Page 116 - Electromagnetics
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The Fourier integral theorem implies that the integrand is zero. Then, expanding out
                                                   ˜
                        the ρ derivatives, we find that E z (ρ, ω) obeys the ordinary differential equation
                                                    2 ˜      ˜
                                                   d E z  1 dE z   2 ˜
                                                        +       + k E z = 0
                                                    dρ 2  ρ dρ
                        where k = ω/v. This is merely Bessel’s differential equation (A.124). It is a second-order
                        equation with two independent solutions chosen from the list
                                             J 0 (kρ),  Y 0 (kρ),  H  (1) (kρ),  H  (2) (kρ).
                                                               0         0
                        We find that J 0 (kρ) and Y 0 (kρ) are useful for describing standing waves between bound-
                                    (1)         (2)
                        aries while H  (kρ) and H  (kρ) are useful for describing waves propagating in the
                                    0           0
                        ρ-direction. Of these, H  (1) (kρ) represents waves traveling inward while H  (2) (kρ) repre-
                                             0                                           0
                        sents waves traveling outward. Concentrating on the outward traveling wave we find
                        that
                                                            π  (2)
                                                                        ˜
                                          ˜
                                                   ˜
                                         E z (ρ, ω) = A(ω) − j  H 0  (kρ) = A(ω)˜ g(ρ, ω).
                                                            2
                                     ˜
                        Here A(t) ↔ A(ω) is the disturbance waveform, assumed to be a real, causal function.
                        To make E z (ρ, t) real we require that the inverse transform of ˜ g(ρ, ω) be real. This
                        requires the inclusion of the − jπ/2 factor in ˜ g(ρ, ω). Inverting we have
                                                    E z (ρ, t) = A(t) ∗ g(ρ, t)               (2.347)
                                                (2)
                        where g(ρ, t) ↔ (− jπ/2)H  (kρ).
                                                0
                          The inverse transform needed to obtain g(ρ, t) may be found in Campbell [26]:
                                                                               ρ
                                                          π       ρ      U t −

                                                    −1        (2)              v
                                          g(ρ, t) = F  − j  H 0  ω     =         ,
                                                          2        v        2  ρ  2
                                                                           t −
                                                                               v  2
                        where U(t) is the unit step function defined in (A.5). Substituting this into (2.347) and
                        writing the convolution in integral form we have


                                                        ∞         U(t − ρ/v)

                                             E z (ρ, t) =  A(t − t )         dt .
                                                                         2
                                                                    	2
                                                                   t − ρ /v 2
                                                       −∞
                        The change of variable x = t − ρ/v then gives

                                                           ∞

                                                             A(t − x − ρ/v)
                                               E z (ρ, t) =               dx.                 (2.348)
                                                                2
                                                          0    x + 2xρ/v
                        Those interested in the details of the inverse transform should see Chew [33].
                          As an example, consider a lossless medium with µ r = 1, 
 r = 81, and a waveform
                                                  A(t) = E 0 [U(t) − U(t − τ)]
                        where τ = 2 µs. This situation is the same as that in the plane wave example above,
                        except that the pulse waveform begins at t = 0. Substituting for A(t) into (2.348) and
                        using the integral
                                                    dx          	√    √

                                                 √ √       = 2ln   x +  x + a
                                                   x x + a


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