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The maximum transmission efficiency of an sinusoidal AM signal is ________.
1. 21.68%
2. 33.33%
3. 58.88%
4. 65.55%  

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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:

\({P_t} = {P_c}\;\left( {1 + \frac{{{\mu ^2}}}{2}} \right)\),

\(\eta = \frac{{{P_c}{\mu ^2}}}{{2\left( {{P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)} \right)}}\)

\(\eta = \frac{{{P_c}{\mu ^2}}}{{{P_c}\left( {2 + {\mu ^2}} \right)}} = \frac{{{\mu ^2}}}{{2 + {\mu ^2}}} \times 100\)

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