airMAX - How are decibels (dB) used in radio systems?

Overview


Readers will learn how to calculate real-world values from decibels (abstract representations). This is particularly important in the RF world where decibels are used in link budgets.

What are decibels?

Decibels are a convenient way of expressing very large or very small numbers. They take an exponential curve and make it look like a straight line. By itself, a decibel means nothing. Decibels are simply used when representing something. Anything! Here are just a few ways decibels can represent "real-world" values:

  • "dBm", a power ratio. 0 "dBm" references to 1mW
  • "dBp", a population ratio. 0 "dBp" references to 1 person
  • "dBo", a molecular ratio. 0 "dBo" references to 1 oxygen molecule

Do you notice the pattern? Zero "decibels of anything" = 1 of anything.


How are decibels used?

Decibels represent ratios composed of two sides (decibel side & "real-world" side). Whatever operation you perform on the decibel side, you must perform the relative function on the real-world side (and vice-versa). There are rules that must be followed when working with either decibels or "real-world" values:

  • The decibel side has two functions: addition & subtraction.
  • The real-world side has two functions: multiplication & division.
  • Every time decibels are added, you must multiply the real-world figure.
  • And every time decibels are subtracted, you must divide the real-world figure.
  • That's it!

There is a concept known as "the rule of 3's & 10's" which allows you to arrive at any "real-world" number by adding/subtracting 3 and/or 10 decibels.

  • +3dB = x2 real-world number
  • -3dB = ÷2 real-world number
  • +10dB = x10 real-world number
  • -10dB = ÷10 real-world number

So if you're trying to arrive at 30dB of something (maybe 30dBm or 30dBo or 30dBz), you will multiple the real-world figure by 1000! (10dB + 10dB + 10dB = x10 x10 x10).

Examples


A few examples are provided for practice.

A radio transmitter is outputting 13dBm. How many milliwatts is this?

Where do you start when calculating decibels? When dealing with decibels, always start at ZERO. After all, 0dB = 1 of anything, right?

We start at 0dBm. How many milliwatts do we have total?
0dBm = 1mW. Now we need to get from 0dBm to 13dBm. According to the "rule of 3's & 10's," we can't add "13dBm" (it's not the rule of 13's). But, if we add by 10dBm then add by 3dBm, we have added a total of 13dBm, using just 3's & 10's! 0dBm = 1mW +3dBm = x2 _____________ 3dBm = 2mW (we're not done yet) 3dBm = 2mW +10dBm = x10 _____________ 13dBm = 20mW So, 13dBm equals 20mW.
Now try arriving at 69dBm—get ready, it's a really big number! Always start at ZERO!
0dBm = 1mW
Now what would be easiest? We could add by 3 all day and finally arrive at 69dBm, or we could add by 10 to get to 69dBm more quickly.
0dBm = 1mW
+10dBm = x10
_____________
10dBm = 10mW
10dBm = 10mW
+10dBm = x10
_____________
20dBm = 100mW
20dBm = 100mW
+10dBm = x10
_____________
30dBm = 1,000mW
30dBm = 1000mW
+10dBm = x10
_____________
40dBm = 10,000mW
40dBm = 10,000mW
+10dBm = x10
_____________
50dBm = 100,000mW
50dBm = 100,000mW
+10dBm = x10
_____________
60dBm = 1,000,000mW (really big number!)
Pause!
We're almost there. Now if we add 10dBm to 60dBm, we'll arrive at 70dBm. And since we can't subtract 1dBm from 70dBm (only 3's & 10's!), we will have to add by 3dBm now:
60dBm = 1,000,000mW
+3dBm = x2
_____________
63dBm = 2,000,000mW
63dBm = 2,000,000mW
+3dBm = x2
_____________
66dBm = 4,000,000mW
66dBm = 4,000,000mW
+3dBm = x2
_____________
69dBm = 8,000,000mW
So, 69dBm equals 8,000,000mW.

How did we arrive at these answers? In both examples, we started at ZERO decibels and used only 3's & 10's to arrive at our answer. We've proven that it's very easy to arrive at a number that is divisible by 3 or 10. What about numbers like 7dBm? Or 8dBm? Or 17dBm?

How many milliwatts does a 17dBm transmitter produce? . . .Start at ZERO!
0dBm = 1mW
+10dBm = x10
_____________
10dBm = 10mW

Pause! We're still not at 17dBm. But if we add by 10, we'll be at 20dBm. And if we add by 3 twice, we'll arrive at 16dBm. And we can't add by 1! Only 3's & 10's. This is where subtraction becomes important. If we add by 10, we arrive at 20dBm. But, we can subtract 3 from 20, and we arrive at 17dBm! Just remember to divide the "real-world side" whenever you subtract from the decibel side. 10dBm = 10mW +10dBm = x10 _____________ 20dBm = 100mW 20dBm = 100mW -3dBm = ÷2 _____________ 17dBm = 50mW

So, 17dBm = 50mW. There are times when to arrive at the answer, you'll need to add as well as subtract to arrive at the final answer. Here's another example:

21dBm = ??? mW? Start at ZERO.
0dBm = 1mW
+10dBm = x10
_____________
10dBm = 10mW
10dBm = 10mW
+10dBm = x10
_____________
20dBm = 100mW
20dBm = 100mW
+10dBm = x10
_____________
30dBm = 1,000mW
30dBm = 1,000mW
-3dBm = ÷2
_____________
27dBm = 500mW
27dBm = 500mW
-3dBm = ÷2
_____________
24dBm = 250mW
24dBm = 250mW
-3dBm = ÷2
_____________
21dBm = 125mW
So, 21dBm = 125mW.

General concepts to remember:

  • +3dB...........x2
  • +10dB.........x10
  • -3dB............÷2
  • -10dB..........÷10
  • Always start at ZERO!
  • Also, when calculating a decibel with "minus" sign in front (ex. -85dBm), understand that this is NOT a negative value! It simply represents a fraction! (-85dBm = .0000000032mW)
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