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Computational Electromagnetics computational electromagnetics High frequency rigorous methods IE DE MoM FDTD TLM field based current based GO/GTD PO/PTD TD FD TD FD VM FEM

# Computational Electromagnetics

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computational electromagnetics. rigorous methods. High frequency. DE. current based. IE. VM. field based. TD. FD. TD. FD. PO/PTD. FDTD TLM. FEM. MoM. GO/GTD. Computational Electromagnetics. Computational Electromagnetics. - PowerPoint PPT Presentation

### Text of Computational Electromagnetics

Computational Electromagnetics

computationalelectromagnetics

High frequencyrigorous methods

IE DE

MoMFDTDTLM

field basedcurrent based

GO/GTD PO/PTD

TD FD TD FD

VM

FEM

Computational Electromagnetics

Electromagnetic problems are mostly described by three methods:

Differential Equations (DE) Finite difference (FD, FDTD)Integral Equations (IE) Method of Moments (MoM)Minimization of a functional (VM) Finite Element (FEM)

Theoreticaleffort

less more

Computationaleffort

more less

Numerical Differentiation“FINITE DIFFERENCES”

Introduction to differentiation

• Conventional Calculus

– The operation of diff. of a function is a well-defined procedure

– The operations highly depend on the form of the function involved

– Many different types of rules are needed for different functions

– For some complex function it can be very difficult to find closed form solutions

• Numerical differentiation

– Is a technique for approximating the derivative of functions by employing only arithmetic operations (e.g., addition, subtraction, multiplication, and division)

– Commonly known as “finite differences”

Taylor SeriesProblem: For a smooth function f(x),

Given: Values of f(xi) and its derivatives at xi

Find out: Value of f(x) in terms of f(xi), f(xi), f(xi), ….

x

yf(x)

f(xi)

xi

Taylor’s TheoremIf the function f and its n+1 derivatives are continuous on an interval containing xi and x, then the value of the function f at x is given by

nn

ii

n

ii

ii

iii

Rxxn

xf

xxxf

xxxf

xxxfxfxf

)(!

)(...

)(!3

)()(

!2

)(''))((')()(

)(

3)3(

2

Finite Difference Approximationsof the First Derivative using the Taylor Series

(forward difference)

x

yf(x)

f(xi)

xi xi+1

f(xi+1)

h

Assume we can expand a function f(x) into a Taylor Series about the point xi+1

nn

iii

n

iii

iii

iiiii

Rxxn

xf

xxxf

xxxf

xxxfxfxf

)(!

)(...

)(!3

)()(

!2

)(''))((')()(

1

)(

31

)3(2

111

h

Finite Difference Approximationsof the First Derivative using the Taylor Series (forward

difference)Assume we can expand a function f(x) into a Taylor Series about the point xi+1

ni

nii

iii hn

xfh

xfh

xfhxfxfxf

!

)(

!3

)(

!2

)(")(')()(

)(3

)3(2

1

h

xfxfxf iii

)()()(' 1

Ignore all of these terms

1)(

2)3(

1

!

)(

!3

)(

!2

)(")()()(' ni

niiii

i hn

xfh

xfh

xf

h

xfxfxf

Finite Difference Approximationsof the First Derivative using the Taylor

Series (forward difference)

h

xfxfxf iii

)()()(' 1

x

yf(x)

f(xi)

xi xi+1

f(xi+1)

h

Finite Difference Approximationsof the First Derivative using the

forward difference: What is the error?

)()()(

)(' 1 hOh

xfxfxf iii

The first term we ignored is of power h1. This is defined as first order accurate.

1)(

2)3(

1

!

)(

!3

)(

!2

)(")()()(' ni

niiii

i hn

xfh

xfh

xf

h

xfxfxf

)()('

)()( 1

hOh

fxf

xfxff

ii

iii

First forwarddifference

Finite Difference Approximationsof the First Derivative using the Taylor

Series (backward difference)

x

yf(x)

f(xi-1)

xi-1 xi

f(xi)

h

Assume we can expand a function f(x) into a Taylor Series about the point xi-1

nn

iii

n

iii

iii

iiiii

Rxxn

xf

xxxf

xxxf

xxxfxfxf

)(!

)(...

)(!3

)()(

!2

)(''))((')()(

1

)(

31

)3(2

111

-h

Finite Difference Approximationsof the First Derivative using the Taylor Series

(backward difference)

ni

nii

iii hn

xfh

xfh

xfhxfxfxf

!

)(

!3

)(

!2

)(")(')()(

)(3

)3(2

1

Ignore all of these terms

1)(

2)3(

1

!

)(

!3

)(

!2

)(")()()(' ni

niiii

i hn

xfh

xfh

xf

h

xfxfxf

)()()(

)(' 1 hOh

xfxfxf iii

)()('

)()( 1

hOh

fxf

xfxff

ii

iii

First backwarddifference

Finite Difference Approximationsof the First Derivative using the Taylor

Series (backward difference)

x

yf(x)

f(xi-1)

xi-1 xi

f(xi)

h

)()()(

)(' 1 hOh

xfxfxf iii

Finite Difference Approximationsof the Second Derivative using the Taylor Series

(forward difference)

y

x

f(x)

f(xi)

xi xi+1

f(xi+1)

h

xi+2

f(xi+2)

ni

nii

iii hn

xfh

xfh

xfhxfxfxf

!

)(

!3

)(

!2

)(")(')()(

)(3

)3(2

1

nni

nii

iii hn

xfh

xfh

xfhxfxfxf 2

!

)(8

!3

)(4

!2

)("2)(')()(

)(3

)3(2

2

(1)

(2)

(2)-2* (1)

)()()(2)(

)(" )3(2

112i

iiii xhf

h

xfxfxfxf

Finite Difference Approximationsof the Second Derivative using the Taylor Series

(forward difference)

y

x

f(x)

f(xi)

xi xi+1

f(xi+1)

h

xi+2

f(xi+2)

)()()(2)(

)(" )3(2

112i

iiii xhf

h

xfxfxfxf

)()(

)()("22

2

hOh

fhO

h

fxf iii

)(2

2

hOh

f

dx

fd in

xx

n

i

Recursive formula forany order derivative

Higher Order Finite Difference Approximations

)()()(2)(

)(" )3(2

112i

iiii xhf

h

xfxfxfxf

1)(

2)3(

1

!

)(

!3

)(

!2

)(")()()(' ni

niiii

i hn

xfh

xfh

xf

h

xfxfxf

1)(

2)3(

)3(12

1

!

)(

!3

)(

!2

...)()()(2)(

)()()('

nin

i

iiii

iii

hn

xfh

xf

hxhf

hxfxfxf

h

xfxfxf

...)('''32

)(3)(4)()('

212

xf

h

h

xfxfxfxf iiii

)(2

)(3)(4)()(' 212 hO

h

xfxfxfxf iiii

Centered Difference Approximation

)(2

)()()(' 211 hO

h

xfxfxf iii

3

)3(2

1 !3

)(

!2

)(")(')()( h

xfh

xfhxfxfxf ii

iii

3

)3(2

1 !3

)(

!2

)(")(')()( h

xfh

xfhxfxfxf ii

iii

(1)

(2)

(1)-(2) 3

)3(

11 !3

)(2)('2)()( h

xfhxfxfxf i

iii

2)3(

11

!3

)(2

2

)()()(' h

xf

h

xfxfxf iiii

Finite Difference Approximationsof the First Derivative using the Taylor

Series (central difference)

x

yf(x)

f(xi-1)

xi-1 xi

f(xi)

h

xi+1

f(xi+1)

)(2

)()()(' 211 hO

h

xfxfxf iii

Second Derivative Centered Difference Approximation (central

difference)

)()()(2)(

)( 22

11 hOh

xfxfxfxf iiii

3

)3(2

1 !3

)(

!2

)(")(')()( h

xfh

xfhxfxfxf ii

iii

3

)3(2

1 !3

)(

!2

)(")(')()( h

xfh

xfhxfxfxf ii

iii

(1)

(2)

(1)+(2) 4

)4(2

11 !4

)(2)()(2)()( h

xfhxfxfxfxf i

iiii

2)4(

211

!4

)(2

)()(2)()( h

xf

h

xfxfxfxf iiiii

Using Taylor Series Expansions we found the following finite-differences

equations

)()()(

)(' 1 hOh

xfxfxf iii

FORWARD DIFFERENCE

)()()(

)(' 1 hOh

xfxfxf iii

BACKWARD DIFFERENCE

)(2

)()()(' 211 hO

h

xfxfxf iii

CENTRAL DIFFERENCE

)()()(2)(

)( 22

11 hOh

xfxfxfxf iiii

CENTRAL DIFFERENCE

Forward finite-difference formulas

Centered finite difference formulas

Finite Difference Approx. Partial DerivativesProblem: Given a function u(x,y) of two independent

variables how do we determine the derivative numerically (or more precisely PARTIAL DERIVATIVES) of u(x,y)

?),(

?),(

?),(

?),(

?),( 2

2

2

2

2

yx

yxUor

y

yxUor

x

yxUor

y

yxUor

x

yxU

Pretty much the same way

STEP #1: Discretize (or sample) U(x,y) on a 2D grid of evenly spaced points in the x-y plane

x axis

y axis

xi xi+1xi-1 xi+2

yj

yj+1

yj-1

yj-2

u(xi,yj) u(xi+1,yj)

u(xi,yj-1)

u(xi,yj+1)

u(xi-1,yj)

u(xi-1,yj+1)

u(xi-1,yj-1)

u(xi-1,yj-2) u(xi,yj-2)

u(xi+1,yj-1)

u(xi+1,yj-2)

u(xi+1,yj+1)

u(xi+2,yj)

u(xi+2,yj-1)

u(xi+2,yj-2)

u(xi+2,yj+1)

2D GRID

x axis

y axis

i i+1i-1 i+2

j

j+1

j-1

j-2

ui,j ui+1,jui-1,j

ui,j-1

ui,j+1

SHORT HAND NOTATION

Partial First Derivatives

Problem: FIND ?),(

?),(

y

yxuor

x

yxu

recall:

h

xfxfxf iii 2

)()()(' 11

Partial First Derivatives

Problem: FIND ?),(

?),(

y

yxuor

x

yxu

x

yxuyxu

x

yxu jijiji

2

),(),(),( 11

x

y

y

yxuyxu

y

yxu jijiji

2

),(),(),( 11

These are central difference formulas

Are these the only formulaswe could use?

Could we use forward or backwarddifference formulas?

Partial First Derivatives: short hand notation

Problem: FIND ?),(

?),(

y

yxuor

x

yxu

x

uu

x

u jijiji

2,1,1,

x

y y

uu

y

u jijiji

21,1,,

Partial Second DerivativesProblem: FIND ?

),(?

),(2

2

2

2

y

yxuor

x

yxu

recall:

211 )()(2)(

)(h

xfxfxfxf iiii

Partial Second Derivatives

Problem: FIND

2

11

2

2 ),(),(2),(),(

x

yxuyxuyxu

x

yxu jijijiji

x

y

?),(

?),(

2

2

2

2

y

yxuor

x

yxu

2

11

2

2 ),(),(2),(),(

y

yxuyxuyxu

y

yxu jijijiji

Partial Second Derivatives: short hand notation

Problem: FIND

2

,1,1,1

2

,2 2

x

uuu

x

u jijijiji

x

y

?),(

?),(

2

2

2

2

y

yxuor

x

yxu

2

1,,1,

2

,2 2

y

uuu

y

u jijijiji

FINITE DIFFERENCE ELECTROSTATICS

Electrostatics deals with voltages and charges that do no vary as a functionof time.

/),,(),,(2 zyxzyx Poisson’s equation

0),,(2 zyx Laplace’s equation

Where, is the electrical potential (voltage), is the charge density and is the permittivity.

E

o

1

2

3

FINITE DIFFERENCE ELECTROSTATICS: Example

0),(2 yx

Find(x,y) inside the box due to the voltages applied to its boundary. Thenfind the electric field strength in the box.

E

Electrostatic Example using FD

Problem: FIND

2

,1,,1

2

,2 2

xxjijijiji

x

y

0),(),(

2

2

2

2

y

yx

x

yx

2

1,,1,

2

,2 2

yyjijijiji

Electrostatic Example using FD

Problem: FIND

022

2

,1,,1

2

1,,1,

xyjijijijijiji

0),(),(

2

2

2

2

y

yx

x

yx

If x = y

jijijijiji

jijijijiji

jijijijijiji

,1,11,1,,

,,1,11,1,

,1,,11,,1,

4

1

04

022

Electrostatic Example using FD

Problem: FIND 0),(),(

2

2

2

2

y

yx

x

yx

jijijijiji ,1,11,1,, 4

1

Iterative solution technique:(1) Discretize domain into a grid of points(2) Set boundary values to the fixed boundary values(3) Set all interior nodes to some initial value (guess at it!)(4) Solve the FD equation at all interior nodes(5) Go back to step #4 until the solution stops changing(6) DONE

Electrostatic Example using FD

MATLAB CODE EXAMPLE

=0

=0

=0

=0

FINITE DIFFERENCE Waveguide TM modes: Example

0),(),( 22 yxkyx t

Where for TM modes

2222222tzzzt kkkkkk

zjkz

zeyxzyxE ),(),,(~

and

If tk then kz becomes imaginary and the mode does notpropagate.

=0

=0

=0

=0

FINITE DIFFERENCE Waveguide TM modes: Example

0),(),( 22 yxkyx t

Goal: Find all permissible values of kt and the correspondingmode shape ((x,y)) for that mode.

Waveguide Example using FD

Problem: FIND

2

,1,,1

2

,2 2

xxjijijiji

x

y

0),(),(),(

2

2

2

2

yxky

yx

x

yxz

2

1,,1,

2

,2 2

yyjijijiji

Waveguide Example using FD

Problem: FIND

022

,2

2

,1,,1

2

1,,1,

jizjijijijijiji k

xy

0),(),(),(

2

2

2

2

yxky

yx

x

yxz

If x = y=h

0)4(

022

,22

,1,11,1,

,22

,1,,11,,1,

jizjijijiji

jizjijijijijiji

kh

kh

Waveguide Example using FD

0)4( ,22

,1,11,1, jizjijijiji kh

=0

=0 =0

=0

N

...2

1

let where N is the number of interior nodes (i.e. not on a boundary)

If we now apply the FD equation at all interior nodes we canform a matrix equation 0)( 22 IhkA

Where I is the identity matrix

Waveguide Example using FD

0)4( ,22

,1,11,1, jizjijijiji kh

is an eigenvalue equation usually cast in the form

The eigenvalues will provide the permissible values for the transverse wavenumberkt and the eigenvectors are the corresponding mode shapes ((x,y))

0)( 22 IhkA z

A

Waveguide Example using FD

0)4( ,22

,1,11,1, jizjijijiji kh

1 2

3 4

=0 =0

=0

=0

=0=0

=0

=0

h

W

W

Waveguide Example using FD

jijijijijiji ,,,1,11,1, 4

1 2

3 4

=0 =0

=0

=0

=0=0

=0

=0

h

W

WNode #1: 1123 400 Node #2:

2214 400 Node #3:

3341 400 Node #4: 4432 400

4

3

2

1

4

3

2

1

4110

1401

1041

0114

Solve using the Matlab “eig” function

Waveguide Example using FD

0)4( ,22

,1,11,1, jizjijijiji kh

is an eigenvalue equation usually cast in the form

The eigenvalues will provide the permissible values for the transverse wavenumberkt and the eigenvectors are the corresponding mode shapes ((x,y))

HOMEWORK: WRITE A MATLAB PROGRAM THAT CALCULATES THE TRANSVERSE WAVENUMBERS AND MODE SHAPES FOR A RECTANGULAR WAVEGUIDE FOR TM MODES

0)( 22 IhkA z

A

Finite Difference Time Domain Method(FDTD)

•Broadband response with a single excitation.•3D models easily.•Frequency dependent materials accommodated.•Most parameters can be generated e.g.

Scattered fieldsantenna pattersRCS S-parameters etc…..

How does it work?

Based on the 2 Maxwell curl equations in derivative form. Theseare linearized by central finite differencing. We only considernearest neighbor interactions because all the fields are advancedtemporally in discrete time steps over spatial cells.

embedding of an antennain a FDTD space lattice(note that the whole volumeis meshed!)

ie we sample in space & time

closed & open problems

FDTD is especially suitable for computing transients for closedgeometries. When an open geometry is required, such as whenwe are dealing with an antenna, we need a boundary condition to simulate infinity. In this case the FDTD requires an absorbing boundarycondition (ABC) at the grid truncation.

This means there is no reflection from the boundary where themesh ends.

FDTD: The Basic Algorithm• Maxwell’s Equations in the TIME Domain:

t

EEHX

t

HEX

Equate Vector Components: Six E and H-Field Equations

x

E

y

E

t

H

z

E

x

E

t

H

y

E

z

E

t

H

yxz

xzy

zyx

1

1

1

zxyz

yzxy

xyzx

Ey

H

x

H

t

E

Ex

H

z

H

t

E

Ez

H

y

H

t

E

1

1

1

2-D Equations: Assume that all fields are uniform in y

direction (i.e. d/dy = 0)

zyz

xyx

xzy

Ex

H

t

E

Ez

H

t

E

z

E

x

E

t

H

1

1

1

yzxy

yz

yx

Ex

H

z

H

t

E

x

E

t

H

z

E

t

H

1

1

1

2D - TE 2D - TM

1-D Equations: Assume that all fields are uniform in y

and x directions (i.e. d/dy =d/dx= 0)

xyx

xy

Ez

H

t

E

z

E

t

H

1

1

yxy

yx

Ez

H

t

E

z

E

t

H

1

1

1D - TE 1D - TM

Discretize Objects in Space using Cartesian Grid

2D Discretization

Z

0z z Z( , )xE z t

1D Discretization

xx

zz

3D Discretization

Define Locations of Field Components:

FDTD Cell called Yee Cell

• Finite-Difference

– Space is divided into small cells

One Cell: (dx)(dy)(dz)

– E and H components are distributed in space around the Yee cell (note: field components are not collocated)

FDTD: Yee, K. S.: Numerical solution of initial boundary value problems involving Maxwell's equations in isotropic media. IEEE Transactions on Antennas

Propagation, Vol. AP-14, pp. 302-307, 1966.

Replace Continuous Derivatives with Differences

• Derivatives in time and space are approximated as DIFFERENCES

)(2

)('

:

'

211' hErrorh

ffxff

curveofslopefx

f

iiii

Solution then evolves by time-marching difference equations

– Time is Discretized• One Time Step: dt

– E and H fields are distributed in time

– This is called a “leap-frog” scheme.

1-D FDTD

Assuming that field values can only vary in the z-direction (i.e. all spatial derivatives in x and z direction are zero),

Maxwell’s Equations reduce to:

z

E

t

Hxy

1

x

yx Ez

H

t

E

1

z

z), (z)

Ex

Hy

1-D FDTD – Staggered Grid in Space

3

2zn

xE

1

2zn

1

2zn

3

2zn

2zn 1zn zn 1zn

1

2tn

yH tn

3

2zn

xE

1

2zn

1

2zn

3

2zn 1

2tn

Time plane

Interleaving of the Ex and Hy field components in space and time in the 1-D FDTD formulation

1-D FDTD

z

E

t

Hxy

1

x

yx Ez

H

t

E

1

Replace all continuous derivatives with finite differences

t

iEiE

t

E nx

nxx

)2/1()2/1( 2/12/1

t

iHiH

t

H ny

nyy

)()(1

z

iEiE

z

E nx

nxx

)2/1()2/1( 2/12/1

z

iHiH

z

H ny

nyy

)1()(

2

)2/1()2/1( 2/12/1

iEiEE

nx

nx

x

1-D FDTD

Substitution of difference equations in above yields:

z

E

t

Hxy

1

x

yx Ez

H

t

E

1

2

)2/1()2/1(

)1()(1)2/1()2/1(

2/12/1

2/12/1

iEiE

z

iHiH

t

iEiEnx

nx

ny

ny

nx

nx

z

iEiE

t

iHiH nx

nx

ny

ny )2/1()2/1(1)()( 2/12/11

1-D FDTD

After some simple algebra:

z

E

t

Hxy

1

x

yx Ez

H

t

E

1

)1()(2

2)2/1(

2

2)2/1( 2/12/1

iHiHtz

tiE

t

tiE n

yny

nx

nx

)2/1()2/1()()( 2/12/11

iEiEz

tiHiH n

xnx

ny

ny

66

1-D FDTD Algorithm – Flow Chart

( , 1/ 2) ( , 1/ 2) ( , )e

ˆ ˆ ˆt t tn n n n n ny y yE E tJ

Start

Stop

1t tn n

t tn N

1tn

Compute 1-D Faraday’s FDTD equation: For all nodes n inside the simulation region:

Electric current density excitation: For all excitation nodes n:

No Yes

( , 1/ 2)ˆ 0tn nyE

Boundary condition: For all PEC boundary nodes n:

( , ) ( , 1) ( , 1/ 2) ( 1, 1/ 2)t t t tn n n n n n n ny y x xH H t E E

Compute 1-D Ampère-Maxwell’s FDTD equation: For all nodes n inside the simulation region:

( , 1/ 2) ( , 1/ 2) ( 1, ) ( , ) t t t tn n n n n n n n

x x y yE E t H H

FDTD Solution of 1-D Maxwell’s Equations (EXAMPLE)

FDTD equations

(0, ) 0

( , ) 0x

x

E tt

E Z t

Initial conditions

Boundary condition for a perfectly electrically conducting (PEC) material

initial-boundary-value problem

Z

0z z Z

( , ) 0xE Z t (0, ) 0xE t ( , )xE z t

)()1(2

2)2/1(

2

2)2/1( 2/12/1 iHiH

tz

tiE

t

tiE n

yny

nx

nx

)2/1()2/1()()( 2/12/11

iEiEz

tiHiH n

xnx

ny

ny

2))((1 )( oiix

o

oy e

Z

EiH

For all values of i

2))41((2

1)2

1( oiix

ox eEiE

FDTD Solution of the First Two 1-D Scalar Maxwell’s Equations

Excitation pulse: RC2(t) – Time Domain

Excitation pulse: RC2(f) – Frequency Domain

Ma

gn

tiu

de

|R

C2(f)

|A

mp

litu

de

RC

2(t)

FDTD Solution of 1-D Maxwell’s Equations

FDTD Solution of 1-D Maxwell’s Equations

SOME OPEN QUESTIONS??

• How do we determine what t and z should be?• How do we implement real sources?• How do we simulate open boundaries?• How accurate is the solution?

Potential Source of Error: Numerical Dispersion

• In implementing any numerical technique, it is essential to understand the origin and nature of errors that are introduced as a result of the computational approach.

• In the FDTD errors arise from the discretization of the computational space and finite difference approximations to Maxwell’s equations.

• One particularly important source of error is call numerical dispersion.

• To see this consider the continuous wave equation in ID:

Which has a solution of the form we have looked at before:

02

22

2

2

x

Uc

t

U

)(, kxwtjetxU

If we substitute this expression into the wave equation, we get:

•From this we see that the wave number is linearly proportional to the frequency.

•This is more widely known as the phase velocity, when expressed as p = w/k p=c

•To determine how the FDTD effects this relationship, we repeat the same procedure using the difference equations.

ckw

jkcjw

ejkcejw kxwtjkxwtj

222

)(22)(2

)()(

)(

•Represent the second order difference equation (for both time and space) in terms of:

•Doing the same for the time derivative, we get:

0

222

11

211

tc

UUU

x

UUU ni

ni

ni

ni

ni

ni

211

2

2 2

x

UUU

x

Un

i

n

i

n

i

Let’s now look at the discrete wave equation

12

1121 22

n

ini

ni

ni

nin

i UUx

UUUtcU

•Now defining the solution from in difference form:

•Substituting this into the difference form of the wave equation we get:

)(

)~

(

kxwtj

xiktwnjni

e

eU

)( kxwtje

xiktnwjxiktwnj

xiktwnjxiktwnjxiktwnj

xiktnwj

ee

eeex

tc

e

~)1()

~(

))1(~

(~

)1(~

2

~)1(

2

2

)~

( xiktwnje Factor out:

xiktnwjxiktwnj

xiktwnjxiktwnjxiktwnj

xiktnwj

ee

eeex

tc

e

~)1()

~(

))1(~

(~

)1(~

2

~)1(

2

2

)~

( xiktwnje Factor out:

tjwxkjxkjtwj eeex

tce

22~~

2

We get:

tjwxkjxkjtjw eeex

tce

22~~

2

ckw

xkx

tctw

11~

cos)cos(2

1122

~~2

xkjxkjtjwtjw ee

x

tcee

Group time and space terms and divide both sides by 2

Use Euler’s Identity:2

)cos(jxjx ee

x

(FDTD)

(continuous)

•This is a nonlinear equation that represents the relationship between the wave number, and the frequency, w. It is a function of the time step t, spatial step x and frequency w

•This means that as a simulated wave propagate through the solution space it undergoes phase errors because the numerical wave either slows down or accelerates relative to the actual wave propagation in physical space.

•To examine the implication we consider 3 special cases:

Case 1

Consider Δt and Δx 0

•In this case, use the first two terms of the Taylor series expansion of terms

k~

cwk

xkx

tctw

xkx

tctw

~

~

2

11~

12

1

22

222

Same as in real space

Case 2

Use the relation cΔt = Δx

•This is often referred to as the magic time step!

•Plugging in we get:

xktw

xkx

tctw

~coscos

11~

coscos2

ckw

ckt

xkw

~

~~

xtc

xktw ~

Implies that

No dispersion

Unfortunately the relation cΔt = Δx is unstable!

Case 3

•In this case we consider the general case where there is no assumed relation between Δt and Δx.

•Therefore we need to solve for a general expression that can tell us the effect of numerical dispersion for a given relationship.

•To do this, we solve in terms of the wave number.

•Using this equation we can determine the numerical wave number for any relation between Δt and Δx for a given sampling rate.

•Substitute in the appropriate parameters and determine the effects of discretizing the computational space.

1cos1cos1~

21 tw

tc

x

xk

10,2

xx

tc

x

12

cos41cos1~ 1 xk

xk

Where we have used k= w/c, use the sampling note of

Now define numerical phase velocity as

xx

kwx

6364.08042.0cos

1~ 1

cpVk

wp 987.0

~~

•Over a distance of 10λ, which is in this case in no computational cells

•The numerical wave would only propagate 98.73 cells phase error of 45.72º

•If we repeat this analysis for

•Now the phase error at wλ is 11.19º

•The conclusion to draw from this is that the larger the solution space, the greater the phase error is going to be.

(MATLAB DEMO)

cpx 9968.020

Stability

•As stated before, the FDTD method approximates M.E.’s as a set of completed difference equations.

•As such this method is useful only when the solution of the difference equations are convergent and stable.

•Convergence means that as time stepping continues, the solution of the difference equations asymptomatically approach the solution given by Maxwell’s equations.

•On the other hand, stability is basically stated as a set of condition under which the error generated does not grow in an unbounded fashion.

In order for the FDTD equations to remain stable the following relationship (called the Courant stability criteria must be met)

•This is the stability condition for 1D case

•For higher dimensional cases we have the form

,where

D=1, 2, 3 depending on the number of dimensions used

xtc

D

xtc

(MATLAB DEMO)

•So far we have considered a sample 1D FDTD formulation, that physically represents the propagation of a plane wave.

•More often than not, EM BVP’s cannot be accurately represented as a 1D problem.

2D FDTD Formulation

•In the formulation of the two-dimensional FDTD, one can assume either TE ( electric field is to the plane of incidence, or TM; magnetic field is to the plane of incidence).

•This results from the fact that in 2D neither the fields nor the object contain any variations in the z-direction,consequently at

0z

So in this context, Maxwell’s Equations can be reduced to:

z

x

zy

zx

zxyz

EonpolarizatiTE

E

t

H

y

E

t

Ht

HE

Ey

H

x

H

t

EE

t

EH

,

1

1

1

For TM polarization we have:

fieldHpolTM

x

E

y

E

t

H

x

HE

t

y

HE

t

z

yxz

zy

zx

.,

1

1

1

Applying the central difference expression, for TE mode, we get:

21,2

1,,21,2

1),(~

),(~

),(~

),1(~

,21,2

1

)1,(~

),(~

21,2

1,

21

21

21

21

21

21

21

21

1

jiHjiHjiHjiHCbjiECajiE

jiEjiEjiHjiH

jiEjiEjiHjiH

nx

nx

ny

ny

nz

nz

nz

nz

ny

ny

nz

nz

nx

nx

For the TM case, we have:

21

21

21

21

21

211

21

21

21

21

21

211

,,),(~

),(~

,,),(~

),(~

)21,1(

~)2

1,(~

),21(

~)1,2

1(~

21,2

12

1,21

21

21

21

21

21

21

jiHjiHCbjiECajiE

jiHjiHCbjiECajiE

jiEjiEjiEjiEjiHjiH

nz

nz

ny

ny

nz

nz

nx

nx

ny

ny

nx

nx

nx

nx

Using same procedure as for the 1D case we obtain:

Numerical Dispersion 2D case

)sin(~~

),cos(~~

2

)sin(~

sin2

)cos(~

sin2

sin

2

~

sin1

2

~sin

1

2sin

1

22

2

222

kkkk

kskst

tc

s

yk

y

xk

x

t

tc

yx

yx

Plot of the normalised phase velocity vs. travelling wave angle.

Numerical Dispersion 2D case

Writing a 2D FDTD Program

1. Defining physical constants

2. Define program constants

magnitude of electric field

: total # of time steps

: # of nodes in the x, y direction

3. If using a pulse source, define Ez and Hx, or we can implement a hard source.

4. Start time marching nΔt, time value n = 1 : Nt

5. Increment the spatial index

First apply the absorbing boundary conditions

Apply the ABC along the edges of the computational region

yx

t

NN

N

x

t

E

,

0

),(,,, yxyx

(MATLAB DEMO)

Yee’s 3D Mesh

Transition to 3D

The 1D formulation involved 2 degrees of freedom (one field quantity E or H and one time variable). In 3D we have 4 degreesof freedom (3 in space and 1 in time). The following notation issomewhat standard

, , , ,i j k i x j y k z where the increment is the lattice space increments and i,j,k areintegers. Any field may be represented as

, ,( , , , ) ni j ku i y j y k z n t u

Fall 2007 [email protected] 95

1 1

1 1S C

D dS H dlt

The Yee Cell seen in a more Macroscopic view –Faraday's and Ampere’s law

General 3D Formation

•So far we have considered the 1D and 2D formulations.

•Now we will do the 3D formulation.

•Recall that Maxwell’s curl equations are:

Et

E

Jt

DH

t

H

tE

(i, j, k)

kji ,,(i, j, k)

•For simplicity, consider a linear and isotropic medium.

•In Cartesian coordinates, the curl equations reduce to:

t

H

y

E

x

Ez

t

H

x

E

z

Ey

t

H

z

E

y

Ex

zxy

yzx

xyz

~

~

~

Ei j k

/x /y /z

Ex Ey Ez

xyz H

tz

E

y

Et

ˆ

Also, for Ampere’s Law:

zzxy

yyzx

xxyz

Et

E

y

H

x

H

Et

E

x

H

z

H

Et

E

z

H

y

H

3D Discretization

Given a function f(x, y, z, t) we write it in discrete form as:

f(x, y, z, t) = f(iΔx, jΔy, kΔz, nΔt)

=

In general, we let: Δx = Δy = Δz =

),,( kjif n

3D formulation1

1

1

yx zx

y xzy

yxzz

EH EH

t z y

H EEH

t x z

EEHH

t y x

1

1

1

yx zx

y x zy

y xzz

HE HE

t y z

E H HE

t z x

H HEE

t x y

These six coupled equations form the basis for the FDTD algorithm.

3D Discretization

Given a function f(x, y, z, t) we write it in discrete form as:

f(x, y, z, t) = f(iΔx, jΔy, kΔz, nΔt)

=

In general, we let: Δx = Δy = Δz =

),,( kjif n

FDTD Discretization of 3D Curl Equations

z

kjiEkjiE

x

kjiEkjiE

kjikjiHkjiH

y

kjiEkjiE

z

kjiEkjiE

kjikjiHkjiH

nx

nx

nz

nz

r

nx

nx

nz

nz

ny

ny

r

nx

nx

,,1,,,,,,1

,,

1,,,,

,,,,,,,,

,,

1,,,,

21

21

21

21

21

21

0

21

21

21

21

21

21

212

12

12

1

21

21

0

21

21

21

21

21

21

21

21

2

,,,,,,

,,,,,,,,

,,

1,,,,

,,,,1,,,1,

,,

1,,,,

21

211

21

21

21

21

21

21

21

21

21

21

0

21

211

21

21

21

21

21

21

0

21

21

21

21

21

21

21

21

21

21

kjiEkjiEkji

z

kjiHkjiH

y

kjiHkjiH

kjit

kjiEkjiE

x

kjiEkjiE

y

kjiEkjiE

kjit

kjiHkjiH

nx

nx

ny

ny

nz

nz

r

nx

nx

ny

ny

nx

nx

r

nz

nz

Where we have used time interpolation

2

,,,,,,

,,,,,,,,

,,

1,,,,

2

,,,,,,

,,,,,,,,

,,

1,,,,

21

211

21

21

21

21

21

21

21

21

21

21

0

21

211

21

211

21

21

21

21

21

21

21

21

21

21

0

21

211

21

212

12

1

21

21

21

21

kjiEkjiEzkji

y

kjiHkjiH

x

kjiHkjiH

kjit

kjiEkjiE

kjiEkjiEkji

x

kjiHkjiH

z

kjiHkjiH

kjit

kjiEkjiE

nz

ny

nx

nx

ny

ny

r

nz

nz

ny

ny

nz

nz

nx

nx

r

ny

ny

We can simplify these expressions if we set:

And we can define the material constants:

kjimmRm

RamCb

mCa

tR

tR

tR

r

mmR

mmR

b

a

r

r

,,,

,

1

1

0

002

0

zyxr ;1

Substituting in we get:

21

21

21

21

21

21

21

21

21

211

21

21

21

21

21

21

21

21

21

211

21

21

21

21

21

21

21

21

21

211

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

,,,,,,,,,,~

,,~

,,,,,,,,,,~

,,~

,,,,,,,,,,~

,,~

,,1~

,,~

,,~

,1,~

,,,,

1,,~

,,~

,,~

,,1~

,,,,

,1,~

,,~

,,~

1,,~

,,,,

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

21

kjiHkjiHkjiHkjiHmCbkjiEmCakjiE

kjiHkjiHkjiHkjiHmCbkjiEmCakjiE

kjiHkjiHkjiHkjiHmCbkjiEmCakjiE

kjiEkjiEkjiEkjiEkjiHkjiH

kjiEkjiEkjiEkjiEkjiHkjiH

kjiEkjiEkjiEkjiEkjiHkjiH

nx

nx

ny

ny

nz

nz

nz

nz

nz

nx

ny

ny

ny

ny

nz

nz

nx

nx

ny

ny

nx

nx

nz

nz

nz

nx

nz

nz

ny

ny

nz

nz

ny

ny

nx

nx

ABC’s

•In general EM analysis of scattering structures often requires the solution of “open region” problems.

•As a result of limited computational resources, it becomes necessary to truncate the computational domain in such a way as to make it appear infinite.

•This is achieved by enclosing the structure in a suitable output boundary that absorbs all outward traveling waves

ABC

•In this lecture we will describe some of the more common ABC’s, such as:

•Bayliss-Turkel annihilation operators

•Enqquist-Majda one way operators

•Mur ABC

•Liao’s extrapolation method

•And the PML, perfectly matched layer

•The PML method which was introduced in 1994 by Berenger, represents one of the most significant advances in FDTD development, since it conception I 1966, by Kane Yee.

•The PML produces back reflection ~ over a very broad range of incident 86 1010

PMC Region

Early ABC’s

•When Yee first introduced the FDTD method, he used PEC boundary conditions.

•This technique is not very useful in a general sense.

•It wasn’t until the 70’s when several alternative ABC’s were introduced.

•However, these early ABC’s suffered from large back reflections, which limited the efficacy of the FDTD method.

Extrapolation from Interior Node (Taylor 1969)

•If we consider the electric field located on a 2D boundary

•Since and are determined in the FDTD calculations, and are therefore known values, we can solve this equation for the exterior node.

•This equation is only effective on normally incident waves and degrades rapidly when the incident wave is off-normal.

),( jiE n

Wave

y, (j)

x, (i)

)3,(iE n

)2,(iE n

)1,(iE n

2)3,()1,(

)2,(inEinE

inE

)2,(iE n )3,(iE n

)1,(iE n

)3,()2,(2)1,( iEiEiE nnn

Extending this method to 2D:

This approach applies to a wide range of incident field angles, but due to the nature of the averaging process often gives rise to significant non-physical back reflections.

3

)2,1(1)2,(1)2,1(1

)1,(in

xEin

xEin

xE

inx

E

j=1

j=2

j=3

i i+1i-1

Dissapative Medium

•In this technique a lossy medium is used to surround the FDTD computational region.

•The idea is that as that waves propagate into this medium, they are dissipated before they can undergo back reflection

•The problem is that there is often an input impedance between the FDTD region and the long media.

•The back reflection results from the ratio of electrical-conductivity and magnetic-conductivity parameters, respectively.

xand

Linear

0x1x

•Here the are assumed to be free space.

•Therefore the input impedance is:

which is not necessarily equal to Zo.

•To overcome this the conductivity values can be implemented in a non-uniform fashion.

•That is the can have a linear or quadratic profile.

•This tends to require relatively thick medium layers - computational requirements!

00 and

*0

0

jjz

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