3 ms·
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.
by gbh444g 5y 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.
- hereforphone 5y agoYou'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.
- gbh444g 5y agoIt 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.