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1、20.07.2020,1,Chapter 3: Mobile Radio Propagation:Large-Scale Path Loss,20.07.2020,2,3.1 Introduction to Radio wave Propagation Small-scale and large-scale fading,20.07.2020,3,3.2 Free Space Propagation Model,In free space, the received power is predicted by Firiis Equ.,Pr(d): Received power with a d
2、istance d between Tx and Rx Pt: Transmitted power Gt: Transmitting antenna gain Gr: Receive antenna gain : The wavelength in meters. d: distance in meters L: The miscellaneous losses L (L=1) are usually due to transmission line attenuation, filter losses, and antenna losses in the communication syst
3、em. L=1 indicates no loss in the system hardware.,20.07.2020,4,Reflection: 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:occurs
4、 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 Mechanisms,20.07.2020,5,reflection（反射）at large obstacles, Scattering (散射)at small
5、 obstacles, diffraction (衍射)at edges,20.07.2020,6,EIRP find the coverage distance d. Transmit Power: Pt=50W=47dBm Pr(d0)=-24.5dBm PL(dB)=40log(d/d0)=-24.5-(-100)=75.5db If n=4,log(d/d0)=75.5/40=1.8875,d=7718m,Example 1,20.07.2020,14,The model in Equation (3.11) does not consider the fact that the su
6、rrounding 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 value predicted by Equation (3.11).,Log-normal Shadowing,20.07.2020,15,Log-normal Shadowing,20.07.2020,16,Dete
7、rmination of Percentage of Coverage Area,20.07.2020,17,U(r) as a function of probability of signal above threshold on the cell boundary.,20.07.2020,18,Example 2,A local average signal strength field measurements , the measured data fit a distant-dependent mean power law model having a log-normal dis
8、tribution 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 than -25dBm, define the standard deviation, ,for the path loss model at d=10m.,20.07.2020,19,Four receiv
9、ed 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 for these measurements follows the model in Equation (3.12.a), where d0 = 100 m: (a) find the minimum mean square
10、 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) predict the likelihood that the received signal level at 2 km will be greater than -60 dBm; and (e) predict the
11、percentage of area within a 2 km radius cell that receives signals greater than -60 dBm, given the result in (d).,Example 3,20.07.2020,20,The value of n which minimizes the mean square error can be obtained by equating the derivative of J(n) to zero, and then solving for n. (a)Using Equation (3.11),
12、 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 received power at a distance di, and let be the estimate for pi using the path loss model of Equation (3.10). T
13、he sum of squared errors between the measured and estimated values is given by,Setting this equal to zero, the value of n is obtained as n = 4.4.,20.07.2020,21,(b)The sample variance 2 = J(n)/4 at n = 4.4 can be obtained as follows.,therefore = 6.17 dB, which is a biased estimate.,20.07.2020,22,(c)T
14、he 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 greater than -60 dBm, then 92% of the cell area receives coverage above 60dbm,20.07.2020,23,3.5 Outdoor Propagation
15、 Models,Okumura Model(150-1920MHz,1km-100km) Hata Model(150-1500MHz,1km-20km) Egli Model(40-400MHz,0-64km),20.07.2020,24,not provide any analytical explanation its slow response to rapid changes in terrain,Okumura Model,20.07.2020,25,Okumura median attenuation and correction,20.07.2020,26,Find the m
16、edian 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 carrier frequency of 900 MHz, find the power at the receiver (assume a unity gain receiving antenna).,Example 4,20.07.2020,27,HATA mode
17、l &COST 231 extension,20.07.2020,28,Example 5,In the suburban of a large city, d = 10 km, hte = 200 m, hre = 2 m , carrier frequency of 900 MHz, using HATA s model find the path loss.,20.07.2020,29,3.6 Indoor propagation models,20.07.2020,30,Feature of Indoor Radio Channel,The distances covered are
18、much smaller, and the variability of the environment is 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 buildi
19、ng type. Indoor radio propagation is dominated by the same mechanisms as outdoor: reflection, diffraction, and scattering. However, conditions are much more variable.,20.07.2020,31,Path attenuation factors,Partition Losses in the same floor Partition Losses between Floors(floor attenuation factors,
20、FAF),20.07.2020,32,Log-distance Path Loss Model,In door path 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 identi
21、cal in form to the log-normal shadowing model of outdoor path attenuation model.,20.07.2020,33,Attenuation Factor Model,Where 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 FAF,20.07.2020,34,Si
22、gnal Penetration into buildings,RF penetration 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 t
23、he ground floor of a building at frequencies 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.,20.07.2020,35,Ray Tracing
24、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 and graphical information system (GIS) database. SISP models
25、 support ray tracing as a means of deterministically modeling any indoor or outdoor propagation environment. Through the use of building databases, which may be drawn or digitized using standard graphical software packages, wireless system designers are able to include accurate representations of bu
26、ilding and terrain features.,20.07.2020,36,Exercises,1. If a transmitter produces 50W of power, express the transmit power in units of (a) dBm, and (b) dBW. If 50 W is applied to a unity gain antenna with a 900 MHz carrier frequency, find the received power in dBm at a free space distance of 100 m f
27、rom the antenna. What is Pr(10 km)? Assume unity gain for the receiver antenna. 2.If the base stations use 20 W transmitter powers and 10 dBi gain omnidirectional antennas, determine the cell coverage distance d. Let n = 4 and the standard deviation of 8 dB hold as the path loss model for each cell in the city. Also assume that a required signal level of -90 dBm must be provided for 90% of the coverage area in each cell. Assume d0 = 1 km .(Q(0.7)=0.24,f=900MHz),