Introduction to Synthetic-Aperture Radar

Kaitlyn MullerGraduate Workshop on Inverse Problems

Colorado State University

August 11, 2016

1

Synthetic-Aperture Radar Forward Model

Recall the model for EM wave propagation in the frequencydomain is the Helmholtz equation

(∇2 + k2)E in(ω, x) = J(ω, x)

We obtain the following solution using the Green’s function andthe Far-Field approximation

E in(ω, x) =

∫G (ω, x − y)J(ω, y)dy ≈ e ik|x−x0|

4π|x − x0|F (k , x − x0)

where x0 is the antenna center and

F (k , x − x0) =

∫e−ik(x−x0·y)J(ω, y)dy

2

SAR Forward Model continued

Inserting this expression for the incident field into the Born approximated model for thescattered field yields

E scB (ω, x) =

∫G(ω, x − y)k2V (y)E in(ω, y)dy

=

∫e ik|x−y|

4π|x − y |k2V (y)

e ik|y−x0|

4π|y − x0|F (k, y − x0)dy

We assume a monostatic system, i.e. the transmit/receive antennas are collocated

To model antenna reception, we can imagine an array antenna centered at x0:

Srec(ω) ∝∫

z∈antennaE sc(ω, z)W (ω, z)dz

=

∫z∈antenna

∫e ik|z−y|

4π|z − y |k2V (y)E in(ω, y)dyW (ω, z)dz

where W is a weighting function.

3

SAR Forward Model continued

We use the Far-field expansion assuming |y − x0| |x0 − z | to obtain

Srec(ω) ∝∫

e ik|y−x0|

4π|y − x0|k2V (y)E in(ω, y)

∫e−ik(y−x0)·(z−x0)W (ω, z)dzdy

Note we define the last integral above as F rec(k, y − x0).

Inserting the incident field expression we express the received signal as

Srec(ω) =

∫e2ik|x0−y|A(y , x0, ω)V (y)dy

where

A(y , x0, ω) =k2F (k, y − x0)F rec(k, y − x0)

(4π|x0 − y |)2

4

SAR Data

Another item to model: the moving platform

Let γ = antenna path, for a pulsed system the antenna position attime tn is γ(tn)

We assume a continuum model with s replacing n, it is calledslow-time (in contrast to fast-time t)

We replace x0 with γ(s) to obtain the data equation

D(ω, s) = F [V ](ω, s) =

∫e2ik|γ(s)−y |A(y , s, ω)V (y)dy

d(t, s) =

∫e−iω(t−2|γ(s)−y |/c)A(y , s, ω)dωV (y)dy

Goal of SAR: Determine V from d or D

5

Working towards inversion

If we assume A and γ are known then the data depends on 2parameters ⇒ we expect to form a 2D image

To this end we assume V is supported on a known surface, forsimplicity

V (x) = V (x)δ(x3)

where x = (x1, x2) now.

We seek to invert d = F [V ]

We will look at the filtered adjoint as a possible inversion method asit maps functions of (ω, s) to functions of x

Note our data is in the form of a Fourier integral operator assumingA satisfies ‘symbol’ estimates and the phase can be written as a true‘phase’ function ⇒ an inverse will be another FIO

6

FIOsDefinition. A FIO P of order m is defined as

P[u](x) =

∫e iΦ(x ,y ,ξ)p(x , y , ξ)u(y)dydξ

where p(x , y , ξ) ∈ C∞(X × Y × Rn) satisfies the estimate: for everycompact set K ⊂ X × Y and for every multi-index α, β, γ there is aconstant C = C (K , α, β, γ) such that

|∂αξ ∂βx ∂yγp(x , y , ξ)| ≤ C (1 + |ξ|)m−|α|

for all x , y ∈ K and for all ξ ∈ Rn. Also Φ must be a phase function, i.e.if

Φ is positively homogeneous of degree 1 in ξ, i.e.Φ(x , y , rξ) = rΦ(x , y , ξ) for all r > 0

(∂xΦ, ∂ξΦ) and (∂yΦ, ∂ξΦ) do not vanish for all(x , y , ξ) ∈ X × Y × Rn

7

Formal Adjoint of Forward Operator

The adjoint F† is an operator such that

〈f ,F [g ]〉ω,s = 〈F†[f ], g〉x∫f (ω, s)(Fg)∗(ω, s)dωds =

∫(F†f )(x)g∗(x)dx

Inserting the forward operator we have that the LHS is

∫f (ω, s)

(∫e−2ik|γ(s)−x|A∗(x , s, ω)g∗(x)dx

)dωds

=

∫ (∫e−2ik|γ(s)−x|A∗(x , s, ω)f (ω, s)dωds

)g∗(x)dx

We conclude the adjoint is given by

(F†f )(x) =

∫e−2ik|γ(s)−x|A∗(x , s, ω)f (ω, s)dωds

8

The Imaging Operator

We look for an imaging operator of the form

I (z) = B[D](z) =

∫e−2ik|γ(s)−z |Q(z , s, ω)D(ω, s)dωds

I (z) = B[d ](z) =

∫e iω(t−2|γ(s)−z |/c)Q(z , s, ω)dωd(t, s)dtds

Note if we let Q = 1 and assume infinite bandwidth we have

I (z) =

∫δ(t − 2|γ(s)− z |/c)d(t, s)dsdt

=

∫d(2|γ(s)− z |/c , s)ds

We see at each s we backproject the data to all locations that are atthe correct travel time and then sum the contributions coherently.

9

SAR PSFSubbing the data expression into the image we obtain

I (z) =

∫ ∫e−2ik(rs,z−rs,z )QAdωdsV (x)dx

=

∫K(z , x)V (x)dx

where rs,z = |γ(s)− z | and

K(z , x) =

∫e−2ik(rs,z−rs,z )QAdωds

is the point-spread function for SAR.

We aim for K(z , x) = δ(z − x)

We will use the method of stationary phase to analyze how close K is to δ

The stationary phase theorem gives an approximate formula for the large β behavior ofoscillatory integrals of the form ∫

e iβφ(x)a(x)dnx .

10

Stationary Phase

Theorem. We assume (i). a is a smooth function of compactsupport and (ii). φ has only nondegenerate critical points. Then asβ →∞∫

e iβφ(x)a(x)dnx =∑

x0:∇φ(x0)=0

(2π

β)n/2a(x0)

e iβφ(x0)e i(π/4)sgn(D2φ(x0))√| detD2φ(x0)|

+ O(β−n/2−1)

where sgn is the number of positive eigenvalues minus the number ofnegative eigenvalues.

11

Back to the SAR image

We seek to apply this theorem to the SAR image

We introduce a large parameter β via the change of variables k = βk ′ (ω = βω′)

K(z , x) =

∫e−2iβk′(rs,z−rs,x )QA

dω′

βds

We find the main contributions to the integral are from the critical points given by

0 =∂φ

∂ω′∝ rs,z − rs,x

0 =∂φ

∂s∝ Rs,z · γ′(s)− Rs,x · γ′(s)

The first equation says the range from z to γ(s) = range from x to γ(s)

The second equation says the down-range velocity (which gives rise to the Doppler shift)must be the same for z and xWe assume a side-looking radar system so we obtain only the critical point z = x

12

PSF analysisIn the neighborhood of the critical point we force the phase to look like the phase of adelta function (z − x) · ξ using the following formula

f (z)− f (x) =

∫ 1

0

d

dµf (x + µ(z − x))dµ = (z − x) ·

∫ 1

0∇f∣∣∣∣x+µ(z−x)

dµ

with f (x) = −2krs,x .

We therefore rewrite the phase of the PSF as

2k(rs,z − rs,x ) = (z − x) · Ξ(x , z , s, ω)

where

Ξ(x , z , s, ω) =

∫ 1

0∇f∣∣∣∣x+µ(z−x)

dµ.

We make the Stolt change of variables (s, ω)→ ξ = Ξ in the PSF to obtain

K(z , x) =

∫e i(z−x)·ξQ(z , ξ)A(x , ξ)

∣∣∣∣∂(s, ω)

∂ξ

∣∣∣∣dξWe note K is the form of the kernel of a pseudodifferential operator, a special case of anFIO.

13

Determining QAfter using symbol calculus from FIO theory we can see how tospecify Q such that our PSF is as close as possible to a deltafunction,

K (z , x) =

∫e i(z−x)·ξQ(z , ξ)A(z , ξ)

∣∣∣∣∂(s, ω)

∂ξ

∣∣∣∣dξWe choose

Q(z , s, ω) =

χ(z , s, ω)

∣∣∣∣ ∂ξ∂(s,ω)

∣∣∣∣A(z , s, ω)

=

χ(z , s, ω)A∗(z , s, ω)

∣∣∣∣ ∂ξ∂(s,ω)

∣∣∣∣|A(z , s, ω)|2

where χ = 0 whenever A = 0 and in the second line we have notedthat we can think of matched filtering as a part of this filteringprocess

14

Resolution AnalysisAfter applying our ‘optimal’ filter we have the leading order PSF

K(z , x) =

∫Ωe i(z−x)·ξdξ

where Ω is called the data collection manifold.

If Ω = R2 we have the ideal δ function PSF ⇒ the size of Ω determines how close we getto the ideal.

It can be shown that ξ ≈ 2kP[Rs,z ] where P projects a vector onto its first twocomponents

If we assume a straight flight path and flat earth then the data collection manifold is asector of an annulus

If we also assume a narrow beam this sector can be approximately by a rectangle ⇒ wecan factor the PSF integral

K(z , x) ≈∫

e i(z1−x1)ξ1dξ1

∫e i(z2−x2)ξ2dξ2

Note we use the convention that [−b, b] in Fourier space corresponds to resolution 2π/bbecause: ∫ b

−be iρrdρ =

2 sin(br)

r= 2b sinc(br)

15

Example Data Collection Manifold

16

Along-track Resolution

We first consider the ξ2 integral and the limits of integration

We assume a dipole antenna of length L moving along a straightflight path γ(s) = (0, s, h)

Consider two points on the ground whose coordinates differ only inthe 2nd coordinate ⇒ (z − x) · ξ = (z2 − x2)ξ2

Note that

ξ2 ≈2k(z2 − γ2(s))

R

where R = |z − γ(s)|

We need to determine when ξ2 is in Ω which depends on bandwidthand the antenna beam pattern

17

Along-track Resolution continued

The interval of s values over which z2 is in the beam is 2 max |z2 − s|which is the width of the antenna footprint which can be shown tobe R(2λ/L)

⇒ max |ξ2| ≈k

R

2λR

L=

4π

L

Therefore the ξ2 integral is∫ 4π/L

−4π/Le i(z2−x2)ξ2dξ2 =

8π

Lsinc

(4π

L(z2 − x2)

)

Looking at the mainlobe we see the resolution is 2π/(4π/L) = L/2⇒ resolution is better for small antennas.

18

Aside: Dipole Antenna Beam

We assume constant current along a wire antenna which is short compared to λ

We parametrize the line the antenna lies on

x(s) = se

with −L/2 ≤ s ≤ L/2 and denote current density by I e

Then the radiation vector is

F (k, x) =

∫ L/2

−L/2e−ik x·(se)I eds = −LI e sinc(kLx · e/2)

To find the beamwidth we consider the width of the mainlobe

kLx · e/2 = kL sin(θ)/2 ≈ kLθ/2 = 2π

where θ is measured from the normal.

We find the beamwidth is 4πk/L = 2λ/L

Using arclength we find the antenna footprint at the range R to be 2λRL

19

Range Resoluton

Now we consider the limits of the ξ1 integral

ξ1 ≈2k(z1 − γ1(s))

R

The minimum and maximum values are given by:

2kmaxz1/R and 2kminz1/R

where R =√

z21 + h2 and we let sin(ψ) = z1/R

The ξ1 integral is therefore∫ 2kmax sin(ψ)

2kmin sin(ψ)

e i(z1−x1)ξ1dξ1 =∝ e2ik0(z1−x1) sin(ψ) sinc((z1 − x1)∆k sin(ψ))

where k0 = (kmax + kmin)/2 and ∆k = kmax − kmin

Looking at the mainlobe we see the resolution is 2π/(∆k sin(ψ)) ⇒resolution is better for large bandwidth and ψ close to π/2 (larger range)

20

Microlocal Analysis: Study of Singularities

For objects in space their shape is determined by the boundariesbetween different materials

This leads to the expectation that V (x) will have jumpdiscontinuities (a type of singularity)

We want these discontinuities or edges in the image tocorrespond to the actual boundaries of the object

21

Formalizing singularities

Definition. The point (x0, ξ) is not in the wavefront set of f (denotedWF (f )) if for some smooth cutoff function φ with φ(x0) 6= 0 the Fouriertransform ∫

f (x)φ(x)e iξ·xdx

decays rapidly in a neighborhood of the direction ξ.

In other words: to determine if (x0, ξ) ∈WF (f ) we

1 localize around x0

2 Fourier transform

3 Look at decay in direction ξ

22

Example - point scatterer

We model a point scatterer as V (x) = δ(x)

Now looking for the wavefront set of V : first consider localizingaround various points

Note at any x 6= 0 we have φV = 0 ⇒ F [φV ] decays rapidly

⇒ x 6= 0⇒ (x , ξ) /∈WF (V )

However if x = 0 then φV = δ ⇒ F [φV ] = constant, i.e. no decayin any direction

⇒WF (V ) = (0, ξ) : ξ 6= 0

23

Example - wall/edge

We model an edge as V (x) = δ(x · ν) ∝∫e ix ·(ρν)dρ

Note if we localize around any point such that x · ν 6= 0 then φV = 0

We consider now points in the support of the delta function:

F [φV ] =

∫δ(ξ − ρν)dρ = 1

if ξ ∝ ν and 0 otherwise

⇒WF (V ) = (x , ν) : x · ν = 0

24

Singularities and Pseudodifferential Operators

Recall the SAR image

I (z) =

∫K (z , x)V (x)dx = (BF)[V ](z)

If K is of the form

K (z , x) =

∫e i(z−x)·ξa(x , z , ξ)dξ

and a satisfies certain symbol estimates, then K = BF is apseudodifferential operator

Every ΨDO has the pseudolocal property:

WF (Ku) ⊆WF (u)

25

ExampleConsider the wall again

V (x) = δ(x · ν) ∝∫

e ix·(ρν)dρ

Now let K act on V∫K(z , x)δ(x · ν)dx ∝

∫ ∫K(z , x)e ix·(ρν)dρdx

=

∫ ∫ ∫e i(z−x)·ξχ(z , x , ξ)e ix·(ρν)dξdxdρ

We now perform a change of variables ξ = ρξ and apply stationary phase to the x and ξintegrals

Note our phase is φ = ρ[(z − x) · ξ + x · ν] which has critical points z = x and ξ = ν

Therefore ∫K(z , x)δ(x · ν)dx ∝

∫e iz·(ρν)χ(z , z , ρν)dρ+ smoother terms

≈ V (z)

We have singularities appearing at the correct location with the correct orientation in theimage.

26

Polarimetric Synthetic-Aperture Inversion

Problem Description

I Design imaging algorithm specifically for extended targets (curve-like, edges), ortargets with anisotropic scattering behavior

I Develop algorithm for a polarimetric SAR system: system with orthogonallypolarized antennas, capable of transmitting both polarizations and receiving onboth polarizations

Method

I Develop forward model which incorporates directional scattering behavior

I Use backprojection for imaging and design “optimal” filter

I Model target and clutter as stochastic processes

27

Background: Polarimetric SAR

Polarimetric radar includes information about the polarization state of the EM fields andmaintains the vector nature of the fields, hence we return to Maxwell’s equations for theforward problem.

∇× E(ω, x) = iωB(ω, x)

∇×H(ω, x) = J(ω, x)− iD(ω, x)

∇ ·D(ω, x) = ρ(x)

∇ · B(ω, x) = 0

Also we obtain multiple data sets ⇒ more information

Currently polarimetric radar, in particular polSAR, is used mainly on distributed/naturalscatterers for land classification type applications, also some weather applications

Past work has not proven its utility for man-made targets

28

Background: Polarization StateConsider the simplest solution to the time-harmonic Maxwell’s equations:monochromatic plane wave

E (r) = Ee ik·r

E = Ehh + Ev v

Polarization describes behavior of the field vector in time

Pictures from: http : //en.wikipedia.org/wiki/Polarization %28waves%29

29

Background: Polarimetric Scattering

The fully polarimetric wave incident on the scatterer has a defined polarization state andpropagation direction:

E i = E ihh + E i

v v

Wave interacts with scatterer & is received on the antenna in the far-field of the scatterer:

E s = E sh h + E s

v v

Scattering process thought of as transformation of incident wave into scattered waveperformed by the scatterer:(

E sh

E sv

)=

e ikr

r

(Sh,h Sh,v

Sv ,h Sv ,v

)(E ih

E iv

)

30

Modeling Anisotropic Scattering

We model the antennas and the scatterers (target & clutter) as dipoles

Each dipole has an associated orientation

e(x) = [cos θ(x), sin θ(x), 0]T

Scatterers also have an associated scattering strength ρ(x)

31

Forward Model

To obtain expression for the incident field use method of potentials solution of Maxwell’sequations & the Far - Field approximation

Ea(k, x) = Rx,s ×(Rx,s × ea

) e ikRx,s

4πRx,sF a(kRx,s · ea)Pa(k)

where Rx,s = x − γ(s), Rx,s = |Rx,s |, and

F (k cos θ) = sinc

(k

2cos θ

)

32

Method of potential solution to Maxwell’s

equationsNote ∇ · B = 0 ⇒ B can be expressed as the curl of another field

B = ∇× A

where A is called the vector potential

We insert this into the first of Maxwell’s equations to obtain

∇× [E(x , ω)− iωA(x , ω)] = 0

Note: a vector field whose curl is zero can be written as the gradient of a potential

E − iωA = −∇Φ

where Φ is called the scalar potential

If we assume the free-space constitutive relations and use the expressions above in the2nd and 3rd Maxwell’s equations we obtain

1

µ0∇× (∇× A) = J − iωε0(−∇Φ + iωA)

ε0∇ · (−∇Φ + iωA) = ρ

33

Method of potential solution continued

Note: if one adds the gradient of any scalar field, say ψ, to A the physical magneticinduction field with not change because ∇× ψ = 0

Note: If one adds iωψ to Φ then E will remain unchanged

Therefore the transformations

A→ A +∇ψΦ→ Φ + iωψ

do not effect E and H

This is called a gauge transformation and it tells us we must add an additional constraintto our system

We will use the Lorenz gauge

∇ · A− iωε0µ0Φ = 0

34

Method of potential solution continued

Rewriting our expression for A using the triple product identity gives us

∇2A + ω2ε0µ0A = µ0J −∇(iωε0µ0Φ− (∇ · A))

Note the expression in parantheses is exactly the Lorenz gauge so our equation simplifiesto

(∇2 + k2)A = µ0J

It may also be shown that(∇2 + k2)Φ = −ρ/ε0

We already know how to solve the Helmholtz equation so we have

A(x , ω) = µ0

∫e ik|x−y|

4π|x − y |J(y , ω)dy

Φ(x , ω) =1

ε0

∫e ik|x−y|

4π|x − y |ρ(y , ω)dy

35

Finding the incident field

We first apply the Far-field approximation to A and then find E via the expression (whichresults from using the Lorenz gauge in the expression for E)

E = iω[A + k−2∇(∇ · A)]

Approximating all derivatives above using the far field assumptions results in

E = iωε0µ0e ik|x|

4π|x |[F − x(x · F )]

= iωε0µ0e ik|x|

4π|x |[x × (x × F )]

where

F (k, x) =

∫e−ikx·yJ(y , ω)dy

36

Back to the forward model

Recall the incident field is

Ea(k, x) = Rx,s ×(Rx,s × ea

) e ikRx,s

4πRx,sF a(kRx,s · ea)Pa(k)

Current on the receiving antenna is given by:

Da,b(k, s) ∝ eb · E(k,γb(s))

=

∫ρsc (x)

e2ikRx,s

16π2R2x,s

(F sc (kRx,s · esc ))2F b(kRx,s · eb)F a(kRx,s · ea)

Pa(k) esc ·[Rx,s ×

(Rx,s × ea

)]eb ·

[Rx,s ×

(Rx,s × esc

)]dx

37

Forward Model continuedIf we transmit and receive on all possible antenna pairs we obtain the data vector:

D(k, s) =

∫e2ikRx,s ρsc (x)

Aa,a 0 0 0

0 Aa,b 0 00 0 Ab,a 00 0 0 Ab,b

(R⊥a ⊗ R⊥a ) : (esc ⊗ esc )

(R⊥a ⊗ R⊥b ) : (esc ⊗ esc )

(R⊥b ⊗ R⊥a ) : (esc ⊗ esc )

(R⊥b ⊗ R⊥b ) : (esc ⊗ esc )

dx

where Ai,j = 116π2R2

x,s(F sc (kRx,s · esc ))2F i (kRx,s · ei )F j (kRx,s · ej )Pi (k) and

R⊥ = R × R × e

Consider the vectors R⊥a and R⊥b :

R ×(R × e

)= R

(R · e

)− e

(R · R

)= −

[e − R

(R · e

)]= −P⊥

Re

Typically in polarimetric SAR it is assumed that:

e = R ×(R × e

)Our formulation holds in general when the beamwidth is broad, for example inlow-frequency systems

38

Linearized Forward ModelWe define a scattering vector

S(θsc ) =

cos2 θsc

cos θsc sin θscsin θsc cos θsc

sin2 θsc

Then we are able to rewrite the data expression as follows

D(k, s) =

∫e2ikRx,s

Aa,ax

2a Aa,axaya Aa,axaya Aa,ay

2a

Aa,bxaxb Aa,bxayb Aa,byaxb Aa,byaybAb,axaxb Ab,ayaxb Ab,axayb Ab,aybybAb,bx

2b Ab,bxbyb Ab,bxbyb Ab,by

2b

ρsc (x)

cos2 θsc

cos θsc sin θscsin θsc cos θsc

sin2 θsc

dx

where R⊥a = (xa, ya, za) and R⊥b = (xb, yb, zb).

To model the directional scattering behavior of an extended target we assume:

FT (kRx,s · eT ) =

1 if Rx,s · eT ≈ 0

0 otherwise

To remove nonlinearity arising from the radiation pattern of clutter scatterers we makethe following assumption:

FC (kRx,s · eC ) = 1∀k, s, x

39

Total Forward Model

We now combine the target & clutter model to obtain the full data expression:

D(k, s) = DT (k, s) + DC (k, s) + n(k, s) = FT [T ](k, s) + FC [C ](k, s) + n(k, s)

=

∫e2ikRx,s AT (k, s, x)T (x)dx +

∫e2ikRx,s AC (k, s, x)C(x)dx + n(k, s)

where n is a 4× 1 additive noise vector, and C(x) = ρC (x)S(θC ) & T (x) = ρT (x)S(θT )

In addition we assume that E [T ] = µ and C & n are 0 mean. In addition all are2nd-order with the following covariance matrix definitions:

CTl,k (x, x′) = E [(Tl (x)− µl (x))(Tk (x′)− µk (x′))]

RCl,k (x, x′) = E [Cl (x)Ck (x′)]

Snl,k (k, s; k′, s′) = E [nl (k, s)nk (k′, s′)]

where here l = aa, ab, ba, bb and k = aa, ab, ba, bb.

Also note we assume that T , C , and n are all mutually statistically independent.

40

Image Formation

Use FBP based reconstruction method:

I (z) = (KD)(z) =

∫e−i2kRz,sQ(z , s, k)D(k, s)dkds

Define the error process E(z) = I (z)− T (z)

Find Q by minimizing MSE

J (Q) =

∫E [|E(z)|2]dz =

∫E [(E(z)†(E(z))]dz

41

LemmaLet D be given as above, where the amplitudes AT and AC satisfy symbol estimates and let I be given by (1). Assume Sn isgiven by (1) and define ST , SC , and M as follows:

CT (x, x′) =

∫e−ix·ζe ix

′·ζ′ST (ζ, ζ′)dζdζ′ (1)

RC (x, x′) =

∫e−ix·ζe ix

′·ζ′SC (ζ, ζ′)dζdζ′ (2)

µ(x)µ†(x′) =

∫e−ix·ζe ix

′·ζ′M(ζ, ζ′)dζdζ′ (3)

where the integrations are defined element-wise. Then

J (Q) = JT (Q) + B(Q) + JC (Q) + Jn(Q), (4)

and the value of the leading-order singularities of each term in (4) are

JT (Q) ≈∫

e ix·(ζ′−ζ) tr

[ Q(x, ζ′)AT (x, ζ′)η(x, x, ζ′)− χ

Ω(x, ζ′)

†Q(x, ζ′)AT (x, ζ′)η(x, x, ζ′)− χ

Ω(x, ζ′)

ST (ζ, ζ′)

]dxdζdζ′, (5)

JC (Q) ≈∫

e ix·(ζ′−ζ) tr

[ Q(x, ζ′)AC (x, ζ′)η(x, x, ζ′)

†Q(x, ζ′)AC (x, ζ′)η(x, x, ζ′)

SC (ζ, ζ′)

]dxdζdζ′, (6)

42

Lemma continued

Lemma

B(Q) ≈∫

e ix·(ζ′−ζ) tr

[ Q(x, ζ′)AT (x, ζ′)η(x, x, ζ′)− χ

Ω(x, ζ′)

†Q(x, ζ′)AT (x, ζ′)η(x, x, ζ′)− χ

Ω(x, ζ′)

M(ζ, ζ′)

]dxdζdζ′, (7)

Jn(Q) =

∫tr

[Q†(z, ξ)Q(z, ξ)S

n(ξ)

]η(z, z, ξ)dξdz. (8)

Here “leading order” is in the microlocal sense meaning that the higher-order terms are smoother than the leading-order one,and ≈ indicates that we are taking the expected L2 norm of the leading-order term.

43

Steps of Proof

We rewrite each term of the MSE such that the phase resembles that of a delta function:

K(FTT )(z) ∼∫

e i(x−z)·ξQ(z , s(ξ), k(ξ))AT (x , s(ξ), k(ξ))T (x)η(x , z , ξ)dxdξ

We next use symbol calculus (from microlocal analysis) to obtain the leading order termsfor each part of the MSE

The linearity of the expectation operator allows use to rewrite each expression in terms ofthe covariance matrices for T , C , and n

Finally we introduce the spectral density functions and carry out the x and ξ integrations

Note that for the noise process we assume stationarity which simplifies it without usingthe symbol calculus

44

Theorem1 Under the same assumptions as the above Lemma, any filter Q satisfying a symbol estimate and also minimizing the

leading-order mean-square error J (Q) must be a solution of the following integral equation ∀r and ∀q:

(∫ηe ix·(ζ

′−ζ)[

(QATη − χ

Ω)(ST + M)(AT )† + (QAC

η)SC (AC )†]dζ′)

(r,q)

+

(QS

nη

)(r,q)

= 0 (9)

Here “leading order” is in the microlocal sense, meaning that the higher-order terms are smoother than theleading-order one.

2 If, in addition, we make the stationarity assumptions

ST (ζ, ζ′) = ST (ζ)δ(ζ − ζ′), (10)

SC (ζ, ζ′) = SC (ζ)δ(ζ − ζ′), (11)

then the filter Q minimizing the leading-order total error variance V(Q) is given by

Q† =

[|η|2(AT (ST )†(AT )† + AC (SC )†(AC )†) + η(Sn)†

]−1ηAT (ST )†χ†

Ω(12)

if the matrix in brackets is indeed invertible.

45

Steps of Proof

Find the variation of J with resect to Q:

0 =d

dε

∣∣∣∣ε=0

JT (Q + εQε) +

d

dε

∣∣∣∣ε=0

JC (Q + εQε) +

d

dε

∣∣∣∣ε=0

Jn(Q + εQε)

+d

dε

∣∣∣∣ε=0

B(Q + εQε)

Use linear algebra and “tricks of the trace” to simplify

For part (ii). simply assume T and C are stationary and ignore the bias term to arrive atthe algebraic expression for Q

46

Numerical Results - Experiment Parameters

Assume straight flight path of the form: γ(s) = [x0, s, z0]

Assume target is a straight line (20 pixels in length) in center of scene with a givenorientation

Antenna polarizations example: ea = [1, 0, 0] and eb = [0, 1, 0]

Frequency range: 1-1.5 GHz

Scene = 50 m2, 100 by 100 pixels

We define the clutter process C(x) = ρC (x)S(θC (x)) where ρC (x) are i.i.d complexGaussian and θC (x) are i.i.d uniform between 0 and π.

We will be comparing the results of our model and imaging scheme with standardpolSAR imaging which assumes R⊥ = e

47

Numerical Results - Example One

Figure: HH component of target vector T , eT = [1, 0, 0].

48

Numerical Results - Reconstructions of HH

component

Left is standard polSAR image reconstruction, right is reconstructed with our imagingalgorithm, SCR=20dB

49

MSE and Image SCR

Scene SCR Image SCR (standard) Image SCR (coupled)

-20 0.3881 0.2695-10 1.2272 0.85220 3.8809 2.6949

10 12.2723 8.521920 38.8085 26.9486

50

Numerical Results - Example Two

Figure: HV component of target vector T , eT = [1/√

(2), 1/√

(2), 0]

51

Numerical Results - Reconstructions of HV

component

Left is standard polSAR image reconstruction, right is reconstructed with our imagingalgorithm, SCR=-20dB

52

MSE and Image SCR

Scene SCR Image SCR (standard) Image SCR (coupled)

-20 0.1883 0.6475-10 0.5955 2.04760 1.8831 6.475

10 4.3572 14.353920 13.7786 45.3912

53

Day 2 Conclusions

In SAR the data is of the form of an FIO applied to the unknown function V that wewish to reconstruct

We design an approximate FIO imaging operator to invert the data and reconstruct V

The composition of forward and imaging operators is a ΨDO and hence preservessingularities in the process of mapping V to I

In addition this microlocal reconstruction technique aims to choose a filter such that thePSF is as close as possible to ideal, giving good resolution in the reconstructed image

Polarimetric SAR attempts to provide a more physical model for the propagation of EMwaves

In order to make a linear model which is easily inverted we must make simplifyingassumptions on scatterers which hopefully can be improved

We can incorporate statistical considerations into microlocal techniques easily to handlenoise and random clutter present in the scene of interest and the measured data

There are still plenty of problems to work on in radar which require mathematics!

54