Question

What would be the percentage power saving when the carrier and one of the side bands were suppressed in an AM wave modulated to a depth of 75% ?

• A. 89.1 %
• B. 9.1 %
• C. 7.81 %
• D. 100 %

Answer 1

Answer: (A)

Here, modulation index, $\mu =\frac{75}{100}=\frac{3}{4},$
Power produced by the AM transmitter
As $P_t=P_c\left(1+\frac{{\mu }^2}{2}\right)=P_c\left(1+\frac{9}{32}\right)=P_c\times \frac{41}{32}$
$P_{SB}=P_c\frac{{\mu }^2}{4}=P_c\times \frac{9}{16\times 4}=\frac{p}{64}{P}_c$
On suppressing, power saved is
$=P_c+\frac{9}{64}{P}_c=\frac{73}{64}{P}_c$
$\therefore$ Percentage saving $=\frac{\frac{73}{64}{P}_c}{P_t}\times 100$
$=\frac{\frac{76}{64}{P}_c}{\frac{41}{32}{P}_c}\times 100=\frac{73\times 32}{64\times 41}\times 100=89.1\%$