Lecture #6 Implementation of One‐ Dimensional FDTDemlab.utep.edu/ee5390fdtd/Lecture 6 --...

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Lecture 6 Slide 1

EE 5303

Electromagnetic Analysis Using Finite‐Difference Time‐Domain

Lecture #6

Implementation of One‐Dimensional FDTD These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited

Lecture Outline

Lecture 6 Slide 2

• Review of Lecture 5

• Sequence of Code Development

• FDTD Implementation

– Numerical boundary conditions

– Grid resolution

– Courant stability condition

– Perfect 1D boundary condition

– Sources

– Total number of iterations

– Revised FDTD Algorithm

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Lecture 6 Slide 3

Review of Lecture #5

Representing Functions on a Grid

Slide 4Lecture 6

A grid is constructed by dividing space into discrete

cells

Example physical

(continuous) 2D function

Function is known only at discrete points

Representation of what is

actually stored in memory

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3D Grids

Lecture 6 Slide 5

10, 10, 15x y zN N N

A three‐dimensional grid looks like this:

A unit cell from the grid looks like this:

x y

z

Lecture 6 Slide 6

Yee Cell for 1D, 2D, and 3D Grids

xy

z

xEyE

zE

xHyH

zH

3D Yee Grid2D Yee Grids1D Yee Grid

zE

xHyH

xy

xy

zHyE

xE

Ez Mode

Hz Mode

z

xE

yH

yExH

Ey Mode

Ex Mode

z

Benefits• Implicitly satisfies divergence equations• Naturally handles physical boundary conditions• Elegant approximation of the curl equations using finite‐differences

Consequences• Field components are in physically different locations• Field components may reside in different materials even if they are in the same unit cell• Field components will be out of phase

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Formulation of Update Equations (1 of 4)

0

0

H H

Slide 7Lecture 6

We normalized the magnetic field,

0 0

r rH EE H

c t c t

Assume linear, isotropic, and non‐dispersive materials and expand the curl equations.

0

0

0

y xx xz

yy yx z

y x zz z

H EH

y z c t

EH H

z x c t

H H E

x y c t

0

0

0

y xx xz

yy yx z

y x zz z

E HE

y z c t

HE E

z x c t

E E H

x y c t

Lecture 6 Slide 8


Finite‐Difference Equation for Hx Finite‐Difference Equation for Hy Finite‐Difference Equation for Hz

Finite‐Difference Equation for Ex Finite‐Difference Equation for Ey Finite‐Difference Equation for Ez

2 2

, , 1 , ,, 1, , ,

, , , ,, ,

0

t t

i j k i j ki j k i j k

t t t t

i j k i j ki

xj k

t t

z

xxx

y yz

H H

E E

y z

c t

E E

2 2

, , 1 , , 1, , , ,

, , , ,, ,

0

t t

i j k i j k i j k i j k

t t t t

i j k i j ki j

x x z z

y yk

ty yt

z x

E

c

H

E

H

t

E E

2 2

1, , , , , 1, , ,

, , , ,, ,

0

t t


t t t t

i j k i j ki

y y x x

zj k

t tzzz

x y

c

E E E

H

t

E

H

2 2 2 2

, , , 1, , , , , 1

, , , ,, ,

0

t t t t


t t t t

i j k i j ki j k

xx tx

z

t

z y

x t

y

y z

E E

H H

c

H

t

H

2 2 2 2

, , , , 1 , , 1, ,

, , , ,, ,

0

t t t t


t t t t

i j k i j ki j k

yy ty

x

t

x z

y t

z

z x

E E

H H

c

H

t

H

2 2 2 2

, , 1, , , , , 1,

, , , ,, ,

0

t t t t


t t t t

i j k

y y x

i j ki j k

zz t t tz

x

z

H H H H

x y

E E

c t

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Slide 9Lecture 6

We assumed the problem was uniform in the x and y directions.

0x y

Maxwell’s equations separated into two sets of equations.

0

0

0

0

yy yx

y xx x

y xx x

yy yx

HE

z c t

H E

z c t

E H

z c t

EH

z c t

2 2

2 2

2 2

2 2

1

0

0

1

0

1

1

0

t t

t t

t t

t t

k ky y

k ky

k kx x

k kx x

k ky

kt tyyt t

kt t xx t t t

kt t yy t t t

kt tt t x

y

k kx x

k k

y

k

xy y xk

x

z c t

z c t

z c t

z c

E E

E

H H

t

E

E E

E E

H H

H H

H H


Slide 10Lecture 6

We derived update equations by solving the finite‐difference equations for the future time value.

2 2

2 2

1

1

t t

t t

k ky y

k k

k kx x

k k

k t tHxt t

t tkEyy yt t

x

t

x

H mz

m

E E

E

H

H HE

z

0

0

kEy k

yy

kHx k

xx

c tm

c tm

We arrived at the following FDTD algorithm.

0 0 k kEy Hxk k

yy xx

c t c tm m

2 2

1

t t

k t tHx

k k

kx tx

y

t

yk mHE

HE

z

2 2

1t tt tk

Eyt t t

kx

k ky y

kx

E mEH

z

H

Loop over z

Loop over z

Includes:• Basic FDTD engine

Excludes:• Dirichlet BC’s• Calculate source parameters• Simple soft source• Perfectly absorbing BC’s• TF/SF source• Fourier transforms• Reflectance/Transmittance• Calculate grid parameters• Incorporate device

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Lecture 6 Slide 11

Sequence of Code Development

Lecture 6 Slide 12

Step 1 – Basic FDTD Algorithm

• Basic update equations

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Lecture 6 Slide 13

Step 2 – Add Simple Soft Source

• Basic update equations• Add a soft source

Lecture 6 Slide 14

Step 3 – Add Absorbing Boundary

• Basic update equations• Add a soft source• Add perfect boundary

condition

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Lecture 6 Slide 15

Step 4 – Add TF/SF


condition• Incorporate TF/SF “one‐

way” source

Lecture 6 Slide 16

Step 5 – Move Source & Add T/R



way” source• Move position of source• Calculate transmittance

and reflectance

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Lecture 6 Slide 17

Step 6 – Add Device (Complete Algorithm)



way” source• Move position of source• Calculate transmittance

and reflectance• Add a real device

Lecture 6 Slide 18

Summary of Code Development Sequence

Step 1 – Implement basic FDTD algorithm Step 2 – Add the source

Step 3 – Add absorbing boundary Step 4 – Add “one‐way” source

Step 5 – Calculate transmittance and reflectance

Step 6 – Add a device

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Lecture 6 Slide 19

NumericalBoundary Conditions

Lecture 6 Slide 20

A Problem at the Boundary of the Grid

2 2

1

t t

k t tHx

k k

kx tx

y

t

yk mHE

HE

z

We must implement the update equations for every point in the grid.

1

What do we do at ?

does not exist.zNy

z

E

k N

2 2

1t tt tk

Eyt t t

kx

k ky y

kx

E mEH

z

H

0

What do we do at 1?

does not exist.xH

k

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Lecture 6 Slide 21

Dirichlet Boundary Condition

2 2

2 2

1

0

t t

z

t t

z

z z

k t tHx zt t

N tHx

k kyk k

x x

N Nx x

y

zt t

y

N

Em k N

z

m k N

H H

H Hz

E

E

2 2

2

1

1 1 1

1

1

1 0

t t

t

k ky y

y

t tkEyt t t

t

Eyt t t

k k

y

x x

x

H H

H

m kz

m k

E E

Ez

E

One easy thing to do is assume the fields outside the grid are zero. This is called a Dirichlet boundary condition.

We modify the update equations as follows.

Lecture 6 Slide 22

Equations MATLAB Code

2 2

2 2

1

0

t t

z z z z

t t

k k k k kx x Hx y y zt tt t

N N N Nx x Hx y ztt t

H H m E E k N

H H m E k N

1 1 1 1

2

1

2 2

1

1

0ty y Ey xt t t t

k k k k kt ty y Ey x xt t t t t

E E m H k

E E m H H k

Update Equations

%% MAIN FDTD LOOP%for T = 1 : STEPS

% Update H from E (Dirichlet Boundary Conditions)for nz = 1 : Nz-1

Hx(nz) = Hx(nz) + mHx(nz)*(Ey(nz+1) - Ey(nz))/dz;endHx(Nz) = Hx(Nz) + mHx(Nz)*(0 - Ey(Nz))/dz;

% Update E from H (Dirichlet Boundary Conditions)Ey(1) = Ey(1) + mEy(1)*(Hx(1) – 0)/dz;for nz = 2 : Nz

Ey(nz) = Ey(nz) + mEy(nz)*(Hx(nz) - Hx(nz-1))/dz;end

end

DO NOT use ‘if’ statements to implement boundary conditions.

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Lecture 6 Slide 23

Periodic Boundary Condition

2 2

2 2

1

1

t t

t t

z

k t tHx zt t

k t

k

tHx zt t

ky y

N

k kx x

k k y

x x

y

m k Nz

m k Nz

H

E

E E

E

H

H H

2 2

2 2

1

1

1

1

z

t t

t t

k ky y

k ky y

t tkEyt t

N

t

t tk

k k

Ey

x

t

x x

x

t t

m kz

m kz

EH

E

EH

E

H

H

Some devices are periodic along a particular direction. When this is the case, the field is also periodic.

We modify the update equations as follows.

z

perio

dic

perio

dic

Lecture 6 Slide 24

Grid Resolution

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Lecture 6 Slide 25

Consideration #1: Wavelength

The grid resolution must be sufficient to resolve the shortest wavelength.

N Comments

10 to 20 Low contrast dielectrics

20 to 30 High contrast dielectrics

40 to 60 Most metallic structures

100 to 200 Plasmonic devices

0min

max max

c

f n nmax is the largest refractive index found

anywhere in the grid. fmax is the maximum frequency in your simulation.

First, you must determine the smallest wavelength:

min 10NN

Second, you resolve the wave with at least 10 cells.

Lecture 6 Slide 26

Consideration #2: Mechanical Features

The grid resolution must be sufficient to resolve the smallest mechanical features of the device.

mind

Unit cell of a diamond lattice

1dN 4dN 1dN 1dN

mind

First, you must determine the smallest feature:

min 1d dd

dN

N

Second, you must divide this by 1 to 4.

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Calculating the Initial Grid Resolution

Must resolve the minimum wavelength

min 10NN

x

Must resolve the minimum structural dimension

min 1d dd

dN

N

Set the initial grid resolution to the smallest number computed above

min ,x y d

y

Slide 27Lecture 6

Lecture 14 Slide 28

Resolving Critical Dimensions (1 of 3)

We have not yet considered the actual dimensions of the device we wish to simulate.

This means we likely cannot resolve the exact dimensions of a device.

Not an exact fit.We cannot fill a fraction of a cell.

x

yd

Suppose we wish to place a device of length d onto a grid.

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Lecture 14 Slide 29


To fix this, we first calculate how many cells N are needed to resolve the most important dimension. In this case, let this be d.

x

yd

x

dN

10.5 cellsN

Second, we determine how many cells we wish to exactly resolve d. We do this by rounding N up to the nearest integer.

ceilx

dN

11 cellsN

Lecture 14 Slide 30


Third, we adjust the value of x to represent the dimension d exactly.

x

yd

x

d

N

11 cellsN

We call this step “snapping” the grid to a critical dimension.

Unfortunately, using a uniform grid, we can only do this for one dimension per axis.

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“Snap” Grid to Critical Dimensions

Decide what dimensions along each axis are critical

Compute how many grid cells comprise dc, and round UP

ceil

ceil

x x x

y y y

M d

M d

Typically this is a lattice constant or grating period along x Typically this is a layer thickness along y

and x yd d

initial grid

critical dim

ensio

n

Slide 31Lecture 6

Adjust grid resolution to fit this dimension in grid EXACTLY

x x x

y y y

d M

d M

ad

justed

grid

critical dim

ensio

n

decrease

Lecture 6 Slide 32

Courant Stability Condition

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Lecture 6 Slide 33

Numerical Propagation Through Grid

During a single iteration, a disturbance in the electric field at one point can only be felt by the immediately adjacent magnetic fields. It takes at least two time steps before that disturbance is felt by an adjacent electric field. This is simply due to how the update equations are implemented during a single iteration.

z

Lecture 6 Slide 34

Physical Propagation Through Space

Electromagnetic waves propagate at the speed of light in a vacuum. Inside a material, they propagate at a reduced speed.

z

00 299792458 m s refractive index

cv c n

n

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Lecture 6 Slide 35

A Limit on t

zNumerical distance covered in one time step:

Physical distance covered in one time step: 0c t n

Over a time duration of one time step t, an electromagnetic disturbance will travel:

Because of the numerical algorithm, it is not possible for a disturbance to travel farther than one unit cell in a single time step.

We need to make sure that a physical wave would not propagate farther than a single unit cell in one time step.

0c tz

n

This places an upper limit on the time step.

0

n zt

c

n should be set to the smallest refractive index found anywhere

in the grid. Usually this is just made to be 1.0 and dropped from the equation.

Lecture 6 Slide 36

The Courant Stability Condition

Refractive index is greater than or equal to one, so our condition on tcan be written more simply as:

0

zt

c

For 2D or 3D grids, the condition can be generalized as

0 2 2 2

1

1 1 1t

cx y z

The Courant stability condition

Sort of a worst case.n=1 produces the fastest possible physical wave.

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Lecture 6 Slide 37

Practical Implementation of the Stability Condition

The stability condition will be most restrictive along the shortest dimension of the grid unit cell.

min min , ,x y z

This can be generalized to account for special cases.

min min

02

nt

c

1. Your grid is filled with dielectric and travels slower

everywhere.2. Your model includes dispersive materials with

refractive index less than one.

A good equation to ensure stability and accuracy on any grid is then

min

02t

c

Note: A factor of 0.5 was included here as a safety margin.

Lecture 6 Slide 38

Time Step for Our 1D Grid

bc

02

n zt

c

nbc = refractive index at the grid boundaries.

This means a wave will travel the distance of one grid cell in exactly two time steps.

It is a necessary condition for the perfect boundary condition we will soon implement.

This implies that we cannot have different materials at the two boundaries using this boundary condition.

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Lecture 6 Slide 39

Perfect 1D Boundary Condition

z

Lecture 6 Slide 40

The Problem

The finite‐difference equation here requires knowledge of the electric field outside of the grid.

2 2

1z

t t

z

z

z k t tHxt

N Nx x

Ny y

t

NE EH

zH m

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Lecture 6 Slide 41

Implementing the Perfect Boundary Condition

Then…

1

2

z z

t t t

N Ny yE E

2 time steps later

z

If and only if…

• the field is only travelling outward at the boundaries,• the materials at the boundaries are linear, homogeneous, isotropic and non‐dispersive,

• The refractive index at both boundaries is nbc,• t = nbcz/(2c0) exactly.

Lecture 6 Slide 42

Visualizing the Perfect Boundary Condition

0 0 0 0 0 0 0 0 0 0 0 0 3 4 3 2 1 0T = 9 0 0

0 0 0 0 0 0 0 0 0 0 0 0 2.5 3.5 3.5 2.5 1.5 0.5T = 10 0 0

0 0 0 0 0 0 0 0 0 0 0 0 2 3 4 3 2 1T = 11 0 0

0 0 0 0 0 0 0 0 0 0 0 0 1.5 2.5 3.5 3.5 2.5 1.5T = 12 0.5 0

0 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 3 2T = 13 1 0

0 0 0 0 0 0 0 0 0 0 0 0 0.5 1.5 2.5 3.5 3.5 2.5T = 14 1.5 0.5

0 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0T = 3 0 0

0 0 0 0 0 0 0 0 0 0 0 0 2.5 1.5 0.5 0 0 0T = 4 0 0

0 0 0 0 0 0 0 0 0 0 0 0 3 2 1 0 0 0T = 5 0 0

0 0 0 0 0 0 0 0 0 0 0 0 3.5 2.5 1.5 0.5 0 0T = 6 0 0

0 0 0 0 0 0 0 0 0 0 0 0 4 3 2 1 0 0T = 7 0 0

0 0 0 0 0 0 0 0 0 0 0 0 3.5 3.5 2.5 1.5 0.5 0T = 8 0 0

0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0T = 1 0 0

0 0 0 0 0 0 0 0 0 0 0 0 1.5 0.5 0 0 0 0T = 2 0 0

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Lecture 6 Slide 43

Summary of the 1D Perfect Boundary Condition

Conditions• Waves at the boundaries are only travelling outward,• Materials at the boundaries are linear, homogeneous, isotropic and non‐dispersive,• The refractive index is the same at both boundaries and is nbc,• Time step is chosen so physical waves travel 1 cell in exactly two time stepst = nbcz/(2c0).

Implementation at z‐Low BoundaryAt the z‐low boundary, we need only modify the E‐field update equation.

23

3 2 2 1 11 1

1

1

t

y

tkEy

x

x yt t t

hh Eh h h h m

zH E

H

Implementation at z‐High BoundaryAt the z‐high boundary, we need only modify the H‐field update equation.

2 2

3

3 2 2 1 1 z

z z

t t

z k tHx

N Nx xt t

NyN

y

ee

EEe He He e m

z

Lecture 6 Slide 44

Sources

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Lecture 6 Slide 45

The Gaussian Pulse

In FDTD, we typically use short pulses as sources. This approximates an impulse function so we can excite a single simulation with a broad range of frequencies at the same time.

1 e

0t 0t 0t

2

0expt t

g t

Lecture 6 Slide 46

Frequency Content of Gaussian Pulse

The Fourier transform of a Gaussian pulse is another Gaussian function.

2 2

2 2maxmax

1exp G exp

t fg t f

ff

max

1f

The frequency content of the Gaussian pulse extends from DC up to above fmax. The frequency fmax is actually the 1/e point of the frequency spectrum.

maxf

max

1 1

e f

max

1

f

Frequency

G f

0

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Lecture 6 Slide 47

Designing the Pulse Source (1 of 2)

Step 1: Determine the maximum frequency of interest in your simulation.

maxf

Step 2: Compute the pulse duration to have sufficient energy at fmax.

maxmax

1 1 f

f

max

0.5

f

Step 3: You may need to reduce your time step. Your Gaussian pulse should be resolved by at least 10 to 20 time steps.

t

tN

Typically, you determine a first t based on the Courant

stability condition, then determine a second t based on resolving the maximum frequency, and finally go with the smallest value of the two t’s.

All of this should be satisfied automatically if t = nz/(2c0).10tN

Lecture 6 Slide 48

Designing the Pulse Source (2 of 2)

Step 4: Compute the delay time t0

0 6t The pulse source must start at zero and gradually increase. NO STEP FUNCTIONS!!

WRONG!!The step function at the beginning will induce very large field gradients.

0 0t

t

STILL WRONG!!While better, this source still starts with a step function that will produce large field gradients.

0t

t

CORRECT!!This source “eases” into the Gaussian source.

0 6t

t

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Visualizing the Arrays

Lecture 6 49

z Usually several hundred points.

Ey, Hx, ER, UR, mEy, and mHx are stored in arrays of length Nz.

tUsually several

thousand time steps.

gE is stored in an array of length STEPS.

Lecture 6 Slide 50

Two Ways to Incorporate a Simple Source

Source #1: Simple Hard SourceThe simple hard source is the easiest to implement. After updating the field across the entire grid, one field component at one point on the grid is overwritten with the source. This approach injects power into the model, but the source point behaves like a perfect electric conductor or perfect magnetic conductor and will scatter waves.

2

and/or t H Ek kt tt yx

kk g EH g OVERWRITE

Not usually a practical source.We won’t use it.

Source #2: Simple Soft SourceThe simple soft source is better than the hard source because it is transparent to scattered waves passing through it. After updating the field across the entire grid, the source function is added to one field component at one point on the grid. This approach injects power into the model in both directions. It is great for testing boundary conditions.

2 2

and/or t t

k kx x H Ek kt t

ky t tt

kyt

H H E gEg

ADD TORarely used.Use this until we learn TF/SF.

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Movie of Source #1 – Simple Hard Source

Lecture 6 51

Movie of Source #2 – Simple Soft Source

Lecture 6 52

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Movie of Source #3 – TF/SF Soft Source

Lecture 6 53

Lecture 6 Slide 54

Total Number of Iterations

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Considerations for Estimating the Total Number of Iterations

Lecture 6 55

The total number of iterations depends heavily on the device being modeled and what properties of it are being calculated.

Device Considerations1. Highly resonant devices typically require more iterations.2. Purely scattering devices typically require very few iterations.3. More iterations are needed the more times the waves bounce

around in the grid.

Information Considerations1. Calculating abrupt spectral shapes requires many iterations.2. Calculating fine frequency resolution requires many iterations.3. Calculating only the approximate position of resonances often

requires fewer iterations. Great for hunting resonances!

Lecture 6 Slide 56

A Rule of Thumb

How long does it take a wave to propagate across the grid (worst case)?

maxprop

0

distancetime

velocityzn N z

tc

Simulation time T must include the entire pulse of duration .12T

Simulation time should allow for 5 bounces.

prop5T t

A rule‐of‐thumb for total simulation time is then

prop12 5T t Note: For highly resonant devices, this will NOT be enough time!

Given the time step t, the total number of iterations is then

STEPS roundT

t

This must be an integer quantity.

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Lecture 6 Slide 57

Revised FDTD Algorithm

Lecture 6 Slide 58

Revised FDTD Algorithm

Finished!

Compute Grid Resolution

2 2

2 23

1t t

z z z z

t t


N N N Nx x Hx y ztt t

H H m E E z k N

H H m E z k Ne

min minmin ,

round

d

c

c

dz

N N

N d z

z d N

Update H from E

no

Compute Time Step bc 02t n z c

Compute Source

max 0

2

0

0.5 5

exp

f t

t tg t

Compute Update Coeff’s0 0 k k

Ey Hxk kyy xx

c t c tm m

Initialize Fields0k k

y xE H

1 1 1 1

2

1

2

3

2

1

1

ty y Ey xt t t t


E E m H z k

E E m H H z k

h

Update E from H

src srck ky y tt t t

E E g

Inject Source

Visualize fields

Done?yes

2

13 2 2 1 1, ,

tx th h h h h H

Record H‐Field Boundary Term

3 2 2 1 1, , Nzy t t

e e e e e E

Record E‐Field Boundary Term

z

z

Loop over time

t

Build DeviceERyy and URxx

Initialize Boundary Terms

3 2 1 3 2 1 0h h h e e e

Includes:• Basic FDTD engine• Dirichlet BC’s• Calculate source parameters• Simple soft source• Perfectly absorbing BC’s

Excludes:• TF/SF source• Fourier transforms• Reflectance/transmittance• Calculate grid parameters• Incorporate device

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Lecture 6 Slide 59

Equations MATLAB Code

min minmin ,

round

d

c

c

dz

N N

N d z

z d N

Grid Resolution % COMPUTE DEFAULT GRID RESOLUTIONdz1 = min(LAMBDA)/nmax/NRES;dz2 = dmin/NDRES;dz = min(dz1,dz2);

% SNAP GRID TO CRITIAL DIMENSIONSN = ceil(dc/dz);dz = dc/N;

2 2

2 2 2

1

t t

z z z z

t t

z


N N N Nx x H

Nyx yt ztt t t

H H m E E k N

H H m E kE N

1

2 2

1 1 1 1

22

1

2

1

1t


ty y Ey xt t x tt tt

E E m H H k

E H kHE m

Update Equations

%% MAIN FDTD LOOP%for T = 1 : STEPS

% Update H from E (Perfect Boundary Conditions)for nz = 1 : Nz-1

Hx(nz) = Hx(nz) + mHx(nz)*(Ey(nz+1) - Ey(nz))/dz;end Hx(Nz) = Hx(Nz) + mHx(Nz)*(E3 - Ey(Nz))/dz;H3=H2; H2=H1; H1=Hx(1);

% Update E from H (Perfect Boundary Conditions)Ey(1) = Ey(1) + mEy(1)*(Hx(1) – H3)/dz;for nz = 2 : Nz

Ey(nz) = Ey(nz) + mEy(nz)*(Hx(nz) - Hx(nz-1))/dz;endE3=E2; E2=E1; E1=Ey(Nz);

% Inject Soft SourceEy(nzsrc) = Ey(nzsrc) + g(T);

end

src srck ky y tt t t

E E g

Lecture #6 Implementation of One‐ Dimensional FDTDemlab.utep.edu/ee5390fdtd/Lecture 6 --...

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Transcript of Lecture #6 Implementation of One‐ Dimensional FDTDemlab.utep.edu/ee5390fdtd/Lecture 6 --...