time-domain model for transducer array
TRANSCRIPT
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Technical Report 1654May 1994
Time-Domain Model fora HydroacousticTransducer Array
George W. BenthienDon Barach
uG061994U'" 4 ( '94-24924:"• ° lull. M uiii li i
94 8 05 101
TAPProvd for public release; disri"on is unlimited.
Technical Report 1654May 1994
Time-Domain Model for a HydroacousticTransducer Array
George W. BenthienDon Barach
NAVAL COMMAND, CONTROL ANDOCEAN SURVEILLANCE CENTER
RDT&E DIVISIONSan Diego, California 92152-5001
K. E. EVANS, CAPT, USN R. T. SHEARERCommanding Officer Executive Director
ADMINISTRATIVE INFORMATION
The work was accomplished during the period 1 July 1993 to 31 March 1994 under thesponsorship of the Space and Naval Warfare Systems Command (SPAWAR PMW 182-32).
Released by Under authority ofE. F. Rynne, Jr., Head J. M. Holzmann, HeadAcoustic Analysis Branch Acoustic Systems and
Technology Division
ACKNOWLEDGMENT
The authors gratefully acknowledge Bob DeLaCroix of Hydroacoustics, Inc., for severalhelpful discussions concerning their modeling of the HLF-6A transducer and for allowing usaccess to their computer code which we used to compare with our results. We also gratefullyacknowledge some helpful suggestions by Dr. Steven Black of the Naval Research Laborato-ry/Underwater Sound Reference Detachment.
SM
EXECUTIVE SUMMARY
OBJECTIVES
To develop a mathematical model for a transducer array which can handle nonlineardevices such as the HLF-6A hydraulic transducers. The model should be capable ofhandling both sinusoidal and nonsinusoidal electrical inputs.
APPROACH
Since the model must be able to handle nonlinear devices it is necessary to formulatethe model in the time domain rather than in the frequency domain. The internal com-ponents of the transducers were modeled using standard hydraulic flow equations whichare in the form of nonlinear ordinary differential equations. The equations are nonlinearsince the pressure drop across a hydraulic valve is proportional to the square of the flowthrough the valve. The acoustic radiation interaction impedances were obtained in thefrequency domain using the boundary element program CHIEF. The frequency-domaininteraction impedances were then transformed to the time domain using an FFT routineto produce interaction impulse responses. The radiation impedance relations in the fre-quency domain become convolution relations in the time domain. When these radiationconvolution integral relations axe combined with the model for the transducer internalcomponents, the result is a coupled set of nonlinear differential-integral equations thatis solved numerically by an explicit time-stepping procedure similar to a second-orderRunge-Kutta method.
RESULTS
Our time-domain model was first applied to an array of simple linear transducers forwhich the solution could be obtained by conventional frequency-domain methods. Theagreement between the results of our time-domain model and the frequency-domain resultswas excellent. A comparison was also made between our model and a model developed byHydroacoustics, Inc. for a single transducer element. Again the agreement was quite good.Finally, predicted results using our time-domain model were compared with experimentalresults for a three-element close-packed array. The agreement was reasonably good inboth level and trends.
iii
CONTENTS
EXECUTIVE SUMMARY iii
1 INTRODUCTION 1
2 LINEAR MODEL 2
2.1 BASIC EQUATIONS .............................. 2
2.2 SOLUTION PROCEDURE .......................... 7
2.3 RESULTS .................................... 8
3 HYDROACOUSTIC MODEL 16
3.1 FIRST-STAGE SERVOVALVE ........................ 16
3.2 SECOND-STAGE VALVE ........................... 20
3.3 PRESSURE COMPENSATION ........................ 24
3.4 RADIATING DISKS AND DRIVE CAVITY ................. 26
3.5 SUMMARY OF EQUATIONS ......................... 28
3.6 RESULTS .................................... 29
4 CONCLUSIONS AND RECOMMENDATIONS 44
5 REFERENCES 45
DT f C. IAAooesloD cl.
c n
Jiat n . .. .--.
LIST OF FIGURES
1 Simple linear transducer element ............................. 2
2 Mutual velocity impulse coefficient for a pair of elements (dl/ = 2)...... 4
3 Self-acceleration impulse response (d/s = 2) ...................... 5
4 Mutual acceleration impulse response A12 (d/s = 2) ................. 5
5 Mutual acceleration impulse response A1s (d/s = 2) ................. 6
6 Three-element array configuration ............................ 9
7 Velocity magnitude of outer element of three-element array .......... . 11
8 Velocity phase of outer element of three-element array .............. 11
9 Velocity magnitude of middle element of three-element array ........ .12
10 Velocity phase of middle element of three-element array ............. 12
11 Time response of a single element in the free field (s/A = 0.110) ....... .13
12 Middle element of three-element array, Q=oo .................... 13
13 Middle element of three-element array, Q=100 ................... 14
14 Middle element of three-element array, Q=20 .................... 14
15 Middle element of three-element array, Q=10 .................... 15
16 Idealized hydroacoustic transducer ........................... 17
17 First-stage servovalve and associated drive circuit ................. 18
18 Second-stage valve ...................................... 21
19 Supply port geometry .................................... 22
20 Single element, -6 dB drive level ............................ 32
vi
21 Single element, -3 dB drive level ............................ 32
22 Single element, 0 dB drive level ............................. 33
23 Single element, 3 dB drive level .............................. 33
24 Single element, -6 dB drive level, HAl load model ................ 34
25 Single element, -3 dB drive level, HAl load model ................ 34
26 Single element, 0 dB drive level, HAl load model .................. 35
27 Single element, 3 dB drive level, HAl load model .................. 35
28 NRaD impedance magnitude comparison ....................... 36
29 NRaD impedance phase comparison .......................... 36
30 HAl impedance magnitude comparison ........................ 37
31 HAI impedance phase comparison ............................ 37
32 Impedance magnitude, NRaD vs HAI, frequency-domain model ....... .38
33 Impedance phase, NRaD vs HAI, frequency-domain model .......... .38
34 Impedance magnitude, NRa.D vs HAI, time-domain model .......... .39
35 Impedance phase, NRlaD vs HAI, time-domain model ............. .39
36 Source level for three-element array, experiment vs computed ........ .40
37 Normalized source level, single element vs array .................. 40
38 Velocity of middle element of three-element array at freq=FL ........ .41
39 Drive cavity pressure of middle element of three-element array at freq=FL. 41
40 Velocity comparison, middle element of 1 x 6 array vs single element. ... 42
41 Drive cavity pressure comparison, middle element of 1 x 6 array vs singleelement ............................................. 42
vii
42 Normalized source level, single element vs six-element array ........... 43
LIST OF TABLES
I Normalized variables ........ .............................. 9
viii
1 INTRODUCTION
This report contains a description of a mathematical model for a hydroacoustic trans-ducer array. Although the model was developed with the HLF-6A transducer in mind, itcan be applied to other hydroacoustic transducer arrays. This model is based partly onprevious modelling work for the HLF-4 transducer [1], but involves new work relating toacoustic radiation loading. Previous models have attempted to couple frequency-domainmodels for the acoustic field with time-domain models for the internal components ofthe transducer. In this report we present a complete time-domain model for the coupledtransducer/acoustic-field problem which is applicable to both single elements and arrays.This model can also handle nonsinusoidal signals, which is important since linear superpo-sition is not valid for nonlinear devices. The acoustic radiation model involves obtainingcertain interaction coefficients in the frequency domain with a boundary element programcalled CHIEF [2,3] and then transforming these coefficients to the time domain by usingFourier transforms. When these coefficients are combined with a model for the internalcomponents, the result is a coupled set of differential-integral equations that can be solvedby an explicit time-stepping procedure with prescribed initial conditions.
Since the model for acoustic radiation loading is new, it was first applied to a simplelinear transducer array for which the solution could also be obtained by the standardfrequency-domain approach. Section 2 contains a description of the linear model thatincludes the new acoustic radiation approach. Results of the time-domain model are com-pared with corresponding frequency-domain results in order to validate the new acousticradiation approach. The agreement was excellent. Section 3 contains a description ofthe models for the various components of the HLF-6A hydraulic transducer, such as thefirst-stage servovalve, the second-stage valve, the drive cavity, the radiating disks, thepressure compensation device, and the radiation loading. Although the equations for thistransducer are more complicated than those for the simple linear transducer, the formof the equations and the general solution procedure are essentially the same. Computedresults for a single HLF-6A transducer in water are compared with computed results forthe Hydroacoustics, Inc. (HAI) model when several drive levels are used and when theinput signal is sinusoidal in time. The agreement is quite good for all drive levels, butthere are some small departures at the higher drive levels (less than 1 dB), which weattribute to differences in the radiation models. Results are also presented for a proposedsix-element array. Since the interelement spacing for this array is quite large, the perfor-mance of each element is not greatly different from that of a single element in the freefield. Predicted results using our time-domain model are also compared with experimen-tal results for a three-element array which is fairly closely packed. The comparison isreasonably good both in level and in trends. Section 4 is a summary of our conclusionsand recommendations.
2 LINEAR MODEL
In this section a time-domain radiation model will be developed and applied to a simplelineax transducer. The resulting linear model will be used to validate the time-domainapproach to the acoustic radiation loading which will later be applied to the nonlinearHLF-6A hydroacoustic transducer. The simple linear transducer is shown in figure 1. Itconsists of two rigid circular pistons of mass M placed symmetrically at the ends of arigid cylindrical housing. A spring of stiffness K restrains each piston and each piston isdriven by the same prescribed force F.
Subsection 2.1 gives the mass/spring relations for this simple linear transducer as wellas a time-domain formulation for acoustic loading on the transducer. The result is asystem of coupled differential-integral equations with specified initial conditions. Thesolution procedure for solving this system is described in subsection 2.2. In subsection 2.3the results for a three-element array are compared with those from a frequency-domainmodel after the initial transients have died out, in order to validate the time-domainformulation.
... PISTON
IFCYIDIA: •HOUSING
- ---------------------- SYMMETRYPLANE
Figure 1. Simple linear transducer element.
2.1 BASIC EQUATIONS
The differential equation describing this simple device is
Mii + Ku = F - Fra'd(1
where Fr" is the acoustic radiation force on the piston and u is the displacement of thepiston. The acoustic radiation force is given by
Fr•d =L pdA (2)
2
where p is the acoustic pressure and A is the surface of the piston in contact with thewater. Now consider a planar array of N of these elements. The radiation force on them-th element of an array can be expressed as a linear function of the piston normalvelocities. In the frequency domain this relation takes the form
NF.Gd(w) = , zm.(w)v.(w) (3)
where F," is the radiation force on the m-th element, Zmn is the mutual radiationimpedance between the m-th and n-th elements, and Vn is the velocity of the ,n-th piston.These acoustic interaction impedances z,, are obtained by using the boundary elementprogram CHIEF, which is based on the Helmholtz integral equation
pOC) =Gi( )b- (C,) + iwpv ()G(C,a)] dS (4)
where S is the radiating surface (including all array elements), v is the normal velocity, pis the fluid density, and G is the free-space Green's function. To obtain zn the integralequation 4 is solved for p when all the piston velocities are set to zero except for vn, whichis set to 1. The resulting pressure is integrated over the m-th piston, yielding Zmn.
When the interaction relation in equation 3 is transformed into the time domain viaa Fourier transform, it takes the form
N tF~rd(t) = r-'.(t) + E Cmn(r)Vn(t - r)di (5)
where Cmn is the Fourier transform of zmn for m 5 n, rg' is the real, high-frequencyasymptotic limit of zmm, and Cm, is the Fourier transform of zmm - r•. It turns out thatfor numerical reasons it is better to work with accelerations than with velocities. Theconversion of the interaction equations 3 and 5 from velocities to accelerations gives thefrequency-domain relation
N
Fmd (W) = Mi (w) a(W) (6)n=1
and the time-domain relation
N P
F~r."(t) Z jo ]1 /Ln(Tr)Cn(t - 7-) Lir (7)
where mmn(w) = zmn(w)/iw and
mn(t) = Cmn (r)d + m3•=n (8)
= fjCmm (r) dr +r<0 m=n()
3
Figure 2 is a plot of the impulse interaction coefficient (12 appearing in equation 5 versusnormalized time. This plot illustrates the reason for expressing the convolution integralsin terms of acceleration instead of velocity. The fact that the functions C,,,t (m # n)alternate in sign over a very short interval gives rise to subtraction errors when theconvolution integrals are evaluated. Figures 3-5 show the mutual impulse functions A11,A12, and A13 appearing in equation 7 versus normalized time. Notice that these functionsare significantly nonzero only over a small time interval (nominally the time for sound totravel across both the interazting elements). To obtain the mutual impulse functions un,,,the CHIEF [3] program was used to calculate the mutual radiation impedances zn over abroad frequency range. The mutual impedances were divided by iw and transformed to thetime domain by using an FFT routine. The self-impulse function /Unn has a discontinuityat t = 0. To eliminate the resulting high-frequency oscillation in the computed Am,, weuse the fact that the even extension of a time function has a Fourier transform which istwice the real part of the original transform. Thus, can be computed by taking theFFT of twice the real part of z,,,/iw.
X10-11.0
0.8 f
0.6
0.4
0.2.) 0.0'
0 0.0
e -0.2
-0.4
-0.6
-0.8
-1.0
-1.2 0 8
0 1 2 3 4 5 6 7 8
TIME/(s/c)
Figure 2. Mutual velocity impulse coefficient for a pair of elements (d/s = 2).
4
x 10
1.0
0.8
0.6
-, 0.4
0.2
0.0
! !
0 1 2 3 4 5 6 7 8
TIME/(s/c)
Figure 3. Self-acceleration impulse response (dls - 2).
xlO-11.0
0.8
0.6
- 0.4
0.2
0.0
-0.2 I i I I I0 1 2 3 4 5 6 7 8
TIME/(s/c)
Figure 4. Mutual acceleration impulse response /12 (dfs = 2).
5
xlO-11.0
0.8
0.6
-. 0.4
0.2
0.0
-020 1 2 3 4 5 6 7 8
x1O
TIME/(s/c)
Figure 5. Mutual acceleration impulse response s13 (d/s = 2).
Combining the acoustic interaction equation 7 with the element equation 1 gives thefollowing coupled system of differential and integral equations:
N t~
Mam,+ Kum = FmZ]PmfEt(7)an,(t -1) dr (l0a)
'.m = am (lob)ilmn = V, m= 1,...,N (10c)
where Fm is the prescribed force on the m-th element. Notice that there is a set ofequations like 10 for each element in the array and that these sets of equations are cou-pled through the convolution integrals in 10a that involve histories of all the elementaccelerations. It will be assumed that the elements are not moving before t = 0 and thus
M(O) = vm(0) = am(O) = 0 m = 1,...,N (11)
The differential-integral equations 10 and the initial conditions 11 describe the problemto be solved numerically.
6
2.2 SOLUTION PROCEDURE
In this section a solution procedure will be described for the coupled set of integral-differential equations 10. As with systems of ordinary differential equations, the solutionis advanced successively by time steps of length 6 beginning at the initial time t = 0. Todescribe the method it is only necessary to specify how the solution is advanced from atime t, = rb to the next time 4.+1 = t, + 6.
It will be convenient to first write the equation 10a in a somewhat different form.When t is in the interval [tv, t 1+J] the integrals in 10a can be split as follows:
mn(r) an(t -r) dr = I m(r)an(t - r) dr + I.Am.(r)an(t - r)dT (12)
An element in the array does not feel the effects of its neighboring elements immediatelysince there is a time delay required for sound to tiavel between the elements. Thus,•,n(t) = 0 for t < din/c, where dmin is the minimum distance between any two elementsand c is the sound speed. It will be assumed in what follows that the time step 6 is smallerthan d .. n/a With this assumption
0 l mn(r)an(t - r) dr = 0 m n (13)
since t - t, 6 b < d/c. When m = n the first integral in equation 12 can be approximatedusing the trapezoidal rule as follows:
t-"trminm(r)an (t - T) dr "- I[ IMM(0)a.(t) + Amm(t - t.)am (4)] (t - 4) (14)
It follows from equations 12 to 14 that
N t
>II jAmn(7jan(t - T) cir = [m()al)+Immt- t,)am(t4)] (t - 4v) +
E j mn(r)an(t - r) dr (15)n---1 tr
Substituting 15 into equation 10a and solving for am yields
am(t) = M I'(O)( - ) Fm(t) - Kum(t) - !Amm(t - tt)am(tv)(t - 4.) -
N t
E I"m.(r)an(t - r) dr (16)
Combining 16 with the last two equations of 10, we obtain
7
+ 1 2?,,m(t) -- M + ½m,.(O)(t - t.) F,(t) KU,(t) - lim,(t - tr)G,(tt)(t - t7) -
N ,
E It-tr tn.(,r)aii(t - 'r) drJ (17a)
V=i(t) (17b)
Notice that the right-hand side of equation 17 involves only the two unknown functionsurn, vmr, and past values of an which are known at this stage.
The systems of equations 17 can be solved in the same manner as a system of ordinarydifferential equations. The only modification necessary is to update the accelerations anat the end of each step by using equation 16. It should be pointed out that a time stepmust be completed for each element in the array before the next time step is addressed.It is necessary to store the time histories of the accelerations for each element in the arrayand to update these histories following each time step. Since the /p 's are essentiallyzero outside a small time interval, only the relevant portions of the time histories need tobe saved.
The results to be presented later in this paper were obtained using a second-orderRunge-Kutta method known as Heun's method [4]. For a system of ordinary differentialequations
•(t) f f(Y (t),1 0 (18a)S= YO (18b)
This solution procedure advances the solution Yn at time tn to the solution yR+, at time
tn+1 by using
Y,+1 = , + [2 [in + f(Y. + , , + ,,+ 1 )]6 (19)
The algorithm given by equation 19 together with the formulas for updating accelerationscan be used to solve the differential-integral equations given in equation 17.
2.3 RESULTS
In this section we will present computed results for a single element and for a 1 x 3array as shown in figure 6.
The results will be checked against those obtained from an independent frequency-domainanalysis. With the dimensionless variables defined in table 1, equation 10 can be writtenas follows:
8
d I- d
Figure 6. Three-element array configuration.
Md. + Kfim F. -j E -,fn() df (20a)
di-T = n .10b)
di-m = j' m=1,...,N (20c)
Table 1. Normalized variables.
- " time P(O) - F(sf/c) drive force
1 1i-= -u(si/c) displacement pQ) =-20(11/c) A
f(ý-= -v(sf/c) velocity =M mass
a() = 5a(sf/c) acceleration I = K stiffnesspc28
The dimensionless parameters in equation 20 were chosen to make the low-frequencyradiation mass of the circular piston equal to the mass of the piston and to make thepiston diameter one-tenth of a wavelength at the resonance frequency. This leads to thedimensionless parameters
K 1 = 2=r2 (21)3 75
9
The dimensionless forcing functions F, were taken to have the form
Ip.1 sin(wt)
- -. 1- sin ~2w- (22)
where A is the acoustic wavelength. In the numerical examples the force amplitudesIF. I/pC2 3 2 were taken to be 1. The spacing d between elements for the array examplewas taken to be 2s.
Figures 7-10 show a comparison of the frequency response of the outer and middleelements of the 1 x 3 array when both a frequency-domain approach and a time-domainapproach are used. The agreement is very good. Figure 11 shows the solution of thedifferential-integral equations 10 for a single element. The steady-state normalized ve-locity amplitude for this computed solution and the value obtained from a steady-statefrequency-domain analysis of this problem were both 15.5. Figure 12 shows the computedtime response for the middle element in the 1 x 3 array. Due to the large number ofoscillations in the given time interval only the envelope of the response is discernible. Ittakes a very long time for the response to reach steady state (about 300 times as long asit takes sound to travel across the array).
Figures 13-15 show the effect of adding an increasing amount of damping resistanceto the linear transducer model. The Q values given are for the element operating in avacuum and are thus a measure of the internal resistance. As the internal losses increase,the level of the response decreases and the time to reach steady state decreases. The effectof losses is important since hydroacoustic transducers typically have very high internallosses.
10
xlOi ANALYTIC COMPUTED
4 .0 :
3.5
1.5
1.0 -
0.5 [
0.2 0.4 0.6 0.8 1.0 1.2 !.4 1.6 1.8x1O',
Figure 7. Velocity magnitude of outer element of three-element array.
xioi ANALYTIC COMPUTED
10
6-
4 I I I , I . I2 0.4 0.6 0 1.0 1.2 1.4 1.6 1,8
0 10
sIX
Figure 8. Velocity phase of outer element of three-element array.
i I
xlOi ANALYTIC COMPUTED
4
3 -
I 2
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8X10-1sfk
Figure 9. Velocity magnitude of middle element of three-element array
xlo 2 ANALYTIC COMPUTED2.0
1.5
o1.0
0.5L
-1.0
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8X10-1
sIX
Figure 10. Velocity phase of middle element of three-element array.
12
X1O1
1.5
1.0
*0.5
Q0.
4).5
-1.0
-1.5
2.0 ~ O0.0 0.5 1.0 1*5 2.0 2.5
dIOPTIME /(s/c)
Figure 11. Time response of a single element in the free field (s/A =0.110).
2C10 0
15
10
5
80
-5
-10
-150 L0.0 0.5 1.0 1.5 2.0 Z5 3.0
X10'TIME/(s/c)
Figure 12. Middle element of three-element array, Q=oo.
13
xlO°
10
.5
-10
0.0 0.5 1.0 1.5 2.0 2.5 3.0
x103
TIME/(s /c)
Figure 13. Middle element of three-element array, Q=100.
x10 0
15
10
0
00
-10
0.0 0.5 1.0 1.5 2.0 2-5 3.0
x103
TIMEI(slc)
Figure 14. Middle element of three-element array, Q=20.
14
X160
15
10
5
10
-15
0.0 0.5 1.0 1.5 2.0 2.5 3.0
TIME/(s/c)
Figure 15. Middle element of three-element array, Q=10.
15
3 HYDROACOUSTIC MODEL
Figure 16 is an idealized representation of the internal components of a hydroacoustictransducer like the HLF-6A. The transducer consists of two metallic radiating disks whichare clamped around their circumference to a cylindrical housing. The disks are drivenat their centers by two drive pistons, which in turn are driven by the pressure in an oil-filled drive cavity. The drive cavity pressure is modulated by a sliding valve that controlsthe flow into and out of the cavity. This valve is driven by a hydraulic servovalve. Theservovalve is called the first-stage valve and the main valve is called the second-stage valve.As the second-stage valve slides back and forth, the drive cavity is alternately coupled tothe supply and return ports of a hydraulic pump. The pump is designed to maintain aconstant pressure at the supply port. The supply pressure is much higher than the returnpressure.
This section contains descriptions of the mathematical models for the main compo-nents of the HLF-6A transducer. The equations for the first-stage servovalve are given insubsection 3.1 and the equations for the second-stage valve are given in subsection 3.2.The equations for the added stiffness due to the pressure compensation mechanism aregiven in subsection 3.3. The mathematical model for the radiating disks, drive cavity, andthe acoustic loading is presented in subsection 3.4. The model for the acoustic loading isa straightforward extension of the model previously described for the simple linear trans-ducer. A summary of the differential-integral equations for an array of hydroacousticelements is presented in subsection 3.5. Subsection 3.6 presents results obtained fromthis model for both single elements and arrays. Single-element results are compared withthose of an existing HAI model for various drive levels. Results for a proposed six-elementarray are compared with those of a single element. Model results for a close-packed three-element array are compared with experimental results.
3.1 FIRST-STAGE SERVOVALVE
This section describes a mathematical model for the first-stage servovalve used in theHLF-6A and its associated drive circuit. A diagram of the servovalve and drive circuit isshown in figure 17.
The internal components of the servovalve (enclosed by a dotted box in figure 17) were
not modeled in detail. Instead an empirical relationship of the form
S= K..1 (23)
was used to relate the drive current I and the servovalve displacement x,,. We will nextshow how to relate the servovalve displacement to the servovalve output flow Q. and the
16
2"ATING DISK
Figure ~ ~ ~ DIV 16 daie hdo Psic ransdcr
17 V 1 M _ '
ARMATURE W:SERVO VALVE
FLAPPER
'P PSNOZZLE NOZZLE
FEEDBACK
WIREP ps Pr P'r P's
SERVOVALVESPOOL
Qp
Ks F P PSSPRING' " _ MAIN "-] ISPRIN*•G
VALVE
Figure 17. First-stage servovalve and associated drive circuit.
18
main valve displacement x,. Let us define P1 and P2 as follows:
P= P. ifX, <0 { P, ifz. f<0 (24)
Then the flow Q, through the servovalve satisfies the equations
P1- I = P2 - 2i (25)
where I. is the sign of Qt, and fi is a constant that depends on the geometry of the valveand the density of the fluid. Thus
ItQ,, > 0 (26)
The flow Q. also satisfies the orifice equations
P~P.P ~ 2 VtceQ (27)
where a,1 depends on the geometry of the two identical output orifices and on the densityof the fluid. The flow Qp through the shunt orifice satisfies
P. - .' = I.a 2Q2 (28)
where I. is the sign of Qp and a 2 is a constant that depends on the geometry of the shuntorifice and on the density of the fluid. The flows Q., Q., and Qp also satisfy the balanceequation
QV = Q, + Q. (29)
It follows from equation 29 that
IQ, = I.(QM- Q.) _Ž 0 (30)
Addition of the equations 25, 27, and 28 yields the equatieu
P1 - P2 = I.a 2Q2 + 21val Q2 + 24,-PQv (31)
Combining equations 29 and 31 gives
P1 - P2 = I,0 2(Q.- Q.)2 + 21tcQl + 21v (32)III
or
rI(A 2 + 21ct, + 21.) Q,2 - (21.c 2Q.)Q0, + (I°a 2Q.2 - PA + P2) 0 (33)
The quadratic equation 33 has the solutions
Q, = -B + I, /B 2 -4AC (34)2A
19
where I, = 1 and
A = I. + 21 (ai +-) (35a)
B = -21.Q 2Q. (35b)
C = a2 Q~P.2 - +P 2 (35c)
Since Q. must be real, we must have
B 2 - 4AC >_ 0 (36)
As we are ignoring the compliance of the end cavities, the flow Q. is related to thedisplacement zv by
Qe = Aviv (37)
From equations 28, 29, and 37 it follows that the driving force F, on the second-stagevalve spool is given by
F, = (P. - P)A,- I=aA,,(Q,, - Aviv) (38)
However, Q. can be determined by solving the quadratic equation 33, given i. and henceQ.. The quantities , I,., and I, are determined so that the conditions 26, 30, and 36 aresatisfied. Thus, we have shown that
F,= f(x i.,,t) (39)
3.2 SECOND-STAGE VALVE
The second-stage valve is pictured in figure 18. The valve spool is shown in the neutralposition (x,, = 0). The overlap between the valve spool and the valve sleeve on each sideof the drive cavity opening when the valve spool is in the neutral position is called thevalve overlap (wol).
The equations describing the flow through the supply (return) port of the valve intothe drive cavity are different when the supply (return) port is open and when it is closed.When the supply (return) port is open, the flow at low frequencies is given approximatelyby the flow through an orifice whose area is changing slowly with time. It is shown ina previous paper [1,Appendix D] that the pressure drop across the open supply port isgiven by
P 2 - = Ph QIQ.I (40)2 17A 2Pwhere P1 is the pressure on the upstream side of the supply port, P 2 is the pressure onthe drive cavity side of the supply port, AP is the area of the supply port opening, and
20
os Or
S~t1
I I~WOL 4DRIVE
CAVITY
Figure 18. Second-stage valve.
q is the ratio of the flow stream area at the point of greatest contraction to the valveopening area. The pressure P1 is different from the supply pressure P. since there arepassages and fittings between the pump and the valve. Similarly, the pressure drop acrossthe return port when it is open is given by
P h QIQT I (41)2 j2A2(
where P3 is the pressure on the downstream side of the return port and Ap, is the area ofthe return port opening.
We next derive expressions for the areas A.., A,p as a function of valve position. Thesupply port opening is pictured in figure 19. The rounded corners shown on the valvesleeve and the valve spool are greatly exaggerated (rad=0.002 in.). When the supply portis open we have
X, = V(zt8 - wol +"-2. rad2 + (rcl + 2. rad)2 - 2. rad (42)
where z., is the displacement of the valve from neutral position (a positive increase inz., tends to open the supply port). The area, A., of the supply port opening is givenapproximately by
Asp -- ,DDX, (rcl < D.) (43)
where D, is the diameter of the valve spool.
Similarly, when the return port is open we have
X, = /(-xi. - wol + 2. rad2 + (rcl + 2. rad)2 - 2. rad (44)
andAlP "-* rD.X2 (rcl A). (45)
21
- -- - - -VALVE SPOOL
rcl Iv
IVALVE SLEEVE T.DIECW
V9- wof
Figure 19. Supply port geometry.
Combining equations 40 and 43 results in
P1- A = Ph QoIQo (46)
Similarly, combining equations 41 and 45 results in
A- P3 = Ph Q, IQr I (47)2 t 27 2 D 2 z 2
The pressure drops P. - P1 and P 2 - PD axe given by
PP - P, = L°9° + phQ.IQ.I (48)2172A,2
and P2 - PD= LoQs + PhQ, QI (49)
where L, is the supply flow path inertance, Lo. is the supply output flow path inertance, A.is the supply flow path equivalent area, and A,. is the supply output flow path equivalentarea. The pressure drops PD - P2 and P3 - P, are given by
P3 - P, = LQ0 + PhQT IQvI (50)
2v72A2
and PnD - P2 = L,&oQr + PhQwIQT l (51)
where L, is the return flow path inertance, Lo is the return output flow path inertance, A,is the return flow path equivalent area, and A,. is the return output flow path equivalentarea.
22
Combining equations 40, 48, and 49 yields
(r r~)D (_ 1 1 )2-D D X2 Q.IQoI (52)
when the supply port is open. Similarly
(L' + L")Q" = PD - P- ,I2 XA + ";j2 + T2DH )QQI (53)21 D X2 Q, IQ,[ 83
when the return port is open.
When the supply port is closed the leakage flow is given by [1,p.15]:
Q (54)
where12 p1=L 27 2l7r2D (rcl) 2
RL = rD1,(rcl)3 ' ko = Ph ZL = wol - . (55)
and #u is the shear viscosity coefficient of the fluid. Similarly, when the return port isclosed
dosed = Sgn(PD - Pr) [-14g& + V/1k4RL + k!IPD - Pr ] (56)
where now ZL = z,, + wol.
The equation of motion for the second-stage valve spool can be written in the form
M, oi , o + 2K:,,o = F, + Fjg, (57)
or equivalently1
61, = - (Ft, + FfI, - 2K.:,o) (58a)iv, = v1. (58b)
where z.. is the lineal valve displacement, MA,, is the mass of the valve spool, Ko is thestiffness of each of the two restraining springs, F. is the driving force on the spool givenin equation 33, and Ff 1m is the axial force on the spool due to the flow through the valve.The force Fp0o. is given approximately by [5]:
F 1, = -sgn(z,,)phQU cos0 (59)
where ph is the density of the hydraulic fluid, Q is the flow through the valve (supply orreturn), U is the velocity of the jet at the point of maximum contraction (vena contracia),
23
and coo - 0.36 (9 is the angle of the flow relative to the axis). Since Q = qUA, whereA is the area of the valve opening, equation 59 can be written
F1. = -0.36 sgn(x,.)-0- (80)
The area of the valve opening is given approximately by
A - rDx, (61)
Thus, equation 60 can be written
Ff.,,= -0.36 sgn (z..).'p.'
Since the flow through the closed port is small, we can write approximately
FC -0.36 sgn(z.) , + Q!) (63)
3.3 PRESSURE COMPENSATION
In order to allow the transducer to operate at large depths the interior of the transduceris filled with a pressurized gas. The pressure of the gas is adjusted to be equal to thehydrostatic pressure in the water at the operating depth. Thus, the pressure Po of theinterior gas is adjusted so that
Po = P=tvn + pgD (64)
where Pa.m is atmospheric pressure, p is the density of the water, g is the acceleration ofgravity, and D is the operating depth.
In addition to providing depth compensation, the pressurized gas also presents a dy-namic stiffness to each of the radiating disks. By symmetry it is only necessary to considerone of the radiating disks and half of the interior volume. When the disk vibrates theinterior volume changes which in turn produces a change in pressure. It will be assumedthat the pressurized gas obeys the ideal gas law
PVI = PoV 7=1 .4 (65)
where -y is the adiabatic gas constant, P is the pressure, V is the volume of gas, Po is theinitial pressure charge, and Vo is the initial volume of gas. Let P and V/ be the changesin pressure and volume produced by the vibration of the disk, i.e.
P = Po+-P v = vo+ +, (66)
24
Combining equations 65 and 66, we obtain
( + T) (1 + i' 1 (67)
Assuming that P < Po and f/ Vo, it follows from equation 67 that
(+ P ) (1+4i -Yf/ - (68)or
(69)
If u is the normal displacement of the disk, then the volume change V is given by
f =L udA (70)
The displacement u can be approximated by
u = uO1 I + u20 2 + u303 + u40 4 (71)
where 0i, 02, q, and 04 are the dominant mode shapes of the disk (normalized to beunity at the disk center). Combining equations 69-71, we obtain
= -7P 0 ,A4 , (72)
where a,• is defined by
a11 = •L•ndA (73)
The finite-element modal force component Fm• is given by
FmLP!Ai•mdA -"PAam (74)
Thus
Fm• = '/ Po A2 4
*,fo 4
= - ctkm u (75)
where
;im ff7PA2
F P- - (76)
25
Let P be the fraction of the interior volume that is filled with gas. Then
V0 = Ah (77)
where A is the area of the disk and h is the half height of the cylindrical housing. Sub-stituting equation 77 into equation 76, we get
SPoA78)
Equation 75 is the desired dynamic stiffness relation.
3.4 RADIATING DISKS AND DRIVE CAVITY
The radiating disks were modeled by using the finite-element method. Four symmetricmodes were used to represent the motion of the disks. The modes were normalized so thatthe normal component of displacement at the disk center was unity. In terms of thesemodes the equation of motion for the m-th element can be written
Mifi'(t) + Kiut(t) = P,(t)Ap - F,,(t) i = 1,2,3,4 (79)
where Mi is the modal mass of the i-th mode, K. is the modal stiffness of the i-th mode,U' is the i-th modal displacement of the m-th element, Pm is the drive cavity pressure ofthe m-th element, Ap is the area of the drive piston, and F, is the i-th modal componentof the acoustic radiation force on the m-th element. This equation can be written in thealternative form
'(t) + K=u4 (t) Pn(t)Ap - Fm(t) (80a)tm(t) = 4(t) (80b)S(t) = vt,1() i= 1,2,3,4 (80c)
Equation 7 for the radiation force in the linear model can be extended to this problem asfollows:
4 N tF,(t ='j p(r)a-n(t - -r) dr i = 1,2,3,4 (81)
j=1 nf=10
Combining equations 80 and 81, we obtain
4 4•NMia,,(t) + Kiu' (t) = Pm(t)Ap - F, kj (t) -E i ,(-r)aj (t - -r) dr (82a)
jui j=1 n.1
i)m(t) = a,,(t) (82b)i&'(t) = vu(t) i= 1,2,3,4 (82c)
26
In order to solve the differential-integral equations for this problem, we will rewriteequations 82 as was done in the linear problem. Let t be in the interval t, < t 5 t,+s.Then the integrals on the right-hand side of equation 82 can be split as follows:
4,(r) a,'. (t-,r)d-r = J 'j (,-r)a(t-,r)d- + J,•,.C(-)a•,(t-,r)d'r (83)
Since an element does not feel the effects of its neighboring elements immediately, itfollows that j,0 C(,)a,(t - -) d-r = 0 m 96 n (84)
When m = n the first integral in equation 83 can be approximated by using the trapezoidalrule as follows:
j0 2s~1 Ta( - ) T -~ I4,(O)aGln(t) + Im(t - tr)a()]85
Inserting equations 83-85 into equation 82 yields
Mi,.(t) + (L,• ,,(O)aj (t) Pm(t)Ap - Kjui,( - kijui,,(t) -j=1 j=l
E ,I uJ ~(T)aj(t -,r) cir-j=l =1 t
(5Vn l',,(i - t7)aiCt,) (86a)j-1
tL(t - a,(t) (86b)vi•t:) = ~(v2t) i = 1,2,3,4 (86c)
The three equations in 86 can be solved for the as in terms of the remaining variables.
The pressure Pm in the drive cavity satisfies the equation
CPmn(t) = net flow into the cavity = Qn(t) - Apv'(t) (87)
where C is the compliance of the cavity, vu is the center velocity of element m, and Q'is the flow into the drive cavity of the m-th element through the second-stage valve. Sincethe modes were normalized to unity at the center, the velocity v:' can be written
4V,() = v,,(t) (88)
The net flow Qn is given by
Qi(t) = Q'(t) - Q (t) (89)
where Q. and Qr are the supply and return flows for the m-th element. Substitutingequations 88 and 89 into equation 87 produces
Pm(t) = r - Q(,t) - A _ C (90)7=1
27
3.5 SUMMARY OF EQUATIONS
In this subsection we will present a summary of the differential-integral equations for ahydroacoustic array obtained in previous subsections. The equations will be given for them-th element of an array. There will be a system of equations like this for each elementin the array.
Radiating disk equations:
i
6Vm = am
• = v. i = 1,2,3,4
where* 4 4 N t
Mi,,,, + Kiu,, PmAP E k,,&m - E E/ ( T0 A.o,)al.(t r)- drj=1 j=1 n==1
Drive cavity equation:
Valve spool equations:
1,1 = -- (F, + FFi - 2K.x,.)
MW,zt,. = Vt,.
where
F,, = f(t,, ,,t)+2 Q2Ffo -m -0. 36sgn (m.,) Ph r.--)
twDtxz,
Valve flow equations:
(L. + L.)Q. = P. - PD + + QIQ. supply port open
28
(L.+L,.o -- PDAP - + 2)+ 2D--- QJQI return port open
Q*=sgn(P.-PD) V4gRL+ /-k-RL+k4IP. -PDI] supply port leakage
Qr= 89n (PD - P,) A4 RL + /fjlcAtRr + k4-1PD - Pr return port leakcage
The initial conditions for ui and vi (i = 1,2,3,4), z., v,, Q1, and Qr are taken tobe zero. The initial condition on the drive cavity pressure Pn is taken to be (P. + P')/2.The solution procedure as outlined for the linear problem is applicable to the solution ofthis set of differential-integral equations.
3.6 RESULTS
This subsection contains computation results from the time-domain model for bothsingle elements and arrays of HLF-6A transducers. The following normalizations havebeen employed in the figures:
1. Frequency has been shifted by a value designated as FL.
2. Time has been divided by a/c, where a is the diameter of an element and c is the
sound speed.
3. All source levels have had the single-element target source level subtracted.
4. Array source levels also have had 20 loglo(N) subtracted, where N is the number ofelements.
5. Velocities have been divided by the sound speed c.
6. Drive cavity pressures have been divided by the supply pressure.
Figures 20-23 show a comparison of results from the NRaD model and the HAI modelfor sinusoidal drives of four different levels and a number of different frequencies. At eachdrive frequency the fundamental component of the time response is plotted. A drive levelof 0 dB corresponds to half the maximum displacement of the servovalve. For all drive
29
levels the agreement is quite good, but there axe small differences (less than 1 dB) at thehigher drive levels. HAI models the load (disk+pressure compensation+radia.tion) by amass, stiffness, resistance, and shape factor for each drive frequency. The values of theseparameters are chosen to match the drive point impedance at the drive frequency. Thisload is not correct at harmonics of the drive frequency produced by the nonlinearitiesof the device. In addition this approach is not applicable to nonsinusoidal drive signals.Figures 24-27 show the same comparison as above except that the NRaD model wasmodified to use the HAI loads. For all drive levels the results are virtually identical. Thisleads us to believe that the differences in figures 20-23 are due to differences in the waythe load was modeled.
The next eight figures contain comparisons related to the drive point impedance ofthe load. Figures 28 and 29 show a comparison of the magnitude and phase of the drivepoint impedance between the NRaD time-domain model and the NRaD frequency-domainmodel. In the time-domain model only the fundamental components were used to form theimpedance. The agreement is quite good over the entire frequency band. Figures 30 and31 show the same comparison for the HAI time-domain and frequency-domain models.Again the agreement is excellent. Figures 32 and 33 show a comparison of the drive pointimpedance magnitude and phase between the NRlaD frequency-domain model and theHAI frequency-domain model. The agreement is excellent. Figures 34 and 35 show thesame comparison for the time-domain models. Again the agreement is good. Since theNRaD and HAI models produce virtually the same drive point impedance at the load,it appears that any differences in results must be due to higher harmonics of the drivefrequency.
The next four figures show results for a three-element line array in which the center-to-center spacing is twice the element diameter. Figure 36 shows a comparison of normalizedsource level computed with the NRaD time-domain model and the experimental resultsobtained at Lake Seneca by Naval Research Laboratory/Underwater Sound ReferenceDetachment (NRL/USRD).' The comparison is reasonably good both in levels and intrends, but there appears to be a small frequency shift in the peaks. Figure 37 shows acomparison of model results for a single element and for the three-element array. Thiscomparison indicates that at this element spacing the array source level can not be ob-tained by adding 20 loglo(3) to the single-element source level. Figures 38 and 39 showthe time response of the center velocity and drive cavity pressure for the middle elementof the three-element array. The fact that the transient behavior dies out very quickly isdue to inherently high losses in the second-stage valve flow.
The next three figures are for a proposed six-element line array in which the center-to-center spacing is 9.26 times the element diameter. Figures 40 and 41 show a comparisonof the time response of the center velocity and drive cavity pressure for a single element
'Data supplied by Steven Black of NRL/USRD in a private communication
30
versus a middle element of the six-element array. At this large element spacing the resultsare virtually identical. Figure 42 shows that the source level for this six-element arraycan be obtained by adding 20 log,0(6) to the single-element source level.
31
HAl NRaD
-B--
(2 -5
0S-5
z-5 5 15 5 35 45 55 65 75 85 95 105
xlO0
FREQUENCY-FL
Figure 20. Single element, -6 dB drive level.
xiOo HAl NLaD
0 -10
-5 5 is 25 35 45 55 65 75 85 95 105
FREQUENCY-FL
Figure 21. Single element, -3 dB drive level.
32
o HAl NRaDxd ................................ ,,
&n• -5
1 -10
0z
-15 1I I I I I
-5 5 1 .5 35 45 55 65 75 85 95 105x0O°
FREQUENCY-FL
Figure 22. Single element, 0 dB drive level.
xl~o HAI NRaD
ri 5
S-10
0z-15 1 L L I I I I I IL
-5 5 15 25 35 45 55 65 75 85 95 105x1O°
FREQUENCY-FL
Figure 23. Single element, 3 dB drive level.
33
x100 HAl N RDwod
0
...........5
0
U
0ci -I0
S -10
0z
-5 5 15 25 35 45 55 65 75 85 95 105X10 0
FREQUENCY-FL
Figure 24. Single element, -6 dB drive level, HAI load model.
,lO0 HAI NRaDmod
X1 .................
-~ 0
U
0U/ -5
S -10
0z-15 - L "L I IL
-5 5 15 25 35 45 55 65 75 85 95 105x100
FREQUENCY-FL
Figure 25. Single element, -3 dB drive level, HAI load model.
34
x- 0° HAl NRaDmod
5
01
0CA -5
N
i -10
"-5 5 15 25 35 45 55 65 75 85 95 105X09FREQUENCY-FL
Figure 26. Single element, 0 dB drive level, HAI load model.
xl0o HAl NRaDmod
5 .......
S -10
- 1 5 . _ l I I I I I
5 5 15 25 35 45 55 65 75 85 95 105
FREQUENCY-FL x~
Figure 27. Single element, 3 dB drive level, HAI load model.
35
I NRaD-TIME NRaD-FREQ
8
7
6
43I I I
-5O 0 50 100 150 20 250xlOp
FREQUENCY-FL
Figure 28. NRaD impedance magnitude comparison.
x1o HNRD.D-TME NRaD-FREQ90
.
~'60
30
S 0
~-30
-50 0 50 100 150 2 M0
FREQUENCY-FL
Figure 29. NRaD impedance phase comparison.
36
xI0' HAl-TIME HAI-FREQ8
7
I) 4
-50 0 50 I00 150 200 250xlO°
FREQUENCY-FL
Figure 30. HAl impedance magnitude comparison.
xldO HAI-TIME HAI-FREQ90
S60
S 30
W0
Z-30
-90 I I I
-50 0 50 100 150 200 250
FREQUENCY-FL
Figure 31. HAl impedance phase comparison.
37
xlO' HAl NRaD
a I,
X10 5 .................... o
7
16
S 5
41
3 -L L I-
-s0 0 s0 100 150o0 250~
FREQUENCY-FL
Figure 32. Impedance magnitude, NRaD vs HAI, frequency-domain model.
xlo HAI NRsD901
I I I ,
-' 60
30
~ 0
FPEQUENCY-FL
Figure 33. Impedance phase, NI~aD vs HAI, frequency-domain model.
38
S. . .. • . a uu l Il I 0. a i
x10' HAl NRaD
8
V
7
S6
S 5
j~~ 43I I I i I
-5O 0 5O 100 1L0 20025 Xlo°
FREQUENCY-FL
Figure 34. Impedance magnitude, NRaD vs HAI, time-domain model.
xi 0o HAl NRaD
- OIII I I
90
-00-30
0 60
-30 0 01015 0 5
x1O°
FREQUENCY-FL
Figure 35. Impedance phase, NRaD vs HAI, time-domain model.
39
219 EXPM~MENT COMPUTED
a sI
0 ..... .... ..... ....... ..... .... . ... .....
-5
i -10
-15
-2 -1
-30
-35
0Z I 5 9 I
"--0 0 50 100 150 200 2.5 300x1O0
FREQUENCY-FL
Figure 36. Source level for three-element array, experiment vs computed.
x100 SINGLE ARRAY
-5
S-10
S -15
o -20U) -25
-30
-40
-50 0 50 100 150 200 250 300xlO°
FREQUENCY-FL
Figure 37. Normalized source level, single element vs array.
40
x1o"6
4
-4
-6I I i I0.0 0.5 1.0 1.5 2.0 2.5
X102
TIME/(s/c)
Figure 38. Velocity of middle element of three-element array at freq=FL.
6 -SIXlIlgI
U 5
4
3
°0.o o1.0 ,.5 2.0 Mo"TIME/(s/c)
Figure 39. Drive cavity pressure of middle element of three-element array at freq=FL.
41
-40 SI LE . .ARRAY
4
S 0
* -2
-4
0.0 0.5 1.0 1.5 2.0 2.5
TIME/(s/C)
Figure 40. Velocity comparison, middle element of 1 x 6 array vs single element.Sx10-1 SINGLE ARRAY
9
U 5o o3
*2
0 ..5 1.0 13 -L.0 2.512.
"TIvME/(s /c)
Figure 41. Drive cavity pressure comparison, middle element of I x 6 array vs singleelement.
42
x1 0 SINGLE ARRAY
-10
-5
S-10
S-15
o -2•0 0 00 10 00 0 0
_2D
S-25
-30O
o 3 -0
-405
"-50 0 50 1'00 1150 t0 2500
FREQUENCY-FL
Figure 42. Normalized source level, single element vs six-element array.
43
__ __ __ __ _ I | i - ! i-
4 CONCLUSIONS AND RECOMMENDATIONS
A mathematical model has been derived for an array of hydroacoustic transducersand applied to the HLF-6A. The major new feature of this model is the time-domainapproach to the array acoustic interaction problem. This approach has been validated foran array of simple linear transducers by comparison with a frequency-domain approach.This linear problem was also used to study the effects of internal damping on the timerequired to reach steady state. It was found that when the damping was small the timeto reach steady state could be very long compared with that of a single element. Asthe damping was increased, the time to reach steady state decreased dramatically andthe steady state level became lower. Results for a single HLF-6A transducer have beencompared with corresponding results from HAI's computer model. The agreement is ingeneral very good. Slight differences are attributable to differences in the way acousticradiation loading is handled. The results of our time-domain model were also comparedwith experimental results for a close-packed three-element array. The comparison showedthe same general trends and levels, but appeared to be slightly shifted in frequency. Aproposed, wide-spaced, six-element array was also modeled. At this proposed spacing theinteraction effects were minimal. In both the three- and six-element arrays the time toreach steady state was about the same as for a single element since the internal dampingof the valve is large.
In the future we would like to investigate an alternative approach in which the in-teractions are modeled at the drive point impedance level instead of at the radiationimpedance level. This would avoid the difficulty of choosing which modes to incorporatein the model. We would also like to further modularize our computer code so that otherdifferential equation solvers could be employed.
44
5 REFERENCES
1. Benthien, G., "A Mathematical Model For A Hydroacoustic Transducer," NavalOcean Systems Center, San Diego, CA, Technical Report 588, March 1982.
2. Schenck, H., Improved Integral Formulation for Acoustic Radiation Problemc, J.Acoust. Soc. Am., vol. 44, 1968, pp. 41-58.
3. Benthien, G.W., D. Barach, and D. Gillette, "CHIEF Users Maaual," Naval OceanSystems Center, San Diego, CA, Technical Document 970, September 1988.
4. Golub, G.H. and J.M. Ortega, Scientific Computing and Differential Equations,Academic Press, San Diego, CA, 1992, pp. 26-30.
5. Blackburn, J.F., Fluid Power Control, Blackburn, J.F., Reethof, G., and Shearer,J.L. eds., Technology Press of M.I.T. and John Wiley & Sons, New York, NY, 1960,p. 300.
45
REPORT DOCUMENTATION PAGE Form APPOvOMB No. 0704-0188
"Pulic repao"g burden for lfts coictioin of rfotmntoali is eansaied to average I hour per reeporae. ircidnlg the time for revneeiing instrucions. searching ean data wiNces. gatherlng andma'•tintig f tM needed. Mid &oWnoplsenng "id reviewng the colleotior of informator. Send commernts regardig fis burden estimae or ay cow aspect of Its cliebon at infmrmation. nclwiingS fggeeboti for redu1cing lls dUrdeM . to Werlaeort Hadquwtars Serice.• , Directorlte tor Informatin Operatns end Reports, 1215 Jefferson Davs lga. Suft 1204. ninn. VA 20-4302.aid to hie OMfce of Managemen and Budge. Pepaeirori Reditclion Protect (0704-0188), Wesangoi. DC 20503
1. AGENCY USE ONLY (Leave bnh) 2. REPORT DATE 3. REPORT TYPE AND DATES COVERED
May 1994 Final: 1 Jul 1993 to 31 Mar 1994
4. TITLE AND SUBTITLE 5. FUNDING NUMBERS
TIME-DOMAIN MODEL FOR A HYDROACOUSTIC TRANSDUCER ARRAY PE 0603747NDN309127
6. AUTHOR(S)
George W. Benthien and Don Barach
7. PERFORMING ORGANIZATION NAME(S) AND ADORESS(ES) 8. PERFORMING ORGANIZATIONREPORT NUMBER
Naval Command, Control and Ocean Surveillance Center (NCCOSC)RDT&E Division TR 1654San Diego, CA 92152-5000
9. SPONSORING/MONITORING AGENCY NAME(S) AND ADDRESS(ES) 10. SPONSORING/MONITORING
AGENCY REPORT NUMBERSpace and Naval Warfare Systems CommandSPAWAR PMW 182-32Washington, DC 20363-5200
11. SUPPLEMENTARY NOTES
12a. DISTRIBUTIONIAVAILABIUTY STATEMENT 12b. DISTRIBUTION CODE
Approved for public release; distribution is unlimited.
13. ABSTRACT (Maxinum 200 worO
This report contains a description of a mathematical model for a hydroacoustic transducer array. Previous models have attempted tocouple frequency-domain models for the acoustic field with time-domain models for the internal components of the transducer. In thisreport, we present a complete time-domain model for the coupled transducer/acoustic-field problem that is applicable to both single ele-ments and arrays. This model can also handle nonsinusoidal signals, which is important since linear superposition is not valid for nonlineardevices. The acoustic radiation model involves obtaining certain interaction coefficients in the frequency domain and then transformingthese coefficients to the time domain by using Fourier transforms. When these coefficients are combined with a model for the internal com-ponents, the result is a coupled set of differential-integral equations that can be solved by an explicit time-stepping procedure. For pur-poses of validation, the time-domain acoustic-radiation model was applied to a simple linear trandsucer array for which the solution couldalso be obtained by the standard frequency-domain approach. The agreement between the two approaches was excellent. The time-domainmodel was also compared with experimental data for a three-element close-packed array. The agreement was reasonably good in both leveland trends.
14. SUBJECT TERMS 15. NUMBER OF PAGES
acoustic interactions transducer 57nonlinear array transient 16. PRICE CODEtime-domain
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