Question

A 1 kW signal is transmitted using a communication channel which provides attenuation at the rate of - 2 dB per km. If the communication channel has a total length of 5 km, the power of the signal received is

[Gain in dB = 10 log$\left( \frac{P_{0}}{P_{i}} \right)$]

• A. 900 W
• B. 100 W
• C. 990 W
• D. 1010 W

Answer 1

Answer: (B)

100 W

Answer 2

: Loss suffered in the communication channel

$= ( - 2dB/km)5km = - 10dB$

$\therefore 10\log(p_{o}/p_{i}) = - 10or\log(p_{o}/p_{i}) = - 1$

Or $(p_{o}/p_{i}) = 10^{- 1} = \frac{1}{10}$

Or $p_{o} = p_{i}/10 = \frac{1kW}{10} = 100W$