I found some multiple choices in a well known book . The problem is that I don't get the answers in some and I want to do so.

Question 1:

The signal x(t) with Fourier transform $X(j\omega) = u(\omega)- u(\omega -\omega_0 )$ can undergo impulse-train sampling without aliasing, provided that the sampling period $T <\frac{2\pi}{\omega_0}$.

Answer: True

Okay here's what I think: $x(t)$ should be sampled with at least:

\begin{align} ωs &> 2B \\ &> 2(B_1-B_2)\\ &>2(\omega_0 - 0) \\ &> 2\omega_0 \\ \frac{2\pi}{T_s} &> 2\omega_0 \\ T_s&<\frac{\pi}{\omega_0} \end{align}

What do I miss?

Question 2: enter image description here

Answer: $50\pi$

For $x(t)$, $\omega_{\text{max}} = 50π + 50π = 100π$ and so $ω_s > 2\omega_{\text{max}} = 200π$. So , if we actually sample with $\omega_s<\omega_{\text{Nyquist}} = 200π$ , then there is no chance to avoid aliasing for all $X(j\omega)$ spectrum but we can gain a proper copy of it as long as $\omega_0 < \frac{\omega_s}{2} = 75π$. Again what do I miss?

  • 2
    $\begingroup$ Have you considered accepting any answers to your previous questions? $\endgroup$
    – Matt L.
    Dec 19, 2020 at 12:45

1 Answer 1


Question 1

Perhaps the easiest way to understand this is to draw $X(j\omega)$: Plot of X(j\omega)

With the corresponding aliases after sampling in blue: Plot of X(j\omega) with aliases \frac{\omega_s}{2\pi} \sum_{k=-\infty}^{\infty} X(j(\omega-k\omega_s))

So long as the blue rectangle do no overlap you would not have aliasing. At this point it should be more clear that you need $$ \omega_s \gt \omega_0 \\ T_s \lt \frac{2\pi}{\omega_0} \\ $$

Question 2

Again, plotting $G(j\omega)$ (red), together with $75X(j\omega)$ and an alias image after sampling by $\omega_s$ (dashed blue) can help visualize the situation: enter image description here

Indeed, because $\omega_{\max} > \frac{1}{2}\omega_s$ you have some aliasing. However the question is not whether you can avoid aliasing, but rather to find the value $\omega_0$ below which the aliasing does not affect the spectrum. In the above graph you may notice that the red and dashed blue curves overlap up to $50\pi$. So $\omega_0 = 50\pi$.


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