much has been said about convolution of continuous time signals. But what about frequency convolution?

For instance, consider the function:

$$f(t) = \text{sinc}(t)=\frac{\sin x}{x}$$

I take the Fourier of f(t):

$$\mathscr{F}\Big\{f(t)\Big\} = \mathscr{F} \Big\{ \text{sinc}(t) \Big\}$$

$$F(\Omega) = \text{rect}(\Omega)$$

where: $$\text{rect}(\Omega) = \begin{cases}1 & |\Omega|< 1/2 \\ 0 &|\Omega| >1/2\end{cases}$$

Then perform convolution of $F(\Omega)$ with itself in the frequency domain:

$$G(\Omega) = F(\Omega) * F(\Omega)$$

$$G(\Omega) = \text{rect}(\Omega) * \text{rect}(\Omega)$$

$$G(\Omega) = \int \limits_{-\infty}^{\infty} \text{rect}(u) ~\text{rect}(u-\Omega)~ du$$

answer according to wolfram alpha

its a triangle between -1 and 1.

I would have thought that graphically you perform this convolution by sliding $\text{rect}(u-\Omega)$ across $\text{rect}(u)$ starting at $u=-1/2$ and ending at $u = 1/2 + 1 = 3/2$. since the width of $\text{rect}(u-\Omega)$ is 1 in length.

but this contradicts the $\Omega$ range given by wolfram alpha of -1 to 1.

what gives?

  • $\begingroup$ i guess i'm wrong, you start convolution at $u=-1$ and end at $u=1$ $\endgroup$
    – pico
    May 2, 2020 at 21:34
  • $\begingroup$ I am confused about your use of $t = 1/2$ since you are in the frequency domain. If you meant $\Omega = 1/2$, wouldn't it be $\Omega = -1/2$ since $|\Omega|<1/2$? $\endgroup$ May 2, 2020 at 21:35
  • $\begingroup$ I think, my mistake was thinking the right edge of $\text{rect}(u-\Omega)$ was u, its actually u+1/2. the center of $\text{rect}(u-\Omega)$ is u. $\endgroup$
    – pico
    May 2, 2020 at 21:38
  • 3
    $\begingroup$ So whether it is frequency domain or time domain it really doesn't matter what the independent variable is (time, frequency, x, u etc). Convolution is convolution. $\endgroup$ May 2, 2020 at 21:40


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