Concept:
The total AM Power is given by:
\({P_t} = {P_c} + \frac{{{P_c}{\mu ^2}}}{2}\)
Expanding the above, we can write:
\( = \begin{array}{*{20}{c}} {{P_c}}\\ \downarrow \\ {Carrier}\\ {Power} \end{array}\begin{array}{*{20}{c}} + \\ {}\\ {}\\ {} \end{array}\;\begin{array}{*{20}{c}} {\frac{{{P_c}{\mu ^2}}}{4}}\\ \downarrow \\ {USB}\\ {Power} \end{array}\begin{array}{*{20}{c}} + \\ {}\\ {}\\ {} \end{array}\;\begin{array}{*{20}{c}} {\frac{{{P_c}{\mu ^2}}}{4}}\\ \downarrow \\ {LSB}\\ {Power} \end{array}\)
% Power saving is defined calculated as:
\(\% P = \frac{{Power\;saved}}{{Total\;power}} \times 100\)
When the carrier and one side-band is suppressed, the percent of power-saving will be:
\( = \frac{{{P_c} + \frac{{{P_c}{\mu ^2}}}{4}}}{{{P_c}\left[ {1 + \frac{{{\mu ^2}}}{2}} \right]}} \times 100\)
\( = \frac{1}{2}\left[ {\frac{{4 + {\mu ^2}}}{{2 + {\mu ^2}}}} \right] \times 100\)
Calculation:
Given:
\(\mu = \frac{{50}}{{100}} = 0.5\)
% Power saved will be:
\(\% P = \frac{{4 + {{0.5}^2}}}{{2\left( {2 + {{0.5}^2}} \right)}}\)
= 94.44%