Notes 17

Polarization of Plane Waves

1

ECE 3317 Applied Electromagnetic Waves

Prof. David R. Jackson Fall 2020

Polarization

The polarization of a plane wave refers to the direction of the electric field vector in the time domain.

We assume that the wave is traveling in the positive z direction.

2

( , )z tE

( ),z tS

x

y

z

Consider a plane wave with both x and y components

ˆ ˆ( ) ( ) jkzx yE z x E y E e−= +

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

Assume: xj

y

E aE be β

= =

=

real number

Polarization (cont.)

Phasor domain:

3

x

y

xE

yE

Time Domain:

( ) ( )

( ) ( )

Re cos

Re cos

j tx

j j ty

ae a t

be e b t

ω

β ω

ω

ω β

= =

= = +

E

E

Depending on b/a and β, three different cases arise:

4

Linear polarization Circular polarization Elliptical polarization

Polarization (cont.)

x

y

( )tE

0z =

Power Density: *12

S E H= ×

( )*

1 ˆ ˆ ˆ ˆ2

y xx y

E ES x E y E x yη η

= + × − +

, yxy x

EEH Hη η

= = −

( )221 ˆ2 x yS z E Eη

= +

Assume lossless medium (η is real):

5

21 ˆ2

S z Eη

=or

From Faraday’s law:

Polarization (cont.)

Hence

At z = 0:

( )ˆ ˆ cosx a y b tω= ±E

coscos

x

y

a tb t

ωω

== ±

EE

Linear Polarization

6

( )( )

cos

cosx

y

a t

b t

ω

ω β

=

= +

E

E

0β β π= =or

0ββ π

+ =− =

sign :sign :

(shown for β = 0)

Linear Polarization (cont.)

7

This is simply a “tilted” plane wave.

( )ˆ ˆ cosx a y b tω= ±E

( )tE

x

y x′y′

a

b

φ

1tan ba

φ − = ±

At z = 0:

2 2 2 2 2 2 2 2cos sinx y a t a t aω ω= + = + =E E E

coscos( / 2) sin

x

y

a ta t a t

ωω π ω

== ± =

EE

Circular Polarization

8

/ 2b a β π= = ±and

( )( )

cos

cosx

y

a t

b t

ω

ω β

=

= +

E

E

Note: The top sign is always for β = +π /2.

( ) 1 1tan tan ( tan )y

x

t tφ ω− − = =

EE

( )t tφ ω=

Circular Polarization (cont.)

9

( )tE

x

y

a

( )tφ

( ) ( )d tt t

dtφ

φ ω ω= ⇒ =

IEEE convention

10

Circular Polarization (cont.)

( )( )

cos

cosx

y

a t

b t

ω

ω β

=

= +

E

E

RHCP

LHCP

( )tE

x

y

a

( )tφ/ 2β π=

/ 2β π= −

/ 2β π= ±

Note: The mechanical angular

velocity is the same as the electrical radian frequency ω.

There is opposite rotation in space and time.

11

Examine how the field varies in both space and time:

Rotation in space vs. rotation in time

( )ˆ ˆ( ) j jkzE z x a y be eβ −= +

( ) ( )ˆ ˆ( , ) cos cosz t x a t kz y b t kzω ω β= − + − +E

Phasor domain

Time domain

Circular Polarization (cont.)

Notice that the rotation in space matches the left hand!

12

A snapshot of the electric field vector, showing the vector at different points.

Circular Polarization (cont.)

RHCP

z

13

Animation of LHCP wave

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

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

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 buildings Propagation through the ionosphere

Receive antenna

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). The linear antenna “throws away” half of the incident signal.

14

Circular Polarization (cont.)

z

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 antennas rotated by 90o, and fed 90o out of phase.

15

Circular Polarization (cont.)

Antenna 1

1xV =1 90oyV j= − = ∠ −

Antenna 2

+

-

+

-

/ 2β π= −

x

y

Realization of method 1 using a 90o delay line

16

Circular Polarization (cont.)

2 1 / 2l lβ β π− =

ANT01 inZ Z=

Antenna 1

Antenna 2

Feed line

Signal

02 01 / 2Z Z=Power splitter :

02Z

x

y

01Z

01Z 1l

2l

17

An array of CP antennas

Circular Polarization (cont.)

18

The two antennas can realized by using two different modes of a single microstrip or dielectric resonator antenna.

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

19

Circular Polarization (cont.)

20

Helical antennas on a GPS satellite

Circular Polarization (cont.)

21

Other Helical antennas

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. (The proof is omitted.)

22

Circular Polarization (cont.)

Note: It does not matter how the receive antenna is rotated

about the z axis.

RHCP wave

RHCP Antenna

2 1 / 2l lβ β π− =

Antenna 1

Antenna 2

Feed line

Output signal

02 01 / 2Z Z=Power combiner :

02Z

x

y

01Z

01Z 1l

2l

z

The two signals add in phase.

23

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/LP, receive antenna is LP/CP

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.

Circular Polarization (cont.)

Includes all other cases that are not linear or circular

Elliptic Polarization

The tip of the electric field vector stays on an ellipse.

24

( )tE

x

y

( )tφ

(This is proved in Appendix A.)

00

β ππ β

< <− < <

LHEPRHEP

Elliptic Polarization (cont.)

Rotation Property

25

LHEP

RHEP

( )tE

x

y

( )tφ

(This is proved in Appendix B.)

( )( )

cos

cosx

y

a t

b t

ω

ω β

=

= +

E

E

Rotation Rule

Here we give a simple graphical method for determining the type of polarization (left-handed or right handed).

26

First, we review the concept of leading and lagging sinusoidal waves.

27

Two phasors: A and B

0 β π< <

0π β− < <

Note: We can always assume that the phasor A is on the real axis (zero degrees phase)

without loss of generality, since it is only the phase difference between the two phasors

that is important.

Rotation Rule (cont.)

Re

Im

β A

BB Aleads

Re

Im

A

B

B Alags

β

0j

j

A aeB be β

=

=

( ) 0a β π< < LHEP

Observation:

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

28

Now consider the case of a plane wave.

Rotation Rule (cont.)

Re

Im

βxE

yEy xE Eleads

LHEP

( )tE

x

y

( )tφ

( ) 0b π β− < < RHEP

29

Rotation Rule (cont.)

Observation:

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

RHEP

( )tE

x

y

( )tφ

Re

Im

xE

yE

y xE Elags

β

Rotation Rule:

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

30

The rule works in both cases, so we can call it a general rule:

Rotation Rule (cont.)

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

What is this wave’s polarization?

Example

31

Rotation Rule (cont.)

x

y

z

xE

zE

Example (cont.)

Therefore, in time the wave rotates from the z axis to the x axis.

32

Rotation Rule (cont.)

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

Re

Im

xE

zE z xE Eleads

LHEP or LHCP

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

33

Rotation Rule (cont.)

Example (cont.)

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

x

y

z

xE

zERotation

LHEP

34

Rotation Rule (cont.)

Example (cont.)

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

x

y

z

xE

zERotation

AR 1ABCD

= = >major axisminor axis

Axial Ratio (AR) and Tilt Angle (τ)

35

Note: In dB we have ( )dB 10AR 20log AR=

( )tE

τ x

y

A

B

C

D

τ = tilt angle

o o

LHEP

RHEP

45 45

AR cot

0 :0 :

sin 2 sin 2 sinξ

ξ

ξξ

ξ γ β

− ≤ ≤ +

=

><

=where

Axial Ratio and Handedness

tan 2 tan 2 cosτ γ β=

Tilt Angle

Note: The tilt angle τ is ambiguous by the

addition of ± 90o.

1

o

tan

0 90

ba

γ

γ

− ≡

≤ ≤

Axial Ratio (AR) and Tilt Angle (τ ) Formulas

36

tan 2 tan 2 cosτ γ β=Tilt Angle:

The title angle is zero or 90o if:

/ 2β π= ±

Note on Tilt Angle

37

( )tE

x

y

Example

38

Relabel the coordinate system: x xz yy z

→→→ −

Find the axial ratio and tilt angle.

[ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

LHEP

x

y

z

xE

zE

Example (cont.)

39

1ˆ ˆ(1)2

jkzjE x y ej

− += + −

Normalize:

( )1.249ˆ ˆ(1) 0.6324 j jkzE x y e e− = +

or

ˆ ˆ(1 ) (2 ) jkzE y j x j e− = + + −

LHEP

x

z

y

xE

zE

40

71.565oy xE Eβ = ∠ − ∠ =

( ) [ ]1 1tan tan 0.632 0.564 radba

γ − − = = =

Hence

( )1.249ˆ ˆ(1) 0.6324 j jkzE x y e e− = +

o

10.63241.249[rad] 71.565

abβ

==

= =

Example (cont.)

LHEP

x

z

y

xE

zE

41

o16.845τ =

o29.499ξ =

LHEP

Results:

AR 1.768=

o71.565β =

[ ]0.564 radγ =

Example (cont.)

o o

LHEP

RHEP

45 45

AR cot

0 :0 :

sin 2 sin 2 sinξ

ξ

ξξ

ξ γ β

− ≤ ≤ +

=

><

=where

42

o16.845τ =

AR 1.768=

o

o o

16.84516.845 90

τ

τ

=

= +

Note: We are not sure which choice is correct:

We can make a quick time-domain sketch to be sure.

Example (cont.)

( )tE

x

y

τ

LHEP

43

( )( )

cos

cosx

y

a t

b t

ω

ω β

=

= +

E

E

Example (cont.)

o

10.63241.249[rad] 71.565

abβ

==

= =

o

AR 1.76816.845τ

=

=

Given:

Results:

Summary

( )tE

x

y

τ

LHEP

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

or

Squaring both sides, we have

Here we give a proof that the tip of the electric field vector must stay on an ellipse.

44

Appendix A

This is in the form of a quadratic expression:

( )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

or

Collecting terms, we have

45

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

Appendix A (cont.)

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

(determines the type of curve)

46

Discriminant:

Appendix A (cont.)

coscos( )

cos cos sin sin

x

y

a tb tb t b t

ωω β

ω β ω β

== +

= −

EE

Here we give a proof of the rotation property.

47

00

β ππ β

< <− < <

LHEPRHEP

Appendix B

LHEP

RHEP

( )tE

x

y

( )tφ

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 = −

Hence

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

( )

( )

0 0

0 0

ddtddt

a

b

φβ π

φπ β

< < <

− < < >

LHEP

RHEP (proof complete)

48

Appendix B (cont.)