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gbh444g | 4 years ago

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.

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hereforphone|4 years ago

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?

gbh444g|4 years ago

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.