0
$\begingroup$

Using the formulation of the short-time Fourier transform where the signal slides behind the fixed window. When looking at it from the perspective of the "filtering view", I don't understand why the window needs to be flipped. Here's my slides from class showing the flipped window (2nd slide):

enter image description here

$\endgroup$
  • 1
    $\begingroup$ The filter-bank point of view is that $y_f(n) = X(n,f) = x \ast w_f(n)$ where $w_f(n) = w(n) e^{2i \pi f n}$ is some band-pass filter around the frequency $f$, while the short-time Fourier transform point of view is that $Y_n(f) = X(n,f) = FT[x(.) w(n-.)]$ where $w(n-.)$ isolates 20ms of the signal around the $n$th time sample $\endgroup$ – reuns Nov 7 '16 at 3:36
  • 1
    $\begingroup$ In the first case you look at one row of the STFT, in the other case at one column $\endgroup$ – reuns Nov 7 '16 at 3:38
  • $\begingroup$ But what I'm wondering is from the filter bank perspective why does the second slide flip the window to be w[-n] rather than w[n]] is it required/more useful to do that for some reason? $\endgroup$ – Austin Nov 7 '16 at 3:56
  • 1
    $\begingroup$ Because usually the window is not symmetric and has its support on $[0,N-1]$, for example the Hanning window $\tilde{w}(n) = (1-\cos(\pi n / N)) 1_{n \in [0,N-1]}$ in your notation and $w(n) = (1-\cos(\pi n / N)) 1_{n \in [-N+1,0]}$ in mine $\endgroup$ – reuns Nov 7 '16 at 4:08
  • 1
    $\begingroup$ And note how from the filtering point of view, the filter isn't causal with such a window (the most convenient when dealing with a numerical signal already recorded) $\endgroup$ – reuns Nov 7 '16 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.