# DTFT of window function applied to input signal

$$x[n] = \cos(\omega_1n) + \cos(\omega_2n)$$

$$w[n] = 1/N$$ for $$0 \leq n < N, 0$$ for everything else

Find the DTFT of $$y[n]=x[n]w[n]$$ expressed by the DTFT of $$w[n]$$, $$W(\omega)$$

I was thinking that multiplication in time domain is convolution in frequency domain. But $$X(\omega)$$ is 0 except for two frequencies so the convolution integral is 0? Which can't be correct? What to do here?

It seems you are mixing discrete and continuous notations and it's unclear which it is for time and frequency. You use the term DTFT which would indicate both are discrete but you use the frequency index $$\omega$$ which would be continuous in the frequency domain.

Your time domain equation doesn't make any sense, since your left side depends on $$n$$ which doesn't show up in the right hand side at all. The way it's written $$x[n]$$ is simply a constant. I'm not just nit-picky, the exact way you write & define this can make a big difference.

I was thinking that multiplication in time domain is convolution in frequency domain.

Correct

But X(ω) is 0 except for two frequencies

Maybe. $$X(\omega)$$ is, but $$X[k]$$ may or may not be depending on the exact relationships of $$N$$, $$\omega_1$$, $$\omega_2$$ and the sample rate.

so the convolution integral is 0?

Nope, it's not.

If the frequency domain is continuous $$X(\omega)$$ is non-zero in a special way, so that the convolution integral is perfectly well defined and non-zero. If it is discrete, the convolution is a sum, not an integral.

• to me, the question seems completely clear and unambiguous and homework. all three. Commented Dec 21, 2021 at 3:55