Imaging of Dielectric Objects by BCS-TE

UNIVERSITY

OF TRENTO

DIPARTIMENTO DI INGEGNERIA E SCIENZA DELL’INFORMAZIONE

38123 Povo – Trento (Italy), Via Sommarive 14

http://www.disi.unitn.it

IMAGING OF DIELECTRIC OBJECTS BY BCS-TE

L. Poli, G. Oliveri, P. Rocca, and A. Massa

June 2013

Contents

I Problem Statement

3

1 Mathematical Formulation 4

1.1 2D TE Inverse Scattering Formulation . . . 4

1.2 Inverse CS Problem - Scattering TE Planewave . . . 6

1.2.1 Single-Task BCS . . . 6 1.2.2 Multi-Task BCS . . . 8 1.3 Legenda . . . 9

II Numerical Validation

10

3 XY-MT-BCS (Exploiting Correlation between x-y Field Components) 50 3.1 XY-MT-BCS - Calibration . . . 51

3.2 XY-MT-BCS - Basic Tests . . . 55

4 MV-MT-BCS (Exploiting Correlation between Views) 91 4.1 MV-MT-BCS - Calibration . . . 92

4.2 MV-MT-BCS - Basic Tests . . . 96

4.3 MV-MT-BCS - Advanced Tests . . . 133

5 MV-XY-MT-BCS (Exploiting Correlation between Views and Field Components) 168 5.1 MV-XY-MT-BCS - Calibration . . . 169

5.2 MV-XY-MT-BCS - Basic Tests . . . 173

5.3 MV-XY-MT-BCS - Advanced Tests . . . 210

6 Single/Double Field Components 245 7 Comparison between BCS techniques 261 7.1 Basic Tests . . . 261

7.2 Advanced Tests . . . 270

7.3 Test Varying the Number of Views . . . 273

7.4 Test Varying the Number of Measurement Points . . . 278

7.5 Test Varying the Number of Views/Measurement Points . . . 283

Part I

1 Mathematical Formulation

1.1 2D TE Inverse Scattering Formulation

Si consideri un cilindro dielettrico di sezione arbitraria innitamente esteso lungo l'asse z parallelo all'asse del cilidro stesso, sul quale incide un onda di tipo trasverso elettrico TE, ossia un'onda elettromagnetica carat-terizzata da un vettore di campo elettrico perpendicolare all'asse del cilindro. Dato il tipo di illuminazione,

il campo incidente elettrico Einc sarà privo della componente lungo l'asse z:

Einc(x, y) = Eincx (x, y)x + Eb

y

inc(x, y)yb (1)

dovebxeybrappresentano i versori paralleli rispettivamente all' asse x e all' asse y del sistema di riferimento.

Si assume che il cilindro dielettrico abbia la stessa permeabilita' magnetica del vuoto (µ = µ0); si assume

che il materiale dielettrico sia lineare e isotropo, ma che potrebbe essere non omogeneo lungo le coordinate trasverse:

˜

ε = ˜ε (x, y) (2)

dove ε rappresenta la permettivita' dielettrica complessa del materiale.

Sia Etot il campo elettrico totale, ovvero il campo il campo generato da una sorgente in presenza del

cilindro dielettrico; il campo scatterato Escatt e' denito come la dierenza tra il campo totale Etot e il

campo incidente Einc:

Etot(x, y) = Einc(x, y) + Escatt(x, y) (3)

Dato il tipo di illuminazione abbiamo che:

Etot(x, y) = Etotx (x, y)x + Eb y tot(x, y) ˆy (4) Escatt(x, y) = Escattx (x, y)bx + E y scatt(x, y) ˆy (5)

Dal principio di equivalenza possiamo considerare il campo scatterato come generato da una corrente equivalente che irradia in spazio libero, dove la densita' di corrente e' denita come:

Jeq(x, y) = τ (x, y) Etot(x, y) (6)

che in questo caso puo' essere denita anche come:

Jeq(x, y) = Jeqx (x, y)x + Jb

y

eq(x, y) ˆy (7)

Si puo' dimostrare che:

Escatt(x, y) =

Z

s

G (x, y/x0, y0) τ (x0, y0) Etot(x0, y0) dx0dy0 (8)

Sostituendo la (8) nella (3) si ottiene Z

s

τ (x0, y0) Etot(x0, y0) G (x, y/x0, y0) dx0dy0= Etot(x, y) − Einc(x, y) (9)

La (9) e' un' equazione vettoriale che, dalla (6) puo' essere riscritta come Z

s

Jeq(x, y) G (x, y/x0, y0) dx0dy0= Etot(x, y) − Einc(x, y) (10)

Le componenti scalari dell'equazione (10) sono ricavabili osservando che

G (x, y/x0, y0) = Gx_bx + Gyy_b J = J_eqxx + J_{b eq}yy_b (11)

Jeq· G = Jeqxx + Jb y eqby



Gxx + G_b y_by (13)

che per simmetria della diadica di Green G si puo' scrivere anche

G · Jeq=  Gxx + G_b y_by J_eqx_bx + J_eqy_by
₍₁₄₎ Quindi si ricava Jeq· G = JeqxG xx_{+ J}y eqG xy
b x + J_eqxGyx+ J_eqyGyy
b y (15)

Noto questo e' possibile riscrivere la (10) nelle sue componenti scalari come segue Z s J_eqxGxx+ J_eqyGxy
dx0_dy0 _{= E}x tot(x, y) − E x inc(x, y) (16) Z s J_eqxGyx+ J_eqyGyy
dx0dy0= E_toty (x, y) − E_incy (x, y) (17)

La (16) e la (20) rappresentano le equazioni fondamentali dello scattering inverso, le quali costituiscono

un sistema di due equazioni in due incognite, rappresentate da Jx

eq e Jeqy. Per poter risolvere con tecniche

numeriche vengono discretizzate secondo la procedura di Richmond [4]. Il dominio di indagine e' suddiviso in

N celle rettangolari (D(n)

ind)su cui sono denite altrettante funzioni base ad impulso rettangolare; si assume

che le celle siano sucientemente piccole anche' sia la costante dielettrica (e quindi la funzione oggetto

τ (xn, yn)) che l' intensita' del campo elettrico Etotx,v(xn, yn)e Etoty,v(xn, yn) siano uniformi all' interno delle

stesse. In forma discretizzata:

E_totx,v(x, y) ' N X n=1 E_totx,v(xn, yn) ψn(x, y) (18) Etoty,v(x, y) ' N X n=1 Etoty,v(xn, yn) ψn(x, y) (19) τ (x, y) ' N X n=1 τ (xn, yn) ψn(x, y) (20) dove ψn(x, y) = ( 1 (x, y) ∈ D_ind(n) 0 (x, y) /∈ D_ind(n) ) (21)

Approssimando le celle rettangolari D(n)

ind con equivalenti celle circolari si ottiene una notevole

sempli-cazione delle espressioni. Il raggio della cella n-esima e' dato da:

an=

r

dxndyn

π (22)

Elaborando questa procedura, utilizzando delle funzioni di test impulsive ed introducendo inoltre M punti di misura, si giunge al seguente sistema algebrico

E_scattx,v (xv_m, y_mv) = N X

n=1

τ (xn, yn) Etotx,v(xn, yn) Gxx,ext_2d (ρnm) +

τ (xn, yn) Etoty,v(xn, yn) Gxy,ext2d (ρnm)

E_scatty,v (xv_m, y_mv) = N X n=1 τ (xn, yn) E x,v tot (xn, yn) G yx,ext 2d (ρnm) +

τ (xn, yn) Etoty,v(xn, yn) Gyy,ext_2d (ρnm) m = 1, ..., M (24)

dove Gxx,ext_2d (ρnm) =        1 −jπk0anH1(2)(k0an) 4 se ρnm≤ an  −jπanJ1(k0an) 2ρ3 nm  h k0ρnm(ym− yn) 2 H₀(2)(k0ρnm)+ h (xm− xn) 2 − (ym− yn) 2i · H₁(2)(kρnm) i se ρnm> an (25) Gxy,ext_2d (ρnm) = Gyx,ext_2d (ρnm) =        0 se ρnm≤ an  −jπanJ1(k0an) 2ρ3 nm  (xm− xn) (ym− yn) h 2H₁(2)(k0ρnm)− k0ρnmH (2) 0 (k0ρnm) i se ρnm> an (26) Gyy,ext_2d (ρnm) =             1 − jπk0anH1(2)(k0an) 4  se ρnm≤ an  −jπanJ1(kan) 2ρ3 nm  h k0ρnm(xm− xn) 2 H₀(2)(k0ρnm)+ h (ym− yn)2− (xm− xn)2 i · H₁(2)(k0ρnm) se ρnm> an (27)

dove J1 e' la funzione di Bessel di ordine 1, H

(2)

0 e' la funzione di Hankel di seconda specie e di ordine 0,

H₁(2) e' la funzione di Hankel di seconda specie di ordine 1, (xm,ym)sono le coordinate del generico m-esimo

punto di misura, (xn, yn)sono le coordinate del centro della generica n-esima cella del dominio di indagine

discretizzato, dove ρmn= r h (xm− xn)2+ (ym− yn)2 i (28) k0= ω √ ε0µ0 (29)

1.2 Inverse CS Problem - Scattering TE Planewave

1.2.1 Single-Task BCS

La tecnica del Bayesian Compressive Sampling permette di risolvere un problema lineare del tipo: dato

y = A · xtrovare x ∈ CM tale che x ∈ CM e x è sparso. Considerando la tecnica single-task BCS, possiamo

denire y =         Escattx,v (x1, y1) ... E_scattx,v (xM, yM) E_scatty,v (x1, y1) ... E_scatty,v (xM, yM)         (30) di dimensione 2M × 1;

A =        

Gxx,ext_2d (ρ11) ... Gxx,ext_2d (ρ1N) Gxy,ext_2d (ρ11) ... Gxy,ext_2d (ρ1N)

... ... ... ... ... ...

Gxx,ext_2d (ρM 1) ... Gxx,ext_2d (ρM N) Gxy,ext_2d (ρM 1) ... Gxy,ext_2d (ρM N)

Gyx,ext_2d (ρ11) ... Gyx,ext_2d (ρ1N) Gyy,ext_2d (ρ11) ... Gyy,ext_2d (ρ1N)

... ... ... ... ... ...

Gyx,ext_2d (ρM 1) ... Gyx,ext2d (ρM N) Gyy,ext2d (ρM 1) ... Gyy,ext2d (ρM N)

        (31) di dimensione 2M × 2N; x =         Jx,v eq (x1, y1) ... Jx,v eq (xN, yN) Jy,v eq (xN, yN) ... Jeqy,v(xN, yN)         (32) di dimensione 2N × 1.

Analogamente a quanto visto per le eq. (23) e (24), possiamo scrivere le equazioni di stato

τ (xn, yn) E x,v inc(xn, yn) = Jeqx,v(xn, yn) − N X q=1 n τ (xq, yq) E x,v totG xx,int 2d (ρnq) τ (xq, yq) E y,v totG xy,int 2d (ρnq) o n = 1, ..., N (33)

τ (xn, yn) Ey,vinc(xn, yn) = Jeqy,v(xn, yn) −

N X q=1 n τ (xq, yq) Etotx,vG yx,int 2d (ρnq) τ (xq, yq) E y,v totG yy,int 2d (ρnq) o n = 1, ..., N (34) dove Gxx,int_2d (ρqn) =         −jπaqJ1(kaq) 2ρ3 qn  h k0ρqn(yn− yq)2H (2) 0 (k0ρqn)+ h (xn− xq)2− (yn− yq)2 i · H₁(2)(k0ρqn) se 1 −jπk0aqH₁(2)(k0aq) 4 se q = n q 6= n (35) Gxy,int_2d (ρqn) = Gyx,int2d (ρqn) =             −jπaqJ1(k0aq) 2ρ3 qn  h (xn− xq) (yn− yq) h 2H₁(2)(k0ρqn)− k0ρqnH (2) 0 (k0ρqn) ii se q 6= n 0 se q = n (36) Gyy,int_2d (ρqn) =             −jπaqJ1(kaq) 2ρ3 qn  h k0ρqn(xn− xq) 2 H₀(2)(k0ρqn)+ h (yn− yq) 2 − (xn− xq) 2i · H₁(2)(k0ρqn) se 1 − jπk0aqH1(2)(k0aq) 4 se q = n q 6= n (37)

E_scattx,v (xn, yn) = N X

n=1

Jx,v

eq (xq, yq) Gxx,ext_2d (ρnm) + Jeqy,v(xq, yq) Gxy,ext_2d (ρnm) n = 1, ..., N (38)

E_scatty,v (xn, yn) = N X

n=1

Jx,v

eq (xq, yq) Gyx,ext_2d (ρnm) + Jeqy,v(xq, yq) Gyy,ext_2d (ρnm) n = 1, ..., N (39)

Quindi, calcoliamo il campo totale interno:

Ex,v_tot(xn, yn) = Escattx,v (xn, yn) + Eintx,v(xn, yn) n = 1, ..., N (40)

E_toty,v(xn, yn) = Escatty,v (xn, yn) + Einty,v(xn, yn) n = 1, ..., N (41)

e per nire: τ (xn, yn) = 1 2V V X v=1 ( Jx,v eq (xn, yn) E_totx,v(xn, yn) +J y,v eq (xn, yn) Ey,v_tot(xn, yn) ) (42) 1.2.2 Multi-Task BCS

Considerando invece la tecnica multi-task BCS, possiamo scomporre il problema in due problemi lineari, sfruttando la correlazione tra le componenti x-y di campo elettrico

( y0= A0· x0 y00= A00· x00 (43) y0=   E_scattx,v (x1, y1) ... E_scattx,v (xM, yM)  , y00=   Ey,v_scatt(x1, y1) ... E_scatty,v (xM, yM)   (44) di dimensione M × 1; A0=   Gxx,ext_2d (ρ11) ... G xx,ext 2d (ρ1N) G xy,ext 2d (ρ11) ... G xy,ext 2d (ρ1N) ... ... ... ... ... ... Gxx,ext_2d (ρM 1) ... G xx,ext 2d (ρM N) G xy,ext 2d (ρM 1) ... G xy,ext 2d (ρM N)   (45) A00=   Gyx,ext_2d (ρ11) ... G yx,ext 2d (ρ1N) G yy,ext 2d (ρ11) ... G yy,ext 2d (ρ1N) ... ... ... ... ... ... Gyx,ext_2d (ρM 1) ... G yx,ext 2d (ρM N) G yy,ext 2d (ρM 1) ... G yy,ext 2d (ρM N)   (46) di dimensione M × 2N, e inne x0 =         Jeqxx,v(x1, y1) ... J_eqxx,v(xN, yN) Jxy,v eq (x1, y1) ... Jxy,v eq (xN, yN)         , x00=         Jeqyx,v(x1, y1) ... J_eqyx,v(xN, yN) Jyy,v eq (x1, y1) ... Jyy,v eq (xN, yN)         (47) di dimensione N × 1.

Escatty,v (xn, yn) = N X

n=1

Jyx,v

eq (xq, yq) Gyx,ext2d (ρnm) + Jeqyy,v(xq, yq) Gyy,ext2d (ρnm)

n = 1, ..., N (49)

Quindi, calcoliamo il campo totale interno:

Ex,v_tot(xn, yn) = E

x,v

scatt(xn, yn) + E

x,v

int(xn, yn) n = 1, ..., N (50)

E_toty,v(xn, yn) = Escatty,v (xn, yn) + Einty,v(xn, yn) n = 1, ..., N (51)

e per nire: τ (xn, yn) = 1 4V V X v=1 ( Jxx,v eq (xn, yn) E_totx,v(xn, yn) +J xy,v eq (xn, yn) E_toty,v(xn, yn) +J yx,v eq (xn, yn) E_totx,v(xn, yn) +J yy,v eq (xn, yn) Ey,v_tot(xn, yn) ) (52)

1.3 Legenda

• ST-BCS: versione Single-Task BCS che considera entrambe le componenti x-y di campo elettrico,

sec-ondo l'implementazione esposta nella sotto-sezione 1.2.1;

• X-ST-BCS: versione Single-Task BCS che considera esclusivamente la componente x di campo elettrico;

• Y-ST-BCS: versione Single-Task BCS che considera esclusivamente la componente y di campo elettrico;

• XY-MT-BCS: versione Multi-Task BCS che considera entrambe le componenti x-y di campo elettrico,

secondo l'implementazione esposta nella sotto-sezione 1.2.2, e ne sfrutta la correlazione;

• MV-MT-BCS: versione Multi-Task BCS che considera entrambe le componenti x-y di campo elettrico,

secondo l'implementazione esposta nella sotto-sezione 1.2.1, sfruttando però la correlazione tra le viste e non tra le 2 componenti di campo;

• MV-X-MT-BCS: versione Multi-Task BCS che considera esclusivamente la componente x di campo

elettrico, sfruttando la correlazione tra le viste;

• MV-Y-MT-BCS: versione Multi-Task BCS che considera esclusivamente la componente y di campo

elettrico, sfruttando la correlazione tra le viste;

• MV-XY-MT-BCS: versione Multi-Task BCS che considera entrambe le componenti x-y di campo

elet-trico, secondo l'implementazione esposta nella sotto-sezione 1.2.2 sfruttando sia la correlazione tra le 2 componenti di campo che la correlazione tra le viste;

Part II

Numerical Validation

2.1 ST-BCS - Calibration

TEST CASE: Square Cylinder L = 0.16λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Square cylinder of side λ

6 = 0.1667

• σ = 0[S/m] BCS parameters:

• Initial estimate of the noise: n0=∈1.0 × 10−6, 2.0 × 10−6, 5.0 × 10−6, 1.0 × 10−5, 2.0 × 10−5, 5.0 × 10−5,

1.0 × 10−4, 2.0 × 10−4, 5.0 × 10−4, 1.0 × 10−3, 2.0 × 10−3, 5.0 × 10−3, 1.0 × 10−2

RESULTS: Calibration 10-4 10-3 10-2 10-1 10-6 10-5 10-4 10-3 10-2 ξtot [arbitraty unit] n₀2 selected value Noiseless SNR = 20dB SNR = 10dB SNR = 5dB average (a) 10-2 10-1 100 10-6 10-5 10-4 10-3 10-2 ξint [arbitraty unit] n₀2 Noiseless SNR = 20dB SNR = 10dB SNR = 5dB average (b) 10-6 10-5 10-4 10-3 10-2 10-6 10-5 10-4 10-3 10-2 ξext [arbitraty unit] n₀2 Noiseless SNR = 20dB SNR = 10dB SNR = 5dB average (c)

Figure 1. Behaviour of error gures as a function of the initial estimate of the noise n0, for dierent SNR

2.2 ST-BCS - Basic Tests

TEST CASE: Square Cylinder L = 0.16λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• σ = 0[S/m] BCS parameters:

• Initial estimate of the noise: n0= 5.0 × 10−4

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Re[ τ(x,y)] (d) (e)

Figure 2. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d)

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (d) (e)

Figure 3. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d)

RESULTS: εr= 5.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Re[ τ(x,y)] (d) (e)

Figure 4. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d)

RESULTS: Error Figures 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-5 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 5. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Two Square Cylinders L = 0.16λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Two square cylinders of side λ

6 = 0.1667

• εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0}(one square), εr= 1.5(one square)

• σ = 0[S/m]

BCS parameters:

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 Re[ τ(x,y)] (d) (e)

Figure 6. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d)

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (d) (e)

Figure 7. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d)

RESULTS: εr= 5.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (d) (e)

Figure 8. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d)

RESULTS: Error Figures 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 9. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Square Cylinder L = 0.33λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Square cylinder of side λ

3 = 0.33

• εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0}

• σ = 0[S/m]

BCS parameters:

• Initial estimate of the noise: n0= 5.0 × 10−4

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (d) (e)

Figure 10. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Re[ τ(x,y)] (d) (e)

Figure 11. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Re[ τ(x,y)] (d) (e)

Figure 12. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 13. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Cross-Shaped Cylinder

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object: • Cross-shaped cylinder • εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0} • σ = 0[S/m] BCS parameters:

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re[ τ(x,y)] (d) (e)

Figure 14. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (d) (e)

Figure 15. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (d) (e)

Figure 16. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 17. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: L-Shaped Cylinder

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude A = 1 • Frequency: 300 MHz (λ = 1) Object: • L-shaped cylinder • εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0} • σ = 0[S/m] BCS parameters:

• Initial estimate of the noise: n0= 5.0 × 10−4

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] (d) (e)

Figure 18. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] (d) (e)

Figure 19. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] (d) (e)

Figure 20. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-4 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 21. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Inhomogeneous L-Shaped Cylinder

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Inhomogeneous L-shaped cylinder

• εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0}

• σ = 0[S/m]

BCS parameters:

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] (d) (e)

Figure 22. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Re[ τ(x,y)] (d) (e)

Figure 23. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (d) (e)

Figure 24. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 25. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Line-Shaped Cylinder L = 0.5λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Inhomogeneous L-shaped cylinder

• εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0}

• σ = 0[S/m]

BCS parameters:

• Initial estimate of the noise: n0= 5.0 × 10−4

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (d) (e)

Figure 26. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Re[ τ(x,y)] (d) (e)

Figure 27. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 Re[ τ(x,y)] (d) (e)

Figure 28. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-4 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 29. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

3 XY-MT-BCS (Exploiting Correlation between x-y Field

Compo-nents)

3.1 XY-MT-BCS - Calibration

TEST CASE: Square Cylinder

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Square cylinder of side λ

6 = 0.1667

• σ = 0[S/m] BCS parameters:

• Gamma prior on noise variance parameter: a ∈ 1 × 10−1_{, 2 × 10}−1_{, 5 × 10}−1_{1 × 10}0_{, 2 × 10}0_{, 5 ×}

100_{,1 × 10}+1_{, 2 × 10}+1_{, 5 × 10}+1_{, 1 × 10}+2_{, 2 × 10}+2_{, 5 × 10}+2_{, 1 × 10}+3_{, 2 × 10}+3_{, 5 × 10}+3_{, 1 × 10}+4

• Gamma prior on noise variance parameter: b ∈ 1 × 10+0_{, 5 × 10}−1_{, 2 × 10}−1_{, 1 × 10}−1_{, 5 × 10}−2_{, 2 ×}

10−21 × 10−2, 5 × 10−3, 2 × 10−3, 1 × 10−3, 5 × 10−4, 2 × 10−4, 1 × 10−4

RESULTS: Calibration 10-1 100 101 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-4 10-3 10-2 Noiseless a b 10-4 10-3 10-2 ξtot [arbitrary unit] 10-1 100 101 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-2 10-1 100 Noiseless a b 10-2 10-1 100 ξint [arbitrary unit] 10-1 100 101 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-6 10-5 10-4 10-3 Noiseless a b 10-6 10-5 10-4 10-3 ξext [arbitrary unit] (a) (b) (c) 10-1 10 0 10 1 10 2 10 3 104 10-4 10-3 10-2 10-1 100 10-4 10-3 10-2 SNR = 20 [dB] a b 10-4 10-3 10-2 ξtot [arbitrary unit] 10-1 10 0 10 1 10 2 10 3 104 10-4 10-3 10-2 10-1 100 10-2 10-1 100 SNR = 20 [dB] a b 10-2 10-1 100 ξint [arbitrary unit] 10-1 10 0 10 1 10 2 10 3 104 10-4 10-3 10-2 10-1 100 10-6 10-5 10-4 10-3 SNR = 20 [dB] a b 10-6 10-5 10-4 10-3 ξext [arbitrary unit] (d) (e) (f ) 10-1 100 101 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-4 10-3 10-2 SNR = 10 [dB] a b 10-4 10-3 10-2 ξtot [arbitrary unit] 10-1 100 101 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-1 100 SNR = 10 [dB] a b 10-1 100 ξint [arbitrary unit] 10-1 100 101 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-6 10-5 10-4 10-3 10-2 SNR = 10 [dB] a b 10-6 10-5 10-4 10-3 10-2 ξext [arbitrary unit] (g) (h) (i) 10-1 10 0 10 1 10 2 10 3 104 10-4 10-3 10-2 10-1 100 10-4 10-3 10-2 10-1 SNR = 5 [dB] a b 10-4 10-3 10-2 10-1 ξtot [arbitrary unit] 10-1 10 0 10 1 10 2 10 3 104 10-4 10-3 10-2 10-1 100 10-2 10-1 100 SNR = 5 [dB] a b 10-2 10-1 100 ξint [arbitrary unit] 10-1 10 0 10 1 10 2 10 3 104 10-4 10-3 10-2 10-1 100 10-5 10-4 10-3 10-2 SNR = 5 [dB] a b 10-5 10-4 10-3 10-2 ξext [arbitrary unit] (l) (m) (n)

Figure 30. Behaviour of error gures as a function of the initial estimate of the Gamma prior on the noise

variance parameters a and b, for dierent SNR values: (a), (d), (g) and (l) total error ξtot, (b), (e), (h)

and (m) internal error ξint, (c), (f ), (i) and (n) external error ξext, for (a), (b) and (c) Noiseless case, (d),

RESULTS: Calibration 10-1 10 0 10 1 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-4 10-3 10-2 Average a b 10-4 10-3 10-2 ξtot [arbitrary unit] (a) 10-1 10 0 10 1 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-2 10-1 100 Average a b 10-2 10-1 100 ξint [arbitrary unit] (b) 10-1 10 0 10 1 10 2 10 3 10 4 10-4 10-3 10-2 10-1 100 10-5 10-4 10-3 10-2 Average a b 10-5 10-4 10-3 10-2 ξext [arbitrary unit] (c)

Figure 31. Averaged behaviour of error gures as a function of the initial estimate of the Gamma prior on

3.2 XY-MT-BCS - Basic Tests

TEST CASE: Square Cylinder L = 0.16λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Square cylinder of side λ

6 = 0.1667

• σ = 0[S/m] BCS parameters:

• Gamma prior on noise variance parameters: a = 2 × 10+2_{, b = 1 × 10}−1

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 Re[ τ(x,y)] (d) (e)

Figure 32. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Re[ τ(x,y)] (d) (e)

Figure 33. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 5.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Re[ τ(x,y)] (d) (e)

Figure 34. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-6 10-5 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 35. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Two Square Cylinders L = 0.16λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Two square cylinders of side λ

6 = 0.1667

• εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0}(one square), εr= 1.5(one square)

• σ = 0[S/m]

BCS parameters:

• Gamma prior on noise variance parameters: a = 2 × 10+2_{, b = 1 × 10}−1

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 Re[ τ(x,y)] (d) (e)

Figure 36. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Re[ τ(x,y)] (d) (e)

Figure 37. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 5.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Re[ τ(x,y)] (d) (e)

Figure 38. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-5 10-4 10-3 10-2 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 39. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Square Cylinder L = 0.33λ

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object:

• Square cylinder of side λ

3 = 0.33

• εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0}

• σ = 0[S/m]

BCS parameters:

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] (d) (e)

Figure 40. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] (d) (e)

Figure 41. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Re[ τ(x,y)] (d) (e)

Figure 42. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 43. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: Cross-Shaped Cylinder

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude: A = 1 • Frequency: 300 MHz (λ = 1) Object: • Cross-shaped cylinder • εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0} • σ = 0[S/m] BCS parameters:

• Gamma prior on noise variance parameters: a = 2 × 10+2_{, b = 1 × 10}−1

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] (d) (e)

Figure 44. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Re[ τ(x,y)] (d) (e)

Figure 45. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (d) (e)

Figure 46. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-4 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 47. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)

TEST CASE: L-Shaped Cylinder

Goal: show the performances of BCS when dealing with a sparse scatterer

• Number of Views: V

• Number of Measurements: M

• Number of Cells for the Inversion: N

• Number of Cells for the Direct solver: D

• Side of the investigation domain: L

Test Case Description Direct solver:

• Square domain divided in√D ×√D cells

• Domain side: L = 3λ

• D = 1296(discretization for the direct solver: < λ/10)

Investigation domain:

• Square domain divided in√N ×√N cells

• L = 3λ • 2ka = 2 ×2π λ × L√2 2 = 6π √ 2 = 26.65 • #DOF = (2ka)₂ 2 = (2×2πλ× L√2 2 ) 2 2 = 4π 2 L λ
2 = 4π2_{× 9 ≈ 355.3}

• N scelto in modo da essere vicino a #DOF : N = 324 (18 × 18)

Measurement domain:

• Measurement points taken on a circle of radius ρ = 3λ

• Full-aspect measurements • M ≈ 2ka → M = 27 Sources: • Plane waves • V ≈ 2ka → V = 27 • Amplitude A = 1 • Frequency: 300 MHz (λ = 1) Object: • L-shaped cylinder • εr∈ {1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0} • σ = 0[S/m] BCS parameters:

RESULTS: εr= 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 Re[ τ(x,y)] (d) (e)

Figure 48. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 2.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 Re[ τ(x,y)] (d) (e)

Figure 49. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: εr= 3.0 -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.5 1 1.5 2 2.5 3 3.5 4 Re[ τ(x,y)] (a) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Re[ τ(x,y)] (b) (c) -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Re[ τ(x,y)] -1.5 -1 -0.5 0 0.5 1 1.5 x/λ -1.5 -1 -0.5 0 0.5 1 1.5 y/ λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Re[ τ(x,y)] (d) (e)

Figure 50. Actual object (a) and BCS reconstructed object for (b) Noiseless case, (c) SNR = 20 [dB] , (d) SNR = 10 [dB] , (e) SNR = 5 [dB].

RESULTS: Error Figures 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξtot [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (a) 10-1 100 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (b) 10-4 10-3 10-2 10-1 0.5 1 1.5 2 2.5 3 3.5 4 ξint [arbitraty unit] Re[τ] Noiseless SNR = 20dB SNR = 10dB SNR = 5dB (c)

Figure 51. Behaviour of error gures as a function of εr, for dierent SNR values: (a) total error ξtot, (b)