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SDSC (ISRO) Technical Assistant (Electronics): Previous Paper 2018 (Held On: 8 April 2018)

Option 2 : 0 dBm

ISRO VSSC Technical Assistant Mechanical held on 09/06/2019

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80 Questions
320 Marks
120 Mins

**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**

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

P(dBm) = 10log_{10}(1)

P(dBm) = 0