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The Method of Moments Lf = g where L is a linear operator
replace f by a set of basis functions fi is in the domain of L A set of testing functions (or weighting functions) wi in the range of L is needed to determine ai
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An inner product is defined as:
Due to the linearity of the operator L, we have each weighting function yields one equation
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Matrix Equation
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Method of Moments is the n-th moment of f , when xn is replaced by wn, we continue to call the integral a moment of f . it is also called the Method of Weighted residuals note that is only an approximation to the exact f a residual R can be defined as these weighted residuals are forced to be zero by adjusting the values of a’s this procedure can also be called the method of projection
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Method of Moments an adjoint operator can be defined as
if L= La ,then this operator is self-adjoint and the domain of La is that of L if wi is chosen such that wi =fi, this is called the Galerkin’s method and the resulting matrix is a symmetric one
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Example consider the solution is
the basis function chosen is in the form of fn =(x-xn+1), fn is in the domain of L and also satisfies the boundary conditions
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Apply Galerkin’s Method
exact solution is obtained when N=3
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Basis and Testing Functions
the basis functions are defined over the entire domain of the problem, i.e., and they are called entire-domain basis functions the testing function wi (= fi) is also entire-domain functions if wi is chosen as d(x-xm) for different values of xm between 0 and 1, the testing function is called a delta testing this procedure is also called the point-matching method or the collocation method the advantage of the point matching method is that no integration is necessary, however, the accuracy of the method depends on the locations and the number of testing points
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Delta Testing
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Subdomain Functions P(x-x2) 0 x1 x2 x3 x4 x5 1 0 x1 x2 x3 x4 x5 1 Since the operator involves a second order derivative, we have to choose a function that allows us to carry out the derivative 0 x1 x2 x3 x4 x5 1 the testing function has to be at least a pulse (P) triangular (T) basis function can also be used as the testing function.
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Extended Operator an extended operator extends the domain of L to include function f whose second derivative does not exist but whose first derivative exists let w = T and f =P, the second term on the right-hand-side vanishes in each subregion the extended operator can also be used to include functions that do not satisfy the boundary condition
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Extended Operator in the original operator consider
suppose f does not satisfy boundary condition, i.e., it does not vanish at x = 0, 1
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the equation can be rearranged
Extended Operator the equation can be rearranged
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Extended Operator this implies Lea = Le self adjoint, even though the original boundary conditions are not met one can choose xn as the basis function as oppose to x – xn+1 in the original operator n 4, exact solution is obtained
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Integral Equation Method for Scattering from Infinite Conducting Cylinder
x y S PEC E, H the tangential electric field on the conducting surface is zero Js only exists on the surface S with the conductor removed
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Equivalent Problem with the PEC Cylinder Removed
x y PEC E, H mo, eo Jz at the original location of the conductor radiates in a homogeneous medium normal incidence TM-to-z wave, the field components are Ez, Hx and Hy for a 2-D problem, the field is invariant in z
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Formulation of an Integral Equation for 2D Problem
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Discretization of the Integral Equation
Pn = 1 if t is in cell n = 0 otherwise jn’s are unknowns to be determined delta testing
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Matrix Equation matrix [Z] is referred as the impedance matrix and [Ei] is the excitation vector the point (xm,ym) represents the center of the m-th segment in the model
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Discretization of the Integral Equation
if the cells are small compared to a wavelength, we can approximate the integral as for the diagonal term, i.e., when m = n, the matrix element cannot be approximated as such as a singularity exists when Rmn=0 for small argument, the Hankel function can be replaced by a power series expansion g=
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Approximation of the Matrix Elements
the radius of curvature of each cell is large enough so that each cell may be considered flat
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Some Considerations we need wn and (xn, yn), choose one-tenth of a wavelength for each segment approximating the cylinder the phase center (xn, yn) should be on the scatterer and the total circumference of the cylinder model should be scaled to that of the original scatterer contour model errors: precise location of phase center and flat strip model of the smooth surface discretization error: Jz(t) represented by pulse and point matched only at the phase center approximations: integrand assumed to be constant on the segment for the off-diagonal term numerical errors: round off and calculation of Hankel functions
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Increase the Accuracy of the Solution
improve the self-term evaluation by removing the singularity quadrature can be used to evaluate the 1st and 2nd integrals 3rd integral can be evaluated analytically electric-field integral equation (EFIE)
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TE-Wave Scattering from Conducting Cylinders: MFIE Discretized with Pulse Basis Functions and Delta Testing Functions for the TE case, we have non-zero Hz, Ex and Ey components only on PEC x y Hi PEC
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MFIE Formulation where and
S+ denotes a surface an infinitesimal distance outside the cylinder unit tangent vector can be defined as with the angle W depicted below W(t) Wn wn (xn, yn)
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Pulse Basis and Delta Testing
Pn = 1 if t is in cell n = 0 otherwise
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Derivation of the Off-Diagonal Terms
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Derivation of the Self Terms
the limiting procedure is carried out for flat cells and produces Zmm=-0.5 regardless of the cell size the sole contribution to the integral arises from the immediate neighborhood of the singularity of the Green’s function, since electric currents on a flat plane produce no tangential magnetic field elsewhere on the same plane Even if the cells of the cylindrical model are slightly curved, the diagonal matrix elements are approximately given by Zmm=-0.5
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Rigorous Consideration
when y’not equals to 0 but x equals to zero, Hs equals to zero and hence, only at x = 0 and y = 0, the integral is non-zero first term in the integral goes to zero as it is antisymmetric small x
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Bistatic Cross Section for TE Polarization
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