Q1. The output power of an amplifier is 14dB higher than the input power of 120mW. What is the output power of the power amplifier?

Given that P dB = 10log (Pw/Pref)

Pw = Power in watt
Pref= reference power in watt.

My answer : Pdb = 10log(120mw/1) = -9.21 dB . Since it said that output power of an amplifier is 14dB higher than the input power , output power of the power amplifier = -9.21 + 14 = 4.79dB.

Q2. What is the EIRP of a 50 watt transmitter and a gain of 15dB?

Given that the formula for EIRP is EIRP = Pt * Gt

So EIRP = 15dB * 50w = 750dBw

Am I correct for both questions?

 

Am I correct for both questions?

The short answer is NO.

dB = 10 log (power out / power in)

You already know the dB and the input power. What is the output power?

You should know that approximately for every 3 dB of gain, the power doubles. You can narrow down a quick approximation to "check" your answers.

 

So the answer is 3014mW for the first question?

dB = 10 log (power out / power in)
14dB = 10 log (Pout/ 120mW)
Pout = 3014mW?

 

How did you arrive at that number? Did you just plug them in the power out position until you reached 14 dB?

 

log(x/y)=log(x)-log(y)

 

So the answer is 3014mW for the first question?

dB = 10 log (power out / power in)
14dB = 10 log (Pout/ 120mW)
Pout = 3014mW?

That's correct.

As a check on whether the answer is reasonable, you can look at 14 dB as being 10 dB + 3 dB + 1 dB.

An increase by 10 dB is a factor of 10 -- so you go from 120 mW to 1200 mW

An increase by 3 dB is a factor of 2 -- so you go from 1200 mW to 2400 mW

This puts you close enough that you can determine that an answer of 3000 mW is within reason.

An increase of 1 dB is an increase of 25% -- so you go from 2400 mW to 3000 mW and now you are well within the rounding associated with the 3 dB and 1 dB rules of thumb.