I have a signal $x(t)$ for which I want to find the Nyquist frequency : $$ x(t) = \frac{\sin{\pi t/2}}{\pi t/2} \ast \sum^\infty_{n=-\infty}\delta(t-10n)$$

I am trying to solve this in the time domain like so : $$ \begin{split} x(t) &= \frac{\sin{\pi t/2}}{\pi t/2} \ast \sum^\infty_{n=-\infty}\delta(t-10n)\\ &=\frac{\sin{\pi t/2}}{\pi t/2} \ast[...\delta(t+20)+\delta(t+10)+\delta(t)+\delta(t-10)+\delta(t-20)+...]\\ &=\frac{\sin{\pi (t+20)/2}}{\pi t/2} +\frac{\sin{\pi (t+10)/2}}{\pi t/2} +\frac{\sin{\pi t/2}}{\pi t/2} +\frac{\sin{\pi (t-10)/2}}{\pi t/2} +\frac{\sin{\pi (t-20)/2}}{\pi t/2} +...\end{split}$$

Now shifting will not change the frequency so :

$$\begin{split} \omega_m &= \pi/2\\ 2\pi f_m &= \pi/2\\ f_m &= 1/4\\ F_{\text{Nyquist}} &= 2\times 1/4 = 0.5\end{split}$$

But when I solved this quesiton in the frequency domain I obtained $0.4$ which is the correct answer.

What is the mistake in this method?

  • $\begingroup$ Consider the Fourier transform of $x(t)$, knowing that the Fourier transform of $sinc(t/2)$ is bandlimited in the frequency domain. $\endgroup$ – Andy Walls Sep 23 '17 at 19:54

The primary problem in your time domain analysis is the assumption that if both $x(t)$ and $x(t-d)$ have the same Nyquist sampling rate so will $x(t)+x(t-d)$.

You can see that the summation may alter the signal in such a way that resulting signal's bandwidth may be different than the individual ones.

As an example consider the composite signal $x(t) = x_a(t) + x_b(t)$ where $x_a(t)$ is a low-pass signal with individual Nyquist sampling rate of $w_a$ and $x_b(t) = \sin(w_0 t)$, is a high frequency sine wave with a Nyquist sampling rate of $2w_0$. Therefore both $x(t)$ and $x(t-d)$ will have the same Nyquist sampling rate of $2 w_0$.

Then we have $x(t-d) = x_a(t-d) + \sin(w_0(t-d)) = x_a(t-d) + \sin(w_0 t - w_0 d)$. Therefore the sum of those two composite signals, $x(t) + x(t-d)$, will be $$x(t)+x(t-d) = x_a(t) + x_a(t-d) + \sin(w_0 t) + \sin(w_0 t - w_0 d)$$

Now if $d$ is chosen such that $w_0 d = \pi$ then $\sin(w_0 t) + \sin(w_0 t - \pi) = 0$, for all $t$, therefore $x(t) + x(t-d) = x_a(t) + x_a(t-d)$. There will be cancellation of the high frequency terms so that the Nyquist sampling rate of the sum will be $w_a$ (assuming the sum of low-pass signals will not further alter it). Hence the sum will have a different Nyquist sampling rate eventhough $x(t)$ and $x(t-d)$ individually have the same Nyquist sampling rate.


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