prof. david r. jackson dept. of ece fall 2013 notes 21 ece 6340 intermediate em waves 1
TRANSCRIPT
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Prof. David R. JacksonDept. of ECE
Fall 2013
Notes 21
ECE 6340 Intermediate EM Waves
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Reflection from Slab
0
0
0
0
0
0
ˆ
ˆ
ˆ
yz
yz
yz
jk yi jk z
jk yr jk z
jk yt jk z
E x E e e
E x E e e
E x E T e e
0
0 0
sin
cos
y i
z i
k k
k k
Notes:
1T (2) The origin is the reference plane for T.
TEz
y
z
qi
Ei
r d
T
(1)
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Reflection from Slab (cont.)
Three methods:
1) Plane-wave bounce method (interface reflections)
2) Steady-state wave representation
3) Transverse equivalent network (TEN)
Find the reflection coefficient .
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Method #1
Define interface plane-wave reflection and transmission coefficients:
,r r
0
0
1
2
2T
1T
Plane-wave bounce method (interface reflections)
1
1 1sin sini n
1 r rn
5
01 001
01 00
TE TE
TE TE
Z Z
Z Z
,r r
0
0
1
2
2T
1T
1 1
2 2
1
1
T
T
2 1
Interface Reflections (cont.)
00 012
00 01
TE TE
TE TE
Z Z
Z Z
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Plane-Wave Bounce Diagram
1 0
1 0
2 1 0
22 1 0
22 2 1 0
:
:
:
:
:
j
j
j
j
A T E
B T E e
C T E e
D T E e
E T T E e
1z y z yk d k
r
0
0
1 0E 21 2 2 0
jTT E e
z
D
0E 3 41 2 2 0
jTT E e
d
B
A
C
D
q1
Ey
1 1 1coszk k
1tand
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Bounce Diagram (cont.)
At z = 0:
( 2 )21 0 1 2 2 0
( 4 )3 41 2 2 0 ......
y y
y
jk y jk yr jx
jk yj
E E e TT E e e
TT E e e
So we have
2 430 1 1 2 2 1 2 2 ......y z z
jk y j jrxE E e TT e TT e
2( 2 ) 2 22 z yy y y y zj pjk y p jk y j p jk y j pj pe e e e e e e
Note that (for p an integer)
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Bounce Diagram (cont.)
Hence
2 431 1 2 2 1 2 2
2 421 1 2 1 1 2 1 2
2 421 1 2 1 2 2
2 2 42 41 1 2 2 2
2 221 1 2 2
0
......
......
(1 ......)
1 1 ......
1
z z
z z
z z
z z z
z z
j j
j j
j j
j j j
j jn n
n
TT e TT e
TT e TT e
TT e TT e
TT e e e
TT e e
2 1 Recall :
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Bounce Diagram (cont.)
Hence
0
1, 1,
1n
n
z zz
21 1 2 22
1
11
1z
z
jj
TT ee
2 221 1 2 1
0
1 z znj j
n
TT e e
Next, use
or
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Method # 2
1
2
3
0 0
1 1
0
0 0
0
yz z
yz z
yz
jk yjk z jk zx
jk yjk z jk zx
jk yjk zx
E E e E e e
E Ae Be e
E T E e e
4 unknowns: , T, A, B
Steady-State Wave Representation
4 equations: Ex and Hy must match at both interfaces.
#1
#2
#3
Three regionsy
zSteady-state waves
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Method # 3
Transverse Equivalent Network (TEN)
,
,
y
y
jk y
x
jk y
y
E x z V z e
H x z I z e
E0
z
E0
G
E0 Td
00TEZ01
TEZ00TEZ
I
V+
-
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TEN (cont.)
E0
z
E0
G
d
00TEZ01
TEZ00TEZ
E0
z
E0
G
E0 Td
00TEZ01
TEZ00TEZ
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00 01 101
01 00 1
tan( )
tan( )
TE TETE TE zL TE TE
z
Z jZ k dZ Z
Z jZ k d
Equivalent circuit:
TELZ
00TEZ
E0
z
E0
G
d
00TEZ01
TEZ00TEZ
TEN (cont.)
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We then have
00
00
TE TELTE TEL
Z Z
Z Z
E0
E0
G
TELZ00
TEZ
TEN (cont.)
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1 1
1 1
2 2
2
z z
z z
jk z d jk z d
jk z d jk z d
V z Ae A e
A e e
Region 2:
Find the transmission coefficient T.
E0
z
E0
G
E0 Td
00TEZ01
TEZ00TEZ
#1 #2 #3
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TEN (cont.)
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1 1
2 2z zjk z d jk z dV z A e e
At z = 0: 1 12 1 2 00 0 1z zjk d jk dV V A e e E
Hence 1 1
0
2
1z zjk d jk d
A Ee e
E0
z
E0
G
E0 Td
00TEZ01
TEZ00TEZ
#1 #2 #3
2
TEN (cont.)
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We then have 2 21V d V d A
1 1
2 2z zjk z d jk z dV z A e e
E0
z
E0
G
E0 Td
00TEZ01
TEZ00TEZ
#1 #2 #3
2
TEN (cont.)
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Also 03 0
zjk zV z E T e
0
0
1zjk dT V d e
E
Hence
TEN (cont.)
0
3zjk z dV z V d e Region 3:
E0
z
E0E0T d
00TEZ01
TEZ00TEZ
#1 #2 #3
2
21V d A
1 10
2
1z zjk d jk d
A Ee e
where
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0
1 12
2
11 z
z z
jk d
jk d jk dT e
e e
Final result:
TEN (cont.)
E0
z
E0
G
E0 Td
00TEZ01
TEZ00TEZ
#1 #2 #3
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