Page 439 - Electromagnetics
P. 439

˜ eq


                                               ˜ a h (θ, φ, ω) =  J (r ,ω)e jkˆ r·r    dS ,    (6.49)
                                                              sm
                                                            S
                        are the directional weighting functions.
                          To compute the fields from the potentials we must apply the curl operator. So we
                        must evaluate
                                        e             e                    e
                                         − jkr         − jkr             
  − jkr
                                  ∇×        V(θ, φ) =      ∇× V(θ, φ) +∇         × V(θ, φ).
                                         r              r                   r
                        The curl of V is proportional to 1/r in spherical coordinates, hence the first term on the
                                               2
                        right is proportional to 1/r . Since we are interested in the far-zone fields, this term can
                        be discarded in favor of 1/r-type terms. Using
                                       e            1 + jkr  e          e
                                      
  − jkr     
          − jkr      − jkr
                                    ∇        =−ˆ r                ≈−ˆ r jk   ,    kr   1,
                                         r             r      r           r
                        we have
                                               e                      e
                                                 − jkr                 − jkr
                                          ∇×        V(θ, φ) ≈− jkˆ r ×    V(θ, φ) .
                                                 r                     r
                        Using this approximation we also establish
                                        e                2       e               2  e
                                         − jkr                     − jkr           − jkr
                              ∇× ∇×         V(θ, φ) ≈−k ˆ r × ˆ r ×   V(θ, φ) = k     V T (θ, φ)
                                         r                         r                r
                        where V T = V − ˆ r(ˆ r · V) is the vector component of V transverse to the r-direction.
                          With these formulas we can approximate (6.46) and (6.47) as
                                                                   jk
                                           ˜            ˜                 ˜
                                          E(r,ω) =− jωA eT (r,ω) +     ˆ r × A h (r,ω),        (6.50)
                                                                   c
                                                                  ˜   (ω)
                                                                   jk
                                           ˜            ˜                 ˜
                                          H(r,ω) =− jωA hT (r,ω) −    ˆ r × A e (r,ω).
                                                                  ˜ µ(ω)
                        Note that
                                                                  jk
                                                 ˜          ˜              ˜
                                             ˆ r × E =− jωˆ r × A eT +  ˆ r × ˆ r × A h .
                                                                  ˜   c
                                                              ˜
                                                       ˜
                                          ˜
                                 ˜
                        Since ˆ r × A eT = ˆ r × A e and ˆ r × ˆ r × A h =−A hT ,wehave

                                                ˜          ˜     jk    ˜      ˜
                                             ˆ r × E = η − jωA hT −  ˆ r × A e = ηH.
                                                                 ˜ µ
                        Thus
                                                                  ˜
                                                              ˆ r × E
                                                          ˜
                                                         H =
                                                                η
                        and the electromagnetic field in the far zone is a TEM spherical wave, as expected.
                        Example of fields produced by equivalent sources: an aperture antenna. As
                        an example of calculating the fields in a bounded region from equivalent sources, let
                        us find the far-zone field in free space produced by a rectangular waveguide opening
                        into a perfectly-conducting ground screen of infinite extent as shown in Figure 6.6. For
                        simplicity assume the waveguide propagates a pure TE 10 mode, and that all higher-order


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