Fast Fourier Transforms [2 ed.] 0849371635, 9780849371639

Table of contents :
Cover
Half Title
Title Page
Copyright Page
Dedication
Table of Contents
1: Basic Aspects of Fourier Series
1.1 Definition of Fourier Series
1.2 Examples of Fourier Series
1.3 Fourier Series of Real Functions
1.4 Pointwise Convergence of Fourier Series
1.5 Further Aspects of Convergence of Fourier Series
1.6 Fourier Sine Series and Cosine Series
1.7 Convergence of Fourier Sine and Cosine Series
References
Exercises
2: The Discrete Fourier Transform (DFT)
2.1 Derivation of the DFT
2.2 Basic Properties of the DFT
2.3 Relation of the DFT to Fourier Coefficients
2.4 Relation of the DFT to Sampled Fourier Series
2.5 Discrete Sine and Cosine Transform
References
Exercises
3: The Fast Fourier Transform (FFT)
3.1 Decimation in Time, Radix 2, FFT
3.2 Bit Reversal
3.3 Rotations in FFTs
3.4 Computation of Sines and Tangents
3.5 Computing Two Real FFTs Simultaneously
3.6 Computing a Real FFT
3.7 Fast Sine and Cosine Transforms
3.8 Inversion of Discrete Sine and Cosine Transforms
3.9 Inversion of the FFT of a Real Sequence
References
Exercises
4: Some Applications of Fourier Series
4.1 Heat Equation
4.2 The Wave Equation
4.3 Schrodinger’s Equation for a Free Particle
4.4 Filters Used in Signal Processing
4.5 Designing Filters
4.6 Convolution and Point Spread Functions
4.7 Discrete Convolutions Using FFTs
4.8 Kernels for Some Common Filters
4.9 Convergence of Filtered Fourier Series
4.10 Further Analysis of Fourier Series Partial Sums
References
Exercises
5: Fourier Transforms
5.1 Introduction
5.2 Properties of Fourier Transforms
5.3 Inversion of Fourier Transforms
5.4 The Relation between Fourier Transforms and DFTs
5.5 Convolution — an Introduction
5.6 The Convolution Theorem
5.7 An Application of Convolution in Quantum Mechanics
5.8 Filtering, Frequency Detection, and Removal of Noise
5.9 Poisson Summation
5.10 Summation Kernels Arising from Poisson Summation
5.11 The Sampling Theorem
5.12 Aliasing
5.13 Comparison of Three Kernels
5.14 Sine and Cosine Transforms
References
Exercises
6: Fourier Optics
6.1 Introduction—Diffraction and Coherency of Light
6.2 Fresnel Diffraction
6.3 Fraunhofer Diffraction
6.4 Circular Apertures
6.5 Interference
6.6 Diffraction Gratings
6.7 Spectral Analysis with Diffraction Gratings
6.8 The Phase Transformation Induced by a Thin Lens
6.9 Imaging with a Single Lens
6.10 Imaging with Coherent Light
6.11 Fourier Transforming Property of a Lens
6.12 Imaging with Incoherent Light
References
Exercises
A: User’s Manual for Fourier Analysis Software
B: Some Computer Programs
C: The Schwarz Inequality
D: Solutions to Odd-Numbered Exercises
Bibliography
Index

Citation preview

Fast Fourier Transforms

Second E dition

Studies in Advanced Mathematics Series Editor STEVEN G. KRANTZ Washington University in St. Louis

Editorial Board R. Michael Beals

Gerald B. Eolland

Rutgers University

University of Washington

Dennis de Turck

William Helton

University of Pennsylvania

University of California at San Diego

Ronald DeVore

Norberto Salinas

University of South Carolina

University of Kansas

Lawrence C. Evans

Michael E. Taylor

University of California at Berkeley

University of North Carolina

Titles Included in the Series Steven R. Bell, The Cauchy Transform, Potential Theory, and Conformal Mapping John J. Benedetto, Harmonic Analysis and Applications John J. Benedetto and Michael W. Frazier, Wavelets: Mathematics and Applications Albert Boggess, CR Manifolds and the Tangential Cauchy-Riemann Complex Goong Chen and Jianxin Zhou), Vibration and Damping in Distributed Systems, Vol. 1: Analysis, Estimation, Attenuation, and Design. Vol. 2: WKB and Wave Methods, Visualization, and Experimentation Carl C. Cowen and Barbara D. MacCluer, Composition Operators on Spaces of Analytic Functions John P. D ’Angelo, Several Complex Variables and the Geometry of Real Hypersurfaces Lawrence C. Evans and Ronald F. Gariepy, Measure Theory and Fine Properties of Functions Gerald B. Folland, A Course in Abstract Harmonic Analysis José García-Cuerva, Eugenio Hernández, Fernando Soria, and José-Luis Torrea, Fourier Analysis and Partial Differential Equations Peter B. Gilkey, Invariance Theory, the Heat Equation, and the Atiyah-Singer Index Theorem, 2nd Edition Alfred Gray, Differential Geometry and Its Applications with Mathematica, 2nd Edition Eugenio Hernández and Guido Weiss, A First Course in Wavelets Steven G. Krantz, Partial Differential Equations and Complex Analysis Steven G. Krantz, Real Analysis and Foundations Clark Robinson, Dynamical Systems: Stability, Symbolic Dynamics, and Chaos John Ryan, Clifford Algebras in Analysis and Related Topics Xavier Saint Raymond, Elementary Introduction to the Theory of Pseudodifferential Operators Robert Strichartz, A Guide to Distribution Theory and Fourier Transforms André Unterberger and Harald Upmeier, Pseudodifferential Analysis on Symmetric Cones James S. Walker, Fast Fourier Transforms, 2nd Edition Gilbert G. Walter, Wavelets and Other Orthogonal Systems with Applications Kehe Zhu, An Introduction to Operator Algebras

Fast

Fourier

Transfoiuis Second Edition

James S. Walker University of Wisconsin Eau Claire

CRC Press Boca Raton London New York Washington, D.C.

Published in 1996 by CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 1996 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group No claim to original U.S. Government works 10 9 8 7 6 5 4 international Standard Book Number-10: 0-8493-7163-5 (Flardcover) International Standard Book Number-13: 978-0-8493-7163-9 (Hardcover) Library of Congress catalog number: 96-16852 This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Pubiication Data Catalog record is available from the Library of Congress

Visit the Tayior & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http ://www.crcpress.com

To my wife, and my mother, and the memory o f my father.

Acknowledgments

I would like to take this opportunity to thank Steve Krantz for first suggesting this project to me; it has turned out to be a most rewarding endeavor. I would also like to thank Bob Stern at CRC Press for encouraging me to write a second edition. My students have played a major role in helping me to write this book (especially in spurring me on to write better code for the software). If I were to thank all of them by name the list would be too long; let me just say a warm thank you to all of them for their continuing interest. Finally, I want to express my deep appreciation for my dear wife Ching. She has helped me with the design of FAS and shown me how to eliminate some bugs in the program. She also has been a source of inspiration to me. Without her love and support (including her delicious lunch boxes) I would not have been able to complete this work.

Vll

About the Author

Jam es S. W alker received his doctorate from the University of Illinois at Chicago in 1982. Since then he has been teaching in the Department of Mathematics at the University of Wisconsin - Eau Claire. In addition to his book Fast Fourier TransformSy he is also the author of Fourier Analysis (published by Oxford University Press). He has also published papers in the fields of Fourier analysis and complex variables. Dr. Walker is a member of the American Mathematical Society.

Contents

Preface

Xlll

1

Basic Aspects of Fourier Series 1.1 Definition of Fourier Series .......................................................... 1.2 Examples of Fourier S eries.............................................................. 1.3 Fourier Series of Real F unctions.................................................... 1.4 Pointwise Convergence of Fourier Series ................................... 1.5 Further Aspects of Convergence of Fourier S e r ie s ...................... 1.6 Fourier Sine Series and Cosine S e r i e s .......................................... 1.7 Convergence of Fourier Sine and Cosine S e r i e s .......................... R eferences.................................................................................................. Exercises.....................................................................................................

1 1 5 9 13 15 21 25 30 30

2

The Discrete Fourier Transform(DFT) 2.1 Derivation of the D F T ..................................................................... 2.2 Basic Properties of the D F T ........................................................... 2.3 Relation of the DFT to Fourier C oefficients................................. 2.4 Relation of the DFT to Sampled Fourier Series .......................... 2.5 Discrete Sine and Cosine T ra n sfo rm s .......................................... R eferences............................................................................................... . Exercises.....................................................................................................

35 35 37 40 43 45 48 49

3

The Fast Fourier Transform (FFT) 3.1 Decimation in Time, Radix 2, F F T ................................................. 3.2 Bit R e v e r s a l..................................................................................... 3.3 Rotations in F F T s ........................................................................... 3.4 Computation of Sines and T a n g e n ts .............................................. 3.5 Computing Two Real FFTs S im ultaneously................................. 3.6 Computing a Real F F T ..................................................................... 3.7 Fast Sine and Cosine Transforms.................................................... 3.8 Inversion of Discrete Sine and Cosine T ransform s.......................

53 53 58 63

66 69 71 72 76

IX

FAST FOURIER TRANSFORMS

3.9 Inversion of the FFT of a Real Sequence........................................... R eferences................................................................. Exercises.........................................................................................................

81

Some Applications of Fourier Series 4.1 Heat E q u a tio n .................................................................................. 4.2 The Wave Equation ............................................................................ 4.3 Schrodinger’s Equation for a Free P a rtic le ........................................ 4.4 Filters Used in Signal Processing .................................................... 4.5 Designing F ilters............................................................................... 4.6 Convolution and Point Spread Functions.......................................... 4.7 Discrete Convolutions Using F F T s ................................................. 4.8 Kernels for Some Common F ilte rs.................................................... 4.9 Convergence of Filtered Fourier S eries............................................. 4.10 Further Analysis of Fourier Series Partial S u m s ............................. R eferences..................................................................................................... Exercises........................................................................................................

85 85 91 96 104 110 116 123 126 131 138 141 141

83 83

Fourier Transforms 149 5.1 In tro d u c tio n ..................................................................................... 149 5.2 Properties of Fourier Transform s.................................................... 155 5.3 Inversion of Fourier T ran sfo rm s.......................................... 160 5.4 The Relation between Fourier Transforms and DFTs ................ 165 5.5 Convolution — an In tro d u c tio n .................................................... 170 5.6 The Convolution T h eo rem .............................................................. 176 5.7 An Application of Convolution in Quantum M e c h a n ic s.................. 184 5.8 Filtering, Frequency Detection, and Removalof N o ise.................. 190 5.9 Poisson S u m m a tio n ........................................... ............................ 202 5.10 Summation Kernels Arising from Poisson Sununation................. 205 5.11 The Sampling T h e o r e m .................................................................. 212 5.12 A lia s in g .................................................................................... 219 5.13 Comparison of Three K e r n e ls ........................................................ 226 5.14 Sine and Cosine T ransform s........................................................... 230 R eferences.................................................................................................... 234 Exercises........................................................................................................ 234 Fourier Optics 6.1 Introduction—Diffraction and Coherency of Light. . . . . . . . 6.2 Fresnel D iffraction.............................................. .. .......................... 6.3 Fraunhofer D iffraction..................................................................... 6.4 Circular A p ertu res........................................................................... 6.5 In te rfe re n c e ..................................................................................... 6.6 Diffraction G ra tin g s ........................................................................ 6.7 Spectral Analysis with Diffraction G ra tin g s .................................

247 247 256 262 267 273 275 281

CONTENTS

XI

6.8

The Phase Transformation Induced by a Thin L e n s .................... 6.9 Imaging with a Single L e n s ........................................................... 6.10 Imaging with Coherent L ight........................................................... 6.11 Fourier Transforming Property of a L e n s....................................... 6.12 Imaging with Incoherent L ig h t........................................................ R eferences.................................................................................................. Exercises ......................................................................................................

285 289 293 301 307 314 314

A

User’s Manual for Fourier AnalysisSoftware

331

B

Some Computer Programs

375

C The Schwarz Inequality

389

D Solutions to Odd-Numbered Exercises

391

Bibliography

433

Index

437

Preface

This book is an introduction to the Fast Fourier Transform (FFT) and some of its applications. Since its creation in the mid-1960s, the field of computerized Fourier analysis, based on the FFT, has been near the heart of a revolution in scientific understanding made possible through the aid of the digital computer. No single book can describe all the features of this new science. I have tried to show how the FFT arises from classical Fourier analysis and describe how the FFT is used to do Fourier analysis. This book is now in its second edition. As with the first edition, I have tried to write for as wide an audience as I could, given the mathematical requirements needed to appreciate the basic definitions of Fourier analysis. I hope the text will continue to be helpful to students of electrical engineering, optical engineering, physics, physical chemistry, and mathematics. As an additional aid to study, in this second edition I have included solutions for all of the odd-numbered exercises. The mathematical prerequisites for reading the book are a solid understanding of calculus and, for some of the applications, differential equations. To understand the proofs in Sections 4.9 and 5.9 to 5.12 a first course in advanced calculus would be ideal, although a determined reader could gain the necessary knowledge by pursuing a few references (for example, [Kr] or [Ru,2]). I have generally tried to keep the mathematical proofs to a bare minimum. Readers who wish to pursue the pure mathematical theory might begin with the following reference: James S. Walker, Fourier Analysis, Oxford University Press, Oxford, 1988. This text is referred to as [Wa] in this book. Since I wrote [Wa] for the purpose of explaining the mathematical theory of classical Fourier analysis, I saw no need to reproduce its pure mathematical arguments in this book. What makes this book unique among the vast literature on the FFT is that it comes with computer software {FAS) for doing Fourier analysis on any PC (using MS-DOS version 5.0 and up). Using FAS, readers will be able to generate imme­ diately on their computer screens images of Fourier series, sine and cosine series, Fourier transforms, convolutions, and all the other aspects of Fourier analysis de­ scribed in the text. This second edition includes a new version of this software. This new version provides extensive online help, mouse support, and vastly improved

Xlll

XIV

FAST FOURIER TRANSFORMS

function compiling and screen printing. The online help for creating functions should be particularly helpful for students, since it contains every formula used in the text (except the exercises), and these formulas can be entered into FAS by simply clicking on them with the mouse. Here is a summary of the main topics covered. In Chapter 1, I discuss the principal features of Fourier series. The emphasis here is on using FAS to gain a firm grasp of the main aspects of Fourier series, as well as Fourier sine and cosine series. In Chapter 2, the digital (discrete) version of Fourier analysis is introduced, showing how the Fourier series can be put into a discrete form intended for computer computation. An efficient means for carrying out this computer computation is called a Fast Fourier Transform or FFT. Chapter 3 is an introduction to one type of FFT. Although there are a great many FFT algorithms, I discuss just one. Any readers who are familiar with Bracewell’s book on the Hartley transform, [Br], will recognize the debt I owe to him and his former colleague at Stanford, the late Oscar Buneman. The work done by Buneman significantly improved the original FFT algorithms. Besides the basic FFT algorithm, I also describe the standard method for computing a real FFT and for computing discrete sine and discrete cosine transforms efficiently. Since the purpose of the FFT is to do Fourier analysis. Chapter 4 consists of applications of Fourier series. The classic Fourier series solutions to the heat and wave equations are described with the principal emphasis being on using FAS to study the time evolution of these solutions. Using FAS we can create computer animations of vibrating strings and of the changing temperature of a thin wire. These animations enliven the purely mathematical solutions of the wave and heat equations. Another area where the computer allows one to do much more than the typical textbook discussion is the famous particle in a box problem, in one dimension, from quantum mechanics. Section 4.3 shows how FAS allows one to study the time evolution of a potential-free quantum mechanical particle. This problem, which has important applications to electron diffraction, is too often treated in a sterile fashion with too much emphasis on stationary states. Many students are left with the impression that the stationary states are the only states. The rest of Chapter 4 describes aspects of Fourier series that are important in signal processing, in particular, emphasizing the concept of filtering of Fourier series. FAS makes it possible for students to see with their own eyes why banning and Hamming filters, as well as other types of filters, are used. The subtle notion of point spread functions (kernels) should be easier to learn with the aid of FAS, too. In Chapter 5 , 1 take up the other main area of Fourier analysis, Fourier trans­ forms. Here, FAS is used to aid in understanding the fundamentals of Fourier transforms, Fourier inversion, and convolution. In particular, the difficult concept of convolution integrals should be more accessible with the help of FAS. A stu­ dent will be able to see very quickly what the convolution of two functions looks like. The examples I have treated (heat equation, Laplace’s equation, potential-free Schrödinger equation) will show the practical importance of convolutions. I have

PREFACE

XV

made a point of showing that FAS can be used to implement the classic convolution solutions to these problems in a practical way. The second half of Chapter 5 (Sec­ tions 5.9 to 5.13) covers some very important, but sometimes difficult, material on Poisson summation and sampling theoiy, stressing the close connection between these two topics. It would be hard to overestimate the importance of sampling theory in modern electrical engineering and communication systems design. FAS should help make some of the principles of sampling theory easier to understand. The book concludes with a chapter on Fourier optics. I think that it is important to describe in as much detail as possible one major area of application of comput­ erized Fourier analysis. In Chapter 6, the topics of Fresnel diffraction, diffraction from circular apertures, interference, diffraction gratings, Fourier transforming properties of a lens, and imaging with a single lens, are all discussed. Fourier op­ tics is an essential element of modem optics. It plays a vital role in understanding diffraction and imaging. And these two areas, through their roles in crystallogra­ phy, chemical analysis, and optical and electron microscopy, underlie a great deal of what we have learned about the physical world during the last century. James S. Walker Eau Claire, Wisconsin February 1996

Basic Aspects o f Fourier Series

This chapter is a summary of the basic theory of Fourier series. Some of the deeper theorems are only quoted; their proofs can be found in [Wa], listed in this book’s bibliography. Besides their importance in applications, Fourier series provide a foundation for understanding the FFT.

1.1

Definition of Fourier Series

To understand the definition of Fourier series we will begin with the essential idea: representing a wave form in terms of frequency as opposed to time. We will denote time by x rather than t. Suppose our wave form is described by 4 cos27Tvjc, which has frequency v. Using Euler's identity z= COS0 + i sin 0 ( 1. 1) we can write 4 cos litv x in complex exponential form 4cos2;rv;c = where the complex exponentials have amplitudes of 2 and frequencies of v and —v. See Figure 1.1. (To see how Figure 1.1(b) was produced, consult Exercise 1.3.) Or, suppose our wave form is described by 6 sin 2n vjc, which also has frequency V. Using Euler’s identity (1.1) again, we obtain 6sin27rv;c = where the complex exponentials have complex amplitudes of 3/ and —3i and frequencies of and v. See Figure 1.2. (To see how Figure 1.2(b) was produced, consult Exercise 1.4.)

1. BASIC ASPECTS OF FOURIER SERIES

H— ^--- h-

y interual: [ -8,

81

V increment =

1.6

H----^--- h

(b ) X interual: t -45, 451 V interual: [ -5, Z51

X increment = y increment =

FIGURE 1.1 Frequency representation of 4 cos(2ttvx), v = 9. (a) Graph in x domain (time or space), (b) Graph in frequency domain.

H--- 1--- h-

y internal: [ -10,

103

y increment =

(b) X interual: [ -45, y interual: [ -30,

H--- 1----h

451 301

X increment = y increment =

FIGURE 1.2 Frequency representation of 6 sin(27r vjc), v = 9. (a) Graph in x domain (time or space), (b) Graph in frequency domain.

These two examples show how the waves cos l i t vx and sin ln v x can be ex­ pressed in frequency terms and distinguished from each other using complex ex­ ponentials. Also, Figures 1.1(b) and 1.2(b) show how, in a certain sense, the frequency representation of these waves is simpler. The basic idea in Fourier series is to express a periodic wave as a sum of complex exponentials all of which have the same period. This is made feasible by the following property of complex exponentials having the same period.

LI, DEFINITION OF FOURIER SERIES

THEOREM L I : ORTHOGONALITY OF COMPLEX EXPONENTIALS For all integers m and n, the complex exponentials j period P satisfy the following orthogonality relation J_

0

ilnmx/ P - iln n x / P

/*

P Jo

PROOF have

ll

form ^ n form = n.

We will prove the theorem for the case of P = 2it. For this case, we

1 n27t — / Jo

1 ^27T dx = — / , i(m—n)x dx 2lt Jo 1

= — / 27t J o

(cos (m —n)x dx

i H-----/ sin (m — n )x d x . 27t Jo

(1.2)

If m ^ n, then by calculus we get r»27T

-2n rJo

COS (m

—n)x dx =

sin (m — n)x 2n 2n{m —n)

=

0

and similarly i — / 2n Jo

sin (m — n)x dx - 0.

That takes care ofthe case when m ^ n. Ifm = n, then sin (m—n)x = sinO = 0 and cos (m —n)x — cosO = 1. Hence (1.2) becomes I p^ . 1 p^ --- / ¿"'Xe-'nx dx = --- / IJjc = 1. 2 tc j q 2jt j q

This proves the theorem.

I

Now, suppose g is a periodic function, period P, and g is expanded in a series

1. BASIC ASPECTS OF FOURIER SERIES

of complex exponentials having the same period six) -

^

c„e'

liznx/P

(1.3)

We will show that a reasonable formula for the general coefficient can be obtained using the orthogonality theorem just proved. If we multiply both sides of (1.3) by { I j (being careful to change the name of the index in the series) and integrate term by term, then we get

jlnmx/ P^—ilTtnx/ P

P Jo

m^oo

P Jo

— Pji because of Theorem 1.1. Based on this result, we make the following definition. DEFINITIO N 1,1: I f the function g has period P, then the Fourier coefficients {Cn} fo r g are defined by g{x)e —iln n xIP dx

fo r all integers n. Using these coefficients [cn], the Fourier series fo r g is defined by the right side o f the following correspondence i2nnx/P n=~oo

REMARK 1.1: (a) We have used the correspondence symbol ~ since we have not yet discussed the conditions of validity of Equation (1.3); we will do that in Section 1.4. (b) Note that when g is a real valued function, then c-n — c* where c* is the complex conjugate of Cn. (c) The Fourier coefficients of g can also be found from the formula

T J-ip

1,2. EXAMPLES OF FOURIER SERIES

1.2 Examples of Fourier Series In this section we will cover some basic examples of Fourier series expansions. Example 1.1: Expand the function g(jc) = jc in a Fourier series, period I n , using the interval [-7T, 7T].

SOLUTION

For n = 0, we have

1 Co

nit xd x = 0.

2n

For n 7^ 0, we split the integral for integrate by parts, obtaining

27TJr— j;

Cn — j _

^ Ir

2n

into real and imaginary terms and then

dx

^

^r

X cos nx dx — —— /

2n J^y,

X sin nx dx =

Hence, f ' n

'

(1.4)

where the prime on the sum indicates that the w = 0 term is omitted. If we group the terms for n and —n, then we can rewrite Formula (1.4) in the real form ~ 2 ( - l) " + i s ~ 2^ ------ sinnx. n—\1

^

In Figure 1.3, we show the graphs of the partial sums C r \ 2(—1)”“^^ . S m \x ) = > ------------ smnjc

n=\1

^

for M = 1,2, 4, and 8. The number M denotes the number o f harmonics in the partial sum.

L BASIC ASPECTS OF FOURIER SERIES

(b) partial suiUj Z haraonics X interual: [ -S.2B, 6.Z81 X increment = V interual: [ -4, 41 Y increment = .8

(c) partial sum, 4 harmonics X interual: [ -6.Z8, 6.Z81 X increment = Y interual: [ -4, 41 Y increment = .8

(d) partial sum, 8 harmonics X interual: t -6.Z8, 6.Z81 X increment = Y interual: [ -4, 41 Y increment = .8

FIGURE 1.3 Graphs of some partial sums of the Fourier series for g(x) = x, period 2tc, using the interval [—tt, tt] for computing Fourier series coefficients.

Notice that in Figure 1.3(d) the graph of 5g, on the interval [—n, tt] that we used to calculate the Fourier series, is a wavy (or wiggly) approximation to the original function g(x) = x. Outside the original interval, the graph of is approximating the periodic extension of g(x) = x. See Figure 1.4. I

Example 1,2: Expand the function g (x) = x^ —2x in a Fourier series, period 2, using the interval [ 0 , 2].

SOLUTION

For cq we obtain cq = (1 /2) Jq

— I x d x — —2/3. For n / 0,

1.2. EXAMPLES OF FOURIER SERIES

X interual: [ -6.Z8, 6.281 X increnent = Y interual: [ -4, 41 Y increment = .8

1.256

X interual: [ -6.28, 6.281 X increment = Y interval: [ -5, 51 Y increment = 1

1.256

FIGURE 1.4 (a) The periodic extension, period 27T, of the function g(jc) = x. (b) The graph of the partial sum Ss together with the graph from (a).

SOLUTION we obtain

— 2 x d x = —2/3. For n 7^ 0,

For cq we obtain cq = (1/2)

1 Cn = ^ j

= 2- Joi (x

- 2 x ) e - ' ’'”^ d x

1

2

— 2x) cos m tx dx — - I (x^ — 2x) sin njtx dx = -^ r2 Jo n^TC

after integrating by parts twice. Thus,

2

2

Grouping terms for n and —n, we obtain the real form of the Fourier series for g

g^ _

2

3

^

4

_j_ \ “

^ cos mtx.

n=\

In Figure 1.5, we have graphed the partial sums

2 A 4 S m M = - - + } -T T -^cosm tx A • ^ n=l

1. BASIC ASPECTS OF FOURIER SERIES

(d) periodic extension and 8 harmonic sun X interual: [ -4, 41 X increment = .8 V interual: [ -1.5, .51 V increment = .2

FIGURE 1.5 Graphs illustrating Example 1.2.

for M = 2, 4, and 8 harmonics, and the graph of the periodic extension of g with the graph of superimposed on it. I Example 1.3: Expand the function g(jc) = [ - 1/ 2 , 1/ 2]. SOLUTION obtain

For

cq

+ x in a Fourier series, period 1, over the interval

we obtain

cq

= / 1 1/2

+ x d x = 1/12. For

n 2 (x^ + x) cos In n x dx — i I (x + x) sin Ijtn x dx J-\

0, we

1.3. FOURIER SERIES OF REAL FUNCTIONS 1

(-If

i + — nn

after integrating by parts twice. Thus,

1 12

'

i: n= -oo

( - 1)”

2

1 i + n^iT^ nn

A2nnx

Grouping terms for n and —n, we obtain the real form of the Fourier series for g

1

CXJ

’ cos 2 nnx

n=\

sin 2nnx nn

In Figure 1.6, we show the graphs of the partial sums iyi ^ s m {^)

^co^2nnx

~ 22 n=\

sin 2n nx nn

for M = 3,12, and 36 harmonics, and the graph of the periodic extension o ix ^ F x superimposed on 535. I

1.3

Fourier Series of Real Functions

In each of the examples in the previous section we computed Fourier series for real valued functions. We were able in each case to express those Fourier series in real forms, series involving real valued sines and cosines. We will now show that this is possible in general. Suppose that g is a real valued function. Then the Fourier coefficients of g over the interval [0, P] are given by

Cn = I

p Jo

(1.5)

If we take complex conjugates of both sides of (1.5), then by keeping in mind that g(x) and X are real we can write

1. BASIC ASPECTS OF FOURIER SERIES

10

(a) partial sum« 3 harmonics X interual: [ -2.5, 2.51 X increment = .5 y interual: [ -.5, .751 If increment = .125

(b) partial sum, 12 harmonics X interual: [ -2.5, 2.51 X increment = .5 y interual: t -.5, 11 y increment = .15

(c) partial sum, 36 harmonics X interual: t -2.5, 2.51 X increment = .5 y interual: 1 -.5, 11 y increment = .15

(d) periodic extension and 36 harmonic sum X interual: [ -2.5, 2.51 X increment = .5 y interual: [ -.5, 11 y increment = .15

FIGURE 1.6 Graphs Ulustrating Example 1.3.

^ Jo Thus, whenever g is real valued, we have — c—n .

(1.6)

11

1.3. FOURIER SERIES OF REAL FUNCTIONS

Using (1.6) we can cast the Fourier series for g into a real form by grouping the and terms as follows

g -

00

OO ^

- CO + £

j

n=l

OO

= co + J 2 n=l ^

= co + 2 2 Re

r

1

n=l

where Re means taking the real part of a complex number. Thus, we have (1.7)

g ~ c o + X ] 2Re n=l

Now, we write Cn more explicitly as a complex number

^«=4 F Jo =4 i F Jo

1

F

~4 i r Jo

In n x r

i

2 ^ " - 2 ^" A„ and Bn by

where we 2

= —/

F Jo

^ ^

1 = — /

In n x ,

g(x)cos— — i/x, ^

F Jo

Using Cn = (\!2)A n — {ij2)Bn and in (1.7) we obtain 00

g ^ CQ

f

(1.8)

F

_ oos{2nnx!P) + i sin(2nnx/P )

2nnx

cos —^ +Z!p” ^ n=l

^ ^ . 27Tnx , g { x ) ^ m — — dx.

2nnx ] + Bn Sin —^ J .

12

i. BASIC ASPECTS OF FOURIER SERIES

For notational uniformity, we define (1/2) Aq to be equal to following formulas: 1

Ij t nx

i

cq.

We then have the

In n x 1

\ g~ +XIM«c o s -------- h Bn Sin —-— P J x

Ao

n=l ^

^

^

2

/

2^71it X

g(jc) cos —- — dx ,

(n = 0 , 1, 2 , . . . )

. In n x , ^(-^) sm —— i/x ,

(n = 1, 2, 3 ,...) .

(1.9)

^

Jo

The formulas in (1.9) define the real form of the Fourier series for g, using the period interval [0, P]. Similar formulas hold if the interval is [—P /2 , P /2]. In particular, the coefficients A„ and P„ would then be given by

2

In n x g(x) COS —^ d x ,

An = — J

2 f2^ 2nnx ^n = — g ( x ) s m - — dx, P J- hP ^

(n = 0, 1, 2 , .. .)

(n = 1, 2, 3 ,...) .

(1.10)

If we define the partial sum S m having M harmonics by M

S m {x ) =

L

( 1. 11)

£ n = -M

then, by the same reasoning as above, it follows that S m has a real form given by o . ^

1.

M

i .

27rnA:

S m {x ) = - A o + 2 ^ M n COS —^ n=l ^

„ . 2jTnx 1 + B„ s m -------\ . p i

( 1. 12)

REMARK 1.2: The convention of using 1/2 as a factor on Aq in the constant term in the real form of Fourier series is troublesome. But, it is a common conven­ tion since it simplifies the form of (1.9). It is also the convention adopted in [Wa], to which we make frequent reference. I

1.4. POINTWISE CONVERGENCE OEFOURIER SERIES

1.4

13

Pointwise Convergence of Fourier Series

In this section we will examine the validity of equating a periodic function with its Fourier series. Our approach to this problem will be to examine the limit of partial sums of the Fourier series at any given point. To be more specific, we examine under what conditions we can have = g(x)

(1.13)

where the left side of (1.13) is the Fourier series, period P, for g. A formal definition of (1.13) is lim Sm M = g(x) where S m is the partial sum, containing M harmonics, of the Fourier series for g [see Formula (1.11)]. The types of functions that we will be interested in are mostly covered by the following definition. D EFINITIO N 1,2: A function g on a finite interval [a, b] is piecewise continu­ ous on [a, b\ if the interval [a, b\ can be divided into a finite number o f subintervals on each o f which g is continuous. I f g is piecewise continuous on every finite in­ terval, then g is called piecewise continuous. Example 1,4: (a) The function .. =

f x + 1 f o r —l < x < 0 for0

55

+ W ^h I

m

(3.7)

Diagram (3.7) is called a butterfly. There are N /2 butterflies for this stage of the FFT and each butterfly requires one multiplication, times Figure 3.1 illustrates the butterflies needed for this stage when N = \6.

Ho -W H j

- W^H2

hJ

+ W^H4

H? Hg + + w"^H7 HX

- H,

HY

- W H

H§

- W ^ H

H hJ

-

-

- w'^h) First reduction in FFT algorithm for N — \6.

The splitting of [Hk\ into two half-size DFTs, {H^} and {H^}, can be repeated on {H^} and [H^] themselves. We get

56

3. THE FAST FOURIER TRANSFORM (FFT)

H i = Hl ^ + { W ^ f H l \

H l ^ , ^ = Hl^ - { W^ Ÿ Hl ^

(3.8)

for A: = 0, 1, . . . , N/ A - 1. In (3.8), { H ^ } is the A^/4-point DFT of [ho, h 4, h s , . . . , hN-4), {^(tM is the At/4-point DFT of [hz, h^, hM-i}, [Hl^] is the iV/4-pointDFTof {/ii, h s , . . . , h N - 3}, {H/M is the iV/4-point DFT of [hs, /17, . . . , hN-i}- Or, in terms of the binary expansions of the indices of hj, we have { H ^ } = DFT[h..,oo}, { f f f } = D F T [ h , .10}, {///«} = D F T [h,..01}, { //" } = D FT { h. . n } . Notice that there is a reversal of the last two digits {bits) in the binary expansions of the indices j in [hj] for which we calculate the A^/4-point DFTs In Figure 3.2 we have diagrammed the two stages of the FFT that we have described. If we continue with this process of halving the order of the DFTs, then after R = log2 N stages we reach a point where we are performing N one-point DFTs. A one-point DFT of a number hj is just the identity hj — h j . Moreover, since the process of indexing the smaller size DFTs by superscripts written by reversing the order o f the bits in the indices of {hj} will continue, by the time the one-point stage is reached all bits in the binary expansion of j will be arranged in reverse order. See Figure 3.3. Therefore, to begin the FFT calculation one must first rearrange [hj} so it is listed in bit reverse order. In Figure 3.4 we have shown the calculations needed for performing an 8-point FFT. The result, of course, matches with the definition of an 8-point DFT. In the exercises the reader is asked to perform a similar calculation of a 16-point FFT. Ultimately, this work is essential to understanding the FFT algorithm. The FFT results in an enormous savings in computing time. For each of the log2 N stages there are N ¡2 multiplications, hence there are {N/2) log2 N mul­ tiplications needed for the FFT. This takes much less time than the (A — 1)^ multiplications needed for a direct DFT calculation. When N = 1024, the FFT requires 5120 multiplications, while the DFT requires 1,046, 529 multiplications. Therefore, for N = 1024, the FFT results in a savings, in multiplication time, by

57

3.1. DECIMATION IN TIME, RAD IX 2, EFT

l5 + W^H2 /A h5 + w’ hJ W'Hj . w/5 It 1

+ w'^h)

- W^H2

-

- W^Hj -

FIGURE 3.2 Two stages of FFT, N = 16.

a factor of almost 200. Actually, even more savings can be achieved, since multi­ plication by i (or —i) does not need to be programmed as multiplication, and we do not need to count as multiplications any butterflies that only involve additions and subtractions (see Exercise 3.1). Similar dramatic savings hold in regard to the number of additions needed for the FFT; we leave it as an exercise for the reader to calculate how many additions are needed for the FFT vs. the DFT. Throughout the remainder of the book we shall concentrate on counting multiplications, leaving the addition counts to the reader as exercises. In the next few sections we shall analyze in detail the following components of the FFT: (a) bit reversal reordering of the initial data (b) rotations involved in butterfly computations (c) computation of the sines and tangents needed to perform (b).

3. THE FAST FOURIER TRANSFORM (FFT)

58 hooo. .0 = DFTihooo oJ ^’lOO. .0 “ ^010 .0 “

. ooo) "

.0^ oi

DFT{h__ool :::rDFT(h ,ooi DFT{h ol

rDFT{h _oio) '

:DFT{h ,io)

rDFT{h _ooi) > DFT{h _o,)

rDFT{h ,ioi)

?DFT{h on) DFT{h l',ll..l

= DFT(h,„..,l

— DFT(h,,„)

Bit Reversal Ordering of(hj)

FIGURE 3.3 Reductions of DFTs and bit reversal. The indices are in binary form.

REMARK 3.1: In this section we have described a decimation in time FFT. The other major type of FFT is known as a decimation infrequency FFT. This type of FFT is described in [Wa, Chapter 7.13]. I

3.2

Bit Reversal

In this section we shall examine in detail the first component of the FFT, bit reversal reordering of the initial data. We shall describe two methods of perform­ ing this bit reversal permutation. These methods are Buneman 's algorithm, and another, faster, algorithm. Buneman’s algorithm is the simplest method for performing the bit reversal

3.2. BIT REVERSAL

59

-ho+h^ +hj+hg

► h o + h 4 + h 2 + h j + h i + h3 + h5 + h7

jho-h^+w2(h2-hg)

► h o - h 4 + W ^ 2 " ‘'6>'*‘ '^^N-j

-

hj]

sin ^

.

(3.80a)

We leave it as an exercise for the reader to show that H g^Rk,

(^A: = 0, 1, . . . , ^ A T - 1^

(3.81)

and

N -l

H If =

hi

J

cos ^—

(3.82)

j=0

where {Rk} and [Ik] are the real and imaginary parts of the FFT of { f j } using weight W = Computer procedures, called COSTRAN and SINETRAN, which perform the algorithms described in this section can be found in Appendix B.

3.8

Inversion of Discrete Sine and Cosine Transforms

In this section we will describe how to invert DSTs and DCTs, thereby recovering the original sequences before they were transformed. First, we show that a DST has the nice property that it is its own inverse (up to multiplication by a constant).

3.8. IN VERSIO N OF DISCRETE SINE AN D COSINE TRANSFORMS

11

More precisely, we will show that the inversion of a DST [H^] of the sequence {hj} is given by N-\

U =

=/i/,

N

it=i

1,...,

N -1).

(3.83)

Formula (3.83) follows from N -\ n m k . 7tjk 2 ^sm N k=\

kN

if m = j

0

if m ^ j

(3.84)

We will prove (3.84) in a moment. First, we show that (3.83) follows from it. The left side of (3.83) can be rewritten as O N-l

-N T ^ k=\

N-l

sm ^ ^ hjn si

7tmk

N-l

sm

N

m=l

2 J . 7tmk . 7zjk — > s m ------sm N N

L ^=1

which, by applying (3.84), yields (3.83). Now, to prove (3.84), we note that for fixed j and m, the sequence nmk . n j k \ 2N-Ì s m ----->

is even about N. Hence, from Formula (3.68), using N in place of N / 2 and k in place of 7 , and noting that the middle element sin(7rmA^/A^) sin(7r 7W/A^) equals 0 , we have

. 7tmk . n j k 1 . 7tmk . n j k > s m ------sm ------ = - > s m ------sm —— N M J ^ N N k=0 k=i ^

1 ^ ^ ^ . iTzmk . iT tjk 2 E sm 2N sm 2N it=0 1

2N-:

l Ek=0 cos 1 - i E A=0

2n{m — j) k ~W

l7i{m + j) k

cos •

2N

(3.85)

3. THE FAST FOURIER TRANSFORM (FFT)

78

Moreover, 2 N- 1

E

cos

k=0

[ 2N-1

271 (m — j) k ~2N

2n(m-j )kl 2N

k=0

0

for m ^ j for m = j

2N

(3.86)

where the last equality follows from Formula (2.7), Chapter 2, with 2N in place of N , [The argument being similar to the one used to prove Theorem 2.1(c) in Chapter 2.] Similarly, we have 2N-1

E

lTt{m + j) k = 0. 2N

C O S ---------------

k=0

(3.87)

Substituting from (3.86) and (3.87) back into (3.85) we obtain Formula (3.84). There is also an inversion procedure for DCTs. It is, however, more complicated than the inversion procedure for DSTs. First, we need to prove the following formula:

7tjk Ttmk > cos - r r - cos ■ N N k=0

N

for m = 7 = 0

-N

for m = 7 7^ 0

I — ^ ( —l)-^"^'”

form 7^ 7*.

(3.88)

Since, for fixed j and m, the sequence {cos(7T7A:/A^) cos(7tmk/N)}^^^ ^ is even about N, we have ^

Ttjk Ttmk 1^^^ 7tjk Ttmk 1 1 _ cos — c o s------= - V cos — c o s ------- + ---------(-1)-'+ '” . (3.89) M

k=0

M

9

N

k=0

9

9,

^

^

^

That (3.89) implies (3.88) is shown by a similar argument to the one used above to show (3.84). The details are left to the reader as an exercise. From (3.88) we find that

2 N k=0

2

, m=0

^ jk N

COS —

k=0

COS

Ttmk ~ w

3.8. INVERSION OF DISCRETE SINE AND COSINE TRANSFORMS 1 v^iV-1 hnr-jf EZZl h A - i r

2^0 + ir

79

for j = 0 (3.90)

hj + ^

for j ^ 0 .

^ E ;:I¿

Consequently, since . N-l

1. iV-1 N-l

m=0

N m=0

.N -l

m=0

we obtain from (3.90) hj + ^

r r C ___

N k=Q ^ ^

cos

for i even ^ 0

odd hm

fo r 7 odd

hj + ^ T l~ ^ ^ n h m

N

2^0 +

2 ^r^N-l

(3.91)

for j = 0 .

n JI, M Z^m odd ’ ^

Using (3.91) it is possible to recover [hj] from [H^} provided the sums

2

2 ^2 -

w Z!

m odd

Z

are known. To find these unknown sums, we introduce the sequence hj defined by

2 %=

E

A:=0

»i

n jk N ’

( j = 0 , 1, . . . , 7V- 1) .

COS ■

Then, from (3.91) we obtain N-2

N-l

and

(3.92)

= ^0 + Z

Z

J even

j=0

A^-l A^-1 J lh j = '£ h j. j odd

(3.93)

j=0

From (3.92) and (3.93) it follows that N-2

ho= Z j even

N -l

~ Z j odd

^J-

(3.94)

3, THE FAST FOURIER TRANSFORM (FFT)

80

Hence, since (3.91) implies

ho = 2ho + — ^

h

m odd

we get N-2

N-l

(3.95)

s \ = h o —2 J

j

even

odd

Furthermore, it also follows from (3.93), and the original definitions of ^2, that

2

(3.96) j

Odd

Thus, we can invert {H^} to recover {hj} by the following steps: Step 1. Form the sequence [hj] defined by N-l

=

O ' = 0 , 1 , . . . , V - 1).

^=0 Step 2. Calculate the quantities s\ and ^2 by 'N-2

N-l

J

j

s\ = ho — 2 even

odd

2 J odd

Step 3, The sequence [hj]^^Q is then obtained by for; = 0 h j^

and

hj -

51

for j even and 7^ 0

hj -

52

for j odd.

3.9. INVERSION OF THE EFT OF A REAL SEQUENCE

3.9

81

Inversion of the FFT of a Real Sequence

In this section we show how the symmetry property (3.57) of the FFT of a real sequence can be used to reduce by about one half the computations involved in inverting the FFT. Although there are other methods of doing this than the method described below, this method is of interest because it uses no extra sine nor tangent values other than the sets and { T w h i c h were used for performing the FFT itself. We will assume in this section that W = Let Rk and Ik stand for the real and imaginary parts of the FFT {Fjt}. That is, Fk = R k ^ i I k .

=

(3.97)

N -l)

where Fk is defined by N -l 7=0

Then, by the formula for DFT inversion fj = J } 8 j ’

(3.98)

U = 0 ,h ...,N -l) ^ ^

where

(3.99)

gj = E Fkiw^'^r. ^=0 Using W =

and (3.97) we have

;= 0

= El

' h ) ^cos

- i sin

•

(3.100)

Since {fj} consists of real numbers, so does [gj] because of (3.98). Consequently, only the real part of the sum in (3.100) is non-zero. Therefore, we must have ^ „ 2njk ^ , gj = ^ R k cos — + E N k=0 k=0

. 2njk N

(3.101)

3. THE FAST FOURIER TRANSFORM (FFT)

82

Because of (3.57) we have RN-k = Rk,

iN -k^-h ,

N -1).

(3.102)

It follows from (3.102) that, for each fixed j , the sequences 2jrjk ]‘

Rk cos —— [ ^ ik=i

and

Ik sin ■

N

\k=i

are even about N /2. From Formula (3.68) we obtain N -l

J^Rk k=l

COS

^N-l = R i A - i y + 2 Y , Rk cos ~Y~ k=l

2n jk

Ijtjk

(3.103)

and N -l

.

.

2n jk

27tjk

(3.104)

= 2 Y h sm E ' * s m -----N N k=l k=\ Using (3.103) and (3.104) in (3.101) yields - N —\

gj = Rq -\-

-\-2 ^ ^

k=\

Formula (3.105) shows that ated from an A/^/2-point FCT of

.

- N —\

/?^ cosy^+ 2 ^ 2^

k=l

and consequently

Ikún-^

(3.105)

2^

can be gener­

{0, 2Ri, 2 R 2 , . . . , 2R¡ N-l^ and an A/^/2-point FST of {2/i, 2 / 2 ,. .. , 2 /i^ _ i} . These FCTs and FSTs are even and odd about N /2, respectively. Taking this symmetry into account. Formula (3.105) applies to y = 0, 1 , . . . , — 1. As we noted above, an interesting feature of (3.105) is that to compute the FCTs and FSTs, of order N /2, by the methods of Section 3.7, one needs only the same sines {S{m)}^^^Q and tangents [T(m)}^^^Q generated to calculate the FFT Therefore, inverting the FFT using (3.105) requires only these same sines and tangents. This is a useful memory savings.

83

EXERCISES

A computer procedure, called InvRFFT, which carries out the algorithm de­ scribed in this section can be found in Appendix B.

References For further discussion of DFTs and FFTs, see [Bri], [El-R], [Nu], [Op-S], [PFTV], [Ra], and [Ra-G]. An important set of foundational papers is collected in [Ra-R]. FFT programs in various computer languages can be found in [PFTV]. Exercises Section 1 3.1 Make a diagram, like the one in Figure 3.4, for the 16-point FFT. How many multiplications are required? 3.2 Construct tables of bit reversed numbers for V = 4, 8,16, and 32. 3.3 Suppose that N = 4^. Derive an FFT based on successive divisions of N by 4. [Remark: This is called a radix 4 FFT.] Section 2 3.4 Construct tables of base 4 digit reversals for N = 4 and N = 16. [This is needed because of Exercise 3.3.] 3.5 Describe the analog of Buneman’s algorithm for digit reversal, base 4. [If n = 13201 (base 4), then the digit reversal of n is 10 2 3 1.] 3.6 Repeat Exercise 3.5 for base r, where r is an integer greater than 1. 3.7 Generalize the second algorithm described in this section, using base 4. 3.8 Generalize the second algorithm described in this section, but use base r. Section 3 3.9 Show that S{N/2 —m) = S{m) for m = 0, 1 ,..., N/4. 3.10 Show that, for w = 0, 1 ,..., N/4 C(m) = S

^

~

~

( 4^ ~ N/4

Using these identities and Formula (3.34), show that the values of (T(m))^^Q can be computed using the values (S(m)}^f^. 3.11 What sines and tangents are needed for performing rotations in a radix 4 FFT (N = 4^)7

3. THE FAST FOURIER TRANSFORM (FFT)

84

Section 4 3.12 Use the first three stages of Buneman’s method, along with a hand calculator, to find sin(27Tj/32) for y = 0, 1 , . . . , 8. Check your results by computing the same values directly on a calculator. 3.13 Develop the fourth stage of Buneman’s method to find sin(27ry764) for y = 0, 1, . .. , 16. 3.14 Suppose that {F^} is the A^-point DFT of [fj], defined by A^-l 7=0

Prove that, for each y = 0, 1 ,...,

—1, ~N-l

fj —

(3.106) k=0 N/4

3.15 Explain why Formula (3.106) in Exercise 3.14 shows that only the values {5(y)} ,• 7=0 xN/4 and {TU)}j=o needed for inverting an FFT, when that FFT used weight W = ^i2niN^ (See Sample Program 2 in Appendix B.) 3.16 Prove (3.68) and (3.69). 3.17 Show that (3.81) and (3.82) are true. Section 8 3.18 Show that (3.88) follows from (3.89). 3.19 Prove that, for m. A: = 0, 1, . .. , A —1, N-l

E

(n + ~){m + \ )n (n + \)(k + cos---------------- — cos — N N

N

if m = k

ifm ^k. (3.107) Prove also that a formula like (3.107) holds if sines are used instead of cosines. n= 0

3.20 Using the results of Exercise 3.19, formulate a definition of an A/^-point DCT that is (up to a constant multiple) its own inverse. Do the same for sines instead of cosines. How would you program a fast version of this discrete cosine (sine) transform?

Some Applications o f Fourier Series

In this chapter we will describe computer modeling of solutions to a few funda­ mental equations from mathematical physics. We shall discuss one-dimensional versions of the heat equation, the wave equation, and the Schrödinger wave equa­ tion for an enclosed, freely moving particle. We shall also examine some of the modifications of Fourier series that are used in signal processing.

4.1

Heat Equation

In this section we shall describe computer modeling of a simple problem in heat conduction. Let the function m(x , t), where 0 < x < L and i > 0, be interpreted as the temperature of a thin insulated rod, with a temperature of 0 at the ends, and let the function g(x) be interpreted as the initial temperature m(jc, 0) = g(jc). In [Wa, Chapter 3.1] it is shown that du

) d^u

dt

^ dx^

m(0,

0 = 0,

m(x ,0)

= g(x)

(heat equation) u{ L, t) = 0

(boundary conditions) (initial condition).

(4.1)

The constant a^ is called the diffusion constant for the material of the rod. Its units are commonly cm^/sec. Copper, for example, has a diffusion constant of 1.14, while granite has a diffusion constant of 0.011. Problem (4.1) is a classic problem in mathematical physics. It was first solved by Fourier (see [Fo]). Its derivation and solution by Fourier series is described in [Wa], [Ch-B], and [We]. A second interpretation of Problem (4.1) is that the function u describes the

85

86

4, SOME APPLICATIONS OF FOURIER SERIES

relative concentration of a solvent within a surrounding solute. The function g(x) describes the initial relative concentration. This interpretation is described in many physical chemistry textbooks. Usually, the method of separation of variables is used to solve (4.1). Since this method is so well known, we will describe another method. We begin by expanding m(jc, i) in a Fourier series in the spatial variable x. Since all of the functions in the system [sm(ri7tx/L)}^i satisfy the zero boundary conditions at x = 0 and jc = L, we expand m(x , i) in a Fourier sine series u(x, t) = ^

Bn{t) sin

n=\

(

=

2

nnx "zT ’

m(jc, 0

nnx \ sin —^ Jjc i .

(4.2)

To find the functions Bn(t) we substitute into the heat equation in (4.1) and, dif­ ferentiating term by term, get 00

^

OO n . . nnx \ —\ / n n a \ ^ ^ ^ ^ • n n x B„(t) sin — = 2 2 - ( ~ )

n—l

(4.3)

n=l

Expressing (4.3) in terms of one sine series yields

E

,

/f n n a \ ^ nnx ) B„{t) sm ----- = 0.

(4.4)

Since the sine coefficients of the 0-function are all 0 we infer from (4.4) that , /nna\'^ B'^it) -b ( — j Bn{t) = 0,

(n = 1, 2, 3 ,...) .

(4.5)

Substituting i = 0 into the formula for Bn{t) in (4.2) we also have Bn{^) = “ /* L Jo

m(x ,0)

sm

nnx

dx,

(n = 1, 2, 3 ,...) .

(4.6)

The right side of (4.6) is a constant for each n\ it represents the sine coefficients of the initial temperature w(x, 0). If we define bn by 2 nnx ^n — — I u{x, 0) sin —— dx, L Jo L

(n = 1, 2, 3 ,...)

(4.7)

87

4.1. HEAT EQUATION

then (4.6) becomes Bn(0) = bn,

(^ = 1 , 2 , 3 , . . . ) .

(4.8)

For each n, the solution to the differential equation (4.5) with initial condition (4.8) is —h ^-(nnalLyt (4.9) Combining (4.9) with the first equation in (4.2) yields the following Fourier sine series solution to Problem (4.1), oo

mtx ' \ b n si n -

u(xj) =

(4.10a)

n=l ^

where m(x , 0)

bn sm si

=

nitx

(4.10b)

n=\

In other words, the temperature m(jc, t) fo r t > 0 is obtained by putting damping factors -{n n a /L yt (Gaussian damping factor) on the coefficients o f the Fourier sine series fo r the initial temperature u(x,0). Example 4.1: Suppose that a^ = 1.14 cm^/sec, the diffusion constant of copper, and L = 10 cm. Graph w(x, i) for i = 0.1, 0.5, 1.0, and 2.0 sec; given an initial temperature of 0 for 0 < X < 4 and 6 < x < 10 m(x , 0) = for 4 < X < 5 20x —80 for 5 < X < 6 . 120 - 20x SOLUTION In Figure 4.1 (b) we show the graph of the 25 harmonic partial sum for the Fourier sine series of w(x, 0) using 512 points. We will use this partial sum as an approximation of w(x, 0). Although the graph of this partial sum appears noticeably different than the graph of the initial function w(x, 0) in Figure 4.1(a), we shall see below that it will still provide good approximations for the positive values of t that we are using. Now, for the times given, we have damping factors ^ Using FAS, these factors can be put as coefficients on the terms o f the 25 harmonic partial sum by choosing to do a filter and then choosing Gauss as the type o f filter. For

4. SOME APPLICATIONS OF FOURIER SERIES

88

X interual: [ B, y interval: [ -5,

101 Z51

X increaent = V increaent =

X interual: [ B, y interval: ( -5,

1 3

101 251

X increaent = y increaent =

1 3

FIGURE 4.1 (a) Graph of the initial temperature u{x,0) in Example 4.1. (b) Graph of the 25 harmonic Fourier sine series partial sum for this initial temperature. i = 0.1 the damping constant T asked for by FAS is T = a^O.l = 0.114. And, for t = 0.5, 1.0, and 2.0 sec, we use T = 0.57, 1.14, and 2.28, respectively. The graphs of m(jc, t) for these values of t are shown in Figure 4.2. We will now show that using a 25 harmonic partial sum gives good approxima­ tions to m(jc, 0 for the times i = 0.1, 0.5, 1.0, and 2.0 that we used above. First, we observe that the sine coefficients bn satisfy 1 \K\ = \-z /

^ Jo

nnx 1 u ( x , 0 ) s i n - — dx\ < - 1

5 Jo

10

\u(x,0)\dx.

Thinking of the integral of |w(x, 0)| as an area, we get \ bn\ t) sin —j — ax

(4.17)

and substituting this series for y{x, t) into the wave equation in (4.16) and into the initial conditions in (4.16), we obtain 00

[^n (0 + {ncn/ L)^Bn{t)^ sin

= 0

n=l

_ 2

BniO)

~ L Jo

BL(0) = — I

nnx f i x ) sin - ^ —dx

g(x) sin

(n = 1, 2, 3 , . . . )

dx

(n = 1, 2, 3, . . . ) .

(4.18)

Letting an and bn stand for the iz* sine coefficients of / and g, respectively, we obtain from (4.18) B'Jit) + (nc7T/LfB„(t) = 0,

Bn(0) =a„,

B'„{0) = b„

(4.19)

for « = 1, 2, 3 , ___For each n the solution to (4.19) is sin (n c 7 T /L )i

Bnit) = an cos (n cnl L) t + bn-

ncn/L

(4.20)

Using the notation smnv sine V = --------, nv

, . ^ ( sine 0 = 1 )

(4.21)

we can express (4.20) as Bn(t) = an cos (ncn/L)t -h bnt sine inc/L)t.

(4.22)

93

4.2. THE WAVE EQUATION

Combining (4.22) with (4.17) we have found the following series solution to Prob­ lem (4.16):

y(x, t) = ^

[ Un COS

{nc7tf L)t +

bnt

sinc {nc/L)t^ sin

riTTX

n=\

2

I Jo

bn

nnx y( x, 0) s in '-^^^ dx, ~T

2 C^dy

= iL

. nnx , 0) s m ----- dx.

(4.23)

JLj

Example 4.2: Suppose that L = 50 cm, c = 100 cm/sec, the initial position i s / ( x ) = 0.1x(50— x), and the initial velocity g is zero. Graph y(x, i) for the times t = 0.1, 0.2, and 0.3 sec. SOLUTION Since g(x) = 0 for 0 < x < L, it follows that 2, 3, — Using a 200 harmonic partial sum for y ( x , t ) yields 200

y(x, t) ^

cos(nnct/L) sin

nnx

= 0 forn = 1,

(4.24)

n=l

where L = 50 and c = 100. The sine series partial sum in (4.24) is obtained by applying coefficients cos{nnct/L),

(n = 1, 2, 3 , . . . )

(4.25)

to the terms of the 200 harmonic partial sum of the sine series for y(x, 0) = /( x ) . This can be done by first calculating the sine series partial sum, using 1024 points. Then, after pressing y in response to the question about using a filter, one presses the Tab key to display the second list of filter choices. The filter needed in this case is the Cos filter. You will be asked for the value of c, called the wave constant, which is 100 for this example. Then you will be asked for the value of i, called the time constant. After performing these steps and entering the three values of t for the time constant, you will obtain graphs like the ones shown in Figure 4.3(a). I The approximation in (4.24) is a good one. Calculating the sine coefficients [an)

94

4. SOME APPLICATIONS OF FOURIER SERIES

V interval: ( -80.

801

V increment =

X interval: [ 0. 501 Y interval: [ -40, 601

16

X increment = 5 Y increment = 10

FIGURE 4.3 (a) Graphs o f the string positions for Example 4.2. (b) Graphs of the string positions for Example 4 3 .

for f ( x ) = 0.1x(50 —x), we get

1000 —

[l-(-l)" ],

(4.26)

(« = 1 , 2 , 3 , . . . ) .

Therefore, 200

y(x, t) —

rmx cos(n7Tct/L) sin — ^ L 1

an cos(n7Cct/L) sin

njtx

«=201

^

^

2000

0.

95

4.2. THE WAVE EQUATION

REMARK 4.1: Since Formula (4.26) gives us an exact expression for the Fourier sine series coefficients, it is possible to graph exact partial sums (instead of the Fast Sine Transform [FST] approximations used in the example above). The gain in accuracy, however, is insignificant (see Exercise 4.12) and certainly does not justify the huge increase in time needed to calculate sums o f200 terms for every point plotted. Of course, for most functions we cannot obtain exact expressions for the Fourier sine series coefficients so we must use FST approximations. I Example 4.3: Suppose that L and c have the same values as in the previous example, but that the initial velocity is described by 20x(50 —x) while the initial position is zero. Graph y(x, t) for times t = 0.001, 0.002, and 0.003 sec.

SOLUTION In this case, the coefficients [an} are all 0. Again using 200 harmonics in a partial sum for y { x ,t ) we have 200

y{x, t) ^

[t sine {net/L)} bn sin

nixx

(4.27)

n=\

In this example, one computes the filter factors t sine (nct/L),

{n = 1, 2, 3, . . . ) .

(4.28)

One chooses the filter Sine on the second filter menu (press the Tab key when the first filter menu appears). You are asked to enter the wave constant, which is e = 100 as in the previous example, and the time constants (0.001, 0.002, and 0.003). This is how one obtains the graphs shown in Figure 4.3(b). I The approximation in (4.27) is more accurate than the previous example. Since the values of bn are 200 times the values of an in the previous example (since g = 2 0 0 / where / is as in the previous example), we obtain

^

X

T

.

y(x, t) - 2 ^ t sme — bn sm —— X-/ 1-1 «=1i

E

«=201

200 sin (ncTTi/L) 1000[1 nenfL

(—1)"] nnx ------- sm ------

4, SOME APPLICATIONS OF FOURIER SERIES

96 ^ < > it=101

200000 , -TT---- — < 4.34 X 10 (2k - 1)47t4

.

Therefore, the approximation in (4.27) involves an error of no more than 4.34 x 10” ^ for 0 < a: < 50 and all time i > 0. REMARK 4.2: The filtered Fourier sine series partial sums described in this section can be used to produce simulations of vibrating strings in the form of computer animations. For an example of such a computer animation, view the graphbook VIBRSTR which can be found on the computer disk accompanying this book. I

4.3

Schrodinger’s Equation for a Free Particle

In this section we shall discuss a problem involving the one-dimensional Schrodinger’s equation: - fi dir _ d^ir -hV ir. i dt 2m dx^ We shall assume that the potential term, V, equals 0. Hence, the equation simplifies to dir ifi d^ir dt 2m dx^ We shall now describe computer modeling of the filtered Fourier series solution to the following problem: dir dt

ifi drir 2m dx^

ir(x, 0 = 0 , ir(x, 0) =

/ ( jc) ,

(Schrodinger’s equation) for jc < 0 or jc > L for 0
0 . Later, we will show that i-L

f Jo

m 2=

(4.30)

\ir(x, t ) f d x = 1

provided that the given initial function / satisfies rL

ll/lii =

f Jo

(4.31)

\ f (x )\ ^ dx = 1.

Formulas (4.30) and (4.31) are the normalization conditions that are imposed on the function -[¡r whose evolution from / is governed by Schrodinger’s equation. The constant fi is Planck’s constant divided by I n . The value of ^ is 1.054 x 10“ ^^ erg-sec. By comparison, an electron has mass 0.911 x 10“ ^^ g. Because of the absence of a potential term in Schrodinger’s equation in (4.29), the particle is said to move freely. The boundary conditions in (4.29) say that the particle is constrained to move along the jc-axis between 0 and L. Problem (4.29) is a one-dimensional version of the particle in a box problem. To find the solution 'll/ to (4.29) we proceed as in the previous two sections. We expand \lr{x,t) and f ( x ) in a Fourier sine series in x: f { x , t ) = ^ C „ ( 0 sin

njtx

, / t)Xsm . y/{x,

J 1 dx

n—l

Ä . nnx f i x ) = y c „ sm ——

r 2 p . nnx , \c„ = f i x ) sm dx

(4.32)

^ I L Jo Substituting the series for ^ (x , t) into Schrodinger’s equation in (4.29), differen­ tiating term by term, and combining the two series into one, yields . nnx sm ----- = 0.

ifl / njt \ ^ _

fy-i/ / \

(4.33)

Setting t = 0 in the formula for Cn(t) in (4.32) yields [by comparison to the formula for c„] Cn(0) =

Cn,

(n = 0, ±1, ±2, ...).

(4.34)

Setting each coefficient in (4.33) equal to 0, we get

C'nit)

—ifi / n jnr \ 2^ ( — j Cn(t), 2m

( n = 0 , ±1, ± 2 , . . . ) .

(4.35)

98

4. SOME APPLICATIONS OF FOURIER SERIES

Solving (4.35), subject to the initial condition (4.34), for each n yields C„(t) =

(n = 0, ± 1 , ± 2 , . . . ) .

(4.36)

Combining (4.36) with (4.32), we have our solution to (4.29): CX)

f i x , t) = 2 ^

^

rmx sm ■

(4.37a)

n=l

where 00

Y—V . riTtX ir(,x,0) = 2_^c„ sin —

(4.37b)

n—l

Our solution to (4.29) expresses ^ ( x , t ) as slfiltered Fourier sine series obtained from the Fourier series for the initial function ‘i[r(x,0). The filter factors are the complex, time-dependent exponentials given in Equation (4.36). Example 4,4: Suppose that the initial function V^^(x, 0) is given by 1A(^,0) =

0.5

for 30 < X < 34 for 0 < jc < 30 or 34 < X < 64

(see Figure 4.4) and the particle has a mass of 0.911 X 10 Graph \ilr(x, i)P for t = 0.05 and 0.15 sec.

g (mass of an electron).

SOLUTION The initial function \lr(x,0) is real valued. Consequently, the Fourier series for xlr(x,t) can be split into real and imaginary parts ^ (x , t) = ir^(x, t) — i ' ^ \ x , t)

(4.38)

where //?/ *

X r ^ / n 7T\2 1 . nn. = ¿ - “ 4 s It ) T " ™ - l «=1

n=l

““

)'

(4.39)

4.3. SCHRÖDINGER ’S EQUATION FOR A FREE PARTICLE

X interual: t 0, 64] Y interval: [ -.Z, .81

99

X increment = 6.4 Y increment = .1

FIGURE 4.4 G raph of the initial function in Example 4.4.

FAS can be used to graph partial sum approximations to and approximations to approximate the probability density function

, and use these

(4.40)

To graph a 2000 harmonic partial sum approximation to'ilr^ix, t), call it one first computes a 2000 harmonic partial sum for 0), using FAS (8192 points were used for these examples). Then, one chooses the FresCos filter from the second filter menu. The filter procedure asks you to input the value of f ijlm, which it calls the wave constant. For an electron, the wave constant has a value of 0.578. The FresCos procedure then asks you for the time constant, for which you enter one of the values of t. FAS then computes filter factors

cos

and graphs

fi 2m

(n = 0, ±1, ± 2 , . . . )

the 2000 harmonic partial sum approximation to

To graph the 2000 harmonic partial sum approximation to \lr \x ,t) , call it 52000 0» you choose FresSin from the filter menu, and then enter the same wave constant and the same time constants as for . If you graph 5'2000 ^ *^2000 simultaneously for t = 0.05 and 0.15 sec, then you obtain graphs like the ones shown in Figure 4.5. In Figure 4.6 we show graphs of l-SaoooiJc, t)\^ - [ S § ^ ( x , t ) f + [5|ooo(-^> O f

4. SOME APPLICATIONS OF FOURIER SERIES

100

(a) t = .05 X interual: [ B, 64] if interual: [ -.2, .81

(b) t = .15 X interual: [ 0, 641 if interual: [ -.2, .81

X increment = 6.4 if increment = .1

X increment = 6.4 if increment = .1

FIGURE 4.5 Approximations of the real and imaginary parts of i/r (x, t) in Example 4.4 for t = 0.05 and 0.15 sec. which is the 2000 harmonic approximation to These graphs were ob­ tained by choosing Graphs on the display menu and then entering the formula f(x) =gl(x)

a

2 + g2(x) A 2

This sequence of steps is performed after both S^qqq and S^qqq are displayed si­ multaneously.

(a) t = .05 X interual: [ 24, if interual: [ -.1,

401 .41

X increment = if increment =

1.6 .05

(b) t = .15 X interual: 1 24, if interual: [ -.1,

401 .41

X increment = if increment =

1.6 .K

FIGURE 4.6 Approximations of \ir{x, i)P in Example 4.4 for t = 0.05 and 0.15 sec. The similarity between these approximations of | i)P and one-dimensional Fresnel diffraction patterns from optics will be obvious to anyone familiar with such

4.3. SCHRÖDINGER ’S EQUATION FOR A FREE PARTICLE

(al t = .05 X interual: [ 24, Y interval: I -.1,

401 .41

X increaent = Y increaent =

(b) t = .15 X interval: t 24, Y interval: I -.1,

1.6 .05

FIGURE 4.7 Comparison between |V^(jc, 0)p and \ir(x,

401 .41

101

X increment = Y increment =

1.6 .K

ft>r t = 0.05 and 0.15 sec.

diffraction patterns. (See Chapter 6, Section 6.2.) It is interesting to compare these approximations of |^(jc, with the graph of the initial distribution 0)p; see Figure 4.7. I It is interesting to see what happens if the initial probability density is narrowed, as in our next example.

Example 4.5: Suppose that the initial function tA(jc, 0) is

0)

= i ^ lo

31.75 < X < 32.25 f o r 0 < x < 31.75 and32. 25 < X < 64.

Graph approximations to |i/r(x, and for the same mass, too.

for the same times as in the previous example

SOLUTION The graphs of | S2000P I »0 P) are shown in Figure 4.8. The similarity between the graphs for t = 0.05 and 0.15 sec and Fraunhofer diffrac­ tion patterns from optics will be obvious to anyone familiar with such diffraction patterns. (See Chapter 6, Section 6.3.) I We close this section by showing that, given (4.31), Equation (4.30) must hold. And, we also show that the 2000 harmonic partial sum approximations used in the

4. SOME APPLICATIONS OF FOURIER SERIES

102

(a) t = .05 X interual: [ 24, y interual: I -.1,

40] 1.4]

(b) t = .15 X interual: [ 24, Y interual: t -.1,

X increnent = 1.6 Y increment = .15

FIGURE 4.8 Approximations of \ir(x,

40] 1.4]

X increwnt = 1.6 Y increnent = .15

in Example 4.5 for t — 0.05 and 0.15 sec.

last two examples are fairly good if the 2-Norm r*L 11^112 =

^

(4.41)

|g(x)|^dxj

is used to measure the magnitudes of the differences between functions. We need the following theorem, whose proof is beyond the scope of this text. THEOREM 4.1: The series riTtx an sm si ■ n=l

is a Fourier sine series fo r afunction g, where ||g ||2 < oo, ifandonlyifY^^=\ converges. Moreover, when ||gll2 < oo, then ParsevaVs equality holds L ^ ^ T ,\a n \^ ^

\g { x )\^ d x ^ \\g \\l

and we have the completeness relation

lim 11^-5^112= lim

M->oo

M-^oo

T QQ 7 I ] \^n\

=

0

n>M

where Sm is the M-harmonic partial sum o f the Fourier sine series fo r g.

103

4.3. SCHRÖDINGER ’S EQUATION FOR A FREE PARTICLE

Theorem 4.1 is a corollary of the following theorem (proved in [Ru, Chapter 4]). In this theorem, l|g ||2 stands for [/f'^ \g{x)\^ . THEOREM 4.2: COMPLETENESS OF FOURIER SERIES The series ^ n=—(X>

is a Fourier series fo r a function g, where \\gW2 < 00, if and only i f Y l n= ^ - ^-oo converges. Moreover, when ||g ||2 < 00, then ParsevaVs equality holds ^ 2L

W nf n=—cx)

pLr*L / \ g { x ) f d x = llalli J —L

and we have the completeness relation

K| lim l l g - 5 M l l 2 = lim 2L ^ M-^OO M-a^oo \n\>M

=0

where S m is the M-harmonic partial sum o f the Fourier series fo r g. Using Theorem 4.1 we have, based on (4.31) and the second equation in (4.32), (4.42) n=\

However, we also have ; fi /n7r\2.

\ F2mÍL) ^Cn\ = \Cn\ so (4.42) implies T

^

^

0

) ^Cn\ ,

(4.43)

n=l

Using Theorem 4.1 again, we conclude from (4.43) that the series in the first equation in (4.32) is the Fourier series for i f ( x , t ) and (4.30) holds. Theorem 4.1 can also be used to show that the 2000 harmonic partial sums in the examples above are good approximations. Letting Sm {^, 0 stand for the Mharmonic partial sum of the series in (4.37a), it follows that \jr{x,t) — 5m (^, t) has the Fourier sine series expansion ^ r • Ä «TT

\¡r{x, t) - S m (x , 0 = X I n>M

2n

sin

nnx

(4.44)

4. SOME APPLICATIONS OF FOURIER SERIES

104

and, for i = 0 in particular,

ilr(x, 0) — S m (x , 0) = ^

c„ sin

nnx

(4.45)

n>M

By Theorem 4.1 applied to (4.44) and (4.45) we obtain, by the same method as above, j U { x , 0 ) - S m {x ,0)\\2 = _

I

oo E

\n\>M

1^.1"

= \\f{x, t) - Sm (x , t)\\ 2 .

(4.46)

Equation (4.46) says that the 2-Norm measure o f the magnitude o f the difference between \j/(x,t) and SM(x,t) is a constantfo r all i > 0. Using the choices Graphs (on the display menu) and Norm diff (entering 2 for the Power), one can estimate the 2-Norm magnitude of the difference between the initial function 0) and its M-harmonic Fourier series S m {x , 0) when both graphs are displayed. When M = 2000, the magnitude of the 2-Norm difference obtained for Example 4.4 is approximately 0.0039 while for Example 4.5 it is approximately 0.11. In both cases, 52000(-^, 0 is a reasonable approximation, in terms of 2-Norm, to i)This is especially true if the 2-Norm difference is thought of as measuring the distance between two vectors of length 1 (corresponding to an angular, or more precisely, chordal, distance).

4.4

Filters Used in Signal Processing

In each of the preceding sections an initial Fourier sine series was filtered by multiplying its coefficients by filter factors. These factors were time dependent. In this section we will give an introduction to some of the common time-independent filters, most of which are used in signal processing. Cesàro Filter Cesàro filtering is also known as the method of arithmetic means. To motivate this procedure, consider Figure 4.9. In parts (b) to (d) of Figure 4.9, we can see how the successive partial sums of the Fourier series seem to interlace around the graph of the step function. By interlace we mean not only that the partial sums oscillate about the function, but that at most points they also change from being

4.4. FILTERS USED IN SIGNAL PROCESSING

105

H— I— I— h

(a) Step function X interual: t -1, Y interual: I -Z,

X increment = Y increment =

(c) 20 and 21 harmonics X interual: [ 0, 11 X increment = .1 Y interual: [ .5, 1.51 Y increment =

(b) 16 and 17 harmonics X interual: [ 0, 11 X increment = .1 Y interual: [ .5, 1.51 Y increment = .1

(d) 24 and 25 harmonics X interual: [ 0, 11 X increment = .1 Y interual: [ .5, 1.51 Y increment = .1

FIGURE 4.9 Behavior of Fourier series partial sums for a step function.

above the step function to below the step function (or from below to above) as one passes from one partial sum to the next. It makes sense then to form an average, an arithmetic mean, of partial sums in order to better approximate the function. D EFINITIO N 4.1: Given a function with Fourier series partial sums {Sn the arithmetic meariy or Cesaro filtered Fourier series using M harmonics, is denoted by gm where

gm

= — [So + Si + . . . + S m - i ] .

106

4. SOME APPLICATIONS OF FOURIER SERIES

By replacing Sk by YTj=-k

we get M -l (4.47) k=0 _ j = - k

By fixing a value of n, for some n = 0, ±1, ±2, . . ± M, and counting how often Cn appears in the sums in (4.47), we obtain M

CTMix)= Y , n = -M

0"

innx/L

(4.48)

If we compare gm in (4.48) to S m , where M

Suix)^

(4.49)

E n = -M

we see that gm is obtained from S m by multiplying the coefficients in S m by iht filter factors {1 — These factors are often called convergence factors since they will usually help improve the pointwise convergence of the C es^o filtered Fourier series to the original function. One form of this improved convergence is suppression o f Gibbs’ phenomenon (see Figure 4.10). Another aspect of C es^o filtering is that the filter coefficients are near 0 for \n\ near M,

(a) Unfilteredj 40 harmonics X interval: [ -1, 1] X increment = V interval: [ -2, 21 V increment =

.2 .4

(b) Cesaro filtering, 40 harmonics X interval: I -1, 13 X increment = Y interval: [ -2, 23 Y inesrement =

.2 .4

FIGURE 4.10 Suppression of Gibbs’ phenomenon by Cesaro filtering. consequently the higher frequency harmonics in Sm ^re damped down in g m - This is a common feature of many of the filtering procedures used in signal processing.

4 A. FESTERS USED IN SIGNAL PROCESSING

107

de la Vallée Poussin Filter Another filter that is closely related to the Cesàro filter is the de la Vallée Poussin filter (dlVP filter, for short). To motivate the use of the dlVP filter, we note that for small values of n the Fourier series partial sum Sn is often not a good approximation to the original function. For example, graphs of the 2 and 3 harmonic Fourier series partial sums for the step function shown in Figure 4.11 are not close to that

X interual: [ -1, Y interual: [ -2,

11 21

X incment = Y increnent =

.2

A

F IG U R E 4.il Graphs of 2 and 3 harmonic Fourier series partial sums for a step function. step function at all. Putting this another way, notice that in Figure 4.9, we chose partial sums that were intertwined about the step function and that this intertwining becomes denser with larger number o f harmonics. Therefore, it might be more advantageous to average only the upper half of the partial sums. If we have an even number of harmonics, say 2M, then we define V2M by 1

V2M = — M

2M -1

Sk.

(4.50)

k=M

This function V2m is called the dlVP filtered partial sum using 2M harmonics. We now show that

V2m {x ) =

2M Y.

(4.51)

rt= -2 M

where

if ln| < M

Vn =

if M < |n| < 2 M .

(4.52)

4. SOME APPLICATIONS OF FOURIER SERIES

108

First, we rewrite (4.50) using the definition of Sk'. 2M -1

1

(4.53) k=M L j = - k

Second, we find the coefficient of are two cases to consider.

for each n = 0, ±1, . . . , ±2M . There

Case 1. (|«| < M). For this case appears in each term in brackets in (4.53) for ^ = M to 2M — 1. Therefore, the exponential has coefficient

M Hence

2M -1

1 2M-1 ■

E

k^M

—

fe=M

= 1 as in (4.52).

Case 2. {M < \n\ < 2M). For this case, appears in only those terms in the brackets in (4.53) where k = \n\, , 2M — 1. Therefore, the coefficient of ^i7tnx/L js 2 M -1

ME

M k=\n\

1 2 M -1

2 M - |n|

- E 1 If M k=\n\ .

—

"m

Hence, Vn = 2(1 - \n/2M\) as in (4.52). In FAS, the dlVP filter is found on the second filter menu. FAS will calculate the dlVP filtered Fourier series M Vm {x ) =

(4.54)

E n = -M

where if21n| < M Vn =

i i M „ =

—

j

(4.70)

for the unknown function B n {t). For instance, suppose that we have a string of length L = 10 and that a force of 0.2p sin( M. We now give a name to this important function V m D EFINITIO N 4.2: Given a sequence {F„} o f filter coefficients, the function V m >defined in (4.91), is called the point spread function (PSF) or kernel fo r the filter process. We will now describe how FAS plots filtered Fourier series and how it plots PSFs. This method is similar to the method of approximating sampled Fourier series that we described in Chapter 2, section 2.4. First, since / * V m has period 2L, it can be sampled over the interval [0, 2L]. Replacing x in Formula (4.90) by xj = 2 L j / N for 7 = 0, 1 ,..., — 1, we get M ilnnj/N

f * P m (-^7) —

(7 = 0 , 1 , . . . , i V - 1 ) .

^ ^ ^nFftC

(4.92)

n = —M

With some index shifting we can view (4.92) as an A/^-point DFT [provided M < (l/2)N]. Splitting the sum in (4.92) into two sums, one over positive indices and the other over negative indices, we get M

f * P

m

(x j ) =

Y

M

+Y c - n F - „ e ~ ' I n n j / N

(4.93)

n=l

B=0

Hence, because ^ i2;rn7/iv _ ^z27t(A/^ n)j/N^ M

M i2n(N-n)j/N

f * VM iXj) =

+ Y ^ -n F -n e‘ n=0

(4.94)

n—l

Now, recall that the Fourier series coefficients of / are approximated by a DFT (see Remark 2.1 [a] in Chapter 2). Denoting this DFT by we have (for

4.6. CONVOLUTION AND POINT SPREAD FUNCTIONS

121

M < N/ S) Cn^G n/N

(4.95)

c - „ ^ G n -n lN .

and

Using (4.95), we rewrite (4.94) as the following approximation:

1

^

il n n jf N

n=0 1 M + ^ E

.

(4.96)

«=1 If we define the sequence {//„} by g „f „ / n

Hn =

0 GnF„-N/ N

for n = 0 , 1, . . . , M for n = M + 1,. . . , for n = —M,

1

(4.97)

then we can express (4.96) as N -l

(4.98) n=0 Formula (4.98) shows how we can approximate / * V m at the points X j=j^,

(j = 0 , I , . . . , N - 1 )

by doing an FFT, with weight sequence {Hn} defined in (4.97). The value / * V m (x n ) = / * 7^m (2E) can be obtained by periodicity: /* P

m (2L)

= /* P

m (0).

By connecting all of these values by line segments, / * V m M is approximated for every x in the interval [0, 2L]. (When the interval used is [—L, L], then FAS makes use of periodicity. It just converts to [0, 2L], does the calculations described above, and converts back to [—L, L].) Using the method just described, it is also possible to graph the P SF V m - All we need for this is a sequence {A n U)} whose FFT {G„} is the constant sequence [N], since then (4.97) will reduce to the Fourier coefficients of V m - If we define A n by 'N if j = 0 (4.99) A n U) i f ; = l , 2, . . . , ^ - l « = lo

122

4. SOME APPLICATIONS OF FOURIER SERIES

then the -point DFT of A i s the constant sequence {N}^~q . The function A n is called a discrete delta function. The function A n can be graphed by FAS as follows. Suppose, for example, that N = 1024. You then enter the function

f i x ) = 10245(jc)

and A 1024 will be graphed (see Figure 4.16[a]). If you then produce an M harmonic filtered Fourier series, you will obtain graphs of V m - See Figure 4.16(b) to (d) for graphs of V 40 for Cesàro, banning, and Hamming filtering.

H— I--- h

H--- h— H

H— !— ^— h

Ca) Discrete delta function (10Z4 points) X interval: 1 -1< 11 X increment = .2 y interval: 1 -100. 14001 V increment =

(c) banning PSF, 40 harmonics X interval: t -3.14, 3.141 X increment = y interval: [ -5, 451 y increment = 5

150

.6Z8

FIGURE 4.16 Graphs of some point spread functions.

I— H--- 1----h

(b) Cesaro PSF. 40 harmonics X interval: t -3.14, 3.141 X increment = y interval: 1 -5, 451 y increment = 5

(d) Hamming PSF, 40 harmonics X interval: [ -3.14, 3.141 X increment = y interval: [ -5, 451 y increment = 5

.6ZB

4.7. DISCRETE CONVOLUTIONS USING EFTS

4.7

123

Discrete Convolutions Using FFTs

In the previous section we defined the concept of convolution of two functions. In this section we describe the method by which the computer can be used to approximate the convolution

/ * g (x ) = ^

y ^ f ( s ) g ( x - s) d s

(4.100)

where / and g both have period 2 L. The method used to approximate (4.100) is very similar to the method described at the end of Section 4.6. It basically consists of multiplying Fourier coefficients for / and g and then doing an inverse FFT. Dividing the interval [—L, L] into N subintervals of equal length 2 L / N with left endpoints

and using a left-endpoint sum for the integral in (4.100) yields j

N -l

f * g(x) ^

f i S m ) g ( x - Sm)

(4.101)

m=0

Since Asm = 2 L / N for each m, we have N —- l1 iV

/ * g{x) ^

f ^ ^ ' « ^ s ( x - Sm)-

(4.102)

m=0

Now, since g has period 2L, if we extend the sequence {5^} beyond the interval [—L , L] using the formula 2L Sm — - L - ^ m — , N

(m = 0 , ± 1, ± 2 , . . . )

then we will have for all integers j and m

(4.103)

124

4. SOME APPLICATIONS OF FOURIER SERIES

where xj is defined by Xj = J

2L W ’

( 7 = 0 , ±1, ± 2 ,...) .

Using (4.103), we can rewrite (4.102) with xj in place of x: N -l

f * g(Xj)

(4.104)

^ XI

m=0

The right side of (4.104) is a discrete convolution (multiplied by l / N ) . D EFINITIO N 4.3: Let [uj] and [vj] be two sequences o f numbers, each having period N. The cyclic convolution o f {uj} and {vj} is the sequence {u * Vj ] defined by N -l ■^ Vj

— ^ ^

m=0

For example, Formula (4.104) shows that the sequence { / * g(xy)} is approxi­ mated by 1/A^ times the cyclic convolution of the sequences {f(sj)} and {g(sj)}. A direct calculation of m * uy for j = 0 , 1, ..., — 1 would seem to require 4A^^ real multiplications (if complex-valued sequences are used). The following theorem allows this number to be reduced to (9/2) N log2 N + 6N real multipli­ cations by taking advantage of the FFT. The FFT provides a very efficient way of performing cyclic convolutions. THEOREM 4.4: DISCRETE CONVOLUTION I f [uj] and {uy} are sequences o f period N with DFTs {Uk} and { then the DFT of { u * i;y} is {UkVk}. PROOF

If we put W =

respectively,

, then the DFT of {u * i;y} is N -l

Y^u*Vj f=o Replacing u * vj by the sum that defines it, we have upon rearranging sums N -l

;=0

N -l

=;=0E

^N -l ,m = 0

125

4.7. DISCRETE CONVOLUTIONS USING EFTS N -l m=0

Replacing

A^-1 L;=o

m)k^mk

by N -l

yields

N -l

N -l

J ^ U ^ V j W j ^ = J 2 ^ m W ‘mk L;=o

m=0

j=0

N -m -l

N -l

E

= J^U m W ^^

(4.105)

_ p=—m

m=0

Now, the last sum in brackets can be rewritten as follows: N -m -l

N -m -l

vpw p’^ = p= -m

-1

E

+ E

p=0

p--m

N -m -l

=

N -l

E

^ p - n w ^^'

E

p=0

N)k

p = N —m

Because [vp] has period N, we have Vp-N = Vp for all p. Also, W ^ = 1. Therefore, N -m -l

Y p = -m

N -l

vpWP’^ = Y ^ p W ’'' = Vk. p=0

Hence (4.105) becomes A^-l

N -l

Y u * V j w j ’^ = E j=0

= UkVk

m=0

which proves that the DFT of {u ^ vj] is {UkVk}.

I

Using Theorem 4.4 we can efficiently compute the cyclic convolution {u * First, compute the FFTs of and which are {Uk}^~Q and Then, multiply each element of the FFTs, obtaining {UkVk}^~Q. Finally, take the inverse FFT of {UkVk}j^~Q and divide by N. Since the two FFTs and the inverse FFT each take (3/ 2 ) N log2 N real multiplications, and multiplying each element of the two FFTs takes 4 N real multiplications, we find that the whole

4. SOME APPLICATIONS OF FOURIER SERIES

126

process takes (9/ 2)A^log2 + 6A^ real multiplications. This indicates how effi­ cient the FFT method is, as compared to a direct computation, which would take 4N^ real multiplications. For instance, when N = 1024, the FFT method takes 80 times less multiplications than a direct computation.

4.8

Kernels for Some Common Filters

In this section we will examine kernels (PSFs) for the Cesàro, dlVP, banning, and Hamming filters. To begin our discussion we must first examine the kernel for an unfiltered Fourier series, (cn = L

^ n--oo

i

dx\

\

'

for a function / having period I n . For simplicity, we shall assume that allfunctions discussed in this section have period In ; this results in no loss of generality. The partial sum S m for the Fourier series above is M S m {x ) = ^ c„e'"

(4.106)

n= -M

This partial sum can be thought of as a filtering of a Fourier series, in which case its kernel D m is defined by M D m {x ) =

(4.107)

Y . n=—M

Using this kernel, we have S m {x ) = f * D m (x )

= ^ y*

f (s) DM( x - s)ds.

(4.108)

A closed form expression for D m can be found. First, we split the sum for D m in (4.107) into two sums M d m {x )

=

M

+ n—O

■ n=\

(4.109)

127

4.8. KERNELS FOR SOME COMMON FILTERS

Using Formula (2.7) from Chapter 2, with M in place of N — 1, in place of r for the first sum in (4.109), and e~'^ in place of r for the second sum in (4.109), we obtain \ _

D m {x ) =

^ - i x _ ^-i{M + \)x

1

+ -

\ — e~

_ ^ i{ M + \)x _j_ ^iM x _ ^-i{M-\-l)x

^-iM x

2 cos M x — 2 cos (M + l)jc 2 —2 cos X

(4.110)

Thus, D

m

{x )

cos M x — cos (M 4-1):^ 1 —cos X

(4.111)

Formula (4.111) can be further simplified using trigonometric identities. First, since / 1 1\ cos M x = cos I M + - — - I V 2 2; it follows that / 1 / . 1 cos M x = cos I M + - \ x cos -X + sm I M + - x sm - x \ 2J 2 \ 2) 2 and, similarly. / 1 / . 1 cos {M + l)x = cos I M + - I X cos -,x —sin i M + - I x sm

Hence, by subtraction cos M x — cos (M + l)x = 2 sin

(4.112)

For M = 0, Formula (4.112) becomes 7. i 1 —cosx = 2 sin - X .

2

(4.113)

128

4, SOME APPLICATIONS OF FOURIER SERIES

Combining Formulas (4.11 1) to (4.113) yields the following closed form for D m : D m (x )

=

sin (M + ^)jc sm ^jc

(4.114)

Formula (4.114) is the classic expression for Dirichlefs kernel D m - Another expression that is more amenable to graphing with FAS is D m {x ) = (2M + 1) sine [(M + 0 .5)x / 7t]/ sine (O.Sjc/ tt)

(4.114a)

where sine is the function defined in (4.21). Graphs of Dirichlet’s kernel using Formula (4.114a) are shown in Figure 4.17.

(a) 10 h a m o n i c s X interval: [ X I --3 3 .1 4 , 3.141 X increwmt = ¥ interval: [ -10. 301 ¥ increMnit = 4

(b) 21 haraonics

.628

x interval: [ -3 .1 4 . 3.141 X increMnt = ¥ interval: i -2 8 . 601 ¥ increment = 8

.628

FIGURE 4.17 Dirichlet’s kernel. To derive the kernel for C es^o filtering, we use Definition 4.1 and For­ mula (4.108) to obtain

g m {x )

=

1

M -l ^ / * I^kix)

■/■(iE Dkix)

(4.115)

Denoting the kernel for C es^o filtering by C m , we have from (4.115) M -l C m {x ) = — ^ Dkix) *=o

(4.116)

m ( jt + ^ ) . Use this formula in combination with (4.114a) to draw graphs of Hamming’s kernel for M = 8, 16, and 32 harmonics. Compare these graphs with the ones obtained as solutions to Exercise 4.31(d). 4.41 Using (4.124), graph Ces^o’s kernel for M = 8, 16, and 32 harmonics. Compare these graphs with the ones obtained as solutions to Exercise 4.31(a). 4.42 Using FAS, graph the banning and Hamming filtered Fourier series, using 32 har­ monics, of _ i1 for0 < x: < 1 for-l 0 . 5

(5.3)

then

rect(w)

/

0.5

-ilnux dx =

-0.5

—i l n u

a—ilnux x=0.5 —iln u

x= —0.5

nu

The transform of rect(jc) is therefore sinc(M). Using the notation/(.x) — to denote the Fourier transform operation, we have rect(jc) — ^ sinc(M).

See Figure 5.1. Example 5.2: Suppose that / is defined by

f(x) =

e 0

for JC > 0 for JC < 0.

f { u)

(5.4)

5,1. INTRODUCTION

H— ^— !— h

X interual: [ -16, y interual: [ -.5,

163 1.51

H--- 1— 1— h

X increment = 3.2 Y increment = .2

X interual: [ -16, Y interual: 1 -.5,

161 1.51

X increment = 3.2 Y increment = .2

FIGURE 5.1 (a) Graph of the rect function, (b) Graph of its Fourier transform.

then -27tx{\+iu) -27TX — H ttux

/(« )

dx =

—2jr{\ + iu)

x=0

1 2;r(l + iu) since e

q

^

Similarly, as the reader may show, the function for X > 0 for X < 0 has Fourier transform g(u) =

I n (I — iu)

Example 5 3 : This example will show that -271 \x\

1

1

7T 1 +

(5.5)

5. FOURIER TRANSFORMS

152

SOLUTION

We need to evaluate the Fourier transform integral

/

OO

^-27T\x\^-i2nux

-OO

Splitting the integral into a sum of two integrals, we have

/

OO

poo

^ - 2n\x\^-i 2nux

/

*00

^ - 2nx ^

- i 2 t tu x

t/0

^ j

pO

^ 2jtx^-i 2nux

—c

= /( « ) + !(« ) where / and g are the functions defined in Example 5.2. Using the results of Example 5.2 for / and g, we have n

e-'^^We-i'^Ttux

^ ------ 1------ ^ ------- 1------ ^

J_o o

27t ( 1 + /

X interual: [ -5j 5] X increment = 1 y interval: [ -.5, 1.51 V increment =

FIGURE 5.2 (a) Graph of

m)

2 n ( l — iu)

Example 5.4: Find the Fourier transform of

Mi

w^

X interual: [ -5, 51 X increment = 1 V interval: I -.5, 1.51 V increment =

.2

(b) Graph of its Fourier transform.

A(^)

7t 1 +

- 1-^1

for |jcl < 1 for \x\ > 1.

.2

153

5.1. INTRODUCTION

SOLUTION Since A = 0 outside of the interval [- 1 ,1 ] and A is an even function, we have A(m ) =

j (I —\x\)cosItvuxdx —i j (I —\x\)sinlnuxdx x)cos27tuxdx.

= 2 f\l- . Jo

Performing an integration by parts with this last integral, we obtain A(u)

= —1

7TU J o

sin^TTM

Thus, A(x)

. sin ljTuxdx =

/ SI

coslnu

1—

27T^M^

= sinc^(M).

I

sinc^(M).

Example 5.5: We will calculate the Fourier transform of Fourier transform yields /( « )

f(x) =

f - [

J —OO

/

. The definition of

(5.6)

dx.

= /'

If we differentiate both sides of (5.6) with respect to under the integral sign, we get

du =

2

u, then by differentiating

^

oo

“

i27zxe-^’^ \ - ‘^^“^ dx

-00

(5.7) =

dx.

154

5. FOURIER TRANSFORMS

(Note: the method of differentiating under the integral sign is a common technique. For further discussion, see [Wa, Chapter 6.2].) Integrating by parts in the last integral in (5.7), we obtain

du

= le

POO

x-^oo ■nx^2 x ^ —oo

J —OO

= —2n u Ir ^ —n x ^ ^ —ilixux dx. J —a Thus, = —i Tt uf i u) .

du

(5.8)

Solving (5.8) for the unknown function / , we get f ( u ) = f(Qf)e-

(5.9)

It remains to find /(O ), where dx.

/(O ) == /

(5.10)

- f -OO

Squaring both sides of (5.10), and replacing the dummy variable x by y in one of the integrals, we have OO

POO

/ -Cl

dx /

-OO

dy

J —OO

g-^(^^+y^)dxdy.

(5.11)

Changing to polar coordinates, we get

/

OO

-OO

POO

/

J —oo

^-7T(x^+y^) d x d y = f

r2n POO f e“ ” '' r dr dO

Jo

= Jo r

Jo

[L27T -

r=0

de = 1.

155

5.2, PROPERTIES OF FOURIER TRANSFORMS

Thus, on returning to (5.11), we see that /(0 )^ = 1. Since /(O ) > 0, as we can see from (5.10), we must have /(O ) = 1. Using this fact in (5.9) we have (5.12) Formula (5.12) expresses the remarkable fact that e (as a function of u).

5.2

Fourier transforms to itself

Properties of Fourier Transforms

This section covers some important properties of Fourier transforms, for exam­ ple, how they behave with respect to shifting, scaling, and differentiation. We begin with the following theorem. TH EO REM S.!: The Fourier transform operation f f has the following properties: (a) Linearity: For all constants a and b, a f + hg — ^ a f -h bg. (b) Scaling: For each positive constant p, f

(?)

T

pfipu)

f{px)

and

-M

(c) Shifting: For each real constant c, f i x - c)

f ( u ) e —il n c u

(d) Modulation: For each real constant c. f i x ) e ' ^ ”^^

f i u - c).

PROOF The proof of (a) is straightforward, so we leave the details to the reader. To prove (b), we make the change of variables s = jc/p in the following Fourier transform integral:

156

5. FOURIER TRANSFORMS

f

^ ^

J°° f

dx = p

/

dips)

OO

f( s ) e ~ ‘'^’^^P"^‘’ ds = p f i p u ) .

-OO

Thus, f ( x / p ) — > p f ( p u ) .

Substituting 1/p in place of p, it follows that

f(px) ( l/p ) /( w /p ) and (b) is verified. To prove (c) we make the change of variable ^ = x —c in the following Fourier transform integral:

/

f i x - c)

OO

^oo

fix -

dx = /

ds

J—oo

-OO

/

OO

fis)e~'^^"Us

= f i u ) e ■1271cu

-OO

Thus, (c) holds. To prove (d), we note that ^i27rcx^-i27TMx _ ^ - i 2n{u-c)x^ hence f ( x ) e ,i2ncx

^

J—OO and (d) holds.

I

Here are some examples of Theorem 5.1. Example 5.6: (a) Find the Fourier transform of (1! p)e~^^^f

SOLUTION

Using scaling, linearity, and e

for each p > 0.

e

P

(b) Find the Fourier transform of rect(x —4) + rect(jc + 4).

we have

157

5.2. PROPERTIES OF FOURIER TRANSFORMS

SOLUTION

Using shifting and linearity, we have rect(u)e

rect(jc —4) + rect(x + 4) Since rect(w) = sine (u) and e

+ rect(w)^^^^^“.

= 2 cos Sttm, we have

rect(x —4) + rect(x + 4) — ^ 2 sine (u) cos(87rw).

(c) Find the Fourier transform of e

cos Gnx.

SOLUTION Rewriting cos 6jtx as (1/2)^^^^^^ + ( 1/ 2)^“ ^^^^^ and then using linearity and modulation, we get

Fience, e ^^'^'cosdTTx

I k 1 + (w —3)^

(d) Find the Fourier transform of e SOLUTION

1 2:zr 1 + (w + 3)^

+

for all p > 0.

Using the scaling property, we have ^ -2

t t p \x

lU ^ V p 1 + ( I )'^J

\

u p

Hence, ^-2np\x\

T

I

P

7t

-\-U^

ip > 0).

I Another useful property of Fourier transforms is the conversion of differentiation into multiplication by the transform variable. To be more precise, we have the following theorem.

158

5. FOURIER TRANSFORMS

TH EO REM S.!: (a) Suppose that ||/ ||i and ||/ '||i are finite and that f is continuous, then , T f (x) — i l n u f i u ) , (b) Suppose that l |/ ||i and \\xf(x)\\i are finite, then f is differentiable and T i xf(x) f\u). 2n

PROOF

(a) Using integration by parts, we have JC->00 oo

J —OO

■I

-\-i27tu I

f(x)e

dx.

By the Fundamental Theorem of Calculus, we have f(x) = / q f'(s)ds + /(O). Consequently, the limits lim;c-^oo f ( x) and lim^^-cx) f(x) exist and are finite. If either of these limits were not 0 , then

/

oo

poo

f(x)dx= I

Jo

-oo

pO

f(x)dx+ I

f(x)dx

J —oo

would not converge (since at least one of the integrals on the right side would diverge). Therefore, both limits are 0 and

/'(x )

i2nu i

f(x)e~^^^^^dx = i27tu f(u)

J —oo

and (a) is proved. To prove (b) we differentiate under the integral sign as follows f\u) = — / du

= /'

f ( x ) e -^ ^ -- ^’ dx

-i 2n x f { x ) e

dx.

But, this shows that - ilnxfix)

(5.13)

f'iu).

Multiplying both sides of (5.13) by i ¡ I ti yields (b).

I

5.2, PROPERTIES OF FOURIER TRANSFORMS

159

Example 5 J : (a) Using Theorem 5.2(b), we have ^ ^ r I n du I

xe — TTX^ hence

T

—

J

-lue

Thus, xe transforms into —i times itself (as a function of w). (b) Using the previous example, and Theorem 5.2(b), 2 T

i d --------—m e I n du I

J

Hence, 2n

(c) From (b), it follows that

Thus, \x^ — ll{An)]e of u). REMARK 5.1:

Fourier transforms into —1 times itself (as a function

Bringing absolute values inside the integral sign and using \^— i27tUX\ __ j|

we obtain l/ (M )l


0): dF _ 2 ^ F dt ^ dx^

(heat equation, diffusion constant a^)

(5.67a)

F( x , 0) = f ( x )

(initial condition).

(5.67b)

SOLUTION F(u, t) by

We proceed similarly to the solution of (5.42a) and (5.42b). Define

/

oo

F{x,

dx.

(5.68)

F {u ,t)e ‘^^"^ du.

(5.69)

-o o

Hence, by Fourier inversion, F ( x ,o =

r

j —(

By differentiating under the integral sign, we have 3 F( x , t )

9 3^F{x, t) ------------ =

dF{u, t) Iilnux du dt

f

(5.70)

J-OO Substituting the integrals from (5.70) back into (5.67a) and rearranging terms into one integral we have

/:[

—

at

+ (27ta)^u^F(u, t) e‘^^“^ d u ^ 0 .

(5.71)

5.5. THE CONVOLUTION THEOREM

179

Since the Fourier inverse of the function in brackets in (5.71) is the 0-function, we conclude that the function in brackets is also the 0-function. Therefore, dF( u, t ) , 2 -f {iTvaYu'^Fiu, t) = 0. St

(5.72)

Now, by putting i = 0 in (5.68), we conclude from (5.67b) that F( u, 0) = f ( u) ,

(5.73)

Hence, our transformed problem is 3F(u, t) = —{2nd)^u^F(u,t) St F ( m,0) = / ( w).

(5.74a)

(5.74b)

Fixing a value of u turns (5.74a) into an ordinary differential equation dP dt

= —(27ta)^u^F

which has solution =

(5.75)

Putting i = 0 in (5.75) and using (5.74b) yields A = f (u). Hence,(5.69) becomes

/ OO -CX)

Now, we can apply the convolution theorem (Theorem 5.7) to express F ( x , t ) as a convolution. In (5.76) we have a Fourier inversion of a product of two transforms. T These transforms are / — ^ / and

1 (Ana^ty^ Thus, by Theorem 5.7, our solution F{x, t) to (5.67a) and (5.67b) is F ( x ,0 = / * t H{x)

(5.77)

5. FOURIER TRANSFORMS

180

where ___ ^—7tx^/(4jTa^t) , H( x) ----------^ (4Tca^t)2

(5.78)

I It is easy to calculate computer approximations to (5.77) using FAS. For example, suppose th at/(jc) = 20 rect(x/2) anda^ = 1. We will use for our second function i __ g - x ^ / m (47tt)2

for t = 0.5, 1.0, and 2.0. If we use 1024 points and the interval [—8, 8], then we obtain the graphs shown in Figure 5.10. These graphs can be interpreted as the evolution of temperature, from the initial temperature f { x ) = 20 rect(x/2), through a homogeneous insulated rod (or a thin wire). Here is another application of convolution. Example 5,12: In probability theory, a probability density function (p.d.f.) is a non-negative func­ tion / satisfying f ( x ) d x = 1. For example, f ( x ) = { X j ^ ) e ~ ^ is a p.d.f. For a p.d.f. there is associated a cumulative distribution function (distribution, for short) defined by F(x) = f ( s ) ds. Graph the distribution F(x) for the p.d.f. fix) = SOLUTION We leave it as an exercise for the reader to show that F(x) / * 7f(x) where H is the Heaviside function defined by U(x)

i;

if X < 0 if X > 0.

(5.79)

By doing a convolution of f ( x ) = ( l l ^ ) e ~ ^ with H (x), over [—16, 16] using 1024 points, one obtains the graph shown in Figure 5.11(a). For reasons which we will explain below, there is an undesirable behavior at the ends of the interval [—16, 16] (i.e., the graph obviously decreases near ±16, but F(x) decreases, 2 because its derivative is ( X j ^ ' ) e ~ ^ , which is always positive). Nevertheless, if we change the x-interval to, say, [—10, 10], then we get the graph shown in Figure 5.11(b), and this graph is a good approximation to the distribution F(x) over the interval [—10, 10]. (See Exercise 5.28 for a treatment of the accuracy of this approximation.) I

5.(5. THE CONVOLUTION THEOREM

H— I— I---- h

(a) t = 0 X interual: [ ¥ interual: [ -

(c) t = 1.0 X interual: [ -8, ¥ interual: [ -5,

81

H----1--- h

251

X increment = 1.6 Y increment = 3

81 251

X increment = 1.6 ¥ increment = 3

(b) t = 0.5 X interual: [ -8, ¥ interual: C -5,

(d) t = 2.0 X interual: [ -8, ¥ interual: [ -5,

FIGURE 5.10 Graph of a solution F( x , t ) to (5.67a) and (5.67b) for t

81 251

81 251

X increment = 1 .6 ¥ increment : 3

X increment = 1.6 ¥ increment = 3

0,0.5,1.0, and 2.0.

REMARK 5.4: As we saw in Example 5.12, the computer approximation of the convolution of (Xf with l~L{x) has some undesirable behavior at the ends of the interval [—16, 16]. The reason for this is that, as we showed in Section 4.7 of Chapter 4, an FFT algorithm is used for approximating the periodic (or cyclic) convolution / * g(x) = - T f ( s ) g ( x - s ) d s 2l J _ l

(5.80)

where f and g are periodic functions with period 2L (in the example above, L = 16). What is graphed in Figure 5.11(a) is 2L = 32 times the periodic convolution in (5.80), where / and g are the periodic extensions of the restrictions

182

5. FOURIER TRANSFORMS

X interual: [ -16j 161 y interual: [ -2, 21

X increaent = 3.2 V increaent = .4

(b) X interual: [ -10, 101 Y interual: [ -2, 21

X increment = 2 Y increaent = .4

FIGURE 5.11 (a) FAS computed convolution of (1 fy/7r)e~^ and the Heaviside function over [—16,16] using 1024 points, (b) Approximate distribution for the probability 2 density function (p.d.f.) .

(a) X interual: [ -16, y interual: [ -.5,

161 1.51

X increaent = 3.2 y increaent = .2

(b) X interual: [ -16, y interual: [ -.5,

161 1.51

X increment = 3.2 y increment = .2

FIGURE 5.12 (a) Graph of components of cyclic (periodic) convolution of (1 and the Heaviside function over [—16,16] for x = 15.75. (b) Graph of the same components for the convolution over R for jc = 15.75.

to [—16,16] of ( l l y / 7t)e~^ and H (x). To see why the troublesome behavior near the endpoints ±16 occurs, we show in Figure 5.12(a) what the components of (5.80) look like for x = 15.75. From Figure 5.12(a) it is clear that there will only be a partial overlap of ( I /^ / tt)^“ ^ and not the more complete overlap which the non-periodic convolution would have [as shown in Figure 5.12(b)]. However, as

5.6. THE CONVOLUTION THEOREM

X intcrual: [ -16, Y interual: [ -.5,

161 1.51

X interual: I -16, Y interual: [ -.5,

X increnetit = 3.2 Y increment = .2

161 1.51

X increment = 3.2 Y increment = .2

FIGURE 5.13 ^ (a) Graph of components of cyclic (periodic) convolution of and the Heaviside function over [—16, 16] for x = 9.75. (b) Graph of the same components for the convolution over R for x = 9.75.

we show in Figure 5.13, the periodic and non-periodic overlaps will be much closer when X = 9.75 (due to ( X j being so near 0 when |jc| > 6.25). I As a final application of the convolution theorem, we shall sketch the proof of Parseval’s equalities. THEOREM 5.9: PARSEVAUS EQUALITIES T I f f — > f , then

/ oo

f

\f(u)\'^du.

(5.81a)

f ( u) g*( u)du.

(5.81b)

-OO

.F Furthermore, if g — ^ g, then

/

oo

-oo

poo

f(x)g^(x)dx= / J —oo

PROOF If we put g equal to / , then (5.81b) implies (5.81a). Therefore, it remains to prove (5.81b). The convolution theorem (Theorem 5.7) also holds for

5. FOURIER TRANSFORMS

184

the inverse Fourier transform (by a simple change of variable). Therefore, we have

f

dx = ( / * g*){u)

/ OO ^

^ f { v ) p ‘( u - v ) d v .

(5.82)

-OO

Furthermore, g*(v)

dx

= J— r *oo OO

a

\ * g{x)e‘^^^^ d x \

Hence we can rewrite (5.82) as

/

poo

OO

f{x )g * {x )e -'^ ^ “^ dx = /

f {v)S*(v - u) dv.

(5.83)

j — OO

-OO

Putting M= 0 in (5.83) yields

/

poo

OO

f(x)g*(x)dx=

and the theorem is proved.

5.7

f ( v ) g*( v ) dv J—OO

-OO

I

An Application of Convolution in Quantum Mechanics

In this section we will discuss an example from quantum mechanics which makes use of the convolution theorem. Recall from the discussion at the beginning of section 4.3, that the potential-free version of the one-dimensional Schrodinger’s equation can be expressed as dxlr

iti

dt

2m dx'^ '

5.7. CONVOLUTION IN QUANTUM MECHANICS

185

We now show how to use the convolution theorem to solve a problem involving this version of Schrodinger’s equation. Example 5.13: Draw graphs that approximate |V^p, where ifi 2m dx^

satisfies

(Schrodinger’s equation)

(5.84a)

(initial condition)

(5.84b)

i r ( x , 0) = f ( x )

for the function f ( x ) = rect(jc) and times i = 0.1 and 0.25 sec. Assume that m is the mass of an electron. SOLUTION We will make use of the solution found for Example 5.11. To make use of this solution, we let stand for the constant factor on the right side of Equation (5.84a). Thus, we can express the problem above as follows: ,

( a^ = — \ \ 2m /

f i x , 0) = f i x ) .

(5.85a)

(5.85b)

Therefore, using the Solution (5.76) found for (5.67a) and (5.67b), but now replac­ ing by ill/ 2m, we get

/ OO fix -OO

Furthermore, instead of (5.77) and (5.78), we have ^ { x j ) = / * tP{x)

(5.87)

where

1

tF{x) = ------------------ ^

(5.88)

[4jr(//i/2m)i]2 We will now show how to approximate |V^(x, i)P using Formula (5.86). If we choose L sufficiently large, then we can approximate the integral by the

5. FOURIER TRANSFORMS

186

integral

obtaining

(5.89)

Now, if f ( x ) = rect(x), then f ( u ) = sine (u). Substituting sine (u) in plaee of f ( u ) in (5.89) and writing terms of its real and imaginary parts, we get

V^(x, 0

j

sine (u) jeos ^^(27rw )^i^ —i sin , 2m ) \ 2m

du.

|

(5.90) The integral in (5.90) is a filtered Fourier transform over [—L, L], using a pos. ex­ ponent. The filter funetion has real part eos[(/z/2m)(27TM)^i] and imaginary part —sin[(:^/2m)(27rw)^i]. Sinee (5.90) deseribes a Fourier transform operation, it follows that IV^(x, i)P is a power spectrum. Therefore, to graph | we perform the following steps. First, we ehoose to do the Fourier transform proeedure and ehoose Complex for the type of funetion. Seeond, we ehoose an interval [—L, L] and ehoose the number of points. For this example, we shall use L = 32 and 4096 points. Third, we speeify the real and imaginary parts of the funetion to be transformed. In this ease, we enter f { x ) = sine (x) and f { x ) = 0, respeetively. Fourth, we ehoose to do a Power spectrum with pos. exponent. And, we press y when asked if we want to apply a filter and ehoose to do a Complex filter. For the real part of the filter, we enter the following formula {Note: for an eleetron fi!2m = 0.578): f { x ) = eos[0.578(27Tx) A 2i] \i=0.1

(5.91)

and for the imaginary part of the filter we enter the following formula: f { x ) = —sin[0.578(2:;rjr) A 2 i] \i=0.1.

(5.92)

After transforming, a graph of |V^(jc, 0.1)p is produeed [see Figure 5.14(a)]. Repeating this proeedure for t = 0.25, we obtain the graph of |V^(x, 0.25)p, whieh is shown in Figure 5.14(b). I Let’s now examine a eouple of theoretieal points. First, we will prove the following theorem. This theorem says that if the initial state V^(x, 0) = f { x ) generates a p.d.f. |/ ( x ) p , then eaeh subsequent state il/(x,t) also generates a

187

5.7. CONVOLUTION IN QUANTUM MECHANICS

(a) t = 0.1 X interval: I -10. Y interval: i -.5,

101 1.51

FIGURE 5.14 G raphs of

(b) t = 0.25 X interval: [ -10, Y interval: I -.5.

X increment = 2 Y increment = .2

103 1.53

X increment = 2 Y increment = .2

OP for Example 5.13.

THEOREM 5.10: Suppose xj/ satisfies (5.84a) and (5.84b). Then, fo r all t-values, oo /*oo

/

I \f (x)\ dx.

\ir(x, t ) f d x

In particular, i f j ^ ^ \ f ( x ) \ ^ d x = 1, then

(5.93)

\xl/(x, t)\^ dx = I fo r all t-values.

PROOF To prove (5.93) we use Parseval’s Equality (5.81a) twice. Looking at (5.86), we see that it is the Fourier inversion integral corresponding to the fol­ lowing Fourier transform operation (for each^jc^J i-value):

Hence, by Parseval’s Equality (5.81a),

/ OQ -OO

poo 7 —0 0

- r

\f(u)rdu.

(5.94)

We used the fact that \e^^\^ = 1 when

• oo.

1 4 7 r( ft/ 2 m )i

/ (

-

)

(5.95)

The approximation in (5.95) is an important one. It shows, for instance, that the complicated derivation of |V^(x, i)P given in Example 5.13 can be replaced by the simpler operation shown on the right side of (5.95). That is, for large enough values, we only need to calculate a scaled version of sinc^. [For instance, \ir(x, 0.25)1^ ^ (1/1.816) sinc^(x/1.816), and this approximation appears to be valid based on Figure 5.14(b).] It is worth noting that is a then both sides o f (5.95) are p.d.fs. That i)P is a p.d.f. follows from Theorem 5.10. That (l/[47r(^/2m )i])|/(jc/(47r(]^/2m )0)P is a p.d.f. can be proved by Parseval’s Equality (5.81a) and a change of variables (we leave this to the reader as an exercise). Formula (5.95) is analagous to Fraunhofer diffraction in optics [see Chapter 6, section 6.3, especially formula (6.60)]. If m equals the mass of an electron, for instance, then Formula (5.95) says that electrons diffract in the same way as light waves do when the light undergoes Fraunhofer diffraction. To prove (5.95) we rewrite (5.87) and (5.88), obtaining f(xj) =

1

r f f(s)e [4jT(ih/2m)t]'^ •'-00

-n(x—sr/(47i:tifi/2m)

ds.

(5.96)

To simplify notation, we define q(t) by q(t) = 47ti{hf2m)t Later, we shall make use of the fact that l / q(t ) ^ 0 a.s t exponential inside the integral in (5.96), we get ^ - 7 i { x - s f /q{t) _

(5.97) oo. If we expand the

lq{t)^-Tis^¡q{t)^iTtxs!q{t)^

(5.98)

Now, suppose that for some positive number P, f(s)=0,

for|i|>5 P

(5.99)

which certainly is true for the function f ( s ) = rect(5) in Example 5.13. If we now let i ^ oo, we will have for all |^| < as soon as t is sufficiently large -7is^/q{t)

^ gO ^ j

(5.100)

5.7. CONVOLUTION IN QUANTUM MECHANICS

hence

189

-7T(x-s)^/q(t) ^ ^-7ZX^¡qit)^litxslqit)

(5.101)

Using (5.101) and (5.99) back in (5.96), and also using the definition of q(t), we have

(t \2 JJ-hi q{ty^ hP

q{ty J-jp Factoring e

outside of this last integral yields ^-TCX" -TTX^fqit) q(t)

n \P

q{t )-2

J-\P

And, using (5.99) again, we have ^-nx^lqit) poo ds. q{t )2

J-oo

Replacing q(t) by 47ti(fi/ 2m)t inside the integral yields p -n x^ lq if)

poo

q{t)2

J-oo

xlr(x, t) ^ -------- p - /

Thus,

/(^)^-^-27r.[x/(4;r(/^/2m)0]

^-nx^lqit)

^

\ An{ f i / 2m)t

q{t)h And, since

/

= 1 for all real (¡>, we have

Therefore, we obtain \f{x,tW

\q{t)\

f ( — i— ) \ 47t(fi/ 2m)t J

\

5. FOURIER TRANSFORMS

190

And replacing

by ATc{fi/lm)t yields

\ir{x, o r

1

A n(fi/ 2m )t

f

)

— \47t(fi/2m)t J

for sufficiently large t. Thus, we have demonstrated (5.95). We obtained this asymptotic form for i)p using the assumption in (5.99). But, the argument above still applies if f ( s ) 0 as 1^1 —^ oc rapidly enough (see, for example. Exercise 5.34). Actually, if we interpret the approximation in (5.95) as meaning a small difference in 2-Norm, then (5.95) holds for all functions / for which H/II2 is finite, but we will omit the proof.

5.8

Filtering, Frequency Detection, and Removal of Noise

In this section, we introduce the concept of filtering a function before Fourier transforming it. This technique is widely used in signal processing. Let’s begin with an example. Example 5.14: Suppose the function f { x ) = cos(0.87Tjc) is Fourier transformed over the interval [>-32, 32] using 1024 points. The result is shown in Figure 5.15(b). Notice that the frequency 0.4 that characterizes f ( x ) is clearly marked by a peak at the point 0.4. However, there is a very noticeable oscillation about the jc-axis away from this peak. This phenomenon is known as leakage. Leakage occurs because the FFT method yields approximations of (see Section 5.4): fe=512 -512

The frequency 0.4, however, does not equal k / 6A for any integer k. This failure to realize a precise frequency location results in leakage. That is the explanation on the frequency side. On the other hand, from the time (jc-variable) side, if we look at the graph of /(jc) in Figure 5.15(a) we see that it does not fit very well in the window determined by —32 < jc < 32. In fact, its periodic extension beyond [—32, 32] no longer represents the actualfiinction / ( jc). See Figure 5.16(a). When this occurs, it is said that the effect o f the window becomes visible. In fact, we can see that the pattern of the transform in Figure 5.15(b) looks a lot like the transform of rect (jc/64).

5.8. HLTERING, FREQUENCY DETECTION, NOISE REMOVAL

(a) cos(0.8nx) X interual: [ -32. 32] y interual: [ -2, 21

X increment = 6.4 Y increment = .4

(c) cos(tix) X interual: [ -32, 321 Y interual: [ -2, 21

X increment = 6.4 Y increment = .4

191

(b) transform of cos(0.8lrx) X interual: [ 0, 11 X increment = Y interual: 1 -10, 401 Y increment =

(d) transform of cos(nx) X interual: [ 0, 11 X increment = .1 Y interual: 1 -10, 401 Y increment = 5

FIGURE 5.15 Fourier transforms of two cosine factions over the interval [—32, 32] using 1024 points.

To see this last point more clearly, it helps to look at the function g(x) = cosTtx [see Figure 5.15(c)] for which leakage does not occur. As we can see in Figure 5.15(d), the graph of g(u) has a peak centered on the frequency 0.5 and no leakage. For this example, A:/64 = 0.5 has the solution A: = 32 so the frequency 0.5 belongs to the set {k/ 6A ]^ } ^ ^ 2' the other hand, as we show in Figure 5.16(b), the periodic extension of g(x) beyond [—32, 32] still represents the function g(jc). And, in this case, no window effects are visible. Based on these considerations, a standard method of reducing leakage is to multiply the original function by a function that damps down to 0 at the edges of the X-interval and whose transform does not exhibit the oscillations that the transform o f rect (x/64) does. For example, if we transform the function h{x) = c o s (0.87 t x )[0.5 + 0.5 c o s (7T x /32)], then the transform h shown in Fig­ ure 5.17(b) results. We say that h(u) is a banning filtered Fourier transform of

5. FOURIER TRANSFORMS

192

(a) periodic extension of cos(0.8nx) X interual: [ - M , 641 X increment = 12.8 Y interual: t -2, 21 Y increment = .4

(b) periodic extension of cos(tix) X interual: i -64, 641 X increment = 12.8 Y interual: t -2, 21 Y increment = .4

FIGURE 5.16 Periodic extensions, period 64, of two cosine functions (defined initiaily on [-3 2 , 32]).

-

_____ 1 1—- / 1 l_ L __ 1 1_____ 1 1_____ 1 1_____ 1 1' - 1_____ 1 [_____ 1 1_____ 1 1_____

(a) cos(8.8nx), banning filtered X Interual: t -32, 321 X increment = 6.4 Y interual: [ -2, 21 Y increment = .4

(b) banning filtered transform X interual: [ 8, 11 X increment = .1 Y interual: t -12, 241 Y increment = 3.6

FIGURE 5.17 A banning filtered Fourier transform.

f ( x ) . Comparing Figures 5.15(b) and 5.17(b) we can see that by banning filtering we have significantly reduced leakage. A filtered Fourier transform is defined as follows. D EFINITION 5.4: IfWfWi is finite and F is a bounded, continuous function, then the filtered Fourier transform o f f (using the filter function F) is

f

F i s ) f i s ) e - ‘^^“' ds.

5.8. FILTERING, FREQUENCY DETECTION, NOISE REMOVAL

193

When we use FAS to graph filtered Fourier transforms, we employ the following approximation (if [—L, L] is the chosen interval):

/

rL

OO

J

F (s )f( s)e~‘^^“^ds.

(5.102)

-OO

When using the interval [—L, L], some filters that are often used are Filter name FilterfunctioHy F{s)

Cesaro banning Hamming Parzen Welch

1 - \s/L\ 0.5 + 0.5 cos{ns/L ) 0.54 + 0.46cos(jr5/L) \-[NI{NF\)]\s/L\ 1 - [ N / i N + l)]2(i/L )2

{Note: the parameter N in the Parzen and Welch filters denotes the number of points used by FAS.) It is also possible to use FAS to design your own filter function. For instance, suppose you want to apply the filter F{s) = You would then press y when FAS asks you if you want to apply a filter. Then, select User from the filter menu, and enter the formula f { x ) = exp[—abs (jc)]. A graph of the filter will be drawn. Then FAS will draw a graph of the product of this filter with your original function. This product function will then be Fourier transformed to obtain the filtered Fourier transform. NOISE SUPPRESSION One application of filtering is noise suppression in signal processing. In this case, the filter function multiplies the Fourier transform of the signal and then an inverse Fourier transform is performed. Let’s look at an example. Example 5.15: Suppose a pulsed signal having a single fundamental frequency is transmitted, say a signal of the form ^(x) = (8 cos IOOttx)^ (5.103) for —0.5 < X < 0 .5 . See Figure 5.18(a). There are many situations where such signals are transmitted, such as sonar, or radar, or when a single bit of data is transmitted by a modem. The signal in (5.103) has a fundamental frequency of 50 and amplitude 8. Its Fourier transform is shown in Figure 5.18(b). Notice that the transform has two prominent spikes located at ±50, corresponding to the fundamental frequency of 50 in the signal. Notice also that the transform is localized at ± 50 in the sense that it is essentially 0 except near the points ±50.

194

5. FOURIER TRANSFORMS

H----^--- 1--- h

H---1— ^--- h

Y interual: [ -13,

131

Y increnent =

2.6

X interoal: [ -512, 5121 X increnent = Y interual: I -2, 23 Y increnent = .4

102.4

FIGURE 5.18 (a) Transmitted signal, (b) Fourier transform of the signal, real and imaginary parts.

Now, when the signal s(x) is received it may be corrupted by noise. Letting n{x) stand for such noise, we can represent the received signal / ( x ) as f ( x ) = (cos

+ n{x).

(5.104)

See Figure 5.19(a) for a graph of such a noisy signal. The term (cos IOOttjc)e~

H— ^— h

X interual: [ -.5, .51 Y interual: [ -3, 31

X increnent = .1 Y increnent = .6

X interual: t -512, Y interual: C -.02,

H— I— h

5121 .081

X increnent = Y increnent =

102.4 .01

FIGURE 5.19 (a) Received, noisy signal, (b) Power spectrum of noisy signal; the spikes are located at ±50. represents the desired portion of the received signal; it has been obscured by the undesired noise n{x). This amplitude of the desired signal is also 8 times smaller

5.8. FILTERING, FREQUENCY DETECTION, NOISE REMOVAL

195

than the amplitude of the original signal (such diminution of amplitude is common in signal transmission). Its fundamental frequency, however, is still 50 (here we are assuming that neither frequency shifting nor phase shifting has occurred; those possibilities are treated in the exercises). The noise term n(x) that we used in Formula (5.104) is a simulation of white noise. For further discussion of this type of noise, and a description of how we created the noise term in (5.104), see the subsection entitled WHITE NOISE below. Now, looking at the signal shown in Figure 5.19(a), it is hard to imagine how one could be certain that the transmitted pulse has been received. If, however, we compute the power spectrum |/(w )p of the received signal, then we produce the graph shown in Figure 5.19(b). Notice that there are two spikes standing out clearly from the rest of the graph. These spikes are located at the positions ±50, they tell us that the received signal contains a fundamental frequency o f 50. Having identified that there is a frequency of 50 present in the noisy signal we can try to reconstruct the original signal (this is not always necessary, sometimes it is only necessary to detect the presence of the fundamental frequency). To reconstruct the original signal, we shall perform a filtering operation on the Fourier transform of the noisy signal. This transform is shown in Figure 5.20(a). First, we choose to do a complex Fourier transform (choosing a pos. exponent so that we are doing

(a) X interual: [ V interval: I

12

, 512]

.25,

.25]

X increment = Y increment =

102.4 .05

Cb) .5] X increment = .1 X interual: [ -.5, Y increment = Y interual: I -1.25, 1.25]

FIGURE 5.20 (a) Transform of noisy signal, real and imaginary parts, (b) Reconstructed signal; compare with Figure 5.18(a). an inverse Fourier transform). Second, we select the two graphs of the real and imaginary parts of the complex function shown in Figure 5.20(a) to be the real and imaginary parts of the complex function to be transformed. Finally, we choose to apply a real, user-created filter function, and enter the formula

f i x ) = rect[(x - 50)/8] ± rect[(x ± 50)/8]

(5.105)

5. FOURIER TRANSFORMS

196

for the filter function. This filter function equals 1 over two relatively small intervals centered at ±50 and is 0 elsewhere. Hence, multiplying it with the FFT shown in Figure 5.20(a) produces a product that matches the FFT near the two spikes, but equals 0 elsewhere. This product resembles the transform of the original signal. In Figure 5.20(b) we show the graph of the filtered inverse FFT. Considering how much noise was originally present, it is an excellent reconstruction o f the original signal (divided by a factor of 8); the noisy background has been almost entirely removed. The method described above, which we shall call Fourier transform filtering, is summarized as follows: FOURIER TRANSFORM FILTERING PROCEDURE Step 1. Measure noisy signal. Step 2. Calculate power spectrum of noisy signal. Step 3. From the power spectrum identify characteristic frequencies of the signal. If signal reconstruction is desired, then the following additional steps may be needed: Step 4. Compute FFT of noisy signal. Step 5. Multiply FFT by a suitable filter. Step 6 . Compute inverse FFT. In the exercises you will be asked to apply this method to several noisy signals (included with the disk accompanying this book) in order to determine the charac­ teristic frequencies of the signal pulses that they contain and attempt to reconstruct these signal pulses. W HITE NOISE In Example 5.15 we used a noise term n(x) to simulate random noise in a received signal. We will now describe what we mean by such noise. Perhaps the simplest way of defining a noise signal is to use some type of random number generator. Thus, it would be possible to define a noise term n(x) by the formula f ( x ) = ¿?ran(x)

(5.106)

where is a positive constant and ran(x) is a random number generator [provided with FAS; see the User’s Manual in Appendix A for further discussion of ran(x)]. This will produce a graph containing random fluctuations varying between ±Z?

5.8. FILTERING, FREQUENCY DETECTION, NOISE REMOVAL

197

with a mean of 0. There is, however, another way of generating noise which corresponds more closely to the kind of noise that is actually encountered in signal transmission. This is the noise that we have referred to as white noise. We will now explain what white noise is, and show how we constructed the simulation of white noise that we used in Example 5.15. Let’s assume that the noise n(x) occurs over a fixed time interval of transmission, such as —0.5 < jc < 0.5. Many different signals could be transmitted over time intervals of length 1 and in every case the random noise that occurs will have a different form. Consequently, we really need to examine an ensemble of noise functions, say {wa:(jc) } ^ j . White noise is noise for which the average value of the power spectra from the ensemble is constant at each frequency. That is, the following limit holds: 1 ^ lim - T \ h k { u ) \ ^ K (5.107) N^OO N k=l ^ where is a positive constant. In other words, on average, every frequency is represented with equal power. Clearly, this is not a physically realizable condition. All real signals will have the power damp down to zero as the frequency tends toward infinity. Hence, we shall assume that (5.107) holds only over some finite range of frequencies. Another type of noise is Gaussian noise. For Gaussian noise, we replace (5.107) by 1

^

2

lim — ' y\ nk(u)\ ' ^ = K e - ‘=" N^OO N k=l ^

(5.108)

where K and c are positive constants. Gaussian noise is closer to a physically realizable condition. (For a good discussion of noise in random signals, see [Bas, Chapter 7].) We will now describe how to use FAS to create a simulation of white noise. First, since a noise function n(x) is real valued, its transform h(u) must satisfy h{—u) = n* (m). That is, if Re h is the real part of h and Im h is the imaginary part of h, then Re/5(—w) = Reh(u), I m h ( - u ) = —l mh(u). (5.109) In other words, the real part of h is an even function and the imaginary part of h is an odd function. The equations in (5.109) tell us that we are only free to define h(u) for w > 0 since the values of h( —u) are then determined by these equations. Now, with this fact in mind, the following five-step procedure can be used in FAS to create a simulation of white noise: Step 1. Enter the function f { x ) = a ran(x) for some positive constant a over the interval [0, L]. This produces the real part of the transform of the noise for positive frequencies.

198

5. FOURIER TRANSFORMS

Step 2. Enter the function f ( x ) = bran(x) for some positive constant b over the interval [0, L]. This produces the imaginary part of the transform of the noise for positive frequencies. Step 3. Enter the formula f ( x ) = gl[abs(x)] over the interval [—L, L]. This produces an even function, which is the real part of the transform of the noise function. Step 4. Enter the formula f { x ) = sign(x)g2[abs(x)] over the interval [—L, L]. This produces an odd function, which is the imaginary part of the transform of the noise function. Step 5. Perform a complex, inverse Fourier transform of the real and imaginary parts of the transform created in steps 3 and 4. Press n when asked if you want to apply a filter. The resulting inverse transformed function is the noise function n(x). It will be defined over the interval [—M /(4L ), M /(4L)], where M equals the number of points used. For example, to produce the noise function n(x) used in Example 5.15, we chose M = 1024 points, L = 512, and b = c = 0.01. This produced a noise function defined over [—0.5, 0.5]. We can check that (5.107) does hold (at least approximately) for the procedure described above. In Figure 5.21 we show graphs of the power spectrum averages 1 ^ f 1 - y [.01 ran(jc)]2 + [.01 Tm(x)f

(5.110)

for N = 1000 and N = 30, 000. As you can see, the average appears to be approx­ imately constant for the larger value of A. In fact, if rounding error is neglected, the average in (5.110) will converge as A -> oo to the constant 2(0.01)^ V, where V is the variance of the random number generator ran(x). Thus, the procedure described above will produce a noise function n(x) that simulates a member of the white noise ensemble {rik(x)}^^ of random functions. We say “simulates” because of the following two limitations of the procedure: (1) the frequencies used were restricted to the interval [—L, L]; (2) only a discrete, finite set of frequencies was used. In regard to the first limitation, we mentioned previously that arbitrarily large frequencies of constant power are not physically realizable. (There are some who would say, in fact, that we are simulating pink noise, i.e., noise in the lower frequencies [the “red” end of the spectrum] are of equal power, while noise in the higher frequencies is nonexistent. We shall, however, stick to the more commonly used expression, white noise.) In regard to the second limitation, this limitation occurs with all of the examples throughout this text, since we have consistently been using finite sets of data to model continuous phenomena.

5.8. FILTERING, FREQUENCY DETECTION, NOISE REMOVAL

(a) X interval : [ 3j 5121 X increment = 51.Z Y interval: [ ■ .0001, .O00Z1 y increment =

(b) X interval : [ Y interval: I ■

199

5121 X increment = 51.2 iOl, .00021 Y increment =

FIGURE 5.21 (a) Average of 1000 random power spectrums. (b) Average of 30,000 random power spectrums.

To model Gaussian noise, step 5 in the procedure above is modified slightly. A filtering operation is applied in this step, using a real valued Gaussian filter function f ( x ) = exp(—cjc A 2) for some positive constant c. FURTHER EXAMPLES OF NOISE SUPPRESSION We close this section with two more examples of noise suppression and frequency detection. First, we show how the Fourier transform filtering procedure can be used to recover a succession of pulses that represents a sequence of bits in a binary signal. Second, we illustrate a relatively new method of frequency detection called stochastic resonance. Example 5,16: Suppose that we wish to transmit the following sequence of eight bits: 101 1 0 1 0 1 . Assuming that the time interval of transmission is —4 < x < 4, then this sequence could be represented by the following signal:

5(x) = (cos 20nx ) [ e -107T(x+3.5r + e

-107t(jc+ 1.5)2

+ e

-107t(a:+0.5)2

This is a train of pulses all having fundamental frequency 10. If the reader graphs this signal it will be apparent that the bit 1 is represented by the presence o f a pulse between two integer values o f time and the bit 0 is represented by the absence o f such a pulse. Now, suppose that the signal in (5.111) is received with white noise added to it. See Figure 5.22(a). After applying the Fourier transform filtering

200

5. FOURIER TRANSFORMS

(a) X interval: [ -4, y interval: [ -2,

41 Z1

X increment = Y increment =

.8 .4

Y interval: [ -Z,

Z1

Y increment =

.4

FIGURE 5.22 (a) Sequence of pulses obscured by noise, (b) Noise removed by Fourier trans­ form filtering; the pulse sequence 1 0 1 1 0 1 0 1 is now apparent.

procedure, we obtain the signal shown in Figure 5.22(b). Clearly, we can read off from this signal the sequence of bits 1 0 1 1 0 1 0 1 , which was the originally transmitted sequence. The topic of noise removal is a fundamental one in signal processing. In addi­ tion to the classic method of Fourier transform filtering illustrated in the previous examples, many other interesting methods have been discovered. We close this section with a description of one of these methods, called stochastic resonance. Example 5,17: STOCHASTIC RESONANCE To illustrate this method of noise removal, consider the following noisy signal: g(x) =

c o s (507T x

)

+ lOran(x).

( 5 .1 1 2 )

This signal consists of a random noise term, lOran(x), of amplitude 1 0 added to a single-frequency term, c o s (507T x ) , of amplitude 1 and frequency 25. The noise has a ten times greater amplitude than the cosine term, hence the cosine is completely obscured by the noise. Moreover, as we can see from Figure 5.23(a), the power spectrum of g does not reveal any easily discernible pair of spikes that would identify the frequency of the cosine term (hence, the Fourier transform filtering procedure cannot be used here). The method of stochastic resonance consists in multiplying the signal g by a filter function of the form F(x) = [ \ g ( x ) \ > c ]

( 5 .1 1 3 )

where c is a parameter that is greater than or equal to 0. The function [ |g(x) | > c]

5.8. FILTERING, FREQUENCY DETECTION, NOISE REMOVAL

X interval: [ -64, 641 V interval: [ -5, 251

X interval: [ -64, 641 Y interval: [ -5, 251

X increnent = 1Z.8 Y increment = 3

201

X increment = 12.t Y increment = 3

FIGURE 5.23 (a) Power spectrum of noisy signal, (b) Power spectrum of filtered signal; the two spikes are located at x = ±25.

equals 1 if |g(x)| > c and equals 0 if |g(x)| < c. Thus, the filtered function is

g(x)

if \ g ( x ) \ > c (5.114)

[\g(x)\ > c]g( x) if\g{x)\ < c.

In other words, the filtered function corresponds to registering the signal g(x) only when its intensity |g(x) | is greater than or equal to the threshold value c. It is worth noting that many natural systems have this property of only responding when the stimulus is above a certain threshold. In FAS Formula (5.114) is implemented by entering the user-created filter function f ( x ) = [abs(gl(x)) > c] where g l stands for the graph of the signal g (the first graph displayed on screen). The phenomenon of stochastic resonance occurs when there is a range of values of c, located somewhere between c = 0 and c = ||g||sup (the maximum value of g), which will enhance the detection of the underlying frequency of the obscured signal. In Figure 5.23(b) we show the power spectrum \F{u)\^ of the filtered function for the signal g defined in (5.112), using a value of c = 9.5. As you can see, the filtered power spectrum shows two prominent spikes. These spikes are located at x = ±25, hence we have identified the frequency 25 that was previously obscured by the noise. See [Mo-W] for an excellent, nontechnical description of stochastic resonance. This article also describes how some animals’ nervous systems make use of stochastic resonance.

5. FOURIER TRANSFORMS

202

5.9

Poisson Summation

In this section, we will explore the relationship between Fourier series and Fourier transforms. This relationship is governed by a process known as Poisson summation. Suppose first that / is a function that is 0 outside of the interval [—L, L]. Then the Fourier transform of / reduces to

/(« )

-L

(5.115)

Substituting n/2L for w in (5.115) and multiplying by 1/2L we get

Formula (5.116) says that the Fourier series coefficients o f f ( x ) over the interval [—L, L] are given by {{\/2L) f ( n/ 2L) } ^ __^ where f is the Fourier transform o f / . Or, expressed another way. rmx/L

(5.117)

n=—oo

is the Fourier series expansion, period 2L, of a function / that is 0 outside the interval [—L, L]. Formula (5.117) can be generalized beyond the case of a function that is 0 outside of a finite interval. One version of this is the following theorem. THEOREM 5.11: POISSON SUMMATION Suppose that ||/ ||i is finite. I f f p is defined by

fp(x) =

E

then / p is a periodic function, period 2L, and has Fourier series

i: y O i V

innx/L

(5.118)

203

5.9. POISSON SUMMATION

REMARK 5.5: (a) The function / p is called the periodization of / . (b) If / is 0 outside of [—L, L], then / p is just the periodic extension of / and (5.118) reduces to (5.117). I PROOF OF THEOREM Clearly, / p has period 2L. To prove that the Fourier series expansion in (5.118) is correct, let Cn stand for the Fourier series coeffi­ cient of / p , then

f { x — 2kL)e -innx/L

oo

r ^

dx

pL

k=—oo L

Now, substituting x -h 2kL in place of x, we get ou

1

k^oo

1

f°°

r2kL+L

p Z K L - \- L

J 2 k L -L

■i2n{n/2L)x

2L'^ \ 2L )

Thus, the Fourier series coefficient of / p equals { l / 2 L) f { n / 2 L ) . This says that (5.118) holds and our theorem is proved. I There is an important corollary to this theorem, which we will also call Poisson summation. THEOREM 5.12: POISSON SUMMATION Suppose that f is continuous and ||/ ||i is finite. If f{x-2nL)

5. FOURIER TRANSFORMS

204

converges uniformly fo r |jc| < L and

converges, then 00-1 irntx/L

(5.119)

PROOF By Theorem 5.11, /p (x ) = J2T=-oo ~ of (5.119) as its Fourier series. Since the sum defining /p (x ) converges uniformly and each term is a continuous function, a well-known theorem of advanced calculus says that f p ( x ) is a continuous function. Also, since

converges, the Weierstrass M-test says that the Fourier series 00

1

converges uniformly to a continuous function. Moreover, we know that this con­ tinuous function has the same Fourier coefficients as f p( x) . These two functions must therefore be the same. Thus, (5.119) holds. I We close this section with a simple example of Poisson summation. In the next section we will examine an important application. Example 5.18: Since p/ ( p^ + x^)

jj-^-27tp\u\

.ir-eo />‘ + (. X- 11)2

p > 0, we have by Theorem 5.12

=E

jj.^-27tp\n\^i2nKx ^

(5.120)

205

5.10. SUMMATION KERNELS/POISSON SUMMATION

If, for example, we put x = 0 and p = 1 in (5.120) we obtain

,

00

00

Tve—27T\n\

— — rvo ^

(5.121)

rt— —

It is interesting that by using geometric series one can determine the precise sums of the right-hand sides of (5.120) and (5.121). We leave it as an exercise for the reader to show that Y „¿T 'oo

_____ - _____ = — + (x — n )^

1

;r(l — 2e

1 +

_;r(l+e-2^) -2

5.10

(5.122) cos

2;rx

(5.123)

tt

Summation Kernels Arising from Poisson Summation

In this section we will show how a summation kernel (see Section 4.9, Chapter 4) can be analyzed using Poisson summation. As an example, we will show that the banning kernel is a summation kernel. For simplicity of notation we shall begin by assuming that we are dealing with Fourier series with period 1. Using the functional notation for filter coefficients introduced in Section 4.5 of Chapter 4, let’s suppose that the point spread function (PSF) V m is defined by M

V m {x ) =

£

2nnx

(5.124)

n=—M

where F is an even, continuous function on [—1, 1]. Extending F to be 0 outside of the interval [—1, 1], we can rewrite (5.124) as

P«W =

oo E

Notice that F is still an even function.

F { ^ ) e ‘2nnx

(5.125)

5. FOURIER TRANSFORMS

206

We will now show that Poisson summation can be applied to (5.125) so that OO

V m {x ) =

^

MF[M{x-n)\.

(5.126)

Moreover, we can relax the requirement that F be 0 outside of [—1, 1]. Here is the theorem. THEOREM 5.13: Suppose that F is even and continuous and that F and its Fourier transform F satisfy |F (x)| < A (l + |x|)- -l—a

(5.127)

| F ( m) 1 < B ( 1 + |m|)-* -^

(5.128)

fo r some positive constants A, a, B, and OO

/ y] \

^

Then, fo r all x in R, OO

.

Y i MF[M{x-n)].

(5.129)

REMARK 5.6: Notice that when F is continuous and is 0 outside of [—1, 1], then Inequality (5.127) is definitely satisfied for some A and a. I

PROOF OF THEOREM

By Fourier inversion, we have F{x) = F { - x ) = F{x),

so F is the Fourier transform of F. Inequality (5.127) ensures that converges. In fact, by (5.127), we have

i:

\

/^ ) I

(1 + |n /M |)l+ “

and the sum on the right side converges fo r each fixed M (by comparison with

5.10. SUMMATION KERNELS/POISSON SUMMATION

207

^ | n | ' “ ). Inequality (5.128) also ensures that

^

MF[M(x-n)]

converges uniformly for all x in [—1/2, 1/2]. In fact, for |n| > 1, BM (1 + M |x -^1)1+^ ‘

M F [M{x - n)]

(5.130)

Since \n\ > 1, and x is in [—1 /2,1/2], it follows that \ x - n \ > -\n \

(5.131)

by a simple consideration of distance on the real line. Thus, for \n\ > 1 BM

M F [M{x - n)]

(l + f |n|)l+^

(5.132)

Since BM 1^1

(l +

f

BM

|n|)l+ ^

converges (by comparison with ^ that

(l +

(5.133)

f n ) ‘+ ^

^ ^), it follows from the Weierstrass M-test

oo

^

MF[M{x-n)]

7 1 = —OO

converges uniformly for all x in [—1/2, 1/2]. We have now verified all the conditions for Poisson summation to be applied to the series Yl^=-oo F(n/M)e^^^^^ which defines V m M - Since we showed above that F = F , we have by change of scale M F (Mx)

T

(\ M - )) •

Consequently, Theorem 5.12 yields (for 2L = 1)

5. FOURIER TRANSFORMS

208 °° MF[M(x-n)]^

°° Y

/ n\ F [ - ) e ilnnx ‘

I

which is the same as (5.129).

Based on Theorem 5.13 we can prove the following theorem, which tells us when the PSF V m in (5.125) will be a summation kernel. THEOREM 5.14: Suppose that F is even and continuous, thatF(0) = 1, andthat (5.127) and(5.128) are valid. Then, the kernel V m defined by ilnnx

is a summation kernel over [—1/2, 1/2]. PROOF We need to check the three properties (a) to (c) in Definition 4.4 from section 4.9 of Chapter 4 (for L = 1/2). (a)

By integrating term by term /

I

oo

VM(x)dx^

^i

Y, n = —oo

"2

*'“ 2

SO (a) holds.

(b) We know from Theorem 5.13 that (5.129) is true. Making use of (5.129), and also (5.132) and (5.133), we have oo

\Vm {x )\
i

MF(Mx)\+Y'^ tí

BM

+

(5.134)

209

5.10. SUMMATION KERNELS/POISSON SUMMATION

Since the numerical series BM

converges, let’s denote its sum by the constant S. Then, by (5.128) applied to \M F (M x)\ we can rewrite (5.134) as W m (x )\ < B

M + 5. (1 + M 1 jc1)1+^

Consequently,

\ V m

(x ) \

M

dx < B

dx -|-

(1 + M|jcl)‘+^

Sdx

dt “h S

M /2 (1 +

DO

Li

1 dt

L o o (l + |i|) ‘+^

IB S =■ —-— h S. P

Putting C equal 25/^0 + »S' we see that (b) holds. (c) We now assume that some 5 > 0 has been given and 5 < |x| < 1/2. Applying (5.134), and using (5.128) to bound \M F{M x)\, we obtain

\ V m

{x )\

M

< B

+y]^‘ ^ t

BM

M

< B

1 w

jt +]8

+ 2 - / -;iT + r n=\

Since the series g n=\

B2^+^ t\+P

(5.135)

5. FOURIER TRANSFORMS

210

converges, the quantity in brackets in (5.135) is a finite constant; let’s call it D. Then (5.135) becomes D w

(5 < 1^1
oo in (5.136) we see that the right side of the inequality tends to 0. Hence, for any given € > 0, if M is taken sufficiently large we will have D / m P < 6 and then \V m (x )\ < e ,

so (c) is true.

(S < \ x \ < ^ )

I

Using Theorem 5.14, we can easily establish the following theorem, which can be used for demonstrating that many of the PSFs discussed in Chapter 4 are summation kernels. THEOREM 5.15: Suppose that F is even and continuous^ that F (0) = 1, and that (5.127) and (5.128) are valid. Then, the kernel V m defined by

^

f

( ^ ) / innx/L

is a summation kernel over [—L, L]. PROOF

If we make the change of variable, x = IL t, then oo

F „ ( 2 L ,) =

Y.

il n n t

'" ( s ) " “

is a summation kernel over [—1/2, 1/2] (because of Theorem 5.14). By making the reverse change of variable, t = x l(2 L ), it follows that V m ( x ) is a summation kernel over [—L, L]. I As an application of this theory we will now show that the banning kernel is a summation kernel.

5.10. SUMMATION KERNELS/POISSON SUMMATION

211

Example 5.19: The banning kernel, defined by V m (x ) =

lrt| 1.

(5.138)

It is easy to see that this function F is continuous and even, and that F(0) = 1. Since F is 0 outside of [—1,1] and continuous, we also have (5.127). In particular, if we take A = 4 and a = 1 it is easy to check that (5.127) holds (just draw graphs of F{x) and 4[1 + \x\]~^). We also have that F{x) = 0.5 rect (jc/2) + 0.25 rect {x/2) ^ITTX

^-17TX

]

(5.139)

“ - 0 ] + 0.5 sine 1^2 ^M +

j . (5.140)

so, by scaling and modulation properties, F(u) = sine (2 m) + 0.5 sine K

Replacing sine v by (sm 7tv)/{nv) and simplifying yields F(u) =

— sin(27rM)

~ sine (2 m)

(27Tm)(4 m^ “ 1)

4 m^ — 1

(5.141)

Since I sine ul < 1 for all u in R, it follows from (5.141) that (5.128) holds when ^ = 1 and B is chosen sufficiently large (since lim|M|->oo(l + 1m|)^/(4 m^ — 1) = 1/4). Because (5.127) and (5.128) are valid, it follows from Theorem 5.15 that the banning kernel is a summation kernel. I

5. FOURIER TRANSFORMS

212

Example 5.20: The kernel V m defined by

V m {x ) =

Y .

{ n / M r ^ilnnx

(5.142)

is a summation kernel over [—1/2,1/2]. This is because the function F{x) — e-7zx satisfies all the requirements of Theorem 5.14. REMARK 5.7: The summation kernel defined in (5.142) is related to the theta function kernel, which plays an important role in the theory of heat conduction. See [Wa, Chapter 4.7]. I

5.11

The Sampling Theorem

This section describes a very important concept in signal processing, known as the sampling theorem. The theory of Poisson summation and the theory of sampling are closely related. We begin with the definition of those functions to which the sampling theory applies. D EFINITIO N 5.5: A function f is called hand limited if its Fourier transform / is 0 outside o f a finite interval [—L, L]. The sampling theorem says that a band limited function can be recovered from its samples

provided / is 0 outside of the interval [—L, L]. This theorem has many important applications. For instance, the construction of compact disc players uses sampling theory (see [Mo]). In this case, the frequencies that human ears are sensitive to lie in a finite range (20 to 20, 000 Hz) and, consequently, the recorded music can be sampled effectively. In fact, there is software available now which allows a PC to be used for analyzing and synthesizing sampled music. Another application of sampling theory is to the sending of multiple telephone messages along a single cable. By sampling the messages (at points separated in time by 1/2L units) gaps are created within which other sampled messages can be sent. Using a fiber optic cable, upwards of 25,000 simultaneous messages can be transmitted. Besides

213

5.11. THE SAMPLING THEOREM

illustrating the wonder of fiber optics, this result also illustrates the power of the sampling theorem. Here is the sampling theorem. THEOREM 5.16: Suppose that f is band limited. I f f is 0 outside of [ —L, L], then 00 fix) =

yn \ f ( ^ ) sine (2Lx - n).

(5.143)

PROOF We will prove a very general version of the sampling theorem which yields (5.143) as a special case. Since / is 0 outside of [—L , L] we can periodically extend it. In fact, if we use / p to denote the periodic extension of / with period 2L, then / p (m) =

^

/ ( m - 2nL).

(5.144)

By Poisson summation, we know that the Fourier series for / p is [use / in place o f / i n (5.118)] 1 -( ^injtulL ^

Since f { x ) = / ( —x), we have innu/L

(5.145)

Now, suppose we multiply (5.145) by a function W(u), known in sampling theory as a window function, which satisfies W{u) =

1 0

when f ( u ) ^ 0 when \u\ > L

(5.146)

and is bounded for all other w-values. For such a window function, we have f p( u) W( u) = f ( u ) W( u ) = f ( u )

(5.147)

because f ( u ) = 0 for |w| > L. For example, if we take

W{u) =

1 0

when |w| < L when Iwl > L

(5.148)

5. FOURIER TRANSFORMS

214

then (5.146) and (5.147) hold. Multiplying both sides of (5.145) by W(u), and using (5.147), we get

/(.) ~

E

¿ / ( i ^ ) » '( . y

« = —OO

^

in n u /L

(5.149)

/

Multiplying both sides of (5.149) by and integrating with respect to u from —L to L, we obtain the following equality:

I-l

nir'oo

\2LJ2LJ^i

We will explain why there is equality in (5.150) in Remark 5.8 below. But first, let’s just assume that the equality is valid. If we define S(jc) by W(M)e'2^“^ du Six) - _L ~ ^ J -L

(5.151)

then (5.150) becomes (5.152)

Since / is 0 outside of [—L, L], we have

/ r*LL

POO f {u)e^^^' *^du= /

du = f i x )

(5.153)

J-OQ

-L

by Fourier inversion. Consequently, (5.152) becomes

Substituting —n in place of n, and noting that the sum in (5.154) is over all integers, we obtain 00

n—~OQ

/ n. \

/

n. \

\Z /./

\

Z/./

(5.155)

5.11. THE SAMPLING THEOREM

215

Formula (5.155) describes a very general sampling theorem. The data specific to the function / is the set of sample values

The reconstruction kernel S is defined by Formula (5.151). If we take for W the function in (5.148), then we have

S(x)

(2Lx)

2L J_

and (5.155) reduces to (5.143).

I

REMARK 5.8: In the proof above we did not explain why (5.150) holds. We will now prove the following more precise version of that equality:

du

/ : = lim

Y

/ f— ) — V 2 L ;2 li_ i

(5 156)

^

To prove (5.156) we use f ( u ) = f {u) W{ u) from (5.147), and we have

J-L

\2 L /2 L ;_ l

du

,innu/L

f

\W{u)\du.

(5.157)

Now, using the symbol S]^(u) in place of the M-harmonic Fourier series partial

5. FOURIER TRANSFORMS

216

sum for / , we obtain from (5.157), 27t{x+n/2L)u

f i u ) - s{^(u)

\ W(u)\du < f - s L

du

l|W ||2 (5.158)

the last inequality being the well-known Schwarz inequality (see Appendix C) where the 2-Norms are taken over the interval [—L, L]. If we now use the completeness relation from Theorem 4.2 in Chapter 4, we have (since / is bounded by ||/ ||i its 2-Norm over [—L, L] is finite): lim

M->oo

f-s,

=

0.

(5.159)

Comparing (5.159) with the last inequality in (5.158) we see that (5.156) holds. Formula (5.156) justifies using (5.150) in the proof of the sampling theorem. Moreover, using the definition of S in (5.151), and Formula (5.153), in Inequal­ ity (5.158), we have also shown that for all x in R

/ « -

l|W ||2

< \\f-SU n= -M

II

(5.160)

II2

Inequality (5.160) provides a useful estimate for the magnitude of the difference between / and a partial sum of the reconstruction series for / given in (5.155). It also shows that the sampling series converges uniformly over the whole real line.

I REMARK 5.9: For a band limited function / , a frequently used value of L is the smallest possible value for which f { u ) = 0 when |w| > L. In this case, one must use the window defined in (5.148), and the sampling series must be the one in (5.143). The sampling rate, 2L samples/unit-length, is called the Nyquist rate when this smallest value of L is used. I Example 5.21: Suppose f { x ) sinc^(2x — 1) -|- 3 sinc^(2x + 1). Find the Nyquist sampling rate for / . Approximate f { x ) to within ±0.001 for all x-values over the interval [—4,4], using a partial sum of the Nyquist sampling series for / .

217

5.11. THE SAMPLING THEOREM

SOLUTION Since sine erties imply that

f(u)

A ( m), the linearity, scaling, and shifting prop­

1a ( ! » ) « - + 1 2

•

3 . ^2

G")

Since A(u/ 2) = 0 outside of the interval [—2, 2], it follows that f ( u ) = 0 outside of [—2, 2]. Moreover, [—2, 2] is the smallest interval centered at 0 outside of which f ( u ) = 0 (this is because (1 + (3/2)^*^“ = 2 at w = 2, hence continuity implies that f ( u ) 7^ 0 for w slightly less than 2). Therefore, the Nyquist sampling rate for / is 4. A partial sum of the Nyquist sampling series for / is M

Sm {x ) =

£

f ( j

sine (4jc —k).

k= -M

This sum can be graphed by FAS using the following formula: f ( x ) = sumk{( sine (0.5k — 1) A 2 + 3 sine (0.5k + 1) A 2) sine (4x — k)}.

where sumk is the function in FAS for summing over k. A little experimentation (using 512 points, 1024 points, and 2048 points) shows that M = 20 yields a graph whose Sup-Norm difference from / over [—4,4] is approximately 2.8 x 10“ ^. Thus, we can safely conclude that S2o(^) approximates f ( x ) to within ±0.001 for all x-values in the interval [—4,4]. (Note: although the sampling theorem applies to all x-values in the interval [—4,4], while FAS only performs its calculations for a finite set of values, we can still guage the validity of our approximations by successively doubling the number of points and seeing if the Sup-Norm differences remain relatively constant.) I As we noted above, the sampling theorem is frequently applied to the recording of musical signals for compact discs and in telephone transmission of voice signals. The basic scheme for both of these applications consists of the following three steps: Step 1. Record samples o f the signal and convert to digital data. This is called analog to digital conversion (or A/D conversion).

218

5. FOURIER TRANSFORMS

Step 2. Transmit digital data. For telephone signals this is done along copper wire or fiber optic cable. For CD players, this step involves the production and distribution of the compact discs. Step 3. Reconstruction o f the analog signal from the digital data. This step involves using a digital to analog converter (or D/A converter), an electronic device that produces a partial sum of the sampling series as an output. The D/A converter has the reconstruction kernel built into its circuitry; it only needs the digital data obtained from the original analog signal in order to produce its output. A few observations need to be made about this procedure. First, in Step 1, when the digitization of the samples {f {k/ 2L) ] is performed, only a finite number of bits can be used. That is, each value [ f ( k/ 2L) } is rounded off to the closest value that can be represented with this finite number of bits. The error produced by this rounding is called quantization error. We will not discuss quantization error any further. We shall assume that it is small enough to be ignored and simply refer to {f {k/ 2L)} as the transmitted data. We shall also not discuss any details of Step 2, the transmission of the digitized data. There are many important issues that are of concern (e.g., error correction, optimal compression of the data, and elimination of noise), but, for reasons of space, we shall leave these issues aside. Second, in regard to Step 3, the reconstruction of the analog signal, the result of Example 5.21 and other examples like it should be examined carefully. If, in Example 5.21, we view the interval—4 < < 4 as the time interval of transmission (and reception) of the sampled values of the signal / , then using M = 20 would involve several samples [f{k/A) for k = ±20, ..., ±17] that are not part o f the transmitted values. If we can only use samples over the interval [—4, 4], then the largest value of M that we can choose is M = 16. When we use M = 16, then the Sup-Norm difference between / and the reconstruction 5i6 over the interval [—4,4] is approximately 2.4 x 10“ ^. Although this is not as small a value as required in the example, it still produces an acceptable approximation of the signal / . See Figure 5.24. Third, when a sampling series partial sum

Sm {^)

is used for the D/A conversion, it is important to consider the response time of the re­ construction kernel S. In Figure 5.25 we show three reconstruction kernels (which we shall describe further in Section 5.13). The kernel S3 shown in Figure 5.25(c) appears to be essentially 0 for |x| > 0.5. Therefore, we could ignore those terms f {k/ 2L) S^{x —k/ 2L) in (5.161) for which |x —k/2L\ > 0.5. The time length of 0.5 is called the response time for the kernel. The kernel Si(x) = sine (4x)

5.12. ALIASIN G

(a) X interual: C ~4, y interval: [ -1,

X increnent = y increnent =

(b) X interval: f -4, y interval: I -1,

41 41

X increnent = y increnent =

.8 .5

FIGURE 5.24 (a) Graph of the signal f ( x ) = sinc^(2jc — 1) + 3 sinc^(2jc + 1). (b) Graph of the Nyquist sampling series partial sum S\e» The ratio |1/ — 5i6l|2/ll/ll2 is about 9.76 x 10“ ^ (this gives quantitative support to the visual similarity between / and S\e).

shown in Figure 5.25(a) has a much longer response time. A D/A converter that uses this sine function kernel would have to wait for most (if not all) of the data before it could begin producing output. On the other hand, a D/A converter that uses the kernel S 3 would have to wait for a much shorter set of data to be received (about 1/16 of the total data) before it begins producing output. We will examine these points in more detail in Section 5.13.

5.12

Aliasing

The sampling theorem proved in the previous section applies to band limited functions only. Not all functions are band limited, however. In this section we will discuss what happens if / is not band limited. It is still possible to obtain an approximate sampling reconstruction, if some error is allowed to exist between the original function / and its sampling series. To begin, we will call a function almost band limited if there are positive con­ stants A and a for which \ f { u ) \ < A [ l + \uW -1—a holds for all u in R.

(5.162)

5. FOURIER TRANSFORMS

220

(a) X interual: [ -4, 41 X increment = .8 V interual: [ -.5, 1.51 V increment = .2

1

^

i

V ^

-t

1

1 '

(c) X interual: C -4, 41 X increment = .8 Y interual: 1 -.5, 1.51 Y increment =

(b) X interual: [ -4, 41 X increment = V interual: [ -.5, 1.51 V increment =

FIGURE 5.25 (a) The kernel Si(x) = sine (4x). (b) The kernel S2(x) = sine ( 8x). (c) The kernel SsCx) = 0.75 sine (6x) sine (2x).

Now, suppose that / is almost band limited but, not band limited. If we look again at (5.144), we see that / p is no longer the periodie extension of / . For any given L, it is the periodization of / having period 2 L that we defined in Remark 5.5(a). Nevertheless, by the Poisson summation theorem (Theorem 5.11), we still have Formula (5.145). If we use the window function

W(u) =

1

if |w| < L if\u\ > L

(5.163)

then, multiplying (5.145) by W(u) and integrating over —L < u < L , wq obtain a new form of (5.150). Namely,

221

5.12. ALIASING

du

¿ j m c Since the window W is the same one as in (5.148), we obtain

/ -LL /p(M)e'2^“ d u = ~L

^ / n \ J2 f ( ^ ) yi—-on

L]

(5.173)

223

5.12. ALIASING

Furthermore, using (5.162) we can bound the integral in (5.173), obtaining \f(x) - Af(x)\
0 as L -> oo, Inequality (5.174) shows that for an almost band limitedfunction f we can always replace f with its alias A / and have negligible error when L is taken sufficiently large. Or, put another way, when the sampling rate o f l L samples per unit length is large enough, there will be negligible error between f and its sampling series. This last result holds because (5.168) and (5.174) give

f(x)~

/ ( ^ ^ sine (2 Lx - «)

^ n=—oo

4A 1 T (L + 1)«

(5.175)

There are, of course, practical limitations. It is not always possible to sample at theoretically desired sampling rates. If we use other sampling series, based on different kernels, we can obtain results that are similar to those described above. For instance, let’s require that the window W satisfy the following three conditions: IW(«)1 < 1 for all M-values W{u) = 0

if |w| > L

(5.176a) (5.176b)

W(m) = 1 if |m| < L j l .

(5.176c)

There are many examples of such windows (e.g., see Example 5.22 below). We define the alias function A f for such a window, by Af(x) =

J

f p( u) W( u) e‘^^“^ du.

(5.177)

It then follows that (5.178)

where S(x)

=± f 2Lj_

du

(5.179)

5. FOURIER TRANSFORMS

224

is the kernel that we defined in Section 5.11. We leave it as an exercise for the reader to show that pL+2nL | / ( m) - f ( u ) W{ u + 2nL)e‘'^^"^’‘\du / n=—OG'x~ L-\-2fiL 00

\ f ( x)

■Afix)\
0 and

5.33 Graph approximations to for t = 0.25, 0.5, 1.0, and 2.0, fi/2m = 0.578, and f {x) = Check Formula (5.95) for i = 1.0 and 2.0. 5.34 Show that if f ( x) = e f^tF(x) =

, then 1

-Ttx^/[l+4jvi(fi/2m)t]

(5.206)

[1 + Ani(fi/2m)t]'2 Using (5.206), show that (5.95) holds for sufficiently large t when f ( x) = e

.

Section 8 5.35 Can you give a physical interpretation of why the filtered Fourier transform method of noise suppression works? 5.36 Given the function cos IG tvx + 0.8 cos 157tjc —0.6 cos 187Tx , perform a filtered Fourier transform on the interval [—16, 16] using 1024 points. Use Parzen, banning, Hamming, and Welch filters. 5.37 Repeat Exercise 5.36, but use sin I67tx + 0.8 cos Snx —0.3 sin I27tx. 5.38 Repeat Exercise 5.36, but use the constant function/(x) = 1. Compare the shapes of the peaks of the filtered transforms in Exercise 5.36 with the shape of each peak in this exercise. Explain the similarity. 5.39 The disk accompanying this book contains the function data file, signall, which is a received signal that has been corrupted by noise. The originally transmitted signal was a pulse of amplitude 1 containing a single fundamental frequency. Determine this frequency and reconstruct, at least approximately, the original signal. (Use the interval [—16, 16] and 1024 points.) 5.40 The disk accompanying this book contains the function data file, signall, which is a received signal that has been corrupted by noise. The originally transmitted signal was a pulse of amplitude 1 containing two fundamental frequencies. Determine these frequencies. (Use the interval [—16, 16] and 1024 points.)

5. FOURIER TRANSFORMS

240

5.41 The disk accompanying this book contains the function data file, signal!, which is a received signal that has been corrupted by noise. The originally transmitted signal was a sequence of pulses all having amplitude 1 and a single fundamental frequency. These pulses represent a sequence of bits. Determine this sequence. Repeat this exercise for the signals signal4 and signal!. (All these signals are defined over the interval [—16, 16] using 1024 points.) 5.42 The disk accompanying this book contains the function data file, signal!, which is a received signal that has been corrupted by noise. The originally transmitted signal was a pulse of amplitude 1 containing a single fundamental frequency equal to 10. Suppose that the received signal was a radar signal received after bouncing off an object. Determine the fundamental frequency in the received signal. Is the object moving away from or towards the transmitter? (Use the interval [—16, 16] and 1024 points.) REMARK 5.10: Exercise 5.42 illustrates the basic idea behind Doppler radar. See also Exercise 5.48. 5.43 The disk accompanying this book contains the function data file, signal?, which is a received signal that has been corrupted by noise. Use the method of stochastic resonance to find the fundamental frequency of the signal. (Use the interval [—1, 1] and 1024 points.) 5.44 Repeat Exercise 5.43 for the signal signals. 5.45 The disk accompanying this book contains the function data file, signal9, which is a noisy signal of the form f ( x) =

cos

(27T

vjc

—(j))e' -\Qnx^ Fn{x)

defined over the interval [—0.5, 0.5] using 1024 points. Using the Fourier transform filtering method, determine the fundamental frequency v in the received signal. Observe that the power spectrum | / ( m)|^ does not indicate the value of the phase shift (j). If the original signal was 5(x) = 4 cos(507rx)^” ^®^^ , then try to estimate the phase shift. 5.46 Repeat Exercise 5.45, but use the signal contained in the file signal 10. 5.47 The disk accompanying this book contains the signals signalll, signalll, ..., signal!!, all of which are defined over the interval [—1, 1] using 2048 points. Using either of the two methods described in the text, determine a single characteristic frequency for each signal. 5.48 A signal pulse has been transmitted by a radar device with a frequency of 75 for a time interval of length 1. The disk accompanying this text contains the function data files signally, signal18, signal19, and signallO. These signals are the readings of a receiver for successive time intervals of length 1. Which of these signals contains the returning echo of the transmitted signal? Is the object that reflected the pulse moving away from or toward the receiver?

EXERCISES

241

Section 9

5.49 Show that (5.122) and (5.123) are true. Also, find the exact sums of oo

oo

E ,Y + n

and n=l

n=0

5.50 Using Poisson summation, show that for i > 0 1

ix-nŸI'lt

(5.207)

n=—oo The identity (5.207) is known as JacobVs identity. 5.51 Approximate YlT=-oo for t = 0.001, using a filtered Fourier series partial sum. Using FAS^ graph this approximation over [—1/2, 1/2] with 1024 points. 5.52 Show that YlT=-oo

cos Innx > 0 for all x GR and all i > 0. for all a > 0.

5.53 Find the exact value of YlT=i Section 10

5.54 Use Theorem 5.15 to give another proof that the Cesko kernel is a summation kernel over [—tt, tt]. 5.55 Show that the dlVP kernel Vm is a summation kernel over [—tt, tt]. [The kernel Vm is defined in (4.54) and (4.55) in Chapter 4.] 5.56 Suppose that F is continuous and even, F(0) = 1, F = 0 outside of [—1, 1] and the derivatives F' and F" are both continuous. Show that (5.127) and (5.128) hold for this function F, hence V m M = Y1^=-m F(«/M)£*"^ is a summation kernel. [Hint: use the inequality in Exercise 5.6 to bound |F(m)|.] 5.57 Given a positive constant p > 0, the Riesz kernel is defined by M r «— 1/ «n=— M

^-ip -j

Use the result of Exercise 5.56 to show that the Riesz kernel is a summation kernel when p > 2. REMARK 5.11: It is known that the Riesz kernel is a summation kernel when p > 0. See [St-W, Chapter 7]. I

5. FOURIER TRANSFORMS

242

Section 11

5.58 Show that the function/(x) = sinc^(2x) cos(47rx) has aNyquistrateof 8. Using FAS, graph the approximations to /(x) given by the sampling series partial sums Sm (x )

sine (8x —k) k=-M

V /

for M = 32, 48, and 64. Use 1024 points over the interval [—4, 4]. Using Sup-Norm and 2-Norm differences, estimate the accuracy of these partial sums as approximations of / . 5.59 Determine the Nyquist sampling rates of the following functions. Also compute the Sup-Norm differences between these functions and the indicated partial sums of their Nyquist sampling series over the given intervals. (a) f (x) = 2 sine ^(4x) H- sine ^(4x —12) + sine ^(4x -1-12); compare with the Nyquist partial sum S32 (x) over [—4, 4]. (b) /(x) = sinc^(8x) — sinc^(8x —2) — sinc^(8x -f- 2); compare with the Nyquist partial sum 5'i6(:r) over [—1, 1]. (c) /(x) = sinc^(2x), compare with the Nyquist partial sum 5'32(x) over [-4,4]. 5.60 Suppose that f(u) = 0 for lw| > L. Explain why, if g(x) = 2L sine (2Lx), then / * g = / . Using FAS, check this result for f ( x ) = sinc^(x) over the interval [—8,8]. Use 1024 points. 5.61 Show that {sine (2Lx —n)}^_^ satisfies the following orthogonality relation: 2L

f

sine (2Lx —m) sine (2Lx —n)dx

0 if m ^ n ifm =n.

i;

{Hint: use Parseval’s equalities.) Conclude that if / is band limited, with / = 0 outside of [—L, L], then /

/(x) sine (2Lx—n) dx

f

^ sm2Ln(x —n/2L) ^ f { x ) ----------------------dx. 7 ( —n/2L) t

x

5.62 The disk that accompanies this book contains a function data file called samples1. Retrieve these data using an interval of [—4, 4] and 1024 points. The non-zero function values are samples [f{k /^)}^^}^2 of ^ signal / , using a sampling rate of 8. Assuming that the Nyquist rate of / is 8, reconstruct / using a sinc-function sampling series. 5.63 The disk that accompanies this book contains a function data file called samples!. Retrieve these data using an interval of [—4,4] and 1024 points. The non-zero function values are samples ^ signal / , using a sampling rate of 8. Assuming that the Nyquist rate of / is 4, reconstruct / in the following two ways: (a) using the Nyquist kernel Si(x) = sine (4x), (b) using the kernel S2 (x) = sine (8x) at twice the Nyquist rate. Find the Sup-Normdifference between these two reconstructions.

EXERCISES

243

Section 12 5.64 For afunction / satisfying (5.162), andkernel S whose window W satisfies (5.176a) to (5.176c), show that the following inequality holds:

n= -M

\W{u)\^ du^ ^ and

where 1]WII2 =

is the M-harmonic partial sum of the

Fourier series for /p . Using this result, show that 4A X T ;^ + ||/ p - S a (L/2 + l>

/w- Î j i à H - i ) n =-M

m |L > /2I.

5.65 Using FAS, graph the approximations to f (x) = 0.5 exp(—0.25;ta: 2) given by a

M

Sm (.x ) = ^

f(.k) sine (x - k)

k= -M

for M = 8, 16, and 32. Use the interval [—8, 8] and 1024 points. Find the SupNorm differences between / and Sm for each M. [Note: if —8 < jc < 8 defines the allowed x-values, then only M = 8 is a valid choice for M.] 5.66 Repeat Exercise 5.65, but use /(x) = (1/7

t

)[1/(1 H-

x

a

2)].

5.67 Suppose /(x) = 0.5exp(—0.257rx^)cos^(7rx/4) for ~2 < x < 2 and /(x) = 0 for |x I >2. Using FAS graph 4L^

/ it \ j

• 4 /W =

- k)

for L = 4, 8, and 16, over the interval [—2, 2]. Estimate the Sup-Norm differences between / and A f for each L. Repeat these computations, but use the kernel S defined in Formula (5.185). 5.68 Suppose that f (u) = 0 for \u\ < L jl and for |m| > L. Show that / can be reconstructed using the samples {/(^/^)}£_oo reconstruction kernel S(x) = sine (Lx) [2 (L7 ) —1]. {Note: The sampling rate here is half the Nyquist rate.) cos

tx

5.69 Repeat Example 5.23, but use the function f ix)

cos

M

r

(57

tx

/8)

if |x |< 4 if lx| >4.

5. FOURIER TRANSFORMS

244

5.70 Given that the window W is defined by ^1 if \u\ < Lj l W{u) = ' sin^(ttw/L) if L/ 2 < |m| < L 0 if \u\ > L.

Show that W satisfies (5.176a) to (5.176c) and that its kernel S satisfies S(x) = -[2 sine {Lx) + 4 sine {2Lx) + sine {Lx —1) + sine {Lx + 1) 8

—2 sine (2Ljc—2) —2 sine {2Lx + 2)]. Hint: you may want to use the trigonometric identity 2 sin^ 0 — I —cos (2^). 5.71 Using the functions f\ and /2 in Example 5.23, compute additional columns for Tables 5.1 and 5.2 using the kernel S in Exercise 5.70. Section 13 5.72 Compile a table like Table 5.3, but use the function f{x) =

-O.IttIxI smnx 0

if |x| < 4 if |x| > 4.

over the interval [—4,4]. Use a sampling rate of 8 for Si and a sampling rate of 16 for both S2 and S3 . 5.73 Repeat Exercise 5.72, but use the function f{x)

e

sin^ :7rx

0

if 1^1 < 4 if |jc| > 4.

over the interval [—4,4]. 5.74 Repeat Exercise 5.72, but use the function f{x)

= I EL-3(-l)'sinc"(2x-20 0

if |jc| < 4 if |x| > 4.

over the interval [—4,4]. 5.75 Add a fourth column to Table 5.3 in the text, but use the kernel S given in Exer­ cise 5.70 with 2L = 8 . Does this kernel S appear to have a shorter response time than the kernel S3 (S2 , SO? 5.76 Fast evaluation of sampling series, (a) Show that if = k/(2L), then the sampling series partial sum Sm in Formula (5.161) satisfies M

SMiXj) = ^ 2 f(Xk)S(Xj-k).

(5.208)

k = -M

(b) Show that the sum in (5.208) can be approximated using FFTs, for N = 2^ points {xj} from a given interval [—

EXERCISES

245

5.77 The method involving FFTs introduced in the previous exercise is programmed into FAS in the procedure entitled Interpolation (sampling) in the Series menu (see the User’s Manual in Appendix A for more details). Use this FAS procedure to graph sampling series partial sums given the following data (in each case use 1024 points), and find the relative 2-Norm error between the function and its FAS computed sampling series partial sum (the relative 2-Norm error equals \\f — 5^112/ II/II2 where is the M5-computed sampling series). (a) Interval, [-4,4]. Kernel, sinc(4x). Function, exp(-(x/2)^^). Sampling increment, 1/(2L) = 0.25. Order of partial sum, M = 16. (b) Interval, [-4,4]. Kernel, sinc(8x). Function, exp(-(x/2)^^). Sampling increment, 1/(2L) = 0.125. Order of partial sum, M = 32. (c) Interval, [—4, 4]. Kernel, 0.75 sine (6x) sine (2x). Function, exp(—(x/2)^^). Sampling increment, 1/(2L) = 0.125. Order of partial sum, M = 32. (d) Interval, [—4,4]. Kernel, sinc(x). Function, exp(—(x/3)^^). Sampling increment, 1/(2L) = 1.0. Order of partial sum, M = 4. (e) Interval, [—4, 4]. Kernel, 0.875 sine (lx) sine (x). Function, exp(—(x/3)^^). Sampling increment, 1/(2L) = 0.125. Order of partial sum, M — 32. Calculate the explicit sampling series partial sums for each of the four cases above, and find the relative 2-Norm error between these explicit sums and the M5-computed sums (the relative 2-Norm error equals || 5m — II2/ II5'm II2 where 5^ is the M5-computed sum and 5m is the explicitly computed sum). For which of these kernels does the fast method provide acceptable results (that is, a small rel­ ative 2-Norm error between the explicit partial sum and the M5-computed partial sum)? 5.78 Periodic reconstruction kernels, (a) Over the interval [—4, 4], using 1024 points, plot the following partial sum of a sampling series: f(x) = sumk[exp(—(a/2)

A

20) sine (4i)] \k=—16,16 \a=k/4 \ z

\ t —u—w/2 \u=v—wgn(v/w) \v=z+w/2\w=S

= x —

k/4

(5.209)

(Note: this formula can be loaded into FAS using the formula file perjsser included with the disk accompanying this book.) The kernel in the sampling series in (5.209) is the periodic extension of sine (4x), period 4 (where sine (4x) is graphed over [—4,4]). (b) Find the Sup-Norm difference between the function graphed in (a) and the graph of the M5-computed sampling series partial sum 5^^ obtained in part (a) of Exercise 5.77. (Your answer should be approximately 7.42 x 10“^"^, which shows that the two sums are virtually identical.) (c) Compute similar partial sums using periodic kernels and compare them with the FAS computed sampling series partial sums obtained in parts (b) to (e) of Exercise 5.77. In every case, the two sums should be virtually identical. 5.79 Explain the results of Exercise 5.78.

5. FOURIER TRANSFORMS

246

Section 14 5.80 Using FAS, graph the Fourier sine and cosine transforms of the following functions (use the interval [0,16] and 512 points). (a) rect (jc - 2) (b) 1/[4 + (x -6)2] (c)

cos 12;rjc

(d)

sin 12 ttx

5.81 Explain why a Fourier sine transform and a Fourier cosine transform can be ap­ proximated by a fast sine transform and a fast cosine transform, respectively.

Fourier Optics

In this chapter, we will discuss some applications of Fourier analysis to optics. We will show how FAS can be used to analyze some important problems in optics, such as the behavior of diffraction gratings and imaging with lenses.

6.1 Introduction—^Diffraction and Coherency of Light In this section we will derive the fundamental formula of diffraction theory. Although we do not have space for a completely rigorous discussion (see [Go] or [Bo-W]), our treatment should capture the main ideas. If the presentation below is found too obscure, some readers might wish to take Formula (6.22) fo r granted (it is a classical formula in diffraction theory); however, in that case, do read Definition 6.1 where the notation used throughout this chapter is given. We shall assume that unpolarized light of wavelength À traveling from a point P to a point Q along a curve C can be described as follows. The light amplitude V^(2, 0 at the point Q is given by it(Q,t)

X\C\

( 6 . 1)

where V^(P, t) is the light amplitude at P , |C| is the length of the curve C, and J q r)(s) ds is the line integral (using the arc-length differential ds) of the index of refraction r]{s) along the curve C. Since s represents arc length and r}{s) is the index of refraction, fQT](s) ds is often called the optical length of C (as opposed to the length of C which equals J q 1 ds). We should note that we are omitting a time delay tQ in (6.1), where tQ is the time it takes light to travel from P to Q along the curve C. In other words, to be absolutely precise, instead of t/t(P , t) in Formula (6.1) we should have V^(P, t — tQ). We omit the time delay tQ for the sake of simplicity. At the distances we shall consider, the speed of light is so great that tQ ^ 0.

247

6. FOURIER OPTICS

248

Although Formula (6.1) may appear strange, it will allow us to quickly derive the fundamental formulas of optical diffraction and imaging. Formula (6.1) is related to Fermat’s principle and to the Feynman path integral method in optics (which is a more rigorous development of the approach described here). A couple of examples should help to clarify the meaning of Formula (6.1). First, though, we need to adopt some convenient notation. To enable us to write our formulas compactly, we shall adopt the following vector notation. D EFINITIO N 6.1: A vector x will have two components x = (x, y) and the point (x, y, D) will be denoted by (x, D). When we want to express integrals like

/

oo

poo

/

f(x,y)dxdy

or

-oo J —oo

/

oo

poo

I

g(u,v)dudv

-oo J —oo

then we will write

f

f i x ) dx

JR2

or

/ g(u) du. JR2

And, we will write |u — x| for the distance [(« —x)^ + (tJ — and X.

between u

Example 6.1: Suppose that light of wavelength X is shone onto an opaque screen with a single point P = (x, 0) cut out of it. See Figure 6.1. Assume that the index of refraction (u. D)

FIGURE 6.1 Diffraction from a single point. rj is a. constant, say ?/ = 1 for simplicity. Then the light at a point Q = (u, D), on a parallel plane D units in front of the screen, will be caused by the light traveling

6.1. DIFFRACTION AND COHERENCY OF LIGHT

249

along the straight line segment C from (x, 0) to (u, D). (See, however, Remark 6.1 below.) Since f ^ l d s = |C|, we obtain from (6.1) that l//'(X, 0, i) ;27r|P|

( 6 .2)

Now, we define the intensity /(u , D) of the light at (u, D) by -T

t)\^dt.

/ ( u , £ ) ) = lim T^OO T

fi

(6.3)

Equation (6.3) is a mathematical idealization (since we cannot actually observe T -> oo). Consequently, we shall use instead 1

Hu,D) = \xlr{u,D,t)\^dt. 1 Jo

(6.4)

where T is taken so large that the limit in (6.3) can be assumed to be attained. For example, T might be the time involved in taking a photograph, which is huge relative to the fluctuations of visible light. Using (6.2) in (6.4), we obtain (since \e^^\ = 1 for all real 0) /(u , D) =

/(X, 0)

X2|Cp*

(6.5)

Formula (6.5) says that the light intensity decreases in inverse proportion to the square o f the distance, |C|, from (x, 0) to (u, D). This result is consistent with light being an electromagnetic phenomenon, and it shows the purpose of dividing by ICl in Formula (6.1). [The reason for also dividing by X will only be clear after our work in the next section.] REMARK 6.1: In the example above we made the assumption that only light traveling along the straight line segment C from (x, 0) to (u, D) contributes to the light V^(u, D, t) at (u, D). This assumption is consistent with classical geometrical optics, since the light rays are straight in a medium with constant index of refraction. However, a more rigorous approach like Feynman’s path integral method would assume that the photons of light can travel all possible paths C from (x, 0) to (u, D). Then, an integral of quantities similar to those in (6.2) would be formed over all paths C from (x, 0) to (u, D). We will not pursue this approach here; because when a long time average is taken (as we did above) the results of this

6. FOURIER OPTICS

250

more rigorous method end up reducing to formulas like the ones described in this section. For a fascinating nontechnical discussion of Feynman’s method, see [Fe].

I This last example showed why the factor 1/|C | is in Formula (6.1). Our next example will show why the phase factor,

is in Formula (6.1). Example 6,2: Suppose that light of wavelength X is shone onto an opaque screen with two points cut out of it. Let’s say these points are at (x, 0) and (z, 0). Assuming again that we have a constant index of refraction ^ = 1, we now must sum the contributions of the light from (x, 0) and (z, 0). If we let Cx denote the line segment from (x, 0) to (u, D), then (6.2) generalizes to

^

XICxI

XlCzl

( 6.6)

We will now calculate the intensity I (u, D) using Formula (6.4). First, we observe that I,/

r. . m 2

\ir(u, D, 01 =

llA(X,0,i)l2

,o...+ A2|Cx |'

-l“2Re

|l/r(z,0,i)P

^^|Czl

> ( x , 0, 0 ,0 ji^nC vi-iC ^n' . , ^ 2 |C x |.|C z |

(6.7)

Since our next step will be to compute l / T times the integral f j dt of each side of (6.7), the following definition will be useful. The coherency function F(x, z) is defined by the identity

r(x , z)V /(x , 0)/(z, 0) = I

f

f i x , 0, 0V^*(z, 0, t) dt.

(6 .8 )

^ 70

{Note: in the special case that either /(x , 0) = 0 or /(z , 0) = 0, we define F(x, z) = 0.) With this definition in mind, we now use (6.4) to calculate the

6.1. DIFFRACTION AND COHERENCY OF LIGHT

251

intensity I (u, D) from (6.7). We obtain

/(u , D) =

7(x,0) _^ 7(2,0) A.2|Cx |2

X2|Cz P

rr(x, z)V7 (X,0)7 (2,0) ^

L

A2lCx|-|Cz|

(6.9)

Formula (6.9) is very complicated, reflecting the fact that we placed no constraints upon the nature of the light radiating onto the opaque screen (other than it is of one wavelength A,). There are two fundamental constraints that can be used to simplify (6.9). The first is called spatial incoherency (incoherencyy for short). The light will be called incoherent if the coherency function F satisfies F(x, z ) = 0

[Incoherency].

fo rx ^ z

( 6 . 10)

In this case, (6.9) simplifies to

7(u, D) =

7(x,0) , 7(z,0) .o +

A2|Cx P

X2|Cz |2

( 6 . 11)

Thus, in the case o f incoherent illumination, the intensity at (u, D) is the sum o f the intensities that would result from the points (x, 0) and (z, 0) individually. The second form of constraint is called spatial coherency {coherency, for short). It is coherency that we will deal with most frequently in this chapter. In order to define coherency, we first note that F(x, z) satisfies the following inequality: l F ( x ,z ) |< l .

(6. 12)

This inequality is proved using the Schwarz inequality (see Exercise 6.7). In­ equality (6.12) says that F(x, z) cannot have values of magnitude greater than 1. Coherency consists in F(x, z) attaining this maximum value of 1. That is, the light will be said to be coherent if F(x, z) satisfies F(x, z) = 1

[Coherency].

If the light is coherent, then using (6.13) in (6.9) yields

(6.13)

6. FOURIER OPTICS

252

I{n,D) -

/(x ,0 ) /(z ,0 ) + X2|Cx|2 A2|C z |2 2 V /(x ,0 )/(z , 0)

^ A2|Cx |- |C z I

2;r -(|Cxl — |Cz|)

(6.14)

Thus, in the case of coherent illumination, we see from this formula that the intensity 7(u, D) contains an oscillatory term: 2 V /(x ,0 )/(z ,0 ) ^"|Cxl • |Cz| As the point

■cos

'2n

T

(|Cxl - ICzl)

(u, D) varies, this term will vary between 2 V /(x , 0 )/(z, 0) A^ICxI • ICzl

and - 2 V /( x , 0)7 (z, 0) A^ICxI • ICzl thus creating interference fringes superimposed upon the background intensity described by /(x , 0) 7(z, 0) A2|Cx P

A2|Cz P '

A reader who desires to graph examples of Formulas (6.11) and (6.14) might begin by looking in the Textbook formulas entry in the on-line help for the function creation procedure of FAS. Much experimental work has verified that all types of illumination will satisfy the coherency condition (at least to a good approximation) provided the points (x, 0) and (z, 0) are sufficiently close (how close is called the coherency interval of the light). Some light has such a small coherency interval that it can be assumed incoherent, while other kinds of light satisfy the coherency condition (6.13) over a wide range. For example, in Figure 6.2 we show a diagram illustrating how coherent light can be created. If light from a strong source is channeled through a point drilled in an opaque screen (a pinhole filter), and then collimated by a lens sitting at a focal distance from the point in the screen, then the light immediately behind the collimating lens is coherent. (This will be shown in Section 6.8; see Exercise 6.49). Often a laser is used as the initial source, since the light from the laser is nearly coherent to begin with.

6.L DIFFRACTION AND COHERENCY OF LIGHT

Focusing Lens

Pinhole Filter

Monochromatic Filter, Wavelength k

253

Collimating Lens

Focal Length

Coherent Light

FIGURE 6.2 Creation of coherent, monochromatic light. We shall now determine the intensity / (u, D) generated from an aperture in an opaque screen that is illuminated by coherent light. See Figure 6.3. In this case, we

FIGURE 6.3 Diffraction from an aperture. generalize (6.2) and (6.6), by forming an integral that consists of superpositioning all contributions along light rays Cx from (x, 0) to (u, D). Thus, we have f{u ,D ,t)= i JR2

>^|Cxl

(6.15)

In (6.15) we have introduced the aperture function A(x) to account for the finite extent of the aperture. In particular, we assume that A(x) = 0, where

for |x| > R

(6.16)

is a finite positive number which marks off the extent of the aperture. The

6, FOURIER OPTICS

254

type of aperture function that we will typically look at has the form A(x) =

1 0

if (x, 0) is in the aperture if (x, 0) is not in the aperture.

(6.17)

But, other types of aperture functions could be used to take into account varying amounts of light transmitted through the aperture as well as other effects. Now, we calculate the intensity /(u , D) using (6.4) and (6.15):

f

/(u,£>) = 1 ; f i n , D , t ) f *in, D , t ) d t T Jo

-LL

V ^|Cxl

MCz\

j \ i x , o , t ) r ( ^ , o , t ) d t ^ d \d z .

(6.18)

Using the coherency assumption, that F(x, z) = 1 over the extent o f the aperture, the last expression in brackets in (6.18) equals y/l{x, 0)7 (z, 0). Hence, I(u,D)=i f JR 2

- ! l ^ e ' x l C x l ^ ^ e - ' x l C z l y / ( x , O)/(z, 0 ) d x d z . MCx\ MCz\

The integrals then separate out and we obtain

/( u

i5)vZ55>,-¥iCz, J .

rr UR2

Substituting

MCx\ ^|C xI

J UR2

MCz\

\

Xin place of zin the second integral above, we get I in , D) =

/* \JR^

A(x) V / ( x, 0 ) ^,2.|C ^I^^ ^|Cxl

2

(6.19)

255

6.1. DIFFRACTION AND COHERENCY OF LIGHT

If we define the transmittance function T (x) through the aperture by r(x ) = A (x)V /(x, 0) then (6.19) can be rewritten as ( 6 .20)

7(u, D) = | i U r 2 >-|Cx I

For simplicity, we shall usually assume that the light intensity / (x, 0) is a constant over the aperture. In fact, to simplify as much as possible, we shall assume that I (x, 0) = 1

for X in the aperture.

( 6 .21)

Using (6.21), we find that T (x) equals A(x), hence Formula (6.20) becomes

7(u, D) = \ f

A(x)

'x lC x I j,

\Jk 2 A.|Cxl

( 6 .22)

Formula (6.22) is a classic formula in diffraction theory. For most of the rest of this chapter we will examine (6.22) for various types of apertures. REMARK 6.2: A rigorous treatment of our subject (see [Go, Chapter 3] or [BoW]) would include an obliquity factor in the integrand of (6.22). However, for the type of illumination illustrated in Figure 6.2, this obliquity factor is approximately 1. This approximation by 1 is a standard approximation in diffraction theory. I We close this section by introducing a convenient notation for the integral on the right side of Formula (6.8), since it will appear frequently in subsequent formulas. D EFINITIO N 6,2:

The cross-correlation function y(x, z) is defined by

y(x, z ) =

1 t lim — / \l/(x,0,t)'il/''(z,0,t)dt. T^oo T Jo

(6.23a)

For calculational purposes, however, we shall use 1 rrT

f i x , 0, t ) f * {z , 0, t ) d t

(6.23b)

5. FOURIER OPTICS

256

where T is the same large time value used to calculate the intensity in For­ mula (6A). Based on Definition 6.2 we can simplify the definition of the coherency function r to the following: r ( x ,. ) = - ^ = ^ 4 = , V y(x, x)y(z, z)

6.2

(6.24)

Fresnel Diffraction

In this section we will discuss Fresnel diffraction, which is one of the funda­ mental types of diffraction in optics. Fresnel diffraction is based on an integral approximating the one in Formula (6.22) for moderate distances D from the aper­ ture. Formula (6.22) says that the intensity /(u , D) is given by

|JR2 >^|Cxl

(6.25)

Since Cx is the line segment from (x, 0) to (u, D) we have |Cxl = [|u -x |2 + £»2j5

(6.26)

We will now show that A.|Cxl in the denominator inside the integral in (6.25) can be replaced by XD, To show this we look at the ratio A|Cxl = [l + |u XD

(6.27)

and we assume that lu ~ x |/D ¿5 small enough that [ l + |u - x |V z > ^ ] " ^ 1.

(6.28)

In fact, the approximation in (6.28) will hold to within 95% accuracy provided lu - x l/D < 0.32.

(6.29)

For a small aperture (relative to D), this inequality is not too restrictive on what u-values we allow in (6.25).

6.2, FRESNEL DIFFRACTION

257

Because of (6.28) and (6.27) we can say that A,|Cx| ^ AD and then (6.25) simplifies to 2

f A(x)e^^^ AD 7 r 2

-i—

/(u , D)

dx

(6.30)

We will now show how to approximate the phase factor in the integrand in (6.30). Here we have to approach the approximation differently than above, since for most light (e.g., visible light) the reciprocal of the wavelength, 1/A, is enormous. Consequently, In D I'k is a terrible approximation of 2;r|Cxl/A. To obtain a useful approximation we multiply (6.27) by I ttDIX, obtaining

^ IC x I = ^

[ l + |u - xlV£>2]

(6.31)

Using the series expansion from calculus (1 + a)2 = 1 + - a —

+ . ..

(6.32)

iu - x r + .

(6.33)

to rewrite the right side of (6.31), we have 2n

In D

7t

T

Now, let’s assume that |u —x |/D is small enough that tt|u —x|"^/4AD^, and higher power terms, can be neglected in (6.33). Then (6.33) simplifies to 27T.^ .

2ttD

7t

.

.9

(6.34)

From (6.34) we obtain e'x iC x I

g i^ D ^ ^ \u - x \'

(6.35)

Substituting the right side of (6.35) into (6.30), and factoring the constant outside the integral, we have

/(u , D)

f

Jr 2

A(x)e^n^“

dx

(6.36)

6. FOURIER OPTICS

258

We have now established two things. First, since \e^ ^

/(u , D)

— f .A (x)cS I«-^l XD JR2

|^ = 1 we have 2

(6.37)

dx

Formula (6.37) is the Fresnel diffraction formula. Second, all of our approximations remain valid for the integral describing t//"(u , D, t) in (6.15). Thus, from (6.15) we obtain

f

V^(u, D, t) ^ ------A(x)V^(x, 0, XD JR2

dx.

(6.38)

Formula (6.38) will prove useful later, when we discuss imaging by a lens. Here is an example of how FAS can be used to analyze the Fresnel diffraction formula (6.37). Example 6.3: Suppose that coherent light of wavelength k = 5 x 10“ "^mm (greenish-blue) light is shined onto a square aperture with aperture function A(x) = A(x, y) given by A (v,y) =

1 if |x| < 1 mm and |y| < 1 mm 0 if |x| > 1 mm or |y| > 1 mm.

Graph 7(u, D) for D = 100 mm, 200 mm, and 300 mm. SOLUTION

It is not too difficult to see that A{x, y) = rect

rect

.

(6.39)

For u = (w, i;) we have

(6.40)

Thus, using (6.39) and (6.40) in the integral in (6.37), we see that the integral over separates into iterated integrals from —oo to oo in each variable x and y.

6.2. FRESNEL DIFFRACTION

259

Therefore, we obtain

/(u , D) =

1 /x\ — re c t(-), D'>2 J-OO o■Dr^

(6.41) 0^12 (A D ) 2

J-OO

'^2/

Because of (6.41) we see that it suffices to graph I\ (u, D), defined by 2

7i(w,D) =

(6.42)

i- r J-OO r - ( f^ )2 /' (XD)2

since the second factor in (6.41) is just I\ (f, D), which is obtained from 1\ (w, D) by changing variables from x to y and u i o v . In particular, we have 7(u, D) = 7i(w ,D )7i(i;,D ).

(6.43)

Now, the integral in (6.42) (6.44) (AD)2 J-OO

^2/

can be expressed in real and imaginary parts as

cos — (u — x r AD^ ^

dx

, sin — (u — x)^ AD (AD) ^2

dx.

(6.45)

Therefore, we can graph the real and imaginary parts of the integral in (6.44) by do­ ing two convolutions. First, we convolve rect(x/2) with (AD)~^/^ c o s [(7 t / A D ) x ^] to produce the real part. Second, we convolve rect(x/2) with (AD)“ ^/^ sin[(7T/AD)x^] to produce the imaginary part. If g l stands for the real part and g2 stands for the imaginary part, then g l(x ) A 2 H- g2(x) A 2 will create the intensity 7i (m, D ) in Formula (6.42). We show graphs of I\(u, D) for A = 5 x 10” "^ and

6. FOURIER OPTICS

260

H----h-M-

H----h-H----h

(a) aperturej D = 0 mm X interual: [ -2, 21 X increment = A V interual: [ -.Z, 1.81 V increment =

.2

(c) D = 200 mm X interual: [ -Z. 21 X increment = A Y interual: I -.2, 1.81 Y increment =

,2

(b) D = 100 mm X interual: 1 -2, 21 X increment = .4 Y interual: [ -.2, 1.81 Y increment = .2

(d) D = 300 mm X interual: 1 -2, 21 X increment = A Y interual: [ -.Z, 1.81 Y increment =

.2

FIGURE 6.4 G raphs of the intensity function I\ (u, D) from Example 6.3.

D = 100, 200, and 300 in Figure 6.4. These graphs were produced over an initial interval of [—4,4] using 4096 points. By changing variables in (6.42) from x t o y and from u to v, we obtain I\ (v, D), the other factor in Formula (6.43) for /(u , D). The intensity function /(u , D) then generates a diffraction pattern like the one shown in Figure 6.5. We see in Figure 6.5 the classic checkerboard (or plaid) pattern of a Fresnel diffraction pattern from a square aperture. I

REMARK 6.3: (a) It is clear from Figure 6.5 that the Fresnel diffraction pattern represents a distorted image of the original aperture, (b) It is also worth noting the similarity between Fresnel diffraction and the electron diffraction that

261

6.2. FRESNEL DIFFRACTION

FIGURE 6.5 Fresnel diffraction from a square aperture; computer-generated graph.

we described in Section 5.7 of Chapter 5. In fact, if we put XD = Aittifi/lm)

(6.46)

then the integral in (6.44) differs from ( / * tF){u) in Formula (5.87) of Chap­ ter 5, for f { x ) = rect(jc/2), by only a factor of Since | / = 1, we can ignore this difference, which does not occur anyway when we graph the expression in (6.42). Equation (6.46) represents more than just a mathematical relationship. Because electrons (like all particles in quantum mechanics) have wavelengths, these particles diffract through apertures in much the same way as light does. In fact, everything we discuss in this chapter on coherency, diffraction, and imag­ ing applies to electron optics, as in electron microscopy, and electron diffraction. Furthermore, neutron diffraction spectroscopy uses similar ideas. I As we can see from the example above, FAS can be used to graph the individual factors of the intensity function /(u , D) whenever the aperture function factors in the following way: A(x) = A i ( x ) A 2(y),

(6.47)

When this happens, we have

I(u,

V,

D) = I\(u, D)l2(v, D)

(6.48)

6. FOURIER OPTICS

262

where 1

poo

-

h{u,D) =

dx

(6.49)

dy

(6.50)

(XD)2 J - cc

h i v , D) =

^ Í

D^2 J~00 J~ (XD)2

In such a situation, we shall call the aperture function A separable. There are a number of separable aperture functions that are important cases for study, for example, aperture functions for rectangular apertures or aperture functions that are Gaussian exponentials. These examples, and the case of edge diffraction, are treated in the exercises.

6.3

Fraunhofer Diffraction

In this section we begin our discussion of Fraunhofer diffraction, which plays a very important role in applications of diffraction theory. This type of diffraction occurs as a limiting case of Fresnel diffraction, either when the distance from the aperture is very large or when the dimensions of the aperture are very small. Fraunhofer diffraction theory is essential for understanding the diffraction of X-rays from crystals. X-ray diffraction photographs from crystals are used to determine the underlying crystal structure. For example, the structure of DNA as well as many protein structures have been determined from diffraction patterns. Another important application is Fraunhofer diffraction from diffraction gratings. This diffraction plays a role in physical chemistry, where it is used to identify chemical compounds. It is also used in astronomy, where it provides a way of identifying the chemical constituents of stars. We will now show how Fraunhofer diffraction arises as a limiting case of Fresnel diffraction. From Formula (6.37) we have for the diffracted light intensity

/(u , D)

- ÍJr 2

dx

(6.51)

Expanding the exponential in (6.51), we get gS|U-X|2 ^

(6.52)

263

6.3. FRAUNHOFER DIFFRACTION

Recall the assumption in (6.16) that for |x| > R.

A(x) = 0,

(6.53)

We will show that 1,

for lx| < R

(6.54)

(far-field case)

(6.54a)

(small aperture case).

(6.54b)

wing two cases:

D

oo

R ^O

For the case of (6.54a), we observe that, for \x\ < R, as D ^

7t

^

n

7VR^

-

(6.55)

0.

0 < — xr < —

~kD

oo

XD

Hence, JO

1

so (6.54) holds. On the other hand, for the case of (6.54b) we still have (6.55)

provided R is sufficiently small (and D stays fixed). Thus, (6.54) holds again. Now, assuming that (6.54) holds, we obtain from (6.52) that for |x| < R.

(6.56)

Since we have (6.53), the integral in (6.51) is actually taken over only those x for which |x| < R. So we can use (6.56) to simplify the integral in (6.51). Substituting the right side of (6.56) into the integrand in (6.51), and factoring the exponential involving |u p outside the integral, we obtain

/(u, D)

e^D lui"

XD

L

A ( x ) e - ‘w ^ - ^ d x

(6.57)

Since the complex exponential involving |u p has magnitude 1, Formula (6.57) becomes 2

/(u , D)

A(x)e

(6.58)

6. FOURIER OPTICS

264

Formula (6.58) is the Fraunhofer diffraction formula. The integral in (6.58) can be viewed as a two-dimensional Fourier transform. The two-dimensional Fourier transform of A(x) is denoted by A(u), where A(u) is defined by A(u) =

f

JR2

A(x)e

(6.59)

Using Formula (6.59), we can rewrite Formula (6.58) as 2

/(u , D)

(6.60)

which shows that the intensity function from Fraunhofer diffraction is the modulussquared o f the (scaled) Fourier transform o f the aperture function. Example 6,4: Discuss Fraunhofer diffraction from a rectangular aperture, having aperture func­ tion for X = ( x ,y ) A(x) =

1 0

if |x| < 0.1 mm and |y| < .05 mm if |x| > 0.1 mm or |y| > .05 mm.

SOLUTION It is not too hard to see that A(x, y) = rect(5jc) rect(lOy). Sub­ stituting this expression for A(x, y) into (6.59) and separating into two integrals of X and y, we obtain [for u = (u,v) and u • x = wjc + i;y] A(u,v)=

f

JR2

rcct(5x)

rect(5x)^

dxdy

i

dx I

rect(10y)e iln vy dy

Thus, from (6.60) we obtain 7( m, V, D) ^

^ sinc^ sinc^ ( —-— ^ . (50A.D)2 \5XD/ \ 10X d J

(6.61) ^ ^

6.3. FRAUNHOFER DIFFRACTION

265

In Figure 6.6(a) we show a computer generated graph of I(u, v, D); notice how well it corresponds to the photograph of such a diffraction pattern in Figure 6.6(b). From (6.61) it is easy to calculate where there is zero intensity in the diffrac­ tion pattern. There will be zero intensity whenever sine ^[w/(5AD)] = 0 or sinc^[i;/(10AD)] = 0. Hence the equations for zero intensity are u = ±5AD, dilOAD,. ..

and

i; = iblOAD, ± 20A D ,. . . .

Notice that these equations define lines of zero intensity in the u-v plane, and they are located at positions that are proportional to AD. I Example 6.5: Describe the Fraunhofer diffraction pattern from a vertical slit.

SOLUTION A vertical slit is a very thin rectangular aperture. Hence, its aperture function is A(x, y) = rect

rect

(6.62)

where the height b is many times larger than the width a. The transform of A is A(u, v) = ab sine (au) sine (bv). Therefore, the intensity function for the diffrac­ tion pattern is . ^

0/ au\

(ad)

.

'

(6.63)

A computer-generated graph of such a function is shown in Figure 6.7. The diffrac­ tion pattern has zero intensity at the lines . AD 2AD 3AD u = ± ---- , ± -------, ± -------, a a a

AD 2AD 3AD V ==±— , ± —— , ± - — , b b b (6.64) Since b is much larger than a, the pattern is visible mainly along the w-axis. There is a strip of high intensity along the w-axis (of width IXDjb), marked off by spots of low intensity around the points u = EXDj a, ±2XD!a, ± 3 X D ! a , ___ I and

266

6. FOURIER OPTICS

FIGURE 6 .6 Fraunhofer diffraction by a rectangular aperture, (a) Computer-generated graph of the intensity I (u, v, D) (using a logarithmic intensity scale and set­ ting XD = 1). (b) Photograph of a Fraunhofer diffraction pattern from a rectangular aperture. (From Walker, J.S., Fourier Analysis^ Oxford Univer­ sity Press, Oxford, 1988. With permission.)

267

6A, CIRCULAR APERTURES

FIGURE 6.7 Diffraction pattern from a vertical slit (negative image).

6.4

Circular Apertures

In this section we will describe Fraunhofer diffraction from circular apertures. The diffraction integral for a circular aperture is usually described in terms of Bessel functions (see [Wa, Chapter 7.6]). The advantage of the method described below is that no special knowledge of Bessel functions is needed. For a circular aperture, the aperture function is if |x| < R if 1x1 > R

A(x)

(6.65)

where R is the radius of the aperture. If we use (6.65) in Formula (6.59) we have A(u) =

f

( 6 .66)

J{X:\X\

\AD \= €.

From all these equalities it is not hard to deduce that

\CD\ = R 2 - [ r I - \

u\^Y

\BC\ = e - |AB| - \CD\.

( 6 . 120 )

We now make the approximations .1

2

2Ri

24

( 6 . 121)

288

6. FOURIER OPTICS

which are similar to the approximations we made in discussing Fresnel diffraction in Section 6.2. These approximations will certainly be valid when R\ and R 2 are relatively large in comparison to the radius of the lens aperture. Using (6.121) in (6.120), we obtain ur ¿K\

ZiV2

Employing (6.122), we have \AB\ + rj\BC\ + \CD\

e r j ( I - v) (\AB\ + \CD\) "1

1 1 lup

Before we make use of this result, we make the following definition. For a double convex lens with spherical surfaces having radii of curvature R\ and R 2, we define the focal length f via the formula (6.124)

From (6.124), we can express (6.123) as \A B \-\-n\B C \ + \ C D \ ^ e r ] -

2/

(6.125)

Combining (6.125) with (6.119), we have (6.126) Substituting the right side of (6.126) in place of the last factor in (6.116) yields

1

x\c\

(6.127)

XD

As a final simplification, we define S to be the constant ^ (D + e »?), and then (6.127) becomes X|C|

XD

(6.128) ^ ^

6.9. IMAGING WITH A SINGLE LENS

289

Using (6.128) we can express the light lens as

D + e, t ) at the exit plane of the

^ (u , D + €,t)

f

Jr 2

A(x)i/f(x, 0,

*1"^ i(x£ V

i*(u). (6.129)

The aperture function A(x) was introduced in Section 6.1. The factor P(u) is the pupil function o f the lens, whose main purpose is to express the fact that the lens has a finite extent. Most often, we will assume that P(u) is the simplest type of aperture function, i.e., if u lies inside the lens aperture if u lies outside the lens aperture.

P(u)

(6.130)

Formula (6.129) consists of two parts. The first part is the integral

—

f

XD J r 2

A (x)\lr(x, 0,t)e>^D^^

dx

(6.131)

which describes Fresnel diffraction from the aperture to the entrance plane o f the lens. The second part. P (u )

(6.132)

describes the phase transformation due to the lens. After the light described by (6.131) has passed through the lens, it has been multiplied by the phase factor ^~i7z\n\^/xf Yens pupil function P(u). The pupil function P(u) in Formula (6.130) describes a lens having no aberra­ tions (this is also called the diffraction-limited case). The pupil function plays the role of an aperture function when light diffracts through the lens aperture. When there are no aberrations, then the only limitation to perfect imaging is the finite ex­ tent of the lens, through which light from the original aperture must diffract. (Since we are dealing with monochromatic light, we are ignoring chromatic aberration, which affects all simple lenses.) In subsequent sections, we will examine the profound implications of For­ mula (6.129).

6.9

Imaging with a Single Lens

In this section, we will derive the general formula for imaging with a single lens. This derivation shows that there is a wave optical reason for the classic lens

6. FOURIER OPTICS

290

equation. In subsequent sections we will discuss two specific cases of the imaging formula, the cases of coherent and incoherent illumination, which are the two fundamental types of illumination. Lens -7

/

Object Plane

Exit Plane of Lens

(X. D + A + e)l

\ \

- (u, D + e)

1

(g ? Aperture (Object)

Observation | Plane j

/ _______

Illumination, -------- L_ — n -------- 1 Wavelength X

L

------------- - A --------------------

FIGURE 6.22 Geometry of single lens imaging. Suppose that we have the setup shown in Figure 6.22, a lens situated D units in front of the aperture and an observation plane located A units behind the lens. From Formula (6.129) we have that the light emergent from the exit plane of the lens is described by J8 V^(u, D + 6, 0

XD

f

dxe

A(x)t/f(x, 0,

F (u).

(6.133)

Our formulas in this section will be very unwieldy unless we do something to abbreviate them. Accordingly, we adopt the following conventions. DEFINITIO N 6.3: We will write f instead o f f ^ 2, cill integrals will be over R^. Instead o f ir(x, 0, t) we will just write j/(x, t), the vector x being used to indicate (x, 0). Similarly, we shall write V''(u, 0 place q/“i/^(ii, D + 6, t), and j / ( X, t ) in place o f \lr (X, D A , t). The function xl/'(X,t) represents the light at the observation plane. With these conventions. Formula (6.133) simplifies to

'^{vi,t) ^ - — i A(x)\l/(x,t)e>^D\^ aD J

dxe

F (u).

(6.134)

Applying the Fresnel diffraction formula (6.38) to V^(u, i) instead of A (x)i/r(x, 0, t).

6.9. IMAGING WITH A SINGLE LENS

291

we can express the light f ( X , t) at the observation plane by I

ir{X, t) ^

n

du.

/ V^(u, AA J

(6.135)

Our next step will be to substitute the right side of (6.134) into the integrand in (6.135). To simplify notation we define E(x, u, X) by E(x, u, X) =

V

(6.136)

This quantity is what we get if we combine all the complex exponentials involved in (6.134) and (6.135). Now, substituting the right side of (6.134) into the integrand in (6.135), and combining all the complex exponentials so that we get £(x, u, X), we obtain

if

t)

A(x)xlr(x, t)E(x, u, X )F(u) dx d u .

(6.137)

To save further space, we define c b y c = ^ A + i5, so that

ir(X, t)

X^DA

//

A(x)V^(x, t)E(x, u, X ) P ( u ) d x d u .

(6.138)

Our next step is to simplify the expression for E(x, u, X) in (6.136), which will help to make (6.138) more meaningful. If we expand the quantities |u —x p and | X - u p in (6.136) and recombine exponentials we obtain E(x , u , X ) = e T [ i + i - } ] l u i % - ‘T “ -(§ + ? )^ i% |x P ^ i||X p _

(6.139)

The first exponential factor in (6.139) will be completely eliminated if we assume that 3

+ i-A = 0 .

(6.140)

Formula (6.140) is the classic imaging equation for a thin lens. Let’s assume that (6.140) holds. Then the first factor in (6.139) is just Hence (6.139) becomes

E{x, u, X) = ^

-« ¥ « -(i+ ? )3 ix P 3 ix p _

(6.141)

6. FOURIER OPTICS

292

Substituting the right side of (6.141) in place of E( x, u, X) in (6.138), and factoring the complex exponential involving |X p outside the integrals, we get

k^DA n

p

/ / A{x)\lr(x,t)P(n)e

_27Tj|^i I ^ v

I

^

(6.142)

We also interchanged the integrals to obtain (6.142). Since

Jpin)e

V

+

we rewrite (6.142) as

j

f O i , t)

A(x)i/r(x, t)P

dx. (6.143)

Finally, suppose an observation of the intensity I (X) defined by -T /(X )

- 'r i:

IV^(X, t)Y-dt

is made (for some large value of T). Then we obtain from (6.143) 7(X)

...h ill X i A(z)^^^'^'^P (

XD

+ — M XA/ l

k (x ,

z)dxdz.

(6.144)

where y(x, z) is the cross-correlation function defined in (6.23b). Formula (6.144) is the general equation fo r imaging with monochromatic light. It is extremely complicated; and that is because we have made no assumptions about the initial illuminating light other than its being unpolarized and having a single wavelength A,. In the next section we will discuss one method of simplifying (6.144). This method makes the assumption that we are dealing with coherent illumination. In Section 6.12 we will discuss the other basic method of simplifying (6.144), which makes the assumption that we are dealing with incoherent light.

293

6.10. IMAGING WITH COHERENT LIGHT

6.10

Imaging with Coherent Light

In the previous section we derived the general formula for imaging. In this section we will discuss one of the two fundamental assumptions, coherency of illumination, that can be used to simplify that formula. We assume that the light illuminating the aperture in Figure 6.22 is coherent. That is, the coherency function F(x, z) satisfies the following condition. F(x, z) = 1

[Coherency]

(6.145)

for all points x and z within the aperture. Hence, based on the definition of the cross-correlation function y(x, z) in (6.23b), along with Formula (6.8), we have y (x ,z) = V /(x , 0 )V /(z, 0)

(6.146)

for all X and z within the aperture. For simplicity, we shall again assume, as we did in Section 6.1, that the light intensity is constant over the aperture. In fact, to simplify as much as possible let’s assume that /(x , 0) = 1 and 7(z, 0) = 1 for all X and z within the aperture. We then find that Formula (6.146) simplifies to y (x ,z) = 1

(6.147)

for all X and z within the aperture. For x or z lying outside the aperture, we would have either A(x) = 0 or A(z) = 0, so we are free to use (6.147) for all x and z in the imaging Equation (6.144). Doing so, we find that the two integrals in (6.144) separate into integrals over x and z, as follows:

Substituting X in place of z in the second integral above, we get

/(X )

.4 /

/tt IVi2 ^

2

(— \X D

— ) dx Xa )

(6.148)

294

6. FOURIER OPTICS

To simplify (6.148) further, we make the small object assumption that, for all x lying in the object’s aperture, |x| is small enough that Ì7L|^|2

1,

for X within the object’s aperture.

(6.149)

Formula (6.149) is a natural approximation to make in the field of microscopy, for exaimple. Since A(x) = 0 for x outside the aperture, we can use (6.149) for all x in (6.148). Then, (6.148) simplifies to

/(X)

—

f A{x)P (— ^ \X D

X^DA J

+ — ) dx XA )

(6.150)

As a final simplification, we define the degree o f magnification M by A M = —. D

(6.151)

We then make the substitution x = —s / M (and dx = dsj M^), so that (6.150) becomes

/(X ) W i — A I — ) — ^ P ^ \J M \ M J (A.A)2

( — (X - s) 1 ds 7

(6.152)

The integral in (6.152) is a two-dimensional convolution. The function

-M 7 \-M Ì ).

{magnified, inverted object)

(6.153)

{point spread function).

(6.154)

is convolved with

(x a )2

f— x ), j

The interpretation of the function in (6.153) as a magnified, inverted object is fairly clear (the factor l / M on the outside is to preserve the 2-Norm squared, which measures total energy). In the following four examples, we will show how FAS can be used to understand the role of the function in (6.154), which is called the point spread function (or PSF).

295

6.10. IMAGING WITH COHERENT LIGHT

Example 6,9: Suppose that coherent light having wavelength A = 5 x 10“ ^ mm illuminates an edge having aperture function At \ A{x, j ) =

10

ifx>0 .. „ if X < 0

„

and this edge is imaged by a lens with a square aperture of side length 40 mm and focal length 1 m. Assume that D = 2 m, and A = 2 m. Describe the image that results. SOLUTION We will use (6.152) to describe the image (even though, strictly speaking, the small object condition is not satisfied). By the scaling property. P(AAx)

(A.A)2^ ( a.a ^ )

(6.155)

where, for this example, since A.A = 1, P(XAx) = P{ x, y) =

1 0

if |x| < 20 and |y| < 20 if |x:| > 20 or lyl > 20

= rect(^)rectm . V40/ V40/

(6.156)

It follows that

(kAy

P ( — X I = 40 sine (40X) 40 sine (40F). \iA /

(6.157)

When this last function is convolved with A ( —X, —F), the integral separates into integrals over X and Y separately. Since the integral along the Y -direction will always be 40 sine (40F) dY = rect (

/ -0 0

,

V 4 0 / l3 .= o

= 1 we only need to eonvolve 40 sine (40X) with

’

—i 1 10

f or X > 0 for X < 0

{Fourier inversion)

6. FOURIER OPTICS

296

to obtain a representation of the image (Note: by convolving with the original edge function, rather than its inversion through the origin, we follow the standard procedure in optics.) Using FAS to perform this convolution, we get the results shown in Figure 6.23. In Figure 6.23(a) we have the graph of 40 sine (40Z), and

(a) PSF X interual: [ -1, 1] Y interual: I -18, 521

X increment = Y increment =

H----^----1----h Cb) edge X interual: [ -10, Y interual: t -.1,

10] 1.41

X increment = 2 Y increment = .15

(c) image of edge X interual: [ -.5, Y interual: [ -.2,

.51 1.31

.1 X increment = Y increment = .15

FIGURE 6.23 Imaging of an edge using coherent light. in Figure 6.23(b) we have the graph of the edge. The resulting image, obtained by convolving these two functions and then squaring, is shown in Figure 6.23(c). To obtain these figures we used an initial interval of [—10,10] and 4096 points. The graph shown in Figure 6.23(c) is a good prediction of the effects of coherent imaging. See [Go, p. 133] for a nice photograph illustrating this. In particular, the oscillation and Gibbs’ phenomenon at the edge boundary which appear in Figure 6.23(c) are plainly visible in that photograph (the Gibbs’ effect produces a bright band near the edge boundary). These are serious defects of coherent imaging (they can be mitigated, however, through image processing; see Exercise 6.63). A

6.10. IMAGING WITH COHERENT LIGHT

297

more subtle defect involves the displacement of the edge boundary in the image. If the image edge boundary is located at the point of half of maximum intensity, then (because of the Gibbs’ effect) the boundary is located slightly to the right of the true object edge boundary. I Example 6,10: Suppose that coherent light having wavelength A. = 5 x 10~^ mm illuminates a small square with aperture function A(x, y)

1 0

if |jc| < .001 nun and |y| < .001 mm if |x| > .001 mm or |y| > .001 mm

and this square is imaged by a lens having focal length of 1/11 m with a square aperture of side length 20 mm. Assume that Z) = 0.1 m, and A = 1 m. Describe the image. SOLUTION

10

In this example, the magnified inverted object is defined by

V io ’ lo ;

if |x| < .01 nun and |y| < .01 nun if |x| > .01 nun or |y| > .01 mm

jo

0.1 rect(50x) rect(50y). The PSF is the Fourier transform of (since A.A = 0.5) P(0.5jc,0.5y)

1 0

if |x| < 20 and |y| < 2 0 if |jc| > 20 or |y| > 20

= rect(^)rectm . V40/ V40/ Therefore, the convolution in (6.152) splits into two identical convolutions in x and y. The convolution with respect to x is

f

rect(50^)40 sine [40(x —s)] ds.

In Figure 6.24 we show details of this convolution process. In Figure 6.24(a) we show the graph of 40sinc(40x), which corresponds to the PSF along the x-direction, while in Figure 6.24(b) we show the the graph of \/O re c t(5 0 x ), which corresponds to the magnified, inverted image. The convolution of these two

298

6. FOURIER OPTICS

(a) PSF X interval: [ -1, 11 Y interval: [ -15, 451

H--- 1--- 1--- h

(b) point-like object X interval: [ -1, 11 Y interval: [ -.1, .41

X increment = .2 Y increment = 6

H--- ^----h

(c) X interval: [ -1, 11 Y interval: [ -.1, .31

X increment = .2 Y increment = .05

X increment = .2 Y increment = .04

FIGURE 6.24 Coherent image of a point-like object.

corresponds to the magnified, inverted image. The convolution of these two func­ tions, which is shown in Figure 6.24(c), is essentially just a reproduction of the PSF in Figure 6.24(a). For this reason, the function

XA

/

is called the point spread function. The effect of the PSF is to spread out a point, in this case a very small square, or pixel The convolution in (6.152) can be interpreted as a summation of the spreads of all the pixels composing the original image. (Note: the actual image will be the square of the convolution shown in Figure 6.24(c). We wanted to emphasize here, however, the reproduction of the PSF by imaging a small point-like object.) I

6.10. IMAGING WITH COHERENT LIGHT

299

REM ARK 6.8: These examples show some of the sensitivities of coherent imaging. Coherent imaging spreads out the images of very tiny features of an object; it also tends to brighten the edges of objects. For instance, dust particles will appear as ringed objects due to their tracing out of the PSF of the imaging lens (see Exercise 6.60). false details include enhancement and fringing of edges (as in our first example), and interference patterns due to the interference of nearby point spreads (see Exercise 6.61). I

Our next two examples illustrate methods of image enhancement that are based on modifying the transforms o f objects. These are just two examples taken from a wide variety of examples that belong to the category known as spatial filtering of images (i.e., modifying transforms of objects in order to enhance their images). Example 6,11: In the field of microscopy there is a method known as dark field imaging. This method is used when intense background light obscures the details in an image. An opaque object (aperture stop) is placed over the center of the lens pupil. This aperture stop blocks out much of the background light and allows the previously obscured details to emerge. In this example we describe a one-dimensional model for dark field imaging. In Figure 6.25(a) we show the graph of the following function (one-dimensional object): A(x) = ex p (-(x /3 ) A 10)[I + .02 c o s (67T x )] (6.158) Its transform is shown in Figure 6.25(b). The central spike is due mostly to the transform of exp(—(x/3) A 10) which forms a kind of bright background on which is superimposed the small oscillations of 0.02cos(6;rx). The dark field method employs a pupil function of the form [see Figure 6.25(c)]:

P df (^A x ) =

0 10 0

if|x |> 4 0 if 2.5 < |x| < 4 0 if|x| 4 0 ifl^ l < 4 0

(6.160)

6. FOURIER OPTICS

300

(a) object X interval: [ -4. 41 X increment = .8 y interval: [ -.5, 1.51 y increment = .2

H— ^— h

(b) transform of object function X interval: t -5, 51 X increment = y interval: 1 -3, 71 y increment =

1 1

H--- 1--- h (d) dark field image X interval: 1 -4, 41 X increment = y interval: 1 -.05, .11 y increment =

(c) dark field pupil function X interval: 1 -128, 1281 X increment = 25.6 y interval: 1 -1, 141 V increment = 1.5

FIGURE 6.25 D ark field imaging.

is shown in Figure 6.26. In order to measure the visibility of the oscillations in the two images, one calculates the relative contrast between the peaks and valleys by contrast =

fmax

fir

+ In

(6.161)

For the bright field image in Figure 6.26(a) we get a contrast of 0.04, or 4%. While for the dark field image the contrast is about 1, or 100%. Thus, the dark field method has improved the image contrast by a factor of about 25. There are problems associated with dark field imaging. As the reader might have observed, in the previous example the contrast was dramatically improved with dark field imaging, but the frequency o f the oscillations was increased by a factor o f 2. This is Sifalse detail in the dark field image. It is this problem of false

6.11. FOURIER TRANSFORMING PROPERTY OF A LENS

301

(b) dark field image X interval: [ -4, 41 X increment = Y interval: [ -.05, .11 Y increment =

(a) bright field image X interval: 1 - 4 , 41 X increment = Y interval: [ -.5, 1.51 Y increment =

FIGURE 6.26 Comparison of bright field and dark field imaging of the object shown in Figure 6.25(a).

details which makes the interpretation of dark field images troublesome. A method which alleviates some of the problems with false details is gray field imaging. In gray field imaging the aperture stop partially transmits a small fraction of the light at the center of the transform, thus providing a gray background in the image (against which more details of the object are visible). Example 6.12: Suppose that the object described by (6.158) is imaged, but now a gray field pupil function of the form P gf (^A jc) =

0 10 1

if |x| > 40 if 2.5 < |x| if |x| < 2.5

40

(6.162)

is used [see Figure 6.27(a)]. In this case, the gray field image shown in Figure 6.27(b) results. The false detail no longer appears. Also, the contrast is about 0.4, or 40%, which is a tenfold improvement over the contrast in the bright field image.

6.11

Fourier Transforming Property of a Lens

As we saw in Section 6.10, with the dark field and gray field imaging examples, sometimes images can be enhanced by manipulating the transform of the object.

6. FOURIER OPTICS

302

H--- h

-l-M

(a) Gray field pupil function X interual: 1 -128j 1281 X increment = 25.6 y interual: [ -2, 131 'i increment = 1.5

(b) Gray field image X interual: [ -4, 41 X increment = .8 Y interual: [ -.5, 1.51 Y increment = .2

FIGURE 6.27 Gray field pupil function and gray field image of the object shown in Fig­ ure 6.25(a).

This requires, however, knowledge of the transform. While computer approxima­ tions can sometimes be used, it is an amazing fact that when coherent illumination is used, nature produces the transform o f the object automatically in the back focal plane o f the imaging lens! [See Formula (6.181).] In this section we will derive this important property mathematically. If we put A = / in Formula (6.139), then we have

E(x , u , X ) =

^

^

(6.163)

Substituting the right side of (6.163) into Formula (6.138) and rearranging the integrals, we obtain

ixp

e^^e^f X ^D f

V^(X, t)

j

j

d u J x .(6.164)

We now show how to simplify the integral

(6.165)

303

6.11. FOURIER TRANSFORMING PROPERTY OF A LENS

which is part of (6.164). First, we observe that ( D \ „-(x + -x )

2

/ D \ = |u p - 2u • ( x + ^ x ) 1+ X + y X

(6.166)

It then follows from (6.166) that (6.167)

Hence (6.165) can be written as — ( XD J

P i u ) e w W \ - ‘^ ^ i ^ + 7 ^ ) dn

x+^ -fXf

I p(u)e^l“-(^+7X)r^u

(6.168)

The integral in brackets is a Fresnel integral, like the one we discussed in Sec­ tion 6.2. In fact, as we pointed out in Remark 6.3, it represents a distorted image of the lens aperture. For example, suppose one uses a square lens aperture with pupil function

P(x, 7) =

I¿1

if \x\ < 20 and \y\ < 20 if |jc| > 20 or I > 20.

(6.169)

Then, we show in Figure 6.28(a) the graph of the real and imaginary parts of

—

f P( u) ew\ ^ - ^ \ ^ du

XD J

along one axis (using XD = 1). Notice that there is some similarity to the pupil function defined in (6.169). This similarity is even greater near the origin; see Figure 6.28(b). In fact, for x near the origin, we have

-XDi j P (u )e ^ l“ "*l^ Ju Ri 0.7166 + 0.6976i =

{b = 0.112).

304

(5. FOURIER OPTICS

H--- 1— ^— h

V interual: [ -.75,

1.251

If increment =

H--- ^ ^ ----h

X interual: [ -10, 10] X increment = 2 If interual: [ -.75, 1.251 V increment =

.2

.2

FIGURE 6.28 Fresnel diffraction from the pupil function defined in (6.169). (a) Graphs of the real and imaginary parts; (b) real and imaginary parts near the origin.

Hence, since x close to the origin implies P(x) = 1, we can make the following approximation: du ^ e'^P(x).

(6.170)

Notice that (6.170) will also hold for x outside the lens aperture, since P(x) is then 0 and [as you can see in Figure 6.28(a)] the left side of (6.170) is approximately 0. We shall assume that (6.170) holds for other pupil functions as well.^ Hence, replacing x by x + (Z )//)X , we have the following approximation:

1 xb

■j

P{u)e>-D l " “ ( ^ + 7 ^ ) l du ^

+

jX j .

(6.171)

Substituting the right side of (6.171) into (6.168) we get

T j P(u)

X) (6.172)

Using the right side of (6.172) in (6.164) we get (writing

For a good example, use FAS to view the graphbook CH6SEC11.

in place of

6.11. FOURIER TRANSFORMING PROPERTY OF A LENS

305

in |Y|2

/

jyt i'vi2

A(x)V^(x,

^

DvP / D \ f ^ \ P \ x - \ - j X j dx, (6.173)

11C I ^

I

Expanding |x + D X //P , and factoring an exponential involving |X p outside the integral, we can rewrite (6.173) as

V

. i i (■ -? )« '■

/

; 2;r V Y ‘ ^-^d x. A { x } f ( x , t )P ( x + 5 x ) e - 'W

(6.174)

From (6.174) we obtain, ««'ng i/ie coherency assumption (6.147) as we did in Section 6.10,

7 (X )« i|^

j

A(x )P ^ + y X j e

(6.175)

The presence of the pupil factor P (x + D X / f ) in (6.175) is known as vignetting. The object A(x) may not fully appear in the integrand if the translated pupil (translated by —D X / f ) does not completely contain the object. One way of minimizing vignetting is to assume that the object described by the function A is very small relative to the lens pupil size. Here we are also assuming that the pupil function has the form described in in (6.130). Let R a stand for the radius o f the object A; in other words, R a is the smallest positive number for which A(x) = 0 whenever |x| < R a - Similarly, let Rp denote the radius of the lens aperture. Our smallness assumption is that R a P p, in which case A(x)P

+ j X ^ = A(x),

for |X| < ^ ( R p - R a ) ^ ^ R p .

(6.176)

Using (6.176), Formula (6.175) simplifies to

/(X)

for |X| < ^ (R p - R a ) ^ ^ R p . A/

(6.177)

6. FOURIER OPTICS

306

Formula (6.177) shows that the intensity in the backfocal plane o f the lens is the modulus-squared o f the (scaled) Fourier transform o f the objectfunction (provided one does not stray too far from the origin). REMARK 6.9:

If we let D -> 0, then (6.175) becomes / ( X ) w | ^ J A(x)F(x)e ‘^ f ^ ' ^ d x

(6.178)

Formula (6.178) does not suffer from vignetting. In fact, we only need to assume that R a < Rp (not much less, just less) in order to obtain from (6.178) (6.179)

/(X )

without the restrictions on X that are in (6.177). (Note: we also obtained For­ mula (6.179) in Exercise 6.53.) The practice of placing an object within the en­ trance aperture of a lens (D = 0 and R a < Rp) is a very common practice, precisely because of the absence of vignetting effects. (See [HTW] for some beautiful examples.) I Finally, we shall show that when an object is placed at a focal length in front of the lens, then an exact Fourier transform is obtained in the back focal plane. To see this, we note that when the object is placed at a distance of D = / units in front of the imaging lens. Formula (6.174) simplifies to ir(X,

A { x ) f { x , t )P( x + X )e -‘v ^ ' ^ d x .

If we make the smallness assumption R a ^(X , 0

Rp again, then we get

A(x)\l/(x,t)e ^ f ^ ^ d x ,

It is also true that when \X\ > Rp if/(X, t) = 0,

(6.180)

for \X\ < Rp — R a ^ Rp-

R a , then A( x) P( x -h X) = 0. Hence for |X| ~> Rp

R a ^ Rp.

Since Rp — R a ^ Rp ^ Rp -I R a , ^ ^ combine these last two results by writing

fi x,

f A(x)fix,

P(X).

(6.181)

6.12. IMAGING WITH INCOHERENT LIGHT

307

Formula (6.181) shows that, except for the pupil factor P(X), the light ir{X, t) at the back focal plane o f the lens is precisely the (scaled) Fourier transform o f A(x)ilr(x, t). This result is the basis for two-lens imaging. A second lens is used to perform a second Fourier transform, thereby obtaining by Fourier inversion an inverted image of the object (see Exercise 6.67). Furthermore, because of (6.181), when modifications of the transform of the object function A are required (as in gray field imaging or Schlieren imaging, for instance), then these modifications can be made in the back focal plane o f the first lens. For example, in gray field imaging, the partially transmitting material can be placed over the origin of the back focal plane of the first lens (instead of over the center of the lens aperture). Since the basic theory and results are quite similar to what we described in Section 6.10, we will not discuss two-lens imaging. The interested reader can find good discussions in [Go] and [li]. Modifying the transform of an object in order to change the image is known as spatial filtering. For more details see [Go], [li], [Pi], [Mi], and [Me].

6.12

Imaging with Incoherent Light

The second main type of imaging is imaging under incoherent illumination. In Section 6.1 we gave one definition of incoherent illumination; see (6.10). Com­ paring this formula with Formula (6.24) we see that incoherency is also defined by y (x, z) = 0

for X ^ z.

(6.182)

From Formulas (6.23b) and (6.4) we also have y(x, x) = /(x , 0).

(6.183)

Thus, if we write /(x ) in place of 7(x, 0), the cross-correlation function y(x, z) must satisfy the following two conditions: (6.184a)

y (x ,x ) = 7(x) y(x, z) = 0

fo r x /z .

(6.184b)

These conditions, however, lead to trouble if we use them in the imaging Equa­ tion (6.144). In fact, (6.184b) implies that the inner integral in (6.144) has an integrand that equals 0 except at the one point x that equals z, hence (6.144) re-

6. FOURIER OPTICS

308

duces to

' ®

( i w

— —ÌT dz. XD^ X )\

/ “"

k

Thus, /(X ) ^ 0, which is not a very useful result. To extricate ourselves from this problem, we must more carefully formulate the notion of incoherency. We will examine what happens if we view the cross­ correlation function y in Formulas (6.184a) and (6.184b) as a limit. For example, if y(x, z) is defined by the following limit: y(x, z ) =

lim I(x)e €■^0+

(6.185)

then (6.184a) and (6.184b) are satisfied. The functions have an important property. If a function g(z) is continuous at a point x, and bounded for all values of z, then

j

g(z)/(x)^“ ^ '^ “ ^*^/^^ dz ^ €^I(x)g(x)

(6.186)

provided e is close enough to 0. For example, in Figure 6.29(a) we show the graph of for x restricted to the jc-axis, z = (0, 0), 6 = 0.05, and I (x) equal to the constant 400. When this function is integrated as a function of z against a function g(z), like the one shown in Figure 6.29(b), a convolution is being performed. And, as you can see from Figure 6.29(c), the function g is closely approximated. This is what Formula (6.186) says should happen when 6 = 0.05 and I (x) = 400. We will demonstrate (6.186) at the end of this section. For now, let’s just assume that (6.186) holds. Because of the limit in (6.185), let’s also assume that e is so close to 0 that y(x, z), too. We can then rewrite (6.186) as

/ giz)y{x, z ) d z

e^/(x)^(x).

(6.187)

Formula (6.187) is our new definition o f incoherency. That is, illumination is incoherent if the approximation in (6.187) can be assumed to hold for each bounded function g that is continuous at x. If we apply (6.187) to the z-integral in the imaging formula (6.144), replacing g(z) by the function of z in the integrand of (6.144), we obtain

309

6.12. IMAGING WITH INCOHERENT LIGHT

X interual: [ -4, V interual: [ -3,

(b) test object X interual: [ -4, Y interual: [ -3.

41 51

X increment = Y increment =

41 Z71

X increment = .8 if increment = 3

(c) conuolution X interual: I -4, Y interual: 1 -3,

.8 .8

41 51

X increment = Y increment =

.8 .8

FIGURE 6.29 Illustration of Formula (6.186).

y(x, z) dz

e l (x) A (x)eSlxi p

(kD + ■k A

(6.188)

Substituting the right side of (6.188) in place of the z-integral in the imaging equation (6.144), we obtain for the intensity /(X ) at the observation plane

/(X ) «i

---- 2 i

(A 2 £ )A )2 7

1

(— + — ) \X D

A.A J

I{ x)dx.

(6.189)

6. FOURIER OPTICS

310

Notice that the complex exponential factor does not appear in (6.189), since ^ = 1.

Unlike coherent illumination, no small object approximation is needed when in­ coherent illumination is used. For simplicity, we shall assume that the intensity /(x ) is uniform over the aper­ ture, say, I (x) = B where is a positive constant. Then, writing k in place of we can rewrite (6.189) as dx.

(6.190)

We will generally ignore the constant /c, since we can treat the rest of For­ mula (6.190) very efficiently without worrying about the value of k . (See Re­ mark 6.10 below.) As we did for the coherent case, we substitute x = —s / M and dx = ds/M^, where M = A/Z) is the degree of magnification, and we obtain from (6.190) ds.

(6.191)

The integral in (6.191) is a two-dimensional convolution. As opposed to the coherent case, the function (6.192) is convolved with . / I

\ 2

(6.193)

(AA)2

The function in (6.192) is a magnified, inverted, form of |A p which corresponds to the intensity of the aperture function (object). The function in (6.193) is the PSF fo r incoherent illumination. Notice that this PSF is the modulus-squared of the PSF for coherent illumination. The function in (6.193) is the power spectrum of the function P(A.Ax). The operations involved are

P(ÀAx)

—^ P (— x) (ÀA)2

Và A

)

—^ P f— x) (ÀA)2

\-k ^

(6.194)

)

When using FAS there is a choice for computing a power spectrum automatically; it is part of the Fourier transform procedure.

6.12. IMAGING WITH INCOHERENT LIGHT

311

The procedures for computing images under incoherent illumination are quite similar to those described in section 6.10, the main difference being that (6.194) is used in place of (6.155) to compute PSFs. We will limit ourselves to one example (some others are described in the exercises). Example 6.13: Suppose that incoherent light having wavelength A = 5 x 10” ^ mm illuminates an edge defined by Ai \ ifx> 0 10 if jc < 0 and this edge is imaged by a lens with a square aperture of side length 40 mm and focal length 1 m. Assume that D = 2 m and A = 2 m. Describe the image that results. SOLUTION Since |A p = A, the only difference between this example and Example 6.9 is that the PSF is now {note: AA = 1) — (XA)2^

x)

J

= [40 sine i 40X)Y [40 sine (407)]^.

In this case, using ParsevaVs equality [see (5.81a), Chapter 5]

f

/

oo

[reet()'/40)]^rfy

-OO

r»20 r = / I dy = 40. J-2 0

Therefore, we will convolve

/ m = jj

for X > 0 for X < 0

with g(X) = 40 [40 sine (40X)]^. The result is shown in Figure 6.30(b); it was obtained using FAS with [—4,4] as the interval and 8192 points. It is very interesting to compare this result for incoherent imaging of an edge with the result for coherent imaging from Example 6.9. For incoherent imaging there is no oscillation or Gibbs’ effect at the boundary of the edge. The absence o f these defects is a major advantage that incoherent imaging has over coherent imaging. There is also a more subtle advantage involving the location of the edge boundary. If the edge boundary is marked by half of maximum intensity, then

312

H--- ^--- 1--- h

H---- ^— I— h (a) edge X interual: l -10, Y interual: t -.1,

10] 1.41

(b) image X interval: [ -.5, .51 Y interual: [ -4, 461

X increnent = Z Y increment = .15

X increment = Y increment =

FIGURE 6.30 Imaging of an edge using incoherent light.

the edge boundary in incoherent imaging nearly equals the edge boundary of the object. (This near equality can be proved theoretically; we leave this proof to the reader.) For coherent imaging, we saw in Example 6.9 that the edge boundary in the image is displaced by a small amount. I Our remarks at the end of the example above show some of the nice features that an incoherent image of an edge has in comparison to the coherent image of an edge. This one example does not, however, tell the whole story. We will continue to compare these two types of imaging in the exercises. It is also important to note that processing o f images is more difficult with incoherent light (there is no Fourier transform in the focal plane).

REMARK 6.10: There is a small problem that occurs in the previous example. Notice that the/

ds =

and 6 ^ 0

(6.197a)

for each € > 0.

(6.197b)

The approximation in (6.197a) follows from 0 < e < e and e~^/^ ^ 0as 6 -> 0+. Formula (6.197b) holds because the integral on its left side splits into a product of two integrals, both of which are equal to dx. dx = e

Making a change of variable, we have

dx. Since

dx = \ (see Example 5.5 in Chapter 5), we have Now, by making the change of variable z = x + s, ¿/z = ¿/s, we have

j

g (z)e

=

j

^

(6. 198)

g(x-\-s)e

Using (6.197a), we then have

f g(x +

Js ^

J

f

g(x +

Js.

(6.199)

J{S:\S\ 1 mm or \y\ > 2 mm

when A = 5 X 10 ^ mm, D = 200 mm, 300 mm, 400 mm. (Your graphs for 400 mm should look like those in Figure 6.31.) 6.10 Sketch the diffraction pattern when the aperture function is A(jc, y) = Ai (jc) rect(y) where A\ (x) is the function given in Exercise 6.8. Use the same values of X and D as in Exercise 6.8. 6.11 Graph the intensity function I\ (u, D) when the aperture function is X —5 X mm, D = 100 mm, 200 mm, 400 mm.

and

6.12 Edge Diffraction. Suppose that A\ (x) in Formula (6.49) is given by if 0 < X < 16 w ^ i 1 ifO ^ iW = l o i f --16 i < X < 0. Using FAS, plot approximations to 7i (m, D) in Formula (6.49). Use 8192 points, interval [—16, 16], A, = 5 x 10“^ mm, and D = 200 mm. Confirm that I\{u, D) has the form shown in Figure 6.32. 6.13 Repeat Exercise 6.12 for A. = 6 x 10“"^ mm and A. = 7 x 10“^ while keeping D = 200 mm. Compare your graphs to the result from Exercise 6.12, shown in Figure 6.32. What is the effect of increasing the wavelength? 6.14 Fresnel integrals. A Fresnel cosine integral F^ and Fresnel sine integral F^ are defined by F^(x) =

j

cos

J i,

F^{x) =

j

sin

(6 .2 0 6 )

EXERCISES

317

X interval: [ -.5, Y interval: I -.5,

1.51 1.51

X increment = Ÿ increment =

.2 .2

FIGURE 6.32 Fresnel diffraction by an edge. where is a positive constant, (a) Show that y = F^(x) solves the following initial-value problem: y' = cos(7tx^/p), y(0) = 0. And, show that y = F^(x) solves y' = sin(7tx^/P), y(0) = 0. (b) Use the results in (a) to graph F^(x) and F ^(x), for ^ = 2, over the interval [0, 10.24] using 1024 points [this is done similarly to the method described in Exercise 5.28]. (c) Consult a table (e.g., [A-S, page 321]) and verify that the FAS computed values for F^ and F^ are all accurate to 6 decimal places, (d) Show that if f ( x ) — rect(x/2), then the intensity I\ (u, D) in (6.42) can be expressed as h ( u, D )

1

XD

|[

+ 1) - F¿^(« ■ 1)]^ + [ f ^^ (m + 1) - F s" ° ( m

1)]H . (6.207)

6.15 (a) Using (6.207) in Exercise 6.14, graph the intensity I\ over the interval [—4,4] using 4096 points, letting X = 5 x 10“"^ mm and D = 300 mm. [Note: the functions that you use in formula (6.207) will have to be the odd extensions of the solutions to the initial-value problems defined in Exercise 6.14(a) for 0 < x < 5 {not 0 < X < 4).] (b) Compute the intensity h by the convolution method described in Example 6.3, using the interval [—4,4] and 4096 points, (c) Find the Sup-Norm difference between the two versions of I\ found in parts (a) and (b) over the interval [—2, 2] {not over the interval [—4, 4]). REMARK 6.12: The point of Exercises 6.14 and 6.15 is to show that (for rect functions, anyway) the Fresnel diffraction pattern can be computed accurately using Fresnel integrals, and that the convolution method that we have described gives similar results. The advantage of the convolution method is that it applies to a much wider variety of apertures than just rectangular ones (it applies to all the separable aperture functions, as we pointed out at the end of Section 6.2).

318

6. FOURIER OPTICS

Section 3 6.16 Show that the two-dimensional Fourier transform defined in (6.59) satisfies the following properties [see Theorem 5.1 in Chapter 5]. (a) Linearity. For all complex numbers a and p aA(x)

-h

pB(x)

q: A

( u ) -f-

PB(u).

(b) Scaling. If p is a positive constant, then A

^

p^A(pu)

and

A(px)

(c) Shifting. For each c in R^, A(x —c)

A(u)^ -ilnC-U

(d) Modulation. For each c in R^, - L A(u - c). 6.17 Symmetry through the origin. Most aperture functions are real valued (for an exception, see Exercise 6.27). Prove that if A(x) is a real-valued aperture function, then the intensity function, 7(u, D), for its diffraction pattern satisfies 7 (-u , D) = 7(u, D).

(6.208)

Formula (6.208) says that the Fraunhofer diffraction pattern from an aperture, with real-valued aperture function, is symmetric about the origin (unchanged after reflection through the origin). 6.18 Translational symmetry. Show that if an aperture is shifted by c = (c, d), then its diffraction pattern is unchanged. 6.19 Rotational symmetry. Suppose that an aperture is symmetrical in the sense that a rotation by an angle 0, about the origin in the plane of the aperture, leaves the aperture unchanged. Give a simple physical explanation for why the diffraction pattern from such an aperture possesses the same rotational symmetry. 6.20 The diffraction pattern from an aperture shaped like an equilateral triangle pos­ sesses six-fold rotational symmetry (i.e., a 60° rotation about the origin leaves the diffraction pattern unchanged). See Figure 6.33. The equilateral triangle, however, possesses only threefold rotational symmetry. Explain why the diffraction pattern has more symmetry than the aperture. 6.21 Abbe'sformula. Suppose 7^ is a bounded region in the plane where Green’s identity

319

EXERCISES

FIGURE 6.33 Equilateral triangle and its diffraction pattern.

holds. [Here, dlZ denotes the boundary of the region, rj denotes the unit normal pointing outward from the boundary, and ds denotes the arc-length differential along the boundary (traveling in a direction so that the unit normal points to the right).] Show that for all u, except u = (0, 0), -i27tU-\

I '

dX :

6. FOURIER OPTICS

320

REMARK 6.13: Formula (6.210) is called Abbe’s formula. It shows that if TZ is a region where Green’s identity holds, then Fraunhofer diffraction is generated from the boundary oflZ. Green’s identity holds for a wide variety of regions, such as circular discs, annuli, rectangles, triangles, and many other more complicated regions. I 6.22 The diffraction pattern from an equilateral triangle is shown in Figure 6.33. Using Formula (6.210) from Exercise 6.21, derive the intensity function 7(u, D) for an equilateral triangular aperture. In particular, show why there are arms of higher intensity in the pattern at right angles to the three sides of the triangle.. Section 4 6.23 By choosing the jc-interval appropriately, estimate the first three zeroes of d{p) where a{x) is given in (6.81). And, estimate the values of the two local extreme values of d{p) lying to the right of the origin. 6.24 Graph the radial dependence of I (u, for a circular aperture of radius /? > 0. You should be able to verify Figure 6.34.

X Interual: [ 0. 41 X incre»ent = ••» ¥ Intepual: [ -.5, 3.51 V lncr«*ent = .4

FIGURE 6.34 Radial dependence of the square root of intensity / aperture of radius R.

for diffraction from a circular

6.25 Diffraction from annuli. Suppose that coherent light of wavelength X illuminates an annular aperture, with aperture function A(x)

h;

if R\ < |x| < Ri if R\ > |x| or |x| > /?2.

Graph A(p) and the radial dependencies of the intensity functions /(u , D) for the following radii:

321

EXERCISES

(a) R i = 1 ,R 2 = 2 (b) Ri = 1,R2 = 1.1 (c) Ri = l ,/?2 = 1.05. 6.26

Alternate derivation of (6.76). Let TZ stand for a disc of radius R centered at the origin, (a) Use Formula (6.210) in Exercise 6.21 to show that ___ î _ O-Ttp)^

L

¡X:|X| 2 1.11 (a) See Figure D.3. (b) See Figure D.4.

(d) See Figure D.5.

Section 3

1.13 View the GraphBook CH1SEC3 in FAS . Section 4

1.15 (a) For 0 < x < 1, the F. series converges to For x = 0, the F. series converges to [e + l]/2. See Figure D.6. (c) For x = 0, the F. series converges to 1, for 0 < X < 1 the F. series converges to 0, for x = 1 the F. series converges to 1/2, for 1 < X < 2 the F. series converges to 1, for x = 2 the F. series converges to 3/2, for 2 < X < 3 the F. series converges to 2, and for x = 3 the F. series converges to 1. See Figure D.7. (e) For —ir¡2 < x < tt the F. series converges to sinx, for x = d=7r/2 the F. series converges to 0. See Figure D.8. 1.17 On [—0.1, 0.1], 41 harmonics. On [0.1, 0.3], 9 harmonics. On [0.3, 0.5], 7 harmonics. Section 5

1.19 View the GraphBook CH1SEC5 in FAS . 1.21 The periodic extension of |x| is continuous and has a piecewise continuous derivative [equal to —1 on the interval (—1,0) and equal to 1 on the interval (0,1)] hence Theorem 1.3 is applicable. The F. series for |x| on [—1,1] is 0.5 — (4/ tt^) + 1)7tx]/(2/: + 1)^. Consequently,

I k l - S2M+\{X) I =

4

^

cos(2A: +

1)7T x

E

7T^ k=M +l {2k + 1)2

Jm

(2x + 1)^

dx =

1

7t^2M + l

It follows that sup I |x| —S2 m +i (x ) I < .01 if 2M + 1 > 21. Thus, 21 har­ monics are needed. Using FAS the Sup-Norm difference between abs (x) and

A P P E N D IX D . SO LU TIO N S TO O D D -N U M B E R E D EXERCISE S

393

a 21-harmonic Fourier series partial sum is found to be about 9 x 10 which confirms this estimate. A simple explanation for why Sm (^) is furthest from g(x) at x = 0 is that ^'(0) fails to exist, but Sm M is infinitely differentiable at x = 0 . Section 6

1.23 1 . (a) cosine series: - sin 3 + /

nnx ——

(—1)'^“*'^6 sin 3 -----0

n=l

^ 2n7T(l —(—l)'^ cos3) . nnx sine series: > . ---------^ ^2^ 2 _ 9 n=l

2

(c)

00

4 (l-h 3 (-lf)

cosine series: — h

nnx

---------- r— 7 ;--------- COS -------

n=l

Sine series les:

(e)

4n^n^(~ ir+ ^ n=l L

- Ì)

sm

nnx

1 ^—\ 2 . ^nn . nn nnx cosine series: — h / — (sm -------- sm — ) cos ^ ^ riT T ^ 3 n=l ^ nn sine series:

2 nn 2nn . nnx 7 — (cos-------cos------) sm ----^ nrr ^ ^ ^ n=l

1.25 If g(x) =

sin(nnx/L), then

m nx nnx 2 . nnx 2 „ - / g(x) sm —— dx = ~ S B , , sm ------sin------ ax L L L Jo L L Jo ^ —^

1.27 If g{x) = Ao/2 +

2

, m nx nnx sm / S I ------sm ------ ax = Bn. L L ^ Jo

cosinTtx/L), then

2 çL 2 — y g{x)dx = — /\

°° 1I ^ 2 N nnx , cos----- ax = An -A TiM o ddxx + ' ^ A n L Jo 2. Jo

and

A P P E N D IX D. SO L U T IO N S TO O D D -N U M BERED EXERCISES

394

2

nnx , cos----- ax

m tx

— / g{x)co%—~ d x ^ Jo L

L Jo

0 rL

00 +

1.29 (a) See Figure D.9.

mixx nnx , cos------cos------ ax — An

(c) See Figure D.IO.

Section 7

1.31 View the GraphBook CH1SEC7 in FAS . CHAPTER 2 Section 1

2.1 If = 1 for 7 = 0 , . . . , V —1, then Fk = N \ik is divisible by N and = 0 if k is not divisible by N (in particular, Fq = N and Fk = Ofork = 1, . . . , V —1). 2.3 F k ^ i+ 2 Y ,] = iC 0 s (n jk /S ) 2.5 Using the result of Exercise 2.4, we have 1

N -l

/

px

In jk lN

(D .l)

N hence by averaging (D.l) and Formula (2.3) we obtain N -l 2 n jk/N

Ck

(D.2)

The expression in brackets in (D.2) is the DFT of { [g(0) + g(P)]/2, g {P /N ) , . . . , g[{N-\)p/m i Note: Formula (D.2) can also be obtained by using the Trapezoidal Rule to approximate the integral for Ck in Formula (2.1). Section 2

2.7 G2k -

^

= Hi,.

Section 3

2.9 Aliasing begins with k = 512 (to see this, it helps to look at the case of i^ = 513). Section 4

2.11 Sup-Norm difference = 7.196 x 10“^. 2.13 For explicit partial sum u se/(x ) = 0.5-h(2/7r)sumk(sin((2A:-hl)jc)/(2^-h 1)) \^=0,17. The Sup-Norm difference is 3.516 x 10“^.

395

A P P E N D IX D . SO L U T IO N S TO O D D -N U M BERED EXERCISES

Section 5 2.15 For explicit partial sum use f ( x ) = —32sumk[sm(a7tx/2)/ (an) 3] \a=2k + 1 \k=0,20. The Sup-Norm difference is 1.467 x 10~^^.

A

2.17 For explicit partial sum use/(jc) = (—4/7r)sumk((—1) a A: sin(A:7rx/2)/A:) \A:=1,41. The Sup-Norm difference is 7.5 x 10“^. 2.19 For explicit partial sum use f ( x ) = (2/;r)sumk ((1 — cos(2A:7r/3)) sin(A:7rjc/3)/A:) \A:=1,41. The Sup-Norm difference is 6.8 x 10”^. 2.21 For Exercise 2.17 the Sup-Norm difference is 4.7 x 10~^. For Exercise 2.19 the Sup-Norm difference is 1.71 x 10~^. CHAPTERS Section 1 3.1 The 16-point FFT is as follows (Note: -> /îO + ^ 8 + ^ 4 + ^12 /ZQ 4-/^8 ho

z

h\2 hi h\0

he

-> ho~h% -> h4 + h ii —> h4 - h\2 «2 + « 1 0 -> h 2 ~ h io ^6+^14

h\A

hi

hg

hs h\3

h3 hn hi

-> -> ->

h\5

h e ~ h \4 h\ + hg h\ — hg h s + h i3 h s -h \3 ^3+^11

h3 -h\\ hi +hi3 hi - h\s

= h o -h ^ + iQ i4 -h i2 ) = = /îQ + ^8 “ ^4 - ^12 h o ~ h % - i(h4 - h\2) = -> h i ■¥ h\o + + h\4 = -» h i - h i Q - \ - i { h ^ - h i 4 ) = —> h i 4-h\o - h^ - hi4 = ^ 2 -^ 1 0 - ^ 14) = -> h\ -h hg 4- h^ h \ 3 = hi - hg + i{hs - h i 3 ) = hi -\- hg — h$ — h i 3 = hi —hg —iQi^ —hi3 ) =

= - 1 and 80 81 81 83 84 85 86 81

= i):

s o + «4 Si + W'^S5 S 2 + 'S 6 -> S3 + W®S7 SO - S 4 SI - W^gs —>• S2 - S3 - M 'S? ^8 ^8 + 811 gg + W^gi3 89 810 810 + ^^14 ^11 + W^gis 811 ^3 + ^11 + ^7 + ^15 = ^12 —> 8% - 811 -> ^3 “ ^11 + + ^ 15 ) = ^13 -> gg - W'^gi3 ^3 + ^11 + ^7 + ^15 = 814 -> 810 - Î8U —> h^-hii- i{hq - hi3) = 8\5 -> S ll -

= = = = = = = = = =

qo q\ qi q3 q4 qs qe qi qs qg

->

-> -> —>

= q\o = qn = q \2

= q\3 = q\4 = q\5

= Ho = Hi qi + W'^qio = »2 q3 + W'^qii = ■»3 = H4 ^4 + iqn qs + = Hs = He «6 + qq + = Hj qo-qs

QO+QS q\ +

= Hg q\ - ^ q g q i - W^qio = »10

q3 - ^^qu = »11 q4 - ^q\2 = »12 qs W^qi3 = »13 qe q i - w 'q i s

= »14 = »15

Note: the scheme above requires 10 multiplications (since multiplications by i can be programmed without using multiplications). This is a significant improvement over the (1/2)16 log2 16 = 32 multiplications described in the text. The programs for FFTs in Appendix B take advantage of this increased efficiency. If N = 1024,

396

A P P E N D IX D. SO L U T IO N S TO O D D -N U M BERED EXERCISES

then this more efficient FFT requires only 3586 multiplications, a savings by a factor of almost 300 over a direct DFT. 3.3 For N — 4^, the AT-point DFT Hk =

= J2f=o

decomposes into (D.3)

W* +

where ^iv-i H k= E h 4 j(w y ^ , j=o

iiv-i H i = Y , h A j+ x iw y ’^ j=0 jN -l

\N -l

H i=

E

h4j+2(W Y^,

;=o

H l=

E

^4,+3(W V'

7=0

are the A/^/4-point DFTs of the subsequences of [hj] whose base 4 expansions of their indices end in 0, 1, 2, and 3, respectively. Based on (D.3) and the fact that these A^/4-point DFTs each have period N/A, and using the relations = /, = —1, and — —i ( when W = we obtain (for A: = 0, 1, ...,7V/4-l): Hk -

+ (H/W*) + ( H ^ w y + ( H ^ W y

Hk+^N = tik + ^k+\N

- {h I

Hk+lN = Hk -

- ( H iw y - m i w y w

’‘) + { H ^ W y - ( H l w y T^u/^k\ + UH^W^'^).

(D.4)

The computations in (D.4) constitute the basic computations that are needed for a 4^-point FFT (they are analogous to the butterflies used for 2^-point FFTs). The 4^-point FFT is implemented by continuing the decomposition described above. It requires R = log4 N stages. Since only the quantities and for A: = 1, 2, . . . , N /4 — 1 in the first line of (D.4) require complex multiplications [their values are stored and retrieved from memory in order to perform the computations in the second, third, and fourth lines of (D.4)], each stage of the 4^-point FFT requires 3A^/4 —3 complex multiplications. Actually, to be more precise, since the first stage of the FFT uses weight = /, it

A P P E N D IX D. SO LU TIO N S TO O D D -N U M B E R E D EXERCISES

397

does not require multiplications (see the 16-point FFT diagrammed below). Thus a 4^-point FFT requires (3A^/4 —3)(log4 N — Ì) multiplications. For example, if N = 1024 = 4^, then 3060 complex multiplications are needed [instead of 1, 046, 529 multiplications for a direct DFT, or 3586 multiplications for a power of 2 FFT (see the solution of Exercise 3.1 above)]. Note: FAS does not have code for radix 4 FFTs, since that would require two separate codings for N = 4^ and N = 2^ (when P is odd). To illustrate the radix 4 FFT described above, here is how it works for the case of N = 16 = 4^. The left column below is the base 4 digit reversal indexing of {hj}jl^Q. For checking these computations the reader should use the relations 4

_

i,

=

- l ,a n d

ho h4

^0 + ^ 4 + ^ 8 + ^ 1 2 ho + ih^. —hg —ih\2

h% h\2 h\ hs hg

ho ho

hl3 hi he hlO hl4 hi hi hn h\5

-> —>

—> ->

hi h\ h\ hi h2 hi hi h^,

—>

hi

'■ = —i .

= = - h 4 + h^ - h i 2 = - ih4 - /zg + ih\2 = + ^ 5 + ^ 9 + ^13 = + //Z 5 - h g - i h i 2 = + hg - h i 2 = — ih^ - h g - \ - ih\^ = + h ^ + h i o + hi4 = + ih^ —h\Q — ih\4 = -h ^ + h io - h \ 4 = — ih(^ —h\o + ih\4 = -\-hii + / ^ i 5 = + i h j —h i i — ihi^ = - h q + h io - / z i5 = — ih j —h\o ihi^ =

80 81 81 81 84 85 86 81 8S 89

g o + ^ 4 + ^8 +

8l

^11

811

= Ho + ^13^^

->

g - ì + g l W ^ + g n ' < V ^ + 8ì5'iV'^ ^ 0 + ^^4 - g 8 - 1811 —>

gl + /g 5 W -g 9 W ^ -ig i3 W ^

->

g2 +

—> —>■ —>

^10

811 811 814 815

+ g 5 ^ +

i86^^ - i8 \4 ^ ^ g i + igiW^ - g n V ^ - i g i i ' ^ ' ^

^ 0 - -?4 + ^8 - ^12 g l - g 5 W ■VggW'^ - g l 3 ^ ^

= «2 = fl-3 = «4

= = = = =

HS H6 Hj Hg

g l - g è ' ^ ^ + gw '^'^-gìA ^/^

= HW = fin

184 - 8 S + ^811 81 - ig sW - ggW'^ + / g i 3 82 - igeW^ - gioW^* + igi4W(' g 2 - i g i W ^ - g n W ^ + igisw'^

= «12

—> go ->

= H\

= «13 = «14 = «15

This scheme requires 9 multiplications, which is a slight improvement over the 10 multiplications needed for the radix 2 FFT diagrammed in the solution of Exer­ cise 3.1. Section 2 3.5 If N = 4^, there are R stages. The first stage begins with the numbers 0,1, 2, and 3. To proceed from one stage to the next, you multiply the preceding numbers by 4 to get the first quarter numbers of the next stage, then you add 1 to the first quarter numbers to get the second quarter numbers, then add 2 to the first quarter numbers to get the third quarter numbers, then add 3 to the first quarter numbers to get the fourth quarter numbers. After R = log^ N stages, this process generates the digit reversal permutation n - > P a^ (^ ). Y o u then swap each hn with /zp ^ („). 3.7 If A = 4^ where R is even, then Equation (3.21) remains the same (only Pm is the base 4 digit reversal permutation for M = ^/N). If R is odd, then (3.25) is replaced by

A P P E N D IX D , SO LU TIO N S TO O D D -N U M BERED EXERCISES

398

n = Pm (K)(4M) + L

PN(n) = PM( L) ( 4M) 4- K

n+M

Pjsi{n + M) = Pat(w) + M

n + 2M n

for

PN(n + 2M) =

3M — ^ Pat(w+ 3M) = Pn W

+

= 1, 2 , . . . , M - 1 and L = 0, 1, . . . ,

+ 2M +

3M

- 1, and M = VF74.

Section 3

3.9 S(N/ 2 —m) = sm[27t(N/2 —m)/N] = sin(7r —27rm/A/^) = sin(27tm/N) = S(m) 3.11 {S(n)}^^Q~^ and { T (if = i is used to reduce multiplications by for N/ 4 < k < N/ 2 — 1, to multiplications by Section 4

3.13 Define Z4 by Z4 = (2 +

SiJ-N) = 64

then

+ Z4

3 S(^N) + S(^N) S i — N) = 64 Z4

5 5(^JV) + 5( t^AT) S{— N) = 64 Z4

7 5(|iV) + 5(4A^) S(— Af) = — ^ 64 Z4

.(9 64

S ( à N ) + S{^,N) Z4

^ Si ^, N) + S { ^ N ) Z4

64

Z4

64 15.,.

5(iAr) + 5(^AT) Z4

All other sine values have been previously computed. 3.15 The FFT, Yjk=o

requires only the sines {S(j)}jlQ and the tangents

{T(j)}^lo' Therefore, the computation required by formula (3.106) needs only these same values in order to invert an FFT.

A P P E N D IX D . SO LU TIO N S TO O D D -N U M B E R E D EXERCISE S

399

3.17 We have

N-l

i TUjk

since {aj cos

Rk = h o S j cos N ;=i

,

iTtjk

ho ^ 2 ^ - h j cos . . 2 N ;=1 N-l

27r/A:

= 7=0 E hj cos-----J

Thus

=

^

^ - h N - j cos ^ 2 ;=i

is odd

IjTjk

^ 2k

And

N-l

. 2;r7A: A

h = Y l ;=i

since {sj sin

is odd

. 7^ 2717*/: Î , . 77T . 27T7^ > hrsi-i sin — sin--------- > hj sin — sin-------7=1

7=1

N -l

= -E

, J7T , I j T j k I h j Sin - — s i n --------

7=1

N

^ hj jcos

Hence

= Hg_^ + h .

N

- cos

since {sin ^ sin

is odd

j = //2i+i -

c

400

A P P E N D IX D. SO L U T IO N S TO O D D -N U M BERED EXERCISES

3.19 We have

{ n i ) ( m + h)jt

{n + \ ) { k - \ - \ ) n

y cos ------ r:---------— COS------- ^ ------- -— N N «=0

1

= ^ I ]{ c o s

in + \){i^ + ^ + i)7T

N

n=0

in + i ) ( m — k)n

+ COS -

1

iV N— -l1

N

J{m-hk-\-l)7z/{2N) in{m+k-\-l)n/N

n=0 N -l + ^ ^i{m-k)7t/(2N)^in{m-k)7T/N^ n=0

1

— -Re 2 ^

1 _ J{m-\-k-\-l)Tü

________ 1 _ ^/(m+it+l)7r/A^ 1 _ J { m -k )n

^________ I -he‘iim-k)7t/{2N) 2 ^_ ^i{m—k)7T/N

Now, if m and k have opposite parity (i.e., one is odd and one is even), then 1 _ ^i(m+k+i)jr _ Q other hand, if they have the same parity, then 1 — ^i{m-k)7T _ Q ^ ^ same parity, then m + A: + 1 is odd. Let’s say m + A: + 1 ^ 2 7 + 1 for some integer j. In that case, so

2 _ ^i(m+k+l)7T 2 _ ^/(2;+1)7t/N

2 — ^i{m-\-k-\-l)7t/N

J(2;+l)7r/(2iV). 2

2(2 —^-i(2;+l);r/iV^ -

2

cos[(2 ; H-1 )7t/A^]

401

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

. 2sin[(2; + l)jr/(2Af)] = i 1 -c o s [ ( 2 ; + 1 ) 7t/ N ] ' Thus, when m and k have the same parity, I — gi(m+k+l)n

Re {e«'(m+^+l)^/(2iV)_;^__ f 1 —e‘

= 0.

Similarly, when m and k have opposite parity 1 _ pi{m-k)n

Re{^-(m-^):r/(2V) 1

^

^

q

\ _ ^ i{m -k)n/N ^

^

Thus, i f m ^ k ^ * (« +

^cos-

n=0

j){m

+ j)jr (n +

N

j)(k

cos

+

\)n

N

=

0.

While if m = k N -l

^cos

(« + i)(m + i)n-

N

n=0

( n -{■ l ) ( k + ^ ) y r

cos -

1

1

= 2^-

------------ — = 2^^® N n=0

This demonstrates (3.107). A similar argument can be used to demonstrate that ^ ^ n=0

. (n + ^)(m + l ) 7T . {n + \ ) {k + h n sm — ^ ^ o,« ^ ^ sm N N

iiv

if m = A:

0

ifm ^k.

which is the analog of formula (3.107), when sines are used instead of cosines. CHAPTER 4

Section 1 4.1

(a) See Figure D. 11.

4.3

To demonstrate (4.159), we have

\bn\ =

(b) See Figure D. 12.

2 f" . nnx , , ax --— / f W sm ----L L Jo

~ ljo

2 _ 7 /

L Jo

nnx \ f ( x ) \ \ s m — \dx

L

402

A P P E N D IX D, SO LU TIO N S TO O D D -N U M B E R E D EX ERCISE S

To demonstrate (4.160), we have M

flTÎX

«=1

-in n a lL ft n=M +\

n=M+l

< L\\f\\^e~^M -¥\)H na/L)hY^^-(2M +y)k{TtalL)h

^

ÙÔ g-(M+lŸ{na/L)h

2 “ L

1 - e-(2M+3)(7ra/L)2f '

4.5 Using inequality (4.160) from Exercise 4.3, we have ^-AianlDh u { x j ) — b\e

4.7

sin — Li

(a) See Figure D. 13.

0 as i ^ oo.

(d) See Figure D. 14.

Section 2 4.9 (a) See Figure D.15. The string is vibrating twice as fast when c — 200. 4.11 We have ncTtt . n n x /ncTtŸ „ ytt = - y — J K cos ------ s m ------

= c

o [

—

j

ncTtt . K cos —I— sm

htcx

]=

c^yxx-

The rest of the verifications are just as easy. See Figure D.16 for a graph of this solution when n = 3, K = —1, L = 10, c = 100, and t = 1.01, 1.02, and 1.03. Section 3 4.15 The initial distributions can be interpreted as probabilities of electrons passing through two slits. Hence the subsequent distributions illustrate the interference ef­ fects observed in the classic double-slit effect in quantum mechanics. For example, see Figure D.17 (the solution of Exercise 4.14(a) for t = 0.3) where interference fringes are clearly visible. As the distance between the rectangles increases, the frequency of the interference fringes undergoes a proportional increase.

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

403

4.17 For Example 4.4 and t = 0.15, see Figure D.18. For Example 4.5 and i = 0.15, see Figure D. 19. Letting mn stand for the mass of a neutron and tHq stand for the mass of an electron, we have = 1836 Mq. Hence, the neutron diffraction pattern for the time in = 1836 ìq will be identical to the electron diffraction pattern at time ìq. 4.19 See the solution of 4.17.

Section 4 4.21 Cesàro: 147 harmonics. dlVP: 29 harmonics, banning: 59 harmonics. Hamming: 56 harmonics.

Section 5 4.23 For co = 29.997T, see Figure D.20. The graphs have a similar form to the graphs for Exercise 4.11 for « = 3, Â' = —1, L = 10, and c = 100 (see Fig­ ure D.16). (The graphs for Exercise 4.11 illustrate the motion of what is called a fundamental harmonic. See [Wa, Chapter 3.2] for further discussion of funda­ mental harmonics.) 4.25 \ico ^ then the filter function F (jc) in (4.80) has a sharp peak at X = m / M . Thus, the m}^ harmonic of j(x , t) is amplified the most. Hence, the expression for y in (4.77) becomes y{x, t)

COm sm cot — (OÛnCL>mt~\

lioi For (o

mnx

Km sin ■

J

(Om, L’Hospital’s rule yields y(x, t)

^ sin (Omt — COmt COS COmt 2coi

Km sin

mnx

(D.5)

In Exercise 4.23, for example, com = co3 = 307t, so the dominant term in (D.5) (when i ^ 1) is —(i/607r)(cos30;ri)F3 sin(0.37rx). And this domi­ nant term (again for i ^ 1) is similar to a multiple of the fundamental harmonic cos(307r/) sin(0.37rx) (as we saw above in the solution of Exercise 4.23). 4.27 See Figure D.21. 4.29 See Figure D.22.

Section 6 4.31 View the GraphBook CH4SEC6 in FAS . 4.33 For banning filtering and 20 harmonics, the graphs for parts (a) and (b) are both shown in Figure D.23; their Sup-Norm difference is 1.02 x 10~^^. For 40 harmonics, their Sup-Norm difference is 1.57 x 10“ ^^. For Hamming filtering, using 20 harmonics, the Sup-Norm difference is 1.15 x 10“ ^^, and using 40 har­ monics, the Sup-Norm difference is 1.61 x 10~^^. For dlVP filtering, using 20

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

404

harmonics, the Sup-Norm difference is 1.398 x 10 the Sup-Norm difference is 1.97 x 10“^^.

and using 40 harmonics,

4.35 We have 1 f{s)g(x - s ) d s ^ — I L

1

f i x - v)g(v) i - d v )

Jx+ L

1

= g * fix)

Section 7 4.37 (9/ 4) N l o g 2 A^+29A^/4—4 real multiplications. Here is how this answer was obtained. As shown in Section 3.5 of Chapter 3, the two initial FFTs can be per­ formed simultaneously; this requires (3/2)A log2 N multiplications. Multiplying the transforms requires 4N real multiplications. If you implement the method of Chapter 3, Section 3.9, to perform the inverse FFT, this requires 2 (A /2 —1) = A —2 initial multiplications to produce the two initial sequences. Then, the FFTs can be done simultaneously, requiring (3/4)A log2(A /2) = (3/ 4) N \ o g 2 N — 3 N/ 4 multiplications. Finally, 2(A —1) = 2 N —2 multiplications are needed to produce the two starting values for each recursion. Dividing every computed value by N requires an additional N multiplications. The total is (9/4) A log2 N + 29A /4 —4 real multiplications. When N = 1024, this method is 34 times faster than a direct computation (which takes multiplications). Section

8

4.39 The graphs for hanning’s kernel (8 harmonics) using both methods are shown in Figure D.24. The Sup-Norm difference between the two graphs is 7.84 x 10“ ^^. For 16 harmonics, the Sup-Norm difference is 2.57 x 10~^^. For 32 harmonics, the Sup-Norm difference is 9.19 x 10~^^. 4.41 The graphs for C es^o’s kernel (8 harmonics) using both methods are shown in Figure D.25. The Sup-Norm difference between the two graphs is 8.37 x 10“ ^^. For 16 harmonics, the Sup-Norm difference is 2.79 x 10~^^. For 32 harmonics, the Sup-Norm difference is 1.022 x 10“ ^^ 4.43 View the GraphBook CH4SEC8 in FAS . Section 9 4.45 The graph of / * I\^ is shown in Figure D.26. The Sup-Norm difference between / * and / over the interval [—0.5, 0.5] is 4.307 x 10“ ^. The Sup-Norm

405

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

difference between / * I 22 and / over [—0.5, 0.5] is 1.196 x 10 The Sup-Norm difference between / * 1^4 and / over [—0.5, 0.5] is 3.092 x 10"^. 4.47 Since D

m

(x )

= [sin(M +

l / 2 ) x ] [ l / s m ( x / 2 ) ]

and 1/ sin(jc/2) > 1 on the intervals (0, tt] and [—tt, 0), it follows that \ D ( )\ > 1 whenever sin (M + l/2 )x = ±1 (which happens for x = (2A:+1)7t/ ( 2 M + 1), k = 0, ±1, dz2, ..., ±M). Therefore, property (c) in Definition 4.4 does not hold. It can also be shown that property (b) does not hold (i.e., that limM-^00 \DM(x)\dx = 00), but we will omit the proof. Section 10 m

x

4.49 Using the results in Exercise 4.48, we have t2M ^ 2tc/ ( 2 M + 1/2). Since (2M + l/2 )/(M + 1/2) -> 2 as M ^ 00, it follows that t2M ^ xm for large values of M.

4.51 For V50 , Vioo. and 1^200. we obtain the estimates shown in Table D.l. T able D .l 50 1 00 200

Maximum Values of some Filtered Partial Sums harmonics 1.0715 ^50 ^8.3602 X 10-2 V s o fe o ) harmonics •^100 i 4.1417 X 10-2 Vioo(a:ioo) 1.0713 harmonics •^200 i 2.0709 X 10-2 1^200fe o o ) 1.0714

These results indicate that, as M Xm ^ 0 and

00, Vm (x m )

1-07... > 1 .

Thus, the amount of overshooting appears to be approximately 0.07. CHAPTER 5 Section 1 5.1 (a) (3/2) sine (m/ 2), (b) 2sinc (w) c o s (87T w ), ( c ) 2/[1 + ( 2 7 t u ) \ (d) 4sinc[8(w — 1)] +4sinc[8(w + 1)], (e) . 5.3 (a) Exact transform is (1/3) sinc^(w/3). See Figure D.27 for graphs of the exact and M^S-computed transforms. The Sup-Norm difference between the two trans­ forms, over [—16, 16], is 1.34 x 10~^. (b) Exact transform is (1/3) sine (u/3). Sup-Norm difference between exact transform and M5'-computed transform, over [—16, 16],is 1.66x 10“ ^. (c) Exact transform i s —/7rw/(l+7r^M^)^. Sup-Norm difference between exact transform and M5-computed transform, over [—16, 16], is 7.87 X 10“ ^. (d) Exact transform is —/(sin47rw)( sine u). Sup-Norm differ­ ence between exact transform and M5-computed transform is 1.776 x 10“ ^. Section 2 5.5 Applying (5.14) to f " in place of / , we get \f"{u)\ < ||/ " ||i . And, by Theorem 5.2(a) applied twice, we have f " (i27cu)^ f ( u ) = Thus I/(w) I < IIf " II1 and the desired inequality must hold.

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

406

5.7

We have ¿[m3 _ L u ] e ~ ^ u ^

(a)

(b)

2tt

xe

-2n\x\

^ 7T^(1 +

(c)

(sin Sttjc) rect(x/6) — ^ 3/ sine [6(m + 4)] —3i sine [6(m —4)]

(d)

e '^^eosÔTTx

,

(e)

1

1

1 + [iTtiu - 3)2]

1 + [2jt{u + 3)2]

_ |; ,+ 1 |

4 C 0 S (2 7 T M )

+ g-

1 + (27Tm)2

Section 3 5.9 If / is the funetion defined in (5.21), then 11/ — 0.1873, 11/ - 53^112 « 0.1327, \\f -

II2 ^ 0.2625, H/ —5^^ II2 ^

« .0939.

5.11 Letting x — xm yield the first maximum of S ^ ( x ) to the right of x = 0, we obtain the results in Table D.2.

Table D.2 Maximum Values of 5 ^ . M = 8

^8

.0625

M = 16

X\6 Ri .03125

M = 32

X32

M = 64

X64 « .00781

.0156

(X8>

2.678

S{^(xu) ^ 3.14 4 (^ 3 2 )

3.41

‘^¿(■^64)

3.55

These results indieate that, as M ^ oo, S ^ ( x m ) overshoots the limiting value of / ( 0 + ) = 7T. In other words, there is a Gibbs’ phenomenon. Section 4 5.13 If 1024 points are used, over an interval of [—16, 16], then the Sup-Norm difference between the M5-computed transform and the exact transform is 5.57 x 10~^. Over the interval [—4,4], the Sup-Norm difference is 3.33 x 10“ ^. 5.15 If 1024 points are used, over an interval of [—16, 16], then the Sup-Norm difference between the M^-computed transform and the exact transform is 2.887 x 10~^^. Over the interval [—4,4], the Sup-Norm difference is 2.887 x 10“ ^^.

407

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

Section 5 5.17

rx-0.5 , rect(v)dv = f ^ I i ^ s O d v ^ 0

If x < 0 , then X - 0 . 5 < -0 .5 so

f Tect(v)dv =

A similar calculation gives 0 if x > 2. If 0 < x < 1, then f l o s ^ 1 ifu = .X. If 1 < jc < 2, then 5.19

rect(u) dv =

^ \ d v = 2 — x.

(a) We have /»x+0.5 2

/ * yP{x) = T ' 1

+ (x —s Ÿ

7 -0 .5 ^

ds Jx—0.5 ^

yl dv

yl

x+0.5 a:—0.5

= iT a n -.

_ iT a „ -. ( i ^

(b) The Sup-Norm differences are as follows: j = y = 1.0: 9.89 X 10-4, y = 2.0: 5.84 x 10-4.

)

.

0.5: 1.495 x 10 y ^ ^

(l/jr) —^)/3^] (^) The Sup-Norm differences are as follows: j = 0.5: 3.51 x 1 0 -^ j = 1.0: 1.626 x 1 0 -^ j = 2.0: 2.336 x 101-3 5.21 A * 3;P for y = 0.25 is shown in Figure D.28. / * f { x ) = 1/(1 -h x^) is shown in Figure D.29. 5.23

for y =z 0.25 and

The Fourier transform version of the solution is r f (w)(cos 27tut)e^^^^^ du. j —i

W{x

Expressing c o sl j tu t as 0.5^^^^"^ + 0.5^-^^^“^ combining exponentials and ap­ plying Fourier inversion, we obtain W ( x , t ) = ( l / 2 ) f ( x - { - t ) - \ - ( l / 2 ) f ( x — t) which solves the given problem (provided / is twice differentiable). A graph of W(x, 1.6) for f ( x ) = is shown in Figure D.30. Section 6 5.25

(b) See Figure D.31.

5.27 W chavC i/i* r H

(d) See Figure D.32.

(f) See Figure D.33.

g - i ^ n a f u h ^ - ( i n a f u ^ r ^ ^-(2;ra)2«2(i+r)_

Therefore,

we have ^—{iTzaŸu^it+x)

(t+z)H{x).

It follows that t H * xH — (t+x)H. A similar argument shows that y P =t= y P = (y+r)P-

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

408

5.29

(a) We have

20

20 TQCt(s/2) t H ( x — s ) ds

= /'

\l\nt

(x+i)/v^ 20 e ^ ds Jx-\

J{){X-

y/A n t (x+ 1 )/ n/4 î

= 10

\— i [ V ^ Jo

e“ " d s ^

2 J (x -

e ^ ds

= 10(erf[(x + 1)/V 4i] - erf[(x - 1)/V 4f]). (c) For t = 0.5, the Sup-Norm difference is 2.518 x 10 For t = 1.0, the Sup-Norm difference is 4.469 x lO“ '*. For t = 1.5, the Sup-Norm difference is 4.56 X 10-'*. Section 7 5.31

See Figure D.34 for graphs of the FAS approximation of | P and of

1 {Aixfi/ 2m)t

^ i f ( \ A n f i t l 2m J

where f ( u ) = 0.5 sine (w/4) and t = 0.1. The 1-Norm difference between the graphs is approximately 4 x 10“ ^. For i = 0.15 the 1-Norm difference is ap­ proximately 1.94 X 10“ ^ and for t — 0.2 the 1-Norm difference is approximately 1.22 X 10“ ^. (It is best to use 1-Norms since we are comparing p.d.f.s.) 5.33

See Figure D.35 for graphs of the FAS approximation of

(A7tfil2m)t

and of

” ) f ( \ 47t f it / 2m J

where /(w ) = y/2e and i = 1.0. The 1-Norm difference between them is approximately 3.54 x 10“ ^. For t = 2.0 the 1-Norm difference is approximately 9.09 X 1 0 -^ Section 8 5.35 Noise suppression will work well when Step 5 (multiplying the FFT by a filter) produces a good approximation of the transform of the original signal. The examples in the text are good illustrations of when this occurs. The transform of

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

409

the noise function is widely spread out and of low amplitude in comparison to the transform of the signal which is very narrow and highly concentrated near one frequency. Hence, when the filter is applied, most of the transform of the signal is retained, while very little of the transform of the noise is allowed through. 5.37 The banning filtered transform is shown in Figure D.36. 5.39 The frequency is 2. For further details, including a reconstruction of the original signal, view the GraphBook CH5SEC8 in FAS . 5.41 signal!: 1 0 1 1 1 0 0 1 0 1 0 0 0 1 1 1, signaM: 0 0 0 1 1 0 1 0 0 0 0 1 1 1 0 0, signal!: 1 0 1 0 0 0 1 1 1 0 1 0 1 0 0 1 . To see a reconstruction of the pulse train for signal!, view GraphBook CH5SEC8 in FAS . 5.43 The frequency is 25. 5.45 The frequency is 27. The phase shift is 0.2. 5.47 The frequencies are 7.* 2S, signall2: !9, signal!!: 14.5, signal!4: 19, signal!!: 30.5, signal!!: 40. Section 9 5.49 Using geometric series formulas we can rewrite the right side of (5.120) as follows:

n=0

n=0

^e-2yrp^-i27tx TC + ][ _^ —Inp^ilnx ^

7r(l —

^2 _ ^ ~ 2 n p _ ^ —2yTp^~i27TX^

7T(1 -

1 —2e~'^^P cos 2ixx + e~^”P Using this result in (5.120) yields (5.122). Formula (5.123) follows from (5.122) by putting p — \ and x = 0. Also, using some algebra on (5.123) we obtain -2n

“

n—\

1+

2I

\ — e~ 2jt

- 1

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

410

and _ 1 r ^(1 + e

^

4 -; 1 + «2 “ 2 [ 1 - ^-2^ n=0

1

J

5.51 See Figure D.37. This is a graph of Se4(x) = YlnL-64 ^“ 2«^7r^( 00i)^i27rnx which is an excellent approximation of the full series. This graph of S ^ ( x ) was generated using a 64-harmonic, filtered Fourier series partial sum for the function f ( x ) = l024S(x). 5.53 The sum is

7T 1 + ^“ 27Tfl 2a I —

j

2^2

Section 10 where

5.55 Vm = E n=- M F{ n/ M) e

1 F(x) =

2( 1 - \ x \ )

.0

if k l < 2 k< \x\< l if 1x 1 > 1.

Clearly, F(x) is even and continuous, and since it is 0 outside of [—1, 1] it satis­ fies (5.127). Furthermore, \ _ i (COSTTW —COS2jrM)/(7TM)2 ” l3/2

if W7^ 0 ifw = 0

It follows, since F is continuous at m = 0 and \F(u)\ < 2/ ( 7Tm)2 for u ^ 0, that (5.128) holds. Consequently, by Theorem 5.15, Vm is a summation kernel. 5.57 V m ( x ) = T . n = - M F{ n/ M) e '”^ where F( a :) = tect(x/2). Clearly, F(0) = 1; also F(x) is continuous and is 0 outside of [—1,1]. When p > 2, we have F '( jc) = —2p x ( l —x^)^“ ^rect(jc/2) F"(x) = { - 2 p { \ - x ' ^ y - ' ^ + p i p - \ ) { 2 x f { \ - x ' ^ Y - ' ^ ] x & a { x / 2 ) . Hence F' and F ” are both continuous when p > 2. Thus, F satisfies all the requirements given in Exercise 5.56, so the Riesz kernel is a summation kernel when p > 2 .

Section 11 5.59 (a) Nyquist rate: 8. 11/— 53211sup ^ 2.09 x 10“ ^. (b) Nyquist rate: 16. 11/ - '^lellsup ^ 6.74 X lO“ '*. ( c ) Nyquist rate: 8. | | / - SnWsup ^ 1-38 x 10“ "^.

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

411

5.61 By Parseval’s equalities:

/

OO

2L I

sine si (2Lx — m) sinc(2Ljc —n) dx

-OO

=

à

£

( s ) ‘' - ' “ “ " ‘ à / I L ^+i27tnx/2L

2L£,

_

(è ) 1 0

if m = n ifm^jézn.

The last equality holding because of the orthogonality of the complex exponentials [ei2^nx/ 2L^ If / is band limited with / = 0 outside of [ - L , L], then OO

fix) =

^

/ ( ^ ) sinc(2Lx - m).

Multiplying this equation by 2L sinc(2Lx —n) and integrating from —cx) to oo, we obtain 2L

f { x ) sinc(2Lx — n ) d x =

Em =-oo / ( ^ ) 2 ^ / £ o sinc(2L.x - m) sinc(2Lx - n ) d x = f ( ¿ ) . 5.63 Graphs of both reconstructions are shown in Figure D.38. The Sup-Norm difference between the two graphs is 1.87 x 10” ^. Section 12 5.65 Graph of shown in Figure D.39. |1/ — ^gUsup ^ 6.39 x 10~^, | | / — ^lellsup « 6.39 X 10-3, ^ 6.39 x IQ-^. 5.67 For the kernel sinc(32x), the graph of (i.e. L = 16) is shown in Figure D.40. Also, | | / - Siellsup 8.58 x 10"^ 11/ - S32IUUP « 2.09 x 10"^ and 11/— IIsup 5.11x10-^. For the kernel S we have | | / —5i6|| sup 1.02x10-3, 11/ - 532llsup 2.72 X 1 0 -^ and | | / - Seallsup ^ 6.93 x 1 0 " / 5.69 See Table D.3. 5.71 See Table D.4. Section 13 5.73 See Table D.5.

412

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

Table D.3 Comparison of Two Sampling Series for the Function / 2L, M Sup-Norm diff., kernel S\ Sup-Norm diff., kernel S2

4 ,16

2.02 1.10 5.68 2.97

2.16 X 10-^ 1.31 x 10-4 7.08 X 10-5 3.68 X 10-5

8, 32 16,64 32,128

X X X X

10-4 10-4 10-5 10-5

Table D.4 New Columns for Tables 5.1 and 5.2, Section 12 2L,M Function f \ Function /2 4, 16 8, 32 16,64 32,128

1.77

X

8.90 4.46 2.24

X X X

10-2 10-5 10-5 10-5

1 28 2.46

X 1 0 -2 X

10“ 5

5 83 X 1 0 -4

1.43

X

10-4

Table D.5 Relative 2-Norm Errors for Three Kernels Time Span, r Kernel Si Kernel S3 Kernel S2 1.72 X 10-4 3.68 X 10-5 1.77 X 10-5 4.0 1.72 X 10-4 2.33 X 10-5 3.5 4.69 X 10-5 1.73 X 10-4 3.0 6.36 X 10-5 3.10 X 10-5 8.29 X 10-5 4.08 X 10-5 1.79 X 10-4 2.5 1.735 X 10-4 1.16 X 10-2 5.67 X 10-5 2.0 1.74 X 10-4 1.6 8 X 1 0 -2 8.25 X 10-5 1.5 1.32 X 10-2 1.89 X 10-4 2.69 X 10-2 1.0 5.12 X 10-4 5.57 X 10-2 2.76 X 10-2 0.5

5.75 The entries for column three are as follows: 4.0: 4.79 x 10

3.5: 4.71 x

10-^, 3.0: 4.75 x 10“ ^, 2.5: 4.81 x 10"^, 2.0: 5.10 x lO^^, 1.5: 7.99 x lO“ '*, 1.0: 1.88x 10“ ^,0.5: 1.57x 10“ ^. The kernel S appears to have a slightly longer response time than S 3. It does have a shorter response time than both S2 and S i.

5.77 Letting stand for the MS-computed partial sum, and S m stand for the explicitly computed partial sum, the following approximations of relative 2-Norms were obtained:

413

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

(a) l l / - 5 f g ||2 / ||/ ||2 = 2.887x 10- 2,

I|5f6-5l6ll2/l|5l6ll2 = 1.445x10-2

(b) Il/-5f2ll2/ll/ll2 = 7.438x10-2,

I|5f6-532||2/I1532ll2 = 7.050x10-2

(c) ||/-53p2ll2/ll/l|2 = 5.854x 10-2,

I|5f2-^32ll2/ll532ll2 = 6.215x10-2

(d) 11/- 5 f | | 2 / l l / l l 2 = 1.578 x 10 - 1

lISfg-54||2/1|54||2 = 5.800

X

10-2

l l /- 5| 2ll 2/ ll /l l2 = 3.360x10-I|5f2-532ll2/l|532ll2 = 3.005x10-2.

(e )

The two kernels for case (c) and case (e) give acceptable results (the relative 2Norm error \\S^2 ~ ‘^32ll2/ll‘S'32ll2 is reasonably small for both of these cases). For case (b), the relative 2-Norm error is not as small, but probably acceptable using the rule of thumb described in the text. 5.79 To see that formula (5.209) uses a periodic reconstruction kernel, you should graph the reconstruction kernel over [—12, 12] using the formula f ( x ) = sine ( 4 t ) \ t = u —w / 2 \ u = v —wgn(v/ w) \ v = x 4 - w / 2 \ w = S You will see that this graph consists of three repetitions of the graph of sine (4x) (initially defined over [—4, 4]). The method derived in Exercise 5.76 uses two FFTs to compute a discrete convolution. As described in Section 4.7 of Chapter 4, this presumes periodicity. In particular, the sequences f ( xk ) and S(xk) should be periodic with respect to their indices. Therefore, if we use a periodic function f ( x ) and a periodic kernel S(x), then the method of evaluating (5.208) using two FFTs produces a sampling series sum with a periodic kernel. Since the sums in Exercise 5.77 only involve terms where the sample points kl (2L) belong to the interval [—4,4], where f ( x ) equals its periodic extension, the M5-computed sampling series partial sums must match the partial sums using the periodically extended kernels computed in Exercise 5.78. (The very tiny Sup-Norm differences are attributable to differences in rounding error between the two methods.) Section 14 5.81 Approximate the integral Riemann sums to obtain f s(u)

by

Q N -l

N E/ =0

for some positive i2. Then, use

.

^

sm

TtujQ

N

Drop the j = 0 term (since it is 0) and put u = k/ Q, obtaining A^-l fs

414

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

which shows that f s ( k / Q) can be approximated by Q / N times a DST. This DST is then performed using a fast sine transform. Similarly, a fast cosine transform can be used to approximate a cosine transform. CHAPTER 6 Section 1 6.1 |V^(jc,0, O p = Therefore, T(x, z) = ( l / T )

= 1, hence /(x , 0) = 1. Similarly, /(z , 0) = 1. dt = H / T ) f J l d t = l.

6.3 Let 0 = (0,0) and let the point source be located at (0, —D). Then, by Example 6.1, the light at (x, 0) on the screen is given by V r ( o , - £ > .f yx [ w^ +g ^] ^ A-v/ | x |2 + Hence, setting 5(i) equal to

—D, t), we are done.

6.5 (a) Let C(x) = 1 and C(z) = 1. (b) For 6.1: C(x) = 1 = C(z). For 6.2: C(x) = e'(2:rA)a-X ^ ^iQ.n/X)Sk-Z_ Pqj. (-( x) = gi(2;rA)[D2+|X|2]5 and C(z) = gi(23r/^)[o^+|zp]2

Replace the integral in brackets in (6.18) by

C (x )C * (z )7 /(x ,0 )v '/(z,0 ) and separate integrals. 6.7

By the Schwarz inequality. T

I

t

I

t

- T ^ I i x , 0)V/(z,0). Therefore, |T(x, z)| < 1. Section 2 6.9 6.11

The graphs are similar to those shown in Figure 6.31. The graph of I\ (w, 200) is shown in Figure D.41.

6.13 The graphs are similar to the one in Figure 6.32. Increasing the wavelength decreases the number of oscillations per unit length. 6.15 (a) The graph of I\ is similar to the graph in Figure 6.4(d). It was obtained as follows: (1) was graphed over the interval [0, 5] by entering the formula f i x ) = C0S(7TJC A 2/0.15) \Diff \y=0

415

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

(2)

was graphed over [0, 5] by entering the formula f { x ) = sin(;rx

A

2/0.15) \D iff \y=0

(3) I\ was graphed over the interval [—4,4] by entering the formula: f i x ) = (1/0.15) {[sign(a)gl(abs(a)) - sign(Z?)gl(abs(Z?))]

A

2

+ [sign(a)g2(abs(a)) - sign(Z?)g2(abs(Z?))]

A

2}

\a=x-\-l \b=x —l (b) The graph of I\ using the convolution method is shown in Figure 6.4(d). (c) The Sup-Norm difference between the two graphs of I\ over [—2, 2] is 1.47 x 10~^. Section 3 6.17 A ( - u ) = / r 2 dx = ( / r 2 Aix)e~^^^^-^ d x f . Thus, A (-u ) = [A(u)]*. It follows that |A(—u ) p = |A(u)p, hence / ( —u, D) = / (u, D). 6.19 If an aperture is rotated by an angle 9, then the diffraction pattern must also rotate through the same angle 6 . Consequently, if the aperture is unchanged by the rotation, then the diffraction pattern is unchanged as well. 6.21

We have -/27TU-X

Therefore, if u 7^ (0, 0), we obtain

32 dx^

+

-i2nU -X

dy^J _(27t)^|u I^

Applying Green’s identity to the second integral, we obtain i27rU-X Jn

9/? L(27r)2|u|2

—

(27T)2 |uP

which proves formula (6.210).

i

f

h n 9>? L

J

ds

dx.

A P P E N D IX D . SO L U T IO N S TO O D D -N U M BERED EXERCISES

416

Section 4 6.23 The first three zeroes of a(p) are p = 0.61, 1.12, and 1.62. The first two local extrema to the right of the origin are a(p) = —0.416 at p = 0.82 (a local min.), and a(p) = 0.202 at p = 1.34 (a local max.). 6.25 The graph of A(p) for R\ = 1 and /?2 = 2 is shown in Figure D.42. The graph of the intensities I\ (u, D) have radial forms that are similar to the graph shown in Figure 6.9. Section 5 627 When the mica is placed over the right aperture, the aperture func­ tion is A(jc, y) = A q(x -1- 8/2, };) + iAo(x — 8/2, y). Therefore, A( u, v ) = Ao(u, + ie -inSu ]. Hence, I { u , V, D) = I q ( u , V, D)

2 4“ 2 sin

/

27t 8

IT

u

d"

It follows that the diffraction pattern looks like the pattern for a single aperture, but with dark vertical interference fringes at w = {XD/ 8){—l/A -t- k) for each integer k. Notice that the interference fringes are displaced slightly to the right of the origin. By a similar analysis, if the mica is placed over the right aperture, then the dark interference fringes are located at w = {XD/ 8)[\/A -f k] for each integer k. Hence, the fringes are displaced slightly to the left of the origin. Thus, we can see that the diffraction patterns differ, depending on whether the mica is placed over the left or right aperture. 6.29 A(x) Ao(u) -tD ) 4 cos^(27TC • u/AD). Since c • u = au + bv, this gives the desired intensity function. Since cos^ = Owhen^ = ±(2/:-f-l)7r/2 for each integer A, putting^ = 2;rc-u/AZ)leads to c • u = ±XD{2k + l)/4. Since c is a normal vector for this last set of equations, the dark fringes must be perpendicular to c. For c = (8/2, 0), the distance between successive dark fringes is XD /8 [this follows from Formula (6.90)]. By rotation (see the solution to Exercise 6.19), it follows that the distance between the dark fringes is XD /8 for every c that satisfies |c| = 8/2. Section

6

6.31 The instrument function for (a) is shown in Figure D.43. 6.33 The instrument function for (a) is shown in Figure D.44. Section 7 6.35 There is a broadening of the spectral line widths, hence there will be less resolution of distinct wavelengths. 6.37 The error in the central slit position seems to cause a slight decrease in the intensity of the off-center spikes, but no apparent loss of resolution (since the spikes seem to be of the same width as they were for the non-misaligned grating in Example 6.8).

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

417

6.39 By shifting and linearity (see Exercise 6.16), we have M -lN -l

A(u) m=0 «=0 M-1

N -l

m =0

n=0

= Ao(u)

Y,

Substituting a •u (and c •u) in place of Su in the derivation of Equation (6.97) from Equation (6.94), we obtain

7(u, D) = 7o(u, D)

úvr(7tM2í • u/ XD)

úv^ { tcN c • u/ XD)

sin^(7ra • \x/XD)

sin"^(7rc • u/ XD)

and have maximum values of and whenever a • u = mXD and c • u = nXD, respectively. The equations in (6.212) are just another way of expressing a • u = mXD and c • u = nXD. The two equations in (6.212) describe two collections of parallel lines. The lines in one collection are described by the equation a •u = mXD, hence are all perpendicular to a since a is a normal vector for these equations. Likewise, the lines in the second collection are all perpendicular to c. Because a and c are not parallel to each other, the lines in these two collections intersect at a grid of points [the simultaneous solutions of the two equations in (6.212)]. Since, even for very moderate sizes of M and N, the functions and are sharply peaked at their maxima, the diffraction pattern appears as a collection of high-intensity dots, each dot having intensity M^N^Io(u, v, D). 6.41 The instrument function in (6.110) has a sharp peak at m = NXD, and its first zero to the right of this peak is at m = NXD + XDIfi. Hence, by Rayleigh’s criterion, for < X2 we must have NX\ D X \ D/ p < NXiD. Dividing out by D and subtracting NX\ yields X\ / p < N{Xi — >.1) from which (6.111) follows. 6.43 First-order spectrum: l / M, second-order spectrum: 1 / ( 2 M ) , . . . , A:*-order spectrum: \ / ( kM ) . Note: these are the same separations as for a vertical slit grating; the relative separations are determined by the structure factor, not the form factor. 6.45 The instrument functions both look similar to the one shown in Figure 6.19. But, for N = 100, the spectral lines are narrower (by a factor of 2) than those for 77 = 50. 6.47 Using the criterion that the intensity is approximately zero if it is less than 1% of the peak intensity, the half-width of the first-order spectral line is approximately 0.29.

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

418

Section 8 6.49 If we use formula (6.203) for the light entering the lens, since D = / the focal length of the lens, we obtain the following expression for the light exiting the lens: f ( x , €, t)

e ~ ' ' P{x).

Using the usual Fresnel approximations, we then have

P(x).

f(x,e,t)

Consequently, 1

^ (W

S(t)e‘^ ^

S*it)e~‘^ f -P{x)P*(z) dt

P( x)P*i ^)l : r

\S{t)\^dt.

i Jo

Letting K equal [I/ (Xf )^](l / T) \S(t)\^ dt, this gives y(x, z) ^ /cP(x)P*(z). Hence, using Formula (6.24) we obtain F(x, z) =

y(x, Z)

_ f Pj x)

f P(z) Y

V T M V ÿ M ^ VI^(x)lAl^(z)l/ ‘

If P is the pupil function defined in (6.130), then we obtain F(x, z) ^ 1 whenever X and z are in the aperture. Hence, the light is coherent across the aperture. Furthermore, since /(x , e) = y(x, x) we have /(x , 6) ^ /c|P(x)|^ = is the pupil function in (6.130) and x is in the aperture. 6.51 By the phase transformation due to the lens, we obtain V^(u, 6, t) ^ ^^^A(u)V^(u, 0,

F (u).

By Fresnel diffraction, we then have Jb

V^(X, A + 6, 0 ^ — f îA(u, 6, À A JR2 Combining these two formulas yields (6.214).

du.

k,

when P

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

419

6.53 If A = / , then g ^ |X - u |% ^ |u p ^

= ^ i7 lX l% ^ X .u _

in | V | 2

Consequently, after factoring e^f' this factor using | • p, we obtain

outside the integral in (6.216) and eliminating

f

/ ( X , / + €)

A(u)e

-Xu

-/ JR2

6.55 If the lens has pupil function V( X) and focal length / , then V^(X, / + 6, i) becomes V^(X,/ + e , i ) e ^ '^ 'V ( X ) after passing through the lens. The factor formula (6.218) in Exercise 6.54, yielding formula (6.219).

canceled in

Section 9 6.57 Including fourth-power approximations will change the approximations in (6.121) to the following:

L '

j

2/?i

8/?J

r«2 |„|2-|5 _ |U|2 |U|4

With these approximations, (6.122) becomes lup

lup

\BC\ = e - \ A B \ - \ C D \ .

1U|2

lul'^

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

420

Consequently, (6.125) changes to \ AB\ -bn\ BC\- \- \CD\

where form

k

is a. constant. The term a:|u |^ produces an extra factor in (6.132) of the where a = 2k /X.

Section 10 6.59 The image in each case looks similar to the image shown in Figure 6.23(c). As k increases, the oscillations in the image decrease in frequency (due to widening oftheP 5F ). 6.61 The graph of the image is shown in Figure D.45. There is ringing due to the dust particles and there are interference effects. 6.63 The image from the Gaussian filtered pupil is shown in Figure D.46. This image is a significant improvement over the image obtained in Example 6.9. There is no Gibbs’ effect (i.e., there is no extra brightness near the edge; the edge would appear blurred due to a very gradual change in intensity). There is also a complete absence of extraneous oscillation. The image edge (as measured by half of maxi­ mum intensity) is closer to the true edge; in fact it appears to be precisely equal to the true edge. This image of an edge is not as good as can be obtained with inco­ herent illumination [see Figure 6.30(b)]. Nevertheless, it is a major improvement over the coherent image using an unmodified pupil [see Figure 6.23(c)]. (In fact, for some of the applications described in [Mi-T], it is a very effective substitute for an incoherent image.) 6.65 p = .04, ^ 480% improvement, p = .08, ^ 430% improvement, p = 0.16, ^ 300% improvement, p = 0.2, 250% improvement, p — 0.25, 200% improvement. False details appear for p > 0.2. Raising the value of P qf for |x 1 < 3 causes a decrease in contrast improvement (over the bright field image), but increases the value of p at which false details appear. Section 11 6.67 Let the coordinates of the back focal plane be specified by (x /, / + F ), where the subscript I stands for image. Then, by repeating the derivation of (6.181) we obtain îA(X, t)e

ilTt V

Y,

d X V ( x i)

where V ( x i ) is the pupil function of the second lens. Substituting the expression from (6.181) for V^(X, t) into this formula, we get

/ [ / M x . ) n x , t ) e ^ ^ ' ^ d x ^ P ( X ) e ^ '^ - ^ ' dXVixi).

APPENDIX a SOLUTIONS TO ODD-NUMBERED EXERCISES

421

Computing ( 1 / r ) Jq dt = 7(x/) in the usual way, using the coherency assumption F(x, z) = 1, we obtain

/(X /)

/

dXV{ X[ )

( /

Assuming that P is a pupil function like the one defined in (6.130), we will have \V\^ = V. And, the innermost integral is a Fourier transform of the aperture function A. Hence /(x/)

-i2n V

P{X)e >-F I

VV

Y

dX

V{x,).

}

Using the convolution theorem (Theorem 5.7 in Chapter 5), we obtain {Note: the transform of P{ —x) is PCX)]: V{xi).

7(x/)

Making some changes of variables, we then have dVL ylF2 The integral in the formula above is analogous to formula (6.152), hence a mag­ nified, inverted image is formed. The magnification factor is F / f and the PSF is 1

H \ -x )f JSection 12 6.69 The PSF is graphed in Figure D.47. There is much less oscillation in the incoherent PSF, it has no negative values, and is more narrowly concentrated at the origin. The incoherent images of tiny point-like objects would look like tiny point-like objects (although slightly wider), while the coherent images would look like ripples spreading out from a central point (like when a pebble is dropped in a pond). 6.71 The image is shown in Figure D.48. It is clearly a better image than the one shown in Figure D.45. This superiority is due to the lack of ringing and the lack of interference effects in the incoherent image. 6.73 When c = 17, or 19, the relative contrast is about 1 for coherent illumination. For incoherent illumination it is about 0.577 fore = 17, and about 0.524 fore = 19.

422

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

When c = 21, 23, or 25, the relative contrast is 0 for coherent illumination. For incoherent illumination it is about 0.476 for c = 21, about 0.426 for c = 23, and about 0.376 for c = 25. The advantage of incoherent illumination is that it produces some relative contrast for each of the given frequencies. For c = 17 and 19, coherent illumination exhibits almost twice as much contrast as incoherent illumination. 6.75 The PSF (see Example 6.13) is [40sinc(40X)]^[40sinc(407)]^. As we showed in that example, f^^[40sinc(40X)]^ d X = 40. Consequently, the nor­ malized PSF is 40 sinc2(40Z) 40 smc^(40Y). Furthermore, because /f^ 4 0 sin c ^ (4 0 X )iiX = 1 and /f^ 4 0 s in c ^ ( 4 0 F ) d y = 1, it makes sense to define the normalized PSF along the X-direction to be 40sinc^(40X). Using this normalized PSF the image of the edge function given in Example 6.13 has exactly the same form as the image in Figure 6.30(b). The difference, however, is that the intensity (T-scale) is now the same one needed for the original object (e.g., a 7-interval of [—2, 2] works well for both the object and the image). 6.77 The coherent PSF has a transform of P( XAu) = rect(w/20). The ob­ ject function Ak(x) = exp(—(x/3)^^)(l + c os kn x) has a transform of Ak(u) = d{u) + 0.5d(u + k/2) + 0.5d(u — k/2), where d(u) is the transform of a(x) = exp(—(jc/3)^^). The transform d(u) essentially consists of a narrow spike at the origin. Hence Ak{u) consists of three spikes at m = 0 and u = ± k /2 . The trans­ form of the coherent image is Ak(u)P(kAu) = Ajc(u) rect(w/20). When k = 17, and 19, then k /2 = 8.5 and 9.5. Hence, all three spikes are unchanged by the multiplication by rect(M/20), since rect(M/20) = 1 for m = 0, w = ±8.5, and u = ±9.5. Thus, the coherent image will be nearly identical to the object, and the relative contrast will be 1. When k = 21, or 23, or 25, however, rect(M/20) = 0 for u = ±.k/2. Hence Ajt(w)rect(M/20) ^ a(w), so the image is approximately a{x) = exp(—(jc/3)^^) which has a relative contrast of 0. If we let G(w) stand for the transform of the coherent PSF, then G(u) = P ( k A u ) = rect(M/20). It follows that the transform of the normalized incoherent PSF is G * G(u) _ ^ ^ \ G * G(0) "" ^ Vm / * (It is necessary to divide by G * G(0) so that we obtain a value of 1 at w = 0 , since the integral of the normalized incoherent PSF is 1.) From this last formula, it follows that the transform of the normalized incoherent image is Ik(u)A(u/20) where Ik(x) = exp(—(x/3)^^)(l ± coskitx). Hence

ft(»)A ( ¿ ) = m

+

(

“ - j ) A ( ¿ ) + 5« (» + 0

A(¿) .

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

423

For k = 17, 19, 21, 23, and 25, the values of A ( m/ 20) are 0.575, 0.525, 0.475, 0.425, and 0.375, which accord nicely with the contrasts obtained for Exercise 6.73.

424

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

Figure D.l. Exercise 1.3. X interval: [ -45, 451 X increnent = y interval: [ -30, 381 Y increMsnt =

S 6

Figure D.2. Exercise 1.5(a). X interval: [ -Z5, 251 X increment = y interval: [ -45, 451 Y increment =

5 9

Figure D.3. Exercise 1.11(a). X interval: I -1.57, 1.571 X increment = .314 y interval: [ -.5, 1.5) Y increment = .2

Figure D.4. Exercise 1.11(b). X interval: ( B, 31 X increment = y interval: [ -5, 151 Y increment =

Figure D.5. Exercise 1.11(d). X interval: [ -.5, .51 X increment = .1 y interval: I -2, 21 Y increment = .4

Figure D.6. Exercise 1.15(a). X interval: ( -3.5, 3.51 X increment = y interval: ( -2, 51 Y increment = .7

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

Figure D.7. Exercise 1.15(c). X interual: 1 -5. 51 X increwsnt = Ÿ interuul: t -3. 71 ¥ increMsnt =

Figure D.9. Exercise 1.29(a). X interual: [ 0, 31 X increnent = ¥ interual: [ -Z, 21 ¥ incnMent =

1 1

.3

A

Figure D.ll. Exercise 4.1(a). X interual: [ 8, 101 X increnent = 1 ¥ interual: [ -188, 14081 ¥ increnent =

150

425

Figure D.8. Exercise 1.15(e). X interual: t -6« 61 X increment = V interual: [ -6, 61 ¥ increwsnt =

1.2 1.2

Figure D.IB. Exercise 1.29(c). X interual: [ 8, 21 X increaent = ¥ interual: [ -4, 41 ¥ increnent =

.2 .8

Figure D.12. Exercise 4.1(b). X interual: [ 8, 161 X increnent = 1.6 ¥ interual: [ -10, 901 ¥ increnent = 10

426

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

Figure D.13. Exercise 4.7(a). X interual : [ 9, 181 X increnent = 1 y interval: [ -109. 1400] V increnent =

150

Figure D.16. Exercise 4.11. X interual: [ 8. 10] X increnent = 1 y interual: [ -2. 2] y increnent = .4

Figure D.15. Exercise 4.9(a). X interval: [ 0. 10] X increnent = 1 y interval: I -.12. .13] y increnent =

Figure D.17. Exercise 4.14(a). X interval: [ 24. 40] X increnent = y interval: 1 -.1. .41 y increnent =

Figure D.14. Exercise 4.7(d). X interval: I 0, 401 X increnent = 4 y interval: [ -30. 1301 y increnent = 16

1.6 .05

Figure D.18. Exercise 4.17. X interval: [ 28. 36] X increnent = y interval: 1 -.1. .41 y increnent =

.8 .85

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

421

Figure B.19. Exercise 4.17. X interval: [ 2S, 36] X increaent = .8 Y interval: t -1, 31 Y increment = .4

Figure D.2B. Exercise 4.23. X interval: 1 8. 10] X increment = 1 Y interval: 1 -.8003, .00031 Y increment =

Figure D.21. Exercise 4.27. X interval: C 0. 101 X increment = 1 Y interval: I -.0003, .00031 Y increment =

Figure D.22. Exercise 4.29. X interval: [ 0, 101 X increment = 1 Y interval: [ -.001, .0011 Y increment =

Figure D.23. Exercise 4.33. X interval: C -3.14, 3.141 X Increment = Y interval: C -1, 21 Y increment = .3

Figure D.24. Exercise 4.39. X interval: f -3.14, 3.141 X increment = Y interval: I -1. 91 Y increment = 1

.628

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

428

Figure D.25. Exercise 4.41. X interual: t -3.14, 3.141 X incre*:nt = Ï interual: I -1, 91 Ÿ increment = 1

1r ^ 1

T

T

1— 1

Figure D.Z7. Exercise 5.3(a). X interval: [ -16, 161 X increment = Y interval: 1 -.1, .41 Y increment =

Figure D.Z6. Exercise 4.45. X interval: [ -.5, .51 X increment = .1 Ÿ interval: C -.5, 1.51 Y increment = .Z

1 1

3.2 .05

Figure D.29. Exercise 5.21. X interual: t -8, 81 X increment = 1.6 Y interval: [ -.5, 1.51 Y increment = .2

Figure D.2B. Exercise 5.21. X interval: [ -8, 81 X increment = 1.6 Y interval: [ -.25, .751 Y increment = .1

Figure D.30. Exercise 5.23. X interval: [ -16, 161 X increment = 3.2 Y interval: 1 -.25, .751 Y increment = .1

429

APPENDIXD. SOLUTIONS TO ODD-NUMBERED EXERCISES

Figure D.31. Exercise 5.25(b). X interval: [ -10, 10] x increment = 2 y interval: [ -3, 171 y increment = 2

Figure D.32. Exercise 5.25(d). X interval: [ -10, 10] X increment = 2 y interval: [ -2, 18] y increment = 2

1

^

.— L

T

T

1

Figure D.33. Exercise 5.25(f). X interval: t -10, 10] X increment = 2 y interval: [ -.5, 1.51 y increment = .2

Figure D.34. Exercise 5.31. X interval: [ -16, 16] X increment = y interval: t -.1, .4] y increment =

3.2 .05

Figure D.35. Exercise 5.33. X interval: [ -16, 161 X increment = y interval: [ -.1, .41 y increment =

Figure D.36. Exercise 5.37. X interval: I -10, 10] X increment = y interval: ( -10, 10] y increment =

2 2

3.2 .05

430

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

Figure D.37. Exercise 5.51. X interval: t -.5, .51 X increnent = .1 V interval: 1 -1, 191 Y increnent = Z

Figure D.38. Exercise 5.63. X interval: [ -4, 41 X increnent = Y interval: I -5, 51 Y increnent =

Figure D.39. Exercise 5.65. X interval: [ -8, 81 X Increnent = 1.6 Y interval: [ -.1, .91 Y increnent = .1

Figure D.4B. Exercise 5.67. X interval: I -2, 21 X increnent = .4 Y interval: C -.25, .751 Y increnent =

Figure D.41. Exercise 6.11. X interval: t -4, 41 X increnent = .8 Y interval: 1 -.5, 1.51 Y increnent =

Figure D.42. Exercise 6.25. X interval: 1 0. 51 X increnent = .5 Y interval: t -5. 151 Y increnent = 2

.8 1

.1

APPENDIX D. SOLUTIONS TO ODD-NUMBERED EXERCISES

Figure D.43. Exercise 6.31. X interual: [ -4, 41 X increaent = .8 y interual: [ -IBB, 6BB1 V increw:nt =

7B

431

Figure D.44. Exercise 6.33. X interual: ( -4, 41 X increnent = .8 y interual: 1 -IBB, 15BB1 y increment =

16B

Figure D.45. Exercise 6.61. X interual: [ -1. 11 X increment = .2 y interual: [ -.5, 1.51 y increment =

Figure D.46. Exercise 6.63. X interual: [ -.5. .51 X increment = .1 y interual: [ -.2, 1.31 y increment = .15

Figure D.47. Exercise 6.69. X interual: [ -1, 11 X Increment = .2 y Interual: [ -IBB, 24BB1 y increment =

Figure D.48. Exercise 6.71. X interual: I -1, 11 X increment = .2 y interual: [ -5, 451 y increment = 5

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Index

2, radix, 53 4, radix, 83

band limited function, almost, 219 basic aspects of Fourier series, 1 basic properties of the DFT, 37 Bessel functions, 267 binary expansions, 54,59 bit reversal, 58 bit reversal permutation, 62 bit reverse order, 56 bit reversed image, 60 BITREV, 63, 376 blurring, image, 330 boundary conditions, zero, 86 boundary, image edge, 297 bright background, 299 Buneman’s algorithm, 58, 64 Buneman’s method, 66, 69 butterfly calculation, 63 butterfly diagram, 55

A/D conversion, 217 aberration, chromatic, 289 aberration, spherical, 327 Airy pattern, 270 Airy rings, 270 algorithm, Buneman’s, 58, 64 alias of the function, 43, 221 aliasing, 219 almost band limited function, 219 A1t\Anim, 373 analysis software, Fourier, 331 analysis, spectral, 275 animations, xiv, 96, 142, 373 annuli, diffraction from, 320 aperture function, 253 aperture, circular, 267 aperture, square, 260 apertures, linear array of, 275 applications ofFourier series, xiv, 85 approximating convolutions, 174 array theorem, 323 array, parallelogram, 324 assumption, small object, 294 assumption, smallness, 305

cable, fiber optic, 212 cable, single, 212 calculation, butterfly, 63 calculation, speed of, 50 G esso ’s kernel, 130 chemistry, 234 chromatic aberration, 289 circular apertures, 267 classic lens equation, 290 coefficients, filter, 93 coherence, generalized, 315 coherency function, 250

background, bright, 299 band limited, 212

437

438

coherency interval, 252 coherency of illumination, 293 coherency, spatial, 251 coherent light, creation of, 325 coherent light, imaging with, 293 common filters, kernels for some, 126 communication theory, 149 commutativity, 174 compact disks, 218 comparison of three kernels, 226 comparison operations, 339 completeness of Fourier series, 103, 164 completeness relation, 102, 216 complex exponentials, 2 complex exponentials, orthogonality of, 3 complex FFT, 69 complex filter, 186 computation of sines and tangents, 66 computation time, 69 computation time, reduction in, 65 computed transform, 170 computing a real FFT, 71 computing time, savings in, 56 computing two real FFTs simultane­ ously, 69 conditions, normalization, 97 conditions, zero boundary, 86 constant damping, 91 diffusion, 85 Planck’s, 97 separation, 279 sequence, 49 time, 93, 99 wave, 93, 99 constructive interference, 274 continuity, uniform, 137 continuous function, piecewise, 13, 14,21 continuous, piecewise, 13

INDEX

contrast, relative, 300 convergence of filtered Fourier se­ ries, 131 convergence of Fourier cosine series, pointwise, 27 convergence of Fourier series, 15 convergence of Fourier series, pointwise, 13 convergence of Fourier sine series, 25 convergence of Fourier sine series, pointwise, 25 convergence, uniform, 15 conversion, A/D, 217 converter, D/A, 218 convex lens, double, 287 CONVOLUTION, 358 convolution, xv, 119, 170, 173 approximating, 174 cyclic, 124, 181 discrete, 124 graphing, 174 in quantum mechanics, 184 periodic, 181 theorem, 177, 178 theorem, Fourier series, 117 two-dimensional, 294 using FFTs, discrete, 123 correlation, pair, 360 COSINE SERIES, 351 cosine series, definition of Fourier, 24 cosine series, Fourier, 21 cosine transform, discrete, 46 cosine transform, Fourier, 231 cosine transforms, sine and, 230 cosines, orthogonality of, 34 COSTRAN, 76 creation of coherent light, 325 criterion, Rayleigh, 281 criterion. Sparrow, 282 cross-correlation function, 255 cumulative distribution function, 180 cumulative truncation error, 342

INDEX

cyclic convolution, 124,181

D/A converter, 218 damping constant, 91 damping factor, Gaussian, 87 dark field imaging, 299 DCT, 46, 72 de la Vallée Poussin filter, 107 decimation in frequency FFT, 58 decimation in time FFT, 53 definition of Fourier cosine series, 24 definition of Fourier series, 1 definition of Fourier sine series, 23 definition of the Fourier transform, 150 degree of magnification, 294 delta function, discrete, 122 density function, probability, 96 derivative, left-hand, 14 derivative, right-hand, 14 designing filters, 110 destructive interference, 274 detail, false, 301 DFT, 35, 46, 57, 70, 83, 165 DFT to Fourier coefficients, relation of the, 40 DFT to sampled Fourier series, rela­ tion of the, 43 DFT, basic properties of the, 37 DFT, derivation of the, 35 DFT, inverse, 37, 39 DFT, relation between Fourier trans­ forms and, 165 diagram, butterfly, 55 DIFFEQ, 341 difference equations, 344 differentiable function, Weirstrass nowhere, 20 differentiation, 155 diffraction, xv and coherency of light, 247 edge, 262

439

electron, xiv, 97, 184, 261 formula, Fraunhofer, 264 formula, Fresnel, 258 Fraunhofer, 188, 262 Fresnel, 256 from annuli, 320 grating, slit, 278 gratings, 247, 275 gratings, spectral analysis with, 281 limited, 289 patterns, Fraunhofer, 101 patterns, Fresnel, 100 spectroscopy, neutron, 261 diffusion constant, 85 Dirichlet’s kernel, 128 Dirichlet’s problem, 170 discrete convolution, 124 discrete convolutions using FFTs, 123 discrete cosine transform, 46 discrete delta function, 122 discrete Fourier transform, 35 discrete sine and cosine transforms, 45 discrete sine and cosine transforms, inversion of, 76 discrete sine transform, 46 discs, compact, 218 display menu, 331 distorted image, 260 distribution function, cumulative, 180 dlVP filter, 107 double convex lens, 287 draw graphs, 361 DST, 46, 72, 90 DST, inverse, 91

edge boundary, image, 297 edge diffraction, 262 electron diffraction, xiv, 97, 184, 261 electron optics, 261

440

enclosed freely moving particle, 85 equality, Parseval’s, 40, 102, 183, 187,311 equation, classic lens, 290 equation, heat, 85, 178 equation, inhomogeneous wave, 112 equation, Schrodinger’s, 97, 185 equations, difference, 344 ERASE SCREEN, 362 error, cumulative truncation, 342 error, maximum truncation, 342 error, quantization, 218 error, truncation, 342 Euler’s identity, 1 even symmetry, 21 examples of Fourier series, 5 examples of noise suppression, 199 exponentials, complex, 2 exponentials, orthogonality of com­ plex, 3 extension, periodic, 8, 182 extension, odd, 25, 233

factor filter, 95 form, 277 Gaussian damping, 87 interference, 274 obliquity, 255 phase, 250 shape, 277 structure, 277 weight, 66 false detail, 301 FAS, xiv, 32, 34, 37, 48, 50, 66, 87,99,108,119,130,145, 174, 197,331 computer software, xiii function creation procedure of, 252 graphing by, 277 installing, 374 trace procedure of, 138

INDEX

user’s manual, 331 FAS.EXE, 331 fast cosine transform, 53 fast Fourier transform, xiii, 53 fast sine and cosine transforms, 72 fast sine transform, 53 FCT, 53,72 Fermat principle, 248 FFT,xiii, 1,53,57,83,121,170 algorithm, xiv, 53 complex, 69 computing a real, 71 decimation in frequency, 58 decimation in time, 53 discrete convolutions using, 123 of a real sequence, inversion of the, 81 rotations in, 63 simultaneously, computing two real, 69 fiber optic cable, 212 field imaging, dark, 299 field imaging, gray, 301 filter coefficients, 93 complex, 186 de la Vallée Poussin, 107 designing, 110 dlVP, 107 factors, 95 FresCos, 99 FresSin, 99 function, 193 Hamming, 109, 193 banning, 109, 192 kernels for some common, 126 Lanczos, 145 pinhole, 252 process, 120 used in signal processing, 104 filtered Fourier series, convergence of, 131 Fourier sine series, 98

INDEX

Fourier transform, 192 Hamming, 109 banning, 109 inverse FTT, 196 filtering, 190 filtering, spatial, 299, 307 finite geometric series, 36 form factor, 277 FORMULA, 333 formula, Fraunhofer diffraction, 264 formula, Fresnel diffraction, 258 Fourier analysis software, 331 coefficients, relation of the DFT to, 40 cosine, 35 cosine series, 21 cosine series, definition of, 24 cosine series, pointwise conver­ gence of, 27 cosine transform, 231 inversion, 178, 206 optics, X V , 247 sine, 35 sine series, 21, 86 sine series, convergence of, 25 sine series, definition of, 23 sine series, filtered, 98 sine series, pointwise convergence of, 25 sine transform, 231 transforming property of a lens, 301 FOURIER SERIES, 350 Fourier series, 4, 12, 32 applications of, xiv, 85 basic aspects, 1 completeness of, 103, 164 convergence of, 15 convergence of filtered, 131 convolution theorem, 117 definition of, 1 examples of, 5 of real functions, 9

441

pointwise convergence of, 13 relation of the DFT to sampled, 43 Fourier transform, xiv, 149,186,273 and DFTs, relation between, 165 definition of the, 150 discrete, 35 fast, xiii, 53 filtered, 192 inverse, 160 inversion of, 160 properties of, 155 spectroscopy, 234 FOURIER TRANSFORMS, 355 Fraunhofer diffraction, 188,262 Fraunhofer diffraction formula, 264 Fraunhofer diffraction patterns, 101 free particle, 96 frequency of the grating, 282 FresCos filter, 99 Fresnel diffraction, 256 Fresnel diffraction formula, 258 Fresnel diffraction patterns, 100 Fresnel integral, 303 FresSin filter, 99 fringes, interference, 252, 275 FST, 53,72 FTT, filtered inverse, 196 function alias of the, 43,221 almost band limited, 219 aperture, 253 Bessel, 267 coherency, 250 cross-correlation, 255 cumulative distribution, 180 discrete delta, 122 filter, 193 Fourier series of real, 9 Heaviside, 180 instrument, 278 Lipschitz, 20 of graphs, 338 of the lens, pupil, 289

442

INDEX

periodic, 4 piecewise continuous, 13, 14,

holographic grating, 282

21 point spread, 110, 120, 294 probability density, 96 transmittance, 255 unit, 279 Weirstrass nowhere differentiable, 20 window, 213, 220

Gaussian damping factor, 87 Gaussian noise, 197, 199 generalized coherence, 315 geometric series, finite, 36 GET DATA, 347 Gibbs phenomenon, 15, 30, 131, 139, 164, 296 Gibbs phenomenon, suppression of, 106 GRAPH, 364 graphbook, 361 graphing convolutions, 174 graphs procedure, draw, 32 graphs, draw, 361 graphs, functions of, 338 grating, diffraction, 247, 275 grating, frequency of the, 282 grating, holographic, 282 grating, slit diffraction, 278 gray field imaging, 301

half plane, upper, 170 Hamming filtered, 109 Hamming filters, 109, 193 Hamming kernel, 131 banning filtered, 109, 192 banning filters, 109, 192 banning kernel, 130, 205, 211 harmonics, number of, 5 heat equation, 85, 178 Heaviside function, 180

identity, Euler’s, 1 illumination, coherency of, 293 illumination, incoherent, 307 illumination, non-uniform, 323 illumination, plane wave, 314 image blurring, 330 image edge boundary, 297 image processing, 296 image, bit reversed, 60 image, distorted, 260 image, processing of, 312 imaging, xv dark field, 299 gray field, 301 two lens, 307 with a single lens, 289 with coherent light, 293 with incoherent light, 307 with lenses, 247 impulse kernel, point, 134 incoherency, spatial, 251 incoherency, time, 281 incoherent illumination, 307 incoherent light, imaging with, 307 index of refraction, 249 inequality, Schwarz, 216,251,389 inhomogeneous wave equation, 112 initial setup, 332 installing FAS, 374 instrument function, 278 insulated rod, thin, 85 integral, Fresnel, 303 interference factor, 274 interference fringes, 252, 275 interference phenomenon, 273 interference, constructive, 274 interference, destructive, 274 INTERPOLATION SERIES, 352 interval, coherency, 252 inverse DFT, 37, 39 inverse DST, 91

INDEX

inverse Fourier transform, 160 inverse FTT, filtered, 196 inversion, Fourier, 178, 206 inversion of discrete sine and cosine transforms, 76 inversion of Fourier transforms, 160 inversion of the FFT of a real se­ quence, 81 inversion theorem, 160 InvRFFT, 83 iteration, 340

kernel, xiv, 120 G esso ’s, 130 comparison of three, 226 Dirichlet’s, 128 for some common filters, 126 Hamming, 131 banning, 130,205,211 point impulse, 134 reconstruction, 215 summation, 132, 212

Lanczos filter, 145 laser, 252 leakage, 190 leakage, no, 191 left-hand derivative, 14 left-hand limit, 14 lemma, Riemann-Lebesgue, 160 lens, double convex, 287 lens equation, classic, 290 lens, Fourier transforming property of a, 301 lens imaging, two, 307 lens, imaging with a single, 289 lenses, imaging with, 247 light, creation of coherent, 325 light, diffraction and coherency of, 247 light, imaging with coherent, 293 light, imaging with incoherent, 307

443

light, monochromatic, 285 light, visible, 257 limit, left-hand, 14 limited, diffraction, 289 limits, 344 linear array of apertures, 275 linearity, 37, 155, 318 Lipschitz function, 20 logical operations, 340 magnification, degree of, 294 manual FAS, user’s, 331 maximum truncation error, 342 menu, display, 331 messages, multiple telephone, 212 method, Buneman’s, 66, 69 method, Schlieren, 328 modulation, 155,318 monochromatic light, 285 multiple telephone messages, 212 music, sampled, 212 neutron diffraction spectroscopy, 261 NEXT, 332 no leakage, 191 noise suppression, 193 noise suppression, examples of, 199 noise, Gaussian, 197 noise, white, 195, 196 non-uniform illumination, 323 notation, vector, 248 number of harmonics, 5 numbers, even-indexed, 75 numbers, odd-indexed, 75 Nyquistrate, 216 object assumption, small, 294 obliquity factor, 255 odd extension, 25, 233 odd symmetry, 21 operations, comparison, 339 operations, logical, 340

444

optic cable, fiber, 212 optics, electron, 261 optics, Fourier, xv, 247 order, bit reverse, 56 order spectrum, second, 281 orthogonality of complex exponen­ tials, 3 orthogonality of cosines, 34 orthogonality of sines, 34 orthogonality relation, 3 oscillatory term, 252

pair correlation, 360 parallelogram array, 324 parameters, 337 Parseval equality, 40,102,183,187, 311 partial sums, 5 particle, free, 96 particle in a box, 97 particle in a box problem, xiv, 97 pattern. Airy, 270 patterns, Fraunhofer diffraction, 101 patterns, Fresnel diffraction, 100 p.d.f., 180 periodic convolution, 181 periodic extension, 8,182 periodic function, 4 periodicity, 37, 40, 44 periodization, 203 permutation, bit reversal, 62 phase factor, 250 phase transformation induced by a thin lens, 285 phenomenon, Gibbs, 15, 30, 131, 139, 164, 296 phenomenon, interference, 273 piecewise continuous, 13 piecewise continuous function, 13, 14,21 pinhole filter, 252 Planck’s constant, 97 plane, upper half, 170

INDEX

plane wave illumination, 314 point impulse kernel, 134 point spread function, xiv, 110,120, 294 pointwise convergence of Fourier co­ sine series, 27 pointwise convergence of Fourier se­ ries, 13 pointwise convergence of Fourier sine series, 25 Poisson summation, xv, 149, 202, 204, 212 Poisson summation, summation ker­ nels arising from, 205 power spectrum, 186,195,310 principle, Fermat, 248 PRINT, 368 probability density function, 96,180 problem, Dirichlet’s, 170 problem, particle in a box, xiv, 97 process, filter, 120 processing, filters used in signal, 104 processing, image, 296 processing of images, 312 processing, signal, xiv, 85,149,190, 193 products, sums and, 336 program, installing the, 374 properties of Fourier transform, 155 properties of the DFT, basic, 37 property of a lens, Fourier transform­ ing, 301 property, scaling, 157 PSF, 205,298,310 pulsed signal, 193 pupil function of the lens, 289 pupil, translated, 305

quantization error, 218 quantum mechanics, xiv, 96 quantum mechanics, convolution in, 184

INDEX

radix 2, 53 radix 4, 83 ran(x), 196, 334 random number, 196, 334 rate, Nyquist, 216 ratio, signal-to-noise, 228 Rayleigh criterion, 281 real FFT, computing a, 71 real functions, Fourier series of, 9 real sequence, inversion of the FFT of a, 81 REALFFT, 72 reconstruction kernel, 215 RECURSIVE, 344 reduction in computation time, 65 refraction, index of, 249 relation between Fourier transforms andDFTs, 165 relation, completeness, 102, 216 relation of the DFT to Fourier coef­ ficients, 40 relation of the DFT to sampled Fourier series, 43 relation, orthogonality, 3 relative contrast, 300 relative separation, 282 REM, 371 resolution, 282 resonance, stochastic, 199 response time, 218, 226 reversal, bit, 58, 62 Riemann-Lebesgue lemma, 160 right-hand derivative, 14 rings. Airy, 270 rod, thin insulated, 85 rotational symmetry, 318 rotations in FFTs, 63

sampled music, 212 sampling series, 219, 352 sampling theorem, 212, 390 sampling theory, xv, 149, 230 scaling, 155, 318

445

scaling property, 157 Schlieren method, 328 Schrodinger’s equation, 185 Schwarz inequality, 216,251,389 second order spectrum, 281 SELECT GRAPH, 349 separation constant, 279 separation, relative, 282 SERIES, 349 series applications of Fourier, xiv, 85 completeness of Fourier, 164 convergence of Fourier, 15 convergence of Fourier sine, 25 definition of Fourier, 1 definition of Fourier sine, 23 filtered Fourier sine, 98 finite geometric, 36 Fourier, 4,12, 32 Fourier cosine, 21 Fourier sine, 21,86 modifications of Fourier, 85 sampling, 219, 352 setup, initial, 332 shape factor, 277 shifting, 155,318 signal processing, xiv, 85, 149, 190, 193 signal, pulsed, 193 signal-to-noise ratio, 228 simulation of white noise, 197 simultaneously, computing two real FFTs, 69 sine function, 334 sine and cosine transforms, 230 sine, Fourier, 35 SINE SERIES, 351 sine series, Fourier, 21, 86 SINE TRANSFORM, 357 sine transform, discrete, 46 sine transform, Fourier, 231 sines and tangents, computation of,

66 sines, orthogonality of, 34

446

SINESTANS, 69 SINETRAN, 76 single cable, 212 single lens, imaging with a, 289 slit diffraction grating, 278 slit, vertical, 265 small object assumption, 294 smallness assumption, 305 software, Fourier analysis, 331 Sparrow criterion, 282 spatial coherency, 251 spatial filtering, 299, 307 spatial incoherency, 251 spatial variable, 86 spectral analysis, 275 spectral analysis with diffraction grat­ ings, 281 spectroscopy, Fourier transform, 234 spectroscopy, neutron diffraction, 261 spectrum, power, 186,195, 310 spectrum, second order, 281 spherical aberration, 327 spread function, point, xiv, 110,120, 294 stochastic resonance, 199 string, vibrating, 91 structure factor, 277 SUB BITREV, 375 SUB COSTRAN, 375 SUB FTBFLY, 375 SUB InvRFFT, 375 SUB REALFFT, 375 SUB SINESTANS, 375 SUB SINETRAN, 375 substitutions, 338 summation kernel, 132, 212 summation kernels arising from Pois­ son summation, 205 summation, Poisson, xv, 149, 202, 204, 212 sums and products, 336 sums, partial, 5 sup-norm difference, 50 Sup-Norms, 365

INDEX

suppression, noise, 193 suppression of Gibbs’ phenomenon, 106 symmetry, even, 21 symmetry, odd, 21 symmetry, rotational, 318 symmetry through the origin, 318 symmetry, translational, 318

tangents, computation of sines and,

66 telephone messages, multiple, 212 theorem, array, 323 theorem, convolution, 177, 178 theorem, Fourier series convolution, 117 theorem, inversion, 160 theorem, sampling, 212, 390 theory, communication, 149 theory, sampling, xv, 149, 230 thin insulated rod, 85 thin lens, phase transformation in­ duced by a, 285 time constant, 93, 99 time, decimation in, 53 time incoherency, 281 time, response, 218 TRACE, 372 transform computed, 170 discrete cosine, 46 discrete Fourier, 35 discrete sine, 46 discrete sine and cosine, 45 exact, 168, 233 fast cosine, 53 fast Fourier, xiii, 53 fast sine, 53 fast sine and cosine, 72 filtered Fourier, 192 Fourier, xiv, 149, 186, 273 Fourier cosine, 231 Fourier sine, 231

INDEX

inverse Fourier, 160 inversion of discrete sine and cosine, 76 inversion of Fourier, 160 properties of Fourier, 155 sine and cosine, 230 spectroscopy, Fourier, 234 TRANSFORMS, 355 translated pupil, 305 translational symmetry, 318 transmittance function, 255 truncation error, 342 truncation error, cumulative, 342 truncation error, maximum, 342 two lens imaging, 307 two-dimensional convolution, 294

uniform convergence, 15 unit function, 279 upper half plane, 170 user’s manual FAS, 331

447

vector notation, 248 vertical slit, 265 vibrating string, 91 view graphbook, 361 vignetting, 305 wave constant, 93, 99 wave equation, inhomogeneous, 112 wave illumination, plane, 314 wavelength dependent, 281 Weirstrass nowhere differentiable func­ tion, 20 white noise, 195, 196 white noise, simulation of, 197 window function, 213, 220 X-ray, 262 zero boundary conditions, 86