Page 340 - Electromagnetics

erroneously chosen < k i we would not have properly enclosed the pole and would have
obtained an incorrect inverse transform.
Now that we know how to represent the Fourier transform pair, let us apply the
˜
transform to solve (4.394). Our hope is that by representing ψ in terms of a spatial
Fourier integral we will make the equation easier to solve. We have
∞+ j
1 x − jk x x
˜
2
2
(∇ + k ) ψ (k x , y,ω)e dk x = 0.
t
2π
−∞+ j
Differentiation under the integral sign with subsequent application of the Fourier integral
˜
theorem implies that ψ must obey the second-order harmonic differential equation
2
d 2 x
˜
+ k y ψ (k x , y,ω) = 0
dy 2
2
2
2
where we have defined the dependent parameter k y = k yr + jk yi through k + k = k .
x y
Two independent solutions to the differential equation are e ∓ jk y y and thus
˜
ψ(k x , y,ω) = A(k x ,ω)e ∓ jk y y .
Substituting this into the inversion integral, we have the solution to the Helmholtz equa-
tion:
∞+ j
1
˜ − jk x x ∓ jk y y
ψ(x, y,ω) = A(k x ,ω)e e dk x . (4.400)
2π
−∞+ j
If we define the wave vector k = ˆ xk x ± ˆ yk y , we can also write the solution in the form
∞+ j
1
˜ − jk·ρ
ψ(x, y,ω) = A(k x ,ω)e dk x (4.401)
2π
−∞+ j
where ρ = ˆ xx + ˆ yy is the two-dimensional position vector.
The solution (4.401) has an important physical interpretation. The exponential term
looks exactly like a plane wave with its wave vector lying in the xy-plane. For lossy media
the plane wave is nonuniform, and the surfaces of constant phase may not be aligned
with the surfaces of constant amplitude (see § 4.11.4). For the special case of a lossless
medium we have k i → 0 and can let → 0 as long as > k i . As we perform the inverse
2
transform integral over k x from −∞ to ∞ we will encounter both the condition k > k 2
x
2
2
2
2
and k ≤ k .For k ≤ k we have
x x
√
2
2
− jk x x ∓ jk y y − jk x x ∓ j k −k x y
e e = e e
where we choose the upper sign for y > 0 and the lower sign for y < 0 to ensure that the
waves propagate in the ±y-direction, respectively. Thus, in this regime the exponential
represents a propagating wave that travels into the half-plane y > 0 along a direction
which depends on k x , making an angle ξ with the x-axis as shown in Figure 4.28. For k x in
[−k, k], every possible wave direction is covered, and thus we may think of the inversion
integral as constructing the solution to the two-dimensional Helmholtz equation from a
continuous superposition of plane waves. The amplitude of each plane wave component
is given by A(k x ,ω), which is often called the angular spectrum of the plane waves and
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