Can anyone actually explain to me how this mathematics question was does as I had it in an engineering test?/?
The power gain of an amplifier relates input power, Pi, to output power, Po, by;
Power gain= 10log(small 10) Po/Pi decibles (dB)
a) As part of the assesment of a new system you have been asked to find the power gain when Po=15W and Pi=300mW
B) you also have to find out what the output power will be if input power is 400mW and the power gain is 20dB.
Please may someone explain this to me step by step as it completely mindblows me, thanks to anyone who can help me it will be much appreciated.
Thanks everyone :)
Answers:
For Part A)
First of all i'm assuming that when you say small 10 you mean base 10.
Just substitute Po and Pi in the equation:
power gain = 10*log(15/.3) = 19.99 dB
B)
Since you know the power gain and Pi you can solve for Po:
20 = 10*log(Po/.4)
2 = log(Po/.4)
10^2 = Po/.4
Po = 100*.4 = 40
Therefore
Po = 40W
I dont have a clue
Doing this problem just takes a few calculations:
A)
you have all the components to solve this problem, thus
power gain = 10*log(base 10)(15/0.300) [make sure to convert mW to W]
=10*(1.698) = aprroximately 17 dB
B)
Here, you have to solve for P_o
The equation setup is as follows: 20 = 10*log(base 10)(P_o/0.400)
First, divide by 2 on both side: 2 = log(base 10)(P_o/0.400)
Now take the inverse of a log(base 10), which is '10^'
10^2 = P_o/0.400 ---------> 100 = P_0/0.400
Multiplying by 0.4, you get P_o = 40 W
Hope this helps
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