MOB Subject 2 – "Link budget " practical
Master of Computer Science 1 - MOB Mobile Computing1/4 Baey, Fladenmuller – Subject 2MOB Subject 2 – "Link budget " practicalWhat is the maximum propagation range you can get between a transmitter and a receiver? Usingone spreadsheet, you will try to understand the impacts of different parameters implied in the linkbudget calculation.Create an excel table like the one below. You have to generate automatically the answers in thewhite 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 issufficient.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 isverified:IF: Returns one value if a condition you specify evaluates to TRUE and another value if itevaluates to FALSE.SyntaxIF(logical_test,value_if_true,value_if_false)Transmitter Receiver Unit13 14,77 dBmPx: transmitter power19,95 29,99 mW-79 -85 dBmSx min with 11 MbpsThreshold of reception12,59 3,16 pW(10-12W)-89 -94Access pointSx max with 1 MbpsThreshold of reception1,26 0,40Cable L: loss 3 1Antenna G: gain of antenna 7 6 dBi17 19,77 dBmTx = Px – L + G Tx: EIRP50,12 94,84 mWMarginMax distance at 11 MbpsDistanceMax distance at 1 MbpsBased on this spreadsheetCase 1: Transmitter and Receiver of Cisco traditional cards: Master of Computer Science 1 - MOB Mobile Computing2/4 Baey, Fladenmuller – Subject 2Output power of the amplifier: 20 dBm Antenna Gain: 0 dBiWe get:2010010--Margin-Lr-GrysensitivitEIRP20103104,2420log-Margin-Lr-GrysensitivitEIRP591020103104,2420log-Margin-Lr-GrysensitivitEIRP1010100103104,2420log1059105910soandMax distance at 11 Mbps is 0,560kmMax distance at 1 Mbps is 1,68kmCase 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 mCable loss: 0,2 dB/mGain: 11 dBiMax distance at 11 Mbps is 1,778kmMax distance at 1 Mbps is 5,02kmCase 3: The transmitter has the characteristics of Case 1 and the receiver has those of Case 2.Max distance at 11 Mbps is 1,778kmMax distance at 1 Mbps is 5,02kmCase 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 thesame worst case 1.Max distance at 11 Mbps is 0,560kmMax distance at 1 Mbps is 1,68km1. What do you observe?If one considers that Case 3 is the uplink, the Case 4 corresponds to the downlink. We thereforeasymmetry in scope.2. Can one add a “Pringles antenna” (home made antenna from a crips box) to the transmitterwhen 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 usean antenna with strong gain or to increase the transmitter power? Master of Computer Science 1 - MOB Mobile Computing3/4 Baey, Fladenmuller – Subject 2The two solutions are the same EIRP but it is preferable to increase the gain of the antenna as thiswill 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 aproblem .Other comparisonsSubsequently, one will consider that one operates with the maximum value of the EIRP and theconnection between the transmitter and the receiver is in LOS. The Friis formula which gives theattenuation 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 whichcan 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 – Marginand Px –L + Ge = EIRP = 20 dBmand Gr = 0 dBi.Thus: 20 log10 (4π / 0,000125) = 100Substitutes: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)/20where Margin = 0We 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. Onesupposes 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 willvary 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 Computing4/4 Baey, Fladenmuller – Subject 2 . log10(4π / 0, 125 ) + 20 log10 (d)] + Gr – Margin⇒ 20 log10 (d) = EIRP – 100 + Gr – Margin – Sensitivity⇒ d = 10 ^ (EIRP – 100 –1 0 – Margin – Sensitivity) /20 where. 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