EE750 Advanced Engineering Electromagnetics Lecture 14mbakr/ee750/Lecture14_2003.pdf · M....
Transcript of EE750 Advanced Engineering Electromagnetics Lecture 14mbakr/ee750/Lecture14_2003.pdf · M....
2EE750, 2003, Dr. Mohamed Bakr
Applications of MoM
• Example on static problems
• Example on 2D scattering problems
• Wire Antennas and scatterers
References
R.F. Harrington, “Field Computation by Moment Methods”
C.A. Balanis, “Advanced Engineering Electroamgnetics”
M. Sadiku,”Numerical Techniques in Electromagnetics”
S.M. Rao et al., “Electromagnetic scattering by surfaces ofarbitrary shape”
3EE750, 2003, Dr. Mohamed Bakr
A Charged Conducting Plate
• Find the charge distribution and capacitance of a metalicplate of dimensions 2a×2a whose potential is Φ = Vo
z
x
y
2a
2a
Φ = Vo
4EE750, 2003, Dr. Mohamed Bakr
A Charged Conducting Plate (Cont’d)
• The potential and charge satisfy for the unbounded medium
•••• The well-known solution for this problem is
• As the plate is assumed to be in the xy plane we may alsowrite
εqΦ ev−=∇ 2
zdydxdqGΦV
ev′′′∫∫∫ ′′=
′)(),()( rrrr
rrr
r ′=′′′∫∫∫′
=′
-RzdydxdR
qΦV
ev ,4
)()(
πε
zyyxxRydxdR
qzyxΦ
a
a-
a
a-
es 222 )()(,4
)(),,( +′−+′−=′′∫ ∫
′=
πεr
5EE750, 2003, Dr. Mohamed Bakr
A Charged Conducting Plate (Cont’d)
• We divide the conducting plate into N square subsectionsand define the subsectional basis function
∆
=otherwise,0
subsectionththe,on1 nSf
n
n
• We then expand the unknown surface charge density interms of the subsectional basis functions
( ) ( )yyxxRR
qqLV
a
a
a
a-
eses
′−+′−=∫ ∫==−
22o ,
4)(
πε
ydxdR
fydxd
R
fV
a
a
a
a-
n
nn
a
a
a
a-
nnn
∫ ∫ ′′∑=′′∫ ∫∑
=−− πεαπε
α
44o
6EE750, 2003, Dr. Mohamed Bakr
A Charged Conducting Plate (Cont’d)
• But as the nth basis function is nonzero only over the nthsubsection we may write
ydxdR
VS nn
n ∫∫ ′′∑=∆ πεα
4
1o (one equation in N unknowns)
• We utilize point matching by enforcing the above equation atthe centers of each subsection
Nm
xyxxRydxdR
VS n
mmmmn
n
,,2,1
)()(,4
1 22o
L=
∫∫ ′−+′−=′′∑=∆ πεα
• Alternatively, NmlVn
mn n ,,2,1,o L=∑= α
∫∫ ′′=∆S n m
mn ydxdR
l πε4
1
7EE750, 2003, Dr. Mohamed Bakr
A Charged Conducting Plate (Cont’d)
• It follows that the coefficients αn are obtained by solving
=
V
V
V
lll
lll
lll
NNNNN
N
N
o
o
o
2
1
21
22212
12111
MMMMMM
L
L
α
αα
• Postprocessing: The capacitance of the conducting plate isapproximated by
V
S
V
qC
N
nnn
t
o
1
o
∑ ∆== =
α
8EE750, 2003, Dr. Mohamed Bakr
A Charged Conducting Plate (Cont’d)
The charge distribution along the width of the plate
Harrington, FieldComputation byMoment Methods
9EE750, 2003, Dr. Mohamed Bakr
Scattering Problems
conductor
nEi
Es
• An incident wave generates surface currents that in turngenerate a scattered field such that
0=+× )( EEn si (zero total tangential electric field)
• In a scattering problem it is required to determine the surfacecurrents. Es is obtained as a byproduct
10EE750, 2003, Dr. Mohamed Bakr
Scattering by a Conducting Cylinder of a TM Wave
•••• Incident field has only z direction
x
y
n
ρ′ ρE i
z
aE zizE=
•••• Fields are dependent on x and y directions. It follows that wecan solve this problem as a 2D problem
11EE750, 2003, Dr. Mohamed Bakr
Scattering by a Conducting Cylinder (Cont’d)
•••• Starting with Maxwell’s equations
,)( HE ωµj−=×∇ EJH ωεj+=×∇ )(
•••• For the case J=Jz we have
JjEkE zzz ωµ=+∇ 22
•••• The corresponding Green’s function is obtained by setting
to obtain)()( yyxxJ z ′−′−= δδ
)(4
),( 2o ρρρρ ′−−=′ kH
kG
η
•••• The scattered electric field is thus given by
CdkHJk
EC
zsz ′∫ ′−′−=
′)()(
4)( 2
o ρρρρ η
(We consider only the z component)
12EE750, 2003, Dr. Mohamed Bakr
Scattering by a Conducting Cylinder (Cont’d)
•••• For the problem at hand we must have for allpoints on the surface of the cylinder
EE sz
iz −=
•••• It follows that we have
C,CdkHJk
EC
ziz ′∈∀′∫ ′−′=
′ρρρρρ )()(
4)( 2
oη
The only unknown in this equation is Jz
•••• We expand Jz in terms of the subsectional basis functions
∆
=otherwise,0
subsectionththe,on1 nCf
n
n ∑==
N
nnnz fJ
1α
13EE750, 2003, Dr. Mohamed Bakr
Scattering by a Conducting Cylinder (Cont’d)
•••• It follows that we have
C,CdkHfk
EC
n
N
nn
iz ′∈∀′∫ ′−∑=
′=ρρρρ )(
4)( 2
o1α
η
C,CdkHk
ECn
N
nn
iz ′∈∀′∫ ′−∑=
∆=ρρρρ )(
4)( 2
o1α
η
(one equation in N unknowns)
•••• We utilize point matching to enforce the above equation at
the centers of the subsections Nmyx mmm ,,2,1),,( L==ρ
N,m,CdkHk
ECn
m
N
nnm
iz L,2,1)(
4)( 2
o1
=′∫ ′−∑=∆=
ρρρ αη
(N equation in N unknowns)
14EE750, 2003, Dr. Mohamed Bakr
Scattering by a Conducting Cylinder (Cont’d)
For a uniform plane wave incident at an angle φi we have
eysine xcoskjE jii
iz
rk .)( =+= φφ
Harrington, FieldComputation byMoment Methods
15EE750, 2003, Dr. Mohamed Bakr
Pocklington’s Integral Equation
•••• The target is to determine the current distribution andconsequently the scattered field due to an incident field for afinite-diameter wire
x
y
z
l/2
l/2
Ei
2a
16EE750, 2003, Dr. Mohamed Bakr
Pocklington’s Integral Equation (Cont’d)
•••• The main relation for this scatterer is)()( aEaE s
ziz =−== ρρ
•••• The equations governing the scattered field are)).()(/( AAE ∇∇−−= ωµεω jj
•••• We need only the z component of the field
)()(2
22
z
AA
j-E
zz
sz ∂
∂+= βωµε
r Az
j-E z
sz )()(
2
22
∂∂+= β
ωµεr
•••• The z component of the magnetic vector potential is
zdadR
ezJsdR
eJA
l/
l/-
R-j
z
R-j
Szz ′′∫ ∫ ′′=′′∫∫= φφ
πµ
πµ π ββ 2
2
2
0
),(4
)(4
)( rr
rr ′= -R
17EE750, 2003, Dr. Mohamed Bakr
Pocklington’s Integral Equation (Cont’d)
•••• If the wire is thin, Jz is not a function of φzdd
RezJaA
l/
l/-
R-j
zz ′∫
∫ ′′=
2
2
2
0 42
1)(2)(
π β
φππ
πµr
)(zI z ′ ),( zG ′r
•••• The distance R in cylindrical coordinate is
aa
r
ρzz ′−
R
)()(2 222 zzacosaR ′−+′−−+= φφρρ•••• For observation points on the wire surface we have
)()(22 222 zzcosaaR ′−+′−−= φφ
)()2
(4 222 zzsinaR ′−+′−= φφ
18EE750, 2003, Dr. Mohamed Bakr
•••• But as , we may write
Pocklington’s Integral Equation (Cont’d)
•••• But as Az has a φ symmetry, we may writeAz(ρ=a,z,φ) = Az(ρ=a,z,0)
,),()(),(2/
2/∫ ′′′=
−
l
lzz zdzzGzIzaA µ )()
2(4 222 zzsinaR ′−+
′= φ
•••• The scattered field at the wire surface is thus given by
zdzzGzIz
j-zaE
l
lz
sz ′′∫ ′
∂∂+=
−),()()(),(
2/
2/2
22β
ωε),(),( zaEzaE s
ziz −=
zdzzGz
zIzaEj-l
lz
iz
′′∂∂+∫ ′=
−),()()(),(
2
22
2/
2/
βωε
Pocklington’s integral equation (only Iz is not known)
19EE750, 2003, Dr. Mohamed Bakr
Solution of Pocklington’s Integral equation
∆
•••• Divide the wire into N non overlapping segments
•••• Expand the unknown current in terms of the
basis functions )()(1
zuIzI n
N
nnz ∑=
=
<<
=+
otherwise,0
,1 2/12/1 zzzu
n-n
n
•••• For pulse functions we have
•••• For triangular functions we have
<<∆−−∆
= +
otherwise0
, 11 zzzzz
u n-n
n
n
20EE750, 2003, Dr. Mohamed Bakr
Solution of Pocklington’s Equation (Cont’d)
•••• It follows that
zdzzGz
zuIzaEj- n
l
l
N
nn
iz
′′∂∂+′∫ ∑=
− =),())((),(
2
22
2/
2/ 1βωε
∫ ′′∂∂+′∑=
−=
2/
2/2
22
1),())((),(
l
ln
N
nn
iz zdzzG
zzuIzaEj- βωε
∫ ′′∂∂+∑=
= ln
N
nn
iz zdzzG
zIzaEj- ),()(),(
2
22
1βωε
using a pulse function
)()(1
zGIzE n
N
nn
iz ∑=
=One equation in N unknowns
21EE750, 2003, Dr. Mohamed Bakr
Solution of Pocklington’s Equation (Cont’d)
•••• Enforcing this equation at the center of each segment, we get
N equations in N unknowns
NmzGIzE mn
N
nnm
iz ,2,,1),()(
1L=∑=
=
=
)(
)(
)(
)()()(
)()()(
)()()(
2
1
2
1
21
22221
11211
zE
zE
zE
I
I
I
zGzGzG
zGzGzG
zGzGzG
Niz
iz
iz
NNNNN
N
N
MMMMMM
L
L