In a 100% modulated AM signal with a carrier power of 100 W, what is the power in the lower sideband?

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RRB JE ECE 22 Apr 2025 Shift 2 CBT 2 Official Paper

Answer 1

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)

The total power is the sum of the carrier power and the sideband power, i.e.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

Calculation:

Given: Pc = 100 W and μ = 1

We can write:

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

\(= 100 \times \frac{1^2}{2}\)

\(P_s=50~W\)

The total sideband power = 50 W

power in upper sideband + power in lower side band = 50 W

power in upper sideband = power in lower sideband = 25 W