FM numerical: s(t) = 9cos(70000t + 16sin2000t)

Given: $m(t) = 3\cos(2000t)$, $c(t) = 9\cos(70000t)$, $s(t) = 9\cos(70000t + 16\sin 2000t)$.

So $A_c = 9$ V, $\omega_m = 2000$ rad/s, $\beta = 16$.

$$f_m = \frac{2000}{2\pi} = 318.31\ \text{Hz}, \qquad f_c = \frac{70000}{2\pi} = 11.14\ \text{kHz}$$

Total modulated signal power

FM has constant amplitude, so (normalised to 1 Ω):

$$P = \frac{A_c^2}{2} = \frac{9^2}{2} = 40.5\ \text{W}$$

Modulation index

From the expression directly:

$$\beta = 16$$

Peak frequency deviation

$$\Delta f = \beta f_m = 16 \times 318.31 = 5092.96\ \text{Hz} \approx 5.093\ \text{kHz}$$

(In rad/s: $\Delta\omega = 16 \times 2000 = 32000$ rad/s. The frequency sensitivity is $k_f = \Delta f/A_m = 5092.96/3 = 1697.65$ Hz/V.)

Bandwidth (Carson's rule)

$$B_T = 2(\Delta f + f_m) = 2(5092.96 + 318.31) = 10822.5\ \text{Hz} \approx 10.82\ \text{kHz}$$

(Equivalently $2(\beta+1)\omega_m = 2 \times 17 \times 2000 = 68000$ rad/s.)

Answer: $P = 40.5$ W, $\beta = 16$, $\Delta f = 5.093$ kHz, $B_T \approx 10.82$ kHz.