AM numerical: s(t) = 5[1 + 0.6cos(6280t)]cos(2π10⁴t)

Compare with $s(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct$:

$A_c = 5$ V, $\mu = 0.6$, $\omega_m = 6280$ rad/s so $f_m = \dfrac{6280}{2\pi} = 999.5 \approx 1$ kHz, and $f_c = 10^4$ Hz = 10 kHz.

(a) Modulation depth

$$\mu = 0.6 \quad (60\%)$$

(b) Maximum and minimum amplitude

$$\begin{aligned} A_{max} &= A_c(1 + \mu) = 5 \times 1.6 = 8\ \text{V} \\ A_{min} &= A_c(1 - \mu) = 5 \times 0.4 = 2\ \text{V} \end{aligned}$$

(c) Frequency components and amplitudes

$$s(t) = 5\cos2\pi(10^4)t + \frac{0.6\times5}{2}\cos2\pi(10^4 + 10^3)t + \frac{0.6\times5}{2}\cos2\pi(10^4 - 10^3)t$$

(Using exact $f_m = 999.5$ Hz: 9.0005 kHz and 10.9995 kHz.)

(d) Power across 1 kΩ

$$\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{25}{2000} = 12.5\ \text{mW} \\ P_{USB} = P_{LSB} &= \frac{(1.5)^2}{2\times1000} = 1.125\ \text{mW} \\ P_t &= 12.5 + 2(1.125) = 14.75\ \text{mW} \end{aligned}$$

Check: $P_c(1 + \mu^2/2) = 12.5 \times 1.18 = 14.75$ mW.

Answer: $\mu = 0.6$; 8 V and 2 V; 10 kHz (5 V), 9 kHz and 11 kHz (1.5 V each); $P_t = 14.75$ mW.