1、2 2、Time dispersionTime dispersion3 3、Impairments of the Wireless ChannelImpairments of the Wireless Channel 1 1、Four Basic Propagation MechanismFour Basic Propagation MechanismLarge-scale path loss and shadowingSmall-scale fadinglRadio Propagation Model“.characterization of radio wave propagation a
2、s a function of frequency,distance and other conditions.”l General ray-tracing method(simulation)Multiple ray-tracing with propagation effects(reflection,diffraction,LOS path,scattering,etc)Might consider building material(steel,concrete,brick,etc)l Empirical methodOn-site measurementCurve-fittingl
3、Statistical method IllustrationDistance inwavelengthSignalPowerT-R distanceTXRXRXLarge-scale free-space lossShadowing lossSmall-scale fadingl The path loss is dependent on the distance between the transmitter and receiver l It also depends on the propagation environmentTerrain characteristicsWavelen
4、gthTransmitting and receiving antenna heightsl A path model is important for determination of the base station(cell)coverage area2.4.2 Propagation Over Smooth Plane2.4.2 Propagation Over Smooth Plane2.4.4 Log-Distance Path Loss with Shadowing2.4.4 Log-Distance Path Loss with Shadowing2.4.1 Free Spac
5、e Propagation2.4.1 Free Space Propagation2.4.5 Indoor/Outdoor Path Loss Model2.4.5 Indoor/Outdoor Path Loss Model2.4.3 Diffraction and Fresnel Zones2.4.3 Diffraction and Fresnel Zones2.4.6 Radio Cell Coverage 2.4.6 Radio Cell Coverage l The Friis Equationl Free-Space Path Lossl No interference,no ob
6、structionsl There is no loss of energy in free space,but there is attenuation due to the spreading of the waves24rttrPPGGdWhere:Pt=total power radiated by an isotropic source Gt=transmitting antenna gain Gr=receiving antenna gain d=distance between transmitting and receiving antennas =wavelength of
7、the carrier signal=c/fc fc=carrier frequencyWhat troubles some people when they see this question is that the path loss also increases as the square of the frequency.While it is true that absorption of RF by various materials(buildings,trees,water vapor,etc.)tends to increase with frequency,remember
8、 we are talking about“free space”here.Effective area of an isotropic antenna However,from electromagnetic theory,we note that the effective area of an isotropic antenna in any direction is given by:A=2/(4 )1.Free space power flux density(W/m2)power radiated over surface area of sphereS Pt/(4 d2)By c
9、overing some of this area,receivers antenna“catches”some of this flux.2.Effective area of an isotropic antenna The power received by an antenna of effective area A is given by:PR=SAHowever,from electromagnetic theory,we note that the effective area of an isotropic antenna in any direction is given b
10、y:A=2/(4 )ThusPR=SA=Pt 2/(4 d)23.Directional Radiation While the isotropic antenna is a useful illustrative device,most antennas are not isotropic.Instead,they have gain that is a function of the azimuth angle and the elevation angle.When nonisotropic antennas are used,we get24rttrPPGGdl d D and d ,
11、where d is the distance between the transmitting antenna and receiving antenna D is the largest linear dimension of antenna is the carrier wavelength l ddf22DdfAssumes far-field24rttrPPGGdEIRPGPtt22()10log()(4)/20log()4pcLddBdcfdBd ()20log()20log()147.56()pcLdfddBThe term(4d/)2 is known as the free-
12、space path loss denoted by Lp(d),which is What does“dB”mean?l dB stands for deciBel or 1/10 of a Bell The Bel is a dimensionless unit for expressing ratios and gains on a log scaleGains add rather than multiplyEasier to handle large dynamic ranges)log()(log(10log10PP121210dB12PPPP-202468100.050.10.1
13、50.20.250.30.350.40.45-2024681010-510-410-310-210-1What does“dB”mean?lEx:Attenuation from transmitter to receiverPT=100,PR=10attenuation is ratio of PT to PRPT/PRdB=10 log(PT/PR)=10 log(10)=10 dBlUseful numbers:1/2dB -3 dB1/1000dB=-30 dBWhat does“dB”mean?l dB can express ratios,but what about absolu
14、te quantities?l Similar units reference an absolute quantity against a defined reference.n mWdBm=n/mWdB n WdBW=n/WdB Ex:1 mWdBW=？dBWExample 2.2Transmitter power:50W.Antenna:unit gain.Carrier frequency:900 MHz.Distance:d=100m.Find:1)dBm value of the transmitter,2)dBm value of the received power,3)Pr(
15、d=10 Km)=?(mWdBm)1)01)Transmit power:101 mlog4W7 dBm.ttPP23(dBm)10222)3.5 10 mW.10log24.5 dBm(4)trrtrrGGPPPPd 101003)(10km)(100m)20log10000 24.5 dBm40 dB 64.5 dBm.rrPP Relative distanceExample 2.3In door WLAN with fc=900 MHz.find the required transmit power in order to guarantee minimum received pow
16、er of 10 uW with distance up to 10 m.What if fc=5 GHz?Assume unit antenna gains.21,0.33 m,10 m,10 W.F41.45W1.61 dBW.rom trrctrtrdPcGPPGdfGGAt 5 GHz,0.06 m.So 43.9 W16.42 dBW.tPFree space propagation does not apply in a mobile environment and the propagation path loss depends not only on the distance
17、 and wavelength,but also on the antenna heights of the mobile station and the base station,and the local terrain characteristics such as buildings and hills.A simple two-path model can be used to illustrate the effect of transmitting and receiving antenna heights.Free Space PropagationFriis Equation
18、Free-Space Path LossWhat does“dB”mean?24rttrPPGGdPropagation Over Smooth Planel Two-path model where:transmit power.:received signal power :transmit antenna height :receive antenna height :transmit-receiver distance ,:trtrtrPPhhdG G antenna gainsthrh1d2dd),(ttGP),(rrGPthrh1d2dd221()trddhhThe propaga
19、tion distance of the direct path is：222()trddhhThe propagation distance of the reflected path is：22222122()()1()1()trtrtrtrdddhhdhhdhhhhdddGiven that dhthr,and1aa211a122 It can be derived that dh2hdrt Given that dhthr,we have d1 d and d2d.However,since the carrier wavelength is very small compared w
20、ith d,a slight change in the propagation distance can cause a significant change in the received signal carrier phase.4t rhhd l How is two-ray model obtained?22jf2rttrjexpe1)4(GGPPf1d(d)Taken into account the phase difference,the received signal power isWhere f and f are the amplitude attenuation an
21、d carrier phase shift introduced by the reflection.If f 1 and f=for ht,hr,the angel is small,and cancels out of the equation,leaving it to be essentially frequency independent It shows an inverse fourth-power law,rather than the inverse-square law of free-space propagation.This points to a far more
22、rapid attenuation of the power received.It shows the effect of the transmit and receive antenna heights on propagation losses.The dependence on antenna height makes intuitive sense.And the corresponding path loss is:)dB()dlog(40)hlog(20hlog20)(Lrtp dCritical distancel Define Critical distanceIf ddc,
23、then signal power falls off with d-4.If d hr This plot can be separated into three segments:l For small distance(ddc the signal components only combine destructively.Example 2.4 If two-ray model is suitable for urban micro-cell with ht=10 m,hr=3 m,and fc=2 GHz.Find dc.What if for indoor microcell wi
24、th ht=3 m,hr=2 m?Find the power loss for dc.44 10 3Urban:800 m.3 2044 3 2Indoor:160 m/.3 20trctrchhdhhd 222244222244103Urban:1 10.000000002286.6 dB80032Indoor:1 10.00000005572 6 dB160trrtrttrrtrth hPGGPdh hPGGPd .Compared with the free space model?-96dB-83dBHow about d=3Km?This model is a simple but
25、 very good ray-tracing modelAccurate in predicting large-scale signal strength over large distance with tall towersUsed widely in prediction cellular propagation lossEffective when a single ground reflection dominates the multipath propagation effectl Free-Space Path Lossl Two-Path Modell Compare the free space path loss model with the two path model.
