Page 438 - Electromagnetics

There is one situation in which it is relatively easy to use the Schelkunoff equivalence.
Consider a perfectly conducting ground screen with an aperture in it, as shown in Figure
6.5. We assume that the aperture has been illuminated in some way by an electromagnetic
wave produced by sources in region 1 so that there are both fields within the aperture
and electric current flowing on the region-2 side of the screen due to diffraction from
the edges of the aperture. We wish to compute the fields in region 2. We can create an
equivalent problem by placing a planar surface S 0 adjacent to the screen, but slightly
offset into region 2, and then closing the surface at infinity so that all of the screen plus
region 1 is excluded. Then we replace region 1 with homogeneous space and place on S 0
eq
˜
˜
˜ ˜
˜
˜
eq
the equivalent currents J s = ˆ n × H, J ms =−ˆ n × E, where H and E are the fields on S 0 in
eq
˜
the original problem. We note that over the portion of S 0 adjacent to the screen J ms = 0
˜
˜
eq
since ˆ n × E = 0, but that J s = 0. From the equivalent currents we can compute the
fields in region 2 using the potential functions. However, it is often difficult to determine
˜
eq
J s over the conducting surface. If we apply Schelkunoff’s equivalence, we can formulate
a second equivalent problem in which we place into region 1 a perfect conductor. Then
eq
eq
˜
˜
we have the equivalent source currents J s and J ms adjacent and tangential to a perfect
conductor. By the image theorem of § 5.1.1 we can replace this problem by yet another
eq
eq
˜
˜
equivalent problem in which the conductor is replaced by the images of J s and J ms in
eq
˜
homogeneous space. Since the image of the tangential electric current J s is oppositely
directed, the fields of the electric current and its image cancel. Since the image of the
eq
˜
magnetic current is in the same direction as J ms , the fields produced by the magnetic
eq
˜
current and its image add. We also note that J ms is nonzero only over the aperture
˜
(since ˆ n × E = 0 on the screen), and thus the field in region 1 can be found from
1
˜ ˜
E(r,ω) =− ∇× A h (r,ω),
c
˜ (ω)
where

˜ c ˜
A h (r,ω) = ˜ (ω)[−2ˆ n × E ap (r ,ω)]G(r|r ; ω) dS
S 0
˜
and E ap is the electric field in the aperture in the original problem. We shall present an
example in the next section.
6.3.5 Far-zone fields produced by equivalent sources
The equivalence principle is useful for analyzing antennas with complicated source
distributions. The sources may be excluded using a surface S, and then a knowledge
of the fields over S (found, for example, by estimation or measurement) can be used to
compute the fields external to the antenna. Here we describe how to compute these fields
in the far zone.
eq
eq
˜
˜
˜
˜
Given that J s = ˆ n × H and J ms =−ˆ n × E are the equivalent sources on S,we may
compute the fields using the potentials (6.43) and (6.45). Using (6.20) these can be

approximated in the far zone (r r , kr 1) as
e − jkr
˜
A e (r,ω) = ˜µ(ω) ˜ a e (θ, φ, ω),
4πr
e − jkr
c
˜
A h (r,ω) = ˜ (ω) ˜ a h (θ, φ, ω), (6.48)
4πr
where

˜ eq

˜ a e (θ, φ, ω) = J (r ,ω)e jkˆ r·r dS ,
s
S
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