corrcoef & xcorr

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Salvatore Turino
Salvatore Turino 2011-11-10
Hello i've two complex functions ( size 1x1x2501) and i need to do a correlation between these (cross-correlation).
i've tried to use this command:
r=corrcoef(Hmimo_tb(1,:)',Hmimo_tb1(1,:)','coeff');
where Hmimo_tb and Hmimo_tb1 are my two signals in which the only difference is the fact that they have been measured in different positions. The difference betweeen these two signals is max equal to 1.5e-13, so they are only affected by noise.
i obtain as result:
ans =
1.0000 1.0000 + 0.0000i 1.0000 - 0.0000i 1.0000
the function that i'm going to correlate are complex but the 0.0000i leave me some doubts.... Another doubt is the fact that the the signals are not equal in fact as i've told before there is a difference of 1.5e-13 that is not reported on the secondary diagonal why?
what are the difference between corrcoef and xcorr?

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采纳的回答

Wayne King
Wayne King 2011-11-10
Salvatore, corrcoef() is not the cross correlation sequence. It does not shift one vector with respect to the other.
x = cos(pi/4*n);
y = cos(pi/4*n-(3*pi)/4);
[r,p] = corrcoef(x,y);
But
[c,lags] = xcorr(y,x,'coeff');
[maxcorr,I] = max(c);
lags(I)
You see that if you allow for shifts then y and x are perfectly correlated and that happens at lag 3, which makes perfect sense since the frequency of x and y is pi/4 radians/sample and y is shifted (3*pi)/4 radians.
Now, note for
lags(length(x))
c(length(x))
This is exactly equal to r in [r,p] = corrcoef(x,y);
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Wayne King
Wayne King 2011-11-10
It can mean a phase shift. It depends on the nature of the signals whether it is more natural to view it as a phase shift or just a delay. If the signals are sine waves, I think it is more natural to think of it as a phase shift. Have you tried to understand my examples??? I've shown you a number of example where you find the delay in by the peak in the cross correlation.
Salvatore Turino
Salvatore Turino 2011-11-10
ok so i understand that the results different from the lag-0 have a non-clear meaning or bytheway what i want to study deeper is if for example the xcorr between M1(measure 1) and M2 gives in the lags=1000 c=0.002 how much shifted is respect the c=0.003 in the lags=1000 of the xcorr between M1 and M3?

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更多回答（2 个）

Walter Roberson
Walter Roberson 2011-11-10
0.0000i implies that there is a non-zero complex component which is too small to be represented using your current display format (which is probably "format short f")
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Walter Roberson
Walter Roberson 2011-11-10
With values that small, it could indicate round-off error.
Salvatore Turino
Salvatore Turino 2011-11-10
mmm well consider that also in the positive value there is a round-off error in my opinion in surplus because the two measurements are not equal (but near to be). does exist a way to not have this approximation?

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Salvatore Turino
Salvatore Turino 2011-11-11
Wayne i've tried your code
x = cos(pi/4*n);
y = cos(pi/4*n-(3*pi)/4);
[r,p] = corrcoef(x,y);
[c,lags] = xcorr(y,x,'coeff');
[maxcorr,I] = max(c);
lags(I)
but i'm on trouble. i've set n=0:1:100 and as you say i have as result 3. you say that those functions are correlated at lags 3 but watching the "c" if lags 3 correspond to c(:,3) i have this result: -0.0139333076031825
so why do you say that they are perfectly correlated?
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Wayne King
Wayne King 2011-11-11
Salvatore, you keep making this mistake. c(3) is not at lag three. You are forgetting about the negative lags. If you enter lags(3) for the example you have above, you see that c(3) is the value of the cross correlation sequence at lag -98. c(104) is the cross correlation sequece at lag 3. That value is very close to 1.

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