Transcript No Slide Title

Mobile Radio Propagation
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Mobile radio channel is an important factor in
wireless systems.
Wired channels are stationary and
predictable, while radio channels are random
and have complex models.
Modeling of radio channels is done in
statistical fashion based on receiver
measurements.
Types of propagation models
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Large scale propagation models
 To predict the average signal strength at a
given distance from the transmitter
 Controlled by signal decay with distance
Small scale or fading models.
 To predict the signal strength at close
distance to a particular location
 Controlled by multipath and Doppler
effects.
Radio signal pattern
Received Power (dBm)
-30
3
-40
-50
-60
-70
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18 20 22 24 26
T-R Separation (meters)
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Measured signal parameters

Electrical Field (Volts/m)
Magnitude E = IEI
Vector
Direction E = xEx + yEy + zEz
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Power (Watts or dBm)
Power is scalar quantity and easier to measure.
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Relation between Watts and dBm
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5
P (dBm) = 10 log10 [ P (mW)]
P(mW)
P(dBm)
10
10
1
0
10-1
-10
10-2
-20
10-6
-60
Physical propagation models
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Free Space Propagation
 Transmitter/receiver have clear LOS path
Reflection
 Wave reaches receiver after reflection off
surfaces larger than wavelength
Diffraction
 Wave reaches receiver by bending at sharp
edges (peaks) or curved surfaces (earth).
Scattering
 Wave reaches receiver after bouncing off
objects smaller than wavelength (snow, rain).
Free Space Propagation
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Transmitter and receiver have clear,
unobstructed LOS path between them.
(Courtesy: webbroadband.blogspot.com)
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Friis transmission equation
Pr = Pt Gt Gr 2
(4)2 d2 L
Pt = Transmitted Power (W)
Pr = Received Power (W)
Gt = Transmitter antenna gain
Gr = Receiver antenna gain
L = System loss factor
 Due to line losses, but not due to propagation
 L 1
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Antenna Gain
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Power Gain of antenna
G = 4Ae / 2,
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Ae is effective aperture area of antenna
Wavelength  = c / f (Hz)
= 3 • 108 / f , meters
Relation between Electric field and Power
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Received power
Pr = IErI2 2 Gr
4 
Impedance of medium:  =  / 
For air or vacuum:
 = (4 • 10-7) /(8.85 • 10-12 )
= 377 
Example
If the received power is Pr = 7 • 10 -10 W,
antenna gain Gr = 2 and transmitting frequency
is 900 MHz, determine the electric field strength
at the receiver.
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Solution
f = 900 MHz = >
 = (3 • 108) / (900 • 106) = 0.33 m
From field-power equation:
IErI = [(Pr •  • 4) / ( 2 • Gr)]1/2
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= [(7 • 10 -10 • 377 • 4) / (0.332 • 2)]1/2
= 0.0039 V/m
Example
A transmitter produces 50W of power.
If this power is applied to a unity gain antenna
with 900 MHz carrier frequency, find the received
power at a LOS distance of 100 m from the
antenna. What is the received power at 10 km?
Assume unity gain for the receiver antenna.
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Solution
Pr = Pt Gt Gr 2
(4)2 d2 L
Pt = 50 W, Gt = 1, Gr = 1, L = 1, d = 100 m
 = (3 • 108) / (900 • 106) = 0.33 m
Solving, Pr = 3.5 • 10-6 W
Pr (10 km) = Pr (100 m) • (100/10000)2
= 3.5 • 10-6 • (1/100)2
= 3.5 • 10-10 W
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Electric Properties of Material Bodies
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Fundamental constants
Permittivity  = 0 r , Farads/m
Permeability  = 0 r ,Henries/m
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Conductivity , Siemens/m
 Types of materials
 Dielectrics – allow EM waves to pass
 Conductors – block EM waves
 Metamaterials – bend EM waves
Reflection at dielectric boundaries
Er =  : Reflection coefficient
Ei
Et = T = 1 +  : Transmission coefficient
Ei
Er
E
i
i
i = r
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r
Et
Vertical Polarization
Hi
||=
Ei
i
Er
Hr
1, 1, 1
r
2, 2, 2
t
Et
2 sint - 1 sini
2 sint + 1 sini
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Horizontal Polarization
Hi
Ei
i
Er
Hr
r 1, 1, 1
2, 2, 2
t
Et
T = 2 sint - 1 sini
2 sint + 1 sini
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Reflection from Perfect Conductor (ET =0)
Ei
i
Er
r
Et
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Vert. polarization
Horiz. polarization
i = r
Ei = Er
i = r
Ei = - Er
Ground Reflection (2-Ray Model)
T (transmitter)
Ei
ELOS
ht
R (receiver)
Er=Eg
i
0
d
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ETOT = ELOS +Eg
hr
Field Equations
d = several kms
ht = 50-100m
ETOT= ELOS + Eg
ETOT(d) =
4E0 d 0 ht hr
d 2
For d > 20hthr / 
Received power Pr=
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2
Pt Gt Gr ht hr
d4
2
Example
A mobile is located 5 km away from a
base station, and uses a vertical /4
monopole antenna with a gain of 2.55dB
to receive cellular radio signals.
The electric field at 1 km from the
transmitter is measured to be 10-3 V/m.
The carrier frequency used is 900 MHz.
(a) Find the length and gain of the
receiving antenna.
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Example
A mobile is located 5 km away from a
base station, and uses a vertical /4
monopole antenna with a gain of 2.55dB
to receive cellular radio signals.
The electric field at 1 km from the transmitter
is measured to be 10-3 V/m.
The carrier frequency used is 900 MHz.
(b) Find the received power at the mobile
using the 2-way ground model assuming the
height of the transmitting antenna is 50 m and
receiving antenna is 1.5 m above the ground.
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Solution:
d0 = 1 km
E0 = 10 -3 V/m
ht =
50 m
hr = 1.5 m
d = 5 km
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(a)
f = 900 MHz
 = (3 • 108) / (900 • 106) = 0.33 m
Length of receiving antenna,
L =  / 4 = 0.33/4 = 0.0833 m = 8.33 cm
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(b)
Gain of antenna = 2.55 dB = > 1.8
4E0 d 0 ht hr
Er (d) =
d 2
= 2 • 10-3 • 1 • 103 • 2 • 50 • 1.5
(5 • 103)2 • 0.333
= 113.1 • 10-6 V/m
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Pr (d)
= I Er I2 2 Gr

4
= (113.1 • 10-6) 2 • (0.333) 2 • 1.8
377
4
= 5.4 • 10-13 W
= -92.68 dBm
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Diffraction
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Diffraction allows radio signals
to propagate around the curved
surface or propagate behind
obstructions.
Based on Huygen’s principle of
wave propagation.
Knife-edge Diffraction Geometry
(a) T is transmitter and R is receiver,
with an infinite knife-edge obstruction
blocking the line-of-sight path.
T
d1
ht
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 h


hobs
R
d2
hr
Knife-edge Diffraction Geometry
(b) T & R are not the same height...
T
ht
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
d1
h
h’

d2

R
hr
Knife-edge Diffraction Geometry
...If  and  are small and h<<d1 and d2,
then h & h’ are virtually identical and
the geometry may be redrawn as in (c).
T
ht
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
d1
h
h’

d2

R
hr
Knife-edge Diffraction Geometry
(c) Equivalent where the smallest height
(in this case hr ) is subtracted from all
other heights.
T
ht - hr

hobs-hr
d1
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

d2
R
Assumptions
h << d1, d2
h >> 
Excess path length
2
h
(d1  d 2 )

2d1d 2

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h ( d1  d 2 )
2 d1d 2
...Assumptions
h << d1, d2
h >> 
Phase difference
 = 2/
= 2  h2 (d1 + d2 )
2  d 1 d2
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Diffraction Parameter
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v=
2
=
2( d1  d 2 )
d1d 2

h
Three Cases
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Case I: h > 0
Case II: h = 0
Case III: h < 0
Case I: h > 0
 and  are positive since h is positive.

h
T
d1
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R
d2
Case II: h = 0
 and  equal 0, since h equals 0.
T
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d1
d2
R
Case III: h < 0
 and  are negative, since h is negative.
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The electric field strength of
the diffracted wave is given by:
Ed = F(v) • Eo
where Eo is the free space field strength in
the absence of both ground and knife edge.
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Approximate Value of
Fresnel Integral F(v):
Gd (dB) = 20 log IF(v)I
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v Range Gd (dB)
v -1
0
-1v 0
20 log (0.5 – 0.62 v)
0v1
20 log (0.5 e-0.95v )
1 v 2.4 20 log (0.4 –
v2.4
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0.1184  (0.38  0.1v) 2 )
20 log (0.225 / v)
Example
Compute the diffraction loss
between the transmitter and
receiver assuming:
 = 1/3 m
d1 = 1 km
d2 = 1 km
h = 25 m
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Solution:
Given   = 1/3 m
d1 = 1km
d2 = 1km
h = 25m
V = h 2( d1  d 2 )
d1d 2
=
25
= 2.74
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2(1000 1000)
(0.3)(1000)(1000)
Using the table,
Gd (dB) = 20 log (0.225/2.74)
= -22 dB
Loss = 22 dB
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Scattering
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When a radio wave impinges
on a rough surface, the
reflected energy is spread out
or diffused in all directions.
Ex., lampposts and foliage.
The scattered field increases
the strength of the signal at
the receiver.
Radar Cross Section (RCS) Model
RCS (Radar Cross Section) =
Power density of scattered wave
in direction of receiver
Power density of radio wave incident
on the scattering object
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Radar Cross Section (RCS) Model
PR = PT • GT • 2 • RCS
(4)3 • dT 2 • dR 2
Where,
PT = Transmitted Power
GT = Gain of Transmitting antenna
dT = Distance of scattering object
from Transmitter
dR = Distance of scattering object
from Receiver
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Practical Link Budget
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Most radio propagation models
are derived using a combination
of analytical and empirical
models.
Empirical approach is based on
fitting curves or analytical
expressions that recreate a set
of measured data.
...Practical Link Budget
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Advantages of empirical models;
Takes into account all propagation
factors, both known and unknown.
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Disadvantages:
New models need to be measured for
different environment or frequency.
Log-Distance Path Model
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Over many years, some classical
propagation models have been
developed, which are used to predict
large-scale coverage for mobile
communication system design.
T
d0
R
PT
PR(d0)
PR(d)
...Log-Distance Path Model
Path loss at d0 = PT/P(d0) = K(d0)n = PL(d0)
Path loss at d = PT/P(d) = K(d)n = PL(d)
PL(d) / PL(d0) = (d/d0)n
PL(d) [dB] = PL(d0) [dB] + 10n log10 (d/d0)
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Received Power in Log-distance model
PR(d) [dbm] = Pt [dbm] – PL(d) [db]
n -> path loss exponent
d0 -> reference distance close to transmitter
Environment
Free space
Urban area cellular radio
LOS in building
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n
2
2.7 – 3.5
1.6 – 1.8
Log-Normal Shadowing
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Log-distance path loss normal gives
only the average value of path loss.
Surrounding environment may be
vastly different at two locations
having the same T – R separation d.
Log-Normal Shadowing
More accurate model includes a random
variable to account for change in
environment.
PL(d) [db] = PL(d) + X
= PL(d0) + 10n log (d / d0) + X
X -> Zero mean Gaussian
random variable (dB)
 -> Standard deviation (dB)
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Received Power in
Log-Normal Shadowing Model
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PR(d) [dbm] = PT[dbm] – PL(d) [db]
Values of n and  are computed from
measured data.
Linear regression method which
minimizes the difference between
measured and estimated path
Estimated over a wide range of
measurement locations and T – R
separations.
...Received Power in
Log-Normal Shadowing Model

Probability [ PR (d) >  ] = Q 


 PR (d ) 



PR (d )  
Probability [ PR (d) <  ] = Q 


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



Calculation of Q Function
2
x
x
  x /2
Q(z) = Q function = 1 x ee
2 dx
2 z z
Q(-z) = 1- Q(z)
2
Q(0) = 1/ 2
Q(z) obtained from Appendix F,
Table F.1, page 647
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Calculation of Q Function
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Example
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Four received power measurements
were taken at the distances of
100m, 200m, 1 km and 3 km from a
transmitter. These measured values
are given in the following table.
The path loss equation model for other
measurements follows log normal
shadowing model where
d0 = 100 m.
Example
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 at 2 km will be greater than
–60 dBm.
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Solution:
T-R distance
100 m
200 m
1 km
3 km
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Measured Power
0 dBm
- 20 dBm
- 35 dBm
- 70 dBm
Let Pi be the average received power
at distance di
Pi (d) = Pi (d0) – 10n log (d /100)
d = d0 = 100m = > P0= 0 dBm
A.
d1= 200 m, P1= -3n,
d2= 1 km, P3= -10n,
d3= 3 km, P4= -14.77n
Mean square error J =  (P – Pi)2
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= (0 – 0)2 + [-20 – (-3n)] 2
+ [-35 – (-10n)] 2 + [-70 – (-14.77n)] 2
= 6525 – 2887.8n + 327.153n2
Minimum value = > dJ(n) / dn
= 654.306n – 2887.8 = 0  n = 4.4
B.
Variance 2 = J / 4 = ( P – Pi)2 / 4
= (0+ 0) + (-20 +13.2)2 + (-35 + 44)2 + (-70 + 64.988)2
4
= 152.36 / 4 = 38.09
  = 6.17 dB
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C.
Pi (d = 2 km)
= 0 – 10(4.4) log (2000/100)
= -57.24 dBm
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D.
Probability that the received signal
will be greater than –60 dBm is:
_____
PR = [PR(d) > -60 dBm] = Q [(- PR (d)) /  ]
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= Q [(-60 + 57.24) / 6.17 ]
= Q [- 0.4473]
= 1 – Q [0.4473]
= 1 – 0.326
= 0.674 = > 67.4%
% of Coverage Area
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Given a circular
coverage area
of radius R...
In the area A,
the received power
PR  
The area A is
defined as
U()
r
R
Area A
Calculation of Coverage Area U()
U () = (1 /  R2 ) ƒ Prob [PR (R) >  ] dA
_____
Where Prob [PR (R) > ] = Q [  - PR (R) /  ]
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Final Equation for U()
1
U ( )  [1  erf ( a )  e
2
where :
(1 2 ab)
b2
1  ab
(1  erf (
))]
b
  Pt  PL( d 0 )  10n log(
a
b

R
)
d0
2
10n log e
 2
The error function erf(z) =
2

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z
e
 x2
0
 1  2Q(
2z )
Alternate method: Use Fig. 4.18 (p 143)
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Example
For the previous problem,
predict the percentage of area
with a 2 km radius cell that
receives signals greater than
–60 dBm.
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Solution
From solution to previous example,
Prob [PR (R) > ] = 0.674
=>( / n) = 6.17 / 4.4
= 1.402
From Figure 4.18,
Fraction of total area = 0.92 => 92%
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Other Propagation Models
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Outdoor propagation models
 Longley Rice model:
point-to-point communication systems
(40MHz–100MHz)
 Okumara’s model:
widely used in urban areas
(150 MHz – 300 MHz)
 Hata model:
graphical path loss
(150 MHz – 1500 MHz)
Other Propagation Models
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Indoor propagation models
 Log-distance path loss
model
 Ericsson multiple breakdown
model