22-Elec-A3 Signals and Communications · May 2018

Question 1 of 5: Fourier Series of a Rectangular Pulse Train — Harmonic Amplitude, Power and Duty-Cycle Optimisation

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

Question 1: Fourier Series of a Rectangular Pulse Train — Harmonic Amplitude, Power and Duty-Cycle Optimisation (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 third-harmonic term of the Fourier series at $d = 0.25$; the average power of the whole pulse train; the duty cycle that maximises the third-harmonic amplitude; and the average power carried by the third harmonic at that optimum.

Approach. Expand the pulse train as a trigonometric Fourier series — whose coefficients are the familiar $\operatorname{sinc}$ samples of the single-pulse spectrum — read off the $n = 3$ term, obtain the total power directly from the definition of mean square, then maximise the third-harmonic coefficient analytically over $d$ and convert that amplitude into power.

1. Write the Fourier series of the pulse train. Take the pulse centred on $t = 0$ so the waveform is even and only cosines appear. With $\tau = dT$ the exponential coefficients are $$c_n = \frac{1}{T}\int_{-\tau/2}^{\tau/2} A\,e^{-j2\pi n t/T}\,dt = \frac{A\tau}{T}\cdot\frac{\sin(\pi n \tau/T)}{\pi n \tau/T} = A\,d\;\operatorname{sinc}(nd),$$ where $\operatorname{sinc}(x) \equiv \sin(\pi x)/(\pi x)$. Because $c_{-n} = c_n$ here, the one-sided (trigonometric) series is $$x(t) = A d + \sum_{n=1}^{\infty} a_n \cos\!\left(\frac{2\pi n t}{T}\right), \qquad a_n = 2 A d \operatorname{sinc}(nd).$$ The DC term $Ad$ is simply the time average of the waveform, as it must be.
2. Extract the third harmonic and evaluate it at $d = 0.25$. Setting $n = 3$, $$a_3 = 2 A d \operatorname{sinc}(3d) = \frac{2A\sin(3\pi d)}{3\pi},$$ which is the general expression asked for "in terms of the signal parameters". Substituting $d = 0.25$ gives $\sin(0.75\pi) = 0.70711$ and $3\pi = 9.42478$, so $a_3 = 2A(0.70711)/9.42478 = 0.15005\,A$. The third harmonic therefore sits at $3/T$ hertz with amplitude $$\boxed{\,x_3(t) = 0.1501\,A\,\cos\!\left(\frac{6\pi t}{T}\right)\ \text{V}\,}$$ (with the pulse centred on the origin; if the pulse instead starts at $t=0$ the same amplitude appears with a phase lag of $3\pi d = 135^\circ$, which does not change any of the numbers below).
3. Compute the average power of the whole pulse train. Power is the mean square, and the waveform is $A$ for a fraction $d$ of every period and zero for the rest, so the integral collapses to a single rectangle: $$P_x = \frac{1}{T}\int_{0}^{T} x^2(t)\,dt = \frac{1}{T}\left(A^2 \cdot dT\right) = A^2 d.$$ With $d = 0.25$, $$\boxed{\,P_x = 0.25\,A^{2}\ \text{W (into 1 }\Omega)\,}$$ As a check, Parseval's theorem gives the same figure from the spectrum: $P_x = c_0^2 + 2\sum_{n\ge1}|c_n|^2 = A^2d^2\left[1 + 2\sum \operatorname{sinc}^2(nd)\right]$, which converges to $A^2 d$ — a useful reminder that the sinc-squared series carries all the power the time domain says it must.
4. Maximise the third-harmonic amplitude over duty cycle. From step 2 the amplitude depends on $d$ only through $\sin(3\pi d)$: $$|a_3(d)| = \frac{2A}{3\pi}\,\bigl|\sin(3\pi d)\bigr| .$$ Note that the $d$ multiplying the sinc has cancelled against the $d$ in its denominator, so this is a pure sine in $d$ — the maximum is reached whenever $|\sin(3\pi d)| = 1$, i.e. $3\pi d = \pi/2 + k\pi$: $$d = \tfrac{1}{6},\ \tfrac{1}{2},\ \tfrac{5}{6}.$$ All three give the same magnitude, $$\boxed{\,d = \tfrac{1}{6} = 16.67\%\ \Rightarrow\ |a_3|_{\max} = \frac{2A}{3\pi} = 0.2122\,A\,}$$ The smallest of the three, $d = 1/6$, is the natural design answer because it is the one that also keeps the pulse narrow (the $d = 1/2$ square wave and the $d = 5/6$ nearly-DC waveform give the same third-harmonic amplitude but very different DC and low-order content). Compared with the given 25% duty cycle the improvement is $0.2122/0.1501 = 1.414$, i.e. 3.01 dB.
5. Convert that amplitude into average power. A real sinusoid of amplitude $C$ carries mean-square value $C^2/2$, so the power in the third harmonic at the optimum duty cycle is $$P_3 = \frac{|a_3|_{\max}^2}{2} = \frac{1}{2}\left(\frac{2A}{3\pi}\right)^{2} = \frac{2A^{2}}{9\pi^{2}}.$$ Evaluating, $9\pi^2 = 88.826$, hence $$\boxed{\,P_3 = 0.02252\,A^{2}\ \text{W}\,}$$ At $d = 1/6$ the total power is $A^2/6 = 0.1667A^2$, so the third harmonic carries 13.5% of the signal's power — a large share for a third harmonic, which is precisely why $d = 1/6$ is the classic choice when a pulse train is used to drive a tripler.