# how to compute the signal passing from the low pass filter?

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?

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?

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 ?

• 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. – Rima Feb 6 at 13:34
• That's Correct. – Hilmar Feb 6 at 17:32

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 $$h[n]$$.