Which of these signals have same Nyquist rate as $x(t)$?

Question

The options are:

1. $x^2(t)$
2. $x(2t)$
3. Time derivative of $x(t)$
4. Convolution of $x(t)$ with itself

I guess, the Nyquist rate remains the same as long as bandwidth of $X(w)$ remains the same, as Nyquist rate = 2 * bandwidth.

The Fourier transform of the fourth option is $X^2(w)$. I think this has the same bandwidth as $X{w}$, because the zero terms outside of the bandwidth give zero squares. So this is correct

The third option: Derivative leads to multiplication by jw. So, again, jw(X(w)) is zero outside of the bandwidth of $X(w)$. So this is correct

Second option: i think it's incorrect, because scaling time domain also scales w domain, which changes badnwidth.

First option: This leads to convolution in the w domain. I think convolution does not preserve bandwidth. Because, say, for some k within the bandwidth, the term $X(k)X(w-k)$ goes outisde the bandwidth.

Is my work right?

Answer 1

Your understanding is right.

With the usual Fourier transform pair notation :

$$x(t) \longleftrightarrow X(\omega)$$

and assuming bandwidth of $x(t)$ is $W$, then the four cases will be

• 1-) $x^2(t) \longleftrightarrow X(\omega) \star X(\omega) \implies Bandwidth = 2W$
• 2-) $x(2t) \longleftrightarrow \frac{1}{2}X(\omega/2) ~~~~~ ~\implies Bandwidth = 2W$
• 3-) $ x'(t) \longleftrightarrow j\omega X(\omega) ~~~~~ ~~~~~\implies Bandwidth = W$
• 4-) $ x(t)\star x(t) \longleftrightarrow X^2(\omega) ~~~\implies Bandwidth = W$

You can see that, cases 3 and 4 has the same bandwidth as $x(t)$ and therefore their Nyquist sampling frequencies will also be the same.