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So, I have a noisy signal in time domain, $f(t) = t \eta(t)$ where $\eta(t)$ is white noise with variance $\sigma$ and mean zero, and that it has the property $\langle \eta(t)\eta(t') \rangle = \sigma^2 \delta(t-t')$ ($\langle \ldots \rangle$ denotes ensemble averaging). What I want to do is that I want to calculate the square magnitude of the spectrum in frequency domain upon ensemble averaging, $\langle |z(\omega)|^2 \rangle$. To do this, I follow Dave's answer in https://physics.stackexchange.com/questions/53739/magnitude-of-the-fourier-transform-of-white-noise. So, it will be $$ \langle |z(\omega)|^2 \rangle = \iint e^{i\omega t} e^{-i\omega t'}t't \langle \eta(t)\eta(t') \rangle dt dt' $$ Using the property stated above, $\langle \eta(t)\eta(t') \rangle = \sigma^2 \delta(t-t')$, I can obtain $$ \langle |z(\omega)|^2 \rangle = \sigma^2 \int t^2 dt $$. Thus, the ensemble averaged square magnitude of my function is independent of $\omega$.

Now I perform the same calculation using MATLAB and I obtained the following pictures. enter image description here

It seems like when I take ensemble average of the lower curve, it will be dependent on $\omega$, which contradicts my calculation. Can someone please point out my mistake(s)?

clear all;
clc;

n = 2048;
ft = rand(1,n)*(2+2)-2 + 0;
t = linspace(-10,10,n);
Dt = 1/(t(3)-t(1));

fw = ifft(fft(t.*ft));
w = (-length(t)/2:length(t)/2-1)*2*pi*Dt/length(t);
subplot(211)
plot(t,t.*ft)
xlabel('t')
ylabel('f(t)')

subplot(212)
plot(w,abs(fw).^2)
xlabel('\omega')
ylabel('|z(\omega)|^2')

That's my code. By the way I think you are using normal distribution for the noise, I used white noise type.

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nougako
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  • You are taking the FT of $t\eta(t)$, not of $\eta(t)$. That factor of $t$ definitely makes the FT dependent on frequency, even if $\eta$ does not. In fact, I would say you are getting the exact result you should be getting. – bob.sacamento Dec 01 '15 at 19:42
  • Of course the FT of $t\eta(t)$ (and also $\eta(t)$ alone) will depend on $\omega$. But what I calculate is the ensemble average of the squared magnitude of the spectrum which, according to my possibly wrong calculation, does not depend on frequency. – nougako Dec 01 '15 at 19:58
  • Just wondering: You seem to have moved on from this question. Did my answer get you what you needed? If so, please accept it as the answer (click the check mark / tick beside it). – Peter K. Dec 02 '15 at 22:08

2 Answers2

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After thinking more about this, I am pretty sure that what you really want to do, and what your MATLAB probably in fact is doing, is study the a function like

$\displaystyle f(t)=\int^tdt^\prime (t-t^\prime)\eta(t^\prime)$

with $\eta(t)$ having the correlation property you've already mentioned. This will give an FT that is definitely frequency dependent and, I think, looking like your MATLAB results.

bob.sacamento
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Looking at your code, this line looks odd:

fw = ifft(fft(t.*ft));

Why are you immediately taking the inverse FFT? Surely you just want:

fw = abs(fft(t.*ft));

???

This is not really an answer, more a request for more information.

I try to repeat your example in R and I cannot get the FFT you get. What I get is plotted below. Can you share your matlab code? I'm wondering what it's doing.

enter image description here

R Code Below

#27432

T2 <- 1000

f <- rnorm(2*T2+1) * seq(-T,T,1)

par(mfrow=c(2,1))
plot(f, type="l", col="blue")
plot(abs(fft(f)), type="l", col="blue")
Peter K.
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  • I know my previous answer is not an answer but I want to respond to your answer by sharing my code as you requested. Can I write a code in a comment? By the way, yeah you are write. That ifft should have been ifftshift, my bad. – nougako Dec 01 '15 at 21:01
  • @nougako: I've edited your question and added it there. Have a look at my updated answer. – Peter K. Dec 01 '15 at 21:02
  • By the way, I think the ensemble average of my $f(t)$ is approximately equal to the time average. Based on this, can I say that $f(t)$ is an ergodic signal? – nougako Dec 01 '15 at 21:11
  • @nougako : I'm uncertain if $f(t)$ is ergodic, because of the time dependence on the variance. – Peter K. Dec 02 '15 at 15:52