Multiple choice on sampling and aliasing 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 $T &lt;\frac{2\pi}{\omega_0}$.
Answer: True
Okay here's what I think: $x(t)$ should be sampled with at least:
\begin{align}
ωs &amp;&gt; 2B \\
&amp;&gt; 2(B_1-B_2)\\
&amp;&gt;2(\omega_0 - 0) \\
&amp;&gt; 2\omega_0 \\  
\frac{2\pi}{T_s} &amp;&gt; 2\omega_0 \\
 T_s&amp;&lt;\frac{\pi}{\omega_0} 
\end{align}
What do I miss?
Question 2:

Answer: $50\pi$
For $x(t)$, $\omega_{\text{max}} = 50π + 50π = 100π$ and so $ω_s &gt; 2\omega_{\text{max}} = 200π$.
So , if we actually sample with $\omega_s&lt;\omega_{\text{Nyquist}} = 200π$ , 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 $\omega_0 &lt; \frac{\omega_s}{2} = 75π$. Again what do I miss?
Question 1
Perhaps the easiest way to understand this is to draw $X(j\omega)$:

With the corresponding aliases after sampling in blue:

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:

Indeed, because $\omega_{\max} &gt; \frac{1}{2}\omega_s$ 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 $\omega_0 = 50\pi$.