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)
Plane Wave Impinging on a Dielectric Interface
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]
Plane Wave Impinging on a Dielectric Interface
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)
Plane Wave Impinging on a Dielectric Interface
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
Graphical Representation of Phase Matching Conditions
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°
Graphical Representation of Phase Matching Conditions
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]
Case I: E-field Perpendicular to Plane of Incidence
z
x
ki
HiEi
ktkr
HrEr HtEt
θ
Magnitude of Reflected and Transmitted Waves
of region 1
4-30
The reflected wave is given by
[4.13]
[4.14]
Case I: E-field Perpendicular to Plane of Incidence
z
x
ki
HiEi
ktkr
HrEr HtEt
θ
Magnitude of Reflected and Transmitted Waves
of region 1
4-31
The transmitted wave is given by
[4.15]
[4.16]
Case I: E-field Perpendicular to Plane of Incidence
z
x
ki
HiEi
ktkr
HrEr HtEt
θ
Magnitude of Reflected and Transmitted Waves
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
Case I: E-field Perpendicular to Plane of Incidence
At z=0
At z=0
z
x
ki
HiEi
ktkr
HrEr HtEt
θ
Magnitude of Reflected and Transmitted Waves
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
Magnitude of Reflected and Transmitted Waves
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
Magnitude of Reflected and Transmitted Waves
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
Magnitude of Reflected and Transmitted Waves
4-37
The incident wave is given by
[4.28]
[4.29]
Case II: E-field Parallel to Plane of Incidence
(fig. 4.9)
z
x
ki
Hi
Ei
ktkrHr
Er Ht
Et
θθt
Magnitude of Reflected and Transmitted Waves
4-38
The reflected wave is given by
[4.30]
[4.31]
Case II: E-field Parallel to Plane of Incidence
(fig. 4.9)
z
x
ki
Hi
Ei
ktkrHr
Er Ht
Et
θθt
Magnitude of Reflected and Transmitted Waves
4-39
The transmitted wave is given by
[4.32]
[4.33]
Case II: E-field Parallel to Plane of Incidence
(fig. 4.9)
z
x
ki
Hi
Ei
ktkrHr
Er Ht
Et
θθt
Magnitude of Reflected and Transmitted Waves
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]
Case II: E-field Parallel to Plane of Incidence
(fig. 4.9)
z
x
ki
Hi
Ei
ktkrHr
Er Ht
Et
θθt
Magnitude of Reflected and Transmitted Waves
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)
Normal Incidence of a Plane Wave on a Perfect Conductor
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
Normal Incidence of a Plane Wave on a Perfect Conductor
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 = −
Oblique Incidence of a Perpendicularly Polarized Plane Wave on a Perfect Conductor
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
Oblique Incidence of a Perpendicularly Polarized Plane Wave on a Perfect Conductor
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
Oblique Incidence of a Perpendicularly Polarized Plane Wave on a Perfect Conductor
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
Oblique Incidence of a Perpendicularly Polarized Plane Wave on a Perfect Conductor
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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