View
229
Download
0
Category
Preview:
Citation preview
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
Surface Equivalent Problem 1
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
Surface Equivalent Problem 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 /'
Surface Equivalent Problem 2
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
Combined Field Equation for PEC Cylinder
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
Combined Field Equation for PEC Cylinder
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
Uniqueness of the Solution
*
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
Uniqueness of the Solution
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
Recommended