Linear amplifier with a gain of 30 dB is fed with 1.0 μW power. The output power of the amplifier

Concept:

The gain in dB is defined as:

\(P(dB)=10~log_{10}\frac{P_{out}}{P_{in}}\)

Power in dBm is expressed as:

\(P(dBm)=10log_{10}(\frac{P}{1m})\)

P = Power to be expressed in dBm

Calculation:

Given

Pin = 1 μW

P(dB) = 30 dB 

Putting these values in the above equation, we get:

\(30=10~log_{10}\frac{P_{out}}{1~\mu W}\)

\(3=log_{10}\frac{P_{out}}{1~\mu W}\)

Taking Antilog on both sides, we get:

\(10^3=\frac{P_{out}}{1~\mu W}\)

P_out = 1 mW

Convert output power to dBm

\(P(dBm)=10log_{10}(\frac{1mW}{1mW})\)

P(dBm) = 10log₁₀(1)

P(dBm) = 0 dB