Question 1 of 5: Fourier Series of a Pulse Train Through a Filter Cascade

Question 1: Fourier Series of a Pulse Train Through a Filter Cascade (20 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.

Find. The exponential Fourier series coefficients of the pulse train, the time-domain signal that emerges from the two cascaded filters, and its average power.

Approach. Take the Fourier coefficients of the rectangular train from the standard sinc result, multiply each one by the RC filter's frequency response evaluated at that harmonic, discard every harmonic above the ideal filter's cut-off, and sum the surviving power by Parseval's theorem.

1. Fix the pulse geometry and the harmonic grid. The duty cycle sets the pulse width directly, and the period sets the spacing of the spectral lines: $$\tau = dT = 0.20 \times 100\ \mu\text{s} = 20\ \mu\text{s}, \qquad f_0 = \frac{1}{T} = \frac{1}{100\times10^{-6}} = 10\ \text{kHz}.$$ Every spectral line therefore sits at a multiple of 10 kHz.
2. Part (a) — evaluate the exponential Fourier coefficients. For a train of rectangular pulses of height $A$ and width $\tau$ centred on the origin, $$c_n = \frac{1}{T}\int_{-\tau/2}^{\tau/2} A\,e^{-jn\omega_0 t}\,dt = \frac{A\tau}{T}\cdot\frac{\sin(n\pi\tau/T)}{n\pi\tau/T} = A\,d\,\operatorname{sinc}(nd),$$ where $\operatorname{sinc}(x) = \sin(\pi x)/(\pi x)$. Substituting $A = 2$ V and $d = 0.2$, $$\boxed{\,c_n = 0.4\,\operatorname{sinc}(0.2n)\,}$$ Because the pulse is drawn symmetrically about $t = 0$, the signal is even and every coefficient is real; there is no phase to report beyond the sign.
3. Tabulate the first few lines. Evaluating the boxed expression:

   $n$	0	1	2	3	4	5
   $f = nf_0$ (kHz)	0	10	20	30	40	50
   $c_n$ (V)	0.4000	0.3742	0.3027	0.2018	0.0935	0

   The $n = 5$ line vanishes because $d = 1/5$ places the first null of the sinc envelope exactly on the fifth harmonic. In one-sided (trigonometric) form the same result reads $x(t) = 0.4 + \sum_{n\ge1} 2c_n\cos(2\pi n f_0 t)$, so the fundamental has amplitude $2c_1 = 0.748$ V.

4. Evaluate the RC filter at each harmonic. A first-order low-pass section has $$H(f) = \frac{1}{1 + j2\pi f\,RC}.$$ The choice $RC = T = 1/f_0$ is what makes this problem clean: at the $n$th harmonic $2\pi (n f_0)(RC) = 2\pi n$ exactly, so $$\boxed{\,H_n = \frac{1}{1 + j2\pi n}\,}$$ independent of the numerical value of $T$. Hence $H_0 = 1$, $|H_1| = 1/\sqrt{1+4\pi^2} = 0.15717$ at $\angle -80.96^\circ$, and $|H_2| = 1/\sqrt{1+16\pi^2} = 0.079327$ at $\angle -85.45^\circ$. The attenuation is severe: even the fundamental is cut to about one sixth.
5. Apply the ideal low-pass filter. The second stage passes everything with $|f| \le B = 25$ kHz and removes the rest. The harmonics sit at 0, 10, 20, 30, … kHz, so the lines at $n = 0, 1, 2$ survive and every line from $n = 3$ (30 kHz) upward is deleted. Because the ideal filter has unit gain in its passband, the surviving coefficients are simply $Y_n = c_n H_n$: $$Y_0 = 0.4, \qquad |Y_1| = 0.3742 \times 0.15717 = 0.058812, \qquad |Y_2| = 0.3027 \times 0.079327 = 0.024014.$$
6. Part (b) — assemble the output waveform. Converting to one-sided amplitudes ($2|Y_n|$) and carrying the RC phase lags, $$\boxed{\,y(t) = 0.4 + 0.1176\cos\!\left(2\pi\times10^{4}t - 80.96^\circ\right) + 0.0480\cos\!\left(2\pi\times2\times10^{4}t - 85.45^\circ\right)\ \text{V}\,}$$ The dc term dominates by a factor of more than three, which is why the output looks like a slightly rippled constant rather than a pulse train.
7. Part (c) — compute the average power by Parseval's theorem. For a periodic signal, power is the sum of the squared magnitudes of all its exponential coefficients, positive and negative frequencies together: $$P_y = |Y_0|^2 + 2\sum_{n=1}^{2}|Y_n|^2 = 0.16 + 2\left(0.058812^2 + 0.024014^2\right) = 0.16 + 0.008071,$$ $$\boxed{\,P_y = 0.1681\ \text{W into 1 }\Omega\,}$$

As a physical check, the input power is $P_x = A^2 d = 4 \times 0.2 = 0.8$ W, so the cascade delivers only 21% of the incoming power — consistent with two low-pass stages acting on a signal whose energy is spread across many harmonics.