large-scale-propagation models in wireless communications.ppt

propagation models for wireless communications

1. 1
Mobile Radio Propagation:
Large-Scale Path Loss

2. 2
Introduction to Radio Wave Propagation
Reflection
–Large buildings, earth surface
Diffraction
–Obstacles with dimensions in order of wavelength
Scattering
–Foliage, lamp posts, street signs, walking pedestrian, etc.
transmitted
signal
received
signal
T
s

max

3. 3
Large-scale propagation models
large-scale propagation models
characterize signal strength over large
T-R separation distances
small-scale or fading models:
characterize the rapid fluctuations of the
received signal strength over very short
travel distances or short time durations

4. 4
0 20 40 60 80 100 120
-1
0
1
0 20 40 60 80 100 120
-1
0
1
0 20 40 60 80 100 120
-1
0
1
0 20 40 60 80 100 120
-2
0
2
First Path
Echo path
(case 1)
Echo
path
(case 2)
Constructive
addition
(case 1)
Destructive
addition
(case 2)
Multipath Fading

5. 5
Large-Scale & Small-Scall Fading

6. 6
Large-Scale & Small-Scall Fading (Contd.)
The distance between small scale fades is on the
order of /2

7. 7
Path Loss

8. 8
Propagation Models
Free Space Propagation Model - LOS path exists
between T-R
May applicable for satellite communication or
microwave LOS links
Frii’s free space equation:
-Pt : Transmitted power
-Pr : Received power
-Gt : Transmitter gain
-Gr: Receiver gain
-d: Distance of T-R separation
-L: System loss factor L1
: Wavelength in meter
Ld
GGP
dP
rtt
r 22
2
)4(
)(




9. 9
Antenna Gain
Relationship between antenna gain and effective
area
•G = antenna gain
•A
e = effective area
•f = carrier frequency
•c = speed of light (3 * 10
8
m/s)
 = carrier wavelength
2
2
2
44
c
AfA
G
ee





10. 10
Propagation Models (Contd.)
Path Loss – difference (in dB) between the effective
transmitted power and the received power, and may
or may not include the effect of the antenna gains
Path loss for the free space model when antenna
gains included
PL(dB) = 10 log(P
t/P
r)
= -10 log(Gt Gr 
2
/ (4)
2
d
2
L)
Path loss for the free space model when antenna
gains excluded
PL(dB) = 10 log(P
t/P
r)
= -10 log(
2
/ (4)
2
d
2
L)

11. 11
Fraunhofer distance

2
2D
d
f

Where D is the largest physical linear dimension
of the antenna. Additionally, to be in the far-field
region, d, must satisfy

ff dDd and

12. 12
Propagation Models (Contd.)
Modified free space equation
Pr(d) = Pr(d
0)(d
0/d)
2
d  d
0  d
f
Modified free space equation in dB form
Pr(d) dBm = 10 log[Pr(d
0
)/0.001W] + 20 log(d
0
/d)
where P
r
(d
0
) is in units of watts.

d
f is Fraunhofer distance which complies:
d
f
=2D
2
/
where D is the largest physical linear dimension of the
antenna
In practice, reference distance is chosen to be 1m
(indoor) and 100m or 1km(outdoor) for low-gain
antenna system in 1-2 GHz region.

13. 13
Example (link budget)
Free Space Loss Path
Frequency 0.9000GHz
ERP 50.0000Watts
ERP in dBm 46.9897dBm
Transmission Line Loss 0.0000dB
Tx Antenna Gain 0.0000dBi
Path Length 0.1500Km
Free Space Path Loss 75.0484dB
Rx Antenna Gain 0.0000dBi
Rx Transmission Line Loss 0.0000dB
Rx Signal Strength -28.0587dBm
Rx Threshold (sensitivity) -85.0000dBm
Fade Margin 56.9413dB
RF Link Budget Calculator

14. 14
Relating Power to Electric Field
P
d
= EIRP / 4d
2
= P
t
G
t
/ 4d
2


In free space, the power flux density P
d (in W/m
2
) is given by
Or in another form
P
d
= E
2
/ R
fs
= E
2
/  W/m
2

where R
fs
is the intrinsic impedance of free
space given by =120  = 377 , then
P
d
= E
2
/ 120 W/m
2

15. 15
Relating Power to Electric Field (Contd.)
P
r
(d) = P
d
A
e
= A
e
(E
2
/ 120 )
At the end of receiving antenna
Or when L=1, which means no hardware losses are taken into
consideration
where Ae is the effective aperture of the receiving antenna
P
r(d) = P
t G
t G
r 
2
/ (4)
2
d
2

16. 16
Large-scale Path Loss (Part 2)
The three basic Propagation Mechanisms
Reflection
Diffraction
Scattering

17. 17
Reflection, Diffraction and Scattering
Reflection occurs when a propagating
electromagnetic wave impinges upon an object which
has very large dimensions when compared to the
wavelength of the propagating wave.
Diffraction occurs when the radio path between the
transmitter and receiver is obstructed by a surface
that has sharp irregularities (edges).
Scattering occurs 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.

18. 18
Fresnel Reflection Coefficient (Γ)
It gives the relationship between the electric field
intensity of the reflected and transmitted waves to
the incident wave in the medium of origin.
•The Reflection Coefficient is a function of the material
properties
• It depends on
Wave Polarization
Angle of Incidence
Frequency of the propagating wave
Reflection

19. 19
Reflection from Dielectrics

20. 20
•
The behavior for arbitrary directions of polarization is illustrated through
the two distinct cases in the figure
Case 1
• The E - field polarization is parallel with the plane of incidence
i.e. the E - field has a vertical polarization, or normal component
with respect to the reflecting surface
Case 2
• The E - field polarization is perpendicular to the plane of incidence
i.e. the E - field is parallel to the reflecting surface ( normal to the
page and pointing out of it towards the reader)

21. 21
•The dielectric constant ε of a perfect (lossless)
dielectric is given by
ε = ε
0 ε
r
where ε
r is the relative permittivity
and ε
0 = 8.85 * 10
-12
F/ m
• The dielectric constant ε for a power
absorbing, lossy dielectric is
ε = ε
0 ε
r - j ε’
where ε’ = σ / 2π f

22. 22
•

23. 23
•
In the case when the first medium is free space and μ
1
=

μ
2

the Reflection coefficients for the two cases of vertical and
horizontal polarization can be simplified to

irir
irir


2
2
||
cossin
cossin



iri
iri


2
2
cossin
cossin





24. 24
Brewster Angle
It is the angle at which no reflection occurs in the medium of
origin
It occurs when the incident angle θ
B is such that the Reflection
Coefficient Γ
| | = 0
For the case when the first medium is free space and the
second medium has a relative permittivity ε
r
, the above
equation can be expressed as
21
1
)sin(





B
1
1
)sin(
2



r
r
B




25. 25
Ground Reflection (Two- Ray) Model

26. 26
Whenever 

321 20
3
20
rthhhh
d 
The received E-field can be approximated
mV
d
k
d
hh
d
dE
dE
rt
TOT
/
22
)(
2
00



The power received at distance d is given by
4
22
d
hh
GGPP
rt
rttr
For large T- R distances so received
power falls off to the 4
th
power of d, or at 40 db/
decade
rthhd

28. 28
•This power loss is much more than that in free
space
•At large values of d, the received power and path
loss become independent of frequency.
• The path loss for the 2- ray model in db
PL (db) = 40 log d – ( 10 log G
t + 10 log G
r +
20 log h
t + 20 log h
r )

29. 29
Diffraction
Phenomena: Radio signal can propagate around the
curved surface of the earth, beyond the horizon and
behind obstructions.
Huygens's principle
The field strength of a diffracted wave in the shadowed
region is the vector sum of the electric field components
of all the secondary wavelets in the space around the
obstacles.

30. 30
Fresnel Zone Geometry
The wave propagating from the transmitter to the receiver via
the top of the screen travels a longer distance than if a direct
line-of-sight path exists.

31. 31
Fresnel Zone Geometry(Cont’d)
Angle ,
Fresnel-Kirchoff diffraction parameter
Normalizing ,

32. 32
Fresnel Zone Geometry(Cont’d)
The concentric circles on the plane are
Fresnel Zones.

33. 33
The radius of the nth Fresnel zone circle
The excess total path length traversed by a ray
passing through each circle is

34. 34

35. 35
Consider a receiver at point R, located in the
shadowed region.
The electric field strength Ed,
where E0 is the free space field strength

36. 36
Graphical representation of
The diffraction gain:

37. 37
Lee’s approximate solution:

38. 38
Multiple Knife-edge Diffraction

39. 39
Scattering:
When does Scattering occur?
When the medium through which the wave travels
consists of objects with dimensions that are small
compared to wavelength
The number of obstacles per unit volume is large
Large-scale Path Loss (part 4)
How are these waves produced:
By rough surfaces, small objects or by other irregularities
in the channel
Normally street signs, lamp posts, trees induce scattering
in mobile communication system

40. 40
Rayleigh Criterion:
Surface roughness is tested using the Rayleigh
criterion,its given by
 h
c= /8sin
i
 where,
 
i is the angle of incidence
 h
c
is the critical height of surface protuberance
 for a given 
i
The surface is considered smooth if the minimum to
maximum protuberance h <= h
c
and rough if h>

h
c

41. 41
Radar cross section model:
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 square meters.
RT
2
TTR
20logd -20logd - )30log(4-
]RCS[dBm)20log((dBi)G(dBm)P(dBm)P


Bistatic radar equation

42. 42
Practical link budget design using path loss models
Log –distance Path Loss Model


















0
0
0
log10)()(
or
)(
d
d
ndPLdBPL
d
d
dPL
n
n is the path loss exponent which indicates the rate at
which the path loss increases with distance,

43. 43

44. 44
Log-normal Shadowing:

X
d
d
ndPLXdPLdBdPL 









0
0
log10)()(])[(
X
σ
is the Zero –mean Gaussian distributed random
variable with standard deviation σ(also in dB)
PL(d) is a random variable with a normal distribution.Define























2
z
erf1
2
1
dx
2
exp
2π
1
Q(z)
z
2
x
 












)(
)(Pr
dP
QdP
r
r
The probability that the received signal level will exceed a
certain value γ can be calculated from the cumulative density
function as

45. 45
   







2
00
22
)(Pr
1
)(Pr
1
)(
R
rr
rdrdrP
R
dArP
R
U
Determination of Percentage of Coverage Area



























b
erf
b
U
1
1
1
exp1
2
1
)(
2


46. 46

47. 47

48. 48

49. Radio Propagation Mechanism
As a mobile moves through a coverage area, these
three mechanism have an impact on the
instantaneous received signal strength.
•If a mobile does have a clear line-of-sight path to
the base station, then diffraction and scattering will
not dominate the propagation.
•If a mobile is at street level with out LOS, then
diffraction and scattering will probably dominate
the propagation.
49