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When 80% AM is transmitted total power is 10kW. What is the power saved if the carrier and one side band are suppressed?

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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}$$\big )$
$\Rightarrow P_c = \Large\frac{\normalsize 10}{1 + \Large\frac{0.8^2}{2}}$$ = 7.58\;kW$
$\Rightarrow$ Power Saved $= P_c \big (1 + \large\frac{m^2}{4} $$\big )$
$\quad \quad\quad \;\quad\quad = 7.58 \big (1+\large\frac{0.8^4}{4}$$\big) $$ = 8.78\;kW$
answered Jul 12, 2014 by balaji.thirumalai

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