Time and Frequency Domain Solutions of EM Problems Using by B. H. Jung, T. K. Sarkar, Y. Zhang, Z. Ji, M. Yuan, M.

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By B. H. Jung, T. K. Sarkar, Y. Zhang, Z. Ji, M. Yuan, M. Salazar-Palma, S. M. Rao, S. W. Ting, Z. Mei, A. De

Numerical strategies of electromagnetic box difficulties is a space of paramount curiosity in academia, and govt. This booklet presents a compendium of answer innovations facing critical equations bobbing up in electromagnetic box difficulties in time and frequency domain names. Written by way of major researchers within the box, it files the authors' special space/time separation process utilizing Laguerre polynomials. a variety of examples that illustrate some of the methodologies and simple computing device codes make this quantity hugely available for engineers, researchers, and scientists.

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Additional resources for Time and Frequency Domain Solutions of EM Problems Using Integral Equations and a Hybrid Methodology (Wiley Series in Microwave and Optical Engineering)

Example text

In short, it is seen by relating (yN — y, W^ ) = 0 in Eq. 120) to the theory of functional approximations that (a) The weighting function W* must be in the range of the operator or, more generally, in the domain of the adjoint operator. (b) Because the weighting functions are orthogonal to the error of the approximation, the set {Wk} should span yN. (c) As N —» oo, yN -» y. Therefore, the weighting function should be able to represent the excitation y in the limit. If the weighting functions do not satisfy all of these criterions, then a meaningful solution may not be obtained by the method of moments.

Therefore, by definition Лф,=А,ф. 102) where A,- is the eigenvalue corresponding to the eigenvectors φ, and ξ,. We represent the solution x as follows: Σ α /Ψί ;=i (1-103) where a, are the unknowns to be solved for. We substitute Eqs. , (φ ξ„) = 0 forA:^ m ), we can multiply both sides of Eq. 104) by the left-hand eigenvector ξ* and integrate the product over the domain of the operator SA and represent it in the form of inner product as in Eq. 9), that is, with Л(Ьк>£,к)=(уАк); «г в* = т 7 Г Т \ л к \Φ*,ς*/ а (1Л05) Once the unknown coefficients are given by Eq.

115). This results in the following: oo x = x (z) = a0 + Σ {aicos (iz) + bisin (iz)) /=i (1-117) where a, and bh are the unknown constants to be solved for. Now the total solution given by Eq. 117) has to satisfy the boundary conditions of Eq. 114). Observe that Eq. 117) is the classical Fourier series solution, and hence, it is complete in the interval [0,2л-]. Now if we solve the problem again and choose the weighting functions to be same as the basis functions, then we find the solution still to be given by Eq.

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