cad 2011 slides

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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


Resources

Core Resources

Hammond P and Sykulski J K:Engineering Electromagnetism

Oxford University Press 1994

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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


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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Example

I

Ia

a

a2a

slicestubes

2a


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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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:)(


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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Tubes and Slices

I I

2

RRR

ave


Tubes and Slices

Constructionlines

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Tubes and Slices

Coarse

tube/slicelines


Tubes and Slices

Improvedtube/slicelines

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Tubes and Slices

Finite

elementsolution


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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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

...


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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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


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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The finite-difference method

02

boundarytheoneverywheredefined

n

or

Equation specified inside:

A unique solution exists!



i,j

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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



i,j

0:2'2

2

2

2

2


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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i,j

0:2'2

2

2

22


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



0:2'2

2

2

2

2


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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The finite-difference method Example

x

y

2 2

2

2

4 4

V= 100 V

V= 0



x

y V= 100 V V= 100 V

V= 0

V= 0

y

x

0

x

V

0

y

V

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x

y V= 100 V

V= 0

y

x

tube

tube

slice

slice



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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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



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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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



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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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



V= 0


introduce an array v[i,j]

sweep the nodes in i (1 toM) andj (1 to N)

i

j



3,2,12,1, 41

iiiji VVVV

1004

11,1,,1,

jMjMjMji VVVV



identify the nodes next to a boundary

and apply relevant formulae

1004

11,,1,1,

NiNiNiji VVVV

M

N

11

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V= 0



i

j







M

N

11

apply five-point formula to all other nodes 1,1,,1,1,

4

1 jijijijiji VVVVV



V= 0




i

j







M

N

11

apply five-point formula to all other nodes Can we accelerate

the convergence?

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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



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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V= 0



i

j







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



V= 0




i

j







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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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


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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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


The FD method for wave problem

t

x

time

space

t= 0

x = 0 x =L

Boundary


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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The FD method for diffusion problem

t

x

time

space

t= 0

x = 0 x =L

Boundary


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.


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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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


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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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


The FD method for 2D problemsExplicit scheme computationally laborious as very small t needed.

x

yt

time

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The FD method for 2D problems

Crank- Nicolson MN simultaneous equations need to ne solved on each time step.

x

yt

time


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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The pros and cons of the FD scheme


SOR natural for a five point scheme writing a code relatively straightforward




FD method: Alternative formulations

Other issues:

material boundaries and interfaces conducting and magnetic materials

current carrying conductors

induced eddy currents magnetic non-linearity


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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FD method: Alternative formulationsApply Stokes theorem

x

y

1

2

3

4

0



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



x

y

1

2

3

4

0



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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x

y

1

2

3

4

0



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



x

y

1

2

3

4

0



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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Assembling all the pieces

and expressing in terms of A0:

where


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


FD method: Alternative formulationsGenerally:

04321

0443322110

Qj

IAAAAA

where


SWSENWNE

SWSENWNEQ 0

and

1

1h

SN SENE

2

2h

WENWNE

3

3h

SN SWNW

4

4h

WE SWSE

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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.


The finite element methodThe pros and cons of the FD scheme


SOR natural for a five point scheme

writing a code relatively straightforward




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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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


Coarse mesh

-6-4

-20

24

6

-6-4

-20

24

60

5

10

15

20


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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



2D mesh

8/3/2019 CAD 2011 Slides

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3D meshes



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.

8/3/2019 CAD 2011 Slides

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...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


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

8/3/2019 CAD 2011 Slides

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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



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

8/3/2019 CAD 2011 Slides

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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 )(

,



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

8/3/2019 CAD 2011 Slides

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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



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 )(

8/3/2019 CAD 2011 Slides

40/74 4



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


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

8/3/2019 CAD 2011 Slides

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x

y

V= 0 V= 100

1 2

3

11

1

2

2

2 3

3

3

A

B CD

1

234

56

Node


1 1 0 ?

2 1 1 ?

3 2 1 100

4 0 1 0

5 2 0 100

6 0 0 0


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



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


1 1 0 ?

2 1 1 ?

3 2 1 100

4 0 1 0

5 2 0 100

6 0 0 0


A 6 1 2

B 6 4 2

C 1 2 3

D 1 5 3


For example, for element A:

01

11

10

2112

1331

3223

yyxx

yyxx

yyxx

and the element area A=0.5.

8/3/2019 CAD 2011 Slides

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x

y

V= 0 V= 100

1 2

3

11

1

2

2

2 3

3

3

A

B CD

1

234

56

Node


1 1 0 ?

2 1 1 ?

3 2 1 100

4 0 1 0

5 2 0 100

6 0 0 0


A 6 1 2

B 6 4 2

C 1 2 3

D 1 5 3


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



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


1 1 0 ?

2 1 1 ?

3 2 1 100

4 0 1 0

5 2 0 100

6 0 0 0


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

8/3/2019 CAD 2011 Slides

43/74 4



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


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



x

y

V= 0 V= 100

1 2

3

11

1

2

2

2 3

3

3

A

B CD

1

2 34

56


For element A:

2

10000

2

1000000

000000

000000

00002

1

2

121000

211

N

222126

121116

626166

NNN

NNN

NNN

NA


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

8/3/2019 CAD 2011 Slides

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x

y

V= 0 V= 100

1 2

3

11

1

2

2

2 3

3

3

A

B CD

1

234

56


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


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



x

y

V= 0 V= 100

1 2

3

11

1

2

2

2 3

3

3

A

B CD

1

2 34

56


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


A 6 1 2

B 6 4 2

C 1 2 3

D 1 5 3

8/3/2019 CAD 2011 Slides

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x

y

V= 0 V= 100

1 2

3

11

1

2

2

2 3

3

3

A

B CD

1

234

56


A 6 1 2

B 6 4 2

C 1 2 3

D 1 5 3


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



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

8/3/2019 CAD 2011 Slides

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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



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

8/3/2019 CAD 2011 Slides

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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


]][[][][1

pfpfff VNNV

Final equation:

Formal solution:



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


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

8/3/2019 CAD 2011 Slides

48/74 4


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!


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

8/3/2019 CAD 2011 Slides

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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)



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.

8/3/2019 CAD 2011 Slides

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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


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

8/3/2019 CAD 2011 Slides

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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


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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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



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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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


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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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


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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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



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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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



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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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


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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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


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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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 !


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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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


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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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


Solving Ax = b SOR = 1 opt = 1.885

Iteration count: 4

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Prescribed accuracy achieved

in 106iterations.

Iteration count: 106



Iteration count: 690


in 106iterations.


in 690 iterations.

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Solving Ax = b SORAssume a linear initial solution: Final solution:


in51 iterations(instead of106when a zero

solution was initially used).


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%



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Solving Ax = b Conjugate Gradient MethodConjugate Gradient Method

Minimise which leads to a condition: bxxAxx TTf 21 0 bxAf



Minimise which leads to a condition: bxxAxx TTf 2

10 bxAf

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Minimise which leads to a condition: bxxAxx TTf 21 0 bxAf


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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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


Coinrecognitionsensor

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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.


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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Local

Global

Start



Global

Local




DirectSearch

IndirectSearch

Simulatedannealing

Evolutionstrategy

Rosenbrocks

methodPatternsearch

Simplexmethod

Powells

conjugatedirection

...

Steepest decent

Newtons method

Variable-metricprocedure

Conjugategradient

Trust regionmethod

...

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Single-minimum objective function

Optimization


Optimization

Multiple-minima objective function

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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


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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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:


+Ms -Ms

Changes of shimming magnet distributionduring optimisation

Convergence

Flux distributions

1 iteration

10 iterations

3 iterations

3 iterations

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Pareto Optimisation

F1

F2

UTOPIA

NADIR

DISTOPIA

Objective

domain

searchspace

POF Pareto Optimal Front

POF

F2max

F2min

F1min

F1max


Designer

Company design sheets and priorknowledge

Optimisation and designprograms

Experts on finite element analysis

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Input data

Output data

Materialslibrary

Induction Motor

Design Routines

FEsoftware

Optimisation

Analytical & Control Software

Industrially

oriented systemfor inductionmotor design

MATLAB

FE software

Excel


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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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

cad 2011 slides

Documents