Angle modulated signal 12 sin(10⁶t + 5 sin10⁴t)

Compare with the standard single-tone angle-modulated wave $v(t) = A_c\sin(\omega_ct + \beta\sin\omega_mt)$:

$A_c = 12$ V, $\omega_c = 10^6$ rad/s, $\omega_m = 10^4$ rad/s, $\beta = 5$.

a) Carrier frequency

$$f_c = \frac{\omega_c}{2\pi} = \frac{10^6}{2\pi} = 159.15\ \text{kHz}$$

b) Modulating frequency

$$f_m = \frac{\omega_m}{2\pi} = \frac{10^4}{2\pi} = 1591.55\ \text{Hz} \approx 1.59\ \text{kHz}$$

c) Modulation index

The peak phase deviation is the coefficient of $\sin 10^4t$:

$$\beta = 5\ \text{rad}$$

(Treated as FM, the peak deviation is $\Delta f = \beta f_m = 5 \times 1591.55 = 7957.7$ Hz.)

d) Power in 100 Ω

The amplitude of an angle-modulated wave is constant, so

$$P = \frac{A_c^2}{2R} = \frac{12^2}{2 \times 100} = \frac{144}{200} = 0.72\ \text{W}$$

Answer: $f_c = 159.15$ kHz, $f_m = 1.59$ kHz, $\beta = 5$, $P = 0.72$ W.