Question 2 of 6: Continuous-time filtering by a discrete-time lowpass filter at three sampling-rate pairs

Question 2: Continuous-time filtering by a discrete-time lowpass filter at three sampling-rate pairs (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. The chain $x_{c}(t) \to \text{C/D}(T_{1}) \to x[n] \to H(e^{j\omega}) \to y[n] \to \text{D/C}(T_{2}) \to y_{c}(t)$, with an ideal discrete-time lowpass filter of cutoff $\omega_{c} = \pi/5$ and a triangular input spectrum $X_{c}(j\Omega)$ of unit height that falls linearly to zero at $\Omega_{0} = 2\pi\times 5\times 10^{3}\ \text{rad/s}$ (a 5 kHz bandwidth).

Find. Labelled sketches of $X(e^{j\omega})$, $Y(e^{j\omega})$ and $Y_{c}(j\Omega)$ for each of the three rate pairs, including every amplitude and every band edge.

Approach. Use the three standard relations once and then tabulate: C/D sampling replicates and rescales the spectrum, the ideal filter truncates it, and D/C reconstruction rescales the frequency axis by $T_{2}$. Only two derived quantities decide each sketch — the normalised band edge $\omega_{0} = \Omega_{0}T_{1}$ and the ratio $T_{2}/T_{1}$.

1. Write down the three governing relations. Sampling at $T_{1}$ gives $$X\!\left(e^{j\omega}\right) = \frac{1}{T_{1}} \sum_{k=-\infty}^{\infty} X_{c}\!\left(j\frac{\omega - 2\pi k}{T_{1}}\right),$$ so the triangle is replicated every $2\pi$, scaled in height by $1/T_{1}$, and its edge lands at the normalised frequency $\omega_{0} = \Omega_{0}T_{1}$. Filtering gives $Y(e^{j\omega}) = H(e^{j\omega})X(e^{j\omega})$, which simply truncates each replica to $|\omega| \lt \omega_{c}$. Ideal bandlimited reconstruction at $T_{2}$ gives $$Y_{c}(j\Omega) = T_{2}\,Y\!\left(e^{j\Omega T_{2}}\right), \qquad |\Omega| \lt \pi/T_{2}.$$
2. Note the one general consequence, before any numbers. Chaining the three scalings, whenever there is no aliasing, $$Y_{c}(j\Omega) = \frac{T_{2}}{T_{1}}\, X_{c}\!\left(j\Omega\frac{T_{2}}{T_{1}}\right) \quad\text{on the retained band},$$ whose inverse transform is $$\boxed{\,y_{c}(t) = x_{c,\text{lp}}\!\left(t\,\frac{T_{1}}{T_{2}}\right)\,}$$ a pure time scaling of the lowpass-filtered input. The spectral height is multiplied by $T_{2}/T_{1}$, but the time-domain amplitude is unchanged — the taller spectrum is produced by stretching the waveform, not by amplifying it. Only when $T_{1} = T_{2}$ is the overall system a genuine continuous-time LTI filter of cutoff $\Omega_{c} = \omega_{c}/T$.
3. Case (a): $1/T_{1} = 1/T_{2} = 2\times 10^{4}$. Then $T_{1} = T_{2} = 5\times 10^{-5}\ \text{s}$ and $$\omega_{0} = \Omega_{0}T_{1} = 2\pi(5\times10^{3})(5\times10^{-5}) = \frac{\pi}{2} \lt \pi ,$$ so sampling at 20 kHz on a 5 kHz signal is comfortably above the Nyquist rate and there is no aliasing. The replicas of $X(e^{j\omega})$ are triangles of height $1/T_{1} = 2\times 10^{4}$ and half-width $\pi/2$, centred at every multiple of $2\pi$. Since $\omega_{c} = \pi/5 \lt \omega_{0} = \pi/2$, the filter cuts into the triangle; the retained edge height is $$\frac{1}{T_{1}}\left(1 - \frac{\omega_{c}}{\omega_{0}}\right) = 2\times10^{4}\left(1 - \tfrac{2}{5}\right) = 1.2\times 10^{4}.$$ Reconstructing with $T_{2} = T_{1}$ leaves the peak at $T_{2}/T_{1} = 1$ and the band edge at $\Omega = \omega_{c}/T_{2} = 2\pi\times 2\times 10^{3}$, i.e. $$\boxed{Y_{c}(j\Omega)\ \text{is}\ X_{c}\ \text{ideally lowpassed at}\ 2\ \text{kHz},\ \text{peak}=1,\ \text{edge}=0.6 }$$

Case (a) is the benchmark: because the two rates agree, the dashed box really is a continuous-time lowpass filter, and its cutoff is $\Omega_{c} = \omega_{c}/T = (\pi/5)(2\times10^{4})/\pi = 2\ \text{kHz}$, i.e. one tenth of the sampling rate. The next two cases break that equivalence.

4. Case (b): $1/T_{1} = 4\times10^{4}$, $1/T_{2} = 10^{4}$. Now $T_{1} = 2.5\times10^{-5}$, $T_{2} = 10^{-4}$, and $$\omega_{0} = 2\pi(5\times10^{3})(2.5\times10^{-5}) = \frac{\pi}{4} \lt \pi \quad\text{(no aliasing).}$$ The replicas have height $1/T_{1} = 4\times10^{4}$ and half-width $\pi/4$; the filter at $\pi/5$ again clips them, leaving edge height $4\times10^{4}(1 - \tfrac{4}{5}) = 8\times 10^{3}$. Reconstruction at the four-times-longer $T_{2}$ scales the peak to $T_{2}/T_{1} = 4$ and compresses the frequency axis by four, so the output band edge is $\omega_{c}/T_{2} = 2\pi\times 10^{3}$: $$\boxed{\ \text{peak } Y_{c} = 4,\quad \text{band edge } 1\ \text{kHz}, \quad y_{c}(t) = x_{c,\text{lp}}(t/4)\ }$$ Here the discrete-time cutoff maps back to $\omega_{c}/T_{1} = 2\pi\times 4\ \text{kHz}$ on the original signal, so the output is the 4 kHz-lowpassed input played back four times more slowly — which is why its band edge lands at $4/4 = 1\ \text{kHz}$. Its peak value in time is unchanged.

5. Case (c): $1/T_{1} = 10^{4}$, $1/T_{2} = 3\times10^{4}$. Here $T_{1} = 10^{-4}$ and $$\omega_{0} = 2\pi(5\times10^{3})(10^{-4}) = \pi \quad\text{exactly.}$$ This is critical sampling: $1/T_{1} = 10^{4} = 2\times(5\times10^{3})$ is precisely the Nyquist rate. The replicas therefore meet at $\omega = \pm\pi$ — but the triangle has already fallen to zero there, so the replicas touch at zero amplitude and there is no aliasing error. $X(e^{j\omega})$ is a continuous triangular wave of peak $1/T_{1} = 10^{4}$ reaching zero at every odd multiple of $\pi$. The filter clips it at $\pi/5$, leaving edge height $10^{4}(1 - \tfrac{1}{5}) = 8\times10^{3}$. Reconstruction at the three-times-shorter $T_{2} = 1/(3\times10^{4})$ scales the peak by $T_{2}/T_{1} = 1/3$ and expands the frequency axis by three, so the band edge sits at $\omega_{c}/T_{2} = 2\pi\times 3\times10^{3}$: $$\boxed{\ \text{peak } Y_{c} = \tfrac{1}{3},\quad \text{edge height } \tfrac{4}{15},\quad \text{band edge } 3\ \text{kHz},\quad y_{c}(t) = x_{c,\text{lp}}(3t)\ }$$

The three cases separate two effects that students routinely conflate. What the discrete-time filter removes depends only on $\omega_{c}$ compared with $\omega_{0} = \Omega_{0}T_{1}$, so it is set by the input rate alone; here $\omega_{c} = \pi/5$ is below $\omega_{0}$ in all three cases and the triangle is always clipped. What happens to the time axis depends only on the ratio $T_{2}/T_{1}$, and it is a stretch, not a gain.