Question

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 __________.

Hint:

Decibels $(\text{dB})$ are a logarithmic unit. Every $10 \text{ dB}$ loss reduces power by a factor of 10. A $100 \text{ dB}$ loss is a factor of $10^{10}$ reduction.

Answer 1

Correct Answer: 8

Step 1: Total loss $= -5 \text{ dB/km} \times 20 \text{ km} = -100 \text{ dB}$.
Step 2: $dB = 10 \log_{10}(P_{out}/P_{in}) \Rightarrow -100 = 10 \log_{10}(P_{out}/100 \text{ W})$.
Step 3: $-10 = \log_{10}(P_{out}/100) \Rightarrow 10^{-10} = P_{out}/100$.
Step 4: $P_{out} = 100 \times 10^{-10} = 10^{-8} \text{ W}$.
Correction: If input is $0.1 \text{ kW} = 100 \text{ W}$, output is $10^{-8}$. If output is $10^{-x}$, $x=8$. (Note: Often in these problems, attenuation refers to current/voltage or the power values are different. Based on the logic of standard dB loss, $x$ would be 8.)