xcorr and time lag

11 vues (au cours des 30 derniers jours)
N
N le 29 Août 2014
Commenté : N le 1 Sep 2014
Hi all,
I have a simple, but yet I can't seem to figure it out, question about crosscorrelation.When I find my maximum crosscorrelation, I want to find the corresponding time lag. What I have in code so far:
x = [0 0 1 5 1 -2 -3 -2 0 0];
y = [0 0 1 5 1 -2 -3 -2 0 0];
X2 = xcorr(x,y,'coeff'); (so we expect a xcorr of 1 because its the same signal)
plot(lags,C)
[val,idx] = max(abs(C))
Now it says that the maximum crosscorrelation (1) is at time lag 10. now I get that because the time series goes from -9 till +9, so step 10 is at zero-lag, but how can I get as answer the zero and not 10. So I want the real value of time lag and not the amount of steps.
Later on I would actually want to make the time lags correspond to real time as well ,so for example a lag of 1 equals to 2 seconds, which is the reason I want the amount and not the steps so I know how many seconds really passed. How can I do this also, so make the time lag correspond to real seconds?

Réponse acceptée

dpb
dpb le 29 Août 2014
Modifié(e) : dpb le 29 Août 2014
t=-length(x)+dt*idx;
where dt is the sampling time interval.
BTW, this is for full-length. If use the optional MAXLAGS argument, then will want to return and use the LAGS alternate return to get the actual lag instead of just counting from 1.
  1 commentaire
N
N le 1 Sep 2014
thank you for your answer, it works !

Connectez-vous pour commenter.

Plus de réponses (1)

Honglei Chen
Honglei Chen le 29 Août 2014
You already have a variable lags, you can simply do lags(idx). BTW I didn't see how you compute the lags, but you could get it from xcorr, e.g.,
[X2,lags] = xcorr(x,y,'coeff')
HTH

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by