Lesson 3
Radio Frequency Components,
Measurements, and Mathematics
Dr. Tahseen Al-Doori
Objectives
 The Components of RF Communications
– Transmitter
– Receiver
– Antenna
– Isotropic Radiator
– Intentional Radiator (IR)
– Equivalent Isotropically Radiated Power (EIRP)
Dr. Tahseen Al-Doori
Objectives
 Units of Power and Comparison
–
–
–
–
–
–
Watt
Milliwatt
Decibel (dB)
dBi
dBd
dBm
 RF Mathematics
 System Operating Margin (SOM)/Link Budget
 Fade Margin
 Inverse Square Law
Dr. Tahseen Al-Doori
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
Dr. Tahseen Al-Doori
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
 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
 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
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
 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
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
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
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
 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
 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
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
Units of Power and Comparison
 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.
 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
 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
Units of Power
Units of Comparison
Watts (W)
Decibal (dB)
Milliwatts (mW)
dBi
dBm
dBd
Dr. Tahseen Al-Doori
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
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
 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
 10^1=10
 10^2=100
 10^3=1000
 10^4=10,000
log10(10)=1
log10(100)=2
log10(1000)=3
log10(10,000)=4
Dr. Tahseen Al-Doori
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
 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
 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
 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.
Dr. Tahseen Al-Doori
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
 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
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
 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.
 When working with 802.11 equipment, it is not
often that you will have an antenna with a dBd
value. 802.11 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
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
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
Exercises





1. convert the following:
400mW ______dBm
14dBm ______mW
7dBW ______dBm
250mW ______dBW
Dr. Tahseen Al-Doori
Calculator
 http://www.terabeam.com/support/calculatio
ns/index.php
Dr. Tahseen Al-Doori
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 1.25
x 0.8
3 dB
x2
x 0.5
10 dB x10
x 0.1
Dr. Tahseen Al-Doori
Example

35dB = 30dB+3dB+3dB-1dB
= 1000x2x2x0.8
=3200mW or 3.2W
 Try it:
1. Convert 24dBm to _______mW
2. Convert 0.5W to _______dBm
3. Convert 47dBm to ______W
Dr. Tahseen Al-Doori
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
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
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 0 + 10
+ 10 + 3. To calculate the mW, you must multiply 1
× 10 × 10 × 2, which calculates to 200 mW.
Dr. Tahseen Al-Doori
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
 8W
Dr. Tahseen Al-Doori
 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
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
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
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
Dr. Tahseen Al-Doori
 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
1Mbps
-94 dBm
2 Mbps
5.5 Mbps
6 Mbps
-93 dBm
-92 dBm
-86 dBm
9 Mbps
-86 dBm
18 Mbps
-86 dBm
36 Mbps
-80 dBm
54 Mbps
-71 dBm
As you can see we are dealing with negative nos.
Therefore, -71 dBm is the highest Rx
Dr. Tahseen Al-Doori
 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
Case 1
 The Diagram below shows simple summary
of gains and losses in an office.
Dr. Tahseen Al-Doori
 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
 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 – 73.98 dB + 2.14 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) = 24.16 dBm
Dr. Tahseen Al-Doori
Case 2
Dr. Tahseen Al-Doori
 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
 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
 Parabolic antenna
+25 dBi
 50' LMR 600 cable
–2.21 dB
 Lightning arrestor–.1 dB
 10' LMR 600 cable
–.44 dB
 Total signal
–65.5 dBm
 So the SOM is
 –65.5 dBm – (–80 dBm) = 14.5 dBm
Dr. Tahseen Al-Doori
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
 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
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
 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
 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
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
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
 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
 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
Dr. Tahseen Al-Doori