$A message signal m(t) is applied to an ideal FM modulator and the resultant FM signal is applied to an ideal demodulator (based on PLL) as shown in the figure below The frequency sensitivity of the Fm modulator (k_{f}) is 5Hz/V and the frequency sensitivity of. the VCO (k_{v}) used in the PLL is 10 Hz/V. If the average power of the message signal m(t) is 20 W, then the average power of the signal y(t) will be equal to$

40 W

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5 W

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20 W

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10 W

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Solution

The correct option is B $5 W$
$For properly designed PL based demodulator, y (y) = \frac{k_{f}}{k_{v}} m (t) = \frac{5}{10} m (t) = \frac{1}{2} m (t) P_{y} = {(\frac{1}{2})}^{2} P_{m} = \frac{1}{4} (20) W = 5 W$

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The correct option is B 5WFor properly designed PL based demodulator,y(y)=kfkvm(t)=510m(t)=12m(t)Py=(12)2Pm=14(20)W=5W

The correct option is B $5 W$
$For properly designed PL based demodulator, y (y) = \frac{k_{f}}{k_{v}} m (t) = \frac{5}{10} m (t) = \frac{1}{2} m (t) P_{y} = {(\frac{1}{2})}^{2} P_{m} = \frac{1}{4} (20) W = 5 W$