alt Hacker NewsTheoretically, it makes sense (go look at any of the diagrams of what a "zero crossing" is online, and it totally does.
The problem is that sign(x[n-1]) != sign(x[n]) describes a place where two successive samples differ in sign, but no sample is actually has a value of zero. Thus, to perform an edit there, if your goal is to avoid a click by truncating with a non-zero sample value, you need to add/assign a value of zero to a sample. This introduces distortion - you are artifically changing the shape of the waveform, which implies the introduction of all kinds of frequency artifacts.
Zero crossings are not computed by finding a minimum between two consecutive samples - that would almost never involve a sign change. And if they are computed by finding the minimum between two consecutive samples that also involves a sign change, there's a very good chance that you'll be long way from your desired cut point, even if you ignore the distortion issue.
It really was a completely misguided idea. If the situation was:
sign(x[n-2) != sign(x[n]) && x[n-1] == 0
> Thus, to perform an edit there, if your goal is to avoid a click by truncating with a non-zero sample value, you need to add/assign a value of zero to a sample.
No, you (the editor, not an algorithm) look at the waveform and see where the amplitude begins to significantly oscillate and place the edit at a reasonable point, like where the signal is near the noise floor and at a point where it crosses zero. There's no zero stuffing.
This kind of thing isn't computed, a human being is looking at the waveform and listening back to choose where to drop the edit point. You don't always get it pop-free but it's much better than an arbitrary point as the sample is rising.
I mean, you could use an algorithm for this. It would be a pair of averaging filters with like a VAD, but with lookahead, picking an arbitrary point some position before activity is detected (peak - noise_floor > threshold)) which could be where avg(x[n-N..n]) ~= noise_floor && sign(x[n]) != (sign(x[n-1]).