Question 6 of 10: AM Signal Parameters, Power Budget and Efficiency

Question 6: AM Signal Parameters, Power Budget and 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. From Figure 5 (positive frequencies only): impulses of weight $1/4$ at 46 kHz, $1/2$ at 48 kHz, $2$ at 50 kHz, $1/2$ at 52 kHz and $1/4$ at 54 kHz. Load resistance normalised to 1 Ω.

[Figure not reproduced: Figure 5 (redrawn): the tall centre line is the carrier at fc = 50 kHz with weight Ac/2 = 2; the sideband pairs sit 2 kHz and 4 kHz away with weights 1/2 and 1/4. See the official exam paper.]

Find. $f_c$, $A_c$ and $m(t)$; the modulation index; the carrier, sideband and total powers with the efficiency; and the largest gain $K$ compatible with envelope detection, with the resulting efficiency.

Approach. Identify the carrier line by its symmetry position, convert each sideband weight back into a message-tone amplitude, then apply the standard AM power formulas; part (d) is governed by the no-overmodulation constraint.

1. Locate the carrier and its amplitude. The spectrum of $[A_c+m(t)]\cos 2\pi f_ct$ is $$\Phi_{AM}(f)=\frac{A_c}{2}\big[\delta(f-f_c)+\delta(f+f_c)\big]+\frac{1}{2}\big[M(f-f_c)+M(f+f_c)\big].$$ The lone impulse sitting at the centre of symmetry is the carrier term, at 50 kHz with weight 2. Hence $f_c = 50$ kHz and $A_c/2 = 2$, giving $$\boxed{f_c = 50\ \text{kHz},\qquad A_c = 4\ \text{V}}$$
2. Convert the sideband weights into tone amplitudes. A message tone $A_k\cos 2\pi f_kt$ contributes impulses of weight $A_k/2$ to $M(f)$, and the AM spectrum carries a further factor $1/2$, so a printed sideband of weight $w_k$ at offset $f_k$ implies $A_k = 4w_k$. The offsets from 50 kHz are 2 kHz (weight $1/2$) and 4 kHz (weight $1/4$), giving $A_1 = 4(1/2) = 2$ V and $A_2 = 4(1/4) = 1$ V: $$\boxed{m(t)=2\cos(2\pi\,2000\,t)+\cos(2\pi\,4000\,t)\ \text{V}}$$
3. Part (b): modulation index. Both cosines reach their maxima together at $t = 0$, so the positive peak of the message is $m_p = A_1 + A_2 = 2 + 1 = 3$ V (a numerical sweep of $m(t)$ over one period confirms the maximum is exactly 3 V at $t = 0$). Therefore $$\mu=\frac{m_p}{A_c}=\frac{3}{4}=\boxed{0.75}$$ Since $\mu \lt 1$ the signal is not overmodulated and the envelope never folds.
4. Part (c): carrier power. The carrier term is a sinusoid of amplitude $A_c$ into 1 Ω: $$P_c=\frac{A_c^2}{2}=\frac{4^2}{2}=8\ \text{W}$$
5. Sideband power. The sideband term is $m(t)\cos 2\pi f_ct$, whose mean square is $\overline{m^2}/2$. For a sum of incoherent tones $\overline{m^2}=\sum A_k^2/2 = 2^2/2 + 1^2/2 = 2.5\ \text{V}^2$, so $$P_s=\frac{\overline{m^2}}{2}=\frac{2.5}{2}=1.25\ \text{W}$$
6. Total power and efficiency. Carrier and sidebands are orthogonal, so the powers add: $$P_{\varphi_{AM}}=P_c+P_s=8+1.25=9.25\ \text{W}$$ and the power efficiency, the fraction of transmitted power that actually carries information, is $$\eta=\frac{P_s}{P_c+P_s}=\frac{1.25}{9.25}=\boxed{13.51\%}$$

Almost 87% of the transmitter output is spent on the carrier, which conveys no information at all. Part (d) asks how much of that waste can be recovered while keeping the receiver simple.

7. Part (d): the envelope-detection constraint. An envelope detector recovers $|A_c + Km(t)|$, which equals $A_c + Km(t)$ — and hence preserves the message shape — only if the bracket never goes negative. The binding case is the most negative excursion of the message, $-m_p$, so the requirement is $A_c - Km_p \ge 0$, i.e. $K \le A_c/m_p$: $$K_{\max}=\frac{A_c}{m_p}=\frac{4}{3}=\boxed{1.333}$$ At this gain $\mu = K_{\max}\,m_p/A_c = 1$ exactly — the boundary of overmodulation, where the envelope just touches zero.
8. Efficiency at $K_{\max}$. Scaling the message by $K$ scales its mean square by $K^2$ while leaving the carrier untouched: $$P_s'=K_{\max}^2\frac{\overline{m^2}}{2}=\left(\frac{4}{3}\right)^2(1.25)=\frac{20}{9}=2.222\ \text{W}$$ so $P_{\text{total}}' = 8 + 2.222 = 10.222$ W and $$\eta'=\frac{2.222}{10.222}=\boxed{21.74\%}$$

Comparison. Raising the message gain to its maximum permissible value lifts the efficiency from 13.51% to 21.74%, an improvement of roughly 1.6 times, at no cost in receiver complexity — the same envelope detector still works. But 21.74% is the ceiling for this particular message: with $\mu$ already at unity there is no further headroom, and pushing $K$ higher would overmodulate, causing the envelope to fold through zero and producing gross distortion that no envelope detector can repair. The residual inefficiency is structural, not a design error: full AM deliberately transmits the carrier so that receivers can stay cheap. Systems that need the power back (DSB-SC, SSB) suppress the carrier and pay for it with coherent detection at every receiver.