Lecture 16

Basic properties of Fourier Transforms

MatLab Code% setupN=256;dt=1.0;tmax=dt*(N-1); t=dt*[0:N-1]; fmax=1/(2.0*dt);df=fmax/(N/2);f=df*[0:N/2,-N/2+1:-1]'; dw=2*pi*df;w=dw*[0:N/2,-N/2+1:-1]';

% a sample function, p(t)w0 = 2*pi*fmax/10;p = sin(w0*t).*exp(-0.25*w0*t);

pt=fft(p); % fourier transformpr=ifft(pt); % inverse transform

p(t)

t

t

ifft(fft(p))

1: Relationship to integral transforms

Integral transforms:

C() = -+

T(t) exp(-it) dt

T(t) = (1/2) -+

C() exp(it) d

Discrete transforms

Ck = n=-N/2N/2 Tn exp(-2ikn/N ) with k=-½N, …, ½N

Tn = N-1k=-N/2N/2 Ck exp(+2ikn/N ) with n=-½N, …, ½N

(this agrees with definition in MatLab HELP)

kk and tn = nt and = 2/(Nt)

so t = 2/N

Ck = C(k) = -+

T(t) exp(-ikt) dt

t n=-N/2N/2 T(tn) exp(-iktn)

= t n=-N/2N/2 Tn exp(-i k nt)

= t n=-N/2N/2 Tn exp(-2i kn )

So forward integral transform is t times discrete forward transform

kk and tn = nt and = 2/(Nt)

so t=2/N and = 1/(Nt)

T(tn) = (1/2) -+

C() exp(itn) d

(/2) k=-N/2N/2 C(k) exp(iktn)

= (1/t) N-1 k=-N/2N/2 Ck exp(i k nt)

= (1/t) N-1 k=-N/2N/2 Ck exp(2i kn )

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

So …

Except for a normalization factor of t,

The discrete transform is just the “Riemann Sum” approximation to the integral transform, with particular choice t=2/N

Properties of the integral transform carry over to the discrete transform (well, more-or-less)

Error Estimates for the DFTAssume uncorrelated, normally-distributed data, dn=Tn, with

variance d2

The matrix G in Gm=d is Gnk=N-1 exp(+2ikn/N )

The problem Gm=d is linear, so the unknowns, mk=Ck, (the coefficients of the complex exponentials) are also normally-distributed.

Since exponentials are orthogonal, GHG=N-1I is diagonaland Cm= d

2 [GHG]-1 = N-1d2I is diagonal, too

Apportioning variance equally between real and imaginary parts of Cm, each has variance 2= N-1d

2/2.The spectrum sm

2= Crm

2+ Cim

2 is the sum of two uncorrelated, normally distributed random variables and is thus

2-distributed.

The 95% value of 2 is about 5.9, so that to be significant, a

peak must exceed 5.9N-1d2/2

examplef(t) = exp(-a2t2)

f() = -+ exp(-a2t2) exp(-it) dt

= 2 0+ exp(-a2t2) cos(t) dt

= exp( -2/4a2 ) / 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 .

t fft( exp(-a2t2) ) with a=0.05

exp( -2/4a2 ) / a with a=0.05

C() is the area under T(t)

2: area under T(t)

0

T(t)

area

C() = -+

T(t) exp(-it) dt

C()

= -+

T(t) exp(0) dt

= -+

T(t) dt = area

area = dt sum(p) = 18.2245

t

t

area = t real(fft(p)) = 18.2245

Multiplying C() by

exp(-it0)

shifts the timeseries T(t) by t0.

3: Time shift

0t

T(t)

0 t0

T(t-t0)

C() = -+

T(t) exp(-it) dt

Transform(shifted) =

-+

T(t-t0) exp(-it) dt =

-+

T(t’) exp{-it’+t0)} dt’ =

exp(-it0) -+

T(t’) exp(-it’) dt’ =

exp(-it0) Transform(unshifted)

t’ = t-t0 so t=t’+t0

and dt’=dtand t’ as t

p(t)

ifft(exp(-it0)fft(p(t))) with t0=50

t

t

Multiplying C() by

i

Gives the transform of dT/dt

4: derivative

transform(dT/dt) =

-+

dT/dt exp(-it) dt

T exp(it |-+

+i -+

T exp(it dt

= -i transform(T)

Integration by parts

u dv = uv - v du

u= exp(it) dv=(dT/dt) dt

du = -i exp(-it) dt

v = dv = T

assuming T exp(it |-+

= 0

p(t)

dp/dt

ifft( i fft(p))

Dividing C() by

i

Gives the transform of T dt

but note that you must set the zero-frequency element manually, since 1/ is undefined at =0. Furthermore, this process is not very stable, since 1/can be very large for small ’s.

5: integral

p(t)

tcumsum(p)

ifft(fft(p)/(i))a bit of drift, presumably

due to round off error

t

t

t

The transform of

the convolution of two timeseries

is the product of their

transforms

6: convolution

transform( f(t)*g(t) ) =

-+

-+

f(t-) g() d exp(it) dt

-+

g() -+

f(t-) exp(it) dt d

-+

g() -+

f(t’) exp{it’+)} dt’ d

-+

g() exp(i) d -+

f(t’) exp(it’) dt’

transform(g(t)) transform(f(t))

t’=t-sot=t’+dt’ = dt

t

t

t

f(t)g(t)

conv(g,f)

ifft(fft(g)fft(f))

7: FFT of a spike at t=0 is a constant

C() = -+

(t) exp(-it) dt = exp(0) = 1

t

p(t) is a spike at t0=0

imag(fft(p))

real(fft(p))

7: FFT of a spike at t=t0 is sinusoidal

C() = -+

(t-t0) exp(-it) dt = exp(-it0)

= cos(t0) - i sin(t0)

t

p(t) is a spike at t0=5

imag(fft(p))=sin(t0)

real(fft(p))=cos(t0)

5

8: FFT of a sinusoid cos(1t) is a pair of spikes at w=±1

Note rule -+

exp(-it) exp(it) dt = (1-2)

cos(t) = ½ { exp(it) + exp(-it) }

-+

cos(t) exp(-it) dt =

½ -+

exp(it) exp(-it) dt + ½ -+

exp(it) exp(-it) dt

= ½{ (-1) + (+1) }

t

p(t)=cos(1t)

imag(fft(p))=0real(fft(p))

±1

Aliasing

the tendency in a digital world for high frequencies to look like low

frequencies

MatLab Script to evaluate cosines at ever increasing frequency

L = 0; % make plots in groups of 10 starting at frequency L+1for k = [1:10]w1= (L+k)*dw;p=cos(w1*t);

subplot(10,1,k);plot(t,p,'b');hold on;axis( [0, tmax, -1.5, 1.5] );

end

t

cos(kt)

k

k

t

cos(kt)

k

k

knyquist

same!

t

cos(kt)

k

kfrequencies seem to be getting lower!

t

cos(kt)

k

kbut now higher again!

in a digital world we have chosen t = 2/N

sin(ntk) = sin(knt) = sin(knt) = sin(2kn/N)cos(ntk) = cos(knt) = cos(knt) = cos(2kn/N)

let m=n+Nsin(mtk) = sin{k(n+N)wt} = sin{2k(n+N)}

= cos(2knsin(2k) sin(2kncos(2k)

cos(mtk) = cos{k(n+N)wt} = cos{2k(n+N)} = cos(2kncos(2k) sin(2knsin(2k)

but sin(2k)=0 and cos(2k)=1 for all integers k, so

sin(mtk)=sin(ntk) and cos(mtk)=cos(ntk)

in a digital world n+N = n

andsince time and frequency play symmetrical roles in exp(-it)

tk+N = tk

Example: cos(0t) with 0=ny/20 sampled with frequency 0=ny/32

Now connect the dots the resulting function has a lower frequency

n+N = n

let n=-m, then

-m = N-m

This is why the fourier coefficientsof ‘negative frequencies’

are placed at the ‘high frequency’ end of the transform vector

they “are” the high frequency coefficients

lets reconsider the example where we computed the Fourier Transform of

p(t) = exp(-a2t2)

this function is non-zero at negative times

0

p(t)

t

you must not just leave out the values of the function at negative times

you’d be leaving out half the information

instead, you must wrap them to large times using the rule

t-m = tN-m

so you must put the negative values at the right and end of the vector, p

N=256;dt=0.5;tmax=dt*(N/2); tmin=dt*(-N/2+1);t=dt*[0:N/2,-N/2+1:-1]'; % note put negative times on the end ...fmax=1/(2.0*dt);df=fmax/(N/2);f=df*[0:N/2,-N/2+1:-1]'; dw=2*pi*df;w=dw*[0:N/2,-N/2+1:-1]';

a=0.05;p = exp( -(a*t).^2 );

pt=fft(p);

0

p(t)

t

0

p(t)

t

Multiplying C() by exp(-it0) 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 …

Remember the Time shift ?

with N=256, trying to shift it 280t just shifts it 280-256=24

24