FM numerical: u(t) = 10cos(100,000t + 10sin5000t)

Given: $m(t) = 2\cos(5000t)$, $c(t) = 10\cos(100000t)$, $u(t) = 10\cos(100000t + 10\sin 5000t)$.

So $A_m = 2$ V, $\omega_m = 5000$ rad/s, $A_c = 10$ V, $\omega_c = 10^5$ rad/s, $\beta = 10$.

$$f_m = \frac{5000}{2\pi} = 795.77\ \text{Hz}, \qquad f_c = \frac{10^5}{2\pi} = 15.915\ \text{kHz}$$

Total modulated signal power

FM has constant amplitude; normalised to 1 Ω:

$$P = \frac{A_c^2}{2} = \frac{10^2}{2} = 50\ \text{W}$$

Modulation index

From the expression: $\beta = 10$.

Peak frequency deviation

$$\Delta f = \beta f_m = 10 \times 795.77 = 7957.7\ \text{Hz} \approx 7.96\ \text{kHz}$$

($\Delta\omega = 10 \times 5000 = 50000$ rad/s; $k_f = \Delta\omega/A_m = 25000$ rad/s/V, i.e. 3978.9 Hz/V.)

Bandwidth (Carson's rule)

$$B_T = 2(\Delta f + f_m) = 2(\beta + 1)f_m = 2 \times 11 \times 795.77 = 17507\ \text{Hz} \approx 17.51\ \text{kHz}$$

(In rad/s: $2 \times 11 \times 5000 = 110000$ rad/s.)

Answer: $P = 50$ W, $\beta = 10$, $\Delta f = 7.96$ kHz, $B_T \approx 17.51$ kHz.