habilitation - chapitre 1 · 2 • c2 and ρ2 are the velocity and density of the second fluid 2 ;...

Chapter IV - The geometric approximation of the transmission through a plane interface

Chapter IV

The geometric approximation of the transmission through a plane interface

I) Introduction

In this chapter, we use the same notations as in the previous chapter:

• c1 and ρ1 are the velocity and density of the first fluid 1 ( ), K c= ρ1 12

• and c2 ρ2 are the velocity and density of the second fluid 2 ; in the case of a solid material,

the notations are and c for the longitudinal and transverse speeds respectively (cl t λ and

μ are the corresponding Lamé’s coefficients),

• and cij ρij are dimensionless coefficients defined by c c cij i j= / and ρ ρ ρij i j= / ,

• f, f x and f y are the temporal and spatial frequencies, related to time t and spatial

coordinates x and y through Fourier transforms.

II) The stationary phase theorem

In this section, we only remind some classical results from the Signal Processing Theory ; we

do not try to prove the formulas, neither want to discuss their validity conditions.31

If is a continuous function at , we have for c small enough: ( )r t t0

( )4 1. a ( ) ( )[ ] ( )10

2 20c j

r t j t t c dt r tπ

exp / .− ≈−∞

+∞

∫

From this expression, we can prove the stationary phase theorem: if is a continuous

function,

( )r t( )ϕ t a continuous function with a single minimum or maximum c such that

( )ϕ ' c = 0 and ( )ϕ ' ' c ≠ 0 , and k > 0:

1


( )4 1. b ( ) ( )[ ] ( ) ( )[ ]

( )( )

( )

( )( )

( )⎪⎪⎩

⎪⎪⎨

⎧

<−−

>×≈∫

∞+

∞−.0'' if ,

''4/exp

,0'' if ,''

4/exp

exp2expc

ckj

cck

j

cjkcrdttjktrϕ

ϕπ

ϕϕπ

ϕπϕ

The general principle of the stationary phase theorem is based of the following idea: the

function varies slowly compared to the phase ( )r t ( )ϕ t ; however, if the phase varies from

0ϕ to πϕ 20 + , the resulting integral vanishes ; as a consequence, only the points

corresponding to a stationary phase (with a zero derivative) have a significant contribution to

the final integral.

Equation (4.1b) can be generalized to the particular case of an integral over the interval [ ]a b, ,

with a phase function ( )ϕ t presenting N different single-order minimums or maximums

, in this interval: c ii , 1≤ ≤ N

( )4 1. c ( ) ( )[ ] ( ) ( )[ ] ( )( )

r t jk t dt r c jk cj

k ca

b

i ii

ii

N

exp expexp /

' ',ϕ π ϕ

ε π

ϕ∫ ∑≈

=2

4

1

where ( )[ ]ii c''sign1 ϕε =±= .

III) Transmission through a plane interface between two fluids

1. Computation of the transmitted field

We are interested in the same problem as in the previous chapter: a plane interface separating

two homogeneous fluids is located in the plane z h= > 0 ; a point-like source is in fluid 1 at

x y z= = = 0 and we want to determine the acoustic field transmitted on the other side of the

interface in fluid 2, in a plane located at a distance d from the interface.

The configuration of interest is illustrated on figure 1.

The basic principle for the computation of the transmitted field is unchanged ; it can be

summarized as follows:

• the incident field is decomposed into plane monochromatic waves using Fourier transforms

over time and space,

2


• then, the reflected and transmitted fields are written using the reflection and transmission

coefficients,

• the reflection and transmission coefficients are obtained from the boundary conditions to

be satisfied in the plane corresponding to the interface,

• we finally return to the real space using inverse Fourier transforms over frequency and

spatial frequencies and . f x f y

Plane interface

Point-like source

h

r

d

Observer

Fluid 1

Fluid 2

z axis

Figure 1: geometry of the problem of interest

This problem has already been addressed in the previous chapter ; thus, we do not give here

the details of the different computation steps. The only difference is that the problem will be

written here in terms of the velocity potentials instead of the acoustic pressures.

The incident field in terms of the velocity potential can be written for monochromatic waves

(frequency f) as follows

( )4 1. a ( ) ( )~ , , ,exp /

,φππi x y z f

j fR cR

=24

1

3


where R x y z= + +2 2 2 . Taking the Fourier transform of (4.1a) over the spatial coordinates

x and y, we obtain using the same notations as in the previous chapter (this expression is only

valid near the interface and for positive values of the frequency f):

( )4 1. b ( ) ( )~ , , , expΦi x yf f z fj

j z=2 1

1νν .

Similarly, the reflected and transmitted fields can be written in terms of the velocity potentials

as follows:

( )4 1. c ( ) ( )

( ) ( )

~ , , , exp

~ , , , exp ,

Φ

Φ

r x y

t x y

f f z fj

R j

f f z fj

T j z

= −

=

⎧

⎨⎪⎪

⎩⎪⎪

2

2

111 1

112 2

νν

νν

,z

where and are the reflection and transmission coefficients, respectively. Using the

boundary conditions to be satisfied in the plane corresponding to the interface (the acoustic

pressure and normal displacement must be continuous), one can obtain:

R11 T12

( )4 1. d ( )

( )[ ] ( ) ( )[ ]R j h

T j h j

112 1 1 2

2 1 1 21

121 1

2 1 1 21 2 12 1 2 1 2

2

2

=−+

=+

− = −

⎧

⎨⎪⎪

⎩⎪⎪

ρ ν ρ νρ ν ρ ν

ν

ρ νρ ν ρ ν

ν ν ν ν ν ν

exp ,

exp , exp .T h

)

The expression of the transmission coefficient is different from the one written in the previous

chapter because the problem is described in terms of the velocity potentials instead of the

acoustic pressures (in fact, the difference reduces to an additional density ratio). The transmitted monochromatic field in fluid 2 can be obtained using an inverse Fourier

transform over the spatial frequencies and : f x f y

(4 2. ( ) ( ) ( ) ( )

( ) ( )

~ , , , , exp exp

exp exp .

φν

ν ν ν ν

π π

t

x y

x y z fj

j h j d

j f x j f y df df

= ×

− −

−∞

+∞

−∞

+∞

∫∫ 22 2

112 1 2 1 2T

x y

4


Due to the cylindrical symmetry of the problem, the transmitted field ( )~ , , ,φt x y z f only

depends on r x y= +2 2 . We now have to evaluate the integral given by equation (4.2) in the

frame of the geometric ray approximation.

2. The ray model approximation

The basic principle of the geometric ray model approximation consists in a simplification of

the integral given in equation (4.2), retaining a single contribution that can be interpreted in

terms of a geometric ray from the source to the observation point and satisfying the Snell-

Descartes’s refraction law at the interface. Among all the plane waves that contribute to the

total transmitted field, we only retain those that correspond to the classical properties of

geometric acoustics.32

For an incident ray, arriving with an incidence θ1 , refracted with an angle θ2 , the transmitted

field is just described by an amplitude and a phase ; this phase is related to the geometric

trajectory of the ray through the relationship

( )4 3. a expcos cos

j fh

cd

c2

1 1 2 2π

θ θ ;+

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

this expression corresponds to the propagation time from the source point to the observation

point along the geometric ray. Figure 2 shows an illustration of this trajectory, in addition to

the different physical parameters of the problem.

The amplitude of the transmitted field is related to the transmission coefficient of the plane

monochromatic wave corresponding to the geometric ray ( )T12 1 2ν ν, ; this transmission

coefficient does not depend on frequency and can be written as follows

( )T12 1 21 1

2 1 1 2 2

1

2

1

2 2ν ν

ρ νρ ν ρ ν ν

νρρ

, .=+

=+

In the frame of the geometric ray approximation, ν1 and ν2 are related to the incidence and

refraction angles through the relationships

[ ]νπ

θii

i

fc

i= ∈2

1 2

cos , , ,

5


such that only depends on c , , (T12 1 2ν ν, ) 1 c2 ρ1 , ρ2 , θ1 and θ2 . In order to take into account

the diffraction and the non-linearity of the Snell-Descartes’s law, it is necessary to add a

correction term to the amplitude of the transmitted field, this correction term insuring the

conservation of energy.

To calculate this correction term, we consider ray tubes corresponding to incidence angles

between θ1 and θ δθ1 + 1 , with the corresponding refraction angles varying from θ2 to

θ δθ2 + 2 . These different parameters are illustrated on figure 2.

δ r

δ ri

δθ 2

θ2

θ1 δθ 1

Plane interface

Point-like source

h

r

d

Observer

Fluid 1

Fluid 2

ri

sin sinθ θ1

1

2

2c c=

Figure 2: computation of the transmitted amplitude using ray tubes Differentiation of the Snell-Descartes’s law yields the following equation:

cos cos

.θδθ

θδθ1

11

2

22c c

=

The incident ray reaches the interface at a horizontal distance r from the source ; then we

have i

( )4 3. b r h

rh

i

i

=

=

⎧⎨⎪

⎩⎪

tan ,

cos.

θ

δθδθ

1

21

1

6


Similarly, the refracted ray reaches the observation point at a horizontal distance r from the

source and we have

( )4 3. c r h d

rh d h

dcc

= +

= + = +⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

tan tan ,

cos cos coscoscos

.

θ θ

δθδθ

θδθ

θθθ

δθ

1 2

21

1 22

2 21

2

1

13

21

The ray tubes are also characterized by an azimutal angle varying from ϕ and ϕ δϕ+

(rotation of figure 2 with respect to the line containing the source and perpendicular to the

plane of the interface) ; due to the symmetry of the problem, this angle remains unchanged

during the refraction from fluid 1 to fluid 2.

The surface of the incident ray tube in the plan of the interface (perpendicular to the

propagation direction) is r ri iδϕδ θ cos 1 ; similarly, the surface of the refracted ray tube in the

plane of the interface is r ri iδϕδ θ cos 2 and becomes r rδϕδ θ cos 2 in the plane parallel to the

interface containing the observation point ; from the interface to the observation point, the

surface of the refracted ray tube has increased with a ratio r r r ri iδ δ / , this the amplitude of the

corresponding ray must also decrease with a ratio given by r r r ri iδ δ / .

Finally, the total amplitude of the transmitted field results from the following three

contributions:

• the amplitude of the incident ray in the plane of the interface, given by cos /θ π1 4 h ,

• the transmission coefficient ( ) ( )T T12 1 2 12 1 2ν ν θ θ, ,≡ ,

• the amplitude ratio r r r ri iδ δ / resulting from the variation of the surface of ray tubes.

Introducing these different contributions in equations (4.3b) and (4.3c), the amplitude of the

transmitted field can be written as follows:

( )4 3. d ( )1

412 1 2 1

2

1

1

2

1 2

2

1

31

32

1 2πθ θ θ

θθ

θθ

T , cos

coscos

coscos

./ /

h dcc

h dcc

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

From equations (4.3a) and (4.3d), we finally obtain32:

7


( )4 3. e ( )( )

~ , , ,, cos exp sec sec

cos sec cos sec./ /φ

π

θ θ θ π θ θ

θ θ θ θt x y z f

j fhc

dc

h dcc

h dcc

≈

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

14

212 1 2 11

12

2

2

11 2

1 22

1

31

32

1 2

T

( sec / cosθ θ= 1 ).

In equation (4.3e), the only frequency-dependant term is the phase term. As a consequence,

this formulation, obtained for a monochromatic source, can be generalized to the case of a

transient excitation described by a function ( )ψ t :

( )4 3. f ( )( )

φπ

θ θ θ ψ θ θ

θ θ θ θt x y z t

thc

dc

h dcc

h dcc

, , ,, cos sec sec

cos sec cos sec./ /≈

− −⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

14

12 1 2 11

12

2

2

11 2

1 2

2

1

31

32

1 2

T

This generalization requires the following assumptions:

• the temporal function ( )ψ t is band-limited, particularly at low frequencies,

• the geometric ray approximation (high frequency approximation) is valid for all

frequencies that are present in the spectrum of ( )ψ t : as a consequence, the approximation

must be valid for the smallest frequency (largest wavelength) of the temporal function ( )ψ t .

In this section, the transmitted field has been calculated using a very intuitive approach of the

problem (geometric ray) ; we are now looking at the same problem, but using some more

rigorous mathematics.

3. Computation of the transmitted field using the stationary phase theorem

We start with equation (4.2) for the transmitted field in fluid 2. Due to their definitions, the

coefficients ν1 and ν2 can be either real (propagative or homogeneous plane waves) or purely

imaginary (evanescent or heterogeneous plane waves). Thus, the expression of the transmitted

field can be decomposed into two parts:

8


• the first part corresponds to an integration over a domain in the ( )f fx y, space limited to a

disk of radius : this first contribution to the transmitted field corresponds to

propagative of homogeneous plane waves in both fluids, (f c/ max ,1 2 )c

)

β

• the second part corresponds to the remaining term, thus to an integration over the full

space, except the previously mentioned disk: this second contribution to the

transmitted field corresponds to evanescent of heterogeneous plane waves in one of the

two fluids.

( f fx y,

The geometric ray approximation is a high frequency approximation, only valid if the fields

can propagate in volumes (this excludes surface waves) on distances greater than a few

wavelengths (one problem is to give a quantitative evaluation of few): these validity

conditions allows to neglect the contribution of the second part of the integral, such that only

propagative or homogeneous plane waves in both fluids effectively contribute to the

transmitted field.32

However, the function to be integrated shows a term that does not depend on frequency (the

transmission coefficient ), and a phase term that varies linearly with frequency. Thus, if

the frequency increases, this phase term show a more and more rapid rotation, justifying the

use of the stationary phase theorem.31

T12

Starting from equation (4.2), we first make the following changes of variables:

x r f f df df f df dy r f f

x r x y r r

y r

= = == =

⎧⎨⎩

cos , cos , .sin , sin ,

α βα β

such that the transmitted field can be rewritten as follows:

( )4 4. a ( ) ( ) ( ) ( ) ( )( )~ , , , , exp exp ,/ max ,

φν

ν ν ν νt r r r

f c cx y z f

jj h j d I f f df≈ ∫ 2 1

12 1 2 1 20

1 2 T

where the function is defined by ( )I fr

( )4 4. b ( ) ( )[ ]I f j rf dr r= − −∫ exp cos .2

0

2π α β

π β

Due to the cylindrical symmetry, the function ( )I fr does not depend on α ; an adequate

choice of this parameter yields

9


( ) ( ) ( )( ) ( ) ( ) ( )

I f I f I f

I f j rf d I f j rf dr r r

r r r r

= +

= − = −

⎧⎨⎪

⎩⎪ −∫ ∫1 2

1 2

2

2 2

3 22 2

,

exp cos , exp cos ./

/

/

/π β β π β

π

π

π

π β

)

The first contribution can be calculated using the stationary phase theorem ; this

computation is quite simple and we only give the final expression

( rfI1

( ) ( )I f e

j rfrfr

j r

r1

4 2≈

−π π/ exp.

Moreover, we also have the relationship ( ) ( )I f I fr2 1=*

r . As an immediate consequence, the

transmitted field given by equation (4.4a) reduces to

( )4 4. c ( ) ( ) ( ) ( )( )

( ) ( )[ ]

~ , , , , exp exp

exp exp .

/max ,

/ /

φν

ν ν ν ν

π ππ π

t r r

f c c

jr

jr

x y z fr

jj h j d f df

e j rf e j rf

≈ ×

− +

∫−

12

2 21

12 1 2 1 20

4 4

1 2 T

Looking at equation (4.4c), we can observe two different phase terms defined by

( )ϕ π π ππ

± = − + − ±f hfc

f dfc

f rfr r r2 2 24

2

12

22

22

2 m .r

The integral in equation (4.4c) being evaluated in the frame of the stationary phase theorem,

we are first interesting in determining the minimums and maximums of these phase terms:

( )

{∂ϕ∂

π π

±

< <

>

=−

−

+−

−

±f

fh

ffc

fd

ffc

frr

r

r

r

r

r

2 22

12

2

0

2

22

2

0

0

1 244 344 1 244 344

.π2

Starting from this expression, we observe that only ( )ϕ+ fr shows a minimum or maximum.

The phase is a monotonous decreasing function and its contribution to the transmitted

field is negligible. It results therefore that equation (4.4c) can be written ( )ϕ− fr

( )4 4. d ( ) ( ) ( )( )( )~ , , , , exp ./max ,

φν

ν ν ϕt r r

f c cx y z f

rj

j f f df≈ +∫1

2 112 1 20

1 2 T r

10


The spatial frequency that make the phase function f r ( )ϕ+ fr stationary satisfies the

following relationship:

( )4 4. e hf

fc

fd

ffc

frr

r

r

r

2

12

22

22

2−

+

−

= .

In terms of plane monochromatic waves, we have

( )4 4. f ffc

fcr = =

11

22sin sin ,θ θ

where θ1 and θ2 are the incidence and refraction angles, respectively ; equation (4.4e) finally

reduces to

( )4 4. g h dtan tan .rθ θ1 2+ =

This relationship corresponds precisely to the geometric interpretation that can be made in the

frame of the ray model, as illustrated by figure 2.

The final step consists now in the computation of the second derivative of the phase ( )ϕ+ fr

with respect to : fr

( )∂ ϕ∂

π

π

2±

− −

− −

= − −⎛⎝⎜

⎞⎠⎟ + −

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− −⎛⎝⎜

⎞⎠⎟ + −

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

ff

hfc

f dfc

f

f hfc

f dfc

f

r

rr r

r r r

2

2

12

21 2 2

22

21 2

22

12

23 2 2

22

23 2

2

2

/ /

/ /

.

If is the frequency that make the phase function fr0 ( )ϕ+ fr stationary, one can obtain after

some calculations steps:

( )4 4. h ( )∂ ϕ

∂π

θθθ

2±

=

= − +⎛⎝⎜

⎞⎠⎟

ff

cf

h dcc

r

r f fr r

213

1

2

1

31

320

2cos

coscos

.

11


We now use the stationary phase theorem to calculate the integral in equation (4.4d) ; it

results from equations (4.4e), (4.4f), (4.4g) and (4.4h) the following formulation of the

transmitted field in fluid 2:

( )4 4. i ( )( )

~ , , ,, cos exp sec sec

cos sec cos sec./ /φ

π

θ θ θ π θ θ

θ θ θ θt x y z f

j fhc

dc

h dcc

h dcc

≈

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

14

212 1 2 11

12

2

2

11 2

1 22

1

31

32

1 2

T

This expression is identical to equation (4.3e) resulting from intuitive principles using ray

tubes.

This proves that the geometric ray approximation is a direct consequence of the stationary

phase theorem, applied to the computation of the transmitted field through the interface.32

4. Numerical computation of the transmitted field

We have developed a software to calculate the transmitted field, first using the complete and

theoretical description, second using the geometric ray approximation.

a – Computation using the theoretical description

This approach consists in a numerical evaluation of the integral given in equation (4.2),

without any particular approximations, except those resulting from a numerical procedure.

This computation can be performed using two different approaches.

The first solution consists in using algorithms based on Fast Fourier Transforms (1D and 2D

transforms) ; this method is efficient in terms of computation time, but can also be complex

due to the algorithm itself (sampling, aliasing, ...)26,33,34.

The second method consists in a classical integration algorithm, like the well-known

Simpson’s rule.26 To do this, we first start with equation (4.2) and change the double integral

into a single integral using the cylindrical symmetry of the problem ; one can obtain21,22:

(4 5. ) ( ) ( ) ( ) ( ) ( )~ , , , , exp exp ,φ πν

ν ν ν ν πt r r rx y z fj

j h j d J rf f df=+∞

∫22

21

12 1 2 1 2 00T

12


where and . For simplicity reasons, we preferred the second

solution, even if the method is quite slower.

r x y2 2= + 2y2f f fr x

2 2= +

b – Computation using the geometric ray model

The computation of the transmitted field using the geometric ray model reduces to the

numerical implementation of equation (4.4i) that gives the amplitude and the phase of the

field in the frame of the ray approximation.

c - Comparison between the two methods

In all the following, we consider the following parameters:

• frequency: f=5 MHz,

• distance between the source and the interface: h=10 mm,

• density and velocity of fluid 1: ρ1 =1,0 and =1500 m/s (c1 λ1 =0,3 mm),

• density and velocity of fluid 2: ρ2 =4,5 and c =6000 m/s (2 λ2 =1,2 mm).

The transmitted field has been calculated using the two methods described above as a

function of r in planes that are parallel to the interface, and for different values of d.

Comparison of the amplitudes: the comparison between the amplitudes of the transmitted

field using the two methods is not a problem ; figure 3 shows the two amplitudes variations

on the same graph (thick curves, left scales), and the relative difference (thin curves, right

scales).

Comparison of the phases: the comparison of phases is more complex, particularly in the

neighborhood of a phase equal to 0 in which case numerical errors can have a significant

contribution ; this kind of comparison requires a high precision that is not necessarily

accessible depending on the computation time that must be reasonable, or depending on the

numerical methods used for solving the problem ; this is the reason why a higher difference

between the two results can be acceptable, compared to the case of the amplitudes ; on figure

3, the relative difference (in modulus) is shown.

Looking at figure 3, we can make the following remarks:

13


• at 10 mm under the interface, the relative difference between the amplitudes of the

transmitted field is less than 0,02 %, while the relative difference between the

corresponding complex values remains less than 1,4 % ; these two limits are acceptable,

particularly if the results have to be compared with experimental measurements where

errors can be more significant,

• at 4 mm under the interface, the relative difference between the amplitudes of the

transmitted field reaches 0,12 % ; while the difference of the corresponding complex

values reaches 3,5 % ; once again, these differences are acceptable,

• at a depth less than 1 mm or 0,5 mm under the interface, the differences increase and

cannot be justified by numerical errors only ; in fact, these differences show that the ray

approximation is not enough to give a correct description of the transmitted field near the

interface ; such a result is not surprising because a depth of 1 mm under the interface

corresponds to less that one wavelength in fluid 2 at the considered frequency.

Finally, the validity of the ray model approximation depends on the precision required for the

computation of the transmitted field ; a depth of about 3 or 4 wavelength under the interface

appears as an acceptable limit, while smaller values generate some significant distortions in

comparison with the theoretical results.

14


0.0 15.0 30.0 45.0 60.0r (mm)

0.00

1.75

3.50

5.25

7.00*10 -4

Amplitude of the transmitted field

0.0000

0.0055

0.0110

0.0165

0.0220Relative difference (%)

0.0 15.0 30.0 45.0 60.0r (mm)

0.0

0.3

0.7

1.0

1.4Relative difference (modulus) between the two solutions (%)

Figure 3a: z=10 mm (8,3λ2 ) Figure 3b: z=10 mm (8,3λ2 )

0.0 10.0 20.0 30.0 40.0r (mm)

0.00

3.25

6.50

9.75

13.00*10 -4


0.00

0.03

0.06

0.09


0.0 10.0 20.0 30.0 40.0r (mm)

0.00

0.88

1.75

2.63

3.50Relative difference (modulus) between the two solutions (%)

Figure 3c: z=4 mm (3,3λ2 ) Figure 3d: z=4 mm (4,3λ2 )

0.0 5.0 10.0 15.0 20.0r (mm)

0.0

6.3

12.5

18.8

25.0*10 -4


0.000

0.675

1.350

2.025


0.0 5.0 10.0 15.0 20.0r (mm)

0

4

8

12

16Relative difference (modulus) between the two solutions (%)

Figure 3e: z=1 mm (0,8λ2 ) Figure 3f: z=1 mm (0,8λ2 )

0.0 2.5 5.0 7.5 10.0r (mm)

0.0

7.0

14.0

21.0

28.0*10 -4


0.0

4.5

9.0

13.5


0.0 2.5 5.0 7.5 10.0r (mm)

0

12

24

36

48Relative difference (modulus) between the two solutions (%)

Figure 3g: z=0,5 mm (0,4λ2 ) Figure 3h: z=0,5 mm (0,4λ2 )

15


IV) Transmission through a fluid/solid interface

We now consider the same problem as above, except that the plane interface separates a fluid

and a solid material. In order to simplify the next developments, we suppose that the source is

located at and the interface is the plane z h= − < 0 z = 0 .

In a solid medium, two kinds of acoustic waves can exist: the longitudinal and transverse

waves.4,28,32 While the propagation in a fluid can be described by a scalar equation (a single

scalar such that acoustic pressure is enough to describe the propagation of the wave), it

becomes vectorial in the solid (in this case, the acoustic field can be described by the

displacement vector at any observation point, for example). This makes the boundary

conditions in the plane of the interface a little more complex.

1. The incident wave in the fluid

Similarly to the previous section, the velocity potential corresponding to the incident wave

can be written near the interface

( ) ( ) ( ),exp2

where,exp,,,~1

11 hjjAzjAfzff iiyxi ν

νν ==Φ

such that the corresponding displacement vector is given by

( )4 6. a ( ) ( ) ( )~ , , , ~ exp ; ; .U f f z fj f

A e A j zff

ff fi x y i

j zi

x y=−

∇ =−⎛

⎝⎜

⎞⎠⎟

12 2

11

1

πν

νπ

ν

If the fluid is not viscous, the xz and yz components of the strength tensor are zero and the

only zz component needs to be calculated:

( )4 6. b ( )( )( ) ( )

~ , , , ,~ , , , ,

~ , , ,~ ~ ~

exp .

,

,

,, , ,

τ

τ

τ∂∂

∂∂

∂∂

π ρ ν

i xz x y

i yz x y

i zz x yi x i y i z

i

f f z f

f f z f

f f z f KU

xU

yU

zj f A j z

=

=

= + +⎛

⎝⎜

⎞

⎠⎟ = −

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

0

0

2 1 1

16


2. The reflected wave in the fluid

The velocity potential corresponding to the reflected wave is written in the following manner

( ) ( )~ , , , expΦr x y if f z f RA j z= − ν1 ,

where R is the reflection coefficient ; it results therefore the following expression of the

displacement vector

( )4 7. a ( ) ( ) ( )~ , , , ~ exp ; ; ,U f f z fj f

RA e RA j zff

ff fr x y i

j zi

x y=−

∇ = −⎛⎝⎜

⎞⎠⎟−1

2 21

11

πν

νπ

ν

and the strength tensor

( )4 7. b

( )( )( ) ( )

~ , , , ,~ , , , ,

~ , , ,~ ~ ~

exp .

,

,

,, , ,

τ

τ

τ∂∂

∂∂

∂∂

π ρ ν

r xz x y

r yz x y

r zz x yr x r y r z

i

f f z f

f f z f

f f z f KU

xU

yU

zj f RA j z

=

=

= + +⎛

⎝⎜

⎞

⎠⎟ = − −

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

0

0

2 1 1

3. The transmitted longitudinal wave in the solid

The displacement vector of a longitudinal wave in a solid is parallel to the direction of

propagation ; we can easily verify that this vector can be written as

( )4 8. a ( ) ( ) ( ), ; 2 ; 2 exp,,,~

lyxlilyxl ffzjATfzffU νππν −=

where is the longitudinal transmission coefficient. Thus, the different components of the

strength tensor are given by:

Tl

17


( )4 8. b

( ) ( )

( ) ( )

( )

~ , , ,~ ~

exp ,

~ , , ,~ ~

exp ,

~ , , ,~

,, ,

,, ,

,,

τ μ∂∂

∂∂

π μν ν

τ μ∂∂

∂∂

π μν ν

τ λ∂

l xz x yl x l z

x l l i l

l yz x yl y l z

y l l i l

l zz x yl x

f f z fU

zU

xj f T A j z

f f z fU

zU

yj f T A j z

f f z fU

= +⎛

⎝⎜

⎞

⎠⎟ =

= +⎛

⎝⎜

⎞

⎠⎟ =

=

4

4

( )∂

∂∂

∂∂

μ∂∂

π μ ν

xU

yU

zU

z

jfc

f f T A j z

l y l z l z

tx y l i l

+ +⎛

⎝⎜

⎞

⎠⎟ +

= − − −⎛⎝⎜

⎞⎠⎟

⎧

⎨

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

~ ~ ~

exp .

, , ,2

82

22

22 2

4. The transmitted transverse wave in the solid

The displacement vector of a transverse wave in a solid is perpendicular to the direction of

propagation ; we can easily verify that this vector can be written as

( )4 9. a ( ) ( ) ( )( )~ , , , exp ; ;, , , ,U f f z f A j z T T f T f Tt x y i t t t x t t y x t x y t y= +ν ν ν π 2 ,

where and T are the transverse transmission coefficients with respect to the x and y

directions. Thus, the different components of the strength tensor are given by:

Tt x, t y,

( )4 9. b

( )( ) ( )[ ]

( )( )

~ , , ,~ ~

exp ,

~ , , ,~ ~

exp

,, ,

, ,

,, ,

,

τ μ∂∂

∂∂

μ ν ν π π

τ μ∂∂

∂∂

μ ν π

t xz x yt x t z

i t t x t x x y t y

t yz x yt y t z

i t x y t x

f f z fU

zU

x

j A j z f T f f T

f f z fU

zU

y

j A j z f f T

= +⎛

⎝⎜

⎞

⎠⎟

= − −

= +⎛

⎝⎜

⎞

⎠⎟

= − +

2 2 2 2

2

4 4

4 ( )[ ]( )

( )( )

ν π

τ λ∂∂

∂∂

∂∂

μ∂∂

πμν ν

t y

t zz x yt x t y t z t z

t i t x t x y t y

f T

f f z fU

xU

yU

zU

z

j A j z f T f T

2 2 24

2

4

−

= + +⎛

⎝⎜

⎞

⎠⎟ +

= +

⎧

⎨

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪

,

,, , , ,

, ,

,

~ , , ,~ ~ ~ ~

exp .

t y

5. The boundary conditions in the plane of the interface

The boundary conditions that must be satisfied in the plane of the interface are:

• the continuity of the normal displacement,

• the continuity of the xz, yz and zz components of the strength tensor.

18


Starting from equations (4.6a), (4.6b), (4.7a), (4.7b), (4.8a), (4.8b), (4.9a) and (4.9b), the four

boundary conditions yield a linear system of 4 equations with 4 unknown variables R, ,

and T :

Tl Tt x,

t y,

( )4 10. a

( )( )

( )

νπ

ν πνπ

ρ πμ μν

π ν ν π π

π ν π

1 1

1

2

22 2

1

2 2 2 2

2

22

2

42

2

4 4 4

4 4

fR T f T f T

f

f Rfc

f f T f T f T f

f T f T f f T

f T f f T

l l x t x y t y

tx y l t x t x y t y

x l l t x t x x y t y

y l l x y

+ − + =

− − −⎛⎝⎜

⎞⎠⎟ + + =

+ − − =

−

, ,

, ,

, ,

,

,

,

( )t x t y t yf T, , .+ − =

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ ν π2 2 24 0

ρ

0

−

We do not give in detail the different steps required for the computation of the different

reflection and transmission coefficients ; we prefer to give the final expressions:

( )4 10. b R

Tfc

Tf f

c

Tf f

clt

t xl x

t

t yl y

t

=−+

=+

⎧

⎨⎪⎪

⎩⎪⎪

= −+

= −+

⎧

⎨⎪⎪

⎩⎪⎪

Δ ΔΔ Δ

ΔΔ Δ

Δ Δ

Δ Δ

1 2

1 2

1

22

0

1 2

1

22

1 2

1

22

1 2

2

1

1

,

,

,

,

,

,

π ρρ

ρ νρ

ρ νρ

with the following notations:

( )4 10. c ( )Δ

Δ Δ

Δ

0

2

22 2

12

02 2

2

2 41

24

1

24

= − −

= + +

=

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

fc

f f

f ffc

tx y

l t x y

l

t

,

,

.

π ν νπ ρ νρ ν

2

6. Computation of the transmitted longitudinal wave

As in the previous section, the transmitted longitudinal wave is computed using the two

methods, first by numerical evaluation of the integrals that define the components of the

displacement vector resulting from the transmitted wave (inverse Fourier transforms), second

using the ray model (stationary phase theorem). We now prefer using an algorithm based on

2D inverse fast Fourier transforms, instead of using classical methods like Simpson’s rule.26

19


In the frame of the ray approximation, the incident and refraction directions are in the same

plane ; thus we can consider that the source and observation points are both located in the

plane (y=0), up to a rotation. ( )xz

Using the stationary phase theorem, we can calculate:

• the geometric ray from the source to the observation point corresponding to the transmitted

longitudinal wave,

• the corresponding incidence angle in the fluid θ1 ,

• the two refraction angles in the solid, θl and θt , corresponding to the transmitted

longitudinal and transverse waves.

The Snell-Descartes’s law is written as follows:

ffc

fc

fc

ffc

fc

fc

xl

lt

t

y

ll

l tt

t

= = =

=

= = =

⎧

⎨

⎪⎪

⎩

⎪⎪

11

11

1

0

2 2 2

sin sin sin ,

,

cos , cos , cos .

θ θ θ

ν π θ ν π θ ν π θ

After some computation steps that we do not present in detail, the displacement vector

resulting from the longitudinal wave can be written in the following way

( )4 11. a ( )( )

~ , , ,, , cos exp sec sec

cos sec cos sec,/ /u x y z f

j fhc

dc

h dcc

h dcc

l

l l tl

l

ll

ll

= ≈

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

01

4

21 11

1

11

1 2

1

31

31 2π

θ θ θ θ π θ θ

θ θ θ θ

T

where is a vector that depends on the three angles (Tl lθ θ θ1, , )t θ1 , θl and θt

( )4 11. b ( ) ( )Tl l t

l

t l l

tt

ll t

l l

c cc

cc

θ θ θρ

ρθ θ θ

θ θ θρ θρ θ

11

2 22

21 1 1

2

2 2 0

2 2 2, ,

cos sin ; ; cos

cos sin sin/ cos/ cos

.=−

+ +

In all the following, we use the following parameters:

• frequency: f=5 MHz,

• distance between the source and the interface: h=10 mm,

• distance between the interface and the observation plane: z=d=10 mm,

• density and velocity of fluid 1: ρ1 =1,0 et c =1500 m/s (1 λ1 =0,3 mm),

20


• density et velocity of solid 2: ρ2 =7,0, =5900 m/s (cl λl =1,18 mm) and =3300 m/s

(

ct

λt =0,66 mm).

The transmitted longitudinal wave is computed as a function of the x coordinate of the

observation point (y=0).

-240.0 -120.0 0.0 120.0 240.0x (mm)

0.0

0.8

1.6

2.4

3.2*10 -5

Modulus of the x component of the displacement

-240 -120 0 120 240x (mm)

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0Modulus ratio

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0Variation of the phase

Figure 4a: component ~,ul x Figure 4b: component ~

,ul x

-240.0 -120.0 0.0 120.0 240.0x (mm)

0.0

2.0

4.0

6.0

8.0*10 -5

Modulus of the z component of the displacement

-240 -120 0 120 240x (mm)

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0Modulus ratio

-1.0

-0.5

0.0

0.5

1.0

1.5


Figure 4c: component ~,ul z Figure 4d: component ~

,ul z

Figures 4a and 4b represent the x component ( )~ , , ,,u x y z d fl x = =0 of the displacement vector

obtained using the two computation methods. The upper curve of figure 4b (left scale)

corresponds to the modulus ratio, and the lower curve to the phase difference divided by 2π

(right scale).

Figures 4c and 4d are similar to figures 4a and 4b, they represent the z component

( )~ , , ,,u x y z d fl z = =0 of the displacement vector.

Looking at these figures, we can make the following remarks:

• figures 4a and 4c show a very good agreement between the two computation methods,

• figures 4b and 4d show some significant differences for the modulus and phase of the

different components of the displacement vector ; anyway, the modulus ratio remains near

21


1 around x=0, and the phase difference is also negligible in this domain ; in fact the

differences increase for observation points far from the source position.

In fact, this observation does not invalidate the ray approximation. Indeed, the differences can

be important in a spatial domain where the different components of the displacement vector

are small, therefore particularly sensible to numerical noise resulting from numerical

algorithms like Fast Fourier Transforms. In this spatial domain, the numerical precision of the

method based on a numerical inverse Fourier transform is small and not really significant ;

these effects, consequences of our numerical approach, explain the graphics represented on

figures 4b and 4d.

Source

Fluid

Solid

Interface

Figure 4e: longitudinal displacement vector in the solid

Figure 4e represents the different geometric rays from the source to different observation

points located in the plane z=10 mm under the interface. For each observation point, we

plotted a vector parallel to the calculated displacement vector, with a length proportional to

the amplitude of this displacement. Looking at this figure, we clearly observe that

• the amplitude of the displacement reaches its maximum at the vertical of the source,

22


• the amplitude of the displacement decreases when the distance between the source and the

observation point increases,

• the direction of the displacement vector is clearly parallel to the geometric ray and this

corresponds to a longitudinal wave in the solid.

7. Computation of the transmitted transverse wave in the solid

As above, the transverse wave can be calculated either by numerical integration or using the

ray approximation. In this case, we obtain the following expression of the displacement:

( )4 12. a ( )( )

~ , , ,, , cos exp sec sec

cos sec cos sec,/ /u x y z f

j fhc

dc

h dcc

h dcc

t

t l tt

t

tt

tt

= ≈

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

01

4

21 11

1

11

1 2

1

31

31 2π

θ θ θ θ π θ θ

θ θ θ θ

T

where is a vector that depends on the three angles (Tt lθ θ θ1, , )t θ1 , θl and θt

( )4 12. b ( ) ( )Tt l t

t

l

l t t

tt

ll t

l l

cc c

ccc

θ θ θρρ

θ θ θ

θ θ θρ θρ θ

11

22

22

21 1 1

2

2 2 0

2 2 2, ,

sin cos ; ; sin

cos sin sin/ cos/ cos

.= −+ +

The vector describing the transverse transmission through the interface can be

complex. The geometric ray approximation allows to insure the Snell-Descartes’s law from

the source to the observation point for a velocity ratio c ; thus the incidence angle

(Tt lθ θ θ1, , )t

ct1 / θ1 can

generate an evanescent or heterogeneous transmitted longitudinal wave

⎪⎩

⎪⎨

⎧

−=−=

>=

.imaginary)(purely 1sinsin1cos

,1sinsin

221

1

lll

ll

jc

c

θθθ

θθ

For the generalization to the transient regime using a temporal excitation function ( )ψ t , it is

necessary to insure the reality of the transmitted displacement vector ; one can find the

following formulation21,28:

23


( )4 12. c

( )( )[ ]

( )[ ]

u x y z tt

hc

dc

h dcc

h dcc

TH thc

dc

h dcc

t

t l tt

t

tt

tt

t l tt

t

t

, , ,, , cos sec sec

cos sec cos sec

, , cos sec sec

cos sec

/ /= ≈ℜ − −

⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

+

ℑ −⎛⎝⎜

⎞⎠⎟

+

01

4

14

1 11

1

11

1 2

1

31

31 2

1 11

1

11

π

θ θ θ θ ψ θ θ

θ θ θ θ

π

θ θ θ θ ψ θ θ

θ

T

T

−

θ θtt

th dcc

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

1 2

1

31

31 2/ /

cos sec,

θ

where ( )TH tψ is the Hilbert transform of the function ( )ψ t defined by the relationship21

( )4 12. d ( )( )

TH tu

t uduψ

πψ

=−−∞

+∞

∫1

.

The numerical parameters used for computation are the same as above.

24


-240 -120 0 120 240x (mm)

0.0

3.0

6.0

9.0

12.0*10 -5

Modulus of the x component of the displacement

-240 -120 0 120 240x (mm)

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0Modulus ratio

-1.0

-0.5

0.0

0.5

1.0

1.5


Figure 5a: component ~,ut x Figure 5b: component ~

,ut x

-240 -120 0 120 240x (mm)

0.0

2.5

5.0

7.5

10.0*10 -5

Modulus of the z component of the displacement

-240 -120 0 120 240x (mm)

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0Modulus ratio

-1.0

-0.5

0.0

0.5

1.0

1.5


Figure 5c: component ~,ut z Figure 5d: component ~

,ut z

Figures 5a, 5b, 5c and 5d represent the x and z components ( )~ , , ,,u x y z d ft x = =0 and

of the displacement vector obtained with the two computation methods,

using the same graphical conventions as above. Looking at these figures, we can make the

following remarks:

(~ , , ,,u x y z d ft z = =0 )

• the comparison of the different curves is acceptable, even if the modulus ratio shows

greater differences than in the case of longitudinal waves ; this is related to the fact that the

computation method based on inverse Fourier transforms seems to be more sensible to

numerical noise and errors than in the previous case,

• we can observe significative differences between the modulus and phases at x=0 ; this

observation point, located at the vertical of the source, corresponds to a zero-transmission

(the transmission coefficients are zero for a normal incidence), and the numerical results

obtained in this case are not significant,

• the modulus and phase show a discontinuity near x ≈ ±9 4, mm (≈ ±2 6, mm in the plane

of the interface), thus corresponding to an incidence angle θ1 of about 14,7°: this

corresponds to the critical angle for longitudinal waves in the solid.

25


Source

Interface

Source

Interface

Figure 5e: transverse displacement vector in

the solid (real part)

Figure 5f: transverse displacement vector in

the solid (imaginary part)

Figures 5e and 5f show the geometric rays from the source to different observation points

located in the plane z=10 mm under the interface. For each observation point, we represent a

vector whose direction and amplitude correspond to the real displacement obtained by

simulation (real part on figure 5e, imaginary part on 5f). Looking at these representations, we

can make the following observations

• the displacement is zero in front of the source,

• the displacement vector is perpendicular to the direction of the geometric ray, this

corresponds to a transverse wave in the solid,

• the imaginary part is zero for incidences less than the critical angle for longitudinal waves

in the solid.

26

habilitation - chapitre 1 · 2 • c2 and ρ2 are the velocity and density of the second fluid 2 ;...

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