EE757 Numerical Techniques in Electromagnetics Lecture 12mbakr/ECE757/Lecture12_2016.pdf · Lecture 12 . EE757, 2016, Dr. Mohamed Bakr 2 1D FEM We consider a 1D differential equation

1

EE757

Numerical Techniques in Electromagnetics

Lecture 12

2 EE757, 2016, Dr. Mohamed Bakr

1D FEM

We consider a 1D differential equation of the form

subject to the

boundary conditions

),0(, L x fdx

d

dx

d

qdx

d p

Lx

,)0(

and are functions associated with the physical

parameters and f is the excitation

Notice that the boundary conditions may be a Dirichlet,

Neuman or mixed Dirichlet and Neuman.


1D FEM (Cont’d)

qdx f-dx dx

dF

Lx

LL2

00

2

2

25.0)(

The functional associated with this problem is

(Prove it)!

Element

Node

1 2 M-1 M

1 2 3 N-2 N-1 N

xe1 x

e2

e

We divide our computational domain into M elements

with a total number of N nodes


1D FEM (Cont’d)

j

e

1

2

1 1 2

2 2 3

3 3 4

M M M+1

We utilize an index table to determine the global index of

each local node

For this simple 1D case, we have N=M+1 and

xx xx ee

ee

121 ,


1D FEM (Cont’d)

We approximate the unknown function by a linear

approximation over each element, i.e.,

eee

ex xbax ,)(

It follows that we have

,11 xbaeeee

xbaeeee

22

Express ae and be in terms of nodes values

e

jj

ej

exNx

2

1

)()(

is the jth interpolation function of the eth element

where and where is

the length of the eth element, i.e.,

)(xNej

l/xxxNeee )()(

21 l/xxxNeee )()(

12

xxleee12

le


The Homogenous Neuman BC Case

For this case ( = q = 0), the functional is given by

LL

dx f-dx dx

dF

00

2

2

5.0)(

M

e

xe

xe

eM

e

xe

xe

dx f-dxdx

dF e

e

1

2

11

2

1

)(

22

5.0)(

M

e

eeFF

1

)()(

F can be written as a sum of subfunctionals

Use elemental expansion


The Homogenous Neuman BC Case (Cont’d)

xe

xe

exe

xe

e

eee dx f-dx

dx

dF

2

1

2

1

2

2

)()( 5.0

The eth subfunctional is thus given by


N i

FF M

ei

ee

i

,,2,1,)()(

1

Notice the coefficient of any node value may be obtained

through summing the associated coefficients obtained by

differentiating each subfunctional

Substituting in the eth subfunctional with the expansion

we get e

jj

ej

exNx

2

1

)()(


The Homogenous Neuman BC Case (Cont’d)

xe

xe

ei

i

e

i

xe

xe

e

j

ej

ei

e

i

e

j

ej

ei

i j

e

i

ee

dxN f-

dx NNdx

Nd

dx

NdF

2

1

2

1

2

1

2

1

2

1

5.0)(


dxNf-dxNNdx

Nd

dx

Nd

d

Fd ei

xe

xe

ej

ei

xe

xe

ej

ei

j

e

je

i

e

2

1

2

1

2

1

Differentiating w.r.t.

Or in matrix form bKeee

e

eF

121222

12

e

i


The homogenous Neuman BC case (Cont’d)

The coefficients of the matrix Ke and the vector be are

dxNN

dx

Nd

dx

NdK

xe

xe

ej

e

i

ej

eie

ij

2

1

dxNfbei

xe

xe

ei

2

1

The process of assembly involves storing the local

elemental components into their proper location in the

system of equations

0

F bK 11

NNNN


An Assembly Example

assuming that we have only 3 elements (4 unknowns)

Element

Node

1 2

1 2 3 4

3

Initialization:

0

0

0

0

,

0000

0000

0000

0000

bK

1st element:

b

b

KK

KK)1(

2

)1(

1)1(

)1(22

)1(

12

)1(

21

)1(

11)1( and, bK

Update the global system to get


An Assembly Example (Cont’d)

0

0,

0000

0000

00

00)1(

2

)1(

1

)1(22

)1(

21

)1(

12

)1(

11

b

b

KK

KK

bK

2nd element:

b

b

KK

KK)2(

2

)2(

1)2(

)2(22

)2(

12

)2(

21

)2(

11)2( and, bK

Update the global system to get

0

,

0000

00

0

00

)2(2

)2(1

)1(2

)1(

1

)2(

22

)2(

21

)2(

12

)2(

11

)1(22

)1(

21

)1(

12

)1(

11

b

bb

b

KK

KKKK

KK

bK


An Assembly Example (Cont’d)

By assembling the 3rd element we finally have the system

of equations

b

bb

bb

b

KK

KKKK

KKKK

KK

)3(2

)3(

1

)2(2

)2(1

)1(2

)1(

1

)3(22

)3(

21

)3(

12

)3(

11

)2(

22

)2(

21

)2(

12

)2(

11

)1(22

)1(

21

)1(

12

)1(

11

,

00

0

0

00

bK


The General Boundary Case

For the case q0 or 0 , the functional is augmented by

the subfunctional

qFLx

b

2

2)(

NNbqF

2

2)(

qF

N

N

b

It follows that only KNN is incremented by and bN is

incremented by q


Dirichlet’s Boundary Conditions

Dirichlet’s boundary conditions are imposed by eliminating the

corresponding nodal values

Example: for the case N=4, we have

b

b

b

b

KKKK

KKKK

KKKK

KKKK

4

3

2

1

44434241

34333231

24232221

41312111

, bK

if 1=p

pKb

pKb

pKb

p

KKK

KKK

KKK

414

313

212

4

3

2

1

444342

343332

242322

0

0

0

0001

K

pKb

pKb

pKb

KKK

KKK

KKK

414

313

212

4

3

2

444342

343332

242322


Example

Determine the power reflected by this inhomogeneous

metal-backed dielectric slab for a uniform incident plane

wave with a z polarized electric field. Both r and r may

vary with x.

y

x

x=0 x=L

,

Ez


Example (Cont’d)

The expression for the incident electric field is

aakrk yoxoincz kk jEyxE sincos),.exp(),( o

)sincosexp(),(ooo ykjxkjEyxE

incz

Notice that all field components must have a variation of

to satisfy field continuity )sinexp( ykj o

The wave equation for this problem is

JEE

j

21

Taking into account that E=Ezaz, /z=0 and J=0, we get

011 2

o

Ek

yyxxzr

rr


Example (Cont’d)

Substituting for we get Ekjy

E

yzo

z

r

sinand0,1

0)sin

(1 2

2o

Ek

dx

Ed

xz

r

rz

r

with the BC Ez(0)=0

x1 x2 x3

1 2

xM+1

M

An analytical solution is obtained by dividing the slab into

layers m=1, 2, , M , where r and r are assumed constant

with values rm and rm, respectively.


Example (Cont’d)

The wave equation in each layer, thus becomes

0)sin

(1 2

2o2

2

Ekxd

Edz

rm

rmz

rm

0)sin( 22

o2

2

Ekxd

Edzrmrm

z

which has the solution

)sinexp())exp()exp(( y kjxjk-BxkjAE oxmmxmmzm

The analytical solution is obtained by enforcing continuity of

the electric and magnetic field components at the layers

interface to get

sin2

o rmrmxm kk


Example (Cont’d)

kk

kk

A

BR

xmrmxmrm

xmrmxmrm

mm

m

mm

11

11

1,,

)2exp()2exp(1

)2exp(11

1,1

1,11 xkj

xkjR

xkjRR mxm

mxmmmm

mxmmmmm

A

BR )(conductor,1

1

11 (Prove it)!


FEM Solution

0)sin

(1 2

2o

Ek

dx

Ed

dx

dz

r

rz

r

with the BC Ez(0)=0

Our problem is given by

A boundary condition at x=L to have a finite computational domain

For L x, we have

)sinexp())cosexp()cosexp((),( oo ykjxkjERxkjEyxE oooz

)sinexp()(),( ykjxEyxE ozz

Differentiating Ez(x) relative to x we get

))cosexp()cosexp((cos)(

oo xkjERxkj Ekj

dx

xEdooo

z


FEM Solution (Cont’d)

Manipulating we get

)(cos)cosexp(cos2)(

o xEkjxkj Ekjdx

xEdzooo

z

)cosexp(cos2)(cos)(

o Lkj EkjxE kjdx

xEdoozo

z

Lx

Utilizing the continuity of the electric and magnetic field we may convert this boundary condition into

)cosexp(cos2)(cos)(1

o

Lkj EkjxE kjdx

xEdoozo

z

rLx


FEM Solution (Cont’d)

Comparing with our 1D FEM formulation we have

)cosexp(cos,cos

)sin

(,/1),(

oooo

22o

LkkE2jq kjμ

k μ xEr

rrz

Once the field is solved, the reflection coefficient is given by

)cosexp(

)cosexp(()(

o

o

LkjE

Lkj ExER

o

oz

It follows that our problem is given by

0)sin

(1 2

2o

Ek

dx

Ed

dx

dz

r

rz

r

with Ez(0)=0

)cosexp(cos2)(cos)(1

o

Lkj EkjxEkjdx

xEdoozo

z

rLx

Documents

EE757 Numerical Techniques in Electromagnetics Lecture 12mbakr/ECE757/Lecture12_2016.pdf · Lecture 12 . EE757, 2016, Dr. Mohamed Bakr 2 1D FEM We consider a 1D differential equation