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An AM signal with a carrier of 1 kW has 200 W in each side band. The percentage of modulation is:
1. 20%
2. 89.4%
3. 49.7%
4. 40%

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Correct Answer - Option 2 : 89.4%

Concept:

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.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

Calculation:

Given: Pc = 1000 W and Ps = 200W

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

\(μ^2=\frac{800}{1000}\)

μ = 0.8944

% Modulation = 89.44%

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