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 x o to x 1 and oriented in the x direction.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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
Further Explanation
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
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
TE Wave Scattering from PEC Strips or Cylinders â€“ EFIE
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
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
magnetic vector potential
electric scalar potential
continuity equation
t
tn-1
tn
tn+1
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
jn
jn/(tn-tn-1)
area = jn
area = jn
-jn/(tn+1-tn)
pulse doublet
Pulse Doublet
corresponds to the slopes of the triangular edges
cancel each other, zero total charge
no fictitious line charge generated
Testing Function
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
Matrix Equation
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
y
e0m0
er(x,y) e0
m0
Ei
x
TM Wave Scattering from Inhomogeneous Dielectric Cylinders â€“ 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
inside scatterer
outside scatterer
Formulation of the Volume Integral Equation
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,
no induced charge
Volume Integral Equation
Pn(x,y)=1 if (x,y) with cell n
0 otherwise
Jz(x,y)=
Final Equation
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
note that when and only appears in the diagonal term
Approximation to Matrix Evaluation
Bistatic Cross Section
note that the sampling rate should be 10/ld where
E2, H2
e0
y
S
E1, H1
e1
x
Scattering from Homogeneous Dielectric Cylinders: Surface Integral Equations Discretized with Pulse Basis and Delta Testing Functions, TM Case
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
E2, H2
J2,M2
y
mo, eo
0
PEC
x
Formulation of the Surface Equivalent Problem 1
radiating in free space
J1,M1
y
E1, H1
m1, e1
0
PEC
x
Formulation of the Surface Equivalent Problem 2
radiating in a homogeneous medium of
E2, H2
J2,M2
y
mo, eo
0
PEC
x
Surface Equivalent Problem 1
for TM case
E2, H2
J2,M2
y
mo, eo
0
PEC
x
Surface Equivalent Problem 1
J1,M1
y
E1, H1
m1, e1
0
PEC
x
Surface Equivalent Problem 2
no source
J1,M1
y
E1, H1
m1, e1
0
x
Surface Equivalent Problem 2
Matrix Equation Using Puse/Delta Functions
Matrix Elements Using Puse/Delta Functions
-0.5
0.5
For the Self Terms
recall that
y
PEC
Hi
x
fi
Hzi=exp(-jk[x cos fi + y sin fi])
Combined Field Equation for PEC Cylinder
1l0 circumference, pulse basis/delta testing
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
Uniqueness of the Solution
Does this produce an external field? The answer is no since
(no source).
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.
Combined Field Equation for PEC Cylinder
then the solution is unique
on the PEC
just inside S
on S
Uniqueness of the Solution
= 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)
Uniqueness of the Solution
If a is real and positive, on S+, otherwise the above expression cannot be zero.
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 a is between 0.2 and 1.