Prof. David R. Jackson Dept. of ECE

Fall 2016

Notes 16

ECE 6340 Intermediate EM Waves

1

Polarization of Waves Consider a plane wave with both x and y components

ˆ ˆ( , , ) ( ) jkzx yE x y z x E y E e−= +

( = .)y xE Eβ −In general, phase of phase of

Assume

Ex

Ey

x

y

0jx

jy

E ae aE be β

= =

=

2

(polar form of complex numbers)

Polarization of Waves (cont.)

Time Domain:

( ) ( )( ) ( )

Re cos

Re cos

j tx

j j ty

a e a t

be e b t

ω

β ω

ω

ω β

= =

= = +

E

E

Depending on b and β, three different cases arise.

At z = 0

x

y

E (t) |E (t)|

φ (t)

3

Linear Polarization

At z = 0:

ˆ ˆ( ) cosx a y b tω= ±E

( )coscos cos

x

y

a tb t b t

ωω β ω

=

= + = ±

E

E

a

b

x

y x´ y´

E

β = 0 or π

(shown for β = 0)

(+ for 0, - for π)

4

Circular Polarization

At z = 0:

2 2 2 2 2 2 2

2

cos sinx y a t a t

a

ω ω= + = +

=

E E E

( )coscos cos( / 2) sin

x

y

a tb t a t a t

ωω β ω π ω

=

= + = ± =

E

E

b = a AND β = ± π /2

5

Circular Polarization (cont.)

( )1 1 1tan tan ( tan ) tan (tan )y

x

t tφ ω ω− − − = = =

EE

tφ ω=

6

so

waveddtφω ω== =angular velocity of waveNote:

a

x

y

( )tE

φ(t)

waveω ω=

/ 2β π= ±

Circular Polarization (cont.)

tφ ω=

IEEE convention

ˆ ˆ( cos sin )a x t y tω ω= E

a

x

y

( )tE

φ(t)

β = +π / 2

β = −π / 2

RHCP

LHCP

7

/ 2β π= ±

Circular Polarization (cont.)

8

( ) ( )( )ˆ ˆcos sina x t kz y t kzω ω= − + −E

( ) ( )( )ˆ ˆcos sina x kz y kz= −E

( )j t kze ω −

Note: The rotation is space is opposite to that in time.

Time and distance dependence:

( ) ( )( )ˆ ˆcos sina x t y tω ω= +E z = 0

t = 0

RHCP

( )t t kzω ω→ −

Circular Polarization (cont.)

RHCP

RHCP

z

z

“Snapshot" of wave

9

Circular Polarization (cont.) Animation of LHCP wave

http://en.wikipedia.org/wiki/Circular_polarization

(Use pptx version in full-screen mode to see motion.)

10

Unit Rotation Vectors

1ˆ ˆ ˆ( )2

1 ˆ ˆ( )2

r x jy

l x jy

= −

= +

RHCP (β = - π / 2)

LHCP (β = + π / 2)

1 ˆˆ ˆ( )2

ˆˆ ˆ( )2

x r l

jy r l

= +

= −

Add:

Subtract:

11

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= +

12

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.

13

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.

14

A more complicated version using four antennas

(omni CP) Note: LHCP is radiated in the negative z direction.

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)

15

Circularly Polarized Antennas (cont.) Method 4: Sequential rotation of antennas is used (shown for N = 4).

I = 1 ∠ 0o

I = 1 ∠ -90o I = 1 ∠ -90o

I = 1 ∠ -180o

I = 1 ∠ -270o

LP elements CP elements

The elements may be either LP or CP.

16

This is an extension of method 2 when used for LP elements.

Using multiple CP elements instead of one CP element often gives better CP in practice (due to symmetry).

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 as sensitive to “multipath” signals that bound off of other objects.

† Satellite transmission often uses CP because of Faraday rotation in the ionosphere. 17

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

18

Pros and Cons of CP (cont.)

LP-LP

Effects of alignment errors

α = alignment angle error

cos2α : The received signal can vary from zero dB to -∞ dB

CP-LP

TX-RX Gain Loss

-3 dB, no matter what the alignment

LP-CP -3 dB, no matter what the alignment

CP-CP 0 dB (polarizations same) or -∞ dB (polarizations opposite), no matter what the alignment

Examples of Polarization

AM Radio (0.540-1.6 MHz): linearly polarized (vertical) (Note 1)

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) (Note 2)

DBS Satellite TV (11.7-12.5 GHz): transmits both LHCP and RHCP (frequency reuse) (Note 3)

GPS (1.574 GHz): RHCP (Note 3)

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.

19

Elliptical Polarization

Includes all other cases

β ≠ 0 or π (not linear)

β ≠ ±π / 2 (not CP)

β = ± π / 2 and b≠ a (not CP)

x

y

( )tE

φ(t)

Ellipse

20

Elliptical Polarization (cont.)

00

β ππ β

< <− < <

LHEPRHEP

x

LHEP

RHEP

y

φ(t)

Ellipse

21

Note: The rotation is not at a constant speed.

Proof of Ellipse Property

cos( )

cos cos sin siny b t

b t b tω β

ω β ω β

= +

= −

E

so 2

cos sin 1x xy b b

a aβ β

= − −

E EE

22 2cos siny x x

b bba a

β β − = − − E E E

2 22 2 2 2 2cos 2 cos siny x x y x

b b bba a a

β β β + − = −

E E E E E

cosx a tω=E

22

Consider the following quadratic form:

( )2

2 2 2 2 2 2cos sin 2 cos sinx y x yb b ba a

β β β β + + − =

E E E E

22 2 2 22 cos sinx x y y

b b ba a

β β + − + = E E E E

2 2x x y yA B C D+ + =E E E E

2 22 2 2 2 2cos 2 cos siny x x y x

b b bba a a

β β β + − = −

E E E E E

Proof of Ellipse Property (cont.)

23

2

2 22

22

4

4 cos 4

4 cos 1

B AC

b ba a

ba

β

β

∆ = −

= −

= −

224 sin 0b

aβ ∆ = − <

Hence, this is an ellipse.

so

Proof of Ellipse Property (cont.)

Discriminant:

0,0,0,

∆ >∆ =∆ <

hyperbolalineellipse

From analytic geometry:

If β = 0 or π we have ∆ = 0, and this is linear polarization. 24

Rotation Property

coscos( )

x

y

a tb t

ωω β

== +

EE

00

β ππ β< <

− < <LHEPRHEP

x

LHEP

RHEP

y

φ(t)

ellipse

25

We now prove the rotation property:

cos cos sin siny b t b tω β ω β= −E

Recall that

Rotation Property (cont.)

Take the derivative:

[ ]tan cos tan siny

x

b ta

φ β ω β= = −EE

( )( )2 2sec sec sind b tdt aφφ ω ω β = −

so

( )( )2 2sin cos secd b tdt aφ β φ ω ω = −

( ) 0 0

( ) 0 0

dadtdbdt

φβ π

φπ β

< < ⇒ <

− < < ⇒ >

LHEP

RHEP (proof complete)

Hence

26

Phasor Picture

( )tE

x

y ( ) 0a β π< < LHEP

Ey leads Ex

Re

β

Ey

Ex

Im

LHEP

Rule:

The electric field vector rotates in time from the leading axis to the lagging axis.

27

A phase angle 0 < φ < π is a leading phase angle (leading with respect to zero degrees). A phase angle -π < φ < 0 is a lagging phase angle (lagging with respect to zero degrees).

Phasor Picture (cont.)

x

y

( )tE

( ) 0b π β− < < RHEP

Rule:

The electric field vector rotates in time from the leading axis to the lagging axis.

Ex

Ey

Im

Re β

Ey lags Ex

RHEP

28

A phase angle 0 < φ < π is a leading phase angle (leading with respect to zero degrees). A phase angle -π < φ < 0 is a lagging phase angle (lagging with respect to zero degrees).

Example

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

y

Ez

z

x Ex

What is this wave’s polarization?

29

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

30

[ ]ˆˆ(1 ) (2 ) jkyE z j x j e= + + −

y

Ez

z

x Ex

LHEP or LHCP

Note: 2x zE E πβ≠ ≠ ± (so this is not LHCP)

Example (cont.)

and

31

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

LHEP

Example (cont.)

32

y

Ez

z

x Ex

Axial Ratio (AR) and Tilt Angle (τ )

x

y ( )tE

τ

A

B C

D

AR 1ABCD

≡ = ≥major axisminor axis

( )dB 10AR 20log AR=

33

Axial Ratio (AR) and Tilt Angle (τ )

( )

( )

o o

0

0

45 45

sin 2 sin 2 sin

AR cot

ξ

ξ

ξ

ξ γ β

ξ

>

<

− ≤ ≤ +

=

=

implies LHEP

implies RHEP

Axial Ratio

tan 2 tan 2 cosτ γ β=

Tilt Angle

Note: The tilt angle τ is ambiguous by the addition of ± 90o.

(We cannot tell the difference between the major and minor axes.)

1tan ba

γ − =

We first calculate γ :

34

0jx

jy

E ae aE be β

= =

=

(-90o < τ < 90o)

( )AR cot ξ=

x

y ( )tE

τ

A

B C

D |ξ |

Axial Ratio (AR) and Tilt Angle (τ) (cont.)

Physical interpretation of the angle |ξ|

35

o

LP CP

0 45ξ< <Note :

Special Case

tan 2 tan 2 cos 00 / 2

τ γ βτ π

= =⇒ = or

Tilt Angle:

The tilt angle is zero or 90o if:

/ 2β π= ±

x

y

( )tE

coscos( / 2)

sin

x

y

a tb t

b t

ωω π

ω

=

= ±

=

EE

22

2 2 1yx

a b+ =

EE

36

37

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 2

r x jy l x jy= − = +

ˆ ˆ RH LHj jRH LH RH LHE r A l A r A e l A eφ φ= + = +

: ( ) RH LHA t A A= +Point E

: ( ) RH LHB t A A= −Point Ex

y ( )tE

A

B

AR RH LH RH LH

RH LH RH LH

A A A AA A A A

+ += =

− −

Hence

38

LHCP/RHCP Ratio (Cont.) Hence

1 /AR

1 /LH RH

LH RH

A AA A

+=

−

AR 1 AR 1AR 1 AR 1

LH

RH

AA

− +=

+ −or

From this we can also solve for the ratio of the CP components, if we know the axial ratio:

1 /AR

1 /LH RH

LH RH

A AA A

+= ± −

or

(We can’t tell which one is correct from knowing only the AR.)

(Use whichever sign gives AR > 0.)

39

z

y

Ez

x Ex

LHEP

Re-label the coordinate system:

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

x xz yy z

→→→ −

Find the axial ratio and tilt angle.

Example

Find the ratio of the CP amplitudes.

40

y

z

Ez

x Ex

LHEP

ˆ ˆ(2 ) (1 ) jkzE x j y j e− = − + +

1.249 o1 0.2 0.6 0.6324 0.6324 71.5652

y j

x

E j j eE j

+= = + = = ∠

−

( )1 1tan tan 0.632 0.564ba

γ − − = = =

radians

Example (cont.)

o/ 0.6324, 71.565b a β⇒ = =

41

tan 2 tan 2 cosτ γ β=

o o45 45

LHEP

RHEP

sin 2 sin 2 sin

AR cot

0 :0 :

ξ

ξ γ β

ξ

ξξ

− ≤ ≤ +

=

=

><

o

o o

16.84516.845 90

τ =

−or

o29.499ξ =

LHEP

Results

AR 1.768=

Example (cont.) Formulas

42

o16.845τ =

AR 1.768=

x

y

( )tE

τ

Example (cont.)

Note: Plotting the ellipse as a function of time will help determine which value is correct for the tilt angle τ.

( ) o

cos

0.6324 cos( 71.565 )x

y

tt

ω

ω

=

= +

E

E

Normalized field (a = 1):

43

AR 1.768=

x

y

( )tE

τ

Example (cont.)

We know that the polarization is LHEP, so the LHCP amplitude must dominate. Hence, we have

AR 1 AR 1AR 1 AR 1

LH

RH

AA

− +=

+ −or

0.2775 3.604LH

RH

AA

= or

Hence

3.604LH

RH

AA

=

Poincaré Sphere

( )AR cot ξ=

0

0

ξ

ξ

>

<

implies LHEP

implies RHEP

/ 2 2θ π ξ= −

x

y ( )tE

τ

A

B C

D |ξ |

44

2φ τ=

x y

z

LHCP

RHCP

Unit sphere

2τ

2ξ

longitude = 2τ

latitude = 2ξ

Poincaré Sphere (cont.)

linear (equator)

LHCP (north pole)

RHCP (south pole) φ = 0 (no tilt)

φ = 90o

(45o tilt)

latitude = 2ξ longitude = 2τ

LHEP

RHEP

ξ = 45o

ξ = 0o

ξ = -45o

x

y

45

Poincaré Sphere (cont.)

View in full-screen mode to watch the “world spinning”.

46

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 AE 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

47

( )* ˆ 0A BE H z× ⋅ =

( ) ( )( )

( ) ( )( )

* *

0

* *

0

1ˆ ˆ ˆ

1 ˆ ˆ ˆ

A B A B

A B B A

E H z E z E z

z E E E E z z

η

η

× ⋅ = × × ⋅

= ⋅ − ⋅ ⋅

( ) ( ) ( )A B C B A C C A B× × = ⋅ − ⋅Note :(orthogonal in the complex power sense)

Appendix Proof of orthogonal property

x

y

z

A

B

* 0A BE E⋅ =

( ) ( )ˆ ˆ ˆ ˆj jxa y be xb y aeβ β++ → −

Hence

48

( ) ( )( ) ( )( ) ( )

:, , ;

, , / 2 ;

, , / 2 ;/ /

A B

b a a b

θ φ π θ φ π

ξ τ ξ τ π

β γ π β π γ

→

→ − +

→ − +

→ + −

→

From examining the polarization equations:

( )( ) ( )( )*ˆ ˆ ˆ ˆ 0j jxa y be xb y ae ab abβ β++ ⋅ − = − =