Question 6 of 10: Standard AM — Recovering the Message, Modulation Index and Power Efficiency

Question 6: Standard AM — Recovering the Message, Modulation Index and Power Efficiency (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. $\varphi_{AM}(t)=[A_c+m(t)]\cos2\pi f_ct$; one-sided line spectrum $\Phi_{AM}(f)$: 46/48/50/52/54 kHz at 1/4, 1/2, 2, 1/2, 1/4.

Find. (a) $f_c,A_c,m(t)$; (b) $\mu$; (c) $P_s,P_c,P_{\varphi_{AM}},\eta$; (d) $K_{max}$ for envelope detectability, and $\eta$ at that $K$.

Approach. Read the carrier and message tones directly off the line spectrum (the tallest line is the carrier, symmetric sideband pairs are the message tones), then use the standard AM power/efficiency formulas and the envelope-detector non-negativity condition $A_c+Km(t)\ge0\ \forall t$.

1. Part (a) — f_c, A_c, m(t). The tallest (centre) line at 50 kHz has height $A_c/2=2\Rightarrow A_c=4$ V, so $f_c=\boxed{50\ \text{kHz}}$, $A_c=\boxed{4\ \text{V}}$. Sidebands at $f_c\pm2$ kHz (48, 52 kHz) have height $B_1/4=1/2\Rightarrow B_1=2$ V at $f_1=2$ kHz; sidebands at $f_c\pm4$ kHz (46, 54 kHz) have height $B_2/4=1/4\Rightarrow B_2=1$ V at $f_2=4$ kHz. So $$m(t)=\boxed{2\cos(2\pi\cdot2000t)+\cos(2\pi\cdot4000t)\ \text{V}}.$$
2. Part (b) — modulation index. $\mu=\max|m(t)|/A_c$. Since $f_2=2f_1$ (harmonically related), both cosines reach their individual peak of $+1$ simultaneously at $t=0$, so $\max m(t)=B_1+B_2=3$ V: $$\mu=\frac{3}{4}=\boxed{0.75}.$$ Note for part (d) that this message is asymmetric. Writing $u=\cos(2\pi\cdot2000t)$, $m=2u+\cos(2\cdot2\pi\cdot2000t)=2u+2u^2-1$, a parabola in $u\in[-1,1]$ whose vertex sits at $u=-\tfrac12$. Its extremes are therefore $$\max m=3\ \text{V (at }u=1),\qquad \min m=-1.5\ \text{V (at }u=-\tfrac12),$$ both confirmed over a full period. The positive and negative peaks differ by a factor of two, which is what makes part (d) interesting.
3. Part (c) — powers and efficiency. With unit load, $$P_c=\frac{A_c^2}{2}=\frac{16}{2}=\boxed{8\ \text{W}}.$$ Expanding $\varphi_{AM}(t)$, the two message tones each split into an upper and lower sideband of amplitude $B_i/2$: $$P_s=2\times\frac{(B_1/2)^2}{2}+2\times\frac{(B_2/2)^2}{2}=2\Big(\frac{1^2}2\Big)+2\Big(\frac{0.5^2}2\Big)=1+0.25=\boxed{1.25\ \text{W}},$$ $$P_{\varphi_{AM}}=P_c+P_s=8+1.25=\boxed{9.25\ \text{W}},\quad \eta=\frac{P_s}{P_{\varphi_{AM}}}=\frac{1.25}{9.25}=\boxed{13.51\%}.$$
4. Part (d) — K_max and revised efficiency. An envelope detector recovers $m(t)$ faithfully precisely when the envelope never goes negative, i.e. $A_c+Km(t)\ge0$ for every $t$. Only the most negative excursion of $m(t)$ can violate that, so the binding constraint is $A_c+K\,(\min m)\ge0$, i.e. $K\le A_c/|\min m(t)|$. From part (b), $\min m(t)=-1.5$ V (not $-3$ V — the message is asymmetric), so $$K_{max}=\frac{A_c}{|\min m(t)|}=\frac{4}{1.5}=\boxed{\frac83\approx2.667}.$$ Sideband power scales as $K^2$ (message amplitudes scale linearly, power quadratically): $$P_s'=K_{max}^2P_s=\Big(\frac{64}{9}\Big)(1.25)=\boxed{8.889\ \text{W}},\quad P_{\varphi_{AM}}'=P_c+P_s'=8+8.889=\boxed{16.889\ \text{W}},$$ $$\eta'=\frac{P_s'}{P_{\varphi_{AM}}'}=\frac{8.889}{16.889}=\boxed{52.63\%}.$$ This is far HIGHER than part (c)'s 13.51%: at $K=1$ the message used only a fraction of the available envelope headroom, and scaling it up to the detection limit moves a great deal of power out of the carrier and into the sidebands. It also comfortably exceeds the familiar single-tone 100%-modulation ceiling of $33.3\%$, which is not a universal bound — that figure assumes a symmetric sinusoid, whose negative peak equals its positive one. Here the negative peak is only half the positive one, so the message can be driven to a much larger mean-square value before the envelope touches zero.