I have a reference signal (s1) of length = 5 and another signal (s2) of length = 25 samples (containing a shifted version of the same 5 sample signal s1).
I want to find the normalised cross correlation between the two signals to calculate the sample distance (delay / lag) between signals s1 and s2.
I pad s1 with zeros (so it is the same length as s2 as required for xcorr 'coeff' option):
s1(numel(s2)) = 0;
and then do:
[R lags]=xcorr(s2,s1,'coeff');
[vm im]=max(R); %max. correlation and index
s1_lag=lags(im);
to find the normalised cross-correlation for lags of -24 to 24 samples.
Since s2 contains a shifted version of s1 I would expect to obtain a maximum correlation value of 1 but maximum correlation is 0.4776 at a lag of 19 samples. I don't understand this?
If I let s1 = s2 and repeat xcorr (now s1 and s2 are identical) I get a maximum correlation value of 1.0 at 0 sample lag as expected.
The delay corresponds to the max peak value (but not necessarily 1 (1 will be the value of xcorr
between the same vector - auto correlation at lag 0)
and two opposite signals will have value of -1).
xcorr
provides normalization scheme. The syntax
xcorr(x,y,'coeff')
divides the output by norm(x)*norm(y)
so that, for auto-correlations, the sample at zero lag is 1.
http://matlab.izmiran.ru/help/toolbox/signal/spectra3.html
The following code computes xcorr for the two signal as OP, with delay 5:
% s1) of length = 5 and another signal (s2) of length = 25
s1=[1 2 3 4 5]
s2=[0 0 0 0 0 s1 1 1 2 3 1 5 2 3 2 4 1 ]
s1(numel(s2)) = 0;
s1
[R lags]=xcorr(s2,s1,'coeff');
[vm im]=max(R) %max. correlation and index
s1_lag=lags(im)
% review the plot
figure
plot(lags,R), hold on,plot(s1_lag,vm,'ro'),ylabel('Amplitude'),xlabel('lag[n]');
title('cross-correlation');
% result
%vm = 0.5756
%im = 31
%s1_lag = 5
The result is:
Max = 0.5756 (need not to be 1, but it is the peak value)
delay = 5 ( match the actual delay between the two signal which is 5)
Update:
For signals of unequal length and padding the shorter with zeros, normalization in xcorr
has no meaning.
You can use xcorr (s1,s2,'none')
or xcorr (s1,s2)
and xcorr
internally pads the shorter signal with zeros for equal length.
You get the the position of peak value which indicates the time offset at which the two signals are the most similar. In our example, using xcorr (s1,s2,'none'), the result is:
vm = 55.0000
s1_lag = 5
In Matlab there is a function named alignsignals
, which can be used with xcorr
as in the following code:
% method 2: align signals and xcorr for the new aligned signals .
%in that case you get max of xcor = 1, delay =0
[Xa,Ya] = alignsignals(s2,s1)
% after aligning signals, take the part of signal Xa with equal lentht of Ya
[R2 lags2]=xcorr(Xa(1:length(Ya)),Ya,'coeff');
[vm2 im2]=max(R2) %max. correlation and index
s1_lag2=lags2(im2)
figure
plot(lags2,R2), hold on,plot(s1_lag2,vm2,'ro'),ylabel('Amplitude'),xlabel('lag[n]');
title('cross-correlation2');
The following plot is the resultant xcorr :
The alligned signals
Xa = 0 0 0 0 0 1 2 3 4 5 1 1 2 3 1 5 2 3 2 4 1
Ya = 0 0 0 0 0 1 2 3 4 5