Modulated signal Z(t) = 50cos{98.6×10⁶×2πt + 15sin(2×10³×2πt)}

Given: $Z(t) = 50\cos\{2\pi(98.6\times10^6)t + 15\sin(2\pi\times 2\times10^3\, t)\}$ V, $R = 50\ \Omega$.

Comparing with $A_c\cos[2\pi f_c t + \beta\sin 2\pi f_m t]$: $A_c = 50$ V, $f_c = 98.6$ MHz, $f_m = 2$ kHz, $\beta = 15$.

a) Type of modulation

The amplitude is constant and the phase varies sinusoidally, so it is angle modulation. If the message is $m(t) = A_m\cos 2\pi f_m t$, the phase term $15\sin 2\pi f_m t$ is proportional to its integral, so it is frequency modulation (FM); the 98.6 MHz carrier also lies in the FM broadcast band. (If the message were a sine, the same wave would be PM; the expression alone cannot tell the two apart.)

b) Frequency deviation

$$f_i(t) = \frac{1}{2\pi}\frac{d\theta}{dt} = f_c + \beta f_m\cos 2\pi f_m t$$ $$\Delta f = \beta f_m = 15 \times 2\ \text{kHz} = 30\ \text{kHz}$$

c) Frequency sensitivity

$k_f = \Delta f / A_m$. The message amplitude is not given, so taking $A_m = 1$ V:

$$k_f = \frac{30\ \text{kHz}}{1\ \text{V}} = 30\ \text{kHz/V}$$

(For any other $A_m$, $k_f = 30/A_m$ kHz/V.)

d) Modulation index

$$\beta = \frac{\Delta f}{f_m} = \frac{30}{2} = 15$$

e) Power delivered to 50 Ω antenna

$$P = \frac{A_c^2}{2R} = \frac{50^2}{2 \times 50} = 25\ \text{W}$$

Answer: FM; $\Delta f = 30$ kHz; $k_f = 30$ kHz/V (for $A_m = 1$ V); $\beta = 15$; $P = 25$ W.