prof. david r. jackson dept. of ece fall 2013 notes 12 ece 6340 intermediate em waves 1
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1
Prof. David R. JacksonDept. of ECE
Fall 2013
Notes 12
ECE 6340 Intermediate EM Waves
2
Wave Impedance
Assume TMz wave traveling in the +z direction
1ˆt tTM
H z EZ
ˆ ˆ ˆTMt tz E Z z z H
ˆTMt tE Z z H
ˆ TMt tz E Z H
TM z z
c
k kZ
k
From previous notes:
so
or
3
-z wave:
z zk k zz z j k zjk z jk ze e e so we can simply substitute
1ˆt tTM
H z EZ
1ˆt tTM
H z EZ
Summary:+ sign: +z wave
- sign: - z wave
Wave Impedance (cont.)
In this formula the wavenumber and the wave impedance are taken to be the same for negative and positive
traveling waves.
4
TEz wave:
TE
z z
kZ
k k
1ˆt tTE
H z EZ
Wave Impedance (cont.)
Note: 2TE TMZ Z
5
Transverse Equivalent Network (TEN)
Denote
(assume TMz wave) Also, write
0( , , ) ( , ) ( )t tE x y z E x y V z
( ) z zjk z jk zV z V e V e
0 ˆ( , ) ( , ) ( , )t TM tE x y e x y x y
ˆ( , , ) ( , ) ( , ) ( )t TM tE x y z e x y x y V z
where Note: V(z) behaves as a voltage function.
Note: We can assume that both the unit vector and the transverse amplitude function are both real (because of the corollary to the “real” theorem).
Hence we have
There is still flexibility here, since there can be any real scaling constant in the definition of t.
6
Transverse Equivalent Network (cont.)
1ˆˆ ( , )
1ˆˆ ( , )
z
z
jk zt TM tTM
jk zTM tTM
H z e x y V eZ
z e x y V eZ
1ˆt tTM
H z EZ
( ) zjk zV z V e
ˆ( , , ) ( , ) ( , ) ( )t TM tE x y z e x y x y V z
For the magnetic field we have
The result for the transverse magnetic field is then
1ˆ ˆt t tTM
H z E z EZ
Now use
7
Transverse Equivalent Network (cont.)
Define
ˆ ˆˆ( , ) ( , )TMTMh x y z e x y
ˆ( , , ) ( , ) ( , ) ( )t TM tH x y z h x y x y I z
and
Note: I(z) behaves as a transmission-line current function.
( )z zjk z jk z
TM TM
V e V eI z
Z Z
We then have
8
Transverse Equivalent Network (cont.)
Summary
ˆ ˆˆ( , ) ( , )TMTMh x y z e x y
ˆ( , , ) ( , ) ( , ) ( )t TM tH x y z h x y x y I z
ˆ( , , ) ( , ) ( , ) ( )t TM tE x y z e x y x y V z
( )z zjk z jk z
TM TM
V e V eI z
Z Z
( ) z zjk z jk zV z V e V e
9
TEN
( )
( )t
t
V z E
I z H
+z-z
zWaveguide
Z0 = ZTM or ZTE + V (z)-
I (z)
zkz = kz of waveguide mode
Transverse Equivalent Network (cont.)
10
Transverse Equivalent Network (cont.)Note on Power
*1
2TENP VI
* *
*
2*
1 1ˆ ˆ
2 21 ˆˆ ˆ( , ) ( , ) ( ) ( , ) ( , ) ( )21 ˆˆ ˆ( , )2
WGz t t
TM t TM t
TMt TM
S E H z E H z
e x y x y V z h x y x y I z z
VI x y e h z
ˆˆ ˆ ˆˆ ˆ ˆ ˆ ˆ 1TM TM TMTMe h z e z e z z z
2*1( , )
2WGz tS VI x yHence
TEN:
WG:
A B C B A C C A B Note :
11
Transverse Equivalent Network (cont.)
The total complex power flowing down the waveguide is then
2*1( , )
2WG WG
z t
S S
P S dS VI x y dS
2( , )WG TEN
t
S
P P x y dS or
If we choose2
( , ) 1t
S
x y dS
WG TENP Pthen
12
At z = 0:
Hence
0 0 0 0t t t tE E H H
(0 ) (0 ) (0 ) (0 )V V I I
Junction:
These conditions are also true at a TL junction.
TEN: Junction
(EM boundary conditions)
+
-
+
-00 0,TM
zZ k 01 1,TMzZ k(0 )V (0 )V
(0 )I (0 )I
z0 r
same cross sectional shape
z = 0
13
Example: Rectangular Waveguide
TE10 mode incident
01 00
01 00
TE TE
TE TE
Z Z
Z Z
1T
TEN:
T1
00 0,TEzZ k 01 1,TE
zZ k
b a
0
r
z
a
z = 0
x
y
#0
#1
14
0 000 2
0 20
0 0 0
2 2
0 0
/
1 1
TE
z
Zk
ka
k
k a k a
101 2
1
1
TEZ
k a
Similarly,
0 376.730313461771 [ ]
Example: Rectangular Waveguide (cont.)
15
Example (Numerical Results)X-band waveguide
f = 10 GHz
2.2225 [cm]
1.0319 [cm]
a
b
Choose 2.2 ( )r teflon
6.749 [GHz] ( )cf air - filled guide
00 01510.25 [ ] 285.18 [ ]TE TEZ Z
0.28295
0.71705T
The results are:
0 154.74 [rad/m]zk 1 276.87 [rad/m]zk
16
Example (cont.)
2 2%
%
100 100 0.28295 8.00
100 8.00 92.0
r
t
P
P
Note:2% 100tP T
This is because the impedances of the two guides are different.
%
%
8.00
92.0
r
t
P
P
(conservation of energy and orthogonality)
0 0z zP P
0 inc refz z zP P P
2ref incz zP P
0z tP P
conservation of energy
othogonality
17
2 22
% 01 012 2
2
0000
( , )2 2
100 1001
( , ) 22
t
tTE TES
t
TEtTES
V Tx y dS
Z ZP
Vx y dS ZZ
Alternative calculation:
% 92.0tP 2% 00
01
100TE
t TE
ZP T
Z
Example (cont.)
18
Example (cont.)
ˆ( , , ) ( , ) ( , ) ( )t TE tH x y z h x y x y I z
ˆ( , , ) ( , ) ( , ) ( )t TE tE x y z e x y x y V z
Fields
1ˆ( , , ) sin zjk zt
xE x y z y Te
a
1
01
1ˆ( , , ) sin zjk z
t TE
xH x y z x Te
a Z
0 0ˆ( , , ) sin z zjk z jk zt
xE x y z y e e
a
0 0
00
1ˆ( , , ) sin z zjk z jk z
t TE
xH x y z x e e
a Z
Air region:
Dielectric region:
ˆ ˆ,
ˆ ˆ,
, sin
TE
TE
t
e x y y
h x y x
xx y
a
0 0
0 0
1
1
00
01
1
1
z z
z z
z
z
jk z jk zair
jk z jk zairTE
jk zdiel
jk zdielTE
V z e e
I z e eZ
V z Te
I z TeZ
b a
0
r
z
a
z = 0
x
y
#0
#1
TL part
19
Matching Elements
A couple of commonly used matching elements:
Capacitive diaphragm
x
y
Inductive post (narrow)x
yTEN
Z0TE Z0
TE Lp
TEN
Z0TE Z0
TE Cd
00 10TE
z
Zk
20
Discontinuity
Rectangular waveguide with a post:
Top view:
Higher-order mode region
z
x
TE10TE10
Post
TE10
ap = radius of post
Assumption: only the TE10 mode can propagate.
21
Discontinuity (cont.)
TEN : Z0TE Z0
TE Lp
Note: The discontinuity is a approximately a shunt load because the tangential electric field of the dominant mode is approximately continuous, while the tangential magnetic field of the dominant mode is not (shown later).
Top view:
The TL discontinuity is chosen to give the same reflected and transmitted
TE10 waves as in the actual waveguide.
00 10
0
2
0
1
TE
z
Zk
k a
Higher-order mode region
z
x
TE10TE10
22
Discontinuity (cont.)
Top view:
Narrow strip model for post (w = 4ap)
Assume center of post is at x = x0
Flat strip model (strip of width w)
In this model (flat strip model) the equivalent circuit is exactly a shunt inductor (proof given later).
z
x
TE10TE10
w
z = 0- z = 0+
x0
23
Discontinuity (cont.)
Top view:
inc scay y yE E E
10
sin zjk zincy
xE e
a
scayE field produced by strip current
(that is, the field scattered by the strip)
z
x
TE10TE10
w
z = 0- z = 0+
x0
24
Discontinuity (cont.)
We assume a field representation in terms of TEm0 waveguide modes (therefore, there is only a y component of the electric field).
Note that TMm0 modes do not exist.
0
0
1
1
, , sin
, , sin
mz
mz
jk zscay m
m
jk zscay m
m
m xE x y z A e
a
m xE x y z A e
a
Transverse fields radiated by the post current (no y variation):
1/220 2
0mz
mk k
a
0
0
1 0
1 0
, , sin
, , sin
mz
mz
jk zsca mx TE
m m
jk zsca mx TE
m m
A m xH x y z e
Z a
A m xH x y z e
Z a
0
1ˆt tTE
m
H z EZ
By symmetry, the scattered field should have no y variation.
25
Discontinuity (cont.)
Top view:
0 0y yE z E z
From boundary conditions:
0 0sca scay yE z E z Hence
z
x
TE10TE10
w
z = 0- z = 0+
x0
10
sin zjk zincy
xE e
a
26
Discontinuity (cont.)
Equate terms of the Fourier series:
1 1
sin sinm mm m
m x m xA A
a a
m m mA A A 1 1 1A A A
10 ˆ sint
xE y V z
a
0 0V V Modeling equation: so
0 0sca scay yE z E z
10 100 0sca scay yE E
10 100 0y yE E
27
Discontinuity (cont.)
0 0V V
Z0TE Z0
TE jXp
This establishes that the circuit model for the flat strip must be a parallel element.
p pX L
28
Discontinuity (cont.)
0 0x x syH z H z J x
00
1sinm m syTE
mm
m xA A J x
Z a
From the Fourier series for the magnetic field, we then have:
Magnetic field:
0
sinsy mm
m xJ x j
a
0
2sin
a
m sy
m xj J x dx
a a
00
12 sinm syTE
mm
m xA J x
Z a
Represent the strip current as
or
29
Discontinuity (cont.)
1 10
2sin sinm mTE
m mm
m x m xA j
Z a a
Therefore
Hence0
2
TEm
m m
ZA j
In order to solve for jm, we need to enforce the condition that Ey = 0 on the strip.
30
Discontinuity (cont.)
0 0 022
0
1/
2 2
2
sy
w wJ x I x x x
wx x
Assume that the strip is narrow, so that a single “Maxwell” function describes accurately the shape of the current on the strip:
0
2sin
a
m sy
m xj J x dx
a a
0
0
/2
0 2/2 2
0
2 1/sin
2
x w
m
x w
m xj I dx
a awx x
Hence
I0 is the unknown
(I0 is the total current on the strip.)
31
Discontinuity (cont.)
Hence
where
0
2m mj I c
a
0
0
/2
2/2 2
0
1/sin
2
x w
m
x w
m xc dx
awx x
Note: cm can be evaluated in closed form (in terms of the Bessel function J0).
00 sin
2m
m xm wc J
a a
(Please see the Appendix.)
32
Discontinuity (cont.)
To solve for I0, enforce the electric field integral equation (EFIE):
0inc scay yE E on strip
10
0
1
, , sin
, , sin
z
mz
jk zincy
jk zscay m
m
xE x y z e
a
m xE x y z A e
a
0 01
sin sin2 2m
m
x m x w wA x x x
a a
Hence
(unit-amplitude incident mode)
0 00,2 2
w wz x x x
33
Discontinuity (cont.)
0 01
sin sin2 2m
m
x m x w wA x x x
a a
00 0
1
sin sin2 2 2
TEm
mm
Zx m x w wj x x x
a a
00 0 0
1
2sin sin
2 2 2
TEm
mm
Zx m x w wc I x x x
a a a
or
or
34
Discontinuity (cont.)
00 0 0
1
2sin sin
2 2 2
TEm
mm
Zx m x w wI c x x x
a a a
This is in the form
0 0 02 2
w wf x I g x x x x
To solve for the unknown I0, we can use the idea of a “testing function.” We multiply both sides by a testing function and then integrate over the strip.
0 0
0 0
/2 /2
0
/2 /2
x w x w
x w x w
T x f x dx I T x g x dx
35
Discontinuity (cont.)
0 0
0 0
/2 /2
00
1/2 /2
2sin sin
2
x w x wTEm
mmx w x w
Zx m xT x dx I c T x dx
a a a
Galerkin’s method: The testing function is the same as the basis function:
22
0
1/
2
T xw
x x
36
Discontinuity (cont.)
0
0
0
0
/2
2/2 2
0
/2
00 2
1 /2 2
0
1/sin
2
2 1/sin
2
2
x w
x w
x wTEm
mm x w
xdx
awx x
Z m xI c dx
a awx x
Hence
37
Discontinuity (cont.)
01 0
1
2
2
TEm
m mm
Zc I c c
a
Hence
10
20
1
22
TEm
mm
cI
Zc
a
10
20
1
TEm m
m
a cI
Z c
so
or
38
Discontinuity (cont.)
Summary
10
20
1
TEm m
m
a cI
Z c
0
2m mj I c
a
0
2
TEm
m m
ZA j
0
0
1
1
, , sin
, , sin
mz
mz
jk zscay m
m
jk zscay m
m
m xE x y z A e
a
m xE x y z A e
a
00 sin
2m
m xm wc J
a a
39
Discontinuity (cont.)
Reflection Coefficient:
1 11 1
1 1inc
A AA A
A
10
10
TEin
TEin
Z Z
Z Z
10
1010
TEpTE
in p TEp
jX ZZ Z jX
jX Z
From these equations, Xp may be found.
1A
Post Reactance:
where
jXp0 10TEZ Z 0 10
TEZ Z
40
Discontinuity (cont.)
Result:
10
1
2TE
pjX Z
101 1 2
TEZA j
1 0 1
2j I c
a
10
20
1
TEm m
m
a cI
Z c
0
0 sin2m
m xm wc J
a a
010 2
0
1
TEZ
k a
41
Appendix
In this appendix we evaluate the cm coefficients .
0
0
/2
2/2 2
0
1/sin
2
x w
m
x w
m xc dx
awx x
Use 0x x x
/20
2/2 2
1 1sin
2
w
m
w
m x xc dx
awx
42
Appendix (cont.)
/20
2/2 2
1 1sin
2
w
m
w
m x xc dx
awx
0/2
2/2 02
sin cos1 1
cos sin2
w
m
w
m xm x
a ac dx
m xm xwx a a
This term integrates to zero (odd function).
or
43
/20
2/2 2
1 1sin cos
2
w
m
w
m x m xc dx
a awx
/20
20 2
cos2
sin
2
w
m
m xm x a
c dxa w
x
Appendix (cont.)
or
44
Next, use the transformation sin2
cos2
wx
wdx d
/20
0
cos sin2 2
sin cos2cos
2
m
m wm x wa
cwa
Appendix (cont.)
45
/20
0
2sin cos sin
2m
m x m wc d
a a
Next, use the following integral identify for the Bessel function:
0
1cos sinnJ z z n d
so that
0
0
1cos sinJ z z d
0
0
1sin cos sin
2m
m x m wc d
a a
Appendix (cont.)
or
46
0
0
1sin cos sin
2m
m x m wc d
a a
00sin
2m
m x m wc J
a a
Hence
Appendix (cont.)
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