This comes from Stephane Mallat's Wavelet Tour text; however, I will phrase my question independently of it. I apologize that this is sort of long-winded.

We have a function $f$ which satisfies the equality:

$$ (f * \bar{f})[n] = \delta[n] \tag{1} $$ which holds for $n \in \mathbb{Z}$. $*$ denotes convolution: $$ (f*g)(t) = \int_{\mathbb{R}}f(t-u)g(u)du $$ and also, $$ \bar{f}(t) := f(-t)^* $$ where $z^*$ denotes the complex conjugate.

We're interested in inspecting the Fourier transform of (1). According to the text, this results in:

$$ \sum_{n=-\infty}^{\infty} |\hat{f}(w+2\pi n)|^2 = 1 \tag{2} $$

where $\hat{f}(w) = \int_{\mathbb{R}} f(t) e^{-iwt} dt$ is the Fourier transform of $f$.

The left-hand side of (2) makes sense: we treat the left-hand side of (1) as continuous in $n$. Taking the Fourier transform yields $|\hat{f}(w)|^2$. However, since we are truly taking the discrete time Fourier transform of the left-hand side of (1), we actually have a periodic summation of this which in turn, yields the left-hand side of (2).

Now, when I try to pull the same trick on the right-hand side of (1), I run into problems: Clearly, $$\hat{\delta}(w)=\int \delta(t) e^{-iwt}dt = 1$$ But since we are sampling on over the integers on the right-hand side of (1), we should have a periodic summation: $$\sum_{n=-\infty}^{\infty} \hat{\delta}(w+2\pi n) = +\infty$$

I've gone astray somewhere, but I can't seem to pinpoint where. Any help?


1 Answer 1


The problem is that $\delta(t)$ and $\delta[n]$ are very different things, and you can't just 'sample' $\delta(t)$ to obtain $\delta[n]$. As you said, you need to take the DTFT of both sides of (1), and the DTFT of $\delta[n]$ is simply

$$\sum_{n=-\infty}^{\infty}\delta[n]e^{-i\omega n}=1$$


$$\delta[n]=\begin{cases}1,\quad n=0\\0,\quad\text{otherwise}\end{cases}$$

  • $\begingroup$ Terrific! Exactly what I was looking for. Your answer got me thinking: what is the "discrete delta" a sampling of then? A box function, $\chi_{[-1/2,1/2]}(t)$, works just fine. It's Fourier transform is of course $2\sin(w/2)/w$. This gives the periodic sum: $\sum \frac{\sin((w+2\pi n)/2)}{w+2\pi n}$ that converges to 1! Of course, your answer is much more direct :) $\endgroup$
    – Chester
    Feb 20, 2015 at 17:48

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