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Exercise Problems · Q12

Q.An AM transmitter radiates 10 kW power at 85 percent modulation.What would be the power saving if the carrier and one of the sidebands are suppressed?

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[!TLDR]

With ma=0.85m_a = 0.85, the carrier is 7.35 kW7.35\ \text{kW} and each sideband 1.33 kW1.33\ \text{kW}; keeping just one sideband saves about 86.77 % of the total power.

First split the 10 kW total into carrier and sideband powers. The carrier power is

Pc=Pt1+ma22=101+0.72252=101.36125=7.35 kWP_c = \dfrac{P_t}{1+\dfrac{m_a^{2}}{2}} = \dfrac{10}{1+\dfrac{0.7225}{2}} = \dfrac{10}{1.36125} = 7.35\ \text{kW}

Each sideband carries

PSB=Pc ma24=7.35×0.72254=1.33 kWP_{SB} = \dfrac{P_c\,m_a^{2}}{4} = \dfrac{7.35\times0.7225}{4} = 1.33\ \text{kW} …

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