Correct Answer - Option 3 : 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
\(P_c=\frac{{A_m}^2}{2}\)
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:
\(P_{s1}= \frac{{{P_c}{u^2}}}{4}\)
Calculation:
Given μ = 20% = 0.2
\(P_c=\frac{{A_m}^2}{2}\)
Total Sideband power is:
\(P_{sb}=\frac{{A_m}^2\mu^2}{4}\)
Assume Am = 1
\(P_{sb}=\frac{0.2^2}{4} \ \times \ 100\)
Psb = 1%
Hence option (3) is the correct answer.