GATE 2017 EC (EC2) – Question 60
An AM transmitter has unmodulated carrier power 5 kW and is modulated by a sinusoidal signal. Its maximum modulation percentage is 50%. If this is reduced to 40%, the maximum unmodulated carrier power (in kW) usable without overloading the transmitter is ________.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 5.19 to 5.23
Explanation
For single-tone AM, total average transmitted power is $P_T=P_C(1+m^2/2)$. The keyed overload model keeps this total average-power rating fixed.
Initially $P_T=5(1+0.5^2/2)=5.625$ kW. With m=0.4, the permissible new carrier power is $P_C'=5.625/(1+0.4^2/2)$.
This is **5.20833 kW**, or about 5.21 kW, within the official 5.19–5.23 interval. A separate peak-envelope-power limit would be a different constraint.