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Interpreting xcorr results compared to corrcoef

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I am having trouble understanding why I am getting different outputs when using coeff and a normalized xcorr, specifically at 0 lags.
Shouldn't 0 lags produce the same r value when using an xcorr compared to coeff since the time series is matched up in both cases?
Note - I am using "coeff" because the two sets of data are equal in length, but the ampltitudes are much different.
Below illustrates the example:
I have two sets of data (A and B) that are 1000 datapoints each.
I use corrcoef(A,B) and then I use xcorr(A,B,500,'coeff").
corrcoef outputs a single value that is rather low such as r = 0.15.
xcorr outputs r values at lags -500 to 500, and they are all higher than r = 0.15.
However my question is, shouldn't r at 0 lag be = 0.15 since the data is already lined up in time?

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Accepted Answer

David Goodmanson
David Goodmanson on 17 Aug 2019
Hi Jonathan,
For the input column vectors, corrcoeff subtracts the mean off of each one and normalizes each to be a unit vector. xcorr with the 'coeff' option normalizes, but doesn't subtract off the mean.
a = (1:1000)';
b = mod((a.^(4/3)),1000);
corrcoef([a b])
ans =
1.0000 0.1745
0.1745 1.0000
xcorr(a,b,0,'coeff') % zero lag only
ans =
0.7865
aa = a-mean(a);
bb = b-mean(b);
xcorr(aa,bb,0,'coeff')
ans =
0.1745 % agrees

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Shuya Zhong
Shuya Zhong on 7 May 2020
Super helpful and solved my doubts! Thanks!
David Goodmanson
David Goodmanson on 7 May 2020
Hello SZ, thanks for the comment. it's good to hear that a blast from the past still might be useful.

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