欢迎来到三一办公! | 帮助中心 三一办公31ppt.com(应用文档模板下载平台)
三一办公
全部分类
  • 办公文档>
  • PPT模板>
  • 建筑/施工/环境>
  • 毕业设计>
  • 工程图纸>
  • 教育教学>
  • 素材源码>
  • 生活休闲>
  • 临时分类>
  • ImageVerifierCode 换一换
    首页 三一办公 > 资源分类 > PPT文档下载  

    《计算电磁学》第十讲ppt课件.ppt

    • 资源ID:1307433       资源大小:1.84MB        全文页数:50页
    • 资源格式: PPT        下载积分:16金币
    快捷下载 游客一键下载
    会员登录下载
    三方登录下载: 微信开放平台登录 QQ登录  
    下载资源需要16金币
    邮箱/手机:
    温馨提示:
    用户名和密码都是您填写的邮箱或者手机号,方便查询和重复下载(系统自动生成)
    支付方式: 支付宝    微信支付   
    验证码:   换一换

    加入VIP免费专享
     
    账号:
    密码:
    验证码:   换一换
      忘记密码?
        
    友情提示
    2、PDF文件下载后,可能会被浏览器默认打开,此种情况可以点击浏览器菜单,保存网页到桌面,就可以正常下载了。
    3、本站不支持迅雷下载,请使用电脑自带的IE浏览器,或者360浏览器、谷歌浏览器下载即可。
    4、本站资源下载后的文档和图纸-无水印,预览文档经过压缩,下载后原文更清晰。
    5、试题试卷类文档,如果标题没有明确说明有答案则都视为没有答案,请知晓。

    《计算电磁学》第十讲ppt课件.ppt

    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 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 transverse 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 component 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 the 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 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,(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 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 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 field 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 (3-5) 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 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 analytical 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 function 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 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 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 according 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 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 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 (3-13), 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 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 the 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 functions.,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 contour 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, 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 .,11/7/2022 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 both 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.,11/7/2022 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 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) 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 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 current.,In this Section, we consider the H-field formulation.,The 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 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. 3-3 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.,11/7/2022 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),11/7/2022 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 discontinuity 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 by 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 uniform 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 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 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 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 scattered 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,

    注意事项

    本文(《计算电磁学》第十讲ppt课件.ppt)为本站会员(牧羊曲112)主动上传,三一办公仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。 若此文所含内容侵犯了您的版权或隐私,请立即通知三一办公(点击联系客服),我们立即给予删除!

    温馨提示:如果因为网速或其他原因下载失败请重新下载,重复下载不扣分。




    备案号:宁ICP备20000045号-2

    经营许可证:宁B2-20210002

    宁公网安备 64010402000987号

    三一办公
    收起
    展开