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?

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

Started by ●March 3, 2006

Reply by ●March 3, 20062006-03-03

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/

Reply by ●March 3, 20062006-03-03

> 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?

Reply by ●March 3, 20062006-03-03

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/

Reply by ●March 3, 20062006-03-03

>> > 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...

Reply by ●March 3, 20062006-03-03

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

Reply by ●March 3, 20062006-03-03