1 EE 543 Theory and Principles of Remote Sensing Topic 3 - Basic EM Theory and Plane Waves

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EE 543Theory and Principles of

Remote Sensing

Topic 3 - Basic EM Theory and Plane Waves

O. Kilic EE543

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Outline• EM Theory Concepts• Maxwell’s Equations

– Notation– Differential Form– Integral Form– Phasor Form

• Wave Equation and Solution (lossless, unbounded, homogeneous medium)

– Derivation of Wave Equation– Solution to the Wave Equation – Separation of Variables– Plane waves

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EM Theory ConceptThe fundamental concept of em theory is that a

current at a point in space is capable of inducing potential and hence currents at another point far away.

J

E, H

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Introduction to EM Theory

• In remote sensing we are interested in the interactions of em waves with the medium and target of interest.

• The existence of propagating em waves can be predicted as a direct consequence of Maxwell’s equations.

• These equations satisfy the relationship between the vector electric field, E and vector magnetic field, H in time and space in a given medium.

• Both E and H are vector functions of space and time; i.e. E (x,y,z;t), H (x,y,z;t.)

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What is an Electromagnetic Field?

• The electric and magnetic fields were originally introduced by means of the force equation.

• In Coulomb’s experiments forces acting between localized charges were observed.

• There, it is found useful to introduce E as the force per unit charge.

• Similarly, in Ampere’s experiments the mutual forces of current carrying loops were studied.

• B is defined as force per unit current.

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Why not use just force?

• Although E and B appear as convenient replacements for forces produced by distributions of charge and current, they have other important aspects.

• First, their introduction decouples conceptually the sources from the test bodies experiencing em forces.

• If the fields E and B from two source distributions are the same at a given point in space, the force acting on a test charge will be the same regardless of how different the sources are.

• This gives E and B meaning in their own right.• Also, em fields can exist in regions of space where there

are no sources.

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Maxwell’s Equations

• Maxwell's equations give expressions for electric and magnetic fields everywhere in space provided that all charge and current sources are defined.

• They represent one of the most elegant and concise ways to state the fundamentals of electricity and magnetism.

• These set of equations describe the relationship between the electric and magnetic fields and sources in the medium.

• Because of their concise statement, they embody a high level of mathematical sophistication.

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Notation: (Time and Position Dependent Field Vectors)

E (x,y,z;t) Electric field intensity (Volts/m)

H (x,y,z;t) Magnetic field intensity (Amperes/m)

D (x,y,z;t) Electric flux density (Coulombs/m2)

B (x,y,z;t) Magnetic flux density (Webers/m2, Tesla)

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Notation: Sources and Medium

J (x,y,z;t) Electric current density (Amperes/m2)

Jd (x,y,z;t) Displacement current density (Amperes/m2)

e Electric charge density (Coulombs/m3)

Permittivity of the medium (Farad/m)

Permeability of the medium (Henry/m)

Conductivity of the medium (Siemens/m)

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Two Forms of Maxwell’s Equations

• Differential form– This is the most widely used form.– They describe the relationship between the electric and

magnetic fields and sources in the medium at a point in space.

• Integral form– Integral form of Maxwell’s equations can be derived

from the differential form by using Stoke’s theorem and Divergence theorem.

– These set of equations describe the field vector relations over an extended region in space.

– They have limited use. Typically, they are applied to solve em boundary value problems with symmetry.

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Maxwell’s Equations – Physical Laws

• Faraday’s Law Changes in magnetic field induce voltage.

• Ampere’s Law Allows us to write all the possible ways that electric currents can make magnetic field. Magnetic field in space around an electric current is proportional to the current source.

• Gauss’ Law for Electricity The electric flux out of any closed surface is proportional to the total charge enclosed within the surface.

• Gauss’ Law for Magnetism The net magnetic flux out of any closed surface is zero.

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Differential Form

-t

BE =

; ;d s ct t

D DH = J J J = J J

vD=

0B=

Faraday’s Law:

Ampere’s Law:

Gauss’ Law:

(1)

(2)

(3)

(4)

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Some Observations (1)How many scalar equations are there in Maxwell’s

equations? Answer = 8 ??

-

ˆ ˆ ˆ ˆˆ ˆ-

ˆ ˆ ˆ ˆˆ ˆ- yx

x y z

y x zz

x

x

y

yz

z

z x

t

x y z x y zt

x y z xy z

yt ttx y

z

BE =

E E E

E E E BE B BE

B B

E

B

Three scalar equations for each curl(3x2 = 6)

1 scalar equation for each divergence (1x2 = 2).

ˆ ˆ ˆ

v

yx zx y z vx y z

x y z

D=

DD DD D D

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Some Observations (2)

But, the divergence equations are related to the curl equations. This is known as the conservation of charge.

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Conservation of Charge

If we take the divergence of both sides of eqn. 2 and use the vector identity:

0H

0J Dt

(5)

Using eqn. (3) in eqn. (5), we obtain the continuity law for current:

vJt

(6)

Flow of current out of a differential volume

Rate of decrease of charge in the volume

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Some Observations (3)

Thus, the divergence equations (eqn. 3, 4) are dependent on the curl equations (eqn. 1, 2).

So, Maxwell’s equations represent 6 independent equations.

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Some Observations (4)

How many variables are there in Maxwell’s equations? Answer:12, three for each component of E, H, D, and B vectors.

Therefore, the set of Maxwell’s equations is

not sufficient to solve for the unknowns.

We need 12-6 = 6 more scalar equations. These are known as constitutive relations.

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Constitutive RelationsConstitutive relations provide information

about the environment in which electromagnetic fields occur; e.g. free space, water, etc.

D= E

B= H9

7

10

364 10

o

o

Free space values.

(7)

(8)

permittivity

permeability

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Duality Principle

• Note the symmetry in Maxwell’s equations

-

E H

H E

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Integral Form of Maxwell’s Equations

Faraday's Law

Ampere's Law

0

C S

C S S

eS V

S

dl dst

dl ds dst

ds dv Q

ds

E B

H D J

D.

B.(9)

Using Stoke’s theorem and Divergence theorem:

HW #2.1 Prove by applying Stoke’s and Divergence theorems to Maxwell’s eqn in differential form

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Time Harmonic Representation - Phasor Form

• In a source free ( ) and lossless ( ) medium characterized by permeability and permittivity , Maxwell’s equations can be written as:

0sJ

0 0cJ

-

;

0

0

t

t

HE =

EH =

D=

B=(10)

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Time Harmonic Fields• We will now assume time harmonic fields; i.e.

fields at a single frequency. We will assume that all field vectors vary sinusoidally with time, at an angular frequency w; i.e.

, , ; ) ( , , )cos( )ox y z t x y z wt E( E

( , , ; ) Re ( , , )

( , , ; ) Re ( , , )

jwt

jwt

x y z t E x y z e

x y z t H x y z e

E

H

In other words:

(11)

Note that the E and H vectors are now complex

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Time Harmonic Fields (2)

• The time derivative in Maxwell’s equations becomes a factor of jw:

jwt

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Phasor Form of Maxwell’s Equations

Maxwell’s equations can then be written in phasor form as:

-

0

0

E jw H

H jw E

D

B

=

=

=

=

Phasor form is dependent on position only. Time dependence is removed.

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The Wave Equation (1) If we take the curl of Maxwell’s first equation:

E jw H

Using the vector identity:

2A A A

0eq J

0

0E

And assuming a source free, i.e. and lossless;

medium:

; ; 0

d d s ct

DH = J J J J = J J

i.e.

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The Wave Equation (2)

2 2( ) ( )E r w E r

2 2k w

2 2( , , ) ( , , ) 0E x y z k E x y z

Define k, which will be known as wave number:

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Wave Equation in Cartesian Coordinates

2 2 22

2 2 2

ˆ ˆ ˆ( , , ) ( , , ) ( , , ) ( , , )

andx y zE x y z x E x y z y E x y z z E x y z

x y z

2 2( , , ) ( , , ) 0E x y z k E x y z

where

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Scalar Form of Wave Equation

2 2( , , ) ( , , ); , , E x y z k E x y z x y z

For each component of the E vector, the wave equation is in the form of:

Denotes different components of E in Cartesian coordinates

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Solution to the Wave Equation – Separation of Variables

( , , ) ( ) ( ) ( ); , ,E x y z f x g y h z x y z

Assume that a solution can be written such that

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Separation of Variables

2 2 22

2 2 2

2 2

22

22

2 2 2

2 2 2 2

22

2 2

22

2

0

divide by fgh:

1 1 1

f g h

1

f

1

g

Let

1

h

x y z

y zx

fk

x

f g hgh fh fg k fgh

x y z

f g hk

x y z

k k k k

hk k

z

g

y

This decomposition is arbitrarily defined at this point. Will depend on the medium and boundary conditions. Determines how the wave propagates along each direction.

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Separation of Variables

2 22 2

2 2

2 22 2

2 2

2 22 2

2 2

The equation can be split into three parts, and

f, g and h can be solved separately.

1

f1

g

1

h

x x

y y

z z

f fk k f

x xg g

k k gy y

h hk k h

z z

+

2

2

2

22

22

2 2

1

h

1

g

1

f

y zx

fk k

h

zxk

g

y

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Possible Solutions to the Wave Equation

1 2

1 1

( ) traveling wave

or

( ) cos( ) sin( ) standing wave

x xjk x jk x

x x

f x Ae A e

f x C k x D k x

Standing wave solutions are appropriate for bounded propagation such as wave guides.

When waves travel in unbounded medium, traveling wave solution is more appropriate.

Energy is transported from one point to the other

HW 2.2: Show that the above are solutions to the wave equation by plugging the solution on the differential eqn on the previous page

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The Traveling Wave• The phasor form of the fields is a mathematical

representation.• The measurable fields are represented in the time domain.

1 1 1;

jkx jE Ae A A e

1 1

( , , ; ) Re ( , , )

ˆRe cos

jwt

jkx jwt

x y z t E x y z e

Ae e x A wt kx

E

Then

Let the solution to the -component of the electric field be:

Traveling in +x direction

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Traveling Wave

-1.5

-1

-0.5

0

0.5

1

1.5

-4 -3 -2 -1 0 1 2 3 4

kx

E(x

,y,z

;t)

wt = 0 wt = PI/2

As time increases, the wave moves along +x direction

( , , ; ) cos x y z t wt kxE

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Standing Wave

1 1 1cos( ); jE C kx C C e

1 1

( , , ; ) Re ( , , )

ˆRe cos( ) cos cos( )

jwt

jwt

x y z t E x y z e

C kx e x C wt kx

E

Then, in time domain:

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-1.5

-1

-0.5

0

0.5

1

1.5

-4 -3 -2 -1 0 1 2 3 4

kx

E(x

,y,z

;t)

wt = 0 wt = PI/2 wt = -PI/2 wt = PIwt = PI/4 wt = 3PI/4 wt = 5PI/8 wt = 3PI/8

Standing WaveStationary nulls and peaks in space as time passes.

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To summarize

• We have shown that Maxwell’s equations describe how em energy travels in a medium

• The E and H fields satisfy the “wave equation”.

• The solution to the wave equation can be in various forms, depending on the medium characteristics

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The Plane Wave Concept

• Plane waves constitute a special set of E and H field components such that E and H are always perpendicular to each other and to the direction of propagation.

• A special case of plane waves is uniform plane waves where E and H have a constant magnitude in the plane that contains them.

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Example 1 (1/5)Assume that the E field lies along the x-axis and is traveling

along the z-direction.

1

-

ˆ ˆ

o

o

jkz jkzoo

o

H Ejw

Ey E e y e

We derive the solution for the H field from the E field using Maxwell’s equation #1:

ˆ jkzoE xE e

Intrinsic impedance; 377 for free space

wave number

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Example 1 (2 of 5)

0

0

E Ex y

H Hx y

22

2

2

2

2

0

0

xx

y

y

Ek E

zH

k Hz

2 2k w

Thus the wave equation (Page 26) simplifies to:

Where as before

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Example 1 (3 of 5)

( , , ) ( )

( , , ) ( )

E x y z E z

H x y z H z

direction of propagation

x

y

z

E, H plane

E and H fields are not functions of x and y, because they lie on x-y plane

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Example 1 (4 of 5)

ˆ cos( )

ˆ cos( )

o

o

o

x E wt kz

Ey wt kz

E

H

phase term

*** The constant phase term is the angle of the complex number Eo

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Wavelength: period in spacek = 2

Example 1 (5 of 5)

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Velocity of Propagation (1/3)

• We observe that the fields progress with time.

• Imagine that we ride along with the wave.

• At what velocity shall we move in order to keep

up with the wave???

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Velocity of Propagation (2/3)

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4

z

Ex

wt = 0 wt = PI/2 wt = PI

constantwt kz a

dz d wt av

dt dt k

wv

k

Constant phase points

E field as a function of different times

ˆ cos( )ox E wt kz E

kz

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Velocity of Propagation (3/3)

1

dz wv

dt k

k w

wv

k

In free space:

9

7

8

1

10 F/m

364 10 H/m

c 3 10 m/s

fs

o o

o

o

v c

Note that the velocity is independent of the frequency of the wave, but a function of the medium properties.

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Example 2

A uniform em wave is traveling at an angle with respect to the z-axis. The E field is in the y-direction. What is the direction of the H field?

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Solution: Example 2

ˆ ˆ ˆcos sine z x

x

zy

E

The E field is along the unit vector:

The direction of propagation is along y.

Because E, H and the direction of propagation are perpendicular to each other, H lies on x-z plane. It should be in the direction parallel to:

ˆ ˆˆ ˆ ˆ/ / sin cosh y e z x

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Plane Wave Characteristics

1ˆ( , , ; ) cos

x y z t A wt kx eE

amplitudefrequency

Wave number, depends on the medium characteristics

phase

Direction of propagation

polarization

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Example 3

Write the expression for an x-polarized electric field that propagates in +z direction at a frequency of 3 GHz in free space with unit amplitude and 60o phase.

1ˆ( , , ; ) cos

x y z t A wt kz eE

=1= 2*3*109

o ow 60o

x

+ z-direction

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Example 4If the electric field intensity of a uniform plane wave

in a dielectric medium where = or and = o is given by:

Determine:• The direction of propagation and

frequency• The velocity• The dielectric constant (i.e. permittivity)• The wavelength

9 ˆ377cos(10 5 )t y z E

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Solution: Example 4 (1/2)

1. +y direction; w = 2f = 109

2. Velocity:

3. Permittivity:

9810

2 10 m/s5

wv

k

81 1 3 10

o r o r r

cv

88 3 10 9

2 10 2.254r

r

v

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Solution: Example 4 (2/2)

4. Wavelength:

2 25k

m

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Example 5Assume that a plane wave propagates along +z-

direction in a boundless and a source free, dielectric medium. If the electric field is given by:

Calculate the magnetic field, H.

ˆ ˆ( ) jkzx oE E z x E e x

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Example 5 - observations

• Note that the phasor form is being used in the notation; i.e. time dependence is suppressed.

• We observe that the direction of propagation is along +z-axis.

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Solution: Example 5 (1/4)

(source free; i.e. 0)E jw H J

From Maxwell’s equations in phasor form, we can write:

ˆ ˆ ˆ

ˆ

y yz x z x

x

E EE E E EE x y z

z y z x x y

Ey

z

Eqn. 1

Eqn. 2

where, in Cartesian coordinates the curl operator is given as:

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Solution: Example 5 (2/4)From eqn. 1 and eqn. 2:

( )1 1ˆ

( )1 1ˆ ˆ( )

ˆ ˆ ˆ

x

jkzjkzo

o

jkz jkz jkzoo o

o oo

E zH E y

jw jw z

E ey y jkE e

jw z jw

Eky E e y e yH ew

E EH

Intrinsic impedance, I = V/R

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Solution: Example 5 (3/4)

• E, H and the direction of propagation are orthogonal to each other.

• Amplitudes of E and H are related to each other through the intrinsic impedance of the medium.

• Note that the free space intrinsic impedance is 377

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Observation

o EH

This answer can be generalized to the following (for plane waves):

where o is the direction of propagation.

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Homework 3.1The magnetic field of a uniform plane wave traveling in free space is given by

1. What is the direction of propagation?2. What is the wave number, k in terms of permittivity, o

and permeability, o?

3. Determine the electric field, E.

ˆ jkzoH xH e

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Plane Waves Along Arbitrary Directions

o

ˆ.( ) jko roE r E e

ˆ ˆ ˆ (position vector)

ˆ unit vector along propagation direction

r xx yy zz

O

E, H plane

o r

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Example 6

• A z-polarized electric field propagating along direction in a

dielectric medium where = 9o, = o.

The frequency is 100 MHz.a) Write the electric field in phasor form and in

time domain. Assume an arbitrary phase and unit amplitude.

b) Calculate the magnetic field, H in phasor form and in time domain.

1ˆ ˆ ˆ

2o x y

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Solution: Example 6 (1/2)

ˆ

ˆ

8

22 2

;

ˆ

1ˆ ˆ ˆ ˆ ˆ ˆ;

2

2 10 9 2

ˆ

2ˆ( ; ) cos( ( ) )

2

o

o

o

jjk r jko ro o o o

j jko r

o o

x yj

o

E E e E e E E e

ze e

o x y r xx yy zz

k w

E ze

z t z wt x y

E

a)=1, unit amplitude

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Solution: Example 6 (2/2)

12

12

2 ( )

2 ( )

ˆ

1ˆ ˆ ˆ

2

1ˆ ˆ

2

o

o

j x y

j x y

o EH

x y ze

y x e

where

1 3773 3

o

o

b)

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Polarization• The alignment of the electric field vector of

a plane wave relative to the direction of propagation defines the polarization.

• Three types:– Linear– Circular– Elliptical (most general form)

Polarization is the locus of the tip of the electric field at a given point as a function of time.

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Linear Polarization• Electric field oscillates

along a straight line as a function of time

• Example: wire antennas

y

x

x

y

E

E

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Example 7

ˆ( ; ) cos( )oz t xE wt kz E

For z = 0 (any position value is fine)

ˆ(0; ) cos( )ot xE wtE

x

t = 0t =

- Eo Eo

y

Linear Polarization: The tip of the E field always stays on x-axis. It oscillates between ±Eo

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Example 8

ˆ ˆ( , ) cos( ) 2 cos( )z t x wt kz y wt kz E

Let z = 0 (any position is fine)

ˆ ˆ(0, ) cos( ) 2 cos( )t x wt y wt E

x

y

t = /2

t = 0

1

2

Linear Polarization

Exo=1 Eyo=2

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Circular Polarization

• Electric field traces a circle as a function of time.

• Generated by two linear components that are 90o out of phase.

• Most satellite antennas are circularly polarized.

y

x

y

x

RHCP

LHCP

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Example 9ˆ ˆ( ; ) cos( ) sin( )z t x wt kz y wt kz

E

2cos( )wt kz Exo=1 Eyo=1

Let z= 0

ˆ ˆ(0; ) cos( ) sin( )t x wt y wt E

x

RHCPy

t=0

t=/2w

t=

t=3/2w

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Elliptical Polarization

• This is the most general form

• Linear and circular cases are special forms of elliptical polarization

• Example: log spiral antennas

y

x

LH

y

x

RH

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Example 10ˆ ˆ( ; ) cos( ) cos( )a bz t xa wt kz yb wt kz

E

or a b a b

y x

bE E

a

ExEy

Linear when

Circular when

Elliptical if no special condition is met.

2 2 2

and 2a b

y x

a b

E E a

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Note:

• We have decomposed E field into two orthogonal components to identify polarization state.

• Examples were x and y components, assuming the wave travels in z-direction.

• For arbitrary propagation directions, the E field can still be decomposed into two components that lie on a plane perpendicular to the direction of propagation.

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Example 11

0.5ˆ ˆ3 4 V/mj zE x j y e

Determine the polarization of this wave.

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Solution: Example 11 (1/2)Note that the field is given in phasor form. We would like to see the trace of the tip of the E field as a function of time. Therefore we need to convert the phasor form to time domain.

20.5

2

( ; ) Re

ˆ ˆRe 3 4 ;

ˆ ˆ3 cos( 0.5 ) 4 cos( 0.5 )

ˆ ˆ3 cos( 0.5 ) 4 sin( 0.5 )

( ; ) 3cos( 0.5 )

( ; ) 4sin( 0.5 )

jwt

jj z jwt

x

y

z t Ee

x j y e e j e

x wt z y wt z

x wt z y wt z

z t wt z

z t wt z

E

E =

E =

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Solution: Example 11 (2/2)

22

(0; ) 3cos(0.5 )

(0; ) 4sin(0.5 )

(0; )(0; )1

9 16

x

y

yx

t z

t z

tt

E =

E =

EE

Let z=0

Elliptical polarization

x yE E

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Example 12

Find the polarization of the following fields:

ˆ ˆ( ) jkzE r jx y e

ˆ ˆ( ) (1 ) (1 ) jkxE r j y j z e

ˆ ˆ( ) (2 ) (3 ) jkyE r j x j z e

ˆ ˆ( ) 2 jkzE r jx j y e

a)

b)

c)

d)

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Solution: Example 12 (1/4) 2ˆ ˆ ˆ ˆ( )

ˆ ˆ( ; ) sin( ) cos( )

j kzjkz jkzE r jx y e xe ye

r t x wt kz y wt kz

E

a)

x

y

zLet kz=0

t=0

t=/2w

t=

t=3/2w

RHCP

Observe that orthogonal components have same amplitude but 90o phase difference.

Circular Polarization

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Solution: Example 12 (2/4)

4 4

4 4

ˆ ˆ( ) (1 ) (1 )

1 2 ; 1 2

ˆ ˆ( ; ) 2 cos( ) 2 cos( )

jkx

j j

E r j y j z e

j e j e

r t y wt kx z wt kx

E

b)

Observe that orthogonal components have same amplitude but 90o phase difference.

y

z

Let kx=0

t=-/4w

t=+/4w

t=3

t=5/4w RHCP

x

Circular Polarization

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Solution: Example 12 (3/4)

c)

1 1

ˆ ˆ( ) (2 ) (3 )

2 5 ; 3 10

1 1tan ; tan

2 3

ˆ ˆ( ; ) 5 cos( ) 10 cos( )

jky

j j

E r j x j z e

j e j e

r t x wt ky z wt ky

E

Observe that orthogonal components have different amplitudes and are out of phase.

Elliptical Polarization

zLet ky=0

t=-/w

t=+/w

x

y

Left Hand

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Solution: Example 12 (4/4)

2

ˆ ˆ( ) 2

ˆ ˆ( ; ) sin( ) 2sin( )

ˆ ˆ2 sin( )

jkz

j

E r jx j y e

j e

r t x wt kz y wt kz

x y wt kz

E

d)

Observe that orthogonal components are in phase. Linear Polarization

x

y

z

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Example 13

Show that an elliptically polarized wave can be decomposed into two circularly polarized waves, one left-handed other right-handed.

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Solution: Example 13

ˆ ˆ ; where , are complex numbersjkzE xa yb e a b

1 1 1 1

1 1 1 1

ˆ ˆ ˆ ˆ

ˆ ˆ

LHCP RHCP

jkz jkz

jkz jkz

E E E

xa jya e xb jyb e

x a b e y ja jb e

1 1

1 1

a a b

b j a b

Elliptically polarized wave in general is of the following form:

1 1

1 1

ˆ ˆ

ˆ ˆ

jkz

LHCP

jkz

RHCP

E xa jya e

E xb jyb e

LHCP and RHCP waves can be written in the following form:

If we let

a b

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Coherence and Polarization

• In the definition of linear, circular and elliptical polarization, we considered only completely polarized plane waves.

• Natural radiation received by an anatenna operating at a frequency w, with a narrow bandwidth, w would be quasi-monochromatic plane wave.

• The received signal can be treated as a single frequency plane wave whose amplitude and phase are slowly varying functions of time.

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Quasi-Monochromatic Waves

( )( )ˆ ˆ( ; ) ( ) ( ) yx

o o

jkz j tjkz j t

x yz t x t e y t e E E E

amplitude and phase are slowly varying functions of time

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Degree of Coherence

1

2

*

22

yx

xy

x y

E E

E E

where <….> denotes the time average.

0

1lim

T

Tdt

T

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Degree of Coherence – Plane Waves

are constant. Thus:,

1

x

o

y

o

jjkzx x

jjkzy y

x y

xy

E E e e

E E e e

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Unpolarized Waves

• An em wave can be unpolarized. For example sunlight or lamp light. Other terminology: randomly polarized, incoherent. A wave containing many linearly polarized waves with the polarization randomly oriented in space.

• A wave can also be partially polarized; such as sky light or light reflected from the surface of an object; i.e. glare.

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Poynting Vector

• As we have seen, a uniform plane wave carries em power.

• The power density is obtained from the Poynting vector.

• The direction of the Poynting vector is in the direction of wave propagation.

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Poynting Vector

* 2

*

W/m

1Re

2where denotes time average

S E H

S E H

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Example 14

Calculate the time average power density for the em wave if the electric field is given by:

ˆ jkzoE xE e

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Solution: Example 14 (1/2)

1ˆ ˆ ˆ;

1ˆ ˆ

ˆ

jkzo

jkzo

H o E o z

z x E e

Eye

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Solution: Example 14 (2/2)

*

2

2

1Re

2

1ˆ ˆRe

2

1ˆ

2

o

o

S E H

Ex y

z E

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Plane Waves in Lossy Media

• Finite conductivity, results in loss

• Ohm’s Law applies:

cJ E

Conductivity, Siemens/mConduction current

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Complex Permittivity

; where s c

s

s

s

H J jw E J J J

J E jw E

J jw j Ew

J jw E

jw

From Ampere’s Law in phasor form:

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Wave Equation for Lossy Media

2 2 0E w E

Attenuation constantPhase constant

12

1

k w

w

k j

jw

Wave number:Loss tangent,

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Example 15 (1/2)

ˆ ;

ˆ ;

ˆ

ˆ ˆ

jkzo

jkz jo

z j zo

j zz j z zo o

E xE e k w j

EH y e e

E xE e e

E EH y e e y e e

Plane wave propagation in lossy media:complex number

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Example 15 (2/2)

ˆ( ; ) cos( )

ˆ( ; ) cos( )zo

zoz t xE wt z

Ez t y w z

e

e t

E

H

Plane wave is traveling along +z-direction and dissipating as it moves.

attenuation propagation

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Attenuation and Skin Depth

Attenuation coefficient, , depends on the conductivity, permittivity and frequency.

12

1

k j

k w jw

Skin depth, is a measure of how far em wave can penetrate a lossy medium

1

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Lossy Media

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Example 16

• Calculate the attenuation rate and skin depth of earth for a uniform plane wave of 10 MHz. Assume the following properties for earth: = o

= 4o

= 10-4

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Solution: Example 16

4 92

7

4

10 36 104.5 10 1

2 10 4

120 30 10 0.00942 2 4 4

1106.1 m

o

o

w

First we check if we can use approximate relations.

Slightly conducting

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References

• http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/waves/u10l1b.html

• Applied Electromagnetism, Liang Chi Shen, Jin Au Kong, PWS