Page 344 - Electromagnetics
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hence
                                                                      ω ˜µ
                                                +          −             ˜
                                               A (k x ,ω) + A (k x ,ω) =−  I(ω).              (4.406)
                                                                       k y
                        Solution of (4.405) and (4.406) gives the angular spectra

                                                                      ω ˜µ
                                                +          −             ˜
                                               A (k x ,ω) = A (k x ,ω) =−  I(ω).
                                                                      2k y
                        Substituting this into the field expressions and combining the cases for y > 0 and y < 0,
                        we find
                                                ∞+ j
                                           ˜
                                        ω ˜µI(ω)     e  − jk y |y|
                           ˜                                − jk x x       ˜    ˜
                          E z (x, y,ω) =−                  e    dk x =− jω ˜µI(ω)G(x, y|0, 0; ω).  (4.407)
                                          2π          2k y
                                               −∞+ j
                             ˜
                        Here G is the spectral representation of the two-dimensional Green’s function first found
                        in § 4.11.7, and is given by
                                                            ∞+ j


                                                        1        e − jk y |y−y |
                                       ˜                                  − jk x (x−x )
                                       G(x, y|x , y ; ω) =               e        dk x .      (4.408)
                                                       2π j         2k y
                                                           −∞+ j
                          By duality we have
                                                ∞+ j
                                         c ˜
                                       ω˜  I m (ω)     e − jk y |y|
                          ˜                                 − jk x x       c ˜
                         H z (x, y,ω) =−                   e     dk x =− jω˜  I m (ω)G(x, y|0, 0; ω) (4.409)
                                          2π           2k y
                                               −∞+ j
                                                ˜
                        for a magnetic line current I m (ω) on the z-axis.
                          Note that since the earlier expression (4.346) should be equivalent to (4.408), we have
                        the well known identity [33]
                                                 ∞+ j
                                              1       e − jk y |y|  − jk x x  (2)
                                                            e    dk x = H 0  (kρ).
                                              π         k y
                                               −∞+ j
                          We have not yet specified the contour appropriate for calculating the inverse transform
                        (4.407). We must be careful because the denominator of (4.407) has branch points at

                               2
                                   2
                        k y =  k − k = 0, or equivalently, k x =±k =±(k r + jk i ). For lossy materials we have
                                   x
                        k i  < 0 and k r  > 0, so the branch points appear as in Figure 4.30. We may take the branch
                        cuts outward from these points, and thus choose the inversion contour to lie between the
                        branch points so that the branch cuts are not traversed. This requires k i < < −k i .It
                        is natural to choose   = 0 and use the real axis as the inversion contour. We must be
                        careful, though, when extending these arguments to the lossless case. If we consider the
                        lossless case to be the limit of the lossy case as k i → 0, we find that the branch points
                        migrate to the real axis and thus lie on the inversion contour. We can eliminate this
                        problem by realizing that the inversion contour may be perturbed without affecting the
                        value of the integral, as long as it is not made to pass through the branch cuts. If we
                        perturb the inversion contour as shown in Figure 4.30, then as k i → 0 the branch points
                        do not fall on the contour.




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