Page 473 - Electromagnetics

We can compute the integrals over K using the complex plane technique. We consider K
to be a complex variable, and note that for dissipative media we have k = k r + jk i , where
k r > 0 and k i < 0. Thus the integrand has poles at K =±k. For the integral involving
e + jK R we close the contour in the upper half-plane using a semicircle of radius and
use Cauchy’s residue theorem. Then at all points on the semicircle the integrand decays
exponentially as →∞, and there is no contribution to the integral from this part of
the contour. The real-line integral is thus equal to 2π j times the residue at K =−k:
∞
e e
jK R − jkR
KdK = 2π j (−k).
−∞ (K − k)(K + k) −2k
For the term involving e − jK R we close in the lower half-plane and again the contribution
from the infinite semicircle vanishes. In this case our contour is clockwise and so the real
line integral is −2π j times the residue at K = k:
e e
∞
− jK R − jkR
KdK =−2π j k.
−∞ (K − k)(K + k) 2k
Thus
e − jkR
˜
G(r|r ; ω) = . (A.49)
4π R
Note that if = 0 then this reduces to
e − jωR/v
˜
G(r|r ; ω) = . (A.50)
4π R
Our last step is to find the temporal Green’s function. Let p = jω. Then we can write
e κ R
˜
G(r|r ; ω) =
4π R
where
1
2
κ =− jk = p + 2 p.
v
We may find the inverse transform using (A.43). Letting x = R, ρ = , and σ = we
find

2 2
δ(t − R/v) 2 − t I 1 t − (R/v) R
− R

G(r|r ; t) = e v + e U t − .
4π R 4πv t − (R/v) 2 v
2
We note that in the case of no dissipation where = 0 this reduces to
δ(t − R/v)

G(r|r ; t) =
4π R
which is the inverse transform of (A.50).
Fourier transform representation of the static Green’s function
In the study of static fields, we shall be interested in the solution to the partial differ-
ential equation
2

∇ G(r|r ) =−δ(r − r ) =−δ(x − x )δ(y − y )δ(z − z ). (A.51)

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