# Convolution theorem for inverse DTFT

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:

$$$$y[n] = (x*h)[n] \iff Y(\omega) = X(\omega)H(\omega), \quad n \in \mathbb{Z}, \omega \in \mathbb{R}$$$$ where $$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.:

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

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 $$$$y[n] = h[n] \cdot x[n] = x[n]$$$$ and by equation (1), $$$$Y(\omega) = X(\omega) * H(\omega) = X(\omega)$$$$ However, it is known that the Fourier transform for a rectangular window (i.e. $$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!

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.
• 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. Oct 21, 2022 at 2:43