《计算电磁学》第十讲ppt课件.ppt

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1、7/31/2022 8:21 PM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 3 Two-dimensional Electromagnetic Fields Dec. 12 , 20117/31/2022 8:21 PMOutline3.1 Transverse Magnetic Fields (横磁场) 3.2 Conducting Cylinders, TM case (导电柱

2、，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） 7/31/2022 8:21 PM3.1 Transverse Magnetic Fields (横磁场) An arbitrary electromagnetic field can be expressed as the sum of a transverse magnetic

3、(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 component of H.

4、 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! 7/31/2022 8:21 PM The time harmonic electromagnetic field ( time variation) satisfies the j teMaxwells equations j EHjHE + J(3-1)(3-2)where J is t

5、he volume distribution of electric currents. For TM fields, assume that , and similarly for . ( , )zzE x yEuJThe Maxwells equations then lead to (3-3)where is the wavenumber ( is the wavelength). 2/k This is a two-dimensional Helmholtz equation (亥姆赫兹方程). 22zzzEk EjJ7/31/2022 8:21 PM Solutions may be

6、 obtained by first finding the 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 xyxyuuxyxyuu(2)04zkEIHk(3-4)where is the intrinsic impedance of free-space and /120 (2)0His the Hankel function （汉克尔函数）of the second k

7、ind, zero order. The of (3-4) is the Greens function for the operator of (3-3). zEThe solution is the superposition of due to all elements of source , or zEzJ ds7/31/2022 8:21 PM(2)0( )( )4zzkEJHkds(3-5)where the integration is over the cross section of the cylinder of currents .zJNew words (生词)：tra

8、nsverse electric 横电的isotropic 各向同性scalar 标量distribution 分布wavenumber 波数electric currents 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7/31/2022 8:21 PM 3.2 Conducting Cylinders, TM case (导电柱，TM情形) Consider a perfectly conducting cyli

9、nder excited by an impressed electric field as represented by Fig. 3-1. The impressed field induces surface currents on theizEzJconducting cylinder, which produce a scattered field (散射场) . szEFig. 3-1. Cross section of a cylinder and coordinate system. 7/31/2022 8:21 PMThe field due to is given by (

10、3-5) specialized to the cylinder surface C. zJThe boundary condition is 0iszzzEEEon C (3-6)Combining (3-5) and (3-6), we have the integral equation(2)0( )( )4izzCkEJHkdl(3-7) : known ( )izE : unknown( )zJ7/31/2022 8:21 PM The simplest numerical solution of (3-7) consists of pulse basis function and

11、point matching procedure. The scatterer contour C is divided into N segments and pulse functions defined as nC1on( )0on all othernnmCfC(3-8)Letting the electric current , substituting it in (3-7), and the resultant1NznnnJfequation at the midpoint of each , we obtain the matrix equation (,)mmxymC mnn

12、mlg(3-9)where the elements of are the coefficients. nn7/31/2022 8:21 PMthe elements of aremg(,)imzmmgExy(3-10)The elements of are mnl(2)220()()4nmnmmCklHkxxyydl(3-11)A solution for the current is given by .1znnmmJflgNo simple analytical expression is available for the integral (3-11). But we can eva

13、luate it by various approximations. The crudest approximation is to treat an element as a filament of current znJCwhen the field point is not on . nC7/31/2022 8:21 PMThat is, (2)220()()4mnnnmnmklC Hkxxyy(3-12)when .mnFor the diagonal elements (对角元素) , the Hankel function has integral nnlsingularity,

14、 and the integral must be evaluated analytically. For this problem, we approximate by a straight line and use the small argument formula nC(2)02( )1log2zHzj (3-13)where is Eulers constant. 1.7817/31/2022 8:21 PMAn evaluation of (3-11) then gives 21log44nnnnk Cklk Cje(3-14)where 2.718e Better approxi

15、mations for this problem 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 anglebetween the wave vector and the x axis is .iThe incident field is ( cossin)iijk xyijzEeek r(3-15)7/31/202

16、2 8:21 PMThis determines the excitation according to (3-10). mgAn approximation evaluation of is given by (3-12) and (3-14). mnlThe solution for is found by matrix inversion in the usual manner. zJWe calculate the radar cross section (雷达散射截面) , which is 2( )( )2siEE (3-16)where is the distant field

17、from . ( )sEzJIt can be found by using the asymptotic expression for in (3-5). (2)0H7/31/2022 8:21 PMThe result is ( cossin )( )( , )sjk xyzCEkKJx y edl(3-17)where (3 /4)1( )8j kKek (3-18) Substituting (3-15) and (3-17) in (3-16), we obtain 22( cossin )( )( ,)4jk xyzCkJx y edl (3-19)7/31/2022 8:21 P

18、Mperfectly 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 雷达散射截面 asymptotic 渐进的New words 生词7/31/2022 8:21 PM 3.3 Various Approximations (各种近似) The accuracy of a solution and the convergence rate

19、depend on the approximation.The solution of Section 3-2 can be improved by more accurate evaluation of the .mnl(1)For the , additional terms can be included in (3-13), but this will not appreciably affect convergence, since (3-14) is exact in the limit . nnl0nC(2) For the terms, , we can expand the

20、integrand of (3-11) in a Taylor mnlmnseries about , and integrate the dominant terms analytically. (,)nnxyThis will give both accuracy and convergence to the exact solution as .N The rate of convergence is almost twice as fast if a piecewise linear approximation to is used instead of the step approx

21、imation. zJ7/31/2022 8:21 PM 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 the subsectional-basis type, the accuracy and convergence of Galerkin solution were about the same as for the point-m

22、atching 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 functions. When we compute them using a computer, we divide each segment into nCsmaller subintervals, and approximate th

23、e integral over each subinterval by (3-12) if nonsingular and by (3-14) if singular. 7/31/2022 8:21 PM Let us explain it in detail. Let Fig. 3-2 (a) represent a small section of the contour of a cylindrical conductor. Fig. 3-2. (a) Section of the contour. (b) Expansion function consisting of three c

24、onstrained pulses.(a)(b)7/31/2022 8:21 PMLet the subintervals , , and be further subdivided as indicated by points a, b, c, and d. 1nCnC1nC Fig. 3-2 (b) shows the same contour straightened out, and an expansion function constructed of three pulses. This three-stepped function approximates a triangle

25、 function, shown dashed. Each represents the field at due to expansion function atmnlzE(,)mmxynf .(,)nnxyWhen , mn2122231122mnmnllll(3-20)where and are given by (3-12) with replaced by and 21l23lnCabCcdC is given by (3-14) with replaced by . 22lnCbcC7/31/2022 8:21 PMFor non-diagonal elements, the pr

26、ocedure is the same, except that (3-12) is used for all since the first point never coincide with the source point. ijl 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 both expansion and testing. Thus, we can cal

27、culate it using approximations (3-12) and (3-14), which is 221221233211133133111248mnmnllllllllll(3-21)where the are the same that appear in (3-20). ijlijlIn the Galerkin solution, the of (3-10) should also be modified to represent mga numerical integration of with the testing function of Fig. 3-4.

28、izE7/31/2022 8:21 PMAccuracy 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7/31/2022 8:21 PM3.4 Transverse Electric Fields (横电场) A two-dimensional TE field in isotropic media has no z component of

29、 E and only a z component of H. The fields can be illustrated by1HAj EA(3-22)(3-23)where the magnetic vector potential A and the electric scalar potential satisfy22k AAJ22qk (3-24)(3-25)7/31/2022 8:21 PMThe electric charge density q is related to J by the equation of continuity (连续性) j q J(3-26)Both

30、 (3-24) and (3-25) are Helmholtz equations, the same as (3-3). Hence, the solutions are of the form (3-5). Defining the two-dimensional Greens function(2)01( , )4GHkj (3-27) we can express solutions to (3-24) and (3-25) in unbounded two-dimensional space as ( ) ( , )JGdsA (3-28)注： unbounded 无界的7/31/

31、2022 8:21 PM 1( ) ( , )qGds (3-29)where the integration is over 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 curren

32、t. In this Section, we consider the H-field formulation. The total field at any point is the sum of the impressed field plus zHizHthe scattered field due to the J on C. szHiszzzHHH(3-30)That is,7/31/2022 8:21 PMThe scattered field is related to its source J by (3-22) and (3-28), or szzCHJGdul(3-31)w

33、here the vector designates the reference direction of J. dlIf the interior of C lies on the left side of , then dlzCJH (3-32)where denotes that is evaluated just external to C. CzHSpecializing (3-30) to , we have CizzCCJHJG d ul(3-33)The is not continuous at C. zHThus, the Greens function G is singu

34、lar. A simple interchange of differentiation and integration is not always possible. 7/31/2022 8:21 PMFig. 3-3 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 , . CzHJ C0zH If the scatterer

35、is a conducting sheet of infinitesimal thickness, it should be treated as the limit of one of finite thickness. 7/31/2022 8:21 PMWe can write (3-33) in general operator notation as ( )izL jH (3-34)where ( )zCCL JJJGdul(3-35)Let us solve (3-35) using the method of moments. The basis function is the p

36、ulse function. The point-matching procedure is used for testing. The current , and the resultant matrix equation is (3-9) with nnJf(,)imzmmgHxy (3-36)7/31/2022 8:21 PM( , )mnmnzlHm n(3-37)where is the Kronecker delta function. mn stands for at on due to unit current density on at . ( , )zHm nzH(,)mm

37、xyCnC(,)nnxyFig. 3-4. Elements of current and local coordinates. Fig. 3-4. represents a typical current element and local coordinatesnJC l .( , )x y7/31/2022 8:21 PMFrom symmetry, and the fact that the discontinuity in is J, we havezH000012xxzzyyHH (3-38)and hence, by (3-37),nn1/ 2l(3-39)If and the

38、field point is distant from , then the sourcenC( , )x ynJC lbehaves as a point source. 20A()4nyCHkj(3-40)and from (3-22)2201()cos()44nznCjHHkCHkjx(3-41)From (3-28)7/31/2022 8:21 PMwhere is the Hankel function of order 1. 21HWe can translate this to an arbitrary origin by replacing by andmn by , wher

39、e coscosn RmnmnR(3-42)is unit vector from the source point to the field point . (,)nnxy(,)mmxyThe result can be used as an approximation for all . mnThus, (3-37) becomes, for mn21()()4mnnmnjlk CHkn R(3-43)The solution is given by . 1nnmmJflg7/31/2022 8:21 PMExample. Consider TE plane-wave scattering

40、 by conducting cylinders. An impressed uniform plane wave from the direction is given by i( cossin)iijk xyizHe(3-44)The are determined from this by (3-36) and the are given by (3-39) andmgmnl(3-43) for a first-order solution.Again the scattering cross section (雷达散射截面) is of interest, given by 2( )(

41、)2siHH (3-45)where is the distant field from J, obtained by using the asymptotic formula ( )sHfor in (3-41) and summing over all elements of source. (2)1H7/31/2022 8:21 PMThe result is ( cossin)( )( ,)eiijk xyszCHKkJ x ydln Rwhere K is given by (3-18). Substituting (3-44) and (3-46) in (3-45), we ob

42、tain 2( cossin)( )( ,)e4iijk xyCkJ x ydl n R(3-46)(3-47)which can be evaluated once J is found. 7/31/2022 8:21 PM 3.6 Alternative Formulation (替代表达） The TM problem was 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 pr

43、oblem 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. iEThe scattered field is produced by produced by transverse currents J on C. sEFor the present problem, these become sj EA(3-48)7/31/2022 8:21 PM(

44、 )( ) ( , )CGdA J l(3-49)11( )( , )CdJGdljdl (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,on0isCEE(3-51)7/31/2022 8:21 PMThank you.7/31/2022 8:21 PMTo be continued.7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM7/31/2022 8:21 PM