prof. david r. jackson dept. of ece fall 2013 notes 16 ece 6340 intermediate em waves 1
TRANSCRIPT
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Prof. David R. JacksonDept. of ECE
Fall 2013
Notes 16
ECE 6340Intermediate EM Waves
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Polarization of WavesConsider a plane wave with both x and y components
ˆ ˆ( , , ) ( ) jkzx yE x y z x E y E e
( = .)y xE E In general, phase of phase of
Assume
Ex
Ey
x
y
0jx
jy
E ae a
E be
2
(polar form of complex numbers)
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Polarization of Waves (cont.)
Time Domain:
Re cos
Re cos
j tx
j j ty
ae a t
be e b t
E
E
Depending on b and b, three different cases arise.
At z = 0
x
y
E (t)|E (t)|
(t)
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Linear Polarization
At z = 0:
ˆ ˆ( ) cosx a y b t E
cos
cos cosx
y
a t
b t b t
E
E
a
b
x
y x´ y´
E
b = 0 or
(shown for b = 0)
(+ for 0, - for )
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Circular Polarization
At z = 0:
2 2 2 2 2 2 2
2
cos sinx y a t a t
a
E E E
cos
cos cos( / 2) sinx
y
a t
b t a t a t
E
E
b = a AND b = p /2
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Circular Polarization (cont.)
1 1tan tan ( tan )y
x
t
E
E
t
6
so
wave
d
dt
angular velocity of waveNote:
a
x
y
( )tE
f(t)
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Circular Polarization (cont.)
t
IEEE convention
ˆ ˆ( cos sin )a x t y t E
a
x
y
( )tE
(t)
b = + p / 2
b = - p / 2
RHCP
LHCP
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Circular Polarization (cont.)
RHCP
RHCP
z
j t kze
(opposite rotation in space and time)
z
"snapshot" of wave
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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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Unit Rotation Vectors
1ˆ ˆ ˆ( )
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ˆ ˆ( )2
r x jy
l x jy
RHCP ( = - / 2)
LHCP ( = + / 2)
1 ˆˆ ˆ( )2
ˆˆ ˆ( )2
x r l
jy r l
Add:
Subtract:
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General Wave Representation
ˆ ˆ
1 1ˆ ˆ( ) ( )
2 2
x y
x y
E x E y E
E r l E j r l
1 1( ) ( )
2 2RH x y LH x yA E j E A E j E
Note: Any polarization can be written as combination of RHCP and LHCP waves.
This could be useful in dealing with CP antennas.
ˆ RH LHE r A l A
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Circularly Polarized Antennas
Method 1: The antenna is fundamentally CP.
A helical antenna for satellite reception is shown here.
The helical antenna is shown radiating above a circular
ground plane.
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Circularly Polarized Antennas (cont.)
+-Vy = -j
z
x
y
Vx = 1
+-
RHCP
Method 2: Two perpendicular antennas are used, fed 90o out of phase.
Two antennas are shown here being fed by a common feed point.
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A more complicated version, using four antennas
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Circularly Polarized Antennas (cont.)
Method 3: A single antenna is used, but two perpendicular modes are excited 90o out of phase.
(a) A square microstrip antenna with two perpendicular modes being excited.
(b) A circular microstrip antenna using a single feed and slots. The feed is located at 45o from the slot axis.
(c) A square patch with two corners chopped off, fed at 45o from the edge.
(c)
(a) (b)
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Circularly Polarized Antennas (cont.)
Method 4: Sequential rotation of antennas is used.
I = 1 0o
I = 1 -90oI = 1 -90o
I = 1 -180o
I = 1 -270o
LP elements CP elements
The elements may be either LP or CP.
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This is an extension of method 2 (when used for LP elements). Using four CP elements instead of one CP element often gives better CP in practice (reduced cross-pol from higher-order modes).
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Pros and Cons of CP
Alignment between transmit and receive antennas is not important. The reception does not depend on rotation of the electric field vector
that might occur†.
When a RHCP bounces off an object, it mainly changes to a LHCP wave. Therefore, a RHCP receive antenna will not be very sensitive to “multipath” signals that bound off of other objects.
† Satellite transmission often uses CP because of Faraday rotation in the ionosphere.
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A CP system is usually more complicated and expensive. Often, simple wire antennas are used for reception of signals in
wireless communications, and hence they are already directly compatible with linear polarization. (Using a CP transmitted signal will result in 50% of the transmitted power being wasted, or a 3 dB drop in the received signal).
Pros
Cons
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Examples of Polarization
AM Radio (0.540-1.6 MHz): linearly polarized (vertical) FM Radio (88-108 MHz): linearly polarized (horizontal) TV (VHF, 54-216 MHz; UHF, 470-698 MHz: linearly polarized
(horizontal) (some transmitters are CP) Cell phone antenna (about 2 GHz, depending on system): linearly
polarized (direction arbitrary) Cell phone base-station antennas (about 2 GHz, depending on
system): often linearly polarized (dual-linear slant 45o) DBS Satellite TV (11.7-12.5 GHz): transmits both LHCP and RHCP
(frequency reuse) GPS (1.574 GHz): RHCP
Notes:
1) Low-frequency waves travel better along the earth when they are polarized vertically instead of horizontally.2) Slant linear is used to switch between whichever polarization is stronger. 3) Satellite transmission often uses CP because of Faraday rotation in the ionosphere.
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Elliptical Polarization
Includes all other cases
b 0 or (not linear)
b p / 2 (not CP)
b = p / 2 and b a (not CP)
x
y
( )tE
(t)
Ellipse
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Elliptical Polarization (cont.)
0
0
LHEP
RHEP
x
LHEP
RHEP
y
(t)
Ellipse
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Note: The rotation is not at a constant speed.
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Proof of Ellipse Property
cos( )
cos cos sin sin
y b t
b t b t
E
so2
cos sin 1x xy b b
a a
E EE
22 2cos siny x x
b bb
a a
E E E
2 22 2 2 2 2cos 2 cos siny x x y x
b b bb
a a a
E E EE E
cosx a tE
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Consider the following quadratic form:
22
2 2 2 2 2 2cos sin 2 cos sinx y x y
b bb
a a
E E EE
22 2 2 22 cos sinx x y y
b bb
a a
E EE E
2 2x x y yA B C D E EE E
2 22 2 2 2 2cos 2 cos siny x x y x
b b bb
a a a
E E EE E
Proof of Ellipse Property (cont.)
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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
Proof of Ellipse Property (cont.)
Discriminant:
0,
0,
0,
hyperbola
line
ellipse
From analytic geometry:
If = 0 or we have = 0, and this is linear polarization.22
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Rotation Property
cos
cos( )x
y
a t
b t
E
E
0
0
LHEP
RHEP
x
LHEP
RHEP
y
(t)
ellipse
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We now prove the rotation property:
cos cos sin siny b t b t E
Recall that
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Rotation Property (cont.)
Take the derivative:
tan cos tan siny
x
bt
a
E
E
2 2sec sec sind b
tdt a
so
2 2sin cos secd b
tdt a
( ) 0 0
( ) 0 0
da
dtd
bdt
LHEP
RHEP (proof complete)
Hence
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Phasor Picture
( )tE
x
y( ) 0a LHEP
Ey leads Ex
Re
b
Ey
Ex
Im
LHEP
Rule:
The electric field vector rotates in time from the leading axis to the lagging axis.
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A phase angle 0 < < is a leading phase angle (leading with respect to zero degrees).
A phase angle - < < 0 is a lagging phase angle (lagging with respect to zero degrees).
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Phasor Picture (cont.)
x
y
( )tE
( ) 0b RHEP
Rule:
The electric field vector rotates in time from the leading axis to the lagging axis.
Ex
Ey
Im
Reb
Ey lags Ex
RHEP
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A phase angle 0 < < is a leading phase angle (leading with respect to zero degrees).
A phase angle - < < 0 is a lagging phase angle (lagging with respect to zero degrees).
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Example
ˆˆ (1 ) (2 ) jkyE z j x j e
y
Ez
z
xEx
What is this wave’s polarization?
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Example (cont.)
ˆˆ (1 ) (2 ) jkyE z j x j e
Therefore, in time, the wave rotates from the z axis to the x axis.
Ez leads Ex Ez
Ex
Im
Re
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ˆˆ(1 ) (2 ) jkyE z j x j e
y
Ez
z
xEx
LHEP or LHCP
Note:2x zE E (so this is not LHCP)
Example (cont.)
and
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ˆˆ (1 ) (2 ) jkyE z j x j e
LHEP
Example (cont.)
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y
Ez
z
xEx
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Axial Ratio (AR) and Tilt Angle ( )
x
y
( )tE
A
B
C
D
1AB
ARCD
major axis
minor axis
dB 1020logAR AR
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Axial Ratio (AR) and Tilt Angle ( )
o o
0
0
45 45
sin 2 sin 2 sin
cotAR
implies LHEP
implies RHEP
Axial Ratio
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
We first calculate :
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0jx
jy
E ae a
E be
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cotAR
x
y
( )tE
A
B
C
D| |
Axial Ratio (AR) and Tilt Angle () (cont.)
Physical interpretation of the angle ||
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o0 45
LP CP
Note :
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Special Case
tan 2 tan 2 cos 0
0 / 2
orTilt Angle:
The title angle is zero or 90o if:
/ 2
x
y
( )tE
cos
cos( / 2)
sin
x
y
a t
b t
b t
E
E
22
2 21yx
a b
EE
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Sometimes it is useful to know the ratio of the LHCP and RHCP wave amplitudes.
LHCP/RHCP Ratio
1 1ˆ ˆ ˆ ˆ ˆ( ) ( )
2 2r x jy l x jy
ˆ ˆ RH LHj jRH LH RH LHE r A l A r A e l A e
: ( ) RH LHA t A A Point E
: ( ) RH LHB t A A Point Ex
y
( )tEA
B
RH LH RH LH
RH LHRH LH
A A A AAR
A AA A
Hence
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LHCP/RHCP Ratio (Cont.)Hence
1 /
1 /LH RH
LH RH
A AAR
A A
1 1
1 1LH
RH
A AR AR
A AR AR
or
From this we can also solve for the ratio of the CP components, if we know the axial ratio:
1 /
1 /LH RH
LH RH
A AAR
A A
or
(We can’t tell which one is correct from knowing only the AR.)
(Use whichever sign gives AR > 0.)
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z
y
Ez
xEx
LHEP
Re-label the coordinate system:
ˆˆE (1 ) (2 ) jkyz j x j e
x x
z y
y z
Find the axial ratio and tilt angle.
Example
Find the ratio of the CP amplitudes.
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y
z
Ez
xEx
LHEP
ˆ ˆE (2 ) (1 ) jkzx j y j e
1.249 o10.2 0.6 0.6324 0.6324 71.565
2y j
x
E jj e
E j
1 1tan tan 0.632 0.564b
a
Example (cont.)
o/ 0.6324, 71.565b a
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tan 2 tan 2 cos
o o45 45
LHEP
RHEP
sin 2 sin 2 sin
cot
0 :
0 :
AR
o o16.845 90
o29.499
LHEP
Results
1.768AR
Example (cont.)
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o16.845
1.768AR
x
y
( )tE
Example (cont.)
Note: Plotting the ellipse as a function of time will help determine which value is correct for the tilt angle .
cos
cos( )x
y
a t
b t
E
E
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1.768AR
x
y
( )tE
Example (cont.)
We know that the polarization is LHEP, so the LHCP amplitude must dominate. Hence, we have
1 1
1 1LH
RH
A AR AR
A AR AR
or
0.2775 3.604LH
RH
A
A or
Hence
3.604LH
RH
A
A
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Poincaré Sphere
cotAR
0
0
implies LHEP
implies RHEP
xy
z2
/ 2 2
LHCP
RHCP
x
y
( )tE
A
B
C
D| |
Unit sphere
2
2
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Poincaré Sphere (cont.)
linear (equator)
LHCP (north pole)
RHCP (south pole) = 0 (no tilt)
= 90o
(45o tilt)
latitude = 2 longitude = 2
LHEP
RHEP
= 45o
= 0o
= -45o
x
y
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Poincaré Sphere (cont.)
View in full-screen mode to watch the “world spinning.”
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Poincaré Sphere (cont.)Two diametrically opposite points on the Poincaré sphere have the property that the electric field vectors (in the phasor domain) are orthogonal in the complex sense. (A proof is given in the Appendix.)
x
y
z
A
B
* 0A BE E
Examples
-
ˆ :
ˆ :
A
B
x
x
E x A
E y B
equator on axis
equator on axis
ˆ ˆ (LHCP) :
ˆ ˆ (RHCP) :
A
B
E x y j A
E x y j B
north pole
south pole
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Appendix
Proof of orthogonal property
x
y
z
A
B
* 0A BE E
ˆ ˆ ˆ ˆj jxa y be xb y ae
Hence
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:
, , ;
, / 2, ;
, , / 2 ;
/ /
A B
b a a b
From examining the polarization equations:
*ˆ ˆ ˆ ˆ 0j jxa y be xb y ae ab ab