cad 2011 slides
TRANSCRIPT
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Computing and CAD IIProfessor Jan K. Sykulski
Computing and CADProfessor Jan Sykulski
FIEEE, FIET, FInstP, FBCS, CEng, CITP
Electrical Power Engineering
School of Electronics & Computer Science
University of Southampton, UK
Computing and CAD IIProfessor Jan K. Sykulski
Resources
Core Resources
Hammond P and Sykulski J K:Engineering Electromagnetism
Oxford University Press 1994
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Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
trR
11
m
m
l
S
lm
S
lm
S
R
m
i
m
i
11
11
S
lm
m
S
lrt
m conductors in parallel:
S
lR
n conductors in series: srR
S
n
l
rs
n
n
S
l
S
nl
R
n
i
1
S
lR
Computing and CAD IIProfessor Jan K. Sykulski
If tubes and slices are in correct position
there is negligible effect as they are not disturbing the field.
If they are not in correct position:
tubes increase resistance giving R+ upper bound
slices decrease resistance giving R lower bound
Only in undisturbed field R = R+
otherwise R < R < R+
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Computing and CAD IIProfessor Jan K. Sykulski
Example
I
Ia
a
a2a
slicestubes
2a
Computing and CAD IIProfessor Jan K. Sykulski
Example
I
Ia
a
a2a
slicestubes
+
2a
depthunitpera
aR 2
2 depthunitper
a
a
a
aR 5.1
2
2a2a
a a
a
a
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Computing and CAD IIProfessor Jan K. Sykulski
Example
I
Ia
a
a2a
)(22
tubesdepthunitpera
aR
)(5.12
slicesdepthunitpera
a
a
aR
2a
depthunitperRR
Rave 75.12
5.12
2
)!(25.075.1: guaranteeddepthunitperRAnswer
depthunitperRprogramelementfiniteafromresultAccurate 73.1:)(
Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
t
is
jij
r
RwhereRIPower
1
1
2
1
1
s
kt
l lsr
RwhereR
VPower
1
1
2
1
1
Tubes:
Slices:
where for each r:
S
l
S
lr
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Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
I I
2
RRR
ave
Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
Constructionlines
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Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
Coarse
tube/slicelines
Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
Improvedtube/slicelines
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Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
Finite
elementsolution
Computing and CAD IIProfessor Jan K. Sykulski
Tubes and Slices
Resistance(per unit depth)
[value material resistivity]
Guaranteedaccuracy
Actual error
Hand calculation 2.7333 22.0 % 12.4 %
TAS coarse 2.4388 9.5 % 0.3 %
TAS refined 2.4324 4.0 % 0.3 %
Finite elements 2.4316 0.9 % -
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Computing and CAD IIProfessor Jan K. Sykulski
Classification ofmethods of fieldmodelling:
analogue methods
analytical solutions
numerical methods
(algebraical computation)
graphical computation
The most popularmethods:
separation of variables
images
analogue techniques
conformal transformations
Laplace transforms
transmission-line
modelling
finite differences
finite elements
boundary elements integral formulations
tubes and slices
...
Computing and CAD IIProfessor Jan K. Sykulski
Review of field equations0
2
2
2
2
2
2
z
V
y
V
x
VLaplaces equation: where V is a function of position V(x,y,z)
often written for convenience as:
In two-dimensional problems (2D): 02
2
2
2
y
V
x
Vwhere V(x,y)
V could be an electrostatic potential or magnetostatic potential.
02 V
Some fields (e.g. a magnetic field) may be described using a vector potential:
02 A where A(x,y,z) = Ax(x,y,z)i + Ay(x,y,z)j + Az(x,y,z)k
02 xA
02 zA
02
yA
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Computing and CAD IIProfessor Jan K. Sykulski
Review of field equations
Laplaces equation: 02 V
02 A
In two-dimensional problems (2D): kkA AAz kJ zJas
where ),( yxA
02 A
02
2
2
2
y
A
x
Aor
Computing and CAD IIProfessor Jan K. Sykulski
Review of field equationsLaplaces equation: 0
2 V
02 A
Poissons equation:
Helmholtz equation:
Diffusion equation:
Wave equation:
fV 2
022 VkV
t
V
hV
2
2 1
2
222
t
VV
FA 2
fVkV 22
homogeneous
non-homogeneous
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method
02
boundarytheoneverywheredefined
n
or
Equation specified inside:
A unique solution exists!
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method
i,j
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method
i,j
0:2'2
2
2
22
yxDinequationsLaplace
...
!3!2,
3
33
,
2
22
,
,,1
jijiji
jijix
x
x
x
xx
...
!3!2,
3
33
,
2
22
,
,,1
jijiji
jijix
x
x
x
xx
Adding:
4,
2
22
,,1,1 2 xOx
x
ji
jijiji
where O{(x)4} represents terms containing fourth and higher order powers ofx.
Neglecting these terms yields:
2,1,,1
,
2
2 2
xx
jijiji
ji
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method
i,j
0:2'2
2
2
2
2
yxDinequationsLaplace
2,1,,1
,
2
2 2
xx
jijiji
ji
Similarly, under the same assumptions, in the y direction:
21,,1,
,2
2 2
yy
jijiji
ji
02121
1,,1,2,1,,12
jijijijijiji
yx
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method
i,j
0:2'2
2
2
22
yxDinequationsLaplace
02121
1,,1,2,1,,12
jijijijijiji
yx
If for convenience we choose a square mesh,
so thatx = y :
04 ,1,1,,1,1 jijijijiji
or
1,1,,1,1,
4
1 jijijijiji
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method
0:2'2
2
2
2
2
yxDinequationsLaplace
02121
1,,1,2,1,,12
jijijijijiji
yx
A five-point computation scheme:
04 ,1,1,,1,1 jijijijiji
or
1,1,,1,1,
4
1 jijijijiji
If for convenience we choose a square mesh,
so thatx = y :
4
1
11
1
i 1 i +1i
j 1
j +1
j
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method Example
x
y
2 2
2
2
4 4
V= 100 V
V= 0
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method Example
x
y V= 100 V V= 100 V
V= 0
V= 0
y
x
0
x
V
0
y
V
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method Example
x
y V= 100 V
V= 0
y
x
tube
tube
slice
slice
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method Example
x
y V= 100 V V= 100 V
V= 0
V= 0
y
x
0
x
V
0
y
V
V= 100 V
V= 0
1 2 3 4
5
2
4Fictitious nodes
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
1 2 3 4
5
2
4
040100122 VVV
040100231 VVV
040100342
VVV
04100100453 VVV
041000544 VVV
100
200
100
100
100
42000
14100
01410
00141
00024
5
4
3
2
1
V
V
V
V
V
Ax=b
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
1 2 3 4
5
2
4
040100122
VVV
040100231 VVV
040100342
VVV
04100100453
VVV
041000544
VVV
1,1,,1,1,41
jijijijiji VVVVV
4
1
11
1
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
1 2 3 4
5
2
4
040100122 VVV
040100231 VVV
040100342
VVV
04100100453 VVV
041000544 VVV
1,1,,1,1,
4
1 jijijijiji VVVVV
10024
121
VV
1004
1312
VVV
1004
1423
VVV
2004
1534
VVV
10024
145
VV
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
1 2 3 4
5
2
4 10024
121
VV
1004
1312
VVV
1004
1423
VVV
2004
1534
VVV
10024
145
VV
v1=0.0; v2=0.0; v3=0.0; v4=0.0; v5=0.0; iter=0;
A simple iterative scheme:
.
.
do {
v1 = 0.25*(2*v2 +100);
v2 = 0.25*(v1 + v3 +100);
v3 = 0.25*(v2 + v4 +100);
v4 = 0.25*(v3 + v5 +200);
v5 = 0.25*(2*v4 +100);
iter += 1;
printf (%3d %6.2f %6.2f %6.2f %6.2f
%6.2f\n,iter,v1,v2,v3,v4,v5):
} while (iter < 20);
.
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
An improved iterative scheme:
introduce an array v[i,j] sweep the nodes in i (1 toM) andj (1 to N)
i
j
identify the Neumann boundary nodes
and apply the formula with fictitious nodes
1,1,,1,
24
1 jijijiji VVVV
1,,1,1, 241 jijijiji VVVV
identify the special nodes
and apply special formulae
N
M1
1
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
An improved iterative scheme:
introduce an array v[i,j]
sweep the nodes in i (1 toM) andj (1 to N)
i
j
identify the Neumann boundary nodes
and apply the formula with fictitious nodes
3,2,12,1, 41
iiiji VVVV
1004
11,1,,1,
jMjMjMji VVVV
identify the special nodes
and apply special formulae
identify the nodes next to a boundary
and apply relevant formulae
1004
11,,1,1,
NiNiNiji VVVV
M
N
11
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
An improved iterative scheme:
introduce an array v[i,j] sweep the nodes in i (1 toM) andj (1 to N)
i
j
identify the Neumann boundary nodes
and apply the formula with fictitious nodes
identify the special nodes
and apply special formulae
identify the nodes next to a boundary
and apply relevant formulae
M
N
11
apply five-point formula to all other nodes 1,1,,1,1,
4
1 jijijijiji VVVVV
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
An improved iterative scheme:
introduce an array v[i,j]
sweep the nodes in i (1 toM) andj (1 to N)
i
j
identify the Neumann boundary nodes
and apply the formula with fictitious nodes
identify the special nodes
and apply special formulae
identify the nodes next to a boundary
and apply relevant formulae
M
N
11
apply five-point formula to all other nodes Can we accelerate
the convergence?
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method Example
i
j i,j
kji
k
ji
k
ji
k
ji
k
jiVVVVV
1,,1
)1(
1,
)1(
,1
)1(
,
4
1
New value availableOld value
New Old
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method Example
i
j i,j
kji
k
ji
k
ji
k
ji
k
ji VVVVV 1,,1)1(
1,
)1(
,1
)1(
,4
1
k
ji
k
ji
k
ji VV ,)1(
,
)1(
,
where k is an iteration count
Let
where is a residual
or)1(
,,
)1(
,
kjik
ji
k
ji VV
Let
)1(,,
)1(,
kji
kji
kji VV 1 < 2
kji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji VVVVVVV ,1,,1)1(
1,
)1(
,1,
)1(
,4
1
kjik jikjik jikjikji VVVVVV 1,,1)1( 1,)1( ,1,)1(,4
11
or
five-point formulaover-relaxation
SORSuccessive Over-Relaxation
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
An improved iterative scheme:
introduce an array v[i,j] sweep the nodes in i (1 toM) andj (1 to N)
i
j
identify the Neumann boundary nodes
and apply the formula with fictitious nodes
identify the special nodes
and apply special formulae
identify the nodes next to a boundary
and apply relevant formulae
M
N
11
apply five-point formula to all other nodes
apply SOR to accelerate convergence
kjik
ji
k
ji
k
ji
k
ji VVVVV 1,,1)1(
1,
)1(
,1
)1(
,4
1
k
ji
k
ji
k
ji VV ,)1(
,
)1(
,
)1(
,,
)1(
,
kjikjikji VV 1 < 2
choose optimum value for
Choosing opt problem dependent
schemes for estimating opt
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
An improved iterative scheme:
introduce an array v[i,j]
sweep the nodes in i (1 toM) andj (1 to N)
i
j
identify the Neumann boundary nodes
and apply the formula with fictitious nodes
identify the special nodes
and apply special formulae
identify the nodes next to a boundary
and apply relevant formulae
M
N
11
apply five-point formula to all other nodes
apply SOR to accelerate convergence
k
ji
k
ji
k
ji
k
ji
k
jiVVVVV
1,,1
)1(
1,
)1(
,1
)1(
, 4
1
k
ji
k
ji
k
jiVV
,
)1(
,
)1(
,
choose optimum value for
monitor errors to terminate iterations%)1..(100
max
max
)1(
,,
gecriterionlocalV
k
jiji
local
%)5.0..(100
1
max
,
)1(
,
gecriterionglobalV
NM ji
k
ji
global
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Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference method ExampleV= 100 V
V= 0
1 2 3 4
5
2
4
100
200
100
100
100
42000
14100
01410
00141
00024
5
4
3
2
1
V
V
V
V
V
Other methods of solvingAx=b exist
SOR particularly easy to implement in FD
Matrix sparsity may be explored
Matrix symmetry may be an issue
The pros and cons of the FD scheme
intuitive and easy to implement
SOR natural for a five point scheme writing a code relatively straightforward
FD grid difficult to fit into practical devices
curved boundaries awkward to handle mesh grading difficult
boundary conditions require special schemes
Computing and CAD IIProfessor Jan K. Sykulski
The finite-difference methodThe FD scheme for diffusion and wave problems (time derivatives)
Diffusion equation in 1D: Wave equation in 1D:t
V
x
V
2
2
The time derivatives can be obtained with the aid of Taylors series:
t
VV
t
V kiki
,1,
2
2
2
2
t
V
x
V
21,,1,
2
2 2
t
VVV
t
V kikiki
tVV
xVVV kikikikiki
,1,
2
,1,,1 2 2
1,,1,
2
,1,,1 22t
VVVx
VVV kikikikikiki
A new value of the potential (at a new time instant k+1)
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Computing and CAD IIProfessor Jan K. Sykulski
The FD method for diffusion problem
t
x
time
space
t= 0
x = 0 x =L
Boundary
and initial conditions specified
tVV
x
VVV kikikikiki
,1,2
,1,,12
FD scheme:
i,k+1
i1,k i+1,ki,k
xt
Computing and CAD IIProfessor Jan K. Sykulski
The FD method for wave problem
t
x
time
space
t= 0
x = 0 x =L
Boundary
and initial conditions specified
FD scheme:
i,k+1
i1,k i+1,ki,k
xt
21,,1,
2
,1,,122
t
VVV
x
VVV kikikikikiki
i,k1
Stability condition:
1
12
x
t
To achieve good accuracy
x must be small.
But this necessitates very small
time steps tand thus the scheme
may be computationally inefficient.
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Computing and CAD IIProfessor Jan K. Sykulski
The FD method for diffusion problem
t
x
time
space
t= 0
x = 0 x =L
Boundary
and initial conditions specified
tVV
x
VVV kikikikiki
,1,2
,1,,12
FD scheme:
i,k+1
i1,k i+1,ki,k
xt
Stability condition:
21
2
x
t
To achieve good accuracy
x must be small.
But this necessitates very small
time steps tand thus the scheme
may be computationally inefficient.
Explicit scheme.
Computing and CAD IIProfessor Jan K. Sykulski
The FD method for diffusion problemCrank & Nicolson (1947) introduced a scheme which is always convergent.
is replaced by the mean of FD representations on time rows kandk+1:2
2
x
V
tVV
x
VVV
x
VVV kikikikikikikiki
,1,2
,1,,1
2
1,11,1,1 22
2
1
so that kikikikikiki rVrVrrVrVVrrV ,1,,11,11,1,1 2222 where 2xt
r
We have 3 unknowns on the left and three known values on the right.
With N nodes for each time row, N simultaneous equations need to be solved at each time step.
Implicit scheme.
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Computing and CAD IIProfessor Jan K. Sykulski
The FD method for diffusion problemA weighted average approximation:
tVV
x
VVVRx
VVVRkikikikikikikiki
,1,
2
,1,,1
2
1,11,1,1 2)1(2
where 0 R 1 R = 0 explicit scheme
R = Crank - Nicolson
R = 1 fully implicit
backward scheme
ImplicitExplicit Fully implicit
Computing and CAD IIProfessor Jan K. Sykulski
The FD method for diffusion problemA weighted average approximation:
Rxt
r212
12
tVV
x
VVVR
x
VVVR
kikikikikikikiki
,1,2
,1,,1
2
1,11,1,1 2)1(
2
where 0 R 1 R = 0 explicit scheme
R = Crank - Nicolson
R = 1 fully implicit
backward scheme
ImplicitExplicit Fully implicit
0 R < stable if
R 1 unconditionally stable
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Computing and CAD IIProfessor Jan K. Sykulski
The FD method for 2D problems
Diffusion equation in 2D:t
V
y
V
x
V
2
2
2
2
i.e. solution on a rectangle.
x
y
x
y
t
time
t
Computing and CAD IIProfessor Jan K. Sykulski
The FD method for 2D problemsExplicit scheme computationally laborious as very small t needed.
x
yt
time
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Computing and CAD IIProfessor Jan K. Sykulski
The FD method for 2D problems
Crank- Nicolson MN simultaneous equations need to ne solved on each time step.
x
yt
time
Computing and CAD IIProfessor Jan K. Sykulski
The FD method for 2D problemsAlternating Direction Implicit (ADI) method 25 times faster than explicit
x
yt
time
7 times faster than Crank-Nicolson
and then the other way round. This process is repeated sequentially.
One second derivative, say in x, replaced by implicitscheme, whereas the other one by explicit.
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Computing and CAD IIProfessor Jan K. Sykulski
The pros and cons of the FD scheme
intuitive and easy to implement
SOR natural for a five point scheme writing a code relatively straightforward
FD grid difficult to fit into practical devices
curved boundaries awkward to handle mesh grading difficult
boundary conditions require special schemes
FD method: Alternative formulations
Other issues:
material boundaries and interfaces conducting and magnetic materials
current carrying conductors
induced eddy currents magnetic non-linearity
Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulationsHelmholtz equation in x,y, coordinates
sJAjy
A
yx
A
x
where reluctivity
1 and(x,y).
x
y
1
2
3
4
0
JNE ,NE,NEJNW ,NW,NW
JSW ,SW,SW JSE ,SE,SE
a
bcd
e
f g h
h1h3
h4
h2
NW
SE
AjJx
A
y
Acurl sz
jior its curl form
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Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulationsApply Stokes theorem
x
y
1
2
3
4
0
JNE ,NE,NEJNW ,NW,NW
JSW ,SW,SW JSE ,SE,SE
a
bcd
e
f g h
h1h3
h4
h2
NW
SE
AjJx
A
y
Acurl sz
ji
lFsF ddcurland perform surface integration over the cell bdfg:
dydxAjJdx
A
y
AN
S
E
W
s
lji
where E=h1/2, N=h2/2, W=h3/2 and S=h4/2.
On ab the contribution to the integral is:
dyx
Ady
x
A
y
A
Ex
N
NE
N
00
jji
and the following approximation may be used:
)(1
01 NOh
AAxA
Ex
and thus:
1
01
0h
AANdy
x
A
y
A NEN
jji
Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulations
x
y
1
2
3
4
0
JNE ,NE,NEJNW ,NW,NW
JSW ,SW,SW JSE ,SE,SE
a
bcd
e
f g h
h1h3
h4
h2
NW
SE
dydxAjJdx
A
y
AN
S
E
W
s
ljiOn ab the contribution
to the integral is:
The complete line integral is therefore:
lji dx
A
y
A
04321 A 44332211 AAAA
where:
1
1h
SN SENE
2
2h
WENWNE
3
3h
SN SWNW
4
4h
WE SWSE
1
01
0h
AANdy
x
A
y
A NEN
jji
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Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulations
x
y
1
2
3
4
0
JNE ,NE,NEJNW ,NW,NW
JSW ,SW,SW JSE ,SE,SE
a
bcd
e
f g h
h1h3
h4
h2
NW
SE
dydxAjJdx
A
y
AN
S
E
W
s
lji
The complete line integral:
For the surface integral of the RHS consider the first quadrant:
dydxAjNEJdydxAjJN E
NENE
N E
s 0 00 0
0
0 0
ANEdydxA
N E
200
0 KOy
Ay
x
AxAA
A double Taylor expansion about node 0yields:
and neglecting higher terms:
4433221104321 AAAAAdx
A
y
A
lji
Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulations
x
y
1
2
3
4
0
JNE ,NE,NEJNW ,NW,NW
JSW ,SW,SW JSE ,SE,SE
a
bcd
e
f g h
h1h3
h4
h2
NW
SE
dydxAjJdx
A
y
AN
S
E
W
s
ljiAssembling all the pieces
and expressing in terms of A0:
04321
0443322110
Qj
IAAAAA
where
SWSENWNE SWJSEJNWJNEJI 0
SWSENWNE
SWSENWNEQ 0
and
1
1h
SN SENE
2
2h
WENWNE
3
3h
SN SWNW
4
4h
WE SWSE
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Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulations
Assembling all the pieces
and expressing in terms of A0:
where
SWSENWNE SWJSEJNWJNEJI 0
SWSENWNE SWSENWNEQ 0
and
11 h
SN SENE
2
2 h
WE NWNE
3
3h
SNSWNW
4
4h
WE SWSE
Assume: h1 = h2 = h3 = h4
J = 0= 0
= constant
1 = 2 = 3= 4
Q0 = 0
I0 = 0
04321
0443322110
Qj
IAAAAA
4
43210
AAAAA
a five point computational scheme
Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulationsGenerally:
04321
0443322110
Qj
IAAAAA
where
SWSENWNE SWJSEJNWJNEJI 0
SWSENWNE
SWSENWNEQ 0
and
1
1h
SN SENE
2
2h
WENWNE
3
3h
SN SWNW
4
4h
WE SWSE
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Computing and CAD IIProfessor Jan K. Sykulski
FD method: Alternative formulationsThree-dimensional FD solution of Laplacian and Poissonian problems
(O) outside
(I) inside(E) east
(W) west
(N) north
(S) south
1
2
3
45
6
0
div B = 0 and H = grad V
div (H) = div ( grad V) = 0
Apply Gauss theorem
to the small volume surrounding node 0.
sFF ddvdiv
04
1
4
1
4
4
1
4
1
4
4
2
1
4
4
2
1
4
)303101
6
06
303101
5
05
316315
0
04
6530
3
03
316315
0
02
6510
1
01
SESWONEONWO
SEISWINEINWI
SEONEOSEINEI
SEOSWOSEISWI
SWONWOSWINWI
NWONEONWINEI
hrhhrhh
VV
hrhhrhh
VV
hhhhhhr
VV
hhhrh
VV
hhhhhhr
VV
hhhrh
VV
A seven-point computational scheme
to calculate V0 in terms of the potentials
in the six surrounding nodes.
Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodThe pros and cons of the FD scheme
intuitive and easy to implement
SOR natural for a five point scheme
writing a code relatively straightforward
FD grid difficult to fit into practical devices
curved boundaries awkward to handle mesh grading difficult
boundary conditions require special schemes
The FE scheme
less intuitive
much more difficult to implement
writing a code challanging
FE mesh easy to fit into practical devices
curved boundaries simple to handle mesh grading possible and effective
boundary conditions naturally satisfied
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method
The similarities and differences between FD and FE methods
both rely on discrete representation of the solution
a grid or mesh of nodes is needed
for better accuracy a fine grid/mesh is required
both lead to a large system of equations Ax=b
solution of this equation is computationally expensive
local and global errors need to be monitored
visualisation and post-processing challenging because of large amounts of data
But:
FD solution defined on a set of nodes only, FE solution described everywhere
FD grid restricted because of parallel lines, FE mesh flexible
boundary conditions in FE naturally satisfiedNote:
if FD grid and FE mesh matching identical final system of equations
Computing and CAD IIProfessor Jan K. Sykulski
Coarse mesh
-6-4
-20
24
6
-6-4
-20
24
60
5
10
15
20
The finite element method
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Computing and CAD IIProfessor Jan K. Sykulski
Mesh refinement
-6-4
-20
24
6
-6-4
-20
24
60
5
10
15
20
-6-4
-20
24
6
-6-4
-20
24
60
5
10
15
20
-6-4
-20
24
6
-6-4
-20
24
60
5
10
15
20
The finite element method
Computing and CAD IIProfessor Jan K. Sykulski
2D mesh
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Computing and CAD IIProfessor Jan K. Sykulski
3D meshes
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method
02
2
2
22
y
V
x
VV
Consider Laplacian equation in 2D for an electrostatic system
where the electric fieldE is given by
jiE
y
V
x
VVVgrad
The stored field energy:
dVdy
V
x
V
dEdW
2
22
2
2
1
2
1
2
1
2
1
DE
The principle of equilibrium requires that the potential distribution must be such as to
minimise the stored field energy. This minimum-energy principle is mathematically equivalent
to our original differential equation in the sense that that a potential distribution which
satisfies Laplaces equation will also minimise the energy, and vice versa.
where integration is carried out over a 2D region, and is thud taken per unit length/depth.
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method
...22
fyexdxycybxaV
Consider a single element and the following approximating polynomial
We choose as many terms as there are nodes in the element:
First order triangle
Second order triangle
Rectangle
1
2
3
1
2
3
4
56
1 2
34
cybxaV
22fyexdxycybxaV
dxycybxaV
Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
cybxaV
In the three vertices (nodes) the potential is
333
222
111
cybxaV
cybxaV
cybxaV
or
c
b
a
yx
yx
yx
V
V
V
33
22
11
3
2
1
1
1
1
and rearranging
3
2
1
1
33
22
11
11
1
VV
V
yxyx
yx
cb
a
and substituting back yields:
3
2
1
1
33
22
11
1
1
1
1
V
V
V
yx
yx
yx
yxV or
3
1
,i
ii yxVV
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
2
1
1
33
22
11
1
1
1
1
VV
V
yxyx
yx
yxV 3
1
,i
ii yxVV
where
yxxxyyyxyxA
2332233212
1
yxxxyyyxyxA
3113311322
1
yxxxyyyxyxA
1221122132
1
where A is the element area
At the vertices:
12
2
2
1,
1231322332111 A
Ayxxxyyyxyx
Ayx
02
1, 2232322332221 yxxxyyyxyx
Ayx
02
1, 3233322332331 yxxxyyyxyx
Ayx
Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
2
1
1
33
22
11
1
1
1
1
V
V
V
yx
yx
yx
yxV
3
1
,i
ii yxVV
where
yxxxyyyxyxA
2332233212
1
yxxxyyyxyxA
3113311322
1
yxxxyyyxyxA
122112213
2
1
where A is the element area
In general
jiyx jji 0,
ji 1
x
y
1
2
3
1
1
Each function vanishes at all vertices but one,
where it assumes the value of one.
23
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
1
,
i
ii yxVV
e
edSVW
2)(
2
1
We can now associate energy with each element,
and remembering that in 2D this energy will be
taken per unit length/depth we write
where integration is performed over the element area.
Using
3
1
,i
ii yxVV we find
3
1
,i
ii yxVV
so that the element energy becomesj
i j
j
e
ii
eVdSVW
3
1
3
1
)(
2
1
which may be written in a compact form VNVWeTe )()(
2
1
where [V] is the vector of vertex values of potential, the superscript T denotes transposition,
and the 33 square element matrix [N](e) is defined bydSN j
e
i
e
ji )(
,
Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
1
,i
ii yxVV
dSN je
i
e
ji )(
,
VNVW eTe )()(2
1
yxxxyyyxyxA
2332233212
1
yxxxyyyxyxA
3113311322
1
yxxxyyyxyxA
1221122132
1
jiji2332
111
2
1xxyy
Ayx
jiji3113
222
2
1xxyy
Ayx
jiji1221
333
2
1xxyy
Ayx
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8/3/2019 CAD 2011 Slides
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
1
,
i
ii yxVV
dSN je
i
e
ji )(
,
VNVW eTe )()( 21
jiji2332
111
2
1xxyy
Ayx
jiji3113
222
2
1xxyy
Ayx
jiji1221
333
2
1xxyy
Ayx
The scalar product of two vectors, say a andb,
in a Cartesian coordinate system in 2D:
jijiba yxyx bbaa
jjijjiii yyxyyxxx babababa
yyxx baba
223
2
32
)(
1,14
1xxyy
AN
e
31231332
)(
2,14
1xxxxyyyy
AN
e
12232132
)(
3,14
1xxxxyyyy
AN
e
23313213
)(
1,24
1xxxxyyyy
AN
e
231
2
13
)(
2,24
1xxyy
AN
e
12312113
)(
3,24
1xxxxyyyy
AN
e
23123221)(
1,341 xxxxyyyyA
N e
31121321
)(
2,34
1xxxxyyyy
AN
e
212
2
21
)(
3,34
1xxyy
AN
e
Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
1
,i
ii yxVV
VNVW eTe )()(2
1
223
2
32
)(
1,14
1xxyy
AN
e
31231332
)(
2,14
1xxxxyyyy
AN
e
12232132
)(
3,14
1xxxxyyyy
AN
e
23313213
)(
1,24
1xxxxyyyy
AN
e
2312
13
)(
2,241 xxyyA
Ne
12312113
)(
3,24
1xxxxyyyy
AN
e
23123221
)(
1,34
1xxxxyyyy
AN
e
31121321
)(
2,34
1xxxxyyyy
AN
e
212
2
21
)(
3,34
1xxyy
AN
e
This completes the specification for an
arbitrary element in the finite-element mesh.
The total energy associated with the entire
region will be found as the sum of individual
element energies:
elementsAlleWW )(
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8/3/2019 CAD 2011 Slides
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element methodFirst order triangular element
1
2
3
3
1
,
i
ii yxVV
VNVW eTe )()( 21
This completes the specification for an
arbitrary element in the finite-element mesh.
The total energy associated with the entire
region will be found as the sum of individual
element energies:
elementsAll
eWW
)(
When assembling elements we notice that
some nodes will be shared between elements.The global matrix must reflect the way in
which individual elements are linked.
12
3
1
4
3
8
5
6
7
2
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
0 2
1
V= 0 V= 100
0
y
V
0
y
V
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
Node
number x-coordinate y-coordinate Potential
1 1 0 ?
2 1 1 ?
3 2 1 100
4 0 1 0
5 2 0 100
6 0 0 0
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
Node
number x-coordinate y-coordinate Potential
1 1 0 ?
2 1 1 ?
3 2 1 100
4 0 1 0
5 2 0 100
6 0 0 0
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
The element matrices may now be determined.
For each element we need the 33 matrix
based on the coordinates of the vertices andthe size of the element.
223
2
32
)(
1,14
1xxyy
AN
e
31231332
)(
2,14
1xxxxyyyy
AN e
12232132
)(
3,14
1xxxxyyyy
AN
e
23313213
)(
1,24
1xxxxyyyy
AN
e
231
2
13
)(
2,24
1xxyy
AN
e
12312113
)(
3,24
1xxxxyyyy
AN
e
23123221)(
1,34
1xxxxyyyy
AN e
31121321
)(
2,34
1xxxxyyyy
AN e
212
2
21
)(
3,34
1xxyy
AN
e
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
Node
number x-coordinate y-coordinate Potential
1 1 0 ?
2 1 1 ?
3 2 1 100
4 0 1 0
5 2 0 100
6 0 0 0
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
The element matrices may now be determined.
For example, for element A:
01
11
10
2112
1331
3223
yyxx
yyxx
yyxx
and the element area A=0.5.
-
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
Node
number x-coordinate y-coordinate Potential
1 1 0 ?
2 1 1 ?
3 2 1 100
4 0 1 0
5 2 0 100
6 0 0 0
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
The element matrices may now be determined.
For example, for element A:
223
2
32
)(
1,14
1xxyy
AN
A
01
1110
2112
1331
3223
yyxx
yyxxyyxx and the element
area A=0.5.
Thus:
2
101
2
1 22
31231332
)(
2,14
1xxxxyyyy
AN
A
2
11011
2
1
and so on giving the matrix for elementA as
2
1
2
10
211
21
02
1
2
1
AN
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
Node
number x-coordinate y-coordinate Potential
1 1 0 ?
2 1 1 ?
3 2 1 100
4 0 1 0
5 2 0 100
6 0 0 0
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
This process needs to be repeated for each
element of the; however, careful inspection of
elementsB, CandD reveals that, in our
example, all four elements matrices are the
same, i.e.:
This is of course a happy coincidence, or
rather a clever use of proportions and local
node numbering, to save some work.
In general the element matrices
are all different !
2
1
2
10
2
11
2
1
02
1
2
1
DCBANNNN
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
A global matrix may now be assembled.
Initially all entries are zero.
000000
000000
000000
000000
000000
000000
N
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
2
1
2
10
2
11
2
1
0
2
1
2
1
DCBANNNN
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
A global matrix may now be assembled.
For element A:
2
10000
2
1000000
000000
000000
00002
1
2
121000
211
N
222126
121116
626166
NNN
NNN
NNN
NA
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
2
1
2
10
2
11
2
1
02
1
2
1
DCBANNNN
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
A global matrix may now be assembled.
For element B:
2
1
2
10
2
100
2
1000000
2
10102
10
000000
002
10
2
1
2
1
2
12
1000
2
11
N
222426
424446
626466
NNN
NNN
NNN
NB
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
2
1
2
10
2
11
2
1
0
2
1
2
1
DCBANNNN
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
A global matrix may now be assembled.
For element C:
2
1
2
10
2
100
2
1000000
2
1010
2
10
0002
1
2
10
002
1
2
11
2
1
2
1
2
1
2
12
1
0002
1
2
1
2
1
1
N
333231
232221
131211
NNN
NNN
NNN
NC
2
1
2
10
2
11
2
1
02
1
2
1
DCBANNNN
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
Element Vertex 1 Vertex 2 Vertex 3
A 6 1 2
B 6 4 2
C 1 2 3
D 1 5 3
A global matrix may now be assembled.
For element D:
2
1
2
10
2
100
2
1
0102
10
2
12
10102
10
02
10
2
1
2
1
2
10
002
1
2
11
2
1
2
1
2
1
2
12
1
2
100
2
1
2
1
2
1
2
11
N
333531
535551
131511
NNN
NNN
NNN
ND
2
1
2
10
2
11
2
1
02
1
2
1
DCBANNNN
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
105.0005.0
0105.005.0
5.00105.00
05.0015.00
005.05.021
5.05.00012
N
2
1
2
10
2
100
2
1
0102
10
2
12
1010
2
10
02
10
2
1
2
1
2
10
002
1
2
112
1
2
1
2
1
2
1
2
1
2
100
2
1
2
1
2
1
2
11
N
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
105.0005.0
0105.005.0
5.00105.00
05.0015.00
005.05.021
5.05.00012
N
To minimise the total energy we must differentiate
with respect to a typical value of Vkand equate to
zero:
VNVW T2
1
0
kV
W
where k refers to node numbers in the global
numbering system. However, some nodes will
have a prescribed potential (e.g. on the
boundary) and we can differentiate only with
respect to free nodes.
0][
][
][][
][][][][
][
p
f
pppf
fpffT
p
T
f
kfkV
V
NN
NNVV
VV
W
Differentiation with respect to the free potentials
results in the following matrix equation:
0][
][]][][[
p
ffpff
V
VNN
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
0][
][]][][[
p
ffpff
V
VNN
105.0005.0
0105.005.0
5.00105.00
05.0015.00
005.05.021
5.05.00012
N
]][[]][[ pfpfff VNVN
]][[][][1
pfpfff VNNV
Formal solution:
In our example:
502
502
21
21
VV
VV
5021
VV
6
5
43
2
1
005.05.0
5.05.000
21
12
V
V
V
V
V
V
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
234
56
105.0005.0
0105.005.0
5.00105.00
05.0015.00
005.05.021
5.05.00012
N
]][[]][[ pfpfff VNVN
]][[][][1
pfpfff VNNV
Final equation:
Formal solution:
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example
x
y
V= 0 V= 100
1 2
3
11
1
2
2
2 3
3
3
A
B CD
1
2 34
56
105.0005.0
0105.005.0
5.00105.00
05.0015.00
005.05.021
5.05.00012
N
]][[]][[ pfpfff VNVN
bxA
Final equation:
This is in the form:
Comments:
[Nff] (orA) may be a very large matrix
b is given by sources (boundary conditions)
A is normally sparse (we cannot see this here)
A is often symmetrical
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example 2Consider a topologically regular mesh:
1
2
3
4
5
21
22
23
24
25
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
All these meshes are topologically identical!
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example 2
1
2
3
4
5
121
22
23
24
25
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20The global matrix may be partitioned as follows:
][][][
][][][
][][][
][
pppfpp
fpfffp
pppfpp
NNN
NNN
NNN
N
The [Nff] sub-matrix is of particular interest:
6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
6 0 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 010 0 0 0 0 0 0 0 0 0 0 0
11 0 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0
13 0 0 0 0 0
14 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0
16 0 0 0 0 0 0 0 0 0 0 0
17 0 0 0 0 0 0 0 0 0
18 0 0 0 0 0 0 0 0 0
19 0 0 0 0 0 0 0 0 0
20 0 0 0 0 0 0 0 0 0 0 0
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Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example 2
1
2
3
4
5
121
22
23
24
25
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20The non-zero elements may be numbered:
6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
6 0 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0
11 0 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0
13 0 0 0 0 0
14 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0
16 0 0 0 0 0 0 0 0 0 0 0
17 0 0 0 0 0 0 0 0 0
18 0 0 0 0 0 0 0 0 0
19 0 0 0 0 0 0 0 0 0
20 0 0 0 0 0 0 0 0 0 0 0
15274553
1426354452
1325344351
1224334250
113241
10234049
922313948
821303847
720293746
62836
519
418
317
216
1
The matrix is symmetrical.
Only 53 values need to be stored(instead of 1515=225)
Computing and CAD IIProfessor Jan K. Sykulski
The finite element method Example 2
1
2
3
4
5
121
22
23
24
25
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20The non-zero elements may be numbered:
15274553
1426354452
1325344351
1224334250
113241
10234049
922313948
821303847
720293746
62836
519
418
317
216
1
For example:
272625242322212019181716
151413121110987654321
4553
354452
344351
334250
3241
4049
313948
303847
293746
2836
+
60 values stored rather than 1515=225.
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Computing and CAD IIProfessor Jan K. Sykulski
The art of sparse matrices Example
1
2
3 4
56
7
8
9
The matrix:
Due to symmetry, half of this 99 matrix needs to be stored and there are 25 non-zero elements.
A simple storage scheme places the non-zero elements in a linear array and complements this
array with two integer arrays which store the corresponding row and column numbers.
11 21 51 22 32 42 52 33 43 44 54 74 84 55 65 75 66 76 96 77 87 97 88 98 99
1 2 5 2 3 4 5 3 4 4 5 7 8 5 6 7 6 7 9 7 8 9 8 9 9
1 1 1 2 2 2 2 3 3 4 4 4 4 5 5 5 6 6 6 7 7 7 8 8 9
elements
row
column
11 21 51 22 32 42 52 33 43 44 54 74 84 55 65 75 66 76 96 77 87 97 88 98 99
1 2 5 2 3 4 5 3 4 4 5 7 8 5 6 7 6 7 9 7 8 9 8 9 9
elements
row
1 4 8 10 14 17 20 23 25 pointers identifying the beginning points of columns.
99989796
888784
77767574
6665
55545251
444342
3332
2221
11
Computing and CAD IIProfessor Jan K. Sykulski
The art of sparse matrices Example
1
2
3 4
56
7
8
9
The matrix:
A profile storage scheme:
11 21 22 32 33 42 43 44 51 52 . 54 55 65 66 74 75 76 77 84 . . 87 88 96 97 98 99
1 1 2 2 1 5 4 4 6
1 3 5 8 13 15 19 24 28
column number of the beginning of each row.
the diagonal element which forms the end of each row.
Strictly speaking the second integer array is not necessary as the ending of a row is always
on the main diagonal. However, the addressing functions become much more complicated.
99989796
888784
77767574
6665
55545251
444342
3332
2221
11
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b
All numerical schemes (except TAS) lead to Ax = b
FD and FE formulations result in very large systems of equations
The A matrix is very sparse
A is often symmetrical
Efficient storage systems are required
Fast and accurate computation is needed
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleElimination method
Gaussian elimination:
addand3
1
addand3
2
222
1132
1223
321
321
321
xxx
xxx
xxx
addand7
4
1874
2177
1223
32
32
321
xx
xx
xxx
Forward elimination
23 x
12 x
4221
2177
1223
3
32
321
x
xx
xxx 31 x Backward substitution
Two steps:
1
2
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleElimination method
Gaussian elimination in matrix notation:
2
11
12
122
321
213
3
2
1
x
x
x Compose anaugmented matrix
and apply elementary
raw transformations:
Elementary Row and Column operations (transformations):1. Interchanging two rows or columns,2. Adding a multiple of one row or column to another,3. Multiplying any row or column by a nonzero element.
2122
11321
12213
:bA
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleElimination method
Gaussian elimination in matrix notation:
2
11
12
122
321
213
3
2
1
x
x
x Compose anaugmented matrix
and apply elementary
raw transformations:
2122
11321
12213
:bA
2122
11321
12213
18740
21770
12213
422100
21770
12213
13
12
)2(3
)1(3
RR
RR
23 47 RR
422100
210210
16802163
422100210210
1890063
2100
1010
3001
23
13
3
212
RR
RR
12 RR
21/1
21/1
63/1
2
1
3
xHence:
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleElimination method
2122
1132112213
18740
2177012213
422100
21770
12213
13
12
)2(3
)1(3
RR
RR
23 47 RR
422100
210210
16802163
422100210210
1890063
23
13
3
212
RR
RR
12 RR
21/1
21/1
63/1
2100
1010
3001
Forward elimination
Backward substitutionNotes: watch for zeros on the diagonal interchanging rows
multiplications lead to very large numbers subtractR1ai1/a11 fromRi, etc
rearrange equations so that largest coefficient is on diagonal at each step pivoting scaling dividing each row by the magnitude of the largest coefficient
so that all coefficients have the same order of magnitude reduce round-off errors
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = bElimination method
Steps:1. Augmentnn matrix into n(n+1) matrix andscale.
2. Change rows so that a11 has the largest magnitude.
3. SubtractR1ai1/a11from rows 2 to n.
4. Repeat 2 and 3 for subsequent rows:
Rjaij/ajj for rows i=j+1 to n.
5. Solvexn=bn/ann6. Solve xn1,xn2,xn3 x2,x1 where
nnnnnn
n
n
b
b
b
x
x
x
aaa
aaa
aaa
2
1
2
1
21
22221
11211
nnnnn
n
n
baaa
baaa
baaa
21
222221
111211
nnn
n
n
ba
baa
baaa
''00
'''0
''''
2222
111211
n
ij
jijiii
i xaba
x
1
'''1
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b A = LUTriangular decomposition
100
101
0
00
2
112
21
2221
11
21
22221
11211
n
n
nnnnnnnn
n
nuuu
lll
lll
aaa
aaaaaa
LUA
nnnn
n
n
lll
ull
uul
21
22221
11211
1
2
1
1
,...,2,1,
j
k
niijkjikijij ulal
njjil
ula
uii
i
k
kjikij
ij ,...,3,2,
1
1
111 ii aljFor
11
1
11
111
a
a
l
auiFor
jjj
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b A = LUTriangular decomposition
100
10
1
0
00
2
112
21
2221
11
21
22221
11211
n
n
nnnnnnnn
n
n
u
uu
lll
ll
l
aaa
aaa
aaa
LUA
nnnn
n
n
lll
ull
uul
21
22221
11211
1
1
,...,2,1,
j
k
niijkjikijij ulal
njjil
ula
uii
i
k
kjikij
ij ,...,3,2,
1
1
The RHS ( ofAx = b ):
nil
blb
bl
bb
ii
i
k
kiki
i ,...,3,2,
'
''
1
1
11
11
Backward substitution:
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleTriangular decomposition
bAx
122
321213
A
2
1112
b
100
1103/23/11
13/42
03/71003
1
1
,...,2,1,
j
k
niijkjikijij ulal
njjil
ula
u ii
i
k
kjikij
ij ,...,3,2,
1
1
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleTriangular decomposition
bAx
122
321
213
A
2
11
12
b
100
110
3/23/11
13/42
03/71
003
2
3
4
'b
nil
blb
bl
bb
ii
i
k
kiki
i ,...,3,2,
'
''
1
1
11
11
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleTriangular decomposition
bAx
122
321213
A
2
1112
b
100
1103/23/11
13/42
03/71003
2
34
'b
2100
3110
43/23/11
2
121313
313/123/243/13/24
3
32
231
x
xx
xxx
Augmenting b to U gives:
Getting b is in fact:
213/42
1103/71
12003
:bL
and forward substitution yields:
21/33/4422'
33/7/41113/7/'111'
43/12'
3
12
1
b
bb
b
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleTriangular decomposition
bAx
122
321
213
A
2
11
12
b
100
110
3/23/11
13/42
03/71
003
2
3
4
'b
2100
3110
43/23/11
2
121313
313/123/243/13/24
3
32
231
x
xx
xxx
213/42
1103/71
12003
:bL 21/33/4422'
33/7/41113/7/'111'
43/12'
3
12
1
b
bb
b
Forward substitution
Backward substitution
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b ExampleTriangular decomposition
bAx
122
321213
A
2
1112
b
100
1103/23/11
13/42
03/71003
2
34
'b
2100
3110
43/23/11
':bU
2
121313
313/123/243/13/24
3
32
231
x
xx
xxx
213/42
1103/71
12003
:bL 21/33/4422'
33/7/41113/7/'111'
43/12'
3
12
1
b
bb
b
First find b from Lb=b byforward substitution, then
Compute x from Ux=b bybackward substitution
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = bCholesky decomposition
If matrix A is symmetric: aij = aji
andpositive definite: TA > 0 for all vectors
(in other words A has all positive eigenvalues)
then a Cholesky decomposition (which is twice as fast a LU) gives:
TLLA
1
1
2
i
kikiiii LaL
NiijLLaL
L
i
k
jkikijii
ji ,...,2,11
1
1
Cholesky decomposition is extremely stable numerically, even without any pivoting.
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Computing and CAD IIProfessor Jan K. Sykulski
Errors ExampleConsider
2
2
01.199.0
99.001.1
y
x with obvious solution x = 1y = 1
Graphically:
x x
y y
?
Ill conditioned problems
Computing and CAD IIProfessor Jan K. Sykulski
Errors Example 2Consider
66.2
35.2
880.0
47.154.083.0
78.156.033.4
53.205.102.3
3
2
1
x
x
x
with a solution
1
2
1
x
Gaussian elimination, using pivoting and three digits rounded gives:
00962.000362.000
65.677.344.10
23.778.156.033.4
66.2
35.2
880.0
x?
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Computing and CAD IIProfessor Jan K. Sykulski
Errors Example 3
Consider
1440.0
8642.0
1441.02161.0
8646.02969.1
2
1
x
x
Let
4870.0
9911.0x be assumed to be a possible solution.
Calculate a residual vector:
8
8
10
10xAbr
The actual solution is:
2
2x and not a single figure in is meaningful!x
The size of the residual vector may not be a sufficient indication of the error in x !
Computing and CAD IIProfessor Jan K. Sykulski
Matrix and vector norms
Anorm is a measure of the magnitude (not size !) of a matrix.
A norm must satisfy the following conditions:
BAAB
BABA
AkkA
AifonlyandifAandA
.4
.3
.2
000.1
Errors Norms
n
j
ijni
aA
11max
2/1
1 1
2
m
i
n
j
ijeaA
For example
maximum row-sum norm
Euclidean norm
Example
547424
745
16A
69.14e
A
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Computing and CAD IIProfessor Jan K. Sykulski
Condition numbers
Errors Condition numbers
1 AAAcond
Let and where is an error and is an approximate solution.
A residual may be defined as
It can be shown that
The relative error can be as great as the relative residual times the condition number!
Let (where ).
It can be shown that
The error can be as large as relative errors in coefficients of A times the condition number!
What can we do? estimate condition number (e.g. LINPACK estimate)
improve the solution through iterative refinement
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b Iterative methodsGaussSeidel iteration
1. Rearrange the rows so that diagonal elements are dominating
2. Assume initial vector of solution
3. Apply
N
ij
kj
ii
iji
j
kj
ii
ij
ii
iki x
a
ax
a
a
a
bx
1
)(1
1
)1(1
SOR iteration 21
1 )(1
1
)1()(1
wherexaxab
axx
N
ij
kjij
i
j
kjiji
ii
ki
ki
Example:
1
1
1
1
1
1
1
1
4111
1411
1141
1114
xgivingx
1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
No of iterations
Error < 10524 18 13 11
min14 18 24 35 55 100
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Computing and CAD IIProfessor Jan K. Sykulski
x
y V= 100 V V= 100 V
V= 0
V= 0
y
x
0
x
V
0
y
V
Solving Ax = b SOR
1. GaussSeidel iterations
( = 1)
2. SOR withopt = 1.885
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b SOR = 1 opt = 1.885
Iteration count: 4
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b SOR = 1 opt = 1.885
Prescribed accuracy achieved
in 106iterations.
Iteration count: 106
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b SOR = 1 opt = 1.885
Iteration count: 690
Prescribed accuracy achieved
in 106iterations.
Prescribed accuracy achieved
in 690 iterations.
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b SORAssume a linear initial solution: Final solution:
Prescribed accuracy achieved
in51 iterations(instead of106when a zero
solution was initially used).
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b SORAssume a linear initial solution: Final solution:
5 iterations needed to keep the local error below 1%
24 iterations needed to keep the local error below 0.2%
51 iterations needed to keep the local error below 0.05%
81 iterations needed to keep the local error below 0.01%
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b Conjugate Gradient MethodConjugate Gradient Method
Minimise which leads to a condition: bxxAxx TTf 21 0 bxAf
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b Conjugate Gradient MethodConjugate Gradient Method
Minimise which leads to a condition: bxxAxx TTf 2
10 bxAf
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Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b Conjugate Gradient MethodConjugate Gradient Method
Minimise which leads to a condition: bxxAxx TTf 21 0 bxAf
Computing and CAD IIProfessor Jan K. Sykulski
Solving Ax = b Comparison of methodsIncomplete Cholesky Conjugate Gradient (ICCG) Method
Comparison of methods
Test problem:
If the coefficient matrix is symmetric, positive definite and sparse:
incomplete Cholesky preconditioning
conjugate gradient iterative solution
Poissons equation over a unit square
uniform mesh of first order triangles with n points
nnodes
mbandwidth
Gausselimination
Gauss-Seidel SOR ICCG
m2n mniter n lognoperation
count (time)
36 8 1 2 1 1 time inarbitrary
units441 23 33 619 123 28
961 33 152 - 536 70
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Computing and CAD IIProfessor Jan K. Sykulski
A typical CAD system for electromagnetics
Pre-processor
Geometrical data
Material propertiesExcitation sourcesBoundary conditions
Mesh
Solver Ax=b
Post-processorField plots, forces,stored energy,losses, errors, etc
Data
files
Resultsfiles
CAD
Designer
From design system Model
Back to design system Solution
Computing and CAD IIProfessor Jan K. Sykulski
Coinrecognitionsensor
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Computing and CAD IIProfessor Jan K. Sykulski
Multiphysics problems
The current and resultant Joule heating in an electric switch contactare modelled as the switch is actuated. Mechanical, thermal andcurrent flow are modelled using direct coupled field elements.
Computing and CAD IIProfessor Jan K. Sykulski
Optimization techniques
Deterministic Stochastic
Always follows same
path from same initial
conditions
Finds local minimum
Fast: 5 to 100 evaluations
Initial conditions do not
determine path of
optimization
Attempt to find globalminimum
Slow: hundreds or thousandsof evaluations
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Computing and CAD IIProfessor Jan K. Sykulski
Local
Global
Start
Optimization techniques
Deterministic Stochastic
Global
Local
Computing and CAD IIProfessor Jan K. Sykulski
Optimization techniques
Deterministic Stochastic
DirectSearch
IndirectSearch
Simulatedannealing
Evolutionstrategy
Rosenbrocks
methodPatternsearch
Simplexmethod
Powells
conjugatedirection
...
Steepest decent
Newtons method
Variable-metricprocedure
Conjugategradient
Trust regionmethod
...
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Computing and CAD IIProfessor Jan K. Sykulski
Single-minimum objective function
Optimization
Computing and CAD IIProfessor Jan K. Sykulski
Optimization
Multiple-minima objective function
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Computing and CAD IIProfessor Jan K. Sykulski
C-core shaped magnet
100351
BBBmaxF i
,iC
DE Differential Evolution
ES Evolution Strategy
GBA Gradient Based Method
Method Starting Optimum n
DE1 9 random 0.0803 720
DE2 13 random 0.0704 881
ES 0.7532 / 0.4344 / 0.6411 0.0642 450
GBA 0.7532 0.0855 188
ES/DE/MQ 0.7532 0.0718 118
Computing and CAD IIProfessor Jan K. Sykulski
Magnetiser
59
1
2,, BBk
kcalculatedkdesiredf
kokdesired 90sinBB ma x, 591 k
Method Starting Optimum n
DE1 11 random 1.235E-5 987
DE2 11 random 5.423E-5 1035
ES 1.457E-3 1.187E-5 433
ES 9.486E-2 1.318E-4 351
GBA 1.457E-3 1.238E-4 41
GBA 9.486E-2 2.433E-4 281
ES/DE/MQ 1.457E-3 1.961E-5 234
ES/DE/MQ 9.486E-2 2.125E-5 206
NF/GA/SQP 6.570E-5 189
Unconstrained optimisation
Method Optimum N
ES/DE/MQ 1.58E-5 246
NF/GA/SQP 4.65E-5 155
Constrained optimisation
NF Neuro-Fuzzy modelling
GA Genetic Algorithm
SQP Sequential Quadratic Programming
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Computing and CAD IIProfessor Jan K. Sykulski
Yoke
MagnetPole piece
Optimized shimming magnet distribution of MRI system
Simplified axi-symmetric model
3 mm
15 mm
Shimming magnet (Br=0.222 T)
n
si
zozi pPyxBBF )(),()( ,245
1
MM
Objective function:
Computing and CAD IIProfessor Jan K. Sykulski
+Ms -Ms
Changes of shimming magnet distributionduring optimisation
Convergence
Flux distributions
1 iteration
10 iterations
3 iterations
3 iterations
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Computing and CAD IIProfessor Jan K. Sykulski
Pareto Optimisation
F1
F2
UTOPIA
NADIR
DISTOPIA
Objective
domain
searchspace
POF Pareto Optimal Front
POF
F2max
F2min
F1min
F1max
Computing and CAD IIProfessor Jan K. Sykulski
Designer
Company design sheets and priorknowledge
Optimisation and designprograms
Experts on finite element analysis
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Computing and CAD IIProfessor Jan K. Sykulski
Input data
Output data
Materialslibrary
Induction Motor
Design Routines
FEsoftware
Optimisation
Analytical & Control Software
Industrially
oriented systemfor inductionmotor design
MATLAB
FE software
Excel
Computing and CAD IIProfessor Jan K. Sykulski
Hierarchical (three-layer) structure
Knowledge base
Approximate solutions
(e.g. equivalent circuits, semi-empirical, design sheets)
Extensive optimisation Large design space
2D finite element, static or steady-state
Constrained optimisation, coupling Medium design space
3D finite-element, transient
Fine tuning of the design Small design space
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Computing and CAD IIProfessor Jan K. Sykulski
Current and future developments
adaptive meshing
reliable error estimation high speed computing
efficient handling of non-linearity and hysteresis
modelling of new types of materials
linear movement and rotation
combined modelling of fields and circuits
coupled problems (em + stress + temperature, etc)
optimisation
integrated design systems
... Significant progress has been made Tremendous effort continues Much still needs to be done
Commercial software