Post on 30-Jan-2016
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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How to calculate Fourier coefficients (continued)
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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How to calculate Fourier coefficients (continued)
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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How to calculate Fourier coefficients (continued)
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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Calculation of Fourier coefficients: examples (continued)
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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Calculation of Fourier coefficients: examples (continued)
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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Exponential form of Fourier Series (continued)
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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Exponential form of Fourier Series (continued)
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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Transition to Fourier integral (continued)
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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Transition to Fourier integral (continued)
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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Convolution and transforms (continued)
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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Convolution and transforms (continued)
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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Convolution and transforms (continued)
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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Convolution and transforms (continued)
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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Convolution and transforms (continued)
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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Fourier analysis: a deeper view (continued)
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
86ControlNumber
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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Fourier analysis: a deeper view (continued)
(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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Fourier analysis: a deeper view (continued)
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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Fourier analysis: a deeper view (continued)
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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Fourier analysis: a deeper view (continued)
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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Fourier analysis: a deeper view (continued)
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
92ControlNumber
Fourier analysis: a deeper view (continued)
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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Fourier analysis: a deeper view (continued)
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