MOB Subject 2 – "Link budget " practical
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Master of Computer Science 1 - MOB Mobile Computing
1/4 Baey, Fladenmuller – Subject 2
MOB Subject 2 – "Link budget " practical
What is the maximum propagation range you can get between a transmitter and a receiver? Using
one spreadsheet, you will try to understand the impacts of different parameters implied in the link
budget calculation.
Create an excel table like the one below. You have to generate automatically the answers in the
white boxes from the ones of the white boxes.
It is supposed that the operating frequency band is 2.4 GHz and that a margin of 10 dB is
sufficient.
The ART, a French standard regulation body, indicates that the EIRP should not exceed 100 mW.
Using the following command, indicate automatically on your spreadsheet if this condition is
verified:
IF: Returns one value if a condition you specify evaluates to TRUE and another value if it
evaluates to FALSE.
Syntax
IF(logical_test,value_if_true,value_if_false)
Transmitter Receiver Unit
13 14,77 dBmPx: transmitter power
19,95 29,99 mW
-79 -85 dBmSx min with 11 Mbps
Threshold of reception
12,59 3,16 pW
(10
-12
W)
-89 -94
Access point
Sx max with 1 Mbps
Threshold of reception
1,26 0,40
Cable L: loss 3 1
Antenna G: gain of antenna 7 6 dBi
17 19,77 dBmTx = Px – L + G Tx: EIRP
50,12 94,84 mW
Margin
Max distance at 11 MbpsDistance
Max distance at 1 Mbps
Based on this spreadsheet
Case 1: Transmitter and Receiver of Cisco traditional cards:
Master of Computer Science 1 - MOB Mobile Computing
2/4 Baey, Fladenmuller – Subject 2
Output power of the amplifier: 20 dBm
Antenna Gain: 0 dBi
We get:
20
10010--Margin-Lr-GrysensitivitEIRP
20
103
104,24
20log-Margin-Lr-GrysensitivitEIRP
5
9
10
20
103
104,24
20log-Margin-Lr-GrysensitivitEIRP
1010
100
103
104,24
20log
10
5
9
10
5
9
10
so
and
Max distance at 11 Mbps is 0,560km
Max distance at 1 Mbps is 1,68km
Case 2: Transmitter and Receiver of Cisco cards with “Pringles antennas” deported
Output power of the amplifier: 10 dBm
Loss in each connector: 0.2 dB (2 connectors are necessary)
Cable length: 3 m
Cable loss: 0,2 dB/m
Gain: 11 dBi
Max distance at 11 Mbps is 1,778km
Max distance at 1 Mbps is 5,02km
Case 3: The transmitter has the characteristics of Case 1 and the receiver has those of Case 2.
Max distance at 11 Mbps is 1,778km
Max distance at 1 Mbps is 5,02km
Case 4: The receiver has the characteristics of Case 1 and the transmitter has those of Case 2.
It gets the same thing as in case 1 because the receiver is identical to case 1 and the issuer has the
same worst case 1.
Max distance at 11 Mbps is 0,560km
Max distance at 1 Mbps is 1,68km
1. What do you observe?
If one considers that Case 3 is the uplink, the Case 4 corresponds to the downlink. We therefore
asymmetry in scope.
2. Can one add a “Pringles antenna” (home made antenna from a crips box) to the transmitter
when using case 1 specifications ? Why?
Not, because it exceeds the limit imposed by the ART.
3. By supposing that one emits with the maximum value of the EIRP, would it be better to use
an antenna with strong gain or to increase the transmitter power?
Master of Computer Science 1 - MOB Mobile Computing
3/4 Baey, Fladenmuller – Subject 2
The two solutions are the same EIRP but it is preferable to increase the gain of the antenna as this
will increase the scope for the reverse link where the antenna acts as a receiver.
4. In which case that will not be checked anymore?
The antennas are more directive and therefore must be able to align. If it does more, it can be a
problem ...
Other comparisons
Subsequently, one will consider that one operates with the maximum value of the EIRP and the
connection between the transmitter and the receiver is in LOS. The Friis formula which gives the
attenuation signal in free space will thus apply. We consider a margin of 10 dB.
1. Trace the curve of the propagation range according to the sensitivity of the receiver which
can vary between -85 dBm and -94 dBm. One will take a null gain in reception.
A
libre
= 20 log
10
(4πd / λ)
= 20 log
10
(4π / λ) + 20 log
10
(d)
= 20 log
10
(4π / 0,125) + 20 log
10
(d)
With λ = 0,000125 km and f = 2,4 GHz, the formular becomes:
Sensitivity = Px – L + Ge – A
libre
+ Gr – Margin
and Px –L + Ge = EIRP = 20 dBm
and Gr = 0 dBi.
Thus: 20 log
10
(4π / 0,000125) = 100
Substitutes:
Sensitivity = EIRP – [20 log
10
(4π / 0,125) + 20 log
10
(d)] + Gr – Margin
⇒ 20 log
10
(d) = EIRP – 100 + Gr – Margin – Sensitivity
⇒ d = 10 ^ (EIRP – 100 –10 – Margin – Sensitivity)/20
where Margin = 0
We get d = 10 ^ (20 - 110 – Sensitivity)/20
⇒ d = 10 ^ (– 90 – Sensitivity) / 20 (km)
2. Distance according to the gain.
a) Express the distance according to the gain of antenna and the sensitivity receiver. One
supposes that the transmitter emits with the maximum value of the EIRP.
d = 10 ^ (EIRP – 100 + Gr – Margin – Sensitivity)/20
⇒d = 10 ^ (20 – 100 + Gr – 10 – Sensitivity)/20
⇒d = 10 ^ (– 90 + Gr – Sensitivity)/20 (km)
b) Trace for each flow, the curve of the distance according to the antenna gain that one will
vary from 0 to 24 dBi. You will take the same gain for the transmitter and the receiver and
"data manufacturer" card for sensitivity associated with each flow.
d = 10 ^ (– 90 + Gr – Sensitivity)/20 (km)
Master of Computer Science 1 - MOB Mobile Computing
4/4 Baey, Fladenmuller – Subject 2
