Question 1 of 10: Sampling and Reconstruction of a Squared-Sinc Signal

Question 1: Sampling and Reconstruction of a Squared-Sinc Signal (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. Input $x(t)=\operatorname{sinc}^2(\pi t)$; sampling interval $T_s = 1/4$ s, so the sampling rate is $f_s = 1/T_s = 4$ Hz. The reconstruction filter is an ideal brick-wall lowpass of gain $A$ and cut-off $f_c$.

Find. The baseband spectrum $X(f)$ and its bandwidth, the Nyquist rate, the sampled spectrum $X_s(f)$, and the pair $(A, f_c)$ that makes the recovered output exactly equal the original signal.

[Figure not reproduced: Figure 1 (redrawn): the sampler multiplies x(t) by the impulse train s(t), and the reconstruction filter H(f) recovers y(t). See the official exam paper.]

Approach. Read $X(f)$ off the supplied transform table, use its bandwidth to get the Nyquist rate, apply the multiplication-by-impulse-train result to get the periodic replication that is $X_s(f)$, and then choose the filter so that it passes exactly one replica with unit net gain.

1. Identify the transform pair for the squared sinc. The table lists $$B\operatorname{sinc}^2(\pi B t)\;\longleftrightarrow\;\Delta\!\left(\frac{f}{2B}\right).$$ Comparing with $x(t)=\operatorname{sinc}^2(\pi t)$ gives $B = 1$, and since $B=1$ the leading coefficient is already correct, so $$\boxed{X(f)=\Delta\!\left(\frac{f}{2}\right)}$$ a triangular pulse of unit height at $f=0$ falling linearly to zero at $f=\pm 1$ Hz.
2. Check the height against the area of x(t). The transform evaluated at the origin must equal the total area of the time signal, $X(0)=\int_{-\infty}^{\infty}\operatorname{sinc}^2(\pi t)\,dt = 1$, which agrees with the unit peak just found. This is the quickest sanity check available on any sketch of a transform.

The triangle vanishes for $|f| \gt 1$ Hz, so $x(t)$ is strictly bandlimited with one-sided bandwidth $B = 1$ Hz. Part (b) follows immediately.

3. Nyquist rate from the bandwidth. Sampling theory requires a rate of at least twice the highest frequency present: $$f_{\text{Nyq}} = 2B = 2\times 1\ \text{Hz} = \boxed{2\ \text{Hz}}$$ The chosen rate $f_s = 4$ Hz is twice this, so the signal is comfortably oversampled and no aliasing can occur.
4. Spectrum of the sampled waveform. Multiplication by an impulse train is convolution by an impulse train in frequency; from the table $s(t)\leftrightarrow f_s\sum_n \delta(f-nf_s)$ with $f_s = 1/T_s$, so $$X_s(f) = X(f) * f_s\sum_{n=-\infty}^{\infty}\delta(f-nf_s) = f_s\sum_{n=-\infty}^{\infty}X(f-nf_s).$$ Substituting $f_s = 4$ Hz, $$\boxed{X_s(f)=4\sum_{n=-\infty}^{\infty}\Delta\!\left(\frac{f-4n}{2}\right)}$$
5. Locate the replicas in the requested window. The replica centres are the multiples of $f_s$, so within $|f| \lt 10$ Hz they sit at $f = 0, \pm 4$ and $\pm 8$ Hz. Each is a triangle of peak $f_s X(0) = 4$ and half-width 1 Hz, so adjacent replicas are separated by a 2 Hz guard band and never overlap.

Because the replicas are disjoint, perfect recovery is possible: the filter must keep the $n=0$ triangle intact, kill the $n=\pm1$ triangles entirely, and undo the factor $f_s$ that the sampling operation introduced.

6. Set the gain from the amplitude condition. Inside the passband $Y(f) = A\,X_s(f) = A f_s X(f)$. Demanding $y(t)=x(t)$, i.e. $Y(f)=X(f)$, requires $A f_s = 1$, hence $$A=\frac{1}{f_s}=\frac{1}{4}=\boxed{0.25}$$ Equivalently $A = T_s$: the gain simply restores the energy lost to the sampling duty cycle.
7. Set the cut-off from the two exclusion conditions. The filter must pass the whole baseband triangle, which demands $f_c \ge B = 1$ Hz, and must reject the nearest edge of the first replica, which begins at $f_s - B = 4 - 1 = 3$ Hz, demanding $f_c \le f_s - B$. Combining, $$\boxed{1\ \text{Hz}\ \le\ f_c\ \le\ 3\ \text{Hz}}$$ Any cut-off in this 2 Hz window works; $f_c = f_s/2 = 2$ Hz is the natural centred choice and gives the largest tolerance to filter imperfection on both sides.