Page 336 - Electromagnetics

−1
Here the spatial frequencytransform variable k z has units of m . The forward and
inverse transform expressions are
∞

z − jk z z
ψ (x, y, k z , t) = ψ(x, y, z, t)e dz, (4.387)
−∞
1 ∞ z jk z z
ψ(x, y, z, t) = ψ (x, y, k z , t)e dk z , (4.388)
2π
−∞
by(A.1) and (A.2).
We interpret (4.388) much as we interpreted the temporal inverse transform (4.2).
Anyvector component of the electromagnetic field can be decomposed into a continuous
z
superposition of elemental spatial terms e jk z z with weighting factors ψ (x, y, k z , t).In
z
this case ψ is the spatial frequency spectrum of ψ. The elemental terms are spatial
sinusoids along z with rapidityof variation described by k z .
z
As with the temporal transform, ψ cannot be arbitrarysince ψ must obeya scalar
wave equation such as (2.327). For instance, for a source-free region of free space we
must have
∞
1 ∂ 1
2 z jk z z
∇ − ψ (x, y, k z , t)e dk z = 0.
2
c ∂t 2 2π −∞
2
2
2
2
Decomposing the Laplacian operator as ∇ =∇ +∂ /∂z and taking the derivatives into
t
the integrand, we have
1 ∞
2 2 1 ∂ 2 z jk z z
∇ − k − ψ (x, y, k z , t) e dk z = 0.
z
t
2
2π c ∂t 2
−∞
Hence
1 ∂
2
2 2 z
∇ − k − ψ (x, y, k z , t) = 0 (4.389)
z
t
2
c ∂t 2
bythe Fourier integral theorem.
The elemental component e jk z z is spatiallysinusoidal and occupies all of space. Because
such an element could onlybe created bya source that spans all of space, it is nonphysical
when taken byitself. Nonetheless it is often used to represent more complicated fields.
If the elemental spatial term is to be used alone, it is best interpreted physically when
combined with a temporal decomposition. That is, we consider a two-dimensional trans-
form, with transforms over both time and space. Then the time-domain representation
of the elemental component is
1 ∞ jk z z jωt
φ(z, t) = e e dω. (4.390)
2π
−∞
Before attempting to compute this transform, we should note that if the elemental term
is to describe an EM field ψ in a source-free region, it must obeythe homogeneous scalar
wave equation. Substituting (4.390) into the homogeneous wave equation we have
2 ∞
1 ∂ 1
2 jk z z jωt
∇ − e e dω = 0.
2
c ∂t 2 2π −∞
Differentiation under the integral sign gives
1 ∞
2 ω 2 jk z z jωt
−k + e e dω = 0
z
2π c 2
−∞
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