Page 465 - Electromagnetics

Using Euler’s identity we can write

1 −e e jθ
π − z sin θ j z cos θ
lim I (k z ) dk z = lim j e dθ.
2 2 jθ
→∞ 2π →∞ 0 e

Thus, as long as z > 0 the integrand will decay exponentially as →∞, and

lim I (k z ) dk z → 0.
→∞

Similarly, I (k z ) dk z → 0 when z < 0 if we close the semicircle in the lower half-plane.

Thus,
∞

I (k z ) dk z =− lim I (k z ) dk z − lim I (k z ) dk z . (A.22)
δ→0 δ→0
−∞ 1 2
The integrals around the poles can also be computed by writing k z in polar coordinates.
Writing k z =−k + δe jθ we find

1 −e ) jδe
0 jz(−k+δe jθ jθ
lim I (k z ) dk z = lim dθ
δ→0 2π δ→0 π (−k + δe jθ − k)(−k + δe jθ + k)
1
1 π e − jkz j − jkz
= jdθ =− e .
2π 0 −2k 4k
jθ
Similarly, using k z = k + δe , we obtain

j jkz
lim I (k z ) dk z = e .
δ→0 4k
2
Substituting these into (A.22) we have
j − jkz j jkz 1
˜ g(z,ω) = e − e = sin kz, (A.23)
4k 4k 2k
valid for z > 0.For z < 0, we close in the lower half-plane instead and get
1
˜ g(z,ω) =− sin kz. (A.24)
2k
Substituting (A.23) and (A.24) into (A.21) we obtain
z
sin k(z − ζ) 1 ∞ sin k(z − ζ)
˜ ˜ ˜
ψ(x, y, z,ω) = S(x, y,ζ,ω) dζ − S(x, y,ζ,ω) dζ
2k 2k 2k
−∞ z
where we have been careful to separate the two cases considered above. To make things
a bit easier when we apply the boundary conditions, let us rewrite the above expression.
Splitting the domain of integration we write

0 sin k(z − ζ) z sin k(z − ζ)
˜ ˜ ˜
ψ(x, y, z,ω) = S(x, y,ζ,ω) dζ + S(x, y,ζ,ω) dζ −
2k k
−∞ 0
∞ sin k(z − ζ)

˜
− S(x, y,ζ,ω) dζ.
0 2k
© 2001 by CRC Press LLC