Figure (EP) shows a communication system. What is the output power when input signals is of 1.01 mW? [ gain in `dB = 10 log_10(P_0//P_t)]`.

Explanation

The distance travelled by the signal is 5 km
Loss suffered in path of transmission = 2 dB/km
So, total loss suffered in 5 km = - `2 xx 5 = - 10` dB
Total amplifier gain = 10 dB + 20 dB = 30 dB
Overall gain in signal = 30 - 10 = 20 dB
According to the question, gain in dB = `10log_(10).(P_(0))/(P_(i))`
`therefore " "20=10 log_(10).(P_(0))/(P_(i))`
or `log_(10) .(P_(0))/(P_(i))=2`
Here, `P_(i) = 1.01` mW and `P_(0)` is the output power.
`therefore" "(P_(0))/(P_(i))=10^(2)=100`
`implies " "P_(0) = P_(i) xx 100 = 1.01 xx 100`
or `P_(0) = 101` mW
Thus, the output power is 101 mW.