Page 475 - Electromagnetics

Here res{ f (k z )} denotes the residues of the function f (k z ). The integrand in (A.53) has

poles of order 1at k z =± jk ρ , k ρ ≥ 0.If z − z > 0 we close in the upper half-plane and
enclose only the pole at k z = jk ρ . Computing the residue using (A.13), we obtain

e − jk x x e − jk y y e −k ρ (z−z )
xy

G (k x , k y , z|r ) = j , z > z .
2 jk ρ

Since z > z this function decays for increasing z, as expected physically. For z−z < 0 we
close in the lower half-plane, enclosing the pole at k z =− jk ρ and incurring an additional
negative sign since our contour is now clockwise. Evaluating the residue we have

e − jk x x e − jk y y e k ρ (z−z )
xy
G (k x , k y , z|r ) =− j , z < z .

−2 jk ρ
We can combine both cases z > z and z < z by using the absolute value function:

e − jk x x e − jk y y e −k ρ |z−z |
xy
G (k x , k y , z|r ) = . (A.54)
2k ρ
Finally, we substitute (A.54) into the inverse transform formula. This gives the Green’s
function representation
1 1 ∞ e −k ρ |z−z |
jk ρ ·(r−r ) 2
G(r|r ) = = e d k ρ , (A.55)
2
4π|r − r | (2π) 2k ρ
−∞
2
where k ρ = ˆ xk x + ˆ yk y , k ρ =|k ρ |, and d k ρ = dk x dk y .
On occasion we may wish to represent the solution of the homogeneous (Laplace)
equation
2
∇ ψ(r) = 0

in terms of a 2-D Fourier transform. In this case we represent ψ as a 2-D inverse transform
and substitute to obtain
1 ∞ 2 xy jk x x jk y y
∇ ψ (k x , k y , z)e e dk x dk y = 0.
(2π) 2 −∞
Carrying out the derivatives and invoking the Fourier integral theorem we find that

2
∂ 2 xy
− k ρ ψ (k x , k y , z) = 0.
∂z 2
Hence

xy
ψ (k x , k y , z) = Ae k ρ z + Be −k ρ z
where A and B are constants with respect to z. Inverse transformation gives

1 ∞ k ρ z −k ρ z jk ρ ·r 2
ψ(r) = A(k ρ )e + B(k ρ )e e d k ρ . (A.56)
(2π) 2
−∞

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