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Angle modulated signal 20cos(6×10⁸t + 7sin1250t)

Short answer

Carrier f_c = 6×10⁸/(2π) ≈ 95.49 MHz, modulating f_m = 1250/(2π) ≈ 198.94 Hz. Modulation index β = 7, maximum deviation Δf = β f_m ≈ 1.39 kHz. Power in 10 Ω = 20²/(2×10) = 20 W.

The angle modulated signal is given by S(t) = 20 cos(6×10⁸t + 7 sin 1250t). Determine: i) The carrier and modulating frequency ii) The modulation index iii) Maximum frequency deviation iv) Power dissipated in 10 Ω resistor

Compare with s(t)=Accos⁡(ωct+βsin⁡ωmt)s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt):

Ac=20A_c = 20 V, ωc=6×108\omega_c = 6 \times 10^8 rad/s, ωm=1250\omega_m = 1250 rad/s, β=7\beta = 7.

i) Carrier and modulating frequency

fc=6×1082π=95.49×106 Hz=95.49 MHzfm=12502π=198.94 Hz\begin{aligned} f_c &= \frac{6 \times 10^8}{2\pi} = 95.49 \times 10^6\ \text{Hz} = 95.49\ \text{MHz} \\ f_m &= \frac{1250}{2\pi} = 198.94\ \text{Hz} \end{aligned}

ii) Modulation index

β=7\beta = 7

iii) Maximum frequency deviation

Δf=βfm=7×198.94=1392.6 Hz≈1.39 kHz\Delta f = \beta f_m = 7 \times 198.94 = 1392.6\ \text{Hz} \approx 1.39\ \text{kHz}

iv) Power in 10 Ω

P=Ac22R=2022×10=20 WP = \frac{A_c^2}{2R} = \frac{20^2}{2 \times 10} = 20\ \text{W}

Answer: fc=95.49f_c = 95.49 MHz, fm=198.94f_m = 198.94 Hz; β=7\beta = 7; Δf=1392.6\Delta f = 1392.6 Hz; P=20P = 20 W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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