topic 5a waveguide introduction & analysis setup...lecture 5a slide 5 waveguide modes the field...
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Lecture 5a Slide 1
EE 4347
Applied Electromagnetics
Topic 5a
Waveguide Introduction & Analysis Setup These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited
Lecture Outline
Lecture 5a Slide 2
• What is a waveguide?
• Governing equations for waveguide analysis.
• Setup for TEM, TE, and TM modes
• Setup for analyzing hybrid modes
• Setup for analyzing slab waveguides
Note, this lecture covers just setting up Maxwell’s equations for solution. This lecture does not attempt to obtain solutions.
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Lecture 5a Slide 3
What is a Waveguide?
Lecture 5a Slide 4
What is a Waveguide?
A waveguide is a structure that confines the propagation of waves to a single path.
They are “pipes” for electromagnetic waves.
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Lecture 5a Slide 5
Waveguide Modes
The field inside a waveguide must obey Maxwell’s equations. This limits what field configurations are possible into a discrete set. Each solution is called a “mode.” Each mode looks different and behaves differently inside the waveguide.
Lecture 5a Slide 6
Slab Vs. Channel Waveguides
Slab waveguides confine energy in only one transverse direction.
Channel waveguides confine energy in both transverse directions.
ConfinementConfinement
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Slide 7
Map of Waveguides (LI Media)
Transmission Lines
• Contains two or more conductors.• No low‐frequency cutoff.
• Confines and transports waves.• Supports higher‐order modes.
• Has TEM mode.• Has TE and TM modes.
stripline
coaxial microstrip
slotline
coplanar
Lecture 5a
“Pipes”
• Has one or less conductors.• Usually what is implied by the label “waveguide.”
Metal Shell Pipes Dielectric Pipes
Inhomogeneous
Homogeneous
• Enclosed by metal.• Does not support TEM mode.• Has a low frequency cutoff.
• Supports TE and TM modes
• Supports TE and TM modes only if one axis is uniform.
• Otherwise supports quasi‐TM and quasi‐TE modes.
rectangular circular
Channel Waveguides
Slab Waveguides
• Composed of a core and a cladding.• Symmetric waveguides have no low‐frequency cutoff.
• Confinement only along one axis.• Supports TE and TM modes.• Interfaces can support surface waves.
• Confinement along two axes.• TE & TM modes only supported in circularly symmetric guides.
dielectric Slab interface
optical Fiber rib
dual‐ridge
no uniform axis(no TE or TM)
Waveguides
Homogeneous Inhomogeneous• Supports only quasi‐(TEM, TE, & TM) modes.
Single‐Ended
Differential
buried parallel plate
coplanar strips
photonic crystal
shielded pairlarge‐area
parallel plate
uniform axis(has TE and TM)
Notes on Transmission Lines
• Contains two or more conductors
• No low frequency cutoff. Works down to DC.
• Supports TEM, TE, and TM modes when the dielectric is homogeneous
• Supports higher‐order modes, not just TEM.
• Serve more as a circuit element than a wave device
• Very compact for low frequency signals.
• Tend to be lossy at very high frequencies (> 10 GHz) due to skin effect and dielectric loss.
Lecture 5a Slide 8
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Notes on Metal Pipe Waveguides
• Contains on a single conductor
• Has a low frequency cutoff below which there is no propagation of waves.
• Supports TE and TM waves only if dielectric is homogeneous.
• Field confined to inside of the waveguide.
• Less lossy for very high frequency waves.
• Prohibitively large size at low frequencies.
Lecture 5a Slide 9
Notes on Dielectric Waveguides
• Does not contain any metals
• Symmetric dielectric waveguides have not low‐frequency cutoff.
• Symmetric waveguides (e.g. slabs & circularly symmetric) support TE and TM modes.
• Most have a low frequency cutoff below which no waves can propagate.
• Hybrid modes still tend to be strongly linearly polarized.
• Optical fibers are dielectric waveguides.
• Field extends outside of the core.
Lecture 5a Slide 10
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Lecture 5a Slide 11
Channel Waveguides for Integrated Optics
Stripe waveguide Diffused waveguide Buried‐strip waveguide
Buried‐rib waveguide Rib waveguide Strip‐loaded waveguide
Lecture 5a Slide 12
Channel Waveguides for Radio Frequencies
Coaxial Cable
Twisted Pair Transmission LineIsolated Wire
Shielded‐Pair Transmission Line
Rectangular Waveguide
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Lecture 5a Slide 13
Channel Waveguides for Printed Circuits
Transmission lines are metallic structures that guide electromagnetic waves from DC to very high frequencies.
Microstrip
Stripline Slot Line
Parallel‐Plate Transmission Line
Coplanar Line
Lecture 5a Slide 14
Structures Supporting Surface Waves
Surface‐Plasmon Polariton (SPP)
Dyakonov Surface Wave Bloch Surface Wave
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Notes on Waveguides
• Waveguides support an infinite number of discrete modes
• The modes have a constant amplitude profile that just accumulates phase as it propages.
• Modes have cutoff frequencies, below which they are not supported and decay very quickly.
Lecture 5a Slide 15
Lecture 5a Slide 16
Governing Equations for Waveguide
Analysis
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Lecture 5a Slide 17
Steps for Waveguide Analysis
1. Draw the waveguide2. 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.
Lecture 5a Slide 18
Various Wave Equations
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
We can solve these equations independently.
wave numberk
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Lecture 5a Slide 19
Expand Maxwell’s Equations
We must analyze waveguides using Maxwell’s equations.
E j H
H j E
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
We have six field components to solve for: Ex, Ey, Ez, Hx, Hy, and Hz.
Yikes!!
Lecture 5a Slide 20
General Form of Solution for Waveguides
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 we can solve the problem by just analyzing the cross section in the x-y plane. This reduces to a two‐dimensional problem.
3D 2D
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Lecture 5a Slide 21
Animation of a Waveguide Mode
Lecture 5a Slide 22
Assume the Form of the Solution
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
If we substitute this solution into our six equations, we get
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 more simple, but we still have six field components to solve for.
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Lecture 5a Slide 23
Setup for TEM, TE, and TM Modes
Lecture 5a Slide 24
Existence Conditions for TEM, TE, and TM ModesTEM modes only exist in transmission lines with two or more conductors embedded in a homogeneous fill.
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.
No TEMSupports TEM
Supports TE and TM No TE or TM
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Lecture 5a Slide 25
Goal of Following Derivation
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
0, 0,0, 2
z zx
c
z zy
c
E HjH
k y x
E HjH
k x y
0, 0,0, 2
0, 0,0, 2
z zx
c
z zy
c
E HjE
k x y
E HjE
k y x
0,
0,
?
?z
z
E
H
We now only need to find Ez and Hz.
2 2
2 2
0
0
z z
z z
E k E
H k H
Lecture 5a Slide 26
Reduce the Number of Terms to Solve (1 of 2)
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1
Eq. 1
Eq. 1
zy x
zx y
y xz
Ej E j H a
y
Ej E j H b
xE E
j H cx y
0,0, 0,
0,0, 0,
0, 0,0,
Eq. 1
Eq. 1
Eq. 1
zy x
zx y
y xz
Hj H j E d
y
Hj H j E e
xH H
j E fx 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 to 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
y xk
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Lecture 5a Slide 27
Reduce the Number of Terms to Solve (2 of 2)
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
x yk
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
x yk
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
y xk
Lecture 5a Slide 28
Reduced Set of Equations
Step 5 – Define the cutoff wave number kc as
2 2 2ck k
We now have all of the transverse field components expressed in terms of the longitudinal components.
0, 0,0, 2
0, 0,0, 2
z zx
c
z zy
c
E HjH
k y x
E HjH
k x y
0, 0,0, 2
0, 0,0, 2
z zx
c
z zy
c
E HjE
k x y
E HjE
k y x
Now all that we have to do is determine E0,z and H0,z. The remaining field components can be calculated from just these two terms.
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Lecture 5a Slide 29
How Do We Find E0,z and H0,z?
Recall that in LHI media, our 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 our solution into the bottom equations above gives
2 20, 0, 0z c zE k E 2 2
0, 0, 0z c zH k H
Lecture 5a Slide 30
Solution Categories
If E0,z = H0,z = 0, we obtain a transverse electromagnetic (TEM) solution because all of the field components are transverse to the direction of propagation. Analysis reduces to an electrostatics problem.
If E0,z = 0 and H0,z ≠ 0, we obtain a transverse electric (TE) solution because the electric field has no longitudinal component.
If E0,z ≠ 0 and H0,z = 0, we obtain a transverse magnetic (TM) solution because the magnetic field has no longitudinal component.
If E0,z ≠ 0 and H0,z ≠ 0, we obtain a hybrid solution.
E0,z H0,z Solution
0 0 TEM
0 TE
0 TM
Hybrid
2 20, 0,
2 20, 0,
0
0
z c z
z c z
E k E
H k H
In LHI media, E0,z and H0,z are solved independently. This means solutions can be obtained in any combination. This is the origin of TEM, TE, TM, and hybrid modes.
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Lecture 5a Slide 31
TEM Analysis (1 of 3)
For TEM waves, we have E0,z = H0,z = 0. Under this condition, Maxwell’s equations reduce to
0,zE
y
0, 0,
0,0,
Eq. 1y x
zx
j E j H a
Ej E
x
0,
0, 0,0,
Eq. 1y
y xz
j H b
E Ej H
x y
Eq. 1c
0,zH
y
0, 0,
0,0,
Eq. 1y x
zx
j H j E d
Hj H
x
0,
0, 0,0,
Eq. 1y
y xz
j E e
H Hj E
x y
Eq. 1 f
0, 0,
0, 0,
0, 0,
Eq. 2
Eq. 2
0 Eq. 2
y x
x y
y x
j E j H a
j E j H b
E Ec
x y
0, 0,
0, 0,
0, 0,
Eq. 2
Eq. 2
0 Eq. 2
y x
x y
y x
j H j E d
j H j E e
H Hf
x y
Lecture 5a Slide 32
TEM Analysis (2 of 3)
0, 0,
0, 0,
0, 0,
Eq. 2
Eq. 2
0 Eq. 2
y x
x y
y x
j E j H a
j E j H b
E Ec
x y
0, 0,
0, 0,
0, 0,
Eq. 2
Eq. 2
0 Eq. 2
y x
x y
y x
j H j E d
j H j E e
H Hf
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).
After dropping E0,z and H0,z, we get
k We see that for TEM
Previously, we defined .2 2 2ck k
If = k, then kc = 0 implying that there is no cutoff frequency for the TEM mode.
In summary, for the TEM mode we have
and 0 no cutoffck k
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Lecture 5a Slide 33
TEM Analysis (3 of 3)
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, 0, 0z c zE k E
2 20, 0,z c zE k E
20,
0
0zE
Lecture 5a Slide 34
Alternate Derivation of TEM Analysis
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. 3
0 Eq. 3
Eq. 3
Eq. 3
E a
D b
D E c
E V d
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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Lecture 5a Slide 35
TE Analysis in LHI Media
We are free to set E0,z = 0 and H0,z ≠ 0. This means we do not have to solve for E0,z.We only have to obtain a solution for H0,z.
2 20, 0, 0z c zE k E 2 2
0, 0, 0z c 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, we can calculate the impedance.
0,2
0,TE
0,0,2
z
x c
zy
c
HjE k y k
ZHjH
k y
0,0, 2
0,0, 2
zx
c
zy
c
HjH
k x
HjH
k y
0,0, 2
0,0, 2
zx
c
zy
c
HjE
k y
HjE
k x
We must still determine by solving the wave equation.
Lecture 5a Slide 36
TM Analysis in LHI Media
We are free to set E0,z ≠ 0 and H0,z = 0. This means we do not have to solve for H0,z.We only have to obtain a solution for E0,z.
2 2 2 20, 0, 0, 0,0 0z c z z c 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
0,0, 2
zx
c
zy
c
EjH
k y
EjH
k x
0,0, 2
0,0, 2
zx
c
zy
c
EjE
k x
EjE
k y
From this, we can calculate the impedance.
0,2
0,TM
0,0,2
z
x c
zy
c
EjE k x
ZEjH k
k x
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Lecture 5a Slide 37
Setup for AnalyzingHybrid Modes
Lecture 5a Slide 38
Hybrid Mode Analysis (1 of 3)
To setup a solution to Maxwell’s equations, we 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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Lecture 5a Slide 39
Hybrid Mode Analysis (2 of 3)
We can write our four remaining equations 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.
Lecture 5a Slide 40
Hybrid Mode Analysis (3 of 3)
Quasi‐LP AnalysisRecognizing that the hybrid modes tend to be strongly linearly polarized, we can make a simplifying approximation that the cross coupling between E0,x and E0,y is weak and can be neglected. Under this condition, our 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
xx xy
yx0, 0,2
0, 0,
0
0x x
y yyy
E E
E E
20, 0,
20, 0,
0
0xx x x
yy y y
E E
E E
Quasi‐LP AnalysisThese final two equations simplify even more for homogeneous dielectrics and for slab waveguides. They are still typically solved numerically on a computer.
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Lecture 5a Slide 41
Setup for Analyzing Slab Waveguides
Lecture 5a Slide 42
Geometry and Solution
z
x
y
0, , j zE x y z E ex
Amplitude Profile
Wave oscillations
phase constant
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Lecture 5a Slide 43
LI Slab Waveguide Analysis (1 of 3)
Given this geometry
0y
0,zE
y
0, 0,
0,0, 0,
0, 0,
Eq. 1
Eq. 1
y x
zx y
y x
j E j H a
Ej E j H b
x
E E
x y
0, Eq. 1zj H c
0,zH
y
0, 0,
0,0, 0,
0, 0,
Eq. 1
Eq. 1
y x
zx y
y x
j H j E d
Hj H j E e
x
H H
x y
0, Eq. 1zj E f
Lecture 5a Slide 44
LI Slab Waveguide Analysis (2 of 3)
We see that Maxwell’s equations have decoupled into two sets of equations.
0,0,
0, 0
0
,
0,
0
,
,
E
Eq. 2
Eq
q. 2
.
2
zx
y x
yz
y
j E j H a
Ej H c
x
Ej E j H b
x
0,
0, 0,
0,0,
0, 0,
Eq. 2
Eq.
q
2
E . 2
z
y x
yz
x y
Hj H j E e
x
j H j E d
Hj E f
x
0,0, 0,
0, 0,
0,0,
Eq. 3
Eq. 3
Eq. 3
zx y
y x
yz
Hj H j E a
xj E j H b
Ej H c
x
0,0, 0,
0, 0,
0,0,
Eq. 3
Eq. 3
Eq. 3
zx y
y x
yz
Ej E j H d
xj H j E e
Hj E f
x
TE Mode (i.e. E0,z = 0) TM Mode (i.e. H0,z = 0)
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Lecture 5a Slide 45
LI Slab Waveguide Analysis (3 of 3)
0, 20,
0, 0,
0,0,
10 Eq. 4
Eq. 4
1 Eq. 4
yc y
x y
yz
Ek E a
x x
H E b
EH c
j x
0, 20,
0, 0,
0,0,
10 Eq. 5
Eq. 5
1 Eq. 5
yc y
x y
yz
Hk H a
x x
E H b
HE c
j x
TE Mode
TM Mode
We can derive wave equations by substituting the 2nd and 3rd equations into the 1st.
Lecture 5a Slide 46
Typical Modes in a Slab Waveguide
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
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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 a single direction without loss of generality.
• Maxwell’s equations rigorously decouple into two distinct modes.
• No approximations are necessary
Lecture 5a Slide 47