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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?

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  • $\begingroup$ 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. $\endgroup$ – MBaz Oct 14 at 15:59
  • $\begingroup$ @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. $\endgroup$ – Ryder Rude Oct 14 at 16:02
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    $\begingroup$ 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. $\endgroup$ – MBaz Oct 14 at 17:49
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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.

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