A signal of 0.1 , kW is transmitted in a cabl | vedclass

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(D) The total attenuation in the cable is given by: $	ext{Total Attenuation} = (	ext{Attenuation per km}) 	imes (	ext{Length}) = -5 \, dB/km 	ime

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$8$

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(D) The total attenuation in the cable is given by: $	ext{Total Attenuation} = (	ext{Attenuation per km}) 	imes (	ext{Length}) = -5 \, dB/km 	imes 20 \, km = -100 \, dB$.The formula for gain in $dB$ is: $	ext{Gain} = 10 \log_{10}(\frac{P_0}{P_i})$.Here,$	ext{Gain} = -100 \, dB$,$P_i = 0.1 \, kW = 100 \, W = 10^2 \, W$.Substituting the values: $-100 = 10 \log_{10}(\frac{P_0}{10^2})$.$-10 = \log_{10}(\frac{P_0}{10^2})$.$10^{-10} = \frac{P_0}{10^2}$.$P_0 = 10^{-10} 	imes 10^2 = 10^{-8} \, W$.Comparing $P_0 = 10^{-8} \, W$ with $10^{-x} \, W$,we get $x = 8$.

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

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