EE757 Numerical Techniques in Electromagnetics … · M. Sadiku,”Numerical Techniques in...

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EE757

Numerical Techniques in Electromagnetics

Lecture 10

2 EE757, 2016, 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 of arbitrary shape”


A Charged Conducting Plate

Find the charge distribution and capacitance of a metalic

plate of dimensions 2a2a whose potential is = Vo

z

x

y

2a

= 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 also

write

qΦ ev

2

zdydxdqGΦV

ev

)(),()( rrrr

rrr

r

-R zdydxdR

qΦ

V

ev ,4

)()(

zyyxxR ydxdR

qzyxΦ

a

a-

a

a-

es 222)()(,

4

)(),,(

r



We divide the conducting plate into N square subsections and

define the subsectional basis function

otherwise,0

subsectionth the,on1

n S f

n

n

We then expand the unknown surface charge density in terms

of the subsectional basis functions

yyxxR R

qqLV

a

a

a

a-

es

es

22

o ,4

)(

ydxdR

fydxd

R

f

Va

a

a

a-

n

nn

a

a

a

a-

nnn

44o



But as the nth basis function is nonzero only over the nth

subsection we may write

ydxdR

VS nn

n

4

1o (one equation in N unknowns)

We utilize point matching by enforcing the above equation at

the centers of each subsection

N m

xyxxR ydxdR

VS n

mmm

mnn

,,2,1

)()(,4

1 22

o

Alternatively, N m lVn

mn n,,2,1,o

S n m

mn ydxdR

l4

1



It follows that the coefficients n are obtained by solving

V

V

V

lll

lll

lll

NNNNN

N

N

o

o

o

2

1

21

22212

12111

Postprocessing: The capacitance of the conducting plate is

approximated by

V

S

V

qC

N

nnn

t

o

1

o



The charge distribution along the width of the plate

Harrington, Field

Computation by

Moment Methods


Scattering Problems

conductor

n

Ei

Es

An incident wave generates surface currents that in turn

generate a scattered field such that

0 )( EEn si (zero total tangential electric field)

In a scattering problem it is required to determine the surface

currents. 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 ρ

Eiz

aE zi

zE

Fields are dependent on x and y directions. It follows that we

can solve this problem as a 2D problem


Scattering by a Conducting Cylinder (Cont’d)

Starting with Maxwell’s equations

,)( HE j EJH j )(

For the case J=Jz we have

JjEkE zzz 22

The corresponding Green’s function is obtained by setting

to obtain )()( yyxxJ z

)(4

),( 2o ρρρρ

kH

kG

The scattered electric field is thus given by

CdkHJk

EC

zsz

)()(4

)( 2o ρρρρ

(We consider only the z component)



For the problem at hand we must have for all

points on the surface of the cylinder EE

sz

iz

It follows that we have

C ,CdkHJk

EC

ziz

ρρρρρ )()(4

)( 2o

The only unknown in this equation is Jz

We expand Jz in terms of the subsectional basis functions

otherwise,0

subsectionth the,on1

n C f

n

n

N

nnnz fJ

1



It follows that we have

C ,CdkHfk

EC

n

N

nn

iz

ρρρρ )(4

)( 2o

1

C ,CdkHk

ECn

N

nn

iz

ρρρρ )(4

)( 2o

1

(one equation in N unknowns)

We utilize point matching to enforce the above equation at

the centers of the subsections N m yx mmm ,,2,1),,( ρ

N , m ,CdkHk

ECn

m

N

nnm

iz ,2,1)(

4)( 2

o1

ρρρ

(N equation in N unknowns)



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

eysine xcoskjEj

iiiz

rk.)(

Harrington, Field

Computation by

Moment Methods


Pocklington’s Integral Equation

The target is to determine the current distribution and

consequently the scattered field due to an incident field for a

finite-diameter wire

x

y

z

l/2

l/2

Ei

2a


Pocklington’s Integral Equation (Cont’d)

The main relation for this scatterer is

)()( aEaEsz

iz

The equations governing the scattered field are

)).()(/( AAE jj

We need only the z component of the field

)()(2

22

z

AA

j-E

zz

sz

r A

z

j-E z

sz )()(

2

22

r

The z component of the magnetic vector potential is

zdadR

ezJsd

R

eJA

l/

l/-

R-j

z

R-j

Szz

2

2

2

0

),(4

)(4

)( rr

rr -R



If the wire is thin, Jz is not a function of

zddR

ezJ aA

l/

l/-

R-j

zz

2

2

2

0 42

1)(2)(

r

)(zI z ),( zG r

The distance R in cylindrical coordinate is

a

r

zz

R

)()(2222

zzacosaR

For observation points on the wire surface we have

)()(22222 zzcosaaR

)()2

(4222 zzsinaR


But as , we may write


But as Az has a symmetry, we may write

Az(=a,z,) = Az(=a,z,0)

,),()(),(2/

2/

l

lzz zdzzGzIzaA )()

2(4

222 zzsinaR

The scattered field at the wire surface is thus given by

zdzzGzIz

j-zaE

l

lz

sz

),()()(),(2/

2/2

22

),(),( zaEzaEsz

iz

zdzzGz

zIzaEj-l

lz

i

z

),()()(),(2

22

2/

2/

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

zzz

un-n

n

For pulse functions we have

For triangular functions we have

otherwise 0

, 11

zzz

zz

u n-n

n

n


Solution of Pocklington’s Equation (Cont’d)

It follows that

zdzzGz

zuIzaEj- n

l

l

N

nn

i

z

),())((),(2

22

2/

2/ 1

2/

2/2

22

1

),())((),(l

ln

N

nn

i

zzdzzG

zzuIzaEj-

ln

N

nn

i

zzdzzG

zIzaEj- ),()(),(

2

22

1

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

N m zGIzE mn

N

nnm

iz ,2,,1),()(

1

)(

)(

)(

)()()(

)()()(

)()()(

2

1

2

1

21

22221

11211

zE

zE

zE

I

I

I

zGzGzG

zGzGzG

zGzGzG

Niz

iz

iz

NNNNN

N

N

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