Page 427 - Engineering Electromagnetics, 8th Edition
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CHAPTER 12   Plane Wave Reflection and Dispersion      409

                     or
                                                         η 2 − η 1
                                              E x10  = E x10
                                                −
                                                      +
                                                         η 2 + η 1
                     The ratio of the amplitudes of the reflected and incident electric fields defines the
                     reflection coefficient, designated by  ,
                                              E  −   η 2 − η 1
                                            =  x10  =       =| |e jφ                  (9)
                                              E x10  η 2 + η 1
                                                +
                     It is evident that as η 1 or η 2 may be complex,   will also be complex, and so we
                     include a reflective phase shift, φ. The interpretation of Eq. (9) is identical to that
                     used with transmission lines [Eq. (73), Chapter 10].
                         The relative amplitude of the transmitted electric field intensity is found by
                     combining (9) and (7) to yield the transmission coefficient, τ,

                                          E  +     2η 2
                                      τ =   x20  =      = 1 +   =|τ|e  jφ i          (10)
                                          E x10  η 1 + η 2
                                            +
                     whose form and interpretation are consistent with the usage in transmission lines
                     [Eq. (75), Chapter 10].
                         Let us see how these results may be applied to several special cases. We first let
                     region 1 be a perfect dielectric and region 2 be a perfect conductor. Then we apply
                     Eq. (48), Chapter 11, with   = σ 2 /ω, obtaining

                                           2

                                                      jωµ 2
                                             η 2 =           = 0
                                                    σ 2 + jω    2
                     in which zero is obtained since σ 2 →∞. Therefore, from (10),
                                                   E  +  = 0
                                                    x20
                     No time-varying fields can exist in the perfect conductor. An alternate way of looking
                     at this is to note that the skin depth is zero.
                         Because η 2 = 0, Eq. (9) shows that
                                                     =−1
                     and
                                                 E  +  =−E  −
                                                  x10
                                                          x10
                         The incident and reflected fields are of equal amplitude, and so all the incident
                     energy is reflected by the perfect conductor. The fact that the two fields are of opposite
                     sign indicates that at the boundary (or at the moment of reflection), the reflected field
                     is shifted in phase by 180 relative to the incident field. The total E field in region 1 is
                                         ◦
                                          E xs1 = E xs1  + E  −
                                                  +
                                                        xs1
                                              = E x10  e − jβ 1 z  − E  +  e jβ 1 z
                                                  +
                                                             x10
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