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:
That “log” really freaks people out,
especially if you've never taken a math class that included logarithms in the curriculum. Logarithms are a way of
multiplying or dividing large numbers by adding smaller ones. Another area of confusion is in the descriptions
of the inverse operation, that is, taking a decibel number and working it
backwards to get to the number that was multiplied or divided by.
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 “10X.”
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 10X key you will get the result of 102 (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 log10. So the log10 of 100 is 2 because
10 to the 2 power is 100. All you really
have to understand is that the 10X 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 10X 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 Pmeasured/Preference 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: 10X. So with 0.7 showing on the calculator press
the 2nd key and log (to access the 10X 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.
