Some Indications on the Analytic Expression of Non-Radiating Currents in 2D TM Imaging Problems

UNIVERSITY

OF TRENTO

DIPARTIMENTO DI INGEGNERIA E SCIENZA DELL’INFORMAZIONE

38050 Povo – Trento (Italy), Via Sommarive 14

http://www.disi.unitn.it

SOME

INDICATIONS

ON

THE

ANALYTIC

EXPRESSION

OF

NON-RADIATING

CURRENTS

IN

2D

TM

IMAGING

PROBLEMS

P. Rocca, M. Donelli, G.L. Gragnani, and A. Massa

May 2008

rents in 2D TM Imaging Problems

Paolo Ro a,* Massimo Donelli,* Member, IEEE, Gian Luigi Gragnani,** and Andrea Massa* Member, IEEE,

*

ELEDIA

Resear hGroup

DepartmentofInformationandCommuni ationTe hnologies-UniversityofTrento ViaSommarive14,38050Trento-Italy

Tel. +390461882057,Fax +390461882093 E-mail: andrea.massaing.unitn.it,

{paolo.ro a,massimo.donelli}disi.unitn.it

**DepartmentofBiophysi alandEle troni Engineering-UniversityofGenoa, ViaOperaPia11/A,16145Genoa,Italy

Tel. +390103532244,Fax +390103532245 E-mail: gragnanidibe.unige.it

rents in 2D TM Imaging Problems

PaoloRo a*,MassimoDonelli*,GianLuigi Gragnani**,andAndreaMassa*

Abstra t

Inthis ontribution, ananalyti alpro edure for building the non-radiating partof the equivalent ur-rent ininverse s atteringproblems is presented. The mathemati alformulation, on erned witha

2D

ongurationand

T M

illumination,isprovidedanddis ussed.

Intheframeworkofinverses atteringte hniques,severalte hniques onsidertheintrodu tionofanequivalent urrentdensity dened into the diele tri domain. A ordingly, the problem is formulatedin terms of an inversesour eonethroughtheintrodu tionofanequivalent urrentinordertolinearizetheoriginalinverse

s attering problem [1℄[2℄[3℄[4℄. Unfortunately, atypi al drawba k ofinverse sour eproblems relies in their non-uniqueness due to thepresen e of non-radiating omponentsof the equivalent urrents. As far asthe

presen e ofnon-radiating urrentsin avolume formulationis on erned,theissuehasbeendis ussedin [5℄

and an approa h is presentedfor the re onstru tionof the non-radiatingpart of theequivalent volumetri

sour e. Moreover, in [6℄, a wide review and some representativeresults on the presen e of non-radiating

urrentsaswellas aboutthe relationshipbetweennon-radiating sour es and s attering obje ts that ould

beinvisibleunderparti ularillumination onditionsarereported. Asamatteroffa t,although ithasbeen

provedin[7℄thataperfe tlyinvisibleobje tdoesnotexist,ontheotherhand,ithasbeenalsodemonstrated

that as atteringobje t anbenons atteringatparti ulardire tionsof illumination[8℄,thustheequivalent

inversesour eproblemin ludenon-radiating omponents(i.e. thenullspa eoftheoperatorisnotempty).

Inthefollowing,apreliminary pro edureforbuildinganon-radiating urrentisreported.

2 Mathemati al Formulation

In the framework of inverse sour e problems, non-radiating urrents,

Q

N R

(¯

r)

, are urrents that produ e

elds that are identi ally zero outside their sour e volume [9℄, hen e they satisfy the following integral

relationship

Z

Z

Z

V

Q

N R

(¯

r

′

)G

0

(|¯

r − ¯

r

′

|)d¯

r

′

≡ 0 ¯

r

′

∈ V, ¯

r /

∈ V

(1) where the

G

0

(|¯

r − ¯

r

′

|)

is thefree-spa e Green's fun tion. In the following, let us onsider a urrent

Q(¯

r)

whosesupportisatwodimensionaldomain

D

. Asfaras

2D

problemswith

T M

illuminationsare on erned,

Q(¯

r)

isnonradiating[

Q(¯

r) = Q

N R

(¯

r)

℄ if

f (¯

ρ) =

Z

D

J

0

(k

0

|¯

ρ − ¯

ρ

′

|)Q(¯

ρ

′

)d¯

ρ

′

≡ 0 ,

∀ ¯

ρ

(2)

J

0

(x)

beingthe( ylindri al) Besselfun tion oftherstkindandzero-thorderand

|¯

ρ| = ρ =

px

2

_{+ y}

2

Itiswellknownthatasetofnonradiating urrents anbedenedforaspheri aldomainin

3D

ase[6℄[10℄.

Con erning the

2D T M

ase,letus onsider a urrentwhose supportis a ir ulardomainof radius

a

, and

havingthefollowingexpression:

Q(ρ) = Q(ρ, ϑ) = [α

1

J

n

(l

1

ρ/a) + α

2

J

n

(l

2

ρ/a)] e

jnϑ

(3)

where

l

1

and

l

2

are two dierent zeroes of

J

n

(x)

and

α

1

and

α

2

are onstants. In order to analyze the

non-radiating ondition(2),letus ompute

f (¯

ρ)

f (¯

ρ) =

Z

D

J

0

(k

0

|¯

ρ − ¯

ρ

′

|)Q(¯

ρ

′

)d¯

ρ

′

D = {¯

ρ : ρ ≤ a}

(4)

Towardsthisend,letus onsiderthat

J

0

(k

0

|¯

ρ − ¯

ρ

′

|) =

+∞

X

µ=−∞

J

µ

(kρ

′

)J

µ

(kρ)e

jµϑ

e

−

jµϑ

′

(5) andsin e

d¯

ρ

′

= ρ

′

dρ

′

dϑ

′

,bysubstituting(5)in(4), weget

R

ρ6a

J

0

(k

0

|¯

ρ − ¯

ρ

′

|)Q(¯

ρ

′

)d¯

ρ

′

=

=

R

2π

0

R

a

0

h

P

+∞

µ=−∞

J

µ

(kρ

′

)J

µ

(kρ)e

jµϑ

e

−

jµϑ

′

i

[α

1

J

n

(l

1

ρ

′

/a) + α

2

J

n

(l

2

ρ

′

/a)] e

jnϑ

′

ρ

′

dρ

′

dϑ

′

=

=

R

2π

0

R

a

0

J

n

(kρ

′

)J

n

(kρ)e

jµϑ

[α

1

J

n

(l

1

ρ

′

/a) + α

2

J

n

(l

2

ρ

′

/a)] ρ

′

dρ

′

dϑ

′

=

= 2πJ

n

(kρ)e

jµϑ

α

1

R

₀

a

J

n

(kρ

′

)J

n

(l

1

ρ

′

/a)ρ

′

dρ

′

+ α

2

R

₀

a

J

n

(kρ

′

)J

n

(l

2

ρ

′

/a)ρ

′

dρ

′



(6)

Moreover,letusre allthat

Z

J

µ

(b

1

x)J

µ

(b

2

x)xdx =

x

b

2

1

− b

2

2

[b

1

J

µ+1

(b

1

x)J

µ

(b

2

x) − b

2

J

µ

(b

1

x)J

µ+1

(b

2

x)] + C

(7) hen e

α

1

Z

a

0

J

n

(kρ

′

)J

n

(l

1

ρ

′

/a)ρ

′

dρ

′

= −α

1

a

2

a

2

_k

2

_{− l}

2

1

l

1

J

n

(ka)J

n+1

(l

1

)

(8) and

α

Z

a

J

(kρ

′

)J

(l

ρ

′

/a)ρ

′

dρ

′

= −α

a

2

l

J

(ka)J

(l

) .

f (¯

ρ) = −2πJ

n

(kρ)e

jµϑ



α

1

a

2

a

2

_k

2

_{− l}

2

1

l

1

J

n

(ka)J

n+1

(l

1

) + α

2

a

2

a

2

_k

2

_{− l}

2

2

l

2

J

n

(ka)J

n+1

(l

2

)



;

(10)

A ordingto(2),

Q(¯

ρ) = Q

N R

(¯

ρ)

if

f (¯

ρ) = 0

,thereforesin ethefollowing ondition

α

2

= −α

1

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

(11)

nulls(10),thenanon-radiating urrentisobtainedbysubstituting (11)in (3),thusobtainingthefollowing

expression

Q

N R

(ρ) = α

1



J

n

(l

1

ρ/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ/a)



e

jnϑ

_.

(12)

2.2 Testing the Non-Radiating Condition

Let us onsider a

2D T M

problem where a ylindri al s attererof ir ular ross-se tion of radius

a

liesin

free-spa e. Ouraimis toverifywhether (1)issatisedwhen onsidering

Q

N R

(ρ)

denedin (12).

Insu h as enario,theintegralin(1)turnsouttobeequalto

Z

Z

ρ6a

H

0

(2)

(k

0

|¯

ρ − ¯

ρ

′

|)Q(¯

ρ

′

)d¯

ρ

′

(13) sin e

G

0

(|¯

ρ − ¯

ρ

′

|) = H

0

(2)

(k

0

|¯

ρ − ¯

ρ

′

|)

,

H

(2)

0

beingtheHankelfun tion ofthese ondkindoforderzero. Thus,

weshould verifythat

Z

Z

ρ6a

H

0

(2)

(k

0

|¯

ρ − ¯

ρ

′

|)Q(¯

ρ

′

)d¯

ρ

′

=











0

ρ > a

6= 0

ρ 6 a

(14) with

Q(¯

ρ) = Q

N R

(¯

ρ) = α

1

h

J

n

(l

1

ρ/a) −

a

2

k

2

−

_l

2

2

a

2

_k

2

−

_l

2

1

l

1

l

2

J

n

+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ/a)

i

e

jnϑ

;

Towardsthisend,letus onsiderthefollowingrelationships

H

0

(2)

(k |¯

ρ − ¯

ρ

′

|) =

+∞

X

µ=−∞

J

µ

(kρ

′

)H

µ

(2)

(kρ)e

jµ(ϑ−ϑ

′

₎

ρ > ρ

′

(15)

H

0

(2)

(k |¯

ρ − ¯

ρ

′

|) =

+∞

X

µ=−∞

J

µ

(kρ)H

µ

(2)

(kρ

′

)e

jµ(ϑ−ϑ

′

)

ρ 6 ρ

′

(16)

Inparti ular,when

ρ > a

Z

Z

ρ6a

"

+∞

X

µ=−∞

J

µ

(kρ

′

)H

µ

(2)

(kρ)e

jµ(ϑ−ϑ

′

₎

#

α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



e

jnϑ

′

ρ

′

dρ

′

dϑ

′

=

=

Z

2π

0

Z

a

0

h

J

n

(kρ

′

)H

n

(2)

(kρ)e

jnϑ

i

α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

dϑ

′

=

= H

(2)

n

(kρ)e

jnϑ

Z

2π

0

dϑ

′

Z

a

0

J

n

(kρ

′

)α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

=

= 2πH

_n

(2)

(kρ)e

jnϑ

Z

a

0

J

n

(kρ

′

)α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

=

= 2πH

n

(2)

(kρ)e

jnϑ



α

1

Z

a

0

J

n

(kρ

′

)J

n

(l

1

ρ

′

/a)ρ

′

dρ

′

−

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

Z

a

0

J

n

(kρ

′

)J

n

(l

2

ρ

′

/a)ρ

′

dρ

′



=

0

(17)

Sin e the twointegrals within the square bra ket in the last rowof (17) are the same of Eq. (6), where

their sumhas beenmade nullby imposing ondition(11), it has been provedthat theeld radiatedby a

non-radiating urrentisnulloutsideitssupport.

Otherwise,theeldinside thesupportofthe urrent(

ρ ≤ a

)isgivenby

Z

Z

ρ6a

"

_+∞

X

µ=−∞

J

µ

(kρ)H

µ

(2)

(kρ

′

)e

jµ(ϑ−ϑ

′

)

#

α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



e

jnϑ

′

ρ

′

dρ

′

dϑ

′

=

=

Z

2π

0

Z

a

0

h

J

n

(kρ)H

n

(2)

(kρ

′

)e

jnϑ

i

_α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

dϑ

′

=

= 2πJ

n

(kρ)e

jnϑ

Z

a

0

H

n

(2)

(kρ

′

)α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

(18) Sin e

H

(2)

n

(x) = J

n

(x) − jY

n

(x)

(19)

where

J

n

(x)

and

Y

n

(x)

are the Bessel fun tions of the rst and se ond kind of order

n

, respe tively, by

substituting (19)into(18)itresultsthat

2πJ

n

(kρ)e

jnϑ

Z

a

0

[J

n

(kρ

′

) − jY

n

(kρ

′

)] α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

=

−j2πJ

n

(kρ)e

jnϑ

Z

a

0

Y

n

(kρ

′

)α

1



J

n

(l

1

ρ

′

/a) −

a

2

_k

2

_{− l}

2

2

a

2

_k

2

_{− l}

2

1

l

1

l

2

J

n+1

(l

1

)

J

n+1

(l

2

)

J

n

(l

2

ρ

′

/a)



ρ

′

dρ

′

6=

0

urrent.

3 Con lusions

Inthispaper,apro edureforindi atingthenon-radiatingpartoftheequivalent urrentininverses attering problems when formulatedin terms of an inverse sour e and on erned with a

2D

onguration and

T M

illumination,ispresented.

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[2℄ S. Caorsi,G.L.Gragnani,and M.Pastorino,Equivalent urrentdensityre onstru tionformi rowave imagingre onstru tion, IEEETrans.Mi rowaveTheoryTe h.,vol.37,no.5,pp.910-916,May1989.

[3℄ T. M.Habashy, E. Y. Chow, and D. G. Dudley, Proleinversionusing therenormalized sour e-type integralequationapproa h,IEEETrans.AntennasPropagat.,vol.38,no.5,pp.668-681,May1990.

[4℄ S. Caorsi, G. L. Gragnani, and M. Pastorino, A multiview mi rowave imaging system for two-dimensional penetrable obje ts, IEEE Trans. Mi rowave Theory Te h., vol. 39, no. 5, pp. 845-851, May1991.

[5℄ T. M. Habashy, M. L.Oristaglio,and A. T. de Hoop, Simultaneous nonlinear re onstru tion of two-dimensionalpermittivityand ondu tivity, RadioS ien e,vol.29,no.4,pp.1101-1118,Jul.-Aug.1994.

[6℄ G. Gbur, Nonradiating sour esand the inverse sour e problem, Ph.D. Thesis, Dept. of Physi s and

Astronomy,UniversityofRo hester,Ro hester, N.Y.,2001.

[7℄ E.WolfandT.Habashy,Invisiblebodiesanduniquenessoftheinverses atteringproblem, J.Modern

Opt.,vol.40, pp.785-792,1993.

[8℄ A.J.Devaney,Nonuniquenessintheinverses atteringproblem, J.Math.Phys.,vol.19,pp.1526-1531,

1978.

[9℄ A. J. Devaney and G. C. Sherman,Nonuniqueness in inversesour e and s attering problems, IEEE

Trans.AntennasPropagat.,vol.AP-30,no.5,Sep.1982.

[10℄ A.Gamliel,K.Kim,A.I.Na hman,andE.Wolf,Anewmethodforspe ifyingnonradiating,