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phase of fft of random autocorrelation is deterministic...why?

Started by Jack March 3, 2006
Hello,

If I generate a random sequence x and calculate:

phase of FFT(autocorrelation of x)

then the phase is the same no matter what...

That puzzles me a little bit....Can anybody explain to
me why the phase is deterministic in this case?



Jack wrote:

> Hello, > > If I generate a random sequence x and calculate: > > phase of FFT(autocorrelation of x) > > then the phase is the same no matter what... > > That puzzles me a little bit....Can anybody explain to > me why the phase is deterministic in this case? > > >
The autocorrelation operation automatically yields a result that has even symmetry around the origin. You'd find that if you arranged the result of the autocorrelation so that it's origin was at sample 1 (time = 0 -- don't you love Matlab?) the fft would be entirely real. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/
> The autocorrelation operation automatically yields a result that has even > symmetry around the origin. You'd find that if you arranged the result of > the autocorrelation so that it's origin was at sample 1 (time = 0 -- don't > you love Matlab?) the fft would be entirely real.
Ok, the fft would be entirely real...but how does that answer my question?
Jack wrote:
>>The autocorrelation operation automatically yields a result that has even >>symmetry around the origin. You'd find that if you arranged the result of >>the autocorrelation so that it's origin was at sample 1 (time = 0 -- don't >>you love Matlab?) the fft would be entirely real. > > > Ok, the fft would be entirely real...but how does that answer my question? > > >
Well, the phase of a real number is either 0 or 180 degrees, no other choices. That's a lot more deterministic than "anything", which is what you'd get from an arbitrary complex number. If you happen to be taking the autocorellation of a positive-definite process then the correctly-centered fft phase will always be 0, forget the 180 degrees. Taking the 'usual' fft with uncentered data will give a phase that is always the same (spinning about at a rate proportional to the offset), which would end up looking pretty darn deterministic. This answer seems entirely adequate to me. Perhaps you should try to restate your question so I know what parts I've left out? -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/
>> > Well, the phase of a real number is either 0 or 180 degrees, no other > choices. That's a lot more deterministic than "anything", which is what > you'd get from an arbitrary complex number. > > If you happen to be taking the autocorellation of a positive-definite > process then the correctly-centered fft phase will always be 0, forget the > 180 degrees. Taking the 'usual' fft with uncentered data will give a > phase that is always the same (spinning about at a rate proportional to > the offset), which would end up looking pretty darn deterministic. >
excellent answer...now I understand...thank you very much...
hmmm...i guess i hit the send button too fast...

you say that the phase of a real number is always either 0 or 180...

but in this case I am calculating the phase of a complex number which
you can see on subplot (1,3,3) in the matlab code below.

try running this matlab code and you will see what I am talking about..

i appreciate the help :-)

clc
close all
clear

N=40;
for k=1:1000
x=randn(1,N);
rxx=xcorr(x);
phase_x=angle(fft(rxx));
subplot(1,3,1)
plot(rxx)
ylim([-20 100]);
subplot(1,3,2)
plot(phase_x)
subplot(1,3,3)
plot(fft(rxx))
drawnow;
t=['Press any key ' num2str(k)];
title(t)
pause()
end



ok...i get it now...the phase has nothing to do with the input but only with
the number of points....stupid me...