Polarimetry of homogeneous half-spacespogudin/workshop/Gilman.pdf · 2018. 7. 24. · Polarimetry...

Polarimetry of homogeneous half-spaces1

M. Gilman, E. Smith, S. Tsynkovspecial thanks to: H. Hong

Department of MathematicsNorth Carolina State University, Raleigh, NC

Workshop on Symbolic-Numeric Methodsfor Differential Equations and Applications

Courant Institute of Mathematical Sciences, New York, July 20, 2018

1Work supported by AFOSR, grants # FA9550-10-1-0092, FA9550-14-1-0218, andFA9550-17-1-0230

M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, 2018 1 / 34

Preliminaries Overview

Overview

Scattering by homogeneous half-spaces:extension of a textbook case: Fresnel reflection coefficients;inverse problem: material properties from scattered waves;quantifier elimination for polarimetry with uniaxial dielectric tensor.

Inhomogeneous half-spaces: vertical stratification, Leontovichboundary conditions, rough surface scattering.


Preliminaries Motivation

Motivation: radar imagingSynthetic aperture radar (SAR) is an established signal processingtechnology that retrieves the “averaged reflectivity”.

https://www.intelligence-airbusds.com/en/

5751-image-detail?img=17281#.WyuwpKknbUI

TerraSAR-X (spaceborne)

SAR scattering model assumes independent point scatterers:

acceptable for detection and tracking;not helpful for reconstruction of scatterer properties (e.g.,aboveground biomass, significant wave height).


Preliminaries Electromagnetic waves

Physical preliminaries

Electric E and magnetic H fields obey the Maxwell’s equations inmaterial:

1c∂H

∂t+ curlE = 0 ,

1c∂D

∂t− curlH = −4π

cj ,

divH = 0,

divD = 0.

The material is characterized by dielectric ε and conductivity σtensors:

D = ε ⋅E , j = σ ⋅E .

In isotropic dielectric with losses:

ε = εI , σ = σI Ô⇒ D = εE , j = σE .

In vacuum:ε = 1, σ = 0Ô⇒ D = E , j = 0 .


Preliminaries Electromagnetic waves

Transverse waves

Homogeneous isotropic perfect dielectric is described byε = εI, where ε is piecewise constant and σ = 0 ; hence

1c∂H

∂t+ curlE = 0 ,

1c∂(εE)∂t

− curlH = 0 , div(εE) = 0

yield, away from discontinuities of ε(r):

∆E = ε

c2∂2

∂t2E

Plane waves: solutions to the Maxwell’s equations of the form

E(r , t) ∼ ei(k ,r)−iωt where√εω = ∣k ∣c, H = c

ωk ×E

(for linear problems, take either real or imaginary part)Transversality: divE = divH = 0Ô⇒ (k ,E) = (k ,H ) = 0.


Preliminaries Refraction-reflection problem

Linear polarizations

Plane waves in vacuum: solutions to the Maxwell’s equations ofthe form (for linear problems, take either real or imaginary part):

E(r , t) ∼ ei(k ,r)−iωt, ω = kc, k = ∣k ∣, H = k ×E/kwith E ⊥ k , H ⊥ k , E ⊥H , ∣E ∣ = ∣H ∣.Planar interface: is characterized by the normal n .If we have a planar interface and k ∦ n , we can define theincidence plane and two linear polarizations:

Horizontal polarization (or E-polarization)

E(r , t) = EA ⋅ ei(k ,r)−iωt, EA = E0n × (k/k)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

vector amplitude ⊥ incidence plane

, H (r , t) = k ×E(r , t)/k.

Vertical polarization (or H-polarization)

H (r , t) =H A ⋅ ei(k ,r)−iωt, H A = H0n × (k/k)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

vector amplitude ⊥ incidence plane

, E(r , t) = −k ×H (r , t)/k.


Preliminaries Refraction-reflection problem

Polarimetry

Any plane wave can be represented as a linear combination of twowaves in basic linear polarizations (with a possible phase shift).We can do so for the incident and reflected waves.The reflecting properties can be fully described by a 2 × 2 matrixrelating amplitudes of scattered and incident waves:

S = [SHH SHVSVH SVV

] , such that EAr = S ⋅EA

i .

If we control EAi and measure EA

r then we can determine S .Entries of S may be complex Ô⇒ max. 8 measurements (only 7 ifthe absolute phase is unavailable).This is how many scatterer parameters we potentially canreconstruct.


Homogeneous half-spaces Isotropic material. Direct problem

Perfect isotropic dielectric: D = εE , j = 0

The simplest setup to test the methodology: the field in eachhalf-space is described by a scalar equation

(ε(z) ∂2

∂t2 −∆)U = 0, ε(z) =⎧⎪⎪⎨⎪⎪⎩

1, z > 0,ε, z < 0,

(*)

where U is any field component.Born approximation (i.e., perturbations, or weak scatterer): δ ≪ 1,

U = U(0) +U(1), ε = 1 + (ε − 1), ∣U(1)∣ ∼ δ∣U(0)∣, ∣ε − 1∣ ∼ δ.

The incident field is U(0); it satisfies (*) everywhere with ε = 1, so

(ε(z) ∂2

∂t2 −∆)U(1) = −(ε(z) − 1) ∂2

∂t2 U(0).

Linearization: throw away (ε − 1)U(1) ∼ δ2.



Solution in two domains

Linearized equation: ( ∂2

∂t2 −∆)U(1) = −(ε(z) − 1) ∂2

∂t2 U(0).

For r = (x, y, z) we take U(0) = u(0)Aei(kir−ωt), where ki = (Ki,0,−qi).The interface is (z = 0). Fourier representation:U(x, y, z, t) = u(z)eiKix−iωt, where U is any field component; so theincident field is u(0)(z) = u(0)Ae−iqiz; scattered field:

( d2

dz2 + q2i )u(1)(z) = −θ(−z)(ε − 1)k2u(0)(z)

= −θ(−z)(ε − 1)k2u(0)Ae−iqiz,

where θ(⋅) is a step function. For z > 0, u(1)(z) is also a planewave; for z < 0, u(1)(z) is a forced oscillation; note the resonance.“General solution without RHS + particular solution with RHS”:

u(1) =⎧⎪⎪⎨⎪⎪⎩

u(0)ABeiqiz, z > 0 (B is the scattering coefficient),

u(0)A(Az +C)e−iqiz, z < 0, A = −−ik2(ε−1)qi

.



Origin of the linearly growing term

The linearly growing term Aze−iqiz is unphysical.This term is an artifact of the Born approximation, in particular —ignoring the change of the vertical wavenumber:

k =√εω

c, q(exact)

t =√εω2

c2 −K2i , q(Born)

t = qi =√

ω2

c2 −K2i .

For z→ 0, the difference between the two periodic solutions is

e−iq(exact)t z − e−iq(Born)

t z = e−iq(Born)t z(e−i(q(exact)

t −q(Born)t )z − 1)

≈ e−iq(Born)t z( − i(q(exact)

t − q(Born)t )z)

≈ e−iq(Born)t z ⋅ (−i)1

2(ω/c)2

qi(ε − 1)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶A

z.

Hence, the transmitted part of the Born solution is valid in thevicinity of the interface (important for interface conditions).



Interface conditions

Needed where divD = div(ε ⋅E) does not exist.Relate the free-space (F) and material (M) sides of the interface

Ex,y∣(F) = Ex,y∣(M) , Hx,y∣(F) = Hx,y∣(M) .

For plane waves U = U(z)eiKix−iωt, the interface conditions reduceto two conditions for a single variable, typically — the componentthat is normal to the incidence plane (xz):

Horizontal polarization, E = (0,Ey,0), H = (Hx,0,Hz):

Ey∣(F) = Ey∣(M) ,dEy

dz∣(F)

= dEy

dz∣(M).

Vertical polarization, H = (0,Hy,0), E = (Ex,0,Ez):

dHy

dz∣(F)

= (ε−1 dHy

dz)∣

(M), Hy∣(F) = Hy∣(M) .



Reflection coefficients

We linearize the Interface conditions and use universal notation ufor Ey and Hy according to the polarization:

both polarizations: u(1)∣(F)

= u(1)∣(M),

horizontal:du(1)

dz∣(F)

= du(1)

dz∣(M),

vertical, linearized:du(1)

dz∣(F)

= du(1)

dz∣(M)

− (ε − 1) du(0)

dz∣z=0

From u(1) =⎧⎪⎪⎨⎪⎪⎩

u(0)ABeiqiz, z > 0,

u(0)A(− −ik2(ε−1)qi

z +C)e−iqiz, z < 0,we obtain:

horizontal: B ≡ SHH = −(ε − 1) k2

4q2i,

vertical: B ≡ SVV = (ε − 1)( 12 −

k2

4q2i) = SHH + ε−1

2 .

Both expressions coincide with the Fresnel formulas as ∣ε − 1∣→ 0.M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, 2018 12 / 34

Homogeneous half-spaces Isotropic material

Summary for isotropic dielectric

Test case of the methodology.Direct problem:

Real-valued reflection coefficients.Obtained correct asymptotics for reflection coefficients.Single scatterer parameter: (ε − 1).No cross-polarized scattering: can satisfy the differential equationsand interface conditions within a single linear polarization.

Inverse problem:

Inversion is straightforward: (ε − 1) = −4q2i

k2SHH.

2 out of 4 entries of the matrix S .The data has only 1 out of 8 degrees of freedom because

SVV

SHH= Q = K2

i − q2i

k2

does not depend on the properties of the scatterer.


Homogeneous half-spaces Isotropic material

Isotropic lossy dielectric

Finite isotropic conductivity: j = σE .For E = Ee−iωt, and similarly for H and D :

curl H = −ikD + 4πcσE = −ik(ε + i

4πωσ)E = −ikε′E .

Use ε′ instead of ε for the case of perfect dielectric.Need σ/ω ≪ 1 to use in Born approximation.Direct problem:

two parameters of the scatterer: (ε − 1) and σ/ω;reflection coefficients are complex;otherwise similar to the perfect dielectric.

Inversion: (ε′ − 1) = −4q2

ik2 SHH (similar to the case of perfect

dielectric). Note that SVV/SHH is independent of the materialproperties Ô⇒ need the absolute phase to detect σ.


Homogeneous half-spaces Anisotropic material

Dielectric tensor

Anisotropic material: Di = εijEj, ε ≠ εI .From energy considerations, ε is symmetric.For a real symmetric matrix ε, choose (x′, y′, z′) such that

Di = diag(εx′ , εy′ , εz′)ij Ej.

Uniaxial model: two (rather than three) independent parameters(i.e., matrix eigenvalues): εx′ = εy′ = ε⊥; εz′ = ε∥; ez′ is theoptical axis. We still require ∣ε⊥ − 1∣ ≪ 1 and ∣ε∥ − 1∣ ≪ 1.Let ez′ = (α,β, γ), α2 + β2 + γ2 = 1, and ζ = ε∥ − ε⊥, then

εxx = ε⊥ + α2ζ, εyy = ε⊥ + β2ζ, εzz = ε⊥ + γ2ζ,

εxy = εyx = αβζ, εxz = εzx = αγζ, εyz = εzy = βγζ.For η = ε−1, we have, up to the first order:

ηxx = 1/εxx, ηyy = 1/εyy, ηzz = 1/εzz,

ηxy = ηyx = −εxy, ηxz = ηzx = −εxz, ηyz = ηzy = −εyz.M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, 2018 15 / 34

Homogeneous half-spaces Anisotropic material. Direct problem

Governing equations

We can reduce the Maxwell’s equations

−dEy

dz= ikHx, −dHy

dz= − ikDx,

dEx

dz− iKiEz = ikHy,

dHx

dz− iKiHz = − ikDy,

iKiEy = ikHz, iKiHy = − ikDz,

to the following system for Ey and Hy:

(ηyyd2

dz2 + k2 −K2i ηyy)Ey = − k(iηxy

ddz+Kiηyz)Hy,

(ηxxd2

dz2 + k2 −K2i ηzz)Hy − 2iηxzKi

dHy

dz= − i

k(iηyzKi − ηxy

ddz

)(d2Ey

dz2 −K2i Ey),

and substitute Ex and Hx expressed via Ey and Hy into

Ex,y∣(F) = Ex,y∣(M) , Hx,y∣(F) = Hx,y∣(M) .


Homogeneous half-spaces Anisotropic material. Direct problem

Calculating scattering coefficients

In Born approximation, the system of two equations decouples ina different way for different polarizations of the incident field.For example, when Ei = (0,E(0)Ae−iqiz,0), then∣Hy∣ ∼ ∣H(1)∣ ≪ ∣E(0)A∣, and in

(ηyyd2

dz2 + k2 −K2i ηyy)Ey = −k(iηxy

ddz+Kiηyz)Hy,

we drop RHS ∼ δ2 and then linearize LHS to obtain

( d2

dz2 + q2i )E(1)y (z) = −θ(−z)(ε − 1)k2E(0)Ae−iqiz + ICsÐ→ SHH,

whereas for Hi = (0,H(0)Ae−iqiz,0) and ∣Ey∣ ∼ ∣E(1)∣ ≪ ∣H(0)A∣ welinearize both sides:

( d2

dz2 + q2i )E(1)y = −k(iηxy

ddz+Kiηyz)H(0)y

+ ICsÐ→ SHV.



Direct and inverse problems

Direct (α,β, γ, ξ, ζ)Ð→ S and inverse S Ð→ (α,β, γ, ξ, ζ) problems:

SHH = −14

k2

q2i(ξ + β2ζ), SHV = 1

4kqi(α + Ki

qiγ)ζβ,

SVV = 14((ξ + α2ζ) −

K2i

q2i(ξ + γ2ζ)), SVH = −1

4kqi(α − Ki

qiγ)ζβ,

1 = α2 + β2 + γ2 (components of ez′),

wherekqi

andKi

qi— parameters, k2 = K2

i + q2i ; ξ = ε⊥ − 1, ζ = ε∥ − ε⊥.

Inverse problem: 4 polynomials of 3rd degree and 1 polynomial of2nd degree.Degrees of freedom vs. orientation of optical axis (Sij ∈ R):

2 if β = 0 because SHV = SVH = 0;3 if (α = 0 and βγ ≠ 0) OR (γ = 0 and αβ ≠ 0) because SHV ± SVH = 0;4 if αβγ ≠ 0.



Solvability of the inverse problem

Proposition: The system of equations can be solved with respectto ε⊥, ε∥, α, β, and γ for the given SHH, SVV, SHV, SVH, and θinc ifand only if

(SVV + VSHH)2 ⩾ 4WSHVSVH, (**)

where

W =q2

i −K2i

k2 = cos2 θinc − sin2 θinc = cos 2θinc.

If α = 0 then SHV = SVH, and for θinc > π/4 we have W < 0 Ô⇒RHS < 0 in (**) Ô⇒ the problem always has a solution.So, (**) puts an additional constraint on the values of the reflectioncoefficients and thus implies a limitation of solvability of thelinearized inverse problem.Derivation of (**) has taken several weeks ...


Homogeneous half-spaces Symbolic computations

Quantifier elimination method rules!Hoon Hong <[email protected]>to me--------Dear Mikhail,

I applied a general method and obtained thefollowing result (in about 3 seconds):........................The following system has a real solution for a,b,g,z,x0 = -s1 - 1/4*k^2*(x+b^2*z),0 = -s2 + 1/4*((x+a^2*z)-K^2*(x+g^2*z)),0 = -s3 - 1/4*k*(a-K*g)*z*b,0 = -s4 + 1/4*k*(a+K*g)*z*b,0 = -1 + a^2 + b^2 + g^2,0 = - k^2 + K^2 + 1

iffK^4*s1^2-2*K^4*s1*s2+K^4*s2^2+4*K^4*s3*s4-2*K^2*s1^2+2*K^2*s2^2+s1^2+2*s1*s2+s2^2-4*s3*s4 > 0



Summary for anisotropic dielectric

Direct problem:Real-valued reflection coefficients.Multiple scatterer parameters.Can produce cross-polarized scattering.

Inverse problem:

All 4 entries of the matrix S .The data has 2 to 4 (out of 8) degrees of freedom.Special condition of solvability (because the material parametersmust be real).If we allow uniaxial conductivity, we can get all 8 degrees offreedom. Separate solvability problems for Re(S) and for Im(S).

Biaxial dielectric — may serve as an example of underdeterminedsystem of equations (not done yet).


Inhomogeneous half-spaces Overview

Inhomogeneous half-spaces

Main deficiency of homogeneous half-space models: onlyspecular reflection, no backscattering.

Backscattering is the primary configuration for remote sensing.Options to obtain backscattered signal:

horizontally inhomogeneous material;horizontally inhomogeneous boundary conditions;horizontally inhomogeneous shape of the interface.


Inhomogeneous half-spaces Horizontally inhomogeneous material

Horizontally inhomogeneous material

For ε = ε(x, y), Maxwell’s equations reduce to

∆E − grad divE = ε

c2∂2

∂t2E , div(εE) = 0.

Plane wave E ∼ ei(k ,r)−iωt is not a solution (try ε = 1 + a sin(2Kix)).Linearization is needed to move forward.

We consider an acoustic problem ∆U = ε

c2∂2

∂t2 U for U(r , t) where

U = U(0) +U(1) and ε = ε(0) + ε(1)(x, y), ∣U(1)∣ ≪ ∣U(0)∣, ∣ε(1)∣ ≪ ε(0),whereas ∣ε(0) − 1∣ is not necessarily small.Linearization yields U(0) as the Fresnel solution for the horizontalpolarization:

U(0) =⎧⎪⎪⎨⎪⎪⎩

U(0)i +U(0)r , z > 0,

U(0)t , z < 0.



Linearization

In Fourier domain:

ε(1)(x, y) = 1(2π)2 ∬ ε(1)(Kx,Ky)ei(Kxx+Kyy)dKxdKy,

we obtain

( d2

dz2 + q2)u(1) = 0, z > 0,

( d2

dz2 + q′2)u(1) = − k2u(0)Aε(1)B e−iq′rz, z < 0,

where u(1) = u(1)(Kx,Ky, z), ε(1)B = ε(1)(Kx −Ki,Ky),

q2 = k2 −K2x −K2

y , q′2 = k2ε(0) −K2x −K2

y , q′2r = k2ε(0) −K2

i .Backscattering for z > 0 means Kx = −Ki, Ky = 0, i.e.,ε(1)B ≡ ε(1)(−2Ki,0). This is the Bragg harmonic of ε(x, y).



Reflection coefficients

General solution:

u(1) = u(0)A ⋅⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

Beiqz, z > 0,Ce−iq′z + A1e−iq′rz, z < 0 and q′ /= q′r,Ce−iq′z + A2ze−iq′rz, z < 0 and q′ = q′r.

Interface conditions: continuity for u(1) and du(1)/dz at z = 0.Reflection coefficients:

B =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2ε(1)Bk2qi

(q′ + q)(q′r + q′)(q′r + qi), q′ ≠ q′r,

−2ε(1)Bk2qi

(q′r + q)2q′r(q′r + qi), q′ = q′r.

B is insensitive to the resonance in the material (i.e., twoexpressions for B coincide if q′ = q′r).



Summary for inhomogeneous material

Scalar problem for inhomogeneous isotropic dielectric.Linearization of ε about ε(0).The scattering coefficient is proportional to ε(1)B .Vector problem (i.e., Maxwell’s equations rather than waveequation) is huge (check [GST, 2017]). It demonstratesdepolarization if Ky ≠ 0. No depolarization for backscattering.Not done yet: inhomogeneous anisotropic material. We expect toobtain depolarization in backscattering.



Polarization: additional information about thetarget

Polarimetry is useful for segmentation/classification ofterrains:

RAMSES (Airborne SAR) https://earth.esa.int/documents/10174/

669756/Urban_Classification_3Drendering.pdf

Polarimetry reveals, e.g.:forested land by high HV;bare ground and grass by high HH-VV;

although the models are still semi-empirical.


Inhomogeneous half-spaces Reflection-only setups

Alternatives for refraction-reflection problem

We can consider reflection-only problem because we are notinterested in the field below the interface.

Advantage: simpler setup.Disadvantage: boundary properties should represent bulkproperties.

Possible options:

Variable boundary conditions on a plane surface (example:Leontovich boundary condition).Homogeneous boundary conditions on a non-plane surface (i.e.,rough surface scattering).Combination of the two inhomogeneities.



Leontovich boundary condition

Let ε = ε(0) = const. In Fourier domain:

u(Kx,Ky, z) = eiωt∬ U(x, y, z, t)e−i(Kxx+Kyy)dx dy,

consider the regular reflection-refraction problem

u(Kx,Ky, z) =⎧⎪⎪⎨⎪⎪⎩

uAi (Kx,Ky)e−iqz + uA

r (Kx,Ky)eiqz, z > 0,uA

t (Kx,Ky)e−iq′z, z < 0,

where q2 = k2 −K2x −K2

y , q′2 = k2ε(0) −K2x −K2

y . If u and ∂u/∂z arecontinuous at z = 0, then we can show that

∂u(Kx,Ky, z)∂z

∣z=+0

= −i√ε(0)k2 − (K2

x +K2y ) ⋅ u(Kx,Ky, z)∣

z=+0.

Hence, in the coordinate domain, the relation between U and∂U/∂z will be non-local.



Leontovich boundary condition, cont’d

For ε(0) ≫ 1, we can simplify√ε(0)k2 − (K2

x +K2y ) ≈

√ε(0)k, so

∂u(Kx,Ky, z)∂z

∣z=+0

= − i√ε(0)k ⋅ u(Kx,Ky, z)∣

z=+0

⇓∂U(x, y, z, t)

∂z∣z=+0

= − i√ε(0)k ⋅U(x, y, z, t)∣

z=+0.

These are local boundary conditions, called Leontovich conditions.Generalization:

∂U(x, y, z, t)∂z

∣z=+0

= −i√ε(x, y)k ⋅U(x, y, z, t)∣

z=+0, ε(x, y) ≫ 1.



Leontovich boundary condition, cont’d

We don’t need a solution for z < 0 because we have boundaryconditions at z = 0.With ε(x, y) = ε(0) + ε(1)(x, y), ∣ε(1)∣ ≪ ε(0), and U = U(0) +U(1),U(0) = uA

i ei(Kix−qiz)−iωt, ∣U(1)∣ ≪ ∣U(0)∣, we can obtain

uAr (Kx,Ky)

uAi

= −ε(1)B

(ε(0))3/2

qi

k.

— this is the reflection coefficient ∝ ε(1)B = ε(1)(Kx −Ki,Ky).

The reflection coefficient coincides with that obtained in thereflection-refraction problem for ε(0) ≫ 1.Need more validation.Possible next step: vector problem and anisotropic ε(1).



Rough surface scattering

Intensely studied topic, important for remote sensing of seasurface (see, e.g., [Beckmann & Spizzichino, 1963], [Bass & Fuks,1979], [Voronovich, 1998], [Bruno et al., 2002]).The simplest setting: Dirichlet problem at z = h(x, y) for the scalar(Helmholtz) equation (strip e−iωt factor):

(∆ + k2)(u(0)i + u(0)r + u(1)) = 0, z > h(x, y),(u(0)i + u(0)r + u(1)) = 0, z = h(x, y).

In Fourier domain:

u(1)(Kx,Ky, z) =∬ u(1)(x, y, z)e−i(Kxx+Kyy)dx dy,

h(Kx,Ky) =∬ h(x, y)e−i(Kxx+Kyy)dx dy,

we have for the first order field:

( d2

dz2 + q2)u(1) = 0, where q2 = k2 −K2x −K2

y .



Rough surface scattering, cont’d

Linearization of the domain and boundary conditions for ∣qih∣ ≪ 1:

u(1)(x, y,0) = − (u(0)i + u(0)r )z=h(x,y), (***)

u(0)i (Kx,Ky)∣z=h(x,y) = ∬ u(0)Ai ei(Kix−qiz)∣z=h(x,y)

e−i(Kxx+Kyy)dx dy

≈ ∬ u(0)Ai eiKix(1 − iqih(x, y))e−i(Kxx+Kyy)dx dy

= u(0)Ai ((2π)2δ(Kx −Ki)δ(Ky) − iqihB),

where hB = h(Kx −Ki,Ky), and similarly for u(0)r , so (***) yields

u(1)(Kx,Ky,0) = u(0)Ai ⋅ 2iqihB.

Using u(1) = u(0)Ai Beiqz we find reflection coefficient to be 2iqihB.Vector problem — no depolarization [Voronovich, 1998].


Summary

This talk dealt with the following topics:Need for physics-based scattering models for inverse problems inradar imaging.Polarimetry and degrees of freedom of the scattering matrix.Half-space models of scatterer: horizontally homogeneous andhorizontally inhomogeneous.Single-domain models: Leontovich boundary condition and roughsurface scattering.

Not mentioned:pulsed/modulated signal, interaction between surface harmonics,speckle, coherence, non-linearities in the scatterer, etc.

THANK YOU!


Backup slides

Degrees of freedom: special cases

2 d.o.f. if β = 0; 3 d.o.f. if α = 0 or γ = 0.

Ki

γ

β

α

qi

x

y

z

c = (α, β, γ)

interface

plane of incidence

k = (Ki, 0,−qi)

material

free space


Backup slides Anisotropic material. Monte-Carlo simulations

Limited solvability: θinc < π/4; α = 0Ô⇒ SHV = SVH



Limited solvability: blow-up



Unlimited solvability: θinc > π/4; α = 0Ô⇒ SHV = SVH


Polarimetry of homogeneous half-spacespogudin/workshop/Gilman.pdf · 2018. 7. 24. · Polarimetry...

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Transcript of Polarimetry of homogeneous half-spacespogudin/workshop/Gilman.pdf · 2018. 7. 24. · Polarimetry...