22-Elec-A3 Signals and Communications · December 2015

Question 3 of 6: Amplitude Modulation — Waveform, Efficiency, Spectrum and Detection

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: 07-Elec-A3 Signals and Communications, December 2015 — 3 hours, closed book, a standard non-programmable calculator permitted. Six questions of equal value (20 marks each); any five constitute a complete paper, and only the first five presented are marked. All six are solved here, since the set is intended as a study resource.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (AM/FM, PCM, frequency conversion); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Fourier series, LTI systems); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (z-transform, difference equations, stability); S. Haykin, Communication Systems, 5th ed. (modulation and detection).

Question 3: Amplitude Modulation — Waveform, Efficiency, Spectrum and Detection (20 marks)

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.

Quantity	Symbol	Value
Modulation index	$a$	0.8
Total average transmitted power	$P_{T}$	2 W
Message waveform	$m_{n}(t)$	triangular, unit peak, zero mean
Message frequency	$f_{m}$	10 kHz
Carrier frequency	$f_{c}$	10 MHz
Mean square of a unit-peak triangle	$\overline{m_{n}^{2}}$	1/3

Find. (a) the complete time-domain expression for $s(t)$ with all constants evaluated; (b) the power efficiency; (c) the line spectrum out to the 4th message harmonic; (d) the envelope and its parameters; (e) an envelope-detector block diagram.

Approach. Standard AM with carrier is $s(t) = A_{c}[1 + a\,m_{n}(t)]\cos(2\pi f_{c}t)$ where $m_{n}$ is normalised to unit peak. The total power splits into a carrier term and a sideband term; setting that total to 2 W fixes $A_{c}$, after which efficiency, spectrum and envelope all follow directly.

Write the power balance. For a zero-mean normalised message, $$P_{T} = \frac{A_{c}^{2}}{2}\left[1 + a^{2}\,\overline{m_{n}^{2}}\right]$$ the first term inside the bracket being the carrier and the second the sidebands.
Evaluate the mean square of the triangular message. A symmetric triangular wave of unit peak has $\overline{m_{n}^{2}} = 1/3$ (compare $1/2$ for a sinusoid). Hence $$1 + a^{2}\overline{m_{n}^{2}} = 1 + \frac{0.64}{3} = 1.21333$$
Solve for the carrier amplitude. $$A_{c}^{2} = \frac{2P_{T}}{1.21333} = \frac{4}{1.21333} = 3.29670 \quad\Longrightarrow\quad A_{c} = 1.8157\ \text{V}$$
State the time-domain signal. With the triangular message expanded as a cosine series of unit peak, $$\boxed{\,s(t) = 1.8157\left[1 + 0.8\,m_{n}(t)\right]\cos\!\left(2\pi \times 10^{7}\,t\right)\ \text{V}\,}$$ $$m_{n}(t) = \frac{8}{\pi^{2}}\sum_{k\ \text{odd}} \frac{1}{k^{2}}\cos\!\left(2\pi k \times 10^{4}\,t\right) = 0.8106\cos(2\pi f_{m}t) + 0.0901\cos(6\pi f_{m}t) + \cdots$$

Figure 3.1 — The AM waveform over two message periods, with the envelope drawn in red. The true carrier-to-message frequency ratio is 1000:1; the carrier is drawn at a much lower ratio so the individual cycles remain visible.

Compute the power efficiency. Efficiency is the fraction of transmitted power carrying information, i.e. the sideband power over the total: $$\eta = \frac{a^{2}\overline{m_{n}^{2}}}{1 + a^{2}\overline{m_{n}^{2}}} = \frac{0.21333}{1.21333} = 0.17582 \quad\Longrightarrow\quad \boxed{\,\eta = 17.58\%\,}$$ In absolute terms the sidebands carry $0.3516$ W and the carrier wastes $1.6484$ W. Even at a modulation index of 0.8, more than four-fifths of the transmitter power goes into a carrier that conveys nothing — the standing objection to AM with carrier, and the reason DSB-SC and SSB exist.
Build the spectrum. Multiplying the message series by the carrier translates each message harmonic to a pair of lines at $f_{c} \pm kf_{m}$ with one-sided amplitude $A_{c}aA_{k}/2$, alongside the carrier line of amplitude $A_{c}$: $$A_{c} = 1.8157\ \text{V};\qquad \frac{A_{c}aA_{1}}{2} = \frac{1.8157 \times 0.8 \times 0.8106}{2} = 0.5887\ \text{V};\qquad \frac{A_{c}aA_{3}}{2} = 0.0654\ \text{V}$$ A triangular wave contains odd harmonics only, so within the stated 4-harmonic limit there are lines at $f_{c}$, $f_{c} \pm 10$ kHz and $f_{c} \pm 30$ kHz, and none at $f_{c} \pm 20$ kHz or $f_{c} \pm 40$ kHz. The occupied bandwidth is $2 \times 3f_{m} = 60$ kHz, spanning 9.97 MHz to 10.03 MHz.

Figure 3.2 — One-sided line spectrum of the AM signal. The dominant line is the carrier, which carries no information; the missing even-order sidebands are a direct signature of the triangular message.

Describe the envelope. Since $a = 0.8 \lt 1$, the bracket $1 + a\,m_{n}(t)$ never changes sign and the envelope is simply $$E(t) = A_{c}\left[1 + 0.8\,m_{n}(t)\right]$$ a triangular wave of the same 10 kHz frequency (period 100 µs), oscillating between $$E_{\max} = A_{c}(1 + a) = 3.2682\ \text{V}, \qquad E_{\min} = A_{c}(1 - a) = 0.3631\ \text{V}$$ with mean level $A_{c} = 1.8157$ V. As a consistency check, $(E_{\max} - E_{\min})/(E_{\max} + E_{\min}) = 0.8 = a$, which is exactly how the modulation index is measured on an oscilloscope.

Figure 3.3 — The envelope alone. Because $E_{\min} \gt 0$, the envelope is an undistorted replica of the message riding on a DC pedestal.

Part (e) asks for the detector. The envelope never collapses to zero, which is the precondition for the simplest possible receiver: a diode that follows the positive peaks, an RC network that holds the charge between carrier cycles but discharges fast enough to track the message, and a series capacitor that strips the DC pedestal.

Figure 3.4 — Envelope detector. The RC time constant must satisfy $1/f_{c} \ll RC \ll 1/f_{m}$; with $f_{c} = 10$ MHz and $f_{m} = 10$ kHz, a value near 1–10 µs sits comfortably between the two limits.

Quantity	Result
Carrier amplitude	$A_{c} = 1.8157$ V
Time-domain signal	$s(t) = 1.8157[1 + 0.8m_{n}(t)]\cos(2\pi\times10^{7}t)$ V
Power efficiency	$\eta = 17.58\%$ (sidebands 0.3516 W, carrier 1.6484 W)
Spectrum lines	Carrier 1.8157 V; $f_{c}\pm f_{m}$: 0.5887 V; $f_{c}\pm3f_{m}$: 0.0654 V; no even-order sidebands
Occupied bandwidth	60 kHz (9.97–10.03 MHz)
Envelope	Triangular, 10 kHz; $E_{\max} = 3.2682$ V, $E_{\min} = 0.3631$ V, mean 1.8157 V
Detector	Diode → RC low-pass → DC block → amplifier

Topic: Conventional amplitude modulation: power budget, efficiency, line spectrum of a non-sinusoidal message, and envelope detection.

Key relations: $s(t) = A_{c}[1 + a\,m_{n}(t)]\cos\omega_{c}t$; $P_{T} = \tfrac{A_{c}^{2}}{2}\left[1 + a^{2}\overline{m_{n}^{2}}\right]$; $\eta = a^{2}\overline{m_{n}^{2}}/(1 + a^{2}\overline{m_{n}^{2}})$; triangular wave $\overline{m_{n}^{2}} = 1/3$ with $A_{k} = 8/(\pi^{2}k^{2})$ for odd $k$; $E_{\max,\min} = A_{c}(1 \pm a)$; detector constraint $1/f_{c} \ll RC \ll 1/f_{m}$.

Why this works: The carrier and the sidebands are orthogonal over any long observation, so their powers add without cross terms — that is what makes the single bracket $\left[1 + a^{2}\overline{m_{n}^{2}}\right]$ exact rather than approximate, and it is why efficiency depends on the message shape through $\overline{m_{n}^{2}}$ and not merely on the index. A triangular message is markedly less efficient than a sinusoid at the same index because its mean square is $1/3$ rather than $1/2$: the waveform spends most of its time away from its peaks. Envelope detection works only because $a \lt 1$ keeps $1 + a\,m_{n}(t)$ positive; the diode then recovers $|s(t)|$'s upper boundary, which is the message plus a constant.

Common pitfall: Using $\overline{m_{n}^{2}} = 1/2$ out of habit — correct for a sinusoid, wrong here — inflates both $A_{c}$ and the efficiency. The second trap is drawing sidebands at every multiple of $f_{m}$: the triangular wave has no even harmonics, so the spectrum has visible gaps at $f_{c} \pm 20$ kHz. Third, "modulation index 0.8" refers to the normalised message; forgetting to normalise the message to unit peak before applying $a$ corrupts every subsequent number.

Source: Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., Ch. 4 (amplitude modulation, power efficiency, envelope detection); Haykin, Communication Systems, 5th ed., Ch. 3.

← Question 2 Next: Question 4 →