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

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?

• Sorry, I messed up my comment.... I was referring to the first option: it's easy to see that squaring expands the bandwdith by looking at a single tone in the time domain. I'll delete my previous comment. – MBaz Oct 14 at 15:59
• @Mbaz I've written about the first option too. Squaring leads to convolution in the w domain, which leads to expansion of bandwidth because the terms in the convolution summation have shifted arguments. So I think the first option is incorrect. – Ryder Rude Oct 14 at 16:02
• It is indeed incorrect. My point is that it's not necessary to go to the frequency domain and convolve to see it; $\sin^2(2\pi f_0 t) = 0.5 +0.5\sin(2\pi 2f_0 t)$ is enough. – MBaz Oct 14 at 17:49

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