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This expression cannot be computed in closed form. For a short dipole we may use (6.33)
                        to approximate the power, but the result is somewhat misleading since the current on a
                                                      ˜
                        short dipole is much smaller than I. A better measure of the strength of the current is
                        its value at the center, or feedpoint, of the dipole. This input current is by (6.26) merely
                               ˜
                        ˜ I 0 (ω) = I(ω) sin(kl). Using this we find
                                                 ˇ 2
                                                                               l
                                                |I 0 | 1  2     π  3    π    
    2
                                                                          ˇ 2
                                         P av ≈ η    (kl)    sin θ dθ = η |I 0 |   .
                                                4π 4      0             3      λ
                        This is exactly 1/4 of the power radiated by a Hertzian dipole of the same length and
                        current amplitude (5.95). The factor of 1/4 comes from the difference between the current
                        of the dipole antenna, which is zero at each end, and the current on the Hertzian dipole,
                        which is constant across the length of the antenna. It is more common to use a dipole
                        antenna that is a half wavelength long (2l = λ/2), since it is then nearly resonant. With
                        this we have through numerical integration the free-space radiated power

                                                   ˇ 2     π  2    π
                                                  |I 0 |  cos  2  cos θ
                                                                               ˇ 2
                                           P av = η 0                 dθ = 36.6|I 0 |
                                                   4π  0      sin θ
                        and the radiation resistance
                                                      2P av     2P av
                                               R r =        2  =     = 73.2  .
                                                     ˇ
                                                                 ˇ 2
                                                    |I(z = 0)|  |I 0 |




                        6.3   Fields in a bounded, source-free region
                          In § 6.2 we considered the first important special case of the Stratton–Chu formula:
                        sources in an unbounded medium. We now consider the second important special case
                        of a bounded, source-free region. This case has important applications to the study of
                        microwave antennas and, in its scalar form, to the study of the diffraction of light.

                        6.3.1   The vector Huygens principle

                          We may derive the formula for a bounded, source-free region of space by specializing
                        the general Stratton–Chu formulas. We assume that all sources of the fields are within
                        the excluded regions and thus set the sources to zero within V . From (6.7)–(6.8) we have


                                         N
                               ˜                    ˜             ˜                ˜

                               E(r,ω) =        (ˆ n × E) ×∇ G + (ˆ n · E)∇ G − jω ˜µ(ˆ n × H)G dS −
                                        n=1  S n
                                         N
                                             1
                                                          ˜


                                      −        c     (dl · H)∇ G,                              (6.34)
                                            jω˜
                                        n=1
                                                  na +  nb
                        and
                                         N
                               ˜                    ˜              ˜          c     ˜

                               H(r,ω) =       (ˆ n × H) ×∇ G + (ˆ n · H)∇ G + jω˜  (ˆ n × E)G dS +
                                        n=1  S n
                        © 2001 by CRC Press LLC
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