Question

In a high-quality coaxial cable, the power drops by a factor of 10 approximately every 5 km. If the original signal power is 0.25 W (=2.5 x 10-1), how far will a signal be transmitted before the power is attenuated to 25 μW? As part of your answer, include a Table showing the signal power vs. distance in 5 km intervals (like Table 1.2 in Section 1.4.4). If optical fibre is used instead of the coaxial cable, briefly explain how you would expect the above calculated distance value to change

Answer 1

Step-by-step explanation:

20 km

Input to the cable

0.25 watts = 250 milliwatts = 2.5 x 10^-1 watts

Output from the cable

25 uW = 2.5 x 10^-5 watts

The ratio of input to output

(2.5 x 10^-1 watts)/(2.5 x 10^-5 watts = 10^4

Since the attenuation is a factor of 10 every 5 km

X km = (5 km)( log [ 10^4]) = (5 km)(4) = 20 km

Answer 2

We know that the:

Given data,

power drops by a factor of 10 approximately every 5 km.

If the original signal power is 0.25 W (=2.5 x 10-1).

Find out,

how far will a signal be transmitted before the power is attenuated to 25 μW?

Let us solve the problem,

Input to the cable

0.25 watts = 250 milliwatts = 2.5 x 10^-1 watts

Output from the cable

25 uW = 2.5 x 10^-5 watts

The ratio of input to output is,

(2.5 x 10^-1 watts)/(2.5 x 10^-5 watts = 10^4)

Since the attenuation is a factor of 10 every 5 km

X km = (5 km)( log [ 10^4])

         = (5 km)(4)

         = 20 km

Hence if optical fibre is used instead of the coaxial cable, the

above calculated distance value to be changed.