ECE 3317
Prof. David R. Jackson
Spring 2013
Notes 17
Polarization of Plane Waves
1
Polarization
The polarization of a plane wave refers to the direction
of the electric field vector in the time domain.
z
S  z, t 
y
E ( z, t )
x
We assume that the wave is traveling in the positive z direction.
2
Polarization (cont.)
Consider a plane wave with both x and y components
y
Ey
x
Ex
Phasor domain:
E( z)  ( xˆ E x  yˆ E y ) e jkz
Assume:
E x  a  real number
E y  be j
(In general,  = phase of E y  phase of Ex )
3
Polarization (cont.)
y
Time Domain:
E (t)
At z = 0:
x
Ex  Re  ae jt   a cos t 
Ey  Re  be j e jt   b cos t   
Depending on b/a and , three different cases arise:
 Linear polarization
 Circular polarization
 Elliptical polarization
4
Polarization (cont.)
Power Density:
1
S  E  H*
2
Hy 
From Faraday’s law:
Ex

Hx  
  Ey
1
S  xˆ E x  yˆ E y    xˆ 
2
 


Ey


 Ex
ˆ

y




*



Assume lossless medium ( is real):
S

1
2
zˆ E x  E y
2
2

or
1
2
S
zˆ E
2
5
Linear Polarization
 0
or
E x  a cos t 

E y  b cos t   
Ex  a cos t
At z = 0:
E y  b cos t
+ sign:  = 0
- sign:  = 
E   xˆ a  yˆ b  cos t
y
x
y
b
E
This is simply a
“tilted” plane wave.
a
(shown for  = 0
x
6
Circular Polarization
b = a AND     /2
E x  a cos t 
E y  b cos t   
At z = 0:
Ex  a cos t
E y  a cos(t   / 2)  a sin t
E  E x2  E y2  a 2 cos2 t  a 2 sin 2 t  a 2
2
7
Circular Polarization (cont.)
y
E (t )
a
 Ey
f  tan 
 Ex
1
f
x

1

tan
( tan t )


f  t
8
Circular Polarization (cont.)
E x  a cos t 
y
E y  b cos t   
IEEE convention
E (t )
a
   / 2
f
LHCP
x
   / 2
RHCP
    /2
f  t
df
 
dt
9
Circular Polarization (cont.)
Rotation in space vs. rotation in time
Examine how the field varies in both space and time:
E( z )   xˆ a  yˆ be j  e  jkz
Phasor domain
E( z, t )  xˆ a cos t  kz   yˆ b cos t  kz   
Time domain
There is opposite rotation in space and time.
10
Circular Polarization (cont.)
A snapshot of the electric field vector, showing the vector at different points.
Notice that the rotation in space
matches the left hand!
RHCP
z
11
Circular Polarization (cont.)
Animation of LHCP wave
(Use pptx version in full-screen mode to see motion.)
http://en.wikipedia.org/wiki/Circular_polarization
12
Circular Polarization (cont.)
Circular polarization is often used in wireless communications to
avoid problems with signal loss due to polarization mismatch.
 Misalignment of transmit and receive antennas
 Reflections off of building
 Propagation through the ionosphere
Receive antenna
z
The receive antenna will always receive a signal,
no matter how it is rotated about the z axis.
However, for the same incident power density, an optimum linearly-polarized wave will
give the maximum output signal from this linearly-polarized antenna (3 dB higher than
from an incident CP wave).
13
Circular Polarization (cont.)
Two ways in which circular polarization can be obtained:
Method 1) Use two identical antenna rotated by 90o, and fed 90o out of phase.
y
Antenna 2
   / 2
Antenna 1
Vy   j  1 90
o
x
+
+
Vx  1
This antenna will radiate a RHCP signal in the positive z direction, and LHCP
in the negative z direction.
14
Circular Polarization (cont.)
Realization with a 90o delay line:
y
 l2   l1   / 2
Antenna 2
Antenna 1
x
Z01
Z01  Z
ANT
in
Z01
l2
Power splitter
l1
Signal
Feed line Z 02
Z02  Z01 / 2
15
Circular Polarization (cont.)
An array of CP antennas
16
Circular Polarization (cont.)
The two antennas are actually two different modes of a microstrip or
dielectric resonator antenna.
17
Circular Polarization (cont.)
Method 2) Use an antenna that inherently radiates circular polarization.
Helical antenna for WLAN communication at 2.4 GHz
http://en.wikipedia.org/wiki/Helical_antenna
18
Circular Polarization (cont.)
Helical antennas on a GPS satellite
19
Circular Polarization (cont.)
Other Helical antennas
20
Circular Polarization (cont.)
An antenna that radiates circular polarization will also receive
circular polarization of the same handedness
(and be blind to the opposite handedness).
y
RHCP Antenna
It does not matter how
the receive antenna is
rotated about the z axis.
Antenna 2
Antenna 1
x
Z0
Z0
l2
z
RHCP wave
l1
Output signal
Z02  Z01 / 2
Power combiner
21
Circular Polarization (cont.)
Summary of possible scenarios
1) Transmit antenna is LP, receive antenna is LP
 Simple, works good if both antennas are aligned.
 The received signal is less if there is a misalignment.
2) Transmit antenna is CP, receive antenna is LP
 Signal can be received no matter what the alignment is.
 The received signal is 3 dB less then for two aligned LP antennas.
3) Transmit antenna is CP, receive antenna is CP
 Signal can be received no matter what the alignment is.
 There is never a loss of signal, no matter what the alignment is.
 The system is now more complicated.
22
Elliptic Polarization
Includes all other cases that are not linear or circular
y
E (t )
f
x
The tip of the electric field vector stays on an ellipse.
23
Elliptic Polarization (cont.)
y
RHEP
f
x
LHEP
0  
Rotation property:
    0
LHEP
RHEP
24
Elliptic Polarization (cont.)
Here we give a proof that the tip of the electric field vector must stay on an ellipse.
Ex  a cos t
Ey  b cos(t   )
 b cos t cos   b sin t sin 
so
or
2

E
E
 
 
Ey  b cos   x   b sin   1   x  
a  

a



2
b

b
2
E y  Ex  cos     sin  b    Ex 2
a

a
Squaring both sides, we have
2
2


b
b
b



 
2
2 
2
2
2
E y  Ex  cos    2 Ex E y  cos    sin  b    Ex 
a

a

a


25
Elliptic Polarization (cont.)
2
2


b
b
b



 
2
2 
2
2
2
E y  Ex  cos    2 Ex E y  cos    sin  b    Ex 
a

a

a


Collecting terms, we have
2


b

b

2 
2
2
2
Ex    cos   sin     E y  2 Ex E y  cos    b 2 sin 2 
a

 a 

or
2
b
 b

Ex    Ex Ey 2 cos    Ey 2  b2 sin 2 
a
 a

2
This is in the form of a quadratic expression:
AEx 2  B Ex Ey  CEy 2  D
26
Elliptic Polarization (cont.)
Discriminant:
  B 2  4 AC
(determines the type of curve)
2
b
b
2
 4   cos   4  
a
a
2
2
b
 4   cos 2   1
a
so
2
b
  4   sin 2   0
a
Hence, this is an ellipse.
27
Elliptic Polarization (cont.)
Here we give a proof of the rotation property.
Ex  a cos t
E y  b cos(t   )
 b cos t cos   b sin t sin 
0  
    0
LHEP
RHEP
y
E (t )
f
x
28
Elliptic Polarization (cont.)
tan f 
Ey
Ex

b
cos   tan t sin  
a
Take the derivative:
sec2 f
df  b 
     sec2 t   sin  
dt  a 
Hence
df
 b  2

2
  sin    cos f sec t  
dt
 a 

(a)
0  
(b)
    0
df
0
dt
df
0
dt
LHEP
RHEP
(proof complete)
29
Rotation Rule
Here we give a simple graphical method for determining the
type of polarization
(left-handed or right handed).
30
Rotation Rule (cont.)
First, we review the concept of leading and lagging sinusoidal waves.
Two phasors: A and B
Im
0  
B leads A
B

A
Re
Im
Note: We can always assume
that the phasor A is on the real
axis (zero degrees phase)
without loss of generality.
    0
B lags A
A
B

Re
31
Rotation Rule (cont.)
Now consider the case of a plane wave
( a) 0    
LHEP
Im
Ey

Ey leads Ex
Ex
Phasor domain
Re
Rule:
The electric field vector
rotates from the leading
axis to the lagging axis
y
E (t )
x
Time domain
32
Rotation Rule (cont.)
(b)
    0
RHEP
Im
Ey lags Ex
Ex

Rule:
The electric field vector
rotates from the leading
axis to the lagging axis
Ey
Re
Phasor domain
y
E (t )
x
Time domain
33
Rotation Rule (cont.)
The rule works in both cases, so we can call it a general rule:
The electric field vector rotates from the leading
axis to the lagging axis.
34
Rotation Rule (cont.)
Example
E   zˆ (1  j)  xˆ (2  j ) e jky
z
Ez
y
x
Ex
What is this wave’s polarization?
35
Rotation Rule (cont.)
Example (cont.)
E   zˆ (1  j)  xˆ (2  j ) e jky
Im
Ez leads Ex
Ez
Re
Ex
Therefore, in time the wave rotates from the z axis to the x axis.
36
Rotation Rule (cont.)
Example (cont.)
E   zˆ (1  j)  xˆ (2  j ) e jky
z
Ez
y
Ex
x
LHEP or LHCP
Note:
Ex  Ez
and
 

2
(so this is not LHCP)
37
Rotation Rule (cont.)
Example (cont.)
E   zˆ (1  j)  xˆ (2  j ) e jky
z
Ez
y
Ex
x
LHEP
38
Axial Ratio (AR) and Tilt Angle ()
y
E (t )
B
C

x
D
A
major axis AB
AR 

1
minor axis CD
Note: In dB we have
ARdB  20log10  AR 
39
Axial Ratio (AR) and Tilt Angle () Formulas
b
  tan  
a
0    90o
1
Axial Ratio and Handedness
sin 2  sin 2 sin 
 45    45
o
Tilt Angle
o
AR  cot 
tan 2  tan 2 cos 
Note: The tilt angle  is ambiguous
by the addition of  90o.
  0:
  0:
LHEP
RHEP
40
Note on Title Angle
Note: The title angle is zero or 90o if:
   / 2
Tilt Angle:
tan 2  tan 2 cos 
y
E (t )
x
41
Example
E   zˆ (1  j)  xˆ (2  j ) e jky
z
LHEP
Find the axial ratio and tilt angle.
Ez
y
Ex
x
Relabel the coordinate system:
xx
z y
y  z
42
Example (cont.)
E   xˆ (2  j )  yˆ (1  j )  e  jkz
y
Ey
LHEP
z
Ex
x
Renormalize:

 1  j    jkz
E   xˆ (1)  yˆ 
 e
 2  j 

or
E   xˆ (1)  yˆ  0.6324 e j1.249  e jkz
43
Example (cont.)
E   xˆ (1)  yˆ  0.6324 e j1.249  e jkz
Hence
y
a 1
b  0.6324
LHEP
Ey
  1.249[rad]  71.565o
z
Ex
x
  E y  Ex  71.565o
b
  tan    tan 1  0.632   0.564
a
1
44
Example (cont.)
tan 2  tan 2 cos 
sin 2  sin 2 sin 
 45    45
o
AR  cot 
  0:
  0:
  0.564
  71.565o
o
Results
  16.845o
  29.499o
LHEP
RHEP
AR  1.768
LHEP
45
Example (cont.)
AR  1.768
y
  16.845o
E (t )

x
Note: We are not sure
which choice is correct:
  16.845o
  16.845o  90o
We can make a quick time-domain sketch to be sure.
46
Example (cont.)
Ex  a cos t 
E y  b cos t   
y
a 1
b  0.6324
E (t )

  1.249[rad]  71.565o
x
47