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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
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)
Some big advantages
•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
FormulaDifferenceCentral
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
Quadratic
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