### Notes 4 - Waveguides part 1 general theory

```ECE 5317-6351
Microwave Engineering
Fall 2011
Prof. David R. Jackson
Dept. of ECE
Notes 4
Waveguides Part 1:
General Theory
1
Waveguide Introduction
In general terms, a waveguide is a devise that confines electromagnetic
energy and channels it from one point to another.
Examples
– Coax
– Printed circuit lines
(e.g. microstrip)
– Optical fiber
– Parallel plate waveguide
– Rectangular waveguide
– Circular waveguide
Note: In microwave engineering, the term “waveguide” is often used to
mean rectangular or circular waveguide (i.e., a hollow pipe of metal).
2
General Solutions for TEM, TE and TM Waves
Assume ejt time dependence and homogeneous source-free materials.
Assume wave propagation in the  z direction
 e
  jk z ,
z
e
y
PEC
jk z z
k z    j
x
E  x , y , z    e t  x , y   zˆ e z  x , y   e
jk z z
J E
transverse
components
H  x , y , z    h t  x , y   zˆ h z  x , y   e
, ,
z
jk z z
3
Helmholtz Equation
  E   j  H
  H  j  E  J
E 
v

H  0
     E   j    H   j 
 j  E  J 
Vector Laplacian definition :
 E     E       E
2
where
2
2
2
2
 E  xˆ   E x   yˆ   E y   zˆ   E z 
4
Helmholtz Equation
    E   j    H   j 
 j  E  J 
     E    E   j   j  E  J
2

 v 
2
2
  


E


 E  j  J

  
 v 
  E    E  j  J    




2
2
 v 
  E    E  j  E    




2
2
Assume Ohm’s
law holds:
J E
5
Helmholtz Equation (cont.)
 v 
 E    E  j  E    

  
2
2
 
 

2
2
  E      j  E    v 
 

  
 
2
2
  E    c E     v 
  
k    c
2
2
 
2
2
  E  k E    v 
  
Next, we examine the term on the right-hand side.
6
Helmholtz Equation (cont.)
 v 
 E  k E   

  
2
  H  j  E  J
      H   j     E      J
2

      H   j     E       E 
 0     E     j 
 E  0
 v  0

state, there can never be any volume
charge density inside of a linear,
homogeneous, isotropic, source-free
region that obeys Ohm’s law.
7
Helmholtz Equation (cont.)
Hence, we have
 Ek E 0
2
2
Helmholtz equation
8
Helmholtz Equation (cont.)
Similarly, for the magnetic field, we have
  H  j  E  J
      H   j    E    J
      H   j    E     E
      H      j     E
     H  
 j  c    E
     H  
 j  c    j  H 
    H    H 
2
 j  c    j  H 
9
Helmholtz Equation (cont.)
Hence, we have
 H k H 0
2
2
Helmholtz equation
10
Helmholtz Equation (cont.)
Summary
 Ek E 0
2
2
 H k H 0
2
2
Helmholtz equations
These equations are valid for a source-free homogeneous isotropic
linear material.
11
Field Representation
Assume a guided wave with a field variation in the z direction
of the form
jk z
e z
Then all six of the field components can be expressed in terms
of these two fundamental ones:
 Ez, H z 
12
Field Representation (cont.)
Types of guided waves:
TEMz
 TEMz: Ez = 0, Hz = 0
 TMz: Ez  0, Hz = 0
 TEz: Ez = 0, Hz  0
 Hybrid: Ez  0, Hz  0
TMz , TEz
Hybrid
w
r
h
Microstrip
13
Field Representation: Proof
Assume a source-free region with a variation e
  H  j c E
  E   j  H
1)
2)
3)
E z
y
 jk z E y   j  H x
jk z E x 
E y
x

E z
E x
y
x
  j  H y
  j  H z
jk z z
4)
5)
6)
H z
y
 jk z H y  j  c E x
jk z H x 
H y
x

H z
H x
y
x
 j  c E y
 j  E z
14
Field Representation: Proof (cont.)
Combining 1) and 5)
 1  
H z 

 jk z 
   jk z H x 
   j  H x
y
x 
 j  c  
E z
E z
y
2

kz 
   j  
Hx
x
j  c 

k z H z
 c
 j  c
E z
y
jk z
H
x
z
 (k  k z ) H x
2
2
2
kc
 Hx 
E z
j 


c
2 
kc 
y
kz


x 
H
kc   k  k
2
2
z

1/ 2
z
Cutoff wave
number
A similar derivation holds for the other three transverse field components.
15
Field Representation (cont.)
Summary
Hx 
E z
j 


c
2 
kc 
y
kz
H z 

x 
E z
H z 
j
H y  2 c
 kz

kc 
x
y 
Ex 
 Ez
H z 
j

k



z

2 
kc 
x
y 
These equations give the
transverse field
components in terms of
longitudinal components,
Ez and Hz.
k    c
2
2
kc   k  k
2
j 
Ey  2 
kc 
kz
H z 
 

y
x 
 Ez
2
z

1/ 2
16
Field Representation (cont.)
Therefore, we only need to solve the Helmholtz equations for the
longitudinal field components (Ez and Hz).
 Ez  k Ez  0
2
2
 Hz k Hz  0
2
2
17
Transverse Electric (TEz) Waves
 Ez  0
The electric field is “transverse”
(perpendicular) to z.
In general, Ex, Ey, Hx, Hy, Hz  0
To find the TEz field solutions (away from any sources), solve
(  k ) H z  0
2
2
2
2
 2


2 
(  k ) H z  0   2 

k Hz  0
2
2
y
z
 x

2
2
18
Transverse Electric (TEz) Waves (cont.)
2
2
 2


2 



k
 2
Hz  0
2
2
y
z
 x

Recall that the field solutions we seek are assumed to
jk z z
vary as
jk
e
 2
2

 
2
2
 2 
 kz  k
2
x
y

k c2

 H z ( x , y , z )  hz ( x , y ) e


 hz  x , y   0


2
 2

2 
  2 

k
h x, y   0
c  z 
2
y
 x

2
 2
 
2
  2 
h
x
,
y


k
hz  x , y 



z
c
2
y 
 x
z
z
kc  k  k z
2
2
2
Solve subject to the appropriate
boundary conditions.
(This is an eigenvalue problem.)
2
T h e e ig e n v a lu e k c is a lw a ys re a l.
19
Transverse Electric (TEz) Waves (cont.)
Once the solution for Hz is obtained,
jk z  H z
 Hx 
H
y
2
kc
x
jk z  H z

2
kc
y
Ex 
Ey 
 j   H z
2
kc
y
j   H z
2
kc
x
For a wave propagating in the positive z direction (top sign):
Ex

Hy
Ey


Hx
kz
For a wave propagating in the negative z direction (bottom sign):

Ex
Hy

Ey
Hx


TE wave impedance
kz
Z TE 

kz
20
Transverse Electric (TEz) Waves (cont.)
Also, for a wave propagating in the positive z direction,
ˆ y  x, y 
e t  x , y   xˆ e x  x , y   ye
ˆ x  xe
ˆ y
zˆ  e t  ye
e x  Z TE h y
ˆ x  yh
ˆ y
 zˆ  e t  Z T E  xh
e y   Z TE hx
 Z TE h t
 ht 
1
Z TE
( zˆ  e t )
Similarly, for a wave propagating in the negative z direction,
ht 
1
Z TE
(  zˆ  e t )
ht  x, y   
1
Z TE
 zˆ  e  x , y  
t
21
Transverse Magnetic (TMz) Waves
 Hz  0
In general, Ex, Ey, Ez ,Hx, Hy  0
To find the TEz field solutions (away from any sources), solve
(  k ) E z  0
2
2
2
2
2




2
2
2 
(  k ) E z  0   2 


k
 Ez  0
2
2
y
z
 x

22
Transverse Magnetic (TMz) Waves (cont.)
 2
2

 
2
2
 2 

k

k
z
2
x
y

2
k
c



 ez  x, y   0


2
 2

2 
  2 
 kc  ez  x, y   0
2
y
 x

kc  k  k z
2
2
2
solve subject to the appropriate
boundary conditions
2
 2
 
2
  2 
e
x
,
y


k
e x, y 


c z 
2  z
y 
 x
(Eigenvalue problem)
23
Transverse Magnetic (TMz) Waves (cont.)
Once the solution for Ez is obtained,
 Hx 
j  c  E z
Ex 
y
2
kc
j  c  E z
Hy  
x
2
kc
Ey 
jk z  E z
2
kc
x
jk z  E z
2
kc
y
For a wave propagating in the positive z direction (top sign):
Ex

Hy
Ey
Hx

kz
 c
For a wave propagating in the negative z direction (bottom sign):

Ex
Hy

Ey
Hx

kz
 c
TM wave impedance
Z TM 
kz
 c
24
Transverse Magnetic (TMz) Waves (cont.)
Also, for a wave propagating in the positive z direction,
ˆ y  x, y 
e t  x , y   xˆ e x  x , y   ye
ˆ x  xe
ˆ y
zˆ  e t  ye
 zˆ  e t  Z T M
ˆ
 xh
e x  Z TM h y
x
ˆ y
 yh

e y   Z TM hx
 Z TM h t
 ht 
1
Z TM
( zˆ  e t )
Similarly, for a wave propagating in the negative z direction,
ht 
1
Z TM
(  zˆ  e t )
ht  x, y   
1
Z TM
 zˆ  e  x , y  
t
25
Transverse ElectroMagnetic (TEM) Waves
 Ez  0 , H z  0
In general, Ex, Ey, Hx, Hy  0
From the previous equations for the transverse field components, all of
them are equal to zero if Ez and Hz are both zero.
Unless

kc  0
2
(see slide 16)
2
2
2
For TEM waves k c  k  k z  0
Hence, we have
kz  k  
 c
26
Transverse ElectroMagnetic (TEM) Waves (cont.)
In a linear, isotropic, homogeneous source-free region,
E  0
In rectangular coordinates, we have
E x
x

E y
y

E z
z
0
Notation:
 t  xˆ
t  E  0
  t   et  x , y  e
 e
 t

x
 yˆ

y
0
   e  x, y    0
  e  x, y    0
jk z z
jk z z
t
t
t
 t   et  x , y 
0
27
Transverse ElectroMagnetic (TEM) Waves (cont.)
Also, for the TEMz mode we have from Faraday’s law (taking the z
component):
zˆ     E   zˆ   j  H
   j  H z
0
Taking the z component of the curl, we have
E y
x

E x
y
Notation:
0
Hence
e y
x

ex
y
 t  xˆ
0

x
 yˆ

y
or
 t  et  x , y   0
 t   et  x , y 
0
28
Transverse ElectroMagnetic (TEM) Waves (cont.)
 t   et  x , y 
0
et  x , y     t   x , y 
 t   et  x , y 
0
 t    t  x, y 
0
Hence
 t   x, y   0
2
29
Transverse ElectroMagnetic (TEM) Waves (cont.)
Since the potential function that describes the electric field in
the cross-sectional plane is two dimensional, we can drop the
“t” subscript if we wish:
   x, y   0
2
   x, y   0
2
Boundary Conditions:
  x , y   V a co n d u cto r " a "
a
b
  x , y   V b co n d u cto r " b "
This is enough to make the potential function unique. Hence,
the potential function is the same for DC as it is for a highfrequency microwave signal.
30
Transverse ElectroMagnetic (TEM) Waves (cont.)
Notes:
 A TEMz mode has an electric field that has exactly the same shape as
a static (DC) field. (A similar proof holds for the magnetic field.)
 This implies that the C and L for the TEMz mode on a transmission
line are independent of frequency.
 This also implies that the voltage drop between the two conductors of
a transmission line carrying a TEMz mode is path independent.
 A TEMz mode requires two or more conductors (a static field cannot
be supported by a single conductor such as a hollow metal pipe.
31
TEM Solution Process
A) Solve Laplace’s equation subject to appropriate
B.C.s.:  2   x , y   0
B) Find the transverse electric field:
C) Find the total electric field:
D) Find the magnetic field:
E  x, y , z   et  x , y  e
H 
1
Z TE M
Z TEM 

kz


k
et  x , y       x , y 

  zˆ  E  ;

 



 jk z z
, kz  k
 z p ro p a g a tin g

32
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