# Invertibility of Time-Dependent Fourier Transform

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

I have the 2nd edition, and there the case $$m=0$$ is considered, so the equation reads (Eq. $$(10.21)$$ on p. 716)
$$x[n] = \frac{1}{2\pi w}\int_0^{2\pi} X[n, \lambda)d\lambda$$
(which of course assumes $$w\neq 0$$). This makes sense because $$e^{j\lambda m}$$ equals $$1$$ for $$m=0$$.
• Indeed, and that is different from 3rd Ed. that I read from. So, actually $$x[n] = \frac{1}{2\pi w}\int_0^{2\pi} X[n, \lambda) d\lambda$$ 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? – GKH Dec 11 '19 at 19:36
• @GKH: Yes, for general $m$ that last equation is correct. – Matt L. Dec 11 '19 at 20:05
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 $$m = 0$$).