Link Calc Example

8/3/2019 Link Calc Example

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2/18/2008 Pietrosemoli 1

Link Budget Calculation

ICTP-ITU School on New Perspectives onWireless Networking 2008

Abdus Salam ICTP

Ermanno Pietrosemoli

Latin American Networking School(Fundacin EsLaRed) ULA

Mrida Venezuela www.eslared.org.ve


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Radio Links Design

Choice of Frequency Path Profiles

Power Budget

Coverage area Site survey


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Path Profile

LOS (Line of sight)

K Factor (Earth Curvature)

Fresnel Zone


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K factor

K = (Apparent Earth Radius)/(Real Earth Radius)

EarthEarth

OpticOptical Horizon Radio Horizon, K= 4/3Radio Horizon, K= 4/3


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Radio trajectory

K = 4/3, 90% of the time, dielectricconstant decreases with altitude

Reach 1/3 beyond the horizon

K = infinity, straight trajectory

K = 2/3, upwards curvature, less reach ,0.6 F1 in critical paths


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Link Profile

Objects on the path

Foliage

Plane surfaces and bodies of water

Fresnel Zones


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First Fresnel Zone

Food Mart

DirectPath=L

FirstFresnelZone

Reflectedpath=L+/2


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Radio Link ClearanceDistancia en km 1ra Zona 0.7 *1ra Zona de Fresnel Curvatura TOTAL

de Fresnel @ 2.4 GHz en metros Terrestre metros

1 5.5 3.9 0.0 3.9

2 7.8 5.5 0.2 5.6

3 9.6 6.7 0.4 7.1

4 11.1 7.7 0.7 8.4

5 12.4 8.7 1.0 9.7

6 13.6 9.5 1.5 11.0

7 14.6 10.2 2.0 12.3

8 15.6 11.0 2.7 13.6

9 16.6 11.6 3.4 15.0

10 17.5 12.2 4.2 16.4

11 18.4 12.8 5.0 17.9

12 19.2 13.4 6.0 19.4

13 19.9 14.0 7.0 21.0

14 20.7 14.5 8.2 22.7

15 21.4 15.0 9.4 24.4

16 22.1 15.5 10.7 26.217 22.8 16.0 12.0 28.0

18 23.5 16.4 13.5 29.9

19 24.1 16.9 15.0 31.9

20 24.7 17.3 16.7 34.0

25 27.7 19.4 26.0 45.4

30 30.3 21.2 37.5 58.7


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Obstructed Line of Sight


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Optical Line of Sight


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Radio Line of Sight


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Fresnel Zone obstruction

Attenuation


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Profiles

How to get the data:

Topographic Maps

GPS

Walking the path withan altimeter

DEMs and appropiatesoftware


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Beyond Line of Sight

Reception is feasible, but with an increased power

budget by diffraction on obstacles

OFDM (Orthogonal Frequency Diversity Modulation)and MIMO (Multiple Input- Multiple Output) based

solutions can make a constructive use of multipath toovercome LOS

Suitable obstacles are abundant in urbanenvironment, much less so in rural areas, trees DO

NOT reflect radio waves, they rather absorb or scatterthem


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Example 1

Find the FSL between two sites 20 km apartin the 2, 4GHz frequency band

Repeat for the 5,7 GHz frequency


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Exercise

Find the received signal level at 10 degrees from the

boresight of a 24 dBi Hyperlink HG2424 antenna fedfrom a Linksys WRT54G Router with 12 meters ofLMR400 cable. The receiving antenna is

omnidirectional, located at 13 km and with a gain of 8dBi at 2, 4GHz operating frequency. The receivingantenna cable is LMR 200 and 7 meters long. Bothantennas are protected by cabling arrestors that

introduce 0,5 dB of additional loss each.The link is meant to attain 11 Mbit/s nominal speed.


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From www.hyperlinktech.com

we find the radiation pattern ofThe antenna. But there are two.

Which one do we choose?

Since the receiving antenna is

Omnidirectional, we will assumethat it is vertically polarised, and

therefore we must use vertical

polarisation at both ends.

For vertical polarisation, we haveabout 8 dB signal drop at 10

Offset, so the effective antenna

gain in this direction is 24-8=16 dBi

Antenna Gain


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Relative and absolute height

GalileoBuilding

Street Level = 60 m

Sea Level

Building heigth = 10 m

Antenna heigth: 3m above the roof

13m above the street, 73 m above sea level

Basement

height: -3m


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FSL

L= 100 +20Log(13km/km) = 122.28 dB

The power reaching the receiving antenna willbe 30,3 dBm -122,3 dB = -92 dBm

Adding the receiving antenna gain, the powerat the antenna terminal will be: -92 + 8 = -84dBm


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Receiver Cable loss

From www.hyperlinktech.com we find the

loss for the LMR 200 cable which is 0,55dB/m, so for 7 m we will have 3,85 dB loss.But we must have 2 connectors at each end,with an estimated loss of 0,2 dB each, plus

one adapter from the RPTN connector of thelinksys to the N male connector of the cable,which has a loss of 0,15 dB, so adding theo,5 dB loss of the lightning arrestor, the total

loss of the cabling will be:3,85+2*0,2+0,15+0,5=4,9 dB, so the inputpower at the receiver will be: -84 dBm 4,9dB = -88,9 dBm


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We can now build a graph of power over distance:

Rx

Gr

Tx

Gt

At Ar

18

dBm

km

Threshold

Margin

14,3

EIRP: 30,3 dBm

-92

FSL= 122.28 dB

-84

-88,9


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Homework

Design a point to multipoint system. Remote units are scattered

in every direction from the base station. The farthest CPEis 8 km away. Base station and remote (CPE) radios are equal,with 15 dBm power output and 85 dBm minimum receiverpower at 2.4 GHz. Base station requires 20 m of coaxial cablebetween the radio and the antenna with 10 dB of attenuation,

wereas the CPEs all use 10m long cables with 5 dB attenuation.

Choose the best suited antenna for the base station and for theCPE, among the one described in the following page.


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Documents

Link Calc Example