Correct Answer - Option 3 : 20%
Concept: power efficiency of AM is defined as \(\left( \eta \right) = \frac{{{\mu ^2}}}{{2 + \mu }} \times 100\% \)
Where μ = modulation index = \(\frac{{{A_m}}}{{{A_c}}}\)
Solution:
\(m\left( t \right) = \frac{1}{2}\cos {\omega _1}\left( t \right) - \frac{1}{2}\sin {\omega _2}t\)
Since in msg signal, two different frequencies ω1 ⋅ ω2 are present So it is a multitone signal.
\({A_{{m_1}}} = \frac{1}{2}\)
\({A_{{m_2}}} = \frac{1}{2}\)
\(s\left( t \right) = \left[ {1 + m\left( t \right)} \right]\cos {\omega _c}t\)
Comparing with standard
AM signal equation \(s\left( t \right) = {A_c}\left[ { + {K_a}m\left( t \right)} \right]\cos {\omega _c}t\)
Ac = 1, Ka = 1
\({\mu _1} = \frac{{{A_{{m_1}}}}}{{{A_c}}} = \frac{1}{2}\)
\({\mu _2} = \frac{{{A_{{m_2}}}}}{{{A_c}}} = \frac{1}{2}\)
\(\mu = \sqrt {\mu _1^2 + \mu _2^2} = \sqrt {{{\frac{1}{2}}^2} + {{\frac{1}{2}}^2}} = \sqrt {\frac{1}{2}} \)
\(\eta = \frac{{{\mu ^2}}}{{2 + \mu }} \times 100\% = \frac{{\frac{1}{2}}}{{2 + \frac{1}{2}}} \times 100\% = \frac{1}{5} \times 100\% \)
η = 20%