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Apr 3, 2024

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Path loss For the following free-space formula: 1. Modify it to accommodate system losses like line and connector losses, fading margin and diversity gain. You will probably find it advantageous to put everything into decibel form. Ans1 : - Pr (dB)=Pt (Db)+Gt(dB)+Gr(dB)+20log 10 ( Lambda 4 Pi r ) –(Ls+Fm+Gd) Here 2. Now, suppose that the path loss coefficient is n = 4. Modify your equation to accommodate this. Ans2 : - The modified Pr equation for the path loss coefficient is Pr (dB)=Pt (Db)+Gt(dB)+Gr(dB)+10nlog 10 ( Lambda 4 Pi r ) –(Ls+Fm+Gd) 3. Suppose that you have a system operating at 1940 MHz ( uplink). The system conforms to the cellular model with path loss coefficient = 4. Calculate the cell radius given the following system data: minimum receivable power for the base station is -110 dBm Pt = 45 dBm (GSM) Transmit antenna is 60⁰ beamwidth antenna with gain = 17.1 dB Receive antenna has a gain of 2.1 dBi. cable is 30m long with losses of 0.2 dB/m. connector losses: 0.2 dB. jumper losses: 0.5 dB.
Path loss Ans3 : - Pt =45dbm Pr=-110 dbm Cable losses=0.2 db/m (for 30m cable =total cable losses =6db) =0.15453 m =154.639 x 10 3 Km Using Free space formula, 40log 10 ( Lambda 4 Pi R ) = -167.5 R= 189.5 m 4. Now, perform the cell radius calculation for path loss coefficient = 2. Comment on the difference. Ans: - Now, path loss coefficient=2 -110=64.2+20*log 10 ( Lambda 4 Pi R )-6.7 20log 10 ( Lambda 4 Pi R ) = -110-64.2+6.7 log 10 ( Lambda 4 Pi R )= 167.5 20 5. To determine path loss, do we need to know the initial and final powers? Ans5: - The initial power is the power with which the signal is transmitted, and the final power is the power that is received after the signal has traveled through the medium. The difference between these two values gives us the path loss, which is a measure of how much signal strength is lost during transmission. This loss can be due to various factors,
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Path loss including free-space loss, absorption, scattering, reflection, and diffraction, as well as system losses like line and connector losses, fading, margin and diversity gain. 6. What is the path loss in dB at 1.9 GHz over 1 m? Over 1 km? Ans: - PL_db =20log 10 (d) + 20log 10 (f) + 20log 10 (4pi/3x10 8) Whereas, d is distance between the transmission and receiver f = 1.9GHz =1.9x10 9 Hz c=3x10 8 Calculating for 1m PL_db = 20log 10 (1) + 20log 10 (1.9x10 9 ) + 20log 10 (4pi/3x10 8) 7. Is the P.L. constant over every linear increment of distance? For example, is the P.L. from 1 to 2 m the same as from 100 to 101 m? Show some examples and explain. Ans: - No, the path loss is not constant over every linear increment of distance. The free- Space path loss formula indicates that pass loss increases logarithmically with distance. This means that the rate of increase in path loss as the distance increases. For example, the path loss from 1 to 2 m will not be the same as from 100 to 101 m because the path loss depends on the distance. As distance doubles, the path loss increases by a constant amount in dB, not linearly.
Path loss 8. Use the table below ( assume constant unknowns ), calculate the path loss coefficient. (d1, r1) = (100m, -97.22dbm) (d2, r2) = (200m, -103.24dbm) Pr= 10*n*log 10 ( d 1 d 2 ) n=2 9. Use the table below to estimate the received power at 2 km from the table. You may need a semi-log graph paper (can be downloaded from the web). Distance from Transmitter (m) Received power (dBm) 100 -97.22 200 -103.24 1000 -117.22 3000 -126.76 Ans: - To calculate the path loss coefficient and estimate the received power at 2 km, %Given distance_values = [100, 200, 1000, 3000]; %Distance from transmitter in meter received_power_values = [-97.22, -103.24, -117.22, -126.76]; % Received Power % Plot the data points %Estimate Received power at 2km received_power_at_2km = interp1(distance_values,received_power_values,2000)
Path loss
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