A signal of 0.1 kW is transmitted in a cable.

Question:

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

Solution:

Power of signal transmitted : $P_{i}=0.1 \mathrm{KW}=100 \mathrm{w}$

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

Total length of path $=20 \mathrm{~km}$

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

Gain in $\mathrm{dB}=10 \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{i}}}$

$-100=10 \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{i}}}$

$\Rightarrow \log _{10} \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{0}}=10$

$\Rightarrow \log _{10} \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{0}}=\log _{10} 10^{10}$

$\Rightarrow \frac{100}{\mathrm{P}_{0}}=10^{10}$

$\Rightarrow \mathrm{P}_{0}=\frac{1}{10^{6}}=10^{-8}$

$\therefore \mathrm{x}=8$

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