Spherical wave scattering by a rigid screen with a soft edge

Applied Acoustics 60 (2000) 353±370 www.elsevier.com/locate/apacoust

Spherical wave scattering by a rigid screen with a soft edge T. Hayat*, S. Asghar Mathematics Department, Quaid-i-Azam University, Islamabad, Pakistan Received 17 November 1997; received in revised form 9 July 1999; accepted 9 August 1999

Abstract An analytic solution for the di€raction by a half-plane is given for a spherical wave incidence in still air as well as in a moving ¯uid. This is important because point source is known as a fundamental radiating device. The half-plane is rigid except for a soft strip in the vicinity of the edge. The di€raction coecient in both cases is strongly dependent upon frequency. The calculations indicate that the ®eld is enhanced as a result of ¯uid motion and this enhancement of the ®eld is independent of the direction of ¯ow. It is found that the magnitude of the far ®eld on the source side of the semi- in®nite plane seems to depend more critically on the length of the soft strip and the wave number. # 2000 Elsevier Science Ltd. All rights reserved. Keywords: Scattering theory; Spherical wave; Moving ¯uid; Rigid screen with soft edge

1. Introduction The study of di€raction by a semi-in®nite plane with mixed boundary conditions is a problem of current interest and arises in the design of the noise reduction by barriers. In particular, noise from motorways, railways and airports can be shielded by a barrier which intercepts the line of hearing from the noise source to the receiver. The acoustic ®eld in the shadow region of a barrier (when transmission through the barrier is negligible) is due to di€raction at the edge alone. An ideal barrier should be such that it is a good attenuator of sound and economical at the same time. The later requirement is not dicult to appreciate when one considers the miles of motorway in heavily built-up areas. One possible economic barrier construction is to have a rigid barrier (hence reducing the transmitted noise) of cheap material which is robust and not necessarily a good attenuator of edge di€racted noise. The barriers having absorbing lining on the surfaces are good attenuators of * Corresponding author. 0003-682X/00/$ - see front matter # 2000 Elsevier Science Ltd. All rights reserved. PII: S0003-682X(99)00042-0

354

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

sound. The provision of a barrier covered completely with an absorbing lining presents several diculties, among them the cost of construction and maintenance. Since di€raction phenomena are governed by the conditions at the di€racting edge, it would be more economic to cover the region only in the immediate vicinity of the edge with sound absorbing material as has been pointed out by Butler [1]. However the most important question is that of the length of absorbing material required to obtain approximately the same noise attenuation as a complete absorbent barrier. It is hoped that the present work will help to answer this question. The presence of an acoustically absorbing lining on a surface is described by an impedance relation between pressure (p) and the normal velocity ¯uctuation on the lining surface [2]: @p ˆ ik p; Re… ˆ 0 c=za † > 0; @n where the sound wave has time harmonic variation eÿi!t (! is the angular frequency); k ˆ !=c is the free space wave number, c is the velocity of sound, 0 is density of the medium, za is the acoustic impedance of the surface, n the normal pointing into the lining, and the complex speci®c admittance of the acoustic lining. We remark that j j ! 0, corresponds to an acoustically hard (or perfectly re¯ecting) surface and j j ! 1 corresponds to an acoustically soft surface (pressure ¯uctuation vanishing on the surface). The limiting case when the edge region on the barrier surface is soft is considered. In case of noise radiated from aero engines and for noise inside wind tunnels, it is necessary to discuss the acoustic di€raction in the presence of a moving ¯uid. If the wavelength of sound is much smaller than the length scale associated with the barrier, the di€raction process is governed to all intents and purposes by the solution to the canonical problem of di€raction by a semi-in®nite rigid plane with a soft edge. Rawlins [3] has presented theoretical work on this model by considering di€raction of sound by a rigid screen with a soft edge. In an other paper, Rawlins [4] derived an approximate boundary condition of an absorption type and utilized to study the problem of di€raction by an acoustically penetrable or an electrically dielectric half-plane. In continuation of this vein of seeking a better understanding of the interaction of acoustic ®elds with half planes, the object of this paper is to obtain the scattering response of a spherical wave by a rigid screen with a soft edge. Di€raction of a spherical wave (point source) by a rigid screen with a soft edge is calculated in still air and the moving ¯uid. The mathematical method used to solve the problem is the Jones method and the Wiener±Hopf technique [5]. Finally the far ®elds in both cases are calculated using asymptotic approximation [6]. 2. Propagation in still air 2.1. Formulation of the problem We consider a semi-in®nite plane occupying a position y ˆ 0; x  0 as shown in Fig. 1. The half-plane is assumed to be in®nitely thin, and over the interval ÿ ` < x < 0

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

355

Fig. 1. Geometry of the problem.

there is a soft substance; the remainder, ÿ 1 < x < ÿ`, of the half plane is rigid. The perturbation velocity u of the irrotational sound wave can be expressed in terms of a total velocity potential t …x; y; z† by u ˆ grad t …x; y; z†. The resulting pressure in the sound ®eld is given by p ˆ ÿ0 @ t =@t. We consider a point source to be located at the position …x0 ; y0 ; z0 †. The wave equation satis®ed by t in the presence of a point source is  2  @ @2 @2 2 ‡ ‡ ‡ k …1† t …x; y; z† ˆ …x ÿ x0 †…y ÿ y0 †…z ÿ z0 †; @x2 @y2 @z2 subject to the following boundary conditions: @ t …x; 0 ; z†=@y ˆ 0 t …x; 0



t …x; 0

; z† ˆ 0

‡

; z† ˆ

…x < ÿ`†;

…2†

…ÿ` < x < 0†; t …x; 0

ÿ

; z†

…3† )

‡

@ t …x; 0 ; z†=@y ˆ @ t …x; 0ÿ ; z†=@y

…x > 0†:

…4†

It is assumed that a solution can be written in the form t …x; y; z†

ˆ

0 …x; y; z†

‡ …x; y; z†;

…5a†

where 0 is the incident wave which accounts for the inhomogeneous source term and …x; y; z† is the solution of the homogenous wave equation (1) that corresponds to the di€racted ®eld. In addition, for the unique solution of the boundary value problem Eqs. (1)±(4) [7], we insist that represents an outward travelling wave as r ˆ …x2 ‡ y2 ‡ z2 †1=2 ! 1

356

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

and also satis®es the edge condition [6] t …x; 0; z†

ˆ O…1† and @ t …x; 0; z†=@y ˆ O…xÿ1=2 †

t …x; 0; z†

ˆ O…1† and @ t …x; 0; z†=@y ˆ O……x ‡ `†ÿ1=2 † as x ! ÿ`:

as x ! 0‡ ;

) …5b†

2.2. Solution of the problem We de®ne the Fourier transform pair by 9 …1 > ÿiksz > t …x; y; z†e dz; t …x; y; s† ˆ > = ÿ1 …1 > > ; t …x; y; s†eiksz ds: > t …x; y; z† ˆ …k=2†

…6†

ÿ1

In Eqs. (6) the transform parameter is taken conveniently to be ks; s is nondimensional. For analytic convenience one can write k ˆ kr ‡ iki …kr ; ki > 0†. The decomposition (6) is common in other ®eld theories as well, for example, Fourier optics [8,9]. Transforming (1) and the boundary conditions (2)±(4) with respect to z by using (6) and then using Eq. (5a), we obtain  2  @ @2 2 2 ‡ ‡ k

…7a† 0 …x; y; s† ˆ eÿiksz0 …x ÿ x0 †…y ÿ y0 †; @x2 @y2 

 @2 @2 2 2 ‡ ‡ k

…x; y; s† ˆ 0; @x2 @y2

…7b†

@…x; 0 ; s†=@y ˆ ÿ@o …x; 0; s†=@y; …x; 0 ; s† ˆ ÿo …x; 0; s†;

…x < ÿ`†;

…ÿ` < x < 0†;

…x; 0‡ ; s† ˆ …x; 0ÿ ; s†; @…x; 0‡ ; s†=@y ˆ @…x; 0ÿ ; s†=@y;

…8† …9†

) …x > 0†;

…10†

where

2 ˆ …1 ÿ s2 †:

…11†

We observe that the mathematical problem formulated in the transformed plane s is the same as in the two-dimensional case [3] for line source except that k2 2 replaces k2 and there is an extra factor exp…ÿikszo † comes in the right-hand side of Eq. (7a). Now, the method of solution of the boundary value problem consisting of Eqs. (7)±(10)

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

357

is essentially the same as used by Rawlins in Ref. [3]. However the detailed calculations are given for the convenience of the readers. The spatial Fourier transform of the velocity potential is de®ned as …1 …x; y; s†eix dx; …12† …; y; s† ˆ ÿ1

and its inverse as …x; y; s† ˆ

1 2

… 1‡i ÿ1‡i

…; y; s†eÿix d;

…13†

where  ˆ  ‡ i. The solution of Eq. (7a) can be written as eÿiksz0 …1† H0 ‰k f…x ÿ x0 †2 ‡ …y ÿ yo †2 g1=2 Š 4i   … eÿiksz0 1‡i exp if…x ÿ xo †2 ‡ i…k2 2 ÿ 2 †1=2 y ÿ y0 d ˆ 4i ÿ1‡i …k2 2 ÿ 2 †1=2

0 …x; y; s† ˆ ÿ

…14†

Introducing x0 ˆ r0 cos 0 ; y0 ˆ r0 sin 0 ; 0 < 0 < ;  ˆ ÿk cos 1 , in Eq. (14) and lettering r0 ! 1, we obtain, using the asymptotic form for the Hankel function 0 …x; y; s† ˆ b…s†eÿik …x cos 0 ‡y sin 0 † ; where b…s† ˆ ÿ

…15†

 1=2 eÿiksz0 2 ei…k r0 ÿ =4† ; 4i k r0

…16†

and 0 is the angle of incidence measured from the x-axis. Applying the transform (12) to (7b) gives …; y; s† ˆ A1 …†eiy ; y > 0;

…17†

ˆ A2 …†eÿiy ; y < 0;

…18†

where  ˆ …k2 2 ÿ 2 †1=2 is de®ned on the cut sheet for which Im…† > 0 when jIm…†j < Im…k †. From Eqs. (17) and (18) eÿi` ÿ …; 0‡ ; s† ‡ 1 …; 0‡ ; s† ‡ ‡ …; 0; s† ˆ A1 …†; 0

0

…19a†

eÿi` ÿ …; 0‡ ; s† ‡ 1 …; 0‡ ; s† ‡ ‡ …; 0; s† ˆ iA1 …†;

…19b†

eÿi` ÿ …; 0ÿ ; s† ‡ 1 …; 0ÿ ; s† ‡ ‡ …; 0; s† ˆ A2 …†;

…20a†

0

0

0

eÿi` ÿ …; 0ÿ ; s† ‡ 1 …; 0ÿ ; s† ‡ ‡ …; 0; s† ˆ ÿiA2 …†;

…20b†

358

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

where …1

‡ …; y; s† ˆ

0

… ÿ`

ÿ …; y; s† ˆ 1 …; y; s† ˆ

…x; y; s†eix dx;

ÿ1

…0

ÿ`

…x; y; s†ei…x‡`† dx;

…x; y; s†eix dx:

9 > > > > > > > > = > > > > > > > > ;

…21†

The primes denote di€erentiation with respect to y, and the ‹ subscripts denote functions which are regular and analytic in the upper …Im…† > ÿIm…k †† and lower …Im…† < Im…k cos 0 †† a-plane. Eliminating A1 …† from (19a) and (19b) and A2 …† from (20a) and (20b) yields the equations 0

0

0

eÿi` ÿ …; 0 ; s† ‡ 1 …; 0 ; s† ‡ ‡ …; 0; s†

…22a†

ˆ i‰eÿi` ÿ …; 0 ; s† ‡ 1 …; 0 ; s† ‡ ‡ …; 0; s†Š:

…22b†

Transforming the boundary conditions (8) and (9) with respect to x we have ÿ …; 0 ; s† ˆ

b…s†k sin 0 eik l cos 0 ; … ÿ k cos 0 †

…23†

1 …; 0 ; s† ˆ

ÿ
ib…s† 1 ÿ eÿi…ÿk cos 0 †` : … ÿ k cos 0 †

…24†

0

Substitution of Eqs. (23) and (24) in Eqs. (22a) and (22b) yield b…s†k sin 0 eÿi…ÿk cos 0 †` 0 0 ‡ 1 …; 0 ; s† ‡ ‡ …; 0; s† … ÿ k cos 0 †  ˆ i eÿi` ÿ …; 0 ; s† ‡

 ÿ
ib…s† 1 ÿ eÿi…ÿk cos 0 †` ‡ ‡ …; 0; s† : … ÿ k cos o †

Adding and subtracting Eqs. (25a) and (25b) gives ^ÿ …† ‡ ei` ^‡ …† ˆ ÿ

…25a†

k sin o b…s†eik ` cos 0 ; …k ‡ †1=2 … ÿ k cos 0

eÿi` ÿ …† ‡ L…† 1 …† ‡ ‡ …† ˆ

ÿ
ib…s† ÿ1 ‡ eÿi…†ÿk cos 0 †` ; … ÿ k cos 0

…25b†

…26†

…27†

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

359

where
…k ÿ †1=2 ÿ ÿ …; 0‡ ; s† ÿ ÿ …; 0ÿ ; s† ; 2i  0  0 1 1 …; 0‡ ; s† ‡ 1 …; 0ÿ ; s 0 ‡ ‡ …; 0; s† ; ^‡ …† ˆ 2 …k ‡ †1=2 ^ÿ …† ˆ

ÿ …† ˆ


1ÿ ÿ …; 0‡ ; s† ‡ ÿ …; 0ÿ ; s† ; 2

1 …† ˆ


iÿ 0 0 1 …; 0‡ ; s† ÿ 1 …; 0ÿ ; s† ; ‡ …† ˆ ‡ …; 0; s†; 2

L…† ˆ …k2 2 ÿ 2 †ÿ1=2 ; L…† ˆ L‡ …†Lÿ …†; L …† ˆ …k  †ÿ1=2 :

…28†

The unknown functions A1;2 …† appearing in Eqs. (17) and (18) have been determined after solving Eqs. (17) and (18) in the Appendix and are given by A1 …† ˆ

 ib…s† …k ‡ k cos 0 †1=2 … ÿ k cos 0 † …k ‡ †1=2

h i  W p …k ‡ †` 0 …k ÿ k cos 0 †1=2 ÿi…ÿk cos 0 †` p ÿ2 e ‡ …k ÿ †1=2 2 …k ‡ † hpi W …k ÿ †` eÿi` 0 ÿ 
 p  S‡ …k † ÿ D‡ …k † ‡ 2 …k ÿ † ÿ
 S‡ …k † ‡ D‡ …k † ; hpi W0 …k ‡ †` ÿ
ib…s†…k ‡ k cos 0 †1=2 p ‡ S‡ …k † ÿ D‡ …k † A2 …† ˆ 1=2 … ÿ k cos 0 †…k ‡ † 2 …k ‡ † hpi W0 …k ÿ †` eÿi` ÿ
p S‡ …k † ‡ D‡ …k † ; ‡ 2 …k ÿ † where hpi ei…k ‡=4† n p po 1 ‡ 2i …k ‡ †`F… …k ‡ †`† …k ‡ †` ˆ p ` ` > 0; arg…k ‡ † 4 ;

W0

…29†

…30†

360

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370 2

F…z1 † ˆ eÿiz1 S‡ …† ˆ

z1

2

eit dt;

hpi S‡ …k † ib…s†…k ‡ k cos 0 †1=2 ‡ …k ‡ †` W 0 …k ‡ †1=2 … ÿ k cos 0 † …k ‡ †1=2 ‡

D‡ …† ˆ

…1

ib…s†…k ÿ k cos 0 †1=2 eik ` cos 0 ; …k ‡ †1=2 … ‡ k cos 0 †

…31†

hpi D‡ …k † ib…s†…k ‡ k cos 0 †1=2 ÿ W0 …k ‡ †` 1=2 1=2 …k ‡ † … ÿ k cos 0 † …k ‡ † ÿ

ib…s†…k ÿ k cos 0 †1=2 eik ` cos 0 ; …k ‡ †1=2 … ‡ k cos 0 †

…32†

and s‡ …k †; D‡ …k † are obtained by outting  ˆ k in expression (31) and (32), and solving for S‡ …k † and D‡ …k †, respectively. The solution of the boundary value problem in the s-plane is now known and is given by t …x; y; s† ˆ 0 ‡ 1 ˆ 0 ‡ 2

… 1‡i ÿ1‡i

1 2

… 1‡i ÿ1‡i

A1 …†eÿix‡iy d; y > 0;

A2 …†eÿixÿiy d; y < 0;

…33†

…34†

where ÿ Im…k † <  < Im…k cos 0 †. 2.3. Asymptotic expressions for the far ®eld For the purpose of the scattered far ®eld expressions (33) and (34) are approximated asymptotically for k r ! 1. This is achieved by using the result. 1 2

… 1‡i

f…† eÿix‡ijyj d ÿ1‡i … ÿ k cos 0 †   f…ÿk cos 1 2jNjF…jNj† p ei…k ‡=4†  k 2k  …cos 1 ‡ cos 0 † exp‰ik …x cos 0 ÿ y sin 0 Š ‡ Sgn…cos 0 †H‰ÿ cos 0 …cos 0 ‡ cos 1 †Š k jsinj0  f…k cos 0 †;

…35†

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

361

as k  ! 1, where x ˆ  cos 1 ; y ˆ  sin 1 ; 0 < 1 < ; F…jNj† ˆ eÿiN

2

…1 jNj

2

eit dt; jNj ˆ

    k  1=2 cos 1 ‡ cos 0 : sin 1 2

The above result is obtained by the saddle point method, modi®ed slightly because of the presence of the pole  ˆ k cos 0 . When N is large the above result (35) can be simpli®ed by using the asymptotic expansion for the Fresnel integral, viz. i ‡ O…jNjÿ3 †; asjNj ! 1; arg…N† < : …36† F…jNj†  2j N j Expression (36) can be used in expression (35) if no pole occurs at  ˆ k cos 0 in the integrand. Thus, upon using the results (35) and (36), for 0 < 0 < , expressions (33) and (34) yield 9 t …r; ; s† ˆ I ‡ RF ‡ 1 for  ÿ 0 <  < ; > > > > > > ˆ I ÿ RF ‡ 1 for  <  ÿ 0 < ; = …37† for 0 <  <  ÿ 0 ; > ˆ I ‡ 1 ; > > > for 0 ÿ  <  < 0; > ˆ I ‡ 2 ; > ; for ÿ  <  < 0 ÿ ; ˆ 2 ; where x = r cos ; y ˆ r sin; ÿ 4  4 , I = incident wave ˆ b…s†eÿik r cos…ÿ0 † , RF = reflected wave ˆ b…s†eÿik r cos…‡0 † . In Eq. (37), 1 and 2 represent the di€racted waves in the illuminated region …0 <  < † and shadow region …ÿ <  < 0†, respectively, and are given by ( 1 eik…r‡r0 †ÿiksz0 D3 …r; r0 ; ; 0 ; s† 1 ˆ 4k …r0 †1=2 …r†1=2 ) eik …R‡r0 †ÿiksz0 D4 …; 0 ; R; r0 ; s† …38† ‡ …R†1=2 ( 1 eik …r‡r0 †ÿiksz0 D3 …r; r0 ; ; 0 ; s† 2 ˆ …r†1=2 4k …r0 †1=2 ) eik …R‡r0 †ÿiksz0 G2 …; 0 ; s† …39† ‡ …R†1=2

362

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

where (

) G1 …; 0 ; s 2 Q F… Q †…1 ‡ cos †1=2 …1 ‡ cos 0 †1=2 ÿ ; D3 …r; r0 ; ; 0 ; s† ˆ b…s† …cos  ‡ cos 0 † (

) G2 …; 0 ; s 2 Q1 F… Q1 †eik ` cos 0 …1 ÿ cos 0 †1=2 ‡ ; D4 …R; r0 ; ; 0 ; s† ˆ b…s† …cos  ‡ cos 0 †‰2…1 ÿ cos †1=2 Šÿ1

G1 …; 0 ; s† ˆ

G2 …; 0 ; s† ˆ  Qˆ

k r 2

 
W0 ‰…k `…1 ÿ cos ††1=2 k ÿ   p   S …k † ÿ D …k † ; ‡ ‡ …2k †1=2 …1 ‡ cos †ÿ1=2 2

 
W0 ‰…k `…1 ÿ cos ††1=2 Š k ÿ   p   …k † ‡ D …k † ; S ‡ ‡ …2k †1=2 …1 ‡ cos †ÿ1=2 2

1=2 

     cos  ‡ cos 0  R 1=2 cos  ‡ cos 0 ; Q1 ˆ ; sin  sin  2

R2 ˆ r2 ‡ `2 ‡ 2r` cos ;  ˆ sgn…† cosÿ1



 ` ‡ r cos  : R

The ®elds 1 …x; y; z† and 2 …x; y; z† can now be calculated by taking the inverse Fourier transform of Eqs. (38) and (39), which will then complete our problem. Thus, the inverse Fourier transform of Eqs. (38) and (39) gives

1

2

( … 1 1 1 eik …r‡r0 †‡iks…zÿz0 † D3 …r; r0 ; ; 0 ; s†ds ˆ 2 p p

8 r0 r ÿ1 ) … 1 1 eik …R‡r0 †‡iks…zÿz0 † D4 …R; r0 ; ; 0 ; s†ds ; ‡ p

R ÿ1 ( … 1 1 1 eik …r‡r0 †‡iks…zÿz0 † D3 …r; r0 ; ; 0 ; s†ds ˆ 2 p p

8 r0 r ÿ1 ) … 1 1 eik …R‡r0 †‡iks…zÿz0 † G2 …; 0 ; s†ds : ‡ p

R ÿ1

…40a†

…40b†

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

363

The integrals appearing in Eq. (40) can be evaluated asymptotically. For that we put r ‡ r0 ˆ r12 sin 12 ; z ÿ z0 ˆ r12 cos 12 and deform the contour by the transformation s ˆ cos…12 ‡ ip1 †; …ÿ1 < p1 < 1†. Hence for large kr we obtain ( 1 D3 …r; r0 ; ; 0 ; 12 †ei…kr12 ÿ =4† p p 1 …x; y; z† ˆ rr12 4 2kr0 ) D4 …R; r0 ; ; 0 ; 12 †ei…kR12 ÿ =4† p ; …41† ‡ RR12 (

D3 …r; r0 ; ; 0 ; 12 †ei…kr12 ÿ =4† p rr12 ) G2 …; 0 ; 12 †ei…kR12 ÿ =4† p ‡ ; RR12

1 p 2 …x; y; z† ˆ 4 2kr0

…42†

where R212 ˆ …R ‡ r0 †2 ‡ …z ÿ z0 †2 ; 12 ˆ sgn…12 † cos

ÿ1

  r12 cos 12 : R12

3. The e€ects of convection 3.1. Formulation of the problem In this section we make an assessment of the e€ects to be expected if the sound is propagating in a moving stream. We consider a small amplitude sound wave on a main stream moving with velocity U parallel to the x-axis and discuss the di€raction of a spherical acoustic wave due to a point source from a rigid screen with a soft edge in a moving ¯uid. The perturbation velocity u1 of the irrotational sound wave can be written in terms of total velocity potential t as u1 ˆ gradt . The resulting pressure in the sound ®eld is given by   @ @ ‡U …43† t: P ˆ ÿ0 @t @x The our problem becomes one of solving the following convective wave equation   2 @ @2 @2 2 @ 2 …1 ÿ M † 2 ‡ 2ikM ‡ 2 ‡ 2 ‡ k t …x; y; z† @x @z @x @y ˆ …x ÿ x0 †…y ÿ y0 †…z ÿ z0 †;

…44†

364

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

subject to the boundary conditions: @t …x; 0 ; z†=@y ˆ 0 t …x; 0 ; z† ˆ 0

…x < ÿ`†;

…45†

…ÿ` < x < 0†;

…46†

t …x; 0‡ ; z† ˆ t …x; 0ÿ ; z†

) …x > 0†;

@t …x; 0‡ ; z†=@y ˆ @t …x; 0ÿ ; z†=@Y

…47†

where M ˆ Uc is the Mach number. 3.2. Solution of the problem We assume that the ¯ow is subsonic i.e. jMj < 1 and making the following substutions x ˆ …1 ÿ M2 †1=2 X; x0 ˆ …1 ÿ M2 †1=2 X0 ; y ˆ Y; y0 ˆ Y0 ; z ˆ Z; z0 ˆ Z0 ; k ˆ …1 ÿ M2 †1=2 K; ` ˆ …1 ÿ M2 †1=2 L; t …x; y; z† ˆ ~ t …X; Y; Z†eÿiKMX : Using these substitution in Eqs. (44)±(47), we get  2  @ @2 @2 2 ~ ‡ ‡ ‡K t …X; Y; Z† @X2 @Y2 @Z2 eiKMX0 …X ÿ X0 †…Y ÿ Y0 †…Z ÿ Z0 †; ˆ …1 ÿ M2 †1=2 @ ~ t …X; 0 ; Z†=@Y ˆ 0 ~ t …X; 0 ; Z† ˆ 0

…48†

…X < ÿL†;

…49†

…ÿL < X < 0†;

…50†

~ t …X; 0‡ ; Z† ˆ ~ t …X; 0ÿ ; Z† @ ~ t …X; 0‡ ; Z†=@Y ˆ @ ~ t …X; 0ÿ ; Z†=@Y

) …X > 0†:

…51†

We note that the mathematical problem is the same as in Section 2 except that an extra factor ‰exp…iKMX0 †…1 ÿ M2 †ÿ1=2 Š comes in the wave equation. Now, following the same method of solution as in Section 2, the di€racted far ®elds 1 and 2 in illuminated and shadow regions, respectively, are given by

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

(

D~ 3 …R1 ; R0 ; ~; ~ 0 ; ~12 †ei…KR~ 12 ÿ =4† …R1 R~ 12 †1=2 )  D~ 4 …R1 ; R0 ; ~ ; ~ 0 ; ~ 12 †ei…KR12 ÿ =4† ; ‡ …R1 R12 †1=2

eÿiKM…XÿX0 † 1 …x; y; z† ˆ p 4 2KR0 …1 ÿ M2 †1=2

…52†

(

D~ 3 …R1 ; R0 ; ~; ~ 0 ; ~12 †ei…KR~ 12 ÿ =4† …R1 R~ 12 †1=2 )  G~ 2 …~ ; ~ 0 ; ~ 12 †ei…KR12 ÿ =4† ; ‡ …R1 R12 †1=2

eÿiKM…XÿX0 † 2 …x; y; z† ˆ p 4 2KR0 …1 ÿ M2 †1=2

365

…53†

where ( D~ 3 …R1 ; R0 ; ~; ~ 0 ; s ˆ

) ~ ~ 2 Q F… Q †…1 ‡ cos ~†1=2 …1 ‡ cos ~ 0 †1=2 ~ ~ ~ G1 …;  0 ; s† ; ÿ …cos ~ ‡ cos ~† b~ …s†

8 9
2 1=2

ieiKMX0 ÿiK…1ÿM † sZ0 ei…KlR0 ÿ =4† ; 2…1 ÿ M2 †1=2 …2KlR0 †1=2

 R12 ˆ R21 ‡ L2 ‡ 2R1 L cos ~

X0 ˆ R0 cos ~ 0 ; Y0 ˆ R0 sin ~ 0 ; …0 4 ~ 0 4 †; X ˆ R1 cos ~; Y ˆ R1 sin ~; l2 ˆ ‰1 ÿ s2 …1 ÿ M2 †Š;

366

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

! ~ ~ ˆ sgn…~† cosÿ1 L ‡ R1 cos  ; R1 R~ 212 ˆ …R1 ‡ R0 †2 ‡ …Z ÿ Z0 †2 ; 

R122 ˆ …R1 ‡ R0 †2 ‡ …Z ÿ Z0 †2 ; ~ 12 ˆ sgn…~12 † cos

ÿ1

! R~ 12 cos 12 ; R12

hpi S~ ‡ …Kl† ib…s†…Kl ‡ Kl cos ~ 0 †1=2 ‡ W …K†l ‡ †L S~ ‡ …† ˆ 0 …Kl ‡ †1=2 … ÿ Kl cos ~ 0 † …Kl ‡ †1=2 ‡

ib…s†…Kl ÿ Kl cos ~ 0 †1=2 eiKlL cos ~0 ; …Kl ‡ †1=2 … ‡ Kl cos ~ 0

hpi D~ ‡ …Kl† ib…s†…Kl ‡ Kl cos ~ 0 †1=2 ‡ W Kl ‡ †L D~ ‡ …† ˆ 0 …Kl ‡ †1=2 … ÿ Kl cos ~ 0 † …Kl ‡ †1=2 ‡

ib…s†…Kl ÿ Kl cos ~ 0 †1=2 eiKlL cos ~0 ; …Kl ‡ †1=2 … ‡ Kl cos ~ 0 †

4. Concluding remarks In this paper we have studied the simplest canonical di€raction problem of Wiener±Hopf type. An exact analytic solution for a spherical wave in still air and in presence of a moving ¯uid has been found. It is observed from Eqs. (52) and (53) that as a result of ¯uid motion the ®eld is increased by a factor …1 ÿ M2 †ÿ1=2 in comparison to still ¯uid. This enhancement of the ®eld is independent of the direction of the ¯ow since the ¯uid velocity U appears as jUj2 in the factor …1 ÿ M2† . It is worth noting that the magnitude of the attenuation for the soft strip is much more than one would expect to achieve in practice with an absorbing strip. This is to be expected because no practical surface is entirely soft. Also, the magnitude of the far ®eld on the source side of the half-plane is dependent on k`, ` representing the length of the soft strip and k the wave number. This is because of the constructive/ destructive interaction of the edge di€racted ®elds with re¯ected waves from the hard …ÿ1 < x < ÿ`† and soft …ÿ` < x < 0† regions, respectively. It is also of interest to note from the ®rst of expressions (37) that the sign of the re¯ected wave corresponds to re¯ection from the rigid part of the barrier. In the second of the expressions (37) the re¯ected wave sign corresponds to re¯ection from the soft region of the barrier.

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

367

Appendix From Eq (26) we have k sin 0 b…s†eik ` cos 0 … ÿ k cos 0 †…k ‡ k cos 0 †1=2   k sin 0 b…s†eik ` cos 0 1 1 ˆ ÿei` ^‡ …† ÿ ÿ : … ÿ k cos 0 † …k ‡ †1=2 …k ‡ k cos 0 †1=2

^ÿ …† ‡

…A1† sides of Eq. (A1). The left-hand side is Let f1 …† de®ne a function equal to both 1 regular, analytic and asymptotic to jjÿ2 as jj ! 1 in Im…† < Im…k cos 0 † and the right-hand side is regular, analytic and asymptotic to jjÿ1 as jj ! 1 in Im…† < ÿIm…k †. Hence, by the usual Wiener±Hopt argument [5], the polynomial representing f1 …† can only be a constant equal to zero. Hence from (A1) and (28) we get ÿ …; 0‡ ; s† ÿ ÿ …; 0ÿ ; s† ˆ

ÿ2ib…s†eik ` cos 0 …k ÿ k cos 0 †1=2 … ÿ k cos 0 †…k ÿ †1=2

…A2†

Now by multiplying Eq. (27) by 1=L‡ …† and using the general decomposition theorem, we obtain  

‡ …† ib…s† 1 1 ‡ ÿ ‡ U‡ …† ‡ V‡ …† L‡ …† … ÿ k cos 0 † L‡ …† L‡ …k cos 0 † ˆ

ib…s† ÿ Uÿ …† ÿ Vÿ …† ÿ Lÿ …† 1 …†: … ÿ k cos 0 †L‡ cos 0

…A3†

The functions U …† and V …† are the decomposition [5] of U…† ˆ

eÿi` ÿ …† ; L‡ …†

V…† ˆ ÿ

ib…s†eÿi…ÿk cos 0 †` : L‡ …†… ÿ k cos 0 †

…A4†

…A5†

Similarly, multiplying Eq. (27) by ei` =Lÿ …† we obtain

ÿ …† ib…s†eik ` cos 0 ÿ ‡ Pÿ …† ÿ Qÿ …† Lÿ …† … ÿ k cos 0 †Lÿ …† ˆ ÿP‡ …† ÿ Q‡ …† ÿ L‡ …† 1 …†ei`:

…A6†

368

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

The functions P …† and Q …† are the decomposition of P…† ˆ

ei` ‡ …† ; Lÿ …†

Q…† ˆ ÿ

…A7†

ib…s†ei` : Lÿ …†… ÿ k cos 0 †

…A8†

Expressions (A3) and (A6) are of the similar form as discussed by Nobel ([5] p. 196±202). Thus, following the procedure of Nobel [5] we obtain from (A3) and (A6) S‡ …† ˆ

ib…s†…k ‡ k cos 0 †1=2 …k ‡ †1=2 … ÿ k cos 0 † … 1‡ia i`…k ÿ†1=2 S …† ‡ 1 e d; ‡ 1=2 … ‡ † 2i…k ‡ † ÿ1‡ia

…A9†

ÿ Im…† < a < Im…k cos 0 †;

D‡ …† ˆ

ib…s†…k ‡ k cos 0 †1=2 …k ‡ †1=2 … ÿ k cos 0 † … 1‡ia i` e …k ÿ †1=2 D‡ …† 1 d; ÿ … ‡ † 2i…k ‡ †1=2 ÿ1‡ia

…A10†

ÿ Im…† < a < Im…k cos 0 †; where S‡ …† ˆ ‡ …† ‡ ÿ …ÿ† ‡

ib…s† ib…s†eik ` cos 0 ‡ ; … ÿ k cos 0 † … ‡ k cos 0 †

…A11†

D‡ …† ˆ ‡ …† ÿ ÿ …ÿ† ‡

ib…s† ib…s†eik ` cos 0 ÿ ; … ÿ k cos 0 † … ‡ k cos 0 †

…A12†

The asterisks denote that a pole is present at  ˆ k cos 0 in the region Im…† > ÿIm…k cos 0 †; otherwise the functions are regular and analytic in this domain. An exact solution of Eqs. (A9) and (A10) would be too dicult to obtain. We therefore obtain an approximate asymptotic solution for the situation k `  1. The physical meaning of this condition is that the soft tip covers a region greater than a wavelength. In expression (A9) the path of integration is moved vertically so that it crosses the pole  ˆ k cos 0 of S‡ …†. However, the path is not allowed to cross the

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

369

branch point  ˆ k . Thus the third term of (A11) will give a residue contribution so that S‡ …† ˆ

ib…s†…k ‡ k cos 0 †1=2 ib…s†…k ÿ k cos 0 †1=2 eik ` cos 0 ‡ 1=2 …k ‡ † … ÿ k cos 0 † …k ‡ †1=2 … ‡ k cos 0 † … 1‡ia i` e …k ÿ †1=2 S‡ …† 1 d; ‡ … ‡ † 2i…k ‡ †1=2 ÿ1‡ia

…A13†

Im…k cos 0 † < a < Im…k †: Similarly, from (A10) we have D‡ …† ˆ

ib…s†…k ‡ k cos 0 †1=2 ib…s†…k ÿ k cos 0 †1=2 eik ` cos 0 ÿ 1=2 …k ‡ † … ÿ k cos 0 † …k ‡ †1=2 … ‡ k cos 0 † … 1‡ia i` e …k ÿ †1=2 D‡ …† 1 d; ÿ … ‡ † 2i…k ‡ †1=2 ÿ1‡ia

…A14†

Im…k cos 0 † < a < Im…k †: Since k `  1, the dominant term of the asymptotic expansion for the integral appearing in (A13) and (A14) comes from the region near  ˆ k so that let … 1‡ia i` p  e k ÿ M‡ …† d I1 ˆ … ‡ † ÿ1‡ia M …k †  ‡ 2i

… 1‡ia

ei` …k ÿ †1=2 d; … ‡ † ÿ1‡ia

…A15†

provided M‡ …k † is well behaved in this region, which it will be provided 0 6 0; , Note that M‡ …k † ˆ S‡ …k † and M‡ …k † ˆ D‡ …k † in Eqs. (A13) and (A14), respectively. The integral (A15) can now be evaluated by wrapping the contour of integration around the branch cut  ˆ k . It can then be shown that I1 

M‡ …K †eik ` 

ˆ M‡ …k †W0

…1 0

eit` t1=2 dt; …t ‡ k ‡ †

hpi …k ‡ †` :

Im  > ÿa; …A16†

Now making use of (A16) in (A13) and (A14) we get expressions (31) and (32). Expressions (31) and (32), on being substituted into Eqs. (A11) and (A12), give expressions for  …†, from which, in conjunction with Eqs. (28) and (A2), one can

370

T. Hayat, S. Asghar / Applied Acoustics 60 (2000) 353±370

produce expressions for ‡ …; 0; s†; ÿ …; 0 ; s†, Substitution of these latter expressions, together with expression (24), into Eqs, (19a) and (20a) yield Eqs. (29) and (30). References [1] [2] [3] [4] [5] [6] [7] [8] [9]

Butler GF. Sound J and Vibr 1974;32:367. Morse PM, Ingard KU. Encyclopedia of Phys.. Berlin: Acoustics, Springer Verlag, 1961. Rawlins AD. Sound J and Vibr 1976;45:53. Rawlins AD. Int J Engg Sci 1977;15:569. Noble B. The Wiener±Hopf technique. London: Pergamon Press, 1958. Jones DS. The theory of electromagnetism. London: Pergamon Press, 1964. Peters AS, Stoker JJ. Commun in Pure and Appl Maths 1954;7:565. Mrozowski M. Arch Elektron Uber 1986;40:195. Lakhtakia A. Arch Elektron Uber 1987;41:195.