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I am measuring the phase shift between two pseudorandom signals using correlation ifft(A.conjugate()*B) and picking out the maximum). That works very well, but takes a lot of time. I roughly know where the peaks are, and don't need the correlation for other shifts.

Is there an adaptive cross-correllation algorithm that would speed up the process if I am only interested in the positions of maxima, and already roughly know their positions? Analog to the sliding DFT?

Maybe running the cross-correlation on a decimated version of data, finding the peaks, interpolating, and then doing a manual correlation around the max?

I already know within 200 samples where the maxima are. The record length is about 200k samples.

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  • $\begingroup$ I was wondering something similar: Is it possible to manipulate the template signal such that the cross-correlation is positive for positive lags and negative for negative lags? So that it can do a "synchronous cross-correlation", only computing the value for a single lag, but that lag is constantly adjusted with a feedback loop to track the match? Are you tracking multiple maxima, though? $\endgroup$ – endolith Apr 21 '14 at 20:56
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    $\begingroup$ Interesting idea. In principle I can choose any binary sequence. I am measuring the phase between two copies of the sequence in my signal. That gives me two peaks. I need to measure that phase as precisely as possible, so I need a sharp central peak, but otherwise the sequence is more or less arbitrary. $\endgroup$ – Dan Apr 21 '14 at 21:05
  • $\begingroup$ Can the two peaks occur at any time? At the same time? $\endgroup$ – endolith Apr 29 '14 at 17:43
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If the limited range is significantly more than log(N) shorter than the total signal of length N, then ordinary cross correlation over that limited range may be faster than FFT-based fast correlation. You could just iterate away from the estimated/predicted lag until you find you have gone past a correlation peak (by a sufficient amount to account for expected noise). But if you go more than log(N)/2 points away from your initial estimate, then perhaps switch back to fast convolution cross correlation.

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  • $\begingroup$ "more than log(N) shorter" could be more clearly worded? $\endgroup$ – endolith Apr 29 '14 at 15:26

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