Correct Answer - Option 2 : 33.33%
Concept:
The transmission efficiency of an AM wave is defined as the percentage of total power contributed by the sidebands.
For a sinusoidal AM signal, it is given by:
\(η=\frac{{{μ ^2}}}{{2 + {μ ^2}}} \times 100\)
μ = Modulation index
The maximum efficiency is obtained for μ = 1, i.e.
\(η_{max}=\frac{{{1}}}{{2 + {1}}} \times 100\)
ηmax = 33.33 %
Derivation:
Mathematically, the efficiency can be expressed as:
\(\eta = \frac{{{P}_{SB}}}{{{P_t}}} \times 100\%\)
For sinusoidal input
PSB = Sideband power given by:
\({P_{SB}} = \frac{{{P_c}\;{\mu ^2}}}{2}\)
Pt = Total power given by:
\({P_t} = {P_c}\;\left( {1 + \frac{{{\mu ^2}}}{2}} \right)\),
\(\eta = \frac{{{P_c}{\mu ^2}}}{{2\left( {{P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)} \right)}}\)
\(\eta = \frac{{{P_c}{\mu ^2}}}{{{P_c}\left( {2 + {\mu ^2}} \right)}} = \frac{{{\mu ^2}}}{{2 + {\mu ^2}}} \times 100\)