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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.
3 EE757, 2016, Dr. Mohamed Bakr
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
4 EE757, 2016, Dr. Mohamed Bakr
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 ,
5 EE757, 2016, Dr. Mohamed Bakr
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
6 EE757, 2016, Dr. Mohamed Bakr
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
7 EE757, 2016, Dr. Mohamed Bakr
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
It follows that we have
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
)()(
8 EE757, 2016, Dr. Mohamed Bakr
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)(
It follows that we have
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
9 EE757, 2016, Dr. Mohamed Bakr
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
10 EE757, 2016, Dr. Mohamed Bakr
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
11 EE757, 2016, Dr. Mohamed Bakr
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
12 EE757, 2016, Dr. Mohamed Bakr
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
13 EE757, 2016, Dr. Mohamed Bakr
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
14 EE757, 2016, Dr. Mohamed Bakr
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
15 EE757, 2016, Dr. Mohamed Bakr
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
16 EE757, 2016, Dr. Mohamed Bakr
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
17 EE757, 2016, Dr. Mohamed Bakr
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.
18 EE757, 2016, Dr. Mohamed Bakr
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
19 EE757, 2016, Dr. Mohamed Bakr
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)!
20 EE757, 2016, Dr. Mohamed Bakr
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
21 EE757, 2016, Dr. Mohamed Bakr
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
22 EE757, 2016, Dr. Mohamed Bakr
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