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FM numerical: 5cos(18849.55t), 9 kHz/V — deviation, swing, Carson

Short answer

Peak frequency deviation is kf*Am = 9*5 = 45 kHz; modulation index is 45/3 = 15; frequency swing is 2*45 = 90 kHz; Carson bandwidth is 2(45+3) = 96 kHz.

A modulating signal m(t) = 5 cos 18849.55t is applied to an FM modulator that has a frequency sensitivity of 9 kHz/V. Compute (i) peak frequency deviation, (ii) modulation index, (iii) frequency swing and (iv) Carson's bandwidth.

Given: m(t)=5cos⁡(18849.55 t)m(t) = 5\cos(18849.55\,t), kf=9k_f = 9 kHz/V.

So Am=5A_m = 5 V and

fm=ωm2π=18849.552π=3000 Hz=3 kHzf_m = \frac{\omega_m}{2\pi} = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}

(i) Peak frequency deviation

Δf=kfAm=9 kHz/V×5 V=45 kHz\Delta f = k_f A_m = 9\ \text{kHz/V} \times 5\ \text{V} = 45\ \text{kHz}

(ii) Modulation index

β=Δffm=45 kHz3 kHz=15\beta = \frac{\Delta f}{f_m} = \frac{45\ \text{kHz}}{3\ \text{kHz}} = 15

(iii) Frequency swing

The carrier swings from fc−Δff_c - \Delta f to fc+Δff_c + \Delta f:

Swing=2Δf=2×45=90 kHz (±45 kHz)\text{Swing} = 2\Delta f = 2 \times 45 = 90\ \text{kHz}\ (\pm 45\ \text{kHz})

(iv) Carson's bandwidth

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

Answer: Δf=45\Delta f = 45 kHz, β=15\beta = 15, swing = 90 kHz, BT=96B_T = 96 kHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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