Question

A 20 MHz carrier is modulated by a 720 kHz modulated signal to a depth of 60%. If the rms value of the amplitude of the carrier signal is 50 V, then the power of the modulated signal is equal to

• A. 2.95 kW
• B. 0.73 kW
• C. 1.475 kW
• D. 5.9 kW

Answer 1

The correct option is A 2.95 kW

$Carrierpower=(A_{max}{)}^2=(50{)}^2=2.5kWTotalpower=P_c(1+\frac{{\mu }^2}{2})=2.5(1+\frac{(0.6{)}^2}{2})=2.95kW$