Can anyone actually explain to me how this mathematics question was does as I had it in an engineering test?/?

Question

Can anyone actually explain to me how this mathematics question was does as I had it in an engineering test but couldn't do it:
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 :)

Answer 1

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

Answer 2

First of all, the concept of power gain is quite simple. It is the ratio of the power out to the power in. That is that Po = Pi * G where G is the power gain.
So, in the example (a) 15 = 0.3 * G and so G = 15/0.3 = 50.

Now decibels (dB).
Let us assume that we have a long cable that we wish to send a signal along. We will need to amplify the signal, probably several times along the length of the cable to compensate for the loss along each section of the cable. To do this we will need to multiply the signal level by a gain for each amplifier and divide the signal level by each loss.
Engineers are lazy. There is no point in being an engineer if you want to do everything the difficult way.
It is easier to add and subtract in your head rather than multiply and divide, so to do this let us express all powers and power ratios as logarithms of the absolute values.

The power gain in (a) would be log50 Bells. As it turns out, a Bell is rather large so we use decibels (dB) so the power gain becomes 10log50 = 16.989 or approx 17dB.

In (b), to get the gain do anti-log(20/10) = anti-log2 = 100
So the answer is 400mW * 100 = 40W.

Note that the log of 2 is about 0.3 and so a gain of 2 comes out as 3dB. A gain of 4 is 6dB and so on.

Answer 3

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