Wireless Communication -...
Transcript of Wireless Communication -...
[1] Communication Signal Processing Lab
Wireless Communication
Chapter 4
Mobile Radio Propagation –
Large-Scale Path Loss
[2] Communication Signal Processing Lab
4.1 Introduction to Radio Wave Propagation
The mechanisms behind electromagnetic wave propagation
can generally be attributed to reflection, diffraction, and
scattering.
High-rise buildings diffraction loss
reflection multipath fading
Large-scale propagation model : local average signal strength
over a measurement track of to .
over distance of 100’s ~ 1000’s meters.
Small-scale propagation model : characterize the rapid
fluctuation of the received signal strength over short distance
or short time durations (secs).
5 40
)( s'
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4.2 Free Space Propagation Model
The free space propagation model is used to predict received
signal strength when the transmitter and receiver have a clear
unobstructed line-of-sight path between them, e.g., satellite
communication systems and microwave line-of-sight radio
links.
The Friis free space equation
d : distance , : the transmitted power, : the received power
: the transmitter antenna gain, : the receiver antenna gain
L : the system loss, : the wavelength
Ld
GGPdP rtt
r 22
2
)4()(
tP rP
tG rG
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The gain of an antenna is related to its aperture
: the effective aperture, related to antenna size
meters.
2
4
eAG
eA
c
c
f
c
2=
8103=c
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L = 1 indicates no loss in the system hardware.
The effective isotropic radiated power (EIRP) is defined as
and represents the maximum radiated power available from a
transmitter in the direction of maximum antenna gain, as
compare to an isotropic radiator.
The effective radiated power (ERP) is compared to a half-
wave dipole antenna (instead of an isotropic antenna).
Since a dipole antenna has a gain of 1.64 (2.15dB above an
isotropic).
ttGPEIRP
)(15.2 dBEIRPERP
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Antenna gains are given in units of dBi (dB gain with respect
to an isotropic antenna) or dBd (w.r.t a half-wave dipole).
Path loss, when antenna gains are included,
when antenna gains are excluded,
])4(
log[10log10)(22
2
d
GG
P
PdBPL rt
r
t
])4(
log[10log10)(22
2
dP
PdBPL
r
t
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The far-field, or Fraunhofer region, is defined as the region
beyond the far-field distance ,
D :the largest physical linear dimension of the antenna.
Additionally
Given
fd
22Dd f
ff dDd ,
020
0 ))(()( ddd
ddPdP rr
fr dddP 00 ),(
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This equation may be expressed in units of dBm or dBW.
For example:
when is in units of watts.
for practical systems using low-gain antennas in the
1-2 GHz region is typically chosen to be
1m in indoor environments
100m or 1km in outdoor environments
)20log(]W001.0
)(10log[dBm )( 00
d
ddPdP r
r
)( 0dPr
0d
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Example 4.1
Find the far-field distance for an antenna with maximum dimension
of 1 m and operating frequency of 900 MHz.
Solution:
Largest dimension of antenna , D = 1 m
f = 900 MHz, = c/f = (3108 m/s) / (900 106 Hz)
df = 2(1)2 / 0.33 = 6 m
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Example 4.2
If a transmitter produces 50 W of power, express the transmit power
in units of (a) dBm, and (b) dBW. If 50W is applied to a unity gain
antenna with a 900MHz carrier frequency, find the received power in
dBm at a free space distance of 100 m from the antenna. What is Pr
(10 km)? Assume unity gain for the receiver antenna.
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Solution:
-64.5dBm40dB-dBm5.24]1000/100log[20)100()km10(
10kmat power receiver The
dBm5.24)mW1010log(3.5mW)(log10)dBm(
mW103.5W)105.3()1()100()4(
)3/1)(1)(1(50
)4(
powerreceiver The
dBW 0.17]50log[10)]W1/()W(log[10)dBW(
(b)
dBm 0.47]1050log[10)]mW1/()mW(log[10)dBm(
(a)
MHz900 frequency,Carrier
W50 power,r Transmitte
3-
3-6
22
2
22
2
3
rr
rr
rttr
tt
tt
c
t
PP
PP
Ld
GGPP
PP
PP
f
P
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4.3 Relating Power to Electric Field
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If a current flows through such an antenna, it launches
electric and magnetic fields that can be expressed as
)/(
2
0
)/(
3
2
220
0
)/(
320
0
}{4
sin
}{4
sin
}1
{2
cos
cdtjc
cdtj
c
c
cdtj
c
r
c
c
c
ed
c
d
j
c
LiH
edj
c
d
c
d
j
c
LiE
edj
c
dc
LiE
0 HHE r
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In far-field region, only the radiated field components of
and
need be considered.
In free space, the power flux density is given by
: the intrinsic impedance of free space
E H
)W/m( 2dP
222
22W/m
44
E
R
E
d
GP
d
EIRPP
fs
ttd
fsR ) (377 120
Ω
|E|Pd
377
2
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The power received at distance d
: the effective aperture of the receiver antenna
W480
||
)4(120
||)(
2
22
22
22
rrtt
eedr
GE
d
GGPA
EAPdP
eA
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If the receiver antenna is modeled as a matched resistive load
to the receiver,
Vant : the open circuit voltage at the antenna.
Rant : the resistance of the matched receiver.
ant
ant
ant
ant
ant
rR
V
R
V
R
VdP
4
]2/[)(
222
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Example 4.3
Assume a receiver is located 10 km from a 50 W transmitter. The
carrier frequency is 900MHz, free space propagation is assumed,
Gt = 1, and Gr = 2 , find (a) the power at receiver, (b) the magnitude
of the E-field at the receiver antenna, (c) the rms voltage applied to
the receiver input assuming that the receiver antenna has a purely real
impedance of 50 and is matched to the receiver.
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Solution:
Transmitter power, Pt = 50W. Carrier frequency, fc = 900MHz
Transmitter antenna gain, Gt = 1. Receiver antenna resistance, Gr = 2.
Receiver antenna resistance = 50.
mV187.050107)(V
(c)
V/m 0039.0)4/(33.02
120107
4/
120)(120)(||
(b)
dBm 61.5dBW5.91
)10000)4(
)3/1(2150log(10)
)4(log(10)(
)(
10
2
10
2
22
2
22
2
antr
r
r
e
r
rttr
RdP
G
dP
A
dPE
d
GGPdP
a
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4.4 The Three Basic Propagation Mechanisms
Reflection – occurs from the surface of the earth or from
building and walls.
Diffraction – occurs when the radio path between the
transmitter and receiver is obstructed by a
surface that has sharp irregularities (edges).
Scattering – are produced by rough surface, small objects, or
by other irregularities in the channel, e.g.,
foliage, street signs, and lamp posts.
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4.5 Reflection
The electric field intensity of the reflected and transmitted
waves may be related to the incident wave through the
Fresnel reflection coefficient ( ).
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4.5.1 Reflection from Dielectric
The behavior for arbitrary directions of polarization can be
studied by considering the two distinct cases.
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In Figure 4.4a, the E-field polarization is parallel with the
plane of incidence. (vertical polarization, normal component,
with respect to the reflecting surface).
In Figure 4.4b, the E-field polarization is perpendicular to
the plane of incidence. (parallel to the reflecting surface).
1, 1, 1 :the permittivity, permeability, and conductance.
For perfect (lossless) dielectric, 0r .
For lossy dielectric ( absorbing power), 0r-j
f
2
'
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The reflection coefficients for the two cases are given by
where i is the intrinsic impedance of the ith medium.
The velocity of an electromagnetic wave is
incidence) of plane the tonormal field-(E sinsin
sinsin
incidence) of planein field-(E sinsin
sinsin
12
12
12
12
ti
ti
i
r
it
it
i
r
E
E
E
E
=
=
i
ii
1v
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Snell’s Law
The boundary conditions from Maxwell’s equations give
and
where is either or , depending on whether the E-field
is in (vertical) or normal (horizontal) to the plane of
incidence.
)θ-90sin)θ-90sin 2211 ti ((
ri θθ
it
ir
EE
EE
)1(
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In the general case of reflection or transmission, the
horizontal and vertical axes of the spatial coordinates may
not coincide with the perpendicular and parallel axes of the
propagation waves.
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The vertical and horizontal field components at a dielectric
boundary may be related by
: the angle between the two sets of axes, as shown in Fig
4.5
where Dxx = x for the case of reflection and
Dxx = Tx = 1+ x for the case of transmission.
cossin
sincosR
E
ERDR
E
EiV
iH
CT
dV
dH
0
0
D
DDC
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Figure 4.6 shows a plot of reflection coefficient for
both horizontal and vertical polarization.
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Example 4.4
Demonstrate that if medium 1 is free space and medium 2 is a
dielectric, both |||| and || approach 1 as j approaches 0 regardless
of r.
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Solution:
This example illustrates that ground may be modeled as a perfect
reflector regardless of polarization or ground dielectric properties.
11
1||
0cos0sin
0cos0sin
11
1||
0cos0sin
0cos0sin
0
2
2
||
2
2
||
r
r
r
r
r
r
rr
rr
i
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4.5.2 Brewster Angle
The Brewster angle is the angle at which no reflection occurs
in the medium of origin
= 0
1
1)θsin(
, if
)θsin(
2
0201
21
1
r
rB
r
B
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The Brewster angle occurs only for vertical (i.e.
parallel) polarization.
Example 4.5
Calculate the Brewster angle for a wave impinging on ground having
a permittivity of r = 4.
Solution:
56.265
1sin
5
1
15
3
1)4(
1)4()sin(
1
2
i
i
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4.5.3 Reflection from Perfect Conductors
Since electromagnetic energy cannot pass through a perfect
conductor, a plane wave incident on a conductor has all of its
energy reflection.
For E-field in plane of incidence,
i = r
Ei = Er
for E-field normal to plane of incidence,
i = r
Ei = Er
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4.6 Ground Reflection (Two-Ray) Model
In a mobile radio channel, a single direct path between the
base station and a mobile is seldom the only physical means
for propagation.
The two-ray ground reflection model shown in Figure 4.7 is a
useful and more accurate propagation model.
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ETOT = ELOS + Eg
If E0 is the free space E-field at a reference distance d0 from
the transmitter, then for d > d0 ,
))(cos(),(
))(cos(),(
)( ))(cos(),(
''
''
00''
'
'
00'
000
c
dt
d
dEtdE
c
dt
d
dEtdE
ddc
dt
d
dEtdE
cg
cLOS
c
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The laws of reflection give
For small values of i (i.e., grazing incidence), as shown in
Example 4.4 = 1, Et = 0.
0θθi
it
i
EE
EE
)1(
g
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Assuming perfect horizontal E-field polarization
using the method of images in next page,
))(cos()1())(cos(),(''
''
00'
'
00
c
dt
d
dE
c
dt
d
dEtdE ccTOT
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c
c
rtrt
rtrt
fc
c
hhdd
hh
dhhdhhdd
2
θ
2θ
))( ( 2
)()(
d
2222'''
=
=
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As d becomes large, the difference between the distances d’
and d’’ becomes very small
ELOS and Eg differ only in phase.
At t = d/c
||||||''
00
'
0000
d
dE
d
dE
d
dE
]1θ[
0cos))(cos(),(
00
''
00'''
00''
d
dE
d
dE
c
dd
d
dE
c
dtdE cTOT
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ETOT(d) decays in an oscillatory fashion, with local maxima
being 6dB greater than the free space value and the local
minima plummeting to dB (although in reality this never
happens).
)2
θsin(2
θcos22
θsin)1θ(cos)(|)(|
00
00
200
d
dE
d
dE
d
dEdETOT
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As d is large
where k is a constant related to E0 .
rtrt
rtTOT
rt
hhhhd
d
k
d
hh
d
dEdE
d
hh
20
3
20 as
V/m 22
)(
implies which 3.02
2
θ
2
θsin
3.02
θ
2
00
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Combining Equations (4.2) ,(4.15) and (4.51), the received
power at distance d can be expressed as
)4120
||(
)120
||4
4120
||
120
||(
2
0
2
0
4
222
0
2
0
222
4
22
d
GPE
d
hhG
dEGEA
EP
d
hhGGPP
tt
rtr
rer
rtrttr
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The receiver power falls off with distance raised to the fourth
power, or at a rate of 40 dB/decade. Path loss is more rapid
than in free space and independent of frequency.
)log20log20log10log10(log40)( rtrt hhGGddBPL
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Example 4.6
A mobile is located 5 km away from a base station and uses a vertical
/4 monopole antenna with a gain of 2.55 dB to receive cellular radio
signals. The E-field at 1 km from the transmitter is measured to be
10-3 V/m. The carrier frequency used for this system is 900 MHz.
(a) Find the length and the effective aperture of the receiving
antenna.
(b) Find the received power at the mobile using the two-ray ground
reflection model assuming the height of the transmitting antenna
is 50 m and the receiving antenna is 1.5 m above ground.
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Solution:
T-R separation distance = 5 km ,
E-field at a distance of 1km=10-3 V/m
Frequency of operation, f = 900MHz
dBm 68.92or dBW 122.68W105.4 )km5(
]4
)333.0(8.1[
377
)101.113()(
V/m101.113])105(333.0
)5.1)(50(2[
105
101102
V/m 22
)(
since )b(
0.016mantenna of aperature Effective
8.33cmm0833.04/333.04/ (a)
m333.010900
103
13-
226
6
33
33-
2
00
2
6
8
dP
d P
d
k
d
hh
d
dEdE
hhd
L
f
c
r
r
rtR
rt
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4.7 Diffraction
Diffraction allows radio signals to propagation behind
obstructions.
Huygen’s principle: all points on a wavefront can be
considered as point sources of the production of secondary
wavelets, and that these wavelets combine to produce a new
wavefront in the direction of propagation.
Diffraction is caused by the propagation of secondary
wavelets into a shadowed region.
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4.7.1 Fresnel Zone Geometry
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Assuming h d1,d2 and h , the excess path length (the
difference between the direct path and the diffracted path ) is
The corresponding phase difference is given by
Figure 4.10c
when tanx x , = +
21
)212 (
2 dd
ddh
21
)212 (
2
22
dd
ddh
21
)21(
dd
ddh
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Define Fresnel-Kirchoff diffraction parameter v as
It is clear that is a function of h and the position of the
obstruction.
Fresnel-Kirchoff represent successive regions where
form) convenient (a 2
)(
2)(2
2
21
21
21
21
v
dd
dd
dd
ddhv
2
n
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These circles are called Fresnel zone. The radius of the nth
Fresnel zone circle
21
21
dd
ddnrn
For d1,d2 rn .
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rn have maximum radius if d1 = d2, i.e., the plane is midway
between transmitter and receiver. Hence , this illustrates how
shadowing is sensitive to the frequency as well as the
location of obstructions with relation to the transmitter and
receiver.
Diffraction loss occurs from the blockage of secondary waves
such that only a portion of the energy is diffracted around an
obstacle.
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As long as 55% of first Fresnel zone is kept clear, then
further Fresnel zone clearance does not significantly alter the
diffraction loss.
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4.7.2 Knife-edge Diffraction Model
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The field strength at point R, located in the shadowed region
(diffraction zone) is a vector sum of the fields due to all of
the secondary Huygen’s sources in the plane above the knife
edge. The electric field strength, Ed, is given by
where E0 is the free space field strength in the absence of
both the ground and the knife edge and F(v) is the complex
Fresnel integral.
v
d dttjj
vFE
E))/2exp((
2
)1()( 2
0
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The diffraction gain due to the presence of a knife edge, as
compared to the free space E-field, is given by
An approximate solution is (provided by Lee)
|)(|log20)( vFdBGd
4.2 )225.0
log(20)(
4.21 ))1.038.0(1184.04.0log(20)(
10 ))95.0exp(5.0log(20)(
01- )62.05.0log(20)(
-1 0)(
2
vv
dBG
vvdBG
vvdBG
vvdBG
vdBG
d
d
d
d
d
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Example 4.7
Compute the diffraction loss for the three cases shown in Figure 4.12.
Assume =1/3 m, d1 =1 km, d2 = 1 km ,and (a) h = 25 m, (b) h = 0,
(c) h = 25 m. Compare your answers using values from Figure4.14,
as well as the approximate solution given by Equation (4.61.a)-
(4.61.e). For each of these cases, identify the Fresnel zone within
which the tip of the obstruction lies.
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Solution:
= 1/3, d1 = 1 km, d2 = 1 km.
75.33333.0
625.022 2/
625.010001000
)10001000(
2
25)(
2
is raysn diffractio anddirect ebetween th differencelength path The
dB. 22 is lossn diffractio The
74.210001000)3/1(
)10001000(225
)(2
,25 )(
2
21
21
2
21
21
ΔnnΔ
mdd
ddhΔ
dd
ddhv
mha
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Solution:
(b) h = 0 m, the Fresnel diffraction parameter v = 0.
From Figure 4.14 the diffraction loss is 6 dB.
For this case h = 0 and = 0, the tip of the obstruction lies
in the middle of the first Fresnel zone.
(c) h = -25 m , the Fresnel diffraction parameter v = -2.74.
From Figure 4.14 the diffraction loss is 1 dB.
Using the numerical approximation in Equation(4.61a), the
diffraction loss is equal to 0 dB.
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Example 4.8
Given the following geometry, determine (a) the loss due to knife-
edge diffraction ,and (b) the height of the obstacle required to induce
6 dB diffraction loss. Assume f = 900 MHz.
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Solution:
.m16.4 ,12000
25
2000
0. loss, dB 6For (b)
dB. 25.5 is lossn diffractio The
24.4)200010000()3/1(
20001000020424.0
rad 0424.02.434
15.2)2000
75(tan
2865.0)10000
2575(tan
m 3
1
10900
103h wavelengtThe (a)
1-
1-
6
8
hh
v
v
f
cλ
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4.7.3 Multiple Knife-edge Diffraction
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4.8 Scattering
Objects such as lamp posts and trees tend to scatter energy in
all directions, thereby providing additional radio energy at a
receiver.
Surface roughness is often tested using the Rayleigh criterion
which defines a critical height hc given by
The surface protuberance hhc the surface is smooth .
hhc the surface is rough.
i
chθsin8
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For rough surfaces, rough = s where the scattering loss
factor s is to account for the diminished reflected wave.
Ament found that
])θsin
(8[])θsin
(8exp[
tomodified Boithias
])θsin
(8exp[
20
2
2
ihihs
ihs
I
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4.8.1 Radar Cross Section Model
Knowledge of the physical location and size of the scattering
object is helpful to predict the received signal strength.
The radar cross section (RCS) of a scattering object is
defined as the ratio of the power density of the signal
scattered in the direction of the receiver to the power density
of the radio wave incident upon the scattering object, and has
units of squared meters.
For urban mobile radio systems, models based on the bistatic
radar equation may be used to compute the received power
due to scattering in the far field,
RT
TTR
dd
RCSGPP
log2020log-)30log(4-
]m dB[)log(20)dBi()dBm()dBm( 2
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For medium and large size buildings located 5-10 km away,
RCS values were found to be in the range of 14.1 to 55.7 dB
m2.
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4.9 Practical Link Budget Design Using Path
Loss Models
4.9.1 Log-distance Path Loss Model
The average large-scale path loss is expressed as
where n is the path loss exponent.
)log(10)( )dB(
or
)()(
0
0
0
d
dndPLPL
d
ddPL n
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In free space, n = 2 when obstructions are present, n 2
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d0 = 1 km, for large coverage cellular systems.
d0 = 100 m or 1 m, for microcellular systems.
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4.9.2 Log-normal Shadowing
The path loss PL(d) at a particular location is random and
distributed log-normally (normal in dB) about the mean
distance-dependent value
where X is a zero-mean Gaussian distributed random
variable (in dB) with standard deviation (also in dB).
])[(][])[(
and
)log(10)()(])[( 0
0
dBdPLdBmPdBmdP
Xd
dndPLXdPLdBdPL
tr
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The log-normal distribution describes the random shadowing
effects, or is refered to as log-normal shadowing.
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Since PL(d) is normal in dB, so is Pr(d). The probability that
is Pr(d) (in dB) will exceed a certain value can be
calculated as
where Q-function or error function are defined as
))(
(])(Pr[
dPQdP r
r
)(1)(
and
)]2
(1[2
1
)2
exp(2
1)(
2
zQzQ
zerf
xzQ
z
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The probability that the received signal level will be below
is given by
))(
(])(Pr[
dPQdP r
r
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4.9.3 Determination of Percentage of Coverage
Area
Since Pr[Pr(r) ] is the probability that the random received
signal at d = r exceeds the threshold within an incremental area
dA, the percentage of useful service area U() (i.e., the percentage
of area with a received signal that is equal or greater than )
2
0 0
2
2
])(Pr[R
1
])(Pr[R
1)(
R
r
r
drdrrP
dArPU
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Using (4.71) and substituting t = a + blog(r/R) , it can be
shown that
If , i.e., a = 0, (4.78) becomes
)])1
(1)[21
exp()(1(2
1)(
2 b
aberf
b
abaerfU
)(RPr
))]1
(1)(1
exp(1[2
1)(
2 berf
bU
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Equation (4.78) may be evaluated for a large number of values of
and n, as shown in
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75% boundary coverage means 75% of the time the signal is
to exceed the threshold at the boundary.
If n = 4, = 8 dB, 75% boundary coverage then the area
coverage = 90%.
If n = 2, = 8 dB, 75% boundary coverage area
coverage = 86%.
It is often useful to compute how the boundary coverage
relates to the percent of area covered within the boundary.
[84] Communication Signal Processing Lab
Example 4.9
Four received power measurements were taken at distance 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(4.69a), 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) Predict the likelihood that the received signal level at 2 km will be
greater than -60 dBm;
(e) Predict the percentage of area within a 2 km radius cell that
receives signals greater than -60 dBm.
[85] Communication Signal Processing Lab
Solution:
4.
.8.2887306.654d
)(d
327.1532887.8-6525
))77.14(70())10(35())3(20(0)-(0)(
.77.14ˆ ,10ˆ ,3ˆ ,0ˆ
m).100/log(10)(ˆ find We )(
)ˆ()(
is valueestimated and measured ebetween th errors squared of sum The
2
2222
4321
0
2
n
nn
nJ
nn
nnnnJ
npnpnpp
dndppa
ppnJ
iii
k
i
ii
[86] Communication Signal Processing Lab
Solution:
dB. 60- above coverage receives area cell the
of 88% that determine toused bemay 4.18 Figureor (4.78)Equation )(
%4.67)17.6
24.5760()
)((]dBm60)([ (d)
dBm.24.57)100/2000log()4.4(100)km2(ˆ
km, 2 )(
.dB17.6 ,dB 09.384/36.152
152.36.
)988.6470()4435()2.1320()00()(
.4at 4/)( )(
22
222
2
e
QdP
Qd P P
dp
dc
nJ
nnJb
rrr
[87] Communication Signal Processing Lab
4.10 Outdoor Propagation Models
4.10.1 Longley-Rice Model
The Longley-Rice model is applicable in the frequency range
from 40 MHz to 100 GHz using two-ray ground reflection
model, Fresnel-kirchoff knife edge model, forward scatter
theory, and modified van der Pol-Bremmer method (predict
far field diffraction).
Also called the ITS irregular terrain model.
It is also available as a computer program for frequencies
between 20MHz and 10GHz.
This model operates in two modes: point-to-point mode and
area mode (depending on whether the terrain path profile is
available or not).
[88] Communication Signal Processing Lab
The urban factor (UF) deals with the additional attenuation
due to urban clutter.
Shortcomings: No correction factors to account for the effects
of buildings and foliage, multipath is not considered.
[89] Communication Signal Processing Lab
4.10.2 Durkin’s Model – A Case Study
A computer simulator for predicting field strength contours
over irregular terrain, that was adopted by the Joint Radio
Committee (JRC) in the U.K..
The execution of the Durkin path loss simulator consists of
two parts:
Access a topographic data base and reconstruct the
ground profile information along the radial joining of the
transmitter to the receiver.
Calculates the expected path loss along that radial.
The assumption is that the receiving antenna receives all of
its energy along the radial (LOS and diffraction models) and
experiences no multipath propagation (somewhat pessimistic).
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[91] Communication Signal Processing Lab
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[93] Communication Signal Processing Lab
[94] Communication Signal Processing Lab
Disadvantages :
Do not predict propagation effects due to foliage, building,
other man-made structures.
Do not account for multipath propagation other than ground
reflection, so additional loss factors are often included.
[95] Communication Signal Processing Lab
4.10.3 Okumura Model
One of the most widely used models in urban areas.
Applicable for frequencies in the range 150 MHz to 1920
MHz, distances of 1 km to 100 km, base station antenna
heights ranging from 30 m to 1000m.
Extensive measurements giving the median attenuation
relative to free space.
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[97] Communication Signal Processing Lab
The model can be expressed as
: the gain due to the type of environment.
AREAretemuF GhGhGdfALdBL )()(),()(50
AREAG
[98] Communication Signal Processing Lab
[99] Communication Signal Processing Lab
G(hte) varies at a rate of 20dB/decade and G(hre) varies at a
rate of 10dB/decade for hre 3m.
3m10m )3
log(20)
3 )3
log(10)
301000m )200
log(20)
rere
re
rere
re
tete
te
hh
hG
mhh
hG
mhh
hG
(
(
(
[100] Communication Signal Processing Lab
Other corrections may also be applied to Okumura’s model.
A standard for system planning in modern land mobile radio
system in Japan.
Fairly good in urban and suburban areas, bit not as good in
rural areas.
Example 4.10
Find the median path los using Okumura’s model for d = 50 km,
hte = 100 m, hre = 10 m in a suburban environment. If the base
station transmitter radiates as EIRP of 1 kW at a carrier frequency of
900 MHz, find the power at the receiver (assume a unity gain
receiving antenna).
[101] Communication Signal Processing Lab
Solution:
dB 155.04
946106435125
)()(),()dB(
dB46.10)3
10log(20)
3log(20)(
dB6)200
100log(20)
200log(20)(
dB9
and
dB 43))km 50MHz(900(
curves Okumura theFrom
dB 5.125])1050()4(
)10900/103(log[10]
)4(log[10
as calculated becan losspath space free The
50
232
268
22
2
-.)--(-.
GhGhGdfALL
hhG
hhG
G
A
d L
L
AREAretemuF
rere
tete
AREA
mu
F
F
[102] Communication Signal Processing Lab
Solution:
-95.04dBm.0dB155.04dB-dBm60
)dB()dB()dBm()(
ispower receivedmedian The
50
rr GLEIRPdP
[103] Communication Signal Processing Lab
4.10.4 Hate Model
The Hate model is an empirical formulation of the graphical
path loss data provided by Okumura.
a(hre) is the correction factor for hre which is a function of the
size of the coverage area.
For small to medium city
dh
hahfurbanL
te
retec
log)log55.69.44(
)(log82.13log16.2655.69)dB)((50
dB )8.0log56.1()7.0log1.1()( crecre fhfha
[104] Communication Signal Processing Lab
For large city
For suburban areas, the Hata model is
For rural areas
MHz300for dB97.4)75.11(log2.3(
MHz300for dB1.1)54.1(log29.8(
2
2
crere
crere
fhha
fhha
)
)
4.5)]28/[log(2)()dB( 25050 cfurbanLL
94.40log33.18)(log78.4)()dB( 25050 cc ffurbanLL
[105] Communication Signal Processing Lab
The Hata model predictions are very closely to that of the
Okumura’s models when d 1 km.
This model is well suited for large cell mobile systems, but
not personal communication system (PCS) which have cells
on the order of 1 km radius.
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4.10.5 PCS Extension to Hata Model
The EURO-COST formed COST-231 working committee to
develop an extended version of the Hata model to 2 GHz.
Mte
retec
Cdh
hahfurbanL
log)log55.69.44(
)(log82.13log9.333.46)(20
[107] Communication Signal Processing Lab
4.10.6 Walfisch and Bertoni Model
A model considers the impact of rooftops and building height
by using diffraction to predict average signal strength at street
level.
The model considers the path loss, S, to be
P0 : free space path loss
Q2 : the reduction in the rooftop signal due to the row of
buildings which immediately shadow the receiver at
street level.
P1 : diffraction loss from the rooftop to the street.
12
0 PQPS
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In dB
L0 : free space loss.
Lrts : rooftop-to-street diffraction and scatter loss.
Lms : multiscreen diffraction loss due to the rows of buildings.
msrts LLLS 0)dB(
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[110] Communication Signal Processing Lab
4.10.7 Wideband PCS Microcell Model
Work by Feuerstein et al. in 1991 used a 20MHz pulse
transmitter at 1900 MHz to measure path loss, outage, and
delay spread in typical microcellular systems in San
Francisco and Oakland.
This work revealed that a two-ray ground reflection model is
a good estimate for oath loss in LOS microcells and a simple
log-distance path loss model holds well for OBS (obstructed)
microcell environments.
[111] Communication Signal Processing Lab
For LOS cases,
fff
f
rtrt
f
ddpdnddn
ddpdndPL
hhhh
d
for log10)/log(10
1for )log(10)(
16)(16
1
)2
()2
)((2)(1
112
11
422222
4222222
[112] Communication Signal Processing Lab
For the OBS case,
1)log(10]dB)[( pdndPL
[113] Communication Signal Processing Lab
4.11 Indoor Propagation Models
The indoor radio channel differs from the traditional mobile
radio channel in two aspects
(1) the distances covered are much smaller,
(2) the variability of environment is much greater.
The propagation within buildings is strongly influenced by
the layout of the building, the construction materials, and the
building type.
[114] Communication Signal Processing Lab
Indoor radio propagation is dominated by the same
mechanisms as outdoor : reflection, diffraction, and scattering.
However, conditions are much more variable. For example,
the interior doors are open or closed; antennas positions (desk
or ceiling).
In general, indoor channels may be classified either as LOS
or obstructed (OBS) with varying degrees of clutter.
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4.11.1 Partition Losses (same floor)
Houses v.s. office buildings.
Hard partitions v.s. Soft partitions.
Data bases of losses for a great number of partitions.
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4.11.2 Partition Losses between Floors
The losses between floors of a building are determined by the
external dimensions, materials of the building, the type of the
construction of the floor, and external surroundings.
Even the number of windows and the presence of tinting.
FAF – floor attenuation factors.
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[120] Communication Signal Processing Lab
The attenuation between one floor of the building is greater
than the incremental attenuation caused by each additional
floor.
After about five or six floor separations, very little additional
path loss is experienced.
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[122] Communication Signal Processing Lab
4.11.3 Log-distance Path Loss Model
where the value of n depends on the surroundings and
building type, and X ~N(0, ).
This equation is identical in form to the log-mormal
shadowing model.
Xd
dndPLPL )log(10)()dB(
0
0
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[124] Communication Signal Processing Lab
4.11.4 Ericsson Multiple Breakpoint Model
The Ericsson model provides a deterministic limit on the
range of path loss at a particular distance.
This model was obtained by measurements in a multiple floor
office building.
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[126] Communication Signal Processing Lab
4.11.5 Attenuation Factor Model
This model provides flexibility and was show to reduce X to
around 4 dB, as compared to 13 dB when only log-distance
model was used.
The attenuation factor model is given by
nSF : the exponent value for the same floor.
FAF : a floor attenuation.
PAF : the partition attenuation factor in a ray joining tx and Rx
in 3D (the primary ray tracing technique).
]dB[]dB[)log(10]dB)[(]dB)[(0
0 PAFFAFd
dndPLdPL SF
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Alternatively, FAF may be replaced by an exponent which
already considers the effects of multiple floor separation
nMF : the multiple floor path loss exponent.
]dB[)log(10)(]dB)[(0
0 PAFd
dndPLdPL MF
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Typical values of n
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[130] Communication Signal Processing Lab
[131] Communication Signal Processing Lab
A modification of Equation (4.94)
: a function of frequency.
]dB[]dB[)log(20]dB)[(]dB)[(0
0 PAFFAFdd
ddPLdPL
[132] Communication Signal Processing Lab
Example 4.11
This example demonstrates how to use Equations (4.94) and (4.95) to
predict the mean path loss 30 m from the transmitter, through three
floors of Office Building 1 (see Table 4.5). Assume that two concrete
block walls are between the transmitter and receiver on the
intermediate floors. From Table 4.5 and 4.7, the mean path loss
exponent for same-floor measurements in a building is n = 3.27, the
mean path loss exponent for three-floor measurements is n = 5.22,
and the average floor attenuation factor is FAF = 24.4 dB for three
floors between the transmitter and receiver. Table 4.3 shows a
concrete block wall has about 13dB of attenuation. Let d0 = 1m.
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Solution:
108.6dB
132)30log(22.510]dB)[m1( ]dB)[m30(
.95)Equation(4 using losspath mean The
130.2dB
1324.24)30log(27.310]dB)[m1( ]dB)[m30(
.94)Equation(4 using losspath mean The
PL PL
PL PL
[134] Communication Signal Processing Lab
4.12 Signal Penetration into Buildings
The signal strength received inside of a building due to an
external transmitter is important for wireless systems that
share frequencies with neighboring buildings or with outdoor
systems.
Signal strength received inside a building increases with
height.
At lower floors, clutter induces greater attenuation.
At higher floors, a LOS path may exist.
[135] Communication Signal Processing Lab
RF penetration is also a function of frequency as well as
height.
Measurements in Liverpool : 16.4 dB 11.6dB 7.6dB
(penetration lose) 441MHz 896.5MHz
1400MHz
Measurements by Turkmani : 14.2dB 13.4dB 12.8 dB
900MHz 1800MHz 2300MHz
Windows v.s. without windows : 6dB less penetration loss.
Height : decrease at a rate 1.9 dB per floor from the ground
level up to 15th floor.
decrease at a rate 2 dB per floor from the ground
level up to 9th floor.
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The percentage of windows, Metallic tints (3dB ~ 30dB
attenuation) the angle of incidence of the transmitted wave.
[137] Communication Signal Processing Lab
4.13 Ray Tracing and Site Specific Modeling
Using computer simulation.
Involve the use of Site Specific (SISP) propagation models
and graphical information system (GIS) databases.
Ray tracing techniques are used with aerial photographs or
satellite photographs which convert into usable 3-D databases
for reflection, diffraction, and scattering models.
For example, the SitePlanner® computer-aided design (CAD).
Some day, these modeling techniques may be downloadable
into wireless phones and used to determine instantaneous air
interface parameters.