Slice Absorption Method (SAM) · 8/26/2019 4 Grid to Matrix Scaling & Memory Slide 7 Typical grid required to model a 3D device. 20 20 100 x y z N N N Number of points in grid:

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Advanced Computation:

Computational Electromagnetics

Slice Absorption Method (SAM)

Outline• Three‐Dimensional FDFD

• Matrix Ordering

• Slice Absorption Method

• Plane Wave Source

• Transparent Boundary Condition

• Field Solution

• Fourier‐Space SAM

• Example Simulations

• Dispersion Analysis Using the SAM

Slide 2

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Slide 3

Three‐Dimensional FDFD

Matrix Form of Maxwell’s Equations

Slide 4

, , 1 , ,, 1, , ,

, , 1 , , 1, , , ,

1, , ,

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,

,

, , 1,

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, ,, ,

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i j k i j ki j k i j ky yz z

i j k i j k i j k

i j

i j

i j kx

i j ky

i j k

kx x z z

i j k i j k

kxx

i j k

i j k i j ky y x

yy

i j kzz z

x

y z

z x

HE EE E

E

x y

H

H

E E E

E E E E

, , , , 1, , , 1,

, , , , 1 , , 1, ,

,

, ,

, ,

, 1, , , , , 1,

, ,

,

, ,,

,

,

i j k i j ki j k i j ky yz z

i j k i j k i j k i j kx x z z

i j k

i j kxx

i j kyy

i j k i j k i ji j kzz

k

i j kx

i j k

y y x

y

i j kz

x

H HH H

H

y z

z x

H H H

H H

y

HE

H

x

E

E

xx

yy

x

y

zy x z

z

z

y

x z

y z

z x

x y

H

H

E E

E E

E E H

xx

yy

zz

z y

x z

y x

x

y

z

y z

z

E

E

E

H H

Hx

yH H

x

H

e ey z xx

e e

z y

x z

x

yz x yy

e ex y zzx zy

D D μ

D D μ

e e

e

D D μ he

h

he

e

h hy z xx

h hz x yy

h hx y

z y

x z

y zz

x

x

y

z

D D ε

D D ε

D D

h h

h h

h h ε

e

e

e

3

4

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Block Matrix Form

Slide 5

e ey z z y xx x

e ez x x z yy y

e ex y y x zz z

h hy z z y xx x

h hz x x z yy y

h hx y y x zz z

D e D e μ h

D e D e μ h

D e D e μ h

D h D h ε e

D h D h ε e

D h D h ε e

e ez y x xx x

e ez x y yy ye ey x z zz z

0 D D e μ 0 0 h

D 0 D e 0 μ 0 h

D D 0 e 0 0 μ h

h hz y x xx x

h hz x y yy yh hy x z zz z

0 D D h ε 0 0 e

D 0 D h 0 ε 0 e

D D 0 h 0 0 ε e

e C e μ h

h C h ε e

E H

H E

We can write our matrix equations in block matrix form.

Three‐Dimensional FDFD

Slide 6

Block Matrix Form

e

h

C e μ h

C h ε e

x

y

z

xx

yy

zz

h hz y

h h hz xh hy x

h

h h

h

μ 0 0

μ 0 μ 0

0 0 μ

0 D D

C D 0 D

D D 0

Matrix Wave Equations

1

1

e h

h e

C ε C μ h 0

C μ C ε e 0

Ωe 0

For 3D analysis, is usually too big to solve by simple means.

x

y

z

xx

yy

zz

e ez y

e e ez xe ey x

e

e e

e

ε 0 0

ε 0 ε 0

0 0 ε

0 D D

C D 0 D

D D 0

5

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Grid to Matrix Scaling & Memory

Slide 7

Typical grid required to model a 3D device.

20

20

100

x

y

z

N

N

N

Number of points in grid:Complex #’s for ex, ey, and ez:

Real #’s for ex, ey, and ez:

Size of matrix 𝛀:Number of complex elements:

Number of real elements:Memory to store full 𝛀:

Size of ex:Memory for ex:

Size of full Dx:Memory for full Dx:

Density of Dx:Memory for sparse Dx:

Size of full Ce:Memory for full Ce:

Density of Ce:Memory for Sparse Ce:

Memory for direct solution:

40,000 points120,000 complex #’s240,000 real floating‐point #’s

120k 120k complex #’s14.4 billion complex #’s28.8 billion real #’s214.6 Gb

40k complex numbers625 kb

40k 40k complex numbers23.8 Gb0.005% non zero elements1.5 Mb

120k 120k complex numbers214.6 Gb0.0033% non zero elements8.2 Mb

110 Gb

Slide 8

Matrix Ordering

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Standard FDFD Order

Slide 9

1h e

Ωe 0

Ω C μ C ε

Matrix Wave Equation

Matrix Order

xx xy xz x

yx yy yz y

zx zy zz z

Ω Ω Ω e 0

Ω Ω Ω e 0

Ω Ω Ω e 0

1,1,1

2,1,1

,1,1

1,2,1

2,2,1

,2,1

, ,1

1,1,2

, ,

x

x

x y

x y z

x

x

Nx

x

x

xNx

N Nx

x

N N Nx

E

E

E

E

E

E

E

E

E

e

1,1,1

2,1,1

,1,1

1,2,1

2,2,1

,2,1

, ,1

1,1,2

, ,

x

x

x y

x y z

y

y

Ny

y

y

yNy

N Ny

y

N N Ny

E

E

E

E

E

E

E

E

E

e

1,1,1

2,1,1

,1,1

1,2,1

2,2,1

,2,1

, ,1

1,1,2

, ,

x

x

x y

x y z

z

z

Nz

z

z

zNz

N Nz

z

N N Nz

E

E

E

E

E

E

E

E

E

e

Raster first along x, then y, and then z.

Reorder Operation

Slide 10

reorder

reorder

Ωe 0

Ω Ω

e e

The matrix equation is reordered in a manner that groups all fields located within the same slices into adjacent rows and/or columns.

We still raster first along x, then y, and then z, but we group the x, y, and z components together instead of separate.

1,1,1

1,1,1

1,1,11,1,1

2,1,1

2,1,1

, ,2,1,1

1,1,1

,1,1

, ,,1,1

1,1,1 ,1,1

, , , ,

, ,

, ,

x y z

x

x y zx

x

x y zx y z

x y z

x y z

x

y

zx

x

yN N Nx

z

y

Nx

N N NN

y yN

z z

N N N N N Nz x

N N Ny

N N Nz

E

E

EE

E

EE EE

EE EE E

E E

E

E

e e

3

3

1 mod 1 1 3

1 mod 1 1 3

pqpq

x y z

x y z

p N N N p p

q N N N q q

Ω Ω

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Visual Interpretation of Reordered Equation

Slide 11

The matrix now has a very important “block tridiagonal” symmetry. Each slice through the grid corresponds to a row in the block matrix equation composed of three square matrices a, b, and c and a column vector f that is usually all zeros.

Slice Data Ai quantifies coupling to the i-1 slice.Bi quantifies coupling of fields within the ith sliceCi quantifies coupling to the i+1 slice.fi is a source condition in the ith slice

A

Ω e

f

Slice Equations

Slide 12

1 1 1 2 1

3 2 3 3 3 4 3

1 2 1 1 1 1

2 1 2 2 2 3 2

2 3 2 2 2 1 2

1

N

N N N N N N N

N N N N N N

N N N N N

B e C e f

A e B e C e f

A e B e C e f

A e B e C e f

A e B e

A e B e C e f

f

We can think of the block tridiagonal matrix equation on the previous slide as the block matrix form of the following set of matrix equations. These “slice equations” relate fields in immediately adjacent slices.

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Slide 13

Slice Absorption Method

Steps to Calculate Slice Data for a Single Slice

Slide 14

Step 1: Build a grid containing three slices where the center slice is the slice for which to calculate the slice data. Simple Dirichlet boundary conditions can be used for the z‐axis boundaries because only Slice 2 is of interest here.

Step 2: Construct the standard 3D‐FDFD matrix for three adjacent slices. The slice of interest should be the middle slice.

1

123 123 123

123

h e

Ω C μ C ε

f 0

Step 3: Reorder the data

123 123

123 123

reorder

reorder

Ω Ω

f f

Step 4: Extract the slice data from the middle row of the block matrix equation.

1 1 1 1

2 2 2 2 2

3 3 3 3

B C 0 e f

A B C e f

0 A B e f

123

123

reordered columns and rows

reordered rows

Ω

f

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Absorbing a Slice (1 of 5)

Slide 15

Three Slice EquationsWe can write the slice equations for three adjacent slices. The center slice is the ith slice.

1 2 1 1 1 1

1

1

1 1 2

1

1 1

i i i i i

i i

i

i i i i i

i i i i i i i

i

A e B e C e f

A e B e C e f

A e B e C e f


Slide 16

We solve the middle equation for ei.

11 1 i i i i i i i

e B f A e C e

1 2 1 1 1 1

1

1

1 1 2

1

1 1

i i i i i

i i

i

i i i i i

i i i i i i i

i

A e B e C e f

A e B e C e f

A e B e C e f

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Slide 17

Substitute New Expression into First and Third EquationWe eliminate ei from these equations by substituting our new expression into the first and third equations.

1 2 1 1 1 1

1 1 1 1 2 1

i i i i i i i

i i i i i i i

A e B e C e f

A e B e C e f

11 1 i i i i i i i

e B f A e C e

1 2 11

1 1

11

1 1 1

1 1 1 1 11 2

i i i i i ii i i i i i

i i i i ii i i i i ii

A e B e C f

A B e C e

B f A e C e

B C e ff A e


Slide 18

Revised Slice Equations for Slice i‐1 and i+1These are the equations after the substitution.

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Slide 19

Revised Slice Equations for Slice i‐1 and i+1We expand the equations and collect the coefficients on common electric field terms. This puts the remaining two slice equations back into their original form, but with revised slice data.

1 1 11 2 1 1 1 1 1 1 1

1 1 11 1 1 1 1 1 2 1 1

i i i i i i i i i i i i i i i

i i i i i i i i i i i i i i i

A e B C B A e C B C e f C B f

A B A e B A B C e C e f A B f

Absorbing a Slice (Summary)

Slide 20

Three slice equations Algebraically eliminate ei

1 2 1 1 1 1

1 1 1 1 1

1 1

2

i i i i i i

i i i i i i i

i i i i i i

i

i

A e B e C e f

A e B e C e f

A e B e C e f1

1 1

11 1 1

1 1

11 1 1

i i i i

i i i i i

i i

i i i i i

A A B A

B B A B C

C C

f f A B f

1 2 1 1 1 1 1

1 1 1 1 1 2 1

i i i i i i i

i i i i i i i

A e B e C e f

A e B e C e f

1 1

11 1 1

11 1

11 1 1

i i

i i i i i

i i i i

i i i i i

A A

B B C B A

C C B C

f f C B f

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Forward Absorption from Top

Slide 21

Two slice equations Algebraically eliminate E1

2 1 2 2 2

1 0

3

1 1 1 1

2

2

A e B e C e

A B e C e f

f

e2 0 2 2 2 3 2 A e B e C e f

12 2 1 1

12 2 2 1 1

2 2

12 2 2 1 1

A A b A

B B A B c

C C

f f A B f

Backward Absorption from Bottom

Slide 22

Two slice equations Algebraically eliminate E1

2 1 2 2 2

1 0

3

1 1 1 1

2

2

A e B e C e

A B e C e f

f

e1 0 1 1 1 3 1 A e B e C e f

1 1

11 1 1 2 2

11 1 2 2

11 1 1 2 2

A A

B B C B A

C C B C

f f C B f

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Slice Absorption Procedure

Slide 23

Using the slice absorption procedure described on the previous slide, we can methodically progress through a large stack of slices, one slice at a time, and reduce the entire stack to just two adjacent slices by “absorbing” all interior slices.

Animation of Slice Absorption Method

Slide 24

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Cascading and Doubling

Slide 25

Given two outer slices describing the unit cell of a periodic structure, we can perform an efficient cascading and doubling procedure to quickly describe large stacks.

Note: This can also be used to efficiently model very thick homogeneous layers in a device.

Slide 26

Plane Wave Source

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Total‐Field/Scattered‐Field Framework

Slide 27

Grid StrategyThe grid strategy for SAM is just like FDFD.

For doubly‐periodic devices, an absorbing boundary is only needed at the z‐axis boundaries.

Spacer Region

Spacer Region

Device Region

Slice Equations at TF/SF Interface

Slide 28

1 0 1 1 1 2

2 1 2 2 2 3

A e B e C e 0

A e B e C e 0

For simplicity, we’ll call the TF/SF interface slices 1 and 2.

scattered‐field

total‐field

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TF/SF Corrections (1 of 2)

Slide 29

1 0 1 1 1 2

2 1 2 2 2 3

A e B e C e 0

A e B e C e 0

1 0 1 1 1 2 2,src A e B e C e f 0

2 1 1,src 2 2 2 3 A e f B e C e 0

Eq. (1) exists in the SF region, but e2 is a TF quantity. The source in slice 2 must be subtracted from it to make it look like a SF quantity.

Eq. (1)

Eq. (2)

Eq. (2) exists in the TF region, but e1 is a SF quantity. The source in slice 1 must be added to it to make it look like a TF quantity.

TF/SF Corrections (2 of 2)

Slide 30

1 0 1 1 1 2 1 2,src

2 1 2 2 2 3 2 1,src

A e B e C e C f

A e B e C e A f

From the above equations, we make the following observations:

• The slice data can be calculated without considering the source or TF/SF framework.

• We must calculate the source function into two adjacent planes.

• The TF/SF source is easily incorporated through the right hand side of the above equations.

1 1 2,src 2 2 1,src f C f f A f

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Calculating the Source Terms

Slide 31

1 0 1 1 1 2 1

2 1 2 2 2 3 2

A e B e C e f

A e B e C e f1 1 2,src

2 2 1,src

f C f

f A f

,inc ,inc

,inc ,inc

,inc ,inc

1,src reorder

x y

x y

x y

j k k

x

j k k

y

j k k

z

p e

p e

p e

x y

x y

x y

f

,inc

2,src 1,srczjk ze f f

Unit Amplitude Source

1p

Slide 32

Transparent Boundary Condition

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Notes on the TBC

• Requires homogeneous materials where TBC is implemented (can be generalized)• Energy can only be propagating in a single direction (ensured by TF/SF)• TBC handles evanescent fields naturally so no spacer layer is needed.• Smaller matrices•More efficient simulations

• FFT operators are slow to construct, but there may be a better way

Slide 33

Matrix Tilt Operator

Slide 34

Given the field in a slice ei, we need a matrix operator T that will remove the phase tilt across the grid giving just the periodic envelope term from Bloch’s theorem.

i ia Te

This is calculated as

reorder

T 0 0

T 0 T 0

0 0 T

,inc ,incdiag x yj k ke

x yT

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Matrix 2D‐FFT Operator

Slide 35

Given the amplitude component ai of the field in a slice, we need another matrix operator Fthat will calculate a 2D FFT in order to calculate the amplitudes of the spatial harmonics in that slice.

i is Fa

The 2D‐FFT is defined as

2

, ,

, 0,1, , , 0,1, ,

mp nqj

M N

p q

x y

H m n h p q e

m p N n p N

From this, we can construct F’.

h F h

The full‐vector 2D‐FFT matrix operator is then

reorder

F 0 0

F 0 F 0

0 0 F

Matrix Phase Operator

Slide 36

Given the spatial harmonics, we need an operator that will add the correct phase to each spatial harmonic to propagate them across one slice.

1i i i s Z s

This is calculated as

reorder

Z 0 0

Z 0 Z 0

0 0 Z

1,1

,

z

z

jk z

jk M N z

e

e

Z

2 2 20 r r,z x yk m n k k m k n

0

0

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Overall Matrix Propagators

Slide 37

We need to calculate a single matrix operator that will calculate the field in an adjacent slice.

1i i i e Pe

1. Remove phase tilt.2. FFT the field to transform to Fourier‐space.3. Add phase to the spatial harmonics.4. Inverse‐FFT the field to transform to real‐space.5. Reincorporate the phase tilt.

1

1 1

i

i

i i

i i

i i

T e

F T e

Z F T e

F Z F T e

T F Z F T e

The overall propagator is therefore

11 1i i i

P T F Z FT FT Z FT

TBC at Top of Grid

Slide 38

The slice equation for the top slice (slice #1) can be written as

1 0 1 1 1 2 1 A e B e C e f

The term e0 does not exist because it resides outside of the grid. Assuming the field at the top of the grid is only moving outward (bottom to top), we can calculate e0 from e1 using the matrix propagator calculated in the reflection region.

0 ref 1e P e

Substituting this into the original slice equation yields

1 1 1 2 1 1 1 1 ref B e C e f B B A P

After this calculation, we no longer need A1.

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TBC at Bottom of Grid

Slide 39

The slice equation for the bottom slice (slice N) can be written as

1 1N N N N N N N A e B e C e f

The term eN+1 does not exist because it resides outside of the grid. Assuming the field at the bottom of the grid is only moving outward (top to bottom), we can calculate eN+1 from eNusing the matrix propagator calculated in the transmission region.

1 trnN N e P e

Substituting this into the original slice equation yields

1 trn N N N N N N N N A e B e f B B C P

After this calculation, we no longer need CN.

Slide 40

Field Solution

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Final Matrix Problem

Slide 41

The slice absorption method algorithm is implemented until only two slices remain. These are usually the reflection and transmission record planes. A final matrix equation is constructed from the remaining two slice equations.

1 0 1 1 1 2 1 A e B e C e f

1 1N N N N N N A e B e C e f

1 1 1 1

N N N N

A B e f

B C e f

Recall that A1and CN are no longer needed because these described coupling to slices from outside of the grid.

Field Solution

Slide 42

The fields in the record planes are calculated as

1

ref 1 1 1 1

trn N N N N

e e B C f

e e A B f

We can recover the vector components of the field in each slice as follows.

,ref

1,ref ref

,ref

reorderx

y

z

e

e e

e

,trn

1,trn trn

,trn

reorderx

y

z

e

e e

e

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Calculating Diffraction Efficiencies

Slide 43

The amplitudes of the spatial harmonics are calculated as

ref ref trn trn s FT e s FT e

Assuming the incident wave is given unit amplitude, the diffraction efficiencies are calculated as usual.

2 ,ref r,inc

DE ref

,inc r,inc

2 ,trn r,trn

DE trn

,inc r,inc

Re ,, ,

Re

Re ,, ,

Re

z

z

z

z

k m nR m n S m n

k

k m nT m n S m n

k

These equations assume the source was given unit amplitude.

inc 1S

Slide 44

Fourier‐Space SAM

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Comparison to Real‐Space SAM

Slide 45

Ω e

f

Formulation of Fourier‐Space SAM

Slide 46

e ey z z y xx x

e ez x x z yy y

e ex y y x zz z

h hy z z y xx x

h hz x x z yy y

h hx y y x zz z

D e D e μ h

D e D e μ h

D e D e μ h

D h D h ε e

D h D h ε e

D h D h ε e

e ey z z y xx x

e ez x x z yy y

e ex y y x zz z

h hy z z y xx x

h hz x x z yy y

h hx y y x zz z

K s D s u

D s K s u

K s K s u

K u D u s

D u K u s

K u K u s

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Slide 47

Example Simulations

Photonic Crystals Fabricated by Multi‐Photon Direct Laser Writing

Slide 48

R. C. Rumpf, A. Tal, S. M. Kuebler, “Rigorous electromagnetic analysis of volumetrically complex media using the slice absorption method,” J. Opt. Soc. Am. A 24(10), 3123‐3134, 2007.

This is a scanning electron microscope image of a photonic crystal designed to operate in the infrared.

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Fourier Transform Infrared Spectrometer

Slide 49

Samples are illuminated by a hollow Gaussian beam produced by a Cassegrain optic.

Reflected infrared energy at near normal reflection angle is collected and measured.


Predicting Accurate Geometry

Slide 50

To predict the geometry of the photonic crystal more accurately, the DLW process was simulated in MATLAB.

Idealized Geometry Realistic Geometry

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Numerical Representation of Device

Slide 51


Results from SAM

Slide 52


There is am important lesson here in terms of the importance of incorporating realistic geometry into your models.

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Metallo‐Dielectric Photonic Crystals

Slide 53

Before copper plating After copper plating

A. Tal, Y.‐S. Chen, H. E. Williams, R. C. Rumpf, S. M. Kuebler, “Fabrication and characterization of three‐dimensional copper metallodielectric photonic crystals,” Optics Express 15(26), 18283‐18293, 2007.

“State‐of‐the‐Art” Simulation of Reflectance

Slide 54

V. Mizeikis, S. Juodkazis, R. Tarozaite, J. Juodkazyte, K. Juodkazis, H. Misawa, “Fabrication and properties of metalo-dielectric photonic crystal structures for infrared spectral region,” Opt. Express 15, 8454-8464 (2007)

Results obtained by Lumerical and Misawa’s research group.

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A Better Simulation of Reflectance

Slide 55

A. Tal, Y. -S. Chen, H. E. Williams, R. C. Rumpf, and S. M. Kuebler, "Fabrication and characterization of three-dimensional copper metallodielectric photonic crystals," Opt. Express 15, 18283-18293 (2007)

Results obtained by UCF/Rumpf team…

Side‐by‐Side Comparison

Slide 56

V. Mizeikis, S. Juodkazis, R. Tarozaite, J. Juodkazyte, K. Juodkazis, H. Misawa, “Fabrication and properties of metalo-dielectric photonic crystal structures for infrared spectral region,” Opt. Express 15, 8454-8464 (2007)

reflectan

ce

A. Tal, Y. -S. Chen, H. E. Williams, R. C. Rumpf, and S. M. Kuebler, "Fabrication and characterization of three-dimensional copper metallodielectricphotonic crystals," Opt. Express 15, 18283-18293 (2007)

Results obtained with Lumerical’sFDTD software

Results obtained by UCF/Rumpf

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What Was Their Big Mistake?

Slide 57

Below 10 m (or so), this photonic crystal is a diffracting structure.

The optical configuration inside the FTIR cuts off the higher order modes. Essentially, it is only the zero‐order diffracted mode that gets detected.

Slide 58

Dispersion AnalysisUsing the SAM

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Starting Point

Slide 59

Using the standard SAM algorithm, it is possible to reduce a unit cell to two slice equations, one at each boundary of the unit cell.

1 0 1 1 1 N A e B e C e 0

1 1N N N N N A e B e C e 0

Derivation of the Eigen‐Value Problem (1 of 3)

Slide 60

Assuming the unit cell is infinitely periodic in the z direction, boundary conditions allow e0

and eN+1 to be written in terms of slices contained in the unit cell.

0z zj

Ne e e

1 1z zj

N e e e

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Slide 61

Substituting the new expressions into the slice equations gives

0z zj

Ne e e1 1

z zjN e e e

1 0 1 1 1 N A e B e C e 0 1 1N N N N N A e B e C e 0

1 1 1 1z zj

N Ne A e B e C e 0 1 1z zj

N N N Ne A e B e C e 0

After some algebraic manipulation to get ejzz on the right‐hand side of the equations, we get

1 1 1 1z zj

N Ne B e C e A e 1 1z z z zj j

N N N Ne e C e A e B e


Slide 62

Our last two equations can be written in block matrix form as

1 1 1 1 11 1 1 1

1 1

z z

z z

z z z z

jjN N

j jN N N N NN N N N

ee

e e

B C e 0 A eB e C e A e

C 0 e A B eC e A e B e

This has the form of a generalized eigen‐value problem

1 1 1 1 z zj

N N N N

v e

B C 0 A eA B x

C 0 A B e

vAx Bx

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Parameter Retrieval

Slide 63

The effective parameters of the unit cell can be determined one of two ways:

1. Field Averaging – The eigen‐vectors give the field values at the boundary slices. From these, the fields in all interior slices can be calculated. After all the fields are known, the field averaging technique can be implemented.

2. T‐Matrix Parameter Retrieval – The effective properties of the unit cell can be retrieved from the eigen‐value following the T‐matrix parameter retrieval method.

eff0

ln 2 branch #z zj

z

v me n m

k

o eff

eff

Re

Im

n n

n

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Documents

Slice Absorption Method (SAM) · 8/26/2019 4 Grid to Matrix Scaling & Memory Slide 7 Typical grid required to model a 3D device. 20 20 100 x y z N N N Number of points in grid: