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(C) The power of a single sideband is given by $P_{sb} = P_c \left( \frac{m_a}{2} \right)^2 = P_c \frac{(0.5)^2}{4} = 0.0625 P_c$.The total power of t

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$94.4$

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(C) The power of a single sideband is given by $P_{sb} = P_c \left( \frac{m_a}{2} ight)^2 = P_c \frac{(0.5)^2}{4} = 0.0625 P_c$.The total power of the $AM$ wave is $P_{total} = P_c \left( 1 + \frac{m_a^2}{2} ight) = P_c \left( 1 + \frac{(0.5)^2}{2} ight) = 1.125 P_c$.If the carrier and one sideband are suppressed,only one sideband remains. The power remaining is $P_{rem} = P_{sb} = 0.0625 P_c$.The reduction in power is $\Delta P = P_{total} - P_{rem} = 1.125 P_c - 0.0625 P_c = 1.0625 P_c$.The percentage reduction in power is $\frac{\Delta P}{P_{total}} 	imes 100 = \frac{1.0625 P_c}{1.125 P_c} 	imes 100 \approx 94.4\%$.

An amplitude modulated wave is modulated to $50\%$. If the carrier and one of the sidebands are suppressed,the percentage reduction in power is ....$\%$

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