FM numerical: 5cos(18849.55t), 9 kHz/V, 25 V, 88.7 MHz (five parts)

Given: $m(t) = 5\cos(18849.55\,t)$, $k_f = 9$ kHz/V, $A_c = 25$ V, $f_c = 88.7$ MHz, $R = 10\ \Omega$.

$$A_m = 5\ \text{V}, \qquad f_m = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}$$

i) Peak frequency deviation

$$\Delta f = k_f A_m = 9 \times 5 = 45\ \text{kHz}$$

ii) Modulation index

$$\beta = \frac{\Delta f}{f_m} = \frac{45}{3} = 15$$

iii) Frequency swing

$$\text{Swing} = 2\Delta f = 90\ \text{kHz}$$

The carrier moves between $88.7 - 0.045 = 88.655$ MHz and $88.745$ MHz.

iv) Carson's bandwidth

$$B_T = 2(\Delta f + f_m) = 2(45 + 3) = 96\ \text{kHz}$$

v) Power in 10 Ω

FM has constant amplitude, so total power equals the carrier power:

$$P = \frac{A_c^2}{2R} = \frac{25^2}{2 \times 10} = \frac{625}{20} = 31.25\ \text{W}$$

Answer: $\Delta f = 45$ kHz, $\beta = 15$, swing = 90 kHz, $B_T = 96$ kHz, $P = 31.25$ W.