Question 1 of 10: Sampling and Reconstruction of a sinc² Signal

Question 1: Sampling and Reconstruction of a sinc² 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. $x(t)=\text{sinc}^2(\pi t)$ multiplies a unit impulse train $s(t)$ of period $T_s=1/4$ s ($f_s=1/T_s=4$ Hz); the recovery filter $H(f)$ is an ideal lowpass filter of gain $A$ and cutoff $f_c$.

Find. (a) $X(f)$; (b) the Nyquist frequency of $x(t)$; (c) $X_s(f)$ for $|f|<10$ Hz; (d) $A$ and the range of $f_c$ for which $y(t)=x(t)$.

Approach. Match $x(t)$ against the Fourier-transform pair table entry $B\,\text{sinc}^2(\pi Bt)\leftrightarrow\Delta(f/2B)$, then apply the impulse-train sampling identity $X_s(f)=f_s\sum_n X(f-nf_s)$ and check whether the replicated triangles overlap, touch, or leave a guard band.

1. Part (a) — spectrum X(f). Matching $x(t)=\text{sinc}^2(\pi t)$ to the table pair $B\,\text{sinc}^2(\pi Bt)\leftrightarrow\Delta(f/2B)$ with $B=1$ gives $$X(f)=\boxed{\Delta(f/2)}$$ Read the width from the paper's own definition of the triangular pulse: $\Delta(u)$ has peak 1 at $u=0$ and falls to zero at $u=\pm\tfrac12$, so $\Delta(f/2)$ reaches zero where $f/2=\pm\tfrac12$, i.e. at $$f=\pm1\ \text{Hz}.$$ $X(f)$ is therefore a triangular pulse of peak height 1 at $f=0$ falling linearly to zero at $f=\pm1$ Hz, and $x(t)$ is exactly bandlimited to $\boxed{W=1\ \text{Hz}}$ (equivalently: the table entry $B\,\text{sinc}^2(\pi Bt)\leftrightarrow\Delta(f/2B)$ has its null at $|f|=B$, and $B=1$ here). A direct numerical Fourier transform of $\text{sinc}^2(\pi t)$ confirms $X(0.5)=0.500$, $X(0.9)=0.100$ and $X(f)=0$ for $|f|\ge1$ Hz.

2. Part (b) — Nyquist frequency. With bandwidth $W=1$ Hz, the Nyquist frequency (the minimum sampling rate that avoids aliasing, $2W$) is $$f_{Nyq}=2W=\boxed{2\ \text{Hz}}.$$
3. Part (c) — X_s(f). The sampling frequency is $f_s=1/T_s=\boxed{4\ \text{Hz}}$, which is twice the Nyquist rate found in (b) ($f_s=2f_{Nyq}=4W$) — the signal is comfortably oversampled. Impulse-train sampling gives $$X_s(f)=f_s\sum_{n=-\infty}^{\infty}X(f-nf_s)=4\sum_{n=-\infty}^{\infty}\Delta\!\left(\frac{f-4n}{2}\right).$$ Each replica is a triangle of height $f_s=4$ and half-width $W=1$ Hz centred on a multiple of $f_s=4$ Hz, so the replicas are well separated: the baseband copy occupies $|f|\le1$ Hz, its nearest neighbours occupy $3\le|f|\le5$ Hz, and the intervals $1<|f|<3$ Hz are empty guard bands 2 Hz wide. Direct numerical summation of the shifted triangles confirms $X_s(f)=0$ throughout $1<|f|<3$ Hz — no overlap, hence no aliasing, and with frequency room to spare. For $|f|<10$ Hz this gives replicas of height $f_s=4$ centred at $f=-8,-4,0,4,8$ Hz.

4. Part (d) — filter gain and cutoff range. For $y(t)=x(t)$ we need $H(f)X_s(f)=X(f)$ over the baseband. Inside the baseband ($|f|\le1$ Hz) $X_s(f)=f_sX(f)$, so an ideal LPF of gain $A$ passing only the baseband must satisfy $A\,f_s=1$, i.e. $$A=\frac{1}{f_s}=\boxed{\frac14}.$$ The cutoff must be high enough to pass the whole baseband triangle and low enough to exclude the nearest image. The baseband copy ends at $W=1$ Hz and the first image begins at $f_s-W=3$ Hz, so any cutoff placed inside that guard band works: $$W\le f_c\le f_s-W\quad\Longrightarrow\quad\boxed{1\ \text{Hz}\le f_c\le3\ \text{Hz}.}$$ Any $f_c<1$ Hz clips part of the baseband triangle's own tail, and any $f_c>3$ Hz admits part of the image centred at $f=4$ Hz. Because $f_s=4W$ here rather than the bare minimum $2W$, this is a genuine 2 Hz-wide interval of admissible cutoffs, not a single point.