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Good morning, i had a question regarding the Nyquist rate : Let's say i have a signal $x(t)$ with Nyquist Rate $w_0$. I need to find the Nyquist rate of the following two signals : $$(x * z)(t) \text{ where } z(t) = \sin(\omega_0 t/3)$$ $$(x * z)(t) \text{ where } z(t) = \cos(\omega_0 t)$$

If I understand, it means $x(t)$ is band limited at $\omega_0/2$. I also know that convolution in the time domain is multiplication in the frequency domain. But I am unsure how to conclude on those questions. I hope you can help ! Thanks a lot

Jdip
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Ravinala
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  • Try drawing the magnitude spectrum of $(x \ast z)(t)$, and see what is its maximum frequency. – MBaz Dec 06 '23 at 14:18

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Convolution in the time domain is multiplication in the frequency domain.

$x(t)$ is band limited at $\omega_0/2$, which means that the Fourier Transform of $x(t)$ only has value from $-\omega_0/2$ to $\omega_0/2$ and zero value otherwise.

$z(t)=\sin(\omega_0 t/3)$ only has value at frequency $-\omega_0/3$ and $\omega_0/3$. So after $z(t)$ convolves $x(t)$, in the frequency domain you will get values only at $-\omega_0/3$ and $\omega_0/3$. So the Nyquist rate equals $2\omega_0/3$.

Similarly, for the second case, $z(t)=\cos(\omega_0t)$, which only has value at frequency $-\omega_0$ and $\omega_0$. Interestingly, you will find you got nothing in the frequency domain of $(x*z)(t)$, which means the Nyquist rate equals $0$ because the Nyquist rate = 2 * highest frequency component and the highest frequency component of the output is $0$.

If you draw a graph of the magnitude frequency response, you will understand what happens clearly.

I hope this will help.

Jdip
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Junchi Wang
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  • Thank you very much, it made it clearer, I did not fully grasp the idea of multiply the two functions in the frequency domain. – Ravinala Dec 07 '23 at 15:02