Page 536 - Engineering Electromagnetics, 8th Edition
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518                ENGINEERING ELECTROMAGNETICS

                                     14.2 ANTENNA SPECIFICATIONS
                                     It is important to fully describe and quantify the radiation from a general antenna. To
                                     do this, we need to be aware of a few new concepts and definitions.
                                        In order to evaluate the radiated power, the time-average Poynting vector must
                                     be found (Eq. (77), Chapter 11). In the present case, this will become
                                                                1
                                                                          ∗     W/m 2                (25)
                                                        < S > =  Re {E θs H } a r
                                                                         φs
                                                                2
                                     Substituting (22) and (23) into (25), we obtain the time-average Poynting vector
                                     magnitude:
                                                                      1     I 0 kd    2
                                                                                   2
                                                       | < S > |= S r =         η sin θ              (26)
                                                                      2  4πr
                                     From this we find the time-average power that crosses the surface of a sphere of radius
                                     r, centered at the antenna:
                                                                         1
                                                 2π     π                     I 0 kd    2    π
                                                         2
                                                                                            3
                                         P r =        S r r sin θdθdφ = 2π          η    sin θ dθ    (27)
                                               φ=0  θ=0                  2     4π      0
                                     The integral is evaluated, and we substitute k = 2π/λ.We will also assume that the
                                                                   .
                                     medium is free space, where η = η 0 = 120π.We finally obtain:
                                                                         I 0 d    2
                                                                    2
                                                            P r = 40π          W                     (28)
                                                                        λ
                                     This is the same average power that would be dissipated in a resistance R rad by
                                     sinusoidal current of amplitude I 0 in the absence of any radiation, where


                                                                     1  2
                                                                 P r =  I R rad                      (29)
                                                                        0
                                                                     2
                                     We call this effective resistance the radiation resistance of the antenna. For the dif-
                                     ferential antenna, this becomes

                                                                                 2
                                                                  2P r      2  d
                                                           R rad =    = 80π                          (30)
                                                                  I 0 2       λ
                                        If, for example, the differential length is 0.01λ, then R rad is about 0.08 
. This
                                     small resistance is probably comparable to the ohmic resistance of a practical antenna
                                     (providing a measure of the power dissipated through heat), and thus the efficiency
                                     of the antenna is likely to be too low. Effective matching to the source also becomes
                                     very difficult to achieve, for the input reactance of an electrically short antenna is
                                     much greater in magnitude than the input resistance R rad .
                                        Evaluating the net power from the antenna, as carried out in (27), involved the
                                     integration of the Poynting vector over a spherical shell of presumed large radius,
                                     such that the antenna appeared as a point source at the sphere center. In view of this,
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