For the most part, natural phenomena behave as exponential functions. The exponents of real numbers are the logarithms, and decibels are directly related to logarithms. This is not so difficult to understand as some believe. I am going to try to show you just how simple they can be and how it is possible to calculate the logarithm of any number without a table of logs or a calculator using only a single number. The first task is to provide an example of a range of numbers over one decade of (10:1), and we will then be able to cover any range. The first digit of the logarithmic value is easily obtained:

1. Deriving the First Digit of the Logarithm: . Since we generally count from one to ten, we will choose the number 10 as the "base" number (B =10), although the logarithm base number is not always decimal. The logarithm of a number N is the number of times the base number is multiplied. For instantce, hoosing three values for N: log(N) = log(10) = 1, log(100) = 2, and log(1000) = 3. We have just calculated three exact log values of N for these simple cases. In fact, calculating the log of N was easier than the process of dividing one number by another using normal methods of arithmetic. Notice that the log number 3 is simply the number of decimal places of N = 1000. But logarithms can also be negative, which is the case when the number is less than one. The dividing point is log (1) = 0.

For example, let's choose a large number, say N = 16777216. Move the decimal point to where only one digit is to the left. In this case, the decimal point is moved seven places to the left, and the first digit of the log of N is 7.0000. This is the first estimate of the logarithm of this very large number. The precise value of the logarithm is log (N) = 7.224719896 = 7 + (log 1.677216) = 7.24719896, and therefore the error is - 0.24719896. The value of this error term is comparatively small (much less than 10), so the value 7 is fairly close to the exact number, and we find that the log value error is only - 3.11%. For many cases, this may be sufficiently accurate. When N is less than one, the decimal point is moved to the right until we have one digit to the left of the decimal point, and the number of places moved is a negative number, and any error will be zero or a positive value. The log of any finite number can be estimated in this manner, but this rough estimate will not always be very accurate. Therefore, we will want to improve the accuracy of the estimate considerably.

2. A Binary/Decimal Division Estimation Method:

The log (N) is plotted below for the following values of N: (a.) N=1.414 (the square root of 2), (b.) N=2.828 (2 times square root of 2), (c.)N=5.656 (4 times square root of 2), (d.) N=8 and (e.) N=10, apply to any decade (only four decades are illustrated). Notice that each decade is the similar.

We need only remember one main number in order to obtain a more accurate estimate of the logarithm of any number (two numbers to remember if you don't already know the square root of two). The points plotted in the graph are the values of N stated above, plus multiples of 2, in each decade. All of the values of the log(N) given above are multiples of either 0.301 or 0.301/2 = 0.1505 (the latter being the log of the square root of 2). So you only need to remember this number:

0.301

For the values N of 1, 2, 4, and 8 the logarithm increases by 0.301 in each case (0, 0.301, 0.602, and 0.903). For 2.828 and 5.656, add half of 0.301 (0.1505) to the previous number. In order to obtain the closest estimate, simply choose the closest value to the number (N) for which you want to find the logarithm. In the example of (1.) above, The value of the error, after moving the decimal point in 1.), is 1.677216. The closest point appears to be 1.4142, so we can choose this as the closest value and add 0.1505 (which is 0.301 divided by 2) to the rough estimate of 7, resulting in 7.1505. Comparing this to the actual value, of log (N) = 7.24719896, results in the reduced error of - 1.32%. For most purposes, this level of accuracy may be quite sufficient. If not, then we can further reduce the error by linear interpolation, which will be described next.

3. A Simple Interpolation Method for Greater Accuracy in the Estimate: In (2.), we simply chose the closest number on the above graph for our estimate of log (N), and the error was -1.32%. We can get even closer by using linear interpolation. Over a comparatively small range of numbers, the logarithms are fairly linear. Recall that N = 16,777,216, and we used the closest number, which is one times the square root of two, which is 14,140.000 (rounded off). If we just observe the points on the graph, 1.677 is about half way between 1.414 and 2, so we can just divide 0.1505 (the log of the square root of two) in half and add it to the previous estimate, resulting in 7.1505 +0.07525 = 7.22575, resulting in an error of only Error = - 0.2968%. In most cases where logarithms are used in engineering and science, this will be sufficient accuracy.

4. The Logarithm of a Binary Number: Now we will choose a different base number B = 2 (binary number system), which in this case the number of times that the number 2 is multiplied. For N = 16, the number two is multiplied four times, N = 2 x 2 x 2 x 2 = 16 = 2^4, and its base 2 logarithm is:log (16) = 4. Similarly, for N = square root of 2 = 1.4142, the (binary) log value is 0.5, whereas the (decimal) log value is 0.301 for N =2.

5. Understanding Decibels: A decibel value refers to a ratio of two numbers. The decibel (abbreviated dB) was originally defined in terms of audio power, and an increase in power level of one decibel was the least amount of audio power that the average human ear could hear. We now use decibels to define signal levels over a very wide range for electrical signals, etc. The most common reference point for zero dB is one milliwatt of power. Once the reference value is known, the absolute power can be expressed by a logarithmic value. The decibel is defined as 10 log (P2/P1). Therefore, an increase in power by 10:1 is 10 log(10) = 10 x 1 = 10 dB. Going in the other direction, an increase of 10 dB is 10 log (P2/P1) = 10, so log (P2/P1) = 1.0, which is 10^1 = 10. Now consider an increase in power of 3 dB, in which case 10 log (P2/P1) = 3, and thus log (P2/P1) = 0.300, which is sufficiently close to our magical number of 0.301 for the logarithmic value that corresponds to an increase of 3 dB. Therefore, 3 dB means an increase in power of 2:1 (it will help to remember this). The term dBm means that the reference level is one milliwatt, so in this case 3 dBm means a power level of 2 milliwatts.

6. A Simple Exercise: In order to test what you have learned here, consider the following problem:

a.)Calculate the relative increase in power for an increase of one dB.

b.) Determine the absolute power level for one dBm.

See the answers.

See a plot of decibels and Q&A.

See more plots and answers.

If you find any difficulties in using the methods outlined here, or if you have suggestions for improving the clarity of the presentation, please email me.

WV