When an 802.11 wireless network is designed, two key components are coverage and performance. A good understanding of RF power, comparison, and RF mathematics can be very helpful during the network design phase.

We will introduce you to an assortment of units of power and units of comparison. It is important to know and understand the different types of units of measurement and how they relate to each other.

Some of the numbers that you will be working with will represent actual units of power and others will represent relative units of comparison. Actual units are ones that represent a known or set value.

To say that a man is 6 feet tall is an example of an actual measurement. Since the man’s height is a known value, in this case feet, you know exactly how tall he is. Relative units are comparative values comparing one item to a similar type of item.

For example, if you wanted to tell someone how tall the man’s wife is using comparative units of measurement, you could say that she is 5/6 his height. You now have a comparative measurement: if you know the actual height of either one, you can then determine how tall the other is.

Comparative units of measurement are useful when working with units of power. We can use these comparative units of power to compare the area that one access point can cover versus another access point.

Using simple mathematics, we can determine things such as how many watts are needed to double the distance of a signal from an access point. List below categorizes the different units.

- Watt (W) - Decibel (dB)
- Milliwatt (mW) - dBi
- dBm - dBd

### Watt

A watt is the basic unit of power, named after James Watt, an eighteenth-century Scottish inventor. One watt is equal to 1 ampere (amp) of current flowing at 1 volt. To give a better explanation of a watt, we will use a modification of the classic water analogy.

Many of you are probably familiar with a piece of equipment known as a power washer. If you are not familiar with it, it is a machine that connects to a water source, such as a garden hose, and allows you to direct a stream of high-pressure water at an object, with the premise that the fast-moving water will clean the object.

The success of a power washer is based upon two components: the pressure that is applied to the water and the volume of water that is used over a period of time, also known as flow.

These two components provide the power of the water stream. If you increase the pressure, you will increase the power of the stream. If you increase the flow of the water, you will also increase the power of the stream.

The power of the stream is equal to the pressure times the flow. A watt is very similar to the output of the power washer. Instead of the pressure generated by the machine, electrical systems have voltage.

Instead of water flow, electrical systems have current, which is measured in amps. So the amount of watts generated is equal to the volts times the amps.

### Milliwatt (mW)

A milliwatt (mW) is also a unit of power. To put it simply, a milliwatt is 1/1,000 of a watt. The reason you need to be concerned with milliwatts is because most of the 802.11 equipment that you will be using transmits at power levels between 1 and 100 mW.

Although regulatory bodies such as the FCC may allow power outputs of as much as 4 watts, only rarely in point-topoint communications, such as in building-to-building environments, would you use 802.11 equipment with more than 250 mW of power.

### Decibel (dB)

The first thing you should know about the decibel (dB) is that it is a unit of comparison, not a unit of power. Therefore, it is used to represent a difference between two values.

In wireless networking, decibels are often used to either compare the power of two transmitters or, more often, compare the difference or loss between the EIRP output of a transmitter and the amount of power received by the receiver.

Decibel is derived from the term bel. Employees at Bell Telephone Laboratories needed a way to represent power losses on telephone lines as power ratios. They defined a bel as the ratio of 10 to 1 between the power of two sounds.

Let’s look at an example: An access point transmits data at 100 mW. Laptop1 receives the signal at a power level of 10 mW, and laptop2 receives the signal at a power level of 1 mW.

The difference between the signal from the access point (100 mW) to laptop1 (10 mW) is 100:10, or a 10:1 ratio, or 1 bel. The difference between the signal from laptop1 (10 mW) to laptop2 (1 mW) is 10:1, also a 10:1 ratio, or 1 bel.

So the power difference between the access point and laptop2 is 2 bels. Bels can be looked at mathematically using logarithms. Not everyone understands or remembers logarithms, so we will review them.

First, we need to look at raising a number to a power. If you take 10 and raise it to the third power (10 3 = y), what you are actually doing is multiplying three 10s (10 × 10 × 10).

If you do the math, you will calculate that y is equal to 1,000. So the completed formula is 10 3 =1,000. When calculating logarithms, you change the formula to 10 y = 1,000. Here you are trying to figure out what power 10 needs to be raised to in order to get to 1,000. You know in this example that the answer is 3.

You can also write this equations as y = log10(1,000) or y = log 10 1,000. So the complete equation is 3 = log10(1,000). Here are some examples of power and log formulas:

10^{1} = 10 | log10(10) = 1 |

10^{2} = 100 | log10(100) = 2 |

10^{3} = 1,000 | log10(1,000) = 3 |

10^{4} = 10,000 | log10(10,000) = 4 |

Now let’s go back and calculate the bels from the access point to the laptop2 example using logarithms. Remember that bels are used to calculate the ratio between two powers. So let’s refer to the power of the access point as P AP and the power of laptop1 as P L1 .

So the formula for this example would be y = log10(P AP /P L1 ). If you plug in the power values, the formula becomes y = log10(100/1), or y = log10(100). So this equation is asking 10 raised to what power equals 100. The answer is 2 bels (10 2 = 100).

OK, so this is supposed to be about decibels and so far we have just covered bels. In certain environments, bels are not exact enough, which is why we use decibels instead. To calculate decibels, all you need to do is multiply bels by 10. So the formulas for bels and decibels are as follows:

bels = log10(P_{1}/P_{2} )

decibels = 10 × log10(P_{1}/P_{2} )

Now let’s go back and calculate the decibels of the access point to laptop2 example. So the formula now is y = 10 × log10(P AP /P L1 ). If you plug in the power values, the formula becomes y = 10 × log10(100/1), or y = 10 × log10(100). So the answer is 20 decibels.

Now that you have learned about decibels, you are probably still wondering why you can’t just work with milliwatts. You can if you want, but since power changes are logarithmic, the differences between values can become extremely large and more difficult to deal with.

It is easier to say that a 100 mW signal decreased by 70 decibels than to say that it decreased to .00001 milliwatts. Table below compares milliwatts and decibel change, using 1 mW as the reference point. Due to the scale of the numbers, you can see why decibels can be easier to work with.

milliwatts | decibel change |

.0001 | –40 |

.001 | –30 |

.01 | –20 |

.1 | –10 |

1 | 0 |

10 | +10 |

100 | +20 |

1,000 | +30 |

10,000 | +40 |

100,000 | +50 |

### dBi

Earlier, we compared an antenna to an isotropic radiator. Theoretically, an isotropic radiator can radiate an equal signal in all directions. An antenna cannot do this due to construction limitations.

In other instances, you do not want an antenna to radiate in all directions because you want to focus the signal of the antenna in a particular direction.

Whichever the case may be, it is important to be able to calculate the radiating power of the antenna so that you can determine how strong a signal is at a certain distance from the antenna.

You may also want to compare the output of one antenna to that of another. The gain or increase of power from an antenna when compared to what an isotropic radiator would generate is known as decibels isotropic (dBi).

Another way of phrasing this is “decibel gain relative to an isotropic radiator.” Since you are comparing two antennas, and since antennas are measured in gain, not power, you can conclude that dBi is a comparative measurement and not a power measurement.

The dBi value is measured at the strongest point or the focus point of the antenna signal. Since antennas always focus their energy more in one direction than another, the dBi value of an antenna is always a positive gain and not a loss.

A common antenna used on access points is the half-wave dipole antenna. The half-wave dipole antenna is a small, typically rubber-encased, general purpose antenna and has a dBi value of 2.14.

### dBd

The antenna industry uses two different dB scales to describe the gain of antennas. The first scale, which you just learned about, is dBi, which is used to describe the gain of an antenna relative to a theoretical isotropic antenna.

The other scale used to describe antenna gain is decibels dipole (dBd), or “decibel gain relative to a dipole antenna.” So a dBd value is the increase in gain of an antenna when it is compared to the signal of a dipole antenna.

Like dBi, since dBd is comparing two antennas, and since antennas are measured in gain, not power, you can also conclude that dBd is a comparative measurement and not a power measurement.

The definition of dBd seems simple enough, but what happens when you want to compare two antennas and one is represented with dBi and the other with dBd? This is actually quite simple.

A standard dipole antenna has a dBi value of 2.14. If an antenna has a value of 3 dBd, this means that it is 3 dB greater than a dipole antenna. Since the value of a dipole antenna is 2.14 dBi, all you need to do is add 3 plus 2.14. So a 3 dBd antenna is equal to a 5.14 dBi.

### dBm

Earlier when you read about bels and decibels, you learned that they measured differences or ratios between two different signals. Regardless of the type of power that was being transmitted, all you really knew was that the one signal was greater or less than the other by a particular number of bels or decibels.

dBm also provides a comparison, but instead of comparing a signal to another signal, it is used to compare a signal to 1 milliwatt of power. dBm means “decibels relative to 1 milliwatt.” So what you are doing is setting dBm to 0 (zero) and equating that to 1 milliwatt of power.

Since dBm is a measurement that is compared to a known value, 1 milliwatt, then dBm is actually a measure of power. You can now state that 0 dBm is equal to 1 milliwatt. Using the formula dBm = 10 × log10(PmW), you can determine that 100 mW of power is equal to 20 dBm.

If you happen to have the dBm value of a device and want to calculate the corresponding milliwatt value, you can do that too. The formula is PmW=log–1(PdBm ÷ 10). Another, easier way of looking at the formula, and an easier way of using it, is PmW = 10 (PdBm ÷ 10).

Just divide the dBm value by 10, and raise 10 to that power. If you are given a value of 20 dBm, to convert it to mW, the formula is PmW = 10(20 ÷ 10) or PmW = 102 = 100 mW. It might seem a little ridiculous to have to deal with both milliwatts and dBm.

If milliwatts are a valid measurement of power, then why not just use them? Why do you have to, or want to, also use dBm? These are good questions that are asked often by students. A very practical reason to use dBm can be shown using the free space path loss formula again.

Following are two free space path loss equations. The first equation calculates the decibel loss of a 2.4GHz signal at 100 meters (.1 kilometer) from the RF source, and the second calculates the decibel loss of a 2.4GHz signal at 200 meters (.2 kilometer) from the RF source:

fspl = 32.4 + (20log10(2400)) + (20log10(.1)) = 80.00422 dB

fspl = 32.4 + (20log10(2400)) + (20log10(.2)) = 86.02482 dB

In this example, by doubling the distance from the RF source, the signal decreased by about 6 dB. If you double the distance between the transmitter and the receiver, the received signal will decrease by 6 dB.

No matter what numbers are chosen, if the distance is doubled, the decibel loss will be 6 dB. This rule also implies that if you increase the EIRP by 6 decibels, the usable distance will double.

This “6 dB rule” is very useful for comparing cell sizes or estimating the coverage of a transmitter. Remember, if you were working with milliwatts, this rule would not be relevant. By converting milliwatts to dBm, you have a more practical way to compare signals.

Using dBm also makes it very easy to calculate the effects of antenna gain on a signal. If a transmitter generates a 20 dBm signal and the antenna adds 6 dBi of gain to the signal, then the power that is radiating from the antenna (EIRP) is equal to the sum of the two numbers, which is 26 dBm.