I apologize for the obscure phrasing of the title of the post...My question is quite simple...

Let us imagine an audio signal (any song will do). We want to apply the Short Time Fourier Transform algorithm, so we break our signal into frames and view each frame through a window. For the sake of the example let us assume the window is a Hanning window.

Many algorithms have a parameter to specify the sample at which to center the Hanning curve in a given frame. Now, let us assume that the dominant signal in that frame happens to fall near one of the the tails of the Hanning window for that frame. The window will then squash the importance of the sample and boost some samples which were not that prominent in the original signal. In essense, we have lost some possibly valuable information.

Would I then be correct in assuming that not only is the number of frames of importance in FT, but also the samples at which the window functions are centered?

Thanks in advance.

  • 2
    $\begingroup$ i know MATLAB calls the window "Hanning", but there is no Dr. Hanning. there is a Julius von Hann, whom the Hann window is named after. and there is a Richard Hamming whom the Hamming window (which is very similar to the Hann window) is named after. $\endgroup$ – robert bristow-johnson Mar 1 '17 at 21:05

The STFT itself is invertible, which means by means of the STFT no information about the original signal is lost.

For common implementations of the discrete STFT, we do not calculate the FFT of each possible shift of the window (i.e. every sample), but instead we have a bigger step size (of a few samples in time domain). This sampled STFT (because we evaluate the STFT(t,f) only at t=k*T, where T is the spacing between the windows) is actually a discrete Gabor Transform (of a 1D signal, most references online refer to Gabor Transform of images). In superficial terms, the Gabor transform is invertible as long as the sampling of the STFT is not too coarse. A necessary condition for invertablity (in continuous time) is that $TF\leq 1$, where $T$ is the sampling distance in time and $F$ is the sampling distance in frequency (since in continuous-time, we have infinite frequency resolution). The invertability becomes the easier, the smaller $TF$, i.e. the finer the sampling of the STFT is done.

So, to answer your question: Despite sounding intuitive that some portion of the signal is lost, if the window is positioned at "unlucky" positions, the STFT does not lose information about the original signal, as long as it is sampled fine enough. So, the signal itself does not determine, if information is lost, but the sampling of the STFT does.


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