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?
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
\(P=\frac{\frac{1}{2}{A_c}^2}{R}=\frac{\frac{1}{2}\times25}{20}=0.625watts\)
Draw and explain the square – law modulator.
Compare PAM, PWM and PPM
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Fig. 14(a).png)
Fig.22(a)