1. 1
3.4 Basic Propagation Mechanisms & Transmission Impairments
(1) Reflection: propagating wave impinges on object with size >> λ
• examples include ground, buildings, walls
(2) Diffraction: transmission path obstructed by objects with edges
• 2nd
ry waves are present throughout space (even behind object)
• gives rise to bending around obstacle (NLOS transmission path)
(3) Scattering propagating wave impinges on object with size < λ
• number of obstacles per unit volume is large (dense)
• examples include rough surfaces, foliage, street signs, lamp posts

2. 2
Models are used to predict received power or path loss (reciprocal)
based on refraction, reflection, scattering
• Large Scale Models
• Fading Models
at high frequencies  diffraction & reflections depend on
• geometry of objects
• EM wave’s, amplitude, phase, & polarization at point of intersection

3. 3
3.5 Reflection: EM wave in 1st
medium impinges on 2nd
medium
• part of the wave is transmitted
• part of the wave is reflected
(1) plane-wave incident on a perfect dielectric (non-conductor)
• part of energy is transmitted (refracted) into 2nd
medium
• part of energy is transmitted (reflected) back into 1st
medium
• assumes no loss of energy from absorption (not practically)
(2) plane-wave incident on a perfect conductor
• all energy is reflected back into the medium
• assumes no loss of energy from absorption (not practically)

4. 4
(3) Γ = Fersnel reflection coefficient relates Electric Field intensity
of reflected & refracted waves to incident wave as a function of:
• material properties,
• polarization of wave
• angle of incidence
• signal frequency
boundary between dielectrics
(reflecting surface)
reflected wave
refracted wave
incident wave

5. 5
(4) Polarization: EM waves are generally polarized
• instantaneous electric field components are in orthogonal
directions
in space represented as either:
(i) sum of 2 spatially orthogonal components (e.g. vertical
& horizontal)
(ii) left-handed or right handed circularly polarized components
• reflected fields from a reflecting surface can be computed using
superposition for any arbitrary polarizationE||
E⊥

6. 6
3.5.1 Reflection from Dielectrics
• assume no loss of energy from absorption
EM wave with E-field incident at ∠θi with boundary between 2
dielectric media
• some energy is reflected into 1st
media at ∠θr
• remaining energy is refracted into 2nd
media at ∠θt
• reflections vary with the polarization of the E-field
plane of incidence
reflecting surface= boundary
between dielectrics
θi
θr
θt
plane of incidence = plane containing incident, reflected, & refracted rays

7. 7
Two distinct cases are used to study arbitrary directions of polarization
(1) Vertical Polarization: (Evi) E-field polarization is
• parallel to the plane of incidence
• normal component to reflecting surface
(2) Horizontal Polarization: (Ehi) E-field polarization is
• perpendicular to the plane of incidence
• parallel component to reflecting surface
plane of incidence
θi
θr
θt
Evi
Ehi
boundary between dielectrics
(reflecting surface)

8. 8
• Ei & Hi = Incident electric and magnetic fields
• Er & Hr = Reflected electric and magnetic fields
• Et = Transmitted (penetrating) electric field
Hi Hr
Ei Er
θi θr
θt
ε1,µ1, σ1
ε2,µ2, σ2
Et
Vertical Polarization: E-field in
the plane of incidence
Hi
HrEi
Er
θi θr
θt
ε1,µ1, σ1
ε2,µ2, σ2
Et
Horizontal Polarization: E-field
normal to plane of incidence

9. 9
(1) EM Parameters of Materials
∀ε = permittivity (dielectric constant): measure of a materials ability
to resist current flow
• µ = permeability: ratio of magnetic induction to magnetic field
intensity
• σ = conductance: ability of a material to conduct electricity,
measured in Ω-1
dielectric constant for perfect dielectric (e.g. perfect reflector of
lossless material) given by
ε0 = 8.85 ×10-12
F/m

10. 10
often permittivity of a material, ε is related to relative permittivity εr
ε = ε0 εr
lossy dielectric materials will absorb power  permittivity described
with complex dielectric constant
(3.18)where ε’ =
fπ
σ
2
(3.17)ε = ε0 εr -jε’
highly conductive materials 
∀εr & σ are generally insensitive to operating frequency
r
f
εε
σ
0
<
• ε0 and εr are generally constant
• σ may be sensitive to operating frequency

11. 11
Material εr σ σ/εrε0 f (Hz)
Poor Ground 4 0.001 2.82 ×107 108
Typical Ground 15 0.005 3.77 ×107 108
Good Ground 25 0.02 9.04 ×107 108
Sea Water 81 5 6.97 ×109 108
Fresh Water 81 0.001 1.39 ×106 108
Brick 4.44 0.001 2.54 ×107
4⋅109
Limestone 7.51 0.028 4.21 ×108
4⋅109
Glass, Corning 707 4 0.00000018 5.08 ×103 106
Glass, Corning 707 4 0.000027 7.62 ×105 108
Glass, Corning 707 4 0.005 1.41 ×108 1010

12. 12
• because of superposition – only 2 orthogonal polarizations need be
considered to solve general reflection problem
Maxwell’s Equation boundary conditions used to derive (3.19-3.23)
Fresnel reflection coefficients for E-field polarization at reflecting
surface boundary
• Γ|| represents coefficient for || E-field polarization
• Γ⊥ represents coefficient for ⊥ E-field polarization
(2) Reflections, Polarized Components & Fresnel Reflection
Coefficients

13. 13
Fersnel reflection coefficients given by
(i) E-field in plane of incidence (vertical polarization)
Γ|| =
it
it
i
r
E
E
θηθη
θηθη
sinsin
sinsin
12
12
+
−
= (3.19)
(ii) E-field not in plane of incidence (horizontal polarization)
Γ⊥ =
ti
ti
i
r
E
E
θηθη
θηθη
sinsin
sinsin
12
12
+
−
= (3.20)
ηi = intrinsic impedance of the ith
medium
• ratio of electric field to magnetic field for uniform plane wave in
ith
medium
• given by ηi = ii εµ

14. 14
velocity of an EM wave given by ( ) 1−
µε
boundary conditions at surface of incidence obey Snell’s Law
( ) ( ) )90sin()90sin( 222111 θεµθεµ −=− (3.21)
θi = θr (3.22)
Er = Γ Ei (3.23a)
Et = (1 + Γ )Ei (3.23b)
Γ is either Γ|| or Γ⊥ depending on polarization
• | Γ | ≈ 1  for a perfect conductor, wave is fully reflected
• | Γ | ≈ 0  for a lossy material, wave is fully refracted








−−= −
)90sin(sin90
2
11
it θ
η
η
θ

15. 15
• radio wave propagating in free space (1st
medium is free space)
• µ1 = µ2
Γ|| =
irir
irir
θεθε
θεθε
2
2
cossin
cossin
−+
−+−
(3.24)
Γ⊥ =
iri
iri
θεθ
θεθ
2
2
cossin
cossin
−+
−−
(3.25)
Simplification of reflection coefficients for vertical and horizontal
polarization assuming:
Elliptically Polarized Waves have both vertical & horizontal components
• waves can be depolarized (broken down) into vertical & horizontal
E-field components
• superposition can be used to determine transmitted & reflected
waves

16. 16
(3) General Case of reflection or transmission
• horizontal & vertical axes of spatial coordinates may not coincide
with || & ⊥ axes of propagating waves
• for wave propagating out of the page  define angle ∠θ
measured counter clock-wise from horizontal axes
spatial horizontal axis
spatial vertical axis
θ
⊥
||
orthogonal components
of propagating wave

17. 17
↔vertical & horizontal
polarized components
components perpendicular
& parallel to plane of incidence
Ei
H , Ei
V Ed
H , Ed
V
• Ed
H , Ed
V = depolarized field components along the horizontal &
vertical axes
• Ei
H , Ei
V = horizontal & vertical polarized components of incident
wave
relationship of vertical & horizontal field components at the dielectric
boundary
Ed
H, Ed
V Ei
H , Ei
V = Time Varying Components of E-field






=





i
v
i
H
C
T
d
v
d
H
E
E
RDR
E
E
(3.26)
- E-field components may be represented by phasors

18. 18
for case of reflection:
• D⊥⊥ = Γ⊥
• D|| || = Γ||
for case of refraction (transmission):
• D⊥⊥ = 1+ Γ⊥
• D|| || = 1+ Γ||
R = 





− θθ
θθ
cossin
sincos
, θ = angle between two sets of axes
DC = 




 ⊥⊥
||||0
0
D
D
R = transformation matrix that maps E-field components
DC = depolarization matrix

19. 19
1.0
0.8
0.6
0.4
0.2
0.0
0 10 20 30 40 50 60 70 80 90
|Γ|||
εr=12
εr=4
angle of incidence (θi)
Brewster Angle (θB)
for εr=12
vertical polarization
(E-field in plane of incidence)
for θi < θB: a larger dielectric constant  smaller Γ|| & smaller Er
for θi > θB: a larger dielectric constant  larger Γ|| & larger Er
Plot of Reflection Coefficients for Parallel Polarization for εr= 12, 4

20. 20
εr=12
εr=4
|Γ⊥|1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0 10 20 30 40 50 60 70 80 90
angle of incidence (θi)
horizontal polarization
(E-field not in plane of
incidence)
for given θi: larger dielectric constant  larger Γ⊥ and larger Er
Plot of Reflection Coefficients for Perpendicular Polarization for εr=
12, 4

21. 21
e.g. let medium 1 = free space & medium 2 = dielectric
• if θi  0o
(wave is parallel to ground)
• then independent of εr, coefficients |Γ⊥|  1 and |Γ|||  1
Γ|| = 1
cos
cos
cossin
cossin
2
2
0
2
2
=
−
−
=
−+
−+−
= ir
ir
irir
irir
i
θε
θε
θεθε
θεθε
θ
Γ⊥ = 1
cos
cos
cossin
cossin
2
2
0
2
2
−=
−
−−
=
−+
−−
= ir
ir
iri
iri
i
θε
θε
θεθ
θεθ
θ
thus, if incident wave grazes the earth
• ground may be modeled as a perfect reflector with |Γ| = 1
• regardless of polarization or ground dielectric properties
• horizontal polarization results in 180° phase shift

22. 22
3.5.2 Brewster Angle = θB
• Brewster angle only occurs for vertical (parallel) polarization
• angle at which no reflection occurs in medium of origin
• occurs when incident angle θi is such that Γ|| = 0  θi = θB
• if 1st
medium = free space & 2nd
medium has relative permittivity εr
then (3.27) can be expressed as
1
1
2
−
−
r
r
ε
ε
sin(θB) = (3.28
)
sin(θB) =
21
1
εε
ε
+
(3.27
)
θB satisfies

23. 23
e.g. 1st
medium = free space
Let εr = 4 
116
14
−
−
sin(θB) = = 0.44
θB = sin-1
(0.44) = 26.6o
Let εr = 15 
115
115
2
−
−
sin(θB) = = 0.25
θB = sin-1
(0.25) = 14.5o

24. 24
3.6 Ground Reflection – 2 Ray Model
Free Space Propagation model is inaccurate for most mobile RF
channels
2 Ray Ground Reflection model considers both LOS path & ground
reflected path
• based on geometric optics
• reasonably accurate for predicting large scale signal strength for
distances of several km
• useful for
- mobile RF systems which use tall towers (> 50m)
- LOS microcell channels in urban environments
Assume
• maximum LOS distances d ≈ 10km
• earth is flat

25. 25
Let E0 = free space E-field (V/m) at distance d0
• Propagating Free Space E-field at distance d > d0 is given by
E(d,t) = 











−
c
d
tw
d
dE
ccos00
(3.33)
• E-field’s envelope at distance d from transmitter given by
|E(d,t)| = E0 d0/d
(1) Determine Total Received E-field (in V/m) ETOT
ELOS
Ei
Er
= Eg
θi θ0
d
ETOT is combination of ELOS & Eg
• ELOS = E-field of LOS component
• Eg = E-field of ground reflected component
• θi = θr

26. 26
d’
d”
ELOS
Ei
Egθi θ0
d
ht
hr
E-field for LOS and reflected wave relative to E0 given by:
and ETOT = ELOS + Eg
ELOS(d’,t) = 











−
c
d
tw
d
dE
c
'
cos
'
00
(3.34)
Eg(d”,t) = 











−
c
d
tw
d
dE
Γ c
"
cos
"
00
(3.35)
assumes LOS & reflected waves arrive at the receiver with
- d’ = distance of LOS wave
- d” = distance of reflected wave

27. 27
From laws of reflection in dielectrics (section 3.5.1)
θi = θ0 (3.36)
Eg = Γ Ei (3.37a)
Et = (1+Γ) Ei (3.37b)
Γ = reflection coefficient for ground
Eg
d’
d”
ELOS
Ei
θi θ0
Et

28. 28
resultant E-field is vector sum of ELOS and Eg
• total E-field Envelope is given by |ETOT| = |ELOS + Eg| (3.38)
• total electric field given by
+











−
c
d
tw
d
dE
c
'
cos
'
00












−−
c
d
tw
d
dE
c
"
cos
"
)1( 00
(3.39)ETOT(d,t) =
Assume
i. perfect horizontal E-field Polarization
ii. perfect ground reflection
iii. small θi (grazing incidence)  Γ ≈ -1 & Et ≈ 0
• reflected wave & incident wave have equal magnitude
• reflected wave is 180o
out of phase with incident wave
• transmitted wave ≈ 0

29. 29
• path difference ∆ = d” – d’ determined from method of images
( ) ( ) 2222
dhhdhh rtrt +−−++∆ = (3-40)
if d >> hr + ht  Taylor series approximations yields (from 3-40)
∆ ≈
d
hh rt2 (3-41)
(2) Compute Phase Difference & Delay Between Two Components
ELOS
d
d’
d”θi θ0
ht
hr
∆h
ht+hr
Ei Eg

30. 30
once ∆ is known  we can compute
• phase difference θ∆ =
c
wc⋅∆
=
∆
λ
π2
(3-42)
• time delay τd =
cfc π
θ
2
∆
=
∆
(3-43)
As d becomes large  ∆ = d”-d’ becomes small
• amplitudes of ELOS & Eg are nearly identical & differ only in phase
"'
000000
d
dE
d
dE
d
dE
≈≈ (3.44)
if Δ = λ/n  θ∆ = 2π/n0 π 2π
λ
Δ

31. 31
(3) Evaluate E-field when reflected path arrives at receiver
( )0cos
"
)1(
'"
cos
'
0000
d
dE
c
dd
w
d
dE
c −+










 −
(3.45)ETOT(d,t)|t=d”/c =
t = d”/creflected path arrives at receiver at






−




 ∆
1cos00
c
w
d
dE
c≈
( )[ ]1cos00
−∆θ
d
dE
=
( )[ ]100
−∠ ∆θ
d
dE
=

32. 32
(3.46)
( )( )∆∆ +−





θθ 22
2
00
sin1cos
d
dE
=( ) ∆∆ 





+−





θθ 2
2
002
2
00
1 sin
d
dE
cos
d
dE
|ETOT(d)|=
=
=











 ∆
2
sin2 00 θ
d
dE
∆−





θcos2200
d
dE
(3.47)
(3.48)
ETOT
"
00
d
dE
'd
dE 00
θ∆
Use phasor diagram to find resultant E-field from combined direct &
ground reflected rays:
(4) Determine exact E-field for 2-ray ground model at distance d

33. 33
As d increases  ETOT(d) decreases in oscillatory manner
• local maxima 6dB > free space value
• local minima ≈ -∞ dB (cancellation)
• once d is large enough  θΔ < π & ETOT(d) falls off asymtotically
with increasing d
-50
-60
-70
-80
-90
-100
-110
-120
-130
-140
101
102
103
104
m
fc = 3GHz
fc = 7GHz
fc = 11GHz
Propagation Loss ht = hr = 1, Gt = Gr = 0dB

34. 34
if d satisfies 3.50  total E-field can be approximated as:
k is a constant related to E0 ht,hr, and λ
rad
d
hh rt
3.0
22
2
1
2
<≈
∆
=∆
λ
π
λ
πθ
(3.49)
d > (3.50)
λλ
π rtrt hhhh 20
3
20
≈this implies
For phase difference, θ∆ < 0.6 radians (34o
) sin(0.5θ∆ ) ≈ θ∆











 ∆
2
2 00 θ
d
dE
|ETOT(d)| ≈
e.g. at 900MHz  if ∆ < 0.03m  total E-field decays with d2
2
00 22
d
k
d
hh
d
dE rt
≈





λ
π
(3.51)ETOT(d) ≈ V/m

35. 35
Received Power at d is related to square of E-field by 3.2, 3.15, & 3.51
Pr(d) = (3.52b)





=
π
λ
ππ 4120
)(
120
)( 222
0 rR
e
GdE
A
dE
Pr(d) = 4
22
d
hh
GGP rt
rtt (3.52a)
• received power falls off at 40dB/decade
• receive power & path loss become independent of frequency
rthhif d >>

36. 36
Path Loss for 2-ray model with antenna gains is expressed as:
• for short Tx-Rx distances use (3.39) to compute total E field
• evaluate (3.42) for θ∆ = π (180o
)  d = 4hthr/λ is where the ground
appears in 1st
Fresnel Zone between Tx & Rx
- 1st
Fresnel distance zone is useful parameter in microcell path
loss models
PL(dB) = 40log d - (10logGt + 10logGr + 20log ht + 20 log hr ) (3.53)
PL =
1
4
22 −






=
d
hh
GG
P
P rt
rt
r
t
• 3.50 must hold