Possible to vectorize xcorr?

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Sebastian Holmqvist
Sebastian Holmqvist el 21 de Mzo. de 2013
Greetings.
Currently my workstation can't keep up when I attempt to calculate the cross correlation for very long signals.
This is a MWE of the code:
x = rand(1,200e3);
for i=500:-1:1
y = i.*x;
C(i,:) = xcorr(x, y, 500);
end
The 500 calls to xcorr seems to kill it. Any way to improve this by vectorizing or otherwise optimize it? Currently it takes >60 sec which isn't practical.
I'm only interested in a small number of lags (around ±500). So I attempted to do a classic implementation of the cross correlation instead of the Fourier implementation in xcorr. My end result wasn't an improvement, but maybe you guys/gals can do one better?
Thanks for any ideas!
  1 comentario
Daniel Shub
Daniel Shub el 3 de Abr. de 2013
I may be missing something, but x and y are scaled versions of each other. You just need to compute the autocorrelation of x once and then set the normalisation factor in the loop.

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Respuestas (1)

Sebastian Holmqvist
Sebastian Holmqvist el 3 de Abr. de 2013
After some time, I realized that x is the same throughout all calculations, which means that I can pre-calculate both it's FFT and it's conjugate before the loop.
This is what I ended up with:
x = rand(1,200e3);
corr_len = 2^nextpow2(2*length(x)-1);
x_fft = fft(x, corr_len);
x_conj = conj(x_fft);
for i=500:-1:1
y = i.*x;
corr = ifft(x_conj.*y);
C(i,:) = [corr(end-500/2:end) corr(1:500/2)];
end

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