Limitation of pulse basis delta testing discretization te wave efie
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Limitation of Pulse Basis/Delta Testing Discretization: TE-Wave EFIE PowerPoint PPT Presentation


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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.

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

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Limitation of pulse basis delta testing discretization te wave efie

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

Further Explanation


Discussions

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


Discussions1

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 efie

TE Wave Scattering from PEC Strips or Cylinders – 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

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

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


Triangular basis function

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


Pulse doublet

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

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

Matrix Equation


Bistatic cross section

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


Limitation of pulse basis delta testing discretization te wave efie

y

e0m0

er(x,y) e0

m0

Ei

x

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

inside scatterer

outside scatterer


Formulation of the volume integral equation

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

Volume Integral Equation

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

0 otherwise

Jz(x,y)=


Final equation

Final Equation


Approximation to matrix evaluation

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


Approximation to matrix evaluation1

note that when and only appears in the diagonal term

Approximation to Matrix Evaluation


Bistatic cross section1

Bistatic Cross Section

note that the sampling rate should be 10/ld where


Limitation of pulse basis delta testing discretization te wave efie

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


Formulation of the surface equivalent problem 1

E2, H2

J2,M2

y

mo, eo

0

PEC

x

Formulation of the Surface Equivalent Problem 1

radiating in free space


Formulation of the surface equivalent problem 2

J1,M1

y

E1, H1

m1, e1

0

PEC

x

Formulation of the Surface Equivalent Problem 2

radiating in a homogeneous medium of


Surface equivalent problem 1

E2, H2

J2,M2

y

mo, eo

0

PEC

x

Surface Equivalent Problem 1

for TM case


Surface equivalent problem 11

E2, H2

J2,M2

y

mo, eo

0

PEC

x

Surface Equivalent Problem 1


Surface equivalent problem 2

J1,M1

y

E1, H1

m1, e1

0

PEC

x

Surface Equivalent Problem 2

no source


Surface equivalent problem 21

J1,M1

y

E1, H1

m1, e1

0

x

Surface Equivalent Problem 2


Matrix equation using puse delta functions

Matrix Equation Using Puse/Delta Functions


Matrix elements using puse delta functions

Matrix Elements Using Puse/Delta Functions


For the self terms

-0.5

0.5

For the Self Terms

recall that


Combined field equation for pec cylinder

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 cylinder1

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

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 cylinder2

Combined Field Equation for PEC Cylinder

then the solution is unique

on the PEC


Uniqueness of the solution1

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 solution2

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.


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