Answer:
Gain (dB)=10 log(O.P./I.P)
Also, the loss(dB) = -gain(dB) = 10 log(I.P/O.P)
which is, 50 = 10 log(I.P./10nW)
and hence, anti-log(5) = I.P/10 nW
And thus the power required by the source is antilog(5)*10nW = 1 mW
Note: Power required by the receiver is the O.P. and the power required from the source is the I.P.
Explanation:
Please check the answer.
