MATLAB Cross Correlation Own code

Question

I am trying to understand code written by my predecessor. Rather than using a xcorr in MATLAB, she did the following. Apparently this seems to working. I would really appreciate if someone could explain, what is happening here. She is saying the pattern is symmetric by calculating the variable sym below, in the code below.

close all hidden

t = 0:0.01:2*pi;
x = sin(t)
plot(x,'k')
mu = mean(x)
sigma = std(x)

y = (x-mu)/(sigma);
hold on
plot(y,'r')

yrev = y(end:-1:1);
hold on 
plot(yrev)
hold on
sym = sum(y.*yrev/length(y))
plot(y.*yrev/length(y),'r*')

You mean even she cannot explain what she has done!? This is the way of calculating covariance when having a single vector as input (hence variance). Check the formulae. — NKN
– NKN, Commented Mar 8, 2017 at 19:29

m7913d · Accepted Answer · 2017-03-08 20:35:05Z

2

sym is the normalised cross-correlation between y and the reverse of y.

If sym is close to one, y is a symmetric function.
If sym is close to zero, y is an asymmetric function
If sym is close to minus one, y is a anti-symmetric function

EDIT: relation with xcorr

You would obtain the same result if you calculate sym as follows:

sym = xcorr(y, yrev, 0, 'coeff')

edited Mar 8, 2017 at 20:35

answered Mar 8, 2017 at 19:50

m7913d

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6 Comments

NKN Over a year ago

your answer does not explain why this outcome is not consistent with the result of xcorr?

m7913d Over a year ago

@NKN I edited my answer to incorporate your comment.

NKN Over a year ago

cool(+1), the lag is 0 and coeff Normalizes the sequence so that the autocorrelations at zero lag equal 1.

m7913d Over a year ago

sym will equal 1 if and only if y is an exact symmetric function

Jesh Kundem Over a year ago

@m7913d would appreciate more clarification. y is the area under curve. Dot product with its reverse, is giving me convolution not correlation. And finally why one should divide by length

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