Question 1 of 6: Sampling rate for an ideal C/D – filter – D/C chain

Question 1: Sampling rate for an ideal C/D – filter – D/C chain (12 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. A continuous-to-discrete converter, an ideal discrete-time lowpass filter and a discrete-to-continuous converter, all sharing one sampling period $T$. The input spectrum $|X_c(j\Omega)|$ is a triangle of unit height at $\Omega = 0$ falling linearly to zero at $|\Omega| = \Omega_0$, with $\Omega_0 = 2\pi(1000)$ rad/s, so the signal is strictly bandlimited to $f_0 = 1000$ Hz. The filter passes $|\omega| \lt \omega_c$ with unit gain and blocks everything else.

Find. (a) the smallest sampling rate that leaves the sampled spectrum free of aliasing, and (b) the smallest sampling rate at which the complete chain reproduces its own input exactly when the filter cutoff is fixed at $\omega_c = \pi/2$.

Approach. Part (a) is the sampling theorem applied to the highest frequency in the triangle; part (b) additionally requires the whole signal band, after it is mapped onto the discrete-time frequency axis by $\omega = \Omega T$, to fall inside the filter passband, which is the tighter of the two conditions.

1. Read the highest frequency off the spectrum. The triangle is identically zero for $|\Omega| \gt \Omega_0$, so $$\Omega_0 = 2\pi(1000)\ \text{rad/s} \qquad \Longrightarrow \qquad f_0 = \frac{\Omega_0}{2\pi} = 1000\ \text{Hz}.$$

   There is no energy above 1000 Hz, so the signal is exactly bandlimited and the sampling theorem applies without qualification.

2. Apply the sampling theorem for part (a). Sampling replicates the spectrum every $\Omega_s = 2\pi/T$, so the replicas are disjoint provided
   $$\Omega_s \ge 2\Omega_0 \qquad \Longleftrightarrow \qquad \frac{1}{T} \ge 2 f_0 = 2(1000).$$

   The smallest admissible rate is therefore

   $$\boxed{\ \frac{1}{T}\Big|_{\min} = 2000\ \text{samples/s}\quad (T_{\max} = 0.5\ \text{ms})\ }$$

   At exactly this rate neighbouring replicas touch at zero amplitude, which is the borderline case the theorem allows because the triangle has no impulse at the band edge.

At 1/T = 2000 Hz the replicas of the triangle just touch: the critical, alias-free case of part (a).
3. Map the signal band onto the discrete-time axis. The sampled sequence has $$X(e^{j\omega}) = \frac{1}{T}\sum_{k=-\infty}^{\infty} X_c\!\left(j\frac{\omega - 2\pi k}{T}\right),$$

   so the baseband copy occupies $|\omega| \le \Omega_0 T$. Raising the sampling rate shrinks that occupied band, because a fixed analogue frequency maps to a smaller $\omega = \Omega T$.

4. Impose the reconstruction condition of part (b). With the cutoff fixed at $\omega_c = \pi/2$, the filter removes nothing — and so $y_c(t) = x_c(t)$ — only if the entire baseband copy lies inside the passband:
   $$\Omega_0 T \le \omega_c = \frac{\pi}{2} \qquad \Longrightarrow \qquad T \le \frac{\pi}{2\,\Omega_0} = \frac{\pi}{2\cdot 2\pi(1000)} = \frac{1}{4000}\ \text{s}.$$

   Inverting the inequality gives the minimum rate

   $$\boxed{\ \frac{1}{T}\Big|_{\min} = 4000\ \text{samples/s}\quad (T_{\max} = 250\ \mu\text{s})\ }$$

   This is twice the Nyquist rate of part (a), so the no-aliasing requirement is automatically satisfied and the passband condition is the binding one.

At 1/T = 4000 Hz the signal band has been squeezed to |w| = pi/2, so the fixed cutoff wc = pi/2 passes all of it and nothing is lost.
5. Confirm that the chain really is an identity. Sampling introduces the factor $1/T$, the filter has unit gain in its passband, and the ideal D/C converter contributes the compensating factor $T$ together with an ideal lowpass at $\pi/T$. Since no replica has been clipped and no baseband component has been removed, $Y_c(j\Omega) = X_c(j\Omega)$ for all $\Omega$, so $y_c(t) = x_c(t)$ exactly. Note also that the overall system is linear and time-invariant for any $\omega_c$ once the input is sampled without aliasing; below the Nyquist rate the chain stops being LTI at all.