Link Budget

Calculation
Training materials for wireless trainers

Wednesday, March 5, 14

This 60 minute talk is about estimating wireless link performance by using link budget calculations. It also introduces the Radio Mobile link
design tool.
Version 1.5 by Rob, @2009-11-23
Version 1.7 by Carlo, @2010-01-13 (in Pune)
Version 1.8 by Rob, @2010-02-28
Version 1.9 by Rob, @2010-03-02
Version 1.10 by Rob, @2010-03-12
Version 1.12 by Carlo, @2011-03-22
Version 1.13 by Carlo, @2011-03-23 - corrected a mistake of 1dB in the Lfs in slides 11 to 14
version 1.15 by Ermanno, some changes in formulas and corrections in slides
 Goals

‣ To be able to calculate
how far we can go with
the equipment we have
‣ To understand why we
need high masts for long
links
‣ To learn about software
that helps to automate
the process of planning
radio links

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Wednesday, March 5, 14
 Free space loss
‣ Signal power is diminished by geometric spreading of the
wavefront, commonly known as Free Space Loss.
‣ The power of the signal is spread over a wave front, the area
of which increases as the distance from the transmitter
increases. Therefore, the power density diminishes.

Figure from http://en.wikipedia.org/wiki/Inverse_square 3

Wednesday, March 5, 14

A light bulb analogy will help understand this. If we watch the quantity of light shed over a piece of paper, we will
notice that this diminishes as we take it further away from a light bulb. This is a purely geometric phenomenon, it
happens even in a vacuum where there is nothing that can absorb the em radiation. That is why it is called free
space loss. An even better term would be “geometric spread loss”
 Free Space Loss (@2.4 GHz)

‣ Using decibels to express the loss and using 2.4 GHz as the
signal frequency, the equation for the Free Space Loss is:

Lfs = 100 + 20*log10(d)

‣ ...where Lfs is expressed in dB and d is in kilometers.

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The log in this formula is in base 10.

It is quite easy to use this simple formula, because the result is: 100dB (for 1 km), plus 20 dB for every time you
multiply the distance by 10... (10 km -> 120dB, 100 km-> 140dB, etc...).
Then if you remember that log(2)=~0.3, you simply add 20*0.3 = 6dB every time you double the distance, e.g.
1km -> 100dB
2km -> 106dB
4km -> 112dB
8km -> 118dB
10km -> 120dB
20km -> 126dB
40km -> 132dB
80km -> 138dB
100km -> 140dB
and so on...
 Free Space Loss (any frequency)

‣ Using decibels to express the loss and at a generic frequency

f, the equation for the Free Space Loss is:

Lfs = 92,45 + 20*log(d) + 20*log(f)

‣ ...where Lfs is expressed in dB, d is in kilometers and f is in

GHz.

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The formula in the previous slide is a particular case of this one, obtained making f=2.4 Ghz in this formula.
381

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This graph show the free space path loss for distances up to 40 km, for both 2.4 and 5.3 GHz.
Hint: if the frequency used is in the range of 5 GHz (instead of 2.4GHz) you have to add 6 dB to the loss to
obtain the right result.
 Power in a wireless system
antenna antenna

Tx Rx
cable path loss cable
radio radio
EIRP

Tx power

Rx power
dBm

Margin

Rx sensitivity

distance

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This graph shows the relative amount of gains and losses as well as the absolute power at each point in a
wireless link.

The transmitter provides some amount of power. A small amount is lost in attenuation between the transmitter
and the antenna. The antenna then focuses the power, providing a gain. At this point, the power is at the
maximum possible value for the link. This power is called EIRP (Equivalent Isotropic Radiated Power). Most
regulators impose a limit on the maximum allowable value of EIRP in a given country.

Then there are free space and environmental losses (which together form the path loss), which increase with the
distance between the link endpoints. The receiving antenna provides some additional gain. Then there is a small
amount of loss between the receiving antenna and the receiving radio.

If the received amount of power at the far end is greater than the receive sensitivity of the radio, by a certain
margin M, then the link is possible. The value of M will determine the reliability of the link, a good starting point
is to have at least 10 dB margin. For critical links, it is better to strive for a 20 dB margin.
 Link budget
‣ The performance of any communication link depends on the quality
of the equipment being used.
‣ Link budget is a way of quantifying the link performance.
‣ The received power in an wireless link is determined by three
factors: transmit power, transmitting antenna gain, and
receiving antenna gain.
‣ If that power, minus the free space loss of the link path, is greater
than the minimum received signal level of the receiving radio,
then a link is possible.
‣ The difference between the minimum received signal level and the
actual received power is called the link margin.
‣ The link margin must be positive, and should be maximized (should
be at least 10dB or more for reliable links).
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Do not confuse the link budget with the cost to obtain the equipment! We are not dealing with money here but
with dB.
The link budget reflects the impact of different variables in the ultimate power that reaches the receiver.
Keep in mind that the receiver sensitivity is strongly dependent on the transmission rate: the higher the
transmission rate
the higher the receiver power required for acceptable performance.
If one cannot obtain an acceptable margin at a given transmission rate it might be required to work at a lower
transmission rate.
The free space loss applies when there is a completely unobstructed path between the transmitter and the
receiver, with clearance of at least 60% of the first Fresnel Zone.
Partial obstruction of the 1st Fresnel Zone or the presence of walls or other objects will cause additional losses
to be added to the free space loss to calculate the total path attenuation.
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Wednesday, March 5, 14

Here is a example of a manufacturer’s data sheet showing the transmit power and receive sensitivity at various
modulations and data rates. A data sheet is required for any serious outdoor work, and some data sheets are
more complete than others. Avoid manufacturers who do not publish the detailed specifications of their
equipment.

Note that the minimum RSL is dependent upon rate, and the 1 Mbps rate is used for maximum range. TX power
is usually also rate dependent, but manufacturers rarely indicate this.

As the received signal drops, devices will automatically reduce their speed to attempt to maintain a stable link.

Note that the manufacturer specifies the frequency range from 2412-2462 MHz. The spectrum usage of each
channel is 20 MHz, and the manufacturer is referring to the center frequency.
 Example link budget calculation
Let’s estimate the feasibility of a 5 km link, with one access
point and one client radio.

The access point is connected to an antenna with 10 dBi

gain, with a transmitting power of 20 dBm and a receive
sensitivity of -89 dBm.

The client is connected to an antenna with 14 dBi gain, with

a transmitting power of 15 dBm and a receive sensitivity of
-82 dBm.

The cables in both systems are short, with a loss of 2dB at

each side at the 2.4 GHz frequency of operation.

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 AP to Client link
antenna antenna
+10 dBi +14 dBi
+20 dBm

Tx Rx
cable path loss cable
radio radio
-2 dB EIRP -114 dB @ 5 km -2 dB

Tx power

???
Rx power
dBm

Margin

Rx sensitivity
-82 dBm
distance

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Here is the graph from the previous slide, with values filled in for the AP and client.
 Link budget: AP to Client link

20 dBm (TX Power AP)

+ 10 dBi (Antenna Gain AP)
- 2 dB (Cable Losses AP)
+ 14 dBi (Antenna Gain Client)
- 2 dB (Cable Losses Client)

40 dB Total Gain
-114 dB (free space loss @5 km)

-74 dBm (expected received signal level)

--82 dBm (sensitivity of Client)

8 dB (link margin)

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The AP to Client link is possible, but below 10 dB. This link could be improved.

Remember that Free Space Loss is different than the link budget. FSL refers to the loss due to geometric spreading of signal in free space,
while the link budget calculation will take the FSL into account to determine whether a link is feasible.

Ask to participants how the link margin can be improved. Comment on the answers; they may suggest use of higher gain antennas,
shorter/better cables, higher TX power, better receiver, or amplifiers. You may point out that the use of amplifiers is not an optimal
solution for different reasons: they are expensive, cause more interference, consume more power, create asymmetry in TX/RX power
(unless used on both ends of a link) while antennas work in both directions, the amplifier is one more point-of-failure, etc…
 Opposite direction: Client to AP
antenna antenna
+14 dBi +10 dBi
+15 dBm

Tx Rx
cable path loss cable
radio radio
-2 dB EIRP -114 dB @ 5 km -2 dB

Tx power

???
Rx power
dBm

Margin

Rx sensitivity
-89 dBm
distance

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This shows the calculation to be performed, but in the opposite direction.

 Link budget: Client to AP link

15 dBm (TX Power Client)

+ 14 dBi (Antenna Gain Client)
- 2 dB (Cable Losses Client)
+ 10 dBi (Antenna Gain AP)
- 2 dB (Cable Losses AP)

35 dB Total Gain
-114 dB (free space loss @5 km)

-79 dBm (expected received signal level)

--89 dBm (sensitivity of AP)

10 dB (link margin)

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Wednesday, March 5, 14

The Client to AP link is better, at 10 dB, even though the transmit power is lower. This link is might work but it
could be improved by using higher gain antennas, more sensitive radios, or more powerful transmitters.
 Fresnel Zone

‣ The First Fresnel Zone is an ellipsoid-shaped volume around

the Line-of-Sight path between transmitter and receiver.
‣ The Fresnel Zone clearance is important to the integrity of
the RF link because it defines a volume around the LOS that
must be clear of any obstacle for the the maximum power to
reach the receiving antenna.
‣ Objects in the Fresnel Zone as trees, hilltops and buildings can
considerably attenuate the received signal, even when there is
an unobstructed line between the TX and RX.

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Fresnel zone calculations are independent of the free space loss. A link with plenty of link budget but a blocked
Fresnel zone can have many problems, or not work at all.

The shape of the Fresnel zone is similar to a cigar, it is a 3D object, so one has to mind also the possible
obstruction of the Fresnel zone by objects in the horizontal plane as well, for instance, two close buildings
adjacent to a clear optical LOS might cause a significant loss.
 Optical and Radio LOS

Optical signals also possess a Fresnel zone, but since

the wavelength is so small, we don’t notice it.

Therefore, clearance of optical LOS does not guarantee

the clearance of RADIO LOS.

Wednesday, March 5, 14
 Line of Sight and Fresnel Zones

a free line-of-sight IS NOT EQUAL TO a free Fresnel Zone

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Simply draw a line between two points, and if nothing is in the way, we have optical line of sight.
But radio waves are not confined to a perfectly straight line, they occupy a volume in space. Fresnel zone theory describes how a
propagating wave can cause interference with itself. If the first Fresnel zone is partially blocked by an obstruction, the signal arriving at the
far end would be diminished.
 Fresnel Zone

‣ The radius of the first Fresnel Zone at a given point between

the transmitter and the receiver can be calculated as:

r = sqrt(λ*(d1*d2)/(d))

‣ ...where r is the radius of the zone in meters, d1 and d2 are

distances from the obstacle to the link end points in meters, d is
the total link distance in meters, and λ is the wavelength in m.
‣ Note that this gives you the radius of the zone, not the height
above ground. To calculate the height above ground, you need to
subtract the result from a line drawn directly between the tops
of the two towers.

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It is not strictly necessary to clear the whole of the first Fresnel zone for an acceptable link.
Standard engineering practice is to clear 60 to 70% of the first Fresnel zone; the resulting loss over the 100%
clearance is negligible.
There is an infinite number of Fresnel zones that surround the first one, but we are not concerned with them
since the clearance of the second Fresnel zone will actually reduce the amount of power reaching the receiver.
So we strive for clearance of just 60% of the first Fresnel zone for optimum performance.
In terms of the frequency in MHz, this equation can be written as:
r = 17.31 * sqrt((d1 * d2) / (f * d)), where the distances are in meters
sqt(x) is the square root of x.

For an optical signal, λ =10^-6 m, whereas at 2.4 GHz is 0.12 m so the Fresnel zone is about 300 times smaller,
that is why we do not see it
 Line of Sight and Fresnel Zones

r rMAX

d1 d2

r = sqrt(λ*d1*d2/d)
rMAX = 1/2* sqrt(λ*d)
where all the dimensions are in meters
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λ = lambda is the wavelength = c/f, where c = 300 000 km/s

or in terms of frequency:
r = 17.31 * sqrt((d1 * d2) / (f * d))
with f in MHz, d in km and r in meters.
 Clearance of the Fresnel Zone and
earth curvature
This table shows the minimum height above flat ground required to clear
60% of the first Fresnel zone for various link distances at 2.4 GHz.

Notice that earth curvature plays a small role at short distances, but
becomes more important as the distance increases.

Distance 1st zone 60% Earth curvature Required

(km) (m) (m) (m) height (m)

1 5.5 3.3  0.0  3.9

5  12.4 7.44 0.4 7.84
10  17.5 10.5 1.5 12
15  21.4 12.84 3.3 16.13
20  24.7 15.82 5.9 21.72
25  27.7 16.62 9.2 25.82
30  30.3 18.18 13.3 32.5

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Comment on the values for one or two cases. This table is computed for a frequency of 2.4 GHz.
Note that the indicated values refer to the midpoint of the trajectory between the transmitter and the receiver. Assuming flat
ground, the indicated value corresponds to the height of the towers needed for this link.
A good figure to remember is that at 30 km distance over flat land one would need towers about 30 m tall in order to clear the
earth curvature and 60 % of the first Fresnel Zone.
 Fresnel Zone

‣ Considering the importance of the Fresnel Zone, it is

important to quantify the degree to which it can be
blocked.

‣ Typically, 20% - 40% Fresnel Zone blockage introduces

little attenuation into the link.

‣ It is better to err to the conservative side allowing no

more than 20% blockage of the Fresnel Zone.

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Ways to improve the clearance: higher antennas, repeaters, higher freq., plan higher margin, relocate the link to
avoid the obstacle, etc.
Keep in mind that the calculations shown refer to normal propagation conditions, which happen most of the
time. For critical links, it is also necessary to consider abnormal propagation conditions in which radio waves
will have a different trajectory, and higher antenna height will be required.
 Radio Mobile

‣ Radio Mobile is a free tool to aid in the design and

simulation of wireless systems.
‣ It can automatically calculate the power budget of a
radio link, calculating the Fresnel zone clearance. It can
use digital maps, GIS (Geographical Information
Systems), or any other digital map, including maps
provided by yourself.
‣ Runs on every version of Windows and can be used
in Linux and Mac through Windows emulators

http://www.cplus.org/rmw/english1.html

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 Radio Mobile
‣ From the same site, you can use radio mobile on-line
‣ You only need a browser, so it operating system
independent
‣ You need to register in the site, choose a user name
and password and start using the program
‣ There are some limitations in the on-line version,
since you are only allowed to use radio ham
frequencies, but you can choose the one closest to
your real frequency and the result will be close
enough.

http://www.cplus.org/rmw/rmonline.html

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 Radio Mobile
‣ Uses Digital terrain Elevation Model for the calculation of
coverage, indicating received signal strength at various point along
the path.
‣ Radio Mobile automatically builds a profile between two points in
the digital map showing the coverage area and 1st Fresnel zone.
‣ Different antenna heights can be tried to achieve optimum
performance.

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This link has a clean line-of-sight and a free Fresnel zone.

 Radio Mobile

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In addition to the land cross-section view, Radio Mobile provides all of the hard numbers relating to the
performance of your link: expected receive signal, link quality, EIRP, etc.
The software will tell you the azimut and elevation angles that the antennas should be pointed at, as well as the
location of the most offending obstacle and which portion of the first Fresnel zone is blocking.

This particular link would not work due to obstacles in the path, indicated by the clearance of -1.3 F1, that is
130% of the first Fresnel zone is blocked. This means that even the portion of the first Fresnel zone ABOVE the
line of sight is blocked.
Nevertheless, the simulation show a positive margin of 16.5 dB. This is because the transmitter power is set at an
illegal value of 10 W and the receiver sensitivity is at an unrealistic value of -107 dBm.
 Thank you for your attention

For more details about the topics presented in

this lecture, please see the book Wireless
Networking in the Developing World,
available as free download in many languages at:
http://wndw.net/

Wednesday, March 5, 14

See Chapter 4 of the book for more detailed information about the material covered in this talk.