AM numerical: S(t) = 10(1 + 0.2cos2π10³t)cos2π10⁶t

Compare with $s(t) = A_c[1 + \mu\cos2\pi f_mt]\cos2\pi f_ct$: $A_c = 10$ V, $\mu = 0.2$, $f_m = 10^3$ Hz, $f_c = 10^6$ Hz. Take $R = 1\ \Omega$.

a) Type of modulation

Carrier term is present and the envelope is $10(1 + 0.2\cos\omega_mt)$: it is DSB-FC (conventional AM), single-tone, with 20% modulation.

b) Frequencies

$$f_m = 1\ \text{kHz}, \qquad f_c = 1\ \text{MHz}$$

c) Bandwidth

$$B = 2f_m = 2 \times 1 = 2\ \text{kHz}$$

d) Powers

$$\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{10^2}{2} = 50\ \text{W} \\ P_{SB} &= P_c\frac{\mu^2}{2} = 50 \times \frac{0.04}{2} = 1\ \text{W}\ (0.5\ \text{W each}) \\ P_t &= P_c + P_{SB} = 51\ \text{W} \end{aligned}$$

e) Efficiency

$$\eta = \frac{P_{SB}}{P_t} = \frac{\mu^2}{2 + \mu^2} = \frac{0.04}{2.04} = 0.0196 = 1.96\%$$

Answer: DSB-FC AM; $f_m$ = 1 kHz, $f_c$ = 1 MHz; $B$ = 2 kHz; $P_c$ = 50 W, $P_t$ = 51 W, $P_{SB}$ = 1 W; $\eta$ = 1.96%.