Question

A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (or gain) of 20dB and the third a loss of 10dB. Assuming a transmitted power level of 400 mW, determine the mean output level of the channel.

Answer 1

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

$P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB$ (1)

so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

$P_{out} = 0.100 W = 100 mW$