Question 3 of 5: DSB and AM Modulation of a Two-Tone 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
16-Elec-A3 — Signals and Communications, May 2017. Closed book (one non-programmable
calculator permitted), 3 hours, 4 pages. Five questions, all compulsory, all of equal value
(20 marks each). All five are solved in full below.
Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication
Systems, 5th ed. (AM/DSB power and efficiency, PCM, FM/Carson); B. P. Lathi, Linear Systems and
Signals, 2nd ed. (Fourier series, filtering, average power); A. V. Oppenheim & A. S. Willsky,
Signals and Systems, 2nd ed. (LTI convolution, BIBO stability); J. G. Proakis & D. G. Manolakis,
Digital Signal Processing, 4th ed. (FIR structures, transfer functions); C. K. Alexander &
M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (first-order RC response).
Question 3: DSB and AM Modulation of a Two-Tone Message (20 marks)
Find. Whether m(t) is periodic and its fundamental period; the scaled DSB-SC signal; the
scaled AM signal and its plot; the AM power efficiency; the envelope-detector output; and the transmission
band.
Approach. Note that $\cos(3\pi f_m t)=\cos(2\pi(1.5f_m)t)$, so the message is a two-tone
signal at 1 kHz and 1.5 kHz. Periodicity follows from the ratio of those frequencies; the modulator
amplitudes follow from the standard power formulas $P_{DSB}=A_c^{2}P_m/2$ and
$P_{AM}=(A_c^{2}/2)(1+k_a^{2}P_m)$.
(a) Periodicity and fundamental period. Rewriting the second term,
$\tfrac12\cos(3\pi f_m t)=\tfrac12\cos\!\left(2\pi(1.5f_m)t\right)$, the two tones are at
$f_a=1$ kHz and $f_b=1.5$ kHz. Their ratio $f_b/f_a=3/2$ is rational, so the sum is periodic,
with fundamental frequency equal to the greatest common divisor of the two:
$$f_{0}=\gcd(1000,\,1500)=500\ \text{Hz}\quad\Longrightarrow\quad \boxed{T_0=\frac{1}{500}=2\ \text{ms}}$$
Equivalently the tones are the 2nd and 3rd harmonics of a 500 Hz fundamental. Over one period the waveform
peaks at $m=1.5$ V at $t=0$ (both cosines in phase) and dips to $-1.2199$ V near $t=0.42$ ms — note the
asymmetry, which will matter in part (c).
Figure Q3.1 - Message m(t) over one fundamental period (2 ms).
(b) DSB-SC signal scaled to 10 W. First the message power, obtained by adding the
powers of the two orthogonal tones:
$$P_m=\frac{1^{2}}{2}+\frac{(1/2)^{2}}{2}=0.5+0.125=0.625\ \text{W}$$
A DSB-SC signal is $s(t)=A_c\,m(t)\cos 2\pi f_c t$, whose average power is $A_c^{2}P_m/2$ because the
carrier multiplication halves the message power. Setting this to 10 W,
$$\frac{A_c^{2}(0.625)}{2}=10\ \Longrightarrow\ A_c^{2}=32\ \Longrightarrow\ A_c=4\sqrt2=5.657$$
$$\boxed{s_{DSB}(t)=4\sqrt{2}\;m(t)\cos\!\left(2\pi\times10^{4}\,t\right)\ \text{V}}$$
(c) AM signal scaled to 10 W. A conventional AM signal is
$s(t)=A_c\left[1+k_a m(t)\right]\cos 2\pi f_c t$ and the modulation index is
$\mu=k_a\,|m(t)|_{\max}$. The message peaks at $m_{\max}=1.5$ V, so
$$k_a=\frac{\mu}{|m|_{\max}}=\frac{0.8}{1.5}=0.5333\ \text{V}^{-1}$$
The envelope $1+k_a m(t)$ then ranges from $1+0.5333(-1.2199)=0.349$ up to $1.8$; it never reaches zero, so
the signal is not over-modulated and the envelope is a faithful copy of the message. Total
AM power splits into carrier plus sidebands:
$$P_{AM}=\frac{A_c^{2}}{2}\left(1+k_a^{2}P_m\right)
=\frac{A_c^{2}}{2}\left(1+0.5333^{2}\times0.625\right)=\frac{A_c^{2}}{2}(1.17778)=10$$
$$A_c^{2}=16.981\ \Longrightarrow\ \boxed{A_c=4.121\ \text{V}},\qquad
s_{AM}(t)=4.121\left[1+0.5333\,m(t)\right]\cos\!\left(2\pi\times10^{4}\,t\right)$$
The plot below shows the modulated carrier with its envelope superimposed; the carrier completes 20 cycles
in one 2 ms message period, and the envelope never crosses zero.
Figure Q3.2 - AM signal with its envelope (red).
(d) Power efficiency. Efficiency is the fraction of transmitted power that carries
information, i.e. resides in the sidebands:
$$\eta=\frac{k_a^{2}P_m}{1+k_a^{2}P_m}=\frac{0.28444\times0.625}{1+0.28444\times0.625}
=\frac{0.17778}{1.17778}$$
$$\boxed{\eta = 0.1509 = 15.09\ \%}$$
Of the 10 W transmitted, 1.509 W is useful sideband power and 8.49 W is spent on the carrier, which conveys
no information. This is the price paid for being able to demodulate with a diode envelope detector instead
of a coherent receiver — DSB-SC, by contrast, is 100 % efficient but needs carrier recovery.
(e) Envelope-detector output. An ideal envelope detector follows the positive envelope
of its input. Because the envelope never goes negative, the output is
$$\boxed{v(t)=A_c\left[1+k_a m(t)\right]=4.121+2.198\,m(t)\ \text{V}}$$
a scaled replica of $m(t)$ riding on a 4.121 V dc pedestal, swinging between 1.440 V and 7.418 V. After the
usual series capacitor (dc block) the recovered message is $2.198\,m(t)$, undistorted. Had $\mu$ exceeded
1.23 (the value that would drive $1+k_am(t)$ negative at the message minimum), the detector would have
rectified the envelope and produced clipping distortion.
Figure Q3.3 - Envelope-detector output, a scaled replica of m(t) on a dc pedestal.
(f) Transmission band. AM is a double-sideband scheme, so the spectrum consists of the
carrier line at 10 kHz plus tone pairs at $f_c\pm 1$ kHz and $f_c\pm 1.5$ kHz. The highest message frequency
is $W=1.5$ kHz, hence
$$B_T = 2W = 2(1.5\ \text{kHz}) = \boxed{3\ \text{kHz}},\qquad f_{\text{centre}}=\boxed{10\ \text{kHz}}$$
so the minimum band is 8.5 kHz to 11.5 kHz. The band is set by the 1.5 kHz tone, not by the
nominal 1 kHz "$f_m$" — a distinction the question plants deliberately by writing the second tone as
$\cos(3\pi f_m t)$.
Figure Q3.4 - One-sided line spectrum of the AM signal.
