FM numerical: m(t) = 20cos(2π2000t), 200 Hz/V

Given: $m(t) = 20\cos(2\pi\, 2000\, t)$ V, $c(t) = 50\cos(2\pi 10^7 t)$ V, $k_f = 200$ Hz/V.

So $A_m = 20$ V, $f_m = 2$ kHz, $A_c = 50$ V, $f_c = 10$ MHz.

a) Peak frequency deviation

$$\Delta f = k_f A_m = 200 \times 20 = 4000\ \text{Hz} = 4\ \text{kHz}$$

b) Modulation index

$$\beta = \frac{\Delta f}{f_m} = \frac{4000}{2000} = 2$$

c) Bandwidth containing over 98% of power

Bessel coefficients for $\beta = 2$:

Fraction of power in carrier + $n$ sideband pairs $= J_0^2 + 2\sum_{k=1}^{n}J_k^2$:

• up to $n = 2$: $0.0501 + 2(0.3326 + 0.1245) = 0.964$ (96.4%, not enough)
• up to $n = 3$: $0.964 + 2(0.0166) = 0.998$ (99.8%)

So 3 pairs of sidebands are needed:

$$B = 2 n f_m = 2 \times 3 \times 2\ \text{kHz} = 12\ \text{kHz}$$

This agrees with Carson's rule: $B_T = 2(\Delta f + f_m) = 2(4 + 2) = 12$ kHz.

d) Total modulated signal power

Taking a 1 Ω load (normalised power):

$$P = \frac{A_c^2}{2} = \frac{50^2}{2} = 1250\ \text{W}$$

(FM power equals unmodulated carrier power, independent of $\beta$.)

Answer: $\Delta f = 4$ kHz, $\beta = 2$, $B = 12$ kHz, $P = 1250$ W (normalised to 1 Ω).