Post on 29-Aug-2020
9/17/2019
1
Advanced Computation:
Computational Electromagnetics
Transfer Matrix Method (TMM)
Outline
• One‐dimensional structures in electromagnetics
• Formulation of 44 matrix equation for 1D structures
• Solution in an LHI layer
• Transfer matrices for multilayer structures
• Stability of transfer matrices
• Formulation of 22 matrix equation for 1D structures
Slide 2
1
2
9/17/2019
2
Slide 3
One‐Dimensional Structures in Electromagnetics
1D Structures
Slide 4
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
y
Region IReflection Region
Region IITransmission Region
3
4
9/17/2019
3
3D 1D Using Homogenization
Slide 5
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.
1 2 3 4 r
Physical Device Effective Medium Approximation in 1D
3D 1D Using Circuit‐Wave Equivalence
Slide 6
r, r,i i in
ii i
i
Z
5
6
9/17/2019
4
Slide 7
Formulation of 44 Matrix Equationfor 1D Structures
Starting Point
Slide 8
0 r
0 r
0 r
yzx
x zy
y xz
HHk E
y z
H Hk E
z x
H Hk E
x y
0 r
0 r
0 r
yzx
x zy
y xz
EEk H
y z
E Ek H
z xE E
k Hx y
Start with Maxwell’s equations in the following form. Here, isotropic materials are assumedand the positive sign convention is used for waves.
0H j H Positive sign convention
7
8
9/17/2019
5
Calculation of the Wave Vector Components
Slide 9
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 r,inc r,inc
0 r,inc r, c
,inc
,inc in
sin cos
sin sin
x
y
x
y
k k
k
k
k k
2 2 2, 0 r, r,z i i i x yk k k k
Layer #i
kx and ky Continuous Throughout Device
Slide 10z
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
9
10
9/17/2019
6
Waves in Homogeneous Media
Slide 11
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 derivatives of these solutions are calculated, 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.
It cannot be said that because the structure is not homogeneous in the z direction.
zz jk zjk
z
Reduction of Maxwell’s Equations to 1D
Slide 12
0 r
0 r
0 r
yy z x
xx z y
x y y x z
dHjk H k E
dz
dHjk H k E
dz
jk H jk H k E
0 r
0 r
0 r
yy z x
xx z y
x y y x 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
0 r
0 r
0 r
yzx
x zy
y xz
HHk E
y z
H Hk E
z x
H Hk E
x y
0 r
0 r
0 r
yzx
x zy
y xz
EEk H
y z
E Ek H
z xE E
k Hx y
11
12
9/17/2019
7
Normalize the Parameters
Slide 13
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
0 r
0 r
0 r
yy z x
xx z y
x y y x z
dHjk H k E
dz
dHjk H k E
dz
jk H jk H k E
0 r
0 r
0 r
yy z x
xx z y
x y y x z
dEjk E k H
dzdE
jk E k Hdz
jk E jk E k H
r
r
r
yy z x
xx z y
x y y x z
dHjk H E
dz
dHjk H E
dz
jk H jk H E
r
r
r
yy z x
xx z y
x y y x z
dEjk E H
dzdE
jk E Hdz
jk E jk E H
Solve for the Longitudinal Components Ez and Hz
Slide 14
r
r
r
yy z x
xx z y
x y y x z
dHjk H E
dz
dHjk H E
dz
jk H jk H E
r
r
r
yy z x
xx z y
x y y x z
dEjk E H
dzdE
jk E Hdz
jk E jk E H
Solve the third and sixth equations for the longitudinal field components Hz and Ez.
r
z x y y x
jH k E k E
r
z x y y x
jE k H k H
13
14
9/17/2019
8
r
r
r
yy z x
xx z y
z x y y x
dHjk H E
dz
dHjk H E
dzj
E k H k H
r
r
r
yy z x
xx z y
z x y y x
dEjk E H
dzdE
jk E Hdz
jH k E k E
Eliminate the Longitudinal Components
Slide 15
Eliminate the longitudinal field terms by substituting them back into the remaining equations.
2r r r
2r r r
yy x x y y x
xx y x y x y
dEk H k k H H
dzdE
k H k k H Hdz
2r r r
2r r r
yy x x y y x
xx y x y x y
dHk E k k E E
dz
dHk E k k E E
dz
Rearrange Maxwell’s Equations
Slide 16
Rearrange the terms and the order of the equations.
2
rr r
2
r r
2
rr r
2
rr r
x yx xx y
y y x yr x y
x yx xx y
y y x yx y
k kdE kH H
dz
dE k k kH H
dz
k kdH kE E
dz
dH k k kE E
dz
2r r r
2r r r
2r r r
2r r r
yy x x y y x
xx y x y x y
yy x x y y x
xx y x y x 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
15
16
9/17/2019
9
Matrix Form of Maxwell’s Equations
Slide 17
The remaining four equations can be written in matrix form as
2
rr r
2
rr r
2
rr r
2
rr r
0 0
0 0
0 0
0 0
x y x
x xy x y
y y
x xx y x
y y
y x y
k k k
E Ek k k
E EdH Hdz k k kH H
k k k
2
rr r
2
rr r
2
rr r
2
rr r
x yx xx y
y y x yx y
x yx xx y
y y x yx y
k kdE kH H
dz
dE k k kH H
dz
k kdH kE E
dz
dH k k kE E
dz
BTW…for Fully Anisotropic Materials
Slide 18
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 𝑒 sign convention.
17
18
9/17/2019
10
Slide 19
Solution in an LHI Layer
Matrix Differential Equation
Slide 20
Maxwell’s equations can now be written as a single matrix differential equation.
d
dz
ψΩψ 0
2
rr r
2
rr r
2
rr r
2
rr r
0 0
0 0
0 0
0 0
x y x
y x yx
y
x x y x
y
y x y
k k k
k k kE z
E zz
H z k k kH z
k k k
ψ Ω
19
20
9/17/2019
11
Solution of the Differential Equation (1 of 3)
Slide 21
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 is the
exponential of a matrix 𝑒𝛀 calculated?
2
rr r
2
rr r
2
rr r
2
rr r
x yx xx y
y y x yx y
x yx xx y
y y x yx y
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
Functions of Matrices (1 of 2)
Slide 22
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.
21
22
9/17/2019
12
Functions of Matrices (2 of 2)
Slide 23
To calculate f(A) correctly, 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 Vf(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);
Solution of the Differential Equation (1 of 2)
Slide 24
So far, the following matrix differential equation had a general solution of
0zdz e
dz
Ωψ
Ωψ 0 ψ ψ
It is now possible to 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
23
24
9/17/2019
13
Solution of the Differential Equation (2 of 2)
Slide 25
The solution to the matrix differential equation is therefore
d
dz
ψΩψ 0 1 0ze λW W ψ
The final solution is then
zdz e
dz
λψ
Ωψ 0 ψ W c
c
0zz e Ωψ ψ
The unknown initial values (0) can be combined with W-1 because that product just leads to another column vector of unknown constants.
Interpretation of the Solution
Slide 26
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.
25
26
9/17/2019
14
Getting a Feel for the Numbers (1 of 2)
Slide 27
For a layer with r = 9.0 and r = 1.0 (i.e. n = 3.0) and a wave at normal incidence, the matrix will be
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
Getting a Feel for the Numbers (2 of 2)
Slide 28
It can be observed that the modes occur as either an Ex‐Hy or an 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.
The refractive index is known (n = 3.0), so the eigen‐values are consistent with what may be expected. 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
inc
r r
cos
3
jn zze e
jn
n
r0
r
r
r
1
3
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.
27
28
9/17/2019
15
Mode 4
Mode 3
Mode 2
Mode 1
Visualizing the Modes
Slide 29
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
0.95
-j0.32
Mode 2
0.95
j0.32
Mode 3
0.95
j0.32 Mode 4
0.95
-j0.32
Slide 30
Transfer Matrices for Multilayer Structures
29
30
9/17/2019
16
Geometry of an Intermediate Layer
Slide 31
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
Field Relations
Slide 32
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
Must include k0 in the exponential to normalize Li-1 because the parameter i-1 expects to multiply a normalized coordinate.
Note: Must equate the field on either side of the interfaces and not the mode coefficients c.
31
32
9/17/2019
17
The Transfer Matrix
Slide 33
The transfer matrix Ti of the ith layer is defined as:
Start with the boundary condition equation from the second interface and rearrange terms…
1i i i c T c
iT0 01
1 1 1 1 i i i ik L k Li 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 W…then read off the transfer matrix.
The Transfer Matrix Method
Slide 34
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.
33
34
9/17/2019
18
The Global Transfer Matrix
Slide 35
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
Slide 36
Stability of Transfer Matrices
35
36
9/17/2019
19
The Multi‐Layer Problem
Slide 37
The diagram below is focused on an arbitrary layer in a stack of multiple layers. Now consider the wave solutions in this ith layer.
Wave Solutions in ith Layer
Slide 38
Recall that the wave number 𝑘 can be purely real (pure oscillation), purely imaginary (pure exponential decay), or complex (decay + oscillation).
k k jk
k jk
k kPure oscillation
Pure decay
Decaying oscillation
37
38
9/17/2019
20
Backward Waves in ith Layer
Slide 39
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.
All Waves are Treated as Forward Waves
Slide 40
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
39
40
9/17/2019
21
TMM is Inherently Unstable
Slide 41
The 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.
It is known that backward waves exist. It is also known that decaying fields exist when a wave is evanescent or propagating in a lossy material.
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.
The Fix
Slide 42
All power in the waves is being treated as forward propagating because it power was distinguished as forward and backward waves.
Clearly, the first part of the fix is to identify 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! A normalized magnetic field is being used.
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
41
42
9/17/2019
22
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
Rearrange Eigen Modes
Slide 43
Now that it is known which eigen‐modes are forward and backward propagating, they can be rearranged 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
Also need to adjust the vertical positions of the eigen‐values so that ’ remains a diagonal matrix.
rearrange modes
New Interpretation of the Matrices
Slide 44
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
The matrices are now partitioned 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.
43
44
9/17/2019
23
Revised Solution to Differential Equation
Slide 45
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, the solution can be written as
z
E E
zH H
ez
e
λ
λ
0W W cψ
W W c0
There are now separate mode coefficients c+ and c- for forward and backward propagating modes, respectively.
Slide 46
Formulation of 22 Matrix Equationfor 1D Structures
45
46
9/17/2019
24
Recall Derivation Up to 44
Slide 47
0 r
0 r
0 r
0 r
0 r
0 r
yzx
x zy
y xz
yzx
x zy
y xz
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 r
0 r
0 r
0 r
0 r
0 r
yy z x
xx z y
x y y x z
yy z x
xx z y
x y y x 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
r
r
r
r
r
r
yy z x
xx z y
x y y x z
yy z x
xx z y
x y y x 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
rr r
2
rr r
2
rr r
2
rr r
x yx xx y
y y x yx y
x yx xx y
y y x yx y
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.
Derivation of Two 22 Matrix Equations
Slide 48
2
rr r
2
rr r
x yx xx y
y y x yx y
k kdE kH H
dz
dE k k kH H
dz
2
rr r
2
rr r
x yx xx y
y y x yx y
k kdH kE E
dz
dH k k kE E
dz
The four equations can be written as two separate matrix equations instead.
2r r
2r r r
1x x y x x
y yy x y
E k k k HdE Hdz k k k
2r r
2r r r
1 x y x xx
yy y 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.
47
48
9/17/2019
25
Standard “PQ” Form
Slide 49
The two matrix equations more compactly by defining auxiliary P and Q matrices.
2r r
2r r r
1x x y x x
y yy x y
E k k k HdE Hdz k k k
2r r
2r r r
1 x y x xx
yy y x y
k k k EHdEHdz k k k
x x
y y
E HdE Hdz
P
xx
yy
EHdEHdz
Q
2r r
2r r r
1 x y x
y x y
k k k
k k k
P
2r r
2r r r
1 x y x
y x y
k k k
k k k
Q
Note: This same “PQ” form will be seen again 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.
Matrix Wave Equation
Slide 50
The two staring matrix equations are
To derive a matrix wave equation, first 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, 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
49
50
9/17/2019
26
Numerical Solution (1 of 3)
Slide 51
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, a second‐order differential equation is solved so the modes we are all propagating in a single direction. They must be explicitly written twice to account for forward and backward waves and thus they are automatically distinguished. In the 4×4 approach, a first‐order differential equation was solved that lumped forward and backward modes together.
Numerical Solution (2 of 3)
Slide 52
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
This relation is used to calculate the matrix exponentials.
51
52
9/17/2019
27
Numerical Solution (3 of 3)
Slide 53
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
Solution for the Magnetic Field (1 of 2)
Slide 54
Since the electric and magnetic fields are coupled and not independent, it should be possible to compute V from W. First, 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.
53
54
9/17/2019
28
Solution for the Magnetic Field (2 of 2)
Slide 55
x z z
y
H zde e
H zdz
λ λVλ c Vλ c
From the previous slide,
xx
yy
E zH zdE zH zdz
Q
Recall
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
Combined Solution for E and H
Slide 56
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 obtained for the 44 approach after the modes were sorted.
55
56
9/17/2019
29
Two Paths to Combined Solution
Slide 57
0 r
0 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
2r r
2r r r
2r r
2r r r
1
1
x y x
y x y
x y x
y 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
57