EE750 Advanced Engineering Electromagnetics Lecture 14mbakr/ee750/Lecture14_2003.pdf · M....

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EE750Advanced Engineering Electromagnetics

Lecture 14

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”


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


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



• 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



• 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



• 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



The charge distribution along the width of the plate

Harrington, FieldComputation byMoment Methods


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


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


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)



•••• 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



•••• 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)



For a uniform plane wave incident at an angle φi we have

eysine xcoskjE jii

iz

rk .)( =+= φφ

Harrington, FieldComputation byMoment Methods


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


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



•••• 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


•••• But as , we may write


•••• 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)


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


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


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

EE750 Advanced Engineering Electromagnetics Lecture 14mbakr/ee750/Lecture14_2003.pdf · M....

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