Radio Frequency Calculus
Last Post: September 21, 2006:

Hi Rafael:
In this example:
50 mW (milliwatts) is a measure of power
3 dB is a pure number representing a loss factor
6 dBi is a pure number representing a gain factor
For the math folks:
3 dB is ten to the power of .3 which is approximately the same as 2.
Dividing by 3 dB is approximately the same as dividing by 2.
6 dBi is ten to the power of .6 which is approximately the same as 4.
Multiplying by 6 dBi is approximately the same as multiplying by 4.
So starting with 50 mW of power, divide by 2 and multiply by 4 to find the final power is 100 mW.
This example demonstrates the popular "rule of 3's". Whenever losing or gaining by a factor of 3 dB (or dBi or dBd), just divide or multiply by 2.
The other popular rule is the "rule of 10's". Whenever losing or gaining by a factor of 10 dB, just divide or multiply by 10.
Many CWNP Program exam questions involving RF math are readily solved using these two rules.
Warning: Some writers insist that decibels are always about comparisons. I find it easier to think of decibels as pure numbers which sometimes involve "comparisons" whatever that means. But I am in the minority. A fun example is that 30 dB of cattle (1000) can trample you into dust while 30 dB of cattle (1/1000th) might make a decent meal! These look like pure numbers to me. If I had to find two things to "compare" each time I used decibels I would get confused a lot.
Some writers insist that numbers represented as decibels are "added". Actually decibels are only a useful notation when the numbers they represent are factors in a multiplication/division problem such as problems involving RF gains and loses. Decibel numbers are so very convenient as factors because when multiplying two of them one merely adds their values, just as we do with exponents of a common base. For example 10 dB x 20 dB equals 30 dB. Some are content to write 10 dB + 20 dB equals 30 dB. I find this to be confusing as well.
Treating decibels as pure numbers helps me make sense of this restatement of the above problem:
17 dB milliwatts / 3 dB * 6 dBi = 20 dB milliwatts
Some writers show not two but four rules of 10's and 3's, two for the gains and two for the loses. I am content with two rules but have to be careful which are the loses (divisions) and which are the gains (multiplications).
I hope this helps. Thanks. /criss
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