ee750 advanced engineering electromagnetics lecture 17mbakr/ee750/lecture17_2003.pdf · advanced...

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EE750Advanced Engineering Electromagnetics

Lecture 17

2EE750, 2003, Dr. Mohamed Bakr

2D FEM

• We consider a 2D differential equation of the form

subject to

Ω∈=+

∂∂

∂∂−

∂∂

∂∂− ),(, yxf

yyxxyx βϕϕ

αϕ

α

Γ= 1onpϕ

• αx, αy and β are functions associated with the physicalparameters and f is the excitation

• Notice that the boundary conditions may be a Dirichlet,Neuman or mixed Dirichlet and Neuman.

Γ=+

∂∂+

∂∂

2on. qyx

yx γϕϕα

ϕα nji

where Γ= Γ1∪ Γ 2 is the contour enclosing the domain Ω andn is the unit outward normal


2D FEM (Cont’d)

∫ Γ

−+Ω∫∫

Ω∫∫

+

∂∂+

∂∂=

ΓΩ

Ω

2

2

2

22

2

5.0)(

dqdf-

dyx

F yx

ϕϕγϕ

ϕβϕα

ϕαϕ

• The functional associated with this problem is

(Prove it)!


4654

2533

2452

1421

321i

e

2D FEM Analysis

• Each node has both a local and aglobal index

e

1 2

3• The computational domain is dividedinto triangles (elements)

• A connectivity array n(i,e), i=1, 2, 3 and e=1, 2, …, Mstores the global indices of the nodes

5

1

23 4

1

2

3

4

5 6


2D FEM Analysis (Cont’d)

• We assume that there are Ms line segments on Γ2

Γ2

s=1

s=2

s=3

• We store the index array ns(i,s), i=1, 2 and s=1, 2, …, Msof global indices of nodes on Γ2


Input Data to the 2D FEM Analysis

• The coordinates of the nodes ri=(xi,yi), i=1, 2, …, N, whereN is the total number of nodes

• The values of αx, αy, β and f for each element

• The value of p for each node residing on Γ1

• The value of γ and q for each segment with nodes on Γ2

• The two arrays n(i,e), i=1, 2, 3 and e=1, 2, …, M andns(i,s), i=1, 2 and s=1, 2, …, Ms


Elemental Interpolation

• Over the eth element we utilize the linear approximationΩ∈++= e

eeee yxycxbayx ),(,),(ϕ• The three nodes of the eth element must satisfy the linear

interpolation relationycxba

eeeeee ,111 ++=ϕ ycxbaeeeeee ,222 ++=ϕ

ycxbaeeeeee333 ++=ϕ

• Solving for ae, be and ce we obtain

where

ϕϕ ej

j

ej

e yxNyx ∑==

3

1),(),(

3,2,1),(2

1),( jycxba

AyxN e

jej

ej

e

ej =++=

xxcyybxyyxa eeeeeeeeeee23132132321 ,, −=−=−=




Elemental Interpolation (Cont’d)

• Ae is the area of the eth element and is given by

yx

yx

yx

Aee

ee

ee

e

33

22

11

1

1

1

2

1=

• The interpolation functions satisfy

≠

===

ji

jiyxN ij

ej

ej

ei

0,

,1),( δ


The Homogenous Neuman BC case

• We first consider the case (γ = q = 0)

• The functional is expressed as a sum of elementalsubfunctions

∑==

M

e

eeFF1

)()( ϕϕwhere

∫∫ ΩΩ∫∫ ++=ΩΩ

e

e

e

ee

y

e

xee df-d

dy

d

dx

dF ϕϕβϕ

αϕ

αϕ )(5.0)(2

22

• Substituting with the linear interpolation expression, wewrite


The Homogenous Neuman BC case

Ω∫∫ ∑−Ω∫∫ ∑ ∑+

Ω∫∫ +∑ ∑=

Ω =Ω = =

= =Ω

e i

ei

ei

ei j

ej

ej

ei

ei

ej

ej

eie

iyej

ej

ei

i j

eix

ee

dNfdNN

ddyNd

dyNd

dxNd

dxNd

F

e3

1

3

1

3

1

3

1

3

15.0)(

ϕϕϕβ

ϕϕαϕϕαϕ

∫∫ Ω

Ω∫∫ ++∑=

∂∂

Ω

Ω=

e

ei

ej

ei

e

ej

ei

y

ej

ei

xj

eje

i

e

dfN-

dNNdyNd

dyNd

dxNd

dxNdF βααϕ

ϕ3

1

i=1, 2, 3

[ ] [ ] [ ]bK eeee

eF131333

13

××××

−=

∂∂

ϕϕϕϕϕϕϕϕ


The Homogenous Neuman BC case (Cont’d)

Ω∫∫ ++=Ω

dNNdyNd

dyNd

dxNd

dxNd

K ej

ei

e

ej

ei

y

ej

ei

xeij βαα

i=1, 2, 3 and j=1, 2, 3∫∫ Ω=Ωe

ei

ei dfNb

• If αx, αy , β and f are taken as constants over each element,and utilizing the property

( ) ( ) ( )!!!

!!!2321 nml

nmlAdNNN e

e ne m

e

e l

++=Ω∫∫

Ω

( ) )1(124

1δβαα ij

eeej

ei

ey

ej

ei

ex

e

eij

Accbb

AK +++=

we get

fAb

eeei

3=


The Process of Assembly

• The process of assembly involves storing the localelemental components into their proper location in theglobal system of equations

0=∂∂ϕϕϕϕF [ ] [ ] [ ]bK 11 ××× = NNNN ϕϕϕϕ

• The element is added toK eij K ejnein ),(),,(

• The element is added tobei b ein ),(


An Assembly Example

5

1

23 4

1

2

3

4

5 64654

2533

2452

1421

321i

e

• There are six nodes ℜ∈ℜ∈ ×× 1666 and bK• We initialize K and b with zeros

• Evaluate K(1) and b(1) an add them to the proper locationsto get


An Assembly Example (Cont’d)

=

=

0

0

0,

000000

000000

000

000000

000

000

)1(2

)1(1

)1(3

)1(22

)1(21

)1(32

)1(12

)1(11

)1(31

)1(23

)1(13

)1(33

b

b

b

KKK

KKK

KKK

bK

• Evaluate K(2) and b(2) an add them to the proper locations

+

+

=

++

++

=

0

0,

000000

000

00

000000

00

000

)2(1

)2(2

)1(2

)2(3

)1(1

)1(3

)2(11

)2(12

)2(31

)2(21

)2(22

)1(22

)2(32

)1(21

)1(32

)2(13

)2(23

)1(12

)2(33

)1(11

)1(31

)1(23

)1(13

)1(33

b

bb

bb

b

KKK

KKKKKK

KKKKKK

KKK

bK

• Repeat the same steps for all elements


Incorporating a Boundary Condition of the 3rd Kind

• In this case γ and q are not zeroes• The extra subfunctional is added to

the functional F∫ Γ

−=

Γ2

2

2)( dqF b ϕϕγϕ

• Because Γ2 is comprised of Ms line segments, we may write

)()(1∑==

M s

s

ssbb FF ϕϕ

• We approximate the function ϕ over the segment s by thelinear expression ξξϕϕ =−=∑=

=NNN sss

jj

sj

s21

2

1,1,

• ξ is the normalized distance from node 1 to node 2

sΓ2

ξ=1

ξ=0


Incorporating a 3rd Kind BC (Cont’d)

∫ Γ

−=

s

sb dqF ϕϕγϕ 2

2)(

Use the expansion

∫ Γ

∑∑ ∑=== =s

si

i

si

sj

i j

sj

si

si

sb dNq-NNF ϕϕϕγϕ

2

1

2

1

2

12)(

Differentiate and use d Γ=ls dξ

21,2

1

1

0

1

0

,idlNq-dlNNF ss

ij

ssj

si

sjs

i

sb =∑ ∫ ∫=

∂∂

=ξξγϕ

ϕ

[ ] [ ] [ ]bK ssss

sbF

121222 ××× −=∂∂

ϕϕϕϕϕϕϕϕ

in matrix form

21and21,,1

0

1

0

,j,idlNqbdlNNK ssi

si

ssj

si

sij ==∫=∫= ξξγ

• Assembly is then applied to store these coefficients


The Dirichlet Boundary Condition

• The Dirichlet boundary conditions are imposed byeliminating the known nodes by substituting for theirvalues

=

b

b

KK

KK

u

p

u

p

uupu

uppp

ϕϕϕϕϕϕϕϕ

( )ϕϕϕϕϕϕϕϕ ppuuuuu KbK −=

Original system

Reduced system


An Example: A Shielded Microstrip Line

ϕ=0εo

εr

ϕ=1

εo

εr

ϕ=1

∂ϕ/∂n ϕ=0

• The microstrip is kept at potential ϕ=1 while the externalshielding box is kept at potential ϕ=0

• Symmetry may be employed to reduce the computationaldomain by one half

• The governing BVP is


An Example: A Shielded Microstrip Line (Cont’d)

ερϕ

εϕ

εo

crr

yyxx=

∂∂

∂∂−

∂∂

∂∂−

with ϕ = 0 on the outer conductor, ϕ =1 on the microstripand ∂ϕ / ∂n=0 on the plane of symmetry

• It follows that we have αx= αy=εr, β =0, f = 0

• The electric field is obtained through E=-∇ ϕ. But ϕ overeach element is approximated by

,),(),(3

1ϕϕ e

jj

ej

e yxNyx ∑==

3,2,1),(2

1),( jycxba

AyxN e

jej

ej

e

ej =++=

∑ +−=∂∂−−

∂∂−=

=

3

1)(

2

1j

ej

ej

ej

e

cbAyx

ϕϕϕjijiE Over the eth element



The Finite Element Method in Electromagnetics, Jianming Jin



The equi-potential lines

The Finite Element Methodin Electromagnetics,Jianming Jin

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