Periodic Pulse Train Through Ideal Low-Pass and High-Pass Filters: 17-Phys-B4 Signals and Communications, December 2016, Q1

Question 1: Periodic Pulse Train Through Ideal Low-Pass and High-Pass Filters (1/6 of paper)

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. A period-$T$ square wave read off the figure: $x(t)=+2\text{ V}$ for $|t|<T/8$ (mod $T$) and $x(t)=-2\text{ V}$ for $T/8<|t|<T/2$ (mod $T$) — a duty-cycle-$1/4$ pulse centred on each multiple of $T$, confirmed by the printed $T/8$ and $7T/8$ tick marks bracketing the high pulse. Ideal LPF cut-off 25 kHz for parts (a)–(d); ideal HPF cut-off 8 kHz for part (e); $T=0.1$ ms for parts (b)–(e).

Find. (a) average power of $x(t)$; (b) LPF output $y(t)$; (c) average power of $y(t)$; (d) bandwidth of $y(t)$ as a baseband signal; (e) HPF output (plot).

x(t): a ±2 V square wave, duty cycle 1/4, high pulse centred at each multiple of $T$.

Approach. $x(t)$ is a fixed-amplitude two-level wave so part (a) is immediate from the definition of average power; write $x(t)$'s trigonometric Fourier series (period $T$, duty cycle $1/4$), keep only the harmonics an ideal filter of the stated cut-off would pass, and use Parseval's theorem for the power questions.

Part (a) — average power of $x(t)$. $x(t)=\pm2\text{ V}$ at every instant, so $x(t)^2=4$ identically and $$P_x=\frac{1}{T}\int_0^T x(t)^2\,dt=\boxed{4\text{ W}}$$ independent of the duty cycle — only the two-level amplitude matters.
Setting up the Fourier series. Write $x(t)=4p(t)-2$, where $p(t)$ is a unit-height rectangular pulse train of duty cycle $d=1/4$ centred at $t=0$ (so $p=1$ on $x$'s high interval, $p=0$ on its low interval, and $4(1)-2=2$, $4(0)-2=-2$ recovers $x(t)$). The standard pulse-train coefficients are $c_n=d\,\text{sinc}(nd)$ with $\text{sinc}(u)=\sin(\pi u)/(\pi u)$, giving the trigonometric series $$x(t)=a_0+\sum_{n=1}^\infty a_n\cos\!\Big(\frac{2\pi n t}{T}\Big),\qquad a_0=4c_0-2=\boxed{-1},\qquad a_n=8c_n=\frac{8\sin(n\pi/4)}{n\pi}$$ Evaluating: $a_1=\dfrac{4\sqrt2}{\pi}\approx1.801$, $a_2=\dfrac{4}{\pi}\approx1.273$, $a_3=\dfrac{4\sqrt2}{3\pi}\approx0.600$, $a_4=0$ — each checked against a direct numerical Fourier integral of $x(t)$.
Part (b) — $y(t)$ for $T=0.1$ ms. The fundamental is $f_0=1/T=10\text{ kHz}$, so the harmonics sit at $0,10,20,30,\dots$ kHz. A 25 kHz ideal LPF passes $n=0,1,2$ (up to 20 kHz) and blocks $n\ge3$ (30 kHz and above), so $$y(t)=\boxed{-1+\frac{4\sqrt2}{\pi}\cos\!\big(2\pi\!\cdot\!10^4t\big) +\frac{4}{\pi}\cos\!\big(2\pi\!\cdot\!2\!\cdot\!10^4t\big)}$$

y(t): the DC + first + second harmonic of x(t) that a 25 kHz ideal LPF retains (T = 0.1 ms).

Part (c) — average power of $y(t)$. By Parseval's theorem, sinusoids at distinct nonzero frequencies are orthogonal over a common period, so the average power is just the sum of each term's own power: $$P_y=a_0^2+\frac{a_1^2}{2}+\frac{a_2^2}{2} =1+\frac{16}{\pi^2}+\frac{8}{\pi^2}=\boxed{1+\frac{24}{\pi^2}\approx3.432\text{ W}}$$ confirmed by numerically averaging $y(t)^2$ over one period.
Part (d) — bandwidth of $y(t)$. $y(t)$ is a real baseband signal containing only DC, 10 kHz and 20 kHz components, so its highest-frequency content sets the bandwidth: $$\boxed{BW_y=20\text{ kHz}}$$
Part (e) — HPF output. With $T=0.1$ ms the fundamental $f_0=10\text{ kHz}$ already exceeds the 8 kHz HPF cut-off, so every harmonic $n\ge1$ passes and only the DC term ($n=0$) is removed: $$y_{HP}(t)=x(t)-a_0=\boxed{x(t)+1}$$ i.e. the same duty-cycle-$1/4$ square wave, shifted up by 1 V so its two levels become $+3\text{ V}$ and $-1\text{ V}$.

HPF output = x(t) + 1: same square wave, DC removed, levels now +3 V / −1 V.

Question 1 — final results
Quantity	Result
Average power of $x(t)$	$4\text{ W}$
$y(t)$, $T=0.1$ ms	$-1+\dfrac{4\sqrt2}{\pi}\cos(2\pi\cdot10^4t)+\dfrac{4}{\pi}\cos(2\pi\cdot2\cdot10^4t)$
Average power of $y(t)$	$1+24/\pi^2\approx3.432\text{ W}$
Bandwidth of $y(t)$	$20\text{ kHz}$
HPF output	$x(t)+1$ (levels $+3\text{ V}/-1\text{ V}$)

Concept & Theory

Topic. Trigonometric Fourier series of a periodic pulse train and ideal brick-wall filtering.

Key relations. Rectangular pulse-train coefficients $c_n=d\,\text{sinc}(nd)$; trigonometric form $a_0=c_0,\ a_n=2c_n$ (scaled here by the $4p(t)-2$ mapping); Parseval average power $P=a_0^2+\sum_n a_n^2/2$; an ideal filter simply zeroes the Fourier coefficients outside its passband.

Why this works. Once a periodic signal is expressed as a sum of harmonically-related cosines, an ideal LPF/HPF acts purely as bookkeeping — it keeps or discards whole terms according to each term's frequency, with no distortion of the terms it keeps; orthogonality of those terms is exactly what lets Parseval's theorem turn a power integral into a simple sum of squares.

Common pitfall. Forgetting that the DC term ($a_0^2$, not $a_0^2/2$) contributes fully to average power while every AC term only contributes half its amplitude squared; and assuming a high-pass filter removes "high-frequency energy of $x(t)$" rather than checking numerically that, because the fundamental (10 kHz) already clears the 8 kHz cut-off, only the DC term is actually removed.

Source. A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed., §3.3–3.6 (Fourier series of periodic signals, filtering); S. Haykin and M. Moher, Communication Systems, 5th ed., Ch. 2 (power spectral density and ideal filtering).

Question 1 of 6: Periodic Pulse Train Through Ideal Low-Pass and High-Pass Filters

Question 1: Periodic Pulse Train Through Ideal Low-Pass and High-Pass Filters (1/6 of paper)