plane electromagnetic waves

Electromagnetics Plane Waves

Plane Electromagnetic Waves

Plane waves in lossless mediaPlane waves in lossy media

Group velocity

C-N Kuo, Winter 2004 1

Group velocityFlow of electromagnetic power

Poynting vectorIncidence at conducting boundaryIncidence at dielectric boundary

Incidence at multiple dielectric interfaces


Energy Transmission

• Energy stored in the static electric and magnetic fields.

• Electromagnetic waves carry with the electromagnetic power.

• Energy is transported through space (media) to distant receiving points by electromagnetic waves.


electromagnetic waves.

01

2

2

22 =

∂∂−∇

t

E

cE


Maxwell’s Equation

Law Circuital sAmpere'

Law sFaraday' t

-E

ceSignifican Form Integral Form alDifferenti

•∂+=•∂+=×∇

Φ−=•∂∂=×∇

∫ ∫

∫c

dsD

IdlHD

JH

dt

ddlE

B


charge Magnetic Isolated No 0 0

Law sGauss'

Law Circuital sAmpere'

=•=•∇

=•=•∇

•∂∂+=•

∂∂+=×∇

∫

∫

∫ ∫

S

S

c S

dsBB

QdsDD

dst

DIdlH

t

DJH

ρ


t

HE

∂∂−=×∇ µ

t

EH

∂∂=×∇ ε

For a source free medium, EM relations:

0=•∇ E 0=•∇ H

)( 2∂−=×∇∂−=×∇×∇ EHE µεµ


01

/1c

0

)(

)(

2

2

22

2

22

22

2

=∂∂−∇⇒=

=∂∂−∇⇒

−∇=∇−•∇∇=×∇×∇∂∂−=

∂×∇∂−=×∇×∇

t

E

cEif

t

EE

EEEE

t

E

t

HE

µε

µε

µεµ

01

01

2

2

22

2

2

22

=∂

∂−∇

=∂∂−∇

t

H

cH

t

E

cE


Time-Harmonic Fields

– Arbitrary periodic time functions can be expanded into Fourier series of harmonic sinusoidal components; and transient nonperiodic functions can be expressed as Fourier Integral

– Maxwell’s equations are linear differential equations, sinusoidal time variations of source functions of a given frequency will produce sinusoidal variations of E and H with the same frequency in the steady state. Thus, for source functions with an arbitrary time dependence, electrodynamics fields can be source functions.


arbitrary time dependence, electrodynamics fields can be source functions.

0

free) source(0

]),,(Re[),,,(

=•∇

==•∇

=×∇−=×∇

=

H

E

EjH

HjE

ezyxEtzyxE tj

ερ

ωεωµ

ω


Source-Free Fields• For free-space (simple, non-conducting source-free medium), Helmholtz’s

equation

– where free-space wavenumber

022 =+∇ EkE o

rr

022 =+∇ HkH o

rr

sec)/(1031

where 8 mcc

koo

ooo ×≅===εµ

ωεµω


• In Cartesian coordinates, each field component follows this equation. Consider the E field component in the x direction.

c ooεµ

00),,()( 2

2

2 waveplane uniform2

2

2

2

2

2

2

=+∂

∂ →=+∂∂+

∂∂+

∂∂

xox

xo Ekz

EzyxEk

zyx )(),,( zEzyxE xx =

→=o

okλπ2

oo k

πλ 2=

Consider uniform plane wave with z-dependence


Plane Waves in Lossless Media

• Solutions of wave equation are traveling waves (unique solution).

zjko

zjko

xxx

oo eEeE

zEzEzE+−−+

−+

+=

+= )()()(


• Constants are determined by B.C.

• Instantaneous fields

)cos(),( zktEtzE oox −= ++ ω

+z direction -z direction


Traveling Waves

• At successive times the curve effectively travels in the positive z direction.• For a point of a particular phase (fixed phase),

constzkt o =−ω


• Phase velocity in free space the velocity of propagation of an equi-phase front.

ckdt

dzu

ooop ====

εµω 1


Magnetic Field

• Associated magnetic field can be determined by curl E.– Consider only Ex component exists.

)ˆˆˆ(

00)(

00

ˆˆˆ

+++

+

++−=∂∂=×∇ zzyyxxo

x

zyx

HaHaHaj

zEz

aaa

E ωµ


• Intrinsic impedance for free-space.

00)(x zE

+−+−++

+ ==∂∂

−=

∂∂

−= x

o

zjko

o

ozjko

o

x

oy EeE

keE

zjz

zE

jH oo

ηωµωµωµ1

)(1)(1

)(1

)( zEzH xo

y−− −=

ηnote:

)( 377120 Ω≈≅== πεµωµη

o

o

o

oo k

(real number)


Example 8-1

• Uniform plane wave propagates in a lossless simple medium in the +z-direction.

– Write the instantaneous expression for E for any t and z.

– Write the instantaneous expression for H.


for H.– Determine the locations where Ex is

a positive maximum when t=10^-8(s)

xxEaE ˆ=r

0,1,4 === σµε rr

8

1&10at t )/(10

,100

8-4max, ===

=

− zmVE

MHzf

x


Example 8-1 (cont’)• wavenumber and wavelength.

• Expression for E field.

)/(3

44

103

102

8

8

mradc

k rro

ππεµωµεω =××===

+−== − ψωr

)(2

32m

k== πλ


)]8

1(

3

4102cos[10ˆ

)63

4102cos(10ˆ

)(68

08

10102 @ maximum

)cos(10ˆˆ

84

84

8

4

−−×=

+−×=

==⇒=+−××

+−==

−

−

−

zta

ztaE

radk

k

kztaEaE

x

x

xxx

ππ

πππ

πψψπ

ψω

r

r


Example 8-1 (cont’)

• Expression for H field.

)]8

1(

3

4102cos[

60

10ˆ

ˆˆ),(

84

−−×=

==

−

zta

EaHatzH

x

yyyy

πππ

ηr

)(60 Ω== πε

ηηr

o


• At t=10^-8 (sec) maximum of cosine function occurs at

)]8

(3

102cos[60

ˆ −−×= ztax ππ

πππ nzt 2)8

1(

3

4)10(102 48 ±=−−=×× −

0,1,2n 8

13

2

3

8

13K=±=±=⇒ λnnz


Transverse Electromagnetic Waves

• A transverse electromagnetic (TEM) wave is a wave for which the electric and magnetic field vectors lie in a plane which is transverse or perpendicular to the axis of propagation.

• Phasor expression for a general form.Er

Hr

jkzoeEzE −=rr

)(


H

22222 where),,( kkkkeeeEzyxE zyxzjkyjkxjk

ozyx ==++= −−− µεω

rr

022 =+∇ EkErr

0),,()( 22

2

2

2

2

2

=+∂∂+

∂∂+

∂∂

zyxEkzyx

r


Constant Phase Plane

• Define wavenumber vector.

nzzyyxx akkakakak ˆˆˆˆ =++=r

rrrrrrr

•−•−

zayaxaR zyx ˆˆˆ ++=r


• Constant phase plane.

Rajko

Rkjo

neEeERErrr

rrrr

•−•− == ˆ)(

constantˆ =•=• RakRk n

rrr


Characteristic of Plane-wave Solution

• In charge-free region,

• Since

0)(0 =∇•⇒=•∇ •− Rjko eEErr

zjkyjkxjkzzyyxx

zjkyjkxjkzyx

Rjk

eeekakakaj

eeez

ay

ax

ae

zyx

zyx

−−−

−−−•−

++−=

∂∂+

∂∂+

∂∂=∇

)ˆˆˆ(

) )(ˆˆˆ()(


Rjkneajk •−−= ˆ

0ˆ0)ˆ( =•⇒=−• •−no

Rjkno aEeajkE

rr E field is transverse to the direction of propagation.


TEM Waves

• Uniform plane wave is a TEM wave with E and H field perpendicular to each other.

• Express H field in terms of E field.

)(ˆ1

)(1

)( REaRERH n

rrrrrr

×=×∇−=ηωµ

Rajkon

neEaRHr

rrr

•−×= ˆ)ˆ(1

)(η


• Express E field in terms of H field.

)(ˆ)()( REaREj

RH n ×=×∇−=ηωµ on eEaRH ×= )ˆ()(

η

RkjoeHRH

rrrrr

•−=)(

)(ˆ)(1

)( RHaRHj

RE n

rrrrrr

×−=×∇= ηωε


Polarization• The polarization of a uniform plane wave describes the time-varying behavior

of the electric field intensity vector at a given point in space.

• Linear polarization.– Field is fixed in one direction (in space quadrature, but in time phase).

• Circular polarization.– Superposition of two linearly polarized waves (both in space and time quadrature).


– Superposition of two linearly polarized waves (both in space and time quadrature).

• General expression.

90by leads assume(*ˆˆ

)(ˆ)(ˆ)(0

122010

21

(z)E(z)EejEaeEa

zEazEazEjkz

yjkz

x

yx

−− −=

+=r

)2

cos(ˆ)cos(ˆ

)](ˆ)(ˆRe[),(

2010

21

πωω

ω

−−+−=

+= −

kztEakztEa

ezEazEatzE

yx

tjyx

r

time harmonic field

instantaneous field


Circular Polarization

• Observe field at z=0.

)sin(ˆ)cos(ˆ

),0(ˆ),0(ˆ),0(

2010

21

tEatEa

tEatEatE

yx

yx

ωω +=

+=r

angular velocity ω


angular velocity ω


Circular Polarization (cont’)

• Right-hand (positive circularly polarized wave)– Finger of the right hand follow the direction of the rotation of E, the thumb points to

the direction of wave propagation.– Counterclockwise direction.

jkzjkz

yx

ejEaeEa

zEazEazE−− −=

+= 21

ˆˆ

)(ˆ)(ˆ)(r


• Left-hand (negative circularly polarized wave)– Clockwise direction.

jkzy

jkzx ejEaeEa −− −= 2010 ˆˆ

jkzy

jkzx

yx

ejEaeEa

zEazEazE−− +=

+=

2010

21

ˆˆ

)(ˆ)(ˆ)(r


Linear Polarization

• If E2(z) and E1(z) are in space quadrature, but in time phase.

tEaEatE yx ωcos)ˆˆ(),0( 2010 +=r


• General expression.

• Note: the reception of transmitted signal must be coincident with the polarization.

)cos(ˆ)cos(ˆ),( 2010 kztEakztEatzE yx −+−= ωωr


Example 8-3

• Decomposition of linearly polarized plane wave into a right-hand circularly polarized wave and a left-hand circularly polarized wave of equal amplitude.

jkzyx

jkzyx

jkzx

eajaE

eajaE

eEazE

−−

−

++−=

=

)ˆˆ(2

)ˆˆ(2

ˆ)(

00

0

r


22

right-hand circular polarization left-hand circular polarization


Plane Waves in Lossy Media

• In a source-free lossy medium, the homogeneous vector Helmholtz’s equation:

022 =+∇ EkE c

ωεσµεωεεµωµεω

jjk cc +=−== 1)"'(


• Define propagation constant

ωεσµεω

εεµεω

µεωγ

jjjj

jjk cc

+=−=

==

1'

"1'

βαγ j+=

attenuation constant

phase constant


Low-Loss Dielectrics

• Low-loss dielectric is a good but imperfect insulator with a nonzero equivalent conductivity.

])'

"(

8

1

'2

"1[' 2

εε

εεµεωβαγ +−≅+= jjj

(Np/m) '2

"

εµωεα ≅


(rad/m) ])'

"(

8

11['

(Np/m) '2

2

εεµεωβ

εα

+≅

≅

)'2

"1(

')

'

"1(

'2/1

εε

εµ

εε

εµη jjc +≅−= −

complex impedance impliesE and H fields are not in timephase.

])(8

11[

1 2'

"

' εε

µεβω −≅=pu


Good Conductors

• A good conductor is a medium for which 1>>ωεσ

ωµσωµσωεσµεω

ωεσµεωγ

2

11

jj

jj

jj

+==≈+=

σµσπβα , ff ∝==


σµσπβα , ff ∝==

σα

σµπ

σωµ

εµη )1()1( j

fj

j

cc +=+=≅=

magnetic field intensity lags behind the electric field intensity by 45 deg.

µσω

βω 2≅=pu


Skin Depth

• At very high frequencies the attenuation constant tends to be very large for a good conductor. The amplitude of a wave will be attenuated by the factor

• The distance through which the amplitude of a traveling plane wave

ze α−


• The distance through which the amplitude of a traveling plane wave decreases by a factor of , or 0.368, is called the skin depth, or the depth of penetration of a conductor.

1−e

(m) 11

µσπαδ

f==

(m) 2

1

πλ

βδ == δ

1−epenetration depth

conductor interface


Skin Depth of Various Materials



Current Distribution

• At high frequencies, electric current distribution is no longer uniform due to the effect of skin depth.

• Cross-sectional view of a good conductor.


high frequency


Ionized Gases

• Gases become ionized due to high energy excitation, e.g., ionosphere.

• The electron and ion densities in the individual ionized layers of the ionosphere are essentially equal called plasmas.

• When electromagnetic waves passing through the ionosphere, the electrons


• When electromagnetic waves passing through the ionosphere, the electrons are accelerated by the electric fields more than positive ions.

• Assume the intensity of plasmas is low and ignore the collisions between the electrons and the gas atoms.


Ionized Gases

• An electron of charge –e and mass m in a time-harmonic electric field E in the x-direction at an angular frequency ω experiences a force –eE.

• Such a displacement gives rise to an electric dipole moment:

Exxx

E2

22

2

ωω

m

em

dt

dme =⇒−==−

xp e−=2Ne


• ωp is called plasma angular frequency

)( )1(

)1(

0

2

20

02

2

00

εω

ωω

ε

εωεε

m

NeE

Em

NePED

pp =−=

−=+=

)1()1(

2

1

2

2

2

02

2

0

0

2

f

f

m

Nef

ppp

pp

−=−=

==

εωω

εε

εππω


Ionized Medium

• Consider wave propagation in an ionized medium:

• At ω=ω , β=0. no propagation.

22 )/(1)/(1 ωωωµεωβ ppo cff −=−=


• At ω=ωp, β=0. no propagation.

• At ω=ωp, propagation is possible.

cc

up

p ≥−

==2)/(1 ωωβ

ω

ccdd

u pg ≤−== 2)/(1/

1 ωωωβ

2cuu gp =•


Wave Velocity

• Phase velocity is defined as the velocity of propagation of an equi-phase wavefront.• Group velocity is defined as the velocity propagation of the wave-packet envelope.

ωβ ddug /

1=


ωββω

/

1==pu


Time-Harmonic Traveling Waves

• Consider a wave packet that consists of two traveling waves having equal amplitude and slightly different angular frequencies.

)cos()cos(2

])()cos[(])()cos[(),(

ztztE

ztEztEtzE

ooo

oooooo

βωβωββωωββωω

−∆−∆=∆−−∆−+∆+−∆+=


• Phase velocity:

• Group velocity:

amplitude of slow variation with an angular frequency ∆ω

ωββωβω

∆∆=

∆∆==⇒=∆−∆

/

1.

dt

dzuconstzt g

βωω ===

kdt

dzup ωβ dd

ug /

1=


ω−β Diagram

• For plane waves in a lossless medium, phase constant β is a linear function of ω. As a consequence, phase velocity is a constant.

Fast wave region


• In general, phase constant is not a linear function of ω.

• Dispersion: waves of the component frequencies travel with different phase velocities, causing a distortion in the signal wave shape.

Slow wave region


Ionized Medium

• Consider wave propagation in an ionized medium:

• At ω=ω , β=0. no propagation.

22 )/(1)/(1 ωωωµεωβ ppo cff −=−=


• At ω=ωp, β=0. no propagation.

• At ω=ωp, propagation is possible.

cc

up

p ≥−

==2)/(1 ωωβ

ω

ccdd

u pg ≤−== 2)/(1/

1 ωωωβ

2cuu gp =•


General Relation

• General relation between the group and the phase velocities.

ωωω

ωωβ

d

du

uuud

d

d

d p

ppp2

1)( −==

ωω

d

du

u

uu

p

p

pg

−=

1


• Three possible cases:– No dispersion (independent of frequency)

– Normal dispersion

– Anomalous dispersion

0=ωd

dup

0<ωd

dup

pg uu =

pg uu <

0>ωd

dup

pg uu >


Flow of Electromagnetic Power• Relation between the rate of energy transfer and the electric and magnetic

field intensities associated with a traveling electromagnetic wave.

t

DJH

t

BE

∂∂+=×∇

∂∂−=×∇

r

rr

r

r

t

DEJE

t

BHHEEHHE

∂∂⋅−⋅−

∂∂⋅−=×∇⋅−×∇⋅=×⋅∇

r

rrr

r

rrrrrrr

)()()(


• In a simple medium,

tt ∂∂

2

2

2

)2

1(

)(

)2

1(

)(

2

1)(

EJE

Ett

EE

t

DE

Htt

HH

t

HH

t

BH

σ

εε

µµµ

=⋅∂∂=

∂∂⋅=

∂∂⋅

∂∂=

∂⋅∂=

∂∂⋅=

∂∂⋅

rr

r

r

r

r

rrr

r

r

r


Flow of Electromagnetic Power

• The integral form

222 )2

1

2

1(

)(

EHEt

t

DEJE

t

BHHE

σµε −+∂∂−=

∂∂⋅−⋅−

∂∂⋅−=×⋅∇

r

rrr

r

rrr


∫∫∫ −+∂∂−=⋅×

VVSdvEdvHE

tsdHE 222 )

2

1

2

1()( σµεr

rr

Time-rate of change of the energy stored in the electric and magnetic fields. rate of decrease of the electric and magnetic energies stored

The Ohmic power dissipated in the volume


Poynting Vector

• To be consistent with the law of conservation of energy, the LHS must equal to the power leaving the volume through the surface.

• Poynting vector is defined to represent the power flow per unit area (leaving the enclosed volume)

)(W/m 2HEΡrrr

×= in the direction normal


• Poynting’s theorem

)(W/m 2HEΡ ×=

∫∫∫ ++∂∂=⋅−

VV meSdvPdvww

tsdP σ)(r

r

in the direction normal to both E and H

this term vanishes if the region of concern is loseless

this term vanishes in a static situation


Example 8-7

• Find the Poynting vector on the surface of a long, straight conducting wire (of radius b and conductivity σ) that carries a direct current I.

• DC situation; uniform current distribution over the cross-sectional area.

surface on the ˆˆ ˆI

aHI

aJ

EI

aJ =⇒===r

r

rr


• To verify Poynting theorem.

surface on the 2

ˆˆ ˆ22 b

aHb

aEb

aJ zz πσπσπ φ=⇒===

32

2

2ˆ

b

IaHEΡ r σπ

−=×=rrr

RIb

Ibb

IdsaPsdP

S rS

22

232

2

)(22

ˆ ===⋅−=⋅− ∫∫ σππ

σπl

l

rr

r


• Where the previous equation, the resistance of a straight wire, R=l/σS, has been used. The above result affirms that the negative surface integral of the Poynting vector is exactly equal to the I2R ohmic power loss in


exactly equal to the I2R ohmic power loss in the conducting wire.


Instantaneous Power Densities• Care with nonlinear operations on phasor expressions.

)cos(||

ˆ])(Re[),(

)cos(ˆ])(Re[),(

ηαω

ω

θβωη

βω

−−==

−==

− zteE

aezHtzH

ztEaezEtzE

zoy

tj

oxtj

rr

rr

])()(Re[])(Re[])(Re[ tjtjtj ezHzEezHezE ωωωrrrr

×≠×


• Note:

• Instantaneous power density.

])()(Re[])(Re[])(Re[ tjtjtj ezHzEezHezE ωωωrrrr

×≠×

])(Re[])(Re[),(),(),( tjtj ezHezEtzHtzEtzP ωωrrrrr

×=×=


Instantaneous Power Densities• Power density vector.

)]22cos([cosˆ

)cos()cos(ˆ

])(Re[])(Re[),(

22

22

α

ηα

ωω

θβωθ

θβωβωη

−−+=

−−−=

×=

−

−

zteE

a

ztzteE

a

ezHezEtzP

zo

zoz

tjtjrrr


• On the other hand.

)]22cos([cosˆ 2ηη

α θβωθη

−−+= − zteE

a zoz

)cos(ˆ])()(Re[ 22

ηαω θβω

η−−=× − zte

EaezHzE zo

ztj

rr

not the same!


Average Power Densities

• As far as the power transmitted by an electromagnetic wave is concerned, the average value is a more significant quantity than the instantaneous value.

• Make use of this identity.

ηα θ

ηcos

2ˆ),(

1)( 2

2

0

zoz

T

av eE

adttzPT

zP −== ∫rr


• Make use of this identity.

])()()()(Re[2

1

])(Re[])(Re[),(

2* tj

tjtj

ezHzEzHzE

ezHezEtzP

ω

ωω

rrrr

rrr

×+×=

×=

]Re[2

1)(

2

1)(

2

1]Re[]Re[ *** BABABBAABA ×+×=+×+=×

)]()(Re[2

1),(

1)( *

0zHzEdttzP

TzP

T

av

rrrr

×== ∫


Bounded Region

• In practice, waves propagate in bounded regions where several media with different constitutive parameters are present.

• When an electromagnetic wave


traveling in one medium impinges on another medium with a different intrinsic impedance, it experiences a reflection.

• Two cases will be considered: – normal incidence– oblique incidence.

(lossless) (perfect conductor)


Normal Incidence

• Assume the phasors of the incident fields inside medium 1.

zjioyi

zjioxi

eE

azH

eEazE

1

1

1

ˆ)(

ˆ)(

β

β

η−

−

=

=take z=0 Eio is the magnitude of field intensity @ z=0.

)()()( zHzEzP ×=


• Inside medium 2 (PEC), both electric and magnetic fields vanish. no wave is transmitted across the boundary into the z>0 region.

• The incident wave is reflected, giving rise to a reflected wave (Er, Hr).

)()()( zHzEzP iii ×=

zjroxr eEazE 1ˆ)( β+=


Normal Incidence

• The total field in medium 1.

• Boundary condition (continuity of the tangential components) at z=0 demands that

)(ˆ)()()( 111

zjro

zjioxri eEeEazEzEzE ββ +− +=+=

iororoiox EEEEEaE −=⇒==+= 0)0()(ˆ)0( 21


• The magnetic field intensity Hr of the reflected wave is related to Er:

iororoiox EEEEEaE −=⇒==+= 0)0()(ˆ)0( 21

zEjaeeEazE ioxzjzj

iox 11 sin2ˆ)(ˆ)( 11 βββ −=−= +−

zjiy

zjry

rzrnrr

eE

aeEa

zEazEazH

11

1

00

1

11

1

)(1

)(1

)(

ββ

ηη

ηη

×=×−=

×−=×=


• No average power is associated with the total eletromagnetic wave in medium 1, since are E1(z) in H1(z) phase quadrature.

• Instantaneous electric and magnetic field intensities:

zE

azHzHzH iyri 1

1

01 cos2)()()( β

η=+=

( ) tzEaezEtzE ioxtj ωβω sinsin2ˆ)(Re),( 111 ==


( ) tzE

aezHtzH iox

tj ωβη

ω coscos2ˆ)(Re),( 11

11 ==

),(H of zero),,(E of maxi. ........0,1,2,....n

,4

1)-(2nzor ,1)-(2n z While

),(H of maxi. .......0,1,2,....n

),(E of zero ,2

-nzor ,-n z While

11

1

1

11

tztz

tz

tz

=

+=+=

=

==

λπβ

λπβ


Standing Wave



Oblique Incidence at a Plane Conducting Boundary

• Perpendicular polarization, Ei is perpendicular to the plane of incidence. (θi, angle of incidence).

izixni aaa θθ cossin +=)cossin(

0011),( iini zxj

iyRaj

iyi eEaeEazxE θθββ +−•− ==


[ ]

[ ] )cossin(

1

0

1

00

1sincos

),(1

),(

),(

ii zxjizix

i

inii

iyiyi

eaaE

zxEazxH

eEaeEazxE

θθβθθη

η+−+−=

×=

==


• For the reflected wave:

.reflection of angle theis where

cossin

r

rzrxnr aaa

θθθ −=

zxjryr

xExExE

eEazxE rr θθβ

+==

= −−

)0,()0,()0,(

:0zat

),( )cossin(0

1


• Snell’s law of reflection: the angle of reflection equals to the angle of incidence.

riri

xjr

xjiy

ri

EE

eEeEa

xExExEri

θθ

θβθβ

=−=→

=+=

+=−−

and

0)(

)0,()0,()0,(

00

sin0

sin0

1

11


)cossin(0

1),( ii zxjiyr eEazxE θθβ −−−=

[ ]

[ ] )cossin(

1

0

1

1sincos

),(1

),(

ii zxjizix

i

rnrr

eaaE

zxEazxH

θθβθθη

η−−−−=

×=

E


i

iii

xjiiy

xjzjzjiy

ri

ezEja

eeeEa

zxEzxEzxE

θβ

θβθβθβ

θβ sin10

sincoscos0

1

1

111

)cossin(2

)(

),(),(),(

−

−−

−=

−=

+=

])cossin(

sin)coscos(

cos[2),(

sin1

sin1

1

01

1

1

i

i

xji

izxj

i

ixi

ez

jaez

aE

zxH

θβ

θβ

θβθθβ

θη

−

− +

−=


• Along the z-direction, E1y and H1x maintain standing –wave patterns. No average power is propagated.

• Along the x-direction, E1y and H1z are in both time and space phase and propagate with a phase velocity.

• The propagating wave in x-direction is a nonuniform plane wave because its amplitude varies with x.

ixx

iixx

uu

θλ

βπλ

θθβω

βω

sin

2

sinsin

1

11

1

111

==

===


varies with x.• When

• E1=0 for all x since

• A transverse electric (TE) wave (E1x=0) would bounce back and forth between the conducting planes and propagate in the x-direction.

.....................3,2,1

,cos2

cos1

1

=

−==

m

mzz ii πθλπθβ

0)cossin( 1 =iz θβ


:OA2 toequalh wavelengtguide a has

waveguide-parallel in the wave travelingThe

cos2

2'

"

1

1

1

==

==

θλ

βπλ

i

bOA

OA


axis x along gpropagatin waveno ,0

sinsin

OA'2OA2

:OA2 toequalh wavelengtguide a has

i

11"

"

=

>===

θ

λθ

λθ

λii

g


H-Polarization Incidence

• Incidence wave

• Reflective wave

[ ])cossin(

1

0

)cossin(0

1

1

),(

sincos),(

ii

ii

zxjiyi

zxjizixii

eE

azxH

eaaEzxE

θθβ

θθβ

η

θθ+−

+−

=

−=


[ ])cossin(

1

0

)cossin(0

1

1

),(

sincos),(

rr

ii

zxjryr

zxjrzrxrr

eE

azxH

eaaEzxE

θθβ

θθβ

η

θθ−−

−−

−=

+=

riri

xjrr

xjii

rxix

EE

eEeE

xExEri

θθθθ θβθβ

=−=→=+

=+=

−−

and

0)cos()cos(

0)0,()0,(

:0zat

00

sin0

sin0

11


i

iii

iii

xjiiziixi

xjzjzjiiz

xjzjzjiix

ri

ezazjaE

eeeEa

eeeEa

zxEzxEzxE

θβ

θβθβθβ

θβθβθβ

θβθθβθθθ

sin110

sincoscos0

sincoscos0

1

1

111

111

)]coscos(sin)cossin(cos[2

)(sin

)(cos

),(),(),(

−

−−

−−

+−=

−−

−=

+=

ri zxHzxHzxH1 ),(),(),( +=


ixji

iy

ri

ezE

a θβθβη

sin1

1

0

1

1)coscos(2 −=


• Along the z-direction, E1x and H1y maintain standing –wave patterns. No average power is propagated.

• Along the x-direction, E1z and H1y are in both time and space phase and propagate with a phase velocity.

• The propagating wave in x-direction is a nonuniform plane wave because its amplitude ix

x

iixx

uu

θλ

βπλ

θθβω

βω

sin

2

sinsin

1

11

1

111

==

===


• The propagating wave in x-direction is a nonuniform plane wave because its amplitude varies with x.

• When

• E1x=0 for all x since

• A transverse magnetic (TM) wave (H1x=0) would bounce back and forth between the conducting planes and propagate in the x-direction.

.....................3,2,1

,cos2

cos1

1

=

−==

m

mzz ii πθλπθβ

0)cossin( 1 =iz θβ


Normal Incidence at Plane Dielectric Boundary

• Incidence Wave

• Reflective Wave

zjioyi

zjioxi

eE

azH

eEazE

1

1

ˆ)(

ˆ)(

β

β

η−

−

=

=


• Transmitted Wave

zjroy

rzr

zjroxr

eE

azE

azH

eEazE

1

1

11

ˆ)(

ˆ)(

ˆ)(

β

β

ηη−=−=

=

)0(,ˆ)(

ˆ)(

ˆ)(

2

2

22

===×=

=

−

−

zEEeE

azE

azH

eEazE

ttozjto

yt

zt

zjtoxr

β

β

ηη


• B.C. :Tangential components of the electric and magnetic field intensities must be continuous:

000)0()0()0( tritri EEEEEE =+→=+

2

000

1

)(1

)0()0()0(ηη

tritri

EEEHHH =+→=+

012

120 ir EE

ηηηηη

+−=


• Definitions:

012

20

2it EE

ηηη+

=

12

12

0

0

12

2

0

0

t coefficien Reflection

2t coefficienon Transmissi

ηηηη

ηηητ

+−==Γ

+==

i

r

i

t

E

E

E

E

τ 1=+Γ


• Γ and τ may themselves be complex in the general case. A complex Γ or τsimply means that a phase shift is introduced at the interface upon reflection or transmission:

• Example: if medium 2 is a perfect conductor, η2=0. Then, Γ =-1 and τ=0. Thus, Er0=-Ei0., and Et0=0. The incident wave will be totally reflected, and a standing wave will be produced in medium 1.

• If medium 2 is not a perfect conductor, partial reflection will result.

)()()( zEzEzE +=


wavestanding:2

wave traveling:

)]sin2([

)]()1[(

)(

)()()(

0

0

10

0

0

1

1

111

11

i

i

zjix

zjzjzjix

zjzjix

ri

ΓE

E

zjeEa

eeeEa

eeEa

zEzEzE

τβτ β

βββ

ββ

Γ+=

−Γ+Γ+=

Γ+=

+=

−

−−

−


• Γ >0, (η2 > η1):– The maximum value of E1(z) is Ei0(1+Γ) occurs when 2β1zmax=-2nπ (n=0,1,2,3…)– zmax= -nπ/ β1=-nλ1/2, n=0,1,2,3……– The minimum value of E1(z) is Ei0(1-Γ) occurs when 2β1zmin=-(2n+1)π

(n=0,1,2,3…)– zmax= -(2n+1)π/ 2β1=-(2n+1)λ1/4, n=0,1,2,3……

• Γ <0, (η2 < η1):

)1()( 11 201

zjzjix eeEazE ββ Γ+= − )1()( 11 2

1

01

zjzjiy ee

EazH ββ

ηΓ−= −


2 1

– The maximum value of E1(z) is Ei0(1-Γ) occurs when 2β1zmax=(2n+1)π(n=0,1,2,3…)

– zmax=-(2n+1)π/ 2β1=-(2n+1)λ1/4, n=0,1,2,3……– The minimum value of E1(z) is Ei0(1+Γ) occurs when 2β1zmin=-2nπ (n=0,1,2,3…)– zmax= -nπ/ β1=-nλ1/2, n=0,1,2,3……

• In a dissipationless medium, Γ is real; and H1(z) will be a minimum at locations where E1(z) is a maximum, and vice versa.

• In medium 2:zj

iyzj

ix eEazHeEazE 220

2202 )(,)( ββ

ηττ −− ==


Standing Wave Ratio

• Definition: the ratio of the maximum value to the minimum value of the electric field intensity of a standing wave.

1

1

1

min

max

−=Γ↔

Γ−Γ+

==

S

E

ES


• While the value of Γ ranges from –1 to +1, the value of S ranges from 1 to ∝.• The standing wave ratio in decibels is 20 log10 S.

– S=2→20log102=6.02 dB. Γ =1/3– SWR:2dB →S=1.26, Γ =0.015

1

1

+−=Γ↔

S

S


Normal Incidence at Multiple Dielectric Interfaces

• X-polarized incident field:

• The reflected field in medium 1:– The field reflected from the interface at z=0 as

the incident wave impinges on it

)( 11001

zjr

zjix eEeEaE ββ += −


the incident wave impinges on it– The field transmitted back to medium 1 from

medium 2 after a first reflection from the interface at z=d

– The field transmitted back into medium 1 from medium 2 after a second reflection at z=d

– And so on………– How to determine it?


• Er0 can be solved base the following procedure:– Write down the electric and magnetic field intensity vectors in all three regions– Apply the B.C. to solve the equations.

• )( 11001

zjr

zjix eEeEaE ββ += −

)(1

1100

11

zjr

zjiy eEeEaH ββ

η−= −

zjzj eEeEaE 22 )( ββ −−+ += )()( d,zAt

)0()0(

)0()0( 0,zAt

)continuouscomponent l(tangentia .

21

21

dEdE

HH

EE

sBC

=====


zjy

zjx

zjzjy

zjzjx

eEaH

eEaE

eEeEaH

eEeEaE

3

3

22

22

32

2

33

222

2

222

1

)(1

)(

β

β

ββ

ββ

η

η

−+

−+

−−+

−

=

=

−=

+=

)()(

)()( d,zAt

32

32

dHdH

dEdE

===


Wave Impedance of The Total Field

• Wave Impedance of the total field: the ratio of the total electric field intensity to the total magnetic field intensity at any plane parallel to the plane boundary.

• For instance, a z-dependent uniform plane wave as shown in the previous figure, we write, in general:

)( Total

)( Total)(

zH

zEzZ x=


• For previous case (normal incidence)

)( Total)(

zHzZ

y

=

zjzj

zjzj

zjzji

zjzji

y

x

ee

ee

eeE

eeE

zH

zEzZ

11

11

11

11

1

1

0

0

1

11

)(

)(

)(

)()(

ββ

ββ

ββ

ββ

η

η

Γ−Γ+=

Γ−

Γ+==

−

−

−

−


• Which correctly reduces to η when η2=η . In that case there is no

22

12

1211

111211

1

11 sincos

sincos

)(

)()(

11

11

ηηηη

βηβηβηβηηη ββ

ββ

+−=Γ

++=

Γ−Γ+=

−−=−

−=

−

−

ljl

ljl

ee

ee

lH

lElZ

lZ

ljlj

ljlj

y

x


• Which correctly reduces to η1 when η2=η1. In that case there is no discontinuity at z=0; hence there is no reflected wave and the total-field wave importance is the same as the intrinsic impedance of the medium.

• If the plane boundary is perfectly conducting, η2 =0 and Γ=-1, and

Which is the same as the input impedance of a transmission line of length l that has a characteristic impedance η1 and terminates in a short circuit.

ljlZ 111 tan)( βη=−


Impedance Transformation with Multiple Dielectrics

• The total field in medium 2 is the result of multiple reflections of the two boundary planes at z=0 and z=d; but it can be grouped into a wave traveling in the +z direction and another traveling in the –z direction. The wave impedance of the total field in medium 2 at the left-hand interface z=0 can be found from the right side by replacing η2 by η3 , η1 by η2 , β1 by β2 , and l by d. thus,

djdZ 2223 sincos

)0(βηβηη +=


• The effective reflection coefficient at z=0 for the incident wave in medium 1 is

• Γ0 differs from Γ only in that has been replaced by Z2(0)

djd

djdZ

2322

222322 sincos

sincos)0(

βηβηβηβηη

++=

12

12

0

0

0

00 )0(

)0(

ηη

+−=−==Γ

Z

Z

H

H

E

E

i

r

i

r


Example

• A dielectric layer of thickness d and intrinsic impedance η2 is placed between media 1 and 3 having intrinsic impedances η1 and η3 , respectively. Determine d and η2 such that no reflection occurs when a uniform plane wave in medium 1 impinges normally on the interface with medium2.

• Sol:

)sincos()sincos(

)0(or ,0,0 120

+=+==Γ=

djddjd

Zz

βηβηηβηβηηη


......3,2,1,04

)12(dor 2

)12(

0cosor

sinsin and

coscos

)sincos()sincos(

22

213

23122

2

2123

2322122232

=

+=+=→

==→=

=→+=+

n

nnd

d

dd

dd

djddjd

λπβ

βηηβηηβη

βηβηβηβηηβηβηη


• When η1= η3 , we require d=nλ2/2, n=0,1,2…. That the thickness of the

......3,2,1,02

dor

0inor

sinsin hand,other On the

22

2132

23122

2

=

==→

===→=

n

nnd

ds

dd

λπβ

βηηηβηηβη


• When η1= η3 , we require d=nλ2/2, n=0,1,2…. That the thickness of the dielectric layer be a multiple of a half-wavelength in the dielectric at the operating frequency. Such a dielectric layer is referred to as a half-wave dielectric windows. Since λ2 =up2/f=1/f√µε, where f is the operating frequency, a half-wave dielectric window is a narrow-band device.

• When η1≠ η3 . We require η2 = √ η1 η3, and d=(2n+1)λ2/4, n=0,1,2….• When media 1 and 3 are different, η2 should be the geometric mean of η1 and

η3 , and d should be an odd multiple of a quarter wavelength in the dielectric layer operating frequency in order to eliminate reflection. Under these conditions the dielectric layer acts like a quarter-wave impedance transformer.


Oblique Incidence at a Plane Dielectric Boundary

Since incidence. of plane with thely,respective

wave,ed transmittandreflected, incident,

theof phase)constant of (surface wavefronts

theofon intersecti theare BO',A'O' AO, Line


.reflection of law sSnell' iswhich

incidence, of angle the toequal is reflection of angle The

sin'sin' ,velocity

phase same with the1 mediumin propagate

wavesreflected theandincident both the

Since incidence. of plane with thely,respective

1

ir

irp OOOO'AOOA'u

θθθθ

=→

=→=


2 and 1 mediafor refraction of indices theare and where

sin

sin

sin

sin

'

'

'

'

:2 mediumIn

21

2

1

2

1

1

2

1

2

12

nn

n

n

u

u

u

u

OO

OO

AO

OB

u

AO

u

OB

p

p

i

t

p

p

i

t

pp

===

==→=

ββ

θθ

θθ


2 and 1 mediafor refraction of indices theare and where 21 nn

• Snell’s law of refraction: at an interface between two dielectric media, the ratio of the sine of the angle of refraction (transmission) in medium2 to the sine of the angle of incidence in medium1 is equal to the inverse ratio of indices of refraction n1/n2.

2

1

2

1

2

1

2

1

021

sin

sin

media, cnonmagnetiFor

:2 mediumIn

ηη

εε

εε

θθ

µµµ

====

==

n

n

r

r

i

t


Total Reflection

)(sinsin

angle. critical thecalled is )2

:reflection totalof

threshold the toingcorrespond(which angle incidence :

sin2

:Setting

2121

1

2

n

nc

t

c

ct

−− ==

=

=→=

εεθ

πθ

θεεθπθ


)(sinsin11 nc ==

εθ

1sinsin1cos

1sinsin

If

?

2

2

12

2

1

−±=−=

>=→

>

itt

it

ci

j θεεθθ

θεεθ

θθ


ixixjze

c

zx-jβR a-jβ

tttztxnt

nt

βββe

ee

H Eaaa

a

x

ttnt

θεεθ

εεα

θθ

θθ

βα

θθ

sin,1sin,

For

spatiallyvary andBoth ,cossin

r unit vecto the2, mediumIn

2

122

2

2

122

i

)cossin(

22

22

=−=→

>=

+=

−−

+•


• The positive sign has been abandoned because it would lead to the impossible result of an increasing field as z increase.

• For θi> θc an evanescent wave exists along the interface (in the x-direction), which is attenuated exponentially in medium 2 in the normal direction (z-direction). This wave is tightly bound to the interface and is called a surface wave. Obviously, it is a nonuniform plane wave. No power is transmitted into medium 2 under these conditions.


Example

• The permittivity of water at optical frequencies is 1.75ε0. It is found that an isotropic light source at a distance d under water yields an illuminated circular area of a radius 5 m. Determine d.

nw 32.175.1 water ofindex refractive The ==


mO'P

d

n

mO'P

oc

o

wc

32.42.49tan

5

tan Thus,

2.49)32.1

1(sin)

1(sin

5 area, dilluminate of radius The

11

===

===

=

−−

θ

θ


Example

• A dielectric rod or fiber of a transparent material can be used to guide light or an electromagnetic wave under the conditions of total internal reflection. Determine the minimum dielectric constant of the guiding medium so that a wave incident on one end at any angle will be confined within the rod until it emerges from the other end.


satisfied. becan quartz and glass

22least at be tomedium guiding theofconstant dielectric the

sin1 :requireswhich

1sin

11sin

1sin

,sincos2

,sinsin

1

21

11

02

11

1

1

=→

+≥

=≥−→=

≥→−=

≥

n

ir

rr

ri

ri

r

t

ctt

c

θε

εεεθ

εθ

εθ

θθθπθ

θθ


Perpendicular Polarization• Incidence wave

• Reflective wave

)cossin(

1

0

)cossin(0

1

1

)sincos(),(

),(

ii

ii

zxjizix

ii

zxjiyi

eaaE

zxH

eEazxE

θθβ

θθβ

θθη

+−

+−

+−=

=


• Transmitted wave

)cossin(

1

0

)cossin(0

1

1

)sincos(),(

),(

rr

rr

zxjrzrx

rr

zxjryr

eaaE

zxH

eEazxE

θθβ

θθβ

θθη

−−

−−

+=

=

)cossin(

2

0

)cossin(0

2

2

)sincos(),(

),(

tt

rr

zxjtztx

tt

zxjtyt

eaaE

zxH

eEazxE

θθβ

θθβ

θθη

+−

+−

+−=

=


• B.C.: tangential components of E and H be continuous at the boundary z=0

tri xjt

xjr

xji

tri

eEeEeE

EEEθβθβθβ sin

0sin

0sin

0211

)0()0()0(−−− =+→

=+

tri xjtt

xjrr

xjii

tri

eEeEeE

HHH

θβθβθβ θη

θθη

sin0

2

sin0

sin0

1

211 cos1

)coscos(1

)0()0()0(

−−− −=+−→

=+


• For all x, all three exponential factors that are functions of x must be equal (phase matching)

• Snell’s law of reflection and refraction

ηη 21

trixxx θβθβθβ sinsinsin

211==

2

1

2

1

1

2

sin

sin

n

n

u

u

p

p

i

t ===ββ

θθ

riθθ =


ttiritri

tritri

EEEEEE

HHHEEE

θη

θη

cos1

cos)(1

and

)0()0()0( and )0()0()0(

02

001

000 =−=+→

=+=+

it

ii

ii

i

r

E

Eηη

θη

θη

θηθηθηθη coscos

coscos

coscos

12

12

12

12

0

0

+

−=

+−==Γ⊥


• When θi=0, making θr= θt=0 these expressions reduce to those for normal incidence

it

iiiEθ

ηθ

ηθηθηcoscos

coscos 12120 ++

it

i

ii

i

i

t

E

E

θη

θη

θηθηθη

θητ

coscos

cos2

coscos

cos2

12

2

12

2

0

0

+=

+==⊥


• If medium 2 is a perfect conductor, η2 =0. We have

• Now, we inquire whether there is a combination of η1 , η2 and θi, which makesfor no reflection. Denoting this particular θi by θB⊥

⊥⊥ =Γ+ τ1

)0(,0,,1 000 ==−=−=Γ ⊥⊥ tir EEE τ

0=Γ⊥


for no reflection. Denoting this particular θi by θB⊥0=Γ⊥

larperpendicu of case for the reflection no of angleBrewster thecalled is angle The

)(1

1sin

sin1sin1cos

sincos

221

12212

222

212

12

⊥

⊥

⊥

−−=

−=−=→

=

B

B

itt

tB

n

n

θµµ

εµεµθ

θθθ

θηθη


• For nonmagnetic media, µ1= µ2 = µ0→ θB⊥ does not exist.• In the case of ε1= ε2, µ1≠ µ2 :

)(1

1sin

21 µµθ

+=⊥B



Parallel Polarization

• Incidence wave

• Reflective wave

)cossin(

1

0

)cossin(0

1

1

),(

)sincos(),(

ii

ii

zxjiyi

zxjizixii

eE

azxH

eaaEzxE

θθβ

θθβ

η

θθ+−

+−

=

−=


• Transmitted wave

)cossin(

1

0

)cossin(0

1

1

),(

)sincos(),(

rr

rr

zxjryr

zxjrzrxrr

eE

azxH

eaaEzxE

θθβ

θθβ

η

θθ−−

−−

−=

+=

)cossin(

2

0

)cossin(0

2

2

),(

)sincos(),(

tt

rr

zxjtyt

zxjtztxtt

eE

azxH

eaaEzxE

θθβ

θθβ

η

θθ+−

+−

=

−=


• B.C.: tangential components of E and H be continuous at the boundary z=0, also lead to snell’s law of reflection and refraction:

1

)(1

,coscos)(

)0()0()0( and )0()0()0(

02

001

000 trittiri

tritri

EEEEEE

HHHEEE

ηηθθ =−=+→

=+=+

itrE θηθη coscos 120 −==Γ


it

it

i

r

E

E

θηθηθηθη

coscos

coscos

12

12

0

0// +

−==Γ

it

i

i

t

E

E

θηθηθητcoscos

cos2

12

2

0

0// +

==

=Γ+

i

t

θθτ

cos

cos1 ////


• If medium 2 is a perfect conductor, η2 =0. We have

• Now, we inquire whether there is a combination of η , η and θ , which makes

vanish.conductor surface on the field E total theofcomponent l tangentia themaking

,0,1 //// =−=Γ τ

sunglass. Polaroid 0except allfor ii

2

//

2 →=Γ>Γ⊥ θθ


• Now, we inquire whether there is a combination of η1 , η2 and θi, which makesfor no reflection. Denoting this particular θi by θB⊥0// =Γ

parallel of case for the reflection no of angleBrewster thecalled is angle The

)(1

1sin

sincos

//

221

2112//

2

//12

B

B

Bt

θεε

εµεµθ

θηθη

−−=

=


• µ1= µ2

)(1

1sin

21

// εεθ

+=B

== −−

1

21

1

21// tantan

n

nB ε

εθ


• Because of the difference in the formulas for Brewster angles for perpendicular and parallel polarizations, it is possible to separate these two types of polarization in an unpolarized wave.

plane electromagnetic waves

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