MATLAB: Doubt about xcorr function

xcorr

Hello!

I have question about the xcorr function.

I have a vector with random bits 0s and 1s.

I do the autocorrelation of this vector and the output is the following:

However, shouldn't the autocorrelation for a random vector trace be like this:

sorry for my paint skills..

What is that space below the plot??? Is there any way to take it out?

Many thanks, João

Best Answer

You may want to use the 'unbiased' option in xcorr so the normalization is on the number of points used to compute the correlation. In addition, you may want to specify a maxlag because at the edge, there simply are not enough samples to get an accurate estimate. Here is a small example:

[c,l] = xcorr(rand91,1024)>0.5,500,'unbiased')
plot(l,c)

HTH

Related Solutions

MATLAB: Xcorr

The frequency of your inputs is 1 Hz, or 2*pi radians per second

The phase lag is pi/10 radians, which corresponds to

pi/10*(1/(2*pi)) seconds. That is 5 sample points at your sampling frequency.

Look at c for c(101:110) which corresponds to lags 0 to 10. See that the maximum value occurs at lag 5, which corresonds to 0.05 seconds as expected.

MATLAB: Xcorr

C is the answer you get by dong the correlation and lags is the corresponding and what xcorr(A,B) does is really moving A around and see whether it matches up with B.

Let's use a simple example, A = [1 1] and B = [1 1]

In this case, to have non-zero correlation output, A can be moved back and force by one sample and still have some overlap with B.

So at lag = -1, you can imagine that

A: 1 1
B:   1 1

so C at this point is 1. Similarly for lag = 0, you have

A: 1 1
B: 1 1

so C at this point is 2. If you do this for every position, you get

C = [1 2 1] 
lag = [-1 0 1]

i.e., if the lag is -1 (A move to the left), the correlation is 1, if A does not move, the correlation is 2, ...

From this result, A and B matches best at where the maximum of C occurs, which is lag = 0.

HTH

Best Answer

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MATLAB: Xcorr

MATLAB: Xcorr

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