1- Wave Reflection and Transmission III
- General Relation for E and H
2- Review
Transmission Line
Examples: Optical Fiber Metal Waveguide
Page 379
Examples: Coaxial Line Two-Wire Line Parallel-Plate Line
Transverse Electromagnetic (TEM)
0&0 zz HE
Non Transverse Electromagnetic Transverse Electric (TE) Transverse Magnetic (TM)
Ez = 0 H z ¹ 0
H z = 0 Ez ¹ 0
- Below 30 GHz coaxial cables are mostly used
Disadvantages of coaxial cable for high frequency power transmission:
Dielectric Losses ↑
PowerDf cableTEM-mode constraint for fabrication
Time varying electric field lines extending between the inner conductor of a coaxial cable and inside surface of the guide can excite an EM wave in the waveguide
Page 379
For transmission of high power high frequency (5-100 GHz) metal waveguides are used.
t
DJH
t
BE
B
D
0
EjJH
H
HjE
E v
~~~0
~
~~/~~
linear, isotropic, and homogeneous medium:
Time variation is a sinusoidal function:
tjezyxEetzyxE ),,(~
);,,(
HB
ED
Assumptions:
0~ v
Charge free
medium
EJ~~
jcom p lex
EjH
H
HjE
E
complex
~~0
~
~~0
~
EjJH
H
HjE
E
~~~0
~
~~0
~
complex 22
Propagation Constant
0~~
0~~
22
22
HH
EE
lossless medium
0
jcom p lex
Wavenumber
222 k0
~~0
~~
22
22
HkH
EkE
Plane Wave
Cartesian Coordinate
Propagation in z-direction 0~
0~
~
0~
~0
~
0~
~
0~
~
2
2
2
2
2
2
2
2
2
2
2
2
z
y
y
xx
z
y
y
xx
H
Hkdz
Hd
Hkdz
Hd
E
Ekdz
Ed
Ekdz
Ed
Solution:
)cos(ˆ)cos(ˆ)(~
),( yyxx
tj kztaykztaxezEetzE
)cos(ˆ)cos(ˆ)(~
),( yyxx
tj kztbykztbxezHetzH
EzH~
ˆ1~
fuk
p
1
Intrinsic Impedance
Objective: Drive expressions for E and H for the TE and TM modes
Maxwell Equations
Solution Procedure
)~
,~
(~
),~
,~
(~
),~
,~
(~
),~
,~
(~
zzyzzxzzyzzx HEHHEHHEEHEE
Wave Equations
Use of boundary condition to find a general solution
EjH
HjE~~
~~
Maxwell Equations
y
H
x
E
k
jH
x
H
y
E
k
jH
x
H
y
E
k
jE
y
H
x
E
k
jE
zz
c
y
zz
c
x
zz
c
y
zz
c
x
~~~
~~~
~~~
~~~
2
2
2
2
zyx
zyx
HzHyHxH
EzEyExE
~ˆ
~ˆ
~ˆ
~
~ˆ
~ˆ
~ˆ
~
222 kkc
Cutoff Wavenumber
...
~
,~
~
...~
,),(~),,(~
z
EEj
z
E
EeyxezyxE
k
y
xx
y
zj
xx
x
E
k
jH
y
E
k
jH
y
E
k
jE
x
E
k
jE
z
c
y
z
c
x
z
c
y
z
c
x
~~
~~
~~
~~
2
2
2
2
y
H
k
jH
x
H
k
jH
x
H
k
jE
y
H
k
jE
z
c
y
z
c
x
z
c
y
z
c
x
~~
~~
~~
~~
2
2
2
2
TE mode Ez=0 TM mode Hz=0
zjzj
zz eb
yn
a
xmEeeE
sinsin~~
00
~zH
TM mode
zj
c
y
zj
c
x
zj
c
y
zj
c
x
eb
yn
a
xmE
a
m
k
jH
eb
yn
a
xmE
b
n
k
jH
eb
yn
a
xmE
b
n
k
jE
eb
yn
a
xmE
a
m
k
jE
sincos~
cossin~
cossin~
sincos~
02
02
02
02
Each combination of the integers m and n represents a variable solution, or a mode, denoted TMnm..
Using the Boundary Conditions:
x
E
k
jH
y
E
k
jH
y
E
k
jE
x
E
k
jE
z
c
y
z
c
x
z
c
y
z
c
x
~~
~~
~~
~~
2
2
2
2
zjzj
zz eb
yn
a
xmHehH
sinsin
~~0
0~zE
TE mode
Page 385
zj
c
y
zj
c
x
zj
c
y
zj
c
x
eb
yn
a
xmH
b
n
k
jH
eb
yn
a
xmH
a
m
k
jH
eb
yn
a
xmH
a
m
k
jE
eb
yn
a
xmH
b
n
k
jE
sincos~
cossin~
cossin~
sincos~
02
02
02
02
y
H
k
jH
x
H
k
jH
x
H
k
jE
y
H
k
jE
z
c
y
z
c
x
z
c
y
z
c
x
~~
~~
~~
~~
2
2
2
2
222
2
0
d
p
b
n
a
muf
p
mnp
,...3,2,1/ pdp
TE mode: 10& pnm
TM mode: 01& pnm
Page 392
Quantization of β:
Design a cavity with f101=12.6 GHz
Assume: a=b=d, m=1, n=0, p=1, and up0=c
Hza
f2
1023 8
101
a=1.68 cm
EM Modes Applet
d
a
b
1- Wave Reflection and Transmission III
- General Relation for E and H
2- Review
Chapter 6: Example 6.2, 6.3, 6.5, 6.7 and 6.8 Chapter 7: Example 7.1 Exercise 7.3, 7.5, 7.6 and 7.9 Problem 7.26 Chapte 8:
Example 8.5 Exercise 8.7 and 8.8 Problem 8.3
Differential Form Integral Form
Gauss’s Law
Faraday’s Law
Gauss’s Law for
Magnetism
Ampere’s Law
vD
t
BE
S
QSdD
Sdt
BldE
SC
0 B
0S SdB
t
DJH
Sdt
DJldH
SC
Page 256
t
BE
Sdt
BldE
SC
tSdB
tldE
SC
Magnetic Flux
tSdB
tldEV
SCemf
Electromotive
Force
tVtVs cos)( 0
t
DJH
Sdt
DJldH
SC
PED
0
t
P
t
E
t
DJD
0Displacement
Current
Displacement Field
zjkii
i
zjkii
eE
yzE
zzH
eExzE
1
1
1
0
1
0
ˆ)(
~
ˆ)(~
ˆ)(~
Incident wave
zjktt
t
zjktt
eE
yzE
zzH
eExzE
2
2
2
0
2
0
ˆ)(
~
ˆ)(~
ˆ)(~
Transmitted wave
zjkrr
r
zjkrr
eE
yzE
zzH
eExzE
1
1
1
0
1
0
ˆ)(
~
)ˆ()(~
ˆ)(~
Reflected wave
Wave Reflection and Transmission at Normal Incident
iit
iir
EEE
EEE
00
12
20
00
12
120
2
iE0is a known quantity
Locus
)cos(ˆ)cos(ˆ)(~
),( yyxx
tj kztaykztaxezEetzE
20 0
sin2sin)2tan( 0
x
y
a
a0tan0
4/4/
22
0
0
0
Left polarization
Linear polarization
Right polarization
Find:
Ψ0 Auxiliary angle
Ellipticity angle
circularor oo 4545
0
0
0
Left circular polarization
Linear polarization
Right circular polarization
Page 275
*~~
2
1HEeSav
)/(2
~
ˆ 2
2
mWE
zSav
)/(cos2
ˆ)( 22
2
0mWe
EzzS z
c
av
90o
Page 337
view
x
z y
It is more convenient to first decompose the incident wave:
),(),(),( ||||
iiiiii HEHEHE ),(&),( ||||
rrrr HEHEdetermine Add the components
t
r
i r
.
ik
iH||
iE||
t
tk tH||
tE||
rkrH||
rE||
z
x
iE
iE||
E
E
E n
i
i r
x
k
iH
iE
t
tk
tH
tE
rk
rH
rE
z
ik
iH
iE
kHE ˆˆˆ
For normal incidence, Γ and are independent of polarization. This is not the case when .0i
t
i
it
i
i
t
it
it
i
r
E
E
E
E
cos
cos1
coscos
cos2
coscos
coscos
||
12
2
0||
0||
||
12
12
0||
0||
||
1coscos
cos2
coscos
coscos
12
2
0
0
12
12
0
0
ti
i
i
t
ti
ti
i
r
E
E
E
E
Parallel Polarization Perpendicular Polarization
The Brewster angle ΘB is defined as the incidence angle Θi at which the reflection coefficient Γ=0.
Perpendicular component Parallel component
22/1
12211
1
)/(1sin
B
1
21
|| tan
B
does not exist for nonmagnetic materials
For nonmagnetic materials, the Brewster angle exists
only for parallel polarization and depend on ε2/ε1
0.5-MHz antenna
Normally
incident plane
mVE /3000
mS
r
r
/4
1
72
d=?
Minimum Signal
Amplitude required: )/(01.0 mV
0.5-MHz antenna
Normally
incident plane
mVE /3000
mS
r
r
/4
1
72
d=?
Minimum Signal
Amplitude required: )/(01.0 mV
dit eEE 2
0
2000
Sea water is a good conductor
Use the equations for a good
conductor
dit eEE 2
0
d=7.54 m
Brewster 1
?2 Parallel polarized
plane wave
Polyester
6.21 rr
Air
)/(10 mVE
=50o
?
Parallel polarized beam
?iP
Compare the polarization states of each of the following pairs of plane waves:
Compare the polarization states of each of the following pairs of plane waves:
Find:
Ψ0 Auxiliary angle
γ Rotation angle
Χ Ellipticity angle
sin2sin)2tan( 0
x
y
a
a0tan
0
Example 7.3: Determine the polarization state of a plane wave with electric field )45sin(4ˆ)30cos(3ˆ),( oo kztykztxtzE
20 0
sin2sin)2tan( 0
x
y
a
a0tan0
4/4/
22
0
0
0
Left circular polarization
Linear polarization
Right circular polarization
Find:
Ψ0 Auxiliary angle
Ellipticity angle
Example 7.3: Determine the polarization state of a plane wave with electric field )45sin(4ˆ)30cos(3ˆ),( oo kztykztxtzE
Example 7.3: Determine the polarization state of a plane wave with electric field
)45sin(4ˆ)30cos(3ˆ),( oo kztykztxtzE
22
4/4/
auxiliary angle
)45cos(4ˆ)30cos(3ˆ),( oo kztykztxtzE
E(z, t) = x3cos(wt -kz+30o)+ y4cos(wt -kz+135o)
x
y
a
a0tan
x
y
a
a1
0 tan
20 0
o1.533
4tan 1
0
89.0105cos2.106tancos2tan)2tan( 0 oo
oo 105sin2.106sinsin2sin)2tan( 0
oo or 2.698.20 Since cosδ<0 Sign of γ the same as cosδ; see the rule on page 328
o0.34;0cos0
;0cos0
if
if
0
0
0
Left circular polarization
Linear polarization
Right circular polarization
Page 275
o0.34
Wave traveling out of the page
1r 4r
mVeyE zxji /20ˆ~ 43
90o
view
x
z y
t
rn
iiE
1r
0
4
zat
r
mVeyE zxji /20ˆ~ 43
a) The polarization of the incident wave
b) The angle of incident
)cossin(
01ˆ
~ii zxjkii eEyE
Equation (8.48a)
90o
view x
z y
t
r
E
E
E n
i
mVeyE zxji /20ˆ~ 43
c) The time-domain expressions for the reflected electric and magnetic fields.
1r 4r
1r 4r
mVeyE zxji /20ˆ~ 43
d) The time-domain expressions for the transmitted electric and magnetic fields.
mVeyE zxji /20ˆ~ 43
e) The average power density carried by the wave in the dielectric medium.
1r 4r