Page 440 - Electromagnetics

Figure 6.6: Aperture antenna consisting of a rectangular waveguide opening into a con-
ducting ground screen of infinite extent.

modes excited when the guided wave is reflected at the aperture may be ignored. Thus
the electric field in the aperture S 0 is
π
˜
E a (x, y) = ˆ yE 0 cos x .
a
We may compute the far-zone field using the Schelkunoff equivalence principle of § 6.3.4.
+
We exclude the region z < 0 using a planar surface S which we close at infinity. We
then fill the region z < 0 with a perfect conductor. By the image theory the equivalent
electric sources on S cancel while the equivalent magnetic sources double. Since the
˜
only nonzero magnetic sources are on S 0 (since ˆ n × E = 0 on the screen), we have the
equivalent problem of the source
π

˜ eq =−2ˆ n × E a = 2ˆ xE 0 cos x
˜
J
ms
a
on S 0 in free space, where the equivalence holds for z > 0.
We may find the far-zone field created by this equivalent current by first computing
the directional weighting function (6.49). Since

ˆ r · r = ˆ r · (x ˆ x + y ˆ y) = x sin θ cos φ + y sin θ sin φ,
we find that
b/2 a/2 π

e
˜ a h (θ, φ, ω) = ˆ x2E 0 cos x e jkx sin θ cos φ jky sin θ sin φ dx dy
−b/2 −a/2 a
cos π X sin πY
= ˆ x4π E 0 ab
2
π − 4(π X) 2 πY
where
a b
X = sin θ cos φ, Y = sin θ sin φ.
λ λ
Here λ is the free-space wavelength. By (6.50) the electric field is
˜ jk 0 ˜
E = ˆ r × A h
0
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