MOB Subject 2 – "Link budget " practical
Trang 1MOB 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
Px: transmitter power
Sx min with 11 Mbps
(10-12W)
Access point
Sx max with 1 Mbps Threshold of reception 1,26 0,40
Tx = Px – L + G Tx: EIRP
Margin Max distance at 11 Mbps
Distance
Max distance at 1 Mbps
Based on this spreadsheet
Case 1: Transmitter and Receiver of Cisco traditional cards:
Trang 2Output power of the amplifier: 20 dBm
Antenna Gain: 0 dBi
We get:
20
100 10 -Margin -Lr -Gr y sensitivit EIRP 20
10 3 10 4 , 2 4 20log -Margin -Lr -Gr y sensitivit
EIRP
5
9 10
20
10 3 10 4 , 2 4 20log -Margin -Lr -Gr y
sensitivit
EIRP
10 10
100 10
3
10 4 , 2 4 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?
Trang 3The 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
Alibre = 20 log10 (4πd / λ)
= 20 log10 (4π / λ) + 20 log10 (d)
= 20 log10 (4π / 0,125) + 20 log10 (d)
With λ = 0,000125 km and f = 2,4 GHz, the formular becomes:
Sensitivity = Px – L + Ge – Alibre + Gr – Margin
and Px –L + Ge = EIRP = 20 dBm
and Gr = 0 dBi
Thus: 20 log10 (4π / 0,000125) = 100
Substitutes:
Sensitivity = EIRP – [20 log10(4π / 0,125) + 20 log10 (d)] + Gr – Margin
⇒ 20 log10 (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)