MC

Tutorial- 4Mobile Propagation Models

Fundamentals ofCellular and Wireless Networks

Lecture ID: ET- IDA-113/114

20.05.2012 , v11

Prof. W. AdiPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringA base station is transmitting 100 W at 1200 MHz with an antenna gain of 10 dB.Compute the received power in dBm at the mobile site assuming a two- ray propagation model for the geometry shown below. The mobile antenna gain is 10 dB. L=220 kmBase stationmobile5m30m1 km19 kmBase stationmobile55m 5m30mtower2. Compute the received power in dBm if a 55m tower is built at a distance of 1 km far from the base station. Use the knife-edge diffraction modelProblem 4.1:Page : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering Solution 4.1:

1. Two Ray (Ground Reflection) Model

Wavelength (in m) = c/f = 0.25m (c= 3 . 108 m/s)Gt = transmitter antenna gain = 10 dB = 10.Gr = receiver antenna gain = 10 dB = 10.ht = transmitter height = 30mhr = receiver height = 5md = transmitter-receiver separation distance = 20,000mf = transmitter frequency = 1200 MHz()()()dBmPmWxWxPPdPrrrr52.581040625.11040625.1000,205301010100h h G G P6942242r2trtt-=====--Check ?Page : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering Solution 4.1 (Contd):2. Knife-Edge Diffraction Model

Received power after the tower: Pr(dBm) = Pr(dBm) + Gd = -58.06 dBm 20.56 = -73.62 dBmTR55 m5 m30 m1 km19 kmThe wavelength = c/f = 300/1200 = 0.25 m TR50 m25 m1 km19 kmLos or two ray?

Page : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering

Problem 4-2:

Solution 4-2:tPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering

Solution 4-2:Vant=VRant=50Pr10-10Pr = -68 dBm= 106.8= 1.58 10-7 mw = 1.58 10-10 watt Pr = Rant . i2= Rant . (V/2 Rant )2Pr = V2 / 4Rant

iR=Rant=50ReceiverAntennaPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringProblem 4-3:A police care driving at a speed of 150 km/h on the highway is tracking a carwith mobile station transmitting at carrier frequency of 900 MHz. The police receiver was able to measure a Doppler frequency shift of -75 Hz. What is the speed of the tracked car?When would the police radar lose connection to the tracked car if the radar receiver sensitivity is 90 dBm and the first tracked signal power was 70 dBm at a distance of 100 m from the tracked car.Assume that the following long distance path loss model is valid for this propagation: Pr(d) = Pr(d0) (d0/d)2 Assume also that both cars have a constant speed all the time.Police RadarVp=150 Km/hMobile station

V=?100mPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringSolution 4-3:1. = 300/900 = 1/3 m, Doppler frequency shift f =v cos / , =0 degree Relative speed is v = f = 1/3 x 75 = 25 m/s. 25 m/s = 25 x 10 3 x 3600 = 90 km/h As it is Moving away from the police car with a speed of 90 km/h The followed car has then the speed of 150 + 90 = 240 km/h 2. Pr (d) = Pr (d0) (d0/d)2 , 10 log Pr(d)= 10 log Pr(d0) + 20 log d0 20 log d 10 log Pr(d0) = -70 dBm, d0= 100 m -90 dBm = -70 + 20 log 100 - 20 log d -60 = - 20 log d => d = 1000 mTime required to lose connection = 1000-100 m / 25 m/s = 36 seconds

Page : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringA base station is transmitting Pt Watt at 900 MHz with an antenna gain of 10 dB.Compute the transmitter power Pt such that a mobile receiver at a distance of 20km with a sensitivity of 70 dBm can operate adequately assuming a free space propagation model for the geometry shown below. The mobile antenna gain is 10 dB. Assume the system loss factor L=2.2. Compute the base station transmitter power Pt such that if a 85m tower is built at a distance of 2 km from the base station, the mobile station would receive adequately. Use the knife-edge diffraction model and take the geometry from the illustration below.20 km kmBase stationMobile station

2 km 18 kmBase stationMobile station5m30m tower

85m Problem 4-4:Page : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringSolution 4-4:1. For LOS model the following hold = 300/900 = 1/3 m Gt = Gr = 10, d = 20000, L=2 Pr (d) = -70 dBm= 10 log Pr(mw) = 10 7 mw = 10 10 w

2. The diffraction loss according to knife edge model:

Thus the required Pt power is 1.138 watt x 102.33 = 243.3 watt Or Pt = 30.5 dBm (- 23.3 dB) = 53.8 dBm

Pr (d) =Pt Gt Gr 2(4)2 d2 LPt =Gt Gr 2Pr (d) (4)2 d2 LPt =10 x 10 x (1/3)2 10 10 (4)2 200002 x 2= 1.138 watt = 1138 mw = 30.5 dBm Diffraction loss =()()Gddddvradradradd3.28225.0log20000=3.28,2033.0000182000 x2 x0315.020315.0004.00275.0004.0004.0000,1880tan0275.00275.0000,23085tan2121-23.27dB = 10- 2.33 ===+==+=+======-=lagbaggbbabg305552 km18 kmPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringA car driving through the high-way at a speed of 36 km/h detected the signal from a base stationfirst at point A and lost the connection at point C. The received power when passing point Bwas 60 dBm. Given: The frequency used is 1500 MHz Receiver sensitivity = -80 dBm The line of sight propagation model is valid for the wave propagationCompute the distance ACCompute the Doppler frequency shift at A, B and CV=36 km/hBase Station

CB-60 dBmA1 kmSProblem 4-5:Page : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering1. The distances SA and SC are equal as the power received at both points is 80dBm at a distance d=SA=SC. For line of sight propagation model: Pr (d) = Pr (d0) (d0/d)2 , Converting into dB:10 log Pr(d)= 10 log Pr(d0) + 20 log d0 20 log d d0= 1k m, 10 log Pr(d0) = -60 dBm SC = d, Power at point C should be 90 dBm -80 dBm = -60 + 20 log 1 - 20 log d -80+60 = - 20 log d => log d =1 => d = 10 km d=AS= 10 km BC =( 102 - 12 ) 1/2= 9.95 km AC = 2 x 9,95= 19,9 km 2. Doppler shift at B = 0 Hz as = 90, cos 90=0 = 300/1500 = 0.2 m v = 36 km/h = Doppler shifts at C and A are equal in value = |f| = ( v / ) cos = ( 36000/3600)/0.2 x (9,95/10) 50 Hz Doppler shift at A is +50 Hz, as Cos is positive Doppler shift at C is -50 Hz, as Cos is a negative valueC-80 dBmB-60 dBmA-80 dBmV=36 km/hBase Station

1 kmScos (180-) = -cos is negative!dd=10 KMSolution=909,95 KMPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering