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4/27/2015
1
Lecture 4 Slide 1
EE 4395/5390 – Special Topics
Computational Electromagnetics
Lecture #4
Transfer Matrix Method
These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited
InstructorDr. Raymond Rumpf(915) 747‐[email protected]
Outline
• Maxwell’s equations for 1D structures
• Solution to Maxwell’s equations in a homogeneous layer
• Multilayer structures
• TMM is an inherently unstable method
Lecture 4 Slide 2
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Lecture 4 Slide 3
Maxwell’s Equations for 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
Lecture 4 Slide 6
3D 1D Using Circuit‐Wave Equivalence
1d 2d 3d 4d 5d 6d
1
1N
2
2N
3
3N
4
4N
5
5N
6
6N
in
inN
L
LN
1d2d
3d4d
5d6d
1Z2Z
3Z4Z
5Z6Z
trn LZ Z
refZ
r rN
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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.
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.
,inc ,inc
,inc ,inc
sin cos
sin sin
x r r
y r r
k
k
2 2, , ,z i r i r i x yk k k
Layer #i
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Lecture 4 Slide 9
Waves in Homogeneous Media
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
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.
When we take derivatives of these solutions, we see that
We cannot say that because the structure is not homogeneous in the z direction.
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.
zz jk zjkz
Lecture 4 Slide 10
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 can be ordinary.
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Lecture 4 Slide 11
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 parameters 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
Lecture 4 Slide 12
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.
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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 13
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
Lecture 4 Slide 14
Rearrange Maxwell’s Equations
Here we simply change the order that we our previous 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
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Lecture 4 Slide 15
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
Lecture 4 Slide 16
BTW…for 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
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Lecture 4 Slide 17
Solution to Maxwell’s Equations in a
Homogeneous Layer
Lecture 4 Slide 18
Matrix Differential Equation
Maxwell’s equations can now be written as a single matrix differential equation.
d
dz ψ
Ωψ 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
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Lecture 4 Slide 19
Solution of the Differential Equation (1 of 3)
The matrix differential equation is
d
dz ψ
Ωψ 0
This is actually a set of four coupled differential equations. The system of four equations can be solved as a single matrix equation as follows.
0zz e Ωψ ψ
This is easy to write, but how do we compute the exponential of a matrix?
Lecture 4 Slide 20
Functions of Matrices
It is sometimes necessary to evaluate the function of a matrix.
?f A
It is NOT correct to calculate the function of every element in the matrix A individually. A different technique must be used.
To do this, we first calculate the eigen‐vectors and eigen‐values of the matrix A.
?
eigen-vector matrix of
eigen-value matrix of
f
A
W AA
λ A
The function of the matrix is then evaluated as
1f f A W λ W This is very easy to evaluate because is a diagonal matrix so the function only has to be performed individually on the diagonal elements.
1
2
3
4
0 0 0
0 0 0
0 0 0
0 0 0
λ
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Lecture 4 Slide 21
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 22
Solution of the Differential Equation (2 of 2)
The solution to the matrix differential equation is therefore
d
dz ψ
Ωψ 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 23
Interpretation of the Solution
zz e λψ W 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 energy is in each mode.
Lecture 4 Slide 24
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 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 25
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 refractive index of the material.
Lecture 4 Slide 26
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 27
Multilayer Structures
Lecture 4 Slide 28
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
zi is a local z‐coordinate inside the ith layer that starts at zero at the layer’s left side.
+z
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Lecture 4 Slide 29
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, not the mode coefficients c.
Lecture 4 Slide 30
The Transfer Matrix
The transfer matrix Ti of the ith
layer is defined as:
After some algebra, the transfer matrix is computed as
1i i i c T c
iT
011
i ik Li i ie
λT W W
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Lecture 4 Slide 31
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 32
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 33
TMM is an Inherently Unstable Method
Lecture 4 Slide 34
The Multi‐Layer Problem
The figure below is focused on an arbitrary layer in a stack of multiple layers. We will be examining the wave solutions in the ith layer.
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Lecture 4 Slide 35
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 k
Lecture 4 Slide 36
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 37
All Waves are Treated as Forward Waves
The true 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 38
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 energy 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 39
The Fix
We are treating all energy 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
z x y y x
y xz x y
z x y y x
E H
E H E H
H HE j E j
jE H E H
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 40
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.
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Lecture 4 Slide 41
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 42
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.
We will pick up here next lecture.
