Page 463 - Electromagnetics

1 ∞ jωt
˜
S(x, y, z, t) = S(x, y, z,ω)e dω.
2π
−∞
Substituting into (A.16), passing the derivatives through the integral, calculating the
derivatives, and combining the inverse transforms, we obtain

1 ∞ ∂ 2 2 jωt
˜
˜
+ k ψ(x, y, z,ω) − S(x, y, z,ω) e dω = 0
2π ∂z 2
−∞
where k = ω/c. By the Fourier integral theorem
2
∂ 2
˜
˜
+ k ψ(x, y, z,ω) − S(x, y, z,ω) = 0. (A.19)
∂z 2
We have thus converted a partial differential equation into an ordinary differential equa-
tion. A spatial transform on z will now convert the ordinary differential equation into
an algebraic equation. We write
1 ∞
˜ ˜ z jk z z
ψ(x, y, z,ω) = ψ (x, y, k z ,ω)e dk z ,
2π
−∞
1 ∞
˜ ˜ z jk z z
S(x, y, z,ω) = S (x, y, k z ,ω)e dk z ,
2π −∞
in (A.19), pass the derivatives through the integral sign, compute the derivatives, and
set the integrand to zero to get
˜ z
2 2 z
˜
(k − k )ψ (x, y, k z ,ω) − S (x, y, k z ,ω) = 0;
z
hence
˜ z
S (x, y, k z ,ω)
˜ z
ψ (x, y, k z ,ω) =− . (A.20)
(k z − k)(k z + k)
The price we pay for such an easy solution is that we must now perform a two-
˜ z
dimensional Fourier inversion to obtain ψ(x, y, z, t) from ψ (x, y, k z ,ω). It turns out to
be easiest to perform the spatial inverse transform first, so let us examine
1 ∞
˜ ˜ z jk z z
ψ(x, y, z,ω) = ψ (x, y, k z ,ω)e dk z .
2π
−∞
By (A.20) we have
1 ∞ −1
˜ ˜ z jk z z
ψ(x, y, z,ω) = [S (x, y, k z ,ω)] e dk z ,
2π −∞ (k z − k)(k z + k)
where the integrand involves a product of two functions. With
−1
z
˜ g (k z ,ω) = ,
(k z − k)(k z + k)
the convolution theorem gives

∞

˜ ˜
ψ(x, y, z,ω) = S(x, y,ζ,ω)˜ g(z − ζ, ω) dζ (A.21)
−∞

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