# Find continuous signal given a condition on its samples

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

$$x(nT)=\delta[n]\tag{1}$$

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.

• 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 $|\omega|\ge\omega_m$? – S.H.W Jul 7 '20 at 11:39
• @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 $f_{max}$. – Matt L. Jul 7 '20 at 13:36
• 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? – S.H.W Jul 7 '20 at 18:45
• @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. – Matt L. Jul 7 '20 at 19:04
• @S.H.W: that's right. – Matt L. Jul 7 '20 at 21:04