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?
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?
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.