In an AM with m=1, the power in one sideband is:

Question

Question

This question was previously asked in

RRB ALP Mechanic Radio & TV 23 Jan 2019 Official Paper (Shift 3)

Answer 1

The correct option is 1

Concept:

The power of a transmitted AM wave is given as:

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

\(P_t = {P_c} + P_c\frac{μ ^2}{2}\)

Power in the carrier = Pc

Power in both the sidebands is given by:

\(P_s= \frac{{{P_c}{μ^2}}}{2}\)

Since the power is distributed equally to the left and to the right side of the sideband, the power in one of the sidebands is given by:

\(P_{s1}= \frac{{{P_c}{u^2}}}{4}\) ............(1)

Explanation:

Given modulation index is 1

Putting modulation index =1 in equation 1 we have

\(P_{s1}= \frac{{{P_c}{1^2}}}{4} = \frac{{{P_c}{}}}{4}\)

Hence the power in the sideband is one-fourth of the carrier power.