Question:

Published on: 20 June, 2022

**Find the carrier and modulating frequency, the modulation index and the maximum deviation of the FM wave presented by the voltage equation \(v=5sin\left(6\times{10}^8t+5sin1000t\right)\). What power will this FM wave dissipate across a 20Ω resistor?**

Answer:

Given that \(A_c=5V\), Resistance \(R=20\mathrm{\Omega}\)

Carrier frequency \(f_c=\frac{6\times{10}^8}{2\pi}=95.45MHz\)

Modulating signal frequency \(f_m=\frac{1000}{2\pi}=159.10Hz\)

Modulation index \(\beta=5\)

Maximum deviation \(∆f=\beta f_{m}=5\times159.10=795.45Hz\)

Power dissipated by the load \(20\mathrm{\Omega}\) is

\(P=\frac{\frac{1}{2}{A_c}^2}{R}=\frac{\frac{1}{2}\times25}{20}=0.625watts\)

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