$x(t)$ can be exactly reconstructed from its samples at $\omega_s = 10 \textrm{ rad/sec}$. My conclusion is that the maximum frequency component in $x(t)$ is $5\textrm{ rad/sec}$. But I'm being told that in fact $x(t)$ has maximum frequency component $7 \textrm{ rad/sec}$. How can this be possible?
$\begingroup$
$\endgroup$
-
$\begingroup$ is x(t) affected by noise? or could it be a modulated signal with the higher component corresponding to the carrier frequency? $\endgroup$ – Behind The Sciences Apr 26 '16 at 6:03
$\begingroup$
$\endgroup$
$x(t)$ must be a band pass signal. Under certain conditions on the sampling frequency and its relation to the lower and upper band edges of the signal, $x(t)$ can be sampled at a frequency that is lower than twice its maximum frequency, without introducing aliasing. Have a look at this article and at this answer to a related question.
For the numbers given in your question, if the lower band edge $\omega_l$ of $x(t)$ satisfies $5<\omega_l<7$ then sampling with $\omega_s=10$ will allow reconstruction of the signal without aliasing.
