Simple Power Question

Thread Starter

KaiL

Joined Aug 30, 2014
69
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?
 
Last edited:

JoeJester

Joined Apr 26, 2005
4,126
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.
 
Last edited:

Thread Starter

KaiL

Joined Aug 30, 2014
69
So the answer is 3014mW for the first question?

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

WBahn

Joined Mar 31, 2012
24,974
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.
 
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