Let $x(t)$ be band-limited with $B = \omega_m$. Sampling gives us $$x(nT_s) = \begin{cases} 1, & n = 0 \\ 0, & n \not = 0 \end{cases}$$ And $\omega_s = 2\omega_m = \frac{2\pi}{T_s}$. Find signal $x(t)$.

My try: The first problem is about the definition for band-limited signal. It means $X(j\omega) = 0$ for $|\omega|\gt\omega_m$ or $X(j\omega) = 0$ for $|\omega|\ge\omega_m$? The sampling theorem requires that if $X(j\omega) = 0$ for $|\omega|\gt\omega_m$ then $\omega_s \gt 2\omega_m$ to avoid aliasing. So in this case is it possible to find interpolating functions other than $$x(t) = \begin{cases} \frac{\sin(\omega_mt)}{\omega_mt}, & t \not= 0 \\ 1, & t = 0 \end{cases}$$which I've found using ideal low pass filter? I mean with the given information is $x(t)$ necessarily unique?


The condition


is called the Nyquist criterion for zero intersymbol interference (ISI). It is important for the design of transmit pulses in digital communication systems.

Condition $(1)$ can be expressed in the frequency domain as

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}X\left(\omega-\frac{2\pi k}{T}\right)=1\tag{2}$$

where $X(\omega)$ is the Fourier transform of $x(t)$. From $(2)$ we see that the shifted spectra need to add up to a constant. This is only possible if the maximum frequency of $x(t)$ satisfies $\omega_m\ge \pi/T$. For the minimum value $\omega_m=\pi/T$, the shifted spectra don't overlap and, consequently, $x(t)$ must be an ideal low pass signal with a flat spectrum. This is the only solution and it corresponds to the solution you came up with. If $\omega_m>\pi/T$, there are infinitely many solutions to $(2)$. One well-known example are the raised-cosine pulses.

  • $\begingroup$ Thank a lot for your clear explanation. What's the exact definition for $\omega_m$? $X(j\omega) = 0$ for $|\omega|\gt\omega_m$ or $X(j\omega) = 0$ for $|\omega|\ge\omega_m$? $\endgroup$ – S.H.W Jul 7 '20 at 11:39
  • $\begingroup$ @S.H.W: It's $f_s>2f_{max}$, but the difference between $>$ and $\ge$ is only relevant if there happens to be a sinusoid exactly at $f_{max}$. $\endgroup$ – Matt L. Jul 7 '20 at 13:36
  • $\begingroup$ Sorry but it seems I've misunderstood some matters. I've read sampling theorem which says that "if $x(t)$ is band-limited(i.e. $X(j\omega) = 0$ for $|\omega| \gt \omega_m$) we should use sampling rate $\omega_s \gt 2\omega_m$ to avoid aliasing." I thought it's $\gt 2\omega_m$ instead of $\ge 2\omega_m$ for not losing $\omega = \pm 2\omega_m$ and seems it's not the case. Could you explain this point, please? $\endgroup$ – S.H.W Jul 7 '20 at 18:45
  • $\begingroup$ @S.H.W: To make things easy (and correct) just use $>$, as we both said. In practice this is not relevant because you always choose the sampling rate with some margin. All practical anti-aliasing filters have a certain roll-off so you need to account for a certain transition band width. $\endgroup$ – Matt L. Jul 7 '20 at 19:04
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    $\begingroup$ @S.H.W: that's right. $\endgroup$ – Matt L. Jul 7 '20 at 21:04

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