1. Radio Frequency Components, Measurements, and Mathematics
Lesson 3
Radio Frequency Components, Measurements, and Mathematics
Dr. Tahseen Al-Doori

2. Objectives The Components of RF Communications Transmitter Receiver
Antenna
Isotropic Radiator
Intentional Radiator (IR)
Equivalent Isotropically Radiated Power (EIRP)
Dr. Tahseen Al-Doori

3. Objectives Units of Power and Comparison RF Mathematics
Watt
Milliwatt
Decibel (dB)
dBi
dBd
dBm
RF Mathematics
System Operating Margin (SOM)/Link Budget
Fade Margin
Inverse Square Law
Dr. Tahseen Al-Doori

4. RF Components
Many components contribute to the successful transmission and reception of an RF signal. Figure 1 shows the key components that will be covered in this section. In addition to understanding the function of the components, it is important to understand how the strength of the signal is specifically affected by each of the components.
Dr. Tahseen Al-Doori

5. Dr. Tahseen Al-Doori

6. Transmitter
The transmitter is the initial component in the creation of the wireless medium. The computer hands the data off to the transmitter, and it is the transmitter’s job to begin the RF communication.
The transmitter will take the data provided and modify the AC signal using a modulation technique to encode the data into the signal. This modulated AC signal is now a carrier signal, containing the data to be transmitted. The carrier signal is then transported either directly to the antenna or through a cable to the antenna.
Dr. Tahseen Al-Doori

7. In addition to generating a signal at a specific frequency, the transmitter is responsible for determining the amplitude, or what is more commonly referred to as the power level, of the signal.
The higher the amplitude of the wave, the more powerful the wave is and the further it will travel. The power levels that the transmitter is allowed to generate are determined by the local regulatory body, such as the Federal Communications Commission (FCC) in the United States.
Dr. Tahseen Al-Doori

8. Although we are explaining the transmitter and receiver separately, and although functionally they are different components, typically they are one device that is referred to as a transceiver (transmitter/receiver).
Typical wireless devices that have transceivers built into them are access points, bridges, and client adapters.
Dr. Tahseen Al-Doori

9. Antenna An antenna provides two functions in a communication system.
When connected to the transmitter, it collects the AC signal that it receives from the transmitter and directs, or radiates, the RF waves away from the antenna in a pattern specific to the antenna type.
When connected to the receiver, it takes the RF waves that it receives through the air and directs the AC signal to the receiver. The receiver converts the AC signal to bits and bytes. As you will see later, the signal that is received is much less than the signal that is generated.
This signal loss is analogous to two people trying to talk to each other from opposite ends of a football field. Due to distance alone (free space), the yelling from one end of the field may be heard as barely louder than a whisper on the other end.
Dr. Tahseen Al-Doori

10. The signal of an antenna is usually compared or referenced to an isotropic radiator .
An isotropic radiator is a point source that radiates signal equally in all directions. The sun is probably one of the best examples of an isotropic radiator. It generates equal amounts of energy in all directions.
Unfortunately, it is not possible to manufacture an antenna that is a perfect isotropic radiator. The structure of the antenna itself influences the output of the antenna, similar to the way the structure of a lightbulb affects the bulb’s ability to emit light equally in all directions.
Dr. Tahseen Al-Doori

11. There are two ways to increase the power output from an antenna.
The first is to generate more power at the transmitter, as stated in the previous section. The other is to direct, or focus, the RF signal that is radiating from the antenna. This is similar to how you can focus light from a flashlight. If you remove the lens from the flashlight, the bulb is typically not very bright and radiates in almost all directions. To make the light brighter, you could use more powerful batteries, or you could put the lens back on. The lens is not actually creating more light. It is focusing the light that was radiating in all different directions into a narrow area.
Some antennas radiate waves as the bulb without the lens does, while some radiate focused waves as the flashlight with the lens does.
Dr. Tahseen Al-Doori

12. Receiver The receiver is the final component in the wireless medium.
The receiver takes the carrier signal that is received from the antenna and translates the modulated signals into 1s and 0s. It then takes this data and passes it to the computer to be processed. The job of the receiver is not always an easy one. The signal that is received is a much less powerful signal than what was transmitted due to the distance it has traveled and the effects of free space path loss.
The signal is also often altered due to interference from other RF sources and multipath.
Dr. Tahseen Al-Doori

13. Intentional Radiator (IR)
intentional radiator (IR) is “a device that intentionally generates and emits radio frequency energy by radiation or induction.”
Basically, it’s something that is specifically designed to generate RF as opposed to something that generates RF as a byproduct of its main function, such as a motor that incidentally generates RF noise.
Dr. Tahseen Al-Doori

14. Regulatory bodies such as the FCC limit the amount of power that is allowed to be generated by an IR.
The IR consists of all the components from the transmitter to the antenna but not including the antenna, as see in Fig 1
The power output of the IR is thus the sum of all the components from the transmitter to the antenna, again not including the antenna.
Dr. Tahseen Al-Doori

15. The components making up the IR include the transmitter, all cables and connectors, and any other equipment (grounding, lightning arrestors, amplifiers, attenuators, etc.) between the transmitter and the antenna.
The power of the IR is measured at the connecter that provides the input to the antenna. Since this is the point where the IR is measured and regulated, we often refer to this point alone as the IR. Using the flashlight analogy, the IR is all of the components up to the lightbulb socket but not the bulb and lens. This is the raw power, or signal, that is provided, and now the bulb and lens can focus the signal.
Dr. Tahseen Al-Doori

16. Equivalent Isotropically Radiated Power (EIRP)
Equivalent isotropically radiated power (EIRP) is the highest RF signal strength that is transmitted from a particular antenna.
To understand this better, think of our flashlight example for a moment. Let’s assume that the bulb without the lens generates 1 watt of power. When you put the lens on the flashlight, it focuses that 1 watt of light. If you were to look at the light now, it would appear much brighter. If you were to measure the brightest point of the light that was being generated by the flashlight, due to the effects of the lens, it may be equal to the brightness of an 8 watt bulb. So by focusing the light, you are able to make the equivalent isotropically radiated power of the focused bulb equal to 8 watts.
Dr. Tahseen Al-Doori

17. Units of Power and Comparison
When an 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.
Here, I 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.
Dr. Tahseen Al-Doori

18. Comparative units of measurement are useful when working with units of power.
As you will see later, 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.
Dr. Tahseen Al-Doori

19. Units of Power Units of Comparison Watts (W) Decibal (dB)
Milliwatts (mW)
dBi
dBm
dBd
Dr. Tahseen Al-Doori

20. 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.
Dr. Tahseen Al-Doori

21. 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.
Dr. Tahseen Al-Doori

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

23. 10^1=10 log10(10)=1 10^2=100 log10(100)=2 10^3=1000 log10(1000)=3
Dr. Tahseen Al-Doori

24. Example
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 Pap and the power of laptop1 as PL1 . So the formula for this example would be y = log10(Pap/PL1). 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 (102 = 100).
Dr. Tahseen Al-Doori

25. To calculate decibels, all you need to do is multiply bels by 10
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(P1/P2)
decibels = 10 × log10(P1/P2)
Dr. Tahseen Al-Doori

26. If you plug in the power values, the formula becomes
Now let’s go back and calculate the decibels of the access point to laptop2 example.
So the formula now is
y = 10 × log10(Pap/PL1).
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.
Dr. Tahseen Al-Doori

27. 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 milliwatts.
Dr. Tahseen Al-Doori

28. dBi
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.
Dr. Tahseen Al-Doori

29. Any time you see dBi, think antenna gain.
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.
 Any time you see dBi, think antenna gain.
Dr. Tahseen Al-Doori

30. 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.
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?
Dr. Tahseen Al-Doori

31. This is actually quite simple.
A standard dipole antenna has a dBi value of 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 So a 3 dBd antenna is equal to a 5.14 dBi.
When working with equipment, it is not often that you will have an antenna with a dBd value antennas typically are measured using dBi. On the rare occasion that you do run into one, just add 2.14 and you now know the antenna’s dBi value.
Dr. Tahseen Al-Doori

32. dBm
dBm 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.
Dr. Tahseen Al-Doori

33. RF Mathematics To convert: mW to dBm use this formula: dBm=10Log(mW)
dBm to mW use this formula
mW=10^(dBm/10)
dBi to dBd
dBd=dBi-2.14
mW to dBW
dBW=10Log(mW/1)
Dr. Tahseen Al-Doori

34. Exercises 1. convert the following: 400mW ______dBm 14dBm ______mW
7dBW ______dBm
250mW ______dBW
Dr. Tahseen Al-Doori

35. Calculator http://www.terabeam.com/support/calculations/index.php
Dr. Tahseen Al-Doori

36. Decibels: Do them in your head
The 1,3,10 Rule for converting between whole decibels and actual powers without a calculator!
dB Increase Decrease
1 dB x x 0.8
3 dB x2 x 0.5
10 dB x10 x 0.1
Dr. Tahseen Al-Doori

37. Example 35dB = 30dB+3dB+3dB-1dB = 1000x2x2x0.8 =3200mW or 3.2W Try it:
Convert 24dBm to _______mW
Convert 0.5W to _______dBm
Convert 47dBm to ______W
Dr. Tahseen Al-Doori

38. To add or not to add In mW you just add and subtract normally.
In dBm you convert to mW then add then convert to dBm.
37 dBm + 37 dBm=? dBm
Dr. Tahseen Al-Doori

39. Q1 2 dBd is equal to how many dBi? 5 dBi 4.41 dBi 4.14 dBi
The value cannot be calculated.
To convert any dBd value to dBi, simply add 2.14 to the dBd value.
Dr. Tahseen Al-Doori

40. Q2 23 dBm is equal to how many mW? 200 mW 14 mW 20 mW 23 mW 400 mW
To convert to mW, first calculate how many 10s and 3s are needed to add up to 23, which is To calculate the mW, you must multiply 1 × 10 × 10 × 2, which calculates to 200 mW.
Dr. Tahseen Al-Doori

41. Q3
A wireless bridge is configured to transmit at 100 mW. The antenna cable and connectors produce a 3 dB loss and are connected to a 16 dBi antenna. What is the EIRP?
20 mW
30 dBm
2,000 mW
36 dBm
8 W
Dr. Tahseen Al-Doori

42. Multiplying by two 10s will accomplish this.
To reach 100 mW, you can use 10s and 2s and multiplication and division.
Multiplying by two 10s will accomplish this.
This means that on the dBm side, you must add two 10s, which equals 20 dBm.
Then subtract the 3 dB of cable loss for a dBm of 17.
Since you subtracted 3 from the dBm side, you must divide the 100 mW by 2, giving you a value of 50 mW.
Now add in the 16 dBi by adding a 10 and two 3s to the dBm column, giving a total dBm of 33.
Since you added a 10 and two 3s, you must multiply the mW column by 10 and two 2s, giving a total mW of 2,000, or 2 W.
Dr. Tahseen Al-Doori

43. Q4
Which of the following are valid calculations when using the rule of 10s and 3s? (Choose all that apply.)
Multiply dBm by 3 and add 2 to mW.
Add 10 to dBm and multiply mW by 10.
Add 3 to dBm and multiply mW by 3.
Subtract 10 from dBm and divide mW by 10.
Divide dBm by 10 and subtract 10 from mW.
Dr. Tahseen Al-Doori

44. The only valid calculations are as follows:
Add 3 to dBm and multiply mW by 2.
Subtract 3 from dBm and divide mW by 2.
Add 10 to dBm and multiply mW by 10.
Subtract 10 from dBm and divide mW by 10.
Dr. Tahseen Al-Doori

45. System Operating Margin (SOM)/ Link Budget
SOM is the calculation of the amount of RF signal that is received minus the amount of signal required by the receiver.
The figure on next slide shows all the components and their effects on the SOM, and this is called receiver sensitivity.
Manufacturers determine the receiver sensitivity for each speed supported by the wireless card.
Dr. Tahseen Al-Doori

46. Dr. Tahseen Al-Doori

47. The lower No. the weaker the signal the more reliable the card.
Different speeds use different modulation techniques and encoding methods and some encoding methods are more susceptible to corruption.
The manufacturers carries out tests to determine the least signal required by the receiver to be acceptable as a proper signal.
They look at the figures above and below the threshold. Then give the Rx of the card.
The lower No. the weaker the signal the more reliable the card.
Dr. Tahseen Al-Doori

48. 1Mbps -94 dBm 2 Mbps -93 dBm 5.5 Mbps -92 dBm 6 Mbps -86 dBm 9 Mbps
As you can see we are dealing with negative nos.
Therefore, -71 dBm is the highest Rx
Dr. Tahseen Al-Doori

49. Typically, the faster the speed, the higher the receiver sensitivity.
Have you noticed that we are dealing with negative nos., and so far we have been working with positive dBm.
WHY?
Dr. Tahseen Al-Doori

50. Case 1
The Diagram below shows simple summary of gains and losses in an office.
Dr. Tahseen Al-Doori

51. Until now we have worked primarily with calculating the IR and EIRP.
It is the effect of free space path loss that makes the values negative, as you will see in the calculations based upon the earlier diagram.
Dr. Tahseen Al-Doori

52. The link budget is equal to the received signal minus the receive sensitivity.
In this example, the received signal is the sum of all components, which is
20 dBm + 5 dBi – dB dBi = –46.84 dBm
If the receive sensitivity of the laptop’s radio is –71 dBm, then the link budget is
–46.84 dBm – (–71 dBm) = dBm
Dr. Tahseen Al-Doori

53. Case 2
Dr. Tahseen Al-Doori

54. In this case, the two antennas are 10 kilometers apart.
Let’s look at the SOM of a point-to-point wireless network, as seen in the Figure .
In this case, the two antennas are 10 kilometers apart.
In addition to the effects of the antennas and cables, there are also lightning arrestors.
Assume that the receiver sensitivity is –80 dBm.
Dr. Tahseen Al-Doori

55. Lightning arrestor –.1 dB 50' LMR 600 cable –2.21 dB
In this configuration, the calculation for the link budget is as follows:
Transceiver 10 dBm
10' LMR 600 cable –.44 dB
Lightning arrestor –.1 dB
50' LMR 600 cable –2.21 dB
Parabolic antenna +25 dBi
FSPL –120 dB
Total signal –65.5 dBm
So the SOM is
–65.5 dBm – (–80 dBm) = 14.5 dBm
Dr. Tahseen Al-Doori

56. Fade Margin
Fade margin is a level of desired signal above what is required. Also called the comfort zone.
If a receiver has a receive sensitivity of –80 dBm, then as long as the signal received is greater than –80 dBm, the transmission will be successful.
The problem is that the signal being received fluctuates due to many outside influences.
Dr. Tahseen Al-Doori

57. In order to accommodate for the fluctuation, it is a common practice to add 10 to 20 dBs to the receive sensitivity value. The additional value that is added is known as the fade margin.
Dr. Tahseen Al-Doori

58. Example
Receiver has a sensitivity of –80 dBm and a signal is typically received at –76 dBm. Then under normal circumstances, this communication is successful.
However, due to outside influences, the signal may fluctuate by ± 5 dBm. This means that most of the time, the communication is successful, but on those occasions that the signal has fluctuated to –81 dBm, the communication will be unsuccessful.
Dr. Tahseen Al-Doori

59. By adding a fade margin of 10 dBm, you are now stating that for your needs, the receive sensitivity is –70 dBm, and you will plan your network so that the received signal is greater than –70 dBm.
If the received signal fluctuates, you have already built in some padding, in this case 10 dBm.
Dr. Tahseen Al-Doori

60. If you look back at the diagram and added a fade margin of 10 dBm to the receive sensitivity of –80 dBm, then the amount of signal required for the link would be –70 dBm.
Since the signal is calculated to be received at –65.5 dBm, you will have a successful communication.
However if you chose a fade margin of 15 dBm, the amount of signal required would be –65 dBm, and based upon the configuration in the diagram, you would not have enough signal to satisfy the link budget plus the 15 dBm fade margin.
Dr. Tahseen Al-Doori

61. Conclusion of Fade Margin
Since RF communications can be affected by many outside influences, it is common to have a fade margin to provide a level of link reliability. By increasing the fade margin, you are essentially increasing the reliability of the link.
Dr. Tahseen Al-Doori

62. Inverse Square Law
Earlier we learned about the 6 dB rule, which states that a +6 dB change in signal will double the usable distance of a signal and a –6 dB change in signal will halve the usable distance of a signal.
This rule and these numbers are actually based upon the Inverse Square Law, originally developed by Isaac Newton. This law states that the change in power is equal to 1 divided by the square of the change in distance.
Dr. Tahseen Al-Doori

63. So the power at 2 feet is 4 mW × 1/4, which is equal to 1 mW.
What this means is that if you are receiving a signal at a certain power level and a certain distance (D) and you were to double the distance (2 × D), then the new power level will change by 1 / (2 × D)^2.
If at a distance of 1 feet (call this D) you were receiving a signal of 4 mW, then at a distance of 2 feet (2 × D) the power would change by 1 / 2^2, which is 1/4.
So the power at 2 feet is 4 mW × 1/4, which is equal to 1 mW.
Dr. Tahseen Al-Doori

64. To see how this relates to the 6 dB rule, using the rule of 10s and 3s, consider that to change from 4 mW to 1 mW, you would need to divide the mW column by 2 twice.
This would require you to subtract 3 twice from the dBm column, giving you a –6 dBm change caused by the doubling of the distance of the signal.
Dr. Tahseen Al-Doori

65. Dr. Tahseen Al-Doori