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Non-destructive testing of dielectric layers with defects

C Tasdemir O Mudanyali S Yıldız O Semerci and A Yapar

Istanbul Technical University Electrical and Electronics Engineering Faculty Maslak 34469 Istanbul Turkey

caglatasdemirgmailcom mudanyaliituedutr yildizseldagmailcom semerciogituedutr yaparaituedutr

Abstract A microwave imaging method for non-destructive testing of dielectric surfaces beyond a layered media is presented The method is based on the analytical continuation of the measured data to the surface under test through a special representation of the scattered field in terms of Fourier transform and Taylor expansion Then the problem is reduced to the solution of a coupled system of non-linear integral equations which is solved iteratively via the Newton method with regularization in the least square sense Numerical simulations show that defects having sizes in the order of λ200 can be successfully recovered through the presented algorithm

1 Introduction Non-destructive testing (NDT) of materials is an important subject in the inverse scattering theory due to its wide range of practical applications such as but not limited to automotive industry medicine aerospace engineering construction etc In a typical NDT problem the structure under test is excited by a certain type of field or wave and the reaction is measured on a region (usually non-contact to the structure) to extract the desired properties of the material under test According to the physical configuration number of methodologies have been developed such as magnetic particle method eddy current method ultrasonics visual-optical methods infrared thermography acoustics electromagnetics etc [1-5] In electromagnetic applications microwave signals are capable of penetrating inside the dielectric media allowing the inspection of the surfaces which are not reachable or not tend to be destructed to test These applications are very important especially in the areas of detection of the mechanical damages irregularities or cracks on coated surfaces of vehicles or on dielectric surfaces beyond layered media

In this study a method to determine the location and the shape of damages irregularities cracks etc on a dielectric surface separating two lossy dielectric media beyond a dielectric half space is presented For the sake of simplicity we consider surfaces having variation only in one space dimension A single illumination of plane electromagnetic wave with a fixed frequency is used for excitation and the scattered field measurements are performed on a line parallel to the boundary of the upper half space The method is based on a special representation of the scattered field in each region where the Fourier transform and Taylor expansion are used together Then the problem is reduced to the solution of a non-linear algebraic equation which is solved iteratively by the use of classical Newton Method The presented method is tested by some numerical simulations and satisfactory results are obtained

6th International Conference on Inverse Problems in Engineering Theory and Practice IOP PublishingJournal of Physics Conference Series 135 (2008) 012096 doi1010881742-65961351012096

ccopy 2008 IOP Publishing Ltd 1

2 NDT Method Consider the problem illustrated in Figure 1

Figure 1 In this configuration 0Γ is the destructed surface which is lying between two layers with

electromagnetic parameters 1 1 1 ε micro σ and 2 2 2 ε micro σ respectively The upper half space is assumed to be free space The surface under test can be a flat or a rough one which is represented by a single-valued and continuous function ( )2 1x f x= 0Γ is assumed to be locally rough ie ( )1f x differs

from zero over a finite interval which has a length of 0L The main aim of the non-destructive testing problem considered here is to reconstruct the possible defects 1 2 D D on the surface through a set of scattered electromagnetic field measurements performed in the accessible domain 2x hgt where h represents the boundary between the layers above the surface under test The measurements have also been performed in the same region For the sake of simplicity it will be assumed that the incident field is a TM polarized time-harmonic plane wave whose electric field vector is given by ( )( )1 200 i iE u x x= with ( ) ( )0 1 0 2 0cos sin

1 2 ik x xiu x x e φ φminus += where 0φ is the incident angle

while 0 0 0 0k ω ε micro= Due to the homogenity in the 3x direction the total and the scattered field

vectors will have only 3x components and the problem is reduced to a scalar one in terms of the total

field function ( )u x which satisfies the Helmholtz equation

2 0u k u∆ + = (21)

where 2

0 2

21 2 1

22 2

2

1

( )

( )

( )k x hk h x f xk x f x

k x =

gt

gt gt gt

` (22)

D1 D2

D3 D4 D6

DN

x1

Measurement Line

φ0

Incident Field

C O A T I N G S

ℓ

Γ0

ε0 micro0 0σ

ε1 micro1 1σ

ε2 micro2 2σ

x2

D5

h 1B

2B

3B

β

αSurface under Test

0


2

In order to formulate the problem in an appropriate way we first decompose the total field as

1

2 1

3 1

2

2

2

( )

( )

( ) ( )( ) ( )

( )

s i

s

s

f xf x

u x u x x hu x u x h x

u x x

gt lt

+ gt

= gt

(23)

where the functions 1

su 2su and 3

su are the contributions of the defects andor the roughness of the

surface to the total field in the regions 2x hgt ( )( )2 1 x f x hisin and ( )2 1x f xlt respectively The boundary conditions imposed on the total field yield

1 2 2 s i su u u x h+ =on = (24)

( )1 22

2 2

s i su ux hu

x x+

= =part part

part parton (25)

2 3 2 1( )s su u x f x=on = (26)

322 1

2 2( )

ss

x f xuux x

= =partpart

part parton (27)

under the appropriate radiation condition for 2 x rarr infin

In the following we will give a special representation of the scattered field by the use of Fourier Transform and Taylor expansion To this aim let us first define the Fourier Transform of 1

su as

11 2 1 2 1 2ˆ ( ) ( ) i xs su x u x x e dx x hνν

infinminus

minusinfin

= gtint (28)

The Fourier transform of the reduced wave equation for 1

su yields

21

0 1 222

ˆ ˆ 0s

sd u u x hdx

γminus = gt (29)

where 2 20 0( ) kγ ν ν= minus is the square root function defined in the complex cut ν -plane as

0 0(0) ikγ = minus The solution of (29) can be given as

0 21 2 2ˆ ( ) ( ) xsu x A e x hγν ν minus= gt (210)

by taking the radiation condition into account Here ( )A ν is the unknown spectral coefficient Then

by applying the inverse Fourier Transform one can express 1su as


3

( ) ( ) ( )1 0 21 2

1 2

i x v xsu x A e d x hν γν νπ

infinminus

minusinfin

= gtint (211)

The same procedure can be easily applied for the scattered fields in the regions ( )2 x hβisin and

2x αlt in which ( )( )1max f xβ ge and ( )( )1min f xα le (see Figure 1) where there is no

discontinuity in the 1x -direction Thus

( ) ( ) ( ) ( ) ( )( )1 2 1 2 12 2

1 2

x x i xsu x B e C e e d x hγ ν γ ν νν ν ν βπ

infinminus

minusinfin

= + lt ltint (212)

( ) ( ) ( )1 2 23 2

1 2

i x xsu x D e d xν γ νν ν απ

infin+

minusinfin

= ltint (213)

where 2 21 1( ) kγ ν ν= minus and 2 2

2 2( ) kγ ν ν= minus with 1 1(0) ikγ = minus and 2 2(0) ikγ = minus while

( ) ( )B Cν ν and ( )D ν are the spectral coefficients to be determined Now assume that the scattered field is measured on a line which is parallel to the layers in the

upper half space ie ( )1 1 su x l l h gt is known for all 1x Risin Inserting 2x l= into (211) we observe

that in agreement with (210) the spectral coefficient ( )A ν can be determined from the Fourier transform via

( ) ( ) ( )01 sA u eγ νν ν= (214)

Since now ( )A ν is known using boundary conditions (24) and (25) on 2x h= one can obtain

the unknown spectral coefficients ( )B ν and ( )C ν related to the region 12 ( ( ) )f x hx isin from the following equations

( ) ( ) ( ) ( ) ( ) ( )( )1 1 11 1 1

1 2

h h i xsiu x h u x h B e C e e dγ ν γ ν νν ν ν

π

infinminus

minusinfin

+ = +int (215)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )1 1 1

22

1 1 2 1 21 1

2 2

1 2

sh hi i x

x hx h

u x x u x xB e C e e d

x xγ ν γ ν νν γ ν ν γ ν ν

π

infinminus

minusinfin==

part part+ = minus +

part part int

(216) The regularized solution of ( )B ν and ( )C ν in the sense of Tikhonov are obtained by discretizing

the equations (215) and (216) To be able to find approximate expressions for the scattered field in the regions ( )( )2 1 x f x βisin

and ( )( )2 1x f xαisin we use the Taylor expansions of the scattered field

( ) ( ) ( ) ( ) ( )2 12 2 1 2

0 2

1

m sMms

mmm

u xu x x R x f x x

m xβ

β β=

part= minus + lt le

partsum (217)


4

( ) ( ) ( ) ( ) ( )3 13 2 2 1

0 2

1

m sNms

Nmm

u xu x x Q x x f x

m xα

α α=

part= minus + le lt

partsum (218)

where the remainder terms are

( ) ( ) ( )2 12 1

2 12

1

x M sM

M M

u xR x x d

M xβ

ξξ ξ

+

+

part= minus

partint (219)

( ) ( ) ( )2 13 1

2 12

1

x N sN

N N

u xQ x x d

N xα

ξξ ξ

+

+

part= minus

partint (220)

The m th order derivatives appering in (217) and (218) can be obtained in the form of

( ) ( ) ( ) ( ) ( ) ( ) ( )( )1 1 12 1

1 12

1 [ ] [ ]2

m si xm m

m

u xB e C e e d

xγ ν β γ ν β νβ

γ ν ν γ ν ν νπ

infinminus

minusinfin

part= minus +

part int (221)

( ) ( ) ( ) ( )( )2 13 12

2

1 [ ] 2

m si xm

m

u xD e e d

xγ ν α να

γ ν ν νπ

infin

minusinfin

part=

part int (222)

This representation of the scattered field together with the boundary conditions (26) and (27)

reduces the problem to a system of nonlinear equations which comprises the spectral coefficient ( )D ν related to the scattered field 3

su and the variation of the rough surface 1( )f x Now we proceed by substituting the pairs (217) and (218) into the boundary conditions (26) and

(27) and by neglecting the remainder terms to obtain

( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2

i x i xD f x e d B f x C f x e dν νν ν ν ν ν ν ν νπ π

infin infin

minusinfin minusinfin

Φ = Φ + Φint int (223)

( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin


Ψ = Ψ + Ψint int (224)

where

( )( ) ( ) ( ) ( )( )1 11 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=

minusΦ = minussum (225)

( )( ) ( ) ( )( ) ( )( )1

111 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminusminus

=

minusΨ = minus

minussum (226)

( )( ) ( ) ( ) ( )( )1 12 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν β=

Φ = minussum (227)

( )( ) ( ) ( )( ) ( )( )1

112 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=

Ψ = minusminussum (228)

( )( ) ( ) ( ) ( )( )2 23 1 1

0

[ ]

mN m

mf x e f x

mγ ν α γ ν

ν α=

Φ = minussum (229)


5

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=


A more compact expression for the system above is given by the following operator equations

( ) ( )1 1( )K D f g fν = (231)

( ) ( )2 2( )K D f g fν = (232) In (231) and (232) 1K and 2K are non-linear operators with respect to 1( )f x while they are linear with respect to ( )D ν

Thus now the problem is reduced to the solution of this non-linear system which can be treated by iterative techniques In the application of the iterative scheme we first choose an initial guess for the unknown surface variation f Using this initial guess it is now easy to solve one of the equations mentioned above to obtain the spectral coefficient ( )D ν Note that since both integral equations given by (231) and (232) are of the first kind one has to apply some regularization techniques Here again Tikhonov regularization is applied Once we have obtained the unknown spectral coefficient from one of the equations say the first one surface variation f can be obtained by solving the other non-linear equation which can also be written in an operator form

( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)

The latter one can be solved iteratively via Newton method To this aim for an initial guess 0f the nonlinear equation is replaced by the linearized one

( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)

that needs to be solved for 0f f f∆ = minus in order to improve an approximate boundary 0Γ given by the function 0f into a new approximation with surface function 0f f+ ∆ In (234) MF prime denotes the Frechet derivative of the operator F with respect to f It can be shown that MF prime reduces the ordinary derivative of MF with respect to f The Newton method consists in iterating this procedure ie in solving

( ) ( )0 0 01 23M MF f f F f iprime ∆ = minus = for 1 1i i if f f+ +∆ = + ∆ (235)

It is obvious that this solution will be sensitive to errors in the derivative of MF in the vicinity of

zeros To obtain a more stable solution the unknown f∆ is expressed in terms of some basis functions ( )1 1n x n Nφ = as a linear combination

( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6

A possible choice of basis functions consists of trigonometric polynomials Then (235) is satisfied in the least squares sense that is the coefficients 1 Na a in (236) are determined such that for a set

of grid points 11 1 Jx x the sum of squares

( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j

M n n Mj nF f x a x F f xφ

= =

+sum sum (237)

is minimized The number of basis functions N in (236) can also be considered as a kind of regularization parameter Choosing N too big leads to instabilities due to the ill-posedness of the underlying inverse problem accordingly too small values of N results in poor approximation quality Hence one has to compromise between stability and accuracy and in this sense N serves as a regularization parameter

3 Numerical Results In this section some numerical results which demonstrate the validity and effectiveness of the method will be presented In all the examples the upper space where the sources and observation points are located is assumed to be free-space and the operating frequency is chosen as 12 GHz 1 random noise is added to the simulated data for the scattered field In the application of least squares solution the basis functions are chosen as combinations of ( )1 0cos 2 nx Lπ and ( )1 0sin 2 nx Lπ

0 1n = plusmn plusmnΝ and the number N is determined by trial and error In the first example the dielectric surface is located above a non-magnetic painting material having

electromagnetic parameters 2 07ε ε= 42 10σ minus= and below a non-magnetic painting material having

electromagnetic parameters 1 04ε ε= 41 10σ minus= The reconstruction of the circular defects on a planar

surface shown in Figure 2 is obtained for the truncation number 5M = in the Taylor expansion for 2 iterations The method determines the locations and the shapes of the defects having depths in the order of 200λ very accurately The results given in the Figure 3 shows that the method can be effectively used for reconstruction of the defects on curved surfaces It shows the reconstruction obtained for the truncation numbers M=5 with 3 iterations

Figure 2 Reconstruction of defects on planar surface


7

Figure 3 Reconstruction of defects on curved surface

4 Conclusion The method presented in [6] is extended to the non-destructive evaluation of the dielectric surfaces beyond a layered media Although the method is developed for only two layers it can be extended easily to the multilayered cases The method is very effective for defects having sizes in order of

200λ for an operating frequency of 12 GHz Future studies are devoted to extend the method for 2D surfaces

References [1] J Mohammadi Non-destructive Test (NDT) Methods Applied Fatigue Reliability Assessment

of Structures J Mohammadi Editor ASCE Publications Reston VA USA [2] SI Ganchev NQaddoumi E Ranu and R Zoughi rdquo Microwave detection optimization of

disbond in layered dielectrics with varying thicknessrdquo IEEE Trans on Instrumentation and Measurement Vol44 No2 pp326- 328 1995

[3] JNadakuduti GChen and RZoughi Semiemprical Electromagnetic Modeling of Crack Dedection and Sizing in Cement-based Materials using Near-Field Microwave Methods IEEE Trans on Instrumantation and Measurement Vol55 No2 pp588-597 2006

[4] AMassa MPastorino ARosani MBenedetti A Microwave Imaging Method for NDENDT based on the SMW Technique for the Electromagnetic Field Prediction IEEE Trans on Instrumantation and Measurement Vol55 No1 pp240-247 2006

[5] MBenedetti MDonelli AMartini MPastorino ARosani AMassa An Innovative Microwave-Imaging Technique for Nondestructive Evaluation Applications to Civil Structures monitoring and Biological Bodies Inspection IEEE Trans on Instrumantation and Measurement Vol55 No6 pp1878-1884 2006

[6] I Akduman R Kress and A Yapar ldquoIterative Reconstruction of Dielectric Rough Surface Profiles at Fixed Frequencyrdquo Inverse Problems Vol22 939-954 2006


8

Non-destructive testing of dielectric layers with defects

C Tasdemir O Mudanyali S Yıldız O Semerci and A Yapar

Istanbul Technical University Electrical and Electronics Engineering Faculty Maslak 34469 Istanbul Turkey

caglatasdemirgmailcom mudanyaliituedutr yildizseldagmailcom semerciogituedutr yaparaituedutr

Abstract A microwave imaging method for non-destructive testing of dielectric surfaces beyond a layered media is presented The method is based on the analytical continuation of the measured data to the surface under test through a special representation of the scattered field in terms of Fourier transform and Taylor expansion Then the problem is reduced to the solution of a coupled system of non-linear integral equations which is solved iteratively via the Newton method with regularization in the least square sense Numerical simulations show that defects having sizes in the order of λ200 can be successfully recovered through the presented algorithm

1 Introduction Non-destructive testing (NDT) of materials is an important subject in the inverse scattering theory due to its wide range of practical applications such as but not limited to automotive industry medicine aerospace engineering construction etc In a typical NDT problem the structure under test is excited by a certain type of field or wave and the reaction is measured on a region (usually non-contact to the structure) to extract the desired properties of the material under test According to the physical configuration number of methodologies have been developed such as magnetic particle method eddy current method ultrasonics visual-optical methods infrared thermography acoustics electromagnetics etc [1-5] In electromagnetic applications microwave signals are capable of penetrating inside the dielectric media allowing the inspection of the surfaces which are not reachable or not tend to be destructed to test These applications are very important especially in the areas of detection of the mechanical damages irregularities or cracks on coated surfaces of vehicles or on dielectric surfaces beyond layered media

In this study a method to determine the location and the shape of damages irregularities cracks etc on a dielectric surface separating two lossy dielectric media beyond a dielectric half space is presented For the sake of simplicity we consider surfaces having variation only in one space dimension A single illumination of plane electromagnetic wave with a fixed frequency is used for excitation and the scattered field measurements are performed on a line parallel to the boundary of the upper half space The method is based on a special representation of the scattered field in each region where the Fourier transform and Taylor expansion are used together Then the problem is reduced to the solution of a non-linear algebraic equation which is solved iteratively by the use of classical Newton Method The presented method is tested by some numerical simulations and satisfactory results are obtained


ccopy 2008 IOP Publishing Ltd 1









2 0u k u∆ + = (21)

where 2

0 2

21 2 1

22 2

2

1

( )

( )


k x =

gt

gt gt gt

` (22)

D1 D2

D3 D4 D6

DN

x1

Measurement Line

φ0

Incident Field

C O A T I N G S

ℓ

Γ0

ε0 micro0 0σ

ε1 micro1 1σ

ε2 micro2 2σ

x2

D5

h 1B

2B

3B

β


0


2


1

2 1

3 1

2

2

2

( )

( )

( ) ( )( ) ( )

( )

s i

s

s

f xf x


u x x

gt lt

+ gt

= gt

(23)


su 2su and 3



1 2 2 s i su u u x h+ =on = (24)

( )1 22

2 2

s i su ux hu

x x+

= =part part

part parton (25)

2 3 2 1( )s su u x f x=on = (26)

322 1

2 2( )

ss

x f xuux x

= =partpart

part parton (27)



su as


infinminus

minusinfin

= gtint (28)


su yields

21

0 1 222

ˆ ˆ 0s

sd u u x hdx

γminus = gt (29)







3

( ) ( ) ( )1 0 21 2

1 2


infinminus

minusinfin

= gtint (211)




( ) ( ) ( ) ( ) ( )( )1 2 1 2 12 2

1 2


infinminus

minusinfin

= + lt ltint (212)

( ) ( ) ( )1 2 23 2

1 2


infin+

minusinfin

= ltint (213)






( ) ( ) ( )01 sA u eγ νν ν= (214)



( ) ( ) ( ) ( ) ( ) ( )( )1 1 11 1 1

1 2


π

infinminus

minusinfin

+ = +int (215)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )1 1 1

22

1 1 2 1 21 1

2 2

1 2

sh hi i x

x hx h



π

infinminus

minusinfin==


part part int




( ) ( ) ( ) ( ) ( )2 12 2 1 2

0 2

1

m sMms

mmm

u xu x x R x f x x

m xβ

β β=

part= minus + lt le

partsum (217)


4

( ) ( ) ( ) ( ) ( )3 13 2 2 1

0 2

1

m sNms

Nmm

u xu x x Q x x f x

m xα

α α=

part= minus + le lt

partsum (218)


( ) ( ) ( )2 12 1

2 12

1

x M sM

M M

u xR x x d

M xβ

ξξ ξ

+

+

part= minus

partint (219)

( ) ( ) ( )2 13 1

2 12

1

x N sN

N N

u xQ x x d

N xα

ξξ ξ

+

+

part= minus

partint (220)


( ) ( ) ( ) ( ) ( ) ( ) ( )( )1 1 12 1

1 12

1 [ ] [ ]2

m si xm m

m

u xB e C e e d



infinminus

minusinfin

part= minus +

part int (221)

( ) ( ) ( ) ( )( )2 13 12

2

1 [ ] 2

m si xm

m

u xD e e d

xγ ν α να

γ ν ν νπ

infin

minusinfin

part=

part int (222)





( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



where

( )( ) ( ) ( ) ( )( )1 11 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( )( ) ( )( )1

111 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminusminus

=

minusΨ = minus

minussum (226)

( )( ) ( ) ( ) ( )( )1 12 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν β=

Φ = minussum (227)

( )( ) ( ) ( )( ) ( )( )1

112 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( ) ( )( )2 23 1 1

0

[ ]

mN m

mf x e f x

mγ ν α γ ν

ν α=

Φ = minussum (229)


5

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=



( ) ( )1 1( )K D f g fν = (231)



( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)


( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)





( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8









2 0u k u∆ + = (21)

where 2

0 2

21 2 1

22 2

2

1

( )

( )


k x =

gt

gt gt gt

` (22)

D1 D2

D3 D4 D6

DN

x1

Measurement Line

φ0

Incident Field

C O A T I N G S

ℓ

Γ0

ε0 micro0 0σ

ε1 micro1 1σ

ε2 micro2 2σ

x2

D5

h 1B

2B

3B

β


0


2


1

2 1

3 1

2

2

2

( )

( )

( ) ( )( ) ( )

( )

s i

s

s

f xf x


u x x

gt lt

+ gt

= gt

(23)


su 2su and 3



1 2 2 s i su u u x h+ =on = (24)

( )1 22

2 2

s i su ux hu

x x+

= =part part

part parton (25)

2 3 2 1( )s su u x f x=on = (26)

322 1

2 2( )

ss

x f xuux x

= =partpart

part parton (27)



su as


infinminus

minusinfin

= gtint (28)


su yields

21

0 1 222

ˆ ˆ 0s

sd u u x hdx

γminus = gt (29)







3

( ) ( ) ( )1 0 21 2

1 2


infinminus

minusinfin

= gtint (211)




( ) ( ) ( ) ( ) ( )( )1 2 1 2 12 2

1 2


infinminus

minusinfin

= + lt ltint (212)

( ) ( ) ( )1 2 23 2

1 2


infin+

minusinfin

= ltint (213)






( ) ( ) ( )01 sA u eγ νν ν= (214)



( ) ( ) ( ) ( ) ( ) ( )( )1 1 11 1 1

1 2


π

infinminus

minusinfin

+ = +int (215)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )1 1 1

22

1 1 2 1 21 1

2 2

1 2

sh hi i x

x hx h



π

infinminus

minusinfin==


part part int




( ) ( ) ( ) ( ) ( )2 12 2 1 2

0 2

1

m sMms

mmm

u xu x x R x f x x

m xβ

β β=

part= minus + lt le

partsum (217)


4

( ) ( ) ( ) ( ) ( )3 13 2 2 1

0 2

1

m sNms

Nmm

u xu x x Q x x f x

m xα

α α=

part= minus + le lt

partsum (218)


( ) ( ) ( )2 12 1

2 12

1

x M sM

M M

u xR x x d

M xβ

ξξ ξ

+

+

part= minus

partint (219)

( ) ( ) ( )2 13 1

2 12

1

x N sN

N N

u xQ x x d

N xα

ξξ ξ

+

+

part= minus

partint (220)


( ) ( ) ( ) ( ) ( ) ( ) ( )( )1 1 12 1

1 12

1 [ ] [ ]2

m si xm m

m

u xB e C e e d



infinminus

minusinfin

part= minus +

part int (221)

( ) ( ) ( ) ( )( )2 13 12

2

1 [ ] 2

m si xm

m

u xD e e d

xγ ν α να

γ ν ν νπ

infin

minusinfin

part=

part int (222)





( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



where

( )( ) ( ) ( ) ( )( )1 11 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( )( ) ( )( )1

111 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminusminus

=

minusΨ = minus

minussum (226)

( )( ) ( ) ( ) ( )( )1 12 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν β=

Φ = minussum (227)

( )( ) ( ) ( )( ) ( )( )1

112 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( ) ( )( )2 23 1 1

0

[ ]

mN m

mf x e f x

mγ ν α γ ν

ν α=

Φ = minussum (229)


5

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=



( ) ( )1 1( )K D f g fν = (231)



( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)


( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)





( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8


1

2 1

3 1

2

2

2

( )

( )

( ) ( )( ) ( )

( )

s i

s

s

f xf x


u x x

gt lt

+ gt

= gt

(23)


su 2su and 3



1 2 2 s i su u u x h+ =on = (24)

( )1 22

2 2

s i su ux hu

x x+

= =part part

part parton (25)

2 3 2 1( )s su u x f x=on = (26)

322 1

2 2( )

ss

x f xuux x

= =partpart

part parton (27)



su as


infinminus

minusinfin

= gtint (28)


su yields

21

0 1 222

ˆ ˆ 0s

sd u u x hdx

γminus = gt (29)







3

( ) ( ) ( )1 0 21 2

1 2


infinminus

minusinfin

= gtint (211)




( ) ( ) ( ) ( ) ( )( )1 2 1 2 12 2

1 2


infinminus

minusinfin

= + lt ltint (212)

( ) ( ) ( )1 2 23 2

1 2


infin+

minusinfin

= ltint (213)






( ) ( ) ( )01 sA u eγ νν ν= (214)



( ) ( ) ( ) ( ) ( ) ( )( )1 1 11 1 1

1 2


π

infinminus

minusinfin

+ = +int (215)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )1 1 1

22

1 1 2 1 21 1

2 2

1 2

sh hi i x

x hx h



π

infinminus

minusinfin==


part part int




( ) ( ) ( ) ( ) ( )2 12 2 1 2

0 2

1

m sMms

mmm

u xu x x R x f x x

m xβ

β β=

part= minus + lt le

partsum (217)


4

( ) ( ) ( ) ( ) ( )3 13 2 2 1

0 2

1

m sNms

Nmm

u xu x x Q x x f x

m xα

α α=

part= minus + le lt

partsum (218)


( ) ( ) ( )2 12 1

2 12

1

x M sM

M M

u xR x x d

M xβ

ξξ ξ

+

+

part= minus

partint (219)

( ) ( ) ( )2 13 1

2 12

1

x N sN

N N

u xQ x x d

N xα

ξξ ξ

+

+

part= minus

partint (220)


( ) ( ) ( ) ( ) ( ) ( ) ( )( )1 1 12 1

1 12

1 [ ] [ ]2

m si xm m

m

u xB e C e e d



infinminus

minusinfin

part= minus +

part int (221)

( ) ( ) ( ) ( )( )2 13 12

2

1 [ ] 2

m si xm

m

u xD e e d

xγ ν α να

γ ν ν νπ

infin

minusinfin

part=

part int (222)





( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



where

( )( ) ( ) ( ) ( )( )1 11 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( )( ) ( )( )1

111 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminusminus

=

minusΨ = minus

minussum (226)

( )( ) ( ) ( ) ( )( )1 12 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν β=

Φ = minussum (227)

( )( ) ( ) ( )( ) ( )( )1

112 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( ) ( )( )2 23 1 1

0

[ ]

mN m

mf x e f x

mγ ν α γ ν

ν α=

Φ = minussum (229)


5

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=



( ) ( )1 1( )K D f g fν = (231)



( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)


( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)





( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8

( ) ( ) ( )1 0 21 2

1 2


infinminus

minusinfin

= gtint (211)




( ) ( ) ( ) ( ) ( )( )1 2 1 2 12 2

1 2


infinminus

minusinfin

= + lt ltint (212)

( ) ( ) ( )1 2 23 2

1 2


infin+

minusinfin

= ltint (213)






( ) ( ) ( )01 sA u eγ νν ν= (214)



( ) ( ) ( ) ( ) ( ) ( )( )1 1 11 1 1

1 2


π

infinminus

minusinfin

+ = +int (215)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )1 1 1

22

1 1 2 1 21 1

2 2

1 2

sh hi i x

x hx h



π

infinminus

minusinfin==


part part int




( ) ( ) ( ) ( ) ( )2 12 2 1 2

0 2

1

m sMms

mmm

u xu x x R x f x x

m xβ

β β=

part= minus + lt le

partsum (217)


4

( ) ( ) ( ) ( ) ( )3 13 2 2 1

0 2

1

m sNms

Nmm

u xu x x Q x x f x

m xα

α α=

part= minus + le lt

partsum (218)


( ) ( ) ( )2 12 1

2 12

1

x M sM

M M

u xR x x d

M xβ

ξξ ξ

+

+

part= minus

partint (219)

( ) ( ) ( )2 13 1

2 12

1

x N sN

N N

u xQ x x d

N xα

ξξ ξ

+

+

part= minus

partint (220)


( ) ( ) ( ) ( ) ( ) ( ) ( )( )1 1 12 1

1 12

1 [ ] [ ]2

m si xm m

m

u xB e C e e d



infinminus

minusinfin

part= minus +

part int (221)

( ) ( ) ( ) ( )( )2 13 12

2

1 [ ] 2

m si xm

m

u xD e e d

xγ ν α να

γ ν ν νπ

infin

minusinfin

part=

part int (222)





( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



where

( )( ) ( ) ( ) ( )( )1 11 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( )( ) ( )( )1

111 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminusminus

=

minusΨ = minus

minussum (226)

( )( ) ( ) ( ) ( )( )1 12 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν β=

Φ = minussum (227)

( )( ) ( ) ( )( ) ( )( )1

112 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( ) ( )( )2 23 1 1

0

[ ]

mN m

mf x e f x

mγ ν α γ ν

ν α=

Φ = minussum (229)


5

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=



( ) ( )1 1( )K D f g fν = (231)



( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)


( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)





( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8

( ) ( ) ( ) ( ) ( )3 13 2 2 1

0 2

1

m sNms

Nmm

u xu x x Q x x f x

m xα

α α=

part= minus + le lt

partsum (218)


( ) ( ) ( )2 12 1

2 12

1

x M sM

M M

u xR x x d

M xβ

ξξ ξ

+

+

part= minus

partint (219)

( ) ( ) ( )2 13 1

2 12

1

x N sN

N N

u xQ x x d

N xα

ξξ ξ

+

+

part= minus

partint (220)


( ) ( ) ( ) ( ) ( ) ( ) ( )( )1 1 12 1

1 12

1 [ ] [ ]2

m si xm m

m

u xB e C e e d



infinminus

minusinfin

part= minus +

part int (221)

( ) ( ) ( ) ( )( )2 13 12

2

1 [ ] 2

m si xm

m

u xD e e d

xγ ν α να

γ ν ν νπ

infin

minusinfin

part=

part int (222)





( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



( ) ( )( ) ( ) ( )( ) ( ) ( )( )( )1 13 1 1 1 2 1

1 1 2 2


infin infin



where

( )( ) ( ) ( ) ( )( )1 11 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( )( ) ( )( )1

111 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminusminus

=

minusΨ = minus

minussum (226)

( )( ) ( ) ( ) ( )( )1 12 1 1

0

[ ]

mM m

mf x e f x

mγ ν β γ ν

ν β=

Φ = minussum (227)

( )( ) ( ) ( )( ) ( )( )1

112 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν βminus

=


( )( ) ( ) ( ) ( )( )2 23 1 1

0

[ ]

mN m

mf x e f x

mγ ν α γ ν

ν α=

Φ = minussum (229)


5

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=



( ) ( )1 1( )K D f g fν = (231)



( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)


( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)





( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8

( )( ) ( ) ( )( ) ( )( )2

123 1 1

1

[ ]

1

mM m

mf x e f x

mγ ν β γ ν

ν αminus

=



( ) ( )1 1( )K D f g fν = (231)



( ) ( ) ( )2 2 0MF f K D f g f= minus = (233)


( ) ( )0 0 0M MF f F f fprime+ ∆ = (234)





( ) ( )1 11

N

n nn

f x a xφ=

∆ = sum (236)


6



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8



( )( ) ( ) ( )( )2

1 1 1

1 1

J Nj j j


= =

+sum sum (237)









7












8












8

non-destructive testing of dielectric layers with defects - iopscience

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