Question 5 of 5: Interpolation, band-pass selection and the equivalent continuous-time system

Question 5: Interpolation, band-pass selection and the equivalent continuous-time system (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 processing chain and its data:

Find. The four spectra of part (a), and a closed-form relation between $Y_c(j\Omega)$ and $X_c(j\Omega)$ — and between $y_c(t)$ and $x_c(t)$ — for any input band-limited to $|\Omega| \lt \pi/T$.

[Figure not reproduced: Figure 2 redrawn: ideal C/D at rate T, expansion by 4, the discrete-time filter, and ideal D/C at rate T' = T/4. See the official exam paper.]

[Figure not reproduced: Figure 4 redrawn: the discrete-time filter is an ideal band-pass of gain 4 between pi/4 and 3pi/4. See the official exam paper.]

Approach. Track the spectrum through each block in turn: sampling scales and periodicises, expansion compresses the frequency axis, the band-pass keeps exactly one pair of images, and the D/C converter maps the surviving band back to analogue frequency. Then read the modulation theorem off the final picture.

1. Sampling: $X(e^{j\omega})$. Ideal C/D gives $$X(e^{j\omega}) = \frac{1}{T}\sum_{k=-\infty}^{\infty} X_c\!\left(j\frac{\omega - 2\pi k}{T}\right).$$ The input occupies $|\Omega| \lt \Omega_0 = \pi/T$, so each image occupies $|\omega| \lt \pi$: the signal is sampled exactly at the Nyquist rate and the images just touch at $\omega = \pm\pi$ without overlapping. $X(e^{j\omega})$ is therefore a train of triangles of peak height $1/T$, centred at every multiple of $2\pi$ and reaching zero midway between them.
2. Expansion by four: $X_e(e^{j\omega})$. Inserting three zeros between every sample gives $x_e[n] = x[n/4]$ for $n$ a multiple of 4 and zero otherwise, so $$X_e(e^{j\omega}) = X(e^{j4\omega}).$$ The frequency axis is compressed by four: each triangle shrinks to a half-width of $\pi/4$ and the images now repeat every $2\pi/4 = \pi/2$. Within $|\omega| \le \pi$ there are images centred at $\omega = 0,\ \pm\pi/2$ and $\pm\pi$. The amplitude is unchanged at $1/T$, because expansion inserts zeros rather than redistributing energy.
3. Band-pass selection: $Y_e(e^{j\omega})$. The image centred at $\omega = \pi/2$ spans exactly $$\left[\tfrac{\pi}{2} - \tfrac{\pi}{4},\; \tfrac{\pi}{2} + \tfrac{\pi}{4}\right] = \left[\tfrac{\pi}{4},\; \tfrac{3\pi}{4}\right],$$ which is precisely the pass band of Figure 4. The filter therefore passes the $\pm\pi/2$ images in full, with gain 4, and removes the base-band image at $\omega = 0$ and the image at $\omega = \pm\pi$ entirely. Thus $Y_e(e^{j\omega})$ consists of triangles of peak height $4/T$ centred at $\omega = \pm\pi/2$ (and their $2\pi$-periodic repetitions at $\pm 3\pi/2$, and so on).
4. Reconstruction: $Y_c(j\Omega)$. The ideal D/C converter at period $T'$ keeps one period of the discrete spectrum and rescales it: $$Y_c(j\Omega) = T'\,Y_e\!\left(e^{j\Omega T'}\right),\qquad |\Omega| \lt \frac{\pi}{T'} = \frac{4\pi}{T}.$$ The centre frequency maps as $\Omega = \omega/T' = (\pi/2)(4/T) = 2\pi/T$ and the half-width as $(\pi/4)(4/T) = \pi/T$, while the height becomes $T' \cdot 4/T = (T/4)(4/T) = 1$. Hence $$\boxed{\;Y_c(j\Omega)\ \text{= two triangles of peak 1, centred at}\ \Omega = \pm\frac{2\pi}{T},\ \text{spanning}\ \frac{\pi}{T} \le |\Omega| \le \frac{3\pi}{T}\;}$$ The peak returns to exactly the input peak of 1, which is what the gain of 4 was chosen to achieve.

5. Write the general spectral relation (part b). Nothing in steps 1–4 used the triangular shape — only the band limit $X_c(j\Omega) = 0$ for $|\Omega| \ge \pi/T$, which is what guarantees that the $\pm\pi/2$ images arrive at the filter undistorted and separated. For any such input the chain therefore shifts the base-band spectrum, unchanged in shape and amplitude, to $\pm 2\pi/T$: $$\boxed{\;Y_c(j\Omega) = X_c\!\left(j\left(\Omega - \frac{2\pi}{T}\right)\right) + X_c\!\left(j\left(\Omega + \frac{2\pi}{T}\right)\right)\;}$$
6. Translate to the time domain. A frequency shift by $\pm\Omega_c$ is multiplication by $e^{\pm j\Omega_c t}$, and the two conjugate shifts add to a cosine: $$y_c(t) = x_c(t)e^{\,j2\pi t/T} + x_c(t)e^{-j2\pi t/T},$$ so that $$\boxed{\;y_c(t) = 2\,x_c(t)\cos\!\left(\frac{2\pi t}{T}\right)\;}$$ The whole multirate chain is a double-sideband suppressed-carrier modulator: it lifts $x_c(t)$ onto a carrier at $\Omega_c = 2\pi/T$ (that is, $f_c = 1/T$ Hz, the original sampling frequency) with a gain of two, and does so using nothing but an expander, a fixed band-pass filter and a converter running four times faster.
7. Check the answer three ways. First, dimensional consistency: at $t = 0$ the boxed expression gives $y_c(0) = 2x_c(0)$, and integrating the two unit-peak triangles of $Y_c$ likewise gives twice the area of the single triangle of $X_c$. Second, no aliasing occurs at the D/C converter because the highest frequency present, $3\pi/T$, is below its folding frequency $\pi/T' = 4\pi/T$. Third, the band edges match exactly: had the input occupied more than $|\Omega| \lt \pi/T$, the images at $\omega = 0$ and $\omega = \pm\pi$ would have spilled into the pass band and the clean modulation result would have failed — which is precisely why part (b) states that band limit as a hypothesis.