I am reading the documentation of scipy.signal.stft function. My question: is the 'noverlap' parameter of this function equivalent with STFT hop size?
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1You might want to check out this answer for a reasonably complete description of STFT and how FFT size, overlap %, and Hop size are all related. – robert bristow-johnson Dec 02 '21 at 05:32
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1This answer might also be useful, if you're using a rectangular window and doing fast convolution. – robert bristow-johnson Dec 02 '21 at 05:41
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The hop size $M$ is the number of samples between each successive FFT. The 'noverlap' parameter is the number of samples $n$ to overlap between segments (and will default to $N/2$ where $N$ is the FFT frame size). The relationship between hop size, frame size and overlap is:
$M$ = $N - n$
If there is no overlap then this would simply be the total number of samples in the FFT frame. In certain uses the overlap is given as an "Overlap Factor", where a factor of 2 would refer to a 50% overlap (stated here in case that was a source of the OP's confusion; 'noverlap' is clearly stated in the documentation as the number of samples and does not represent "overlap factor").
Dan Boschen
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For those ending up here, as stated in 'Notes' at the official SciPy documentation:
hop size = nperseg - noverlap
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