Page 271 - Electromagnetics
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and (4.222) as
                                                               ˆ
                                                        ˜
                                                                   ˜
                                                        E =−ηk × H.
                        Here

                                                           ω ˜µ    ˜ µ
                                                       η =     =    c
                                                            k      ˜
                        is the complex intrinsic impedance of the medium.
                          Equations (4.223) and (4.221) show that the electric and magnetic fields and the wave
                        vector are mutually orthogonal. The wave is said to be transverse electromagnetic or
                        TEM to the direction of propagation.

                        The phase and attenuation constants of a uniform plane wave.    For a uniform
                        plane wave we may write
                                                      ˆ       ˆ  ˆ          ˆ
                                                k = k k + jk k = kk = (β − jα)k

                        where k = β and k =−α. Here α is called the attenuation constant and β is the phase

                        constant. Since k is defined through (4.206), we have
                                 2          2   2           2    2  c   2


                                                                                          c
                                                                                    c
                                k = (β − jα) = β − 2 jαβ − α = ω ˜µ˜  = ω ( ˜µ + j ˜µ )(˜  + j ˜  ).
                        Equating real and imaginary parts we have
                                      2   2    2                          2
                                                           c
                                                                                      c
                                                    c
                                                                                c
                                    β − α = ω [ ˜µ ˜  − ˜µ ˜  ],  −2αβ = ω [ ˜µ ˜  + ˜µ ˜  ].

                        We assume the material is passive so that ˜µ ≤ 0, ˜   c    ≤ 0. Letting
                                 2    2   2                             2

                                               c

                                                       c
                                                                                     c
                                                                              c
                                β − α = ω [ ˜µ ˜  − ˜µ ˜  ] = A,  2αβ = ω [| ˜µ |˜  + ˜µ |˜  |] = B,
                        we may solve simultaneously to find that
                                          1                         1
                                       2            2   2        2              2   2
                                     β =     A +   A + B   ,    α =    −A +   A + B   .
                                          2                         2
                                        4
                                                            2
                                                      2
                              2
                                   2
                        Since A + B = ω (˜  c 2  + ˜  c  2 )( ˜µ + ˜µ ),wehave
                                              $
                                              %

                                              % 1         ˜   c  2  ˜ µ   2     ˜ µ ˜  c

                                           ˜
                                   β = ω µ ˜          1 +       1 +      + 1 −          ,     (4.224)
                                              c  &
                                                2         ˜   c 2    ˜ µ  2     ˜ µ ˜   c

                                              $
                                              %

                                              % 1         ˜   c  2  ˜ µ   2     ˜ µ ˜  c

                                           ˜
                                   α = ω µ ˜          1 +       1 +      − 1 −          ,     (4.225)
                                              c  &
                                                2         ˜   c 2    ˜ µ  2     ˜ µ ˜   c

                               c
                        where ˜  and ˜µ are functions of ω.If ˜ (ω) =  , ˜µ(ω) = µ, and ˜σ(ω) = σ are real and
                        frequency independent, then
                                                        $

                                                        %
                                                    √   % 1          σ    2
                                               α = ω µ   &     1 +      − 1 ,                 (4.226)
                                                          2        ω
                                                        $

                                                        %
                                                    √   % 1          σ    2
                                               β = ω µ   &     1 +      + 1 .                 (4.227)
                                                          2        ω
                        © 2001 by CRC Press LLC
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