Correct Answer - Option 4 : 94.4%
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:
\(P_{s1}= \frac{{{P_c}{u^2}}}{4}\)
Calculation:
When the carrier and 1 of the sideband are suppressed, the amount of power saving will be:
\( = {P_c} + \frac{{{P_c}{μ ^2}}}{4}\)
The percentage of power saved will be calculated as:
\(\frac{{{P_c}\left( {1 + \frac{{{μ ^2}}}{4}} \right)}}{{{P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)}} \times 100\)
\(\Rightarrow \frac{{4 + {μ ^2}}}{{4 + 2{μ ^2}}} \times 100\)
With μ = 0.50, the amount of power saving will be:
\(=\frac{{4 + {{\left( {0.5} \right)}^2}}}{{4 + 2{{\left( {0.5} \right)}^2}}} \times 100 \)
\(= \frac{{4.25}}{{4.5}} = 94.44\%\)
