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GATE EC 2021 Official Paper

CT 1: Ratio and Proportion

3536

10 Questions
16 Marks
30 Mins

__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%