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I know STFT is generally applied to non-stationary signals but I tried to apply it to a stationary signal to get a working knowledge.

I created a stationary signal composed of three frequencies as below:

x = 3*cos(2*pi*30*t + phi1) + 2*cos(2*pi*45*t - phi2) + 1*cos(2*pi*70*t + phi3);

I then performed STFT (Short-Time Fourier transform) on this signal using hann window of length = 128.

  1. I have tried ranging the overlapping percentage from 75% to 0% (no overlap) but cannot see and difference in the spectrogram generated. Why could that be?

  2. On varying the length of the window (keeping overlapping percentage same), the bright lines in the spectrogram thickens or thins. Why could that be?

I am using MATLAB's STFT function and documentation can be found here:1

Edit:

I tried a signal with varying frequency over time. As pointed out, decreasing overlap percentage results in coarser time grid.

However, when increasing window size from 64 to 128 keeping overlap percentage same (75% for both), and 128-point FFT, in both cases the STFT is calculated for 65 frequencies (128/2 + 1). And the 64-hann window gives a better result. Does that mean that a smaller window gives better results almost everytime? Obviously, I understand that a smaller window would mean more computation cost.

Another experiment made me realize that keeping everything constant (window size, overlap), and increasing the N in N-DFT gives better results.

Lobster3221
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3 Answers3

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My answers will be intuitive, I hope. There are more rigorous mathematical arguments that can be made, but your example is not stochastic or varying in frequency, so they are not necessary here.

  1. Incomplete but intuitive answer: The spectrum estimate does not appear to change with changing overlap because the spectrum is constant. But see below for more details on block-size, overlap, and time resolution versus frequency resolution.

  2. If you feed a block of $N$ samples to an FFT (the algorithm that outputs a discrete Fourier transform (DFT)), then the output has $N$ "frequency bins". As $N$ grows, you have more frequency bins, and you have finer resolution in frequency a finer grid of frequencies. Your example has just 3 constant frequencies to estimate, so as the frequency resolution grid of frequencies becomes finer, the "weight" will be concentrated in frequency bins closer and closer to the 3 frequencies. As a result, the lines in the spectrogram will be narrower as $N$ increases. You can also change the window; the thickness of the lines might change as you change the window.

Note that as the block-size ($N$) increases, the waiting-time between feeding blocks of samples to the FFT increases. Hence, as you get greater frequency resolution a finer grid of frequencies, you have lesser time resolution a coarser grid of times. Overlapping blocks compensates for that a little bit: as the overlap increases, the waiting-time between FFTs shrinks. The cost is more computation.

I offer the diagram below to address @Lobster3321's comment on this answer. The diagram has been corrected per @OverLordGoldDragon's comment on the location of the insertion of zeros for zero-padding.

The diagram below gives a high-level view of the order of processing. If we had 0% overlap, then we would have a new column in the spectrogram after $N$ samples. With, 50% overlap, on the other hand, we have a new column of the spectrogram after just $N/2$ samples. enter image description here

Joe Mack
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  • It needs pointing that "resolution" in your answer does not refer to the Heisenberg sense; there only the window matters. – OverLordGoldDragon Aug 07 '22 at 19:36
  • @OverLordGoldDragon : I was afraid that I had used an overloaded term. I will change it to something that is less ambiguous. – Joe Mack Aug 07 '22 at 19:38
  • Crossing out what's "wrong but related" or "you might think this", instead of deleting it entirely, is an interesting and I think effective approach. – OverLordGoldDragon Aug 08 '22 at 11:58
  • @JoeMack here, N is the windows size? Or is it the N from N-point DFT? As per the results that I'm seeing, it should be N from the N-point DFT. Please see details in edit. Would be really appreciative of your help. – Lobster3221 Aug 08 '22 at 14:22
  • The window is center-padded, not right-padded. It's also confusing, if not incorrect, to call $M$ an interpolant and $N$ "original", where I presume $M$ is n_fft and $N$ is window length? which also is inconsistent with the rest of the answer. I think there's too much focus on the "windowed DFT" interpretation of STFT; STFT is complex bandpass convolutions, it utilizes no frequencies outside of DFT(x). – OverLordGoldDragon Aug 08 '22 at 22:50
  • @OverLordGoldDragon: The diagram has been corrected and separated from the rest of the answer. It still offers a high-level overview and not a deep dive. (1) The DFT of a zero-padded version of a block does consist of interpolants of the entries of the DFT of the un-padded block. (2) I look at a lot of spectrograms, so my focus is biased toward what is needed to get a good one. Spectral leakage-mitigating windows are applied by default in just about every canned spectrogram routine, and for good reason. – Joe Mack Aug 08 '22 at 23:47
  • you're of course correct, but it matters to distinguish "interpolating" in this sense vs padding x, as latter is a degradation and implies different things. I think you've made it clearer now though. Also correction on my prev comment, it's DFT(x_padded), but it's always x_padded so again n_fft won't "insert" new bins as in the usual padding. Agreed on 2).
    • – OverLordGoldDragon Aug 09 '22 at 01:04