17-Phys-B4 Signals and Communications · December 2016
Question 3 of 6: DSB and AM Modulation of a Staircase Message — Power, Efficiency, Bandwidth
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B4 Communications, National
Examination December 2016 — a three-hour closed-book examination (a standard
non-programmable, no-text-storage calculator is the only aid permitted). The cover
page states any five of the six questions constitute a complete
paper, with only the first five as they appear in the answer book marked; every
question is nonetheless answered in full below so the paper remains a complete
study resource. All six questions carry equal value.
Reference texts. A. V. Oppenheim and A. S. Willsky, Signals
and Systems, 2nd ed. (Fourier series, sampling, z-transform); S. Haykin and
M. Moher, Communication Systems, 5th ed. (AM/DSB/FM modulation, PCM,
mixers and frequency conversion); B. P. Lathi and Z. Ding, Modern Digital and
Analog Communication Systems, 4th ed. (envelope detection, Carson's rule);
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed.
(z-transform stability, partial-fraction inversion).
Question 3: DSB and AM Modulation of a Staircase Message — Power, Efficiency, Bandwidth (1/6 of paper)
Given. $m(t)$ is a period-$\tau$ staircase read off the figure:
over one period it holds $1,2,1,-1,-2,-1$ (in volts) across six equal intervals of
width $\tau/8,\tau/4,\tau/8,\tau/8,\tau/4,\tau/8$, with the two marked
zero-crossings at $t=\tau/2$ and $t=3\tau/2$ pinning down the period and phase
(the crossing steps directly from $+1$ to $-1$). $\tau=1$ ms; DSB carrier $f_c=10$ kHz (b); AM with
$\mu=0.8$ (c–f); neglect harmonics of $m(t)$ beyond the 7th (g).
Find. (a)/(c)/(f) time-domain plots; (b) DSB expression for
$P=10$; (d) AM expression for $P=10$; (e) AM power efficiency; (g) minimum
transmission band.
m(t): period τ, levels
1, 2, 1, −1, −2, −1 V over eighths of τ, matching the
printed τ/2, 3τ/2 zero-crossing ticks.
Approach. Compute $m(t)$'s DC value and mean-square (average
power) directly from its piecewise definition, then apply the standard DSB/AM
average-power formulas to solve for each carrier amplitude; the AM envelope,
efficiency and transmission band follow from the same two numbers plus the
peak value of $m(t)$ and its harmonic content.
Groundwork — DC value, mean square and peak of $m(t)$.
Weighting each level by the fraction of $\tau$ it occupies,
$$\overline{m(t)}=\tfrac18(1)+\tfrac28(2)+\tfrac18(1)+\tfrac18(-1)+\tfrac28(-2)+\tfrac18(-1)=\boxed{0}$$
(the staircase is odd/half-wave symmetric about each zero-crossing, so the DC
term vanishes exactly), and
$$E\big[m(t)^2\big]=\tfrac18(1)^2+\tfrac28(2)^2+\tfrac18(1)^2+\tfrac18(1)^2+\tfrac28(2)^2+\tfrac18(1)^2
=\boxed{2.5}$$
with peak value $m_p=\max|m(t)|=2\text{ V}$. Both numbers are confirmed by a
high-resolution numerical average over one period.
Parts (a) & (b) — DSB signal. A DSB signal has the
form $s(t)=A_c\,m(t)\cos(2\pi f_ct)$, whose average power is
$P_{DSB}=\tfrac12A_c^2E[m^2]$ (the carrier and message are independent of one
another in the time-average sense). Setting $P_{DSB}=10$ with $E[m^2]=2.5$:
$$A_c^2=\frac{10}{0.5\times2.5}=8\ \Rightarrow\ A_c=\boxed{2\sqrt2\approx2.828}$$
$$s_{DSB}(t)=\boxed{2\sqrt2\,m(t)\cos\!\big(2\pi\!\cdot\!10^4t\big)}$$
plotted to scale below (part a) — a rapidly-oscillating 10 kHz carrier
whose envelope tracks $\pm2\sqrt2\,|m(t)|$ and flips sign with $m(t)$.
DSB signal s(t) = 2√2
m(t) cos(2π·10⁴ t), plotted to scale over two periods of τ.
Parts (c) & (d) — AM signal. Standard AM writes
$s(t)=A_c\big[1+\mu\,m(t)/m_p\big]\cos(2\pi f_ct)$; with $\mu=0.8$ and $m_p=2$ the
scale factor is $k=\mu/m_p=0.4$. Its average power is
$$P_{AM}=\frac{A_c^2}{2}E\big[(1+km(t))^2\big]
=\frac{A_c^2}{2}\Big(1+2k\,\overline{m(t)}+k^2E[m^2]\Big)
=\frac{A_c^2}{2}\big(1+0.16\times2.5\big)=\frac{A_c^2}{2}(1.4)$$
using $\overline{m(t)}=0$ from the groundwork step. Setting $P_{AM}=10$:
$$A_c^2=\frac{10}{0.7}=\frac{100}{7}\ \Rightarrow\ A_c=\boxed{\frac{10}{\sqrt7}\approx3.780}$$
$$s_{AM}(t)=\boxed{\frac{10}{\sqrt7}\big[1+0.4\,m(t)\big]\cos\!\big(2\pi\!\cdot\!10^4t\big)}$$
plotted below (part c) with its envelope $A_c[1+0.4m(t)]$ overlaid.
AM signal s(t) (part c),
carrier inside the dashed envelope A_c[1+0.4 m(t)].
Part (e) — power efficiency. Efficiency is the fraction
of total power carried in the sidebands (the only part that carries information):
$$\eta=\frac{P_{sidebands}}{P_{total}}=\frac{k^2E[m^2]}{1+k^2E[m^2]}
=\frac{0.4}{1.4}=\boxed{\frac{2}{7}\approx28.6\%}$$
the remaining $5/7\approx71.4\%$ of the transmitted power is spent on the
unmodulated carrier, which carries no information but is required for simple
envelope detection.
Part (f) — envelope detector output. Since
$\mu=0.8<1$, the AM envelope $1+0.4\,m(t)$ never goes negative (its range is
$1+0.4(-2)=0.2$ up to $1+0.4(2)=1.8$, both positive), so the signal is never
over-modulated and an ideal envelope detector recovers the envelope exactly, with
no rectification distortion:
$$v_{out}(t)=\boxed{A_c\big[1+0.4\,m(t)\big]=\frac{10}{\sqrt7}\big[1+0.4\,m(t)\big]}$$
— the same staircase shape as $m(t)$, scaled and offset, ranging from about
$0.76\text{ V}$ to $6.80\text{ V}$ (a DC-blocking capacitor in a practical
receiver would strip the average and leave a scaled copy of $m(t)$ itself).
Envelope-detector output for
the AM signal of part (c): a scaled, offset copy of m(t).
Part (g) — minimum transmission band. The staircase
$m(t)$ has half-wave odd symmetry ($m(t+\tau/2)=-m(t)$, confirmed by a discrete
Fourier check that every even harmonic coefficient is numerically zero), so only
odd harmonics of $f_{m0}=1/\tau=1\text{ kHz}$ are present; neglecting harmonics
beyond the 7th caps the highest significant message frequency at
$f_{max}=7\times1\text{ kHz}=7\text{ kHz}$. An AM signal occupies $f_c\pm f_{max}$:
$$\big[f_c-f_{max},\,f_c+f_{max}\big]=\boxed{[3\text{ kHz},\ 17\text{ kHz}]}$$
