Question

An AM modulator develops an unmodulated power output of 400W across a 50\(\Omega\) resistive load. The carrier is modulated by a single tone with a modulation index of 0.6. If this AM signal is transmitted, the power developed across the load is

• A. 428 W
• B. 432 W
• C. 472 W
• D. 418 W

Hint:

The total power of an AM signal is dependent on the carrier power and the modulation index as given in the equation: \(P_{total} = P_c \left(1 + \frac{\mu^2}{2}\right)\).

Answer 1

The Correct Option is A

Step 1: We are given the unmodulated power, \( P_c = 400W \) and the modulation index, \( \mu = 0.6 \). The modulated power can be calculated as: \[ P_{total} = P_c \left(1 + \frac{\mu^2}{2}\right) \] Step 2: Substitute values: \[ P_{total} = 400 \left(1 + \frac{0.6^2}{2}\right) = 400 (1 + \frac{0.36}{2}) = 400 (1 + 0.18) = 400 \times 1.18 = 472 W \] Therefore, the power developed across the load is 472W.