Question 4 of 5: Continuous-Time Filtering by a Discrete-Time System — Three Sampling-Rate Cases

Question 4: Continuous-Time Filtering by a Discrete-Time System — Three Sampling-Rate Cases (25 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 $x_c(t) \to \mathrm{C/D}(T_1) \to H(e^{j\omega}) \to \mathrm{D/C}(T_2) \to y_c(t)$, with an ideal discrete-time lowpass filter of unit gain and cut-off $\omega_c = \pi/5$, and a triangular input spectrum of unit peak that falls linearly to zero at $\pm\Omega_0$, $\Omega_0 = \pi \times 10^{5}\ \text{rad/s}$ (i.e. 50 kHz). Three sampling-rate pairs are to be examined.

Find. Labelled sketches of the discrete input spectrum $X(e^{j\omega})$, the filtered spectrum $Y(e^{j\omega})$ and the reconstructed continuous spectrum $Y_c(j\Omega)$, with every height and every band edge marked, for each of the three rate pairs.

[Figure not reproduced: Figure 1 redrawn: the discrete-time filter H(e^jw) sandwiched between an ideal C/D converter (period T1) and an ideal D/C converter (period T2). Only when T1 = T2 does the pair behave as a single continuous-time filter Hc(jOmega). See the official exam paper.]

[Figure not reproduced: Figures 3 and 2 redrawn: the unit-peak triangular input spectrum Xc(jOmega), zero at +/- Omega0 = 10pi x 10^4 rad/s, and the ideal discrete-time lowpass filter of unit gain and cut-off wc = pi/5. See the official exam paper.]

Approach. Apply the three standard relations in order — sampling maps the continuous spectrum to a scaled, frequency-warped periodic replica set; the ideal filter clips each replica; the ideal reconstructor keeps only the baseband and un-warps it with the second sampling period. Whenever T₁ ≠ T₂ the result is a time-scaled version of the filtered signal, not a pure filtering operation.

1. State the three governing relations. Ideal C/D conversion at period $T_1$ gives $$X(e^{j\omega}) = \frac{1}{T_1}\sum_{k=-\infty}^{\infty} X_c\!\left(j\frac{\omega - 2\pi k}{T_1}\right),$$ the filter gives $Y(e^{j\omega}) = H(e^{j\omega})X(e^{j\omega})$, and ideal D/C conversion at period $T_2$ gives $Y_c(j\Omega) = T_2\,Y(e^{j\Omega T_2})$ for $|\Omega| \lt \pi/T_2$ and zero outside. Two consequences are used repeatedly: the triangle edge $\Omega_0$ maps to $\omega_0 = \Omega_0 T_1$, and the replica height is $1/T_1$.
2. Check for aliasing once, in general. Replicas are spaced $2\pi$ apart in $\omega$, so they overlap only if $\omega_0 \gt \pi$, i.e. if $\Omega_0 \gt \pi/T_1$, i.e. if the sampling rate $1/T_1$ is below the Nyquist rate $\Omega_0/\pi = 10^{5}\ \text{Hz}$. All three cases use $1/T_1 \ge 10^{5}$, so $$\boxed{\,\text{no aliasing occurs in any of the three cases}\,}$$ with case (c) sitting exactly at the critical rate, where adjacent replicas just touch at $\omega = \pm\pi$ (with zero amplitude, so no overlap error).

Case (a): $1/T_1 = 1/T_2 = 2\times10^{5}$ Hz. Here $T_1 = T_2 = 5\ \mu\text{s}$, so the cascade really is a continuous-time filter, and no time scaling occurs.

3. Map the input spectrum into $\omega$. The triangle edge maps to $\omega_0 = \Omega_0 T_1 = (\pi\times10^{5})(5\times10^{-6}) = \pi/2$, and the height is $1/T_1 = 2\times10^{5}$. So $X(e^{j\omega})$ is a triangle of peak $2\times10^{5}$, zero at $\omega = \pm\pi/2$, repeated every $2\pi$.
4. Clip with the ideal filter. Since $\omega_c = \pi/5 \lt \omega_0 = \pi/2$, the filter cuts into the triangle. At the band edge the surviving height is $2\times10^{5}\left(1 - \frac{\pi/5}{\pi/2}\right) = 2\times10^{5}(0.6) = 1.2\times10^{5}$, so $Y(e^{j\omega})$ is the same triangle truncated to $|\omega| \le \pi/5$ (and repeated every $2\pi$).
5. Reconstruct. With $T_2 = T_1$ the scale factor $T_2/T_1 = 1$, so the peak returns to $T_2 \cdot 2\times10^{5} = 1$ and the band edge maps back to $\Omega = \omega_c/T_2 = 4\pi\times10^{4}\ \text{rad/s}$ (20 kHz), where the height is $0.6$. Therefore $$\boxed{\,Y_c(j\Omega) = X_c(j\Omega)\ \text{for}\ |\Omega| \le 4\pi\times10^{4},\ \text{zero elsewhere}\,}$$ i.e. the cascade is exactly an ideal continuous-time lowpass filter of cut-off $\Omega_c = \omega_c/T = 4\pi\times10^{4}$ rad/s.

Case (b): $1/T_1 = 4\times10^{5}$ Hz, $1/T_2 = 10^{5}$ Hz. Reconstructing four times more slowly than the signal was sampled stretches the output in time by a factor of four and multiplies its spectral amplitude by the same factor.

6. Sample and filter. Now $T_1 = 2.5\ \mu\text{s}$, so $\omega_0 = (\pi\times10^{5})(2.5\times10^{-6}) = \pi/4$ and the replica height is $1/T_1 = 4\times10^{5}$. Again $\omega_c = \pi/5 \lt \pi/4$, so the filter clips the triangle, leaving a band-edge height of $4\times10^{5}\left(1-\frac{\pi/5}{\pi/4}\right) = 4\times10^{5}(0.2) = 0.8\times10^{5}$.
7. Reconstruct at the slower rate. With $T_2 = 10\ \mu\text{s}$ the amplitude scale is $T_2/T_1 = 4$ and the frequency axis contracts by the same factor: $$\boxed{\,Y_c(j0) = 4, \qquad \text{support } |\Omega| \le \frac{\omega_c}{T_2} = 2\pi\times10^{4}\ \text{rad/s}\,}$$ with height $0.8$ at the band edge. Physically $y_c(t)$ is the filtered waveform played back four times more slowly, $y_c(t) = x_{c,\text{filt}}(t/4)$: the time-domain peak is unchanged, and the spectral peak rises by four precisely because stretching a signal in time compresses and heightens its spectrum. The cascade is therefore not a filter at all here.

Case (c): $1/T_1 = 10^{5}$ Hz, $1/T_2 = 3\times10^{5}$ Hz. The first rate is exactly the Nyquist rate, and the second is three times faster, so time is compressed by a factor of three.

8. Recognise critical sampling. With $T_1 = 10\ \mu\text{s}$, $\omega_0 = (\pi\times10^{5})(10^{-5}) = \pi$: the replicas of the triangle just touch at $\omega = \pm\pi$, where the amplitude is zero. The sampling frequency $\Omega_s = 2\pi/T_1 = 2\pi\times10^{5}$ equals $2\Omega_0$ exactly, so there is no aliasing error, but there is also no guard band. The replica height is $1/T_1 = 10^{5}$, and $X(e^{j\omega})$ becomes a continuous triangular wave rather than a set of isolated triangles.
9. Filter and reconstruct. The filter keeps $|\omega| \le \pi/5$, where the band-edge height is $10^{5}\left(1-\frac{\pi/5}{\pi}\right) = 0.8\times10^{5}$. Reconstruction at $T_2 = 1/3\ \times 10^{-5}$ s scales amplitude by $T_2/T_1 = 1/3$ and expands the frequency axis by three: $$\boxed{\,Y_c(j0) = \tfrac13, \qquad \text{support } |\Omega| \le \frac{\omega_c}{T_2} = 6\pi\times10^{4}\ \text{rad/s}\,}$$ with height $\tfrac13(0.8) = 4/15 \approx 0.267$ at the band edge. Here $y_c(t) = x_{c,\text{filt}}(3t)$: the filtered waveform is played back three times faster, again with its time-domain amplitude unchanged and its spectrum correspondingly wider and lower.