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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
Solution
100 W
Explanation:
Power of the signal transmitted is Pi = 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
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