22-Elec-A3 Signals and Communications · December 2018

Question 1 of 5: Threshold Device Driven by a Cosine — Fourier Series, Filtering and Power

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

Question 1: Threshold Device Driven by a Cosine — Fourier Series, Filtering and Power (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. Input $x(t)=4\cos(2\pi f_0 t)$ with amplitude $A=4$ V and $f_0=120$ Hz, so $T_0 = 1/120\ \text{s} = 8.333$ ms and $\omega_0 = 2\pi f_0$. The threshold device is memoryless: it passes $x(t)$ unchanged wherever $x(t)\ge 2$ V and outputs zero elsewhere. The band-pass filter of part (c) is ideal, with centre frequency 500 Hz and bandwidth 250 Hz.

Find. (a) the fundamental frequency of $y(t)$; (b) its real (cosine) Fourier series; (c) the filter output $z(t)$; and (d) the average powers $P_y$ and $P_z$.

Approach. Locate the conduction interval by solving $4\cos\theta = 2$, integrate the Fourier coefficients only over that interval (the signal is zero everywhere else), then read off which harmonics fall inside the filter passband and apply Parseval's theorem for the powers.

1. Find the conduction angle. Writing $\theta = \omega_0 t$, the device conducts wherever $$4\cos\theta \ge 2 \quad\Longleftrightarrow\quad \cos\theta \ge \tfrac{1}{2},$$ which holds for $|\theta| \le \theta_c$ with $$\theta_c = \arccos\!\left(\tfrac{1}{2}\right) = \boxed{\dfrac{\pi}{3} = 60^\circ}$$ in each period. So $y(t)$ consists of one cosine "cap" of total angular width $2\theta_c = 120^\circ$ per cycle of $x(t)$, as the figure above shows.
2. (a) Fundamental frequency. The threshold is a memoryless non-linearity, so it cannot change the period: $y(t+T_0) = y(t)$ and there is exactly one cap per period, with no repetition at any shorter interval. Hence $$\boxed{f_{0,y} = f_0 = 120\ \text{Hz}}$$ This is the point of the question — a non-linearity generates new harmonics, but it does not move the fundamental, because the output inherits the input's periodicity.
3. Set up the Fourier integrals. $y(t)$ is real and even (the conduction window is symmetric about $t=0$ and the cosine is even), so all sine coefficients vanish and the series is a pure cosine series, $$y(t) = a_0 + \sum_{n=1}^{\infty} a_n\cos(2\pi n f_0 t).$$ Because $y(t)=0$ outside $|\theta| \le \pi/3$, every integral runs only over the cap: $$a_0 = \frac{1}{2\pi}\int_{-\pi/3}^{\pi/3}\! 4\cos\theta\;d\theta, \qquad a_n = \frac{1}{\pi}\int_{-\pi/3}^{\pi/3}\! 4\cos\theta\cos n\theta\;d\theta .$$
4. Evaluate the DC term. $$a_0 = \frac{4}{2\pi}\Big[\sin\theta\Big]_{-\pi/3}^{\pi/3} = \frac{4}{2\pi}\cdot 2\sin\frac{\pi}{3} = \frac{2\sqrt{3}}{\pi} = \boxed{1.1027\ \text{V}}$$
5. Evaluate the harmonic terms. Using $\cos\theta\cos n\theta = \tfrac{1}{2}[\cos(n+1)\theta + \cos(n-1)\theta]$ and integrating, the case $n \ne 1$ gives $$a_n = \frac{4}{\pi}\left[\frac{\sin\!\big((n+1)\frac{\pi}{3}\big)}{n+1} + \frac{\sin\!\big((n-1)\frac{\pi}{3}\big)}{n-1}\right], \qquad n \ne 1 .$$ The case $n=1$ must be taken separately because $\cos(n-1)\theta \to 1$ integrates to $\theta$ rather than a sine: $$a_1 = \frac{4}{\pi}\left[\frac{\sin(2\pi/3)}{2} + \frac{\pi}{3}\right] = \frac{\sqrt{3}}{\pi} + \frac{4}{3} = 1.8847\ \text{V}.$$
6. (b) Tabulate the series. Substituting $n=1,2,\dots$ into the two expressions above: $$\boxed{\,y(t) = 1.1027 + 1.8847\cos\omega_0 t + 1.1027\cos 2\omega_0 t + 0.2757\cos 3\omega_0 t - 0.2205\cos 4\omega_0 t - 0.2757\cos 5\omega_0 t - 0.0630\cos 6\omega_0 t + 0.1378\cos 7\omega_0 t + \cdots}$$ with $\omega_0 = 2\pi(120)$ rad/s. Note that the coefficients change sign from the fourth harmonic onward; a negative $a_n$ simply means that harmonic enters with a phase of $180^\circ$, not that it is weaker than its magnitude suggests.

7. (c) Identify the harmonics inside the passband. The ideal band-pass filter has centre 500 Hz and bandwidth 250 Hz, so it passes $$f \in \left[500 - \tfrac{250}{2},\; 500 + \tfrac{250}{2}\right] = [375,\,625]\ \text{Hz}.$$ The harmonics of $y(t)$ sit at $120n$ Hz: 120, 240, 360, 480, 600, 720, … Of these only $n=4$ (480 Hz) and $n=5$ (600 Hz) fall inside the window — 360 Hz misses the lower edge and 720 Hz clears the upper edge. The DC term is likewise rejected. Therefore $$\boxed{\,z(t) = -0.2205\cos\big(2\pi(480)t\big) - 0.2757\cos\big(2\pi(600)t\big)\ \text{V}}$$ Equivalently, $z(t) = 0.2205\cos(2\pi(480)t + 180^\circ) + 0.2757\cos(2\pi(600)t + 180^\circ)$.
8. (d) Average power of $y(t)$ — directly. Averaging the square over one period, again only over the conduction window: $$P_y = \frac{1}{2\pi}\int_{-\pi/3}^{\pi/3}\! 16\cos^{2}\theta\;d\theta = \frac{16}{2\pi}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{-\pi/3}^{\pi/3} = \frac{8}{\pi}\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)$$ $$\boxed{P_y = 3.769\ \text{W}\ \ (\text{into }1\ \Omega)}$$
9. Cross-check with Parseval's theorem. For a real cosine series the average power is $$P_y = a_0^{2} + \frac{1}{2}\sum_{n=1}^{\infty} a_n^{2}.$$ Summing the coefficients of Step 6 numerically gives $1.2159 + 2.5534 = 3.7693$ W, which matches the closed-form integral to five figures. This is a genuine check, not a restatement: the two routes use different information (the time waveform versus the coefficient list), so agreement confirms both.
10. (d) Average power of $z(t)$. The filter is ideal, so $z(t)$ keeps only the two admitted harmonics and each cosine of amplitude $a_n$ carries power $a_n^{2}/2$: $$P_z = \frac{a_4^{2} + a_5^{2}}{2} = \frac{(-0.2205)^{2} + (-0.2757)^{2}}{2} = \frac{0.04862 + 0.07600}{2}$$ $$\boxed{P_z = 0.0623\ \text{W}}$$ which is only 1.65 % of $P_y$ — almost all of the power of $y(t)$ lives in the DC term and the first two harmonics, which the filter discards.