Question 3 of 6: Amplitude Modulation — Waveform, Efficiency, Spectrum, Envelope, Detector

Question 3: Amplitude Modulation — Waveform, Efficiency, Spectrum, Envelope, Detector (1/6 of paper)

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. Modulation index $\mu=0.8$; average power $P_{\text{avg}}=2$ W; message $m_n(t)$ a unit-peak triangular wave, $f_m=10$ kHz; carrier $f_c=10$ MHz.

Find. (a) $s(t)$ with $A_c$ evaluated numerically, and its plot; (b) power efficiency $\eta$; (c) spectrum lines up to the 4th message harmonic; (d) envelope $e(t)$ and its parameters; (e) an envelope-detector block diagram.

Approach. Use the standard AM power relation $P_{\text{avg}}=\tfrac{A_c^2}{2}\big[1+\mu^2\overline{m_n^2}\big]$ with the known mean-square value of a unit-peak triangular wave to solve for $A_c$; efficiency is the sideband-power fraction of the total; the spectrum follows from the triangular wave's own (odd-harmonic, $1/n^2$) Fourier series.

1. Part (a) — solving for $A_c$ and writing $s(t)$. A unit-peak triangular wave has mean-square value $\overline{m_n^2}=1/3$ (standard result: $\text{rms}=\text{peak}/\sqrt3$). So $$P_{\text{avg}}=\dfrac{A_c^2}{2}\Big(1+\mu^2\cdot\dfrac13\Big)\ \Rightarrow\ A_c=\sqrt{\dfrac{2P_{\text{avg}}}{1+\mu^2/3}}=\sqrt{\dfrac{4}{1.2133}}=\boxed{1.8157\ \text{V}}$$ $$s(t)=\boxed{1.8157\big[1+0.8\,m_n(t)\big]\cos\!\big(2\pi\times10^7t\big)\ \text{V}}$$ with $m_n(t)$ the unit-peak triangular wave at $f_m=10$ kHz. Since $\mu=0.8<1$ there is no over-modulation, so the envelope tracks $m_n(t)$ faithfully (plotted below).
2. Part (b) — power efficiency. The efficiency is the fraction of total power carried in the sidebands (the only part that conveys information): $$\eta=\dfrac{\mu^2\overline{m_n^2}}{1+\mu^2\overline{m_n^2}}=\dfrac{0.8^2/3}{1+0.8^2/3}=\boxed{17.58\%}$$ low efficiency is typical of AM with $\mu<1$: over 82% of the transmitted power is "wasted" carrying the unmodulated carrier itself.
3. Part (c) — spectrum. A triangular wave's Fourier series has only odd harmonics, $c_n=\dfrac{(-1)^{(n-1)/2}8}{\pi^2n^2}$ for odd $n$ (zero for even $n$), so $c_1=0.8106$, $c_3=-0.0901$ ($c_2=c_4=0$ automatically — "beyond the 4th" only removes $n\ge5$, which are already the next odd term $n=5$ onward). Each message harmonic $n$ produces an AM sideband pair at $f_c\pm nf_m$ with amplitude $A_c\mu c_n/2$: $$\text{carrier: }A_c=1.816\text{ V at }f_c;\quad \text{sidebands: }\dfrac{A_c\mu c_1}{2}=\boxed{0.589\text{ V at }f_c\pm f_m},\quad \dfrac{A_c\mu c_3}{2}=\boxed{-0.065\text{ V at }f_c\pm3f_m}$$
4. Part (d) — envelope. Since $\mu<1$, the envelope is simply $e(t)=A_c\big[1+\mu\,m_n(t)\big]$, a triangular wave itself (same shape as $m_n(t)$, scaled and offset), oscillating between $$e_{\min}=A_c(1-\mu)=\boxed{0.363\text{ V}},\qquad e_{\max}=A_c(1+\mu)=\boxed{3.268\text{ V}}$$ at the message rate $f_m=10$ kHz (period 100 μs).

Part (e) — envelope detector. A standard diode-plus-$RC$ envelope detector demodulates $s(t)$ directly: the diode conducts only near each carrier peak, charging $C$ to that peak; between peaks $C$ discharges through $R$. Choosing $RC$ so that $1/f_c\ll RC\ll1/f_m$ (here roughly $100\text{ ns}\ll RC\ll100\ \mu\text{s}$) lets the output track $e(t)=A_c[1+\mu m_n(t)]$ with negligible carrier ripple and no diagonal-clipping distortion (the discharge must stay fast enough to follow the envelope's fastest fall).