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Comment by hereforphone | Hacker News Reader

The phase is dropped but isn't there more going on? Comparing the signal to itself at various lags? I don't see how just dropping the phase accomplishes that.

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Not really. ACF is defined as a convolution of signal X with itself: XX. But FFT turns a convolution into a dot product: FFT[XX] = FFT[X]·FFT[X], or just |FFT[X]|². But what is this really? If X is a sum of A·cos(2πwt+φ) waves, then FFT[X] is a set of A·exp(iφ) complex numbers. What does |FFT[X]|² do? It turns those complex numbers into A². Inversing this FFT gives a sum of A²·cos(2πwt) waves, so in effect ACF has dropped the phases and squared amplitudes. This is also why ACF have this bright vertical line - this is cos(x) functions piling up together.

ACF has a lag argument correct? Also isn't the bright vertical line just the result of the fact that ACF at lag 0 is 1?

It does, but ACF[X] at 0 is the sum of X[i] squares, so when sound gets louder, ACF at 0 also gets higher.

You're probably sick of this conversation by now :). But at least in radio applications I think that acf[0] is normalized so that it's 1 (typically). And again the ACF is calculated at several lag arguments and the sum is used to build the final graph / array.

But you obviously know more about this than me, I'm just putting out what I know. Your paragraph above, I actually copied so I can study it a few times. So thanks.

It sounds this is what I'm doing: taking ACF at equally spaced offsets. Not sure what the sum of ACFs would achieve, but this might turn out a good idea.

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