22-Elec-A3 Signals and Communications · December 2016
Question 1 of 6: Periodic Square Wave through Ideal Low-Pass and High-Pass Filters
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Professional Engineers Ontario / Engineers Canada national examination
07-Elec-A3 — Signals and Communications, December 2016. Three hours, closed book;
a standard non-programmable calculator is the only permitted aid. Six questions of equal value
(20 marks each); any five constitute a complete paper. All six are solved below, because this
set is a study resource rather than an examination script.
Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication
Systems, 4th ed. (Ch. 2–4 Fourier analysis, Ch. 4 amplitude modulation, Ch. 5 angle modulation,
Ch. 6 sampling and pulse-code modulation); A. V. Oppenheim and A. S. Willsky, Signals and
Systems, 2nd ed. (Ch. 3 Fourier series, Ch. 10 the z-transform); J. G. Proakis and D. G.
Manolakis, Digital Signal Processing, 4th ed. (Ch. 3 z-transform, Ch. 7 realisation
structures); S. Haykin and M. Moher, Communication Systems, 5th ed. (Ch. 3 amplitude
modulation, Ch. 4 angle modulation, Ch. 7 digital transmission).
Notation used throughout. The normalised sinc function is
$\operatorname{sinc}(x)=\dfrac{\sin \pi x}{\pi x}$, so that $\operatorname{sinc}(0)=1$ and the zeros
sit at the non-zero integers. Average power of a periodic signal is
$P=\frac{1}{T_0}\int_{T_0}|x(t)|^{2}\,dt$, and the exponential Fourier series is
$x(t)=\sum_{n=-\infty}^{\infty} c_n e^{j n \omega_0 t}$ with
$P=\sum_n |c_n|^{2}$ (Parseval).
Question 1: Periodic Square Wave through Ideal Low-Pass and High-Pass Filters (20 marks)
Given. A two-level periodic waveform of period $T$ that takes the value $+2$ on
$|t| \lt T/8$ and $-2$ on $T/8 \lt t \lt 7T/8$ (read directly from the printed figure: the
high plateau is centred on the origin, and the marked transition instants are $T/8$ and $7T/8$).
The high fraction of the period is therefore $d = \tfrac{T/4}{T} = \tfrac14$. For parts (b)–(e),
$T = 0.1\ \text{ms}$, so the fundamental is $f_0 = 1/T = 10\ \text{kHz}$. Filters are ideal:
a low-pass of bandwidth $25\ \text{kHz}$, and later a high-pass of cut-off $8\ \text{kHz}$.
Find. The average power of $x(t)$; the closed-form low-pass output $y(t)$ and its
average power and baseband bandwidth; and a plot of the high-pass output.
Figure Q1.1 — the periodic input x(t): +2 on |t| < T/8 and −2 on T/8 < t < 7T/8 (duty cycle 1/4, period T).
Approach. Compute the power directly from the two-level nature of the waveform,
then express $x(t)$ as a DC offset plus a scaled rectangular pulse train so its Fourier
coefficients follow from the standard $\operatorname{sinc}$ result; the ideal filters simply keep or
discard whole spectral lines.
Average power of a two-level waveform. Because $|x(t)| = 2$ at every instant,
$x^{2}(t) = 4$ identically and no integration is needed:
$$P_x=\frac{1}{T}\int_{-T/2}^{T/2}x^{2}(t)\,dt=\frac{1}{T}\int_{-T/2}^{T/2}4\,dt=\boxed{P_x = 4\ \text{W (into }1\ \Omega)}$$
This answers part (a). The duty cycle is irrelevant here — it will matter only once the signal
is decomposed into harmonics.
Decompose $x(t)$ into a DC term plus a pulse train. Write
$x(t) = -A + 2A\,p(t)$ with $A = 2$, where $p(t)$ is the unit-height pulse train that equals $1$ on
$|t| \lt T/8$ and $0$ elsewhere. This construction is exact: where $p = 1$ the signal is
$-2+4 = +2$, and where $p = 0$ it is $-2$.
Fourier coefficients of the pulse train. A unit pulse train of duty $d$ has
$c_n^{(p)} = d\operatorname{sinc}(nd)$. With $d = 1/4$, the coefficients of $x(t)$ are therefore
$$c_0 = -A + 2Ad = -2 + 1 = -1,\qquad
c_n = 2Ad\operatorname{sinc}\!\left(\frac{n}{4}\right)=\operatorname{sinc}\!\left(\frac{n}{4}\right),\ n\neq 0 .$$
Evaluating the first few, $c_1 = 0.9003$, $c_2 = 0.6366$, $c_3 = 0.3001$, $c_4 = 0$. The signal is
real and even, so every $c_n$ is real and the phase spectrum is identically zero.
Apply the ideal low-pass filter. With $f_0 = 10\ \text{kHz}$ the spectral lines
sit at $0,\ 10,\ 20,\ 30,\dots\ \text{kHz}$. A bandwidth of $25\ \text{kHz}$ passes $n = 0, 1, 2$ and
rejects $n = 3$ onwards. Combining each conjugate pair into a cosine, $2c_n\cos(n\omega_0 t)$,
$$\boxed{\,y(t) = -1 + 1.8006\cos\!\left(2\pi\times 10^{4} t\right) + 1.2732\cos\!\left(4\pi\times 10^{4} t\right)\,}$$
with $t$ in seconds. This is part (b).
Average power of the filtered signal. Parseval applied to the surviving lines
gives
$$P_y = c_0^{2} + 2c_1^{2} + 2c_2^{2} = 1 + 2(0.9003)^{2} + 2(0.6366)^{2}$$
$$\boxed{P_y = 3.4317\ \text{W}}$$
As a check, summing the full series returns $P_x = 4$ exactly, so the rejected harmonics
($n \ge 3$) carry the remaining $0.568\ \text{W}$, about $14\%$ of the input power.
Baseband bandwidth of $y(t)$. The highest surviving spectral line is the second
harmonic, so as a baseband (low-pass) signal
$$\boxed{B_y = 2f_0 = 20\ \text{kHz}}$$
Note this is set by the signal, not by the filter: the filter's $25\ \text{kHz}$ corner is wider than
the highest line it passes.
Ideal high-pass filter, cut-off 8 kHz. Every harmonic of $x(t)$ lies at
$10\ \text{kHz}$ or above and therefore passes untouched; only the DC term at $0\ \text{Hz}$ is
removed. The output is thus the input with its mean subtracted,
$$z(t) = x(t) - c_0 = x(t) + 1,$$
a square wave of the same timing that alternates between $+3$ (on $|t| \lt T/8$) and $-1$
(elsewhere). Its average power is $P_x - c_0^{2} = 3\ \text{W}$.
The two filtered waveforms are plotted below. The low-pass output is the smooth two-harmonic
approximation of the square wave; the high-pass output retains the sharp edges but has lost its
negative DC pedestal.
Figure Q1.2 — low-pass output y(t) = −1 + 1.8006 cos(ω₀t) + 1.2732 cos(2ω₀t), one and a half periods.
Figure Q1.3 — high-pass output z(t) = x(t) + 1: the DC term is removed, so the levels become +3 and −1.
