chapter 4 reflection and transmission of...

4-1

Chapter 4 Reflection and Transmission

of WavesECE 3317

Dr. Stuart Long

www.ranamok.com

www.bridgat.com

4-2

w

n H1

H2

H3H4

^

1

2

yxl

(fig. 4.1)

Boundary Conditions

-The convention is that is the outward pointing normal at the boundary pointing into region 1

4-3

w

n H1

H2

H3

H4

^

1

2

yxl

(fig. 4.1)

Boundary Conditions

4-4Boundary Conditionsfor H Field

w

n H1

H2

H3

H4

^

1

2

y

xl(fig. 4.1)

www.e-education.psu.edu

4-5

1 2 s-ˆ ( ) × =H H Jn

The discontinuity in tangentialmagnetic field is equal to the

surface current.

w

n H1

H2

H3

H4

^

1

2

yxl

(fig. 4.1)

Boundary Conditionsfor H Field

4-6

1 2ˆ ( ) 0× =E - E n

A similar derivation for yieldsjω× = −E B∇

The tangential electric field is continuous across the boundary

(no magnetic current source)

Boundary ConditionsFor E Field

4-7

s

tan tan

The surface current density only exists on a "perfect" conductor.So if both media have finite conductivities t E H hen and are

both continuous.

Since the

We can now deduce that :

•→

•

J

field cannot exist inside a "perfect" conductorthen the tangential field is zero on the surface

(i.e field is normal to perfect conductor surface.)

→

E -E -

E -

Boundary Conditions

4-8

n̂

w

2

1

A

1n 2n-D D sρ= The discontinuity of the normal D-field is equal to ρs

Boundary ConditionsFor D Field

4-9

1n 2nB - B 0=

The normal B-field is continuous across the boundary

n̂

w

2

1

A

Boundary Conditionsfor B Field

4-10

tan tan norm norm

1t 2t1t 2t

1 2

1t 2t1t 2t

1 2

1n 2n 1 1n 2 2n

1n 2n 1 1n 2 2n

E , H , B , D

D DE E

B BH H

D D

Are all conti

E E

B B

nuo

u

H H

s

ε ε

µ µ

ε ε

µ µ

⇒

⇒

= =

= =

= =

= =

⇒

⇒

Non-Perfect Conductors

1

2

D1

B2

B1 E1

D2 E2

H1

H2 ε2μ2

ε1 μ1

Boundary ConditionsSummary

4-11

1 1

1 1

1 1

1 1

tan tan

norm norm

n s n

t s t

2 2 2

s

2

1

1

= H = B = D = 0

E 0 D 0

B 0 H 0

ˆ D E

ˆ |H | | J | |B |

E

;

; ; | J |

ss

s

ρρ ρ

ε

µ

= =

= =

• = = =

× = = =

⇒

⇒

⇒D

H J

n

n

Perfect Conductors

σ = ∞E, H, B, D=0

n H E^

Boundary ConditionsSummary

1

2

4-12

- Etan is continuous

- Htan is discontinuous by |Js|

- Bnorm is continuous

- Dnorm is discontinuous by |ρs|

- Js and ρs exist only on perfect conductors

- All fields ≡ 0 inside a perfect conductor

- D = ε E

- B = μ H

H

Js

BI

E

Boundary ConditionsConcepts

4-13

Consider a uniform plane wavepropagating in and direction, and with the electric field in

ˆˆ+ +

ˆ the

direction.

z x

y

x k

z

Surfaces of constant phase

E H

(fig. 4.5)

[4.7]

Uniform Plane Wave Propagating in an Arbitrary Direction

k

kx

kz

4-14

Once again we see that H is perpendicular to both E and k

x k

z

Surfaces of constant phase

E H

(fig. 4.5)

4-15

(fig. 4.6)

Plane Wave Impinging on a Dielectric Interface

1μ1ε1

2μ2ε2

krz

krx

ktz

ktx

kzkx

θ

θr θt

x

z

ki

kr kt

ε1= εr1 ε0 ε2= εr2 ε0

[4.13]

[4.11]

[4.15]

Where Reflection Coefficient Transmission Coeffic

ientT

R →→

4-16

Incident Wave Vector

Reflected Wave Vector

Transmitted Wave Ve t

c or

ˆ ˆ

ˆ

ˆ ˆ

ˆ x z

t x t

r

z

x r zk k

k

k

k

k= +

= −

= +t

r

ik

k x

k

x

z

z

x z1

μ1ε12

μ2ε2

krz

krx

ktz

ktx

kzkx

θ

θr θt

x

z

ki

kr kt

(fig. 4.6)


4-17

Remember that Etan is continuous at

the boundary (z=0) thus we have :

-- + rxx txj j xkj x xk kTeee R =

To be true for all values of x

kx = krx = ktxPhase matching condition

The tangential components of the three wave vectors are equal

0 00ˆ ˆ=ˆ rx r tx tz z zxjk x jk x j jk x jk kkz zzjE R T Ee eE e− +− − −− + yy y

[4.19]

[4.17]


4-18

2 21 1

2

2 21 1

2 22

0

0

0

ω µ

ω µ ε

ω µ ε

ε

∇ + =

∇ + =

∇ + =

t

i

rE

E

E

Each wave satisfies the appropriate Maxwell equations therefore the wave

equations become:

In medium 1

In medium 2

1μ1ε1

2μ2ε2

krz

krx

ktz

ktx

kzkx

θ

θr θt

x

z

kI

kr kt

(fig. 4.6)


4-19

2 21 1 1

2

2 2

21 1 1

2 2

2 2

2 22 2 2

r x rz

x

t

z

t x z

k

k

k

kk

k

k

k

k

ω µ ε

ω µ ε

ω µ ε

+ =

=+

=

=

=

+ =

θr

krx

krz

k1

[4.20a]

[4.20b]

[4.21]

θt

ktx

ktz

k2

θkx

kz

k1

4-20

From geometry and an understanding of the phase matching condition we obtain :

1

2

1 sin

sin

sin

r

x t

x

x

t

r

k

k

k

k k

k

θ

θ

θ=

=

= 11 2sins inin sr tk kk θθθ = =

θr= θ

angle of incidence is equal to angle of reflection

krz = kz

Snell’s Law (law of refraction)

k1sin θ = k2sinθt

1μ1ε1

2μ2ε2

krz

krx

ktz

ktx

kzkx

θ

θr θt

x

z

kI

kr kt

(fig. 4.6)

[4.25]

4-21Graphical Representation of Phase Matching Conditions

Radius=k1

k1

4-22

(fig 4.7b)

Radius = k1

k1>k2

θ

θr θt

kx

ki

kr

ktRadius = k2

kz

Note: wave bent away from normal

Graphical Representation of Phase Matching Conditions

For the case where k1 > k2

4-23

Radius = k1

k1>k2

θθr θt

kx

ki

kr kt Radius = k2kz

Note: wave bent away from normal


Another case where k1 > k2(θ increased)

4-24

Radius = k1

k1>k2

θc

θr θt

kx

ki

kr ktRadius = k2

θ=θc

kz

2 1 sink k θ=θt=90°


For the case where k1 > k2at the critical angle

4-25

In medium 2 wave propagates in + direction, but is attenuated in the +z direction . Non-uniform plane wave

also called a surface or effinesant wave

2 2 2 2 2 22 2 -x tz tz xk k k k k k⇒+ = =

x̂

[4.26]

θt

kx

ktz

k2

4-26

[4.27]

x 2 2 1

2

1

If ;

whe

sin

sire n

c c

c

θ θ k k k k θ

kθ

k

= = ⇒ =

=

Critical Angle Angle of incidence above which total

internal reflection occurs. It can only occur when k1>k2.

x

z

c θ

c θ

Critical Angle θc

4-27Magnitude of Reflected and Transmitted Waves

- Depends on polarization of E

- The plane of incidence is defined by the plane formed by the unit normal v vector normal to the boundary and the incident wave vector.

n̂

Case I Perpendicularly PolarizedCase II Parallel Polarized

ik

4-28

z

x

ki

HiEi

ktkr

HrErHt

Et

Case I: E-field Perpendicular to Plane of Incidence

θ

Magnitude of Reflected and Transmitted Waves

4-29

The incident wave is given by

[4.11]

[4.12]


z

x

ki

HiEi

ktkr

HrEr HtEt

θ


of region 1

4-30

The reflected wave is given by

[4.13]

[4.14]


z

x

ki

HiEi

ktkr

HrEr HtEt

θ


of region 1

4-31

The transmitted wave is given by

[4.15]

[4.16]


z

x

ki

HiEi

ktkr

HrEr HtEt

θ


of region 2

4-32

✔ Quick Review

θ

ktz

k1kx

z

x

θt

kx

ktz

k2

4-33

1tan 2 tan

1tan 2

( ) ( )0

( )

1

ta

1

0

2

n

0

1

xx xj k

i r ty y y y

i r

j k xI

I

z

tx

j k xI

I

tz II

x

z

x

x

x

T E e

T

k T

R E e

E E E E E E

H H H

E

k

e

k R

H H

R

H

ω µµ ω ωµ

− − −

= ⇒ ⇒ + =

+ =

+ =

= ⇒ ⇒ + =

−− + =NOTE: At z=0

both Etan and Htanmust be

continuous


At z=0

At z=0

z

x

ki

HiEi

ktkr

HrEr HtEt

θ


4-34

2

2

2

1

2 1

2 1

1

1

2

Using the previou

1 s 2 equations

we can find

I

t

I

z I z I

z

z

Iz t

z tzI

z tz

z

R

k R

T

k T

kT

k

k k

k kRk k

ωµ

µµ

ωµ

µ µµ µ

µ

ωµ−

=

+ =

+

+

− =

−=

+Reflection coefficient for

perpendicularly polarized wave

Transmission coefficient for perpendicularly polarized wave

[4.22]

[4.23]

Case I: E-field Perpendicular to Plane of Incidencez

x

ki

HiEi

ktkr

HrEr HtEt

θ


4-35

1 2 0

For nonmagnetic

materials

equ. 4.23

and 4.23 reduce to:

2

zI

z t

z

z

z

tzI

z t

kk k

k k

kR

kT

µ µ µ

=

−+

+

=

= =

Reflection coefficient for perpendicularly polarized wave

Transmission coefficient for perpendicularly polarized wave

Case I: E-field Perpendicular to Plane of Incidencez

x

ki

HiEi

ktkr

HrEr HtEt

θ


4-36

Case II: E-field Parallel to Plane of Incidence

(fig. 4.9)

z

x

ki

Hi

Ei

ktkr

Hr

Er Ht

Et

θθt


4-37

The incident wave is given by

[4.28]

[4.29]


(fig. 4.9)

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt


4-38

The reflected wave is given by

[4.30]

[4.31]


(fig. 4.9)

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt


4-39

The transmitted wave is given by

[4.32]

[4.33]


(fig. 4.9)

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt


4-40

2

2

1

1

2 1

2

Using Boundary Conditions as previously done for Case I we obtain:

2

z tzIIz tz

zII

z tz

k kR

k k

kT

k kε

ε

εε ε

ε

ε −

=+

=+

Reflection coefficient for parallel polarized wave

Transmission coefficient for parallel polarized wave

[4.34]

[4.35]


(fig. 4.9)

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt


4-41Conditions for No Reflection

1 2 0

Total Transmission

For non-magnetic dielectrics

Case I: For perpendicular polarized

0

0

I z tz

R

R k k

µ µ µ

⇒

=⇒

→

=

= =

1 2 1 2

Since we already know , this is only possible

if , thus we find that for total transmission

to occur both media must be the sa nome ( interfac

e at all)

x txk k

k k ε ε

=

=⇒=

z tzIz tz

k kRk k

−=

+

4-42

2 1Case II: For parallel polarized

along with the phase matching condition we

find that for total transmission to occur the

the angl

0

e of

II z tzR k kε ε= =⇒

1 2

1

inci

dence must be

tanbθεε

−=Brewster Angle or

Polarization AngleAngle of incidence at which the wave is

totally transmitted. It can exist only when incident wave is parallel polarized for non-

magnetic dielectrics .

b θ[4.36]

Conditions for No Reflection

2 1

2 1

z tzII

z tz

k kRk k

ε εε ε

−=

+

4-43

b θ

0=2.25The material is glass with The Brewster angle is 56

ε ε≈ °

(fig 4.10)

4-44Reflection from a Perfect Conductor

Oblique Incidence

Perfect Conductor

z

x

ki

HiEi

krHr

Er

θ

(fig 4.16)

4-45

Perfect conductor

z

x

ki

HiEi

krHr

Er

θ

(fig 4.16)

Reflection from a Perfect Conductor

[4.22]

[4.23]

4-46

Perfect conductor

z

x

ki

HiEi

krHr

Er

θ

(fig 4.16)

Reflection from a Perfect Conductor

[4.34]

[4.35]

4-47Normal Incidence of a Plane Wave on a Perfect Conductor

ki

Hi

Ei

kr

Hr

Er PerfectConductor

z

x

(fig 4.14a)

4-48

0

0

0

0

0

0

ˆ

ˆ

ˆ

ˆ

0

jkz

j

jk

z

z

k

jkz

E e

E eE

E e

eη

η+

−

+

−

= −

=

=

=

= =

r

r

t

i

t

i

E

E

H

E H

H

x

y

x

y

[4.44a]

Perfectconductor

z

x

kr

Hr

Er

kiHi

Ei

[4.44b]

[4.45a]

[4.45b]

Normal Incidence of a Plane Wave on a Perfect Conductor

Etotal= Ei + Er

Htotal= Hi + Hr

4-49

[4.46a]

[4.46b]

Normal Incidence of a Plane Wave on a Perfect Conductor Perfect

conductorz

x

kr

Hr

Er

kiHi

Ei

4-50

[4.47]

[4.49]

[4.48]

Normal Incidence of a Plane Wave on a Perfect Conductor Perfect

conductorz

x

kr

Hr

Er

kiHi

Ei

4-51

2 kz π= − kz π= − 0

0 2E

Efig(4.14b)


Standing-wave pattern of the E field

4-52

3 2

kz π= − 2

kz π= −

0

0

2E

η

Standing-wave pattern of the H field

fig(4.14c)

H


0

4-53

Perfect Conductor

z

x

ki

HiEi

krHr

Er

θ

Note:

where:

sin cos

ˆ ˆˆ ˆ

x zx

x

z

z

k kk

k kk

k

k

θθ

= +

=

=

=

r

i

k zk

xx - z(fig 4.16)

Oblique Incidence of a Perpendicularly Polarized Plane Wave on a Perfect Conductor

4-54

( )

0

0

0

0

ˆ

ˆ

ˆ

ˆ

0

jk x jk z

jk x j

jk x jk z

jk x jk z

k z

x z

x

x z

z

z

x

E e

E

E e

E e

ek

k

η

η

− +

−

− −

+

− −=

×=

= −

×=

=

=

i

r

y

k -

y

y

k y

r

r

t

i

t

i

E

H

H

E

E

H

[4.51a]

[4.51b]

Perfect conductor

z

x

ki

HiEi

krHr

Er

θ

(fig 4.16)

Note: 1IR = −


Has magnitude 1 in the direction of H

4-55

( ) ( )

( ){( ) }

sin0

0

sin

Total fields in medium 1

ˆ -2 sin cos

ˆ 2cos cos cos

ˆ 2 sin sin cos

jkx

jkx

jE kz e

E kz

j kz e

θ

θ

θ

θ θη

θ θ

−

−

=

=

+

y

x -

z -

E

H

Perfect conductor

z

x

ki

HiEi

krHr

Er

θ

(fig 4.16)

Oblique Incidence of a Perpendicularly Polarized Plane Wave on a Perfect

Conductor

4-56

0

0 2E

Standing-wave pattern of the Ey field

fig(4.17)

y E


kz-π/cosθ-2π/cosθ

-λ/cosθ -λ/2cosθ 0 z

4-57

0

02E cos θη

Standing-wave pattern of the Hx fieldx H


kz-π/2cosθ-3π/2cosθ

-3λ/4cosθ -λ/4cosθ 0 z

4-58

Standing-wave pattern of the Hz field

02E sin θη

z H


kz-π/cosθ-2π/cosθ

-λ/cosθ -λ/2cosθ 0 z

4-59Oblique Incidence of a Perpendicularly Polarized

Plane Wave on a Perfect Conductor

Summary

4-60Power Conservation

4-61Power Conservation

Pr

Pt

4-62Example 1

[ ]

0

100 MHz

V1m

f

E

=

= = i

E

1 1 1 0 00

8

1 8

2 2 0

1

21

0

2

2 10 233 10

2 443 3r

fk

m

k m

v

k

πω µ ε ω µ ε

π π

π πω µ ε ω µ ε ε

−

−

= = =

×= =

×

= = = ⋅ =

1 0µ µ= 2 0µ µ=

1 0ε ε= 2 04ε ε=

z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

30i rθ θ= = °

4-63

[ ]

0

100 MHz

V1m

f

E

=

= = i

E

1 2

1

2

1

sin sin

1 sin sin sin30 0.262

sin 0.26 14.48

i t

t i

t

k k

kk

θ θ

θ θ

θ − ⇒

=

= = =

= °

Example 1 Cont.

1 0µ µ= 2 0µ µ=

1 0ε ε= 2 04ε ε=

z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

30i rθ θ= = °

4-64

1 1

1 1

2 2 2 2 22 1 1 1

2 1 1 1

you can check by noting that

sin 0.5

cos 0.866

- 4 sin 1.94

cos 4 cos 2 cos14.5 1.94

x i rx tx

z i rz

tz x i

tz t t

k k k k k

k k k k

k k k k k k

k k k k k

θ

θ

θ

θ θ

= = = =

= = =

= = − =

= = = =

1 0µ µ= 2 0µ µ=

1 0ε ε= 2 04ε ε=

z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

30i rθ θ= = °

Example 1 Cont.

4-65

( )

2 1

1 1

1

2 1

12

2 1 1 1

1

Note:

0.866 1.94 0.3820.866 1.

2 0.8662

2 0.6180.866

94

1+

1.94

0

z zI

z tz z t

I

z tz z tzI

z tz z

z

I

tz

I

k kR

kk kTk k k k k k

k k k kRk k k k

T

k

R

k

µ µµ

µ

µ

µµ

= =

=

=

−=

− −

= =+

+ +

+

= −+

=

+

(-.6 0.118 382)= +

1 0µ µ= 2 0µ µ=

1 0ε ε= 2 04ε ε=

z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

30i rθ θ= = °

Example 1 Cont.

4-66

22 2 20

r1

2 20 0

0 0

2 2 20 0 0

0 0 0

2

20

i

22

1

20 0

0

0

0

2t

2

S = =

2 2

2 2 2

2

S 1 = 1

S

0.14

1

2

22

59

2

2

I I I

I I

E E

E E E

E T

E

E

E

R R

E

R

T

η η

η η η

ηη η

η

η

⇒

⇒

⇒

=

=

=

=

20

0

22 = 0.7 6 9 32I

T Eη

1 0µ µ= 2 0µ µ=

1 0ε ε= 2 04ε ε=

z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

30i rθ θ= = °

0.382

0.618

I

IT

R

=

= −

Example 1 Cont.

4-67

2 20 0

0 0

2 20 0

0 0

2 20 0

0

i

0t t

i

r rS S cos = cos

S S cos = 1 cos3

30

S S cos 0.7639 cos1

0 = 0.8660

4.48 0.7396

=0 0.1264

2 2

2 2.1459

2

2

rz

iz

tz

E E

E E

E E

η η

η η

η η

θ

θ

θ

=

=

= =

=

2 2 20 0 0

0 0i

0tr S 0.76390.1459 S 1 ; ;2 2

S 2

E E Eη η η

===

Example 1 Cont.

4-68

20

0

20

0

20

0

2

0.7396

0.1264

t

0.8660

2

ref

in 2

r

c.

n.

.

a

E

E

E

η

η

η

+

( )see p.99 for general proof

2 2 20 0 0

0 0i

0trS =S = ; ; 2 2

S 0.8660 0.7396 0.1264 2zzz

E E Eη η η

=

Example 1 Cont.

4-69

z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

Example 1 Cont.

4-70

Im

Re

Example 1 Cont.

4-71

0

1.38

z

Standing-wave pattern of the |Ey | field

0.62

-0.87 -1.73 -3.46

2y E

1y E

ytotalE

Example 1 Cont.

4-72

cos 0.866sin 0.5

i

i

θθ

=

=z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

Example 1 Cont.

4-73

0

1

1.2 η

z

Standing-wave pattern of the |Hx | field

-0.87 -1.73 -3.46

2x H 1x H

totalHx

1

0.54 η

Example 1 Cont.

4-74

0

1

0.69 η

z -0.87 -1.73 -3.46

2z H 1z H

1

0.31 η

Standing-wave pattern of the |Hz | field

totalH z

Example 1 Cont.

4-75

2cos 1.94t rθ ε =z

x

ki

HiEi

ktkr

HrEr HtEt

iθ

Example 1 Cont.

4-76

Example 2

[ ]50 Hzf =

1 0µ µ=

2 0µ µ=

1 0ε ε=

z

ki

Hi

Ei

2 081ε ε=sea

water

mho4

mσ =

airAt an Air Seawater boundary, calculate the power density in seawater as compared to that in air for a normally incident wave.

4-77Example 2 Cont.

[ ]50 Hzf =

1 0µ µ=

2 0µ µ=

1 0ε ε=

z

ki

Hi

Ei

2 081ε ε=sea

water

mho4

mσ =

air

4-78Example 2 Cont.

[ ]50 Hzf =

1 0µ µ=

2 0µ µ=

1 0ε ε=

z

ki

HiEi

2 081ε ε=sea

watermho

4m

σ =

air

4-79

z

x

ki

HiEi

kr

Hr

Er

θ

( )

0

0

sin0

0

sin0

0

ˆ

ˆ

ˆˆ 2cos cos cos

2ˆ cos

z

z

jkx

z o

jkx

E kz e

E e

θ

θ

θ θη

θη

=

=

−

=

−

×

×

= × −

=

n

z

z x

y

S

S

S

S

J = H

J = (- ) H

J (- )

J

k

H E

x

z

y

produces currents only indirection

wire grid

ŷ

Direction of Surface Currents

4-80Wave Incident on a “Good Conductor”

z

ki

ktkr

x

θtθi

z

E

4-81Example 3 [ ]300 MHz

parallel polarizedf =

( )Prob. 4.101 0µ µ= 2µ

1 0ε ε= 2 0ε ε=

4530

i r

t

θ θθ

= = °

= °

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt

4-82

[ ] 1

0

1

0

1

1

( sin c

( sin cos )

os )01 1

1

2 ( )2

01 1

1

2 ( )

0

0

0

20

ˆ ˆ( cos ) sin )

2 ˆ ˆ

ˆ ˆ( cos sin )

2 ˆ ˆ( )

( )2

2

i

i

i

ij k x k zi i

jk x z

j k x k zIIi i

jk x zII

R Hk k e

H e

e

R

H

e

k

H

k

θ

θ

θ

θ

θ θωε

η

θ θωε

η

− −

− −

− +

− +

= −

=

=

−

= − −

− −

i

i

r

r

E

E

E

E

x z

x z

x z

x z

( )Prob. 4.10 1 0µ µ= 2µ1 0ε ε= 2 0ε ε=

4530

i r

t

θ θθ

= = °

= °

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt

Example 3 Cont.

4-83

( )

2 21 2

0

( sin )2 2 02 1

2

(0.707 1.335 )0 0

ˆ ˆ( sin )

0.536 ˆ ˆ32

i xj k x k k zIIx i

jk x z

T Hk k k e

H e

θθωε

η

− + −

− +

= − −

= −

t

t

E

E

x z

x z

( )Prob. 4.10

1 0µ µ= 2µ

1 0ε ε= 2 0ε ε=

4530

i r

t

θ θθ

= = °

= °

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt

Example 3 Cont.

4-84

0

0.656

z

0

- 22kπ

0

- 2 kπ

0 0

ExH η

0.758

( )Prob. 4.10Example 3

Cont.

4-85

( )

( )

2 2 2r

21i 0

2 222

2 20 00 0

2 2

t

20 0 00 0 0

2 2 20 0 00 0 0

10

0

2

S 1 = 2 2

2 2 2

2 2

S = 2 2 2

0.005S = = 2

1 1= Re2 2

1 1 2

2

2 II I

II III

I II

I

H H

H H

T T

H

H H

R R

T

H

H

H

H

H R

η η

η η η

η η

η

η

η

η

η

∗

⇒

⇒

⇒

× =

=

=

= =

S E H

0.81 23

( )Prob. 4.10

0.0718 1.0718 ; IIIIR T ==

1 0µ µ= 2µ

1 0ε ε= 2 0ε ε=

4530

i r

t

θ θθ

= = °

= °

z

x

ki

Hi

Ei

ktkrHr

Er Ht

Et

θθt

Example 3 Cont.

4-86

2 20 00 0

2 20 00 0

2 20 00 0t

i

r

i

r

t

0.0052

S

S S cos = cos4

S S cos 0.8123 cos30 0.

5 = 0.0036

S cos = 1 cos4 2 2

2 2

2 2

5 = 0.7071

7035

r

iz

tz

z

H H

H H

H H

η ηθ

ηθ

η

η ηθ

= = =

=

=

( )Prob. 4.10

2 2 20 0 00 r 0t0i S 0.80.0052 S 1 ; ; 12S 2 2

3 2

H H Hη η η ===

Example 3 Cont.

4-87

t2 2 20 0 0

0i 0 0r; ;2 2S 0.7S 0.S 0.7071 0036

2 035z zz H H H

η η η

==

=

( )Prob. 4.10

200

200

200

2

2

0.707

0.703

0.00

1

36

in c.

5

tra

ref

.

.

2

n

H

H

H

η

η

η

+

Example 3 Cont.

4-88

ECE 3317- Chapter 4 Reflection and Transmission of Waves

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chapter 4 reflection and transmission of...

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