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ECE 480
Wireless Systems
Lecture 3
Propagation and Modulation
of RF Waves
1
Circular Polarization
Magnitudes of the x – and y – components of E  z 
are equal
Phase difference is
 
 

2

2
 

2
Left – Hand Circular (LHC) Polarization
Right – Hand Circular (RHC) Polarization
2
Left – Hand Circular (LHC) Polarization
ax  ay  a
 

2

j 

E  z    xˆ a  yˆ a e 2  e  j k z


jkz
ˆ
ˆ
 ax  j y  e
e
j

2
 j
3
jt

E  z,t   Re E  z  e 


 xˆ a cos  t  k z   yˆ a cos   t  k z  
2

 xˆ a cos  t  k z   yˆ a sin  t  k z 
Convert to polar form
E  z,t   E x2  z,t   E y2  z,t  
1
2
1
2
 a 2 cos 2  t  k z   a 2 sin 2  t  k z    a
 E y  z,t  
1
  z,t   tan 

 E x  z,t  
 tan
1
  a sin  t  k z  

    t  k z 
 a cos  t  k z  
4
Linear polarization


E  z,t    a x cos  t  k z 

2


 a y cos  t  k z   
2


= f (z, t)
 ay
  tan 
 ax

1

  constant

 f (z, t)
Circular polarization
E  z,t   a
 f (z, t)
  z,t     t  k z 
= f (z, t)
5
1
2
Back to LHC Polarization
Consider the LHC wave at z = 0
   t
Inclination angle
decreases with time
6
Right – Hand Circular (RHC) Polarization
ax  ay  a
E  z,t   a
 

2
  z,t    t  k z 
7
The direction of polarization is defined in terms of
the rotation of E as a function of time in a fixed
plane orthogonal to the direction of propagation
8
Example RHC Polarized Wave
An RHC polarized plane wave with electric field
modulus of 3 mV/m is traveling in the + y direction in
a dielectric medium with
  4 0
  0
 0
f = 100 MHz
Obtain expressions for
E  y,t  and H  y,t 
9
Solution
The wave is traveling in the + y direction.
Therefore, the field components are in the x and z
directions.
zˆ  xˆ  yˆ
EH
direction of propagation
10
Assign a phase angle of 0 o
to the z component of E  y 
(arbitrary)
The x component of E  y 
will have a phase shift
 

2
Both components have a magnitude of a = 3
E  y   xˆ E x  zˆ E z
j

e  j k y  zˆ a e  j k y
mV
jky
   xˆ j  zˆ  3 e
m
 xˆ a e
2
11
H y  


1

1

3

yˆ  E  y 
y    xˆ j  zˆ  3 e  j k y
 zˆ j
 xˆ  e
  2  f  2   10
k 
 r
8
rad
s
2   10 8 4 4

 
8
3  10
3
c

mA
m
jky
0

120 
 60 
4

rad
m
12
Converting back to the time domain
E  Re E  y  e j  t 
 Re   xˆ j  zˆ  3 e  j k y e j  t 
 3  xˆ sin  t  k y   zˆ cos  t  k y  
mV
m
H  Re H  y  e j  t 
3
jky
jt 
 Re   zˆ j  xˆ  e
e 


1
mA
 xˆ cos  t  k y   zˆ sin  t  k y  

20 
m 13
Elliptical Polarization
Most general case
ax  0
The tip of
ay  0
 0
E traces an ellipse in the x – y plane
Can be left – handed or right - handed
Major axis:
Minor axis:


Rotation Angle

Ellipticity Angle

14
Rotation angle
  cos


2
1
 
ax
a

2
The shape and rotation are defined by the ellipticity angle
1


tan   

 
a
R
4
4
a
R = 1 Circular
R
axial ratio
a
R =  Linear
a
15
tan  0 
ay
ax
0   0  90




tan 2   tan 2 0 cos 
tan 2   sin 2 0 sin 
   45
 0
o
o
o

 
    
2
 2

 
    
4
 4
Circular
Linear
16
Polarization States for Various Combinations of  and 
17
How is the type of polarization determined?
Positive values of  corresponding to sin  > 0
define left – handed rotation
Negative values of  corresponding to sin  < 0
define right – handed rotation
a x and a y are, by definition, > 0
ay
ax
0   0  90
will vary from 0  
Two possible values of
  0 if cos   0
  0 if cos   0
 and 
o
in this range
  0 if sin   0
  0 if sin   0
18
Example: Polarization State
Determine the polarization of a plane wave with an
electric field given by

E  z, t   xˆ 3 cos  t  k z  30
o
  yˆ 3 sin  t  k z  45 
mV
m
o
Solution
Convert the sin term to a cos term by subtracting 90 o



E  z, t   xˆ 3 cos  t  k z  30 o  yˆ 3 cos  t  k z  45 o

mV
m
Convert the – cos term to a + cos term by adding 180 o



E  z, t   xˆ 3 cos  t  k z  30 o  yˆ 3 cos  t  k z  135 o

mV
m
19
Convert to phasor form
E  z   xˆ 3 e
jkz
e
j 30 o
 yˆ 4 e
jkz
e
j 135 o
   y   x  135 o  30 o  105 o
 0  tan
1
 ay

 ax

1
  tan


4
o

53
.
1
 
3

tan 2   tan 2 0 cos 



 tan 106.2 o cos 105 o

   3.442    0.2588 
 0.8900
20
There are two possible solutions for  since the tan
function is positive in both the first and third quadrants

2   tan  1  0.89   41.7 o , 221.7 o  138.3 o

41.7 o
1 
 20.85 o
2
 138.3 o
2 
  69.15 o
2
Which is correct?
cos   0
    69.15
o
21


sin 2   sin 2 0 sin 



 sin 106.2 o sin 105 o

  0.9603   0.9659 
 0.9276
By a similar analysis,
  34.0 o
 The wave is elliptically polarized and the rotation is
left - handed
22
Plane – Wave Propagation in Lossy Media
2 E   2 E  0

2
   2   C    2   '  j  " 

" 

'  
 can be written as
   j
 = attenuation constant
 = phase constant
23
2
2


j






 
  j 2 
2
  2   '  j  2   "
Equate the real and imaginary parts
2
2
2






'


2      "
2

'
  
 2
Solve for  and 

'
  
 2

 1



 1


 

"
 
   1 

 ' 
2
1
2

 

"
 
   1 

 ' 
2

1
2
24
For a uniform plane wave traveling in the + z
direction with an electric field
E  xˆ E x  z 
the wave equation becomes
d 2 E x z
dz
2

2
E x z  0
The solution is
E  z   xˆ E x  z   xˆ E x 0 e   z  xˆ E x 0 e   z e  j  z
H
kˆ  E
c
 yˆ
E x0
c
e
 z

 
" 
c 

1  j 
c
' 
' 

e
1
2
j

25
The magnitude of E x  z  is
E x  z   E x0 e
 z
e
jz
 E x0 e
 z
Decreases exponentially with e -  z
Hy 
Ex
c
also decreases exponentially with e -  z
Define: Skin Depth,  s
s 
1

Distance that a wave must travel before it is
attenuated by
1
 0.3679
e
26
In a perfect dielectric
  0 ,   0 , s  
In a perfect conductor
   ,    , s  0
27
Expressions are valid for any linear, isotropic,
homogeneous medium
"
 10  2
'
"
2
2
10 
 10
'
"
2
10 
'
Low – Loss Dielectric
Quasi – Conductor
(Semiconductor)
Good Conductor
28
Low – Loss Dielectric
Consider
For
1  x 
x  1 ,
1
2
1  x 
1
2
x
 1
2
1
2
" 

  j    ' 1  j   j    '
' 


" 
1  j

2

'


Divide into real and imaginary parts
"
  


2
'
2 
   '   
 - Same as for lossless medium
Np
m
29

 
" 
c 

1  j 
c
' 
' 
x  1 ,

c 
'
1  x 


" 
1  j

2 ' 

c 
1
2

1
2

x
 1
2



" 
1  j

2

'




Same as for the lossless case
30
Good Conductor
 
"
2


  f 
2
     f 
c 

f

j
 1  j 
 1  j 
"


1 j
j 
2
31
Semiconductors – Must use exact solution
32
Example – Plane Wave in Seawater
A uniform plane wave is traveling downward in
the + z direction in seawater, with the x – y plane
denoting the sea surface and z = 0 denoting a
point just below the surface.
The constitutive parameters of seawater are:
r  1,
S
 4
m
 r  80 ,
The magnetic field intensity at z = 0 is given by

H  0,t   yˆ 100 cos 2   10 t  15
a. Determine expressions for
3
o

mA
m
E  z,t  and H  z,t 
b. The depth at which the amplitude of E is 1% of
its value at z = 0
33
Solution
a. The general expressions for the phasor fields are
E  z   xˆ E x 0 e   z e  j  z
H  z   yˆ
E x0
c
e  z e  j  z
"


4



 '     r  0 2  10 3  80  10  9
36 
 9  10 5
 Seawater is a good conductor at 1 KHz
34
   f      10  4   10
3
7
    0.126
Np
 4  0.126
m


j  0.126
j
 
 c  1  j    2 e 4 
 0.044 e 4
 
 4
The general expression for E x0 is
E x0  E x0 e
j 0
j
E  z,t   Re  xˆ E x 0 e 0 e   z e  j  z e j  t 
 x̂ E x 0 e
 0.126 z

cos 2  10 t  0.126 z   0
3

V
m
35
 E e j0

x0
 z
jz
jt 
H  z, t   Re  yˆ
e
e
e



j
4
 0.044 e


 ŷ 22.5 E x 0 e  0.126 z cos 2   10 3 t  0.126 z   0  45 o

A
m
at z = 0:

H  0, t   yˆ 22.5 E x 0 cos 2   10 3 t   0  45 o

A
m
Compare with original expression

H  0,t   yˆ 100 cos 2   10 t  15
22.5 E x 0  100  10
E x0
mV
 4.44
m
3
o

mA
m
3
 0  45 o  15 o
 0  60 o
36




E  z,t   xˆ 4.44 e  0.126 z cos 2  10 3 t  0.126 z  60 o
H  z,t   yˆ 100 e  0.126 z cos 2  10 3 t  0.126 z  15 o
mV
m
mA
m
Note that they are no longer in phase. The electric
field always leads the magnetic field by 45 o.
b. Set the amplitude to 0.01
0.01  e
z
 0.126 z
ln  0.01
 0.126
 36 m
37
Electromagnetic Power Density
Define: Poynting Vector
S  E H
W
m2
Direction of S is in the direction of propagation, k
n̂ 
unit vector
normal to the
surface
Power through a surface, A
P
S
A
ˆ
ndA
 S A cos 
38
Plane Wave in a Lossless Medium
Consider a plane wave traveling in the + z direction
E  z   xˆ E x 0 e  j k z
H  z   yˆ
E x0

ejkz
Want to find the power density vector, S
39
Time – Domain Approach
E  z ,t   xˆ E x 0 cos  t  k z 
H  z ,t   yˆ
E x0

cos  t  k z 
S  z ,t   E  z ,t   H  z ,t 
 ẑ
E x0

2
cos 2  t  k z 
40
Time average of
T
S av
1
  S  z ,t  d t
T 0
2


2
z


ẑ
0
E x0
E x0

S
1 2 

T  

f



2
cos 2  t  k z  d t
2
2
41
Phasor – Domain Approach
S av  Re E  H * 

Ex 0 *
1
jkz
 Re  xˆ E x 0 e
 yˆ
e
2


 ẑ
E x0
jkz



2
2
S av  Re E  H * 
is valid for any media
42
Plane Wave in a Lossy Medium
E  z   xˆ E x  z   yˆ E y  z 


 xˆ E x 0  yˆ E y 0 e   z e  j  z
H z 
1
c


 xˆ E x 0  yˆ E y 0 e   z e  j  z
S av  Re E  H * 


ẑ E x 0
2
 E y0
2
c  c e
2
e
j 
 2 z
 1 
Re 
  c * 


43
S av  z   zˆ
E0
2
2c
e
c  c e
S av  z   zˆ

E0
cos  
j 
2
2c
E 0  E x0
 2 z
2
e  2  z cos  
 E y0
2

1
2
 2 z
e
Note that the average power decays with
44
Homework
The electric field of a plane wave is given by
E  z ,t   xˆ a x cos  t  k z   yˆ a y cos  t  k z   
Identify the polarization state, determine the
polarization angles (, ), and sketch the locus of E (0, t)
for each of the following cases
V
V
ax  3
, ay  4
,   0o
m
m
V
V
ax  3
, ay  4
,   180 o
m
m
V
V
ax  3
, ay  3
,   45 o
m
m
V
V
ax  3
, ay  4
,    135 o
45
m
m
Homework
In a medium characterized by
S
 r  9 ,  r  1 ,   0.1
m
Determine the phase angle by which the
magnetic field leads the electric field
46
Radiation and Antennas
• An antenna may be considered as a transducer that
converts a guided EM wave to a transmitted wave or
an incident wave to a guided EM wave
• Antenna dimensions are generally referred to in
wavelength units
47
48
Reciprocity
• Antenna radiation pattern: The directional function
that characterizes the distribution pattern radiated by
an antenna
• Isotropic antenna: A hypothetical antenna that
radiates equally in all directions
• Used as a reference radiator to compare antennas
• Reciprocal antennas: Antennas that have the same
radiation patterns for transmission as for reception
49
Two aspects of antenna performance
1. Radiation Properties
• Direction of the radiation pattern
• Polarization state of the radiated wave in the TX mode
(Antenna Polarization)
• In the RX mode, the antenna can extract only that
component of the wave whose E – field is parallel
to that of the antennas polarization direction
2. Antenna Impedance
• Pertains to the impedance match between the
antenna and the generator
50
Radiation Sources
Two basic types
1. Current sources
• Dipole and loop antennas
2. Aperture fields
• Horn antennas
51
Far – Field Region
The far – field region is at a distance R where the
wave may be considered to be a plane wave
R ff 
2D

2
D = Maximum effective size of
the antenna
 = Wavelength of the signal
52
Example: Far – Field Distance of an Antenna
A parabolic reflector antenna is 18" in diameter
operates at 12.4 GHz. Find the operating wavelength
and the far – field distance of this antenna.
Solution
c
3  10
 
 0.0242 m
9
f 12.4  10
1m
18" x
 0.457 m
39.37"
8
R ff 
2D

2

2  0.457 
0.0242
2
 17.3 m
53
Antenna Arrays
• Can control the phase and
magnitude of each antenna
individually
• Can steer the direction of
the beam electronically
54
Retarded Potentials
Consider a charge
distribution as shown
 
V R 
1
4 

 v R i 
'
R'
d '
The electric potential
V (R) at a point in
space specified by
the position vector R
is given by
Ri
= position vector of an elemental volume
 '
v
= elemental volume
R'  R  R i
= charge density inside the volume
55
= distance between the volume and the point
If the charge density
is time – varying, the
obvious solution is


V R ,t 
1
4 

 v R i , t 
'
R'
d '
Problem: Does not account for reaction time
Any change in the charge distribution will require a
finite amount of time to change the potential
56
R'
t' 
up
Delay Time
Retarded Scalar Potential


V R ,t 
1

4 
 v  R i , t  t '  
R'
'
d '
Retarded Vector Potential

A  R ,t  
4


J R i ,t  t ' 
'
R'
 d '
Valid under both static and dynamic conditions
57
Time – Harmonic Potentials
In a linear system, the parameters all have the
same functional dependence on time
Consider a sinusoidal time – varying charge
distribution
v 

 
jt 

R i ,t  Re  v R i e


 v R i 
= phasor representation of
 v  R i ,t 
58
 
V R 
1
4 

A R  
4

 v  R i  e  j k R'
R'
'

k
 
J R i e  j k R'
R'
'
d ' V
d '

up
59
In terms of A
H
1

A
  H  j E
1
E
H
j
  E   j H
1
H
E
j
60
The Short (Hertzian) Dipole
Approach: Develop the radiation properties of
a differential antenna and use that model to
predict other configurations
Characteristics of a
Short Dipole
• Current is uniform
over the length
l

50
jt

i  t   I 0 cos  t  Re  I 0 e 
I  I0
A
61
At the point Q
0
A R  
4
k

c


'
J e  j k R'
d '
R'
2

J  zˆ
I0
s
s = cross – sectional area of dipole
d '  s d z
limits of integration
Assume
l
l
 z
2
2
R' ~ R
62
0 e  jkR
A
4 R
l
2
 zˆ I

0
dz
l
2
0
 e  jkR 
 ẑ
I0 l 

4
 R 
e
 jkR
Spherical propagation factor
R
Considers both the magnitude and
phase change wrt R
Change to spherical coordinates
63
R  Range
  Zenith angle
  Azimuth angle
64
ẑ  Rˆ cos   ˆ sin 

I 0 l  e  jkR 
0
A  Rˆ cos   ˆ sin 


4  R 


 Rˆ A R  ˆ A  ˆ A
0 I0 l
 e  jkR 
AR 
cos  

4
 R 
0 I0 l
 e  jkR 
A  
sin  

4
 R 
A  0
65
H
E
1

A
1
j
H
 j
1 
H 
e


 sin 
2
4
 k R  k R  
 1
2 I0 lk 2
j 
 jkR
ER 
0 e


 cos 
2
3
4
  k R 
 k R  
 j
I0 lk 2
1
j 
 jkR
E 
0 e



 sin 
2
3
4
 k R  k R 
 k R  
I0 lk 2
 jkR
H R  H  E  0
66
Electric
Field lines
67
Far – Field Approximation
R  
2 R
kR 
 1

1
k R 
2
,
1
k R 
3
negligible
j I 0 l k e  jkR
E 
sin 
4
R
ER  0
H 
E
0
Independent of 
Proportional to sin 
68
Power Density
S av  Re E  H * 
For the short dipole:
S av  Rˆ S  R , 
  0 k 2 I 02 l 2 
2
S  R ,   
sin


2
2
 32  R 


W
2
 S 0 sin 
m2
69
Define: Normalized Radiation Intensity
F  ,  
S  R , , 
S max
Radiation is maximum when
S max
 

2
(azimuth plane)
  0 k 2 I 02 l 2 
2
 S0  
sin


2
2
 32  R 


2
2
15  I 0  l 

 
R2  
70
F  ,   F    sin 
2
No energy is radiated by the dipole along the direction
of the dipole axis and maximum radiation (F = 1)
occurs in the broadside direction.
71