Question 1 of 6: Sampling rate, effective analogue cut-off and the reconstructed spectrum

Question 1: Sampling rate, effective analogue cut-off and the reconstructed spectrum (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.

[Figure not reproduced: Figure 1.1 — the C/D → discrete-time system → D/C chain of the exam figure. Both converters run on the same sampling period $T$. See the official exam paper.]

Given.

Find. The largest $T$ for which the C/D converter does not alias; the cut-off of the equivalent analogue filter when the chain runs at 24 kHz; and a labelled sketch of $Y_r(j\Omega)$ at that rate.

Approach. Apply the Nyquist condition to the highest frequency present, then use the frequency map $\omega = \Omega T$ to carry the discrete-time cut-off back into analogue frequency, and finally read $Y_r(j\Omega)$ off the effective analogue response $H_{\mathrm{eff}}(j\Omega) = H(e^{j\Omega T})$ for $|\Omega| \lt \pi/T$.

1. Convert the stated bandwidth into a radian frequency. The input is bandlimited to $f_{\max} = 4$ kHz, so the highest radian frequency it contains is $$\Omega_N = 2\pi f_{\max} = 2\pi(4\times 10^{3}) = 8000\pi \;\text{rad/s} \approx 2.513\times 10^{4}\ \text{rad/s}.$$ The exam figure confirms this: the triangle reaches zero exactly at $\Omega = \pm\,2\pi(4\times10^{3})$.
2. Impose the Nyquist condition on the C/D converter. Sampling replicates $X_c(j\Omega)$ at multiples of the sampling frequency $\Omega_s = 2\pi/T$. Adjacent replicas do not overlap provided $\Omega_s \ge 2\Omega_N$, which rearranges to $$\frac{2\pi}{T} \ge 2\Omega_N \quad\Longrightarrow\quad T \le \frac{\pi}{\Omega_N} = \frac{1}{2 f_{\max}}.$$ Substituting $f_{\max} = 4\times10^{3}$ Hz, $$\boxed{\,T_{\max} = \frac{1}{2(4\times10^{3})} = 1.25\times10^{-4}\ \text{s} = 125\ \mu\text{s}\,}$$ equivalently a minimum sampling rate of $1/T_{\max} = 8$ kHz.
3. Map the discrete-time cut-off back to analogue frequency. The C/D converter sets $\omega = \Omega T$, so a discrete-time frequency $\omega_c$ corresponds to the analogue frequency $\Omega_c = \omega_c/T$. With $1/T = 24$ kHz, $$\Omega_c = \frac{\omega_c}{T} = \frac{\pi}{6}\,(24\times10^{3}) = 4000\pi\ \text{rad/s},$$ so the effective continuous-time filter is an ideal lowpass with $$\boxed{\,\Omega_c = 4000\pi\ \text{rad/s} \approx 1.257\times10^{4}\ \text{rad/s}, \qquad f_c = \frac{\Omega_c}{2\pi} = 2\ \text{kHz}.\,}$$ Note that this cut-off is proportional to the sampling rate: doubling $1/T$ would double $f_c$ even though nothing inside the discrete-time system changed.
4. Check that the chain really is alias-free at this rate. Here $1/T = 24$ kHz comfortably exceeds the $8$ kHz found in step 2, so the baseband replica of $X_c(j\Omega)$ stands clear of its neighbours by $2\pi(24-4-4)\times10^{3} = 2\pi(16\times10^{3})$ rad/s. The overall system is therefore LTI in the continuous-time sense and $$Y_r(j\Omega) = H(e^{j\Omega T})\,X_c(j\Omega), \qquad |\Omega| \lt \frac{\pi}{T} = 24000\pi \ \text{rad/s},$$ with $Y_r(j\Omega) = 0$ outside that band because the ideal D/C reconstruction filter removes everything above $\pi/T$.
5. Assemble and label the output spectrum. Since the passband gain is 1, the filter simply truncates the input triangle at $\pm\,\Omega_c$: $$Y_r(j\Omega) = \begin{cases} 1 - \dfrac{|\Omega|}{8000\pi}, & |\Omega| \lt 4000\pi, \\[4pt] 0, & \text{otherwise}. \end{cases}$$ The peak at $\Omega = 0$ is unchanged at 1, and the surviving edge height at $\Omega = \pm\,4000\pi$ is $$1 - \frac{4000\pi}{8000\pi} = \boxed{0.5}$$ so the sketch is the original triangle with its two outer halves cut off vertically at $\pm\,4000\pi$ rad/s.