The Gabor transform (i.e. Short-time Fourier transform with some Gaussian window) can be defined by (see http://en.wikipedia.org/wiki/Gabor_transform) :

$G_x(t_0,\xi) = \int_{-\infty}^\infty e^{-\pi(t-t_0)^2}e^{-2i\pi \xi t}x(t)\,dt$

On this page it is mentioned that the Gabor transform is invertible, and that the original signal can be recovered by the following equation :

$ x(t_0) = \int_{-\infty}^\infty G_x(t_0,\xi) e^{2i\pi t_0 \xi}\,d\xi$

Can somebody help me for a proof of that ?

I thought we would need $G_x(u,\xi)$ for all $\xi$ and $u$ in a wide range in order to reconstruct $x(t_0)$, but here it seems that only $G_x(t_0,\xi)$ for all $\xi$ is needed. Is this true ?


Ok according to a book:

the Windowed Fourier transform is obtained by the formula:

$$Sf(u,\epsilon)=\int_\mathbb{R}f(t)g(t-u)e^{-i\epsilon t}dt$$

where $f$ and $g$ are the signal and window respectively:

and the reconstruction formula is given by:

$$f(t)=\frac{1}{2\pi}\int_\mathbb{R}\int_\mathbb{R}Sf(u,\epsilon)g(t-u)e^{i\epsilon t}d\epsilon du$$.

So I reckon you will need both $u$ and $\epsilon$, to reconstruct $f$.

  • $\begingroup$ yes! I also thought 2 integrations will be required in order to recover $f(t)$, so I don't understand how it could be possible with only $1$. Is there a mistake in Wikipedia here : en.wikipedia.org/wiki/Gabor_transform#Inverse_Gabor_transform ? $\endgroup$ – Basj Nov 15 '13 at 14:57
  • $\begingroup$ just for information, which book is it ? $\endgroup$ – Basj Nov 15 '13 at 14:59
  • $\begingroup$ @Basj: I think there is a mistake there. $\endgroup$ – freak_warrior Nov 15 '13 at 15:42
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    $\begingroup$ @Basj: A wavelet tour of signal processing $\endgroup$ – freak_warrior Nov 15 '13 at 15:42
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    $\begingroup$ Or maybe is there some magical thing that makes that only one integral is needed at the end when the window $g$ is a Gaussian window (Gabor transform) ? $\endgroup$ – Basj Nov 15 '13 at 17:27

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