Before you fully delve into the rule of 10s and 3s, it is important to know that this rule may not give you the exact same answers that you would get if you used the logarithmic formulas. The rule of 10s and 3s provides approximate values, not necessarily exact values.
If you are an engineer creating a product that must conform to RF regulatory guidelines, you will need to use logarithms to calculate the exact values. However, if you are a network designer planning a network for your company, you will fi nd that the rule of 10s and 3s will provide you with the numbers you need to properly plan your network.
This section will take you step-by-step through numerous calculations. All of the calcu- lations will be based on the following four rules of the 10s and 3s:
■ For every 3 dB of gain (relative), double the absolute power (mW).
■ For every 3 dB of loss (relative), halve the absolute power (mW).
■ For every 10 dB of gain (relative), multiply the absolute power (mW) by a factor of 10.
■ For every 10 dB of loss (relative), divide the absolute power (mW) by a factor of 10.
For example, if your access point is confi gured to transmit at 100 mW and the antenna is rated for 3 dBi of passive gain, the amount of power that will radiate out of the antenna (EIRP) will be 200 mW. Following the rule that you just learned, you will see that the 3 dB of gain from the antenna caused the 100 mW signal from the access point to double.
Conversely, if your access point is confi gured to transmit at 100 mW and is attached to a cable that introduces 3 dB of loss, the amount of absolute amplitude at the end of the cable will be 50 mW. Here you can see that the 3 dB of loss from the cable caused the 100 mW signal from the access point to be halved.
In another example, if your access point is confi gured to transmit at 40 mW and the antenna is rated for 10 dBi of passive gain, the amount of power that radiates out of the antenna (EIRP) will be 400 mW. Here you can see that the 10 dB of gain from the antenna caused the 40 mW signal from the access point to increase by a factor of 10. Conversely, if your access point is confi gured to transmit at 40 mW and is attached to a cable that intro- duces 10 dB of loss, the amount of absolute amplitude at the end of the cable will be 4 mW.
Here you can see that the 10 dB of loss from the cable caused the 40 mW signal from the access point to be decreased by a factor of 10.
If you remember these rules, you will be able to quickly perform RF calculations. After reviewing these rules, see Exercise 3.1, which will take you through a step-by-step pro- cedure for using the rule of 10s and 3s. As you work through the step-by-step procedure, remember that dBm is a unit of power and that dB is a unit of change. dB is a value of change that can be applied to dBm. So if you have a +10 dBm signal and it increases by 3 dB, you can add these two numbers together to get a result of +13 dBm signal.
E X E R C I S E 3 . 1
Step-by-Step Use of the Rule of 10s and 3s
1. On a sheet of paper, create two columns. The header of the fi rst column should be dBm, and the header of the second column should be mW.
dBm mW
2. Next to the dBm header, place a + sign and a – sign, and next to the mW header place a × sign and a ÷ sign.
These will help you to remember that all math performed on the dBm column is addi- tion or subtraction and all math performed on the mW column is multiplication or division.
×
dBm mW ÷
+ –
3. To the left of the + and – signs, write the numbers 3 and 10, and to the right of the × and ÷ signs, write the numbers 2 and 10.
Any addition or subtraction to the dBm column can be performed using only the numbers 3 and 10. Any multiplication or division to the mW column can be per- formed using only the numbers 2 and 10.
3 10
2
dBm mW 10
×
÷ +
–
4. If there is a + on the left, there needs to be an x on the right. If there is a – on the left, there needs to be a ÷ on the right.
5. If you are adding or subtracting a 3 on the left, you must be multiplying or dividing by a 2 on the right. If you are adding or subtracting a 10 on the left, you must be mul- tiplying or dividing by a 10 on the right.
6. The last thing you need to do is to put a 0 under the dBm column and a 1 under the mW column.
Remember that the defi nition of dBm is decibels relative to 1 milliwatt. So now the chart shows that 0 dBm is equal to 1 milliwatt.
3 10
2 10
×
dBm mW ÷
+ –
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Before we continue with other examples, it is important to emphasize that a change of ±3 dB equates to a doubling or halving of the power, no matter what power measure- ment is being used. In our usage of the rule of 10s and 3s, we are dealing with milliwatts because that is the typical transmission amplitude measurement used by 802.11 equipment.
However, it is important to remember that a +3 dB increase means a doubling of the power regardless of the power scale used. So a +3 dB increase of 1.21 gigawatts of power would result in 2.42 gigawatts of power.
An animated explanation of the rule of 10s and 3s—as well as explana- tions of each of the following examples—has been created using Microsoft PowerPoint and can be downloaded from this book’s online resource area that can be accessed at www.sybex.com/go/cwna4e. If you do not have PowerPoint on your computer, you can download from Microsoft’s website a PowerPoint Viewer that will allow you to view any PowerPoint file. The names of the PowerPoint files are displayed in the following list.
■ 10s and 3s Template.ppt
■ Rule of 10s and 3s Example 1.ppt
■ Rule of 10s and 3s Example 2.ppt
■ Rule of 10s and 3s Example 3.ppt
■ Rule of 10s and 3s Example 4.ppt
E X E R C I S E 3 . 2
Rule of 10s and 3s, Example 1
In this example, you will begin at 1 mW and double the power three times. In addition to calculating the new power level in milliwatts, you will calculate the power level in dBms.
1. The fi rst thing to do is create the initial chart, as in Exercise 3.1.
3 10
2 10
× dBm ÷
0
mW 1 +
–
E X E R C I S E 3 . 2 ( c o n t i n u e d )
2. Now, you want to double the power for the fi rst time. So to the right of the 1 mW and on the next line, write × 2. Then below the 1, perform the calculation.
×
÷ 3
10
2 10
× 2 +
– dBm
0
mW 1
= 2
3. You are not fi nished yet with this new line. Remember that for whatever is done to one side of the chart, there must be a correlative mathematical equation on the other side. Because you multiplied by 2 on the right side, you must add 3 to the left side.
So you have just calculated that +3 dBm is equal to 2 mW.
3 10
2 10
+ 3
×
÷
× 2 dBm
0
= 3
mW 1
= 2 +
–
4. You have just completed the fi rst doubling of the power. Now, you will double it two more times and perform the necessary mathematical commands. Since this is the fi rst time using this process, all of the steps have been shown using arrows. Future examples will not contain these arrows.
×
÷ 3
10
2 10
× 2
× 2 dBm
0 3
= 6
= 9
mW 1 2
= 4
= 8 +
–
+ 3 + 3
You have just calculated that 4 mW = +6 dBm, and 8 mW = +9 dBm. If you had used the conversion formula for dBm instead of the rule of 10s and 3s, the actual answers would be 4 mW = +6.0206 dBm, and 8 mW = +9.0309 dBm. As you can see, this set of rules is accu- rate but not exact.
E X E R C I S E 3 . 3
Rule of 10s and 3s, Example 2
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E X E R C I S E 3 . 3 ( c o n t i n u e d )
EIRP Data sent
100 mW
–3 dB
+10 dBi
Intentional radiator (IR)
As a reminder, and as seen in the graphic, the IR is the signal up to but not including the antenna, and the EIRP is the signal radiating from the antenna.
1. The fi rst step is to determine whether by using 10 or 2, and × or ÷, you can go from 1 mW to 100 mW.
It is not too diffi cult to realize that multiplying 1 by 10 twice will give you 100. So the bridge is generating 100 mW, or +20 dBm, of power.
+ –
×
÷ 3
10
2 10
× 10
× 10 + 10
+ 10 dBm
0 10 20
mW 1 10 100
2. Next you have the antenna cable, which is introducing –3 dB of loss to the signal.
After you calculate the effect of the –3 dB loss, you know the value of the IR. You can represent the IR as either +17 dBm or 50 mW.
+ –
×
÷ 3
10
2 10
× 10
× 10
÷ 2 + 10
+ 10 – 3
dBm 0 10 20 17
mW 1 10 100 50
E X E R C I S E 3 . 3 ( c o n t i n u e d )
3. Now, all that is left is to calculate the increase of the signal due to the gain from the antenna. Because the gain is 10 dBi, you add 10 to the dBm column and multiply the mW column by 10. This gives you an EIRP of +27 dBm, or 500 mW.
+ –
×
÷ 3
10
2 10
× 10
× 10
÷ 2
× 10 + 10
+ 10 – 3 + 10
dBm 0 10 20 17 27
mW 1 10 100 50 500
So far all of the numbers chosen in the examples have been straightforward, using the values that are part of the template. However, in the real world this will not be the case.
Using a little creativity, you can calculate gain or loss for any integer. Unfortunately, the rule of 10s and 3s does not work for fractional or decimal numbers. For those numbers, you need to use the logarithmic formula.
dB gain or loss is cumulative. If, for example, you had three sections of cable connect- ing the transceiver to the antenna and each section of cable provided 2 dB of loss, all three cables would create 6 dB of loss. Using the rule of 10s and 3s, subtracting 6 dBs is equal to subtracting 3 dBs twice. Decibels are very fl exible. As long as you come up with the total that you need, they don’t care how you do it.
Table 3.1 shows how to calculate all integer dB loss and gain from –10 to +10 by using combinations of just 10s and 3s. Take a moment to look at these values and you will realize that with a little creativity, you can calculate the loss or gain of any integer.
TA B L E 3 .1 dB Loss and gain (–10 through +10)
Loss or gain (dB) Combination of 10s and 3s
–10 –10
–9 –3 –3 –3
–8 –10 –10 +3 +3 +3 +3
–7 –10 +3
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Loss or gain (dB) Combination of 10s and 3s
–3 –3
–2 –3 –3 –3 –3 +10
–1 –10 +3 +3 +3
+1 +10 –3 –3 –3
+2 +3 +3 +3 +3 –10
+3 +3
+4 +10 –3 –3
+5 +10 +10 –3 –3 –3 –3 –3
+6 +3 +3
+7 +10 –3
+8 +10 +10 –3 –3 –3 –3
+9 +3 +3 +3
+10 +10
E X E R C I S E 3 . 4
Rule of 10s and 3s, Example 3
This example is a little more complicated than the previous ones. You have an access point that is transmitting at 50 mW. The signal loss between the access point and the antenna is –1 dB, and the access point is using a +5 dBi antenna. In this example, calculate the IR and the EIRP values.
TA B L E 3 .1 dB Loss and gain (–10 through +10) (continued)
E X E R C I S E 3 . 4 ( c o n t i n u e d )
EIRP Data sent
50 mW
–1 dB
+5 dBi
Intentional radiator (IR)
1. The fi rst step after drawing up the template is to convert the 1 mW to 50 mW. This can be done by multiplying the 1 mW by 10 twice and then dividing by 2.
2. The dBm column then needs to be adjusted by adding 10 twice and subtracting 3.
When the calculations are more complex, it is useful to separate and label the differ- ent sections.
×
÷ 3
10
2 10
× 10
× 10
÷ 2
Transmitter dBm
0 10 20 17
mW 1 10 100 50 +
–
– 3 + 10 + 10
3. The signal loss between the access point and the antenna is –1 dB. Table 3.1 shows that –1 dB can be calculated by subtracting 10 and adding 3 three times.
4. The mW column will need to be adjusted by dividing by 10 and then multiplying by 2 three times. So the IR is either +16 dBm or 40 mW.
3 10
2 10
× 10
× 10
÷ 2
÷ 10
× 2
× 2
× 2
Transmitter
Connector dBm
0 10 20 17
mW 1 10 100 50 7
10 13 16
5 10 20 40 +
–
– 10 + 3 + 3 + 3 + 10 + 10 – 3
×
÷
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E X E R C I S E 3 . 4 ( c o n t i n u e d )
5. The antenna adds a gain of 5 dBi. Table 3.1 shows that +5 dBi can be calculated by adding 10 twice and subtracting 3 fi ve times.
6. The mW column will need to be adjusted by multiplying by 10 twice and dividing by 2 fi ve times. The EIRP is therefore either +21 dBm or 125 mW.
Antenna +
–
×
÷ 3
10
2 10 + 10
+ 10 – 3
Transmitter – 10
+ 3 + 3 + 3
Connector dBm
0 10 20 17
mW 1 10 100 50 7
10 13 16
5 10 20 40
× 10
× 10
÷ 2
÷ 2
÷ 2
÷ 2
÷ 2
× 10
× 10
÷ 2
÷ 10
× 2
× 2
× 2 + 10
+ 10 – 3 – 3 – 3 – 3 – 3
26 36 33 30 27 24 21
400 4000 2000 1000 500 250 125
E X E R C I S E 3 . 5
Rule of 10s and 3s, Example 4
In this example, you have an access point that is providing coverage to a specifi c area of a warehouse via an external directional antenna. The access point is transmitting at 30 mW. The cable and connector between the access point and the antenna creates –3 dB of signal loss. The antenna provides 20 dBi of signal gain. In this example, you will calcu- late the IR and EIRP values.
EIRP Data sent
30 mW
–3 dB
+20 dBi
Intentional radiator (IR)
E X E R C I S E 3 . 5 ( c o n t i n u e d )
It is not always possible to calculate both sides of the chart by using the rule of 10s and 3s. In some cases, no matter what you do, you cannot calculate the mW value by using 10 or 2. This is one of those cases. You cannot set the mW and dBm values to be equal, but you can still calculate the mW values by using the information provided.
1. Instead of creating the template and setting 0 dBm equal to 1 mW, enter the value of the transmitter, in this case 30 mW.
2. In the dBm column, just enter unknown.
Even though you will not know the dBm value, you can still perform all of the neces- sary mathematics.
3 10
2 10
×
÷ +
– dBm
unknown
mW 30
3. The cable and connectors introduce 3 dB of loss, so subtract 3 from the dBm column and divide the mW column by 2. So the output of the IR is 15 mW.
3 10
2 10 – 3 ÷ 2
dBm unknown unknown – 3
mW 30 15
×
÷ +
–
4. The 20 dBi gain from the antenna increases the dBm by 20, so add 10 twice to the dBm column, and multiply the mW column by 10 twice. So the output of the EIRP is 1,500 mW. You can see in the graphic that the 20 dB gain by the antenna and the –3 dB of loss from the cable results in a 17 dB gain from the original dBm. Even though you do not know what the original dBm value is, you can see that it is 17 dB greater.
+ –
×
÷ 3
10
2 10
÷ 2
× 10
× 10 – 3
+ 10 + 10
dBm unknown unknown – 3 unknown + 7 unknown + 17
mW 30 15 150 1,500