Basic properties of Fourier Transforms

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• Lecture 16
• Basic properties of Fourier Transforms

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MatLab Code

• setup
• N256
• dt1.0
• tmaxdt(N-1)
• tdt0N-1
• fmax1/(2.0dt)
• dffmax/(N/2)
• fdf0N/2,-N/21-1'
• dw2pidf
• wdw0N/2,-N/21-1'
• a sample function, p(t)
• w0 2pifmax/10
• p sin(w0t).exp(-0.25w0t)
• ptfft(p) fourier transform
• prifft(pt) inverse transform

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p(t)
t
ifft(fft(p))
t
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1 Relationship to integral transforms

• Integral transforms
• C(w) ?-?? T(t) exp(-iwt) dt
• T(t) (1/2p) ?-?? C(w) exp(iwt) dw
• Discrete transforms
• Ck Sn-N/2N/2 Tn exp(-2pikn/N ) with k-½N, ,
  ½N
• Tn N-1Sk-N/2N/2 Ck exp(2pikn/N ) with n-½N,
  , ½N
• (this agrees with definition in MatLab HELP)

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• wk kDw and tn nDt and Dw 2p/(NDt)
• so DwDt 2p/N
• Ck C(wk) ?-?? T(t) exp(-iwkt) dt
• ? Dt Sn-N/2N/2 T(tn) exp(-iwktn)
• Dt Sn-N/2N/2 Tn exp(-i kDw nDt)
• Dt Sn-N/2N/2 Tn exp(-2pi kn / N )

So forward integral transform is Dt times
discrete forward transform
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• wk kDw and tn nDt and Dw 2p/(NDt)
• so DwDt2p/N and Dw/2p 1/(NDt)
• T(tn) (1/2p) ?-?? C(w) exp(iwtn) dw
• ? (Dw/2p) Sk-N/2N/2 C(wk) exp(iwktn)
• (1/Dt) N-1 Sk-N/2N/2 Ck exp(i kDw nDt)
• (1/Dt) N-1 Sk-N/2N/2 Ck exp(2pi kn / N )

So inverse integral transform is 1/Dt times
inverse discrete transform
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So

• Except for a normalization factor of Dt,
• The discrete transform is just the Riemann Sum
  approximation to the integral transform, with
  particular choice DwDt2p/N
• Properties of the integral transform carry over
  to the discrete transform (well, more-or-less)

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Error Estimates for the DFT

• Assume uncorrelated, normally-distributed data,
  dnTn, with variance sd2
• The matrix G in Gmd is GnkN-1 exp(2pikn/N )
• The problem Gmd is linear, so the unknowns,
  mkCk, (the coefficients of the complex
  exponentials) are also normally-distributed.
• Since exponentials are orthogonal, GHGN-1I is
  diagonal
• and Cm sd2 GHG-1 N-1sd2I is diagonal, too
• Apportioning variance equally between real and
  imaginary parts of Cm, each has variance s2
  N-1sd2/2.
• The spectrum sm2 Crm2 Cim2 is the sum of two
  uncorrelated, normally distributed random
  variables and is thus c22-distributed.
• The 95 value of c22 is about 5.9, so that to be
  significant, a peak must exceed 5.9N-1sd2/2

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example

• f(t) exp(-a2t2)
• f(w) ?-?? exp(-a2t2) exp(-iwt) dt
• 2 ?0? exp(-a2t2) cos(wt) dt
• ?p exp( -w2/4a2 ) / a
• See integral 679 of CRC Math Tables
• Note, by the way, that the Fourier transform of a
  Gaussian is a Gaussian Furthermore, the wider in
  t, the narrower in w.

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Dt fft( exp(-a2t2) ) with a0.05
w
?p exp( -w2/4a2 ) / a with a0.05
w
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2 area under T(t)

• C(w0) is the area under T(t)

area
T(t)
0
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• C(w) ?-?? T(t) exp(-iwt) dt
• C(w0)
• ?-?? T(t) exp(0) dt
• ?-?? T(t) dt area

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area dt ? sum(p) 18.2245
t
area Dt ? real(fft(p)) 18.2245
t
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3 Time shift

• Multiplying C(w) by
• exp(-iwt0)
• shifts the timeseries T(t) by t0.

T(t)
T(t-t0)
t
t0
0
0
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• C(w) ?-?? T(t) exp(-iwt) dt
• Transform(shifted)
• ?-?? T(t-t0) exp(-iwt) dt
• ?-?? T(t) exp-iw(tt0) dt
• exp(-iwt0) ?-?? T(t) exp(-iwt) dt
• exp(-iwt0) Transform(unshifted)

t t-t0 so ttt0 and dtdt and t?? as t??
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p(t)
t
ifft(exp(-iwt0)?fft(p(t))) with t050
t
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4 derivative

• Multiplying C(w) by
• iw
• Gives the transform of dT/dt

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• transform(dT/dt)
• ?-?? dT/dt exp(-iwt) dt
• T exp(iwt) -??
• iw ?-?? T exp(iwt) dt
• -iw transform(T)

Integration by parts ? u dv uv - ? v du u
exp(iwt) dv(dT/dt) dt du -iw exp(-iwt) dt v
?dv T
assuming T exp(iwt) -?? 0
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p(t)
dp/dt
ifft( iw fft(p))
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5 integral

• Dividing C(w) by
• iw
• Gives the transform of ? T dt
• but note that you must set the zero-frequency
  element manually, since 1/w is undefined at w0.
  Furthermore, this process is not very stable,
  since 1/w can be very large for small ws.

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p(t)
t
Dt?cumsum(p)
t
a bit of drift, presumably due to round off error
ifft(fft(p)/(iw))
t
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6 convolution

• The transform of
• the convolution of two timeseries
• is the product of their
• transforms

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• transform( f(t)g(t) )
• ?-?? ?-?? f(t-t) g(t) dt exp(iwt) dt
• ?-?? g(t) ?-?? f(t-t) exp(iwt) dt dt
• ?-?? g(t) ?-?? f(t) expiw(tt) dt dt
• ?-?? g(t) exp(iwt) dt ?-?? f(t) exp(iwt) dt
  
• transform(g(t)) transform(f(t))

tt-t so ttt dt dt
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f(t)
g(t)
t
conv(g,f)
t
ifft(fft(g)?fft(f))
t
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7 FFT of a spike at t0 is a constant

• C(w) ?-?? d(t) exp(-iwt) dt exp(0) 1

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p(t) is a spike at t00
t
real(fft(p))
imag(fft(p))
w
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7 FFT of a spike at tt0 is sinusoidal

• C(w) ?-?? d(t-t0) exp(-iwt) dt exp(-iwt0)
• cos(wt0) - i sin(wt0)

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p(t) is a spike at t05
t
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real(fft(p))cos(wt0)
imag(fft(p))sin(wt0)
w
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8 FFT of a sinusoid cos(w1t) is a pair of spikes
at ww1

• Note rule ?-?? exp(-iw1t) exp(iw2t) dt
  d(w1-w2)
• cos(w1t) ½ exp(iw1t) exp(-iw1t)
• ?-?? cos(w1t) exp(-iwt) dt
• ½ ?-?? exp(iw1t) exp(-iwt) dt ½ ?-??
  exp(iw1t) exp(-iwt) dt
• ½ d(w-w1) d(ww1)

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p(t)cos(w1t)
t
real(fft(p))
imag(fft(p))0
w
w1
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Aliasingthe tendency in a digital world for
high frequencies to look like low frequencies
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MatLab Script to evaluate cosines at ever
increasing frequency

• L 0 make plots in groups of 10 starting at
  frequency L1
• for k 110
• w1 (Lk)dw
• pcos(w1t)
• subplot(10,1,k)
• plot(t,p,'b')
• hold on
• axis( 0, tmax, -1.5, 1.5 )
• end

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cos(kDwt)
k1
k10
t
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same!
cos(kDwt)
k11
k16
nyquist
k21
t
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cos(kDwt)
k21
k31
frequencies seem to be getting lower!
t
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cos(kDwt)
k31
k41
but now higher again!
t
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in a digital world we have chosen DwDt 2p/N

• sin(wntk) sin(kDwnDt) sin(knDwDt)
  sin(2pkn/N)
• cos(wntk) cos(kDwnDt) cos(knDwDt)
  cos(2pkn/N)
• let mnN
• sin(wmtk) sink(nN)DwDt sin2pk(nN)/N
• cos(2pkn/N) sin(2pk) sin(2pkn/N) cos(2pk)
• cos(wmtk) cosk(nN)DwDt cos2pk(nN)/N
• cos(2pkn/N) cos(2pk) - sin(2pkn/N) sin(2pk)
• but sin(2pk)0 and cos(2pk)1 for all integers k,
  so
• sin(wmtk)sin(wntk) and cos(wmtk)cos(wntk)

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in a digital world wnN wn andsince time
and frequency play symmetrical roles in
exp(-iwt) tkN tk
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Example cos(w0t) with w0wny/20 sampled with
frequency w0wny/32
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Now connect the dots the resulting function has a
lower frequency
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wnN wn let n-m, then w-m wN-m
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This is why the fourier coefficientsof negative
frequenciesare placed at the high frequency
end of the transform vectorthey are the high
frequency coefficients
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lets reconsider the example where we computed the
Fourier Transform ofp(t) exp(-a2t2)this
function is non-zero at negative times
p(t)
0
t
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you must not just leave out the values of the
function at negative timesyoud be leaving out
half the informationinstead, you must wrap them
to large times using the rule t-m tN-m
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so you must put the negative values at the right
and end of the vector, p

• N256
• dt0.5
• tmaxdt(N/2)
• tmindt(-N/21)
• tdt0N/2,-N/21-1' note put negative
  times on the end ...
• fmax1/(2.0dt)
• dffmax/(N/2)
• fdf0N/2,-N/21-1'
• dw2pidf
• wdw0N/2,-N/21-1'
• a0.05
• p exp( -(at).2 )
• ptfft(p)

p(t)
0
t
p(t)
0
t
t
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Remember the Time shift ?

• Multiplying C(w) by exp(-iwt0) shifts the
  timeseries T(t) by t0.
• What if we try to shift it by more than the
  length of the time-series?
• It just wraps around

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with N256, trying to shift it 280Dt just shifts
it 280-25624
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