formulation of finite‐difference frequency‐domain

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Advanced Computation:

Computational Electromagnetics

Formulation of Finite‐Difference Frequency‐Domain

Outline

• Periodic boundary conditions

• Matrix wave equations

• Plane wave source by total‐field/scattered‐field

• Formulation summary and examples

• Calculating transmittance and reflectance

Slide 2

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From Last Time

Slide 3

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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

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HE EE E

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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 kx x z z

i j k

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i j kyy

i j k i j k i ji j kzz

k

i j kx

i j k

y y x

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i j kz

x

H HH H

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y z

z x

H H H

H H

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HE

H

x

E

E

xx

yy

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zy x z

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E E

E E

E E H

xx

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E

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H H

Hx

yH H

x

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e e

e e

e e

x

y

zy

y z xx

z x yy

x y

y

zz

z

x z

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D D μ

D D μ

D

h

e

D h

e e

e

e μe

h

h h

h h

h h

y z xx

z x

z

yy

z y

x z

yx y z

y

z

x

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D h h

h e

D eε

D D ε

D Dh ε

h

h e

Sign Convention for FDFD

Slide 4

jkze

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Slide 5

Periodic Boundary Conditions

Problem with the PBC Discussed Until Now

Slide 6

Suppose we wish to simulate a periodic structure with the source wave incident at an angle.

Is the problem still periodic?

field is NOT continuous

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k

Revisiting Bloch’s Theorem

Slide 7

Waves in a periodic structure take on the same symmetry and periodicity as the structure itself.

Actually, waves in a periodic structure obey Bloch’s theorem.

j rE r A r e

Overall E FieldCombines the periodic envelope with the “plane wave” like phase term.

Periodic Envelope, AIt is this part that takes on the same symmetry and period as the structure the wave is in.

“Plane Wave” Phase TermThis has the same form as a plane wave with wave vector . This incorporates an overall “phase tilt.”

We recognize that x = x6 – x1.

Revised Periodic Boundary Condition

Slide 8

Suppose we have the following grid…

We need to relate E1 to E6 in a way that is consistent with Bloch’s theorem.

1

66

1 1

6

j x

j x

E Ae

E A e

x56E E

x

x

We can write expressions for Ei at each point using Bloch’s theorem.

11 1

j xA E e Solve for A1.

6 1

6 1j x xE E e

6 1xjE E e

We now have our revised boundary condition.

66 1

j xE Ae Bloch’s theorem states that A1 = A6.

616 1

j xj xE E e e Replace A1 with expression from above.

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The Final Periodic Boundary Condition

Slide 9

We can think of the grid this way

,inc

1 1 0 0 0

0 1 1 0 01 0 0 1 1 0

0 0 0 1 1

0 0 0 1x x

E

x

jk

x

e

D

For FDFD analysis, inck

56 5 1xjE E

x

E e E

x

x

x

Names for this BCPeriodic BCPseudo‐periodic BCFloquet BCBloch‐Floquet BC

2D Derivative Matrices for 1D Grids With Source Wave at an Oblique Angle

Slide 10

When Nx=1 and Ny>1 ,inc

,ince h

,inc

0

0

x

xx x

x

jk

jk

jk

D D

When Nx>1 and Ny=1e h

e h,inc

and are standard for 1D gridx x

y y yjk

D D

D D I

e h,inc

e h and are standard for 1D grid

x x x

y y

jk D D I

D D

,inc

,ince h

,inc

0

0

y

y

y y

y

jk

jk

jk

D D

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USE SPARSE MATRICES!!!!!!!

Slide 11

WARNING !!The derivative operators will be EXTREMELY large matrices.

For a small grid that is just 100200 points:

Total Number of Points: 20,000Size of Derivate Operators: 20,000 20,000Total Elements in Matrices: 400,000,000Memory to Store One Full Matrix: 6 GbMemory to Store One Sparse Matrix: 1 Mb

NEVER AT ANY POINT should you use FULL MATRICES in the

finite‐difference method. Not even for intermediate steps. NEVER!

Header for yeeder() Function

Slide 12

function [DEX,DEY,DHX,DHY] = yeeder(NGRID,RES,BC,kinc)% YEEDER Construct Yee Grid Derivative Operators on a 2D Grid%% [DEX,DEY,DHX,DHY] = yeeder(NGRID,RES,BC,kinc);%% Note for normalized grid, use this function as follows:%% [DEX,DEY,DHX,DHY] = yeeder(NGRID,k0*RES,BC,kinc/k0);%% Input Arguments% =================% NGRID [Nx Ny] grid size% RES [dx dy] grid resolution of the 1X grid% BC [xbc ybc] boundary conditions% -2: periodic (requires kinc)% 0: Dirichlet% kinc [kx ky] incident wave vector% This argument is only needed for periodic boundaries.%

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Block Diagram for yeeder() Function

Slide 13

NGRID, RES, BC, and optional kinc

If kinc not given, let kinc = [0 0]

DEX = jkx,incINx = 1?

Initialize DEX as sparse

Place two diagonals(1’s and ‐1’s)

Make corrections

PBC?

Insert exp(jkx,incx) terms

no

yes

yesNy = 1?

Initialize DEY as sparse

Place two diagonals(1’s and ‐1’s)

Make corrections

PBC?

Insert exp(jky,incy) terms

no

yes

yes

no

DEY = jky,incI

h e

h e

H

x x

H

y y

D D

D D

CalculateDHX and DHY

Finished

Note:The finite‐difference equation for point (nx,ny) on the grid is located in row mwhere m = (ny – 1)*Nx + nx

no

33 Grid Numerical Example

Slide 14

k0 = 2*pi;theta = 20*degrees;kinc = k0*[sin(theta);cos(theta)];

NGRID = [3 3];RES = [1 1];BC = [-2 -2];[DEX,DEY,DHX,DHY] = yeeder(NGRID,RES,BC,kinc);

DEX =

-1.0000 1.0000 0 0 0 0 0 0 0 0 -1.0000 1.0000 0 0 0 0 0 0

0.9866 + 0.1630i 0 -1.0000 0 0 0 0 0 0 0 0 0 -1.0000 1.0000 0 0 0 0 0 0 0 0 -1.0000 1.0000 0 0 0 0 0 0 0.9866 + 0.1630i 0 -1.0000 0 0 0 0 0 0 0 0 0 -1.0000 1.0000 0 0 0 0 0 0 0 0 -1.0000 1.0000 0 0 0 0 0 0 0.9866 + 0.1630i 0 -1.0000

DEY =

-1.0000 0 0 1.0000 0 0 0 0 0 0 -1.0000 0 0 1.0000 0 0 0 0 0 0 -1.0000 0 0 1.0000 0 0 0 0 0 0 -1.0000 0 0 1.0000 0 0 0 0 0 0 -1.0000 0 0 1.0000 0 0 0 0 0 0 -1.0000 0 0 1.0000

0.4205 - 0.9073i 0 0 0 0 0 -1.0000 0 0 0 0.4205 - 0.9073i 0 0 0 0 0 -1.0000 0 0 0 0.4205 - 0.9073i 0 0 0 0 0 -1.0000

,inc

,inc

inc ,inc ,inc

2.1490 3

5.9043 3

ˆ ˆ ˆ ˆ2.1490 5.9043

3 1 3

3 1 3

0.9866 0.1630

0.4205 0.9073

x x

y y

x x y y x y

x x x

y y y

jk j

jk j

k k a k a a a

N

N

e e i

e e i

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44 Grid Numerical Example (1 of 2)

Slide 15



1 1

1 1

1 1

0.68 0.73 1 0

1 1

1 1

1 1

0.68 0.73 1 0

1 1

1 1

1 1

0.68 0.73 1 0

1 1

1 1

1 1

0.68 0.73 1

ex

i

i

i

i

D

,inc

inc ,inc ,inc

2.1490 4

ˆ ˆ ˆ ˆ2.1490 5.9043

4 1 4

0.6757 0.7372x x

x x y y x y

x x x

jk j

k k a k a a a

N

e e i

44 Grid Numerical Example (2 of 2)

Slide 16



1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

0.05 0.99 1

0.05 0.99 1

0.05 0.99 1

0.05 0.99 1

ey

i

i

i

i

D

,inc

inc ,inc ,inc

5.9043 4

ˆ ˆ ˆ ˆ2.1490 5.9043

4 1 4

0.0551 0.9985y y

x x y y x y

y y y

jk j

k k a k a a a

N

e e i

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Slide 17

Matrix Wave Equations

3D FDFD in Block Matrix Form

Slide 18

e e

e e

e e

h h

h h

h h

y z z y xx x

z x x z yy y

x y y x zz z

y z z y xx x

z x x z yy y

x 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 e

e e

e e

z y x xx x

z x y yy y

y 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 h

h h

h

z y x xx x

z x y yy yh

y 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

rE H

rH E

We can write our matrix equations in block matrix form.

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3D‐FDFD Formulation

Slide 19

Block Matrix Form

er

hr

C e μ h

C h ε e

r

h h

h h h

h h

x

y

z

xx

yy

zz

z y

z x

y 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

1e hr r

1h er r

C ε C μ h 0

C μ C ε e 0

Ae 0

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

For more information on 3D FDFD analysis, see

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, 3123‐3134 (2007).

r

e e

e e e

e e

x

y

z

xx

yy

zz

z y

z x

y 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

3D FDFD Usually Too Big to Solve

Slide 20

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 A:Number of complex elements:

Number of real elements:Memory to store full A:

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

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Recall Diffraction Configurations

Slide 21

Planar Diffraction from a Ruled Grating

Conical Diffraction from a Ruled Grating

Crossed Grating Configurations

Requires a full 3D simulationModel can be reduced to a 2D simulation

• Simplest model• Two independent modes

• Analytically, about the same complexity as 3D

• No independent modes

Advice: Don’t bother developing a 2D code for conical diffraction. Develop a 3D code and use it for 2D.

0 0E H

z z z z

0 z

E Hjk

z z z z

0 0E H

z z z z

z

y

x

Reducing Planar Diffraction to 2D

Slide 22

h h

e

e

x y y x zz z

y z xx x

x z yy y

D h D h ε e

D e μ h

D e μ h

For problems uniform along the z‐direction and propagation restricted to the x-y plane,

e h0 z zz

D D 0

Maxwell’s equation decouple into two distinct modes:

EMode: HMode:

e e

h

h

x y y x zz z

y z xx x

x z yy y

D e D e μ h

D h ε e

D h ε e

e ey z z y D e D e

e

xx x

z x

μ h

D e

e e

e

x y y x zz z

x z yy y

D e D e h

e

μ

D μ h

h hy z z y D h D h

h

xx x

z x

ε e

D hh h

h

x y y x zz z

x z yy y

D h D h ε

D h ε

e

e

x

y

z

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Matrix Wave Equation (EMode)

Slide 23

h h

e

e

x y y x zz z

y z xx x

x z yy y

D h D h ε e

D e μ h

D e μ h

The matrix equations for the Emode were

1 e 1 e x xx y z y yy x z

h μ D e h μ D e Following matrix algebra rules, the second two equations are solve for hx and hy.

h 1 e h 1 e

E z

E x yy x y xx y zz

A e 0

A D μ D D μ D ε

AE = DHX/URyy*DEX + DHY/URxx*DEY + ERzz;

Now substitute these new expressions into the first matrix equation.

h h

h 1 e h 1 e

h 1 e h 1 e

h 1 e h 1 e

h 1 e h 1 e

x y y x zz z

x yy x z y xx y z zz z



x yy x y xx y zz z

D h D h ε e

D μ D e D μ D e ε e

D μ D e D μ D e ε e

D μ D e D μ D e ε e 0

D μ D D μ D ε e 0

Two Matrix Wave Equations

Slide 24

EMode

HMode

h 1 e h 1 eE x yy x y xx y zz

A D μ D D μ D ε

e 1 h e 1 hH x yy x y xx y zz

A D ε D D ε D μ

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TE and TM (Framework #1)

Slide 25

For the Emode, the electric field is always transverse to waves propagating in the x‐y plane.

EMode = TE Mode

For the Hmode, the magnetic field is always transverse to waves propagating in the x‐y plane.

HMode = TM Mode

TE and TM (Framework #2)

Slide 26

We use the labels “TE” and “TM” when we are trying to describe the orientation of a linearly polarized wave relative to a device.

TEaTMa

TE/perpendicular/s – the electric field is polarized perpendicular to the plane of incidence.

TM/parallel/p – the electric field is polarized parallel to the plane of incidence.

In the limit as 90°, we have

EzMode = TM ModeHzMode = TE Mode

This is exactly the opposite of Framework #1 when propagation is restricted to be in the x‐y plane.

n

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Our Matrix Equations Cannot Yet Be Solved

Slide 27

Solution to our matrix wave equation

1 E z z E A e 0 e A 0 0 trivial solution!!

A source bmust be incorporated.

1 E z z E A e b e A b

Slide 28

Plane Wave Source ByTotal‐Field/Scattered‐Field

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The Total‐Field/Scattered‐Field Framework

Slide 29

Problem rows

total‐field

scattered‐field

The Source Field

Slide 30

1,1

src ,inc ,inc

,

exp

x y

x y

N N

f

j k k

f

f X Y

Compute the source as it would exist in a completely homogeneous grid.

incsrc

jk rE r e

Unit amplitude plane wave

Generated by meshgrid()

fsrc = exp(1i*(kxinc*X + kyinc*Y));

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Scattered‐Field Masking Matrix, Q

Slide 31

total‐field

scattered‐field 1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

1

1

0

0

0

Q

0

0

Reshape data from a 2D grid to a 1D array, then insert the 1D array into the center diagonal of a sparse matrix.

Q = diag(sparse(Q(:)));

Note: The TF/SF interface must completely enclose the source in order to completely describe it.

Calculating the Source Vector b (1 of 2)

Slide 32

Quantities that must be subtracted from TF terms to

make them look like SF terms…

totAf

srcfSource field

QSF masking function

tot src f I Q fSource isolated to TF

scat srcf QfSource isolated to SF

Quantities that must be added to SF terms to make them look like TF terms…

scatAf

…but only from SF equations:

totQAf…but only to TF equations:

scatI Q Af

SFTF

SFTF

Combine corrections:

tot scat Ae QAf I Q Af 0 SF

correctionsTF

corrections

Note, we get non‐zero numbers here where ftot or fscat fails to satisfy Maxwell’s equations.

SFTF

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Calculating the Source Vector b (2 of 2)

Slide 33

tot scat Ae QAf I Q Af 0

Our matrix equation incorporating the TF/SF corrections is

These are known quantities that can be brought to the right‐hand side of the equation.

tot scat Ae QAf I Q Af

We now have an equation from which to determine the source vector b.

b

We can write b in terms of just Q, A, and fsrc by substituting in our expressions for ftot and fscat.

tot scat

src src

src

b QAf I Q Af

QA I Q f I Q AQf

QA AQ f

Famous QAAQ Equation

Slide 34

src Ae = b b QA AQ f Quack!

All of the complexities of the total‐field/scattered‐field method are easily implemented using the famous QAAQ (pronounced “quack”) equation.

• Very easy to implement!• TF and SF regions can be arbitrary.• Works regardless of grid (i.e. uniform, nonuniform, unstructured, etc.).• Works regardless of order of accuracy of the finite‐differences.• Should work independent of the numerical method used to construct A.

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Visualizing the Data in b

Slide 35

We see that b is non‐zero only at the interfaces between the total‐field and scattered‐field. This is where the finite‐difference equations contain quantities from both regions and correction terms are needed to make the quantities compatible.

Strictly speaking, the source field fsrc really only needs to be calculated at the points immediately adjacent to the TF/SF interface.

Example Simulation #1

Slide 36

Materials Field Q

Materials Field Q

PML

PML

PML

PML

PML

PML

PML

PML

PML

PML

PML

PML

Scattered‐FieldScattered‐Field


Total‐FieldTotal‐Field


1 1.0n

1 1.0n

2 3.0n

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Example Simulation #2

Slide 37

Materials Field Q

Materials Field Q

PML

PML

PML

PML

PML

PML

PML

PML

PML

PML

PML

PML





1 1.0n

1 1.0n

2 3.0n

Slide 38

Formulation Summaryand Examples

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FDFD Formulation Summary

Slide 39

h 1 e h 1 eE x yy x y xx y zz

A D μ D D μ D ε

e 1 h e 1 hH x yy x y xx y zz

A D ε D D ε D μ

Matrix wave equations

src b QA AQ f

Source vector

FDFD Matrix Problem

EA e = b

HA h = b

Periodic Structures in FDFD

Slide 40

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Cylindrical Source

Slide 41

Modal Sources (i.e. Waveports)

Slide 42

This is a simulation of an anti‐reflective waveguide termination. This required that the fundamental mode be launched into the waveguide as well as calculation of the energy reflected back into the fundamental mode.

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Slide 43

Calculating Transmittance and Reflectance

Process of Calculating Transmittance and Reflectance

Slide 44

1. Perform FDFD simulation2. Extract cross sections of the field on the reflected and

transmitted side.3. Calculate the wave vector components of the spatial

harmonics4. Calculate the complex amplitude of the spatial harmonics5. Calculate the diffraction efficiency of the spatial

harmonics6. Calculate over all reflectance and transmittance7. Calculate energy conservation.

For each frequency of interest…

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Step 1: Perform FDFD Simulation

Slide 45

TF/SF Interface

Periodic BCx

y

Reflection record plane

Transmission record plane

Spacer

Spacer

Step 2: Calculate Wave Vector Components

Slide 46

x Nx points across

Transverse Components

,

2

floor , , 2, 1,0,1,2, , floor2 2

x x incx

x x

mk m k

N Nm

Longitudinal Components

2 20 ref

2 20 trn, n

, ef

tr

ry

xy

xk n k mk m

m nk k k m

Note: transverse wave vector components are the same throughout the entire grid.

Note: the longitudinal components vary throughout the grid and are calculated from the dispersion relation.

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Step 3: Extract Reflected and Transmitted Fields

Slide 47

Step 4: Remove Phase Tilt

Slide 48

Field calculated from FDFD

Phase across grid due to angle of incidence

Field with phase tilt removed.

Bloch amplitude.

,inc

ref refxjk xE x A x e

,inc

trn trnxjk xE x A x e

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2 1 0 1 2FFT M MS S S S S S S

Step 5: Calculate Amplitudes of Spatial Harmonics

Slide 49

, x yj k m x k m y

m

E x y S m e

0S 1S 2S1S2S

Note: This operation is performed for every frequency of interest.

FFT

Step 6: Calculate Diffraction Efficiencies

Slide 50

2

ref ,ref r,inc

DE 2,inc r,incinc

2

trn ,trn r,trn

DE 2,inc r,incinc

Re

Re

Re

Re

y

y

y

y

S m k mR m

kS

S m k mT m

kS

This equation assumes the Emode was solved.

Notes1. This calculation is performed for every frequency of interest.2. is the amplitude of the source. This is usually set equal to 1.0 and dropped from

the above equations.3. We can neglect the negative sign in the equation by also ignoring the negative

sign when calculating .

incS

DER m ,refyk m

2

ref r,inc

DE 2,inc r,incinc

,refRe

Re

y

y

S mR

S

k

k

mm

2 2rref f, 0 e xy k nk k mm

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Step 7: Reflectance and Transmittance

Slide 51

Reflectance is the total fraction of power reflected from a device. Therefore, it is equal to the sum of all the reflected modes.

DE

xN

R R m

DE

xN

T T m

Transmittance is the total fraction of power transmitted through a device. Therefore, it is equal to the sum of all the transmitted modes.

Reflectance and Transmittance on a Decibel Scale

dB 10 dB 1010log 10 logR R T T Be careful NOT to use 20log10!

Step 8: Calculate Power Conservation

Slide 52

Assuming loss or gain is not incorporated into the simulation, the reflectance plus transmittance should equal 100%.

100%R T

It is ALWAYS good practice to calculate this total to check for conservation of power. This may deviate from 100% when:

•The boundary conditions are not working properly and need to be corrected.•Rounding errors are too severe and greater grid resolution is needed.•Simulation incorporates loss or gain in the materials.

The general equation for conservation must also include the absorptance A.

100%A R T

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formulation of finite‐difference frequency‐domain

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