SOLUTION: I am taking an online electronics course and I am stumped on decibel/logarithm section relating to amplifiers. My problems deal in power or voltage gain, and levels. Here are 2 e

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am taking an online electronics course and I am stumped on decibel/logarithm section relating to amplifiers. My problems deal in power or voltage gain, and levels. Here are 2 e      Log On


   



Question 81388: I am taking an online electronics course and I am stumped on decibel/logarithm section relating to amplifiers. My problems deal in power or voltage gain, and levels. Here are 2 examples that I cannot solve. (1) A power amplifier has a gain of 26dB, and an input of 1 volt. What is the voltage level? (2) An amplifier has a voltage gain of 60dB, the input is 10 microvolts, what is the output? I have a couple formulas, and I know the answers, but I can't figure out how to get the answers from the formulas. The answer to problem one is 20, and the answer to problem 2 is 10. Here are the formulas:
Gain= signal out/signal in
dB= 10 x log(base 10) (W out/ W in)
dB= 20 x log(base 10) (V out/ V in)
Output voltage= input level x gain ratio
I need a very detailed expression as to how to solve these problems. I have looked in two textbooks for further explanation, but I cannot find the solutions.
Thank you so much if you can help!!!!

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
To solve these two problems you need only the voltage amplification equation:
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dB+=+20%2ALog+%28V%5B2%5D%2FV%5B1%5D%29
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and a knowledge of a few rules of Logarithms.
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The dB equation defines decibels. It uses the base 10 Logarithms. In this equation,
V%5B2%5D is the output voltage, and V%5B1%5D is the input voltage.
.
Here are a couple of rules of Logarithms that you should be familiar with. Assume that
the base of the Logarithm is b:
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Rule I. …. Log+%28x%5En%29+=+n+%2A+Log+%28x%29
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Rule II. …. Log+%28x%2Fy%29+=+Log%28x%29 – Log%28y%29
.
Rule III. …. Log+%28x%2Ay%29+=+Log+%28x%29 + Log+%28y%29
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Rule IV. …. Log base b of x = y is equivalent to saying b%5Ey+=+x
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Note that in decibel calculations b (the base) is 10. This translates Rule IV to:
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Log (x) = y is equivalent to saying 10%5Ey+=+x
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If you need a further explanation of these rules, you can get it from an Algebra II text,
from elsewhere on this site, or by doing a Google on Logarithms.
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Time to work your first problem by using the voltage form of the decibel equation.
Given the amplifier has a power gain of 26 dB, and an input voltage of 1 volt. Substitute
those two values into the dB equation (26 for dB and 1 for the input voltage) and you get:
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26+=+20%2ALog%28V%5B2%5D%2F1%29
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The plan is to work down to V%5B2%5D, so start by dividing both sides of this equation by
20 to get rid of the 20 on the right side. At the same time note that V%5B2%5D%2F1
is just V%5B2%5D. These simplifications lead to the equation becoming:
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26%2F20+=+Log+%28V%5B2%5D%29
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And after dividing 26 by 20 on the left side the equation reduces to:
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1.3+=+Log+%28V%5B2%5D%29
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Time to apply Rule IV. When you do you get:
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10%5E1.3+=+V%5B2%5D
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Calculator time. Your calculator probably works this way: enter 1.3 then press the 10%5Ex
key and you find the result to be 19.95262315 volts. Close enough for electrical work.
Call it 20 volts.
.
Now to your second problem. If the answer is to be 10, then you probably meant that the
input voltage was 10 millivolts 10%2A10%5E-3+ instead of 10 microvolts 10%2A10%5E-6.
So let’s use 10 millivolts.
.
For the given gain of 60 dB and an input of 10%2A10%5E-3 volts the equation becomes:
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60+=+20%2ALog+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29
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Just as was done in the last problem, divide both sides by 20 to reduce the problem to:
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60%2F20+=+Log+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29
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The left side divides out to a quotient of 3 and the equation becomes:
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3+=+Log+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29
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On the right side apply Rule II (division) to split the term into two separate logarithms:
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3+=+Log+%28V%5B2%5D%29 – Log+%2810%2A10%5E-3%29
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Notice that 10*10^-3 = 10^(-2). Substitute this into the right hand logarithm and get:
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3+=+Log%28V%5B2%5D%29 – Log+%2810%5E%28-2%29%29
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Apply Rule I (exponent rule) to Log%2810%5E-2%29 and get - 2%2ALog+%2810%29. But by Rule IV
you can see that Log base 10 of 10 = 1. (Think 10%5Ey+=+10 which means y must be 1.)
So - 2%2A+Log%2810%29 equals -2*1 equals -2. Substitute this result for – Log%2810%5E-2%29
and the equation becomes:
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3+=+Log%28V%5B2%5D%29 – %28-2%29
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And this simplifies to 3+=+Log%28V%5B2%5D%29+%2B+2
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Subtract 2 from both sides and the equation becomes:
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1+=+Log%28V%5B2%5D%29
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Apply Rule IV to this equation and you get:
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V%5B2%5D+=+10%5E1+=+10
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This agrees with the answer you had for this problem, but recall that 10 millivolts
was used in place of 10 microvolts to get the answer.
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Hope this helps you get a handle on decibel equations.