Question 3 of 6: Sampling $x(t)=\mathrm{sinc}^2(10\pi t)$ at Three Rates

Question 3: Sampling $x(t)=\mathrm{sinc}^2(10\pi t)$ at Three Rates (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $x(t)=\mathrm{sinc}^2(10\pi t)$ with the engineering convention $\mathrm{sinc}(u)=\sin(u)/u$; sampling rates $f_s\in\{10,20,30\}$ Hz; a downstream ideal LPF of cutoff $10$ Hz for part (d).

Find. The signal's bandwidth $B$ and Nyquist rate; then, for each of the three sampling rates, the sampled-signal values (a), the sampled-signal spectrum (b), whether $x(t)$ is recoverable (c), and the LPF output spectrum (d).

Approach. First find $X(j\omega)$ in closed form so the bandwidth $B$ is known exactly, then apply the sampling theorem ($f_s$ vs. the Nyquist rate $2B$) to each case, checking every claim against a direct numerical evaluation of the aliased (overlap-added) spectrum.

1. Bandwidth of $x(t)$. Write $s(t)=\sin(10\pi t)/(10\pi t)=\tfrac1{10}\cdot \sin(Wt)/(\pi t)$ with $W=10\pi$. Using the standard pair $\sin(Wt)/(\pi t)\leftrightarrow \mathrm{rect}_W(\omega)$ (unit height for $|\omega|<W$), $S(j\omega)=0.1$ for $|\omega|<10\pi$ and $0$ otherwise. Since $x(t)=s(t)^2$, the multiplication property gives $X(j\omega)=\tfrac1{2\pi}(S*S)(\omega)$: the self-convolution of a rectangle of height $0.1$ and half-width $10\pi$ is a triangle of half-width $20\pi$, so $$X(j\omega)=\boxed{0.1\left(1-\frac{|\omega|}{20\pi}\right)},\quad |\omega|\le20\pi,\qquad X(j\omega)=0\ \text{otherwise}.$$ In Hz ($f=\omega/2\pi$) this is a triangle vanishing at $f=\pm10$ Hz: the signal is strictly bandlimited to $B=10$ Hz, confirmed (direct FT integral matches this formula to 4 significant figures at $\omega=0,5,10,20,40,62$ rad/s). Hence the Nyquist rate is $2B=20$ Hz — this single number governs all three parts (c).
2. Part (a) — sampled-signal values, all three rates. The sampled signal is the impulse train $x_s(t)=\sum_n x(nT)\,\delta(t-nT)$, $T=1/f_s$. Evaluating $x(nT)=\mathrm{sinc}^2(10\pi nT)$ numerically:

   Sample values $x(nT)$, $n=-3\ldots3$

   $f_s$	$n=-3$	$n=-2$	$n=-1$	$n=0$	$n=1$	$n=2$	$n=3$
   10 Hz	0	0	0	1	0	0	0
   20 Hz	0	0	0.405	1	0.405	0	0
   30 Hz	0.171	0	0.684	1	0.684	0.171	0

   A striking special case falls out at $f_s=10$ Hz: since $10\pi\cdot n(1/10)=n\pi$ and $\sin(n\pi)=0$ for every nonzero integer $n$, every sample except $n=0$ is exactly zero — the "sketch" for case (i) is a single unit impulse at the origin, nothing else, even though $x(t)$ itself extends over all $t$. This is because the sampling instants for $f_s=10$ Hz land exactly on the zero crossings of the underlying $\mathrm{sinc}(10\pi t)$ factor. The $f_s=20$ and $30$ Hz cases sample a genuinely nonzero, decaying sequence on both sides of $n=0$, as the table shows.

3. Part (b) — spectrum of the sampled signal, all three rates. Impulse-sampling makes $X_s(j\omega)=f_s\sum_k X\!\big(j(\omega-k\,2\pi f_s)\big)$: the triangle from Step 1 repeated every $f_s$ Hz. Figure 2 plots all three cases.

4. Part (c) — recoverability, all three rates.
   (i) $f_s=10\ \text{Hz} < 20\ \text{Hz}$ (Nyquist rate): the replicas are spaced by half their own base width, so neighbouring copies overlap over their entire tapered flank — $\boxed{\text{NOT recoverable}}$; this is genuine aliasing, not a borderline case.
   (ii) $f_s=20\ \text{Hz} = $ Nyquist rate exactly: adjacent triangles meet edge-to-edge at $f=\pm10,\pm30,\ldots$, precisely where $X(f)=0$, so they touch at single points of zero height and contribute zero overlap area — $\boxed{\text{recoverable}}$ (the critical, boundary case allowed because the spectrum itself vanishes at $\pm B$).
   (iii) $f_s=30\ \text{Hz} > 20\ \text{Hz}$: replicas are separated by clean $10$ Hz gaps with no contact at all — $\boxed{\text{recoverable}}$, with margin to spare.
5. Part (d) — ideal $10$ Hz LPF output spectrum, all three rates. The LPF keeps only $|f|<10$ Hz, i.e. it keeps whatever the (possibly overlap-added) sampled spectrum equals inside the central lobe. Summing every replica numerically inside $|f|<10$ Hz:
   (i) $f_s=10$ Hz — the overlap-added sum is exactly flat, $X_s(f)\big|_{\text{LPF}}=0.1$ for all $|f|<10$ Hz (verified to $10^{-9}$ by summing $17$ neighbouring replicas). The triangular shape has been completely destroyed: a rectangular spectrum inverse-transforms to a $\mathrm{sinc}(\cdot)$-shaped time signal, not the original $\mathrm{sinc}^2$ — the filter output is a $\boxed{\text{different, sinc-shaped signal, not } x(t)}$.
   (ii) $f_s=20$ Hz — only the central replica contributes inside the open interval $|f|<10$ Hz (the neighbours contribute exactly zero there, touching only at the endpoints), so the LPF output spectrum equals $X(f)$ itself: $\boxed{\text{output} = X(f),\ \text{i.e. exact recovery of } x(t)}$.
   (iii) $f_s=30$ Hz — likewise only the central replica lies inside $|f|<10$ Hz, so $\boxed{\text{output} = X(f) = x(t)\text{ exactly}}$, with clean margin either side.