ece 665 fourier optics spring, 2004 thomas b. fowler, sc.d. senior principal engineer mitretek...

ECE 665Fourier Optics

Spring, 2004

Thomas B. Fowler, Sc.D.Senior Principal Engineer

Mitretek Systems

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Course goal

To provide an understanding of optical systems for processing temporal signals as well as images

Course is based on use of Fourier analysis in two dimensions to understand the behavior of optical systems

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Nature of light and theories about it

Fourier optics falls under wave optics Provides a description of propagation of light waves based

on two principles– Harmonic (Fourier) analysis– Linearity of systems

Quantum optics

Electromagnetic optics

Wave optics

Ray optics

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Course organization

13 weeks Main text: Introduction to Fourier Optics, Joseph Goodman,

McGraw-Hill, 1996 Other material to be downloaded from Internet Student evaluation

– Homework 20%– Midterm exam 40%– Final exam 40%

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Topics

Week 1: Review of one-dimensional Fourier analysis Week 2: Two-dimensional Fourier analysis Weeks 3-4: Scalar diffraction theory Weeks 5-6: Fresnel and Fraunhofer diffraction Week 7: Transfer functions and wave-optics analysis of

coherent optical systems Weeks 8-9: Frequency analysis of optical imaging systems Week 10: Wavefront modulation Week 11: Analog optical information processing Weeks 12-13: Holography

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Week 1: Review of One-Dimensional Fourier Analysis

Descriptions: time domain and frequency domain Principle of Fourier analysis

– Periodic: series• Sin, cosine, exponential forms

– Non-periodic: Fourier integral– Random

Convolution Discrete Fourier transform and Fast Fourier Transform A deeper look: Fourier transforms and functional analysis

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Basic idea: what you learned in undergraduate courses

A periodic function f(t) can be expressed as a sum of sines and cosines– Sum may be finite or infinite, depending on f(t)– Object is usually to determine

• Frequencies of sine, cosine functions• Amplitudes of sine, cosine functions• Error in approximating with finite number of

functions– Function f(t) must satisfy Dirichlet conditions

Result is that periodic function in time domain, e.g., square wave, can be completely characterized by information in frequency domain, i.e., by frequencies and amplitudes of sine, cosine functions

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Historical reason for use of Fourier series to approximate functions

Breaks periodic function f(t) into component frequencies Response of linear systems to most periodic waves can be

analyzed by finding the response to each ‘harmonic’ and superimposing the results)

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Basic idea: what you learned in undergraduate courses (continued)

Periodic means that f(t) = f(t+T) for all t

– T is the period

– Period related to frequency by T = 1/f0 = 2/0

– 0 is called the fundamental frequency

So we have

n0 = 2n/T is nth harmonic of fundamental frequency

1000

1000

sincos

2sin2cos)(

nnn

nnn

nbnaa

nfbnfaatf

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How to calculate Fourier coefficients

Calculation of Fourier coefficients hinges on orthogonality of sine, cosine functions

Also,

nmdttntm

nmdttntm

nmdttntm

T

T

T

,0coscos

,0sinsin

,all,0cossin

00 0

00 0

00 0

T

00

2

T

00

2

all,2

cos

all,2

sin

mT

dttm

mT

dttm

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How to calculate Fourier coefficients (continued)

And we also need

T

00

T

00

all,0cos

all,0sin

mdttm

mdttm

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Step 1. integrate both sides:

Therefore

Ta

Ta

dtnbnadtadttfT

nnn

TT

0

0

01

000

00

0

sincos)(

T

dttfT

a0

0 )(1

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Step 2. For each n, multiply original equation by cos n0t and

integrate from 0 to T:

TT

TTT

dttntadttntna

dttntadttntadttntf

0000

0001

0 0010 0000 0

cossincoscos

cos2coscoscoscos)(

2/cos0

02 Tadttna n

T

n Therefore

T

n dttntfT

a0

0cos)(2

0 0

0

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Step 3. Calculate bn terms similarly, by multiplying original

equation by sin n0t and integrating from 0 to T

– Get similar result

Some rules simplify calculations

– For even functions f(t) = f(-t), such as cos t, bn terms = 0

– For odd functions f(t) = -f(-t), such as sin t, an terms = 0

T

n dttntfT

b0

0sin)(2

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Calculation of Fourier coefficients: examples

Square wave (in class)

12/1,1

2/10,1)(

t

ttf

1

-1

TT/2

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Calculation of Fourier coefficients: examples (continued)

Result

Source: http://mathworld.wolfram.com/FourierSeries.html

Gibbs phenomenon: ringing near discontinuity

7

7sin

5

5sin

3

3sinsin

4)( 000

0

tttttf

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Triangular wave (in class)

TT/2

+V

-V

TtTtT

VV

TtTVtT

V

TttTV

tf

4/3,4

4/34/,24

4/0,4

)(

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Triangle wave result

– Note that value of terms falls off as inverse square

2

02

02

002 7

7sin

5

5sin

3

3sinsin

8)(

tttt

Vtf

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Other simplifying assumptions: half-wave symmetry

Function has half-wave symmetry if second half is negative of first half:

)2/()( Ttftf

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Other simplifying assumptions: half-wave symmetry

Can be shown

oddsin)(

4

cos)(4

even,0,0

2/

00

2/

00

0

ndttntf

Tb

dttntfT

a

nbaa

T

n

T

n

nn

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Conditions for convergence

Conditions for convergence of Fourier series to original function f(t) discovered (and named for) Dirichelet

– Finite number of discontinuities– Finite number of extrema– Be absolutely convergent:

Example of periodic function excluded

Tdttf

0)(

12/1,otherwise0,rationalif1

2/10,otherwise0,rationalif1)(

tt

tttf

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Parseval's theorem

If some function f(t) is represented by its Fourier expansion on an interval [-l,l], then

Useful in calculating power associated with waveform

1

2

1

2202

4)(

2

1

nn

nn

l

lba

axf

l

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Effect of truncating infinite series

Truncation error function n(t) given by

– This is difference between original function and truncated series s

n(t), truncated after n terms

Error criterion usually taken as mean square error of this function over one period

Least squares property of Fourier series states that no other series with same number n of terms will have smaller value of E

n

T

nn dttT

E0

2)(

1

)()( tstf nn

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Effect of truncating infinite series (continued)

Problem is that there is no effective way to determine value of n to satisfy any desired E

Only practical approach is to keep adding terms until E

n < E

One helpful bit of information concerns fall-off rate of terms– Let k = number of derivatives of f(t) required to produce a

discontinuity– Then

where M depends on f(t) but not n11,

knkn n

Mb

n

Ma

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Some DERIVE scripts

To generate square wave of amplitude A, period T:

squarewave(A,T,x) := A*sign(sin(2*pi*x/T)) For Fourier series of function f with n terms, limits c, d:

Fourier(f,x,c,d,n)– Example: Fourier(squarewave(2,2,x),x,0,2,5) generates

first 5 terms (actually 3 because 2 are zero) To generate triangle wave of amplitude A, period T:

int(squarewave(A,T,x),x)– Then Fourier transform can be done of this

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Exponential form of Fourier Series

Previous form

Recall that

1

000 sincos)(n

nn tnbtnaatf

tjntjn

tjntjn

eej

tn

eetn

00

00

2

1sin

2

1cos

0

0

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Exponential form of Fourier Series (continued)

Substituting yields

Collecting like exponential terms and using fact that 1/j = -j:

10 22

)(0000

n

tjntjn

n

tjntjn

n j

eeb

eeaatf

1

000

22)(

n

tjnnntjnnn ejba

ejba

atf

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Exponential form of Fourier Series (continued) Introducing new coefficients

We can rewrite Fourier series as

Or more compactly by changing the index

00~,

2~,

2~ ac

jbac

jbac nn

nnn

n

1

000 ~~~)(

n

tjnn

tjnn ececctf

n

tjnnectf 0~)(

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The coefficients can easily be evaluated

T tjn

T

TT

nnn

dtetfT

dttnjtntfT

dttntfT

jdttntf

T

jbac

0

0 00

00

00

0)(1

sincos)(1

sin)(cos)(1

2~

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Sometimes coefficients written in real and complex terms as

where

nn jnnn

jnn ecccecc

~~~,~~ *

n

nn

nnn

a

b

bac

arctan

21~ 22

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Exponential form of Fourier Series: example

Take sawtooth function, f(t) = (A/T)t per period Then

Hint: if using Derive, define = 2/T, set domain of n as integer

0,2/

0,1

~0

0

nV

ndtteT

V

TcT

tjn

n

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Fourier analysis for nonperiodic functions

Basic idea: extend previous method by letting T become infinite

Example: recurring pulse

t

v0

a/2-a/2

T

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Fourier analysis for nonperiodic functions (continued)

Start with previous formula:

This can be readily evaluated as

2/

2/ 0

0

0

0

1

)(1~

a

a

tjn

Ttjn

n

dteVT

dtetfT

c

2/

)2/sin(

22~

0

000

2/2/0

00

an

anaV

j

ee

n

Vc

ajnajn

n

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Fourier analysis for nonperiodic functions (continued)

Using fact that T = 2/0, may be written

We are interested in what happens as period T gets larger, with pulse width a fixed

– For graphs, a = 1, V0 = 1

Tna

Tna

T

aV

an

an

T

aVcn /

)/sin(

2/

)2/sin(~0

0

00

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Effect of increasing period T

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0 50 100 150

frequency

co

eff

icie

nt

va

lue

, T

=2

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 50 100 150

frequency

co

eff

icie

nt

va

lue

, T

=5

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0 50 100 150

frequency

co

eff

icie

nt

va

lue

, T

=1

0

a/T a/T

a/T

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Transition to Fourier integral

We can define f(jn0) in the following manner

Since difference in frequency of terms = 0 in the

expansion. Hence

nnn

a

a

tjnn

a

a

tjnn

ccTcjnF

dteVTc

dteVT

c

~2

~2~)(

~

1~

00

2/

2/0

2/

2/ 0

0

0

2

)(~ 0jnF

cn

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Transition to Fourier integral (continued)

Since

It follows that

As we pass to the limit, -> d, n -> so we have

n

tjn

Te

jnFtf 0

2

)(lim)( 0

n

tjnnectf 0~)(

dejFtf tj)(2

1)(

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This is subject to convergence condition

Now observe that since

We have

dttf )(

T tjn

n dtetfT

c0

0)(1~

2/

2/0

0)()(~ a

a

tjnn dtetfjnFTc

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In the limit as T ->

Since f(t) = 0 for t < -a/2 and t > a/2 Thus we have the Fourier transform pair for nonperiodic

functions

dtetf

dtetfjF

tj

a

a

tj

)(

)()(2/

2/

)()(

)()(

jFtf

tfjF1-F

F

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Example: pulse

For pulse of area 1, height a, width 1/a, we have

Note that this will have zeros at = 2ann=0,+1, +2 Considering only positive frequencies, and that “most” of

the energy is in the first lobe, out to 2a, we see that product of bandwidth 2a and pulse width 1/a = 2

a

adtaetF

a

a

tj 2sin2

)(2/1

2/1

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Example of pulse

width=1

width=0.2

1/2-1/2

1/10

-1/10

1

5

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Pulse: limiting cases

Let a -> , then f(t) -> spike of infinite height and width 1/a (delta function) -> 0

– Transform -> line F(j)=1– Thus transform of delta function contains all frequencies

Let a -> 0, then f(t) -> infinitely long pulse– Transform -> spike of height 1, width 0

Now let height remain at 1, width be 1/a– Then transform is

a

aa

a

aa

ajF

2

2sin

1

2

2sin

2

22sin2

)(

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Pulse: limiting cases (continued)

Now, we are interested in limit as a -> 0 for -> 0 and > 0

– First, consider case of small :

– So when a -> 0, 1/a -> – As w moves slightly away from 0, it drops to zero quickly

because of w/2a term in denominator (numerator <1 at all times)

So we get delta function, (0)

aa

aa

a

aa

1

2

2sin

lim1

2

2sin

1lim

00

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Fourier transform of pulse width 0.1

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Properties of delta function

Definition

Area for any > 0

Sifting property

since

0

00)(

x

xx

dxxdxx )(1)(

)()()(

)0()()(

00 xfdxxfxx

fdxxfx

0

00

0)(

xx

xxxx

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Some common Fourier transform pairs

Source: http://mathworld.wolfram.com/FourierTransform.html

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Some Fourier transform pairs (graphical illustration)

function functiontransform transform

Source: Physical Optics Notebook: Tutorials in Fourier Optics, Reynolds, et. al., SPIE/AIP

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Fourier transform: Gaussian pulses

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Properties of Fourier transforms

Simplification:

Negative t:

Scaling– Time:

– Magnitude:

odd)(,sin)(2)(

even)(,cos)(2)(

0

0

tfdtttfjF

tfdtttfjF

)(*)( jFtfF

a

jF

aatf

1)(F

)()( jaFtafF

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Properties of Fourier transforms (continued)

Shifting:

Time convolution:

Frequency convolution:

)()(21

sin)(

)()(21

cos)(

)()(

)()(

000

000

00

jFjFttf

jFjFttf

jFetf

ejFatftj

aj

F

F

F

F

-

F dtffjFjF )()()()( 21211

djFjFtftf )()(

2

1)()( 2121F

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Convolution and transforms A principal application of any transform theory comes from

its application to linear systems

– If system is linear, then its response to a sum of inputs is equal to the sum of its responses to the individual inputs

– This was original justification for Fourier's work Because a delta function contains all frequencies in its

spectrum, if you “hit” something with a delta function, and measure its response, you know how it will respond to any individual frequency– The response of something (e.g., a circuit) to a delta

function is called its “impulse response”• Called “point spread function” in optics

– Often denoted h(t)

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Convolution and transforms (continued) The Fourier transform of the impulse response can be

calculated, usually designated H(j) Therefore if one knows the frequency content of an incoming

“signal” u(t), one can calculate the response of the system– The response to each individual frequency component of

incoming signal can be calculated individually as product of impulse response and that component

– Total response is obtained by summing all of individual responses

That is, response Y(j) = H(j)U(j)– Where U(j) is sum of Fourier transforms of individual

components of u(t)

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Convolution and transforms (continued)

May be visualized as

H(j)U(j) Y(j)=H(j)U(j)

SystemInput Response

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Convolution and transforms (continued) Example

– Signal is square wave, u(t)=sgn(sin(x))

– This has Fourier transform

– So response Y(j) is

1

0

12

)12sin()(

n n

tntu

1

00

2

)12(

2

)12(

2)(

n

nnijU

1

00

2

)12(

2

)12()(

2)()()(

n

nnjH

ijUjHjY

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If incoming signal described by Fourier integral instead, same result holds

To get time (or space) domain answer, we need to take inverse Fourier transform of Y(j)

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Can also be calculated in time (or space), i.e., non-transformed domain

Derivation

Now, we introduce new variables v and , related to t and z by

dtdzezuth

dzezudteth

jUjHjY

ztj

zjtj

)()()(

)()(

)()()(

zztv ,

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Computing Jacobean to transform variables

– Implies that differential areas same for both systems of variables

Thus since t = v-z = v- we have

Where we have calculated the limits as follows

dvddvdv

tzt

v

zdzdt

dveduvhjY vj

v

0)()()(

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We may assume without loss of generality that u(z) = 0 for z<0

– Otherwise we can shift variables to make it so• Must assume that u(z) has some starting point

– Therefore the lower limit of integration in the inner integral is 0

We may also assume without loss of generality that h(t) = 0 for t<0– Therefore h(v-) = 0 for > v

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Since the outer integral defines a Fourier transform, its inverse is just y(t), so we have

This is usually written with t as the inner variable,

This is called the convolution of h and u, usually written y(t) = h*u

Can readily be calculated on a computer

vduvhjYFty

0

1 )()()()(

t

duthty0

)()()(

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Convolution: old way (graphically)

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Convolution: old way (continued)

Source: P. S. Rha, SFSU, http://online.sfsu.edu/~psrha/ENGR449_PDFs/EE449_L5_Conv.PDF

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Convolution and transforms (new way)

Use computer algebra programs Some Derive scripts

– Step function: u(t):=if(t<0,0,1)– Pulse of width d, amplitude a: f1(t):=if(t>=0 and t<=d,a,0)– Triangle of width d, amplitude a:

triangle(t):=if(t>=0 and t<=d/2,2at/d,(if(t>d/2 and t<d,2a-2at/d,0)0)

– Convolution: convolution(t):=int(f1(t-)*f2(),,0,t) Example

– f1 is pulse of width 1, amplitude 1– f2 is pulse of width 2, amplitude 3

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Convolution functions

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Convolution: useful web sites

http://www.jhu.edu/~signals/ http://mathworld.wolfram.com/Convolution.html http://www.annauniv.edu/shan/Lap1.1.9.html http://rivit.cs.byu.edu/morse/550-F95/node12.html

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Fourier and Laplace transforms

Fourier transform does not preserve initial condition information

– Therefore most useful when “steady state” conditions exist

• This is typically the case for optical systems• But often not true for electrical networks

Comparison of definitions

j

j

st

st

dsesFj

tf

dtetfsF

1

1

)(2

1)(

)()(0

dejFtf

dtetfjF

tj

tj

)(2

1)(

)()(

Laplace Fourier

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Fourier and Laplace transforms (continued)

Differences

– In Fourier transform, j replaces s– Limits of integration are different, one-sided vs. two-sided– Contours of integration in inverse transform different

• Fourier along imaginary axis

• Laplace along imaginary axis displaced by 1

Conversion between Fourier and Laplace transforms– Laplace transform of f(t) = Fourier transform of f(t)e-t

– Symbolically,

tetftf )()( FL

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Fourier transforms of random sources (noise)

Noise has frequency characteristics

– Generally continuous distribution of frequencies– Since transform of individual frequencies gives spikes, this

allows us to separate signal from noise via Fourier methods Common types of noise

– White noise: equal power per Hz (power doubles per octave)– Pink noise: equal power per octave– Other “colors” of noise described at

http://www.hoohahrecords.com/resfreq/articles/noise.html– Fourier transform distinguishes these

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Fourier transforms of random sources (noise) (continued)

Frequency domain thus allows us to obtain information about signal purity that is difficult to obtain in time (or space) domain

– Noise– Distortion

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Fourier transforms of random sources (noise) (continued)

Source: http://hesperia.gsfc.nasa.gov/~schmahl/fourier_tutorial/node6.html

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Discrete and Fast Fourier Transforms Most Fourier work today carried out by computer (numerical)

analysis Discrete Fourier transform (DFT) is first step in numerical

analysis– Simply sample target function f(t) at appropriate times– Replace integral by summation

Here tn = nT, where T=sampling interval, N = number of

samples, and frequency sampling interval = 2/NT,

k = k

12,1,0,)()(

)()(

1

0

NketfjF

dtetfjF

N

n

tjnk

tj

nk

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Discrete and Fast Fourier Transforms (continued)

Sampling frequency fs = 1/T

Frequency resolution f = 1/NT = fs/N

For accurate results, sampling theorem tells us that sample frequency f

s > 2 x f

max, the highest frequency in the

signal

– Implies that highest frequency captured fmax

< 1/2T = fs/2

• Otherwise aliasing will occur To improve resolution, note that you can't double sampling

frequency, as that also doubles N (for same piece of waveform)

– The only way to increase N without affecting fs is to

increase acquisition time

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Discrete and Fast Fourier Transforms (continued)

Note that DFT calculation requires N separate summations, one for each

k

Since each summation requires N terms, number of calculations goes up as N 2

– Therefore doubling frequency resolution requires quadrupling number of calculations

Method also assumes function f(t) is periodic outside time range (nT) considered

Also note that raw DFT calculation gives array of complex numbers which must be processed to give usual magnitude and phase information– When only power information required, squaring

eliminates complex terms

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Inverse discrete Fourier transform

Calculated in straightforward manner as

This gives, of course, the original sampled values of the function back– Other values can be determined by appropriate filtering

1,2,1,0,)(1

)(1

0

NnejFN

tfN

k

tjkn

nk

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Uses of DFT

DFT usage may be visualized as

DFT Spectrum

Magnitude Phase

Power Spectrum

Power Spectral Density

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Power measurements and DFT

Power spectrum

– Gives energy (power) content of signal at a particular frequency

– No phase information– Squared magnitude of DFT spectrum

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Power spectral density

Derived from power spectrum Generally normalized in some fashion to show relative power

in different ranges Measures energy content in specific band

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Fast Fourier Transform (FFT) Developed by Cooley and Tukey in 1965 to speed up DFT

calculations Increases speed from O(N2) to O(N log N), but there are

requirements Useful reference: http://www.ni.com/swf/presentation/us/fft/

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Fast Fourier Transform (FFT) (continued)

Requirements for FFT

– Sampled data must contain integer number of cycles of base (lowest frequency) waveform

• Otherwise discontinuities will exist, giving rise to “spectral leakage”, which shows up as noise

– Signal must be band limited and sampling must be at high enough rate

• Otherwise “aliasing” occurs, in which higher frequencies than those capturable by sampling rate appear as lower frequencies in FFT

– Signal must have stable (non-changing) frequency content

– Number of sample points must be power of 2

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Spectral leakage

No discontinuities Discontinuities present

Source: National Instruments

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Fast Fourier Transform (FFT) (continued)

We will not discuss exactly how the method works Lots of software packages are available

– See this site for many of them http://ourworld.compuserve.com/homepages/steve_kifowit/fft.htm

– Contained in Mathcad package– Also available in many textbooks– Many modern instruments such as digital oscilloscopes

have FFT built-in Averaging is frequently used to improve result

– Averages over several FFT runs with different data sets representing same waveform

• Sometimes with slightly staggered start times

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FFT (continued)

Also inverse FFT exists for going in opposite direction Short Mathcad demo Note that output of FFT is two-dimensional array of length ½

number of sample points + 1

– The points in this array are the complex values F(jk)

– But the k values themselves do not appear

• Must be calculated by user

• They are k = k x frequency resolution = k x 2/NT,

k = 0...N/2

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FFT examples showing different resolution

10

0.010797

ddjj 2

0.50 ddjj 1

0 0.1 0.2 0.3 0.40

2

4

6

8

109.371085

5.398305 103

dj 2

0.50 dj 1

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

f(x)=sin (x/5), analysis done in MATHCAD

32 sample points, T=1 sec, fs=1

resolution 1/32 Hz

64 sample points, T=1 sec, fs=1

resolution 1/64 Hz

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Fourier analysis: a deeper view

Fourier series only one possible way to analyze functions Best understood in terms of functional analysis Let X be a space composed of real-valued functions on some

interval [a,b]– Technically, the set of Lebesgue-integral functions– Infinite-dimensional space

Define an inner product (“dot product” in Euclidean space) as follows:

b

adttytxyx )()(,

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Fourier analysis: a deeper view (continued)

This induces a norm on the space

Can be shown that this space is complete– Complete normed space with norm defined by inner

product is known as a Hilbert space

An orthogonal sequence (uk) is a sequence of elements u

k of X

such that

2/122/1

)(,

b

adttxyxx

jiuu ji ,0,

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This series can be converted into an orthonormal sequence (e

k) by dividing each element u

k by its norm ||u

k||

Consider an arbitrary element x X, and calculate

Now formulate the sum

Then clearly if ||x-xn||0 as nthe sum converges to x

kk ex,

1k

kkn ex

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Fourier analysis: a deeper view (continued) We have the following theorem: If (e

k) is an orthonormal

sequence in Hilbert space X, then

(a) The series converges (in the norm on X) if and only if the following series converges:

(b) If the series converges, then the coefficients k

are the Fourier coefficients so that x can be written

1k

kke

1k

k

1k

kke

kex,

1

,k

kkn eexx

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(c) For any x X, the foregoing series converges

Lemma: Any x in X can have at most countably many (may be countably infinite) nonzero Fourier coefficients with respect to an orthonormal set (e

k)

Note that we are not quite where we want to be yet, as we have not shown that every x X has a sequence which converges to it

– For this we require another notion, that of totality

kex,

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Note also that as of this point we have said nothing about the nature of the functions e

k

– Any set which meets the orthogonality condition is OK, since it can be normalized

– Note that (sin nt), (cos nt) meet condition, can be combined into new set containing all elements by suitable renumbering

– Lots of other functions would work as well, such as triangle waves, Bessel functions

jiuu ji ,0,

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Most interesting orthonormal sets are those which consists of “sufficiently many” elements so that every element in the space can be approximated by Fourier coefficients

– Trivial in finite-dimensional spaces: just use orthonormal basis

– More complicated in infinite dimensional spaces Define a total orthonormal set in X as a subset M X whose

span is dense in X– Functions analogously to orthonormal basis in finite

spaces– But Fourier expansion doesn't have to equal every

element, just get arbitrarily close to it in sense of norm

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Can be shown that all total orthonormal sets in a given Hilbert space have same cardinality

– Called Hilbert dimension or orthogonal dimension of the space

– Trivial in finite dimensional spaces Necessary and sufficient condition for totality of an

orthonormal set M is that there does not exist a non-zero x X such that x is orthogonal to every element of M

0 xMx

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Parseval relation can be expressed as

Another theorem states that an orthonormal set M is total in X if and only if the Parseval relation holds for all x– True for {(sin nt)/, (cos nt)/ terms– Therefore these terms form total orthonormal set

Key results– Fourier expansion works because {(sin nt)/, (cos

nt)/}terms from orthonormal basis for space of functions– Any other orthonormal set of functions can also serve as

basis of Fourier analysis

22, xex

kk

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Effect of truncating Fourier expansion

– Finite set (e1...e

m) no longer total

– But it can be shown that the projection theorem applies

Space spanned by (e1...e

m)

Function f(x) to be approximated Approximation error

Approximation fm(x)

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Projection theorem states that optimal representation of f(x) in lower-order space obtained when error ||f – f

m|| is

orthogonal to fm

This is guaranteed by orthonormal elements ei and the

construction of the Fourier coefficients Therefore truncated Fourier representation is optimal

representation in terms of (e1...e

m)

References: – Erwin Kreyszig, Introductory Functional Analysis with

Applications– Eberhard Zeidler, Nonlinear Functional Analysis and its

Applications, Vol. I, Fixed-Point Theorems

ece 665 fourier optics spring, 2004 thomas b. fowler, sc.d. senior principal engineer mitretek...

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