22-Elec-A3 Signals and Communications · May 2017

Question 1 of 5: First-Order RC Low-Pass Filter Driven by a Square Wave

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: First-Order RC Low-Pass Filter Driven by a Square Wave (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 Fourier series of x(t); the output y(t) in the time and frequency domains; the 40 dB bandwidth of y(t); and the average power of both the input and the output.

Approach. Expand the even square wave in a cosine Fourier series, multiply each harmonic by the RC filter's frequency response $H(f)=1/(1+jf/f_c)$ evaluated at that harmonic, then read the bandwidth off the resulting line spectrum and obtain both powers from Parseval's theorem.

1. Fix the filter time constant and cut-off frequency. The capacitance is deliberately written as $C=\dfrac{1}{2\pi}\ \text{nF}$ so that the corner frequency comes out round: $$\tau = RC = (10\times10^{3})\left(\frac{10^{-9}}{2\pi}\right) = 1.5915\ \mu\text{s}, \qquad f_c=\frac{1}{2\pi RC}$$ $$\boxed{f_c = 100\ \text{kHz} = 2f_0}$$ The corner therefore sits exactly two harmonics up from the fundamental, so the filter is gentle: it attenuates the fundamental only slightly and rolls the high harmonics off at 20 dB/decade.
2. Fourier series of the input. The wave swings between 0 V and 2 V, so it has a dc level of 1 V and an alternating part of amplitude 1 V. Because it is even with its maximum at the origin, only cosine terms survive and the classical odd-harmonic square-wave series applies: $$\boxed{x(t) = 1 + \frac{4}{\pi}\sum_{n\ \text{odd}}\frac{(-1)^{(n-1)/2}}{n}\cos\!\left(2\pi n f_0 t\right)\ \text{V}}$$ with $f_0=50$ kHz. The first few amplitudes are $a_1=4/\pi=1.2732$, $a_3=-0.4244$, $a_5=0.2546$, $a_7=-0.1819$ V; the alternating signs are what place the maximum at $t=0$.
3. Frequency response of the RC section. With the output taken across the capacitor, $$H(f)=\frac{1/j2\pi fC}{R+1/j2\pi fC}=\frac{1}{1+j f/f_c},\qquad |H| = \frac{1}{\sqrt{1+(f/f_c)^{2}}},\qquad \angle H = -\tan^{-1}\!\frac{f}{f_c}$$ Since $f_c=2f_0$, the $n$-th harmonic sees $f/f_c = n/2$, so $|H_n| = 1/\sqrt{1+n^{2}/4}$ and $\angle H_n = -\tan^{-1}(n/2)$. Numerically:

   n	an (V)	|Hn|	|Yn| (V)	∠Hn	rel. to fundamental
   0 (dc)	1.0000	1.0000	1.0000	0°	—
   1	1.2732	0.8944	1.1388	−26.57°	0 dB
   3	−0.4244	0.5547	0.2354	−56.31°	−13.69 dB
   5	0.2546	0.3714	0.09457	−68.20°	−21.61 dB
   7	−0.1819	0.2747	0.04997	−74.05°	−27.16 dB
   9	0.1415	0.2169	0.03069	−77.47°	−31.39 dB
   11	−0.1157	0.1789	0.02071	−79.69°	−34.81 dB
   13	0.0979	0.1521	0.01489	−81.25°	−37.67 dB
   15	−0.0849	0.1322	0.01122	−82.41°	−40.13 dB
   The dc term passes unattenuated because a capacitor is an open circuit at zero frequency.

4. (a) Output in the time domain. Each harmonic is scaled by $|H_n|$ and delayed by $\angle H_n$: $$\boxed{y(t) = 1 + \sum_{n\ \text{odd}}\frac{4(-1)^{(n-1)/2}}{n\pi\sqrt{1+n^{2}/4}} \cos\!\left(2\pi n f_0 t - \tan^{-1}\frac{n}{2}\right)\ \text{V}}$$ Written out to the seventh harmonic, $$y(t)\approx 1 + 1.1388\cos(\omega_0 t-26.57^\circ) - 0.2354\cos(3\omega_0 t-56.31^\circ) + 0.0946\cos(5\omega_0 t-68.20^\circ) - 0.0500\cos(7\omega_0 t-74.05^\circ)+\cdots$$ with $\omega_0 = 2\pi(50\ \text{kHz})$. The same result can be obtained directly in the time domain by solving the first-order differential equation over each half period; the exponential segments reach 1.9963 V just before the falling edge, which the harmonic sum reproduces.

5. (b) Output in the frequency domain. For a periodic signal the spectrum is a line spectrum. Using the two-sided exponential form $x(t)=\sum_n c_n e^{j2\pi nf_0t}$ with $c_0=1$ and $c_n=a_{|n|}/2$ for odd $n$, $$\boxed{Y(f)=\delta(f)+\sum_{n\ \text{odd}}\frac{2(-1)^{(|n|-1)/2}}{|n|\pi}\, \frac{1}{1+jn/2}\ \delta(f-nf_0)}$$ i.e. an impulse of area 1 at dc plus impulse pairs at $\pm 50,\pm150,\pm250,\ldots$ kHz whose areas are the tabulated $|Y_n|/2$ with the tabulated phases. The one-sided amplitude spectrum is plotted below.

6. (c) 40 dB bandwidth. The "main AC in-band component" is the fundamental, of amplitude 1.1388 V. A component is in band while it is no more than 40 dB below that, i.e. while $$|Y_n| \ge 10^{-40/20}\,(1.1388) = 0.011388\ \text{V}$$ From the table $|Y_{13}|=0.01489$ V ($-37.67$ dB) still qualifies, while $|Y_{15}|=0.01122$ V is $-40.13$ dB and falls outside. The highest in-band harmonic is therefore the 13th: $$\boxed{B = 13 f_0 = 650\ \text{kHz}}$$ measured from dc, since the output is a baseband (low-pass) signal. Note how close the decision is — the 15th harmonic misses the threshold by only 0.13 dB, so the levels must be computed rather than estimated from the $1/n$ envelope alone.
7. (d) Average power of the input. The square wave spends half of each period at 2 V and half at 0 V, so no integration is needed: $$P_x=\frac{1}{T}\int_T x^{2}(t)\,dt=\frac{(2)^{2}+(0)^{2}}{2}=\boxed{P_x = 2\ \text{W}}$$ (that is, 2 W into a 1 Ω reference, or equivalently a mean-square value of 2 V²). Parseval gives the same figure: $1^2+\tfrac12\sum a_n^2 = 1 + \tfrac12(8/\pi^2)\!\sum_{n\,\text{odd}}\!n^{-2}\cdot 2 = 2$.
8. (e) Average power of the output. Parseval's theorem for a periodic signal adds the dc power to half the square of every harmonic amplitude: $$P_y = c_{0}^{2}+\frac{1}{2}\sum_{n\ \text{odd}}|Y_n|^{2} = 1 + \tfrac12\left(1.1388^{2}+0.2354^{2}+0.09457^{2}+\cdots\right)$$ $$P_y = 1 + 0.6484+0.0277+0.00447+0.00125+0.00047+\cdots$$ $$\boxed{P_y = 1.683\ \text{W}}$$ The series converges quickly because the terms fall off as $n^{-4}$ once $n\gg 2$. As an independent check, integrating the exact exponential waveform of Step 4 over one period gives 1.68288 W, agreeing to five figures. About 16 % of the input power is dissipated in the resistor.