8/25/2019

Advanced Computation:
Computational Electromagnetics

PWEM Extras

Outline
• Using 3D PWEM for 2D and 1D Analysis, and using 2D PWEM for 1D
Analysis
• Visualizing the fields
• Formulation of efficient 3D PWEM
• Formulation of efficient 1D PWEM
• PWEM with anisotropic materials
• Incorporating dispersion
• Reduced Bloch mode expansion (RBME) technique
• Generalized symmetry
Slide 2

1
 8/25/2019

Using 3D PWEM for 2D and

1D Analysis, and Using 2D
PWEM for 1D Analysis

Slide 3

Using 3D PWEM for 3D, 2D & 1D Analysis

3D Analysis with 3D PWEM 2D Analysis with 3D PWEM 1D Analysis with 3D PWEM
z z z

y y y
x x x

P  1 along x P  1 along x P  1 along x

Q  1 along y Q  1 along y Q  1 along y
R  1 along z R  1 along z R  1 along z
Conventional 3D PWEM The device is uniform along the z‐ The device is uniform along the x‐
axis so R can be set to 1. and y‐axes so P and Q can both be
set to 1.
Slide 4

2
 8/25/2019

Using 2D PWEM for 2D & 1D Analysis

2D Analysis with 2D PWEM 1D Analysis with 2D PWEM

y y
x x

P  1 along x P  1 along x
Q  1 along y Q  1 along y

Conventional 2D PWEM The device is uniform along the y‐axis so Q can be set to 1.

Slide 5

Visualizing the Fields

Slide 6

3
 8/25/2019

Eigen‐Vectors Contain Field Information

For 2D PWEM, we solved the following eigen‐value problem.

K  
x r
1 1

K x  K y   r  K y s z  k02  r  s z 
V  eigen vectors
λ  eigen values

Recall the field expansion into a plane wave basis

         2 p 2 q
E r     S  p, q  e
p  q 
 jk  p , q  r
k  p, q    
x
xˆ 
y
yˆ

sz The complex amplitudes of the plane waves are

stored in the elements of the eigen‐vectors.

Slide 7

Rearrange Terms
The field expansion is
         2 p 2 q
E r     S  p, q  e
p  q 
 jk  p , q  r
k  p, q    
x
xˆ 
y
yˆ

  
We substitute in the expression for k  p, q  and factor out e
 j  r
.
 2 p 2 q  
       j

xˆ  yˆ  r
 y 
E  r   e  j  r  S  p, q  e 
x

p  q 

This is just an inverse 2D FFT.

We can now write the electric field expression as

    
E  r   e  j  r FFT 1  S  p, q  

Slide 8

4
 8/25/2019

Visualizing the Data

 
S  p, q  S  p, q  on full grid

q q

p p
  
e
 
 j r FFT 1  S  p, q   E r 

Slide 9

Procedure (1 of 2)
Step 1 – Extract eigen‐vector Step 2 – Insert Fourier
and reshape coefficients into large grid.

nxc = ceil(Nx/2);
s = V(:,m);
nx1 = nxc - floor(P/2);
s = reshape(s,P,Q);
nx2 = nxc + floor(P/2);
nyc = ceil(Ny/2);
ny1 = nyc - floor(Q/2);
ny2 = nyc + floor(Q/2);

sf = zeros(Nx,Ny);
sf(nx1:nx2,ny1:ny2) = s; Slide 10

10

5
 8/25/2019

Procedure (2 of 2)
Step 3 – Calculate inverse FFT Step 4 – Calculate phase term.
of S  p, q  .
 
S  p, q  on full grid A  x, y  on full grid

az = ifft2(ifftshift(sf)); phase = exp(-1i*(beta(1)*X + beta(2)*Y));

az = az/max(abs(az(:)));

Step 5 – Calculate overall field.

.* = Ez = phase.*az;

Slide 11

11

Formulation of Efficient 3D
Plane Wave Expansion Method

Slide 12

12

6
 8/25/2019

Notation
Unbolded letters are scalar quantities. a  1.678

 a11 a12 a1N 

a a22  a2 N 
Bold capital letters are matrices, usually square matrices. A   21
   
 
 aM 1 aM 2 aMN 

 x1 
x 
Bold lower‐case letters are column vectors. x 2
  
 
 xM 

e x 
Bold lower‐case letters with a vector sign are column vectors of   a x 
e  e y  a 
column vectors. a y 
 e z 
Slide 13

13

Vector‐Matrix Operations
Definitions
  Ai1 0 
A x  diagonal matrix containing x components of A Ax 
  Ai 2  
A y  diagonal matrix containing y components of A Ai    A   A y 
   
A z  diagonal matrix containing z components of A    A z 
0 AiN 

Vector‐Matrix Operations
 
A  A 2x  A 2y  A 2z  A    A x Ay A z 
 

 0    
A z Ay   A  B  B  A Unless Ax, Ay, Az, Bx, By, and Bz
    
 A    A z  are all diagonal.
  0 A x 
A y    
 Ax 0   A  B   B  A Unless Ax, Ay, Az, Bx, By, and Bz
    are all diagonal.

In our formulation, only [r] and [r] are not diagonal matrices.
Slide 14

14

7
 8/25/2019

Expand the Field Into Two Orthogonal Polarizations

We write the field as the sum of two orthogonal polarizations.

 ˆ
s  P1s1  Pˆ 2s 2
Two polarization vectors are chosen for each spatial harmonic in the plane wave expansion.
These must be orthogonal to the direction of the harmonic. If these are chosen correctly, the
following equations will be satisfied.

k   1
P1  P2     Pˆ1  Pˆ 2  K K
ˆ ˆ
k

k ˆ ˆ  
  P1  P2  K  Pˆ1  Pˆ 2 K
k

k ˆ  
  P2   Pˆ1  K  Pˆ 2  Pˆ1 K
k
Slide 15

15

Recipe for Calculating Orthogonal Polarization Vectors

Given the wave vector of the ith spatial harmonic

ki  k x ,i aˆ x  k y ,i aˆ y  k z ,i aˆ z

We construct another vector that is in a different direction than ki .

vi  4k y ,i aˆ x  2k z ,i aˆ y  3k x ,i aˆ z
We can now calculate two vectors that are perpendicular to the wave vector.
 
ki  vi
pˆ1,i    
ki  vi Any choice of pˆ1,i and pˆ 2,i can be used as long as pˆ1,i  pˆ 2,i  ki .

ki  pˆ1,i Perhaps there exists a better choice than described here.
pˆ 2,i  
ki  pˆ1,i

Slide 16

16

8
 8/25/2019

Wave Equation with Polarization Expansion

We substitute the polarization expansion into our matrix wave equation.

 1   
K   r   K  s  k02  r  s
   

 1 
 
K   r  K  Pˆ1s1  Pˆ 2s 2  k02  r  Pˆ1s1  Pˆ 2s 2
     

Slide 17

17

Projecting a Vector‐Matrix Equation Onto Two Orthogonal

Polarizations
We can project one vector‐matrix onto another using
 
  a  b    a  b   1  
Proj a onto b   2 b Proj a onto b    a 
b
 b b   a
b b

Using this equation, we can project both sides of a vector‐matrix equation onto
our two orthogonal polarizations to get two separate equations.
 
Ax  By One 33 matrix equation

   Pˆ1  Ax   Pˆ1  By

Ax Pˆ1
 By Pˆ1
    
Note:
1 1
Two 11 matrix Pˆ1  Pˆ 2 I
   P2  Ax   Pˆ 2  By
ˆ  equations
Ax Pˆ 2
 By Pˆ 2
    

Slide 18

18

9
 8/25/2019

Project Our Vector‐Matrix Equation Onto Two

Orthogonal Polarizations
We apply the results from the last slide to our vector‐matrix
equation.
 1 
 
K   r   K  Pˆ1s1  Pˆ 2s 2  k02  r  Pˆ1s1  Pˆ 2s 2
     
Project onto Pˆ1
 1 
        
 Pˆ1    K    r   K  Pˆ1s1  Pˆ 2s 2   k02  Pˆ1    r  Pˆ1s1  Pˆ 2s 2
 
We now have two matrix equations.
Project onto Pˆ 2
 1 
        
 Pˆ 2    K   r   K  Pˆ1s1  Pˆ 2s 2   k02  Pˆ 2    r  Pˆ1s1  Pˆ 2s 2
 
Slide 19

19

22 Block Matrix Form

We can put our two previous equations into block matrix form as follows.
 1 
 Pˆ1  K    r   K 
      Pˆ s  Pˆ s   k
1 1 2 2
2
0
 Pˆ1   r 
   Pˆ s  Pˆ s 
1 1 2 2
 1 
 Pˆ 2  K    r   K 
      Pˆ s  Pˆ s   k
1 1 2 2
2 ˆ
   r 
0  P2     Pˆ s  Pˆ s 
1 1 2 2

   1 
  Pˆ   K     1 K  Pˆ ˆ 
 P1   K    r  K  P2   s1 
1   r 
ˆ   
     1
 ˆ     1   1   
P  K    r  K  Pˆ1  Pˆ 2   K    r   K  Pˆ 2  s 2 
  2        
  Pˆ1    r  Pˆ1 ˆ 
 P1    r  P2   s1 
ˆ 
  
 k02   
  Pˆ     Pˆ  Pˆ 2   r  Pˆ 2  s 2 
  2   r  1   

This is a 22 generalized eigen‐value problem. It is smaller than the

33 generalized eigen‐value problem that we derived previously.
Slide 20

20

10
 8/25/2019

Final Form of Generalized Eigen‐Value Problems

Reduced Eigen‐Value Problem for Electric Fields
   
  Pˆ   K     1 K  Pˆ  Pˆ   K     1 K  Pˆ 
s  s    1   r    1  1   r    2
A  1    k02 B  1  A
s 2  s 2   ˆ     1   1 
ˆ 
  P2   K    r  K  P1  P2   K   r  K  P2 
ˆ ˆ 

  Pˆ1    r  Pˆ1  Pˆ1    r  Pˆ 2 

      
B
  Pˆ      Pˆ  Pˆ      Pˆ 
  2   r  1  2   r  2 

Reduced Eigen‐Value Problem for Magnetic Fields

   1 
  Pˆ    K    1 K  Pˆ  Pˆ1   K   r  K  Pˆ 2 
u  u    1   r    1      
A  1   k02 B  1  A
u 2  u 2   ˆ     1   1 
 Pˆ 2   K   r  K  Pˆ 2 
P  K   r  K  Pˆ1
  2        
  Pˆ1    r  Pˆ1  Pˆ1    r  Pˆ 2 
    
B 
  Pˆ      Pˆ  Pˆ 2     r  Pˆ 2 
  2   r  1   

Note that both of these eigen‐value problems are valid for general anisotropic media.
Slide 21

21

Generalized Eigen‐Value Problem

with NO MAGNETIC RESPONSE
Generalized Eigen‐Value Problem for Electric Fields
  Pˆ   K  2  2
 ˆ  Pˆ1   K  Pˆ 2 
s  s    1    P1    
A  1    k02 B  1  A  2  2 
s 2  s 2  ˆ ˆ
  P2   K  P1
ˆ ˆ
 P2   K  P2 
   

  Pˆ1    r  Pˆ1 ˆ 

 P1    r  P2 
ˆ 
  
B
  Pˆ      Pˆ  Pˆ 2    r  Pˆ 2 
  2   r  1    Note that both of these
eigen‐value problems are
Ordinary Eigen‐Value Problem for Magnetic Fields valid for general
   1 
anisotropic media.
  Pˆ    K    1  K  Pˆ  Pˆ1    K   r  K  Pˆ 2 
u  u   1    r    1      
A  1    k02  1  A
u u 2   ˆ     1   1 
ˆ 
 2 P  K   r   K  Pˆ1  P2    K   r  K  P2 
 ˆ   
  2  
This is the most common formulation because it is an ordinary eigen‐value problem.

Slide 22

22

11
 8/25/2019

Recovering 𝐬⃗ from s1 and s2

The generalized eigen‐value problem for the electric fields is
s  s 
A  1    k02 B  1 
s
 2 s 2 

Let the eigen‐vector and eigen‐value matrices be

s  s  W  eigen-vector matrix
A  1    k02 B  1  
s 2  s 2  λ  eigen-value matrix

If we extract a single eigen‐mode, wi and i, we have

s 
w i   i1 
s i 2  These terms are now

ˆ s  Pˆ s . Therefore,
Recall that s  P1 1 2 2
independent of the chosen
polarization vectors.
s x 
 ˆ s 
si  P1si1  Pˆ 2si 2   Pˆ1 Pˆ 2   i1    Pˆ1 Pˆ 2  w i  s y 
s i 2   s z  Slide 23

23

Formulation of Efficient 1D
Plane Wave Expansion Method

Slide 24

24

12
 8/25/2019

Reduction to One Dimension

For the 1D devices where propagation is only in the z‐direction, the wave has no wave vector
components in the x and y directions.
K x  0 and K y  0
Maxwell’s equations reduce to
K z u y  jk0  xx  s x K z s y  jk0   xx  u x
K z u x  jk0  yy  s y K z s x  jk0   yy  u y
0  jk0  zz  s z 0  jk0   zz  u z

Immediately, we see that

sz  0 uz  0

The field will have no components in the longitudinal direction z.

Slide 25

25

Two Modes
Maxwell’s equations have again decoupled into two distinct sets of equations.

K z u y  jk0  xx  s x K z s y  jk0   xx  u x
K z u x  jk0  yy  s y K z s x  jk0   yy  u y
0  jk0  zz  s z 0  jk0   zz  u z

Ex Mode Ey Mode
K z u y  jk0  xx  s x K z u x  jk0  yy  s y
K z s x  jk0   yy  u y K z s y  jk0   xx  u x

Slide 26

26

13
 8/25/2019

Final Eigen‐Value Problems

We can substitute one equation into the other to derive two eigen‐value problems.

Ex Mode Ey Mode
1
K zμ K zs x  k
yy
2
0  xx s x K z μ K z s y  k02  yy  s y
1
xx

1  1 1
  yy  K z s x
1
uy 
jk0  
ux  
jk0
  xx  K z s y

or or
K ε K z u y  k   yy  u y
1
z xx
2
0 K ε K z u x  k02   xx  u x
1
z yy

1 1 1  1
sx  
jk0
 xx  K z u y sy   yy  K z u x
jk0  

Slide 27

27

PWEM with Fully

Anisotropic Materials

Slide 28

28

14
 8/25/2019

Anisotropic 3D PWEM
The matrix equations derived for 3D PWEM are valid for fully anisotropic materials. In the
anisotropic case, however, the material tensors become block matrices composed of
convolution matrices for each tensor element individually.

 xx     
 xy  xz
 
 r   yx 
     
 yy   yz 

 zx       
  zy  zz 
  xx       
 xy  xz
 
 r     yx       
 yy   yz 

  zx       
  zy  zz 

Slide 29

29

Anisotropic 2D PWEM
In general, all field components remain coupled in anisotropic media. Since Maxwell’s
equations will not simplify in this case, there is almost no numerical advantage to
developing a 2D PWEM for anisotropic media. Instead, use your anisotropic 3D PWEM for
2D problems by using only on spatial harmonic in the uniform direction.

z
y
uniform axis
infinite and

P  # harmonics along x
Q  # harmonics along y
x R  # harmonics along z

In this case, you would set R = 1.

Slide 30

30

15
 8/25/2019

Incorporating Dispersion

Slide 31

31

Techniques for Incorporating Dispersion

1. Iterative PWEM
• Can only handle one frequency at a time.

2. Do not use k0 as the eigen‐value

Slide 32

32

16
 8/25/2019

Alternate 2D‐PWEM Formulation (1 of 5)

Recall our starting point for E and H modes
E‐Mode H‐Mode
K x u y  K y u x  jk0  zz  s z K x s y  K y s x  jk0   zz  u z
K y s z  jk0   xx  u x K y u z  jk0  xx  s x
K x s z  jk0   yy  u y K x u z  jk0  yy  s y

We will choose y as our eigen‐value instead of k0.

E‐Mode H‐Mode
  xI  G x  u y    y I  G y  u x  jk0  zz  s z   xI  G x  s y    yI  G y  s x  jk0   zz  u z

 I  G s
y y z  jk0   xx  u x  I  G u
y y z  jk0  xx  s x
   x I  G x  s z  jk0   yy  u y    x I  G x  u z  jk0  yy  s y

G x  G x  components of wave vector expansion

G y  G y  components of wave vector expansion Slide 33

33

Alternate 2D‐PWEM Formulation (2 of 5)

Next we normalize our wave vector terms by dividing by k0.
  k 1K
K   k 1K
K   k 1G
G   k 1G
G x  k01 x  y  k01 y
x 0 x y 0 y x 0 x y 0 y

Our governing equations can now be written as

E‐Mode H‐Mode
  
 x I  G x u y   y I  G
 
y x 
 u  j   s
zz z   
 x I  G x s y   y I  G
  
 s  j   u
y x zz z

  I  G  s
y y z  j   xx  u x   I  G  u
y y z  j  xx  s x
   I  G
x
 s
x z  j   yy  u y    I  G
x
 u
x z  j  yy  s y

Slide 34

34

17
 8/25/2019

Alternate 2D‐PWEM Formulation (3 of 5)

Solving for sy and uy, we get
E‐Mode H‐Mode
  I  G  s  
1 1
u y  j   yy  x x z s y  j  yy  x I  G
 u
x z

We eliminate these terms by substituting these expressions into our

governing equations.
E‐Mode
 
   1  I  G
j x I  G  yy 
x  x x z
 s   I  G
y y x 
 u  j   s
zz  z

 
 y I  G y s z  j   xx  u x

H‐Mode

 
  1  I  G
j x I  G  yy 
x  x x z 
 u   I  G
y 
y x 
 s  j   u
zz  z

 
 y I  G y u z  j  xx  s x
Slide 35

35

Alternate 2D‐PWEM Formulation (4 of 5)

We rearrange these equations to bring y to the right‐hand side.

E‐Mode

y x
 x

 u  j   I  G
G    1  I  G
 yy 
x  x  x 
    s   u
zz  z

y x

j   xx  u x  G s   s
y z y z

H‐Mode

y x
 x

 s  j   I  G
G   1  I  G
 yy 
x  x x 
     u   s
zz  z


y x

j  xx  s x  G u   u
y z y z

Slide 36

36

18
 8/25/2019

Alternate 2D‐PWEM Formulation (5 of 5)

Finally, we write our equations in block matrix form.

E‐Mode
 G


y  
j  x I  G  yy 
x   x x 
     u 
   1  I  G
zz   u 
  x    y  x 
    sz   sz 
 j   xx  G y 

H‐Mode
 G


y 
j  x I  G
 x 
 yy 
  1  I  G
x x 
       s 
zz    s 
  x    y  x 
   u z  u z 
 j  xx  G y 

Slide 37

37

Alternate 3D‐PWEM Formulation (1 of 2)

We choose z to be the eigen‐value.
K y z z 
 u   I  G
 u  j   s
z y xx x  y z z
 s   I  G
K  s  j   u
z y xx x
 z I  G z
 u K
x
 u  j   s
x z  yy  y   I  G  s
z z
 s  j   u
K x
x z  yy  y
K u  K u  j   s

x

y y x zz z
 s K
K x y
 s  j   u
y x zz z

Solve for longitudinal components.

u z   j   zz 
1
 K s
x y
 s
K y x  s z   j  zz 
1
 K u
x y
 u
K y x 
Eliminate longitudinal components.
   1 K
jK x zz
 s  j    K
y x  yy  x zz 
   1 K
 s G
x y
 u   u
z x z x 
   1 K
j K y zz y xx x
   1 K
    s  jK
y zz
 s G
x y 
 u   u
z y z y

z x
  1 K
 s  jK
G x zz
 u  j    K
y x  yy   1 K
x zz
 u   s
x y z x 
 s  jK
G z y 
  1 K
y zz y xx x
  1 K
  j    u  jK
y zz 
 u   s
x y z y Slide 38

38

19
 8/25/2019

Alternate 3D‐PWEM Formulation (2 of 2)

In block matrix form, we have



G z 0   1 K
jK x zz

y 
j   yy   K  1 K
x zz x
 
 s
 x   sx 


0 
G z 
   K
j K y zz y
1
   
xx   jK    K
y zz
1

x  sy  s 
    z  
y



   1 K
jK x zz

y 
j  yy   K
   K
x zz

x
1
 
G z 0  x u 
 u y 
 u
 
x

u y 
 

j K
   K
y zz
1
   
y xx    1 K
 jK y zz

x 0 G
z 

Slide 39

39

Alternate 1D‐PWEM Formulation

For 1D, our eigen‐value problem is
Ex Mode Ey Mode
K z u y  jk0  xx  s x K z u x  jk0  yy  s y
K z s x  jk0   yy  u y K z s y  jk0   xx  u x
 

  
  z I  G z u y  j  xx  s x   
 z I  G z u x  j  yy  s y

  I  G  s
z z x  j   yy  u y 
 z I  G z y
 s  j   u
xx x

 
 
G z s x  j   yy  u y  z s x
 G z s y  j   xx  u x  z s y


 j  xx  s x  G u   u
z y z y
j  yy  s y  G
 u   u
z x z x

 
 G   
j   yy   s x   sx   Gz   j   xx   s y  s 
    z       z  y 
z
 
  j  xx   u  
 j  yy  
G z   y  u y  G z  u x  u x 
Slide 40

40

20
 8/25/2019

Reduced Bloch Mode Expansion

(RBME) Technique

Mahmoud I. Hussein, “Reduced Bloch mode expansion

for periodic media band structure calculations,” Proc. R.
Soc. A 465, pp. 2825‐2848, 2009.

Slide 41

41

The Problem
Suppose we wish to calculate a photonic band
diagram or calculate the full bands throughout
the entire Brillouin zone. a
2 c0
For 2D simulations, typical matrix size is
400×400. For 3D simulations, typical matrix size
is 12,000×12,000. These large matrices must be
solved for each Bloch wave vector of interest.
This is typically 100’s of vectors.

Run time is long and computation is bogged

down.  a
 a 0
0  a x
y  a

42

42

21
 8/25/2019

Step 1: Calculate the Full Solution at Only the Key Points of Symmetry
M 
        
       
       
Γ    v Γ1    v 2    v 3    v  N   
X VΓ 
        
Γ Γ Γ

       
        
       

Notes         
       
Ensure that no redundant points are used 
  1   
v 
 
v 
 
v


or RBME will fail. VX   vX  2   3    N 

    X   X   X  
       
An asymmetric distribution of key points        
       
will lead to asymmetry in the data 
calculated from them.         
       
       
You can improve accuracy by adding   v  N   
VM    v M1    v 2    v 3   
more key points, but this will be slower    M   M   M 
       
and less efficient.        
       

Construct full PWEM Solve full PWEM Append Transformation

Done? Eigen‐Value Problem Eigen‐Value Problem Matrix U
Loop through key points of symmetry
43

43

Step 2: Combine Eigen‐Vector Matrices Using Lowest Order Modes

  Take the M lowest order modes from each

        
        eigen‐vector matrix and construct a new
      
VΓ    v Γ1   v2   v  3    v  N    eigen‐vector matrix U.
    Γ 
 
 Γ 
 
 Γ 
 
       
       

        
       
       
  v  N   
VX    v X1   v 2   v  3  
    X 
 
 X 
 
 X 
 
       
       

        
       
       
  v  N   
VM    v M1   v 2   v  3  
    M  M  
M

                    
         
                
           
  1    M  v 1    v  M    v 1    v M   
U   vΓ  vΓ

      X  X   M  M 
           
           
          

U does not have to be a square matrix. 44

44

22
 8/25/2019

Step 3: Perform Gram‐Schmidt Orthonormalization on the Matrix U

            
           
           
  1    M  v 1    v  M    v 1    v M   
U   vΓ  vΓ

      X  X   M  M 
           
           
          

            
           
           
    v 1    v  M  
U  v 1    v  M    v 1    v  M   
 Γ   Γ   X  X   M  M 
           
           
           


45

45

Gram‐Schmidt Orthonormalization
 
           
           
           
U    v1   v2   v3    v M 1   v M 1  vM 
           
           
             
 

v1
Step 1: v 1 
v1
v 2   v 2  v 1  v 1 This algorithm essentially just
Step 2: v 2 
v 2   v 2  v 1  v 1 removes the components of the
previous vectors and then
v 3   v 3  v 2  v 2   v 3  v 1  v 1
Step 3: v 3  normalizes the amplitude.
v 3   v 3  v 2  v 2   v 3  v 1  v 1


46

46

23
 8/25/2019

Step 4: Construct Eigen‐Value Problem for the Next 

Construct the standard PWEM eigen‐value problem for the next ,

but do not solve it yet.

Ax  λBx
 K x  r 1 K x  K y  r 1 K y E Mode
A 1 1
K x  r  K x  K y  r  K y H Mode
   E Mode
B r
 r  H Mode

47

47

Step 5: Reduce Matrix Size by Expanding into Bloch Modes as the Basis

Perform a linear transformation that expresses the field in terms of the new basis formed
from the reduced set of Bloch modes.

This dramatically reduces the size of the matrices while retaining excellent accuracy
because those Bloch modes more efficiently describe the fields in the structure.

 H AU
A  U 
 H BU
B  U  











  
      
    
     
  
A UH   
  
  
  
  
A U
48

48

24
 8/25/2019

Step 6: Solve the Reduced Eigen‐Value Problem

The reduced eigen‐value problem is solved according to

Ax  λBx  V, λ

The eigen‐values are interpreted the same as in the standard PWEM.

If they are needed, the eigen‐vectors must be transformed back into

a plane wave basis to be compatible with the standard PWEM.

 U
V  UV H

49

49

Step 7: Repeat Procedure for Each Bloch Wave Vector in IBZ

Construct full PWEM Solved Reduced

Done? Transform to U Record
Eigen‐Value Problem Eigen‐Value Problem

Loop through list of ’s

50

50

25
 8/25/2019

Dashboard
Block Diagram of RBME
Build Unit Cell

Build Convolution Matrices

Construct IBZ and Key Points

Generate List of ’s

Construct PWEM Solve PWEM Eigen‐ Append Transformation

Done? Eigen‐Value Problem Value Problem Matrix U
Loop through key points of symmetry

Gram‐Schmidt

Construct PWEM Solved Reduced

Done? Transform to U Record
Eigen‐Value Problem Eigen‐Value Problem

Loop through list of ’s

Draw Band Diagram

51

51

Generalized Symmetry

Slide 52

52

26
 8/25/2019

“Mystery Bands” When Modeling Hexagonal Arrays with Rectangular Symmetry

Unit Cell Reciprocal Lattice

Vectors
a
2 c0
 r  13

r  1

Γ M K Γ

Unit Cell Reciprocal Lattice

Vectors
a Correct physical bands
2 c0
 r  13
r  1
Mystery nonphysical
bands
Γ M K Γ Slide 53

53

Cause of “Mystery Bands”

• Lattice symmetry imposes phase conditions between

adjacent unit cells expressed by the Bloch theorem.
• The higher the symmetry, the tighter the phase
conditions that are imposed.
• There are fewer phase conditions with a rectangular
unit cell so there are additional allowed states.

Slide 54

54

27
 8/25/2019

The Fix
The fix is to construct the convolution matrices for the general
symmetry case.

The standard FFT can no longer be used and this process is more
numerically intensive.

For 2D periodic functions, the expansion is

       
  1  j  pT1  qT2  r
f  x, y     
j pT1  qT2  r

A 
a p,q e a p,q  f x , y e dA
p  q  A

For 3D periodic functions, this is

   
   
  1   j  pT1  qT2  rT3 r
f r     a f r e
j pT1  qT2  rT3  r

p  q  r 

p ,q ,r e a p ,q ,r 
V 
V
dV

These must be numerically evaluated. Slide 55

55

28