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Q.
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% ?
Communication Systems
Solution:
Here, modulation index, $\mu=\frac{75}{100}=\frac{3}{4},$
Power produced by the AM transmitter
As $P_{t}=P_{c}\left(1+\frac{\mu^{2}}{2}\right)=P_{c}\left(1+\frac{9}{32}\right)=P_{c}\times\frac{41}{32}$
$P_{SB}=P_{c} \frac{\mu^{2}}{4}=P_{c}\times\frac{9}{16\times4}=\frac{p}{64}P_{c}$
On suppressing, power saved is
$=P_{c}+\frac{9}{64}P_{c}=\frac{73}{64}P_{c}$
$\therefore $ Percentage saving $=\frac{\frac{73}{64}P_{c}}{P_{t}}\times100$
$=\frac{\frac{76}{64}P_{c}}{\frac{41}{32}P_{c}}\times100=\frac{73\times 32}{64\times 41}\times 100=89.1\%$