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A ` 1 KW ` signal is transmitted using a communication channel which provides attenuatiom 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 ((P_(0))/(P_(i)))]`A. `900 W`B. `100 W`C. `990 W`D. `1010 W` |
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Answer» Correct Answer - B Here `P_(i) = 1 kW = 1000` watt , length of path `= 5 km , P_(0) = ?` Loss of power at a distance of ` 5 km = - 2 xx 5 = - 10 dB`. As gain in `dB = 10 log .(P_(0))/(P_(i)) :. - 10 = log .(P_(0))/(P_(i)) = - 10 log .(P_(i))/(P_(0))` or `log .(P_(i))/(P_(0)) = 1 = log 10 or (P_(i))/(P_(0)) = 10 or P_(0) = (P_(i))/(10) = (1000)/(10) = 100` watt |
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