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Q.
A transmitter supplies $9 KW$ to the aerial when unmoderated. The power radiated when modulated to $60 \%$ is:
Communication Systems
Solution:
We know the relation
$\frac{p_{t}}{p_{c}}=\left[1+\frac{(m)^{2}}{2}\right]$
Given, $m=60 \%=\frac{60}{100}=0.6$
$\frac{p_{t}}{9}=\left(1+\frac{(0.6)^{2}}{2}\right)$
$\Rightarrow p_{t}=9\left[1+\frac{(0.36)}{2}\right]$
$p_{t}=9(1+0.18)=9(1.18)$
$p_{t}=10.62 k W$