Electromagnetic Waves Chapter 1. WHY STUDY ?? In ancient time –Why do paper clip get attracted to rod rubbed with silk –What causes lightening –Why do

Electromagnetic Waves

Chapter 1

Chapter 1

WHY STUDY ?? WHY STUDY ?? • In ancient time

– Why do paper clip get attracted to rod rubbed with silk

– What causes lightening– Why do colors get separated when light beam

passes through a prism

• Modern days

– How we receive TV/radio signals– Why is radio signal good in some corner of the room

but not other– Why do cell phones have signal fluctuation

2

3

Chapter OutlinesChapter OutlinesChapter 1 Electromagnetic Waves

Maxwell’s EquationsAmpere-Maxwell CorrectionEM Wave Fundamental and EquationsEM Wave Propagation in Different MediaEM Wave Reflection and Transmission at

Normal or Oblique IncidenceWaveguide Field EquationsRectangular Waveguides

4

IntroductionIntroduction

In your previous experience in studying electromagnetic, you have learned about and experimented with electrostatics and magnetostatics … concentrating on static, or time invariant electromagnetic fields (EM Fields).

Henceforth, we shall examine situations where electric and magnetic fields are dynamic or time varying !!

5

Where :

• In static EM Fields, electric and magnetic fields

are independent each other, but in dynamic field

both are interdependent.

• Time varying EM Fields, represented by E(x,y,z,t)

and H(x,y,z,t) are of more practical value than

static EM Fields.

• In time varying fields, it usually due to

accelerated charges or time varying currents.

Introduction (Cont’d..)Introduction (Cont’d..)

6

In summary:

Stationary charges electrostatic fields

Steady currents magnetostatic fields

Time varying currents electromagnetic fields or waves

Introduction (Cont’d..)Introduction (Cont’d..)

Maxwell EquationsMaxwell Equations

Below is the generalized forms of Maxwell Equations:

Maxwell Equations

Point or Differential

Form

Integral Form

Gauss’s Law

Gauss’s Law for Magnetic Field

Faraday’s Law

Ampere’s CircuitalLaw

v D

0 B

t

B

E

tc

D

JH

encQd SD

0 SB d

SBLE dt

d

SDSJLH dt

dd c

Maxwell Equations (Cont’d..)Maxwell Equations (Cont’d..)

It is worthwhile to mention other equations that go hand in hand with Maxwell’s equations.

BuEF q

EJ

HB

ED

tv

J

Lorent’z Force Equation

Constitutive Relations

Current Continuity Equation

Circuit - Field relations:Circuit - Field relations:

EJc GVVR

i 1

HB iL

ED t

iLVL

dt

Hmd

VCqe

t

VCi c

c

dt

Ee

Field Relation Circuit Relation

Maxwell Equations (Cont’d..)Maxwell Equations (Cont’d..)

Displacement CurrentDisplacement Current

We recall from Ampere’s Circuital Law for static field,

cJH

EJ c

‘c’ subscript is used to identify it as a conduction current density, which related to electric field Ohm’s Law by:

But divergence of curl of a vector is identically zero,

JH 0

Displacement Current (Cont’d..)Displacement Current (Cont’d..)The current continuity equation,

tc

vJ

dc JJH

We see that the static form of Ampere’s Law is clearly invalid for time varying fields since it violates the law of current continuity.

It was resolved by Maxwell introduction which what we called displacement current density,

Where Jd is the rate of

change of the electric flux density,

td

D

J

Displacement Current (Cont’d..)Displacement Current (Cont’d..)

The insertion of Jd was one of the major contribution

of Maxwell. Without Jd term, electromagnetic wave

propagation (e.g. radio or TV waves) would be impossible.

At low frequencies, Jd is usually neglected

compared with Jc. But at radio frequencies, the two

terms are comparable.

tc

D

JH

Therefore,


By applying the divergence of curl, rearrange, integrate and apply Stoke’s Theorem, we can get the integral form of Ampere’s circuital Law:

dcc iidt

dd

SDSJLH

Do you really understand this displacement current??

Only formula and formula…????


To have clear understanding of displacement current, consider the simple capacitor circuit of figure below.

A sinusoidal voltage source is applied to the capacitor, and from circuit theory we know the voltage is related to the current by the capacitance. i(t) here is the conduction current.


Consider the loop surrounding the plane surface S1. By

static form of Ampere’s Law, the circulation of H must be equal to the current that cuts through the surface. But,

the same current must pass through S2 that passes

between the plates of capacitor.


But, there is no conduction current passes through an ideal capacitor, (where J=0, due to σ=0 for an ideal

dielectric ) flows through S2. This is contradictory in

view of the fact that the same closed path as S1 is used.

But to resolve this conflict, the current passing through S2

must be entirely a displacement current, where it needs to be included in Ampere’s Circuital Law.

122 SSS

d dIt

Qd

tdd SJSDSJLH

So we obtain the same current for either surface though it is conduction current in S1 and displacement current in S2.


The ratio of conduction current magnitude to the displacement current magnitude is called loss tangent, where it is used to measure the quality of the dielectric good dielectric will have very low loss tangent.

Other example for physical meaning:

Ji = current sourceJc= conducted current through resistorJd=displacement current through dielectric material

mi= magnetic current sourcemd=displacement magnetic current

Lossless TEM WavesLossless TEM Waves

Let’s use Maxwell’s equations to study the relationship between the electric and magnetic field components of an electromagnetic wave.

Consider an x-polarized wave propagating in the +z direction in some ideal medium characterized by µ and ε, with σ = 0.

Electromagnetic wave as x-polarized means that the E field vector is always pointing in the x or –x direction. Choose σ = 0 to make medium lossless for simplicity, as given by:

xztEtz aE cos, 0

Lossless TEM Waves (Cont’d..)Lossless TEM Waves (Cont’d..)

A plot of the equation E(z,0) = E0cos(z)ax at 10 MHz

in free space with E0 = 1 V/m.


Upon application of Maxwell equations, we would also find the magnetic field propagates in the +z direction.

But the field is always normal or perpendicular to the electric field vector the wave is said to propagate in a transverse electromagnetic wave mode, or TEM.

TEM Waves has no E field or H field components along the direction of propagation.We can apply Faraday’s Law, to

the propagating electric field equation previously.tt

HB

E

Solve the right hand side and the left hand side to get H !!


yy

zyx

ztEztEz

ztEzyx

aa

aaa

E

sincos

00cos

00

0

dtztE

d

ztEt

y

y

aH

aH

sin

sin

0

0

Where,

CztE

y aH

cos0

We could see that the time varying E is the only source of H, if no conduction current given that can also generate H Thus C must be zero.

Plot of the equation

H(z,0) = (E0/) cos(–

z)ay at 10 MHz in free

space with

E0 = 1 V/m along with

the lighter plot of

E(z,0).

The amplitudes of E

and H are related by

Maxwell equations.



We can apply Ampere’s Circuital Law, to

the propagating magnetic field equation previously.tc

DJH

By considering lossless characteristic, taking the curl of H, equating the equation and then integrate it, we could get :

pu

And propagation velocity relation:

1

puto get

24

Example 1Example 1

Suppose in free space that:

E(z,t) = 5.0 e-2zt ax V/m.

Is the wave lossless?

Find H(z,t).

Solution to Example 1Solution to Example 1

Since the wave has an attenuation term (e-2zt) it is clearly not lossless.

To find H,

2 2

2

5 10

5 0 0

x y z

zt zto y y

zt

e tex y z zte

a a aH

E a a

2 210 10, =zt zt

y yo o

td e dt te dt

H a H a

Therefore,

Solution to Example 1 (Cont’d..)Solution to Example 1 (Cont’d..)

udv uv vdu 2 and .ztu t dv e dt

This integral is solved by parts

where we let

We arrive at:

2 22

10 10

2 4zt zt

yo o

t Ae e

z z m

H a

EM Wave Fundamental & equations

EM Wave Fundamental & equations In free space, the constitutive parameters are σ = 0, µr = 1, εr = 1, so the Ampere’s Law and Faraday’s

Law equations become :

tt

H

E B

E 0

ttc

E

H D

JH 0

If there is some point in space a source of time varying E field, a H field is induced in the surrounding region.

As this H field also changing with time, it in turn induces an E field.

Energy is pass back and forth between E and H fields as they radiate away from the source at the speed of light.

EM Wave Fundamental & equations (Cont’d..)EM Wave Fundamental & equations (Cont’d..)The EM waves radiates spherically, but at a remote distance away from the source they resemble uniform plane wave.

In a uniform plane wave, the E and H fields are orthogonal, or transverse to the direction of propagation (to propagate in TEM mode ).

We will briefly review some of the fundamental features of waves before employing them in the study of electromagnetic.

Consider time harmonic waves, represented by sine waves, rather than transient waves (pulses or step functions), generally as:

xz zteEtz aE cos, 0

The E field is a function of position (z) and time (t). It is always pointing in +x or –x direction x-polarized wave.

EM Wave Fundamental & equations (Cont’d..)EM Wave Fundamental & equations (Cont’d..)


Where,

zeE

E

0

0

fT

12

f 2

Initial amplitude at z = 0

exponential terms attenuation

Angular frequency

Phase constant

Phase shift

And important relations:

2

fdt

dzu p


E(0,t) = Exax = E0cos(t)ax. E(0,t) =Exax = E0cos(t + )ax.

E(z, 0) = E0cos(–z)ax. E(z, 0) = E0e–zcos (–z)ax.


Use Maxwell’s equations to derive formulas governing EM wave propagation.

Consider that the medium is free of any charge, where:

And linear, isotropic, homogeneous, and time invariant (simple media), whereby the Maxwell’s equation can be rewritten as :

0D

0H , 0E

HE ,

EEH

tt


Take curl of both sides of Faraday’s Law,

Consider position derivative acting on a time derivative in a homogeneous material,

Exchange the Faraday’s Law for the curl of H

Invoking a vector identity,

t

HE

HE

t

2

2

tttt

EEE

EE

AAA 2


So now we have :

2

22

tt

EE

EE

2

22

tt

EE

E

But, medium is charge free, so divergence of E is zero,

Helmholtz Wave Equation for E.

This can be broken up into three vector equations ( Ex, Ey and Ez)

35

Task 1 Task 1

By starting with Maxwell’s equations for

simple and charge free media, derive the

Helmholtz Wave Equation for magnetic

field, H


But, with interest on Helmholtz equations for time harmonic fields, the previous Helmholtz wave equation becomes :

ss jj EE 2s

s jt

EE

022 ss EE

because

Generally written in the form:

Where the propagation constant, γ, with real part is attenuation and an imaginary part is phase constant,

jjj )(


The value of α and β in terms of the material’s constitutive parameters:

112

112

2

2


The Helmholtz wave equation for time harmonic magnetic fields, H :

022 ss HH

With loss of generality, we assume that x polarized traveling in the +az direction, where Es has only an x

component with function of z, since for plane wave that the fields do not vary in transverse direction, where in this case is xy plane. Then :

xxss zE aE )(

0)()()()( 2

2

2

2

2

2

2

zE

z

zE

y

zE

x

zExs

xsxsxs

022 zxsE-

By substituting this equation into the Helmholtz wave equation for E field, it becomes:


Hence,

With first and second term is zero :

0)(0)()( 2

2

22

2

2

zE

zzE

z

zExsxs

xs


This is a scalar wave equation, a linear homogeneous differential equation. A possible solution for this equation is :

If we let, then, zsx AeE zsxzsx Ae

z

EAe

z

E 22

2,

0,022 or

So, this equation, becomes:

Which has two solutions,

zsx

zsx

zsx

zsx

eEEAeE

eEEAeE

0

0

,,,0)2(

,,,0)1(

or

or

0)(22

2

zE

zxs

zzxs eEeEzE 00)(


The general solution is linear superposition of these two:

For the represents the E field amplitude of +z

traveling wave at z=0, by reinserting the vector, multiply

by time factor apply Euler’s identity and use the

propagation constant to get :

xz zteEtz aE cos, 0

tje

0E


For the represents the E field amplitude of negative z traveling wave at z=0, and similarly by previous approach to get :

0E

xz zteEtz aE cos, 0

The general instantaneous solution is superposition of these two solutions above.

xzz

s eEeE aE 00or generally as :

xz

xz zteEzteEtz aaE coscos, 00


Evaluating the curl of E to find:

yzz

s eEeE a-E 00

yzz

s ej

Ee

j

EaH

00

We can solve for H ,

If we assume that the wave propagates along +az and Hs has

only an y component, as what we had assume for Es yyss zH aH )(

The magnetic field can be found by applying Faraday’s Law :

ss j HE


We would have been led to the expression,

yzz

s eHeH aH 00

0H

0EBy making comparison, we can find a relationship between

and , or and

0E

0H

j

H

E

0

0intrinsic impedance (in ohms)

njn e

j

j

0

0

H

E 2tan


By plugging in the equation of intrinsic impedance into

equation of magnetic field, we could get:

ynz zte

Etz aH

cos, 0

Where E and H are out of phase by θn at any instant of time

due to the complex intrinsic impedance of the medium. Thus

at any time, E leads H or H lags E by θn . The ratio of

magnitude of conduction current density to displacement

current density in a lossy medium is:

tanjj s

s

d

c

E

E

J

Jloss tangent

EM Wave Fundamental and equations (Cont’d..)EM Wave Fundamental and equations (Cont’d..)

Tan δ is used to determine how lossy a medium is, i.e.

• Good (lossless or perfect) dielectric if

• Good conductor if

,1tan

, 1tan


Basically, we can use Fleming’s Left Hand Rule to

determine the E, H and propagation direction ;

Propagation direction (first finger)

H direction (second finger)

E direction (thumb)


By knowing the EM wave’s direction of propagation,

given as unit vector ap, is the same as the cross

product of Es with unit vector, aE and Hs with unit

vector aH :

SPS

SPS

HaE

EaH

1

And also, a pair of simple formulas can be derived:

SS

SSP HE

HEa

HE

EH

HE

aaa

aaa

aaa

P

P

PIn addition to properties of cross product,


Representation of waves; in (a), the wave travels in the ap

= +az direction and has Es = E0+ e–z ax and Hs = (E0

+/)e–z

ay. In (b), the wave travels in the ap = –az direction and has

Es = E0–ez ax along with Hs = –(E0

–/)ez ay.

50

Example 2 Example 2

Suppose in free space :

H(x,t) = 100 cos(2π x 107t – βx + π/4) az

mA/m. Find E(x,t).

Solution to Example 2 Solution to Example 2

0.100 , , 4

120 0.100 12

j x js z P x

j x j j x js P s x z y

e e

e e e e

H a a a

E a H a a a

We could find:

12 cos yt x E a

So then,

2 22 30 rad m

c f

7 212 cos 2 10

30 4 y

Vx t x

m

E a

Since free space is stated,

and then

52

Example 3 Example 3

Suppose:

E (x,y,t) = 5 cos(π x 106t – 3.0x + 2.0y) az

V/m.

Find :

H (x,y,t).

The direction of propagation, ap


We could find:

3 25 j x j ys ze eE a

Assume nonmagnetic material and therefore have:

3 2 3 210 15j x j y j x j ys s x yj j e e j e e E H a a

3 2 3 2 3 2 3 210 152.53 3.8j x j y j x j y j x j y j x j y

s x y x yo

j je e e e e e e e

j j

H a a a a

6 6 A( , , ) 2.53cos 10 3 2 3.80cos 10 3 2

mx yx y t x t x y x t x y H a a

So that,

Solution to Example 3 (Cont’d..) Solution to Example 3 (Cont’d..)

The direction of propagation :

s sP

s s

E H

aE H

6 4 6 419 12.65j x j y j x j ys s x ye e e e E H a a

Where,

And with the exponential terms canceling in the top

and bottom of the equation for ap, we have:

yyjxj

xyjxj

p aeeaeea 4646 55.083.0

EM Wave Propagation In Different Media

EM Wave Propagation In Different Media

We will consider time harmonic field propagating in different types of media :-

Lossless, Charge-Free

Dielectrics

Conductors

• Lossless, Charge - Free

Charge free, ρv=0, medium has zero conductivity, σ=0.

This is the case where waves traveling in vacuum or free space (free of any charges). Perfect dielectric is also considered as lossless media.

EM Wave Propagation In Different Media (Cont’d..)EM Wave Propagation In Different Media (Cont’d..)

jjj )(

jjjjj 22)0(

, 0

Evaluate the propagation constant,

So,

Where,

Since , the signal does not attenuate as it travels lossless medium.

0

The propagation velocity,

1pu


Evaluate the intrinsic impedance,

j

j

0

For lossless materials, E and H are always in phase. Again,

00

0

r

r

r

r 1200

intrinsic impedance of free space

58

In a lossless, nonmagnetic material with :

εr = 16, and H = 100 cos(ωt – 10y) az mA/m.

Determine : The propagation velocity The angular frequency The instantaneous expression for the

electric field intensity.

Example 4 Example 4


883 10

0.75 1016

p

r

c x mu x

s

The propagation velocity:

The angular frequency:

8 80.75 10 10 7.5 10p

radu x x

s

From given H field :

8( , ) 100cos 7.5 10 10 z

mAy t x t y

m H a

So, the time harmonic H field is:


0.100 ,

1200.100 3

j ys z

j y j ys P s y z x

r

e

e e

H a

E a H a a aWhere,

8( , ) 9.4cos 7.5 10 10 x

Vy t x t y

m E a

Finally, the instantaneous expression for E field is:

• Dielectric

Treating a dielectric as lossless is often a good approximation, but all dielectrics are to some degree lossy finite conductivity, polarization loss etc. With finite conductivity, the E field gives rise to conduction current density results in power dissipation.

Thus, it will give a complex permittivity, complex propagation constant with attenuation constant greater than zero. The intrinsic impedance is also complex, resulting a phase difference between E and H fields.



• Conductor

In any decent conductor, the loss tangent, σ/ωε>>1 or σ>>ωε so that σ ≈ ∞, so that:

11

2,

2

f2

Therefore,

where interior

brackets becomes:


The intrinsic impedance approximated by:

j

j

j

2

1 jj

045)1(2

jej

2pu

f

2

By considering leading to:

and also..

E leads H by 450


A wave from air (free space) penetrates rapidly in a good conductor, with wavelength clearly much shorter.

In a good conductor, the large attenuation means the

penetration depth can be quite small, confining the fields

near the surface or skin, of the conductor skin depth.

A large attenuation means the fields cannot penetrate far into the conductor.

f

11

65

Example 5Example 5

In a nonmagnetic material,

E(z,t) = 10 e-200z cos(2π x 109t - 200z) ax mV/m.

Find H(z,t)


Since α = β, the media is a good metal, with µr =

1 we have:

22

9 7

200, or 10.13

1 10 4 10o

o

Sf

f mx x

45 452 28j je e

We could also find the intrinsic impedance,

Solution to Example 5 (Cont’d)Solution to Example 5 (Cont’d)

So, to calculate H,

1 1 1010 , 10z j z z j z z j z

s x s P s z x ye e e e e e

E a H a E a a a

The instantaneous expression for the magnetic field intensity.

200 9( , ) 360 cos 2 10 200 45zy

mAz t e x t z

m H a

EM Wave Reflection & Transmission at Normal IncidenceEM Wave Reflection & Transmission at Normal Incidence

What happens when a EM wave is incident on a different medium? E.g. Light wave incident with mirror, most of it gets reflected but a portion gets transmitted (rapidly attenuating in the silver backing of the mirror.Consider a plane wave that are normally incident which means the planar boundary separating the two media is perpendicular to the wave’s propagation direction.

Generally, consider a time harmonic x-polarized electric field incident from medium 1 (µr1, εr1, σr1) to

medium 2 (µr2, εr2, σr2)

EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)EM Wave Reflection & Transmission at Normal Incidence (Cont’d..)

xzii zteEtz aE 10 cos),( 1

With incident field:


We have the following sets of equations:

yzjz

tts

xzjztt

s

yzjz

rrs

xzjzrr

s

yzjz

iis

xzjzii

s

eeE

eeE

eeE

eeE

eeE

eeE

aH

aE

aH

aE

aH

aE

22

22

11

11

11

11

2

0

0

1

0

0

1

0

0

Incident Fields

Reflected Fields

Transmitted Fields

tri EEE 000 ,, The E field intensities at z=0

The boundary conditions:

212121

21

tt

tt

HHKHHa

EE

i

riir

E

EEEE

0

0

12

1200

12

120 ,

Applying these boundary conditions at z=0 to get:

i

tiit

E

EEEE

0

0

12

200

12

20

2,

2

Reflection CoefficientTransmissio

n Coefficient


Try this!!


By comparison, 1

Consider a special case when medium 1 is a perfect dielectric (lossless,σ1=0) and medium 2 is a perfect

conductor (σ2= ∞). For this case, showing that the wave is totally reflected fields in perfect conductor

must vanish, so there can be no transmitted wave, E2 = 0.

0 ,1 ,02

The totally reflected wave combines with the incident wave to form a standing wave it stands and does not

travel, it consists of two traveling waves Ei and Er of

equal amplitudes but in opposite directions.


Standing wave pattern for an incident wave in a lossless medium reflecting off a second medium at z=0 where = 0.5.

1

1

min

max

E

ESWR

74

Example 6Example 6

A uniform planar waves is normally incident

from media 1 (z < 0, σ = 0, µr = 1.0, εr = 4.0)

to media 2 (z > 0, σ = 0, µr = 8.0, εr = 2.0).

Calculate the reflection and transmission

coefficients seen by this wave.


The reflection coefficient ;

2 11 2

2 1

120 8; 60 , 120 240

24

This leads to:

240 60 30.60

240 60 5

and the transmission coefficient,

1 1.60

76

Example 7 Example 7

Suppose media 1 (z < 0) is air and media 2 (z

> 0) has εr = 16. The transmitted magnetic

field intensity is known to be:

Ht = 12 cos (ωt - β2z) ay mA/m.

Determine the instantaneous value of the

incident electric field.


We know that,

2 2

2

12t

j z j zt os y y

EmA mAe e

m m

H a a

From transmitted H field, we could find the

transmitted E field,

2t t2 o s

2

30 , so 12 , E 0.36 , and 1.13t

j zox

E mA V Ve

m m m

E a

2 1

2 1

3 21 ; , 1

5 5t i io o oE E E

Since we know the relation between transmitted E

field and incident E field,

12.83, so 2.83t

j zi ioo s x

EE e

E a

Thus, 1( , ) 2.83cos .x

Vz t t z

m E a


EM Wave Reflection & Transmission at Oblique IncidenceEM Wave Reflection & Transmission at Oblique Incidence

A uniform plane waves traveling in the ai

direction is obliquely incident from medium 1 onto medium 2.

Plane of incidence plane containing both a normal to the boundary and the incident’s wave propagation.In figure, the propagation direction is ai and the normal is az,

so the plane incidence is the x z plane. The angle of incidence, reflection and transmission is the angle that makes the field a normal to the boundary.

EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)EM Wave Reflection & Transmission at Oblique Incidence (Cont’d..)

When EM Wave in plane wave form obliquely incident on the boundary, it can be decomposed into:

Perpendicular Polarization, or transverse electric (TE) polarization The E Field is perpendicular or transverse to the plane of incidence.

Parallel Polarization, or transverse magnetic (TM) polarization The E Field is parallel to the plane of incidence, but the H Field is transverse.

We need to decompose into its TE and TM components separately, and once the reflected an the transmitted fields for each polarization determined, it can be recombined for final answer.


• TE Polarization


For TE polarization, the fields are summarized as follows:

ztxtzxj

tts

yzxjtt

s

zrxrzxj

rrs

yzxjrr

s

zixizxj

iis

yzxjii

s

tt

tt

rr

rr

ii

ii

eE

eE

eE

eE

eE

eE

aaH

aE

aaH

aE

aaH

aE

sincos

sincos

sincos

cossin

2

0

cossin0

cossin

1

0

cossin0

cossin

1

0

cossin0

2

2

1

1

1

1

Incident Fields

Reflected Fields

Transmitted Fields


By applying Snell’s Law,

ri i

t

sin

sin

2

1

iTE

i

ti

tir EEE 0012

120 coscos

coscos

We could get:

iTE

i

it

it EEE 0021

20 coscos

cos2

TETE 1

Try to solve this!!!

• TM Polarization



By similar geometric arguments, TM fields are:

yzxj

tts

ztxtzxjtt

s

yzxj

rrs

zrxrzxjrr

s

yzxj

iis

zixizxjii

s

tt

tt

rr

rr

ii

ii

eE

eE

eE

eE

eE

eE

aH

aaE

aH

aaE

aH

aaE

cossin

2

0

cossin0

cossin

1

0

cossin0

cossin

1

0

cossin0

2

2

1

1

1

1

sincos

sincos

sincos

Incident Fields

Reflected Fields

Transmitted Fields


By applying Snell’s Law and employing the boundary conditions, we could get the following expressions relating the field amplitudes:

iTM

i

ti

it EEE 0021

20 coscos

cos2

iTM

i

it

itr EEE 0012

120 coscos

coscos

t

iTMTM

cos

cos1


For TM polarizations, there exists an incidence angle

at which all of the wave is transmitted into the second

medium Brewster Angle, θi = θBA , where:

22

21

21

22

21

22

22 )(

sin

BA

2

11

1sin

r

rBA

When a randomly polarized wave such as light is incident on a material at the Brewster angle, the TM polarized portion is totally transmitted but a TE component is partially reflected.

88

Example 8 Example 8

A 100 MHz TE polarized wave with amplitude 1.0 V/m is obliquely incident from air (z < 0) onto a slab of lossless,

nonmagnetic material with εr = 25 (z > 0).

The angle of incidence is 40. Calculate:(a) the angle of transmission, (b) the reflection and transmission

coefficients,(c) the incident, reflected and transmitted for

E fields.

Solution to Example 8Solution to Example 8

(a) 6

1 28

2 100 102.09 , 10.45 .

3 10r

x rad rad

c x m c m

1

2 2

1 1 1; sin sin 40 ; 7.4

5 5t tr

(b) 1 2

120120 ; 24

25

2 1

2 1

cos cos0.732; 1 0.268

cos cosi t

TE TE TEi t


(c) For incident field:

2.09 sin 40 cos40 1.34 1.601 1j x zi j x j z

s y y

Ve e e

m

E a a

( , ) 1cos 1.34 1.60iy

Vz t t x z

m E a

Thus,

For reflected field:

0.732r io TE oE E

1.34 1.600.732r j x j zs y

Ve e

m E a

Leading to:

Thus,

( , ) 0.732cos 1.34 1.60ry

Vz t t x z

m E a


Finally for transmitted field:

0.268t io TE oE E

2 sin cos 1.35 10.40.268 0.268t tj x zt j x j zs y y

Ve e e

m E a a

To get:

Therefore,

m

Vzxttz y

r aE 4.1035.1cos268.0),(




End

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