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I am currently trying to solve this question.

Let $x[n]=\cos(\frac{\pi}{2}n)$ and $h[n]=\frac{1}{5}\text{sinc}(\frac{n}{5})$. Compute the convolution $y[n]=x[n]∗h[n],$ and write the value of $y[5].$

Hint - Given for the question is to compute the convolution in the frequency domain first.

when I calculated the Fourier transform of $y[n]$, I got

$$Y(e^{jw}) = X(e^{jw}) \times H(e^{jw})$$

$$X(e^{jw}) = (1/2) * [\delta(\omega-\pi/2)+\delta(\omega+\pi/2)]$$

$$H(e^{jw}) = \begin{cases} \text{1,} &\quad \text{if } |\omega| <= \pi/5\\ \text{0,} &\quad \text{otherwise} \\ \end{cases}$$

But what I don't understand is, since the range of $X$ is beyond the range of $H$, how do we multiply the two FT's? I must be misunderstanding the concept somewhere. Could anyone help explain to me how do we find the output?

enter image description here

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Low pass filters get the name because they pass signals at low frequencies and attenuate signals at higher frequencies. The definition of low and high depend on the the cut-off frequency of the filter. In your case, the cut-off frequency is $\frac{\pi}{5}$. Your filter is the ideal low pass filter where signals above the cut-off frequency are completely zeroed out. The input signal is a perfect cosine signal so all of its energy is concentrated at one specific frequency. It happens to be at a frequency above the cut-off frequency, so what should happen?

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Good job so far, you are almost done

since the range of X is beyond the range of H,

X is NOT beyond the range of H. Both functions have the same range and are fully defined from $[-\pi,+\pi]$. You just need to find the value of $H(\omega)$ at $\omega = \pi/2$. Just put $\omega = \pi/2$ into your definition of $H$ and see what number you get.

HINT #2: Your input is a high frequency sine wave and your filter is a low pass filter: what output would you expect? What's the purpose of a low pass filter ?

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  • $\begingroup$ when we put $\omega$ = $\pi$/2 in H, it would give 0, since the value 1 is only between |$\pi/5$|? I assume there should be no output because Y = 0. That means the high frequency is completely blocked by the filter. $\endgroup$ – Rima Feb 6 at 13:34
  • $\begingroup$ That's Correct. $\endgroup$ – Hilmar Feb 6 at 17:32
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Discrete-Time Fourier Transforms of discrete time sequences are defined for $\omega \in (-\infty, \infty)$ and they are $2\pi$ periodic. So, both $X(e^{j\omega})$ and $H(e^{j\omega})$ are defined at all $\omega$'s. For you particular case, $X(e^{j\omega})$ is $0$ except at $\omega = 2n\pi \pm \frac{\pi}{2}$, and $H(e^{j\omega})$ is $0$ except inside interval $\omega \in [-\frac{\pi}{5}, \frac{\pi}{5}] + 2n\pi$.

Hence, $X(e^{j\omega})$ and $H(e^{j\omega})$ never overlap and their product is always $0$. This is mathematical interpretation. Signal Processing interpretation wise, you can say that input sequence $x[n]$ has been filtered out by the low pass filter $h[n]$.

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