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Understanding Logarithms II
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a.) Question by Jacinta:

"What I am confused about is how 3dB corresponds to an increase in power of 2:1. I understand that 0.301 is log 2 but this value is on the RHS of the equal sign (ie why isn't it 1:2) ? I have had a problem to calculate the factor increase if there is a 60dB difference between 2 sounds. so following the steps above:
10 log(P2/P1)=60
log (P2/P1) = 60/10
log (P2/P1) = 6
I don't understand why in your answer above the result from an increase in 10 dB becomes 10^1 (though I understand steps before that). Just translating this result to my problem I would assume that my answer would be 10^6, but wouldn't this be an answer of 1000000? How does log(p2/P1) start to be referred to as 10^ (or at least that is how it appears to me)."

Answer to question: Let's begin with the definition of a logarithm. If we take the expression
x = 10^n, then log x = n. So if we set x = 2, then log 2 = 0.301.
The definition of a decibel is dependent upon what the variable may be. For voltage it 20 log V2/V1, while for power it is 10 log P2/P1. Therefore dB = 10 log P2/P1, and it is on the RHS of the equal sign.
In order to calculate P2/P1, you need to reverse the equation. For dB = 60,
dB = 10 log P2/P1 = 60, and therefore log (P2/P1) = 6. Therefore there are six decades of
power change =1,000,000. This is the great advantage of the use of logarithms when the
numbers become quite large, and this is exactly what happens in electronics and electromagnetism.
It also happens in mathematics when we deal with ratios. Note that the difference between 1,000,012 (the number you might want to seek) and log 1,000,000 is a small fraction of one percent. So othe log of 1,000,000 is usually sufficient. Iff you need the log equivalents of very high degrees of accuracies to many decimal places, then you can take the time to go to your log table.

The base number doesn't need to be 10, but that is the way we count. It can be any number, such as 2. There are various base numbers, such as 2. The base can have a reference number. Ex: Base 10 ref one milliwatt refers to the ratio with respect to one milliwatt.

b.) Solving the problem of calculating a relative increase in power of one dB:

For an increase in power of one dB: 10 log(P2/P!) = 1.0 dB, or log (P2/P1) = 0.10. Therefore the increase in power is 10^0.1. We can utilize a calculator to find the answer, but we want to do it without a calculator or table of logarithms, using the process of known values and linear interpolations. The first step is to locate the known value on the log (N) vs N graph nearest the point log (N) = 0.1. We see that the value of log (1) = 0, which is 0 dB. This point is less than 1.0 dB, and the nearest point above it is log (1.414) = 0.1505, which is 1.505 dB. Therefore, N will lie between 1 and 1.414 (the square root of 2), and log (N) lies between 0 and 0.1505.

We can now utilize linear interpolation to find a better value and increase the accuracy of the estimate. For linear interpolation, the estimated value of N lies along a straight line between the two points 1 and 1.414. We need to decrease the location of the estimate point from 1.505 dB to 1.00 dB, which is a change of

The total spread of the values of log (N) is delta N = (0.1505 - 0) = 0.1505, and this corresponds to a spread of 1.505 dB. Since we want the value of one dB, we have to increase from 0 dB by a factor of 1.0 dB. This is a ratio of 1.0/1.515 = 0.66

The total spread of the values of N is (1.414 -1.0) = 0.414.

Therefore, the estimated value of N lies

0.66 x 0.414 = 0.27327 above the 0 dB point, and

N = 1.0 + 0.27327 = 1.27327 = P2/P1 (since N is a ratio of power level increase).

The actual value is N = 1.25992105 = P2/P1 for a one dB increase in power, and therefore, the power increases by 26% for every dB, as compared to the estimated value of 27.3% . The db calculation error is (1.27327 - 1.25992105) = 0.013349 dB or 1.33%, which is considered to be quite good. Notice that (P2/P1)^3 = 1.25992105^3 = 2, which corresponds to 3 dB for twice the power level.

b.) Solving the problem of calculating the absolute power level of one dBm:

For a signal level of one dBm, the absolute power level is referenced to one miliwatt(zero dBm), so value of the absolute power is

P(dBm) = 1.0 x10^-3 x1.25992105 = 1.26 x 10^-3 = 1.26 mW.

Compare this exact value to the estimated value in (a.) of 1.27 mw This is a difference of only 10 microwatts as compared to the exact value of 1.26 microwatts.

If you found this process to be somewhat difficult, you can try to visualize the estimation process using the graph below. We have shown that there is a 26% change in power for every one dB increment, and that a 3 dB increment represents a doubling of the power.

The curved line is a plot of log(n) vs (n). The estimate point is the circle, and the actual value is the square. The x,y reference points are (1,0) and (1.414,1.5).

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