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1、10/29/20211chapter 3: mobile radio propagation:large-scale path loss10/29/20212 3.1 introduction to radio wave propagation small-scale and large-scale fading10/29/202133.2 free space propagation modelin free space, the received power is predicted by firiis equ.pr d()ptgtgr 242d2lpr(d): received powe

2、r with a distance d between tx and rxpt: transmitted powergt: transmitting antenna gaingr: receive antenna gain : the wavelength in meters.d: distance in metersl: the miscellaneous losses l (l=1) are usually due to transmission line attenuation, filter losses, and antenna losses in the communication

3、 system. l=1 indicates no loss in the system hardware.10/29/20214reflection: occur from the surface of the earth and from buildings and walls.diffraction:occurs when the radio path between the transmitter and receiver is obstructed by a surface that has sharp irregularities (edges). scattering:occur

4、s when the medium through which the wave travels consists of objects with dimensions that are small compared to the wavelength, and where the number of obstacles per unit volume is large. 3.3 the three basic propagation mechanisms10/29/20215 reflection(反射)at large obstacles, scattering (散射)at small

5、obstacles, diffraction (衍射)at edges10/29/20216eirp&erpeirp: effective isotropic radiated power represents the maximum radiated power available from a transmitter in the direction of maximum antenna gain, as compared to an isotropic radiator.erp: effective radiated power erp is used instead of ei

6、rp to denote the maximum radiated power as compared to a half-wave dipole antenna (instead of an isotropic antenna). in practice, antenna gains are given in units of dbi (db gain with respect to an isotropic sourse) or dbd (db gain with respect to a half-wave dipole)ttgpeirp 2.15db10/29/202179dbi an

7、tenna & 3dbi antenna10/29/20218path lossthe path loss, which represents signal attenuation as a positive difference (in db) between the effective transmitted power and the received power. the path loss for the free space model when antenna gains are included is given by quantity measured in db,

8、is defined as thewhen antenna gains are excluded, the antennas are assumed to have unity gain, and path loss is given by)4(log10log10)(222dggppdbplrtrt222()10log10log(4 )trppl dbpd ()32.4420lg20lgpl dbfd(f:mhz,d:km)10/29/20219the friis free space model is only a valid predictor for pr for values of

9、d, which are in the far-field of the transmitting antenna. the far-field of a transmitting antenna is defined as the region beyond the far-field distance df , which is related to the largest linear dimension of the transmitter antenna aperture and the carrier wavelength. the far-field distance is gi

10、ven byto be in the far-field region, d must satisfy22ddffdd ffddandd the far-field region of a transmitting antenna10/29/202110the reference distanceit is clear that equation does not hold for d=0. for this reason, large-scale propagation models use a known received power reference point. the receiv

11、ed power, pr(d), at any distance dd0, may be related to pr at d0.if pr is in units of dbm or dbw, the received power is given by frrddddddpdp0200)()(00( )10log()20log()trpdpl dpl dpd000( )10log()20log() rrfdp dp ddddd10/29/2021113.4 link budge design using path loss model00()() 10 log()dpl dbpl dndl

12、og-distance path loss modelboth theoretical and measurement-based propagation models indicate that average received signal power decreases logarithmically with distance, whether in outdoor or indoor channels. the average large-scale path loss for an arbitrary t-r separation is expressed as a functio

13、n of distance by using path loss exponent n.n is the path loss exponent which indicates the rate at which the path loss increases with distanced0 is the close-in reference distance which is determined d is the t-r separation distance10/29/202112path-loss exponents10/29/202113if a transmitter produce

14、s power:pt=50w, receive sensitivity (minimum usable signal level)is -100dbm.assume d0=100m, with a 900mhz carrier frequency, n=4,gt=gr=1; find the coverage distance d.transmit power: pt=50w=47dbmpr(d0)=-24.5dbmpl(db)=40log(d/d0)=-24.5-(-100)=75.5dbif n=4,log(d/d0)=75.5/40=1.8875,d=7718mexample 1 10/

15、29/202114xddndplxdbpldbdpl)log(10)()()(00the model in equation (3.11) does not consider the fact that the surrounding environmental clutter may be vastly different at two different locations having the same t-r separation. this leads to measured signals which are vastly different than the average va

16、lue predicted by equation (3.11). log-normal shadowing10/29/202115log-normal shadowing 10/29/202116ir0rdadetermination of percentage of coverage area10/29/202117 u(r) as a function of probability of signal above threshold on the cell boundary.10/29/202118example 2 a local average signal strength fie

17、ld measurements , the measured data fit a distant-dependent mean power law model having a log-normal distribution about the mean. assume the mean power law was found to be .if a signal of 1mw was received at d0=1m from the transmitter, and at a distance of 10m, 10% of the measurements were stronger

18、than -25dbm, define the standard deviation, ,for the path loss model at d=10m.3.5( )rp dd10/29/202119four received power measurements were taken at distances of 100 m, 200 m, 1 km, and 3 km from a transmitter. these measured values are given in the following table. it is assumed that the path loss f

19、or these measurements follows the model in equation (3.12.a), where d0 = 100 m: (a) find the minimum mean square error (mmse) estimate for the path loss exponent, n; (b) calculate the standard deviation about the mean value; (c) estimate the received power at d = 2 km using the resulting model; (d)

20、predict the likelihood that the received signal level at 2 km will be greater than -60 dbm; and (e) predict the percentage of area within a 2 km radius cell that receives signals greater than -60 dbm, given the result in (d). example 310/29/202120the value of n which minimizes the mean square error

21、can be obtained by equating the derivative of j(n) to zero, and then solving for n.(a)using equation (3.11), we find = pi(d0)-10nlog(di/ 100 m). recognizing that p(d0) = 0 dbm, we find the following estimates for p, in dbm: the mmse estimate may be found using the following method. let pi be the rec

22、eived power at a distance di, and let be the estimate for pi using the path loss model of equation (3.10). the sum of squared errors between the measured and estimated values is given byip0( /)nd dsetting this equal to zero, the value of n is obtained as n = 4.4.ip10/29/202121(b)the sample variance

23、2 = j(n)/4 at n = 4.4 can be obtained as follows. therefore = 6.17 db, which is a biased estimate. 10/29/202122(c)the estimate of the received power at d = 2 km is(d)the probability that the received signal level will be greater than -60 dbm is(e)67.4% of the users on the boundary receive signals gr

24、eater than-60 dbm, then 92% of the cell area receives coverage above 60dbm 6057.24pr( )60()67.4%6.17rp ddbmq 10/29/2021233.5 outdoor propagation models okumura model(150-1920mhz,1km-100km) hata model(150-1500mhz,1km-20km) egli model(40-400mhz,0-64km)10/29/202124not provide any analytical explanation

25、 its slow response to rapid changes in terrain okumura model 10/29/202125okumura median attenuation and correction10/29/202126find the median path loss using okumuras model for d = 50 km, hte = 100 m, hre = 10 m in a suburban environment. if the base station transmitter radiates an eirp of 1 kw at a

26、 carrier frequency of 900 mhz, find the power at the receiver (assume a unity gain receiving antenna). example 410/29/202127hata model &cost 231 extension10/29/202128example 5in the suburban of a large city, d = 10 km, hte = 200 m, hre = 2 m , carrier frequency of 900 mhz, using hata s model fin

27、d the path loss. 50() 69.55 26.16log900 13.82log200 1.05 (44.9 6.55log200)log10 143.80l urban dbdb250()143.802log(900/28)5.4133.86ldbdb10/29/2021293.6 indoor propagation models10/29/202130feature of indoor radio channel the distances covered are much smaller, and the variability of the environment i

28、s much greater for a much smaller range of t-r separation distances. it has been observed that propagation within buildings is strongly influenced by specific features such as the layout of the building, the construction materials, and the building type. indoor radio propagation is dominated by the

29、same mechanisms as outdoor: reflection, diffraction, and scattering. however, conditions are much more variable. 10/29/202131path attenuation factors partition losses in the same floor partition losses between floors(floor attenuation factors, faf)10/29/202132log-distance path loss model in door pat

30、h loss has been shown by many researchers to obey the distance power law where the value of n depends on the surroundings and building type, and x represents a normal random variable in db having a standard deviation of sigma. this is identical in form to the log-normal shadowing model of outdoor pa

31、th attenuation model.xddndpldbdpl)log(10)()(0010/29/202133attenuation factor modelwhere nsf represents the exponent value for the “same floor” measurement. the path loss on a different floor can be predicted by adding an appropriate value of faf10/29/202134signal penetration into buildings rf penetr

32、ation has been found to be a function of frequency as well as height within the building measurements showed that penetration loss decreases with increasing frequency. specifically, penetration attenuation values of 16.4db, 11.6db,and 7.6db were measured on the ground floor of a building at frequenc

33、ies of 441mhz, 896.5mhz, and 1400mhz, respectly. results showed that building penetration loss decreased at a rate of 1.9db per floor from the ground level up to the fifteenth floor and then began increasing above the fifteen floor.10/29/202135ray tracing and site specific modeling in recent years, the computational and visualization capabilities of computers have accelerated rapidly. new methods for predicting radio signal coverage involve the use of site specific (sisp) propagation models

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