Page 301 - Electromagnetics
P. 301

where s = jω and
                                                         2
                                                     r − sin θ i         σ
                                               D =      2   ,    B =      2  .
                                                     cos θ i            0 cos θ i
                        We can put (4.302) into a better form for inversion. We begin bysubtracting   ⊥∞ , the
                                             ˜
                        high-frequencylimit of   ⊥ . Noting that
                                                                       √
                                                                    1 −  D
                                                     ˜
                                                 lim   ⊥ (ω) =   ⊥∞ =  √ ,
                                                 ω→∞                1 +  D
                        we can form
                                                              √    √             √
                                                               s −   Ds + B   1 −  D
                                        ˜  0    ˜
                                         (ω) =   ⊥ (ω) −   ⊥∞ = √  √        −    √
                                        ⊥
                                                               s +   Ds + B   1 +  D
                                                  √     
 √     √
                                                    D        s −  s + B/D
                                             = 2    √    √    √ √           .
                                                1 +   D    s +  D s + D/B
                        With a bit of algebra this becomes
                                      √                        B  
                                     2 D       s            s +             2B            1
                            ˜ 0                                 D
                             (ω) =−                   1 −         −     √                     .
                             ⊥                   B                                          B
                                     D − 1  s +               s                         s +
                                                D−1                    1 +  D (D − 1)      D−1
                        Now we can apply(C.12), (C.18), and (C.19) to obtain
                                            0       −1    ˜ 0
                                             (t) = F     (ω) = f 1 (t) + f 2 (t) + f 3 (t)    (4.303)
                                            ⊥           ⊥
                        where
                                                       2B          Bt
                                         f 1 (t) =−  √          e  −  D−1 U(t),
                                                 (1 +  D)(D − 1)
                                                     B 2           t  B(t−x)     Bx
                                         f 2 (t) =−√        U(t)   e  −  D−1 I   dx,
                                                   D(D − 1) 2    0          2D
                                                    B         Bt
                                         f 3 (t) = √      I      U(t).
                                                 D(D − 1)    2D

                        Here
                                                   I (x) = e −x  [I 0 (x) + I 1 (x)]

                        where I 0 (x) and I 1 (x) are modified Bessel functions of the first kind. Setting u = Bx/2D
                        we can also write
                                                       √          Bt
                                                    2B D           2D  Bt−2Du
                                            f 2 (t) =−    2  U(t)   e −  D−1 I (u) du.
                                                    (D − 1)      0
                        Polynomial approximations for I (x) maybe found in Abramowitz and Stegun [?], making
                                          0
                        the computation of   (t) straightforward.
                                          ⊥
                          The complete time-domain reflection coefficient is
                                                            √
                                                         1 −  D        0
                                                   ⊥ (t) =  √ δ(t) +   (t).
                                                                       ⊥
                                                         1 +  D


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