# Determine the percentage power saved when carrier and one of the sidebands are suppressed in an AM wave modulated to a depth of 50%.

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Determine the percentage power saved when carrier and one of the sidebands are suppressed in an AM wave modulated to a depth of 50%.
1. 76%
2. 83.3%
3. 55%
4. 94.4%

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Correct Answer - Option 4 : 94.4%

Concept:

The power of a transmitted AM wave is given as:

$P_t = {P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)$

$P_t = {P_c} + P_c\frac{μ ^2}{2}$

Power in the carrier = Pc

Power in both the sidebands is given by:

$P_s= \frac{{{P_c}{μ^2}}}{2}$

Since the power is distributed equally to the left and to the right side of the sideband, the power in one of the sidebands is given by:

$P_{s1}= \frac{{{P_c}{u^2}}}{4}$

Calculation:

When the carrier and 1 of the sideband are suppressed, the amount of power saving will be:

$= {P_c} + \frac{{{P_c}{μ ^2}}}{4}$

The percentage of power saved will be calculated as:

$\frac{{{P_c}\left( {1 + \frac{{{μ ^2}}}{4}} \right)}}{{{P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)}} \times 100$

$\Rightarrow \frac{{4 + {μ ^2}}}{{4 + 2{μ ^2}}} \times 100$

With μ = 0.50, the amount of power saving will be:

$=\frac{{4 + {{\left( {0.5} \right)}^2}}}{{4 + 2{{\left( {0.5} \right)}^2}}} \times 100$

$= \frac{{4.25}}{{4.5}} = 94.44\%$