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A1 KWsignal 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 of5 km , the power of the signal received is [ gain in dB = 10 log ((P_(0))/(P_(i)))] |
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Answer» <P>900W `=(-2dB//km)5km=-10dB` `therefore 10"log"(P_(0)//P_(i))=-10, "or log" (P_(0)//P_(i))=-1` or `(P_(0)//P_(i))=10^(-1)=(1)/(10)` or `P_(0)=P_(i)//10=(1KW)/(10)=100W` |
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