Question

A 1 KW signal is transmitted using a communication channel which provides attenuation at the rate of – 2dB 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 ("P"_0/"P"_"i")]`

Options

• 900 W

• 100 W

• 990 W

• 1010 W

Answer 1

100 W

Explanation:

Power of the signal transmitted is P_i = 1 KW = 1000 W

rate of attenuation of signal is = `-2("dB"/"km")`

Length or total path = 5 km

thus less suffered in the communication channel = (5) (– 2) = – 10 dB and gain in

dB = 10 log `("P"_0/"P"_"i")`

dB = – 10 log `("P"_"i"/"P"_0)`

– 10 = – 10 log `("P"_"i"/"P"_0)`

log' `("P"_"i"/"P"_0) = 1 ("P"_"i"/"P"_0) = 10`

P₀ = `("P"_"i"/"P"_0) = 1000/10` = 100 W