I have noticed that one of the most confusing topics
in ham radio is the decibel. I believe a
lot of the confusion relates to the complex look of the formula:

Decibel calculations are not intuitive. You have to practice and pay attention to
what the calculations mean. The
mechanics of the formula are fairly straightforward and today’s scientific
calculators do all the heavy lifting for you.
You literally just have to push a button to get the answers. Unfortunately, if you don't understand what’s
going on you won’t perform the steps in the correct order.

First you should take a look at your calculator and notice that
the key labeled “log” has a second function above it. That second function is the inverse operation
of “log” and is labeled as “10

^{X}.” You pronounce that “ten to the X” and all that means is you are raising the number 10 to the power of the number in the calculator’s display. If the number in the display is 2, when you press the 10^{X}key you will get the result of 10^{2}(ten squared, or 10 to the second power) which is 100 (10 times 10 is ten squared). Logarithms take that relationship and put it into a different form. In that 10 to the 2nd power equals 100, the base is 10. There are as many bases for logs as there are numbers, but the one on your calculator, and the one that doesn’t specify a base, is the base 10 logarithm, or log_{10}. So the log_{10}of 100 is 2 because 10 to the 2 power is 100. All you really have to understand is that the 10^{X}and log key are inverses of each other.
So try it. Enter 100 into
your calculator and press the log button.
The calculator should now show the number 2. Now press the 2ND button on your
calculator (to use the second function of a key) and press log again. This time because you pressed the 2ND key first you are accessing the 10

^{X}function and should now see 100 on your screen. See? We’re back to our starting point. This is an important point to recognize as we go forward.
Another point of confusion is in which number to put on top in
the P

_{measured}/P_{reference}part of the equation. If you are looking for a gain you put the bigger number on top and divide by the smaller number. E.g., if you have a 50 watt radio and your amplifier puts out 100 watts, what is the gain? Well, in plain numbers your gain is 50 watts. That’s a doubling of your input power. If you plug those numbers into the formula as 100/50 you will get 2. That’s the multiplier. The log of 2 is the power you raise 10 to in order to get 2 as the answer. If you accidentally put the numbers the other way, 50/100, then you’d get 0.5 or 1/2. 100 watts is not half of 50 so you know right away that you did something wrong. So what decibel value represents that doubling of power? Well, with the original answer of 2 in the display (from dividing 100 by 50) press the log key and you should get a number like 0.3010299 etc. We’re almost done at this point, all that’s left is to multiply this answer by 10 to get, rounded off, 3. So a gain of twice our input power is 3dB. If you had instead used the numbers backwards, so 50/100 resulting in 1/2 (0.5) then when you pressed the log key you would have seen -0.3010299, which when multiplied by 10 would give you*minus*3dB!
Eventually most people come to terms with the formula for
calculating dB. The most important thing
to remember is that you are dividing 2 power values (or voltages if you are
working with voltages) to get the ratio of them. Gain is what you have when you divide the
bigger number by the smaller, and loss (minus dB) is what you get when you
divide the smaller by the larger.

Where a lot of people have the most trouble is when given a
decibel value and either asked to calculate the multiplication/division factor,
or given a dB value and a power and asked to calculate either the starting
power or ending power. For instance, if
I tell you that I have an amplifier that provides 7dB of gain no matter how
much power is put into it and ask you to tell me how much power out I’d get if
I fed it with 50 watts on its input how would you calculate that? You do it by working the dB formula
backwards. When you calculate the dB
value of a pair of power numbers, you first divided the 2 powers, pressed the
log key to get the log of that answer, then multiplied by 10 for the final
decibel figure. Here we start with the decibel
figure so the first thing we do is

*divide*that dB number by 10. So we have 7dB divided by 10 equals 0.7 (zero point seven). The second to last operation was pressing log, but when we divide by 10 to get 0.7*that IS the log*of the ratio of the 2 powers! We have to find out what that ratio is by using the inverse of the log function: 10^{X}. So with 0.7 showing on the calculator press the 2nd key and log (to access the 10^{X}function) and you should see 5.011872…, basically just 5. 5 is the ratio of the input to output power. If you are given the input power, as we were, you just multiply that by 5 to get the answer. Here we were given 50 watts in, so 50 times 5 equals 250 watts output! If you were given the output power and asked to calculate the input power to a 7dB amplifier, you would divide the output power by 5. Using the same example, we have a 7dB amplifier with 250 watts output. What’s the input power? It’s just 250 divided by 5 equals 50!
If you

*really*want to get into logarithms and exponents, check out this page: http://www.mathsisfun.com/algebra/exponents-logarithms.html. As always I invite discussion and questions. I really want you to understand this stuff so you can impress all your ham friends with your mastery of decibels.
Very well explained, but I have a question, on second thought let me read your essay again. peter

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