JEE PYQ: Communication System Question 13
Question 13 - 2021 (24 Feb 2021 Shift 2)
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 ______ $[\text{Gain in dB} = 10 \log_{10}(\frac{P_o}{P_i})]$
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Answer: 8
Solution
Total loss $= -5 \times 20 = -100$ dB. $-100 = 10 \log_{10} \frac{P_o}{100}$. $P_o = 10^{-8}$. So $x = 8$.
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