terms of impulses, finding the system response to each impulse, and adding all the responses. It is required to measure the system response to only one signal, the impulse. The reason for the selection of the impulse signal as the basic signal is that it is a simple signal and it is easy to express a given arbitrary signal in terms of impulses.

In the convolution operation, we express the present output as an exclusive function of the present and all past input samples. Let us formulate the convolution operation through an example. Assume that a savings account in a bank yields 10% of interest per year for deposit. If we deposit $1 in the bank now, the money in the account next year will be 1.1$, the

287

288 Convolution and Correlation

i m \ i . i 3

-i) h{2

l.l2

l.l3

-i) h(3

1.1 l.l2

l.l3

-0 n y(n)

0 I

30C 1

1.1 l.l2

l.l3

0 30C

1 1.1 20C

1 1.1 l.l2

1 53C

2 l.l2

0

3 l.l3

0

1 1.1 2

1 3

583 641.3

Fig. 14.1 The linear convolution operation.

year after that it will be 1.12$, etc. The savings account is a process and the rate of growth of money of 1$ is the impulse response. Let us say we open an account with a deposit of $300 and deposit another $200 at the beginning of the next year. The deposits are the inputs. Let us try to find the money, which is the output, in our account for the next three years assuming that we make no other deposits or withdrawals in this period and close the account after three years. At the beginning of this year, the balance is $300 that we have deposited. At the beginning of the next year the balance will be $300 x 1.1 + $200. The year after that the balance will be $300 x l . l 2 + $200 x 1.1. After three years the balance will be $300 x l . l 3 + $200 x l . l2 . The problem of finding the balance can be formulated neatly as shown in Fig. 14.1. The impulse response h(i),i = 0,1,2,3 is {1,1.1, l . l2 , l . l 3 } . The input x(i), i = 0,1,2,3 is {300,200,0,0}. The impulse response when folded about the y-axis, h(0 - i), is { l . l 3 , l . l 2 ,1 .1 ,1}. To find the response j/(0), we simply find the sum of products of all the overlapping samples of x{i) and h(0-i). Then, we shift h(0 — i) to the right by one position to get h(l — i). The sum of products of all the overlapping samples of x(i) and /i(l - i) yields the output y(l). This process can be continued to find the other two outputs. Note that the values of x(i) and h(i) for all values of i not shown in Fig. 14.1 are assumed to be zero. We get the same result if x(i) is folded rather than h(i).

In summary, given the input sequence x(i) and the impulse response of a system h(i), there are four steps to find the output of the system through the convolution operation. 1. Fold one of the sequences about the y-axis. 2. Shift the folded sequence to the point where the output response is to be determined. 3. Multiply the overlapping samples of the two sequences. 4. Add all the products to find the output at that output point.

The Indirect Convolution 289

The linear convolution

The linear convolution operation is defined as follows. Let x(n),n = 0 , 1 , . . . , P - 1 and h(n),n = 0 , 1 , . . . , Q — 1 be two arbitrary sequences. Assume that x(n) — 0, for n < 0 and n > P — 1 and h(n) = 0, for n < 0 and n > Q — 1. Then, the linear convolution of the two sequences is defined as

Min(n,P-l) Min(n,Q-l)

y(n) = 5 3 x{i)h{n -i)= ^2 h(i)x(n - i), i=Max(0,n-Q+l) i=Max(0,n-P+l)

n = 0 , l , . . . , P + Q - 2

It is easy to see that the present output is the sum of products of the two sequences, each other's index running in opposite directions.

14.2 The Indirect Convolution

Implementing the time-domain convolution by computing the DFT of two signals, multiplying the DFTs term by term, and computing the IDFT of the product is known as indirect convolution. The computational complexity of the indirect convolution method, using a fast DFT algorithm, is 0(N log2 N) compared with 0(N2) for the direct convolution.

The circular convolution

Let x(n) and h(n), n = 0 , 1 , . . . , TV — 1 be the samples of one period of periodic signals with period iV. Then, the circular or periodic convolution of the two sequences is defined as

N-l N-l

y(n) = ^2 x(i)h(n - i) = J ^ h(i)x(n -i), n = 0 ,1 , . ..,N- 1 i=0 j=0

The values y(n),n = 0 , 1 , . . . , N — 1 represent samples of one period of the periodic output sequence y(n) with period N. The circular convolution is the same as that of the linear convolution except that the sequences are periodic. The circular convolution of sequences x(n) — {1 ,4 ,2 , -1} and h(n) = {2,3, - 1 , 1 } is illustrated in Fig. 14.2. The samples of one period of the input and output sequences are shown in boldface. The output of one period is {1,14,14,1}.


i Mi) xli) h{0 - i)

h{\-

- 3 3

- 2 - 1

- 1 1

r i - i W h{2-

1 i)

h(3-

3 - 1

1 *)

n yip)

d 2 1 2 3

- 1 1 0 1

1 3 4 1 2 3

- 1 1

14

2 - 1

2 - 1

1 2 3 2

14

3 1

- 1 3

- 1 1 2 3 1

3 - 1

1 3

- 1 3|

Fig. 14.2 The circular convolution operation.

TheDFTofar(n) and h(n) are, respectively, X(k) = { 6 , - l - j " 5 , 0 , - l + j5} and H(k) = {5,3 - j 2 , - 3 , 3 + j2}. The product of the DFTs is X(k)H(k) = {30, - 1 3 - j l 3 , 0 , - 1 3 + ;13}. The inverse of X(k)H(k) is y{n) — {1,14,14,1}. Using the transform method, the output is periodic as the signals are assumed periodic. The superposition sum of the linear convolution output {2 ,11 ,15 ,1 , -1 ,3 , -1} of the two 4-point sequences, periodically placed at intervals of four samples, produces the periodic output affected by aliasing. The last three values of the linear convolution are aligned with the first three values and the corresponding terms added. Note that the fourth output value is correct.

The computation of the linear convolution using the DFT

If we make the period of the input sequences, N, to be at the least P + Q — 1 (seven for the example) by appending the sequences with zeros, then aliasing does not occur and the output samples of one period of the circular convolution correspond to the output of the linear convolution. Let N = P + Q - 1 and

x'(n)

h'(n)

- { - {

x(n) for n = 0 , 1 , . . .,P — 1 0 forn = P,P + l,...,N -1

h(n) for n = 0 , 1 , . . . , Q — 1 0 forn = Q,Q + l , . . . , i V - l

The circular convolution of the sequences of our last example with zero padding, x'{n) = {1 ,4 ,2 , -1 ,0 ,0 ,0} and h'(n) = {2 ,3 , -1 ,1 ,0 ,0 ,0} , is illustrated in Fig. 14.3. The output of one period is {2,11,15,1, —1,3, —1}. T h e D F T o f z » = {1,4,2 , -1 ,0 ,0 ,0 ,0}and/ i ' (n) = {2 ,3 , -1 ,1 ,0 ,0 ,0 ,0} are, respectively, {6,4.5355-j4.1213, —1—j'5, -2.5355-j0.1213,0, -2.5355+

http://-2.5355-j0.1213

The Indirect Convolution 291

- 6 ~ 3 |

- 5 - 1

- 4 1

- 3 0

- 2 C

- 1 c

o ^ y ( 2 -

c c

•i) h'(3-

0 C c

• 0 h'(A-

1 0 0 c

•i) h'{5-

- 1 1 C c c

•i) h'(6-

3 - 1

1 0 c 0

• » )

n y'(n)

c 2 1 2 3

- 1 1 C C c c 2

1 3 4 0 2 3

- 1 1 0 c 1

11

2 - 1

2 C C 2 3

- 1 1 0 2

15

3 1

- 1 C C c 2 3

- 1 1 3 1

4 C C 1 C c 0 2 3

- 1 4

- 1

5 C C

- 1 1 C 0 c 2 3 5 3

6 0 0 3

- 1 1 C c c 2 6

- 1

3 - 1

1 C C c

3 - 1

1 C c

3 - 1

1 C

3 - 1

1 3

- 1 31

Fig. 14.3 The simulation of the linear convolution operation by the circular convolution operation with zero padding.

jO.1213, -1+J5,4.5355 +J4.1213} and {5 ,3 .4142- j l .8284,3- j2 ,0 .5858-j'3.8284,-3,0.5858+ J3.8284,3 +J2,3 .4142+ J1.8284}. Note that we append four zeros to the sequences in order to use efficient algorithms since the length of the sequences become eight, which is a power of two. The product of the DFTs X'(k)H'(k) is {30,7.9497 - j22.364,-13 - J13, -1.9497 + j9.636,0, -1.9497 - J9.636, - 1 3 + jl3,7.9497 + J22.364}. The IDFT of X'(k)H'(k) is {2,11,15,1, - 1 , 3 , - 1 , 0 } . The first seven values of the IDFT correspond to the linear convolution output with the remaining value zero. To summarize, the steps to simulate a linear convolution of two sequences x(n) of length P and h(n) of length Q by a circular convolution are as follows.

(1) Find the smallest N such that N > P + Q-l and N is an integral power of two. The second condition enables us to use power of two DFT algorithms that are the most efficient.

(2) Append both sequences with zeros to get x'(n) and h'{n) of length N.

(3) Compute the DFTs of x'(n) and h'{n) to obtain X'(k) and H'{k).

(4) Find the term by term product of X'{k) and H'(k) to get Y'(fc).

(5) Find the IDFT of Y'(k) to obtain y'{n).

(6) The first P + Q - l samples of y'(n) are the output of the linear convolution of the two sequences x(n) and h(n).


Implementation

In practice, signals are real in most applications. To convolve a single real signal with another real signal, we can use the DFT algorithms described in Chapters 8 and 9. To convolve two real signals x(ri) and y{n) with a single real impulse response h(n) (or a single real signal with two impulse responses), we form a complex signal x'(n) +jy'(n) using the zero padded signals x'{n) and y'{n). Compute the DFT, X'(k) + jY'(k). Compute the DFT, H'(k), of the zero padded impulse response h'(n). Form the term by term product Z(k) = (X'(k) + jY'(k))H'(k). Compute the IDFT of Z(k). The real part gives the convolution of x(n) and h(n) and the imaginary part gives the convolution of y(n) and h(n). This process enables the use of a DFT algorithm for complex data without the necessity of splitting the two individual DFTs of x'{n) and y'(n).

14.3 Overlap-Save M e t h o d

In general, the input signal is very long and, therefore, it may not be possible to manipulate the whole signal or we may not be able to wait too long for the output. In such cases, the convolution operation is carried out over sections of the input signal. For example, consider the impulse response, h(n), of length Q = 4 and the input signal, x(n), of length P = 14 shown in Fig. 14.4. (For illustrative purposes, we use a short length of P = 14. However, the input sequence should be very long to take advantage of this method.) The input signal is extended at the front by Q — 1 = 3 zeros. Remember that the circular convolution without zero padding yields incorrect values at the beginning. We have to choose a block length. Let us say we choose a block length of 8. Then, the

h(n) x(n) 0 2

0 1

0 4

2 7 7 3

1 4 4 0

4 1 1 ~o1 2

3 3 3 0 1

5 5 0 4

9 9

1 1

0 0

2 2

4 4

310 0 2

0 1

0 4

2 2

3 3

8 8

6 6 1 0 1 0 1 0 1 0 1 0 1 0 1

3 I 0 0 2

0 1

0 4 3 1 0 1 0 1 0 | 0 1

y{n) 1141151341441291381401521351131161301391381471481181 0 | 0 [ 0

Fig. 14.4 The overlap-save method of convolution of long sequences.

Overlap-Save Method 293

impulse response is zero padded on the right to make it equal to the block length and we get {2,1,4,3,0,0,0,0}. The DFT of this, with a precision of 2 digits, is 1.25,0.07 - jO.85, -0.25 + jO.25,0.43 + jO.15,0.25, 0.43 - jO.15, -0.25 - jO.25,0.07 + jO.85. The DFT values are divided by, TV = 8, the block length so that we do not need to divide every time an IDFT is computed. Note that, as the impulse response is fixed, we have to compute this DFT only once. Also, as the DFT is conjugate-symmetric, the storage of half of the transform is sufficient.

The input data is divided into overlapping blocks of size 8. There are 4 blocks for the present example. The blocks correspond to the 4 sets of zero padded impulse response shown. Note that at the end also we zero pad the input data so that we get an integral number of blocks. With zero padding, the length of the input data has become 23. We take the first two input blocks and make a set of 8 complex numbers by putting the first block values in real parts and second block values in the imaginary parts. For this example, we get 0 + j ' l , 0 + j 3 , 0 + j 5 , 7 + j 9 , 4 + j l , l + j 0 , 3 + j 2 , 5 + j 4 . The DFT of this, with a precision of 2 digits, is 20 + 25,2.54 + jO.88, - 9 + j6,0.78 - j'2.29, - 6 - j7, -4.54 + J5.12,11 - j l 6 , -14.78 - J3.71. The pointwise multiplication of this DFT with that of the impulse response yields, with a precision of 2 digits, 25 + j31.25,0.94 - j'2.10,0.75 - J3.75, 0.67 - jO.86, -1 .5 - j l .75 , -1.19 + j'2.85, -6.75 + jl.25,2.08 - J12.89. The IDFT (remember the division by 8 required by the IDFT operation has already been done) of this is 20 + j l4 ,29 + ;29,15 + j'29,14 + j'38, 15 + j40,34 + j52,44 + j35,29 + jl3. The first three values are discarded. The last five values of the real part appended by the last five values of the imaginary part are the first ten values of the convolution output, y(n). We combine two blocks to make a complex signal in order to use an algorithm for complex data, rather than taking one block and using an algorithm for real data since the algorithms for complex data are more regular and simpler. In addition, there is no need to split the individual DFTs. We repeat this procedure until all the input values are processed. If we end up with one block at the end, then we still use a complex algorithm by assuming a block with all zeros. An algorithm for real data can be used. This requires having two algorithms and the work saved will be little if the data length is very long. Remember that we have to append Q — 1 zeros to the sequence x(n) in order to produce all output values. Since we prefer to use algorithms that require the data length to be an integral power of 2, we may have to append the data further with zeros. Since, the input blocks are


overlapped by Q — 1 samples to eliminate errors due to circular convolution and saving N — Q + 1 output values, this method is called overlap-save method of indirect convolution.

Implementation

The implementation consists of reading blocks of data, two at a time, computing the DFT, multiplying it with the DFT of the impulse response, computing the IDFT of the product, and storing the valid output. These operations are repeated until the data is exhausted. As we are finding the output of two blocks of length N at a time, the computational complexity for one block is half and it is of computing a complex DFT. The computation of the DFT of the impulse response is carried out once and is ignored in the analysis. A single DFT algorithm is sufficient as it can be used for computing both DFT and IDFT. A table of twiddle factors can be used repeatedly. The multiplication of two DFTs requires 3iV — 4 operations for each block. A block produces N — Q + 1 valid output points. For example, let N = 64. The 2 x 2 PM algorithm for complex data requires 1184 real multiplications and additions. Multiplying the DFTs requires 188 operations. For an impulse response of length 12, 64 — 11 = 53 valid output points are produced. Therefore, the number of operations per output point

i s U84+188 _ 25.9 (approximately 41og2 TV).

As for the choice for block length N, the following considerations must be taken into account. If the block length is long there are two disadvantages: (i) more memory is required and (ii) the number of DFT operations per point increases. On the other hand, if the block length is short then the overlap is more and the number of valid outputs decreases. In general, for efficient implementation, the following condition is required.

Input length > > Block length, N » Impulse response, Q

Table 14.1 shows the number of operations for various block and impulse response lengths. The minimum operation count is shown in boldface. The other major factor is the data transfers which is about 2 log2 N per output point. With the complexity order N log2 N for both arithmetic operations and data transfers, the indirect convolution using fast DFT algorithms is efficient for most cases than direct convolution. As the block sizes are much smaller for any data size, the storage requirements are moderate.

Two-Dimensional Convolution 295

Table 14.1 The number of real arithmetic operations per point for various impulse response and transform block lengths using 2x2 PM DFT algorithm for 1-D convolution by the over-lap save method

N\IR 32 64 128 256 512 1024

12 26.5 25.9 28.1 31.2 34.9 38.8

16 32.7 28.0 29.1 31.7 35.2 38.9

24 61.8 33.5 31.4 32.8 35.8 39.2

32

41.6 33.9 34.0 36.4 39.5

40

54.9 37.0 35.2 37.0 39.9

48

80.7 40.6 36.6 37.6 40.2

56

45.1 38.0 38.3 40.5

64

50.7 39.6 39.0 40.9

72

57.8 41.3 39.7 41.2

80

67.2 43.2 40.4 41.6

88

80.3 45.2 41.2 41.9

14.4 Two-Dimensional Convolution

The 2-D linear convolution of sequences x{n\, n2) , n i , n 2 = 0 , 1 , . . . , P — 1 and /i(rai,n2), n i , n 2 = 0 , 1 , . . . , Q — 1 is given by

Min(ni,P-l) Min(n2 ,P-l)

y(ni,n2) = ^ J ^ a;(ii,t2)/j(ni - i i ,n 2 - »2), i i=Moi (0 ,n i -CJ+l ) i?=Max(0<m— Q+1)

n 1 , n 2 = 0 , l , . . . , P + Q - 2

Consider the 5 x 5 sequence ar(«i,i2) and the 3 x 3 sequence ft^'i, i2) shown in Fig. 14.5. There are the same four steps of the 1-D convolution extended to two dimensions.

2 1 3 2 1

X

1 0 1 0 0

r«i, 4 2 0 1 4

*?) 3 2 1 3 2

3 2 0 5 5

h( 1 3 8

H,i2) 3 5 1

2 6 3

h(ii,-2 6 3

3 5 1

•12)

1 3 8

s( i i , J2W0 - i i ,0 - i-a) x{ii,i2)h{2-iul-i2) 3 6 2

1 5 3

8 3 1 5 1 3 2 1

1 C 1

c c

4 2 0 1 4

3 2 1 3 2

3 2 0 5 5

1 28 1 5 1 3 C 3 311 U

2 C 1 0 1 0

3 5

22 19 31 19

1 C 4 2 5 8 1 35 20 54 11 15

M -3 6 2

* i , -1 5 3

-i2) 8 3 1

16 25 25 24

•u(n\,ni) 11 33 60 47 31 32

17 43 50 33 39 51

20 60 70 45 56 95

15 43 36 35 65 51

22 21 16 40 45

Fig. 14.5 The linear convolution of a 5 x 5 and a 3 x 3 2-D sequences. Origin at upper left-hand corner.


(1) The sequence ft(i'i,Z2) is rotated in the (11,12) plane by 180 degrees about the origin. This operation can be achieved in two steps as shown in Fig. 14.5. Fold the sequence about i\ axis to get h(i\, - i 2 ) . Then, we get h(-ii,—i2) by folding the resulting sequence about 12 axis.

(2) Shift the rotated sequence by an amount (711,712) to get the sequence h(n\ — »i,ri2 — 12)-

(3) Find the products x(ii,i2)h(ni — n , n 2 - 12) of all the overlapping samples.

(4) Sum the products to get the convolution output y(ni ,n 2 ) .

For example, with a shift of 0 — i\, 0 —12, there is only one overlapping pair (1,2). The product of these numbers yields the output y(0,0) = 2. The overlapping samples with a shift of 2 — t i , 1 — i2 is also shown in Fig. 14.5. The process is repeated to get the complete convolution output y (711,712) shown in Fig. 14.5.

Two-dimensional circular convolution

The 2-D circular convolution of periodic sequences z (n i ,n 2 ) , n i , n 2 = 0 , 1 , . . . , N — 1 and /i(ni,7i2), ni,7i2 = 0 , 1 , . . . , N — 1 is given by

JV-l JV-1 y(7ii,n2) = X ] 5 Z x(i2,i2)h(ni -ii,n2 - i2), 7ii,n2 = 0 , 1 , . . . ,N - 1

i1=o i2=o

Fig. 14.6 shows the 5 x 5 sequence x(i i , i2) and the 3 x 3 sequence h(ii, 12) of Fig. 14.5 padded with sufficient zeros so that, by using a power of two DFT algorithm, the linear convolution is correctly simulated.

Overlap-save method

The 5 x 5 sequence x(i\,i2) and the 3 x 3 sequence h(ii,i2) of Fig. 14.5 are zero padded as shown in Fig. 14.7. The outputs of two blocks each of size 4 x 4 are produced at a time. Fig. 14.8(a) shows the combined DFT of the

Two-Dimensional Convolution 297

2 1 3 2 1 0 0 0

1 0 1 0 0 0 0 0

4 2 0 1 4 0 0 0

x' 3 2 1 3 2 0 0 0

[k, 3 2 0 5 5 0 0 0

ii) 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

h!{ii,i2) 1 3 8 0 0 0 0 0

3 5 1 0 0 0 0 0

2 6 3 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Fig. 14.6 The sequences 1(11,12) and h(h,i2) of Fig. 14.5 padded with zeros in order to use the circular convolution to simulate the linear convolution. Origin at upper left-hand corner.

first two blocks,

0+jQ 0 + jO 0 + jO 0 + jO 0 + j O 0 + jO 0 + jO 0 + jO 0+j2 0 + j ' l 2 + j4 l + j 3 0+jl 0 + jO 1 + J 2 0 + j2

the real parts in the upper half and the imaginary parts in the lower half. The DFT of the sequence h'{ii,i2) divided by 16 is shown in Fig. 14.8(b). The division by 16 eliminates the necessity of the division operation each time an ID FT is computed. Figure 14.8(c) shows the product of the two DFTs. The convolution output of the first two blocks is obtained by computing the IDFT (remember the division by 16 required by the IDFT operation has already been done) of the product. The first block output is the real part and the second block output is the imaginary part shown in

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 2 1 3 2 1 0 0 0

x'(i 0 0 1

JL T1 0 0 0 0 0

0 0 4 2 0 1 4 0 0 0

L , » 2

0 0 3 2 1 3 2 0 0 0

) 0 0 3 2 0 5 5 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

h'(ii,i2) 1 3 8 0

3 5 1 0

2 6 3 0

0 0 0 0

Fig. 14.7 The sequences x ( i i , i 2 ) a n d / i ( n , i 2 ) of Fig. 14.5 padded with zeros to convolve using the overlap-save method.


4.00 -8.00 2.00 2.00

15.00 -9.00 5.00

-11.00

-7.00 5.00

-1.00 3.00

-2.00 -2.00 0.00 4.00

2.00 -2.00 0.00 0.00

3.00 -1.00 1.00

-3.00

1.00 1.00

-1.00 -1.00

-4.00 4.00

-2.00 2.00

2.00 -0.38 0.25

-0.38

0.00 -0.88 0.00 0.88

0.06 -0.69 0.44

-0.06

-0.56 0.06 0.06

-0.31

0.88 -0.63 0.38

-0.63

0.00 -0.25 0.00 0.25

0.06 -0.06 0.44

-0.69

0.56 0.31

-0.06 -0.06

(a) (b) 8.00

-4.88 0.50 8.87

30.00 10.38

1.25 5.88

-1.56 -3.31 -0.44 1.06

3.81 1.69

-0.06 -1.19

1.75 1.00 0.00 0.75

2.63 1.13 0.38 1.88

2.31 -1.31 -0.56 0.81

0.31 0.06

-0.81 -1.31

13.00 3.00 7.00

19.00

56.00 16.00 19.00 56.00

3.00 0.00 2.00 6.00

36.00 7.00

13.00 38.00

19.00 8.00 2.00 7.00

51.00 19.00

11.00 33.00

15.00 1.00

7.00 16.00

47.00 18.00

17.00 43.00

(c) (d)

Fig. 14.8 (a) The combined DFT of the first two 4 x 4 blocks of x'(ii, i2) in Fig. 14.7, the real part in the upper half and the imaginary part in the lower half, (b) The DFT of/i'(ii,i2)/16. (c) The product of the two DFTs in (a) and (b). (d) The IDFT of the product in (c) gives the convolution outputs of the first two blocks. The valid outputs are shown in boldface.

Fig. 14.8(d). The valid outputs are shown in boldface. This process is repeated until all the blocks are processed. The complete output of the convolution operation is shown in Fig. 14.5. The optimum block size for various sizes of the sequence h ^ i , ^ ) are given in Table 14.2. The number of arithmetic operations and data transfers per point are, respectively, about 8 log2 N and 4 log2 N.

14.5 Computation of Correlation

The function of the correlation operation is different from that of the convolution. However, the computation is very similar to that of convolution and can be considered as the computation of convolution with slight changes in the procedure. Letar(n), n = 0 , 1 , . . . , P-l and/i(n), n = 0 , 1 , . . . , Q - 1 be

Computation of Correlation 299

Table 14.2 The number of real arithmetic operations per point for various impulse response and transform block lengths using 2 x 2 PM DFT algorithm for 2-D convolution by the over-lap save method

NxNMR 3 x 3 5 x 5 7 x 7 9 x 9 11 x 11 13 x 13 15 x 15

8 x 8 16x16 32x32 64x64

128 x 128

30.0 31.3 36.4 42.6 50.1

42.6 41.8 45.5 51.7

61.4 48.5 48.7 53.4

56.9 52.2 55.2

5 6 . 2 57.1

60.6 5 9 . 1

65.5 6 1 . 1

i

m h(3 + i) 1 21 3

h(2 + i) | 2 h(l + i)

h(0 + i)

- 1 3 2

M-i +

C 2 1 1

- 1 3 2

») h(-2 +

h n Vxhin)

1 a 4

1 - 1

3 2

0 (-3 +

3 1

2 3

2 - 1

2

3 1

- 1

1 - 1

3 2

i) 1 1

1 - 1

3 2 C

11

1 - 1

3 - 1 15

1 - 1 - 2

1

1 - 3 - 2

Fig. 14.9 The linear correlation operation.

two arbitrary sequences. Assume that x(n) = 0, for n < 0 and n > P-1 and h(n) = 0, for n < 0 and n > Q — 1. Then the linear correlation of the two sequences is given by

Min(Q-n-l,P-l)

Vxhin) = J2 x*(i)h(n + i), n = Q-1,Q-2,... ,-P + 1 i=Max(0,—n)

One difference between this operation and the convolution is that there is no folding operation in computing the correlation operation. Another difference is that the correlation operation, in general, is not commutative. A yet another difference is that the first function is conjugated. Otherwise, the correlation operation is similar to convolution as can be seen from Fig. 14.9, which shows the computation of the linear correlation of two sequences {1 ,4 ,2 , -1} and {2 ,3 , -1 ,1} .


5|—4|—3—2|—H q 1] 2| 31 4! 51 6

h?(i) = h'(7-i) hrs (i) = hr'(i — 3) x'U)

hrs'(0 - i) Q _ C _ C _2 _ 3 ^ 1 firs'(1 -t) C C

c C 2 3 - 1

0 0

c

hrs'(2-t) hrs' (3 -1J

/ i rs ' (4- l ) hrs'(b —TJ

hrs'(6 —1)

L(»)

C 0 2 3 - 1

2 3 - 1 1 2 0 C 0

C

- 1 3 2

C C 0 1

0 2 3 - 1 0 C 2 3

C C 2 C

1 3

2 - 1

C 0 1

C

- 1

1 11

C C O

15

- 1 3 C 0 0 0 3 - 1

3 - 1

1-2

2 3 - 1 C 2 3 - 1 C C 2 3 - 1

C O O C C 2 3 - 1

B

Fig. 14.10 The linear correlation operation simulated by the circular convolution operation with zero padding.

The circular correlation

Letz(n), n = 0 , 1 , . . .,N—1 and h(n), n = 0 , 1 , . . .,iV —1 be the samples of one period of periodic signals with period N. Then, the circular or periodic correlation of the two sequences is given by

J V - l

Vxh{n) = Y^ x*(i)h{n + i), n = 0 , 1 , . . . , N - 1 t=0

yxh{n), n = 0 , 1 , . . . , N - 1 gives samples of one period of the periodic output sequence 2/i/i(n) with period N. To compute the linear correlation using the circular correlation, we zero pad the signals exactly the same way as for computing the convolution. In this case, the sequence hr'(i) — h'(N — i) is formed and is circularly right shifted by Q — 1 positions to get hrs'(i) = hr'(i — (Q — 1)). Now, the circular convolution of the sequences x'(i) and hrs'(i) yields the linear correlation output. This computation, shown in Fig. 14.10, corresponds to y'xh{n) = WFT(X'(k)H'*{k)W$~1)k) in the transform domain. The overlap-save method of correlation is similar to that of convolution shown in Fig. 14.4 with the difference mentioned above. That is the formation of H'*(k)W]f~ ' initially instead of H'(k). The linear correlation output of the sequences x(n) and h(n) shown in Fig. 14.4 is

{21,40,26,31,36,52,50,23,25,22,24,25,46,57,38,22,12}

Summary 301

Table 14.3 The linear correlation output, yxh(ni,ri2), of the sequences 1(11,12) and Mil)*z) shown in Fig. 14.5. Origin at upper left-hand corner.

6 15 19 26 21 10 2

5 17 19 26 22 11 3

29 49 53 38 38 31 9

21 49 52 29 36 41 16

44 69 48 43 100 72 20

27 42 36 40 56 49 17

24 25 9 42 55 20 5

The 2-D linear correlation of sequences x(n\, n^), ni,ri2 = 0 ,1 , . . . ,P—1 and / i(ni ,n2) , ni,ri2 = 0 ,1 , . ..,Q — 1 is given by

Min(Q-m-l,P-l) Min(Q-m-l,P-l)

Vxh(ni,n2)= ^2 Y^, x*(ii,i2)h(ni+ii,n2 + i2), ii = Max(0,—m) Z2 = Max(0,—n?)

ni , n2 = Q - 1, Q - 2 , . . . , -P + 1.

The computation of the 2-D correlation is similar to that of the 2-D convolution with the difference mentioned for the computation of 1-D correlation extended to two dimensions. Table 14.3 shows the linear correlation output of the sequences x{ii,i2) and h{ii,ii) shown in Fig. 14.5.

14.6 S u m m a r y

• The convolution operation implemented using the DFT is called the circular convolution and the linear convolution operation is of prime interest in the analysis of LTI systems. In this chapter, the simulation of the linear convolution operation by the circular convolution operation was described. The overlap-save method of convolution was also presented. Use of DFT is more efficient than the direct method for almost all cases except when the impulse response is very short. The implementation of correlation operation is very similar to that of the convolution.


References


Exercises

14.1 Find the linear convolution of the sequences x(n) = {1,2,3} and h(n) = {-2,3} using the DFT.

14.2 Find the circular convolution of the sequences x(n) = {1,2,3,4} and h(n) = {1, - 2 ,1 ,3} using the DFT.

* 14.3 Using the DFT, find the linear convolution of the sequences

x(nx,n2) = 2 1 3 4

-2 1 h(n1,n2) =

2 3 1 4

14.4 Using the DFT, find the circular convolution of the sequences

x(n1,n2) =

3 1 5 - 4 2 1 3 - 1 1 - 2 3 4 2 3 1 3

h{ni,n2)

1 4 2 1

4 5 3 0

2 - 2

1 1

3 - 3

4 5

14.5 Find the linear correlation of the sequences x(n) = {1,2,3} and h(n) = {-2,3} using the DFT.

14.6 Find the circular correlation of the sequences x(n) = {1,2,3,4} and h{n) = {1, -2 ,1 ,3} using the DFT.

* 14.7 Using the DFT, find the linear correlation of the sequences

x(ni,n2) = 2 1 5 3 4 2

- 2 1 - 3 h(ni,n2) =

2 3 1 4

14.8 Using the DFT, find the circular correlation of the sequences

x(ni,n2) =

3 1 5 - 4 2 1 3 - 1 1 - 2 3 4 2 3 1 3

h(ni,n2) =

1 4 2 1

4 5 3 0

2 - 2

1 1

3 - 3

4 5

Chapter 15

Discrete Cosine Transform

For any processing, we prefer a signal to be represented by minimum number of values. This is particularly important for storage and transmission of signals. The more compact the signal is coded, the more is the reduction of storage and bandwidth requirements. Although N values represent a signal in both the time- and frequency-domains, the representation, in general, is more compact in the frequency-domain. The reason is that most of the energy of commonly occurring signals is contained in the lower part of the spectrum. As periodicity is implied in the DFT, there could be large discontinuity between the beginning and end of a period of a signal. This discontinuity represents energy at high frequencies. The discontinuity can be avoided by extending the signal so that it is even-symmetric. A signal so constructed has its energy heavily concentrated in the lower part of the spectrum. Although this procedure is a special case of the DFT, it is called the discrete cosine transform (DCT) and is very widely used in signal and image coding applications.

In Sec. 15.1, we examine the orthogonality property of the sinusoids again to find out how we can define another transform, albeit closely related to the DFT. We present the algorithms for the computation of 1-D and 2-D DCT, respectively, in Sees. 15.2 and 15.3. These algorithms are essentially DFT algorithms with few additional operations.

15.1 Orthogonality Property Revisited

It was stated in Chapter 2 that when the sum of the pointwise products of two discrete signals is zero over a specified interval, the signals are said

303

304 Discrete Cosine Transform

0.3536

0.4619

* -0.1913

(c) (d)

0.3536

-0.3536

0

* -0.

4619

1913

(e)

0.4157

; 0.0975 in

-0.2778 -0.4904

0.4157

0.0975

-0.2778 -0.4904

(9)

Fig. 15.1 The DCT basis functions, with AT = 8.

(h)

to be orthogonal in that interval. The specified interval considered was an integral number of cycles. Now, we consider the orthogonality property of cosines over an integral number of half-cycles. The sum of the N samples of a discrete cosine function shifted by \ sample interval to the left is zero, if summed over an integral number of half-cycles. This is obvious due to the symmetry of this function about the horizontal axis. The sum of the samples of each of the waveforms in Figs. 15.1(b) to (h) is zero.

The 1-D Discrete Cosine Transform 305

Consider the sum of the product of two cosine waveforms

jv-i f N for I = m = 0

E cos ^ ( 2 n + 1)1 cos r^r(2n + l )m = < £ for Z = m ^ 0 n=o ^ 0 otherwise

This result is clear by rewriting the equation as

1 N_1

2 ^ ( c O S 2^T ( 2 n + 1 ) ( ' " m ) + C ° S 2 ^ ( 2 n + 1 ) ( Z + m ) )

n=0

If I 7 m, the functions are cosines, as shown in Figs. 15.1(b) to (h), and their sums over an integral number of half-cycles are zero. If / = m ^ 0, the sum of the second term evaluates to zero (since it is a cosine) while the first term is cos (0) and summed over N terms and divided by two equals y . When I = m = 0, the sum is N. Unlike in the case of the DFT, we get two nonzero constants. With this orthogonal property, we can define a transform with basis functions, Xo(n) = y/ijf) and Xfc(n) — ^/(jf) cos 2^(2n + l)jfe, n = 0 , 1 , . . . , N - 1 and k = 1,2,. . . , N - 1, shown in Fig. 15.1 for N = 8.

15.2 The 1-D Discrete Cosine Transform

The 1-D DCT of a real-valued signal x(n), n = 0 , 1 , . . . , N — 1 is defined as

N-l

Xct{h) = C{k) J2 <n)cos 2^(2" + !)fc. k = 0,l,...,N-l n=0

The 1-D inverse DCT is given by

J V - l

<n)= ^C^Xctik)cos £-{2n + \)k, n = 0 , 1 , . . .,N - 1 2iV" fe=o

where

c(fc) = {$I) t°'k=° f) for* = 1,2,. . . , J V - l

These are the most commonly used definitions. However, it should be noted that there are other forms also. It is assumed that the data length, N, is even.


Computation of the DCT using the DFT

Splitting the summation in the DCT denning equation into that of the even-indexed and odd-indexed input samples, we get

Xct(k) = C W { $ > ( 2 n ) c o s ^ : ( 2 ( 2 n ) + l)fc n=0

+ J^ x(2n + ! ) c o s ^ (2(2n + !) + !)*} n=0

By reversing the order of the computation of the products in the second summation, we get

= C(k){ 5 3 x(2n) cos ^ ( 2 ( 2 n ) + l)fc n=0

JV-1

+ 5 3 ^ ( 2 ^ - 2n - 1) cos ^ ( 2 ( 2 ^ - 2n - 1) + l)Jfe}

Simplifying, we get

f-1 „ j JV-i 2 ? r j

= C(*){ 5 3 *(2n) cos - ^ (n + -)fc + 5 3 x(2iV - 2n - 1) cos - ^ (n + - )*} n=0 n = £

Using the rearrangement of the data

xo(n)~S X{2n)' » = 0 . 1 . - . f - l ^ W - j x ( 2 J V - 2 n - l ) , n = f , f + 1 , . . . , 7 V - 1

and combining the two summations, we get

Xct(k) = C(k) 5 3 ^o(n) cos — (n + ^)k n=0

We illustrate the operation of this equation with a specific example. The computation of XC{(1), with N = 8, involves the sum of products of the basis function, 0.5cos ^ ( n + | ) , and the input data shown, respectively, in Figs. 15.2(a) and (b). The coefRcient can also be computed by the sum of product of the functions shown in Figs. 15.2(c) and (d). As the cosine wave is even-symmetric, a full cycle of the waveform, 0.5cos ^ ( n + | ) , can be


0.4904

0.2778

c " * -0 .0975

-0 .4157

•

0

0.4904

0.2778

1? H -0 .0975

-0 .4157

•

0

bas

•

2

•

•

2

s function

• •

•

4 n

(a)

• •

4 n

(c)

•

6

•

6

4 3.5

3 1?2.5 V 2

1.5 1

0.5

+

0

4 3.5

3 •5-2.5

1.5 1

0.5

+

0

+

+

input data +

2

+

2

+ +

4 n

(b)

+ +

4 n

(d)

+ +

+

6

+

+

+

6

Fig. 15.2 (a) The DCT basis function for the computation of Xct{\), with N = 8. (b) The samples of an arbitrary data, (c) The samples of one cycle of the cosine waveform giving the same sample values as shown in (a) such that the even-indexed samples appear first followed by the odd-indexed samples in reverse order, (d) The data samples shown in (b) rearranged such that the even-indexed samples appear first followed by the odd-indexed samples in reverse order. Now, either the sum of products of samples shown in (a) and (b) or those shown in (c) and (d) yields X c t( l) .

used with the rearrangement of the input data as shown in Fig. 15.2(d). For example with n = 1, cos f (1 + \) = cos ^ ( 7 + \). The input data values shown in Fig. 15.2(b) are rearranged such that the even-indexed values appear in the first half and the odd-indexed values appear in the second half in reverse order. The last equation can be, equivalently, expressed as

J V - l

Xct(k) = C{k)Re{Y^xo(n)e~j¥{n+i)k)} n = 0

i V - 1

= C(k) Re {W?N(J2 xo(n)W%k)},

where the summation in the last expression is the DFT of xo(n).


Example 15.1 Compute the DCT, Xct{k), of x(n) = {2,1,0,3}. Compute x(n) back from Xctik). Solution Form the sequence xo(n) = {2,0,3,1}. Compute the DFT of xo(n), XO(k) — {6, —1 + j , 4, —1 — j}. The twiddle factors are

Wfe = {1,0.92 - jO.38,0.71 - jO.71,0.38 - j'0.92}

Multiplying XO(k), term by term, by W*Q and taking the real parts, we get {6.0,-0.54,2.83,-1.31}. A more efficient procedure for this step is as follows. As the DFT is conjugate-symmetric, if XO(k) = a + jb, then XO(N — k) = a — jb. The twiddle factors are related in the following way. If W*N = c + jd, then W\N~ = —d — jc. Therefore, the products are related.

(a+jb)(c + jd) = (ac-bd)+j(ad + bc) (a — jb)(—d — jc) = — (ad + be) — j (ac — bd)

The imaginary part of the product W%NXO(k) in the first half is the negative of the value of the real part with index N — k. Therefore, multiplication with the twiddle factors over half the range is sufficient. Now,

X^k) = C(fc){6.0, -0.54,2.83, -1.31} = {3.0, -0.38,2.0, -0.92}

The inverse: We trace the steps backwards. Multiply Xctik) by constants

for A; = 0 and J \ for other values of k to get {1.5, -0.27,1.41, -0.65}.

Due to the property mentioned above, the complex signal, W*6XOik), can

be constructed with a scale factor as

{1.5, -0.27 + jO.65,1.41 - j l . 41 , -0.65 + jO.27}

Now, XOik) can be computed with a scale factor as

^ { 1 . 5 , -0.27 + jO.65,1.41 - j l . 41 , -0.65 + jO.27} = {1.5, -0.50 + jO.50,2, -0.50 - jO.50}

These values have been normalized for DCT. For taking the IDFT the first value must be multiplied by JV = 4 and other values by y = 2 (Note that the number of multiplication by constants can be minimized in the implementation of the algorithm.) to get XOik) = {6, - 1 + j , 4 , - 1 — j}. Computing the IDFT of XOik), we get xoin) = {2,0,3,1}. The first half

vT


is the even-indexed values of the input data and the second half, in reverse order, are the odd-indexed values. I

The computational complexity of the DCT is that of computing the DFT of an TV-point real data, in addition to 2N — 3 real multiplications and N — 2 real additions.

15.3 The 2-D Discrete Cosine Transform

The 2-D DCT of a real-valued signal x(ni, n2), n\, n2 = 0 , 1 , . . . , N — 1 is defined as

J V - 1 J V - 1

Xct{kuk2) = C{k1)C{k2) ] T 5 3 x ( n i , n 2 ) c o 8 ^ ( 2 n i + l)*i n i=0 ri2=0 7T

x cos—:(2n2 + l)fe2, h,k2 = 0,1, ...,N - 1

where C(fci) and C(k2) are as defined in the last section. The 2-D inverse DCT is defined as

N-l N-l

x(nun2) = Yl Yl C(fci)C(fc2)Xct(fci,fc2)cos^r(2ni + l)fci ki=0k2=0

7T x c o s — ( 2 n 2 + l)k2, ni,n2 = 0 , 1 , . . . , N - 1

The transform is separable and can be computed by the row-column method. For example, the 2-D DCT can be written as

N - l J V - 1

Xct(kuk2) = C(fc2) £ { £ ( * ! ) ^ x ( n 1 , n 2 ) c o s ^ ( 2 n 1 + l)A;1} ri2=o m=o

x c o s — ( 2 n 2 + l)A;2

Therefore, the algorithm for computing the 2-D DCT is to compute the 1-D DCT of each row of the image followed by the 1-D DCT of each column of the resulting data and vice versa.


Example 15.2 Compute the 2-D DCT of the following matrix of data. The origin is at upper left-hand corner.

2.50 • 1.04 1.50 1.19 .

Computing 1-D DCT of the rows, we get Xct(ki, fo) as

1.98 " 1.07 1.60 2.95 .

The original image can be obtained by computing 1-D inverse DCT of each row of the transform matrix followed by 1-D inverse DCT of each column of the resulting matrix and vice versa. I

15.4 Summary

• In this chapter, algorithms for the computation of the 1-D and 2-D DCT were described. It was pointed out that the DCT is essentially the DFT of an even-extended real signal. Reordering of the input data makes the problem of computing the DCT into a problem of computing the DFT of a real signal with a few additional operations. This approach provides regular, simple, and very efficient

n i n2 -*

1 3 1 1 4 0 4 1 3 2 1 0 2 1 4 3

Solution Computing 1-D DCT of the columns, we get

5.00 0.38 2.00 0.92

3.00 0.77 1.00 1.85

5.00 -1.15

0.00 -2.77

k2 ->•

7.75 1.09 -0.25 -0.90 0.95 -0.52

0.25 -2.02 -0.75 -1.52 1.43 -0.60

Exercises 311

DCT algorithms for practical hardware and software implementations.

References


(2) Gonzalez, R. C. and Woods, P. (1987) Digital Image Processing, Addison Wesley, Reading, Mass.

(3) Jain, A. K. (1989) Fundamentals of Digital Image Processing, Prentice-Hall, New Jersey.

Exercises

15.1 Compute the DCT, Xct(k), of x(n) = {3,4,5,6} using the DFT. Compute x(n) back from Xct(k).

* 15.2 Compute the 2-D DCT of the following matrix of data using the DFT. The origin is at upper left-hand corner.

n2 -¥

1 4 2 3 " 5 6 4 1 2 5 1 3 3 0 4 1

ni

4-

Chapter 16

Discrete Walsh-Hadamard Transform

In computing the DFT, we find the representation of a signal in terms of a set of sinusoids. In the case of discrete Walsh-Hadamard transforms, we represent a signal in terms of a set of orthogonal rectangular waveforms. These transforms represent signals with discontinuities more efficiently and they are useful in image processing tasks. These transforms can be computed using algorithms those are very similar to the DFT algorithms. One of the advantages of this type of transforms is that they do not require multiplication operations in their computation and, hence, require less computational effort than the DFT. The study of the DFT and these transforms provides a contrast in representing a signal by two sets of orthogonal functions, sinusoids and rectangular waveforms.

In Sec. 16.1, the discrete Walsh transform (DWT) and the PM DWT algorithm are presented. In Sec. 16.2, the naturally ordered discrete Hadamard transform (NDHT) and the PM NDHT algorithm are described. In Sec. 16.3, the sequency ordered discrete Hadamard transform (SDHT) and the PM SDHT algorithm are developed.

16.1 The Discrete Walsh Transform

In this chapter, it is assumed that N, the number of samples of the data, is an integral power of two and M = log2 N. The DWT of a data sequence {x(n), n = 0 , 1 , . . . , N - 1} is defined as

N-l M - l

*„(*) = £ x(n) IJ (-ir*"-*-«, ft = 0,1,..., AT-1, n=0 t=0

313

314 Discrete Walsh-Hadamard Transform

where ni is the zth bit in the binary representation of n. The Isb is indicated by the subscript 0. The transform coefficients, Xw(k), are called sequency coefficients. Sequency is defined as, for waveforms with an odd number of zero crossings, 0.5(number of zero crossings+1) and, for waveforms with an even number of zero crossings, 0.5(number of zero crossings). The matrix form of the defining equation, with N = 8, is written as

" MO) " M(i) M2) M3) M4) M5) M 6 )

. M ? )

" 1 1 1 1

1 1 1 1 - •

1 1 - 1 - 1 j

1 1 - 1 - 1 - ]

1 - 1 1 - 1 ]

1 - 1 1 - 1 - ]

1 - 1 - 1 1 ]

1 - 1 - 1 1 - ]

t i l l "

1 - 1 - 1 - 1

L 1 - 1 - 1

L —1 1 1

L - 1 1 - 1

L 1 - 1 1

L - 1 - 1 1

L 1 1 - 1 _

" x(0) " ar(l) x(2)

*(3) x(4) *(5) x{6)

. x(T) .

The DWT basis waveforms, with N = 8, are shown in Fig. 16.1. The DWT basis functions,

A f - l

WN(k,n)= J[{-l)n*k"-^, n,fc = 0 , l , . . . , i V - l ,

can be generated using the bits in the binary representation of n and k, the data and sequency indices. For each value of k, multiplying the corresponding bits of n and k to get a 0 or 1 and finding the product of (-1) to the power of 0 or 1 yields the kernel values. Let k = 3 = Oil and n = 2 = 010. Then, the bit-by-bit product of 110 (remember the bits of k are to be reversed) and 010 is 010. The corresponding kernel value is W8(3,2) = ( -1 ) ° ( -1 ) 1 ( -1 ) ° = - 1 - Since the first row of the kernel matrix is all ones,

N-l

Y,WN(0,n)=N

n=0

Since there are equal number of plus ones and minus ones in all other rows,

J V - l

J2WN(k,n) = 0, Jfc = 1 , 2 , . . . , J V - l n=0

The Discrete Walsh Transform 315

l 1 r fe°°oL

0

•? 11* ci o fe 1 .

0

B- 1 S. 0 fe 1

•

•

_ • _

2

•_ 2

•

.

_*.

_• .

4 n

(a) •

4 n

(o)

•

•

_ • _

6

a— 6

•

_*L

_•.

"c 1

& 1

•

0

"B 1 t* C 0 £ 1

0

"B 1 !£ 0 B£°° 1

•

•

•

—*_

•

2

•

2

•

•

•

4 n

(b)

• • 4

n

(d)

_ • • - .

_ a _

_ • _

•

• . 6

•

6

_•

_«.

•

•

(e)

? 1|« • • • £ 0

OO I

* -1 L — • — • . • •

(9)

B- 1 f cL o -fe -1 L—•—•_

(f)

4 n

(h)

Fig. 16.1 The DWT basis functions, with N = 8.

The Walsh function is orthogonal. That is,

N-l J V - 1 ,

n = 0 n = 0 ^

TV for k = s

fork^s

where <g> represents exclusive-or operation on the bits of k and s yielding a 1 when two bits are different and a 0 when two bits are the same. If both the corresponding bits of the two sequency indices, k and s, are zeros or ones, the products, ( -1)°(-1)° or ( - l ^ - l ) 1 , are all ones, as n varies. This is equivalent to setting the corresponding bit of the resultant sequency index to zero and setting the other bits to one. If k = s, Wjv(k,n)WN(k,n) = Wjv(fc ® k,n) = Wjv(0,n). If k ^ s, since there is at the least one bit different, we get a resultant sequency index that is other than 0, and the sum of the Walsh basis function is zero. For example, k = 6 — 110 and s = 7 = 111 results in W8{Q,n)Ws(7,n) = W8(6® 7,n) = Ws{l,n).


The inverse DWT is given by

1 N-i M-\

^)^E I « 'WlI ( - 1 )" i f e M " I ' i ' n = 0,l,...,iV-l k=0 t=0

As the inverse DWT definition is similar to that of the DWT except for a constant divisor, an algorithm for computing the DWT can be used directly for computing the inverse DWT. The transform pair can be verified using the orthogonal property.

Example 16.1 Let x(n) = {1,3,5,7}. Compute the DWT of x(n). Get back x(n) by computing the inverse DWT of the sequency coefficients. Solution

Xw{0) = 1 + 3 + 5 + 7 = 16 Xw(l) = 1 + 3 - 5 - 7 = - 8 Xw(2) = 1 - 3 + 5 - 7 = - 4 Xw(3) = 1 - 3 - 5 + 7 = 0

The inverse DWT is

x(0) = (16 + (-8) + (-4) + 0)/4 = 1 x(l) = (16 + (-8) - (-4) - 0)/4 = 3 x{2) = (16 - (-8) + (-4) - 0)/4 = 5 ar(3) = ( 1 6 - ( - 8 ) - ( - 4 ) + 0)/4 = 7 I

The PM DWT algorithm

While it is possible to derive the algorithm from the definition, it is much easier to deduce the algorithm from the 2 x 1 PM DIT DFT algorithm due to the similarity in the definitions of the DWT and DFT. Consider the DFT definition, with N = 4, given below.

3

X(fc) = ^x(n )W 4 " f c , it = 0,1,2,3 n=0

Representing the indices in the twiddle factor in terms of binary bits, we get

3

n=0


O

__ V ^ x ( n \ ^ 4 ( ™ i f c i ) j ^ 2 ( " o * i + f c o n i ) ^ l ( n 0 f c o )

n=0 3

rl(n0fco) = 5^ar(n)(-l)(no*1+*oBl)W4 n=0

This definition is the same as that of the DWT if we set all the twiddle factors, except those with powers of (—1), to unity. Therefore, by setting all twiddle factors to unity, except those with powers of ( -1) , we change a DFT algorithm to a DWT algorithm. The PM DWT algorithm is deduced from the 2 x 1 PM DIT DFT algorithm as follows. The input vectors a(ri) are defined as

N N o.(n) = {o0(n),a1(n)} = {(x(n) + x(n + -j),x{n) - x(n + y ) ) } ,

N n = 0,l,...,--l

The output vectors A(k) are defined as

A(k) = {A0(k), At (*)} = {Xw(k),Xw(k + y ) } , k = 0 , 1 , . . . , y - 1

The equations characterizing the PM DWT butterfly are given as

A[*lHh) = 4 r )w-4 r )w 4r+1>(Z) = A[r\h) + A<?\l) A?+1\l) = A[rHh)-A[rHD

The SFG of the PM DWT algorithm, with N = 16, is shown in Fig. 16.2, which is the same as that of the 2 x 1 PM DIT DFT algorithm with the twiddle factors set equal to unity. The number of real addition operations required by the PM DWT algorithm is N log2 N.

Example 16.2 Find the trace of the algorithm shown in Fig. 16.2 for the following input data.

{1,3,1,1,4,0,4,1,3,2,1,0,2,1,4,3}

Solution The trace is shown in Fig. 16.3. The sequency coefficients are


tage 1 stage 2 stage 3

o(7)o

A(0)

A(l)

A(2)

A(3)

A(4)

A{5)

A(6)

A(7)

Fig. 16.2 The SFG of the PM DWT algorithm, with N = 16.

{31, - 1 , -7 ,1,1 ,1 ,11 , - 5 , 9 , 1 , - 9 , - 9 , - 1 , - 1 , - 3 , - 3 }

The 2-D DWT

The 2-D DWT of an N x N image {x(nun2), nun2 = 0 , 1 , . . .,N - 1} is denned as

7V-1 J V - 1 M - l

n i = 0 n 2 = 0

M - l

I n 2 = 0 i = 0

M - l

x JJ ( - l )^) '^ )"- 1 - 1 , fci,fc2=0,l,...,iV-l (=0

Example 16.3 Compute the 2-D DWT of the following matrix of data.


X(o) m x(8) |_3j

*(1) |~3

*(2) m z(10)LU *(3) m a?(ll) I_QJ x{4) x(12) x(5) x(13)

x(6) x(14)

*(7) x(15)

Vector Formation

and Swapping

Stage 1 Output

4 -2

6 2

10 -2

0 -4

20 0

0 0

311 -MO) 9j Xw(&)

rri ^( i ) LU -M9)

yy ^(io) Xw{3) • M i l )

11 X«,(4) l ] ^ ( 1 2 )

rri ^(5) LU XW(U) Til Xw{6) 31 X„,(14)

Xw(7) -M15)

a

Fig. 16.3 The trace of the PM DWT algorithm, with N = 16.

The origin is at upper left-hand corner.

n2 ->

1 3 1 1 4 0 4 1 3 2 1 0 2 1 4 3

Solution Computing the 1-D DWT of the rows, we get

6 9 6

10

2 - 1

4 - 4

- 2 7 2 2

- 2 1 0 0


Computing the 1-D DWT of the columns of the resulting matrix, we get

*1

I " 31

- 1 - 7

1

k2 -»• 1 1

11 - 5

9 1

- 9 - 9

- 1 - 1 - 3 - 3

The original image can be obtained by computing the 1-D row inverse DWTs of the transform matrix followed by the 1-D column inverse DWTs of the resulting matrix and vice versa. I

It can be shown that the 2-D computation is the same as that of 1-D DWT. As such the SFG for computing the 2-D DWT remains the same as in Fig. 16.2 except that the data length is iV2. If data is read row-by-row from the input file, the output would be written column-by-column and vice versa.

16.2 The Naturally Ordered Discrete Hadamard Transform

This transform generates the same sequency coefficients as that of the DWT, but in bit-reversed order. The NDHT of a data sequence {x(n), n = 0 , 1 , . . . , N — 1} is defined as

Xnh(k) = Y, *(n)(-l)^-° U i i > fc = 0 , 1 , . . . , iV - 1, n=0

where ni is the ith bit in the binary representation of n, the lsb is indicated EA f - l ,

„., „^„„. - , . . , , , . . , . . , v - , i=o n i k i ,

n, k = 0 , 1 , . . . , JV - 1 , with N = 8, are shown in Fig. 16.4. The matrix form of the defining equation, with N = 8, is written as

Xnh(0) Xnh(l) Xnh(2)

Xnh(3) Xnh(4)

Xnh{5) Xnh(6)

L Xnh(7)

" 1 1 1 1 1 1 1 - 1 1 - 1 1 - 1 1 1 - 1 - 1 1 1 1 - 1 - 1 1 1 - 1 1 1 1 1 - 1 - 1 1 - 1 1 - 1 - 1 1 1 1 _ 1 - 1 _1 _ 1

1 - 1 - 1 1 - 1 1

1 1 " 1 - 1

- 1 - 1 - 1 1 - 1 - 1 - 1 1

1 1 1 - 1

• ar(0) " x(l) x(2)

x(3) x(4) x(5) x(6)

. «(7) .

The Naturally Ordered Discrete Hadamard Transform 321

I1 Bt"o

0

"e 1 ^ 00

£ 1

• •

0

•? 11* • € o fe 1 1

0

"c 1 t* • € 0 ^ o° 1

0

2

2

•

2

•

2

4 n

(a)

• •

4 n

(c)

•

4 n

(e)

—a • a_ 4

n

(g)

6

6

6

• •

6

"? 1 [• d o— ^ oo 1

fe i L _ 0

"e 1 p e o— £ i L _

0

"B 1 p £ 0 — •. 0° I

~s 1 L _ 0

"c 1 p £ 0 — fe i L

0

_ a _

_ a _

_•_

_•_

•

2

2

•

2

2

•

_a . 4

n

(b)

• •

4 n

(d)

_ • • 4

n

(0 •

•_ 4

n

(h)

_•_

_ i _

•

•

•

6

•

• 6

•

6

•

. t

6

Fig. 16.4 The NDHT basis functions, with AT = 8.

The kernel of NDHT is recursively related in a simpler manner and is, therefore, very easy to generate.

NDHT2 = 1 1 1 - 1

NDHT 2 ) V = N D H T J V N D H T J V

N D H T N - N D H T J V

For example,

N D H T 4

1 1 1 1

1 - 1

1 - 1

1 1

- 1 - 1

1 - 1 - 1

1


By writing the rows of this kernel in bit-reversed order, we can get the DWT kernel. The inverse NDHT is given by

^(») = T7 ] T Xn h(*)(-l)2-«-o "•*••, n = 0 , l , . . . , 7 V - l fc=0

Example 16.4 Let x{n) = {1,3,5,7}. Compute the NDHT of x(n). Get back x(n) by computing the inverse NDHT of the sequency coefficients. Solution

Xnh(0) = 1 + 3 + 5 + 7 = 16 Xnh(l) = 1 - 3 + 5 - 7 = - 4 Xnh(2) = 1 + 3 - 5 - 7 = - 8 Xnh(3) = 1 - 3 - 5 + 7 = 0

The inverse NDHT is

x(0) = (16 + (-4) + (-8) + 0)/4 = 1 x(l) = (16 - (-4) + (-8) - 0)/4 = 3 x(2) = (16 + (-4) - (-8) - 0)/4 = 5 x(3) = (16 - (-4) - (-8) + 0)/4 = 7 I

The PM NDHT algorithm

Since the NDHT produces the same coefficients as that of the DWT, but in bit-reversed order, the PM NDHT algorithm is deduced from the 2 x 1 PM DIF DFT algorithm as follows. The input vectors a(n) are defined as

N N o(n) = {a0(n),oi(n)} = {(x(n) + x(n +—),x(n) - x(n+—))},

N n = 0 , l , . - . , y - l


A(k) = {A0(k), Ax (k)} = {Xnh(2k), Xnh(2k + 1)}, k = 0 ,1, . . . , y - 1

The equations characterizing the PM NDHT butterfly are given as

4 r + 1 )w = <4r)w+4p)(o o[r+1)(ft) = air\h)-a£\l)

a£+1\l) = a[r\h) + aP(l)

a[r+1)(Z) = a^iV-a^Hl)

The Naturally Ordered Discrete Hadamard Transform

stage 1 stage 2 stage 3

323

A(0)

A(l)

A(2)

Fig. 16.5 The SFG of the PM NDHT algorithm, with N = 16.

The SFG of the PM NDHT algorithm, with N = 16, is shown in Fig. 16.5, which is essentially the same as that of the 2 x 1 PM DIF DFT algorithm with the twiddle factors set equal to unity.

Example 16.5 Find the trace of the algorithm shown in Fig. 16.5 for the input data shown in Fig. 16.3. Solution The trace is shown in Fig. 16.6. The sequency coefficients are

{31,9,1, - 1 , - 7 , - 9 , 1 1 , - 3 , - 1 , 1 , 1 , - 1 , 1 , - 9 , - 5 , - 3 } I

The 2-D NDHT

The2-D NDHT of an N x N image {x(m,n2), nun2 = 0 , 1 , . . . ,N - 1} is defined as

N-l N-l

Xnh(kuk2) = Y, E *(ni,n2)(-l) (£ i=; ("i)*(»0«+E^1(»»)'(fe)')> n i=0 ri2=0

kuk2 = 0 , l , . . . , i V - l


Input Vector Formation

Stage 1 Output

Stage 2 Output

Stage 3 Output

x(0) x(8)

*(1) [31 x(9) LU x(2) m ar(10)LU x(3) [ I x(ll)[_0 x(4) x(12) x(5) x(13) x(6) f T | x(14)LU

*(T) m x(15)liJ

E H

4 -2

5 1

li

10 -2

6 4

10 -6

20 0

11 1

-8 4

a

h -l

8 0

4 -2

0 2

0 0

-1 3

PH i

-4 -4

5 -1

E i a

3ll Xnh(0) 9j Xnfe(l)

U Xnh{3) T l Xnh(4) r9j Xnh(5) Til Xnft(6) U X„h(7)

FT] *nft(8) LU *nh(9)

Fl Xnh(W) 1 I ^nh(H)

X„ft(12) Xn7l(13) Xnh{U) X„/>(15)

Fig. 16.6 The trace of the PM NDHT algorithm, with N = 16.

It can be shown that the computation of 2-D NDHT is the same as that of 1-D NDHT. As such the SFG remains the same except that the data length is N2. If data is read row-by-row from the input file, the output would be written row-by-row and vice versa.

Example 16.6 Compute the 2-D NDHT of the matrix of data given in Example 16.3. Solution Computing the 1-D NDHT of the rows, we get

6 9 6 0

- 2 7 2 2

2 - 1

4 - 4

- 2 1 0 0

The Sequency Ordered Discrete Hadamard Transform 325

Computing the 1-D NDHT of the columns of the resulting matrix, we get

fcl

I

k2 ->• " 31 9

- 7 - 9 - 1 1

1 - 9

1 11

1 - 5

- 1 - 3 - 1 - 3

The original image can be obtained by computing the 1-D row inverse NDHTs of the transform matrix followed by the 1-D column inverse NDHTs of the resulting matrix and vice versa. I

16.3 The Sequency Ordered Discrete Hadamard Transform

This transform generates the same sequency coefficients as that of the DWT, but in sequency-order ({0,1,1,2,2,3,3,4} for N = 8). The SDHT of a data sequence {x(n),n = 0 , 1 , . . . , N — 1} is defined as

N-l M_1

Xah(k) = £ x(n)(-l)^i=~0 n ^ ( f c ) , A; = 0 , 1 , . . . , A T - 1 ,

n=0

where rii is the ith bit in the binary representation of n, the lsb is indicated by the subscript 0, and

g0(k) = kM-i 9i {k) = {kM-i + kM-2) mod 2 52(k) = (&M-2 + kMs) mod 2

9M-I (k) = (h + fc0) mod 2

The matrix form of the defining equation, with N = 8, is written as

" X.h(0) ' Xah{\) X8h{2) Xsh(3) Xsh(A) Xsh(5) X8h(6) X8h(7)

1 1 1 1 - 1 - 1 - 1 - 1 1 1 - 1 - 1 - 1 - 1 1 1 1 1 - 1 - 1 1 1 - 1 - 1 1 - 1 - 1 1 1 - 1 - 1 1 1 - 1 - 1 1 - 1 1 1 - 1 1 - 1 1 - 1 - 1 1 - 1 1 1 - 1 1 - 1 1 - 1 1 - 1

• ar(0) "

x(l) x{2) x(3) x(4) x(5) x(6)

. *(7) .


"a 1 O

fe 0 0

B1 1 CI 0 .. 00

& 1

•

0

"a 1 S. 0 ^ 00

=s 1

•

•

2

2

4 it

(a)

4 n

(c)

• •

•

6

•

6

m

•

•

"e 1

5° ° & -1

•

0

"s 1 C 0 .. 00

£ 1

•

0

"a 1 t» !C 0 ^ 00

fe 1

•

•

•

2

2

•

4 n

(b)

•

4 n

(d)

•

a

•

•

6

6

•

<•) "a 1 f £ 0 & -1

(f)

1 M • • •

0

-1 L • . • . • . •.

(g) (h)

Fig. 16.7 The SDHT basis functions, with N = 8.

The SDHT basis waveforms,

WN(k,n) = ( - l ) E " ; 1 "•»<(*), n,fc = 0 , l , . . . , J V - l ,

are shown in Fig. 16.7, with N — 8. It can be seen that the waveforms are sequency ordered and are similar to the DFT basis waveforms. This kernel can be obtained by rearranging the DWT kernel in gray-code order (0,1,3,2,6,7,5,4, for AT = 8). The inverse SDHT is given by

1 N _ 1 M - l x^ = JJ E *.ft(*)(-l)Si-° ni9i{k)> n = 0 , 1 , . . . , iV - 1

k=0

Example 16.7 Let x{n) = {1,3,5,7}. Compute the SDHT of x(n). Get back x(n) by computing the inverse SDHT of the sequency coefficients.

The Sequency Ordered Discrete Hadamard Transform 327

Solution

X8h(0) X8h(l) X8h(2) Xsh(3)

The inverse SDHT is

x(0) = (16 + (-8) + (0) + ( -4)) /4 = 1 x(l) = (i6 + ( - 8 ) - ( 0 ) - ( - 4 ) ) / 4 = 3 x{2) = (16 - (-8) - (0) + ( -4)) /4 = 5 *(3) = ( l 6 - ( - 8 ) + ( 0 ) - ( - 4 ) ) / 4 = 7 I

The PM SDHT algorithm

This algorithm is similar to the PM NDHT algorithm with the following differences: (i) the input vectors are formed with a different pair of elements and they are placed in the bit-reversed order and (ii) the output at the lower node of a butterfly is stored in a different order. The input vectors a(n) are defined as

a(n) = {a0(n),ai(n)} = {x{2n) + x{2n + l),x(2n) - x(2n + 1)}, N

n = 0 , l , . . . , y - l


A(k) = {A0(k), A^k)} = {X8h(2k),Xsh(2k + 1)}, fc = 0 ,1 , . . . , y - 1

The equations characterizing the PM SDHT butterfly are given as

4r+1)(/i) = a£\h)+a£\l) a?+1\h) = 4p)(/»)-4r)(0 air+1\l) = a[r\h)-a[r\l)

af?+1\l) = o[p )(/ i)+a[ r )(Q

The SFG of the PM SDHT algorithm, with N = 16, is shown in Fig. 16.8.

= 1 + 3 + 5 + 7 = 16 = 1 + 3 - 5 - 7 = - 8 = 1 - 3 - 5 + 7 = 0 = 1 - 3 + 5 - 7 = - 4


o(0)

o(4)

o(2)

«(6)

o( l )

o(5)

o(3)

o(7)

stage 1 stage 2 stage 3 > A(0)

A(l )

A(2)

A(3)

•A(4)

A(5)

A(6)

^(7 )

Fig. 16.8 The SFG of the PM SDHT algorithm, with N = 16.

The 2-D SDHT

The 2-D SDHT of an JV x JV image {x(ni ,n2) , ni,n2 = 0 , 1 , . . . , N - 1} is defined as

* .*(*! , to) = £ E x ( n 1 , n 2 ) ( - l ) ( S " ; 1 ( " 0 ^ ( f c 1 ) + E " o 1 ^ ) ' ^ ^ » , n i = 0 n 2 = 0

fcl,fc2=0,l,...,JV-l

Example 16.8 Compute the 2-D SDHT of the matrix of data of Example 16.3. Solution Computing the 1-D SDHT of the rows, we get

6 2 - 2 - 2 9 - 1 1 7 6 4 0 2

1 0 - 4 0 2

Summary 329

Computing the 1-D SDHT of the columns of the resulting matrix, we get

fc2 -»• 31 1 - 1 9 " - 1 1 - 1 1

1 - 5 - 3 - 9 - 7 11 - 3 - 9 .

The original image can be obtained by computing the 1-D row inverse SDHTs of the transform matrix followed by the 1-D column inverse SDHTs of the resulting matrix and vice versa. I

The SFG of the 1-D SDHT algorithm remains the same for the computation of the 2-D SDHT with the same number of data values except for a difference explained below. The vector formation is carried out in the required manner for the row (column) transforms. However, for the column (row) transforms, the vector formation stage occurs after the computation of the row (column) transforms. In this intermediate vector formation stage, the sum and difference values are stored, respectively, as the first and second elements of a vector, at both the upper and lower nodes of a butterfly. If data is read row-by-row from the input file, the output would be written column-by-column and vice versa.

Example 16.9 Find the trace of the algorithm shown in Fig. 16.8: (i) for the 1-D input data shown in Fig. 16.3 and (ii) for the 2-D input data of Example 16.3. Solution (i) The trace is shown in Fig. 16.9. The sequency coefficients are

{31, - 1 , 1 , - 7 , 1 1 , - 5 , 1 , 1 , - 1 , - 1 , - 3 , - 3 , - 9 , - 9 ,1 ,9}

(ii) The trace is shown in Fig. 16.10. The sequency coefficients are the same as that of Example 16.8. I

16.4 Summary

• In this Chapter, 1-D and 2-D DWT, NDHT, and SDHT and the PM algorithms to compute them were described. These transforms, particularly effective for representing waveforms with discontinuities, use a set of orthogonal rectangular waveforms as basis func-

i


Input

Vector Formation

and Swapping

Stage 1 Output

Stage 2 Output

Stage 3 Output

x{0 x(l x(2 x (3 x (4 x(5

x(6 x(7

x(8 x(9

z(10) z ( l l )

*(12) z(13) a;(14) x(15)

_3_

a a a

3 2 1 0

2 1

I 1 1

a

9 -1 10 -4 -2 -2

3 1 8 0

-1 -3

a a a

0 0

-9 5 0 4

a -9 -9 1 9

3T1 Xsh(0)

Gn x8h(i) X8h(2) X8h(3)

mn xsh(4) Lil xsh(5)

m ^ft(6)

n n Jf.fc(8) X,fc(10) X8h(U)

X.fc(12) Jf.h(13) Xgh(14) X,fc(15)

Fig. 16.9 The trace of the 1-D PM SDHT algorithm, with N = 16.

tions. The sequency coefficients, corresponding to the same input data, of these transforms are the same but appear in different order. The PM algorithms for computing these transforms are very similar to those of the PM DFT algorithms. The computation of these transforms is relatively faster since multiplication operation is not required.

References

(1) Gonzalez, R C. and Wintz, P. (1987) Digital Image Processing, Addison Wesley, Reading, Mass.

(2) Sundararajan, D. and Ahmad, M. O. (1998) "Fast Computation of the discrete Walsh and Hadamard Transforms", IEEE Trans. Image Processing, vol. 7, No. 6, pp. 898-904.

Exercises 331

x(0, 0 X(0, 1

z(0,2 x(0,3

x( l ,0 x ( l , l x( l ,2 x( l ,3 z(2,0 z (2 , l x(2,2; a;(2,3

z(3,0 a;(3,1 z(3,2 z(3,3

Vector Formation

and Swapping

Row Transform

Output

Vector Formation

Column Transform

Output

m

5 1

4 1 4 ]

re-4

nr - i

16 -4

1 3

1 -7

nn j

10 -4

-2 -2

0 8

-1 -3

[-b\ 11

-1 -1

0 0

5 -9

-3 -3

9 1

4 0

-9 -9

*8fc(0,0 X8h{\,Q

Xsh%Q Xsh(Z,Q

Xsh{Q,\ X8h(l,\ X s h (2 , l Xsh(3,1 X s / l(0,2 X s f t( l ,2

-X»fc(2,2 -^»fe(3,2

*8ft(0,3 ^sfe(l,3 ^«ft(2,3 ^»/j(3,3

Fig. 16.10 The trace of the 4 x 4 2-D PM SDHT algorithm.

Exercises

16.1 List the values at each stage of computing the inverse DWT of the sequency coefficients shown in Fig. 16.3 using the algorithm shown in Fig. 16.2.

* 16.2 List the values at each stage of computing the inverse NDHT of the sequency coefficients shown in Fig. 16.6 using the algorithm shown in Fig. 16.5.

16.3 List the values at each stage of computing the 1-D inverse SDHT of the sequency coefficients shown in Fig. 16.9 using the algorithm shown in Fig. 16.8.

16.4 List the values at each stage of computing the 2-D inverse SDHT of the sequency coefficients shown in Fig. 16.10 using the algorithm shown in Fig. 16.8.



16.1 Write a program to implement the PM DWT algorithm.

16.2 Write a two stages at a time program to implement the PM DWT algorithm.

16.3 Write a program to implement the PM NDHT algorithm.

16.4 Write a two stages at a time program to implement the PM NDHT algorithm.

16.5 Write a program to implement the PM SDHT algorithm.

16.6 Write a two stages at a time program to implement the PM SDHT algorithm.

16.7 Write a two stages at a time program to implement the 2-D PM SDHT algorithm.

Appendix A

The Complex Numbers

The real number system can be represented by a number line. A real number is the coordinate of a point on the line. Zero is the reference number and the numbers to the right of zero are called positive numbers and those to the left are called negative numbers. With this number system, we can specify a movement along the line. Now, the question is how to specify a movement along any direction, that is a movement in a plane. If we draw another number line at right angles to our original number line, we can specify any movement in the plane with the help of a pair of numbers, one number taken from each line.

Using pairs of ordered numbers, taken from two perpendicular number lines, to represent points in a plane is the complex number system. Although it is not really complex, unfortunately, it is called the complex number system and a number is called complex number. Notwithstanding the fact that each of the pair of numbers is just a real number, the first one is called the real part and the second one is called the imaginary part of the complex number. The plane formed by the two number lines is called the complex plane. The first and the second number lines are called, respectively, the real and imaginary axis of the complex plane. These names are totally misleading and there is nothing complex or imaginary in representing a point in a plane with a pair of real numbers.

We stress this point so much because our principal object in Fourier analysis, the sinusoid, is most efficiently represented by a point moving in a plane. A pair of ordered real numbers is our standard denomination. By using complex numbers, we manipulate sinusoids with manipulation of numbers which is just little bit more involved than real numbers rather

333

334 The Complex Numbers

•-1+/J

,-2+yO

.-2-yi | So

2+/2 ,o+y 1

0+yO

2 - / 2 Jm

- 2 - 1 0 1 2

Fig. A. l The complex plane with some complex numbers.

than using vector algebra or vector diagrams. When the second of the pair of ordered numbers is zero, it is equivalent to a real number. The set of real numbers is a subset of the complex numbers.

Rectangular Form

Figure A.l shows some complex numbers in the complex plane. Remember that the negative symbol '-' is used to represent n radians or 180 degrees (the angle formed by the positive and the negative real axis). Similarly, the symbol j is used to indicate an angle of \ radians or 90 degrees. The rectangular form of a complex number is given by x + jy where x and y are real numbers and j — y/—l since jj = j 2 = —1, a rotation of 180 degrees. The number x is called the real part and y is called the imaginary part of the complex number and j is called the imaginary unit. If the real part of a complex number is zero (the number 0 + j l ) , it is called a pure imaginary number (a number on the imaginary axis). If the imaginary part is zero (the number -2+jO), it is called a pure real number (a number on the real axis).

Let Zi = xi + jyi and z2 — x2 + J2/2- The complex numbers z\ and z2

are equal if and only if x\ — x2 and y\ =yi-Addition

z3 = z\ + z2, where z3 = (xi + x2) + j(yi +1/2)

Subtraction

z3= zi - z2, where z3 = (xi - x2) + j(yi - 2/2)

Polar Form 335

Multiplication

23 = ziz2, where z3 = (xix2 - 2/12/2) + j(xiy2 + x2yi)

Conjugation The conjugate of a complex number z = x+jy, denoted by z*, is defined

asz* = x—jy, that is the imaginary part is negated. The complex conjugate pair 2 + j2 and 2 — j2 is shown in Fig. A.l. z + z* = 2x, z — z* — j2y, and zz* = x2 + y2. The complex conjugate of an expression is equivalent to the expression with every complex quantity conjugated.

(Zl +Z2)* =Z*+Z*, (Zx - Z2)* = Z* - Z*, (Z!Z2)* = Z*Z2.

Division

_ zi _ ziz* _ {xxx2 + 2/12/2) + j(x2yi - y2xi) z3 — — — T — o~, 9

Z2 Z2Z^ X\ + J/2

Polar Form

A point in the complex plane can also be represented by its distance from the origin, called the magnitude (the magnitude is always a positive number), and the angle, called the argument, formed by the line joining the point and the origin, and the positive real axis. The magnitude A and the argument 8 can be obtained from the rectangular form x + jy as

A = +x/(a;2 + 2/2), 9 = tan""1 ^ x

The polar form is written as Aid. As ALQ = AL(Q + 2kir), where k is an integer, the value of the argument 6 such that — it < 6 < 7T is called its principal value. From polar representation, the components of the rectangular form can be derived as x = ^4cos0 and y = AsinO. Using Euler's identity, we get

Aeje = Acos0 + jAsinO

The expression on the left is called the exponential form of the complex number. The multiplication and division operations are relatively easier to carry out in the polar or exponential form.

z3 = zxz2 = Axe^A2ei6* = A1A2e

j^+e^


z2 A2e^ A2

Powers and Roots

Let z = reje. Then, zN = (reje)N = rNejm, where N > 1 is an integer. Since 2kir can be added to the argument of a complex number, where k is an integer, without changing its value and replacing N by jj, we get for z^O

X 1 „• 2fcir + 9 £ W •=. fit Ql N .

Expanding the equation, we get the N distinct roots as

y~z= ^ ( c o s ^ ± f + j s i n ^ ± f ) , fc = 0 , l , . . . , J V - l (A.l)

where %/f is the real positive root. For example, letting r — 1, 6 = TT, and N = 2, with A; = 0, we get

( c o s - + j s i n - ) = 0 + j l

and, with k = 1, we get

. 37T . . 37T.

(cos y + j s i n y ) = 0 - j l

The square of both the roots equals e*-7^ = — 1.

The Roots of Unity

Of particular interest to DFT are the N roots of unity, that is the solutions of the equation xN — 1 = 0. In mathematical literature, the computation of the DFT is referred to as the evaluation of a polynomial of degree N — 1 at N values of the roots of unity. The usual representation of a polynomial is with its coefficients. For example, the representation of the polynomial 2 + 3x with its values at the two roots of unity (1, - 1 ) is (5, - 1 ) , obtained by evaluating the polynomial at 1 and - 1 . In electrical engineering literature, the computation of the DFT is referred to as the evaluation of the ^-transform at JV equally spaced points on the unit circle.

The Complex Exponential, and Cosine and Sine Functions 337

By letting r = 1 and 6 = 0 in Eq. (A.l), we get N complex numbers, called the iVth roots of unity, as

V I = (cos ^ + j sin ^ ) , A = 0 , 1 , . . . , J V - 1

With N = 2, the roots are 1 + jO and - 1 + jO. With N = 3, the roots are 1 + jO, - | + j ^ , and - \ - j ^ . With JV = 4, the roots are 1 + jO, 0 + jl, —1 + jO, and 0 - jl. The first eight roots, with N = 32, are shown in Fig. A.2. By using appropriate signs, the other roots can be obtained. These are the values of

em" = wi2k, fc = o,i, . . . ,3i

The Complex Exponential, and Cosine and Sine Functions

Let us express the complex exponential in the series form. The series representation is similar to that of a real exponential.

6 1 + W";+ 2! 3! 4! ' (r)!

The exponent can be any complex number, but our interest is primarily in the complex exponential with a pure imaginary exponent. One half of the numbers in the series are pure real and the other half pure imaginary. If we group the two sets, we get

•»><i-£+S--)+i(#-£+£--> The two expressions in parenthesis on the right-hand side are series expansions for cosine and sine functions, respectively. Therefore, we get the Euler's identity

eje = cos 6 + j sin d

The Representation of a Real Sinusoid

Figure A.2 shows thirty-two equidistant points of a circle of radius one and center at origin. By definition, the projection of any point on the real axis is cos#, where 8 is the angle formed by the line from the origin to the point and the positive real axis. The line between a point on the circle and the


Imaginary axis k. fl 0.1951+ J0.9808

* • 0.3827 + j0.9239 • 0.5556+ J0.8315

,f 0.7071 + j'0.7071

,]un\ • 0.8315 + J0.5556

| • 0.9239 + jO.3827

cos

»•

I • 0.9808 + j0.1951

J- . i-oopo + JO.OOOO

H t Real axis I * I • I • 1

• I

i . - ' cos(a;n), LJ = ||

-J'l n = 0

8 *

16

24.

*32

Fig. A.2 The first eight roots of unity, with N = 32. The values not shown can be obtained by using appropriate signs for various quadrants. The figure also shows the representation of a cosine waveform as the projection of a rotating vector on the real axis.

The Representation of a Real Sinusoid 339

Imaginary axis

';'0.5

0.5 ^S(UJn)

2 ' 2

Fig. A.3 The representation of a cosine waveform as the sum of two rotating conjugate vectors.

origin is a vector and if we allow it to move it is called a rotating vector, e j 9 . Let the vector move at an angular velocity of u> radians per sample in the counterclockwise direction. If we substitute 6 — wn with u = | f and allow n to vary from 0 to 31, then the plot of the projections of the tip of the vector on the real axis is a cycle of the cosine wave, cos(|f-n) = Re [e?"&"), which is shown in Fig. A.2. By associating a complex constant Ae^ with the complex exponential e J w n , we can get a cosine waveform with arbitrary amplitude and phase shift.

Figure A.3 shows two conjugate vectors with the same amplitude and angular velocity (The angular velocity is negative if the direction of rotation is clockwise.) rotating in opposite directions. The necessity of having two rotating conjugate vectors is to get an exact mathematical expression for a cosine wave rather than describing it as the projection of the tip of a rotating vector on the real axis. The tip of a vector has two projections on the two


axes. By having two conjugate vectors rotating in opposite directions, we can cancel either of the projections and generate a sinusoid in terms of phase-shifted cosine or sine waveform. Although it is an arbitrary choice, we use the phase-shifted cosine to represent a sinusoid in this book.

Appendix B

The Measure of Computational Complexity

An algorithm is a systematic procedure to solve a problem. As there can be many algorithms to solve a problem, a measure is required to evaluate the relative merits of the algorithms. While the ultimate measure is the execution time of the algorithms on the same computer, however, we need a measure that is independent of the computer used although it may be less precise. Such a measure is given by a function that indicates the rate of increase in the number of major operations, such as multiplications, additions, or comparisons, as the number of elements in the input data is increased. For example, if the number of operations required to execute an algorithm is proportional to JV, the data size, then it is referred to have a computational complexity of 0(N). The constant of proportionality and any terms of a lower order are ignored. The growth of the computational complexity of various orders is shown in Fig. B.l. For a large JV, the execution time will be proportional to the computational complexity. For very small JV, due to reduced overhead operations, an algorithm with an higher order of complexity could run faster. When comparing algorithms of the same order of complexity, we have to consider several factors such as run-time, overhead operations, memory required, regularity, simplicity, stability, and numerical accuracy to find out which algorithm is practically better.

The computational complexity of directly evaluating the 1-D DFT is 0(JV2) and that of using fast 1-D DFT algorithms is 0(JVlog2 JV). The computational complexity of directly evaluating the 2-D DFT is 0(JV4). The computational complexity of directly evaluating the 2-D DFT using the row-column method is 0(N3). The computational complexity of evalu-

341

342 The Measure of Computational Complexity

10

10lu

Com

plex

o

-m" J 1 a o •

+ D 0

•

+ D

• X

+

D O

• X

+

D

O

X

+ +

+ +

+ n

A/3* D " o + • o

° „ °

NlBSf , " " _ • * X

log^V

10" 10' 10' 10J

N 10' 10°

Fig. B.l Growth of the computational complexity of various orders.

ating the 2-D DFT using the row-column method along with fast 1-D DFT

algorithms is 0(N2 log2 N).

Appendix C

The Bit-Reversal Algorithm

The Number System

A number consists of i digits in sequence, each digit taking any of r possible values, written as ni-i,rii-2, • • .,rii,no- The symbol r represents the radix or base of the number system. In the familiar decimal number system r = 10, indicating that there are 10 distinct digits used in that number system, namely {0,1,2,3,4,5,6,7,8,9}. The decimal number 253 is equal to 2 x 102 + 5 x 101 + 3 x 10°. If we use only two digits {0,1} (r = 2), then the number system is called radix-2 or binary number system. A digit in the binary number system is called a bit (fiinary digit). The rightmost bit, no, is called the least significant bit (lsb) and the leftmost bit, rij_i, is called the most significant bit (msb). The binary number 11001 is equal to 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 2° = 25 in decimal.

Conversion of a Decimal Number into a Binary Number

One way of converting a decimal number into the corresponding binary number is to find the bits from msb to lsb. Remember that each bit has a weight attached to it depending on its position. The weight attached to bit n0 is 2° = 1 and, in general, the weight of a bit n» is 2*. Given a decimal number, find the largest weight that is equal to or just less than the decimal number. For example, the largest weight that is equal to or just less than 25 is 16. Therefore, we fix m = 1 and subtract 16 from 25 to get 9. We repeat the process again with 9 and continue until the number becomes a 0 or 1. Every weight must be tested in decreasing order. Note that if the

343

344 The Bit-Reversal Algorithm

Table C.l A list of the decimal numbers from 0 to 15 and their bit-reversals.

Decimal

0 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15

Binary ri3n2Tiino

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

Bit-reversed noT»lTl2«3

0000 1000 0100 1100 0010 1010 0110 1110 0001 1001 0101 1101 0011 1011 0111 1111

Bit-reversed decimal

0 8 4

12 2

10 6

14 1 9 5

13 3

11 7

15

weight is greater than the number at any stage, we set the corresponding bit to zero before we test with the next lower weight. This method will be used in the bit reversal algorithm to test the bits in various positions.

The Bit-Reversed Order

If we express a set of decimal numbers in binary form, rewrite the bits of each number in reverse order, and convert the resulting binary sequences to decimal form, we get the bit-reversed order. Decimal numbers from 0 to 15 are shown in the first column of Table C.l. The second column shows the binary representation of the numbers. The third column shows the binary bits of the second column written in reverse order. The fourth column shows the bit-reversed order of the decimal numbers. For example, the binary representation of the decimal number 2 is 0010 and its bit-reversed form is 0100 in binary and 4 in decimal. The in-place DFT algorithms require the input to be in the bit-reversed order or generate the output in that order. Therefore, we need the bit-reversed order of a set of N consecutive integers starting from zero, where JV is a power of two.

The Bit-Reversal Algorithm 345

The Bit-Reversal Algorithm

The complexity of finding the bit-reversed order is reduced by using the similarities in the bit patterns of the numbers. We can easily observe the following similarities in the bit-patterns of the list of binary numbers shown in Table C.l. (1) The bit-reversal of odd numbers in the first half of the list (1,3,5,7) are the even numbers (8,12,10,14) in the second half of the list, because odd numbers have lsbs 1 and since they are in the first half their msbs are zero. Therefore, in the bit-reversed form, they have their msbs 1 and lsbs 0. For example, the bit-reversal of 0001 is 1000. Therefore, the bit-reversal of the even numbers in the second half of the list need not be found. (2) Since the only difference between an even number and the next odd number is in the lsb, lsb 0 for even number and 1 for odd number, once the bit-reversal of an even number is found the bit-reversal of the next odd number is deduced by adding y to the bit-reversal of the even number (setting the msb of the bit-reversal of the even number). For example, the bit-reversal of 4 and 5 are, respectively, 2 and 10. Therefore, the bit-reversals of (1,3,5,7) need not be found directly. With these two observations, the problem of finding the bit-reversal of the set of 16 numbers reduces to finding the bit-reversal of the four even numbers (0,2,4,6) in the first half and those of the four odd numbers (9,11,13,15) in the second half of the list. (3) Notice that the bit-patterns of these numbers in the two halves of the Table C.l are the same except for the lsbs and msbs shown in boldface. The bit-reversals of the odd numbers can be found from those of the even numbers by simply adding ^ + 1 to the bit-reversals of the even numbers. For example, the bit-reversal of 11 is found by adding 9 to the bit-reversal of 2, which is 4, to get 13. Therefore, the list of numbers for which the bit-reversals have to be found directly is further reduced to the first four even numbers (0,2,4,6).

While we can keep on reducing the list by using the similarity in bit patterns, we do not go any further for the following two reasons: (i) program code becomes longer and (ii) coupled with the fact that we need only ^ bit-reversals for an AT-point DFT with vector length two, the problem reduces to the finding of only y bit-reversals directly. The execution time for this operation is a negligible fraction of the execution time of the DFT algorithms.

To find the bit-reversals of the first four even numbers, we use the bit-reversal of the previous even number to find the bit-reversal of the current

346 The Bit-Reversal Algorithm

even number. We can obtain the next even number from an even number by just adding the binary number 0010. This addition results in the change of bits from 1 to 0, starting from the second bit from the right, until the first zero bit is found, which is changed from 0 to 1 leaving the other bits unchanged. Therefore, given the bit-reversal of a number, we have to start from the second bit from the left and change all the 1 bits to zeros and the first zero bit to 1, and leave the rest of the bits unchanged. For example, the bit-reversal of the number 0100 is 0010. The bit-reversal of 0110 is 0110. In the implementation of the algorithm, we use decimal numbers and a 1 or 0 is found by comparing the number with appropriate weights. For the example of finding the bit-reversal of 6, we compare the number 2 (bit-reversal of 4) with 4 and find that the second bit from the left is zero (number is less than the weight). We change that bit to 1 (by adding the weight 4) to get the bit-reversal as 6. The flow chart of the algorithm is shown in Fig. 6.10(d).

Reference

(1) Sundararajan, D., Ahmad, M. O. and Swamy, M. N. S. (1994) "A Fast Bit-Reversal Algorithm", IEEE Trans. Cir. and Sys. II, vol. CAS-41, No.10, pp. 701-703.

Appendix D

Prime-Factor DFT Algorithm

For most applications, we will be using the PM algorithms with a vector length of 2. When N is not an integral power of 2, we use vectors of length such as 6, 10, 14, etc. While the two-point DFT is the fundamental operation in the algorithm, however, we need to compute DFTs of lengths such as 6, 10, and 14, etc., in the vector formation stage. These DFTs are computed efficiently by prime-factor algorithm. The SFG for computing the DFT of six data points is shown in Fig. D.l. This algorithm requires 8 real multiplications and 36 real additions for the computation of a 6-point complex DFT. The trace of the algorithm to compute the DFT of

Fig. D.l The prime-factor DFT algorithm, with N = 6, and its trace.

347

348 Prime-Factor DFT Algorithm

{1,3,5,2,4,6} is also given in Fig. D.l. For computing a 3-point DFT, use the upper half of the SFG without the last stage. The input values x(2) and a;(4) become, respectively, ar(l) and x(2).

Reference

(1) Burrus, C. S. and Parks, T. W. (1985) DFT/FFT and Convolution Algorithms, John Wiley, New York.

Appendix E

Testing of Programs

In this Appendix, we present a variety of test data sets and ways for testing DFT programs. Once a program is written, it must be tested not only with a variety of data but also for different data sizes. For example, a program, written for data lengths of an integral power of 2, must be tested at the least for N = 16, 32, 64, 128, 256, and 512.

Comparing with the Trace of the Algorithm

Using tested programs, make files of DFTs corresponding to random input data of lengths N = 16, 32, 64, 128, 256, and 512. For small DFT lengths, such as 16 and 32, have the trace of the algorithm also.

Once a new program is written, compile and eliminate any syntax errors so that the program is running and gives some output. Compare the output of the program for the random test data with the correct output, starting with TV = 16. If the output is correct, keep testing up to at the least N = 512. Now, we can be reasonably sure that the program is working. Further, we can test for more lengths and use a variety of input signals.

If the program does not work for N = 16, start comparing the trace of the program with that you have already. You should be able to deduct the possible errors in input, vector formation, and the special and general butterflies, and correct the program. Testing for N = 16 is easy as the data size is small and will eliminate most of the errors, if not all. Now test the output for other lengths. If necessary, comparison of traces for N = 32 should eliminate all the errors. Again, we emphasize that testing for lengths longer than 32 is required.

349

350 Testing of Programs

Use of Closed-Form Solutions

We have derived the 1-D and 2-D DFT of several signals in closed-form and several problems have been suggested. The closed-form expressions of these DFTs can be used for testing.

Use of Properties

We found that certain properties produce certain form of output as presented in Chapters 4 and 10. Construct input data sets with those properties and check that the output exhibits the anticipated property. For example, if x(n) is real-valued data, the DFT is hermitian-symmetric. In short, every property can be tested.

Use of Special Inputs

Certain inputs, which produce known outputs, can be used to test programs. For example,

(1) Combination of scaled and delayed impulses (2) Constant input, c (3) Alternating constant input, (—l)"c (4) All zeros input (5) x(n) = e(J '£"2), with N even. The magnitude of the DFT, \X(k)\,

of this signal is equal to y/N for all k. (6) Combination of complex exponentials and sinusoids. In particular,

sinusoids with frequencies that are relatively prime to each other and relatively prime to the DFT length. For example, with N = 16, test with sinusoids with frequency indices 3 and 5.

Use of Certain Operations

(1) Find the DFT of data and compute the IDFT of the transform to get back the data

(2) Carry out the convolution operation using the DFT and check the results with that obtained by the direct method. In a similar way, the correlation operation can also be used.

Number of Operations, Precision, and Execution Time 351

N u m b e r of Opera t ions , Precision, and Execut ion T i m e

The DFT values must be checked for expected precision. Check always that the number of operations required for a given data size is correct with the theoretical values. Check that the execution time is reasonable for the specific computer.

Appendix F

Useful Mathematical Formulas

e±j0 =cos9±jsin9

cos 9 =

. n e»'9-e--»'« sin 9 = —

sin2 9 + cos2 9 = 1

cos 19 = cos2 9 - sin2 9 = 2cos20 - 1 = 1 - 2 sin2 0

sin 29 = 2 sin 0 cos 0

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

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

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

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

sin(-0) = sin(27r - 9) = - sin 6

353

354 Useful Mathematical Formulas

cos(—#) — cos(2?r — 9) = cos#

sin(7r ±9) = =p sin 9

cos(7r ± 9) = — cos#

7T

cos( - ±9) = Tsin0

7T

sin( — ± 0 ) = cos#

c o s ( ^ ± 0 ) = ±s in0

sin(— ±9) = — cos#

sin(A ± B) = sin A cos JB ± cos A sin JB

cos(A ± JB) = cos A cos B =p sin 4 sin B

a mod 6 = o — b\a/b\, & 7 0

sin2 a; = - (1 — cos 2x) £1

sin3 x = - (3 sin a; — sin 3x)

sin4 x = - (cos 4a; - 4 cos 2x + 3)

sin5 x = — (sin 5a; - 5 sin 3a; + 10 sin x) 16

cos2 x — - ( 1 + cos 2a;)

cos3 x — - (3 cos x + cos 3a;)

Useful Mathematical Formulas 355

cos4 x = - (cos 4a; 4- 4 cos 2a; + 3) 8

cos5 x = — (cos 5a; + 5 cos 3a; + 10 cos x) 16

sin 3a; = 3 sin a; — 4 sin3 x

sin 4a; = cos x (4 sin x — 8 sin3 a;)

sin 5x = 5 sin x — 20 sin3 x + 16 sin5 x

cos 3a; = 4 cos3 x — 3 cos x

cos 4a; = 8 cos4 x - 8 cos2 a; + 1

cos 5x = 5 cos a; — 20 cos3 a; + 16 cos5 x

sin x ± sin y = 2 sin —-— cos —-—

x + y x — y cos x + cos y = 2 cos —-— cos —-—

A Ad

. x+y . x-y cos x — cos y = — 2 sin —-— sin —-—

„ i n - i T _ x x3 (l)(3)a;5 (l)(3)(5)a;7

(1) + (2)(3) + (2)(4)(5) + (2)(4)(6)(7) + ' ' ' ' * <

- l ^ • - l

cos a; = — — sin x, x < 1

e - l + (ja;)+ 2! + 3! + 4! + + (r)! +

COfl(af) = l _ * + * _ . . . + ( _ l ) r i ! _ 2! 4! v ' (2r)l

356 Useful Mathematical Formulas

X3 X5 , _ X2r+1

sin(x) = x- — + — + ( - l ) r

3! 5! v ' (2r + l)!

JV-1

i = 0

JV-1 J , _ 1 ci JV\

JV-1

»=0 ^ '

/ udv = uv — I vdu

l'hopital's rule:

If lim f(x) = 0 and lim g(x) = 0, then lim ^ = lim d/^\/ff

Answers to Selected Exercises

2.1.2. A = 7.1, w = | radians/sample, / = ^ = n cycles/sample, period = 12 samples, 0 = | radians.

2.2.3. 3cos( fn+ f ) .

2.3.6. | c o s ( f n ) - 3 ^ s i n ( f n ) .

2.4.5. ^ c o s ( f n + ^ ) .

2.5.3. 5.6283 cos(fn-0.2818)

2.6.4. The function is periodic with a period of 43.

2.7.2. s in(i |^n - f ) , - s i n ( ^ n + f ) , s i n ( ^ n - f )

2.8.1. The fundamental cyclic frequency is ^ . The first sinusoid is the 39th harmonic and the second is the 35th harmonic.

2.9.2. The amplitude of the spectral value at w = f is 2 and the phase is — j . The amplitude of the spectral value at u = — f is 2 and the phase is

4-

The complex spectral coefficient at u = | is \/2 — jy/2 and at u — — | is y/2 + jy/2.

3.4.2. dc=0.25, 3cos(^fn+ £), -0.25cos(7rn).

3.8.2. X(k) = Wf2k, k = 0 , 1 , . . . , 7.

3.11. X{2) = 16(2*1 - i f ) .

3.13. X(fc) = 1 + e - J ^ l l ^ f r * ) (1 ~ ^ ( c o s ( ^ f c ) + j sinffifc)))-

357

358 Answers to Selected Exercises

3.15.5. x(n) = £co6(f jp5n-a?0.

4.2.1. X(k) = {6 + J2, - 1 -j3,-2-jl2,l-j3}. X(23) = X(3) = 1 - j 3 , X( -47) = X( l ) = - 1 - j 3

4.3.2. {2 + j '3, - 1 - j l , - 3 + j l , - 2 + j2} .

4.6.3. Odd half-wave. X(k) = {0, - 2 - j4, 0, - 2 + j4} .

4.7.2. Odd half-wave and odd. X(k) = {0, - 8 , 0 , 8 } .

4.8.1. Odd half-wave and odd. X(k) = {0, - 4 - j2 ,0 ,4 + j2} .

4.12. {13,3,9,23}.

4.15.1. yxh(n) = {10,11,20,31}. yhx = {10,31,20,11}.

4.23.1. The sum of \x{n)\2 is 45. The sum of \X(k)\2/4 is also 45.

5.1.2. The input vectors are {(4 + j 3 , - j l ) , (2 - j 5 , jl)}. The DFT vectors a r e { ( 6 - j 2 , 2 + j 8 ) , ( l - j l , - l - j l ) } . The DFT X(fc) is { 6 - j 2 , l - j l , 2 + j 8 , - l - j l } .

5.5. The input vectors are {(1 + j5 , - 3 - jl), ( - 5 + j2,1)}. The output vectors are {(—4 + jl, 6 + j3) , (—3 — j2, —3)}. When real and imaginary parts are swapped, we get x(n) as {7 — j4, —2 — j'3,3 + j6, — j3}/4.

5.9. The swapped input vectors are {(j'4, - 6 - j2), (2 + jA,-4 + j2), ( - 5 + j 4 , - l ) , ( j 4 , - 8 ) } .

The vectors after Stage 1 are {(2 + j8 , - 2 ) , ( -4 + j2, - 8 - j6), ( - 5 + j 8 , - 5 ) , ( - l + j 8 , - l - j 8 ) } .

The output vectors are { ( - 3 + j l 6 , 7 ) , (0.9497 + j'8.364, -8.9497-j '4.364), ( -2 + j5 , - 2 - j5), (-12.9497 + jO.364, -3.0503 - J12.364)}. To get the IDFT, the real and imaginary parts must be swapped and divided by 8.

5.10. The input vectors are {(3 - j6 , - 1 - j2), (3,1 - j6), (2 - j2,0) , (4 -A-J6)}. The vectors after Stage 1 are {(5 - j 8 ,1 - ji), (7 - j2, - 1 + j2), ( - 1 -j2, - 1 - j2), (-7.7782 - jO.7071,0.7071 - J9.1924)}.

The output vectors are {(12 - j'10, - 2 - jQ), (3 - j 3 , - 1 - ;5) , (-8.7782 -J2.7071,6.7782 - jl.2929), (-10.1924 - J2.7071,8.1924 - jl.2929)}. The output vectors must be swapped to get the DFT values in natural order.


10.7(d). X(14,0) = - 6 4 0 ( \ / 3 - j l )

10.7(f). X(8,6) = 512(V5 - j l ) , X(24,26) = 512(V5 + j l )

10.8(f). x{n1,n2) = ^ cos(ff(2m +3n 2 ) + f )

10.8(1). x{nun2) = ^ e i (? f ( 2 " i+ 2 "=) - f )

10.11. JST(7,21) = X(3 , l ) = - 5 + j l , X ( - 4 , - 3 ) = X(0,1) = 4 + j6 , X ( l , - 3 ) = X ( l , l ) = - 5 + j 3

10.16.

-> k2

24 4 + j6 4 4 - j 6 " 3 + j l - 5 + j 3 - 1 + j l - 5 - j l

1 ^ 6 2 - j 2 2 2 + j2 . 3 - j l - 5 + j l - 1 - j l - 5 - J 3 .

10.18. J V - 1 N-l J V - 1 JV-1

4* E E l*(ni,n2)|2= E E |*(*i,fe>)|2 =896

n i = 0 n 2 = 0 fei =0*2=0

11.1(c). Minimum AT = 14. X{7) = 28

N = 4. X(l) = 4 and A"(3) = 4

JV = 8. Jf (1) = 8 and X(7) = 8

iV = 16. X{7) = 16 and X(9) = 16

JV = 32. X(7) = 32 and X(25) = 32

11.2(b). Xrect{k) = { I , l + j 0 . 6 5 , l + j 2 . 6 1 , l - j 2 . 1 8 , l , l + j 2 . 1 8 , l - j 2 . 6 1 , l - j O . 6 5 }

XtriW = {0 , - j0 .14 , j l .58 , - j l .55 ,0 , j l .55 , - j l .58 , j0 .14}

Xhan{k) = {0 , - j0 .33 , j l .69 , - j l .74 ,0 , j l .74 , - j l .69 , j0 .33}

Xham(.k) = {0.08,0.08 - jO.25,0.08 + jl.76,0.08 - jl.78,0.08,0.08 + j l .78 , 0.08 - j 1.76,0.08+ J0.25}

11.3. Zero-padding is not required since the frequency increment obtainable with the given record length T = 0.1 seconds, —^ = 10 Hz, is lower than the required frequency increment, 20 Hz. Since sampling frequency


must be more than twice the highest frequency, the sampling period Ts

must be less than 2 xg 0 0 0 = 0.0001 seconds. Therefore, the number of samples, N, in the time-domain data must be more than £- = 0 °p01 = 1000. Note that, in practice, typically two times the minimum N is used. In this case, 2000 samples may be used. In addition, N is usually fixed to the next largest power of two, so that a power of two DFT algorithm can be used. The suggested value for N is, therefore, 2048. Note that, with increased JV from the minimum value, we can decrease the frequency increment, or increase the sampling frequency, or both as required.

12.4.

, , A A , . 2?r 1 . 2TT 1 . „2TT x{t) = - - - ( s i n — i + - s i n 2 — i + - s i n 3 — i + •••)

With A = 1 and T = 1, we get

x(t) = (sin(27ri) + - sin2(27rf) + - sin3(27rt) + • • •) 2 7r 2 3

A comparison of the FS and DFT coefficients is shown in Table 18.1.

Table 18.1 Comparison of the exact values of Xc(k),Xa(k), k = 0 , 1 , . . . ,8 (second row) with those obtained from the DFT coefficients with N = 4 (third row), N = 8 (fourth row), and N — 16 (fifth row).

0 0.5

0 0.5

0 0.5

0 0.5

0

1 0

-0.3183 0

-0.25 0

-0.3018 0

-0.3142

2 0

-0.1592 0 0 0

-0.125 0

-0.1509

3 0

-0.1061

0 -0.0518

0 -0.0935

4 0

-0.0796

0 0 0

-0.0625

5 0

-0.0637

0 -0.0418

6 0

-0.0531

0 -0.0259

7 0

-0.0455

0 -0.0124

8 0

-0.0398

0 0

12.11.

2Aw 2A,. ,2ir . ,2-K . 1 . .2TT , , n 2 7 r . x(t) = — + — (sm(—w) cos(—t) + - sin(—2tu) cos(2—t)

1 . , 2 7 T , , 2 - K . +-sin(—3w)cos(3—f) + ---)


With A = 1, T = 1 and w = | , we get

1 2 7T 1 7T 1 37T #(*) = T + —(sin — cos(27rt) + - sin — cos 2(27r£) + - sin —- cos 3(27rf) H )

4 7r 4 2 2 3 4


Table 18.2 Comparison of the exact values of Xc(k),Xa(k), k = 0 , 1 , . . . ,8 (second row) with those obtained from the DFT coefficients with N = 4 (third row), N = 8 (fourth row), and N = 16 (fifth row).

0 0.25

0 0.25

0 0.25

0 0.25

0

1 0.4502

0 0.5 0

0.4268 0

0.4444 0

2 0.3183

0 0.25

0 0.25

0 0.3018

0

3 0.1501

0

0.0732 0

0.1323 0

4 0 0

0 0 0 0

5 -0.09

0

-0.0591 0

6 -0.1061

0

-0.0518 0

7 -0.0643

0

-0.0176 0

8 0 0

0 0

12.18.

Xe,(0,0) = 2.5

Xc(ki,0) = •£-, * i = ± l , ± 2 , . . . Z"KK\

-Yc.(0,*a) = 4-> fe = ±l,±2,. . . 7TK2


13.5.

sin(o; - 10) sin(w + 10) (u) - 10) (w + 10)

Table 18.4 gives a comparison of the exact FT samples and those computed by DFT. 13.8.

wi \ 2 ™=(^y(^)


Table 18.3 Thesecond row ofthe first half of the table shows the exact XC3(ki,0), k\ — 0,1, . . . ,8 . The third, fourth, and fifth rows show those computed by the DFT with 4 x 4, 8 x 8, and 16 x 16 samples, respectively. The second half of the table shows Xc3(0,k2), * 2 = 0 , l , . . . , 8 .

0,0 1,0 2,0 2.5 J0.4775 jO.2387 2.5 j'0.375 0 2.5 jO.4527 jO.1875 2.5 jO.4713 jO.2263

0,0 0,1 0,2 2.5 j'0.3183 jO.1592 2.5 J0.25 0 2.5 jO.3018 jO.125 2.5 jO.3142 jO.1509

3,0 jO.1592

jO.0777 jO.1403

0,3 jO.1061

jO.0518 jO.0935

4,0 5,0 jO.1194 j'0.0955

0 jO.0937 j'0.0626

0,4 0,5 j'0.0796 j'0.0637

0 j"0.0625 jO.0418

6,0 jO.0796

J0.0388

0,6 jO.0531

jO.0259

7,0 8,0 j/0.0682 jO.0597

jO.0186 0

0,7 0,8 jO.0455 j'0.0398

j'0.0124 0

Table 18.4 The first row shows the first 8 samples of Xft(u) computed by DFT with T = 16 and N = 1024. The second row shows the corresponding exact values.

-0.1086 -0.0979 -0.0679 -0.0236 0.0271 0.0749 0.1110 0.1277 -0.1088 -0.0982 -0.0680 -0.0237 0.0270 0.0750 0.1111 0.1278

Table 18.5 gives a comparison of the exact FT samples and those computed

by DFT.

Table 18.5 The first half of the table shows the first 4 x 4 samples of Xft(wi,U2) computed by DFT with Ti,T2 = 4 and NltN2 = 64. The second half shows the corresponding exact values.

1 0.8112 0.4066 0.0907

0.8112 0.6581 0.3298 0.0736

0.4066 0.3298 0.1653 0.0369

0.0907 0.0736 0.0369 0.0082

1 0.8106 0.4053 0.0901 0.8106 0.6570 0.3285 0.0730 0.4053 0.3285 0.1643 0.0365 0.0901 0.0730 0.0365 0.0081


14.3.

14.7.

y(ni ,n2) =

2/(ni,n2) =

4 8

-1 -2

8 18 1

-6

8 26 12 -7

6 26 20 -1

13 25 15 1

21 29 3

-7

15 26 -1 -12

5 12 1

-6

15.2.

Xct(ki,k2) —

11.25 1.33

-2.25 -1.36

1.52

0.09 -2.06 -2.05

-1.75 -1.06 0.75 0.71

-0.90 -2.55 2.21

-3.09

16.2. The values after the vector formation stage are

(30,32), (10,8), (2,0), ( -2 ,0) , ( -6 , - 8 ) , (-18,0), (6,16), (-6,0)

The values after Stage 1 are

(24,36), (-8,28), (8, - 4 ) , ( -8 ,4) , (24,40), (8,8), (16, -16) , (0,0)

The values after Stage 2 are

(32,16), (-16,0), (32,40), (32,24), (40,8), (8,8), (24,56), (8,8)

The output values are

(16,48), (16,16), (64,0), (64,16), (48,32), (16,0), (32,16), (64,48)

The output values are to be divided by 16, the transform length.

Glossary

Algorithm A systematic procedure that provides a solution to a problem. Aliasing The impersonating of high frequency sinusoids as low frequency

sinusoids in the frequency-domain representation of a signal due to sampling the signal, in the time-domain, with a sampling interval that is not small enough. A similar phenomenon occurs in the time-domain.

Angular frequency 2-K times the cyclic frequency. For continuous-time signals, the unit is radians per second. For discrete signals, the unit is radians per sample.

Bit-reversed order The order obtained by reversing the bits of the binary representation of a set of naturally ordered decimal digits.

Complex amplitude A complex number containing the amplitude and phase of a complex sinusoid.

Complex conjugate The complex number obtained by negating the imaginary part of a complex number.

Complex number An ordered pair of real numbers consisting, respectively, of the a;-axis and y-axis components of a vector. The ar-axis component is referred as the real part of the complex number and the y-axis component is referred as the imaginary part.

Complex signal An ordered pair of real signals. Continuity A signal x(t) is said to be continuous at t = to, if it is denned

throughout \t — to\ < St for some positive St and limt_>.t0 x(t) = x(t0).

Continuous-time signal A signal that is defined at all instants of time.

365

366 Glossary

Cosine waveform The projection, on the x-axis from time t = — oo to t = oo, of a point moving with uniform angular velocity in the counterclockwise direction around a circle with center at the origin such that the longest projection on the positive side of the x-axis occurs at time t — 0. The radius of the circle and the angular velocity, respectively, are the amplitude and angular frequency of the waveform.

Cyclic frequency The reciprocal of the period of a periodic signal. For continuous-time signal the unit is Hz (cycles per second). For discrete signals the unit is cycles per sample. Note that the cyclic frequency of a discrete signal is equal to the reciprocal of its period only if the waveform makes one cycle during a period. Otherwise, the cyclic frequency is equal to the ratio of the number of cycles during a period and the period.

Decimation-in-frequency D F T algorithm A type of algorithm in which the given input data is combined to form several smaller groups of input data, the DFT of each of which represents mutually exclusive groups of the frequency coefficients of the given input data.

Decimation-in-time DFT algorithm A type of algorithm in which the frequency coefficients of smaller DFTs of mutually exclusive groups of input data are combined to form the frequency coefficients of the given input data.

Digital signal A signal that is defined only at discrete intervals of time and its amplitude defined only at certain discrete values.

Discrete complex sinusoid x(n) = e-7^*™: With k and N integers, a periodic set of complex numbers having real and imaginary parts, respectively, the sample values of cos(^fcn) and sin(^-/cn).

Discrete Fourier transform A transformation that yields an TV—point frequency-domain sequence corresponding to an N—point time-domain sequence. The JV-point frequency-domain sequence, called the frequency coefficients, represents the time-domain sequence in terms of a set of harmonically related discrete sinusoids.

Discrete signal A signal that is defined only at discrete intervals of time. Fourier series The representation of a continuous-time periodic signal in

terms of a set of infinite sinusoids having harmonically related frequencies.

Glossary 367

Fourier transform The representation of a continuous-time aperiodic signal in terms of a set of infinite sinusoids with continuum of frequencies.

Frequency coefficient The amplitude and phase (or the amplitudes of the cosine and sine components) of a sinusoid at a given frequency.

Frequency-domain signal representation The representation of a signal in terms of its frequency components.

Fundamental harmonic The sinusoidal component of an arbitrary periodic signal whose period is the same as that of the signal.

Harmonic Any of the sinusoidal components of a periodic signal. Harmonic (Fourier) analysis The decomposition of an arbitrary peri

odic waveform into its constituent harmonically related sinusoids. Harmonically related sinusoids A set of sinusoids whose frequencies

are an integral multiple of the frequency of the fundamental harmonic.

Harmonic (Fourier) synthesis The building up of a complex periodic waveform by summing a set of harmonically related sinusoids.

In-place computation The implementation of an algorithm in which the memory required for storing the data values is equal to the size of the input data.

Leakage effect The leakage of energy to adjacent frequencies in the spectrum of a signal due to the selection of an inappropriate record length in the time-domain. A similar phenomenon occurs in the time-domain.

Linear time-invariant system A system which obeys the superposition theorem and the response of which, for a specific input signal, does not change with time.

Periodic sequence A sequence x{n) is periodic with a period N if x(n + aN) = x(n), where a is any integer.

Phase shift The phase shift of a sinusoid is the amount of right or left shift a cosine waveform has to be shifted in order to get the sinusoid.

Picket-fence effect The inability of the DFT to present the continuous spectrum of an aperiodic signal due to the representation of frequency components only at discrete intervals.

P M D F T algorithms A family of fast DFT algorithms in which the computation of a form of the DFT equation, with the input and output quantities represented as vectors, is decomposed into several stages of computation.

368 Glossary

Radix-2 algorithm A type of algorithm in which a problem is split into two smaller independent problems of the same type, each of half the size, recursively from the input or output end. The solution of two smaller problems are combined to form the solution of a larger problem.

SFG The signal-flow graph (SFG) is a pictorial description of an algorithm using nodes and arrows and it is the most effective way of describing DFT algorithms.

Sine function The general form of the sine function, which is even, is s'"^ ' , having a peak value of unity at t — 0 and a waveform similar to an exponentially damped sinusoid at either side of the origin. It has zero crossings at points where t is equal to an integral multiple of 7r.

Sine waveform A cosine waveform with a phase shift of — f radians or -90 degrees.

Sinusoid A cosine or a sine waveform with arbitrary phase shift. A linear combination of the sine and cosine waveforms of the same frequency.

Time-domain signal representation The representation of a signal in terms of its amplitude at instants of time.

Twiddle factor A complex number that is a root of unity. Windows A set of special functions used to modify a truncated time-

domain signal in order to reduce the leakage of energy to adjacent frequencies in its frequency-domain representation. Windows are also used in the frequency-domain.

Uniform convergence For a given tolerance 6 > 0, for all t in the given interval, there exists an N such that \x(t) — xn(t)\ < 5, for n> N, where xn(t) is the series representation of x(t) with the first n terms.

Unit circle The circle with center at the origin and radius one. Zero padding When modeling a finite aperiodic time-domain signal, padding

with zeros at the end makes the signal more closer to the true signal and reduces the frequency increment making the spectrum denser. It can be used in the frequency-domain also. Zero padding is also used to simulate a linear convolution by a circular convolution.

Index

aliasing, 197, 226, 254 folding frequency, 227 folding of frequencies, 229 highest frequency component, 227 reducing the aliasing effect, 231 sampling frequency, 227 sampling rate, 227 sampling theorem, 227 time-domain, 228

antihermitian symmetry, 71

basis functions DCT, 304 DFT, 40 DWT, 314 NDHT, 320 SDHT, 326

bit-reversal, 343 algorithm, 345 bit-reversed order, 128, 344 number system, 343

butterfly computation, 123

complete representation, 58 complex amplitude, 22 complex exponential function, 22, 46,

199, 337 complex numbers, 333

argument, 335 exponential form, 335

imaginary part, 334 imaginary unit, 334 magnitude, 335 operations

addition, 334 conjugation, 335 multiplication, 335 subtraction, 334

polar form, 335 real part, 334 rectangular form, 334 roots, 336

complex plane, 333 computation of a single DFT

coefficient, 130 computational complexity, 341 continuous-time signal, 7 convergence, 251 convolution

circular, 289 frequency-domain, 84 time-domain, 83

linear, 288 simulation by DFT, 290

overlap-save method, 292 convolution, 2-D

circular, 296 spatial frequency-domain, 209 spatial time-domain, 208

linear, 295

369

370 Index

overlap-save method, 296 correlation, 298

circular auto, 85, 210 cross, 85, 209

cosine function, 8, 12, 47, 200, 337

DCT, see discrete cosine transform decimation-in-frequency, 114, 132 decimation-in-time, 108, 132 DFT, see discrete Fourier transform digital signal, 8 Dirichlet conditions, 249 discontinuity, 251 discrete cosine transform

basis functions, 305 orthogonality, 304

computation using the DFT, 306 computational complexity, 309 definition, 305

discrete cosine transform, 2-D, 309 discrete Fourier transform

basis functions, 40 center-zero format, 43 complex exponential, 46 computation with vectors, 101 dc, 45 definition, 37, 38 direct computation, 51 direct implementation, 51 frequency increment, 40 Hann function, 49 impulse, 44 kernel matrix, 43 properties

see properties of the DFT real data

algorithm for complex data, 163

single real data set, 169 two real data sets, 166

real sinusoid, 46 rectangular waveform, 49 vector format, 96, 99, 100

discrete Fourier transform, 2-D center-zero format, 196 complex exponential, 199 computation

algorithms, 212 real data, 217 row-column method, 202

definition, 195 impulse, 197 properties

see properties of the 2-D DFT real sinusoid, 199

discrete signal, 7 discrete Walsh transform, 313

basis functions, 314 definition, 313 orthogonality, 315 PM DWT algorithm, 316 sequency, 314

discrete Walsh transform, 2-D, 318 DWT, see discrete Walsh transform

Euler's identity, 337 even function, 15, 71 even half-wave symmetry, 71

finality of coefficients, 57 Fourier analysis, 32 Fourier Series, 1-D, 249

aliasing effect, 254 approximatation by DFT, 253 continuous-time frequency, 257, 260 convergence, 251 Dirichlet conditions, 249 exponential form, 250 Gibbs phenomenon, 251, 261 sample value at a discontinuity, 257 trigonometric form, 250 waveform reconstruction , 259

Fourier Series, 2-D, 262 Fourier synthesis, 32, 35 Fourier transform, 1-D, 273

approximation by DFT, 277 complex exponential, 276

Index 371

dc signal, 276 definition, 276 impulse, 276 limiting case of the FS, 273 pulse, 275, 277 real sinusoid, 276 relation between FS and FT, 275,

277 signal reconstruction, 281

Fourier transform, 2-D, 282 frequency composition

1-D real signals, 37 2-D real signals, 196

frequency-domain, 10

Gibbs phenomenon, 251, 261

Hamming window, 240 Hann window, 239 hermitian symmetry, 71

IDFT, see inverse discrete Fourier transform

impulse, 9 inverse discrete Fourier transform

center-zero format, 43 computation using DFT, 104, 111 definition, 37, 39 direct computation, 51 vector format, 104

inverse discrete Fourier transform, 2-D center-zero format, 196 definition, 195

inverse discrete Walsh transform definition, 316

inverse naturally ordered discrete Hadamard transform, 322

inverse sequency ordered discrete Hadamard transform, 326

l'hopital's rule, 263, 356 leakage effect, 231

frequency response

DFT, 235 rectangular and Hann

windows, 240 Multiplication of the signal with a

rectangular window, 233 reduction of leakage, 244 spectral resolution, 241 windows, see windows

least squares error, 55-57, 257

naturally ordered discrete Hadamard transform, 320 basis functions, 320 definition, 320 kernel generation, 321 PM NDHT algorithm, 322

naturally ordered discrete Hadamard transform, 2-D, 323

NDHT, see naturally ordered discrete Hadamard transform

odd function, 15, 71 odd half-wave symmetry, 71 orthogonality, 24, 26

complex exponential, 26 cosines over half-cycles, 304 trigonometric functions, 24 Walsh function, 315

overlap-save method, 292, 296

Parseval's theorem, 90, 212 periodicity, 62 phase shift, 11, 12 picket-fence effect, 244

denser spectrum, 245 PM DFT Algorithms, classification,

114 PM DIF DFT algorithms

2 x 1 PM DIF DFT algorithm, 134 butterfly, 134 computational stages, 135

2 x 2 PM DIF DFT algorithm, 154 butterfly, 157 computational stages, 157

372 Index

it x 1 PM DIF DFT algorithms, 132 butterfly, 133 computational stages, 134

PM DIF DFT Algorithms, fundamentals, 112 2 x 1 PM DIF DFT algorithm, 114 compression of data vectors, 113 shift of transform vectors, 112

PM DIF RIDFT algorithms 2 x 1 PM DIF RIDFT algorithms,

180 butterfly, 183 computational stages, 184 special butterflies, 184

2 x 2 PM DIF RIDFT algorithms, 190

butterfly, 190 computational stages, 191 special butterflies, 192

storage of data, 176 PM DIT DFT algorithms

2 x 1 PM DIT DFT algorithm, 125 butterfly, 125 computation of a single DFT

coefficient, 130 computational complexity, 135 computational stages, 126 flow chart description, 141 implementation issues, 148 reordering of the input data,

128 2 x 2 PM DIT DFT algorithm, 151

butterfly, 153 computational complexity, 158 computational stages, 154

6 x 1 PM DIT DFT algorithm, 138 butterfly, 139 computational complexity, 140 computational stages, 140

t i x l P M DIT DFT algorithms, 122

butterfly, 123 computational stages, 124

PM DIT DFT Algorithms,

fundamentals, 106 2 x 1 PM DIT DFT algorithm, 108 shift of data vectors, 106 zero padding of data vectors, 107

PM DIT RDFT algorithms 2 x 1 PM DIT RDFT algorithms,

176 butterfly, 177 computational stages, 178 special butterflies, 178

2 x 2 PM DIT RDFT algorithms, 187

butterfly, 187 computational stages, 187 special butterflies, 189

comparison with DFT algorithms, 193

storage of data, 175 prime-factor DFT algorithm, 139, 347 properties of the 2-D DFT

complex conjugates, 208 convolution, see convolution correlation, see correlation difference, 210 image rotation, 210 linearity, 205 Parseval's theorem, 212 periodicity, 205 reversal property, 207 separable signals, 211 spatial circular shift of a spectrum,

206 spatial circular shift of an image,

206 sum and difference of sequences,

210 symmetry, 207

properties of the DFT circular shift of a spectrum, 66 circular shift of a time sequence, 62 complex conjugates, 81 DFT of overlapping segments, 66 DFT twice in succession, 70 duality, 71

Index 373

linearity, 61 padding the data with zeros, 86

at the end, 86 in between the samples, 89

Parseval's theorem, 90 periodicity, 62 signal defined over a finite range,

80 sum and difference of sequences, 85 symmetry, complex signal, 78

even, 78 even half-wave, 80 odd, 80 odd half-wave, 80

symmetry, imaginary signal, 75 even, 76 even half-wave, 76 odd, 76 odd half-wave, 78

symmetry, real signal, 72 even, 74 even half-wave, 75 odd, 74 odd half-wave, 75

time-reversal, 69

rectangular window, 236 roots of unity, 40, 336 rotating vector, 339 row-column method, 202, 309

sampling frequency, 227 sampling rate, 227 sampling theorem, 227 SDHT, see sequency ordered discrete

Hadamard transform sequency, 314 sequency ordered discrete Hadamard

transform, 325 basis functions, 326 definition, 325 PM SDHT algorithm, 327

sequency ordered discrete Hadamard transform, 2-D, 328

signal continuous-time, 7 digital, 8 discrete, 7, 9

signal representation frequency-domain, 10, 11 time-domain, 7, 10, 11

signal-flow graph, 97 sine function, 240, 252 sine function, 8, 12, 47, 200, 337 sinusoid, 11

amplitude, 11-13 angular frequency, 12 complex, 21, 22 cosine, 8, 12 cyclic frequency, 12 even function, 15 harmonically related, 19 harmonics, 19 highest frequency, 18 odd function, 15 orthogonality, 24, 26 period, 12 periodicity, 17, 18 phase shift, 11-13 phase-shifted cosine, 12 phase-shifted sine, 12 polar form, 12 rectangular form, 14 representation with a rotating

vector, 337 representation with two rotating

vectors, 339 sine, 8, 12 sum of sinusoids, 16, 19, 20, 22

sinusoidal representation of signals advantages, 54

sinusoidal surface, 196 spectral density, 275 spectral resolution, 241 spectrum, 23

amplitude and phase, 23 real and imaginary parts, 23

symmetry

374 Index

antihermitian, 71 even, 71 even half-wave, 71 hermitian, 71 odd, 71 odd half-wave, 71

testing of programs, 349 comparing with the trace, 349 use of closed-form solutions, 350 use of properties, 350 use of special inputs, 350

time-domain, 7, 10 triangular window, 238 twiddle factor, 40

windows, 236 Hamming, 240 Hann, 239 rectangular, 236 triangular, 238

zero padding, 86, 108, 245, 290, 296

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