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FM numerical: m(t) = 5cos(18849.55t), 9 kHz/V

Short answer

f_m=18849.55/(2π)=3 kHz. Δf=k_f A_m=9×5=45 kHz. Modulation index β=Δf/f_m=15. Carrier swing=2Δf=90 kHz. Carson bandwidth=2(Δf+f_m)=96 kHz. Total power in 10 Ω is A_c²/(2R)=25²/20=31.25 W.

A sinusoidal modulating signal m(t) = 5cos(18849.55t) is applied to an FM modulator that has frequency sensitivity of 9 kHz/V. The amplitude of carrier is 25 V and the carrier frequency is 88.7 MHz. Compute a) modulation index b) Carrier frequency swing c) Bandwidth using Carson rule d) total power delivered to 10 Ω resistor.

Given: m(t)=5cos⁡(18849.55t)m(t) = 5\cos(18849.55t), so Am=5A_m = 5 V and

fm=18849.552π=3000 Hz=3 kHzf_m = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}

kf=9k_f = 9 kHz/V, Ac=25A_c = 25 V, fc=88.7f_c = 88.7 MHz, R=10 ΩR = 10\ \Omega.

Peak frequency deviation:

Δf=kfAm=9×5=45 kHz\Delta f = k_fA_m = 9 \times 5 = 45\ \text{kHz}

a) Modulation index

β=Δffm=453=15\beta = \frac{\Delta f}{f_m} = \frac{45}{3} = 15

(wide-band FM, since β≫1\beta \gg 1)

b) Carrier frequency swing

The carrier swings from fc−Δff_c - \Delta f to fc+Δff_c + \Delta f, i.e. from 88.655 MHz to 88.745 MHz:

Carrier swing=2Δf=2×45=90 kHz\text{Carrier swing} = 2\Delta f = 2 \times 45 = 90\ \text{kHz}

c) Bandwidth by Carson's rule

BT=2(Δf+fm)=2(45+3)=96 kHzB_T = 2(\Delta f + f_m) = 2(45 + 3) = 96\ \text{kHz}

d) Total power in 10 Ω

The FM wave has constant amplitude, so its power does not depend on modulation:

P=Ac22R=2522×10=62520=31.25 WP = \frac{A_c^2}{2R} = \frac{25^2}{2 \times 10} = \frac{625}{20} = 31.25\ \text{W}

Answer: β=15\beta = 15; carrier swing = 90 kHz; BT=96B_T = 96 kHz; P=31.25P = 31.25 W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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