Plane Wave Reflection and Transmission (1)

8/17/2019 Plane Wave Reflection and Transmission (1)

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Hon Tat Hui Plane Wave Reflection and Transmission

NUS/ECE EE2011

1

Plane Wave Reflection and Transmission

1 Normal Incidence at a Conducting Boundary

zy

x

Medium 2 – perfect conductor

( )∞=σ2

Medium 1 – air or dielectric medium

( )01 =σ

z = 0

•

( )11,μ ε

r H

r E

r k̂

•

iE

iH ik̂

( ) ( )0

22

== z z HE

Js ρ s+ + +

Surface current density

Surface charge density


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( ) zii e E z

1-j

0ˆ β xE = ( ) 1-j0

1

ˆ zii

E z e

β

η =H y

Incident fields:

Transmitted fields in medium 2:( ) ( ) 022 == z z HE

Boundary conditions,

( ) ( )

( ) ( )

1 1 20

1 1 20

ˆ 0

ˆ

z

s z

z z

z z

=

=

× − =⎡ ⎤⎣ ⎦

× − =⎡ ⎤⎣ ⎦

n E E

n H H J

Reflected electric field:( ) z

r r e E z1 j

0ˆ β += xE ( ) 1 j0

1

1ˆ z

r r z E e

β

η

+= −H y

( ) ( ) ( )

( ) ( ) ( )

1

1

i r

i r

z z z

z z z

= +

= +

E E E

H H H


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( ) ( ) 000 ||2||1 == E E

Tangential

component of E1

Note that the boundary conditions are prescribed andcannot be theoretically derived or proved.

Note that as only one unknown E r 0 is to be found, weonly need the electric field boundary condition.

( ) ( )

( ) ( )

1|| 1 1 0

2|| 1 2 0

ˆ0

ˆ0

z

z

E z

E z

=

=

= ×

= ×

n E

n E

Let,

Then,


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Total electric field in medium 1:( ) ( ) ( )

z

r

z

i

r i

e E e E

z z z

11 j

0

j-

0

1

ˆˆ β β ++=

+=

xx

EEE

At z = 0, using the boundary condition,

( ) 0 00 00||1 =+⇒= r i E E E

00 ir E E −=∴

( ) ( ) ( )

( )( ) z E j

ee E

z z z

i

z z

i

r i

10

j j-

0

1

sin2ˆ

ˆ 11

β

β β

x

x

EEE

−=

−=

+=+

Total electric field:


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Reflected magnetic field :

( ) ( ) ( )

( ) ( )

zi

z

i

r r

e E

e E

z z

1

1

j0

1

j

0

1

1

ˆ1

ˆˆ1

ˆ1

β

β

η

η

η

+

+

=

−×−=

×−=

y

xz

EzH

Reflected electric field :

( ) zir e E z

1 j

0ˆ β +−= xE


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( ) ( ) ( )

( )

( ) z E

ee E

e E e

E

z z z

i

z zi

z

i

zi

r i

1

1

0

j j-

1

0

j

01

j-

1

0

1

cos2ˆ

ˆ

ˆ1

ˆ

11

11

β η

η

η η

β β

β β

y

y

yy

HHH

=

+=

+=

+=

+

+

Total magnetic field in medium 1:


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Instantaneous fields:

Note that both the total electric and total magnetic fields

in medium 1 are standing waves*.

1. They are ⊥ to each other and 90º out of phase.2. The electric field vanishes at z = -nλ /2, n = 0,1,2,

3. The magnetic field vanishes at z =-(λ /4 + nλ /2).

( ) ( ) ( ) ( )t z E e zt z it j ω β ω sinsin2ˆRe, 1011 xEE ==

( ) ( ){ } ( ) ( )t z E e zt z it j ω β η ω coscos2ˆRe, 1

1

011 yHH ==

* See animation “Formation of a Pure Standing Wave” to show the making of a standing wave from

two traveling waves moving in opposite directions

By setting sin( β 1 z)=0 in E1

By setting cos( β 1 z)=0 in H1


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Total electric and magnetic fields in medium 1


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Example 1A uniform plane wave (Ei, Hi) at a frequency of 100 MHz

travels in air in the + x direction. The electric field is

polarised in the y direction. The wave impinges normally ona perfectly conducting plane at x = 0. The amplitude of the

incident electric field is 6×10-3 V/m and its initial phase is

zero.(a) Write phasor and instantaneous expressions for Ei, Hi.

(b) Write phasor and instantaneous expressions for Er , Hr .

(c) Write phasor and instantaneous expressions for E1, H1 in

air.

(d) Determine the position nearest to the conducting plane

where E1 = 0.


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Solutions(a)


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(b)

( x)


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(d) The electric field vanishes at x = -nλ /2, n = 0,1,2,….Except at the boundary

surface (n = 0), the nearest null will be at n = 1. That is, x = - λ /2 = -1.5 m.

(c)


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2 Normal Incidence at a Dielectric Boundary

r H

r E

r k̂

•

iE

iH ik̂

zy

x

Medium 2 – dielectric medium

( )02 =σ

Medium 1 – air or dielectric medium

( )01 =σ

z = 0

•

( )11,μ ε ( )22 ,μ ε

•

t E

t H t k̂


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Incident, reflected, and transmitted fields:( ) z

ii e E z1-j

0ˆ β xE = ( ) zi

i e E

z 1 j-

1

0ˆ β

η yH =

( ) zr r e E z 1 j

0ˆ β xE = ( ) zr r e E z 1

j

1

0ˆ β

η yH −=

( ) zt t

e E z 2-j

0

ˆ β xE = ( )

zt

t

e E

z 2 j-

2

0ˆ β

η yH =

Medium parameters:

222111 , μ ε ω β μ ε ω β ==

2

22

1

11 ,

ε

μ η

ε

μ η ==


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Boundary conditions:

( ) ( )00 ||2||1 E E = ( ) ( )00 ||2||1 H H =

000 t r i E E E =+2

0

1

0

1

0

η η η

t r i E E E =−The boundary conditions lead to:

Solving for E r 0

and E t 0

,

0

12

120 ir E E

η η

η η

+−

=0

12

20

2it E E

η η

η

+=

( ) ( ) ( ) ( ) ( )00 ,000 0||200||1 t r i E E E E E =+=

( ) ( ) ( )

( ) ( )

2

0||2

1

0

1

0||1

00 ,

000

η η η

t r i E H E E

H =−=


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Define:

τ Γ =+1 Note:

1≤Γ

12

12

0

0 t,coefficienReflectionη η

η η Γ

+−

==

i

r

E

E

12

2

0

0 2 t,coefficienonTransmissiη η

η τ

+==

i

t

E

E


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Transmitted

Reflected

Incident

Using Γ and τ , the field expressions in the media can beexpressed in terms of the incident field amplitude E i0:

( ) zii e E z

1-j

0ˆ β xE = ( )

zii e

E z 1 j-

1

0ˆ β

η yH =

( ) zir e E z 1 j0ˆ β Γ= xE ( ) zir e E z1 j

1

0ˆ β η Γ−=

yH

( ) z

it e E z2-j

0ˆ β

τ xE = ( ) zi

t e

E

z2 j-

2

0

ˆ

β

η

τ

yH =


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Reflected power density, S r

( ) ( ){ } ( )

2

20*

1

2

2 0

1

2

1 1ˆRe Re2 2

1

2

incident power density

ir r

i

E z z

E

η

η

⎧ ⎫⎪ ⎪= × = − Γ⎨ ⎬⎪ ⎪⎩ ⎭

= Γ

= Γ ×

E H z

reflected powerof fractiondensity powerincident

density powerreflected2==Γ

Power Density Relationship

density powerincident2

1

1

2

0 =η

i E


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( ) ( ){ }2

2 0*

2

2

2 01

2 1

2 1

2

1 1ˆRe Re

2 2

1

2


i

t t

i

E z z

E

τ η

η τ

η η

η τ

η

⎧ ⎫⎪ ⎪= × = ⎨ ⎬

⎪ ⎪⎩ ⎭

=

= ×

E H z

smitted power tranof fractiondensity powerincident

density powerdtransmitte

2

12 ==η

η τ

Transmitted power density, S t


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Since the media are lossless,

Fraction of reflected power + Fraction of transmitted power = 1

That is,

12

122 =+η

η τ Γ


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( ) ( ){ }

( ) ( ) ( ) ( )

{ }( ) ( ){ } ( ) ( ){ }

( )

*

1 1

* *

* *

2 2

20 0

1 1

2

2 1

2

1Re

2

1Re

21 1

Re Re2 2

2 2

1 incident power density

transmitted power density


i r i r

i i r r

i i

z z

z z z z

z z z z

E E η η

η τ

η

= ×

⎡ ⎤⎡ ⎤= + × +⎣ ⎦ ⎣ ⎦

= × + ×

= − Γ

= − Γ ×

=

= ×

E H

E E H H

E H E H

Total average power density in medium 1, S 1


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Total electric field in medium 1:( ) ( ) ( )

( )( ) ( )[ ]( ) ( )[ ]

( )[ ] z je E

z je E

eee E

ee E

e E e E z z z

zi

z

i

z z z

i

z z

i

z

r

z

ir i

1 j-0

1

j-

0

j- j j-

0

j j-

0

j

0

-j

01

sin2ˆ

sin21ˆ

1ˆ

ˆ

ˆˆ

1

1

111

11

11

β Γ τ

β Γ Γ

Γ Γ

Γ

β

β

β β β

β β

β β

+=

++=

−++=

+=

+=+=

x

x

x

x

xxEEE

travelling wave

(amplitude: )0i E τ

standing wave(amplitude: )

02 i E Γ

The total electric field in medium 1 has local maximum and

minimum values but does not go to zero at any location.

|E1|

z


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See animation “Standing Wave (with Different Reflection

Coefficients)” to show the making of a general standing wave from

two traveling waves with different amplitudes (A1 and A2) moving

in opposite directions.

( )

tcoefficienreflectionA2/A1/

2Aˆ1Aˆ

ˆˆwaveresultantTotal

00

j j-

j

0

-j

01

11

11

===Γ

+=

+==

ir

z z

z

r

z

i

E E

ee

e E e E z

β β

β β

xx

xxE

Note that inside the website you can change the ratio of A2/A1 (i.e.,

the reflection coefficient) to investigate the effect on the resultant

standing wave.

Computer Animation of a Standing Wave


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Total magnetic field in medium 1:( ) ( ) ( )

( )( ) ( )[ ]

( )[ ] ze E

eee E

ee

E

z z z

zi

z z zi

z zi

r i

1

j-

1

0

j- j j-

1

0

j j-

1

0

1

cos2ˆ

1ˆ

ˆ

1

111

11

β Γ τ η

Γ Γ η

Γ η

β

β β β

β β

−=

+−+=

−=

+=

y

y

y

HHH

Note: The total fields in medium 2 are the transmitted

waves and they are pure travelling waves.


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Maxima and minima of EM fields:( ) )( ) ( )

( )[ ]θ β β β θ β

β β

−Γ+=

Γ+=

Γ+=

'2 j-' j

0

' j2- j' j

01

j2-j

01

11

11

11

1ˆ

1ˆ'

1ˆ

z z

i

z z

i

z z

i

ee E

eee E z

ee E z

x

xE

xE

( ) ( )

( ) ( )

( )[ ]θ β β

β θ β

β β

η

η

η

−Γ−=

Γ−=

Γ−=

' j2-

1

' j

0

' j2- j

1

' j

01

j2

1

-j

0

1

1

1

1

1

1

1

1ˆ

1ˆ'

1ˆ

z z

i

z z

i

z z

i

ee E

eee E

z

e

e E

z

y

yH

yH

θ

Γ Γ

j

e= z z −='


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( ) ( )θ β −

Γ+='2-j

01 11' z

i e E zE

( ) ( )θ β

η

−Γ−= ' j2-

1

0

111' z

ie

E zH

E1 achieves a maximum at z’ M when such that:

( ) Γ += 101 i M E ' zE

That is, when:

,2,1,0 ,2'2 1 ==− nn z M π θ β

( )1

'2-j 1 =−θ β M ze

magnitude


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real. bothareandlossless,aremediatheIf 21 η η

2 12 1

2 12 1

0, when

0, when

η η η η

η η η η

> >⎧−Γ = ⎨

< ⎧Γ = Γ ⎨

= <⎩

Therefore,

( )1 2 1

1 2 1

2 ' 2 , when2 ' 2 1 , when

M

M

z n z n

β π η η β π η η

= >⎧⎨ = +


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E1 achieves a minimum at z’m when such that:

( ) Γ −= 101 im E ' zE

Using the same analysis as for the maximum position,

( )1 2 11 2 1

2 ' 2 1 , when

2 ' 2 , whenm

m

z n

z n

β π η η

β π η η

⎧ = + >⎨ =


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Table for positions of the maxima and

minima of the EM field in medium 1

|E1

|max

|E1|min

( )( )Γ

Γ

−=

+=

1

1 :that Note

0min1

0max1

i

i

E

E

E

E

z’m z’ M z’

0


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The ratio of |E1|max to |E1|min is called the standing waveratio S :

Γ Γ

−+== 11

min1

max1

EES

1

1

+−=

S

S Γ


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Example 2

A beam of yellow light with a wavelength of 0.6 μm isnormally incident from air ( z > 0) on to a glass ( z < 0) . If

the glass surface is at the plane z = 0 and the relative

permittivity of glass is 2.25, determine:

(a) the locations of the electric field maxima in medium

1 (air),(b) the fraction of the incident power transmitted into

the glass medium.


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Solutions

(a) We determine the medium parameters,

( )

( )

8.012080

1602

2.012080

12080

Ω80

25.2

120~1

Ω 120

~

12

2

12

12

0

0

2

22

0

0

1

1

1

=+

=+

=

−=+−

=+−

=

=−==

−==

π π

π

η η

η τ

π π

π π

η η

η η Γ

π π

ε ε

μ

ε

μ η

π ε

μ

ε

μ

η

r


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Electric-field magnitude is a maximum at (with Γ < 0):

( )

m6.0with

,...2,1,0 24

'

1

11

μ λ

λ λ

=

=+= nn z M

(b) The fraction of the incident power transmitted into

the glass medium is

( ) 96%or96.02.011ely,Alternativ

96.080

1208.0

2/

2

22

2

2

12

1

2

0

2

2

02

2

2

=−=−=

===⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

Γ

η

η τ

η η τ

avi

av

ii

avi

av

P

P

E E

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Plane Wave Reflection and Transmission (1)

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Transcript of Plane Wave Reflection and Transmission (1)