EE-406: Problem Set 3. Q3

hi, for this problem I don't understand how we get to the given solution. I understand for the first two signals but not the other ones. Could you please elaborate on how we define the cut off frequency please? Wishing you a good weekend.

Problem description Solution

Re: Problem Set 3. Q3

by Tim Tuuva - Saturday, 15 November 2025, 12:41

Hello,

The following figure shows the spectrum of

$|X(\omega)|$ and

$|H(\omega)|$ .

Let us observe for example

$|H(\omega)|$ , you see that it is zero for frequencies below

$-\frac{\pi}{5}$ or above

$\frac{\pi}{5}$ .

This is called an ideal low pass filter.

Its cut-off frequency is

$\omega_c$ =

$\frac{\pi}{5}$ here. Everything is set to zero above this cut-off frequency.

Note if you filter a signal x(t) with h(t). It ends up multiplying in the Fourier domain, hence we get the final answer.

Let me know if it is still unclear