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 $[$ $[$ Gain in $dB=10{\log }_{10}\left(\frac{P_0}{P_i}\right)]$

Answer 1

Answer: 8

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 $dB=10lo{g}_{10}\frac{P_0}{P_i}$
$-100=10lo{g}_{10}\frac{P_0}{P_i}$
$\Rightarrow lo{g}_{10}\frac{P_i}{P_0}=10$
$\Rightarrow lo{g}_{10}\frac{P_i}{P_0}=lo{g}_{10}10^{10}$
$\Rightarrow \frac{100}{P_0}=10^{10}$
$\Rightarrow P_0=\frac{1}{10^8}=10^{-8}$
$\therefore x=8$