Prof. David R. Jackson Notes 17 Polarization of Plane Waves Polarization of Plane Waves ECE 3317 1 Spring 2013.
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1
Prof. David R. Jackson
Notes 17
Polarization of Plane Waves
ECE 3317
Spring 2013
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Polarization
The polarization of a plane wave refers to the direction of the electric field vector in the time domain.
x
y
z
( , )E z t
We assume that the wave is traveling in the positive z direction.
,S z t
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Consider a plane wave with both x and y components
ˆ ˆE( ) ( E E ) jkzx yz x y e
( = E E )y x In general, phase of phase of
Assume:
Ex
Ey
x
y
E
E
x
jy
a
be
real number
Polarization (cont.)
Phasor domain:
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Time Domain:
Re cos
Re cos
j tx
j j ty
E ae a t
E be e b t
Depending on b/a and b, three different cases arise:
x
y
E (t)
At z = 0:
Polarization (cont.)
Linear polarization Circular polarization Elliptical polarization
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Power Density: *1S E H
2
*
E E1ˆ ˆ ˆ ˆS E E
2y x
x yx y x y
H
H
xy
yx
E
E
221ˆS E E
2 x yz
Assume lossless medium ( is real):
21ˆS E
2z
or
Polarization (cont.)
From Faraday’s law:
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At z = 0:
ˆ ˆ cosE x a y b t
cos
cosx
y
E a t
E b t
a
b
x
y x
y
E
b = 0 or
(shown for b = 0)
Linear Polarization
This is simply a “tilted” plane wave.
cos
cos
x
y
E a t
E b t
+ sign: = 0- sign: =
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At z = 0:
2 2 2 2 2 2 2 2cos sinx yE E E a t a t a
cos
cos( / 2) sinx
y
E a t
E a t a t
b = a AND b = p /2
Circular Polarization
cos
cos
x
y
E a t
E b t
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a
x
y
( )E t
f
1 1tan tan ( tan )y
x
Et
E
t
Circular Polarization (cont.)
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t
a
x
y
( )E t
f
b = + p / 2
b = - p / 2
RHCP
LHCP
IEEE convention
Circular Polarization (cont.)
d
dt
cos
cos
x
y
E a t
E b t
b = p /2
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There is opposite rotation in space and time.
Circular Polarization (cont.)
Examine how the field varies in both space and time:
Rotation in space vs. rotation in time
ˆ ˆE( ) j jkzz x a y be e
ˆ ˆ( , ) cos cosE z t x a t kz y b t kz
Phasor domain
Time domain
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RHCP
Notice that the rotation in space matches the left hand!
z
Circular Polarization (cont.)
A snapshot of the electric field vector, showing the vector at different points.
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Circular Polarization (cont.)
Animation of LHCP wave
http://en.wikipedia.org/wiki/Circular_polarization
(Use pptx version in full-screen mode to see motion.)
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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
The receive antenna will always receive a signal, no matter how it is rotated about the z axis.
z
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).
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Circular Polarization (cont.)
Two ways in which circular polarization can be obtained:
This antenna will radiate a RHCP signal in the positive z direction, and LHCP in the negative z direction.
Method 1) Use two identical antenna rotated by 90o, and fed 90o out of phase.
Antenna 1
V 1x oV 1 90y j
Antenna 2
x
y
+
-
+
-
/ 2
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Circular Polarization (cont.)
Realization with a 90o delay line:
2 1 / 2l l
01ANT
inZ Z
Antenna 1
Antenna 2
x
y
l1
l2
Z01
Z01
Feed line
Power splitter
Signal
02 01 / 2Z Z
02Z
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Circular Polarization (cont.)
An array of CP antennas
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Circular Polarization (cont.)
The two antennas are actually two different modes of a microstrip or dielectric resonator antenna.
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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
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Circular Polarization (cont.)
Helical antennas on a GPS satellite
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Circular Polarization (cont.) Other Helical antennas
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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).
It does not matter how the receive antenna is rotated about the z axis.
Antenna 1
Antenna 2
x
y
l1
l2
Z0
Z0
Output signal
Power combinerRHCP wave
02 01 / 2Z Zz
RHCP Antenna
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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.
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Includes all other cases that are not linear or circular
x
y
( )E t
f
Elliptic Polarization
The tip of the electric field vector stays on an ellipse.
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0
0
LHEP
RHEP
x
LHEP
RHEP
y
f
Elliptic Polarization (cont.)
Rotation property:
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cos( )
cos cos sin sin
yE b t
b t b t
so2
cos sin 1x xy
E EE b b
a a
22 2cos siny x x
b bE E b E
a a
2 22 2 2 2 2cos 2 cos siny x x y x
b b bE E E E b E
a a a
cosxE a t
Elliptic Polarization (cont.)
or
Squaring both sides, we have
Here we give a proof that the tip of the electric field vector must stay on an ellipse.
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This is in the form of a quadratic expression:
2
2 2 2 2 2 2cos sin 2 cos sinx y x y
b bE E E E b
a a
22 2 2 22 cos sinx x y y
b bE E E E b
a a
2 2x x y yA E B E E CE D
Elliptic Polarization (cont.)
2 22 2 2 2 2cos 2 cos siny x x y x
b b bE E E E b E
a a a
or
Collecting terms, we have
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2
2 22
22
4
4 cos 4
4 cos 1
B AC
b b
a a
b
a
224 sin 0
b
a
Hence, this is an ellipse.
so
Elliptic Polarization (cont.)
(determines the type of curve)
Discriminant:
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cos
cos( )
cos cos sin sin
x
y
E a t
E b t
b t b t
x
y
( )E t
f
Elliptic Polarization (cont.)
Here we give a proof of the rotation property.
0
0
LHEP
RHEP
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Take the derivative:
tan cos tan siny
x
E bt
E a
2 2sec sec sind b
tdt a
Hence
2 2sin cos secd b
tdt a
( ) 0 0
( ) 0 0
da
dtd
bdt
LHEP
RHEP (proof complete)
Elliptic Polarization (cont.)
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Rotation Rule
Here we give a simple graphical method for determining the type of polarization
(left-handed or right handed).
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First, we review the concept of leading and lagging sinusoidal waves.
B leads A
Reb
B
A
Im
B lags A
Reb B
A
Im
Two phasors: A and B
0
0
Note: We can always assume that the phasor A is on the real axis (zero degrees phase) without loss of generality.
Rotation Rule (cont.)
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( ) 0a LHEP
Rule:
The electric field vector rotates from the leading axis to the lagging axis
Ey leads Ex
Phasor domainRe
b
Ey
Ex
Im
Now consider the case of a plane wave
Rotation Rule (cont.)
( )E t
x
y
Time domain
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( ) 0b RHEP
Rule:
The electric field vector rotates from the leading axis to the lagging axis
Rotation Rule (cont.)
Ex
Ey
Im
Reb
Ey lags Ex
Phasor domain
x
y
( )E t
Time domain
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The electric field vector rotates from the leading axis to the lagging axis.
Rotation Rule (cont.)
The rule works in both cases, so we can call it a general rule:
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ˆˆE (1 ) (2 ) jkyz j x j e
y
Ez
z
xEx
What is this wave’s polarization?
Rotation Rule (cont.)
Example
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Example (cont.)
Therefore, in time the wave rotates from the z axis to the x axis.
Ez leads Ex Ez
Ex
Im
Re
Rotation Rule (cont.)
ˆˆE (1 ) (2 ) jkyz j x j e
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LHEP or LHCP
Note: and 2x zE E (so this is not LHCP)
Example (cont.)
Rotation Rule (cont.)
ˆˆE (1 ) (2 ) jkyz j x j e
z
y
Ez
xEx
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LHEP
Rotation Rule (cont.)
Example (cont.)
ˆˆE (1 ) (2 ) jkyz j x j e
z
y
Ez
xEx
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x
y
( )E t
A
B
C
D
1AB
ARCD
major axis
minor axis
Axial Ratio (AR) and Tilt Angle ()
Note: In dB we have 1020logdBAR AR
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o o45 45
LHEP
RHEP
sin 2 sin 2 sin
cot
0 :
0 :
AR
Axial Ratio and Handedness
tan 2 tan 2 cos
Tilt Angle
Note: The tilt angle is ambiguous by the addition of 90o.
1
o
tan
0 90
b
a
Axial Ratio (AR) and Tilt Angle () Formulas
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tan 2 tan 2 cos Tilt Angle:
Note: The title angle is zero or 90o if:
/ 2
x
y
( )E t
Note on Title Angle
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Example
z
y
Ez
xEx
LHEP
Relabel the coordinate system:
ˆˆE (1 ) (2 ) jkyz j x j e
x x
z y
y z
Find the axial ratio and tilt angle.
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Example (cont.)
LHEP
ˆ ˆE (2 ) (1 ) jkzx j y j e
y
z
Ey
xEx
1ˆ ˆE (1)
2jkzj
x y ej
Renormalize:
1.249ˆ ˆE (1) 0.6324 j jkzx y e e
or
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Example (cont.)
LHEP
oE E 71.565y x
1 1tan tan 0.632 0.564b
a
Hence
1.249ˆ ˆE (1) 0.6324 j jkzx y e e
o
1
0.6324
1.249[rad] 71.565
a
b
y
z
Ey
xEx
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Example (cont.)
tan 2 tan 2 cos
o o45 45
LHEP
RHEP
sin 2 sin 2 sin
cot
0 :
0 :
AR
o16.845
o29.499
LHEP
Results
1.768AR
o71.565
0.564
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Example (cont.)
o16.845
1.768AR
x
y
( )E t
o
o o
16.845
16.845 90
Note: We are not sure which choice is correct:
We can make a quick time-domain sketch to be sure.
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Example (cont.)
x
y
( )E t
cos
cos
x
y
E a t
E b t
o
1
0.6324
1.249[rad] 71.565
a
b
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