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Journal of Sound and Vibration (1996) 197(5), 547–571

EFFICIENT CALCULATIO N OF THE

THR EE-DIMENSION AL SOU ND PR ESSUR EFIELD AR OU ND A N OISE BAR R IER

D. DEcole Nationale des Ponts et Chausse es CERAM , 1 avenue Montaigne ,

93167 Noisy Le Grand Cedex , FR

(Received 4 November 1993, and in nal form 25 March 1996)

A numerical method is presented for calculating the sound pressure around a noisebarrier of constant but arbitrary cross-section. To obtain the exact solution of this problem,the Helmholtz equation must be solved in the three-dimensional domain outside the barrier.

It is shown in this article how to calculate this 3-D sound pressure from solutions of simplerproblems dened on the two-dimensional domain outside a cross-section of the barrier. Thenumerical solution of a large three-dimensional problem is thus avoided and the efciencyof the calculation is considerably improved especially when a whole frequency spectrumis needed. By using the boundary element method to calculate the numerical solutions inthe two-dimensional domains, examples are given of determinations of sound pressureelds created by a point source and by an incoherent line source. In this way the efciencyof barriers of different cross-sections can be compared by using the real sound pressurearound them. From the frequency spectrum at a point, one can then calculate by Fouriertransform the temporal variations of the sound pressure created by a noise source movingin a direction parallel to the barrier.

7 1996 Academic Press Limited

1. INTRODUCTION

A noise barrier is the most common protection used against noise propagation in outdooracoustics. To be able to estimate the efciency of a noise barrier of given geometry, onemust calculate its excess attenuation, dened as the quotient of the sound pressure withthe barrier over the sound pressure in free eld. The calculation of excess attenuations of barriers is thus the main modelling problem and knowledge of acoustic elds around suchobstacles is required.

For a straight wall on a rigid ground, the acoustic eld is well approximated bysemi-analytical formulas derived from the Sommerfeld-MacDonald solution for asemi-innite screen. Maekawa [1] used these results to propose a method which has longbeen used for noise barrier calculations. Pierce [2], Jonasson [3] and Tolstoy [4] improved

the method by solving the diffraction problem for a two-dimensional angle or for apolygonal line.

For more complex shapes, Keller’s geometrical theory of diffraction allows thedetermination of asymptotic behaviors for high frequencies in the shadow zone; see, forinstance, the papers of Kurze and Anderson [5] and Kurze [6]. However more preciseresults for arbitrary shapes and boundary conditions can be derived from numericalsolutions of the Helmholtz equation in the domain outside the noise barrier. They are

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usually obtained from the boundary element method for which one needs only to meshthe boundary of the uid domain. Using such a method Seznec [7] and Hothersall et al .[8, 9] presented numerical results for two-dimensional diffraction problems. In theircalculus the sound pressure is supposed to be created by a coherent line source, the soundpressure of which is given by

p(r)=(i/4)H 0(kr ), (1)

where r is the vector between the sound source and the observation point, r is its norm,k is the wavenumber equal to 2 p f /c, with f the frequency and c the sound speed. H 0 isthe Hankel function of the rst kind. The Helmholtz equation is solved in atwo-dimensional domain which represents the domain outside a cross-section of the wall.

Unfortunately the experimental data and the real sound pressures usually consistof noise radiated from a three-dimensional point source, the sound pressure of whichis given by

p(r) = e ikr /4pr (2)

The two-dimensional models can give only a crude estimate of the real sound eld createdby such a point source. However, Daumas [10] and subsequent authors comparedmeasured insertion losses for sound pressures created by point sources with results of two-dimensional calculations. They found a close agreement between the two values. So2-D calculations give very useful results to predict the performances of 3-D barriers. Onemust notice, however, that the comparison is made on the amplitude of the sound pressureand no result is given for the phase. So the possibility of calculating interference effectsis not clear. The second and more important remark is that these comparisons are madefor situations where the source and the receiver are in the same plane perpendicular to thebarrier. For reception points outside this plane, the 2-D calculations provide no way toestimate the pressure. So the solution of the true three-dimensional problem is desirable

in order to estimate correctly the real attenuation everywhere outside the barrier and tocompare it with experimental results usually obtained by using a rather point-like soundsource like a loudspeaker. Furthermore, knowledge of the pressure in a straight line alongthe barrier allows one to calculate the attenuation for an incoherent line source, which isknown to be a good description of road trafc noise.

The three-dimensional sound pressure can be obtained by numerical methods like theboundary element method or the nite element method. Unfortunately these methodsusually require meshing a two-dimensional surface for the boundary element method ora large part of the three-dimensional domain outside the barrier for the nite elementmethod. So the problem to be solved becomes rapidly very large and only simulations forsmall walls of limited length and for low frequencies are usually done. Using the boundaryelement method Filippi [11] gave results for a noise source, assuming a nite wall with zerothickness and a wavelength of the same order as the dimensions of the wall. Kawai [12]

and Antes [13] made calculations for nite length walls with pressure elds created by pointsources. They discretized a wall of rectangular shape and made calculations for frequenciesabout 100 Hz for walls of some meters in height and width. For instance, for a real barrierwith a length of 10 m and a height of 2 m, the discretization of the rectangular surfaceby a classical 3-D boundary element method with ve nodes per wavelength needsapproximately 500/ l 2 unknowns where l is the wavelength. At 1000 Hz, l = 0·34 m andone must solve a full linear system with about 4 × 10 3 unknowns which leads to possible



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but heavy computations. In comparison, the 2-D solution on a cross-section only requires10/ l 1 30 unknowns for the same frequency.

To avoid such difculties in 3-D calculations, it is proposed here to solve these problemsby using a series of solutions of simpler two-dimensional problems in the frequencydomain, for both real and complex frequencies. To solve the two-dimensional problemsone can use any numerical or analytical method. For the examples described in this articleresults are presented which are obtained from a numerical solution based on the boundaryelement method on a cross-section of the barrier. The meshes are thus one-dimensional,the calculations are two-dimensional but the mathematical transformation presented willgive the three-dimensional sound pressure. Furthermore for a broad band frequencyanalysis it is shown that the whole spectrum for a three-dimensional source can be deducedfrom the two-dimensional spectrum for real and imaginary frequencies.

Then it is shown how this can be used to estimate the attenuation provided by noisebarriers, especially in the case of an incoherent line source. Finally, by Fouriertransformation, the time variations of the pressure are calculated for a source moving ina direction parallel to the barrier.

2. PROBLEM SPECIFICATION

It is supposed that the barrier has a constant cross-section, as for the example presentedin Figure 1, and that it is innite in length. In other words the geometry is unchanged bya translation in the z-direction. The sound pressure is supposed to be created by a singlefrequency point source in the uid domain, with time behavior e −i v t. To calculate thepressure around the barrier one must solve the three-dimensional problem

D pdif + k 2 pdif = 0 in V3, 1 pdif /1n + 1 pinc /1n = 0 on 1V3,

pinc (r) = e ik=r − rs=/4p =r − r s=, 1 pdif /1n − i kp dif = o(1/r ), (3)

where V3 is the exterior 3-D domain outside the barrier with boundary 1V3. The soundpressure pinc and pdif are respectively the incident and the diffracted elds and the sum ptot = pinc + pdif is the total pressure. It is supposed that the boundary of the uid domain1V3, consisting of the ground and the barrier, is rigid. The vector r s is the position of thesound source. The last relation is the radiation condition for outgoing waves at innity.

Figure 1. Structure of constant cross-section.



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According to the limiting absorption principle (see reference [14]), one knows that thesolution is also the limit when n: 0+ of solutions of

D pdif + k 2(n) pdif = 0 in V3, 1 pdif /1n + 1 pinc /1n = 0 on 1V3

pinc (r) = eik(n)=r − rs=

/4p =r − r s=, (4)with pdif $ L 2(V3), k (n) = z k 2 + i n, n q 0 and Im( k (n)) q 0.

To solve this problem by the boundary element method one must descretize theboundary or at least a large part of the wall. The discretization of the ground can beavoided if one uses a Green function with an image source in the ground to take accountof the rigid boundary condition. Nevertheless the dimension of the discrete problem to besolved limits the practical solutions to walls of small lengths and to the low frequencydomain.

To avoid this complex computational scheme, one can start from the following relationin which the eld of a three-dimensional point source in free space is calculated from theelds of two-dimensional sources by Fourier transform (see reference [15], formula 6·616):

eik(n)

z r2 + z2

4pz r 2 + z2= 1

2p g + a

− a

i4 H 0(rz k 2(n) − a 2) eiaz da . (5)

Here 0 E arg (z k 2(n) − a 2) Q p and r = z x 2 + y2 q 0 is the radial distance in the ( x , y)plane between the observation point and the source. For imaginary values of the variableone has, when u q 0, (i/4)H 0(iu)=(1/2 p )K 0(u) where K 0 is the modied Bessel functionof zero order. The function (i/4)H 0(rz k 2(n) − a 2) is in fact the sound pressure of a linesource with a possible extension to the imaginary axis. One can then extend this relationfor point sources to solutions of diffraction problems and calculate the solutions of 3-Dproblems as Fourier transforms of solutions of 2-D problems.

So, one can dene a family of two-dimensional problems depending on a complexparameter m. One must calculate the diffracted qdif , incident qinc or total qtot sound pressures

solutions of the following two-dimensional exterior problems with rigid boundaryconditions on the ground, on the barrier and outgoing wave conditions at innity:

(D + m2)qdif = 0 in V2, 1qdif /1n + 1qinc /1n = 0 on 1V2,

qinc (x , y , m) = (i/4)H 0(rm). (6)

V2 is the 2-D uid domain outside a cross-section of the barrier and 1V2 is its boundary.The outgoing wave condition can be expressed as the existence of a function f with compactsupport such that

qdif (x , m) = g V2

i4 H 0(m=x − z=) f (z) dz. (7)

In the case of a real parameter m one can use the more familiar Sommerfeld radiationcondition

1qdif /1n − i mqdif = o(1/z r ). (8)

This problem has a unique solution for complex msuch that Im( m) e 0 (see reference [14]).



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Now the solution of the three-dimensional problem (4) can be recovered from thesolutions of the previous two-dimensional problems by the formula

p(x , y, z, k (n)) = 12p

g + a

− a

eia zq(x , y , z k 2(n) − a 2) da , (9)

where p and q denote the incident, diffracted or total sound pressure. The solution of thethree-dimensional problem is thus the Fourier transform of two-dimensional solutionswhich are much easier to obtain. The solution of problem (3) can be found as

p(x , y , z, k ) = limn: 0+

12p g + a

− a

eiazq(x , y , z k 2(n) − a 2) da , (10)

Remark . When k Q =a = the solution q(x , y , z k 2 − a 2) is exponentially decreasingbecause

K 0(z) 0 z p /2z e− z (11)

for large values of z. So the integral (9) is in fact exponentially decreasing in a when =a =q kand, generally, only a rather small range in the imaginary complex domain needs to becalculated.

To estimate the gain in computation provided by formula (10), one notices that theboundary element solution of the two-dimensional problem requires only meshing a curveinstead of a surface for the three-dimensional problem. Supposing for instance that the2-D mesh is made of n nodes. For the same precision, the surface mesh of the 3-D problemneeds a number of nodes proportional to n2. If the computational time is proportional tothe power three of the number of nodes, the times required are about n3 in 2-D and n6

in 3-D. One must, however, make several 2-D calculations. To estimate the number of calculations needed, one notices that the 2-D solution oscillates approximately like e ikr

where r is the distance between the source and the receiver. Here rmax is the maximum

distance of interest. To get ve points per period one must sample the wavenumber withthe spacing

Dk = 2 p /5rmax (12)

and the frequency with

D f = c /5rmax , (13)

So the maximum number of calculations to get the 3-D pressure at a frequency f is, inthe real domain,

n = f /D f = 5 frmax /c = 5 rmax /l . (14)

With approximately an equal number of points for the imaginary frequencies, one can

estimate the number of frequency points byn f = 10 rmax /l . (15)

Taking again the example described in the introduction, one can estimate the number of calculations to be (500/ l 2)3 in the 3-D case and (10/ l )3 × 10 rmax /l = 10 4rmax /l 4 in the2-D case. In Table 1 these two values are compared for different frequencies, forrmax =50 m.



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T 1Comparison of the number of operations

Frequency (Hz) 100 500 1000 5000

l (m) 3·40 0·68 0·34 0·072D case 4 × 10 3 2×10 6 4×10 7 2×10 10

3D case 8 × 10 4 109 8×10 10 1015

Although one needs to calculate for several frequencies in the 2-D case, the proposedmethod saves many calculations. The reduction is greater when the frequency is higher.In this example, at 1000 Hz, the 3-D calculation is about 2000 times longer than theevaluation of the 2-D series.

This approach is still more interesting when a complete frequency spectrum over a band[0, f max ] is needed. In the 3-D case one must calculate for each required frequency. Withthe proposed method only two-dimensional solutions over the band [0, f max ] and in alimited part of the complex imaginary frequency axis are needed. The integral (9) has anumerical cost which is negligible compared to the solutions of problems (6). So the 3-Dspectrum is calculated with a computational cost similar to the 2-D one.

3. NUMERICAL SOLUTION OF 2-D PROBLEMS

To solve the two-dimensional problem one can use any method. To illustrate theprevious discussion the following examples will be calculated with the boundary elementmethod. Let Gk(y, x) be the Green function for a half-space with a rigid boundarycondition: that is,

Gk(y, x)=(i/4)H 0(kr ) + (i/4)H 0(kr '), (16)

where H 0 is the Hankel function of rst kind of order zero. The distances are dened by

r = =y − x=and r ' = =y − x '=where x is the source point, x ' the image source with respectto the ground and y the observation point (see Figure 2). The total pressure qtot solutionof equations (6) satises the Kirchoff relation on the boundary

c(y)qtot (y) − g 1V

qtot (x) 1Gk

1nxdx = qinc (y), (17)

Figure 2. Denition of domains and normals.



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with c(y) = u(y)/2p if y$ 1V (12 if y is a smooth point). u(y) is the exterior angle at y and

qinc (y) is the incident pressure created by sources in the exterior domain. For the specialcase in which y is one of the points on the boundary where the barrier intersects the groundsurface one has in fact c(y) = u( y )/p . It is supposed that the normal at the boundary isdirected into Vext . To simplify the expression the rigid boundary condition is used. Relation(17) can be transformed into the more regular expression

qtot (y) − g 1V

(qtot (x) − qtot (y)) 1G0

1nxdx − g 1V

qtot (x)01Gk

1nx− 1G0

1nx1dx = qinc (y); (18)

where G0 is the static Green function given by

G0(y, x)=−(1/2 p ) log r − (1/2 p ) log r '. (19)

To avoid possible problems associated with the singular frequencies for which equation(18) has more than one solution and which are the resonance frequencies of the interiordomain, the Burton and Miller [16] formulation of the integral relation is used. Thisconsists in adding to relation (18) its derivative multiplied by a complex parameter b with

Im( b ) $ 0. They showed that this new equation has a unique solution at every frequency.Lin [17] proved the validity of this approach. Transforming the derivative equation toimprove its regularity one nally obtains

qtot (y) − g 1V

(qtot (x) − qtot (y)) 1G0

1nxdx − g 1V

qtot (x)01Gk

1nx− 1G0

1nx1dx

− b g 1V

(qtot (x) − qtot (y) − 9 tqtot (y) · (x − y)) 12G0

1nx1nydx

− b

g 1V

9 tqtot (y) · nx1G0

1ny

dx − b

g 1V

qtot (x)

012Gk

1nx1ny

− 12G0

1nx1ny

1dx

= qinc (y) + b 1qinc

1ny(y), (20)

where 9 t means the tangential derivative. This relation involves weakly singular terms only.The boundary condition 1q /1n = 0 has been used to remove the corresponding terms inthe equation. More details on the derivation of this relation can be found in reference [18].In the following, the parameter b is chosen as

b = i =k=/max (=k=2, 50); (21)

thus it gives the classical value i/ k at high frequency and a small value at low frequency.The discretization is made by quadratic three nodes elements with the interpolation

r(j ) = N 1(j )r1 + N 2(j )r 2 + N 3(j )r3 (22)

for the geometry where r1, r2 and r3 are the positions of the nodes and N i (j ) are theinterpolation functions dened on [−1, 1]. One has a similar expression for theinterpolation of the pressure.

To solve equation (20) numerically the collocation method is used. To avoid possibleproblems with the non-uniqueness of the normal at the extremity nodes of the elements



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Figure 3. Nodes ( N i ) and collocation points ( C i ) inside an element.

the collocation points are chosen to be inside the element as indicated in Figure 3. In anelement one denes three possible collocation points which have the co-ordinates −0·95,0·05 and 0·95 in the j interval. To get a nal square system one must associate a uniquecollocation point to each node. For the interior node only one choice is possible ( C 2). Foran extremity node the collocation point is chosen among the two nearest collocationpoints. Two choices are possible because there is a collocation point in the two elementshaving this node respectively as rst or third node. The nal choice depends on the globalnumbering of the nodes. Which one is effectively chosen has very little inuence on thenal result.

After the numerical evaluation of the integrals over each element one obtains the globalsystem

AQ = F (23)

where

Q = &q1...qn' and F = &qinc (y1) + b 1qinc /1ny (y1)...

qinc (yn) + b 1qinc /1ny (yn)', (24)

n is the total number of nodes, y1, . . . , yn are the collocation points and q1, . . . , qn thenodal values of the pressure. A results from the assembly of the elementary matrices of

each element.

4. COMPARISON WITH KNOWN 3-D SOLUTIONS

To estimate the precision of the previous methods and their efciency, the numericalresults are compared in two cases for which the 3-D solution can be easily calculated. Themain purpose is to study the inuence of the discretization on the numerical errors. Thesediscretization errors have two origins. A rst part of the error comes from the numericalsolution of the 2-D problems by the boundary element method. This study is classical andnothing new is provided here. Secondly the calculus is based on the evaluation of the 2-Dsolution at discrete frequencies, on nite intervals. To calculate the 3-D pressure for afrequency f , one must calculate the 2-D solution in the interval [0, f ] and for the imaginaryfrequencies. So the precision of the result will depend on the interval chosen in theimaginary axis. With [0, i f max ] denoting this interval, f max must be chosen large enough toallow one to neglect the contribution of frequencies larger than f max in the numericalevaluation of formula (10). Formula (11) can provide an estimate of this upper limit. Toget a solution with an error of order e for points at distance larger than rmin one musthave

e− armin E e: (25)



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Figure 4. Point source in a half-space.

that is,

(c/2prmin ) log (1/ e) E f max . (26)

The solution is also strongly dependent on the number of frequency points chosen in eachinterval. Formula (14) can provide an upper bound on the number of points required inthe real interval [0, f ]. In the imaginary interval, the pressure is more regular and thenumber of points can be smaller.

The rst case is a point source over a rigid plane for which the analytical solution isderived from relation (5) and is given by

eikz r2 + z2

4pz r 2 + z2+ eikz r'2 + z2

4pz r ' 2 + z2= i

8p g + a

− a

(H 0(rz k 2 − a 2) + H 0(r 'z k 2 − a 2)) e iaz da . (27)

r is the distance between the projections of the source and the reception points in the(x , y) plane. r ' is the corresponding distance for the image source. The sound pressure isto be calculated at the three points dened in Figure 4. Their co-ordinates are given inTable 2. Point R1 is a rather difcult case, being very close to the sound source.

In this example the two-dimensional sound pressure is given analytically byq( f ) = (i/4)H 0(2p fr/c) + (i/4)H 0(2p fr '/ c), (28)

with c = 343·5 m/s. This expression is evaluated for discrete frequencies values. To estimatethe inuence of the frequency discretization and the value of f max , four series of calculationswere made with the following values for the frequencies (all in Hz): (D1) f 0 = 0·0001, f 1 = 0·01, f 2 = 0·1, f 3 = 1, f 4 = 5 and between f 5 =10 and f max = f 204 = 2000 in steps of 10;(D2) f 0 = 0·0001, f 1 =0·01, f 2 =0·1, f 3 = 1, f 4 = 5 and between f 5 =10 and f max = f 403 =2000 in steps of 5; (D3) f 0 = 0·0001, f 1 =0·01, f 2 = 0·1, f 3 = 1, f 4 =5 andbetween f 5 =10 and f max = f 404 =4000 in steps of 10; (D4) f 0 = 0·001, f 1 = 0·01, f 2 =0·1,

T 2

Source and reception pointsx (m) y (m) z (m)

Source 0 0·8 0R1 0·1 0·8 0R2 10 1 0R3 10 1 10



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T 3Frequency steps and errors associated with the three points

D f (Hz) e ( f max = 2000 Hz) e ( f max = 4000 Hz)

R1 687 2·6 × 10 −2 6·6×10 −4

R2 6·9 0 0R3 6·9 0 0

f 3 = 1, and between f 4 =2·5 and f max = f 1603 = 4000 in steps of 2.5. For each discretizationthe pressures q( f n ) and q(i f n) were calculated. Between the frequencies f n and f n + 1 thepressure was calculated by a quadratic interpolation on the values of q( f n −1 ), q( f n ) andq( f n + 1 ), thus one could rebuild a continuous approximation of the exact function fromthese discrete frequency points.

The value of D f calculated by formula (13) and the error e, obtained by taking equalityin formula (26), are given in Table 3 for the three calculation points.

The symmetry allows one to calculate

i4p g

amax

0

(H 0(rz k 2 − a 2) + H 0(r 'z k 2 − a 2)) cos (az) da , (29)

where z a 2max − k 2 = 2 p f max /c. This integral is divided into four parts for a respectively

in [0, 0·5k ], [0·5k , k ], [k , 1·5k ] and [1·5k , amax ]. Each part is divided into sub-intervalsand in each interval the numerical integration is made by a Gauss formula with sevenpoints. Changes of variable can be used to improve the numerical integration near a = k .Table 4 gives the pressure at points R1, R2 and R3 for different frequencies and for thefour frequency discretization schemes (D1), (D2), (D3) and (D4). The inuence of themaximum frequency of calculation f max is especially important for points close to the sourcepoint as one can see by examining the results for the point R1. For this point the decreaseof the frequency step has no inuence while the increase in f max clearly improves the result.

T 4Pressure in a half -space

Frequency (Hz) 100 500 1000 2000

R1 analytical 0·734+ 0·155i 0·461+ 0·673i −0·228+ 0·726i −0·717− 0·350iR1 numerical (D1) 0·725 +0·155i 0·452 +0·673i −0·239+ 0·726i −0·842− 0·349iR1 numerical (D2) 0·726 +0·155i 0·452 +0·673i −0·239+ 0·726i −0·842− 0·349iR1 numerical (D3) 0·733 +0·155i 0·460 +0·673i −0·227+ 0·726i −0·716− 0·349iR1 numerical (D4) 0·734 +0·155i 0·461 +0·673i −0·227+ 0·726i −0·716− 0·349i

R2 analytical (10 −2 ) 1·431−0·626i −0·558− 1·041i −0·096+ 0·162i 0·553+ 1·431iR2 numerical (D1) (10 −2 ) 0·986 − 0·689i −0·590− 0·784i −0·068+ 0·147i 0·584 + 1·100iR2 numerical (D2) (10 −2 ) 1·330 − 0·654i −0·589− 1·010i −0·096+ 0·163i 0·576 + 1·400i

R2 numerical (D3) (10 −2 ) 0·986 − 0·689i −0·590− 0·787i −0·066+ 0·146i 0·587 + 1·100iR2 numerical (D4) (10 −2 ) 1·420 − 0·630i −0·563− 1·040i −0·096+ 0·163i 0·557 + 1·430i

R3 analytical (10 −2 ) 0·746+0·829i −0·488− 0·845i −0·287+ 0·500i 0·256+ 0·461iR3 numerical (D1) (10 −2 ) 0·741 + 0·534i −0·530− 0·566i −0·154+ 0·483i 0·362 + 0·377iR3 numerical (D2) (10 −2 ) 0·711 + 0·815i −0·489− 0·826i −0·275+ 0·505i 0·268 + 0·442iR3 numerical (D3) (10 −2 ) 0·741 + 0·534i −0·532− 0·565i −0·154+ 0·484i 0·362 + 0·377iR3 numerical (D4) (10 −2 ) 0·740 + 0·827i −0·488− 0·841i −0·285+ 0·501i 0·257 + 0·460i



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Figure 5. Diffraction by a rigid cylinder.

The results for the points R2 and R3 show that density of the discrete frequencies isessential for points far from the source. On the contrary the value of f max is not important,as expected. The results could be improved by using the fact that, for large distances, thepressure behaves like e ikr and it would be possible to interpolate e −i kr p(r , k ) instead of p(r , k ). This could save many calculations in the 2-D solution especially for points at largedistance for which, according to formula (13), the frequency step must be small.

In the second case, the pressure of a point source diffracted by a rigid cylinder has beenstudied. A solution of this problem is given in the Appendix. It consists mainly of Fouriertransforms of Bessel functions. This solution can be efciently calculated with arbitraryprecision by standard numerical routines. In the present case this formula has beencalculated with high accuracy and it provides a reference solution to which the numericalresults will be compared. So the global efciency of the numerical method (boundaryelement solution followed by a numerical Fourier transform of the interpolated results)can be estimated by comparison with this reference solution. The cylinder has a radius of 1 m and is innite in length. The calculations are for three different points which aredened in Figure 5. Their co-ordinates are given in Table 5.

The source is on the x -axis so the solution is symmetrical with respect to the ( x , z)-planeand is equal to the pressure diffracted by a rigid half-cylinder on a rigid plane. To estimatethe inuence of the frequency discretization two calculations were made with the followingvalues for the frequencies (Hz): (D1) f 0 = 0·001, f 1 = 0·01, f 2 = 0·1, f 3 = 1, f 4 =5 andbetween f 5 =10 and f max = f 204 = 2000 in steps of 10; (D2) f 0 = 0·0001, f 1 =0·01, f 2 =0·1, f 3 = 1, f 4 = 5 and between f 5 =10 and f max = f 403 = 2000 in steps of 5. The circular sectionof the half-cylinder was divided into N + 1 nodes of co-ordinates

N n = (cos ( np /N ), sin (np /N )) (30)

where 0 E n E N . N = 101 was taken to get approximately ve nodes per wavelength atthe maximum frequency. To calculate the numerical solutions of the two-dimensionalproblems a quadratic mesh was used with three node elements as described in Figure 3.

T 5Source and reception points around the cylinder

x (m) y (m) z (m)

Source 2 0 0R1 4 0 0R2 10 0 0R3 10 1 10



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T 6Pressure around a cylinder

Frequency (Hz) 100 500 1000 2000

R1 appendix (10 −2 ) −2·621− 1·187i 3·816− 3·270i 0·912− 4·533i −2·736− 1·890iR1 numerical (D1) (10 −2 ) −2·886− 1·220i 3·740− 3·279i 0·868− 4·519i −2·762− 1·886iR2 numerical (D2) (10 −2 ) −2·618− 1·203i 3·814− 3·275i 0·908− 4·533i −2·767− 1·887i

R2 appendix (10 −2 ) −0·197+ 0·522i −1·071− 0·902i 0·150+ 1·252i −0·783+ 0·018iR2 numerical (D1) (10 −2 ) −0·480 +0·539i −1·066 −0·721i 0·172 +1·097i −0·722 +0·264iR2 numerical (D2) (10 −2 ) −0·228 +0·521i −1·081 −0·876i 0·164 +1·236i −0·777 +0·025i

R3 appendix (10 −2 ) 0·209− 0·526i −0·448− 0·809i −0·515+ 0·724i −0·198− 0·706iR3 numerical (D1) (10 −2 ) −0·141− 0·540i −0·463− 0·660i −0·415+ 0·682i −0·270− 0·639iR3 numerical (D2) (10 −2 ) 0·173 −0·532i −0·451 −0·797i −0·503 +0·730i −0·207 −0·695i

Figure 6. Real part of the pressure at R1. —, Reference; , , calculated.·

Figure 7. As Figure 6 but imaginary part of the pressure.



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Figure 11. As Figure 7 but at R3.

discretizations give an approximation of the solution. The discretization (D2) is clearlybetter than (D1), as expected.

In Figures 6 and 7 the calculated solution is compared with the reference one for thereal and imaginary components of the pressure at point R1 on the frequency band [0, 2000Hz] with the frequency discretization (D2). Figures 8–11 show similar results for pointsR2 and R3. The comparison is quite good except perhaps at very low frequencies where

some differences can be seen.The two preceding comparisons have thus proved the accuracy of the proposed method.The frequency discretization does not need to be very ne except perhaps for very lowfrequencies or for observation points far from the source point. For most practical casesone can calculate the pressure of the three-dimensional problem for frequencies as highas 2000 Hz without difculty, which is not the case if a full three-dimensional problemwere to be solved.

5. APPLICATIONS TO NOISE BARRIERS

5.1.

The excess attenuation is dened as the sound pressure level with a wall divided by the

sound pressure in free eld:

At = 20 log 10 (= pwall =/= p free=). (31)

This formula can be used directly for point sources or coherent line sources for which= p free==(1/4 pr ) or = p free== 1

4=H 0(kr )=respectively.Calculations for incoherent line sources parallel to the barrier will also be made. An

incoherent line source is modelled as uncorrelated three-dimensional point sources locatedon a straight line parallel to the z-axis. The amplitude of such a source as function of the abscissa z can be represented as a random process with the cross-correlationfunction

E(m(z)m*(z + u)) = n(z) d(u), (32)

where n(z) e 0. In free eld the pressure of the whole line at a point x = ( x , y , 0) isgiven by

p(x) = g + a

− a

eikr (z)

4pr (z)m(z) dz (33)



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where r (z) = z x 2 + y2 + z2. The density of acoustic potential energy has the expectation

e = E 0= p(x)=2

4r c2 1= 14r c2 g + a

− a g + a

− a

eikr (z)

4pr (z)e−i kr (z')

4pr (z ') E(m(z)m*(z ') d z dz ' (34)

= 14r c2 g + a

− a

n(z)[4pr (z)]2 dz. (35)

If n is constant one can calculate the energy density for a line in free eld by

e free = 14r c2

pn(4p )2r = n

64pr c2r , (36)

with r = z x 2 + y2. In the presence of a wall, the pressure created at point x = ( x , y, 0)by a unit source located in the plane z = z0 is, with respect to formula (10),

P (x , z0) = 1

2p g + a

− a

e−i az0q(x , y, z k 2 − a 2) da . (37)

For a line source with the amplitude m(z) one has

p(x) = g + a

− a

P (x , z)m(z) dz. (38)

The expectation of the density of acoustic potential energy is thus

ewall =1

4r c2 g + a

− a

=P (x , z)=2n(z) dz. (39)

If the amplitude n is constant

ewall =n

4r c2 g + a

− a

=P (x , z)=2 dz (40)

Using Parseval’s relation yields

ewall =n

8pr c2 g + a

− a

=q(x , y , z k 2 − a 2)=2 da . (41)

This integral is calculated by Gaussian quadratures as for the numerical evaluationof formula (29). Note that this formula needs only the knowledge of the 2-D solutionsand can be efciently calculated. The attenuation provided by the barrier is thengiven by

At = 10 log 10 (ewall /e free ). (42)

5.2.

The previous results can now be used to compare the efciency of noise barriers of different shapes. The attenuations for the four cross-sections of Figures 12–15 were



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Figure 12. Straight wall. Case A.

Figure 13. Left-bent top. Case B.

Figure 14. Right-bent top. Case C.

Figure 15. T-shape wall. Case D.

calculated with a sound source at 50 cm above the ground and 1·9 m to the right of the wall

at point (2, 0·5). This source can be a point source, a coherent or an incoherent line source.The observation point is 2 m behind the wall and 50 cm above the ground at point (−2, 0·5).The two points are on the same cross-section (same z). The origin O is always at the bottomleft of the wall. In the four cases, the thickness of the wall is e = 10 cm and the totalheight H is equal to 2 m. For case D the width of the top is 50 cm. For cases B and D thewall is bent from (3/4) H to H at an angle of 45°. The walls and the ground are assumedrigid.



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Figure 16. Excess attenuation. Case A. —, Coherent line; ·····, point source; – – – , incoherent line.

Figure 17. As Figure 16 but Case B.

Figure 18. As Figure 16 but Case C.



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Figure 19. As Figure 16 but Case D.

Figures 16–19 show that a point source and a coherent line source give almost the sameresult. The attenuation is a complex curve especially for the cases B, C and D because of interference between the different waves diffracted by the barrier and reected from theground.

The comparison between these curves to nd the best shape is not easy. An incoherentline source gives, on the contrary, very different results with an attenuation increasing withfrequency and a much lower average value. This result needs to be modelledthree-dimensionally and cannot easily be deduced from the two-dimensional attenuationfor a coherent line source. There is no interference phenomenon in this case. So theattenuation seems to be less dependent on a particular choice for the frequency or theposition. This could provide a better estimate of the efciency of the wall.

In Figure 20 the attenuations for an incoherent line source for different forms of thetop of the wall are compared. The differences are small but the T-form seems a little betterby some decibels.

In Figure 21 the results presented here are compared with the numerical results of reference [9]. The calculation is made for a source at point (−15, 0, 0) and a receiver atpoint (50, 0, 0). The wall has a T-shape with a width of 20 cm, the top of the T- is 1·5 m

Figure 20. Differences in attenuation for different forms of the top of the wall. —, A; ·····, B; – – – –, C; — — —, D.



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Figure 21. Comparison with reference [9] results. —, Reference [9] results; ·····, coherent line results; – – – –,point source; — — —, incoherent line source.

wide and the global height is 2·72 m. The wall is symmetrical with respect to the y-axis.The comparison of the results shows a good agreement between the results of reference[9], calculated for a coherent line source at the octave center frequencies, and thecalculations here for the 2-D coherent line source and the 3-D point source insertion losses.The insertion loss is dened as the quotient between the pressure without the wall overthe pressure with the wall. The excess attenuation for an incoherent line source is alsogiven, showing a lower attenuation at high frequencies.

6. MOVING SOURCES

The previous results were obtained in the frequency domain, supposing the source washarmonic in time. One can, however, use this to get results for non-stationary cases. More

precisely, the pressure at a point created by a source moving in a direction parallel to thebarrier can be calculated.

6.1.The pressure is the solution of

DP −(1/ c2)12P /1t 2 = − s(t ) d(x − X(t )) (43)

if the source sends a signal with the amplitude s(t ) at the position X(t ) and time t . Thesource is supposed to move along the z-direction, so X(t ) = ( x 0, y0, z(t )). With p denotingthe Fourier transform of P in z and t , one obtains

P (x , y , z, t ) = 1(2p )2 g

+ a

− a g + a

− a

p(x , y, a , v ) eiaz e−i v z da dv ,

p(x , y , a , v ) = g + a

− a g + a

− a

P (x , y , z , t ) e−i az eiv t dz dt . (44)



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Taking the Fourier transform of equation (43) in z and t yields

Dx, y p − a 2 p + k 2 p = − g + a

− a g + a

− a

s(t ) d(x − X(t )) e−i az eiv t dz dt

= − 0g + a

− a

s(t ) e−i az(t) eiv t dt1d(x − x0) d( y − y0)

= − S (a , v ) d(x − x 0) d( y − y0), (45)

with

S (a , v ) = g + a

− a

s(t ) e−i az(t) eiv t dt . (46)

So

p(x , y , a , v ) = S (a , v )q(x , y , z k2

− a2

) (47)where q is the previously dened solution of the two-dimensional problem. The pressureeld is thus

P (x , y , z, t ) = 1(2p )2 g + a

− a g + a

− a

S (a , v )q(x , y , z k 2 − a 2) eiaz e−i v t da dv . (48)

6.2.

For a uniform motion z(t ) = z0 + Ut and with s denoting the Fourier transform of s

S (a , v ) =

g + a

− a

s(t ) e−i a(z0 + Ut ) eiv t dt = e −i az0s(v − aU ). (49)

The speed of the source generates a shift in frequency, a Doppler effect. The pressure isthen

P (x , y, z, t ) = 1(2p )2 g + a

− a g + a

− a

e−i az0s(v − aU )q0x , y , Xv 2

c2 − a 21eiaz e−i v t da dv . (50)

For a harmonic source of unit amplitude pulsating at the angular frequency v 0, one hass(t ) = e −i v 0t. Its Fourier transform is s(v ) = 2 p d(v − v 0) and

P (x , y , z, t ) = 12p e−i v 0t g + a

− a

q(x , y , z (v 0 + aU )2/c2 − a 2) eia(z − z0 − Ut ) da . (51)

The integral is no longer symmetric with respect to 0. One also has

(v 0 + aU )2/c2 − a 2 = 0 (52)

for

a+ = k /(1− M ), a− = − k /(1+ M ), (53)



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where M = U /c is the Mach number and k = v 0/c . The integral (51) is evaluatednumerically for a q 0 over the intervals [0, 0·95 a+ ], [0·95a+ , a+ ], [a+ , 1·05a+ ] and[1·05a+ , amax ] where amax is dened by

a 2max − ( v 0 + amax U )2/c2 = (2 p f max /c)2. (54)

Similarly for a Q 0 the integral is calculated on [0·95 a− , 0], [a− , 0·95a− ], [1·05a− , a− ] and[amin , 1·05a− ] where amin is the negative solution of equation (54). Each interval is dividedinto 500 sub-intervals over which a seven points Gauss formula is used.

6.3.The results are rst compared with those of an analytical solution. In free eld the

retarded potential formula allows one to calculate the pressure created by a point sourcewith a uniform motion and the amplitude s(t ) = e −i v 0t as (see reference [19])

P (x , t ) = e−i v 0(t − =x − y/c)

4p (=x − y=− ( x − y) · U /c) , (55)

where U is the velocity of the source and y is its position at time t − =x − y=/c , which isthe solution of

y = X (t − =x − y=/c), (56)

where X(t ) is the position of the source at time t . To take into account the rigid groundone must add the pressure created by the image source. So

P (x , t ) = e−i v 0(t − =x − y=/c)

4p (=x − y=− ( x − y) · U /c) + e− i v 0(t − =x− y'=/c)

4p (=x − y'=− ( x − y') · U /c) (57)

where

y' = X'( t − =x − y'=/c), (58)

X' being the position of the image source.For the numerical example it is supposed that the source moves along the z-direction

at (2, 0·5) in the ( x , y)-plane with the speed 50 m/s and radiates a harmonic signal of frequency 200 Hz. The pressure is calculated at point (−2, 0·5, 0). In all the examples thesource is supposed to be at z = 0 at time 0. So at this time the source is in front of theobservation point. The source sends the signal s(t )=Re(e −i v 0t). The signal at theobservation point is Re( P (x , y, z, t )) where P is given by formula (57) or (51). P can alsobe written as an amplitude multiplying the source signal as

P (t ) = A(t ) e−i v 0t. (59)

To avoid an over complex gure Re(e iv 0tP (x , y , z, t ))=Re( A(t )) is presented instead of

Re( P (x , y, z, t )). In Figure 22 the analytical and numerical results over the time interval[−0·5 s, 0·5 s] are compared. It can be seen that the numerical results give a very goodestimate of the pressure.

To take more complex examples the straight wall of section 5.2 is again used and thepressure at the observation point behind it is calculated. The source still moves along thez-direction at (2, 0·5) in the ( x , y)-plane. Re( A(t )) with P given by formula (51) is againpresented.



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Figure 22. Source moving in a half-space. —, Analytical; , , numerical.

Figure 23. Pressure for the frequency 200 Hz and the source speed 50 m/s. —, With wall; ·····, without wall.

Figure 24. As Figure 23 but for pressure modulus.



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Figure 25. As Figure 23 but for source speed 100 m/s.

For the results in Figure 23 it is supposed that the source has a speed of 50 m/s andsends a signal of frequency 200 Hz. The time history is presented in the time interval[−0·5 s, 0·5 s] with and without the wall. One notices an important decrease in pressurewhen the wall is added. The behavior is different for the negative and positive timesbecause of the Doppler effect.

Figure 24 presents the modulus of P (x , y , z, t ) over the time interval [−5 s, 5 s] for thecomplex signal s(t ) = e −i v 0t and the same speed of the source. This has the advantage of removing the oscillations and it allows a better estimate of the duration of the noise andof the attenuation provided by the barrier.

For Figure 25 the source is assumed to have a speed of 100 m/s. The duration of thenoise at the reception point is smaller than in Figure 23 and the frequency appearsconsequently higher. A stronger Doppler effect is noticeable in this case. On the contrary

the maximum of the pressure does not seem to be strongly affected by the speed of thesource.

Figure 26. As Figure 23 but for the frequency 800 Hz.



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In Figure 26 the source has a speed of 50 m/s but sends a signal of frequency 800 Hz.The attenuation provided by the barrier is stronger than for a source at 200 Hz.

7. CONCLUSIONThe proposed method allows an efcient determination to be made of excess attenuation

of noise barriers of constant cross-sections. One has to solve only two-dimensionalproblems for real and imaginary frequencies with a very simple one-dimensional mesh. Onecan obtain the pressure eld for a point source and for an incoherent line source.Comparisons can be made between different geometries in this interesting practical case.Furthermore by Fourier transformation one can obtain the time signal at a given pointin space for a noise source moving in a direction parallel to the barrier and can show theinuence of the frequency and of the velocity of the source. It would be interesting toimprove the model to take into account an impedance boundary condition on the barrierand a nite impedance ground.

REFERENCES

1. Z. M 1965 Memoirs of the Faculty of Engineering , Kobe University 11 , 29–53. Noisereduction by screens.

2. A. D. P 1974 Journal of the Acoustical Society of America 55, 941–955. Diffraction of soundaround corners and over wide barriers.

3. H. G. J 1972 Journal of Sound and Vibration 25, 577–585. Diffraction by wedges of niteacoustic impedance with applications to depressed roads.

4. I. T 1989 Journal of the Acoustical Society of America 85, 661–669. Exact, explicitsolutions for diffraction by hard sound barriers and seamonts.

5. U. J. K and G. S. A 1971 Applied Acoustics 4, 35–53. Sound attenuation bybarriers.

6. U. J. K 1974 Journal of the Acoustical Society of America 55, 504–518. Noise reduction bybarriers.

7. R. S 1980 Journal of Sound and Vibration 73, 195–205. Diffraction of sound aroundbarriers: use of the boundary elements technique.

8. D. C. H , S. N. C -W and M. N. H 1991 Journal of Sound and Vibration 146 , 303–322. Efciency of single noise barriers.

9. D. C. H , D. H. C and S. N. C -W 1991 Applied Acoustics 32,269–287. The performance of T-prole and associated noise barriers.

10. A. D 1978 Acustica 40, 213–222. Etude de la diffraction par un e cran mince dispose surle sol.

11. P. J. T. F 1977 Journal of Sound and Vibration 54, 473–500. Layer potentials and acousticdiffraction.

12. Y. K and T. T 1990 Applied Acoustics 31, 101–117. The application of integral equationmethods to the calculation of sound attenuation by barriers.

13. H. A 1991 Boundary element methods in acoustics , 225–260 (chapter 11). Applications inenvironmental noise (Computational mechanics publications). Amsterdam: Elsevier AppliedSciences.

14. C. H. W 1975 Scattering theory for the d ’Alembert equation in exterior domains(volume 442 of Lecture Notes in Mathematics). Berlin: Springer-Verlag.

15. I. S. G and I. M. R 1980 Table of integrals , series and products . New York:Academic Press.

16. A. J. B and G. F. M 1971 Proceedings of the Royal Society of London A323 , 201–210.The application of integral equation methods to the numerical solution of some exteriorboundary-value problems.

17. T. C. L 1984 Journal of Mathematical Analysis and Applications 103 , 565–574. A proof forthe Burton and Miller integral equation approach for the Helmholtz equation.



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18. D. D 1994 Ph .D . Thesis , Ecole Nationale des Ponts et Chaussees . L’acoustique desproble mes couple s uide-structure: application au contro le actif du son.

19. D. S. J 1989 Acoustic and electromagnetic waves . Oxford University Press (SciencePublications).

APPENDIX:POINT SOURCE SCATTERING BY A RIGID CYLINDER

A rigid cylinder of radius b scatters the sound pressure created by a point source locatedat (x 0, y0, z0). The pressure is the solution of

D p + k 2 p = − d(x − x0) d( y − y0) d(z − z0), (60)

with 1 p/1n = 0 on the surface of the cylinder and the outgoing wave condition at innity. p denotes the partial Fourier transform of p in the z-direction:

p(x , y , kz) = g + a

− a

p(x , y, z) e−i kzz dz (61)

The Fourier transform is the solution of Dx, y p + ( k 2 − k 2

z ) p = −e −i kzz0 d(x − x 0)d( y − y0) (62)

The solution of this problem in the domain outside the cylinder is (see reference [19])

p = s+ a

− a

i 4

iH m (mr0)H'm (mb) (Jm(mr )H'm (mb) − H m (mr )J'm (mb)) e−i m(F − F 0) e−i kzz0, (63)

with r0 = z x 20 + y2

0, r = z x 2 + y2 and m= z k 2 − k 2z . F and F 0 are respectively the polar

angles of points ( x , y) and ( x 0, y0) with respect to the x -axis. J m and H m are respectivelythe Bessel and Hankel functions of order m . This formula is valid for r Q r0. In the othercase one must exchange r and r0. The sound pressure solution of (60) is thus

p = i8p s

+ a

− a

e−i m(F − F 0) g + a

− a

H m(mr0)H'm(mb) (Jm (mr )H'm (mb) − H m(mr )J'm (mb)) e ikz(z − z0) dkz . (64)

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