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Electromagnetics:
Electromagnetic Field Theory
Waveguide Analysis Setup
Lecture Outline
•Governing Equations for Waveguide Analysis
•Reduced Set of Equations for Waveguide Analysis
•Analysis Setup• for hybrid modes• for TEM analysis• for TE and TM analysis• for analyzing slab waveugides
Slide 2
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Slide 3
Governing Equations for Waveguide Analysis
Steps for Waveguide Analysis
Slide 4
1. Draw the waveguide.2. Assume a form of the solution. Outer regions must decay exponentially or be equal to
zero.3. Substitute solution into Maxwell’s equations.4. Simplify equations based on the geometry of the waveguide.5. Manipulate equations into a differential equation to solve. This is called the governing
equation.6. Solve the governing equation in each homogeneous region of the waveguide.7. “Connect” the solutions in each region using boundary conditions.8. Calculate the overall field solution.9. Use the field solution to calculate the waveguide parameters such as , Z0, and the
profile of the fields.
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Various Wave Equations
Slide 5
E j H
H j E
1. Maxwell’s Curl Equations
E
E j H Hj
2. Wave Equation in General Media
1 2
H j E
Ej E
j
E E
3. Wave Equation in LHI Media
1 2
2
E E
E E
E
2 2
2 2 0
E k E
E k E
4. Wave Equation Decouples
2 2
2 2
2 2
0
0
0
x x
y y
z z
E k E
E k E
E k E
These equations are solved independently.wave numberk
Expand Maxwell’s Equations
Slide 6
Maxwell’s equations are used to analyze waveguides.
E j H
The two curl equations expand into a set of six coupled partial differential equations.
yzx
x zy
y xz
EEj H
y z
E Ej H
z xE E
j Hx y
yzx
x zy
y xz
HHj E
y z
H Hj E
z xH H
j Ex y
There are six field components to solve for: Ex, Ey, Ez, Hx, Hy, and Hz.
Yikes!!
H j E
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General Form of Solution for Waveguides
Slide 7
A mode in a waveguide has the following general mathematical form.
0, , , j zE x y z E x y e
0 ,E x y
x
y
z
phase constant
complex amplitude,mode shape
accumulation of phase in z direction
j ze
This means the problem can be solved by just analyzing the cross section in the x-y plane. The problem reduces mathematically to two dimensions.
3D 2D
Animation of a Waveguide Mode
Slide 8
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Assume the Form of the Solution
Slide 9
For a waveguide uniform in the z direction, the solution will have the form
0 0, , , , , ,j z j zE x y z E x y e H x y z H x y e
Substituting this solution into the set of six equations gives
0,0, 0,
0,0, 0,
0, 0,0,
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
0,0, 0,
0,0, 0,
0, 0,0,
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
Things are a little simpler, but there are still six field components to solve for.
Slide 10
Reduced Set of Equations for
Waveguide Analysis
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Reducing Number of Terms
Slide 11
0,0, 0,
0,0, 0,
0, 0,0,
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
0,0, 0,
0,0, 0,
0, 0,0,
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
0, 0,0, 2 2
0, 0,0, 2 2
z zx
z zy
E HjH
k y x
E HjH
k x y
0, 0,0, 2 2
0, 0,0, 2 2
z zx
z zy
E HjE
k x y
E HjE
k y x
Now to analyze a waveguide, it is only necessary to solve for E0,z and H0,z.
It is possible to put 𝐸 , , 𝐸 , , 𝐻 , , and 𝐻 , in terms of just 𝐸 , and 𝐻 , .
Reduce the Number of Terms to Solve (1 of 2)
Slide 12
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1a
Eq. 1b
Eq. 1c
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1d
Eq. 1e
Eq. 1f
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
Step 1 – Solve Eq. (1e) for E0,y.
0,0, 0,
1 zy x
HE j H
j x
Step 2 – Substitute this expression into Eq. (1a) to eliminate E0,y.
0, 0,0, 0,
1z zx x
E Hj j H j H
y j x
Step 3 – Recall that k2 = 2 and solve this new expression for H0,x.
0, 0,0, 2 2
z zx
E HjH
k y x
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Reduce the Number of Terms to Solve (2 of 2)
Slide 13
Step 4 – Derive three more similar equations.
Solve Eq. (1d) for E0,x, substitute that expression into Eq. (1b) and solve for H0,y.
0, 0,0, 2 2
z zy
E HjH
k x y
Solve Eq. (1b) for H0,y, substitute that expression into Eq. (1d) and solve for E0,x.
0, 0,0, 2 2
z zx
E HjE
k x y
Solve Eq. (1a) for H0,x, substitute that expression into Eq. (1e) and solve for E0,y.
0, 0,0, 2 2
z zy
E HjE
k y x
Reduced Set of Equations
Slide 14
Step 5 – Define the cutoff wave number kc as
2 2 2ck k
Now all of the transverse field components 𝐸 , , 𝐸 , , 𝐻 , and 𝐻 , are expressed
in terms of just the two longitudinal components 𝐸 , and 𝐻 , .
0, 0,0, 2
c
0, 0,0, 2
c
z zx
z zy
E HjH
k y x
E HjH
k x y
0, 0,0, 2
c
0, 0,0, 2
c
z zx
z zy
E HjE
k x y
E HjE
k y x
Analyzing a waveguide reduces to just solving for E0,z and H0,z. The remaining field components can be calculated directly from these two terms.
This term will have more meaning later on.
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How To Find E0,z and H0,z?
Slide 15
Recall that in LHI media, the wave equation simplified to
1 2
2 2
2 2
2 2
2 2
0
0
0
0
x x
y y
z z
E E
E k E
E k E
E k E
E k E
1 2
2 2
2 2
2 2
2 2
0
0
0
0
x x
y y
z z
H H
H k H
H k H
H k H
H k H
Substituting the solution 𝐸 𝐸 , 𝑒 into the bottom equations above gives
2 20, c 0, 0z zE k E 2 2
0, c 0, 0z zH k H
Solution Categories
Slide 16
Hybrid Mode Analysis
𝐸 , 0 and 𝐻 , 0
Can analyze any waveguide.Usually very complicated.
• No simplifications to the analysis are possible.• Solutions contain all six field components
TEM Analysis
𝐸 , 𝐻 , 0
• Must be a transmission line,• and must have a homogeneous dielectric
• Transverse electromagnetic (TEM) mode• Both electric and magnetic fields are transverse to the direction of propagation.
TE & TM Analysis
𝐸 , 0 or 𝐻 , 0,but not both
• Homogeneous dielectric,• or has a dielectric uniform in one direction
• Transverse electric (TE) mode• 𝐸 , 0 so the electric field is completely transverse to the direction of propagation.
• Transverse magnetic (TM) mode• 𝐻 , 0 so the magnetic field is completely transverse to the direction of propagation.
Slab Waveguide Analysis
• Material properties uniform and infinite along one direction
• Maxwell’s equations simplify and decouple into two independent sets of modes.
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Slide 17
Setup for AnalyzingHybrid Modes
Eliminate 𝐸 , and 𝐻 ,
Slide 18
To setup for hybrid modes, back up to Maxwell’s equations in linear and isotropic media (i.e. can be inhomogeneous).
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1a
Eq. 1b
Eq. 1c
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 2a
Eq. 2b
Eq. 2c
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
Solve Eq. (1c) for H0,z and solve Eq. (2c) for E0,z.
0, 0,0,
1 Eq. 3ay x
z
E EH
j x y
0, 0,
0,
1 Eq. 3by x
z
H HE
j x y
Substitute Eq. (3a) into Eqs. (2a) and (2b), & substitute Eq. (3b) into Eqs. (1a) and (1b).
0, 0,0, 0,
0, 0,0, 0,
1 1 Eq. 4a
1 1 Eq. 4b
y xy x
y xx y
H HH E
x x y
H HH E
y x y
0, 0,0, 0,
0, 0,0, 0,
1 1 Eq. 5a
1 1 Eq. 5b
y xy x
y xx y
E EE H
x x y
E EE H
y x y
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Form a Matrix Equation
Slide 19
The four remaining equations can be written more compactly as
0, 0,
0, 0,
1 1 1 1
Eq. 61 1 1 1
x x
y y
x y x x H E
H E
y y y x
0, 0,
0, 0,
1 1 1 1
Eq. 71 1 1 1
x x
y y
E Hx y x xE H
y y y x
Full Wave Analysis
Solve Eq. (7) for the magnetic field components.
0, 0,
0, 0,
1 1 1 1
1= Eq. 8
1 1 1 1
x x
y y
H Ex y x xH E
y y y x
Solve Eq. (8) into Eq. (6) to arrive at the final wave equation to be solved.
0, 0,2
0, 0,
1 1 1 1 1 1 1 10
01 1 1 1 1 1 1 1
x x
y y
x y x x E Ex y x xE E
y y y x y y y x
Yikes!! This is typically solved numerically on a computer.
Quasi‐LP Analysis
Slide 20
Recognizing that the hybrid modes tend to be strongly linearly polarized, a simplifying approximation can be made. The approximation is that the cross coupling between E0,x and E0,y is weak and can be neglected. Under this condition, the governing equation separates into two independent equations, one for each LP mode.
2 22
2 22
1 1 1 1 1 1
1 1 1 1 1 1
xx
yy
x x y y x y x y
y y x x y x y x
0, 0,2
0, 0,
0
0xx xy x x
yx yy y y
E E
E E
20, 0,
20, 0,
0
0xx x x
yy y y
E E
E E
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Slide 21
Setup for TEM Analysis
Existence Conditions for TEM
Slide 22
TEM modes only exist in transmission lines embedded in a homogeneous fill.
Does Not Support TEMSupports TEM
Does Not Support TEM
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TEM Analysis (1 of 3)
Slide 23
For TEM waves, E0,z = H0,z = 0. Under this condition, Maxwell’s equations reduce to
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1a
Eq. 1b
Eq. 1c
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1d
Eq. 1e
Eq. 1f
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
0, 0,
0, 0,
0, 0,
Eq. 2a
Eq. 2b
0 Eq. 2c
y x
x y
y x
j E j H
j E j H
E E
x y
0, 0,
0, 0,
0, 0,
Eq. 2d
Eq. 2e
0 Eq. 2f
y x
x y
y x
j H j E
j H j E
H H
x y
TEM Analysis (2 of 3)
Slide 24
0, 0,
0, 0,
0, 0,
Eq. 2a
Eq. 2b
0 Eq. 2c
y x
x y
y x
j E j H
j E j H
E E
x y
0, 0,
0, 0,
0, 0,
Eq. 2d
Eq. 2e
0 Eq. 2f
y x
x y
y x
j H j E
j H j E
H H
x y
0, 0, y xH E
Solve Eq. (2d) for H0,y.
0, 0,
2 20, 0,
2 20, 0,
x x
x x
x x
j E j E
E E
E k E
Substitute H0,y into Eq. (2b).
From the previous slide…
k This shows that for TEM analysis
Previously, the cutoff wave number was defined as 𝑘 𝑘 𝛽 .
If = k, then kc = 0 indicating that there is no cutoff frequency for the TEM mode.
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TEM Analysis (3 of 3)
Slide 25
In LHI media, recall that the wave equation was
But for the TEM mode, kc = 0.
The wave equation reduces to Laplace’s equation from electrostatics.
2 20, c 0, 0xy xyE k E
2 20, c 0, 0xy xyE k E
20, 0xyE
Alternate Derivation of TEM Analysis
Slide 26
The TEM mode in a transmission line has no cutoff frequency (kc = 0).This means that it can be analyzed as 0 and the problem reduces to an electrostatics problem.
Derivation
Maxwell’s equations for electrostatics
0 Eq. 3a
0 Eq. 3b
Eq. 3c
Eq. 3d
E
D
D E
E V
Substitute Eq. (3c) into Eq. (3b).
0 Eq. 4E
Substitute Eq. (3d) into Eq. (4).
0V
For isotropic dielectrics
0V
For homogeneous dielectrics
2 0V
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Slide 27
Setup forTE & TM Analysis
Existence Conditions for TE and TM Modes
Slide 28
TE and TM modes only exist in waveguides with a homogeneous fill or in waveguides with a uniform axis like slabs and circularly symmetric guides.
Does Not Support TE or TMSupports TE and TM
Supports TE and TM Does Not Support TE or TM
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TE Analysis in LHI Media
Slide 29
Choose to set E0,z = 0 and H0,z ≠ 0. This means it is only necessary to solve for H0,z.
2 20, c 0, 0z zE k E 2 2
0, c 0, 0z zH k H An added benefit of this solution approach is that H0,z is tangential to all boundaries in a waveguide.
For TE analysis, the other field components are calculated just from H0,z.
From this, the characteristic impedance 𝑍 is
0,2
0, cTE
0,0,2c
z
x
zy
HjE k y k
ZHjH
k y
0,0, 2
c
0,0, 2
c
zx
zy
HjH
k x
HjH
k y
0,0, 2
c
0,0, 2
c
zx
zy
HjE
k y
HjE
k x
is found by solving the wave equation.
TM Analysis in LHI Media
Slide 30
Choose to set E0,z ≠ 0 and H0,z = 0. This means it is only necessary to solve for E0,z.
2 2 2 20, c 0, 0, c 0,0 0z z z zE k E H k H An added benefit of this solution approach is that
E0,z is tangential to all boundaries in a waveguide.
For TM analysis, the other field components are calculated just from E0,z.
0,0, 2
c
0,0, 2
c
zx
zy
EjH
k y
EjH
k x
0,0, 2
c
0,0, 2
c
zx
zy
EjE
k x
EjE
k y
From this, the characteristic impedance 𝑍 is
0,2
0, cTM
0,0,2c
z
x
zy
EjE k x
ZEjH k
k x
is found by solving the wave equation.
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Slide 31
Setup for Analyzing Slab Waveguides
Geometry and Solution
Slide 32
z
x
y
0, , j zE x y z E ex
Amplitude Profile
Wave oscillations
phase constant
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Origin of TE and TM Modes (1 of 2)
Slide 33
Given this geometry
0y
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1a
Eq. 1b
Eq. 1c
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1d
Eq. 1e
Eq. 1f
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
0,0, Eq. 1fy
z
Hj E
x
0,
0, Eq. 1cyz
Ej H
x
Origin of TE and TM Modes (1 of 2)
Slide 34
Given this geometry
0y
0, 0, Eq. 1ay xj E j H 0, 0, Eq. 1dy xj H j E
0,0, 0, Eq. 1bz
x y
Ej E j H
x
0,
0, 0, Eq. 1ezx y
Hj H j E
x
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0,0, Eq. 2fy
z
Hj E
x
0,
0, Eq. 2cyz
Ej H
x
Origin of TE and TM Modes (2 of 2)
Slide 35
0, 0, Eq. 2ay xj E j H 0, 0, Eq. 2dy xj H j E
0,0, 0, Eq. 2bz
x y
Ej E j H
x
0,
0, 0, Eq. 2ezx y
Hj H j E
x
Maxwell’s equations have decoupled into two independent sets of equations.
TE Mode (i.e. E0,z = 0) TM Mode (i.e. H0,z = 0)
Origin of TE and TM Modes (2 of 2)
Slide 36
Maxwell’s equations have decoupled into two independent sets of equations.
TE Mode (i.e. E0,z = 0) TM Mode (i.e. H0,z = 0)
0,0, 0,
0, 0,
0,0,
Eq. 3a
Eq. 3b
Eq. 3c
zx y
y x
yz
Hj H j E
xj E j H
Ej H
x
0,0, 0,
0, 0,
0,0,
Eq. 3d
Eq. 3e
Eq. 3f
zx y
y x
yz
Ej E j H
xj H j E
Hj E
x
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TE Wave Equation
Slide 37
Solve Eq. (3b) for 𝐻 , and solve Eq. (3c) for 𝐻 , .
TE Mode (i.e. E0,z = 0)
0,0, 0,
0, 0,
0,0,
Eq. 3a
Eq. 3b
Eq. 3c
zx y
y x
yz
Hj H j E
xj E j H
Ej H
x
0, 0,
0,0,
Eq. 4a
1 Eq. 4b
x y
yz
H E
EH
j x
Substitute Eq. (4a) and (4b) into Eq. (3a) to obtain an equation that only contains 𝐸 , .
0, 2c 0,
10y
y
Ek E
x x
0,
0, 0,z
x y
Hj H j E
x
0,
0, 0,
1 yy y
Ej E j E
x j x
TM Wave Equation
Slide 38
Solve Eq. (3e) for 𝐸 , and solve Eq. (3f) for 𝐸 , .
TM Mode (i.e. H0,z = 0)
0,0, 0,
0, 0,
0,0,
Eq. 3d
Eq. 3e
Eq. 3f
zx y
y x
yz
Ej E j H
xj H j E
Hj E
x
0, 0,
0,0,
Eq. 5a
1 Eq. 5b
x y
yz
E H
HE
j x
Substitute Eq. (5a) and (5b) into Eq. (3d) to obtain an equation that only contains 𝐻 , .
0, 2c 0,
10y
y
Hk H
x x
0,
0, 0,z
x y
Ej E j H
x
0,
0, 0,
1 yy y
Hj H j H
x j x
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Typical Modes in a Slab Waveguide
Slide 39
TE Modes
TM Modes
ncore = 2.0
nclad = 1.5
ncore = 2.0
nclad = 1.5
01.8
01.8
Effective refractive indices
Effective refractive indices
x
yz
Remarks About Slab Waveguide Analysis
•Waves are confined in only one transverse direction.
•Waves are free to spread out in the uniform transverse direction
•Propagation within the slab can be restricted to the 𝑧direction without loss of generality.
Slide 40
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Summary of This Lecture
• Identify what types of modes are supported• TEM, TE, TM, or hybrid
• Analysis setup
Slide 41
TEM
2 0V
TE
2 20, c 0, 0z zH k H
TM
2 20, c 0, 0z zE k E
Hybrid
0, 0,2
0, 0,
1 1 1 1 1 1 1 10
01 1 1 1 1 1 1 1
x x
y y
x y x x E Ex y x xE E
y y y x y y y x
Slab
0, 2c 0,
0, 2c 0,
1TE: 0
1TM: 0
yy
yy
Ek E
x x
Hk H
x x
41
