Angle modulated voltage 10 sin(10⁸t + 3 sin10⁴t)

Given: $E = 10\sin(10^8 t + 3\sin 10^4 t)$ V, $R = 100\ \Omega$.

Comparing with the standard form $E = A_c \sin(\omega_c t + \beta \sin \omega_m t)$:

$$A_c = 10\ \text{V}, \quad \omega_c = 10^8\ \text{rad/s}, \quad \omega_m = 10^4\ \text{rad/s}, \quad \beta = 3$$

Carrier frequency

$$f_c = \frac{\omega_c}{2\pi} = \frac{10^8}{2\pi} = 15.915 \times 10^6\ \text{Hz} \approx 15.92\ \text{MHz}$$

Modulating frequency

$$f_m = \frac{\omega_m}{2\pi} = \frac{10^4}{2\pi} = 1591.5\ \text{Hz} \approx 1.59\ \text{kHz}$$

Modulation index

$$\beta = 3 \quad (\text{peak deviation } \Delta f = \beta f_m = 3 \times 1591.5 = 4774.6\ \text{Hz})$$

Power in 100 Ω

The amplitude of an angle-modulated wave is constant, so its power equals the unmodulated carrier power:

$$P = \frac{V_{rms}^2}{R} = \frac{(A_c/\sqrt2)^2}{R} = \frac{10^2}{2 \times 100} = 0.5\ \text{W}$$

Answer: $f_c = 15.92$ MHz, $f_m = 1.59$ kHz, $\beta = 3$, $P = 0.5$ W.