Page 469 - Electromagnetics

where
1
2
κ = p + 2 p
v
with p = jω.
We can solve the homogeneous ordinary differential equation (A.34) by inspection:
˜
κz
˜
˜
ψ(x, y, z,ω) = A(x, y,ω)e −κz + B(x, y,ω)e . (A.35)
˜
˜
Here A and B are frequency-domain coefficients to be determined. We can either specify
these coefficients directly, or solve for them by applying specific boundary conditions.
We examine each possibility below.
Solution to wave equation by direct application of boundary conditions. The
solution to the wave equation (A.33) will be unique if we specify functions f (x, y, t) and
g(x, y, t) such that

= f (x, y, t),

ψ(x, y, z, t)
z=0
∂
= g(x, y, t). (A.36)

ψ(x, y, z, t)
∂z z=0
˜
Assuming the Fourier transform pairs f (x, y, t) ↔ f (x, y,ω) and g(x, y, t) ↔ ˜ g(x, y,ω),
we can apply the boundary conditions (A.36) in the frequency domain:

˜ = f (x, y,ω),
˜
ψ(x, y, z,ω)
z=0
∂
˜ = ˜ g(x, y,ω).
∂z ψ(x, y, z,ω) z=0
From these we find
˜
˜
˜
˜
˜
A + B = f , −κ A + κ B = ˜ g,v
or
1 ˜ g 1 ˜ g
˜ ˜ ˜ ˜
A = f − , B = f + .
2 κ 2 κ
Substitution into (A.35) gives
sinh κz
˜
˜
ψ(x, y, z,ω) = f (x, y,ω) cosh κz + ˜ g(x, y,ω)
κ
∂
˜
˜
˜
= f (x, y,ω) Q(x, y, z,ω) + ˜ g(x, y,ω)Q(x, y, z,ω)
∂z
˜
˜
= ψ 1 (x, y, z,ω) + ψ 2 (x, y, z,ω)
˜
˜
where Q = sinh κz/κ. Assuming that Q(x, y, z, t) ↔ Q(x, y, z,ω), we can employ the
convolution theorem to immediately write down ψ(x, y, z, t):
∂
ψ(x, y, z, t) = f (x, y, t) ∗ Q(x, y, z, t) + g(x, y, z, t) ∗ Q(x, y, z, t)
∂z
= ψ 1 (x, y, z, t) + ψ 2 (x, y, z, t). (A.37)
˜
To find ψ we must first compute the inverse transform of Q. Here we resort to a
tabulated result [26]:
√ √
sinh a p + λ p + µ 1 − (µ+λ)t 1
1
2
√ √ ↔ e 2 J 0 (λ − µ) a − t 2 , −a < t < a.
p + λ p + µ 2 2
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