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I am reading Oppenheim & Schafer's (O&S) Discrete Time Signal Processing (2nd or 3rd edition, does not matter) and I find hard to understand a technicality behind the Time-Dependent Fourier Transform (TDFT from now on). Specifically, I have a question regarding the invertibility of the TDFT.

According to the textbook, the TDFT is given by $$X[n, \lambda) = \sum_{m=-\infty}^{+\infty}x[n+m]w[m]e^{-j\lambda m}$$ which can be thought as the DTFT of a windowed signal portion around sample $n$. So, by moving one sample at a time, we have "a collection" of DTFTs of windowed signal portions around sample $n$, and the TDFT is a 3D-function (consider magnitude only) with axis $n$ vs axis $\lambda$.

Then, O&S mentions that the TDFT is invertible iff $w[m]$ has at least one nonzero sample.

If we take the Fourier synthesis equation, we get: $$x[n+m]w[m]=\frac{1}{2\pi}\int_0^{2\pi} X[n,\lambda)e^{j\lambda m}d\lambda, \: \: \: -\infty < m < +\infty$$ which makes perfect sense as it reconstructs the windowed signal around sample $n$ from complex exponentials weighted by the DTFT $X[n, \lambda)$. Dividing by the window: $$x[n+m] = \frac{1}{2\pi w[m]}\int_0^{2\pi} X[n, \lambda)d\lambda$$ if $w[m] \neq 0$.

Here's my question: where did the $e^{j\lambda m}$ term go??

Isn't this equation supposed to recover the full windowed signal $x[n+m]$? The integral looks as a number to me (summing $X[n,\lambda)$ values for all $\lambda$ at index $n$ of the TDFT).

The same equation is in all O&S versions, so I guess it's not a typo and I am missing something very fundamental here :D

Thanks in advance.

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I have the 2nd edition, and there the case $m=0$ is considered, so the equation reads (Eq. $(10.21)$ on p. 716)

$$x[n] = \frac{1}{2\pi w[0]}\int_0^{2\pi} X[n, \lambda)d\lambda$$

(which of course assumes $w[0]\neq 0$). This makes sense because $e^{j\lambda m}$ equals $1$ for $m=0$.

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  • $\begingroup$ Indeed, and that is different from 3rd Ed. that I read from. So, actually $$x[n] = \frac{1}{2\pi w[0]}\int_0^{2\pi} X[n, \lambda) d\lambda$$ is the sample value $x[n]$, the signal's value at the center of the analysis window, and not the signal itself. Keeping the $m$ variable, would that: $$x[n+m] = \frac{1}{2\pi w[m]}\int_0^{2\pi} X[n, \lambda) e^{j\lambda m}d\lambda$$ be correct? $\endgroup$
    – GKH
    Dec 11, 2019 at 19:36
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    $\begingroup$ @GKH: Yes, for general $m$ that last equation is correct. $\endgroup$
    – Matt L.
    Dec 11, 2019 at 20:05
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The same equation is in all O&S versions, so I guess it's not a typo and I am missing something very fundamental here :D

It's a very persistent typo. There's no way you can drop the $e^{j\lambda m}$ term just by dividing by $\omega[m]$ (except for the trivial case where $\lambda m = 2\pi k, k \in \mathbb{I}$, such as when $m = 0$).

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