2 Integral Equations for Dielectric Surfaces

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scattering is in good agreement. The cross-polarized scattering is in excellent agreement. The comparison is excellent in view of the fact that the absolute values are compared and there are no adjustable parameters. Backscattering enhancement is a result of the contributions of the cyclical scattering processes (see 8 of Volume III) which begin at the second order. For co-polarization, the second-order terms can be obscured by the presence of first-order scattering. However, for cross-polarization, the first-order scattering is zero. Thus, the second-order backscattering enhancement is more clearly exhibited as seen in the simulations and the experimental data.

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Integral Equations for Dielectric Surfaces

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Electromagnetic Fields with Electric and Magnetic Sources

In formulating integral equations for electromagnetic wave scattering, a convenient method is through the use of equivalent electric and magnetic currents [Harrington, 1961; Wang, 1991]. For the Maxwell equations with electric sources of current density 1, volume charge density Pv, and surface charge density Ps

= iWtLH (6.2.1) \7 x H = -iWEE + 1 (6.2.2) \7 . EE = Pv (6.2.3) \7 . tLH = 0 (6.2.4) \7 1- iwpv = 0 (6.2.5) \7 s ls - iwps = 0 (6.2.6) where \7 s ls is the surface divergence of 1 s . Vector potential A and scalar

\7 x E

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H=-\7xA tL E = iwA - \7<P

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(6.2.7) (6.2.8)

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Using the gauge condition

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-\7. A = iWE<P tL

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Then

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(6.2.9)

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+ k 2 ) A = -tL1 (\7 2 + k 2 ) <P = _Pv

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(6.2.10)

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(6.2.11)

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63-D WAVE SCATTERING FROM 2-D ROUGH SURFACES

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9 (r,r') =

we have

47fr-r 1- -'I

(6.2.12)

The magnetic field is

\7 x

The electric field is

Jdr'9 (r, r') J (r') <I>(r) = } Jdr'9 (r, r') (r') H= Jdr'g(r,r')J(r') = - Jdr'\7g(r,r') J(r') (6.2.15) E(r) = iWtL Jdr'9 (r, r') J (r') Jdr'9 (r, r') (r') = iWtL [J dr'9 (r, r') J (r') + :2 Jdr'9 (r, r') J (r')] (6.2.16)

A(r) = tL

(6.2.13) (6.2.14)

\7' .

The fields are expressed in terms of electric current or in terms of electric current and electric charge. For the case of equivalent magnetic sources of magnetic current density M, volume charge density m, and surface charge density m s , the Maxwell equations are

\7 x \7 x

E = iwtLH H = -iWEE

(6.2.17) (6.2.18) (6.2.19) (6.2.20) (6.2.21) (6.2.22)

\7. EE

Vector potential analogous to that of the electric source case 1 E

\7. tLH = m \7. M - iwm = 0 \7 s . M s - iwm s = 0 F and scalar potential 1/J can E= -\7xF

be employed. Derivation is

(6.2.23) (6.2.24)

H = -iwF - \71/J

The gauge condition is 1 - \7 . F

= iWtL1/J

(6.2.25)

2.1 Electromagnetic Fields with Electric and Magnetic Sources

Then

(\7 2 + k 2 ) F = -EM (\7 2 + k 2 ) 1/J = _ m

F(r)

(6.2.26) (6.2.27)

0/'(-) = -1 'f/ r

dr' 9

(1",1"') M (1"')

(6.2.28) (6.2.29)

d-' 9 (- -') m (-') r r, r r

The electric field is

E(r)

= - \7 x

dr'9

(1",1"') M (1"') = -

dr'9

dr' \7 9

(1",1"')

(1"')

(6.2.30)

and the magnetic field is

H (1")

dr'g

(r, r') M (1"') - \7

(1",1"') M (1"') +

(1",1"')

dr'9

m :')

= iWE

dr' 9

:2 J

(1",1"') \7' . M (1"')]

(6.2.31)

The fields are expressed entirely in terms of magnetic current or in terms of magnetic current and magnetic charge. When both electric and magnetic sources are present, Eqs. (6.2.16)(6.2.15) and (6.2.30)-(6.2.31) are combined to give E(r) = -

dr' \79

(1",1"')

(1"')

+ iwf.L

H (r) =

dr'9

(1",1"') ] (1"') +

:2 J

\7 \7

dr' 9

(1",1"') \7' . ] (1"')] (6.2.32)

\7' . M

dr' 9

(1",1"') M (7") + (1",1"')

:2 J

dr' 9

(1",1"')

(1"') ]

(6.2.33)

dr'\7g

(1"')

The boundary conditions separating two media of f.L1, surface electric and magnetic sources are

and f.L2,

with

n x E 1 - n x E2 = -M s

nxH 1

(6.2.34) (6.2.35) (6.2.36) (6.2.37)

nxH 2=]s

n E1E1 - n' E2 E 2 = Ps n f.L1H1 - n' f.L2H2 = m s

6 3-D WAVE SCATTERING FROM 2-D ROUGH SURFACES

Region 2

Figure 6.2.1 Scattering of waves by dielectric surface: Physical problem, no surface sources.

Physical Problem and Equivalent Exterior and Interior Problems

Consider an incident wave with electric field E i and magnetic field Hi (Fig. 6.2.1) on an dielectric surface. Let E1, f..L1 be the permittivity and permeability respectively above the surface and E2, f..L2 be the permittivity and permeability respectively below the surface. This is the physical problem. In region 1, the electromagnetic fields are

E 1 = Ei

+ Es

H 1 =H i +H s

(6.2.38) (6.2.39)

In region 2, the electromagnetic fields are E2 and H2. The boundary conditions are

n x E 1 - n x E2 = 0

nxH 1

nxH 2=0

n E1E1

n . E2E2 = 0

n. f..L 1 HI - n. f..L2H 2 = 0

(6.2.40) (6.2.41) (6.2.42) (6.2.43)

Because this is a dielectric surface, there are neither surface currents nor surface charges at the boundary separating regions 1 and 2. Hence we have the boundary condition of (6.2.40)-(6.2.43). The physical problem can be represented by the exterior problem and the interior problem using equivalent surface currents and charges.

2.2.1 Equivalent Exterior Problem, Equivalent Currents and Integral Equations

Consider the equivalent exterior problem A with the same incident electric field E i and magnetic field Hi and same electric field E1 and magnetic field HI in region 1 as in the physical problem. In the equivalent problem A, let

2.2 Physical Problem and Equivalent Exterior and Interior Problems

Rq;o:~:i::t;~ ~

E1, /"1

E 1 =E, +Es

N, + IT,

!"7~

I/!TL

-A -A A A

Surface sources

-A-A M" J s

Region 2

Figure 6.2.2 Equivalent problem A, exterior problem with M s ' J s ' Ps, m s .

there be equivalent sources on the boundary (Fig. 6.2.2)

-M s =nxE 1 -A J s =nxH 1

-A -

(6.2.44) (6.2.45) (6.2.46) (6.2.47)