22-Elec-A3 Signals and Communications · December 2016
Question 3 of 6: DSB and AM Modulation of a Staircase Message
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 3: DSB and AM Modulation of a Staircase Message (20 marks)
Given. A periodic six-step staircase message read from the printed (to-scale)
figure, with $\tau = 1\ \text{ms}$:
Interval (within one period)
$m(t)$
$-\tau \le t \lt -\tau/2$
$+1$
$-\tau/2 \le t \lt +\tau/2$
$+2$
$+\tau/2 \le t \lt \tau$
$+1$
$\tau \le t \lt 3\tau/2$
$-1$
$3\tau/2 \le t \lt 5\tau/2$
$-2$
$5\tau/2 \le t \lt 3\tau$
$-1$
The period is therefore $T_0 = 4\tau = 4\ \text{ms}$, giving a fundamental
$f_0 = 250\ \text{Hz}$; the peak value is $|m|_{\max} = 2$; the carrier frequency is
$f_c = 10\ \text{kHz}$. Prescribed transmitted power is 10 W for both the DSB and the AM case, and
the AM modulation index is $\mu = 0.8$.
Find. Time-domain plots of the DSB and AM waveforms, exact expressions for both
at the prescribed power, the AM power efficiency, the envelope-detector output, and the transmitted
frequency band when harmonics above the seventh are neglected.
[Figure not reproduced: Figure Q3.1 — the staircase message m(t) redrawn from the exam figure: six steps per period, peak 2, period 4τ. See the official exam paper.]
Approach. Characterise the message by its mean, peak and mean-square power; then
apply the standard DSB and AM power formulae to fix the carrier amplitude, and use the message's
symmetry to identify which harmonics actually exist before setting the transmission band.
Message statistics. Over one period the level $\pm 2$ is held for a total of
$2\tau$ and the level $\pm 1$ for a total of $2\tau$, so
$$P_m = \frac{1}{4\tau}\left[(2)^{2}(2\tau) + (1)^{2}(2\tau)\right] = \frac{8\tau + 2\tau}{4\tau} = \boxed{P_m = 2.5}$$
The waveform is even about $t = 0$ and satisfies the half-wave odd condition
$m(t + 2\tau) = -m(t)$; consequently $\overline{m} = 0$ and only odd harmonics of
$f_0 = 250\ \text{Hz}$ are present.
Part (a) — the DSB waveform. Double-sideband suppressed-carrier modulation
forms $s_{\text{DSB}}(t) = A_c\,m(t)\cos(2\pi f_c t)$. Since $f_c = 10\ \text{kHz}$ and
$f_0 = 250\ \text{Hz}$, there are 40 carrier cycles per message period, so each staircase level
appears as a burst of constant-amplitude carrier. The envelope follows $A_c|m(t)|$ and, crucially,
the carrier undergoes a 180° phase reversal wherever $m(t)$ changes sign
(at $t = \tau$ and $t = 3\tau$ within each period). The plot below shows this, drawn with a reduced
carrier-to-message frequency ratio so the individual cycles remain visible.
Figure Q3.2 — DSB-SC waveform (part a). The dashed envelope is ±A₊|m(t)|; note the 180° phase reversal wherever m(t) changes sign. Carrier ratio reduced for clarity.
Part (b) — DSB amplitude for 10 W. The average power of a DSB signal is
$P_{\text{DSB}} = \tfrac12 A_c^{2}P_m$ (the factor $\tfrac12$ is the mean square of the carrier
cosine, valid because $f_c \gg f_0$). Setting this to 10,
$$A_c = \sqrt{\frac{2P_{\text{DSB}}}{P_m}} = \sqrt{\frac{2(10)}{2.5}} = \sqrt{8}$$
$$\boxed{\,s_{\text{DSB}}(t) = 2\sqrt{2}\; m(t)\cos\!\left(2\pi \times 10^{4}\,t\right) = 2.8284\,m(t)\cos\!\left(2\pi \times 10^{4}\,t\right)}$$
Part (c) — AM waveform with $\mu = 0.8$. Standard AM is
$s_{\text{AM}}(t) = A_c\left[1 + k\,m(t)\right]\cos(2\pi f_c t)$ where the modulation index is
$\mu = k|m|_{\max}$. Hence $k = \mu/|m|_{\max} = 0.8/2 = 0.4$, and the normalised envelope
$1 + 0.4\,m(t)$ takes the four values
$$1.8,\quad 1.4,\quad 0.6,\quad 0.2$$
for $m = +2, +1, -1, -2$ respectively. Because the smallest of these is $0.2 \gt 0$, the envelope
never crosses zero and there is no phase reversal — the visual signature that separates
AM from the DSB plot above.
Figure Q3.3 — AM waveform with μ = 0.8 (part c). The envelope never touches zero, so there is no phase reversal. Carrier ratio reduced for clarity.
Part (d) — AM amplitude for 10 W. With $\overline{m} = 0$ the AM power
splits cleanly into carrier and sideband parts:
$$P_{\text{AM}} = \frac{A_c^{2}}{2}\left(1 + k^{2}P_m\right) = \frac{A_c^{2}}{2}\left[1 + (0.4)^{2}(2.5)\right] = \frac{A_c^{2}}{2}(1.4)$$
Setting $P_{\text{AM}} = 10$ gives $A_c^{2} = 100/7$, so
$$\boxed{\,s_{\text{AM}}(t) = 3.7796\left[1 + 0.4\,m(t)\right]\cos\!\left(2\pi \times 10^{4}\,t\right)}$$
Substituting back reproduces $10.000\ \text{W}$, and a direct numerical integration of the shipped
waveform over one message period agrees to three decimals.
Part (e) — power efficiency. Efficiency is the fraction of transmitted power
carried by the information-bearing sidebands:
$$\eta = \frac{k^{2}P_m}{1 + k^{2}P_m} = \frac{0.16 \times 2.5}{1 + 0.16 \times 2.5} = \frac{0.4}{1.4}$$
$$\boxed{\eta = 0.2857 = 28.57\%}$$
The remaining $71.43\%$ (about $7.14\ \text{W}$) sits in the carrier line, which conveys no
information — the price paid for allowing a one-transistor envelope detector at the receiver.
Note that this figure is independent of $A_c$: it depends only on $\mu$ and on the message's
shape through $P_m/|m|_{\max}^{2}$.
Part (f) — envelope-detector output. An ideal envelope detector outputs
$e(t) = A_c\left|1 + k\,m(t)\right|$. Since $1 + km \ge 0.2 \gt 0$ everywhere, the absolute value is
inactive and the recovery is distortionless:
$$e(t) = 3.7796\left[1 + 0.4\,m(t)\right]$$
which is a staircase of the same shape as $m(t)$, riding on a DC pedestal of $3.7796\ \text{V}$ and
spanning $0.756\ \text{V}$ to $6.803\ \text{V}$. After a series capacitor removes the pedestal the
output is $1.5119\,m(t)$ — the message, scaled.
Figure Q3.4 — envelope-detector output (part f): a faithful staircase copy of m(t) on a DC pedestal, spanning 0.756 V to 6.803 V.
Part (g) — transmission band. The message has only odd harmonics of
$f_0 = 250\ \text{Hz}$, so neglecting everything beyond the seventh leaves lines at
$250,\ 750,\ 1250$ and $1750\ \text{Hz}$; the message bandwidth is $W = 7f_0 = 1.75\ \text{kHz}$.
Standard AM is a double-sideband scheme, so the transmitted band extends symmetrically about the
carrier:
$$f_{\min} = f_c - W = 10\,000 - 1750,\qquad f_{\max} = f_c + W = 10\,000 + 1750$$
$$\boxed{\,8.25\ \text{kHz} \le f \le 11.75\ \text{kHz},\quad B_T = 3.5\ \text{kHz}}$$
