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A transmitter puts out a total power of 15 Watts of 16% AM signal. Calculate how much power is contained in each of the sidebands and the fraction of power in the carrier.

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The total power in an AM signal when the carrier power and the percentage of modulation are known is $P_T = P_c \big ( 1 + \large\frac{m^2}{2}$$\big )$
$\Rightarrow 15 = P_c \big ( 1 + \large\frac{0.16^2}{2}$$\big )$
$\Rightarrow P_c = \large\frac{15}{1.013}$$ = 14.81\;W$
The total power in both side bands $ = 15-14.81 = 0.19\;W$
The power in each side band is half of this $ \approx 0.095\;W$
$\Rightarrow$ The fraction of the power in the carrier is $\large\frac{14.81}{15} $$\approx 0.99$
answered Jul 13, 2014 by balaji.thirumalai

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