in trying to understand the convolution theorem for DTFT, I'm faced with the following problem which I can't get my head around.

First, let me state the convolution theorem for the DTFT as follows:

\begin{equation} y[n] = (x*h)[n] \iff Y(\omega) = X(\omega)H(\omega), \quad n \in \mathbb{Z}, \omega \in \mathbb{R} \end{equation} where $x, h, y$ are signals in the time domain and $X,H,Y$ their spectra in the frequency domain.

According to Wikipedia, we also have the converse for the DTFT, i.e.:

\begin{equation} y[n] = h[n]\cdot x[n] \iff Y(\omega) = X(\omega) * H(\omega) \tag{1} \end{equation} where \begin{equation} X(\omega) * H(\omega) = \int_{-\pi}^{\pi} X(\xi)H(\omega - \xi)\,d\xi \end{equation}

Now assume $x[n]$ and $h[n]$ are non-zero within the range $-(N-1)/2 \leq n \leq (N-1)/2$, where $N$ is odd. Moreover, let $h[n] = 1$ within this range such that $h[n]$ can be thought of as a rectangular window.

Now, this implies \begin{equation} y[n] = h[n] \cdot x[n] = x[n] \end{equation} and by equation (1), \begin{equation} Y(\omega) = X(\omega) * H(\omega) = X(\omega) \end{equation} However, it is known that the Fourier transform for a rectangular window (i.e. $H(\omega)$) is an aliased sinc function. It's hard to imagine that the cyclic convolution of $H(\omega)$ with any function $X(\omega)$ leaves $X(\omega)$ unchanged. So, I believe there must be a fault in logic somewhere. Please help illuminate!


1 Answer 1


It may be hard to imagine but there's no fault in your logic. Look at the dual problem: assume that $x(t)$ is a lowpass signal with no frequency components above $\omega_x$. If you filter $x(t)$ with an ideal lowpass filter with cut-off frequency $\omega_c>\omega_x$, the signal $x(t)$ will remain unchanged. Mathematically this means

$$\int_{-\infty}^{\infty}x(t-\tau)\frac{\sin(\omega_c\tau)}{\pi\tau}d\tau = x(t)\tag{1}$$

The validity of Eq. $(1)$ might also be hard to imagine, but we can accept it more easily by considering that its implication in the frequency domain is trivial.

  • 1
    $\begingroup$ Thanks for this. I was thinking about the analogy between my problem and the DFT case where the length of the signal is finite, and $h[n]=1$. It appears in this case, then $H[0]=N$, and $H[k]=0, k\neq 0$. However, the fact that in the DTFT case, $H(\omega)$ is an aliased sinc function rather than an impulse highlights the fact that what I have there is not quite as general as it seems. In particular, the form of $X(\omega)$ must be restricted by the fact that $x[n]=0, |n|>N/2$. It's this restriction that makes the convolution behaves like an identity operator. $\endgroup$
    – Tim Mak
    Oct 21, 2022 at 2:43

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