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 ______.

Answer 1

Step 1. Given data

Power of signal transmitted $P_i=0.1Kw=100w$

Total length of path $=20km$

Rate of attenuation $=-5dB/Km$

Step 2. Calculating the value of $x$

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

We know that Gain is given by

$$\begin{gathered} GainindB=10{\log }_{10}(P_0/P_i) \\ 100=10{\log }_{10}(P_0/P_i) \\ {\log }_{10}(P_0/P_i)=10 \\ {\log }_{10}(P_0/P_i)={\log }_{10}{10}^{10} \\ \frac{100}{P_0}={10}^{10} \\ {P}_0=\frac{1}{{10}^8}={10}^{-8} \end{gathered}$$

Therefore, the value of $x$ is $8$.