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  • A signal of 0.1 kW is transmitted in a cable. The Attenuation of cable is -5 dB per km and cable length is 20 km. The power received at receiver is 10 -x W. The value of x is ______. [Gain in <img alt="dB=10\; log_{10}\left ( \frac{P_{o}}{P_

A signal of 0.1 kW is transmitted in a cable. The Attenuation of cable is -5 dB per km and cable length is 20 km. The power received at receiver is 10 -x W. The value of x is ______. [Gain in dB=10\; log_{10}\left ( \frac{P_{o}}{P_{i}} \right )]
Option: 1 2
Option: 2 4
Option: 3 6
Option: 4 8

Answers (1)

best_answer

Power of signal transmitted: \mathrm{P}_{\mathrm{i}}=0.1 \mathrm{KW}=100 \mathrm{W}
Rate of attenuation =-5 \ \frac{dB}{Km}
Total length of path =20 Km
Total loss suffered =-5 \times 20=-100 \mathrm{~dB}
\\ \text{Gain in} \mathrm{dB}=10 \log _{10}\left(\mathrm{P}_{0} / \mathrm{P}_{\mathrm{i}}\right) \\ \Rightarrow -100=10 \log _{10}\left(\mathrm{P}_{0} / \mathrm{P}_{\mathrm{i}}\right) \\\Rightarrow \log _{10}\left(\mathrm{P}_{\mathrm{i}} / \mathrm{P}_{0}\right)=10\\ \Rightarrow \log _{10}\left(\mathrm{P}_{\mathrm{i}} / \mathrm{P}_{0}\right)=\log _{10} 10^{10}\\\Rightarrow \frac{100}{ \mathrm{P}_{0}}=10^{10} \\ \Rightarrow \mathrm{P}_{0}=\frac{1}{10^{8}}=10^{-8} \\ \Rightarrow \mathrm{x}=8

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