Notes 4 - Waveguides part 1 general theory
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Transcript 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
– Twin lead (twisted pair)
– 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 ejt 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.)
To do this, start with Ampere’s law:
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
In the time-harmonic (sinusoidal steady
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
Ek 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
Ek 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