RF Math Samples
There are two basic types of RF math problems: conversions and effectives.
Conversions deal with converting back and forth between dBm and mW. Remember
the following:
When converting milliwatt values to dBm, express the milliwatt value as factors of 2 and
10.
For example: 80mw = 2 x 2 x 2 x 10. So applying the rules of 10s and 3s, you
substitute +3 for the 2s and +10 for the 10s resulting in +3+3+3+10 = 19dbm
When converting dBm values to milliwatts, express the dBm values as sums of 10 and
3.
For example: 27dbm = 10+10+10-3. Then begin at 1mW and apply the rules of
10s and 3s, multiplying or dividing by 10 or 2 as appropriate, thus resulting in
(1mWx10x10x10)/2 = 500mW.
Convert the values below to either dBm or mW as appropriate:
50mW
30dBm
15dBm
100mW
21dBm
Effectives deal with the effect of adding cables and antenna to an AP. Remember that
cables always introduce loss and antennas always introduce gain. Simply apply the
rules of 10s and 3s to the beginning AP power to arrive at the resulting radiated power.
For example: An AP is set to 35mw of power and is connected to a 12dBi gain
antenna with a cable having 3dBm of loss. Start at 35mW. The cable loss is
3dBm, so divide by 2 resulting in 17.5mw. The antenna gain is 12dBi, so
express 12 with sums of 10s and 3s resulting in 3+3+3+3. Therefore, multiply
17.5mW by 2 four times resulting in a radiated power of 280mW.
Solve these problems for radiated power:
1. An AP has an output power of 65mW and is connected directly to a 21dBi
antenna.
2. An AP has an output power of 50mW and is connected to a 9dBi antenna with a
cable having 6dBm of loss.
3. An AP has an output power of 20dBm and is connected to a 10dBi antenna with
a cable having 3dBm loss.
Answers to Sample Problems
Conversions
50mW= (10 x 10) / 2 = +10+10-3 = 17dBm
30dBm=+10+10+10 = 1x10x10x10 = 1,000mW or 1W
15dBm=+3+3+3+3+3=1x2x2x2x2x2= 32mW
100mW = 10 x 10 = +10+10 = 20dBm
21dBm = +10+10+10-3-3-3 = (((1x10x10x10)/2)/2)/2 = 125mW
Effectives
• 21dBi gain is (((1x10x10x10)/2)/2)/2, which results in
(((65mwx10x10x10)/2)/2)/2 which equals 8,125mW of radiated power.
• Net of antenna gain and cable loss is +3db, so 50mW x 2 = 100mW of
radiated power.
• First convert AP power to mW, resulting in 100mW. 3dBm of cable loss
and 10dBi of gain result in (100mW/2)x10=500mW of radiated power.
Remember that the rules of 10s and 3s are approximations, and there may be more
than one way to arrive at the answer which will result in slightly different answers.