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Influence of the thickness and the attenuation coefficient of a backing on the response of transducers
Ultrasonics
34 (1996) 103-106
Influence of the thickness and the attenuation coefficient of a backing on the response of transducers Jamal Assaad *, Martine Ravez, Christian Bruneel, Jean M. Rouvaen, Fraqis Institut d’Electronique
et de Microelectronique du Nord (IEMN), UMR CNRS 9929, Opto-Acousto-Electronic University of Valenciennes, le Mont Houy BP 311, 59304 - Valenciennes Cedex, France
Haine
Department (OAED),
Abstract
In this paper, the characteristic mechanical impedance of an absorbing medium (backing) is expressed in terms of its thickness THand the tangent of the phase angle (tan 6 = m, 6 is the loss angle). In fact, the backing has been considered as a lossy transmission line. Then, using the model of Mason’s circuit, the electrical impedance and radiated power of an in-air backed transducer will be expressed in terms of tan 6 and Ta. It is shown that the characteristics of such transducers do not change if the thickness is scaled down and the product mT, stays constant. The same properties have been obtained for backed transducer bars using a finite element method (FEM) analysis. Keywords:
Backed transducers; FEM -
1. Introduction
For high-frequency applications in ultrasonic imaging and nondestructive testing, it is often necessary to limit the number of periods in a pulse generated by ultrasonic transducers. Usually, a backing with a high ultrasonic attenuation and with an acoustic impedance matching that of the piezoelectric active transducer element is employed. Most models based on Mason’s circuit consider the backing as a semi-infinite medium without losses. Generally, these models do not take into account the acoustical attenuation coefficient (a). Via the use of a complex elastic tensor, the attenuation coefficient (TV) can be expressed as a function of the loss angle [ 11. Moreover, by considering the absorbing medium as a lossy transmission line the mechanical impedance of this absorbing medium has been expressed in terms of tan 6 and TB. Then, using Mason’s circuit, the characteristics of a backed transducer can be obtained. The FEM is the most accurate [2] method of characterizing backed transducer bars. The aim of this paper is to show how a backed transducer can be modelled by FEM. In fact, the modelling of a backed transducer is
possible but the main drawback is generally the large size of the numerical problem, although with this new formulation the modelling of a backed transducer or a group of backed transducers is possible by scaling the tangent of the phase angle (tan 6 = m), by reducing TR and by keeping the product TBtan 6 constant.
B
C
air
zo A
TB
T
D
MM
Fig. 1. In-water
backed
transducer
IQ
50
100
m (%) * Corresponding author. Fax: + 33-27-14-1 I-89; e-mail: [email protected] 0041-624X/96/$15.00 C 1996 Elsevier Science B.V. All rights reserved PII SOO41-624X(96)00009-1
Fig. 2. The variation
of aTa versus m keeping
WIT, = 1.29 x 10m3.
104
J. Assaad et al. / Wtrasonics 34 (1996) 103-106
2. Mechanical
impedance of backing
A well-known method for generating ultrasonic waves in a propagation medium is the so-called backing method. It relies on the backing of the piezoelectric face (Fig. l), as opposed to the propagation medium. Considering the backing as a lossy transmission line, then the impedance Zz seen at AB (Fig. 1) is given by: Z2 = - jZBl tan&, Te)
that for m less than 30%, the product UT, stays approximately constant. Fig. 3 describes the variations of the relative real part and the relative imaginary part of the characteristic mechanical impedance Z2 as a function of m keeping mT, constant (mT, = 1.29 x 10-j). It is interesting to note that for m < 30%, the imaginary part is practically zero and the real part is about 1 (Z, % Z,).
(1)
where k, is the sound vector, ZBI (ZBI = pc) is the mechanical impedance of the absorbing medium (Fig. 1) with losses. The characteristic impedance Z, of air is very small and supposed equal to zero. The relations between the sound speed c, Young’s modulus (Y) for an absorbing medium is c2 = AY/p. Here, the constant A depends only upon Poisson’s ratio (v), which is supposed to be real and p is the mass density in kg rne3. To take losses into account in the backing, Y and c are to be regarded as complex (c = c’ + jc” and Y = Y’ - jY”). Then the sound vector kB is also complex (kB = k + ja). As v is assumed to be real, then the tangent angle 6 is equal to Y”/Y’ (tan 6 = m = Y”/Y). Then with a few simple calculations the characteristic impedance, as given by Eq. (1) can be expressed as:
Fig. 3. The real part and the imaginary part of the relative acoustical impedance (2,/Z,) of the backing keeping mT, = 1.29 x 10m3.
m = 0.65 %, Tg = 20 cm, ~TB = 2.4
1.5
Z2 = -jZ,J1_jm
tan [;7,(1
+jr)]
r=L
v;=
0.2 ,
4.5
j
1.5
(MHz)
1
3
4.5
frequency
(MHz)
m = lo%, Tg = 1.29 cm 0.2 , 7r
,
1.2
-1+ k
3 frequency
(2)
with
,
‘.*k::~!:i::
/ 1
/
JGS m
(;)L.i(1+?)=2V2(
’
1 +7&J,
(4)
1.5
3 frequency
where V, and V(V2 = AY’/p) are, respectively, the phase velocity of the absorbing medium with and without losses. Z, is the characteristic mechanical impedance of the absorbing medium without losses (Z, = A(PY’)~‘~). Considering longitudinal wave (V = V,) and using Mason’s circuit, the electrical impedance and the acoustical power radiated in the propagation medium can be easily obtained and can be found in many references [ 33.
4.5
1.5
(MHz)
,:
:
4.5 frequenci
4.5 (MHZ)
0.2
;,,,. I r
id
3 frequency
1.5
(MHz)
46 frequeni
(MHz)
3. Numerical results For easier experimentation an aluminum backing has been used with TB= 20 cm. The attenuation coefficient and the longitudinal wave have been measured and respectively found equal to 12 N m-i and 6250 m s-i (V, = 6250 m s-l). Then the products aT, and mT, are respectively found equal to 2.4 and 1.29 x 10m3. Fig. 2 displays the variation of aT, versus m, keeping mT, constant (mTB = 1.29 x 10p3). It is interesting to note
-.1.5
3 frequency
4.5 (MHz)
1.5 frequency
3
4.5
(MHz)
Fig. 4. Variation as a function of m of the relative acoustical (Z,/Z,) of the backing keeping mT, = 1.29 x 10-j.
impedance
:. ..: ,: q J. Assaad et al. I Ultrasonics 34 (I 996) 103-l 06
m=0.65 %, Tg= 20 cm
m =0.65%,TB= 20cm, aT~z2.4
.
5 4
,3
..: .
L
..I...
;.
:
1
m=lo%,Tg A.29 cm
Jjyyyyyids
., ; : :
.j...i__
:
:
2
105
I.::: ..;..<...>..j.
. .._._.I_.
..‘.,.
:
:
:
“::
:: K :
.:‘.: n "0.5 normaked
:.
..
1 frequency
1.5 (f/to)
1 0.5 normalized frequency
$0)
m= lO%,Tg= 1.29 cm
:’
’
.
_,
I
,
:
:
:
:
.:...:
m=20 %,‘I’&.65
0
cm
0.5 1 1.5 normalized frequency (f/fo)
m=dO %, T~zO.32 mm
1 :
..,.......
:
_.
0.5 1 1.5 normalized frequency (f/fo)
:
~
~~1/‘1!
I~~
.:.........
:
:
.:. i,. :::. .:/ w L+LhJ :
:
.:._
1 frequency
-0.5
normalized
1.5 (f/to)
ii
normal&d
frequency
(flro)
m=20%,Tg=0.65cm ‘:.
1 1.5 0.5 normalized frequency (Wfo)
Fig. 6. Variation of the radiated mT, = 1.29 x 10-3.
:
I.
1 1.5 0.5 normalized frequency (f/fe)
3
power
as a function
of m keeping
;
:
.
,,
:_
..:.:. :
‘I.
I,..
._:.._.
:
:
it;
:
:,:, : :
:
1 -0.5 nornmlized frequency
1,’ :
;
..I...
:
:
j
:
:
:.
..,.
1.5 (f/to)
1 nornmlized frequency
-0.5
1.5 (f/6)
impedance of a backed transducer. Fig. 7 describes the real part of the electrical resistance (R) and admittance (G) of an air-backed transducer for different values of m and by keeping mT, equal to 1.29 x 10e3. The comparison is perfect.
10
m=5% m=lO% ,,,,,,, ,,,.,, ,,,,,,,,.
8
.............
6
“0.5
normalized
1 frequency
1.5 (f/fo)
1 0.5 normalized frequency
Fig. 5. The electrical resistance and conductance transducer keeping m’Ia = 1.29 x 10e3.
of an in-air
1.5 (f/fo) backed
Fig. 4 displays the real part and the imaginary part of Z, versus frequencies for different values of m keeping mT, = 1.29 x 10e3. Previous comments are applicable for this figure. Figs. 5 and 6 describe the electrical resistance (R), electrical conductance (G), and the radiated power for a backed lithium niobate plate (V, = 7400 m s-l, p = 4.7 kg rnm3, W>> T, W being the width) whose surface is equal to 15 mm’ and thickness is equal to 1 mm (fO = 3.7 MHz, f0 being the mechanical resonance frequency). For Fig. 6, the transducer has been matched electronically to 50 R using a transformer. In this last example a lithium niobate bar (YZw/36”, [2]) has been considered with W/T = 0.5. There is no simple model to compute the electrical impedance of bars. The FEM has then been used to compute the electrical
8
3 .P E
2
3 FREQUENCY
2
3 FREQUENCY
Fig. 7. The electrical resistance niobate transducer bar keeping
4 (MHz)
4 (MHz)
5
5
and conductance of an in-air lithium mT, = 1.29 x lOmA.
106
J. Assaadet al. / Ultrasonics 34 (1996) 103-106
4. Conclusions
References
The attenuation coefficient of a backing must have a large value. However, generally, this value is not very large and consequently, the thickness of the backing must be quite large. To model a backed transducer by FEM, the main drawback is the large size of the numerical problem. However, by reducing the thickness of the backing and scaling tan 6, the size of the backed domain can be considerably reduced. The modelling of a set of finite backed transducer bars can be now modelled. The resultant pressure of an array including interaction can be also computed using FEM [4].
[ 11 B.N. Capps, Elastomeric materials for acoustical applications, Naval Research Laboratory, Underwater Sound Reference Detachment, Orlando, FL (1989). [2] J. Assaad, C. Bruneel, J.-N. Decarpigny and B. Nongaillard, J. Acoust. Sot. Am. 94 (1993) 2969. [3] B. Auld, Acoustic Fields and Waves in Solids, (Wiley, New York, 1973). [4] J. Assaad and C. Bruneel, Ultrasonics 34 (1996) 107 (this issue).
34 (1996) 103-106
Influence of the thickness and the attenuation coefficient of a backing on the response of transducers Jamal Assaad *, Martine Ravez, Christian Bruneel, Jean M. Rouvaen, Fraqis Institut d’Electronique
et de Microelectronique du Nord (IEMN), UMR CNRS 9929, Opto-Acousto-Electronic University of Valenciennes, le Mont Houy BP 311, 59304 - Valenciennes Cedex, France
Haine
Department (OAED),
Abstract
In this paper, the characteristic mechanical impedance of an absorbing medium (backing) is expressed in terms of its thickness THand the tangent of the phase angle (tan 6 = m, 6 is the loss angle). In fact, the backing has been considered as a lossy transmission line. Then, using the model of Mason’s circuit, the electrical impedance and radiated power of an in-air backed transducer will be expressed in terms of tan 6 and Ta. It is shown that the characteristics of such transducers do not change if the thickness is scaled down and the product mT, stays constant. The same properties have been obtained for backed transducer bars using a finite element method (FEM) analysis. Keywords:
Backed transducers; FEM -
1. Introduction
For high-frequency applications in ultrasonic imaging and nondestructive testing, it is often necessary to limit the number of periods in a pulse generated by ultrasonic transducers. Usually, a backing with a high ultrasonic attenuation and with an acoustic impedance matching that of the piezoelectric active transducer element is employed. Most models based on Mason’s circuit consider the backing as a semi-infinite medium without losses. Generally, these models do not take into account the acoustical attenuation coefficient (a). Via the use of a complex elastic tensor, the attenuation coefficient (TV) can be expressed as a function of the loss angle [ 11. Moreover, by considering the absorbing medium as a lossy transmission line the mechanical impedance of this absorbing medium has been expressed in terms of tan 6 and TB. Then, using Mason’s circuit, the characteristics of a backed transducer can be obtained. The FEM is the most accurate [2] method of characterizing backed transducer bars. The aim of this paper is to show how a backed transducer can be modelled by FEM. In fact, the modelling of a backed transducer is
possible but the main drawback is generally the large size of the numerical problem, although with this new formulation the modelling of a backed transducer or a group of backed transducers is possible by scaling the tangent of the phase angle (tan 6 = m), by reducing TR and by keeping the product TBtan 6 constant.
B
C
air
zo A
TB
T
D
MM
Fig. 1. In-water
backed
transducer
IQ
50
100
m (%) * Corresponding author. Fax: + 33-27-14-1 I-89; e-mail: [email protected] 0041-624X/96/$15.00 C 1996 Elsevier Science B.V. All rights reserved PII SOO41-624X(96)00009-1
Fig. 2. The variation
of aTa versus m keeping
WIT, = 1.29 x 10m3.
104
J. Assaad et al. / Wtrasonics 34 (1996) 103-106
2. Mechanical
impedance of backing
A well-known method for generating ultrasonic waves in a propagation medium is the so-called backing method. It relies on the backing of the piezoelectric face (Fig. l), as opposed to the propagation medium. Considering the backing as a lossy transmission line, then the impedance Zz seen at AB (Fig. 1) is given by: Z2 = - jZBl tan&, Te)
that for m less than 30%, the product UT, stays approximately constant. Fig. 3 describes the variations of the relative real part and the relative imaginary part of the characteristic mechanical impedance Z2 as a function of m keeping mT, constant (mT, = 1.29 x 10-j). It is interesting to note that for m < 30%, the imaginary part is practically zero and the real part is about 1 (Z, % Z,).
(1)
where k, is the sound vector, ZBI (ZBI = pc) is the mechanical impedance of the absorbing medium (Fig. 1) with losses. The characteristic impedance Z, of air is very small and supposed equal to zero. The relations between the sound speed c, Young’s modulus (Y) for an absorbing medium is c2 = AY/p. Here, the constant A depends only upon Poisson’s ratio (v), which is supposed to be real and p is the mass density in kg rne3. To take losses into account in the backing, Y and c are to be regarded as complex (c = c’ + jc” and Y = Y’ - jY”). Then the sound vector kB is also complex (kB = k + ja). As v is assumed to be real, then the tangent angle 6 is equal to Y”/Y’ (tan 6 = m = Y”/Y). Then with a few simple calculations the characteristic impedance, as given by Eq. (1) can be expressed as:
Fig. 3. The real part and the imaginary part of the relative acoustical impedance (2,/Z,) of the backing keeping mT, = 1.29 x 10m3.
m = 0.65 %, Tg = 20 cm, ~TB = 2.4
1.5
Z2 = -jZ,J1_jm
tan [;7,(1
+jr)]
r=L
v;=
0.2 ,
4.5
j
1.5
(MHz)
1
3
4.5
frequency
(MHz)
m = lo%, Tg = 1.29 cm 0.2 , 7r
,
1.2
-1+ k
3 frequency
(2)
with
,
‘.*k::~!:i::
/ 1
/
JGS m
(;)L.i(1+?)=2V2(
’
1 +7&J,
(4)
1.5
3 frequency
where V, and V(V2 = AY’/p) are, respectively, the phase velocity of the absorbing medium with and without losses. Z, is the characteristic mechanical impedance of the absorbing medium without losses (Z, = A(PY’)~‘~). Considering longitudinal wave (V = V,) and using Mason’s circuit, the electrical impedance and the acoustical power radiated in the propagation medium can be easily obtained and can be found in many references [ 33.
4.5
1.5
(MHz)
,:
:
4.5 frequenci
4.5 (MHZ)
0.2
;,,,. I r
id
3 frequency
1.5
(MHz)
46 frequeni
(MHz)
3. Numerical results For easier experimentation an aluminum backing has been used with TB= 20 cm. The attenuation coefficient and the longitudinal wave have been measured and respectively found equal to 12 N m-i and 6250 m s-i (V, = 6250 m s-l). Then the products aT, and mT, are respectively found equal to 2.4 and 1.29 x 10m3. Fig. 2 displays the variation of aT, versus m, keeping mT, constant (mTB = 1.29 x 10p3). It is interesting to note
-.1.5
3 frequency
4.5 (MHz)
1.5 frequency
3
4.5
(MHz)
Fig. 4. Variation as a function of m of the relative acoustical (Z,/Z,) of the backing keeping mT, = 1.29 x 10-j.
impedance
:. ..: ,: q J. Assaad et al. I Ultrasonics 34 (I 996) 103-l 06
m=0.65 %, Tg= 20 cm
m =0.65%,TB= 20cm, aT~z2.4
.
5 4
,3
..: .
L
..I...
;.
:
1
m=lo%,Tg A.29 cm
Jjyyyyyids
., ; : :
.j...i__
:
:
2
105
I.::: ..;..<...>..j.
. .._._.I_.
..‘.,.
:
:
:
“::
:: K :
.:‘.: n "0.5 normaked
:.
..
1 frequency
1.5 (f/to)
1 0.5 normalized frequency
$0)
m= lO%,Tg= 1.29 cm
:’
’
.
_,
I
,
:
:
:
:
.:...:
m=20 %,‘I’&.65
0
cm
0.5 1 1.5 normalized frequency (f/fo)
m=dO %, T~zO.32 mm
1 :
..,.......
:
_.
0.5 1 1.5 normalized frequency (f/fo)
:
~
~~1/‘1!
I~~
.:.........
:
:
.:. i,. :::. .:/ w L+LhJ :
:
.:._
1 frequency
-0.5
normalized
1.5 (f/to)
ii
normal&d
frequency
(flro)
m=20%,Tg=0.65cm ‘:.
1 1.5 0.5 normalized frequency (Wfo)
Fig. 6. Variation of the radiated mT, = 1.29 x 10-3.
:
I.
1 1.5 0.5 normalized frequency (f/fe)
3
power
as a function
of m keeping
;
:
.
,,
:_
..:.:. :
‘I.
I,..
._:.._.
:
:
it;
:
:,:, : :
:
1 -0.5 nornmlized frequency
1,’ :
;
..I...
:
:
j
:
:
:.
..,.
1.5 (f/to)
1 nornmlized frequency
-0.5
1.5 (f/6)
impedance of a backed transducer. Fig. 7 describes the real part of the electrical resistance (R) and admittance (G) of an air-backed transducer for different values of m and by keeping mT, equal to 1.29 x 10e3. The comparison is perfect.
10
m=5% m=lO% ,,,,,,, ,,,.,, ,,,,,,,,.
8
.............
6
“0.5
normalized
1 frequency
1.5 (f/fo)
1 0.5 normalized frequency
Fig. 5. The electrical resistance and conductance transducer keeping m’Ia = 1.29 x 10e3.
of an in-air
1.5 (f/fo) backed
Fig. 4 displays the real part and the imaginary part of Z, versus frequencies for different values of m keeping mT, = 1.29 x 10e3. Previous comments are applicable for this figure. Figs. 5 and 6 describe the electrical resistance (R), electrical conductance (G), and the radiated power for a backed lithium niobate plate (V, = 7400 m s-l, p = 4.7 kg rnm3, W>> T, W being the width) whose surface is equal to 15 mm’ and thickness is equal to 1 mm (fO = 3.7 MHz, f0 being the mechanical resonance frequency). For Fig. 6, the transducer has been matched electronically to 50 R using a transformer. In this last example a lithium niobate bar (YZw/36”, [2]) has been considered with W/T = 0.5. There is no simple model to compute the electrical impedance of bars. The FEM has then been used to compute the electrical
8
3 .P E
2
3 FREQUENCY
2
3 FREQUENCY
Fig. 7. The electrical resistance niobate transducer bar keeping
4 (MHz)
4 (MHz)
5
5
and conductance of an in-air lithium mT, = 1.29 x lOmA.
106
J. Assaadet al. / Ultrasonics 34 (1996) 103-106
4. Conclusions
References
The attenuation coefficient of a backing must have a large value. However, generally, this value is not very large and consequently, the thickness of the backing must be quite large. To model a backed transducer by FEM, the main drawback is the large size of the numerical problem. However, by reducing the thickness of the backing and scaling tan 6, the size of the backed domain can be considerably reduced. The modelling of a set of finite backed transducer bars can be now modelled. The resultant pressure of an array including interaction can be also computed using FEM [4].
[ 11 B.N. Capps, Elastomeric materials for acoustical applications, Naval Research Laboratory, Underwater Sound Reference Detachment, Orlando, FL (1989). [2] J. Assaad, C. Bruneel, J.-N. Decarpigny and B. Nongaillard, J. Acoust. Sot. Am. 94 (1993) 2969. [3] B. Auld, Acoustic Fields and Waves in Solids, (Wiley, New York, 1973). [4] J. Assaad and C. Bruneel, Ultrasonics 34 (1996) 107 (this issue).