For 100% modulation (AM),the useful part of t | vedclass

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(B) For $100\%$ modulation,the modulation index $m_a = 1$.The total power radiated in an $AM$ wave is given by $P_t = P_c(1 + \frac{m_a^2}{2})$,where

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$1/3$ of the total power

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(B) For $100\%$ modulation,the modulation index $m_a = 1$.The total power radiated in an $AM$ wave is given by $P_t = P_c(1 + \frac{m_a^2}{2})$,where $P_c$ is the carrier power.The useful power is the power contained in the sidebands,given by $P_{sb} = P_c(\frac{m_a^2}{2})$.The ratio of useful power to total power is $\frac{P_{sb}}{P_t} = \frac{m_a^2/2}{1 + m_a^2/2} = \frac{m_a^2}{2 + m_a^2}$.Substituting $m_a = 1$:$	ext{Ratio} = \frac{1^2}{2 + 1^2} = \frac{1}{3}$.Thus,the useful part of the total power radiated is $1/3$ of the total power.

For $100\%$ modulation $(AM)$,the useful part of the total power radiated is

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