#### Transcript Notes 8 - Waveguides part 5 TEN

```ECE 5317-6351
Microwave Engineering
Fall 2012
Prof. David R. Jackson
Dept. of ECE
Notes 8
Waveguides Part 5:
Transverse Equivalent Network
(TEN)
1
Waveguide Transmission Line Model
Our goal is to come up with a transmission line model for a waveguide mode.
y
b
x
a
The waveguide mode is not
a TEM mode, but it can be
modeled as a wave on a
transmission line.
z
I
z
+
-
V
2
Waveguide Transmission Line Model (cont.)
For a waveguide mode, voltage and current are not uniquely defined.
y
y
TE10 Mode
A
b
Ey  x 
b
B
x
x
a
z
a
Ey  
j

A
sin
10

kc2 a
a

x  e  jkz z

 j


V  z   VAB  z    E  dr   E y dy   2 A10  b sin 
a
 kc a

A
b
B
0
  jkz z

xe
 V0 sin 

a

x  e  jkz z

The voltage depends on x!
3
Waveguide Transmission Line Model (cont.)
For a waveguide mode, voltage and current are not uniquely defined.
y
x1
TE10 Mode
x2
Hx  x
b
y
b
x
a
a
Hx 
jk z

A
sin
10

kc2 a
a
x2

x  e  jkz z

z
x2
x2
x1
x1
Current on top wall: I  z    J  x  dx   H x  x  dx  
top
sz
x1
Note: If we integrate
around the entire
boundary, we get
zero current.
x
jk z

A
sin
10

kc2 a
a

x  e  jkz z dx

 jk 
 a     
  
  2z A10     cos  x2   cos  x1   e  jkz z
 a 
 kc a
     a 
I   
  
 0 cos  x1   cos  x2   e  jkz z
2  a 
 a 
The current depends on the length of the interval!
4
Waveguide Transmission Line Model (cont.)
Examine the transverse (x, y) fields:
Modal amplitudes

Et ( x, y, z )  et ( x, y ) A e  jk z z  A e  jk z z

H t ( x, y, z )  ht ( x, y ) A e  jk z z  A e  jk z z


The minus sign arises from:
1
h 
( zˆ  et )
Zw

t
Z w  ZTE or ZTM
Wave impedance
5
Waveguide Transmission Line Model (cont.)
Introduce a defined voltage into field equations:
We may use whatever definition
of voltage we wish here.
1 
Et ( x, y, z ) 
et ( x, y) V0 e jkz z  V0e  jkz z 
C1
1 
H t ( x, y, z ) 
ht ( x, y) V0 e  jkz z  V0 e  jkz z 
C1

V
A  0
C1
where
or

V
A  0
C1
V0 V0
C1    
A
A
6
Waveguide Transmission Line Model (cont.)
Introduce a defined current and then from this define a
characteristic impedance:
V
We may use whatever definition
Z0  
of current we wish here.
I

1 
H t ( x, y, z ) 
ht ( x, y ) V0 e jkz z  V0e  jkz z
C1

 V0  jk z z V0  jk z z 
1 
H t ( x, y , z ) 
ht ( x, y ) 
e

e

C2
Z
Z
0
 0

where
C1
C2 
Z0
7
Waveguide Transmission Line Model (cont.)
Summary:
V (z)

1 
Et ( x, y , z ) 
et ( x, y ) V0 e  jk z z  V0 e  jk z z
C1

I (z)
 V0  jk z z V0  jk z z 
1 
H t ( x, y , z ) 
ht ( x , y ) 
e

e

C2
Z
Z
0
 0

V0 V0
C1    
A
A
C1
 Z0
C2
8
Waveguide Transmission Line Model (cont.)
Note on Z0:
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.
V (z)

1 
Et ( x, y , z ) 
et ( x, y ) V0 e  jk z z  V0 e  jk z z
C1

I (z)
 V0  jk z z V0  jk z z 
1 
H t ( x, y , z ) 
ht ( x , y ) 
e

e

C2
Z
Z
0
 0

9
Waveguide Transmission Line Model (cont.)
The transmission-line model is called the transverse equivalent
network (TEN) model of the waveguide
I  z
z
+
V  z
-
Z0 , k z
10
Waveguide Transmission Line Model (cont.)
Power flow down the waveguide:
PWG  z  

WG
P
1
*
E

H
 zˆ dS


t
t

2S
 1
1
V z I* z 
*
2
C
C
 1 2
 1
z  P z  *
 C1C2
TL


*
ˆ
e
(
x
,
y
)

h
(
x
,
y
)

z
dS


t
S  t


S  e ( x, y)  h ( x, y)   zˆ dS 

t
*
t
11
Waveguide Transmission Line Model (cont.)
Set
PWG  z   PTL  z 
Then
C1C2*    et ( x, y )  ht* ( x, y )   zˆ dS
S
12
Waveguide Transmission Line Model (cont.)
Summary of Constants
(for equal power)
C1
 Z0
C2
C1C2*    et ( x, y )  ht* ( x, y )   zˆ dS
S
Once we pick Z0, the constants are determined.
The most common choice:
Z0  Z w
13
Waveguide Transmission Line Model (cont.)
We have two constants (C1 and C2)
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.
14
Example: TE10 Mode of
Rectangular Waveguide
y
Method 1:
 Define voltage
 Define current
(This determines Z0)
b
x
a
Method 2:
 Choose Z0 = ZTE
 Assume power equality
z
15
Example: TE10 Mode (cont.)
y
Method 1
 
E   yˆ A sin  x  e  jk z z
a 
1
 
H t  xˆ A
sin  x  e  jk z z
ZTE
a 

t
b

x
a
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
z
Define:
V   z 
ZTE 
bot
0
 E  a / 2, y, z   dr  E  a / 2, y, z  dy   A b  e
I
k z10
 jk z z
y
top



b
a
 z   J
0
a
top
sz
dx   H
0
A
A  a 
    jkz z
 jk z z
dx  
sin  x  e
dx 
    1  1 e
ZTE
ZTE   
a 
0
a
top
x
16
Example: TE10 Mode (cont.)
V

z   A

b e
y
 jk z z
A  2a   jkz z
I  z 

e
ZTE   

V z
I  z
b

Z0 
x
a
z
Hence
 b 
Z 0  ZTE 

 2a 
Since we have defined both voltage and current,
the characteristic impedance is not arbitrary,
but is determined.
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
ZTE 

k z10
17
Example: TE10 Mode (cont.)
y
V0
A b
C1      b
A
A
b
C1
 b 
 Z 0  ZTE 

C2
 2a 
C
 C2  1
ZTE
1  2a 
 2a 





  b  ZTE   
C1  b
1  2a 
C2 


ZTE   
x
a
z
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
ZTE 

k z10
18
Example: TE10 Mode (cont.)
y
C1  b
1  2a 
C2 


ZTE   
WG
P
b
x
a
 z   P  z  Rp
TL
 1
Rp  
*
C
C
1
2

Rp 
z


*
ˆ
e
(
x
,
y
)

h
(
x
,
y
)

z
dS


t
S  t

 1  ab  
 * 

2
 1  2a    ZTE


b * 

Z



 TE
1
Rp 

 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
ZTE 

k z10
4
19
Example: TE10 Mode (cont.)
y
Method 2
C1C2*    et ( x, y )  ht* ( x, y )   zˆ dS
b
x
S
C1C2*  
S
1
2 x 
sin

 dS
*
ZTE
 a 
1
 *
ZTE

2 x 
sin
0 0  a  dydx
a
z
a b
1  ab 

* 
ZTE
 2 
C1
 Z 0  ZTE
C2
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
ZTE 

k z10
20
Example: TE10 Mode (cont.)
1
*
C1C2  *
ZTE
y
 ab 


 2 
C1
 ZTE
C2
b
x
a
Take the conjugate of the
second one and multiply
the two together.
z
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
Solution:
The solution is
unique to within a
common phase term.
C1 
ab
2
1
C2 
ZTE
ZTE 
ab
2

k z10
21
Example: TE10 Mode (cont.)
y
C1 
C2 
WG
P
ab
2
1
ZTE
ab
2
b
x
a
 z   P  z  Rp
TL
 1
Rp  
*
C
C
 1 2
Rp 
z

ˆ
S  e ( x, y)  h ( x, y)   z dS 

t
*
t
1
ab  1
 *
2  ZTE
 1  ab  
 * 

Z
2



ab  TE

2 
Rp  1
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
ZTE 

k z10
(as expected)
22
Example: Waveguide Discontinuity
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.
y
B
r
x
b
0
a
a = 2.2856 cm
b = 1.016 cm
r = 2.54
f = 10 GHz
a  b  0
z
A
z=0
a
k za
 
 k     158.0  rad / m 
a
k zb
 
  r k     304.1  rad / m 
a
2
2
0
2
2
0
23
Example (cont.)
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
TEN
Z0TEb , kzb
Z0TEa , kza
ZTE 

0
TV0
V
V0
ZTE 
kz
 Choose Z0 = ZTE
 Assume power equality
b
Z 0b  ZTE


k za

k zb
k z10

C1 
Convention:
a
Z 0 a  ZTE


C2 
ab
2
1
ZTE
ab
2
 499.7  
A  1 (since et  x, y  already has 1 [ V / m ])
 259.6  
V0  C1 A 
ab
2
24
Example (cont.)

Va  z   V0 e  jk za z  e  jk za z

Vb  z   V0 Te  jk zb z
V0
Ia  z  
e  jk za z  e  jk za z
Z0a


V0T  jk zb z
Vb  z  
e
Z 0b
Z0TEa , kza
V0
V0
Equivalent reflection problem:
Z0TEa
TEN
Z
TE
0b
Z0TEb , kzb
TV0
Z ob  Z oa

 0.316
Z ob  Z oa
T  1    0.684
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.
25
Example (cont.)

TEN

Va  z  
ab  jk za z
e
  0.316  e  jk za z
2
Vb  z  
ab
 0.684  e jkzb z
2
Ia  z  
ab 1
e  jk za z   0.316  e  jk za z
2 Z0a
Vb  z  
Z0TEa , kza


V0
V0
ab  0.684   jk zb z
e
2
Z 0b
Recall that for the TE10 mode:
V (z)
Et ( x, y , z ) 

1 
et ( x, y ) V0 e  jk z z  V0 e  jk z z
C1

Z0TEb , kzb
TV0
 
et   yˆ sin  x 
a 
1
 
ht  xˆ
sin  x 
ZTE
a 
I (z)
 V0  jk z z V0  jk z z 
1 
H t ( x, y , z ) 
ht ( x , y ) 
e

e

C2
Z
Z
0
 0

26
Example (cont.)
TEN
Z0TEa , kza
V0
V0
Hence, we have that
Eta ( x, y, z ) 
1 
et ( x, y)
C1
Hta ( x, y, z ) 
1 
ht ( x, y )
C2
Etb ( x, y, z ) 

ab  jkza z
e
  0.316  e jkza z
2
TV0


ab 1
e jkza z   0.316 e jkza z
2 Z0 a
1 
et ( x, y)
C1
Z0TEb , kzb

ab
 0.684 e jkzb z
2
1 
Htb ( x, y, z ) 
ht ( x, y )
C2
ab 1
 0.684 e jkzb z
2 Z 0b
27
Example (cont.)
Substituting in, we have
Eta ( x, y, z ) 
Eta ( x, y, z ) 
1 
et ( x, y)
C1
1
ab
2

ab  jkza z
e
  0.316  e jkza z
2

  
ˆ

y
si
n
 x 

a 



ab  jkza z
e
  0.316  e jkza z
2

28
Example (cont.)
Substituting in, we have
Hta ( x, y, z ) 
H ta ( x, y, z ) 
1
1
Z 0TEa
1 
ht ( x, y )
C2

ab 1
e jkza z   0.316 e jkza z
2 Z0 a
 1
  
ˆ
x
s
in

 x 
TE
ab  Z 0 a
 a 
2
a
Z w  ZTE
a
Z0  Z0a  ZTE


ab 1
e jkza z   0.316  e jkza z
TE
2 Z0a

(wave impedance)
(our choice)
29
Example (cont.)
Substituting in, we have
Etb ( x, y, z ) 
Etb ( x, y, z ) 
1
ab
2
1 
et ( x, y)
C1
ab
 0.684 e jkzb z
2

  
ˆ

y
sin
 x 

a 

ab
 0.684  e jkzb z
2
30
Example (cont.)
Substituting in, we have
1 
Htb ( x, y, z ) 
ht ( x, y)
C2
H tb ( x, y, z ) 
1
1
Z 0TEb
ab 1
 jk zb z
0.684
e


2 Z0TE
b
 1
  
 xˆ TE sin  x  
ab  Z 0b
 a 
2
ab 1
 jk zb z
0.684
e


2 Z 0TbE
b
Z w  ZTE
b
Z0  Z0b  ZTE
(our choice)
31
Example (cont.)
Summary of Fields

  
Eta ( x, y, z )    yˆ sin  x   e jkza z   0.316  e jkza z
a 



 1
  
H ta ( x, y, z )   xˆ TE sin  x   e  jkza z   0.316  e  jkza z
 a 
 Z0a


  
Etb ( x, y, z )    yˆ sin  x    0.684  e jkzb z
a 


Z 0 a  499.7   
Z 0b  259.6   
k za  158.0  rad / m 
 1
  
H tb ( x, y, z )   xˆ TE sin  x    0.684  e jkzb z
 a 
 Z 0b
k zb  304.1  rad / m 
32
Example (cont.)
Note: In this problem, Z0 and  are real.
Power Calculations:
Painc
1
1

 Re  V0 I 0*   Re  V0
2

2
2
1  1

TE * 
Z oa  2
1
 2 1
Paref  Re  V0 I 0*   
2
2



1
2
1

Pbtrans  Re  V0 I 0*  1   
2
2

2
ab
2
1
TE
Z oa
2
ab
2
1
2

TE
Z oa
2
ab
2

1
2
1


TE
Z oa

Alternative:
Pbtrans
*


 1


V
1
1

trans trans* 

0
 Re  V0 I 0
  Re  V0 1      TE 1      
2
2


 Z oa
  2

2
ab
2

1
2
1 
TE
Z oa

33
Example (cont.)
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.
Final Results:
Painc  1.161  mW 
Parefl  0.116  mW 
B
y
Pbtrans  1.045  mW 
r
A
z
x
a = 2.2856 cm
b = 1.016 cm
r = 2.54
f = 10 GHz
b
0
z=0
a
a
34
Discontinuities in Waveguide
Rectangular Waveguide
(end view)
Note: Planar discontinuities are
modeled as purely shunt elements.

inductive iris

capacitive iris

resonant iris
The equivalent circuit gives us the correct reflection and transmission of the TE10 mode.
35
Discontinuities in Waveguide (cont.)
Inductive iris in air-filled waveguide
Z 0  Z 0TE 
x
Top view:

TE10
TE10
1

0
k z10
0
  
1 

k
a
 0 
2
T
z
Higher-order mode region
Because the element is a
shunt discontinuity, we have
TEN Model
T  1 
Z0TE
Lp
Z0TE
36
Discontinuities in Waveguide (cont.)
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
37
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