《计算电磁学》第十讲ppt课件.ppt

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1、11/7/2022 4:55 AM,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 3 Two-dimensional Electromagnetic Fields,Dec. 12 , 2011,11/7/2022 4:55 AM,Outline,3.1 Transverse Magnetic Fields (横磁场), 3.2 Conducting Cylinders, TM

2、case (导电柱，TM情形), 3.3 Various Approximations (各种近似),3.4 Transverse Electric Fields (横电场), 3.5 Conducting Cylinders, TE case (导电柱，TE情形), 3.6 Alternative Formulation (替代表达）,11/7/2022 4:55 AM,3.1 Transverse Magnetic Fields (横磁场),An arbitrary electromagnetic field can be expressed as the sum of a transve

3、rse magnetic (TM) part and a transverse electric (TE) part. The TM part has only components of magnetic field H transverse to z, and the TE part has only component of E transverse to z.,For two-dimensional fields in isotropic media, the TM part has only a z component of E and the TE part only a z co

4、mponent of H.,In many cases, the TM and TE parts can be treated separately, reducing the problem to a scalar problem.,In this section, we only consider the TM fields!,11/7/2022 4:55 AM,The time harmonic electromagnetic field ( time variation) satisfies the,Maxwells equations,(3-1),(3-2),where J is t

5、he volume distribution of electric currents.,For TM fields, assume that , and similarly for .,The Maxwells equations then lead to,(3-3),where is the wavenumber ( is the wavelength).,This is a two-dimensional Helmholtz equation (亥姆赫兹方程).,11/7/2022 4:55 AM,Solutions may be obtained by first finding th

6、e field from a two-dimensional point source, that is, a three-dimensional line source.,The field at due to a filament of current I at is,(3-4),where is the intrinsic impedance of free-space and,is the Hankel function （汉克尔函数）of the second kind, zero order.,The of (3-4) is the Greens function for the

7、operator of (3-3).,The solution is the superposition of due to all elements of source , or,11/7/2022 4:55 AM,(3-5),where the integration is over the cross section of the cylinder of currents .,New words (生词)：transverse electric 横电的isotropic 各向同性scalar 标量distribution 分布wavenumber 波数electric currents

8、电流filament细丝,细线zero order 零阶intrinsic impedance 本征阻抗Greens function 格林函数superposition 叠加integration 积分cylinder 柱,11/7/2022 4:55 AM, 3.2 Conducting Cylinders, TM case (导电柱，TM情形),Consider a perfectly conducting cylinder excited by an impressed electric field as represented by Fig. 3-1. The impressed f

9、ield induces surface currents on the,conducting cylinder, which produce a scattered field (散射场) .,Fig. 3-1. Cross section of a cylinder and coordinate system.,11/7/2022 4:55 AM,The field due to is given by (3-5) specialized to the cylinder surface C.,The boundary condition is,on C,(3-6),Combining (3

10、-5) and (3-6), we have the integral equation,(3-7),: known,: unknown,11/7/2022 4:55 AM,The simplest numerical solution of (3-7) consists of pulse basis function and point matching procedure.,The scatterer contour C is divided into N segments and pulse functions defined as,(3-8),Letting the electric

11、current , substituting it in (3-7), and the resultant,equation at the midpoint of each , we obtain the matrix equation,(3-9),where the elements of are the coefficients.,11/7/2022 4:55 AM,the elements of are,(3-10),The elements of are,(3-11),A solution for the current is given by .,No simple analytic

12、al expression is available for the integral (3-11).,But we can evaluate it by various approximations.,The crudest approximation is to treat an element as a filament of current,when the field point is not on .,11/7/2022 4:55 AM,That is,(3-12),when .,For the diagonal elements (对角元素) , the Hankel funct

13、ion has integral,singularity, and the integral must be evaluated analytically.,For this problem, we approximate by a straight line and use the small argument formula,(3-13),where is Eulers constant.,11/7/2022 4:55 AM,An evaluation of (3-11) then gives,(3-14),where,Better approximations for this prob

14、lem will be discussed in Section 3-3.,Example. Consider TM plane-wave scattering by conducting cylinders. In this case, the impressed field is a uniform plane wave. The angle,between the wave vector and the x axis is .,The incident field is,(3-15),11/7/2022 4:55 AM,This determines the excitation acc

15、ording to (3-10).,An approximation evaluation of is given by (3-12) and (3-14).,The solution for is found by matrix inversion in the usual manner.,We calculate the radar cross section (雷达散射截面) , which is,(3-16),where is the distant field from .,It can be found by using the asymptotic expression for

16、in (3-5).,11/7/2022 4:55 AM,The result is,(3-17),where,(3-18),Substituting (3-15) and (3-17) in (3-16), we obtain,(3-19),11/7/2022 4:55 AM,erfectly conducting 完纯导体 induce 感应 scattered field 散射场combine 组合 evaluate 计算 distant field from 远离integral singularity 积分奇异性 straight line 直线radar cross section

17、雷达散射截面 asymptotic 渐进的,New words 生词,11/7/2022 4:55 AM, 3.3 Various Approximations (各种近似),The accuracy of a solution and the convergence rate depend on the approximation.,The solution of Section 3-2 can be improved by more accurate evaluation of the .,For the , additional terms can be included in (3-1

18、3), but this will not,appreciably affect convergence, since (3-14) is exact in the limit .,(2) For the terms, , we can expand the integrand of (3-11) in a Taylor,series about , and integrate the dominant terms analytically.,This will give both accuracy and convergence to the exact solution as .,The

19、rate of convergence is almost twice as fast if a piecewise linear approximation,to is used instead of the step approximation.,11/7/2022 4:55 AM,The solution can be obtained by using the Galerkins procedure, using pulses for both expansion and testing functions.,It was found that, for solutions of th

20、e subsectional-basis type, the accuracy and convergence of Galerkin solution were about the same as for the point-matching solution.,In addition, Galerkins method has great utility in perturbational solutions.,That is, when the solution is represented by only one expansion function or by a few funct

21、ions.,When we compute them using a computer, we divide each segment into,smaller subintervals, and approximate the integral over each subinterval by (3-12),if nonsingular and by (3-14) if singular.,11/7/2022 4:55 AM,Let us explain it in detail. Let Fig. 3-2 (a) represent a small section of the conto

22、ur of a cylindrical conductor.,Fig. 3-2. (a) Section of the contour. (b) Expansion function consisting of three constrained pulses.,(a),(b),11/7/2022 4:55 AM,Let the subintervals , , and be further subdivided as indicated by points a, b, c, and d.,Fig. 3-2 (b) shows the same contour straightened out

23、, and an expansion function constructed of three pulses.,This three-stepped function approximates a triangle function, shown dashed.,Each represents the field at due to expansion function at,.,When ,(3-20),where and are given by (3-12) with replaced by and,is given by (3-14) with replaced by .,11/7/

24、2022 4:55 AM,For non-diagonal elements, the procedure is the same, except that (3-12) is used for,all since the first point never coincide with the source point.,If we wish an approximation to the Galerkin solution, instead of the point-matching solution, the functions of Fig. 3-4 can be used for bo

25、th expansion and testing.,Thus, we can calculate it using approximations (3-12) and (3-14), which is,(3-21),where the are the same that appear in (3-20).,In the Galerkin solution, the of (3-10) should also be modified to represent,a numerical integration of with the testing function of Fig. 3-4.,11/

26、7/2022 4:55 AM,Accuracy 精度Convergence 收敛the rate of convergence 收敛速度appreciably可察觉地,明显地;相当地integrand 被积函数dominant占优势的，主导的，主要的utility实用, 效用perturbational 微扰的dash 虚线,New words生词,11/7/2022 4:55 AM,3.4 Transverse Electric Fields (横电场),A two-dimensional TE field in isotropic media has no z component of E

27、 and only a z component of H.,The fields can be illustrated by,(3-22),(3-23),where the magnetic vector potential A and the electric scalar potential satisfy,(3-24),(3-25),11/7/2022 4:55 AM,The electric charge density q is related to J by the equation of continuity (连续性),(3-26),Both (3-24) and (3-25)

28、 are Helmholtz equations, the same as (3-3).,Hence, the solutions are of the form (3-5).,Defining the two-dimensional Greens function,(3-27),we can express solutions to (3-24) and (3-25) in unbounded two-dimensional space as,(3-28),注： unbounded 无界的,11/7/2022 4:55 AM,(3-29),where the integration is o

29、ver a z=constant cross section of the cylinder., 3.5 Conducting Cylinders, TE case (导电柱，TE情形),The conducting cylinder is illuminated with a TE plane wave. We need to determine the current on the cylinder and the field produced by this current.,In this Section, we consider the H-field formulation.,Th

30、e total field at any point is the sum of the impressed field plus,the scattered field due to the J on C.,(3-30),That is,11/7/2022 4:55 AM,The scattered field is related to its source J by (3-22) and (3-28), or,(3-31),where the vector designates the reference direction of J.,If the interior of C lies

31、 on the left side of , then,(3-32),where denotes that is evaluated just external to C.,Specializing (3-30) to , we have,(3-33),The is not continuous at C.,Thus, the Greens function G is singular.,A simple interchange of differentiation and integration is not always possible.,11/7/2022 4:55 AM,Fig. 3

32、-3 shows an expanded view of the conductor boundary to help clarify these concepts.,Fig. 3-3. Section of cylinder boundary.,At point a on , , and the point b on , .,If the scatterer is a conducting sheet of infinitesimal thickness, it should be treated as the limit of one of finite thickness.,11/7/2

33、022 4:55 AM,We can write (3-33) in general operator notation as,(3-34),where,(3-35),Let us solve (3-35) using the method of moments.,The basis function is the pulse function. The point-matching procedure is used for testing.,The current , and the resultant matrix equation is (3-9) with,(3-36),11/7/2

34、022 4:55 AM,(3-37),where is the Kronecker delta function.,stands for at on due to unit current density on at .,Fig. 3-4. Elements of current and local coordinates.,Fig. 3-4. represents a typical current element and local coordinates,.,11/7/2022 4:55 AM,From symmetry, and the fact that the discontinu

35、ity in is J, we have,(3-38),and hence, by (3-37),(3-39),If and the field point is distant from , then the source,behaves as a point source.,(3-40),and from (3-22),(3-41),From (3-28),11/7/2022 4:55 AM,where is the Hankel function of order 1.,We can translate this to an arbitrary origin by replacing b

36、y and,by , where,(3-42),is unit vector from the source point to the field point .,The result can be used as an approximation for all .,Thus, (3-37) becomes, for,(3-43),The solution is given by .,11/7/2022 4:55 AM,Example. Consider TE plane-wave scattering by conducting cylinders.,An impressed unifor

37、m plane wave from the direction is given by,(3-44),The are determined from this by (3-36) and the are given by (3-39) and,(3-43) for a first-order solution.,Again the scattering cross section (雷达散射截面) is of interest, given by,(3-45),where is the distant field from J, obtained by using the asymptotic

38、 formula,for in (3-41) and summing over all elements of source.,11/7/2022 4:55 AM,The result is,where K is given by (3-18).,Substituting (3-44) and (3-46) in (3-45), we obtain,(3-46),(3-47),which can be evaluated once J is found.,11/7/2022 4:55 AM, 3.6 Alternative Formulation (替代表达）,The TM problem w

39、as treated by an E-field formulation in Section 3-2.,Both cases can be treated either by an E-field method or H-field method.,The TE problem was treated by an H-field formulation in Section 3-5.,Let Fig. 3-1 represent a conducting cylinder excited by an impressed TE field transverse to z.,The scatte

40、red field is produced by produced by transverse currents J on C.,For the present problem, these become,(3-48),11/7/2022 4:55 AM,(3-49),(3-50),where G is given by (3-27). The boundary condition is the tangential component of total E vanishes on the conductor.,That is,(3-51),11/7/2022 4:55 AM,Thank you.,11/7/2022 4:55 AM,To be continued.,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,11/7/2022 4:55 AM,