Page 468 - Electromagnetics

Applying these to (A.25) (and again using Leibnitz’ rule) we have

˜
˜
˜
A(x, y,ω) + B(x, y,ω) = f (x, y,ω),
˜
˜
− jk A(x, y,ω) + jkB(x, y,ω) = ˜ g(x, y,ω),
hence
˜ g(x, y,ω)
˜
˜
2A(x, y,ω) = f (x, y,ω) − c ,
jω
˜ g(x, y,ω)
˜
˜
2B(x, y,ω) = f (x, y,ω) + c .
jω
Finally, substituting these back into (A.25) and expanding the sine function we obtain
the frequency-domain solution that obeys the given boundary conditions:
ω
ω
z
˜
˜
c S(x, y,ζ,ω)e j (z−ζ) S(x, y,ζ,ω)e − j (z−ζ)
c
c
˜
ψ(x, y, z,ω) = − dζ +
2 0 jω jω
ω
˜
ω
1 j z − j z
+ ˜ f (x, y,ω)e c + f (x, y,ω)e c +
2
ω
ω
c ˜ g(x, y,ω)e j z ˜ g(x, y,ω)e − j z
c
c
+ − .
2 jω jω
This is easily inverted using (A.27) to give (A.32).
Fourier transform solution of the 1-D homogeneous wave equation for
dissipative media
Wave propagation in dissipative media can be studied using the one-dimensional wave
equation
2 2
∂ 2 ∂ 1 ∂
− − ψ(x, y, z, t) = S(x, y, z, t). (A.33)
2
2
∂z 2 v ∂t v ∂t 2
This equation is nearly identical to the wave equation for lossless media studied in the
previous section, except for the addition of the ∂ψ/∂t term. This extra term will lead to
important physical consequences regarding the behavior of the wave solutions.
We shall solve (A.33) using the Fourier transform approach of the previous section,
but to keep the solution simple we shall only consider the homogeneous problem. We
begin by writing ψ in terms of its inverse temporal Fourier transform:
1 ∞ jωt
˜
ψ(x, y, z, t) = ψ(x, y, z,ω)e dω.
2π
−∞
Substituting this into the homogeneous version of (A.33) and taking the time derivatives,
we obtain
1 ∞ 2 2 ∂ 2 jωt
˜
( jω) + 2 ( jω) − v ψ(x, y, z,ω)e dω = 0.
2π ∂z 2
−∞
The Fourier integral theorem leads to
2 ˜
∂ ψ(x, y, z,ω) 2
˜
− κ ψ(x, y, z,ω) = 0 (A.34)
∂z 2
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