Topic2_BasicEMTheory

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Topic 3 - Basic EM Theory and Plane Waves

EE 542Antennas & Propagation for Wireless Communications

O. Kilic EE542

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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

• 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)

r

Permittivity of the medium (Farad/m)Relative permittivity (with respect to free space o)

r

Permeability of the medium (Henry/m)Relative permittivity (with respect to free space o)

Conductivity of the medium (Siemens/m)

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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 of Maxwell’s Equations

-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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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.

(5)

(6)

permittivity

permeability

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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=

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Examples of del Operations

• The following examples will show how to take divergence and curl of vector functions

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Example 1

ˆ sin ; sin( ) sin(2 )

Find:

a) the curl of A

b) the divergence of A

c) the gradient of

Let A y x x y

A

A

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Solution 1

ˆ ˆ ˆ ˆ sin

ˆ cos

ˆ ˆ ˆ ˆ sin 0

sin( ) sin(2 ) sin( ) sin(2 ) sin( ) sin(2 )ˆ ˆ ˆ

ˆ ˆcos( ) sin(2 ) 2sin( ) cos(2 )

A x y z y xx y z

z x

A x y z y xx y z

x y x y x yx y z

x y z

x x y y x y

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Example 2

Calculate the magnetic field for the electric field given below. Is this electric field realizable?

ˆ ˆ, , ; ) 5 cos( )

ox y z t x xy yz wtE(

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Solution

-

ˆ ˆ5 cos( )

ˆ ˆ ˆ ˆˆ 5 cos( )

5ˆ ˆ ˆ ˆcos( ){

5ˆ ˆ ˆ ˆ

5ˆ ˆˆ ˆ }

ˆ5 cos( ) -

o

o

o

o

t

x xy yz wt

x y z x xy yz wtx y z

xy zwt x x x y

x xxy z

y x y yy y

xy zz x z y

z y

x wt zt

HE =

H=

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Solution continued

5ˆcos( )

5ˆ cos( )

5ˆsin( )

o

o

o

x wt zt

xz wt

xz wt

w

H

H =

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Solution continued

t

? EH =

To be realizable, the fields must satisfy Maxwell’s equations!

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Solution Continued

5ˆsin( )

5ˆ sin( )

ˆ ˆ5 cos( )

ˆ ˆsin( ) 5

5ˆ ˆ ˆsin( ) sin( ) 5

o

o

o

o

o o

xz wt

w

y wtw

x xy yz wt

t tw wt x xy yz

y wt w wt x xy yzw

H =

E

These fields are NOT realizable. They do not form em fields.

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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.

ˆ, , ; ) ( , , )cos( )

o ox y z t e x y z wtE( E

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Time Harmonics and Phasor Notation

( ) cos( ) sin( )j wte wt j wt

Using Euler’s identity

The time harmonic fields can be written as

ˆ, , ; ) Re ( , , )

ˆRe ( , , )

o

o

j wt

o

j jwt

o

x y z t e x y z e

e x y z e e

E( E

EPhasor notation

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( , , ; ) Re ( , , )

( , , ; ) Re ( , , )

where

ˆ( , , ) ( , , )

ˆ( , , ) ( , , )

o

o

jwt

jwt

j

o

j

o

x y z t E x y z e

x y z t H x y z e

E x y z e x y z e

H x y z h x y z e

E

H

E

H

Note that the E and H vectors are now complex and are known as phasors

Phasor Form

Information on amplitude, direction and phase

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Time Harmonic Fields in Maxwell’s Equations

With the phasor notation, the time derivative in Maxwell’s equations becomes a factor of jw:

( , , ; ) Re ( , , )

( , , ; )Re ( , , )

Re ( , , )

jwt

jwt

jwt

jw

x y z t E x y z e

x y z tE x y z e

t t

jwE x y z e

t

E

E

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Maxwell’s Equations in Phasor Form (1)

( , , ; ) Re ( , , )

( , , ; ) Re ( , , )

Re ( , , )

Re ( , , )Re ( , , )

jwt

jwt

jwt

jwt

jwt

x y z t E x y z e

x y z t H x y z e

H x y z e

E x y z ejwE x y

t

z et t

E

H

H =

E

EH =

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Maxwell’s Equations in Phasor Form (2)

Re ( , , )

Re ( , , )Re ( , , )

Re ( , , ) R

(

e ( , , )

, , ) ( , , )

jwt

jwt

jwt

jwt jwt

H x y z e

E x y z ejwE x y z e

t

t

H x y z

t

H x y z e jwE

jw E

x y

z

z

x y

e

EH =

H =

E

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Phasor Form of Maxwell’s Equations (3)

Maxwell’s equations can thus 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.

-

;

0

0

t

t

HE =

EH =

D=

B=

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Examples on Phasor Form

Determine the phasor form of the following sinusoidal functions:

a) f(x,t)=(5x+3) cos(wt + 30)

b) g(x,z,t) = (3x+z) sin(wt)

c) h(y,z,t) = (2y+5)4z sin(wt + 45)

d) V(t) = 0.5 cos(kz-wt)

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Solutions

30

30

30

( , ) Re ( )

5 3 cos( 30)

5 3 Re

Re 5 3

( ) 5 3 5 3 cos30 sin30

jwt

j wt

j jwt

j

f x t F x e

x wt

x e

x e e

F x x e x j

a)

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Solutions

90

90

90

90

( , , ) (3 ) sin( )

(3 ) cos( 90)

(3 )Re

Re (3 )

Re (3 )

( , ) (3 ) (3 )

j wt

j wt

j jwt

j

g x z t x z wt

x z wt

x z e

x z e

x z e e

G x z x z e j x z

b)

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Solution

( 45)

45 ( )

45

( , , ) (2 5)4 sin( 45)

(2 5)4 cos( 45 90)

(2 5)4 cos( 45)

(2 5)4 Re

Re (2 5)4

( , ) (2 5)4

(2 5)4 cos45 sin45

j wt

j j wt

j

h y z t y z wt

y z wt

y z wt

y z e

y ze e

H y z y ze

y z j

c)

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Solution

( ) 0.5cos( )

0.5Re Re 0.5

0.5

jkz jwt jkz jwt

jkz

v t kz wt

e e e

V e

d)

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Example

• Find the phasor notation of the following vector:

( ) ( )

where

ˆ ˆ( ) 3cos( ) 4sin( ) 8 cos sin

C t E tt

E t wt wt x wt wt z

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Solution

ˆ( ) 3 sin 4 cos

ˆ8 sin( ) cos

Using

sin Re

cos Re

ˆ( ) Re 3 4

ˆRe 8 8

ˆ ˆ3 4 8 8

jwt

jwt

jwt jwt

jwt jwt

EC t w wt w wt x

tw wt w wt z

wt je

wt e

C t w je w e x

w je w e z

C wj w x wj w z

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Example

• Show that the following electric field satisfies Maxwell’s equations.

ˆ jkzoE xE e

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Solution

?

ˆ

1ˆ ˆ

ˆ

1ˆ ;

1 1ˆ

1ˆ( )

ˆ ˆ

jkzo

jkz jkzo o

jkzo

jkzo

jkzo

jkzo

jkz jkzo o

E xE e

wkB E yE e yE e

jw w w

yE e

EH B y e

EE H y e

jw jw

Ejk x e

jw

kxE e xE e

w

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

where

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Laplacian2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

ˆ ˆ ˆ( , , ) ( , , ) ( , , ) ( , , )

ˆ ˆ ˆ( , , ) ( , , ) ( , , )

ˆ

ˆ

ˆ

x y z

x y z

x x x

y y y

z z z

E x y z x E x y z y E x y z z E x y z

x E x y z y E x y z z E x y z

E E Ex

x y z

E E Ey

x y z

E E Ez

x y z

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Scalar Form of Maxwell’s Equations

22

2

ˆ ˆ ˆ( ) ( ) ( )

Ek E

x

E x f x y g x z h x

Let the electric field vary with x only.

and consider only one component of the field; i.e. f(x).

22

2

x

fk f

x

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Possible Solutions to the Scalar Wave Equation

1 2

1 1

( ) traveling wave

or

( ) cos( ) sin( ) standing wave

jkx jkxf x Ae A e

f x C kx D kx

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

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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

EThen

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 electromagnetic 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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Plane Wave Characteristics

1ˆ( , , ; ) cos

x y z t A wt kx eE

amplitudeFrequency (rad/sec)

Wave number, depends on the medium characteristics

phase

Direction of propagation

polarization

1ˆ( , , ; ) cos ;

x y z t A e wt kzE

amplitude phase

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Plane Waves in Phasor Form

1

1

0

ˆ( , , ; ) cos

ˆ( , , )

ˆ

j jkx

jkx

x y z t A wt kx e

E x y z A e e e

E e e

E

Complex amplitude Position dependence

polarization

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Example 1Assume that the E field lies along the x-axis (i.e. x-

polarized) and is traveling along the z-direction.

1

ˆ-

jkzo

o

EH E y e

jw

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

Note the I = V/R analogy in circuit theory.

EH

2 2k w

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Example 1 (2 of 4)

( , , ) ( )

( , , ) ( )

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 (3 of 4)

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

In time domain:

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Wavelength: period in spacek = 2

Example 1 (4 of 4)

w

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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 as shown below. The E field is in the y-direction. What is the direction of the H field?

x

zy

k

E

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Solution: Example 2

ˆ ˆ ˆcos sink z x

x

zy

k

The E field is along y

The direction of propagation is the unit vector

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 ˆ ˆ ˆ ˆ/ / sin cosh k y z x

E

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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

=1W = 2f =

2*3*109

o ow 60o

x

+ z-direction

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Solution 3

1ˆ( , , ; ) cos

x y z t A wt kz eE

=1W = 2f =

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 (1/2)

ˆ ˆo oE EH y y

Intrinsic impedance, I = V/R

E

k

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Solution: Example 5 (2/2)

• 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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Example 6

Sketch the motion of the tip of the vector A(t) as a function of time.

ˆ ˆA x jy

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Solution: Example 6 (1/2)

2

( ) Re

ˆ ˆ ˆ ˆRe Re

ˆ ˆcos( ) cos( )2

ˆ ˆcos( ) sin( )

jwt

j wtjwt jwt jwt

A t Ae

xe jye xe ye

x wt y wt

x wt y wt

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Solution: Example 6 (2/2)

wt = 0x

y

wt = 90o

wt = 180o

wt = 270o

The vector A(t) rotates clockwise wrt z-axis. The tip traces a circle of radius equal to unity with angular frequency w.

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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 wt

E

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 7ˆ 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 8ˆ 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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Example 9

0.5ˆ ˆ3 4 V/mj zE x j y e

Determine the polarization of this wave.

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Solution: Example 9 (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 9 (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 10

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

a)

b)

c)

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Solution: Example 10 (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 10 (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 10 (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 10 (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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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 11

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 11 (1/2)

1ˆ ;

1ˆ ˆ

ˆ

jkzo

jkzo

H z E

z x E e

Eye

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Solution: Example 11 (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 12 (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 12 (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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Field Attenuation in Lossy Medium

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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 13

• 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 13

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

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Homework Assignments

• Due 9/25/08

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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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Homework 3.2

• Find the polarization state of the following plane wave:

ˆ ˆ( ) 2 jkzE r jx j y e

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Homework 3.3

How far must a plane wave of frequency 60 GHz propagate in order for the phase of the wave to be retarded by 180o in a lossless medium with r =1 and r = 3.5?

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1. What is the direction of propagation? Ans: -z2. What is the wave number, k in terms of permittivity, o

and permeability, o?

Ans: free space

3. Determine the electric field, E.

Solution Homework 3.1

o ok w

ˆ jkzoH xH e

k

H

E

ˆ ;jkz oo o o

o

E y H e

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Solution: Homework 3.2

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

Observe that orthogonal components are in phase. Linear Polarization

x

y

z

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Solution 3.3 (1/2)Wavelength: period in spacek = 2

w

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Solution 3.3 (2/2)

8

9

2 3.5

1 3 * 10

2 3.5 2 * 60 * 10 3.5

1

400 3.5

o o

o o

kz

zk w f

f

m