Reciprocity and Symmetry in .NET

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Reciprocity and Symmetry
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5.1 Reciprocity
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When the medium of propagation is reciprocal, the electromagnetic fields obey reciprocity relations [Kong, 1990J. The reciprocity relations give rise to symmetry relations of bistatic scattering coefficients, scattering amplitudes, and dyadic Green's functions. It is also used to derive the relations between emissivity and bistatic scattering coefficients. For a general anisotropic medium with permittivity tensor f, the reciprocal medium relation is that f is symmetric. If the medium is reciprocal, then the electromagnetic fields obey reciprocity relations even though the medium may be inhomogeneous and may have rough boundaries. We consider the electromagnetic fields of a region 0 with permittivity E separated by a rough boundary from region 1 with inhomogeneous permittivity fl(r) (Fig. 2.5.1). The medium 1 is reciprocal so that
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where t denotes the transpose of the dyad. Reciprocity relations relate the electromagnetic fields of different problems.
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Problem (a)
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Consider a source]a placed in region 0 producing fields E a and H a in region o and fields E ia and IlIa in region 1. The Maxwell equations are
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'V x H a = -iWEEa + J a 'V x E a = iWJ1H a 'V x H Ia = -iw~l(r) . E ia 'V
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(2.5.2) (2.5.3) (2.5.4) (2.5.5) (2.5.6) (2.5.7)
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The boundary conditions on rough surfaces S are
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it x H a = it it x E a = it
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Region 0
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Region 1
E (f),' 1
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.- " " ...... '- 8 100
Figure 2.5.1 Reciprocity relation for half-space of inhomogeneous anisotropic medium with rough interface.
Problem (b) Consider a source ]b placed in region O. The Maxwell equations are
'V x Hb = -iWEEb +]b 'V x E b = iwp,Hb 'V x H 1b = -iW~l (r) . E 1b 'V X E 1b = iwp,H 1b
with boundary conditions on surface 8
(2.5.8) (2.5.9) (2.5.10) (2.5.11) (2.5.12) (2.5.13) (2.5.14) (2.5.15) (2.5.16) (2.5.17) (2.5.18)
n x Hb = n x Hib n x E b = n x Elb
'V . (Eb
\7. (Eb
H a) = H a . \7
E b - E b . \7 x H a Eb']a
Substituting (2.5.2) and (2.5.9) in (2.5.14) gives
H a) = iwp,H a . Hb
+ iWEEa . E b -
'V . (Ea X H b) = iw jiBa . H b + iWE Ea' E b - Ea' ] b 'V. (Elb x H Ia) = iWJiB Ia . H Ib + iwElb . fl(r) . Ei a \7 . (E 1a X H Ib) = iwp,H1a . H 1b + iwE 1a . ~1 (r) . E 1b
5.2 Reciprocity Relation for Scattering Amplitudes
1 is symmetric, we have
E 1b' l(r) E1a = E1a . l(r) . E 1b
Thus from (2.5.17)-(2.5.19),
V' . (E 1b x H 1a - E 1a
H 1b) = 0
Integrating (2.5.20) over region 1 and applying divergence theorem to (2.5.20) and the radiation condition over 8 100 gives
dSn (E 1a
H 1b - E 1b X H 1a ) = 0
Similarly, using (2.5.14)-(2.5.16) and performing volume integration over region 0 gives
where is normal pointing into region O. Using the boundary conditions (2.5.6)-(2.5.7) and (2.5.12)-(2.5.18) on 8 and the vector identity of a' (5 x c) = a x b c, the left-hand sides of (2.5.21) and (2.5.22) are equal. Thus,
dV(Ea']b -]a . Eb) = 0
The reciprocity relations can also be applied to scattered field. The total field can be expressed as a sum of incident and scattered fields:
Ea = Ea+ Ea ~ -8 Eb = E b + E b
(2.5.24) (2.5.25)
The previous analysis can be repeated for the case when the half-space medium is absent. Hence
Subtracting (2.5.26) from (2.5.23) gives
dV (~ . ]b - ]
~) =
dV (E: . ]
a . E~) = 0
Reciprocal Relations for Bistatic Scattering Coefficients and
Scattering Amplitudes
To derive the reciprocal relation for the bistatic scattering coefficients, we let ]a and ]b be Hertzian dipoles located at r a and rb, respectively, and
direction ((h, <Pb)
Figure 2.5.2 and (Ob' <Pb)'
Reciprocity relation for bistatic scattering coefficients in direction (Oa, <Pa)
elements pointing in directions 6: and jJ, respectively. Thus,
]a = 6:Il8(r - r a)
(2.5.28) (2.5.29)
= jJIl8(r -
Let]a and ]b be at the far-field region in the directions (Oa, <Pa) and (lh, <Pb), respectively, and at distances r a and rb, respectively, from the target (Fig. 2.5.2). We also let 6: and jJ to be perpendicular to r a and rb, respectively. Then in the vicinity of the target, the incident field will be in the form of plane waves:
. ~ iw jiB ikr -ikf:1' a E = Q--e ae
47l"r a
~ = jJiwj1Jl eikTbe-ikr..r
Substituting of (2.5.28) and (2.5.29) in (2.5.27) gives
E:(rb) . jJ = E:(ra ) &
The direction vectors are
r a = sinOa cos <Pax + sinOa sin <PaY + cosOaz rb = sin Ob cos <PbX + sin Ob sin <PbY + cos 0b Z
(2.5.32) (2.5.33) (2.5.34)
5.2 Reciprocity Relation for Scattering Amplitudes
dipole source with polarization a
(observation point)
Problem (a)
(observation point) dipole source with polarization {3
Problem (b)
Figure 2.5.3 Reciprocity for scattering amplitudes.
If we view 1 a as transmitting the incident field and receiving antenna at rb, then the bistatic scattering coefficient is, on using ki = -ra and ks = fb'