10/1/2015
Instructor
Dr. Raymond Rumpf
(915) 747‐6958
rcrumpf@utep.edu
EE 5320
Computational Electromagnetics
Lecture #5
Transfer Matrix Method
Using Scattering Matrices
 5These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited 
Lecture
Slide 1
Outline
•
•
•
•
•
Recall 44 and 22 formulations
Scattering matrix for a single layer
Multilayer structures
Calculating reflected and transmitted power
Notes on implementation
Lecture 5
Slide 2
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10/1/2015
Recall 44 and 22
Formulations
Lecture 5
Slide 3
Derivation Up to Matrix Form
Start with Maxwell’s
equations from
Lecture 2.
Assume LHI.
Ez E y

 k0  r H x
y
z
Ex Ez

 k0  r H y
z
x
E y Ex

 k0  r H z
x
y
H z H y

 k0  r E x
y
z
H x H z

 k0  r E y
z
x
H y H x

 k0  r E z
x
y
Lecture 5
Assume device is
infinite and uniform
in x and y directions.

 jk x
x
jk y Ez 

 jk y
y
dE y
dz
 k0 r H x
dE x
 jk x E z  k0 r H y
dz
jk x E y  jk y E x  k0  r H z
jk y H z 
dH y
dz
 k 0 r E x
dH x
 jk x H z  k0 r E y
dz

jk x H y  jk y H x  k0 r Ez
Normalize z and
wave vectors kx, ky,
and kz.
z   k0 z
k
kx  x
k0
ky
ky 
k0
Eliminate
longitudinal
components Ez and
Hz by substitution.
k
kz  z
k0
dE y
jky Ez 
 r H x
dz 
dEx
 jkx Ez  r H y
dz 
jkx E y  jky Ex   r H z
dH y
jky H z 
  r Ex
dz 

dH x
 jkx H z   r E y
dz 
jkx H y  jky H x   r Ez
dEx kx ky  
k 2 
H x   r  x  H y

dz 
r
r 

 k

dE y  ky2
k
x y 
Hy
    r  H x 

dz    r
r


dH x kx ky
k 2 
Ex    r  x  E y

dz 
r
r 


dH y  ky2
kx ky
E
    r  Ex 

dz    r
r y

Slide 4
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10/1/2015
Two Paths to Combined Solution
4×4 Matrix
   yz
 kx ky
 
  

 
jkx  yz  zy 
 kx zx 
  yx  yz zx 
  j  ky

  zz
 zz 
 zz 
   zz
  zz  zz 


2



k









y
 Ex 
xz  zx
xz
zx

 xz  k zy





jk
j
k






xx
y
x
y

 

 zz 
 zz 
  zz  zz 
  zz
  Ey  
  zz




2
H



z
k k
x
    
  
   yz   zx 
    x y   yx  yz zx    k x   yy  yz zy 
j
k
k


 y

x
 zz    zz
 zz 
 zz 
 H y    zz
  zz

  k 2
  k k
  

 
  y   xx   xz zx    x y   xy  xz zy 
jky  xz  zx 




 zz    zz
 zz 
   zz
  zz  zz 
Sort Eigen‐Modes


 


  E 
  x 

  Ey 
  H 
  yz  zy 
 x
jkx 


  H y 
  zz  zz 

   xz   zy  
 j  kx
 ky
 
 zz  
  zz
  kˆx2
 
k
  yy  yz zy
 
 zz
zz

 kx ky
 xz  zy
  xy 


 zz
  zz
 W
W   E
 WH
eλ z

 0

e λz
WE 

WH 
Ex
Ey
H x
H
y
0 

 λ  z
e

Anisotropic
Field Solution
Maxwell’s Equations


  E  k0  r  H


  H  k0  r  E
Isotropic or
diagonally
anisotropic
Lecture 5
 W
ψ  z    E
 VH
WE  e

VH   0
λ  z
 W W  e λz
ψ  z   

 V V   0
r  r  kx2 
1  kx ky
 2


 r  k y   r  r
 kx ky 
 r  r  kx2 
1  kx ky
Q
 2


 r  k y   r  r
 kx ky 
0  c  
 

e  λ z  c 
0  c  
 
e  c  
 λz 
P
No sorting! 
PQ Method
Slide 5
Scattering Matrix
for a Single Layer
R. C. Rumpf, “Improved Formulation of Scattering Matrices for Semi‐Analytical
Methods That is Consistent with Convention,” PIERS B, Vol. 35, pp. 241‐261, 2011.
Lecture 5
Slide 6
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Motivation for Scattering Matrices
Scattering matrices offer several important features and
benefits:
• Unconditionally stable method.
• Parameters have physical meaning.
• Parameters correspond to those measured in the lab.
• Can be used to extract dispersion.
• Very memory efficient.
• Can be used to exploit
longitudinal periodicity.
• Mature and proven approach.
• Much greater wealth of
literature available.
Excellent alternatives to S‐matrices do exist!
Lecture 5
Slide 7
Geometry of a Single Layer
Indicates a point that lies on an interface, but associated with a particular side.
Layer i
Medium 1
ψ1
ψ i  0 
ψ1
ψ i  0 
ψ  zi 

i
ψ i  zi 
c1 ci
Medium 2
ψ i  k0 Li 
ψ 2
ψ i  k0 Li 
ψ 2
c2
c2
c1 ci
Li
ψ i  z   field within i th layer
Lecture 5
z
ci  mode coefficients inside i th layer
ci   mode coefficients outside i th layer
Slide 8
4
10/1/2015
Field Relations
Field inside the ith layer:
 Ex ,i  zi  


E y ,i  zi    Wi
ψ i  zi    

 H x ,i  zi    Vi


 H y ,i  zi  
0  ci 
 
e  λi zi  ci 
Wi   e λi zi

 Vi   0
Boundary conditions at the first interface:
ψ1  ψ i  0 
 W1 W1  c1   Wi
 V  V       V
1  c1 
 1
 i
Wi  ci 
 
Vi  ci 
Boundary conditions at the second interface:
ψ i  k0 Li   ψ 2
 Wi
V
 i
Wi  e λ i k0 Li

Vi   0
 ci   W2
   
e λ i k0 Li  ci   V2
W2  c2 
 
V2  c2 
0
Note: k0 has been
incorporated to normalize Li.
Lecture 5
Slide 9
Definition of A Scattering Matrix
c1   S11 S12  c1 
   
 
c2  S 21 S 22  c2 
S 21
c

1
S11
c1
Lecture 5
S11  reflection
S 21  transmission
transmission
reflection
c2
S 22
S12
This is consistent with experimental convention.
c2
Slide 10
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10/1/2015
Derivation of the Scattering Matrix
Solve both boundary condition equations for the intermediate mode


coefficients ci and ci .
1
ci   Wi
   
ci   Vi
ci  e λ i k0 Li
   
c i   0
Wi   W1 W1  c1 
 
Vi   V1 V1  c1 
0
e
 λ i k0 Li
  Wi

  Vi
1
Wi   W2
Vi   V2
W2  c2 
 
V2  c2 
Both of these equations have
 Wi

 Vi
1
Wi   W j

Vi   V j
W j  1  A ij

V j  2  Bij
A ij  Wi1W j  Vi1V j
Bij 
A ij 
Bij  Wi1W j  Vi1V j
We set substitute this result into the first two equations and set them
equal.
1  Ai1
2  Bi1
Bi1  c1  1 e λ i k0 Li
  
Ai1  c1  2  0
0
e
 λ i k0 Li
Bi 2  c2 
 
Ai 2  c2 
  Ai 2

  Bi 2
We write this as two matrix equations and rearrange terms until they
have the form of a scattering matrix.
c1  ? ? c1 
   
 
c2  ? ? c2 
Lecture 5
Slide 11
The Scattering Matrix
The scattering matrix Si of the ith
layer is defined as:

c1 
 i  c1 
  S  
c2 
c2 
i 
S11
S  i

S 21
 
S12

S 22i  
i
i 
After some algebra, the components of
the scattering matrix are computed as

S    A
S    A
S    A
i
 A i1  Xi Bi 2 A i21Xi Bi1
S11
i
12
i1
 Xi Bi 2 A i21Xi Bi1
i
21
i2
 Xi Bi1A i11Xi Bi 2
i
22
Lecture 5
1
i 2  Xi B i1 A i1 X i B i 2
 X B A X A
 X A  B A
 X A  B A
 X B A X A
1
i
i2
1
i2
i
i1
1
Si
 Bi1
i
i2
i2
1
i2
i
i1
i1
1
i1
1
1
i
i1
1
i1
i



Bij  Wi1W j  Vi1V j
Bi 2
Bi1
i 2  Bi 2
A ij  Wi1W j  Vi1V j

X i  e λ i k0 Li
i is the layer number.
j is either 1 or 2 depending on which external medium is being referenced.
Slide 12
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Scattering Matrices in the Literature
For some reason, the computational electromagnetics community has:
(1) deviated from convention, and (2) formulated scattering matrices
inefficiently.
S 22 transmission
c   S11 S12   c 
 
 
 c  S 21 S 22  c 

i 1

i

i

i 1
S12 reflection
S 21
S11
ci1
ci
ci1
ci
Lecture 5
• Here s11 is not reflection.
Instead. it is backward transmission!
• Here s21 is not transmission.
Instead, it is a reflection parameter!
• Scattering matrices can not be interchanged.
• Scattering matrices are not symmetric so they take twice
the memory to store and are more time‐consuming to
calculate.
Slide 13
Limitation of Conventional S‐Matrix Formulation
Note that the elements of a scattering matrix are a function of
materials outside of the layer.
This makes it difficult to interchange scattering matrices arbitrarily.
For example, there are only three unique layers in the multilayer
structure below, yet 20 separate computations of scattering
matrices are needed.
Three unique layers
20 layer stack
Lecture 5
Slide 14
7
10/1/2015
Solution
To get around this, we will surround each layer with an external regions
of zero thickness. This lets us connect the scattering matrices in any
order because they all calculate fields that exist outside of the layers in
the same medium. This will have no effect electromagnetically as long
as we make the external regions have zero thickness.
Gap Medium
Layer i
Gap Medium
Li
Lecture 5
Slide 15
Visualization of the Technique
We calculate the scattering matrices for just the unique layers.
Three unique layers
Then we just manipulate these same three scattering matrices to
“build” the global scattering matrix.
Gaps between the layers are made to have zero thickness so
they have no effect electromagnetically.
Faster! Simpler! Less memory needed!
Lecture 5
Slide 16
8
10/1/2015
Revised Geometry of a Single Layer
same
Gap Medium
Gap Medium
Layer i
ψ1
ψ i  0 
ψ1
ψ i  0 
ψ i  zi 
ψ i  k0 Li  ψ 2
r ,h
r ,h
 r ,h
ψ i  zi 
ψ i  k0 Li  ψ 2
c1 ci
 r ,h
c2
c2
c1 ci
Li
Lecture 5
Slide 17
Calculating Revised Scattering Matrices
The scattering matrix Si of the ith
layer is still defined as:

c1 
 i  c1 

S
 
 
c2 
c2 
S  i 
S  i    11i
S 21
i  
S12

S 22i  
But the elements are calculated as

S    A
i
12
i
i 
S21i   S12
i 
i 
S 22  S11
Lecture 5
 Xi Bi A i1
Si
 X B A X A  B 
X B  X A  B A B 
i 
S11
 A i  Xi Bi A i1Xi Bi
1
i
1
i
i
i
1
i
i
i
i
i
i
1
i
i
i
• Layers are symmetric so the scattering matrix
elements have redundancy.
• Scattering matrix equations are simplified.
• Fewer calculations.
• Less memory storage.
A i  Wi1Wh  Vi1Vh
Bi  Wi1Wh  Vi1Vh
X i  e λ i k0 Li
Slide 18
9
10/1/2015
Layers in TMM are Four‐Port Systems
We have written the scattering matrices as 22 block matrices.
 S11 S12 
S

 21 S 22 
For TMM, this actually expands to a 44 element scattering matrix.
 S11
S
 21
 s11
S12   s21

S 22   s31

 s41
s
S11   11
 s21
s
S 21   31
 s41
s12 
s22 
s32 
s42 
s12
s13
s22
s23
s32
s33
s42
s43
s
S12   13
 s23
s
S 22   33
 s43
s14 
s24 
s34 

s44 
s14 
s24 
s34 
s44 
Each mode provides an I/O mechanism
and there are two modes on each side.
Lecture 5
Slide 19
Scattering Matrices of Lossless Media
If a scattering matrix is composed of materials that have no loss and
no gain, the scattering matrix must conserve energy. That is, all
incident energy must either reflect or transmit.
This implies that the scattering matrix is unitary.
If the scattering matrix is unitary, it must obey the following rules:
S H  S 1
S H S  SS H  S 1S  SS 1  I
Note: If the regions external the layer are different from each other, the scattering
matrices will not be unitary. This is because the field amplitudes will be different even
though the field carries the same amount of power.
Lecture 5
Slide 20
10
10/1/2015
Hints About Stability in These Formulations
• Diagonal elements S11 and S22 tend to be the
largest numbers. Divide by these instead of any
off‐diagonal elements for best numerical stability.
 S11 S12 
S

S
S
22 
 21
• X describes propagation through an entire layer.
Don’t divide by X or your code can become
unstable.
Lecture 5
Slide 21
Multilayer Structures
Lecture 5
Slide 22
11
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Solution Using Scattering Matrices
The scattering matrix method consists of working through the device
one layer at a time and calculating an overall scattering matrix.
S 1
S 2
S  3
S 4
S 5
S  device   S1  S 2   S  3  S 4   S  5
Redheffer star product.
NOT matrix multiplication!
Lecture 5
Slide 23
Redheffer Star Product
Two scattering matrices may be combined into a single scattering
matrix using Redheffer’s star product.
S
 AB 
S
 A
S
S  A 
S A   11A
 
S 21
 B
A 
S12

S22A  
S  B
S B    11B
 
S 21
 B 
S12

S22B 
The combined scattering matrix is then

S11
AB 
S
 AB 
 AB
S11
  AB
 
S 21
 AB 
S12

S22AB 
1
 
 
   
   
 S11
 S12
I  S11 S 22  S11 S 21
A
A
B
A
B
A
1
 AB 
A 
 B  A  
 B
 S12
S12
I  S11 S 22  S12
1
 
 
S21   S21  I  S22 S11
 S 21
AB
B
A
B
A
1
 B 
 A   B
S22AB  S22B  S21B I  S 22A S11
 S 22 S12
R. Redheffer, “Difference equations and functional equations in transmission-line theory,”
Modern Mathematics for the Engineer, Vol. 12, pp. 282-337, McGraw-Hill, New York, 1961.
Lecture 5
Slide 24
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10/1/2015
Derivation of the Redheffer Star Product
We start with the equations for the two adjacent scattering matrices.
A
c1  S11
     A
c2  S 21
A 
S12
c1 

 
A
S22   c2 
 B
c2  S11
      B
c3  S 21
 B 
S12
c2 

 
B
S 22   c3 
We expand these into four matrix equations.
A 
A 
 B 
 B 
Eq. 1
Eq.  3
c1  S11
c1  S12
c2
c2  S11
c2  S12
c3
A 
A 
 B 
 B 


Eq.  2 
Eq.  4 
c2  S 21 c1  S 22 c2
c3  S 21 c2  S 22 c3
We substitute Eq. (2) into Eq. (3) to get an equation with only c2 .
We substitute Eq. (3) into Eq. (2) to get an equation with only c2 .
 I  S S   c
 I  S   S    c
B
11
A
22

2
A
22
B
11

2
 B  A  
 B 
 S11
S 21 c1  S12
c3
A
 A   B
 S 21 c  S 22 S12 c

1

3
Eq.  5 
Eq.  6 
We eliminate c and c by substituting these equations into Eq. (1)
and (4). We then rearrange terms into the form of a scattering matrix.

2
c1  ? ? c1 
   
 
c3  ? ? c3 

2
Overall, this is just algebra. We start with 4 equations and 6
unknowns and reduce it to 2 equations with 4 unknowns.
Lecture 5
Slide 25
Putting it All Together
We have one remaining problem. In the revised framework, the global scattering
matrix places the device in free space. In many applications, we may want
something other than free space outside the device.
We connect the global scattering matrix to the external materials by surrounding it
by “connection” scattering matrices that have zero‐thickness
Device exists within gap medium
S
global 
 S
ref 
 S    S      S     S 





1
2
N
trn 
Device in gap medium
Lecture 5
S device 
Slide 26
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10/1/2015
Reflection/Transmission Side Scattering Matrices
The reflection‐side scattering matrix is
 ref 
1
ref
 ref 
1
ref
S11   A B ref
S12  2 A
s ref
A ref  Wh1Wref  Vh1Vref

S 21ref   0.5 A ref  B ref A ref1 B ref

1
h
r ,I
r ,h
 r ,I
 r ,h
1
h
B ref  W Wref  V Vref
S 22ref   B ref A ref1
lim
L0
The transmission‐side scattering matrix is
 
 B trn A trn1
S11
s trn
trn

 
 0.5 A trn  B trn A trn1 B trn
S12
trn
S21   2A trn1
trn
S22trn    A trn1 B trn

1
h
1
h
1
h
1
h
A trn  W Wtrn  V Vtrn
r ,h
r ,II
 r ,h
 r ,II
B trn  W Wtrn  V Vtrn
lim
L0
Lecture 5
Slide 27
Calculating Transmitted
and Reflected Power
Lecture 5
Slide 28
14
10/1/2015
Recall How to Calculate Source Parameters
Incident Wave Vector
Surface Normal
sin  cos  

kinc  k0 ninc  sin  sin  
 cos  
 aˆ y


aˆTE   nˆ  kinc

 nˆ  k
inc


aˆ  k
ˆaTM  TE inc
aˆTE  kinc
0
nˆ  0
1 
Right‐handed
coordinate system
x
y
Unit Vectors Along Polarizations
z
  0
  0
Composite Polarization Vector

P  pTE aˆTE  pTM aˆTM
In CEM, we usually make

P 1
Lecture 5
Slide 29
Solution Using Scattering Matrices
The external fields (i.e. incident wave, reflected wave, transmitted
wave) are related through the global transfer matrix.
c ref 
 global  cinc 
c   S
 0 
 
 trn 
E
1  x ,inc 
cinc  Wref
E 
 y ,inc 
We get Ex,inc and Ey,inc from the
polarization vector P.  E   P 
x ,inc
x
E   P 
 y ,inc   y 
This matrix equation can be solved to calculate the mode coefficients
of the reflected and transmitted fields.
 global 
c ref  S11


c 
global
 trn  S 21 
 global  
S12
cinc 

 global   0 
S 22   
 global 
c ref  S11
cinc
c trn  S21globalcinc
cinc 
right
cinc
not typically used
Lecture 5
Slide 30
15
10/1/2015
Calculation of Transmitted and Reflected Fields
The procedure described thus far calculated cref and ctrn.
The transmitted and reflected fields are then
inc
 Exref 
 global 
 global  1  Ex 
 ref   Wref c ref  Wref S11 cinc  Wref S11 Wref  inc 
 E y 
 E y 
inc
 Extrn 
 global 
 global  1  Ex 
 trn   Wtrn c trn  Wtrn S 21 cinc  Wtrn S 21 Wref  inc 
 E y 
 E y 
Lecture 5
Slide 31
Calculation of the Longitudinal Components
We are still missing the longitudinal field component Ez on either
size of the layer stack.
These are calculated using Maxwell’s divergence equation.

E  0
 
 
 



E0, x e jk r 
E0, y e jk r 
E0, z e jk r  0
x
y
z

jk x E0, x e

 
jk  r

 jk y E0, y e

 
jk  r

 jk z E0, z e
Note:

   E  0 reduces to

  E  0 when  is homogeneous.
 

 
jk  r
0
k x E0, x  k y E0, y  k z E0, z  0
k z E0, z  k x E0, x  k y E0, y
E0, z  
k x E0, x  k y E0, y
kz
Ezref  
kx E xref  ky E yref
k ref
z
Eztrn
kx Extrn  ky E ytrn

k trn
z
Lecture 5
Slide 32
16
10/1/2015
Calculation of Power Flow
Reflectance is defined as the fraction of power reflected from a device.
 2
Eref
2
2
2
2
R  2
E  Ex  E y  Ez
Einc
Transmittance is defined as the fraction of power transmitted
through a device.
 2
Note: We will derive these
Etrn

k 
formulas in Lecture 7.
T   2 Re  r ,ref z ,trn 
k

Einc
 r ,trn z ,inc 
It is always good practice to check for conservation of energy.
 1 materials have loss
R  T   1 materials have no loss and no gain
 1 materials have gain
Lecture 5
Slide 33
Reflectance and Transmittance on a Decibel Scale
Decibel Scale
PdB  20 log10  A 
PdB  10 log10  P 
How to calculate decibels from an amplitude quantity.
How to calculate decibels from a power quantity.
P  A2
PdB  10 log10  A2   20 log10  A 
Reflectance and Transmittance
Reflectance and transmittance are power quantities, so
RdB  10 log10  R 
TdB  10 log10 T 
Lecture 5
Slide 34
17
10/1/2015
Notes on
Implementation
Lecture 5
Slide 35
Outline
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Step 0 – Define problem
Step 1 – Dashboard
Step 2 – Describe device layers
Step 3 – Compute wave vector components
Step 4 – Compute gap medium parameters
Step 5 – Initialize global scattering matrix
Step 6 – Main loop through layers
Step 7 – Compute reflection side
scattering matrix
Step 8 – Compute transmission side
scattering matrix
Step 9 – Update global scattering matrix
Step 10 – Compute source
Step 11 – Compute reflected and transmitted fields
Step 12 – Compute reflectance and transmittance
Step 13 – Verify conservation of energy
Lecture 5
human does this
computer does the rest
Step 6: Iterate through layers
• Compute P and Q
• Compute eigen‐modes
• Compute layer scattering
matrix
• Update global scattering
matrix
Slide 36
18
10/1/2015
Storing the Problem
How is a the device described and stored for TMM?
We don’t use a grid for this method!
Store the permittivity for each layer in a 1D array.
Store the permeability for each layer in a 1D array.
Store the thickness of each layer in a 1D array.
ER = [ 2.50 , 3.50 , 2.00 ];
UR = [ 1.00 , 1.00 , 1.00 ];
L = [ 0.25 , 0.75 , 0.89 ];
Input arrays for
three layers
We will also need the external materials, and source parameters.
er1, er2, ur1, ur2, theta, phi, pte, ptm, and lam0
Lecture 5
Slide 37
Storing Scattering Matrices
We often talk about the scattering matrix S as a single matrix.
S12 
S
S   11

S 21 S 22 
However, we never actually use the scattering matrix S this way.
We only ever use the individual terms S11, S12, S21, and S22.
So, scattering matrices are actually best stored as the four separate
components of the scattering matrix.
S12 
S
S   11

S 21 S 22 
Lecture 5

S11 , S12 , S 21 , and S 22
Slide 38
19
10/1/2015
Initializing the Global Scattering Matrix
Before we iterate through all the layers, we must initialize the global
scattering matrix as the scattering matrix of “nothing.”
What are the ideal properties of nothing?
1.
Transmits 100% of power with no phase change.
 global 
S12
 S21global   I
2.
Does not reflect.
 global 
S11
 S 22global  0
We therefore initialize our global scattering matrix as
0 I 
S global   

 I 0
This is NOT an identity matrix!
Look at the position of the 0’s and I’s.
Lecture 5
Slide 39
Analytical Expressions for W and 
The dispersion relation with normalized wave vectors is
r  r  kx2  ky2  kz2
Using this relation, we can simplify the matrix equation for 2.
Ω 2  PQ 
1  kx ky

r  r  ky2  r  r
r  r  kx2   kx ky
kx ky
 2

  k y  r  r
r  r  kx2   kz2

  0
kx ky
0 
2
  kz I
2

 k z 
1 0 
I
 identity matrix
0 1 
Since this is a diagonal matrix, we can conclude that
1 0 
WI

0 1 
λ 2  Ω2
 jk
λ z
 0
0 
  jkz I
jkz 
e
λz 
 e jkz z

 0
0 


e jk z z 
We don’t actually have to solve the eigen‐value problem to obtain the eigen‐modes! 
Lecture 5
Slide 40
20
10/1/2015
Calculating the Parameters of the Homogeneous
Gaps
Our analytical solution for a homogeneous layer is
1  kx ky
Qh 

r ,h  ky2  r ,h r ,h
r ,h r ,h  kx2 


kx ky
kz2,h  r ,h r ,h  kx2  ky2
Wh  I
λ h  jkz ,h I
We are free to choose any r and
Vh  Q h Wh λ h1
r that we wish. We also wish to
avoid the case of kz,h = 0. For convenience, we choose
r ,h  1.0
 r ,h  1  kx2  ky2
and
We then have
 kx ky
Qh  
2
  1  kx


1  ky2 

kx ky 

Wh  I
W not even used in TMM.
Vh   jQ h
Lecture 5
Slide 41
Simplifications for TMM in LHI Media
In LHI media,
1 0 
Wi  I  

0 1 
and
Ωi  jkz ,i I
1 0 
I
 identity matrix
0 1 
Now we do not actually have to calculate  because
λ i  Ωi
Given all of this, the eigen‐vectors for the magnetic fields can be calculated as
Vi  Qi Wi λ i1  Qi Ωi1
When calculating scattering matrices, the intermediate matrices Ai and Bi are
A i  Wi1Wh  Vi1Vh  I  Vi1Vh
Bi  Wi1Wh  Vi1Vh  I  Vi1Vh
The fields and mode coefficients are now related through
P
 Px 
1  x 
cinc  Wref
P   P 
 y  y
Lecture 5
 Exref 
 ref   Wref S11cinc  S11cinc
 E y 
 Extrn 
 trn   Wtrn S 21cinc  S 21cinc
 E y 
Slide 42
21
10/1/2015
Calculating Xi = exp(-ik0Li)
Recall the correct answer:
Xi  e
Ωi k0 Li
 e jkz k0 Li

 0



e jk z k0 Li 
0
It is incorrect to use the function exp() because this calculates a
point‐by‐point exponential, not a matrix exponential.
X =
0.0135 + 0.9999i
1.0000
WRONG
X = exp(OMEGA*k0*L);
1.0000
0.0135 + 0.9999i
Approach #2: diag()
Approach #1: expm()
X = expm(OMEGA*k0*L);
X = diag(exp(diag(OMEGA)*k0*L));
X =
0.0135 + 0.9999i
0
X =
0.0135 + 0.9999i
0
0
0.0135 + 0.9999i
0
0.0135 + 0.9999i
Lecture 5
Slide 43
Efficient Calculation of Layer S‐Matrices
There are redundant calculations in the equations for the scattering
matrix elements.

S   S    A
 X B A X A  B 
X B  X A  B A B 
i 
S11
 S 22i   A i  Xi Bi A i1Xi B i
i
12
i
21
i
 Xi B i A
1
i
1
i
1
i
i
i
1
i
i
i
i
i
i
1
i
i
i
This should be calculated as
A i  I  Vi1Vh
Bi  I  Vi1Vh
Xi  e λ i k0 Li
D  A i  Xi B i A i1Xi B i

i 
S11
 S 22i   D1 Xi B i A i1Xi A i  B i

i 
S12
 S 21i   D1Xi A i  B i A i1Bi
Lecture 5


Slide 44
22
10/1/2015
Efficient Star Product
After observing the equations to implement the Redheffer star
product, we see there are some common terms. Calculating these
multiple times is inefficient so we calculate them only once using
intermediate parameters.
S AB   S A  S B 

S11
AB 
 AB
S12
 AB
1
 
 
   
   
 S11
 S12
 I  S11 S 22  S11 S 21
A
A
B
A
 B  A  
 B
I  S11
S 22  S12

 B
 A   B   1  A 
1
 B 
F  S21B I  S22A S11

1
A
 AB 
A
 B  A 
S11
S 21
 S11
 DS11
1
A
 S12
B
A 
 B  A  
D  S12
I  S11 S 22 
 AB 
 B
S12
 DS12
S 21  S 21 I  S 22 S11  S 21
S21AB   FS21A 
1
 B 
 A   B
S22AB  S22B  S21B  I  S 22A S11
 S 22 S12
 B
S22AB   S22B  FS22A S12
Lecture 5
Slide 45
Using the Star Product as an Update
Very often we update our global scattering matrix using a star
product.
When we use this equation as an update, we MUST pay close
attention to the order that we implement the equations so that we
don’t accidentally overwrite a value that we need.
S global  Si   S global 
global 
i
F  S21
global 
reverse order
global 
Lecture 5
S22
 global 
S 21
 global 
S12
 global 
S11
 S22
global 
S22 
i
 global  
I  S22i S11


 FS22S1 2
i
global 
i 
 FS 21
 global 
 DS12
i 
1
 global   i 
 S11  DS11
S 21
 global  
 i   global  
D  S12
I  S11 S 22 
1
global  i 
i

F  S21 I  S22 S11

standard order


D  S12
I  S11
S global  S global  S i 
1
1
 global 
 global 
 i   global 
S11
 S11
 DS11
S 21
 global 
i
S12
 DS12
S21global  FS21global 
i 
S22global  S22i   FS22global S12
Slide 46
23
10/1/2015
Simplified External S‐Matrices in LHI Media
The reflection‐side scattering matrix is
 ref 
1
ref
 ref 
1
ref
S11   A B ref
S12  2 A
s ref
A ref  I  Vh1Vref

S 21ref   0.5 A ref  B ref A ref1 B ref
r ,I
r ,h
 r ,I
 r ,h
1
h
B ref  I  V Vref

S 22ref   B ref A ref1
lim
L 0
The transmission‐side scattering matrix is
 
S11
 B trn A trn1
s trn
trn

 
 0.5 A trn  B trn A trn1 B trn
S12
trn

1
h
A trn  I  V Vtrn
r ,h
r ,II
 r ,h
 r ,II
1
h
B trn  I  V Vtrn
S21   2A trn1
trn
S22trn    A trn1 B trn
lim
L 0
Lecture 5
Slide 47
Block Diagram of TMM Using S‐Matrices
Initialize Global
Scattering Matrix
 kx ky
Qh  
 1  kx2

0 I 
S  global   

 I 0
Calculate Parameters
for Layer i
Calculate Transverse
Wave Vectors
kx  ninc sin  cos 
Calculate Gap Medium Parameters

1  ky2 

kx ky 


Vh   jQ h
ky  ninc sin  sin 
Calculate Scattering Matrix
for Layer i
kz ,i  i i  kx2  ky2
Ai  I  Vi1Vh
i i  kx2 
1  kx ky
Qi   2

i  ky  i i
kx ky 

Ωi  jk z ,i I
Vi  Qi Ωi1
Bi  I  Vi1Vh
A 
 B  A  
D  S12
I  S11 S 22 
 B

i 
S11
 S22i   D1 Xi Bi A i1Xi A i  B i
i 
1

1
i
S12  S 21  D Xi Ai  Bi A B i
Start
S  global  S  global  S i 
Xi  eλ i k0 Li
D  A i  Xi Bi A i1Xi B i
i 
Update Global
Scattering Matrix
1
 A   B   1
F  S 21 I  S 22 S11 

 AB 
A
 B  A 
 S11
 DS11
S11
S 21

 AB 
S12
 B
 DS12
S 21AB   FS21A 
 B
S 22AB   S 22B  FS 22A S12
no
yes
Connect to
External
Regions
S  global   S  ref   S  global 
S  global   S  global   S  trn 
Lecture 5
Loop through all layers
Done?
Finish
Calculate
Source

P  pTE aˆTE  pTM aˆTM

P 1
 Px 
esrc   
 Py 
Calculate Transmitted
and Reflected Fields
Calculate Longitudinal
Field Components
 E ref 
e ref   xref   S11esrc
 E y 
E zref  
 E trn 
e trn   xtrn   S 21esrc
 E y 
E ztrn  
kx Exref  ky E yref
k ref
z
kx E xtrn  ky E ytrn
k trn
z
Calculate Transmittance
and Reflectance

R  Eref
2
 2

T  Etrn  Re  ref
  trn
kztrn 

kzref 
Slide 48
24
10/1/2015
How to Handle Zero Number of Layers
Follow the block diagram!!
Setup your loop this way…
NLAY = length(L);
for nlay = 1 : NLAY
...
end
For zero layers:
ER = [];
UR = [];
L = [];
If NLAY = 0, then the loop will not execute the global scattering
matrix will remain as it was initialized.
0 I 
S  global   

 I 0
Lecture 5
Slide 49
Can TMM Fail?
Yes!
The TMM can fail to give an answer and behave numerically strange
any time kz = 0. This happens at a critical angle when the
transmitted wave is at or near its cutoff.
We fixed this problem in the gap medium, but this can also happen in
any of the layers or in the transmission region.
This happens in any medium where
 r  r  kx2  ky2
Lecture 5
Slide 50
25