《计算电磁学》第八讲ppt课件.ppt

计算电磁学Part II： 矩量法,Dr. Ping DU (杜平),School of Electronic Science and Applied Physics, Hefei University of Technology,E-mail: ,Chapter 1 Deterministic Problems,Nov. 24 , 2011,2,Outline,1.1 Introduction (介绍), 1.2 Formulation of Problems (问题的描述）, 1.3 Method of moments (矩量法）, 1.4 Point Matching(点匹配或点选配), 1.5 Subsectional Bases (子域基）, 1.6 Approximate Operators (近似算子）, 1.7 Extended Operators (扩展算子）,3,1.1 Introduction,Consider equations of the inhomogeneous type (非齐次型),(1-1),where L is an operator (算子), f is the field or response (unknown function to be determined), and g is the source or excitation (known function).,By the term deterministic we mean that the solution to (1-1) is unique. That is, only one f is associated with a given g.,4,Two terminologies: Analysis (分析)& Synthesis (综合),A problem of Analysis involves the determination of f when L and g are given.,2) A problem of Synthesis involves the determination of L when f and g are specified.,Antenna array synthesis 天线阵列综合,Generally speaking, the solution is not unique.,The solution is unique.,Electromagnetic inverse problems 电磁逆问题,Two examples:,5,where and are scalars and * denotes a complex conjugate.,(1-4),(1-3),(1-2),An inner product is a scalar defined to satisfy,1. Inner product (内积),1.2 Formulation of problems,6,2. Operator and its properties,An adjoint operator (伴随算子) and its domain,(1-5),for all f in the domain of L.,If , an operator is self-adjoint (自伴的).,The domain of is that of L.,Properties of the solution depend on properties of the operator. An operator is real if Lf is real whenever f is real.,7,An operator is positive definite if,(1-6),for all in its domain.,3. Solution,If the solution to exists and is unique for all g, then the inverse,operator exists such that,(1-7),If g is known, then (1-7) represents the solution to the original problem.,(1-7) is an inhomogeneous equation for g if f is known. And its solution is .,8,L and are a pair of operators, each of which is the inverse of the other.,Example 1. Given g(x), find f(x) in the interval satisfying,This is a boundary problem for which,The range of L is the space of all functions g in the interval that we wish to consider.,(1-8),(1-9),(1-11),9,The solution to (1-8) is not unique (不唯一) unless appropriate boundary Condition are included. In other words, both the differential operator and its domain are required to define the operator.,Define an inner product for this problem is,(1-11),(1-11) satisfies the postulates (条件) (1-2) to (1-4), as required.,The definition (1-11) is not unique. For example,(1-12),where w(x)0 is an arbitrary weighting function (加权函数), is also an acceptable inner product.,10,However, the adjoint operator depends on the inner product, which can often be chosen to make the operator self-adjoint.,To find the adjoint of a differential operator, we form the left side of (1-5), and integrate by parts (分部积分) to obtain the right side.,For the present problem,(1-12),The last terms are boundary terms, and the domain of may be chosen so that,these vanish.,11,The first boundary terms vanish by (1-9), and the second vanish if,(1-14),It is evident that the adjoint operator to (1-10) for the inner product (1-11) is,(1-15),Since and the domain of is the same as that of L, the operator is,self-adjoint (自伴的).,It can be observed that L is a real operator, since is Lf real when f is real.,12,That L is a positive definite operator shown from (1-6) as follows:,(1-16),Note that L is a positive definite operator even if f is complex.,The inverse operator to L is,(1-17),where G is the Greens function,13,(1-18),That is self-adjoint follows from the proof that L is self-adjoint, since,(1-19),That is positive definite whenever L is positive definite, and vice versa.,14,1.3 Method of Moments,Lets discuss a general procedure for solving linear equations, called the method of moments(矩量法).,Consider the inhomogeneous equation,(1-20),where L is a linear operator, g is known, and f is to be determined.,Let f be expanded in a series of functions in the domain of L as,(1-21),where the are constants. We shall call the expansion functions or basis functions.,15,Substituting (1-21) in (1-20), and using the linearity of L, we have,(1-22),It is assumed that a suitable inner product has been determined for,Define a set of weighting functions (权函数) or testing functions (测试函数),in the range of L.,the problem.,Take the inner product of (1-22) with each .,The result is,(1-23),m=1, 2, 3, ,16,This set of equations can be written in matrix form is,(1-24),where,(1-25),(1-26),17,If the matrix l is nonsingular its inverse exists.,The are then given by,(1-27),and the solution for f is given by (1-21).,For concise expression of this result, define the matrix of functions,(1-28),and write,(1-29),18,This solution may be exact or approximate, depending on the choice of the,and .,The matrix l may be either of infinite order (无限阶) or finite order (有限阶). The former one can be inverted only in special cases, for example, if it is diagonal (对角线的).,If the sets and are finite, the matrix is of finite order, and can be inverted.,Choices of the weighting function and the basis functions are very important. Some factors need to be considered:,(1) accuracy of solution desired,19,(4) realization of a well-conditioned matrix (好条件矩阵).,(3) size of the matrix, and,(2) ease of evaluation of the matrix elements,When analyzing the 3D scattering problem with RWG basis function,the double surface integrals are needed. It is very time consuming.,Characteristic function can be used to calculate the matrix elements.,For a PC with 1GB, the order of the matrix cannot be larger than 5000.,Otherwise, “Out of memory” will appear.,If the condition is bad, the convergence will be very slow.,To address this issues, the preconditioning techniques can be applied.,20,Example 2.,Consider the same equation as in the example of Section 1-2, but with specific source .,Our problem is,(1-30),(1-31),This is a simple boundary-value problem. Its solution is,(1-32),21,This problem can be solved by using the method of moments.,For a power series solution, let us choose,(1-33), so that the series (1-21) is,(1-34),Note that the term x is needed in (1-33), else the will not be in the domain of L.,That is, the boundary condition will not be satisfied.,22,For testing functions, choose,(1-35),The method is that of Galerkin (伽辽金法 ).,Evaluation of the matrices (1-25) and (1-26) for the inner product (1-11) and,is straightforward, and results in,(1-36),(1-37),23,For any fixed N (number of expansion functions), the are given by (1-27) and the approximation to f by (1-34).,For N=1, we have , , and from (1-24),For N=2, the matrix equation (1-24) becomes,(1-38),Solving it, we get,(1-39),24,For N=3, the matrix equation (1-24) becomes,(1-40),Solving it, we get,(1-41),Thus, we obtain, which is the exact solution.,25,Figure 1-1. Solutions using and Galerkins method.,Fig. 1-1shows the relation between the N and the accuracy.,26,1.4 Point Matching (点匹配或点选配),The integration involved in evaluating the of (1-25),is often difficult to perform in problems of practical interest.,A simple way to obtain approximate solutions is to require that equation (1-22) be satisfied at discrete points in the region of interest.,This procedure is called a point-matching method (点选配法）.,In terms of the method of moments, it is equivalent to using the Dirac delta func-tions as testing functions.,Example 3.,Reconsider the problem of Section 1-3, stated by (1-30) and (1-31).,27,Again we choose expansion functions (1-33), so that (1-22) becomes,(1-42),For a point-matching solution, let us take the points,(1-43),which are equispaced in the interval .,Requiring (1-42) to be satisfied at each gives us the matrix equation (1-24), with elements,28,(1-44),(1-45),This result is identical to choosing weighting functions (加权函数),(1-46),where is the Dirac delta function and applying the method of moments with inner product (1-11).,Then lets analyze the accuracy of the point matching method.,29,Consider the solution as N is increased.,For N=1, we have , , and from (1-27) .,For N=2, the matrix equation is,(1-47),Solving it, we get,(1-48),30,For N=3, the exact solution must be obtained again since the exact solution is a linear combination of the and we are applying N independent tests.,Fig. 1-2 plots the relation between the N and the accuracy.,Figure 1-2. Solutions using and point matching method.,31,1.5 Subsectional Bases (分域基函数),Example 4. Again consider the problem of Section 1-3, stated by (1-30) and (1-31).,N equispaced points on the interval are defined by the of (1-43).,A subinterval is defined to be of width 1/(N+1) centered on the . This is shown,in Fig. 1-3(a).,In the above discussion, the basis function is defined over the entire interval.,In fact, the basis function can also be defined over a subsection, which is called the sub-sectional basis function (分域基函数).,It is referred to as entire section basis function (全域基函数).,32,Fig. 1-3. Subsectional bases and functional approximations.,33,A function exists over only one subinterval is the pulse function (脉冲函数).,(1-49),A better behaved function is the triangle function (三角形函数), defined as,(1-50),Note that the pulse function cannot be used as the basis function unless that the extended or approximation operator is applied.,34,A function exists over only one subinterval is the pulse function (脉冲函数).,(1-49),A better behaved function is the triangle function (三角形函数), defined as,(1-50),For the case N=5 the function is shown in Fig. 1-3(d).,35,A linear combination of triangle functions of the form,(1-51),gives a piecewise linear approximation to f, as represented by Fig, 1-2(e).,For , we have,(1-52),where is the Dirac delta function.,36,To follow through the method of moments, let be the basis function.,is chosen as the testing function.,For inner product (1-11), the matrix elements of (1-25) and (1-26) are,(1-53),(1-54),37,1.6 Approximate Operators (近似算子),Example 5. Consider the problem (1-30) and (1-31) by a finite-difference approximation. This involves replacing all derivatives by finite differences.,In complex problems, it is sometimes convenient to approximate the operator to obtain approximate solutions.,For differential operators, the finite-difference approximation has been widely used.,For integral operators, an approximate operator can be obtained by approximating the kernel (核) of the integral operator.,38,(1-56),(1-55),For a given ,For the present problem, consider the interval divided into segments with end points (see Fig. 1-3(a).,For equal to one segment, .,A finite difference approximation to is,39,as for all f in the domain of L.,Apply the method of moments to the approximate equation,(1-57),subject to boundary conditions .,The point-matching procedure at is used.,The matrix elements are,(1-58),(1-59),40,1.7 Extended Operators (扩展算子),Operator: operation (运算) and a domain (定义域),Ways of extended operators:,(1) extend the domain, and,(2) extend the original domain of L.,41,Example 6. Suppose we wish to use pulse functions for an expansion of f in a,moment solution for the operator .,These are not in the original domain of L.,However, for any functions w and f in the original domain,(1-60),obtained from (1-11) by integration by parts (分部积分).,If Lf does not exist, but df/dx does exist, (1-60) can be used to define an extended operator.,42,This extends the domain of L to include functions f whose second derivatives do not exist, but the first derivatives do exist.,It is still assumed that .,To apply the method of moments using the pulse functions and the extended operator, let,(1-61),where P are the pulse functions defined by (1-49).,The testing functions are , where T are the triangle functions defined by (1-50).,43,The matrix elements of l are,(1-62),The matrix elements of g are,(1-63),Example B. Let us extend the original domain of to apply to functions,not satisfying the boundary condition,44,If an extended operator is defined by,(1-64),we have even if the original boundary conditions are not met.,So, the extended operator is self-adjoint(自伴的) regardless of boundary conditions.,Solve this problem using the extended operator.,1) choose the basis functions and testing functions,(1-65),When N3 these functions form a basis for the exact solution (1-32).,45,2) Evaluating the matrices,Using the extended operator for , for N=4 we obtain,(1-66),3) Get the coefficient matrix,Solving (1-66), we have,(1-67),This is the exact solution.,46,Note that if (1-65) are used with the original operator a singular l,matrix results, and hence no solution is obtained.,To illustrate convergence using the extended operator, Fig. 1-4 shows plots of the cases N=2 and N=3, and the exact solution ( ).,When N=2, the matrix equation is,The solution is ,-,(1-68),47,When N=3, the matrix equation is,(1-69),The solution is,48,Fig. 1-4. Extended operator moment solutions using powers of x for expansion and testing.,49,Thank you!,