Page 338 - Electromagnetics

the Fourier transform to represent the spatial dependence. By § 4.11.2 a general two-
dimensional field maybe decomposed into fields TE and TM to the z-direction. In the
˜
˜
TM case H z = 0, and E z obeys the homogeneous scalar Helmholtz equation (4.208).
˜
˜
In the TE case E z = 0, and H z obeys the homogeneous scalar Helmholtz equation.
˜
Since each field component obeys the same equation, we let ψ(x, y,ω) represent either
˜
˜
˜
E z (x, y,ω) or H z (x, y,ω). Then ψ obeys
2
˜
2
(∇ + k )ψ(x, y,ω) = 0 (4.394)
t
c
c 1/2
2
where ∇ is the transverse Laplacian (4.209) and k = ω( ˜µ˜ ) with ˜ the complex
t
permittivity.
˜
We maychoose to represent ψ(x, y,ω) using Fourier transforms over one or both
spatial variables. For application to problems in which boundaryvalues or boundary
conditions are specified at a constant value of a single variable (e.g., over a plane), one
transform suffices. For instance, we mayknow the values of the field in the y = 0 plane
(as we will, for example, when we solve the boundaryvalue problems of § ??). Then
we maytransform over x and leave the y variable intact so that we maysubstitute the
boundaryvalues.
We adopt (4.392) since the result is more readilyinterpreted in terms of propagating
plane waves. Choosing to transform over x we have
∞

˜ x ˜ jk x x
ψ (k x , y,ω) = ψ(x, y,ω)e dx, (4.395)
−∞
1 ∞
˜ x − jk x x
ψ(x, y,ω) = ψ (k x , y,ω)e dk x . (4.396)
2π
−∞
For convenience in computation or interpretation of the inverse transform, we often
regard k x as a complex variable and perturb the inversion contour into the complex k x =
k xr + jk xi plane. The integral is not altered if the contour is not moved past singularities
such as poles or branch points. If the function being transformed has exponential (wave)
behavior, then a pole exists in the complex plane; if we move the inversion contour across
this pole, the inverse transform does not return the original function. We generally
indicate the desire to interpret k x as complex byindicating that the inversion contour is
parallel to the real axis but located in the complex plane at k xi = :
∞+ j
1
˜ ˜ x − jk x x
ψ(x, y,ω) = ψ (k x , y,ω)e dk x . (4.397)
2π
−∞+ j
Additional perturbations of the contour are allowed provided that the contour is not
moved through singularities.
As an example, consider the function

0, x < 0,
u(x) = (4.398)
e − jkx , x > 0,
where k = k r + jk i represents a wavenumber. This function has the form of a plane
wave propagating in the x-direction and is thus relevant to our studies. If the material
through which the wave is propagating is lossy, then k i < 0. The Fourier transform of
the function is
∞
∞ 1
x − jkx jk x x
e
u (k x ) = e e dx = e j(k xr −k r )x −(k xi −k i )x .
0 j(k x − k) 0
© 2001 by CRC Press LLC