# Nyquist Rate - Computing

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

• Try drawing the magnitude spectrum of $(x \ast z)(t)$, and see what is its maximum frequency.
– MBaz
Dec 6, 2023 at 14:18

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.

• Thank you very much, it made it clearer, I did not fully grasp the idea of multiply the two functions in the frequency domain. Dec 7, 2023 at 15:02