Page 349 - Electromagnetics

where φ 0 is the angle between the incident wave vector and the x-axis. By (4.223) we
also have
˜
E 0 (ω)
˜ i jk(x cos φ 0 +y sin φ 0 )
H (r,ω) = (ˆ y cos φ 0 − ˆ x sin φ 0 )e .
η
The scattered fields may be written in terms of the Fourier transform solution to the
Helmholtz equation. It is convenient to use the polar coordinate representation (4.403)
to develop the necessary equations. Thus, for the scattered electric field we can write

˜ s − jkρ cos(φ±ξ)
E (x, y,ω) = f (ξ, ω)e dξ. (4.415)
z
C
By (4.404) the x-component of the magnetic field is
1 ∂E ˜ s 1 ∂
˜ s z − jkx cos ξ ± jky sin ξ
H (x, y,ω) =− =− f (ξ, ω) e e
x
jω ˜µ ∂y jω ˜µ C ∂y
1 − jkρ cos(φ±ξ)
=− (± jk) f (ξ, ω) sin ξe dξ.
jω ˜µ C
To find the angular spectrum f (ξ, ω) and ensure uniqueness of solution, we must apply
the boundary conditions over the entire y = 0 plane. For x > 0 where the conductor
resides, the total tangential electric field must vanish. Setting the sum of the incident
and scattered fields to zero at φ = 0 we have

˜
f (ξ, ω)e − jkx cos ξ dξ =−E 0 e jkx cos φ 0 , x > 0. (4.416)
C
˜ s
To find the boundary condition for x < 0 we note that by symmetry E is even about
z
˜ s
˜ s
y = 0 while H , as the y-derivative of E , is odd. Since no current can be induced in the
x z
y = 0 plane for x < 0, the x-directed scattered magnetic field must be continuous and
thus equal to zero there. Hence our second condition is

f (ξ, ω) sin ξe − jkx cos ξ dξ = 0, x < 0. (4.417)
C
Now that we have developed the two equations that describe f (ξ, ω), it is convenient
to return to a rectangular-coordinate-based spectral integral to analyze them. Writing
−1
ξ = cos (k x /k) we have
d dk x
(k cos ξ) =−k sin ξ =
dξ dξ
and
dk x dk x dk x
dξ =− =− =− .
2
2
k sin ξ k 1 − cos ξ k − k 2 x
Upon substitution of these relations, the inversion contour returns to the real k x axis
(which may then be perturbed by j ). Thus, (4.416) and (4.417) may be written as
∞+ j
f cos

−1 k x
k − jk x x ˜ jk x0 x
2 2 e dk x =−E 0 e , x > 0, (4.418)
k − k x
−∞+ j
∞+ j

k x
f cos −1 e − jk x x dk x = 0, x < 0, (4.419)
k
−∞+ j
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