Question

A signal of $0.1\text{kW}$ is transmitted in a cable. The attenuation of cable is $-5\text{dB}$ per $\text{km}$ and cable length is $20\text{km}$. The power received at receiver is ${10}^{-x}\text{W}$. The value of $x$ is .
[Gain in $dB=10{\log }_{10}\left(\frac{P_0}{P_{\text{i}}}\right)$]

Answer 1

Power of signal transmitted $P_{\text{i}}=0.1\text{kW}=100\text{W}$

Rate of attenuation$=-5\text{dB/Km}$

Total length of path $=20\text{km}$

Total loss suffered$=5\times 20=-100\text{dB}$

Gain in $dB=10{\log }_{10}\left(\frac{P_0}{P_{\text{i}}}\right)$

$\Rightarrow -100=10lo{g}_{10}\left(\frac{P_0}{P_{\text{i}}}\right)$

$\Rightarrow lo{g}_{10}\left(\frac{P_{\text{i}}}{P_0}\right)=10$

$\Rightarrow lo{g}_{10}\left(\frac{P_{\text{i}}}{P_0}\right)=lo{g}_{10}{10}^{10}$

$\Rightarrow \frac{100}{P_0}={10}^{10}$

$\Rightarrow P_0=\frac{1}{{10}^8}={10}^{-8}$
$\Rightarrow x=8$