Post on 12-Jan-2016
1
Prof. David R. JacksonDept. of ECE
Notes 8
ECE 5317-6351 Microwave Engineering
Fall 2012
Waveguides Part 5:Transverse Equivalent Network
(TEN)
2
Waveguide Transmission Line ModelOur goal is to come up with a transmission line model for a waveguide mode.
I
+- V
z
The waveguide mode is not a TEM mode, but it can be modeled as a wave on a transmission line.
z
a
bx
y
3
For a waveguide mode, voltage and current are not uniquely defined.
102sin zjk z
yc
jE A x e
k a a
0
10 02sin sinz z
Bjk z jk z
AB ycA b
jV z V z E dr E dy A b x e V x e
k a a a
The voltage depends on x!
x
y
a
b
A
B
TE10 Mode
yE x
z
a
bx
y
Waveguide Transmission Line Model (cont.)
4
For a waveguide mode, voltage and current are not uniquely defined.
2 2 2
1 1 1
102
10 2 12
01 2
sin
cos cos
cos cos2
z
z
z
x x xjk ztop z
sz xcx x x
jk zz
c
jk z
jkI z J x dx H x dx A x e dx
k a a
jk aA x x e
k a a a
Ix x e
a a
The current depends on the length of the interval!
102sin zjk zz
xc
jkH A x e
k a a
TE10 Mode
x
y
a
b xH x
x1 x2
Current on top wall:
Note: If we integrate around the entire boundary, we get zero current.
z
a
bx
y
Waveguide Transmission Line Model (cont.)
5
( , , ) ( , ) z zjk z jk zt tE x y z e x Ay e eA
1ˆ( )t t
w
h z eZ
( , , ) ( , ) z zjk z jk zt tH x y z h x Ay e eA
Wave impedance
Waveguide Transmission Line Model (cont.)
w TE TMZ Z Z or
Examine the transverse (x, y) fields:
The minus sign arises from:
Modal amplitudes
6
Introduce a defined voltage into field equations:
0 01
1( , , ) ( , ) z zjk z jk z
t tE x y z e x y e eV VC
0 01
1( , , ) ( , ) z zjk z jk z
t tH x y z h x y e eV VC
Waveguide Transmission Line Model (cont.)
0
1
VA
C
where
0 01
V VC
A A
or
0
1
VA
C
We may use whatever definition of voltage we wish here.
7
0 0
0 02
1( , , ) ( , ) z zjk z jk z
t tH x y z h x y eV
eV
ZC Z
Waveguide Transmission Line Model (cont.)
12
0
CC
Z
Introduce a defined current and then from this define a characteristic impedance:
0
VZ
I
0 01
1( , , ) ( , ) z zjk z jk z
t tH x y z h x y e eV VC
where
We may use whatever definition of current we wish here.
8
Summary:
0 01
( )
1( , , ) ( , ) z zjk z j
V
t t
z
k zE x y z e x y e eVC
V
0 0
0
( )
02
1( , , ) ( , ) z zjk z jk z
t
z
t
I
H x yV
z h x y eV
Ze
C Z
Waveguide Transmission Line Model (cont.)
0 01
V VC
A A
10
2
CZ
C
9
Note on Z0:
0 01
( )
1( , , ) ( , ) z zjk z j
V
t t
z
k zE x y z e x y e eVC
V
0 0
0
( )
02
1( , , ) ( , ) z zjk z jk z
t
z
t
I
H x yV
z h x y eV
Ze
C Z
Waveguide Transmission Line Model (cont.)
We can define voltage and current, and this will determine the value of Z0.
Or, we can define voltage and Z0, and this will determine current.
10
The transmission-line model is called the transverse equivalent network (TEN) model of the waveguide
Waveguide Transmission Line Model (cont.)
I z
+- 0 , zZ k
z
V z
11
Power flow down the waveguide:
*
* **
1 2
1ˆ
2
1 1ˆ( , ) ( , )
2
WGt t
S
t t
S
P z E H z dS
V z I z e x y h x y z dSC C
Waveguide Transmission Line Model (cont.)
**
1 2
1ˆ( , ) ( , )WG TL
t t
S
P z P z e x y h x y z dSC C
12
Set
Waveguide Transmission Line Model (cont.)
* *1 2 ˆ( , ) ( , )t t
S
C C e x y h x y z dS
WG TLP z P z
Then
13
Summary of Constants (for equal power)
Waveguide Transmission Line Model (cont.)
* *1 2 ˆ( , ) ( , )t t
S
C C e x y h x y z dS
10
2
CZ
C
The most common choice: 0 wZ Z
Once we pick Z0, the constants are determined.
14
We have two constants (C1 and C2)
Waveguide Transmission Line Model (cont.)
Here are possible constraints we can choose to determine the constants:
We can define the voltage We can define the current We can define the characteristic impedance We can impose the power equality condition
Any two of these are sufficient to determine the constants.
15
Method 1: Define voltage Define current (This determines Z0)
Example: TE10 Mode of Rectangular Waveguide
Method 2: Choose Z0 = ZTE
Assume power equality
z
a
bx
y
16
ˆ sin
1ˆ sin
z
z
jk zt
jk zt
TE
E y A x ea
H x A x eZ a
Define:
0
/ 2, , / 2, , z
botjk z
y
top b
V z E a y z dr E a y z dy A b e
Example: TE10 Mode (cont.)
0 0 0
sin 1 1z z
a a ajk z jk ztop top
sz xTE TE
A A aI z J dx H dx x e dx e
Z a Z
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Method 1
z
a
bx
y
17
zjk zV z A b e
2zjk z
TE
A aI z e
Z
0 2TE
bZ Z
a
Since we have defined both voltage and current, the characteristic impedance is not arbitrary, but is determined.
Hence
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Example: TE10 Mode (cont.)
za
bx
y
0
V zZ
I z
18
10
2 2TE
C bZ Z
C a
01
V A bC b
A A
12
2 1 2
TE TE
C a aC
Z b Z
1
2
1 2
TE
C b
aC
Z
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Example: TE10 Mode (cont.)
za
bx
y
19
1
2
1 2
TE
C b
aC
Z
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Example: TE10 Mode (cont.)
**
1 2
1ˆ( , ) ( , )
WG TLp
p t t
S
P z P z R
R e x y h x y z dSC C
*
*
1 1
21 2p
TE
TE
abR
Zab
Z
4pR
z
a
bx
y
20
* *1 2 ˆ( , ) ( , )t t
S
C C e x y h x y z dS
* 21 2 *
2*
0 0
*
1sin
1sin
1
2
TES
a b
TE
TE
xC C dS
Z a
xdydx
Z a
ab
Z
10
2TE
CZ Z
C 10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Method 2
Example: TE10 Mode (cont.)
za
bx
y
21
*1 2 *
1
2TE
abC C
Z
1
2TE
CZ
C
Solution:
1
2
2
1
2TE
abC
abC
Z
Example: TE10 Mode (cont.)
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Take the conjugate of the second one and multiply the two together.
The solution is unique to within a common phase term.
za
bx
y
22
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
Example: TE10 Mode (cont.)
**
1 2
1ˆ( , ) ( , )
WG TLp
p t t
S
P z P z R
R e x y h x y z dSC C
*
*
1 1
212 2
pTE
TE
abR
Zab abZ
1pR
1
2
2
1
2TE
abC
abC
Z
(as expected)
za
bx
y
23
220
220
158.0 rad / m
304.1 rad / m
za
zb r
k ka
k ka
Example: Waveguide Discontinuity
a = 2.2856 cmb = 1.016 cmr = 2.54f = 10 GHz
b a0
rz
a
z = 0
x
y
A
B
For a 1 [V/m] (field at the center of the guide) incident TE10 mode E-field in guide A, find the TE10 mode fields in both guides, and the reflected and transmitted powers.
0a b
24
TEz
Zk
Example (cont.)
Convention: Choose Z0 = ZTE
Assume power equality
1
2
2
1
2TE
abC
abC
Z
0
0
499.7
259.6
aa TE
za
bb TE
zb
Z Zk
Z Zk
0 1
V / m1 ( , )
2
tA e x y
abV C A
since already has 1 [ ]
TEN
0V 0TV
0V
0 ,TEa zaZ k 0 ,TE
b zbZ k
10
ˆ sin
1ˆ sin
t
tTE
TEz
e y xa
h x xZ a
Zk
25
0
0
0
0
0
0
za za
zb
za za
zb
jk z jk za
jk zb
jk z jk za
a
jk zb
b
V z V e e
V z V Te
VI z e e
Z
V TV z e
Z
Equivalent reflection problem: 0.316ob oa
ob oa
Z Z
Z Z
Example (cont.)
0TEbZ0
TEaZ 1 0.684T
TEN
0V 0TV
0V
0 ,TEa zaZ k 0 ,TE
b zbZ k
Note: The above TL results come from enforcing the continuity of voltage and current at the junction, and hence the tangential electric and magnetic fields are continuous in the WG problem.
26
Recall that for the TE10 mode:
Example (cont.)
0
0
0.3162
0.6842
10.316
2
0.684
2
za za
zb
za za
zb
jk z jk za
jk zb
jk z jk za
a
jk zb
b
abV z e e
abV z e
abI z e e
Z
abV z e
Z
0 01
( )
1( , , ) ( , ) z zjk z j
V
t t
z
k zE x y z e x y e eVC
V
0 0
0
( )
02
1( , , ) ( , ) z zjk z jk z
t
z
t
I
H x yV
z h x y eV
Ze
C Z
ˆ sin
1ˆ sin
t
tTE
e y xa
h x xZ a
TEN
0V 0TV
0V
0 ,TEa zaZ k 0 ,TE
b zbZ k
27
Hence, we have that
Example (cont.)
2 0
11( , , ) ( , 0.31
2) 6za zajk z jk z
ta
a t
abe e
ZH x y z h x y
C
1
0.6841
( , )2
, ( , ) zbjtb
kt
zE x y z e xb
eC
ay
02
10.684
2
1( , , ) ( , ) zbjk z
btb tH x y z h x y e
ZC
ab
1
0.3161
( , , ) ( , )2
za zata
k j zt
j z kabE x y z e x
Cey e
TEN
0V 0TV
0V
0 ,TEa zaZ k 0 ,TE
b zbZ k
28
Substituting in, we have
Example (cont.)
1ˆ 0( , , ) si 3n . 6
2
2
1za zata
jk z jk zE x y z y xa
abe
ae
b
1
0.3161
( , , ) ( , )2
za zata
k j zt
j z kabE x y z e x
Cey e
29
Substituting in, we have
Example (cont.)
0
00
1 1ˆ( , , ) s
10.316
2in
12
za zajk z jk zTEa
ta TEa
TEa
H x y z xa
xZ aab
be e
Z
Z
2 0
11( , , ) ( , 0.31
2) 6za zajk z jk z
ta
a t
abe e
ZH x y z h x y
C
0 0a
a TEZ Z Z
aw TEZ Z
(our choice)
(wave impedance)
30
Substituting in, we have
Example (cont.)
1
0.6841
( , )2
, ( , ) zbjtb
kt
zE x y z e xb
eC
ay
1ˆ( , , ) sin
2
0.6842
zbtb
jk zE x y z y xaa
ae
b
b
31
Substituting in, we have
Example (cont.)
2 0
1( , , ) ( , )
10.684
2zb
tb tjk z
TEb
abe
ZH x y z h x y
C
0
00
1 1ˆ
10.684( , , ) sin
12
2zbjk
tb TEb
TE
Eb
b
zT
H x y z x xZ aab
ae
Z
b
Z
0 0b
b TEZ Z Z
bw TEZ Z
(our choice)
32
Summary of Fields
Example (cont.)
0
1ˆ( , , ) sin 0.684 zbjk z
tb TEb
H x y z x x eZ a
ˆ( , , ) sin 0.684 zbjk ztbE x y z y x e
a
0
1ˆ( , , ) sin 0.316za zajk z jk z
ta TEa
H x y z x x e eZ a
ˆ( , , ) sin 0.316za zajk z jk ztaE x y z y x e e
a
0
0
499.7
259.6
a
b
Z
Z
158.0 rad / m
304.1 rad / m
za
zb
k
k
33
Power Calculations:
22*
0 0 0 *
1 1 1 1 1Re Re
2 2 2 2inc
a TE TEoa oa
abP V I V
Z Z
Example (cont.)
2
2 2*0 0
1 1 1Re
2 2 2ref
a TEoa
abP V I
Z
2
2 2*0 0
1 1 1Re 1 1
2 2 2trans
b TEoa
abP V I
Z
2*
2* 00 0 0
1 1 1 1Re Re 1 1 1
2 2 2 2trans trans trans
b TE TEoa oa
V abP V I V
Z Z
Alternative:
Note: In this problem, Z0 and are real.
34
Final Results:
1.161 mW
0.116 mW
1.045 mW
inca
refla
transb
P
P
P
Example (cont.)
a = 2.2856 cmb = 1.016 cmr = 2.54f = 10 GHz
For a 1 [V/m] incident TE10 mode E-field in guide A (field at the center of the guide) , find the TE10 mode fields in both guide, and the reflected and transmitted powers.
b a0
rz
a
z = 0
x
y
A
B
35
Discontinuities in Waveguide
inductive iris
capacitive iris
resonant iris
Rectangular Waveguide(end view)
Note: Planar discontinuities are modeled as purely shunt elements.
The equivalent circuit gives us the correct reflection and transmission of the TE10 mode.
36
Discontinuities in Waveguide (cont.)Inductive iris in air-filled waveguide
Top view:
z
x
TE10TE10
Higher-order mode region
Z0TE Z0
TE Lp
TEN Model
00 0 10
0
2
0
1
TE
z
Z Zk
k a
1T
Because the element is a shunt discontinuity, we have
T
1
37
Discontinuities in Waveguide (cont.)
Much more information can be found in the following reference:
N. Marcuvitz, Waveguide Handbook, Peter Perigrinus, Ltd. (on behalf of the Institute of Electrical Engineers), 1986.
Equivalent circuits for many types of discontinuities Accurate CAD formula for many of the discontinuities Graphical results for many of the cases Sometimes, measured results