Logs in EquationsDate: 05/20/97 at 12:38:38 From: Hughie Coyle Subject: Calculation of Output Power I have already calculated that a circuit has an attenuation of 60dB's and the input power = 200mW and I need to calculate the mean output power. The formula is as follows : dB = 10log PowerOut/PowerIn -60 = 10log PowerOut/200/1000 The question is: What do I have to do to manipulate this formula in order to find the PowerOut? In order to get PowerOut on its own, I tried taking the natural log of both sides, but the answer I get is incorrect (818mW). Can you please enlighten me as to the correct way of manipulation of logs? Best Regards, H F Coyle Date: 06/24/97 at 14:49:00 From: Doctor Barney Subject: Re: Calculation of Output Power Rev. 1 Hi Hughie, I would be happy to spread some enlightenment. Let's review what a logarithm is. If you have a number B, and you raise it to the power of x, and the result is C, then B^x = C. We say that the log (base B) of C is x: log(C) = x Now, how can you get rid of the log in an equation? Suppose we take our base number, B, and raise it to the power of both sides of that last equation: B^(log(C)) = B^x. But B^x = C, so B^(log(C)) = C. This is sometimes called an inverse logarithm. In plain English, to get rid of the log, raise it to the power of whatever base you are using, and the answer is the thing you were taking the log of in the first place. You can have logarithms of any base you want, but two of the most common are logs with a base of e, or natural logarithms, and logarithms with a base of 10, or common logarithms. A decibel is an engineering unit that was created to make it more convenient to describe ratios of very different power levels. In the formula you site above, which is the definition of a decibel, the log is a base 10 logarithm. (This problem has nothing to do with natural logarithms or with the number e.) As an example of how we might use a decibel, if the ratio of the power of the signal coming out of your stereo amplifier to the power going in is a thousand to one, that's 30 dB: 10 x log(1,000/1) = 30 In the problem you have above, first write the problem like this: -60 = 10log(PowerOut/PowerIn) -60 = 10 x log(PowerOut/200) If you leave the input power in mW, the output power will also be in mW. First isolate the log by dividing both sides of the equation by 10. Then, to get rid of the log, raise 10 to the power of both sides of the equation. Then it should be easy to solve for the output power. Since this circuit has an attenuation of 60 dB, your answer should be much, much smaller than 200 mW. -Doctor Barney, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/