1 Prof. David R. Jackson Dept. of ECE Notes 4 ECE 5317-6351 Microwave Engineering Fall 2011 Waveguides Part 1: General Theory
Dec 17, 2015
1
Prof. David R. JacksonDept. of ECE
Notes 4
ECE 5317-6351 Microwave Engineering
Fall 2011
Waveguides Part 1:General Theory
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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
Waveguide Introduction
Note: In microwave engineering, the term “waveguide” is often used to mean rectangular or circular waveguide (i.e., a hollow pipe of metal).
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General Solutions for TEM, TE and TM Waves
Assume ejt time dependence and homogeneous source-free materials.
Assume wave propagation in the z direction
zjk zze e
ˆ, , , , zjk zt zE x y z e x y z e x y e
, , , ˆ , zjk zt zH x y z h x y h x y ez
,z zjk k j
z
y
x
, ,
PEC
transverse components
J E
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Helmholtz Equation
E j H
H j E J
E j H j j E J
0
vE
H
Vector Laplacian definition :
2E E E
2 2 2 2ˆ ˆ ˆx y zE x E y E z E
where
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Helmholtz Equation
2
2 2
2 2
2 2
v
v
v
E E j j E J
E E j J
E E j J
E E j E
E j H j j E J
Assume Ohm’s law holds:
J E
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2 2 vE E j E
2 2
2 2
2 2
v
vc
v
E j E
E E
E k E
Next, we examine the term on the right-hand side.
Helmholtz Equation (cont.)
2 2ck
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0
0
0v
H j E J
H j E J
H j E E
E j
E
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.)
2 2 vE k E
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Helmholtz equation
2 2 0E k E
Hence, we have
Helmholtz Equation (cont.)
9
2
c
c
c
H j E J
H j E E
H j E
H j E
H j j H
H H j j H
Similarly, for the magnetic field, we have
H j E J
Helmholtz Equation (cont.)
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2 2 0H k H
Hence, we have
Helmholtz equation
Helmholtz Equation (cont.)
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2 2 0H k H
Summary
Helmholtz equations
2 2 0E k E
Helmholtz Equation (cont.)
These equations are valid for a source-free homogeneous isotropic linear material.
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Assume a guided wave with a field variation in the z direction of the form
zjk ze
Field Representation
Then all six of the field components can be expressed in terms of these two fundamental ones:
,z zE H
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Types of guided waves:
Field Representation (cont.)
TEMz: Ez = 0, Hz = 0
TMz: Ez 0, Hz = 0
TEz: Ez = 0, Hz 0
Hybrid: Ez 0, Hz 0
Microstrip
h
w
er
TEMz
TMz , TEz
Hybrid
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Assume a source-free region with a variation zjk ze
E j H
1) zz y x
Ejk E j H
y
2) zz x y
Ejk E j H
x
3) y xz
E Ej H
x y
4) zz y c x
Hjk H j E
y
5) zz x c y
Hjk H j E
x
6) y xz
H Hj E
x y
Field Representation: Proof
cH j E
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Combining 1) and 5)
2
2
2 2
2
1
( )
c
z z z zx
c c
z
z zz z x x
c
z zx c z
c
k
zc z z x
E k H kj H
y x j
E Hj jk k k H
E Hjk jk H j H
y j x
E HjH k
y x
k y x
1/22 2c zk k k
Cutoff wave number
Field Representation: Proof (cont.)
A similar derivation holds for the other three transverse field components.
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2
2
2
2
z zx c z
c
z zy c z
c
z zx z
c
z zy z
c
E HjH k
k y x
E HjH k
k x y
E HjE k
k x y
E HjE k
k y x
Summary
These equations give the transverse field components in terms of longitudinal components, Ez and Hz.
Field Representation (cont.)
2 2ck
1/22 2c zk k k
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Therefore, we only need to solve the Helmholtz equations for the longitudinal field components (Ez and Hz).
Field Representation (cont.)
2 2 0z zH k H
2 2 0z zE k E
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Transverse Electric (TEz) Waves
0zE
In general, Ex, Ey, Hx, Hy, Hz 0
To find the TEz field solutions (away from any sources), solve
2 2( ) 0zk H
2 2 22
2 2 22 2 ) 0( 0 zz k H
x y zk H
The electric field is “transverse” (perpendicular) to z.
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Recall that the field solutions we seek are assumed to vary as zjk ze
( , , ) ( , )jk zz
z zH x y z h x y e
2
2 22 2
2 2, 0z
c
z
k
k k h x yx y
2 2 2c zk k k
2 2
22 2
, 0c zk h x yx y
Solve subject to the appropriate boundary conditions.
2 2 22
2 2 20zk H
x y z
Transverse Electric (TEz) Waves (cont.)
2 2
22 2
, ,z c zh x y k h x yx y
(This is an eigenvalue problem.)2 .ckThe eigenvalue is always real
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Once the solution for Hz is obtained,
2 2
2 2
z z zx x
c c
z z zy y
c c
jk H HjH E
k x k y
jk H HjH E
k y k x
TE wave impedance
TEz
Zk
yx
y x z
EE
H H k
For a wave propagating in the positive z direction (top sign):
yx
y x z
EE
H H k
Transverse Electric (TEz) Waves (cont.)
For a wave propagating in the negative z direction (bottom sign):
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Also, for a wave propagating in the positive z direction,
ˆ ˆˆ
ˆ ˆˆ
1ˆ( )
t x y
t TE x y
TE t
t t
x TE y
y TE
TE
x
z e ye xe
z e Z
e
xh yh
Z h
h z
h
e Z h
e
Z
Z
ˆ ˆ, , ,t x ye x y xe x y ye x y
Similarly, for a wave propagating in the negative z direction,
1ˆ( )t t
TE
h z eZ
Transverse Electric (TEz) Waves (cont.)
1ˆ, ,t t
TE
h x y z e x yZ
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0zH
Transverse Magnetic (TMz) Waves
In general, Ex, Ey, Ez ,Hx, Hy 0
To find the TEz field solutions (away from any sources), solve
2 2( ) 0zk E
2 2 22
2 2 22 2 ) 0( 0 zz k E
x y zk E
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2 2
22 2
, 0c zk e x yx y
solve subject to the appropriate boundary conditions
2
2 22 2
2 2, 0z
c
z
k
k k e x yx y
2 2 2c zk k k
Transverse Magnetic (TMz) Waves (cont.)
2 2
22 2
, ,z c ze x y k e x yx y
(Eigenvalue problem)
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2 2
2 2
c z z zx x
c c
c z z zy y
c c
j E jk EH E
k y k x
j E jk EH E
k x k y
TM wave impedance
zTM
c
kZ
yx z
y x c
EE k
H H
yx z
y x c
EE k
H H
Once the solution for Ez is obtained,
For a wave propagating in the positive z direction (top sign):
For a wave propagating in the negative z direction (bottom sign):
Transverse Magnetic (TMz) Waves (cont.)
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Also, for a wave propagating in the positive z direction,
ˆ ˆˆ
ˆ ˆˆ
1ˆ( )
t x y
t TM x y
TM t
t t
x TM y
y TM
TM
x
z e ye xe
z e Z
e
xh yh
Z h
h z
h
e Z h
e
Z
Z
ˆ ˆ, , ,t x ye x y xe x y ye x y
Similarly, for a wave propagating in the negative z direction,
1ˆ( )t t
TM
h z eZ
1ˆ, ,t t
TM
h x y z e x yZ
Transverse Magnetic (TMz) Waves (cont.)
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Transverse ElectroMagnetic (TEM) Waves0, 0z zE H
From the previous equations for the transverse field components, all of them are equal to zero if Ez and Hz are both zero.
Unless 2 0ck
For TEM waves 2 2 2 0c zk k k
z ck k
In general, Ex, Ey, Hx, Hy 0
(see slide 16)
Hence, we have
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In a linear, isotropic, homogeneous source-free region,
0E
0
, 0
, 0
, 0
z
z
t
jk zt t
jk zt t
t t
E
e x y e
e e x y
e x y
ˆ ˆt x yx y
In rectangular coordinates, we have
Notation:0yx zEE E
x y z
, 0t te x y
Transverse ElectroMagnetic (TEM) Waves (cont.)
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Also, for the TEMz mode we have from Faraday’s law (taking the z component):
ˆ ˆ 0zz E z j H j H
0y xE E
x y
, 0t te x y
0y xe e
x y
, 0t te x y
ˆ ˆt x yx y
Notation:
or
Transverse ElectroMagnetic (TEM) Waves (cont.)
Taking the z component of the curl, we have
Hence
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, 0t te x y
, ,t te x y x y
, 0t te x y , 0t t x y
2 , 0t x y
Hence
Transverse ElectroMagnetic (TEM) Waves (cont.)
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2 , 0x y
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:
Transverse ElectroMagnetic (TEM) Waves (cont.)
Boundary Conditions:
,
,
a
b
a
b
x y V
x y V
conductor " "
conductor " "
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.
2 , 0x y
a b
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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.
Transverse ElectroMagnetic (TEM) Waves (cont.)
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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:
2 , 0x y
1ˆ ;
TEM
H z E zZ
propagating
TEMz
Zk k
, ,te x y x y
, , , ,zjk zt zE x y z e x y e k k