Question 1 of 6: Sampled-data differentiator and the band-limit condition

Question 1: Sampled-data differentiator and the band-limit condition (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. An ideal C/D converter, an ideal discrete-time differentiator and an ideal D/C converter, all sharing the same sampling period.

Find. The reconstructed continuous-time output $y_r(t)$ for each input, and whether the cascade behaves as a true differentiator in each case.

[Figure not reproduced: Figure Q1.1 — the sampled-data chain of the exam figure: an ideal C/D converter, the discrete-time differentiator, and an ideal D/C converter, both converters clocked at the same period T. See the official exam paper.]

Approach. Map each continuous-time frequency to its discrete-time counterpart $\omega_0 = \Omega_0 T$, fold that value into the principal range $-\pi \lt \omega \le \pi$ if it falls outside, evaluate the given frequency response there, and reconstruct the resulting sinusoid at $\Omega = \omega_0 / T$.

1. Fix the folding frequency. The C/D converter maps the continuous frequency $\Omega$ to the discrete frequency $\omega = \Omega T$, and the map is one-to-one only while $|\Omega| \lt \pi/T$. With $T = 1/10$ s, $$\frac{\pi}{T} = 10\pi\ \text{rad/s} \equiv 5\ \text{Hz}.$$ Any input component above 5 Hz will be aliased down into the base band before the differentiator ever sees it. This single number decides both parts of the question.
2. Case (i): place the input frequency. For $x_c(t) = \cos(6\pi t)$ the radian frequency is $\Omega_1 = 6\pi$ rad/s (3 Hz), so $$\omega_1 = \Omega_1 T = 6\pi \cdot \tfrac{1}{10} = 0.6\pi\ \text{rad/sample}.$$ Since $|\omega_1| \le \pi$ there is no aliasing and the sampled sequence is $x[n] = \cos(0.6\pi n)$.
3. Evaluate the differentiator at that frequency. Substituting $\omega_1$ into the given response, $$H(e^{j0.6\pi}) = \frac{j(0.6\pi)}{1/10} = j\,6\pi = 6\pi\,\angle\,{+}90^{\circ},$$ so the gain is $|H| = 6\pi \approx 18.85$ and the phase is a quarter-cycle lead.
4. Form the output sequence and reconstruct it. The sinusoidal-response relation supplied on page 7 of the exam gives $y[n] = |H(e^{j\omega_1})|\cos(\omega_1 n + \angle H(e^{j\omega_1}))$, hence $$y[n] = 6\pi\cos\!\left(0.6\pi n + \tfrac{\pi}{2}\right) = -6\pi\sin(0.6\pi n).$$ The D/C converter turns each discrete frequency $\omega$ back into $\Omega = \omega/T$, so $0.6\pi$ returns to $6\pi$ rad/s and $$\boxed{\,y_r(t) = -6\pi\sin(6\pi t) \approx -18.85\sin(6\pi t)\,}$$
5. Case (ii): place the input frequency. Now $\Omega_2 = 14\pi$ rad/s (7 Hz), so $$\omega_2 = \Omega_2 T = 14\pi \cdot \tfrac{1}{10} = 1.4\pi \gt \pi .$$ The input sits above the 5 Hz folding frequency, so it is aliased. Folding into the principal range, $1.4\pi - 2\pi = -0.6\pi$, and indeed $\cos(1.4\pi n) = \cos(2\pi n - 0.6\pi n) = \cos(0.6\pi n)$: the two inputs produce the identical sequence $x[n]$.
6. Evaluate and reconstruct. Because the differentiator is defined only on $-\pi \le \omega \le \pi$, it responds at $\omega = -0.6\pi$: $$H(e^{-j0.6\pi}) = \frac{j(-0.6\pi)}{1/10} = -j\,6\pi = 6\pi\,\angle\,{-}90^{\circ}.$$ Then $y[n] = 6\pi\cos(-0.6\pi n - \tfrac{\pi}{2}) = -6\pi\sin(0.6\pi n)$, exactly the sequence of case (i), and therefore $$\boxed{\,y_r(t) = -6\pi\sin(6\pi t) \approx -18.85\sin(6\pi t)\,}$$ The 7 Hz input and the 3 Hz input come out of the machine as the same 3 Hz signal.
7. Part (b): compare with a true derivative. For case (i) the exact derivative is $\frac{d}{dt}\cos(6\pi t) = -6\pi\sin(6\pi t)$, which is precisely the boxed answer, so the cascade is behaving as a differentiator. For case (ii) the exact derivative would be $\frac{d}{dt}\cos(14\pi t) = -14\pi\sin(14\pi t)$, an amplitude of $14\pi \approx 43.98$ at 7 Hz, whereas the machine delivers amplitude $6\pi \approx 18.85$ at 3 Hz. Both the frequency and the amplitude are wrong, so the answer to part (b) is: yes for (i), no for (ii).
8. State the condition. The effective continuous-time response of the whole chain is $H_{\text{eff}}(j\Omega) = H(e^{j\Omega T}) = j\Omega$ for $|\Omega| \lt \pi/T$ and zero elsewhere. The cascade is therefore an ideal differentiator only for inputs band-limited to $|\Omega| \lt \pi/T = 10\pi$ rad/s. Nothing is wrong with the discrete-time system; the failure in (ii) is committed by the sampler, before the differentiator is reached, and no discrete-time processing can undo it.