limitation of pulse basis/delta testing discretization: te-wave efie difficulties lie in the...

Limitation of Pulse Basis/Delta Testing Discretization: TE-Wave EFIE

difficulties lie in the behavior of fields produced by the pulse expansion functions

consider the scattered electric field due to a current source extended from xo to x1 and oriented in the x direction

)y()x,x;x(P)y,x(J 10x

)A(j

1AjEs

)kR(Hj4

1*Jk

xj

1E 2

0x2

2

2sx

10

1

0x'x

20x'x

20

x

x

20

2sx |)kR(H

'x|)kR(H

'x4

1'dx)kR(H

4

k)y,x(E

Further Explanation

)y()xx(')xx(')x,x;x(PkJkx

10102

x2

2

2

dx)kR(H'x

)x'x()kR(H)x'x('dx)x'x('x

)kR(H 200

2000

20

0x200

20 |)kR(H

'x)kR(H

1

11

21

0

00

21

x

x

20

2sx R

xx)kR(H

R

xx)kR(H

4

k'dx)kR(H

4

k)y,x(E

1

0

2200 y)xx(R 22

11 y)xx(R

2j)(H 2

1 0

Discussions

electric field produced by the constant pulse of current density Jx(x)

is singular at the edges of the source segment

due to the presence of line charges associated with the discontinuity in current density at the end of the segment as

fictitious line charges give rise to infinite tangential electric field at the two edges of every cell in the model

point matching is done at the center so all the matrix elements are finite and solution can be found, but accuracy may be affected

0J

Discussions

for MFIE with TE polarization, although the transverse electric field is singular, the magnetic field Hz produced by the

pulse is finite along the source segment and therefore the solution is fine

for EFIE with TM polarization, there is no fictitious line charges generated as and therefore the solution is also fine

to avoid generation of fictitious line charges for the EFIE TE case, a smoother basis function is required

for formulation with one derivative, we can use pulse basis and delta testing

for formulation with two derivatives, we need to use triangular basis with delta or pulse testing

0J

TE Wave Scattering from PEC Strips or Cylinders – EFIE

FAj

1AjE

00

s

FAjE e0s

we have studied MFIE for TE polarization but MFIE cannot treat thin structure since is not the same as where the magnetic field on one size is equal to and zero on the other when the equivalent principle is applied

J)HH(n 21

JHn

H

Use of EFIE

EFIE can be used for thin structures but its implementation is more difficult

it is advisable to use basis and testing functions having additional degrees of differentiability to compensate for additional derivatives presence in the TE EFIE

we will consider the use of subsectional triangular basis functions with pulse testing which together provide two degrees of differentiability beyond that of the pulse/delta combination

Formulation of EFIE

total tangential E equals to zero, i.e., tangential scattered E is equal and opposite to tangential incident E

ei tAtjkEt

'dt)kR(Hj4

1)'t(J)'t(t)t(A 2

0t

'dt)kR(Hj4

1)t()t( 2

0e

e

t

JJj t

se

22 )'t(y)t(y)'t(x)t(xR

magnetic vector potential

electric scalar potential

continuity equation

Triangular Basis Function

triangular basis function spans over two current segments for a close structure, we have M basis functions for a M-segment structure

for an open structure, we only have M-1 basis function

two pulses are also used to approximate the triangular basis function

)t,t,t;t(Tj)t(J 1nn1n

N

1nnt

t

tn-1

tn

tn+1

Pulse Doublet

t

J

jkt

J

j

1,

t

Jj ttet

e

t

J t

corresponds to the slopes of the triangular edges

)t,t;t(Ptt

1)t,t;t(P

tt

1

jk

)t(1nn

n1nn1n

N

1n 1nn

e

jn

jn/(tn-tn-1)

-jn/(tn+1-tn)

area = jn

area = jn

pulse doublet

cancel each other, zero total charge

no fictitious line charge generated

Testing Function

)t,t;t(P)t(T 5.0m5.0mm

the choice of pulse testing function permits the analytical treatment of the gradient operator appearing in the mixed potential form of the EFIE according to the equation that

)a(F)b(Fdtdt

dFFdt)t(t)b,a;t(P

b

a

Matrix Equation

mnmn ejZ dtE)t(te5.0m

5.0m

t

t

im

dt'dt)kR(H)t,t,t;'t(T)'t(t)t(t4

kZ 2

01nn1n

t

t

t

tmn

5.0m

5.0m

1n

1n

1n

n

n

1n

t

t 220

n1n

t

t 220

1nn

'dt)kR(Htt

1'dt)kR(H

tt

1{

4

k

1n

n

n

1n

t

t 120

n1n

t

t 120

1nn

}'dt)kR(Htt

1'dt)kR(H

tt

1

25.0m2

5.0m1 )'t(y)t(y)'t(x)t(xR

25.0m2

5.0m2 )'t(y)t(y)'t(x)t(xR

Bistatic Cross Section

further approximate the first integral by approximating the triangle with two pulses and delta testing the equation

the scalar potential contribution dominates when R is made small

bistatic cross section is given by

2N

1n

t

t

}sin)'t(ycos)'t(x(jk1nn1nnTE

1n

1n

'dte)t,t,t;'t(Tj}sin)'t(txcos)'t(ty{4

k)(

TM Wave Scattering from Inhomogeneous Dielectric Cylinders – Volume EFIE Discretization

with Pulse Basis and Delta Testing Functions

convenient to convert the original scattering problem into an equivalent form more amenable to a direct solution

replace the inhomogeneous dielectric scatterer with equivalent induced polarization currents and charges radiating in free space

x

y

Ei

rx,y)

k

HjE 0

EjH 0

HjE 0

EjH 0r

outside scatterer inside scatterer

Formulation of the Volume Integral Equation

ind0r000r JEjE)1(jEjEjH

0E)E()E( r00rr0

inde

rr0r

r

00

1EE)E(

the scatterer is now replaced by induced currents and charges radiating in free space

For TM case, we have Ez, Hx and Hy components only and

therefore, 01

Er

0)y,x(

1

zz

yy

xxEz

rz

no induced charge

Volume Integral Equation

)y,x(E)1(jz)y,x(J zr0

sz

r0

zsz

totalz

iz E

)1(j

JEEE

z0sz AjE

'dy'dx)kR(Hj4

1)'y,'x(Jj

)1(j

JAj

)1(j

JE 2

0z0r0

zz0

r0

ziz

Pn(x,y)=1 if (x,y) with cell n 0 otherwise

Jz(x,y)=

N

1nnn )y,x(Pj

Final Equation

ncell

20z

r0

nN

1nn

iz 'dy'dx)kR(H

j4

1)'y,'x(Jjk

)1)y,x((j

)y,x(Pj)y.x(E

ncell

m20mn 'dy'dx)kR(H

4

kZ

mcell

m20

rmmm 'dy'dx)kR(H

4k

)1(jkZ

2m2

mm 'yy'xxR

Approximation to Matrix Evaluation

approximate the square cell with a circular area of the same surface area so that the integral can be evaluated analytically

cell

2

0'

a

0

20 'd'd')kR(H'dy'dxI

2

210

k

4j)ka(H)k(J

k

a2I,a

)ka(H)k(Jk

a2I,a 2

01

)kR(H)ka(J2

aZ mn

20n1

nmn

Approximation to Matrix Evaluation

)1(jk4

k

k

4j)kR(H

2

aZ

rm2mn

21

mmm

2nm

2nmmn )yy(xxR

)1(jk4

k

k

4j)kR(H

2

aZ

rm2mn

21

mmm

)1(k

j)kR(H2

a

)1(

11

kj)kR(H

2

aZ

rm

rmmn

21

m

rmmn

21

mmm

note that when and only appears in the diagonal term

1)k(J 0 0rm

Bistatic Cross Section

N

1n

)sinydosx(jkn1

nn

2

TMnne)ka(J

k

a2j

4

k)(

note that the sampling rate should be 10/d

where r0d /

Scattering from Homogeneous Dielectric Cylinders: Surface Integral Equations Discretized with Pulse Basis

and Delta Testing Functions, TM Case

x

y S

0E2, H2

E1, H1

1 n

for an inhomogeneous cylinder, we employed the volume integral equation formulation in which a volume discretization is required

for a homogeneous cylinder, it is more convenient to formulate the problem using the surface integral equation approach in which only a surface discretization is necessary

Formulation of the Surface Equivalent Problem 1

x

y

PEC

E2, H2

0

J2,M2

n

nEM,HnJ 2222

k

radiating in free space

Formulation of the Surface Equivalent Problem 2

radiating in a homogeneous medium of

nk

x

y

PEC

E1, H1

0

J1,M1)n(EM,HnJ 1111

11 ,

Surface Equivalent Problem 1

x

y

PEC

E2, H2

0

J2,M2

nk

FAj

1AjE

00

s

0A

for TM case

220s FAjE

220itotal FAjEE

220totali FAjEE

nEMt tt


x

y

PEC

E2, H2

0

J2,M2

nk

2z0tiz FAjME

S

xyz0t

iz y

F

x

FAjME

'drj4

)kR(HJA

20

2

'drj4

)kR(HMF

20

2


nk

x

y

PEC

E1, H1

0

J1,M1

110s FAjE

1s EE

no source

'AA 21

'FF 21

t12 MtnEnE

'drj4

)R'k(HJ'A

20

22

'drj4

)R'k(HM'F

20

22

r00'k r00 /'


nk

x

yE1, H1

0

J1,M1

110i FAjE0

'F'AjE0 220i

n)'F'Aj(nE0 2201

S

x2y2z20t y

'F

x

'F'AjM0

Matrix Equation Using Puse/Delta Functions

M

J

DC

BA

0

E iz

n21

ncell mm

0nn

nncell

20m

iz m'dt)kR(H

R

)'t(yy)'t(cos

R

)'t(xx)'t(sin

j4

km)nm(j'dt)kR(H

j4

jk)t(E

n21

ncell mm

nnn

ncell

20 m'dt)R'k(H

R

)'t(yy)'t(cos

R

)'t(xx)'t(sin

j4

'km)nm(j'dt)R'k(H

j4

''jk0

Matrix Elements Using Puse/Delta Functions

M

J

DC

BA

0

Eiz'dt)kR(H

4

kA

ncell

20mn

'dt)kR(HR

)'t(yy)'t(cos

R

)'t(xx)'t(sin

j4

kB m

21

ncell mm

0mn

'dt)R'k(H4

''kC

ncell

20mn

'dt)R'k(HR

)'t(yy)'t(cos

R

)'t(xx)'t(sin

j4

'kD m

21

ncell mm

mn

For the Self Terms

2

1'dt)kR(H

R

)'t(yy)'t(cos

R

)'t(xx)'t(sin

j4

km

21

ncell mm

0

recall that

2

1

2

11Bmm

2

1)

2

1(1Dmm

0.5-0.5

Combined Field Equation for PEC Cylinder

x

y

Hi

PECi

Hzi=exp(-jk[x cos i + y sin

i])

10 circumference, pulse basis/delta testing

Aj

1AjtEt

00

i

Stisz Az)t(J)t(H

EFIE

MFIE


For the EFIE, the solution will not be unique if has a nonzero solution. Let us consider the problem of finding the resonant frequencies of a cavity by setting

or [A]{x}=0

to determine the eigenvalues of matrix A from which the resonant frequencies can be calculated. The current is then equal to a linear combination of the eigenvectors correspond to each of the eigenvalues and therefore

0)J(LE

0)J(LE

0J

Uniqueness of the Solution

Does this produce an external field? The answer is no since

(no source). 0EE)J(L igentialtan

sgentialtanE

For the magnetic field equation , , just inside S.

If is not zero then

but

which implies that and hence, the field outside S is not unique.

0)J(LH

0Hn s

J s

ernalintsexternal HHnJ

0H sernalint

0H sexternal


it

ist

s EHn)J(E)J(Hn

0J0)J(E)J(Hn st

s

then the solution is unique

Fj

AkAE

0

2s

0

2s

j

FkFAH

0nEF

on the PEC


*

st

sst

s )J(E)J(Hn)J(E)J(Hn

S

*ss

2

st

2

S

st ds)n()J(H)J(EalRe

2ds)J(E)J(H

= 0

= 0

the last term represent real power flowing inside S which is equal to 0 for lossless media and > 0 for lossy media (sourceless)

0)J(H st

0)J(Est

just inside S on S


If is real and positive, on S+, otherwise the above expression cannot be zero.

0)J(H st

Therefore . The solution to the combined field equation is unique at all frequencies.

The combined field equation requires the same number of unknowns but the matrix elements are more complicated to evaluate. Typical value of is between 0.2 and 1.

J0HHn St

St

limitation of pulse basis/delta testing discretization: te-wave efie difficulties lie in the...

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