# how to perform frequency convolution?

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:

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

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

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

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?

• i guess i'm wrong, you start convolution at $u=-1$ and end at $u=1$
– pico
May 2, 2020 at 21:34
• 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 $|\Omega|<1/2$? May 2, 2020 at 21:35
• 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.
– pico
May 2, 2020 at 21:38
• 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. May 2, 2020 at 21:40