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

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...