Prof. David R. Jackson Dept. of ECE Fall 2013 Notes 16 ECE 6340 Intermediate EM Waves 1.

Prof. David R. JacksonDept. of ECE

Fall 2013

Notes 16

ECE 6340Intermediate EM Waves

1

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)

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)

3

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 )

4

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

5

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)

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

7

Circular Polarization (cont.)

RHCP

RHCP

z

j t kze

(opposite rotation in space and time)

z

"snapshot" of wave

8

Circular Polarization (cont.)

Animation of LHCP wave

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

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

9

Unit Rotation Vectors

1ˆ ˆ ˆ( )

21

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:

10

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

11

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.

12

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.

13

A more complicated version, using four antennas

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)

14

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.

15

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).

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.

16

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

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.

17

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

18

Elliptical Polarization (cont.)

0

0

LHEP

RHEP

x

LHEP

RHEP

y

(t)

Ellipse

19

Note: The rotation is not at a constant speed.

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

20

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.)

21

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

Rotation Property

cos

cos( )x

y

a t

b t

E

E

0

0

LHEP

RHEP

x

LHEP

RHEP

y

(t)

ellipse

23

We now prove the rotation property:

cos cos sin siny b t b t E

Recall that

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

24

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.

25

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).

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

26

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).

Example

ˆˆ (1 ) (2 ) jkyE z j x j e

y

Ez

z

xEx

What is this wave’s polarization?

27

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

28

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

29

ˆˆ (1 ) (2 ) jkyE z j x j e

LHEP

Example (cont.)

30

y

Ez

z

xEx

Axial Ratio (AR) and Tilt Angle ( )

x

y

( )tE

A

B

C

D

1AB

ARCD

major axis

minor axis

dB 1020logAR AR

31

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 :

32

0jx

jy

E ae a

E be

cotAR

x

y

( )tE

A

B

C

D| |

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

Physical interpretation of the angle ||

33

o0 45

LP CP

Note :

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

34

35

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

36

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.)

37

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.

38

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

39

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.)

40

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

41

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

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

42

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

43

Poincaré Sphere (cont.)

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

44

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

45

Appendix

Proof of orthogonal property

x

y

z

A

B

* 0A BE E

ˆ ˆ ˆ ˆj jxa y be xb y ae

Hence

46

:

, , ;

, / 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