properties of thin dielectric objects from far field data

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properties of thin dielectric objects from far field data

Noam Zeev

Colaboration: Dr. Fioralba Cakoni

Old Dominion University Department of Mathematics and Statistics

AIP 2009

July 23, 2009

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Outline

1 Direct problem for thin dielectric objects

2 Inverse Problem - far-field

3 Numerical results from far field data

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Outline

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Applications

Important problems in nondestructive testing

detection of flaws in materials in specific (typically thin) areas, as well as the determination of the integrity of thin coatings

In the study of optical devices in communication networks In the detection of thin air pockets inside structures.

we investigate the inverse problem of using far field time harmonic electromagnetic measurements to determine the shape and information about the thickness and physical properties of a thin dielectric film embedded in an inhomogeneous background.

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Direct problem

Assume

1 The obstacle is an infinitely long thin cylinder with an arc as cross section,

2 The electric field is polarized parallel to the cylinder axis.

After factoring out the terme^−iωt, whereωis the frequency, the only component of the total electric fieldusatisfies the

Helmholtz equation

∆u+k²u=0

in the exterior of the cylinder, wherek is the wave number.

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Formulation of the problem

1. LetΓ⊂R²be an oriented piecewise smooth nonintersecting arc .

2. The total fieldu=u^s+uⁱ satisfies,

∆u+k²u =0 in R²\Γ ∂u

∂ν

=0 on Γ

[u]−iλ∂u⁺

∂ν =0 on Γ

r→∞lim

√r ∂u^s

∂r −iku^s

=0 whereλinvolves some physical properties ofΓ.

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Direct problem

In general, givenf ∈H^−1/2(Γ)andh∈H^−1/2(Γ)find v ∈H_loc¹ (R²\Γ)satisfying

∆v +k²n(x)v =0 in R²\Γ ∂v

∂ν

=f on Γ

[v]−iλ∂v⁺

∂ν =h on Γ

r→∞lim

√r ∂v

∂r −ikv

=0

wheren(x)represent the inhomogeneity of the medium.

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Direct problem

Well posedness of the direct problem,i.e.,

1 Existence

2 Uniqueness

3 The solution is continuously dependent of the data.

Approaches:

1 Integral equations - BEM

2 Variational formulation - FEM

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Direct problem - Via integral equations

Theorem The direct scattering problem has a unique solution v which satisfies

kvk_H1(B_R\Γ)≤C

kfk_H−1/2(Γ)+khk_H−1/2(Γ)

x ∈R²\Γ where the positive constantC depends onRbut not onf andh.

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Proof of existence:

LetDbe the region bounded by the extension∂DofΓ. Note that[v]∈H˜^1/2(Γ)and_∂v

∂ν

∈H˜^−1/2(Γ).

LetG(x,y)be the radiating Green’s function of the background medium satisfying

∆G(x,y) +k²n(x)G(x,y) =δ(x−y).

From the Greens representation formula we have

v(x) =









 Z

∂D

∂v(y)

∂νy G(x,y)dsy − Z

∂D

v(y) ∂

∂νyG(x,y)dsy, x ∈D

− Z

∂D

∂v(y)

∂νy

G(x,y)ds_y + Z

∂D

v(y) ∂

∂νy

G(x,y)ds_y, x ∈R²\D where

H˜^1/2(Γ) :={v ∈H^1/2(Γ) :suppv ⊆Γ}

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Using the jumps relations of the single- and double-layer potentials across the boundary∂D, we obtain that the jump[v] satisfies

(i

λI+T_Γ) [v] = i λh+

K_Γ⁰ − I

2

f (∗) where the operatorsK_Γ⁰ : ˜H^−1/2(Γ)→H^−1/2(Γ)and T_Γ: ˜H^1/2(Γ)→H^−1/2(Γ)are defined by

K_Γ⁰ψ (x) :=

Z

Γ

ψ(y) ∂

∂νxG(x,y)dsy forx ∈Γ, (T_Γψ) (x) := ∂

∂ν_x Z

Γ

ψ(y) ∂

∂ν_yG(x,y)dsy forx ∈Γ respectively.

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To solve(∗)we observe that(_λⁱI+T_Γ) : ˜H^1/2(Γ)→H^−1/2(Γ)is a Fredholm operator of index zero. Therefore, it suffices to prove only the injectivity of _λⁱI+T_Γ.

To this end letξ∈H˜^1/2(Γ)satisfies i

λI+T_Γ

ξ =0 and define the following potential

w(x) = Z

Γ

ξ(y) ∂

∂νyG(x,y)ds_y forx ∈R²\Γ.

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ApproachingΓand using the jump relations for double layer potential we obtain

∂w⁺

∂ν = ∂

∂νx

Z

Γ

ξ(y) ∂

∂νy

G(x,y)dS_y =T_Γ(ξ),

[w] =ξ and

∂w

∂ν

=0 Hence we have

[w]−iλ∂w⁺

∂ν =ξ−iλT_Γξ =−iλ i

λI+T_Γ

ξ =0.

Thereforewsatisfies the direct problem withf =h=0, and from uniqueness resultsw =0 inR²\Γwhich finally implies [w] =ξ =0.

Note: knowing[v], we are able to computev⁺,v⁻, ^∂v_∂ν⁺, ^∂v_∂ν⁻.

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Direct Problem- Numerical results

For numerical purposes we need to knowG, we first assume n(x) =1.

Operator

i λI−T_Γ

[u] = i λh Techniques used:

1 Cosine substitution introduced by Yan and Sloan (1988) for T_Γ.

2 We used a Quadrature method developed by Chapko and Kress (for Single layer potential). Adapted by Lars Monch toT_Γ.

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n Reu∞((1,0),(1,0)) Imu∞((1,0),(1,0)) 8 0.071205428169135 0.081916845574100 16 0.242512388888656 0.360640284054226 k=3 32 0.253705871997084 0.346918547805065 64 0.253705867911025 0.346918558878099 16 0.049017043906720 0.282768227362883 k=5 32 0.338614266128542 0.400542203216391 64 0.339344419678081 0.400032127366243 128 0.339344419678121 0.400032127366210 16 0.168887350799880 0.224115244323497 32 0.110221269971698 0.382784494259534 k=10 64 0.482325291178517 0.519975395707543 128 0.482676124016400 0.519608279401000 256 0.482676124016448 0.519608279400957

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Outline

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Inverse Problem - far-field

Inverse problem

We want to detectΓandλfrom the knowledge of the far-field datau∞(ˆx,d), wherexˆandd are on the unit circle. Here we assumen(x) =1 inR².

The scattered field has the asymptotic behavior u^s(x) = e^ikr

√r u∞(ˆx,d) +O(r^−3/2) asr → ∞, whereˆx = _|x|^x .

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Inverse Problem - Uniqueness results

Theorem.AssumeΓ₁andΓ₂are two scattering obstacles with corresponding surface impedanceλ₁andλ₂such that for a fixed wave number the far field patterns coincide for all incident directions d, thenΓ₁= Γ₂andλ₁=λ₂.

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Inverse Problem - far-field

Thefar field operatorF :L²(S)→L²(S)is defined by (Fg)(ˆx) :=

Z

S

u∞(ˆx,d)g(d)ds(d) TheHerglotz wave functionis given by

v_g(x) :=

Z

S

g(d)e^ikx·dds(d)

whereg∈L²(S)is called the kernel ofvg.

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Inverse Problem - far-field

By superposition we have the following relation (Fg) =B(Hg)

whereH:L²(S)→H^−1/2(Γ)is defined by Hg:=iλ∂vg

∂ν

andB:H^−1/2(Γ)→L²(S)takesh∈H^−1/2(Γ)to the far field patternu∞of the solution to the direct problem withf :=0 and h

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Inverse Problem - far-field

Forβ ∈H˜^1/2(Γ)we construct the double layer potential D(β)(x) :=

Z

Γ

β(y) ∂

∂νy

Φ(x,y)ds(y)

whereΦ(x,y)is the fundamental solution of∆v+k²v =0. D has far field patternγFβwhereγ = ^√^e^iπ/4

8πk and Fβ :=

Z

Γ

β(y)∂e^−ikˆ^x·y

∂ν_y ds(y)

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Inverse Problem - far-field

Lemma The compact operatorsF: ˜H^1/2(Γ)→L²(S)and H:L²(S)→H^−1/2(Γ)are injective and have dense range, provided that there does not exist a Herglotz wave function such that its normal derivative vanishes onΓ.

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Inverse Problem - far-field

Fβ =γ⁻¹B(I−iλT_Γ)β

which implies the following factorization of the far field operator Fg=γF(I−iλT_Γ)⁻¹Hg, g∈L²(S)

LemmaFor any piecewise smooth non intersecting arcLand a functionβ_L∈H˜^1/2(L)we defineΦ^L_∞∈L²(S)by

Φ^L_∞(ˆx) :=

Z

L

β_L(y) ∂

∂ν_ye^−ikˆ^x·yds_y.

Then,Φ^L_∞(ˆx)∈Range(F)if and only ifL⊂Γ.

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Proof

IfL⊂Γ, thenΦ^L_∞(ˆx)∈Range(F).

IfL6⊂Γ, assume on the contrary, thatΦ^L_∞(ˆx)∈Range(F), i.e.

there existsβ ∈H˜^1/2(Γ)such that Φ^L_∞(ˆx) =

Z

Γ

β(y) ∂

∂νy

e^−ikˆ^x·yds_y.

Hence by Rellich’s lemma and the unique continuation principle we have that the potentials

Φ^L(x) = Z

L

β_L(y) ∂

∂ν_yΦ(x,y)ds_y and D(x) = Z

Γ

β(y) ∂

∂ν_yΦ(x,y)ds_y coincide inR²\(Γ∪L).

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Proof

Now letx₀∈L,x₀∈/ Γ, and letB(x₀)be a small ball with center atx₀such thatB(x₀)∩Γ =∅. HenceDis analytic inB(x₀) whileΦ^Lhas a singularity atx₀which is a contradiction. This

proves the lemma.

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Inverse Problem - far-field

Theorem.IfF is the far field operator corresponding to the direct scattering problem withuⁱ(x) =e^ikx·d, the following is true:

1 IfL⊂Γthen for every >0 there exists a solution g^L∈L²(S)of the inequality

kFg^L−Φ^L_∞k_L2(S)≤ such thatH_gL

converges to a well defined function in H^−1/2(Γ).

2 IfL6⊂Γthen for every >0 allg^L∈L²(S)satisfying kFg^L−Φ^L_∞k_L2(S)≤

are such that

→0limkg^Lk_L2(S) =∞ and lim

→0kH_gL

k_H−1/2(Γ)=∞

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Outline

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Numerical results from far field data

Having

u∞=

N

X

n=−N

u∞,nexp(inθ)

we then add random noise to the Fourier coefficients ofu∞to obtain the approximate far-field pattern

u∞=

N

X

n=−N

u∞,a,nexp(inθ)

where

u_∞,a,n=u_∞,n(1+χn) withχn a random variable in[−1,1]and=.05.

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Numerical results from far field data

The inversion scheme is based in solving the following ill-posed first kind equation

Z

S

u∞(ˆx,d)g(d)ds(d) = Z

L

β_L(y) ∂

∂ν_ye^−ikˆ^x·ydsy

for an arbitrary open arc L andβ_L∈H˜^1/2(L).

ifL⊂Γwe can find a boundedg ∈L²(S)

ifL6⊂Γall approximates solutions are unbounded.

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Numerical results from far field data

In this study we search for the crack by takingLto be a small segment centered at a sampling pointz with unit normalνz and β_La sequence that converges toδ(x −z).

Hence we want to solve Z

S

u∞(ˆx,d)g_z,ν_z(d)ds(d) =−ikν_z·x eˆ ^−ikˆ^x·z. (∗) Ifz ∈Γandνz coincides with the normal vector toΓatz, then we can find a boundedg_z,ν_z ∈L²(S)that solves(∗). Otherwise all suchgz,νz are unbounded.

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Numerical results from far field data

In order to solve(∗)we use Tikhonov regularization and the Morozov discrepancy principle to deal with the severe ill-posedness of this equation.

Rewriting the above equation as an ill-condition matrix equation Ag_z,ν_z =f_x,ν_z, where ν_z ∈[0, π]

We use the singular value decomposition ofA, i.e.A=UΛV^∗, is then equivalent to solvingΛV^∗g_z,ν_z =U^∗f_z,ν_z.

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Numerical results from far field data

The Tikhonov regularization leads to the problem of solving min_g_z,νz_∈_RⁿkΛV^∗g_z,ν_z −ρz,νzk²_l2+αkg_z,ν_zk²_l2

whereρ_z,ν_z = (ρ₁, ρ₂, ..., ρ_n)^T =U^∗f_z,ν_z whereαis chosen using the Morozov descrepancy principle.

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Numerical results from far field data

The numerical procedure to locate the crack is the following:

we consider a uniform grid of sampling points{z_i}_i=1..N, and for each sampling point we choose a finite number of anglesν_z, and compute for eachz_i, andνi

G(z_i, ν_i) =kf_z,ν_zk_l2/kg_z,ν_zk_l2

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Numerical results from far field data

formula f(t)=[!2+t,2t]

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1

0 1

2

3 4

5

Figure:Original Crack.

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Numerical results from far field data

Figure:Crack reconstructed from far field data, wavenumberk =5.

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Numerical results from far field data

formula f(t)=[t, .5*cos(t*pi/2)+.2*sin(t*pi/2)!0.1cos(3*t*pi/2)]

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1

0 1

2

3

4

5

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Numerical results from far field data

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Numerical results from far field data

formula f(t)=[2sin((t+4/3)*6pi/4)+2, sin((t+4/3)*3pi/4)]

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1 0

1

2

3 4

5

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Numerical results from far field data

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Numerical results from far field data

Having foundgandΓby solving the far field equation, it is now possible to use the boundary conditions

[u] +iλ∂u⁺

∂ν to determineλ. (Zeev’s PhD thesis)

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Numerical results from far field data

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

2.2/(dvg/dnu) for non linear crack lab=2

λ

Figure:λreconstructed from far field data. The solid line represent the exact solution.

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Numerical results - Impedance reconstruction

Goal:Solve the ill posed, nonlinear equation,

F(λ) =u_∞ (∗)

WhereF mapsλ(x)onto the far field patternu∞for a fixed incident fielduⁱ.

Linearizing(∗) Solve

F_λ⁰h+F_λ =u∞

forh, using Newton type methods.

Improve an approximate of the surface impedanceλr into λ˜_r =λ_r +h.

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The scattered field,

u^s =K(I−iλT_Γ)⁻¹Ruⁱ whereRuⁱ =iλ^∂u_∂νⁱ. the corresponding far field,

u∞=K∞(I−iλT_Γ)⁻¹Ruⁱ We computeF_λ⁰h,

F_λ⁰h=K∞ψ_h where

(I−iλT_Γ)ψ_h=−ihT_Γ(I−iλT_Γ)⁻¹Ruⁱ+ih∂uⁱ

∂ν.

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Numerical results - Impedance reconstruction

Leth(x) =PN

k=0a_kb_k(x). we want to solve

N

X

k=0

a_kF_λ⁰b_k +F_λ =u∞

for reala_k. We considered Chebyshev polynomials b_k(x) =cos(kcos⁻¹(x)), and radial basis functions

b_k(x) =e^−c|x−ζ^k^|², withx ∈[−1,1],c >0 andζ_k ∈[−1,1].

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We use a collocation method with respect tomequidistant pointsˆx₁...xˆ_m in the unit circle, i.e. we solve

N

X

k=0

a_k(F_λ⁰b_k)(ˆx_j) =u_∞(ˆx_j)−F_λ(ˆx_j) j =1, ...,m for reala_k.

We apply Tikhonov regularization, then we minimize the following equivalent problem,

C(p,a₀, ...,a_N) :=

p

X

k=0







m

X

i=1

b_ijka_j−c_ik

2+α

N

X

j=0

θjka²_j







whereb_ijk = (F_λ⁰b_j)(ˆx_i;d_k),c_ik =u∞(ˆx_i;d_k)−F_λ(ˆx_i;d_k),αis the regularization parameter andθ_jk are positive weights.

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Numerical results - Impedance reconstruction

We solve the linear system

p

X

k=1 N

X

l=0 m

X

i=1

b¯_ijkb_ilka_l+α

p

X

k=1

θ_jka_j =

p

X

k=1 m

X

i=1

b¯_ijkc_ik j=0, ...,N.

Choseα=2^−k, wherek is the number of iteration.

Usedn=64 quadrature points in the forward solver and n=32 in the backward solver.

Used one incident direction.

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Numerical results from far field data

(a) (b)

Figure:Panel (a) shows a plot ofλ(x)(in red) and the initial guess λ₀(x)(in blue) chosen along the crack. Panel (b) shows a plot ofλ(x) (in red) and lambda reconstructed (in blue) after 10 iterations with Newton type methods. One incident field was used, 5%noise,N=3.

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Numerical results from far field data

(a) (b)

Figure:Panel (a) shows a plot ofλ(x)and the initial guessλ₀(x) chosen along the crack. Panel (b) shows a plot ofλ(x)and lambda reconstructed after 9 iterations with Newton type methods. One incident field was used, 5%noise,N=3.

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Numerical results from far field data

(a) (b)

Figure:Panel (a) shows a plot ofλ(x)and the initial guessλ₀(x) chosen along the crack. Panel (b) shows a plot ofλ(x)and lambda reconstructed after 12 iterations with Newton type methods. One incident field was used, 5%noise,N=3.

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Conclusions:

Have succesfully reconstructed complex thin objects (with no a priori information) using LSM from the knowledge of noisy far-field data.

Reconstructed their physical properties using a Newton type method from the knowledge of noisy far-field data.

This work have been done for Near field data.

N. Zeev and Fioralba Cakoni, The identification of thin objects from far field or near field scattering data, SIAM J.

APPL. MATH., Vol. 69, No. 4, pp. 1024−1042.

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THANK YOU

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Numerical results from far field data

formula f(t)=[!2+t,2t]

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1

0 1

2

3 4

5

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Numerical results from far field data

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1 0

1

2 3

4

5

0.2 0.25 0.3 0.35 0.4 0.45 0.5

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Numerical results from far field data

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Numerical results from far field data

formula f(t)=[t, .5*cos(t*pi/2)+.2*sin(t*pi/2)!0.1cos(3*t*pi/2)]

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1

0 1

2

3

4

5

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Numerical results from far field data

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1

0

1

2

3

4

5 0.15

0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55

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Numerical results from far field data

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Numerical results from far field data

formula f(t)=[2sin((t+4/3)*6pi/4)+2, sin((t+4/3)*3pi/4)]

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1 0

1

2

3 4

5

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Numerical results from far field data

!

!5 !4 !3 !2 !1 0 1 2 3 4 5

!5

!4

!3

!2

!1 0 1 2 3 4 5

0(1 0(15 0(2 0(25

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