Waveguiding Systems and Helmholtz Equation in Microwave Engineering
1
Prof. David R. Jackson
Dept. of ECE
N
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s
6
ECE 5317-6351
ECE 5317-6351
Microwave Engineering
Microwave Engineering
F
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1
9
Waveguiding Structures
Part 1: General Theory
Adapted from notes by
Prof. Jeffery T. Williams
2
In general terms, a
waveguiding system
is a system that confines
electromagnetic energy and channels it from one point to another.
(This is opposed to a wireless system that uses antennas.)
Examples
–
Coax
–
Twin lead (twisted pair)
–
Printed circuit lines
(e.g. microstrip)
–
Parallel plate waveguide
–
Rectangular waveguide
–
Circular waveguide
Waveguide Introduction
Waveguide Introduction
N
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.
Waveguides
–
Optical fiber (dielectric waveguide)
Transmission Lines
3
General Notation for Waveguiding Systems
General Notation for Waveguiding Systems
Assume
e
j
t
time dependence and homogeneous source-free materials.
Assume wave propagation in the
z
direction:
Transverse
(
x
,
y
)
components
N
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:
L
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)
.
Example of waveguiding system (a waveguide)
4
Helmholtz Equation
Helmholtz Equation
(complex)
where
5
Helmholtz Equation
Helmholtz Equation
Vector Laplacian definition:
where
6
Next, we examine the term on the right-hand side.
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
So far we have:
7
To do this, start with Ampere’s law:
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.
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
8
Vector Helmholtz equation
Hence, we have
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
9
Similarly, for the magnetic field, we have
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
10
Hence, we have
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
Vector Helmholtz equation
11
Summary
These equations are valid for a source-free, homogeneous, isotropic,
linear material.
Vector Helmholtz equation
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
12
From the property of the vector Laplacian, we have
Recall:
Scalar Helmholtz equation
Helmholtz Equation (cont.)
Helmholtz Equation (cont.)
13
Assume a
guided wave
with a field variation
F
(
z
)
in the
z
direction of the
form
Field Representation
Field Representation
Then all four of the
transverse
(
x
and
y
) field components can be
expressed in terms of the two longitudinal ones:
(This is a property of any guided wave.)
14
Assume a source-free region with a variation
Field Representation: Proof
Field Representation: Proof
Take
(
x
,
y
,
z
)
components
15
Combining
1)
and
5)
:
Cutoff wavenumber
(real number, as discussed later)
Field Representation: Proof (cont.)
Field Representation: Proof (cont.)
A similar derivation holds
for the other three transverse field components.
16
Summary of Results
These equations give the transverse
field components in terms of the
longitudinal
components,
E
z
and
H
z.
Field Representation (cont.)
Field Representation (cont.)
17
Therefore, we only need to solve the Helmholtz equations for the
longitudinal
field components (
E
z
and
H
z
).
Field Representation (cont.)
Field Representation (cont.)
From table
18
Types of guided waves:
Types of Waveguiding Systems
Types of Waveguiding Systems
TEM
z
:
E
z
=
0
, H
z
=
0
TM
z
:
E
z
0
, H
z
=
0
TE
z
:
E
z
=
0
, H
z
0
Hybrid
:
E
z
0
, H
z
0
TEM
z
TM
z
,
TE
z
Hybrid
Hybrid
19
Waveguides
Waveguides
Two types of modes: TE
z
,
TM
z
We assume that the boundary is PEC.
We assume that the inside is filled with a homogenous isotropic linear
material (could be air)
An example of a
waveguide
(rectangular waveguide)
20
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves
) Waves
In general,
E
x
,
E
y
,
H
x
,
H
y
,
H
z
0
To find the TE
z
field solutions (away from any sources), solve
The electric field is “transverse” (perpendicular) to
z
.
or
21
Recall that the field solutions we seek are assumed to
vary as
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
22
We need to solve the eigenvalue problem subject to the appropriate boundary conditions.
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
(2D Eigenvalue problem)
(A proof of this may be found in the ECE 6340 notes.)
For this type of eigenvalue problem, the eigenvalue is always
real
.
Change notation
23
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
Hence, TE
z
modes exist inside of a waveguide (conducting pipe).
A solution to the eigenvalue problem can always be found for a PEC boundary
(proof omitted).
Neumann boundary condition
(see below)
24
Once the solution for
H
z
is obtained, we use
TE wave impedance
For a wave propagating in the positive
z
direction (top sign):
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
For a wave propagating in the negative
z
direction (bottom sign):
25
For a wave propagating in the positive
z
direction, we also have:
Similarly, for a wave propagating in the negative
z
direction,
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
26
Summarizing both cases, we have
+ sign: wave propagating in the +
z
direction
- sign: wave propagating in the -
z
direction
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
27
Transverse Magnetic (TM
Transverse Magnetic (TM
z
z
) Waves
) Waves
In general,
E
x
,
E
y
,
E
z
,
H
x
,
H
y
0
To find the TE
z
field solutions (away from any sources), solve
or
The magnetic field is “transverse” (perpendicular) to
z
.
28
Transverse Magnetic (TM
Transverse Magnetic (TM
z
z
) Waves (cont.)
) Waves (cont.)
Recall that the field solutions we seek are assumed to
vary as
29
We need to solve the eigenvalue problem subject to the appropriate boundary conditions.
Transverse Electric (TE
Transverse Electric (TE
z
z
) Waves (cont.)
) Waves (cont.)
(Eigenvalue problem)
(A proof of this may be found in the ECE 6340 notes.)
For this type of eigenvalue problem, the eigenvalue is always
real
.
30
Transverse Magnetic (TM
Transverse Magnetic (TM
z
z
) Waves (cont.)
) Waves (cont.)
A solution to the eigenvalue problem can always be found for a PEC boundary
(proof omitted).
Hence, TM
z
modes exist inside of a waveguide (conducting pipe).
Dirichlet boundary condition
(see below)
31
TM wave impedance
Once the solution for
E
z
is obtained, we use
For a wave propagating in the positive
z
direction (top sign):
For a wave propagating in the negative
z
direction (bottom sign):
Transverse Magnetic (TM
Transverse Magnetic (TM
z
z
) Waves (cont.)
) Waves (cont.)
32
For a wave propagating in the positive
z
direction, we also have:
Similarly, for a wave propagating in the negative
z
direction,
Transverse Magnetic (TM
Transverse Magnetic (TM
z
z
) Waves (cont.)
) Waves (cont.)
33
Summarizing both cases, we have
+ sign: wave propagating in the +
z
direction
- sign: wave propagating in the -
z
direction
Transverse Magnetic (TM
Transverse Magnetic (TM
z
z
) Waves (cont.)
) Waves (cont.)
34
Transverse ElectroMagnetic (TEM) Waves
Transverse ElectroMagnetic (TEM) Waves
From the previous table for the transverse field components, all of
them are equal to zero if
E
z
and
H
z
are both zero.
Unless
For TEM waves
In general,
E
x
,
E
y
,
H
x
,
H
y
0
Hence, we have
35
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
From EM boundary conditions, we have:
so
The current flows purely in the
z
direction.
36
In a linear, isotropic, homogeneous source-free region,
In rectangular coordinates, we have
Notation:
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
Hence, we have
37
Also, for the TEM
z
mode, we have from Faraday’s law (taking the
z
component):
Note:
Using the formula for the
z
component of the curl
of
E
, we have:
Hence
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
Hence, we have:
38
Hence
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
39
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:
Boundary Conditions:
This is enough to make the potential function
unique
.
Hence, the potential function is the same for DC as it is for a high-
frequency microwave signal.
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
The field of a TEM mode
does not change shape with frequency
:
it has the same shape as a DC field.
40
Notes:
A TEM
z
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 TEM
z
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 TEM
z
mode is path independent.
A TEM
z
mode requires two or more conductors: a static electric field
cannot exist inside of a waveguide (hollow metal pipe) due to the
Faraday cage effect.
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
41
For a TEM mode, both wave impedances are the same:
Transverse ElectroMagnetic (TEM) Waves (cont.)
Transverse ElectroMagnetic (TEM) Waves (cont.)
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TEM Solution Process
TEM Solution Process
A) Solve Laplace’s equation subject to appropriate B.C.s.:
B) Find the transverse electric field:
C) Find the total electric field:
D) Find the magnetic field:
N
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T
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n
t
h
e
w
a
v
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n
u
m
b
e
r
k
z
=
k
.
43
Note on Lossy Transmission Line
Note on Lossy Transmission Line
A TEM mode can have an arbitrary amount of dielectric loss.
A TEM mode cannot have conductor loss.
Inside conductors:
In practice, a small conductor loss will not change the shape of the fields
too much, and the mode is
approximately
TEM.
If there is conductor loss,
there must be an
E
z
field.
44
If there is only dielectric loss (an exact TEM mode):
Equate real and imaginary parts
Note on Lossy Transmission Line (cont.)
Note on Lossy Transmission Line (cont.)
45
From these two equations we have:
Equations for a TEM mode:
N
o
t
e
:
These formulas were assumed
previously in Notes 3.
Note on Lossy Transmission Line (cont.)
Note on Lossy Transmission Line (cont.)
46
This general formula accounts for
both dielectric and conductor loss
:
Dielectric loss
Conductor loss
When
R
is present the mode is not
exactly
TEM, but we usually ignore this.
Note on Lossy Transmission Line (cont.)
Note on Lossy Transmission Line (cont.)
47
Summary
Summary
T
E
M
M
o
d
e
:
48
Summary (cont.)
Summary (cont.)
T
r
a
n
s
m
i
s
s
i
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l
i
n
e
m
o
d
e
(
a
p
p
r
o
x
i
m
a
t
e
T
E
M
m
o
d
e
)
:
The mode requires
two
conductors.
The mode is purely TEM only when
R
= 0
.
49
Summary (cont.)
Summary (cont.)
T
E
z
M
o
d
e
:
The mode can exist inside of a
single
pipe (waveguide).
50
T
M
z
M
o
d
e
:
Summary (cont.)
Summary (cont.)
The mode can exist inside a
single
pipe (waveguide).
Waveguiding systems are essential in confining and channeling electromagnetic energy, with examples including rectangular and circular waveguides. The general notation for waveguiding systems involves wave propagation and transverse components. The Helmholtz Equation is a key concept in analyzing electromagnetic fields, providing insights into wave behavior in waveguides. By examining the equation and related laws, we gain a deeper understanding of waveguiding principles in microwave engineering.
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Adapted from notes by Prof. Jeffery T. Williams ECE 5317-6351 Microwave Engineering Fall 2019 Prof. David R. Jackson Dept. of ECE Notes 6 Waveguiding Structures Part 1: General Theory 1
Waveguide Introduction In general terms, a waveguiding system is a system that confines electromagnetic energy and channels it from one point to another. (This is opposed to a wireless system that uses antennas.) Examples Parallel plate waveguide Rectangular waveguide Circular waveguide Coax Twin lead (twisted pair) Printed circuit lines (e.g. microstrip) Transmission Lines Waveguides Optical fiber (dielectric 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 Notation for Waveguiding Systems Assume ej t time dependence and homogeneous source-free materials. Assume wave propagation in the z direction: z = = jk z z e e z = + = , j k j z c jk z C ( ) ( ) ( ) = + jk z Example of waveguiding system (a waveguide) z e x y , , , , E x y z e x y e z t z Transverse (x,y) components j c = j ( ) ( ) ( ) = + jk z z h x y , , , , c c H x y z h x y e = j z c t z tan c d Note: Lower case letters denote 2-D fields (the z term is suppressed). c 3
Helmholtz Equation = E j H = E v = + H j E J = 0 H ( ( j ) + = = = E j H ) j j E J ( ) + j E E j ( ) j j E Recall: c c = 2 k E where 2 2 k (complex) c 4
Helmholtz Equation = 2 E k E ( ) 2E Vector Laplacian definition: E E ( ) ( ) ( ) = + + where 2 2 2 2 x y z E E E E x y z ( ) = 2 2 E E k E = 2 2 v E k E + = 2 2 v E k E 5
Helmholtz Equation (cont.) So far we have: + = 2 2 v E k E Next, we examine the term on the right-hand side. 6
Helmholtz Equation (cont.) To do this, start with Ampere s law: = H j E c ( ) ( ) = = H j E c ( ) = = 0 j E c 0 0 E D = 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. 0 v 7
Helmholtz Equation (cont.) Hence, we have + = 2 2 0 E k E Vector Helmholtz equation 8
Helmholtz Equation (cont.) Similarly, for the magnetic field, we have = + H j E J = = = + H H j E E E ( ) + j ( ) j H j E Recall: c c ( ) ( ) ( = ) = H j E c ( )( j ) H j j H c ( ) k H ( )( ) = 2 H H j H c = 2 2 H 9
Helmholtz Equation (cont.) Hence, we have + = 2 2 0 H k H Vector Helmholtz equation 10
Helmholtz Equation (cont.) Summary + = 2 2 0 E k E + = 2 2 0 H k H Vector Helmholtz equation These equations are valid for a source-free, homogeneous, isotropic, linear material. 11
Helmholtz Equation (cont.) From the property of the vector Laplacian, we have + = 2 2 0 E k E z z + = 2 2 0 H k H z z Scalar Helmholtz equation ( ) ( ) ( ) = + + 2 2 2 2 x y z E E E E Recall: x y z 12
Field Representation Assume a guided wave with a field variation F(z) in the z direction of the form ( ) F z e = jk z z (This is a property of any guided wave.) Then all four of the transverse (x and y) field components can be expressed in terms of the two longitudinal ones: ( z E H ) , z 13
Field Representation: Proof jk z Assume a source-free region with a variation e z = = H j E E j H c Take (x,y,z) components H y E y = 4) jk H j E z = 1) jk E j H z z y c x z y x E x H x = 2) jk E j H z = 5) jk H j E z z x y z x c y E x E y H H y = y 3) j H x = y 6) j E x z z x 14
Field Representation: Proof (cont.) Combining 1) and 5): = 1 E y H x = jk jk H j H z z z z x x j c 2 z E y k H x k j H z z z x j c c E y H x = 2 2 z ( ) j jk k k H z z c z x 2 2 z 2 c k k k k c 1 k E y H x Cutoff wavenumber (real number, as discussed later) = H j jk z z x c z 2 c A similar derivation holds for the other three transverse field components. 15
Field Representation (cont.) Summary of Results E y H j = H k z z x c z 2 c k x These equations give the transverse field components in terms of the longitudinal components, Ezand Hz. E x H j = H k z z y c z 2 c k y = 2 2 k c E x H y j = + E k z z = 2 2 z k k k x z 2 c k c E y H x j = + E k z z y z 2 c k 16
Field Representation (cont.) Therefore, we only need to solve the Helmholtz equations for the longitudinal field components (Ez and Hz). + = 2 2 0 E k E z z + = 2 2 0 H k H z z From table , , , E E H H x y x y 17
Types of Waveguiding Systems Types of guided waves: TEMz ( R = TEMz: Ez = 0, Hz = 0 ) 0 TMz: Ez 0, Hz = 0 TEz: Ez = 0, Hz 0 Hybrid: Ez 0, Hz 0 TMz ,TEz ( PEC ) w Hybrid Hybrid r h Microstrip 18
Waveguides We assume that the boundary is PEC. We assume that the inside is filled with a homogenous isotropic linear material (could be air) An example of a waveguide (rectangular waveguide) Two types of modes: TEz ,TMz 19
Transverse Electric (TEz) Waves = 0 z E 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 + = 2 2 ( ) 0 k H z or 2 2 2 + + + = 2 0 k H z 2 2 2 x y z 20
Transverse Electric (TEz) Waves (cont.) 2 2 2 + + + = 2 0 k H z 2 2 2 x y z Recall that the field solutions we seek are assumed to vary as ( ) F z e = jk z z jk z z = ( , , ) H x y z ( , ) h x y e z z 2 2 ( ) + + = 2 c 2 2 z 2 z 2 , 0 k k k k k h x y z 2 2 x y 2 c k 2 2 ( ) + + = 2 c , 0 k h x y z 2 2 x y 2 2 ( ) ( ) + = 2 c , , h x y k h x y z z 2 2 x y 21
Transverse Electric (TEz) Waves (cont.) 2 2 ( ) ( ) + = 2 c , , h x y k h x y z z 2 2 x y Change notation 2 2 ( ) ( ) + = , , x y x y 2 2 x y (2D Eigenvalue problem) ( ) , z h x y k = = eigenvalue = = eigenfunction 2 c For this type of eigenvalue problem, the eigenvalue is always real. (A proof of this may be found in the ECE 6340 notes.) We need to solve the eigenvalue problem subject to the appropriate boundary conditions. 22
Transverse Electric (TEz) Waves (cont.) 2 2 ( ) ( ) Neumann boundary condition (see below) + = 2 c , , h x y k h x y z z 2 2 x y A solution to the eigenvalue problem can always be found for a PEC boundary (proof omitted). Hence, TEz modes exist inside of a waveguide (conducting pipe). ( ) , zh x y z PEC c C c zh n = 0 C Neumann boundary condition 23
Transverse Electric (TEz) Waves (cont.) Once the solution for Hz is obtained, we use jk k jk k H x H y j k H y = = H E z z z x x 2 c 2 c j H x = = H E z z z y y 2 c 2 c k For a wave propagating in the positive z direction (top sign): E H E H = = y x k y x z For a wave propagating in the negative z direction (bottom sign): TE wave impedance E H E H = = y x k y x z Z TE k z 24
Transverse Electric (TEz) Waves (cont.) For a wave propagating in the positive z direction, we also have: ( ) ( ) ( ) = + xe x y ye , , , e x y x y t x y Recall: z e = ye xe t x y = = ( ) e Z h z e = = + yh xh Z x TE Z h y t TE y x e Z h y TE x TE t 1 = z e ( ) h t t Z TE Similarly, for a wave propagating in the negative z direction, 1 = z e ( ) h t t Z TE 25
Transverse Electric (TEz) Waves (cont.) Summarizing both cases, we have 1 ( ) ( ) ( ) = z e x y , , h x y t t Z TE + sign: wave propagating in the + z direction - sign: wave propagating in the - z direction 26
Transverse Magnetic (TMz) Waves = 0 H The magnetic field is transverse (perpendicular) to z. z In general, Ex, Ey, Ez ,Hx, Hy 0 To find the TEz field solutions (away from any sources), solve + = 2 2 ( ) 0 k E z or 2 2 2 + + + = 2 0 k E z 2 2 2 x y z 27
Transverse Magnetic (TMz) Waves (cont.) + + 2 2 2 + = 2 0 k E z 2 2 2 x y z Recall that the field solutions we seek are assumed to vary as ( , , ) z E x y z ( ) = jk z F z e z jk z z = ( , ) e x y e z 2 2 ( ) + + = 2 z 2 , 0 k k e x y 2 c 2 2 z k k k z 2 2 x y 2 c k 2 2 ( ) + + = 2 c , 0 k e x y z 2 2 x y 2 2 ( ) ( ) + = 2 c , , e x y k e x y z z 2 2 x y 28
Transverse Electric (TEz) Waves (cont.) 2 2 ( ) ( ) + = 2 c , , e x y k e x y z z 2 2 x y (Eigenvalue problem) ( ) = , = e x y eigenfunction z 2 c k eigenvalue We need to solve the eigenvalue problem subject to the appropriate boundary conditions. For this type of eigenvalue problem, the eigenvalue is always real. (A proof of this may be found in the ECE 6340 notes.) 29
Transverse Magnetic (TMz) Waves (cont.) 2 2 ( ) ( ) + = 2 c , , e x y k e x y Dirichlet boundary condition (see below) z z 2 2 x y A solution to the eigenvalue problem can always be found for a PEC boundary (proof omitted). Hence, TMz modes exist inside of a waveguide (conducting pipe). ( ) , ze x y z PEC c C c ze = 0 C Dirichlet boundary condition 30
Transverse Magnetic (TMz) Waves (cont.) Once the solution for Ez is obtained, we use j E y jk k jk k E x E y = = H c E z z z x x 2 c j 2 c k E x = = H c E z z z y y 2 c 2 c k For a wave propagating in the positive z direction (top sign): E H E H k = = y x z y x c For a wave propagating in the negative z direction (bottom sign): E H TM wave impedance E H k = = y x z k y x c Z z TM c 31
Transverse Magnetic (TMz) Waves (cont.) For a wave propagating in the positive z direction, we also have: ( ) ( ) ( ) = + xe x y ye , , , e x y x y t x y Recall: = ye xe z e t x y ( ) = = e Z h = = + yh xh z e Z x TM Z y h t TM y x e Z h y TM x TM t 1 = z ( ) h e t t Z TM Similarly, for a wave propagating in the negative z direction, 1 = z e ( ) h t t Z TM 32
Transverse Magnetic (TMz) Waves (cont.) Summarizing both cases, we have 1 ( ) ( ) ( ) = z e x y , , h x y t t Z TM + sign: wave propagating in the + z direction - sign: wave propagating in the - z direction 33
Transverse ElectroMagnetic (TEM) Waves = = 0, 0 E H z z In general, Ex, Ey, Hx, Hy 0 From the previous table for the transverse field components, all of them are equal to zero if Ez and Hz are both zero. E y H j = H k z z ck = x c z 2 Unless 2 c 0 k x E x H j = 2 c 2 2 z = 0 k k k For TEM waves H k z z y c z 2 c k y E x H y j Hence, we have = + E k z z x z 2 c k = = k k z c E y H x j = + E k z z y z 2 c k 34
Transverse ElectroMagnetic (TEM) Waves (cont.) ( ) ( ) ( ) = = jk z jkz , , , , H x y z h x y e h x y e z t From EM boundary conditions, we have: = n sJ H th so n ( ) ( ) e = jkz n , J h x y s t = z J J y s sz x The current flows purely in the z direction. 35
Transverse ElectroMagnetic (TEM) Waves (cont.) In a linear, isotropic, homogeneous source-free region, = 0 E In rectangular coordinates, we have E y E x E z y + + = 0 x z Notation: = + x y = 0 E t x y t ( ) ( ) = jk z , 0 e x y e z t t ( ) ( ) ( ) ( ) = jk z jk z , , 0 e e x y e x y e z z t t t ( ) ( ) = , 0 e x y t t Hence, we have ( ) ( ) = , 0 te x y t 36
Transverse ElectroMagnetic (TEM) Waves (cont.) Also, for the TEMzmode, we have from Faraday s law (taking the z component): ( ) ( z E z j = ) = = 0 H j H z Using the formula for the z component of the curl of E, we have: E x E y Note: y = 0 x = + x y Hence t x y ( ) ( ) ( ) = + e x xe x y ye , , , e x y x y e y y = t x y 0 x e x e y ( ) = y z , e x y x Hence, we have: t t ( ) ( ) = , 0 te x y t 37
Transverse ElectroMagnetic (TEM) Waves (cont.) ( ) ( ) = , 0 te x y t ( ) ( ) = , , e x y x y t t ( ) ( ) ( ) ( ) = = , 0 , 0 te x y x y t t t Hence ( ) = 2 t , 0 x y 38
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: ( ) = 2 , 0 x y ( ) = 2 , 0 x y Boundary Conditions: ( ( ) ) = = , x y conductor " " a a a b , x y conductor " " b b PEC conductors This is enough to make the potential function unique. Hence, the potential function is the same for DC as it is for a high- frequency microwave signal. The field of a TEM mode does not change shape with frequency: it has the same shape as a DC field. 39
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 electric field cannot exist inside of a waveguide (hollow metal pipe) due to the Faraday cage effect. 40
Transverse ElectroMagnetic (TEM) Waves (cont.) For a TEM mode, both wave impedances are the same: = Recall: zk k = = = = = Z TE k k z c c k k = = = = = c Z z TM c c c c Note: is complex for lossy media. 41
TEM Solution Process A) Solve Laplace s equation subject to appropriate B.C.s.: ( ) , x y = 2 0 ( ) ( ) = B) Find the transverse electric field: , , te x y x y ( ) ( ) C) Find the total electric field: = = jk z , , , , E x y z e x y e k k z t z 1 ( ) D) Find the magnetic field: = z ; H E z propagating Note: The only frequency dependence is in the wavenumber kz = k. 42
Note on Lossy Transmission Line A TEM mode can have an arbitrary amount of dielectric loss. A TEM mode cannot have conductor loss. = J E Inside conductors: ( ) I z + + + + + + + - - - - - - - - - - ( ) V z If there is conductor loss, there must be an Ezfield. J E z z In practice, a small conductor loss will not change the shape of the fields too much, and the mode is approximately TEM. 43
Note on Lossy Transmission Line (cont.) If there is only dielectric loss (an exact TEM mode): ( )( ) = + j L G + j C R ( ) = = = jk jk k k z z ( )( ) = j L G + j C ) ( = ) ( )( ( ) = = jk j L G + j C 2 1 tan j 0 0 r r d j c Equate real and imaginary parts ( ) = 1 tan j j 0 0 r r d = LC ( ) 0 0 r r Note denotes real : r r ( )( ) L G ( ) = 2 tan 0 0 r r d 44
Note on Lossy Transmission Line (cont.) Equations for a TEM mode: = LC 0 0 r r ( )( ) L G ( ) = 2 tan 0 0 r r d ( ) ( ) ( ) 2 tan tan tan G C = = = 0 0 2 0 0 0 0 r r d r r d r r d LC LC 0 0 r r From these two equations we have: = LC 0 0 r r Note: These formulas were assumed previously in Notes 3. G C = tan d 45
Note on Lossy Transmission Line (cont.) This general formula accounts for both dielectric and conductor loss: ( )( ) = + j L G + j C R Dielectric loss Conductor loss When R is present the mode is not exactly TEM, but we usually ignore this. 46
Summary TEM Mode: ( ) = 1 tan j = = 0 E H rc r d z z = = r ( ) = = = 0 1 tan k k k k j c rc 0 0 z r rc r r d 1 ( ) = = ( ) ( ) 376.7303 0 = z e x y , , h x y 0 t t 0 47
Summary (cont.) Transmission line mode (approximate TEM mode): The mode requires two conductors. The mode is purely TEM only when R = 0. ( )( ) = = + j L G + j C jk R z = lossless L Z = LC 0 = 0 0 r r 0 0 r r = lossless / C Z 0 0 0 r r G C ( R ) R tan G C = = tan d d L C = z J J lossless Z s sz 0 48
Summary (cont.) TEz Mode: The mode can exist inside of a single pipe (waveguide). = 0, 0 E H z z ( ) = 2 2 c = = k k k 1 tan k k k j 0 0 r rc r r d z kc = real number (depends on geometry and mode number) 1 ( ) ( ) ( ) = z e x y , , h x y = Z t t Z TE k TE z ( ) ( ) k h x y = 2 t 2 c , , h x y z z 49
Summary (cont.) TMz Mode: The mode can exist inside a single pipe (waveguide). = 0, 0 H E z z ( ) = 2 2 c = = k k k 1 tan k k k j 0 0 r rc r r d z kc = real number (depends on geometry and mode number) 1 ( ) k ( ) ( ) = z e x y , , h x y = Z z t t Z TM TM ( ) ( ) k e x y = 2 t 2 c , , e x y z z 50
