Question

A signal of $0.1kW$ is transmitted in a cable. The attenuation of cable is $-5dBperkm$ and cable length is $20km.$ the power received at receiver is $10^{- x}W.$ . The value of $x$ is $\left[\right.$ $\left[\right.$ Gain in $\left.dB =10 \log _{10}\left(\frac{P_{0}}{P_{i}}\right)\right]$

Answer 1

Power of signal transmitted : $P_{i}=0.1Kw=100w$
Rate of attenuation $=-5dB/Km$
Total length of path $=20km$
Total loss suffered $=-5\times 20=-100dB$
Gain in $dB=10log_{10}\frac{P_{0}}{P_{i}}$
$-100=10log_{10}\frac{P_{0}}{P_{i}}$
$\Rightarrow log_{10}\frac{P_{i}}{P_{0}}=10$
$\Rightarrow log_{10}\frac{P_{i}}{P_{0}}=log_{10}10^{10}$
$\Rightarrow \frac{100}{P_{0}}=10^{10}$
$\Rightarrow P_{0}=\frac{1}{10^{8}}=10^{- 8}$
$\therefore x=8$