Answer:
the mean output power level of the channel is 100.475 mW
Explanation:
Given the data in the question;
for the first section; attenuation
16 = 10logāā Ā 400/Pā
Pā = 10.0475 mW
For the second section; Ā amplification
20 = 10logāā Ā pā/10.0475
pā = 1004.75 mW
For the third section; attenuation
10 = 10logāā Ā pā/1004.75 Ā
pā = 100.475 mW
i.e the mean output power level is 100.475 mW
or better still overall attenuation channel = ( 16 - 20) + 10 = 6dB
so
6 = 10logāā Ā 400/Pā
pā = 100.475 mV
the mean output power level of the channel is 100.475 mW