Page 432 - Electromagnetics
P. 432

into (6.24) and carrying out the x and y integrals we have the directional weighting
                        function

                                                        l
                                                          ˜



                                          ˜ a e (θ, φ, ω) =  ˆ zI(ω) sin k(l −|z |)e jkz cos θ  dz .
                                                      −l
                        Writing the sine functions in terms of exponentials we have
                                             ˜
                                            ˆ zI(ω)     jkl     l  jkz (cos θ−1)  − jkl     l  jkz (cos θ+1)




                                ˜ a e (θ, φ, ω) =  e    e         dz − e      e         dz +
                                             2 j       0                    0
                                                  0                  0


                                         + e jkl  e  jkz (cos θ+1)  − e − jkl  e jkz (cos θ−1)  .
                                                −l                 −l
                        Carrying out the integrals and simplifying, we obtain
                                                               ˜
                                                              2I(ω) F(θ, kl)
                                                  ˜ a e (θ,φ,ω) = ˆ z
                                                                k    sin θ
                        where
                                                          cos(kl cos θ) − cos kl
                                                 F(θ, kl) =
                                                                 sin θ
                                                                     ˆ
                        is called the radiation function. Using ˆ z = ˆ r cos θ − θ sin θ we find that
                                                       ˜
                                                      2I(ω)
                                        ˜ a eθ (θ,φ,ω) =−  F(θ, kl),   ˜ a eφ (θ, φ, ω) = 0.
                                                        k
                        Thus we have from (6.23) and (6.21) the electric field
                                                             ˜
                                                           jηI(ω) e − jkr
                                                ˜
                                                E(r,ω) = θ ˆ         F(θ, kl)                  (6.31)
                                                            2π    r
                        and from (6.22) the magnetic field
                                                            ˜
                                                           j I(ω) e − jkr
                                                 ˜
                                                         ˆ
                                                H(r,ω) = φ           F(θ, kl).                 (6.32)
                                                            2π    r
                        We see that the radiation function contains all of the angular dependence of the field
                        and thus describes the pattern of the dipole. When the dipole is short compared to a
                        wavelength we may approximate the radiation function as
                                                      1        2      1   2
                                                  1 − (kl cos θ) − 1 + (kl)   1
                                                                                  2
                                     F(θ, kl   1) ≈   2               2    =   (kl) sin θ.     (6.33)
                                                             sin θ            2
                        So a short dipole antenna has the same pattern as a Hertzian dipole, whose far-zone
                        electric field is (5.93).
                          We may also calculate the radiated power for time-harmonic fields. The time-average
                        Poynting vector for the far-zone fields is, from (6.25),
                                                             ˇ 2
                                                            |I|   2
                                                    S av = ˆ rη  F (θ, kl),
                                                              2 2
                                                           8π r
                        and thus the radiated power is
                                                       ˇ 2     π
                                                       |I|     2
                                                P av = η      F (θ, kl) sin θ dθ.
                                                       4π  0



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