Correct Answer - Option 2 : 66 %
Concept:
The generalized AM expression is represented as:
s(t) = Ac [1 + μa mn (t)] cos ωc t
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.
\({P_s} = P_c\frac{{{μ^2}}}{2}\)
The power in a single sideband will be:
\({P_s} = \frac{1}{2}\times P_c\frac{{{μ^2}}}{2}\)
With \(P_c=\frac{A^2}{2}\), the above can be written as:
\({P_s} = \frac{1}{2}\times \frac{A_c^2}{2}\frac{{{μ^2}}}{2}\)
\({P_s} = \frac{{{A_c^2μ^2}}}{8}\)
\(Power Saved=\frac{P_c}{P_{total}}\) ---(1)
Power Saved = Pc in DSB - SC
\(Power \ Saved=\frac{2}{2 \ + \ μ^2}\)
Analysis:
When μ = 1, the transmitted power will be:
\({P_t} = {P_c}\left( {1 + \frac{{{1^2}}}{2}} \right)=\frac{3}{2}P_c\)
\(Power Saved=\frac{P_c}{P_c(1 \ + \ \frac{μ^2}{2})}\)
As μ = 1
\(Power \ Saved=\frac{2}{2 \ + \ 1^2}\times 100\)
Power Saved = 66 %
