《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 TechnologyE-mail: Chapter 1 Deterministic ProblemsNov. 24 , 20112Outline1.1 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） 31.1 Introduction Consider equations of the inhomogeneous type (非齐次型) ( )L fg(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. 4Two terminologies: Analysis (分析)& Synthesis (综合)1) 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:5where and are scalars and * denotes a complex conjugate. 0if0*,0if0ffff(1-4),fg hf hg h (1-3),f gg f(1-2)An inner product is a scalar defined to satisfy 1. Inner product (内积)1.2 Formulation of problems62. Operator and its propertiesAn adjoint operator (伴随算子) and its domain aL,aLf gf L g(1-5)for all f in the domain of L. If , an operator is self-adjoint (自伴的). aLLThe domain of is that of L.aLProperties of the solution depend on properties of the operator. An operator is real if Lf is real whenever f is real.7An operator is positive definite if 0*,00positive definitefLfpositive semidefinitenegative definite(1-6)for all in its domain. 0f 3. SolutionIf the solution to exists and is unique for all g, then the inverse( )L fgoperator exists such that1L1( )fLg(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 . ( )L fg8L and are a pair of operators, each of which is the inverse of the other. 1LExample 1. Given g(x), find f(x) in the interval satisfying 01x22( )d fg xdx(0)(1)0ffThis is a boundary problem for which 22dLdx The range of L is the space of all functions g in the interval that we wish to consider. 01x(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 10,( ) ( )f gf x g x dx(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, 10( ) ( ) ( )w x f x g x dx(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,12112000121200,d fdf dgdfLf ggdxdxdxdx dxdxd gdgdffdxfgdxdxdx(1-12)The last terms are boundary terms, and the domain of may be chosen so that these vanish. aL11The first boundary terms vanish by (1-9), and the second vanish if (0)(1)0gg(1-14)It is evident that the adjoint operator to (1-10) for the inner product (1-11) is 22adLLdx (1-15)Since and the domain of is the same as that of L, the operator isaLLaLself-adjoint (自伴的).It can be observed that L is a real operator, since is Lf real when f is real. 12That L is a positive definite operator shown from (1-6) as follows:12*11*2000210,d fdfdfdffLffdxdxfdxdx dxdxdfdxdx(1-16)Note that L is a positive definite operator even if f is complex. The inverse operator to L is110( )( , ) ( )LgG x x g x dx(1-17)where G is the Greens function13(1)( , )(1) xxxxG x xx xxx(1-18)That is self-adjoint follows from the proof that L is self-adjoint, since 1L11212,Lffg L g(1-19)That is positive definite whenever L is positive definite, and vice versa. 1L141.3 Method of Moments Lets discuss a general procedure for solving linear equations, called the method of moments(矩量法). Consider the inhomogeneous equation ( )L fg(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 123,fff nnnff(1-21)where the are constants. We shall call the expansion functions or basis functions. nnf15Substituting (1-21) in (1-20), and using the linearity of L, we have ()nnnL fg(1-22)It is assumed that a suitable inner product has been determined for, f gDefine a set of weighting functions (权函数) or testing functions (测试函数), 123,w w w in the range of L.the problem. Take the inner product of (1-22) with each . mwThe result is ,nmnmnwLfwg(1-23)m=1, 2, 3, 16This set of equations can be written in matrix form is mnnmlg(1-24)where11122122,.mnw Lfw Lflw Lfw Lf12n12,mw ggw g(1-25)(1-26)17If the matrix l is nonsingular its inverse exists. 1lThe are then given by n1nnmmlg(1-27)and the solution for f is given by (1-21). For concise expression of this result, define the matrix of functions 123nffff (1-28)and write 1nnnmnmffflg(1-29)18This solution may be exact or approximate, depending on the choice of the nfand . nw 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.20Example 2. Consider the same equation as in the example of Section 1-2, but with specific source . 214gx Our problem is 22214d fxdx (0)(1)0ff(1-30)(1-31)This is a simple boundary-value problem. Its solution is245( )623xxxf x (1-32)21This problem can be solved by using the method of moments.For a power series solution, let us choose 1nnfxx(1-33)1, 2,3,nN, so that the series (1-21) is 11()Nnnnfxx(1-34)Note that the term x is needed in (1-33), else the will not be in the domain of L. nfThat is, the boundary condition will not be satisfied. 22 For testing functions, choose 1nnnwfxx(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 22/Lddx is straightforward, and results in ,1mnmnmnlwLfmn(1-36)(38),2(2)(4)mmmmgwgmm(1-37)23For any fixed N (number of expansion functions), the are given by (1-27) and the approximation to f by (1-34). nFor N=1, we have , , and from (1-24) 111/3l 111/30g 111/10For N=2, the matrix equation (1-24) becomes 12111132301472512(1-38)Solving it, we get 121/107/12(1-39)24For N=3, the matrix equation (1-24) becomes 123113113253014712512513917057(1-40)Solving it, we get 12312013 (1-41)Thus, we obtain 24511623fxxx, which is the exact solution. 25Figure 1-1. Solutions using and Galerkins method.1nnfxxFig. 1-1shows the relation between the N and the accuracy. 261.4 Point Matching (点匹配或点选配)The integration involved in evaluating the of (1-25) ,mnmnlwLfis 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). 27Again we choose expansion functions (1-33), so that (1-22) becomes21221()14Nnnndxxxdx (1-42)For a point-matching solution, let us take the points1mmxN1,2,3,.,mN(1-43)which are equispaced in the interval . 01xRequiring (1-42) to be satisfied at each gives us the matrix equation (1-24), with elements mx281(1)1nmnmln nN(1-44)2141mmgN (1-45)This result is identical to choosing weighting functions (加权函数) ()mmwxx(1-46)where is the Dirac delta function and applying the method of moments with inner product (1-11).( )xThen lets analyze the accuracy of the point matching method.29Consider the solution as N is increased. For N=1, we have , , and from (1-27) . 112l12g 11 For N=2, the matrix equation is 122213/92425/9(1-47)Solving it, we get1211823 (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. nfsFig. 1-2 plots the relation between the N and the accuracy. 00.10.20.30.40.50.60.70.80.9100.050.10.150.20.250.30.35xfExact solutionN=2N=1Figure 1-2. Solutions using and point matching method.1nnfxx311.5 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). mxA subinterval is defined to be of width 1/(N+1) centered on the . This is shownmxin 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).32Fig. 1-3. Subsectional bases and functional approximations.33A function exists over only one subinterval is the pulse function (脉冲函数).11|2(1)( )10|2(1)xNP xxN(1-49)A better behaved function is the triangle function (三角形函数), defined as 11 |(1)|1( )10|1xNxNT xxN(1-50) Note that the pulse function cannot be used as the basis function unless that the extended or approximation operator is applied.34A function exists over only one subinterval is the pulse function (脉冲函数). 11|2(1)( )10|2(1)xNP xxN(1-49)A better behaved function is the triangle function (三角形函数), defined as 11 |(1)|1( )10|1xNxNT xxN(1-50)For the case N=5 the function is shown in Fig. 1-3(d). 2()T xx35A linear combination of triangle functions of the form 1()NnnnfT xx(1-51)gives a piecewise linear approximation to f, as represented by Fig, 1-2(e). For , we have 22/Lddx 11()(1)()2 ()()nnnnLT xxNxxxxxx(1-52)where is the Dirac delta function. ( )x36To follow through the method of moments, let be the basis function. ()nnfT xx is chosen as the testing function. ()mmwP xxFor inner product (1-11), the matrix elements of (1-25) and (1-26) are2(1)(1)101mnNmnlNmnmn 22141311(1)mmgNN(1-53)(1-54)371.6 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. 38211(1)2 ( )11dL fNfxf xfxNN(1-56) 22212()d ff xxf xxf xdxx(1-55)122dfxxfxfxdxxFor a given ,x For the present problem, consider the interval divided into segments with end points (see Fig. 1-3(a). 01x1N nxFor equal to one segment, . 1/(1)xN A finite difference approximation to is 22/Lddx 39 as for all f in the domain of L. dLLN N Apply the method of moments to the approximate equation214dL fx (1-57)subject to boundary conditions . (0)(1)0ffThe point-matching procedure at is used. mxThe matrix elements are 222(1)(1)101mnNmnlNmnmn (1-58)2141mmgN (1-59)401.7 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.41Example 6. Suppose we wish to use pulse functions for an expansion of f in a moment solution for the operator . 22/Lddx These are not in the original domain of L.However, for any functions w and f in the original domain,10,dw dfw Lfdxdx dx(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 .(0)(1)0ffTo apply the method of moments using the pulse functions and the extended operator, let 1()NnnnfP xx(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).()mmwT xx43The matrix elements of l are 2(1)(1)101mnNmnlNmnmn (1-62)The matrix elements of g are 120()(14)mmgT xxxdx(1-63)Example B. Let us extend the original domain of to apply to functions 22/Lddx not satisfying the boundary condition (0)(1)0ff44 If an extended operator is defined by eL1100,edww L fwLfdxfdx(1-64)we have even if the original boundary conditions are not met. ,eew L ff L wSo, 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 functionsnnnfwx1,2,nN(1-65)When N3 these functions form a basis for the exact solution (1-32).452) Evaluating the matricesUsing the extended operator for , for N=4 we obtain ,emnmnlwL f123431234287221723251572111352512224027455735(1-66)3) Get the coefficient matrix Solving (1-66), we have 12345612013(1-67)This is the exact solution. 46Note that if (1-65) are used with the original operator a singular lmatrix results, and hence no solution is obtained.22/Lddx To illustrate convergence using the extended operator, Fig. 1-4 shows plots of the cases N=2 and N=3, and the exact solution ( ). 4N When N=2, the matrix equation is 1231228172315 The solution is ,11310275 -(1-68)47When N=3, the matrix equation is 123312328717232157211132512(1-69)The solution is 10.744444444444442-0.011111111111113-0.7407407407407448Fig. 1-4. Extended operator moment solutions using powers of x for expansion and testing.00.10.20.30.40.50.60.70.80.91-0.1-0.0500.050.10.150.20.250.30.35xfExact solution (N=4)N=2N=349Thank you!