l{3a(fh, <Pb; (}a, <Pa in .NET

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l{3a(fh, <Pb; (}a, <Pa
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On the other hand, if we view lb as transmitting the incident field and receiving antenna at ra' then the bistatic scattering coefficient is, on using ki = -fb and ks = fa,
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Thus, using the reciprocity relation of (2.5.32) in (2.5.35) and (2.5.36) gives cos Oa l{3a ((}b, <Pb; (}a, <Pa
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With slight modification, the method can be used to derive reciprocity relations for multilayered media with rough interfaces separating the layers. Thus, reciprocity holds if the permittivity tensors at all points are symmetric. Reciprocity holds even in the presence of volume and rough surface scattering. We can also use (2.5.28)-(2.5.32) to derive symmetry relations for scattering amplitudes (Fig. 2.5.3). In Problem (a), the incident direction is
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k i = -Ta and the scattered direction is k s = Tb. Note that Ea(r) = ..".----F.E a.
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e ikr =
i3. E:Cfb) =
i3. F((h, <Pb; 7r -
()a, 7r + <Pa) . 0:
(i 7rra
J111 e ikra )
Similarly in Problem (b), the incident direction is direction is ks = a . Thus,
= -Tb and the scattered
-s e = 0: . Eb(r a) = --0: . F(()a, <Pa; 7r
, J11l ()b, 7r + <Pb) . {3 -4-e2Okrb ) 7rTb
(}b, 7f + <Pb) . ~
The relation of reciprocity of (2.5.32) equates (2.5.38) and (2.5.39) sO that
~ . FUh, cPb; 7f - (}a, 7f + cPa) . a =
a. F( (}a, cPa; 7f -
To put (2.5.40) into component forms, we note that Ta, Va, and h a form an orthonormal system, while rb, Vb, and hb form an orthonormal system.
r(()a, <Pa) = sin ()a cos <Pax + sin ()a sin <PaY + cos ()a z v((}a, cPa) = cos (}a cos cPaX + cos (}a sin cPaY - sin (}a z h((}a, <Pa) = - sin <Pax + cos <PaY
Note that
r(7f -
(2.5.41) (2.5.42) (2.5.43)
v( 7r -
(}a, 7r + <Pa) = -r((}a' cPa) (}a, 7r + cPa) = v((}a, <Pa)
(2.5.44) (2.5.45) (2.5.46)
h(7f - (}a, 7f + cPa) = -h((}a' cPa)
In the backward directions, it follows from the definitions of (2.5.41)-(2.5.43)
that vertical polarization keeps the direction while the horizontal polarization vector direction is opposite. Now,
where {3, a = v, h. Let 0: = v(7r - ()a, 7r Then we have Next, let a = h( 7f - (}a, 7r + cPa) (}b, <Pb) in (2.5.44). This gives
+ <Pa)
i3 =
V((}b, <Pb) in (2.5.40).
7r -
!vv(()b, <Pb; 7r - ()a, 7r + <Pa) = !vv(()a, <Pa; 7r - ()b, 7r + <Pb)
= -he(}a, cPa)
b = h((}b, cPb) = -he
i3 =
fhh((}b, cPb; 7r - (}a, 7r + cPa) = fhh ((}a , cPa; 7r - (}b, 7r + cPb)
Finally, let & = h(7r - (}a, 7r + <Pa) = -h(()a, <Pa) and (}b,7r + <Pb) in (2.5.40). This gives
V(()b' <Pb) = v(7r -
5.3 Symmetry Relations for Green's Function
fvh((h, rPb; 7r - (}a, 7r + rPa) = -h((}a, rPa) . F((}a, rPa; 7f - (}b, 7f + rPb) . v( 7r - (}b, 7r + rPb) = - fhv((}a, rPa; 7r - (}b, 7r + rPb) (2.5.50)
Note that because of the change of polarization vectors in the backward direction, there is a negative sign in the reciprocity relation of (2.5.50) but not in (2.5.48) and (2.5.49).
5.3 Symmetry Relations for Dyadic Green's Function
The reciprocal relations of (2.5.23) leads to symmetry relations for the dyadic Green's function. Let the geometrical configuration consist of region 0 and region 1 that are separated by a boundary S (Fig. 2.5.4). Let Glj(r, r') denote dyadic Green's Function, with field point r in region l and source point r' in region j. In (2.5.23), let J a and Jb be in region 0 so that
r dV(E
a Jb -
J a . E b)
Ja =
-.-8(r - r a) J,Wj1,
(2.5.52) (2.5.53)
Substituting (2.5.52)-(2.5.53) in (2.5.51) gives
~. Ea(rb) = & . Eb(ra)
Then by the definition of the dyadic Green's function for r in region 0,
Ea(r) = Goo(r, r a) . & Eb(r) = Goo(r, rb) . ~
Substituting (2.5.55)-(2.5.56) in (2.5.54) gives
(2.5.55) (2.5.56)
~. Goo (rb' r a) . & = & . Goo(ra,rb) . ~
Thus, since & and ~ are arbitrary we obtain
_ -t
Goo (rb' r a) = Goo (ra, rb)
The symmetry relations can readily be extended to the case when the sources are in two different regions. We then consider Problem (c) with source in