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Lecture 4 Slide 1
EE 5337
Computational Electromagnetics
Lecture #4
Transfer Matrix Method
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InstructorDr. Raymond Rumpf(915) 747‐[email protected]
Outline
• Formulation of 44 matrix equation for 1D structures
• Solution in an LHI layer
• Transfer matrices for multilayer structures
• Transfer matrices are unstable
• Formulation of 22 matrix equation for 1D structures
Lecture 4 Slide 2
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Lecture 4 Slide 3
Formulation of 44 Matrix Equationfor 1D Structures
Lecture 4 Slide 4
1D Structures
Sometimes it is possible to describe a physical device using just one dimension. Doing so dramatically reduces the numerical complexity of the problem and is ALWAYS GOOD PRACTICE.
z
x
yRegion IReflection Region
Region IITransmission Region
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Lecture 4 Slide 5
3D 1D Using Homogenization
Many times it is possible to approximate a 3D device in one dimension. It is very good practice to at least perform the initial simulations in 1D and only moving to 3D to verify the final design.
Physical Device Effective Medium Approximation in 1D
1 2 3 4 r
Lecture 4 Slide 6
3D 1D Using Circuit‐Wave Equivalence
r, r,i i in
ii i
i
Z
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Lecture 4 Slide 7
Starting Point
0
0
0
yzr x
x zr y
y xr z
HHk E
y z
H Hk E
z x
H Hk E
x y
0
0
0
yzr x
x zr y
y xr z
EEk H
y z
E Ek H
z xE E
k Hx y
We start with Maxwell’s equations in the following form. Here we have assumed isotropic materials and we will use the positive sign convention for waves.
0H j H Positive sign convention
Lecture 4 Slide 8
Calculation of the Wave Vector Components
The components kx and ky are determined by the incident wave and are continuous throughout the 1D device. The kz component is different in each layer and calculated from the dispersion relation in that layer.
0 ,inc ,inc
0 ,inc ,inc
sin cos
sin sin
x r r
y r r
k k
k k
2 2 2, 0 , ,z i r i r i x yk k k k
Layer #i
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Lecture 4 Slide 9
kx and ky Continuous Throughout Device
z
xinck
kx
refk -kz,air
kx
kz,air 22 2,air 0 airz xk k n k
,inc 0 ai
,inc a
r
0 ir
cos sin
cosz
x
k k n
k k n
kx
kz,11k
22 2,1 0 1z xk k n k
1n
2n
3n
kx
kz,22k
22 2,2 0 2z xk k n k
kx
kz,33k
22 2,3 0 3z xk k n k
kx
kz,airtrnk
22 2 2,trn 0 air ,airz x zk k n k k
Lecture 4 Slide 10
Waves in Homogeneous Media
0 0 0 0 y yx xz zjk y jk yjk x jk xjk z jk zjk r jk rE r E e E e e e H r H e H e e e
A wave propagating in a homogeneous layer is a plane wave. It has the following mathematical form.
0 0 y yx xz zjk y jk yjk x jk xjk z jk z
x x xE r E e e e jk E e e e jk E r jkx x x
When we take derivatives of these solutions, we see that
0 0 y yx xz zjk y jk yjk x jk xjk z jk z
y y yE r E e e e jk E e e e jk E r jky y y
Note: e+jkz sign convention was used for propagation in +z direction.
We cannot say that because the structure is not homogeneous in the z direction.
zz jk zjk
z
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Lecture 4 Slide 11
Reduction of Maxwell’s Eqs. to 1D
0
0
0
yy z r x
xx z r y
x y y x r z
dHjk H k E
dz
dHjk H k E
dz
jk H jk H k E
0
0
0
yy z r x
xx z r y
x y y x r z
dEjk E k H
dzdE
jk E k Hdz
jk E jk E k H
Given that
x yjk jkx y
Maxwell’s equations become
Note: z is the only independent variable left so its derivative is ordinary. d
z dz
Lecture 4 Slide 12
Normalize the Parameters
yy z r x
xx z r y
x y y x r z
dHjk H E
dz
dHjk H E
dz
jk H jk H E
yy z r x
xx z r y
x y y x r z
dEjk E H
dzdE
jk E Hdz
jk E jk E H
We normalize the coordinates (x, y, and z) and wave vector components (kx, ky, and kz) according to
0z k z
Using the normalized parameters, Maxwell’s equations become
0 0 0
yx zx y z
kk kk k k
k k k
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Lecture 4 Slide 13
Solve for the Longitudinal Components Ez and Hz
yy z r x
xx z r y
x y y x r z z x y y xr
dHjk H E
dz
dHjk H E
dzj
jk H jk H E E k H k H
yy z r x
xx z r y
x y y x r z z x y y xr
dEjk E H
dzdE
jk E Hdz
jjk E jk E H H k E k E
We solve the third and sixth equations for the longitudinal field components Hz and Ez.
yy z r x
xx z r y
z x y y xr
dHjk H E
dz
dHjk H E
dzj
E k H k H
yy z r x
xx z r y
z x y y xr
dEjk E H
dzdE
jk E Hdz
jH k E k E
Lecture 4 Slide 14
Eliminate the Longitudinal Components
We eliminate the longitudinal field terms by substituting them back into the remaining equations.
2
2
yy x x y y r r r x
xr x y x y x r r y
dEk H k k H H
dzdE
k H k k H Hdz
2
2
yy x x y y r r r x
xr x y x y x r r y
dHk E k k E E
dz
dHk E k k E E
dz
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Lecture 4 Slide 15
Rearrange Maxwell’s Equations
Here we rearrange the terms and the order of the equations.
2
2
2
2
x yx xx r y
r r
y y x yr x y
r r
x yx xx r y
r r
y y x yr x y
r r
k kdE kH H
dz
dE k k kH H
dz
k kdH kE E
dz
dH k k kE E
dz
2
2
2
2
yy x x y y r r r x
xr x y x y x r r y
yy x x y y r r r x
xr x y x y x r r y
dEk H k k H H
dzdE
k H k k H Hdz
dHk E k k E E
dz
dHk E k k E E
dz
Lecture 4 Slide 16
Matrix Form of Maxwell’s Equations
The remaining four equations can be written in matrix form as
2
2
2
2
0 0
0 0
0 0
0 0
x y xr
r r
x xy x yr
y yr r
x xx y xr
y yr r
y x yr
r r
k k k
E Ek k k
E EdH Hdz k k kH H
k k k
2
2
2
2
x yx xx r y
r r
y y x yr x y
r r
x yx xx r y
r r
y y x yr x y
r r
k kdE kH H
dz
dE k k kH H
dz
k kdH kE E
dz
dH k k kE E
dz
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Lecture 4 Slide 17
BTW…for Fully Anisotropic Materials
2
2
ˆyz yz zy x y yz zx yz zyzx x
y x x yx yyzz zz zz zz zz zz zz zz
zy yxz zx xzxy x y x
zz zz zz zz zzy
x
y
k k kj k k jk
kE jk j k kE
Hz
H
2
2
x y xz zyxz zxx xy
zz zz zz
x y yz zx yz zy yz yz zyx zxyx yy y x x
zz zz zz zz zz zz zz zz
y x yxz zxxx
zz zz zz
k k
k k kj k k jk
k k k
x
y
x
y
xz zy zyxz zx xzxy y x y
zz zz zz zz zz
E
E
H
H
jk j k k
Note: This is for the sign convention.j ze
Lecture 4 Slide 18
Solution in an LHI Layer
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Lecture 4 Slide 19
Matrix Differential Equation
Maxwell’s equations can now be written as a single matrix differential equation.
d
dz
ψΩψ 0
2
2
2
2
0 0
0 0
0 0
0 0
x y xr
r r
y x yxr
y r r
x x y xr
y r r
y x yr
r r
k k k
k k kE z
E zz
H z k k kH z
k k k
ψ Ω
Lecture 4 Slide 20
Solution of the Differential Equation (1 of 3)
The matrix differential equation is
d
dz
ψΩψ 0
This is actually a set of four coupled differential equations.
0zz e Ωψ ψ
This is easy to write, but how do we compute the exponential of a matrix?
2
2
2
2
x yx xx r y
r r
y y x yr x y
r r
x yx xx r y
r r
y y x yr x y
r r
k kdE kH H
dz
dE k k kH H
dz
k kdH kE E
dz
dH k k kE E
dz
The system of four equations can be solved as a single matrix equation as follows.
1
2
3
4
0
0
0
0
zx x
zy y
zx x
zy y
E z e E
E z e E
H z e H
H z e H
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Lecture 4 Slide 21
Functions of Matrices (1 of 2)
It is sometimes necessary to evaluate the function of a matrix.
?f A
It is NOT correct to calculate the function applied to every element in the matrix A individually. A different technique must be used.
11 12 1
21 22 2
1 2
N
N
M M MN
f A f A f A
f A f A f Af
f A f A f A
A
This is more of an array operation than a matrix operation so it is incorrect to perform on a matrix.
Lecture 4 Slide 22
Functions of Matrices (2 of 2)
To calculate f(A) correctly, we first calculate the eigen‐vectors and eigen‐values of the matrix A.
eigen-vector matrix of
eigen-value matrix of
V AA
D A
Given the eigen‐vector matrix V and the eigen‐value matrix D, the function of the matrix is evaluated as
1f f A V D V
f(D) is very easy to evaluate because D is a diagonal matrix so the function only has to be performed individually on the diagonal elements.
1
2
0 0
0 0
0
0 0 0 M
D
D
D
D
1
2
0 0
0 0
0
0 0 0 M
f D
f Df
f D
D
[V,D] = eig(A);
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Lecture 4 Slide 23
Solution of the Differential Equation (1 of 2)
We had the following matrix differential equation and general solution
0zdz e
dz
Ωψ
Ωψ 0 ψ ψ
We can now evaluate the matrix exponential using the eigen‐values and eigen‐vectors of the matrix .
eigen-vector matrix
eigen-value matrix
WΩ
λ
1z ze e Ω λW W
1
2
3
4
0 0 0
0 0 0
0 0 0
0 0 0
z
zz
z
z
e
ee
e
e
λ
Lecture 4 Slide 24
Solution of the Differential Equation (2 of 2)
The solution to the matrix differential equation is therefore
d
dz
ψΩψ 0
1
0
0
z
z
z e
z e
Ω
λ
ψ ψ
ψ W W ψ
We can combine the unknown initial values (0) with W-1 because that product just leads to another column vector of unknown constants.
Our final solution is then
zdz e
dz
λψ
Ωψ 0 ψ W c 1 0c W ψ
c
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Lecture 4 Slide 25
Interpretation of the Solution
zez λWψ c
(z’) – Overall solution which is the sum of all the modes at plane z’.
W – Square matrix who’s column vectors describe the “modes” that can exist in the material. These are essentially pictures of the modes which quantify the relative amplitudes of Ex, Ey, Hx, and Hy.
ez’ – Diagonal matrix describing how the modes propagate. This includes accumulation of phase as well as decaying (loss) or growing (gain) amplitude.
c – Column vector containing the amplitude coefficient of each of the modes. This quantifies how much power is in each mode.
Lecture 4 Slide 26
Getting a Feel for the Numbers (1 of 2)
For a layer with r = 9.0 and r = 1.0 (i.e. n = 3.0) and a wave at normal incidence, we will have
0 0 0 1
0 0 1 0
0 9 0 0
9 0 0 0
Ω
This matrix has the following eigen‐vectors and eigen‐values.
0.32 0.32 0 0
0 0 0.32 0.32
0 0 0.95 0.95
0.95 0.95 0 0
j j
j j
W
3.0 0 0 0
0 3.0 0 0
0 0 3.0 0
0 0 0 3.0
j
j
j
j
λ
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Lecture 4 Slide 27
Getting a Feel for the Numbers (2 of 2)
We see that the modes occur as either an Ex‐Hy or Ey‐Hx pair. This is consistent with plane waves. Due to the normalization, they are 90° out of phase. A sign difference indicates forward and backward waves. Only the relative amplitude difference between E and H is important here.
We know the refractive index (n = 3.0), so the eigen‐values are consistent with what we would expect. The signs correspond to forward and backward waves.
0.32 0.32 0 0
0 0 0.32 0.32
0 0 0.95 0.95
0.95 0.95 0 0
j j
j j
W
3.0 0 0 0
0 3.0 0 0
0 0 3.0 0
0 0 0 3.0
j
j
j
j
λ
inccos
inccos
3
jn zz
r r
e e
jn
n
0
1
3
r
r
r
r
E
H
E
H
The modes in W only contain information about the relative amplitudes of the field components.
The numbers in describe how the modes accumulate phase in the z direction. This is essentially just the complex refractive index of the material.
Lecture 4 Slide 28
Visualizing the Modes
0.32 0.32 0 0
0 0 0.32 0.32
0 0 0.95 0.95
0.95 0.95 0 0
j j
j j
W
3.0 0 0 0
0 3.0 0 0
0 0 3.0 0
0 0 0 3.0
j
j
j
j
λ
Mode 1 Mode 2
Mode 3 Mode 4
Mode 1
Mode 2
Mode 3
Mode 4
0.95 0.95
0.95 0.95
-j0.32
j0.32
j0.32
-j0.32
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Lecture 4 Slide 29
Transfer Matrices for Multilayer Structures
Lecture 4 Slide 30
Geometry of an Intermediate Layer
Layer i Layer i+1Layer i-1
0iψ
1 0 1i ik L ψ
iL
0i ik Lψ
1 0iψ
1iL 1iL
1icic1ic
i izψ
is a local z‐coordinate inside the ith layer that starts at zero at the layer’s left side.
iz0
iz
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Lecture 4 Slide 31
Field Relations
Field inside the ith layer:
,
,
,
,
i i
x i i
y i i zi i i i
x i i
y i i
E z
E zz e
H z
H z
λψ W c
Boundary conditions at the first interface:
Boundary conditions at the second interface:
1 0 1
1 0 1
1 1
0
i i
i i i
k Li i i i
k L
e
λ
ψ ψ
W c Wc
0
0 1
1 1
0
i i
i i i
k Li i i i
k L
e
λ
ψ ψ
W c W c
We need to include k0 in the exponential to normalize Li-1 because the parameter i-1 expects to multiply a normalized coordinate.
Note: We must equate the field on either side of the interfaces and not the mode coefficients c.
Lecture 4 Slide 32
The Transfer Matrix
The transfer matrix Ti of the ith
layer is defined as:
We start with the boundary condition equation from the second interface and rearrange terms.
1i i i c T c
iT
0 011 1 1 1 i i i ik L k L
i i i i i i i ie e λ λW c W c c W W c
011
i ik Li i ie
λT W WWe then read off the transfer
matrix.
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Lecture 4 Slide 33
The Transfer Matrix Method
The transfer matrix method (TMM) consists of working through the device one layer at a time and calculating an overall (global) transfer matrix.
1T 2T 3T 4T 5T
global 5 4 3 2 1 T T T T T T This is standard matrix multiplication.
Reflection Region
Transmission Region
The order of multiplication may seem backwards here, but it is not. Recall the definition of the transfer matrix to have this make sense.
Lecture 4 Slide 34
The Global Transfer Matrix
The transfer matrix so far is not yet the “true” global transfer matrix because it does not connect the reflection region to the transmission region. It only connects the amplitude coefficients of Layer 1 to the amplitude coefficients in the transmission region. This is a result of how we defined the transfer matrix.
ref ref 1 1W c W c
Solving this for c1 yields
trn global 1c T c
The global transfer matrix must connect the amplitude coefficients in the reflection region to the amplitude coefficients in the transmission region. Boundary conditions at the first interface require
The global transfer matrix is derived by substituting this result into the first equation.
11 1 ref ref
c W W c
1trn global 1 ref ref
1global global 1 ref
c T W W c
T T W W1
global 5 4 3 2 1 1 ref T T T T T T W W
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Lecture 4 Slide 35
Transfer Matricesare Unstable
Lecture 4 Slide 36
The Multi‐Layer Problem
The diagram below is focused on an arbitrary layer in a stack of multiple layers. We will be examining the wave solutions in this ith
layer.
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Lecture 4 Slide 37
Wave Solutions in ith Layer
Recall that the wave vector can be purely real (pure oscillation), purely imaginary (pure exponential decay), or complex (decaying oscillation).
k k jk
k jk
k kPure oscillation
Pure decay
Decaying oscillation
Lecture 4 Slide 38
Backward Waves in ith Layer
Due to reflections at the interfaces, there will also be backward traveling waves in each of the layers. These can also have wave vectors that are real, imaginary or complex, so they can oscillate, decay/grow, or both.
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Lecture 4 Slide 39
All Waves are Treated as Forward Waves
The pure transfer matrix method treats all waves as if they are forward propagating. Decaying fields associated with backward waves become exponentially growing fields and quickly become numerically unstable.
,
,
,
,
i i
x i i
y i i zi i i i
x i i
y i i
E z
E zz e
H z
H z
λψ W c
Lecture 4 Slide 40
TMM is Inherently Unstable
Our wave solution was
x
y z
x
y
E z
E zz e
H z
H z
λψ W c
This treats all power as forward propagating.
We know that backward waves exist. We also know that decaying fields exist when a wave is evanescent or propagating in a lossymaterial.
When backward waves are decaying and treated as forward propagating waves, they grow exponentially. This leads to numerically instability.
The TMM is inherently an unstable method because it treats everything as forward propagating.
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Lecture 4 Slide 41
The Fix
We are treating all power as forward propagating because we did not distinguish between forward and backward waves.
Clearly, the first part of the fix is to distinguish between forward and backward propagating waves.
This can be accomplished by calculating the Poynting vector associated with the modes and looking at the sign of the z component. Be careful! We are using a normalized magnetic field.
0 0
0
1
z x y y x
y xz x y
z x y y x
E H
E H E H
H HE E
j j
E H E Hj
0.32 0.32 0 0
0 0 0.32 0.32
0 0 0.95 0.95
0.95 0.95 0 0
i i
i i
W
3.0 0 0 0
0 3.0 0 0
0 0 3.0 0
0 0 0 3.0
i
i
i
i
λ
Lecture 4 Slide 42
Rearrange Eigen Modes
Now that we know which eigen‐modes are forward and backward propagating, we can rearrange the eigen‐vector and eigen‐value matrices to group them together.
0.32 0.32 0 0
0 0 0.32 0.32
0 0 0.95 0.95
0.95 0.95 0 0
i i
i i
W
3.0 0 0 0
0 3.0 0 0
0 0 3.0 0
0 0 0 3.0
i
i
i
i
λ
0.32 0 0.32 0
0 0.32 0 0.32
0 0.95 0 0.95
0.95 0 0.95 0
i i
i i
W
You will also need to adjust the vertical positions of the eigen‐values so that ’ remains a diagonal matrix.
rearrange modes
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Lecture 4 Slide 43
New Interpretation of the Matrices
3.0 0 0 0
0 3.0 0 0
0 0 3.0 0
0 0 0 3.0
i
i
i
i
λ
0.32 0 0.32 0
0 0.32 0 0.32
0 0.95 0 0.95
0.95 0 0.95 0
i i
i i
W
x
y
x
y
E
E
H
H
E E
H H
zz
z
ee
e
λλ
λ
W WW
W W
0
0
We have now partitioned our matrices into forward and backward propagating elements.
x
y
x
y
E
E
H
H
3.0 0 3.0 0
0 3.0 0 3.0
i i
i i
λ λNote: For anisotropic materials, all the eigen‐vectors and eigen‐values are in general unique.
Lecture 4 Slide 44
Revised Solution to Differential Equation
The matrix differential equation and its original solution was
zdz e
dz
λψ
Ωψ 0 ψ W c
After distinguishing between forward and backward propagating waves and grouping them in the matrices, we can write our solution as
z
E E
zH H
ez
e
λ
λ
0W W cψ
W W c0
We now have separate mode coefficients c+ and c- for forward and backward propagating modes, respectively.
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Lecture 4 Slide 45
Formulation of 22 Matrix Equationfor 1D Structures
Lecture 4 Slide 46
Recall Derivation Up to 44
0
0
0
0
0
0
yzr x
x zr y
y xr z
yzr x
x zr y
y xr z
EEk H
y z
E Ek H
z xE E
k Hx y
HHk E
y z
H Hk E
z x
H Hk E
x y
Start with Maxwell’s equations from
Lecture 2.
Assume LHI.
0
0
0
0
0
0
yy z r x
xx z r y
x y y x r z
yy z r x
xx z r y
x y y x r z
dEjk E k H
dzdE
jk E k Hdz
jk E jk E k H
dHjk H k E
dz
dHjk H k E
dz
jk H jk H k E
Assume device is infinite and uniform in x and y directions.
x yjk jkx y
yy z r x
xx z r y
x y y x r z
yy z r x
xx z r y
x y y x r z
dEjk E H
dzdE
jk E Hdz
jk E jk E H
dHjk H E
dz
dHjk H E
dz
jk H jk H E
Normalize z and wave vectors kx, ky,
and kz.
0
0 0 0
yx zx y z
z k z
kk kk k k
k k k
2
2
2
2
x yx xx r y
r r
y y x yr x y
r r
x yx xx r y
r r
y y x yr x y
r r
k kdE kH H
dz
dE k k kH H
dz
k kdH kE E
dz
dH k k kE E
dz
Eliminate longitudinal
components Ez and Hz by substitution.
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Lecture 4 Slide 47
Derivation of Two 22 Matrix Equations
2
2
x yx xx r y
r r
y y x yr x y
r r
k kdE kH H
dz
dE k k kH H
dz
2
2
x yx xx r y
r r
y y x yr x y
r r
k kdH kE E
dz
dH k k kE E
dz
We can write our four equations as two matrix equations.
2
2
1x x y r r x x
y yr y r r x y
E k k k HdE Hdz k k k
2
2
1 x y r r x xx
yy r y r r x y
k k k EHdEHdz k k k
Note: These equations are valid regardless of the sign convention because there is always a k multiplying another k and erasing the sign.
Lecture 4 Slide 48
Standard “PQ” Form
We can write our two matrix equations more compactly as
2
2
1x x y r r x x
y yr y r r x y
E k k k HdE Hdz k k k
2
2
1 x y r r x xx
yy r y r r x y
k k k EHdEHdz k k k
x x
y y
E HdE Hdz
P
xx
yy
EHdEHdz
Q
2
2
1 x y r r x
r y r r x y
k k k
k k k
P
2
2
1 x y r r x
r y r r x y
k k k
k k k
Q
Note: We will see this same “PQ” form again for other methods like MoL, RCWA, and waveguide analysis. TMM, MoL, and RCWA are all implemented the same after P and Q are calculated.
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Lecture 4 Slide 49
Matrix Wave Equation
Our two governing equations are
We can now derive a matrix wave equation. First, we differentiate Eq. (1) with respect to z’.
x x
y y
E HdE Hdz
P
xx
yy
EHdEHdz
QEq. (1) Eq. (2)
22
2
2
0
0x x
y y
E EdE Edz
Ω
Ω PQ
Second, we substitute Eq. (2) into this result.
2
2
x xx x
y yy y
E EH Hd d d d dE EH Hdz dz dz dz dz
P P
2
2
x x
y y
E EdE Edz
P Q
Lecture 4 Slide 50
Numerical Solution (1 of 3)
The system of equations to be solved is
22 2
2
0
0x x
y y
E EdE Edz
Ω Ω PQ
This has the general solution of
x z z
y
E ze e
E z
Ω Ωa a proportionality constant of forward wave
proportionality constant of backward wave
a
a
No mode sorting! Here, we solved a second‐order differential equation where the modes we calculate are all propagating in a single direction. We simply write them twice for forward and backward waves and thus they are automatically distinguished. Before we solved a first‐order differential equation that lumped forward and backward modes together.
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Lecture 4 Slide 51
Numerical Solution (2 of 3)
Recall that
1 1 z z z ze e e e Ω λ Ω λW W W W
So the overall solution can now be written as
1 1x z z
y
E ze e
E z
λ λW W a W W a
2
2 2
Eigen-vector matrix of
Eigen-value matrix of
W Ω
λ Ω
21
1
222
2 NN
zz
zzz
zz
e e
eee
ee
λ
1f f A V D V
We can use this relation to compute the matrix exponentials.
Lecture 4 Slide 52
Numerical Solution (3 of 3)
So the overall solution can now be written as
1 1x z z
y
E ze e
E z
λ λ
cc
W W a W W a
The column vectors a+ and a‐ are proportionality constants that have not yet been determined.
The eigen‐vector matrix W multiplies a+ and a‐ to give another column vector of undetermined constants.
To simplify the math, we combine these products into new column vectors labeled c+ and c‐ .
x z z
y
E ze e
E z
λ λW c W c
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Lecture 4 Slide 53
Solution for the Magnetic Field (1 of 2)
Since the electric and magnetic fields are coupled and not independent, we should be able to compute V from W. First, we differentiate the above solution with respect to z’.
The magnetic field has a solution of the same form, but will have its own eigen‐vector matrix V to describe its modes.
x z z
y
H ze e
H z
λ λV c V c
x z z
y
H zde e
H zdz
λ λVλ c Vλ c
We are free to choose any sign we wish because it can be accounted for in c-. We put the minus sign in the solution here so that both terms in the differentiated equation will be positive. You will see soon why this is desired.
Lecture 4 Slide 54
Solution for the Magnetic Field (2 of 2)
x z z
y
H zde e
H zdz
λ λVλ c Vλ c
We now have
xx
yy
E zH zdE zH zdz
Q
Recall from previous slides that
x z z
y
E ze e
E z
λ λW c W cand
Combining these results leads to
z z z z
z z
e e e e
e e
λ λ λ λ
λ λ
Vλ c Vλ c Q W c W c
QW c QW c
Comparing the terms on the left and right sides of this equation shows that
1 Vλ QW V QWλ
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Lecture 4 Slide 55
Combined Solution for E and H
Electric Field Solution
x z z
y
E ze e
E z
λ λW c W c
amplitude coefficients of forward wave
amplitude coefficients of backward wave
eigen-vector matrix
diagonal eigen-value matrix
c
c
W
λ
Combined Solution
1 x z z
y
H ze e
H z
λ λV c V c V QWλ
x
zy
zx
y
E z
E z ez
H z e
H z
λ
λ
W W 0 cψ
V V 0 c
Magnetic Field Solution
Does this equation look familiar?
This is the same equation we had on for the 44 approach after we sorted the modes.
Lecture 4 Slide 56
Two Paths to Combined Solution
0
0
r
r
E k H
H k E
Maxwell’s Equations Field Solution
2
2
ˆyz yz zy x y yz zx yz zyzx x
y x x yx yyzz zz zz zz zz zz zz zz
zy yxz zx xzxy x y
zz zz zz zz zzy
x
y
k k kj k k jk k
kE jk j k kE
Hz
H
2
2
x y xz zyxz zxxx xy
zz zz zz
x y yz zx yz zy yz yz zyx zxyx yy y x x
zz zz zz zz zz zz zz zz
y x yxz zxxx
zz zz z
k k
k k kj k k jk
k k k
x
y
x
y
xz zy zyxz zx xzxy y x y
z zz zz zz zz zz
E
E
H
H
jk j k k
2
2
2
2
1
1
x y r r x
r y r r x y
x y r r x
r y r r x y
k k k
k k k
k k k
k k k
P
Q
4×4 Matrix Sort Eigen‐Modes
PQ Method
No sorting!
Isotropic or diagonally anisotropic
Anisotropic
E E
H H
zz
z
ee
e
λλ
λ
W WW
W W
0
0
x
y
x
y
E
E
H
H
zE E
zH H
z
z
ez
e
ez
e
λ
λ
λ
λ
0W W cψ
V V c0
W W 0 cψ
V V 0 c