Why is the number of sample frequencies in `scipy.signal.stft()` tied to the hop size?

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This question relates to SciPy's Short-time Fourier Transform function for signal processing.

For some reason I don't understand, the size of the output 'array of sample frequencies' is exactly equal to the hop size. From the documentation:

nperseg : int, optional

    Length of each segment. Defaults to 256.

noverlap : int, optional

    Number of points to overlap between segments. If None, noverlap = nperseg // 2. Defaults to None. When specified, the COLA constraint must be met (see Notes below).


f : ndarray

    Array of sample frequencies.

hop size H = nperseg - noverlap

I'm new to signal processing and Fourier transforms, but as far as I understand a STFT is just chopping an audio file into segments ('time frames') on which you perform a Fourier transform. So if I want to do a STFT on 100 time frames, I'd expect the output to be a matrix of size 100 x F, where F is an array of measured frequencies ('measured' probably isn't the right word here but you know what I mean).

This is kinda what SciPy's implementation does, but the size of f here is what bothers me. It's supposed to be an array describing the different frequencies, like [0Hz 500Hz 1000Hz], and it does, but for some reasons its size exactly the same as the hop size. If the hop size is 700, the number of measured frequencies is 700.

The hop size is the number of samples (i.e. time) between each time frame, and is correctly calculated as H = nperseg - noverlap, but what does this have to do with the frequency array?

Edit: Related to this question

scipy fft
1 Answers

An FFT is an square matrix transform from one orthogonal basis to another of the same dimension. This is because N is the exact number of orthogonal (e.g. that don't interfere with one another) complex sinusoids that fit in a time domain vector of length N.

A longer time vector can contain more frequency information (e.g. it's hard to tell 2 frequencies apart using just 3 sample points, but much easier with 3000 samples, etc.)

You can zero-pad your short time vector of length N to use a longer FFT, but that is identical to interpolating a nice curve between N frequency points, which makes all the FFT results interdependent.

For many purposes (visualization, etc.) an STFT is overlapped, where the adjacent segments share some overlapped data instead of just being end-to-end. This gives better time locality (e.g. the segments can be spaced closer but still be long enough so that each one can provide the frequency resolution required).

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