The propagation of grazing incidence plane waves along a finite impedance boundary is investigated. A solution of the semi-infinite problem, where a harmonic motion, parallel to the boundary, is imposed along a line perpendicular to the boundary, is obtained. This solution consists of quasiplane waves, waves moving parallel to the boundary with amplitude and phase variations perpendicular to the boundary. Several approximations to the full solution are considered

Vanmoorhem, W. K.

Zhong, F. H.

English

NASA Technical Reports Server

N91-16684
ON THE PROPAGATION OF PLANE WAVES ABOVE AN IMPEDANCE SURFACE
F. H. Zhong and W. K. Van Moorhem
Department of Mechanical Engineering
University of Utah
Salt Lake City, Utah 84112
ABSTRACT
The propagation of grazing incidence plane waves along a finite impedance boundary is
investigated. A solution of the semi-infinite problem, where a harmonic motion, parallel to the
boundary, is imposed along a line perpendicular to the boundary, is obtained. This solution consists
of quasiplane waves, waves moving parallel to the boundary with amplitude and phase variations
perpendicular to the boundary. Several approximations to the full solution are considered.
INTRODUCTION
Mathematical modeling of the propagation and reflection of harmonic plane waves above a
finite impedance plane surface is a fundamental topic in acoustics. In the case where the angle
between the normal to the wavefront and the surface is not zero an analytic solution is very easy to
obtain. This solution consists of the incident plane wave propagating toward the surface plus a
reflected plane wave propagating away from the surface at the same magnitude of the angle between its
normal and the surface as the incident wave. The amplitude of the reflected wave is given by a
reflection coefficient that is expressed in terms of incident angle and the specific impedance of the
surface. However at zero incident angle (the wave normal parallel to the surface), complete cancellation
of the incident and reflect waves occurs in this model and a zero solution results. Most acoustic texts
claim that this situation is not possible [1-3].
McAninch[4] recently has investigated a related situation where a plane wave source is
generating waves that would move parallel to a surface if its impedance was infinite but where the
surface impedance is not infinite quasiplanewavesresult. McAninch s investigation, however, uses the
parabolic approximation where only waves traveling in one direction are allowed. This paper
approaches the same problem without the assumption of parabolic approximation.
FORMULATION OF THE PROBLEM
The governing acoustic wave equation for harmonic waves can be put in the form
(V 2 + k 2) _b=0 (1)
where the time dependent part of the potential, e- i co t, has been separated from the spatial part of the
potential, ¢(x,y). When an impedance boundary exists, the solution of equation (1) must also satisfy
the boundary condition
Cy+ T¢= 0 (2)
on y--0. Here the subscript y indicates a partial derivative with respect to y.
13
https://ntrs.nasa.gov/search.jsp?R=19910007371 2020-03-19T19:20:13+00:00Z
For huniquesolution,someextraconstraintsmustbeintroduced.Oneis to assumethatqbwill
not beaffectedby the groundimpedanceasy approachesinfinity, the secondis that the acoustic
pressureatx=0is givenby
¢(O,y)= 1 (3)
EXACT SOLUTION
Weassumethatthesolutionof (1)hastheform of
_b= ei kx + f(x,y) (4)
where e ikx can be considered as a solution without the boundary condition given by (2). Substituting
(4) into (1),(2) and (3) we get a new governing equation and set of boundary conditions
(V 2 +k z )f=0 (5)
fy (x,O) + Tf (x,O) = - Te i k x (6)
and
f (0,y) = 0 (7)
lim f (x,y) = 0 (8)
y--+,,.
Equation (8) results from the first uniqueness condition listed above.
The sine transform,
F 0v,y) = ff (x,y) sin ()Vx) dx
0
(9)
is equivalent to the Fourier transform of an even function and will be applied here. The inverse
transform is given by
x-2 fF (x,y) sin ()vx) d_.f (x,y) =
0
(10)
Applying (9) to (5) and (6) we have
and
k2) F=0
Oy2
Fy (_,,0) + _{F (LO) =- _/
(11)
(12)
14
Here,it is assumedthatk is acomplexnumberwith averysmallpositiveimaghlarypart.
Solving (11)andmakinguseof thegivenboundaryconditions,yields
F (_.,y)= A 0.,0) e- my (13)
A (_.,0) = - T _-
(_.2- k2 ) (T- m )
(14)
where m = "4(_.2-k2). Since the solution is required to remain finite, Re_/(_.2-k 2) >0. Substituting (13)
and (14) into (10) yields the inverse transform of F(_.,y) as
f my
f (x,y) - 2 T _" e- sin (2Vx) d_.
(k2 _k2 ) (?_m)
(15)
For convenience, substitute the identity
sin (_,x) = d _.x. e- i Xx2i (16)
into (15), yielding
7
f (x,y) =- 7-- ( 11 - 12 )
1K
(17)
where
0_ my
11 = _.e- e i _.x d_. (18)
(_,2 - k2 ) (T- m )
0_ _'e- mY
e-i kXd_ "
I2= (_2 k2)(T_m)
(19)
In order to evaluate the above two integrals, introduce the complex variable A = _.+is and define the
contour integrals
j A e-M_y. e iAx dAIcl ( A2 -k-'2 ) (T-M)
(20)
My
ae e-i Ax dA (21)IcIi
(A2-k2) (T-M)
First evaluate the integral Ici where the contour is shown in Figure 1 along with the branch lines
which extend from the imaginary axis to the points A = + k. The value of this contour integral is
determined by the residue within the contour. Writing
7= ot+i 13 {22)
15
it is clear a pole exists within the contour only when _ > 0 (since Re{A 2 - k 2 > 0, y - {AZ-k 2 =0 only
when Re(y) = o_ > 0), and this pole is at A = _/(k 2 + ,,/2). It is easy to determine the residue at this pole
to be
Res(_/k2+_2 )=_C -Ty+iq/-_+"_ x
Y
(23)
and
I1 = Ic1 =- Ic2 - Ic3 - Ic4 - Ic5 - Ic6 + f2 _ i Res (x/k2 (24)
when R + ,,% Ic2 will vanish, while
-i s2a/_+k2 Y sxic 3 = _ s e e- ds (25)
0 (s2+k2)(y-i_s2+k 2 )
k
-i ka/-_-- 7V2 Y e i_vx d_. (26)
IC4=I _.2 - k2 _k2 _. 2 e
( )(y-i - )
_ rti eikx (27)
Ic5 y
k
X
Ic6 =- I _,2_k 2 i _k2 _ _,26 ( )(y+
i k-J'_-- _,2 y eiZ,e x dX, (28)
Substituting the above integrals into (24), yields
k { iL?7x2yeI1 = _2_-_k 2 _k2 _,20 ( ) (y+i - i 2 y }
e
( y-i,Jk2 - k 2 )
e i Z.x d_
+ i s2_+k2 y
s_e_-
o (s2 +k2 ) (y-i,_J+k 2
irt eikx
e-sx ds + --_
)
i 2r_ _/y ei k2_f-_+_
+ y e- x
0
(29)
16
Thevalueof integral12 is much easier to evaluate. We choose a contour in the fourth quadrant, since
there is no pole within the contour. 12 can then be written as
i2 = f s e e- sx ds (30)
0 (s2+k2)(7+i_s2+k 2 )
Here it must be recalled that there is a branch line along the imaginary axis. Subtracting 12 from 11yields
_ i7_ eikx11- ½--q-
+ f _. , e
0 ( _'2 -k2 ) ( Y+ix/-k2-X 2
e e i _.x d_
) (7-i_k2- k 2 )
e i s_--+k z y }
" e "sx ds
) ( ]t+ix/s 2 +k 2 )
+ s e
(s2+k 2 ) ( _,_ i_/s2 + k 2
i2n -TYei_-_--_ x
+ ,,f e or>0
0 c_<0
(31)
By substitution of 7L=_/k2 -t z and s=-i_/k 2 -t 2 , the above two integrals can be combined into one. Thefinal result is
where
f -e ikx +P-K cz>0
f (x,y)--
[ -e ikx -K if<0 (32)
P=2e--tYei kf_-++Tz x
(33)
and
K - 2 y S ( t Cos (ty) - 7Sin (ty)) ei _ x dt
r_ o t('_ +t2) (34)
P is called the surface wave, and it both decays with increasing height y, and also decays with the
distance x due to the imaginary part of 7.
17
ASYMPTOTICEXPANSIONVALID FORSMALL RECEIVERHEIGHTS
1.Soft boundarycase
IntegralK canbeasymptoticallyevaluatedfor largex usingthe saddle-pointmethod. This
methodis discussedby MorseandFeshbach[5]andwill notbediscussedhere.Actually wecanuse
someof conclusionsfrom Wenzel[6] sincewehavethesamefactor",]k2-t2 asoccurredthere.
Thesteepest-descentpathhasbeenshownin [6] to begivenby
T = t + i s (35)
where
s -- t (36)
andt > 0. Again usingtheresiduetheorem,integralK canbe transformedinto the integralL. Note
thati 5'is notin theregionof concernsince[3>0.Thus
I P+L -i5'_ D
K=t
L L -i5'¢ D
(37)
where D is the region between positive real axis and curve s=-t (l+t2/k 2 )-t/2, p is the surface wave
given in (33) and
L = _SDPl T Cos(_+(Ty)-,I.,2_-TS'Sin (Ty) ei k_- Tz X dT
(38)
Substituting (37) into (32), we have
[ P+L -is'e F
K=t
L L - i5'_ F
(39)
The region F in 5' plane is bounded by the curve [3 > 0, 0 < c_=[_(l+ [32/k 2 )-1/2. The region F is
called the surface wave region in the far field (shown in Figure 2), which is same as that of reference
[6]. When 7 e F, we can easily show Re_ / k 2 + 5'2 > k,that means, if the surface wave exists,its
propagation speed is less than the speed of sound in free space. It is also found that Im "J k2 + 5'2
has a close relationship to the quantity (or [3/ k) so that a large imaginary part of 5' and low frequency
of the source can make the surface wave decay very quickly.
get
Expanding each expression in (38) around the saddle point T=0 and integrating each term, we
18
[ + ik y2y2 y3y3
L=lT_i_x eikX,(1-yy)[ 72x (1-Ty+ 2 6 )+O(x -z ) } (40)
Thisasymptoticexpansionis notuniformlyvalid.Theconditionsfor its validity are
and
I?l_-- >> 1
k[2_--'-'fy36 1<<1x
(41)
(42)
Extremely small IT I will not satisfy conditions (41) and (42), so another asymptotic method has to be
developed.
The total solution under the condition of large x can be obtained by substituting (39) into (4)
yielding
fP-L y_ F
_=_-L 7_ F
where P and L are given in (33) and (40).
(43)
If we neglect the surface wave, we can get an explicit
equation for the wave above ground in the far field as
3_
i(k x +--_-- +0)
1 42k )2 )2
*=T_ -k---x ((1-o_y +(13y ) e (44)
where 0 = - arctan (13 y / (1 - o_ y)). Furthermore, if the receiver is on the ground, the above
expression can be written as
20 Log _ = 20 Log a - 10 Log x (45)
with a = (l/IyI)(2k/x) in. This result shows that the acoustic pressure level drops 10dB when the
distance increases 10 times or 3 dB per doubling of distance.
2. Hard boundary case
As mentioned before, the asymptotic expansion given in (40) is not uniformly valid in 7, with
the method failing for small I_. An alternative method is developed in this section which is valid in
the small I_ case. The method is almost the same as that used in evaluating L except the factor 1 /
(_a + T 2) in (38) will not be expanded. After changing variables (38) becomes
L __
Making use of the formula
- _--_ _ X_2 _t2)-I7( 1 7Y) 2x eikX ( 2i---k e-t2dtxi
0
(46)
19
__._K__.e_z2_( 1 t2 2 i z erfc ( -iz ) Im (z) > 0z 2 -t 2)- e- dt = ___K___ez 2
0 2 i z ( erfc ( -iz ) - 2 ) Im (z) < 0
and neglecting the terms of order 72, yields
r
-e ikx [l+y(-_-- -Y)]
2 i_ k
-e ikx [ 1 + 7(_--g----g -y)- 2 ( 1 -YY)]
(47)
(48)
The conditions Im (7/qi ) > 0 and Im (7/_]i ) < 0 can be identified as o_ < [3 and o_ > [3 respectively.
¢z = [3 is the line which divides these two regions in 7 plane. This is exactly the bounding curve o_= [3
(1 + 132/k2) -1_ obtained previously provided that Iyl _ 0. Recognizing this relation,we substitute (48)
into (43) and rewriting surface wave approximately as P = 2 (1 - 7 Y) eikx, finally get the total field
expression as
qb=e ikx 1 +7 _ -Y
the condition for the validity of the above expansion is
lTly<< 1
and
171 << 1
although x can't be small because of the nature of the saddle point method.
(49)
(50)
(51)
ANOTHER ASYMPTOTIC EXPANSION VALID FOR LARGE RECEIVER HEIGHT
The asymptotic expansions obtained above have their limitations in application. For example,
they require the receiver's location to be near the ground. In this section we will derive a asymptotic
expansion which is valid for large R = "_x2+y 2 (except for small y). The idea is similar to that of
Chien and Soroka [7].
By using the identity sin @x)=(ei_.x-e-iXx)/2i and transformation )_=k sin (z), (15) becomes
7 I Tan(z)f (x,y) = -_- 7 + i k Cos ( z )
C
eik(y Cos(z) +x Sin(z)) dz (52)
20
ThecontourCis shownin Fi_mare 3. In order to get an expansion in terms of the variable R = "_/x2+y2,
we transform the Cartesian coordinate system into the polar coordinate system by
and
x = R Sin 0 (53)
y = R Cos 0 (54)
Substituting into (52), yields
Y I Tan(z)f (x,y) = _--_- y+ iCos(z)
C
ik RCos(z- 0)
e dz (55)
the saddle point for the function i k R Cos(z-0) is at z = 0 and the path of steepest descent is found to
be given by
Cos (u- 0) Cosh (v) = 1 (56)
where z = u + i v, by considering Im (i k R Cos (z - 0))=Im (i k R). This path, denoted as C' is
shown in Figure (3). Deforming contour C into C', adding the possible poles (Cos z = i _,/k ), we
have
iT N_ +9f(x,y)=Q+H(-Re(1---_-- Cos0- 1 -- Sin0))Pk 2
(57)
where Q is defined by (54) but with the contour C changed to C', H is Heavyside step function and P
is the surface wave given in (33). The condition for the existence of pole is explained in reference [7],
and will not be repeated here. In the limit of 0 approaching rt/2 the condition for the existence of the
pole in the present case is equivalent to the condition for the existence of the pole in (39).
Q can be evaluated asymptotically with a method similar to that used in evaluating L, i.e. to
expand each term around the saddle point 0 and then integrate them with suitable transformation of the
variable. The result is
X/ 2 yTan0 eikRQ= igkR i(y+iCos0)
1 [ 1 + i k Cos 01 + i---_-ff [ 2 (y+ikCos0) ik 1}(y+ik Cos 0 ) Cos 0 (58)
The conditions for the validity of the above expansion are
and
kR >> 1
Y +iCos0 RCos0
<< 1
(59)
(60)
21
It is clearthat 0 cannotbe too close to n/2. This limitation is complimentaryto the asymptotic
expansionsobtainedpreviously(for smally). Substitutinginto (4) yields
iv "4/
_b=eikX+Q+H(-Re(1---_ - Cos0- 1 --k2 Sin0))P
(61)
In the limit R---)oo, Q and P will va,ish, with the result that only the plane wave term remains.
Equation (38) can be evaluated accurately by numerical methods as well as by asymptotic
expansions. Calculations show that the results match quite well when y is small. Figure 4 gives the
amplitude of acoustic pressure on the ground versus the distance to the receiver obtained by numerical
integration and from (44). Figures 5a, b and c show the amplitude of acoustic pressure versus the
receiver height for several receiver locations as obtained from the asymptotic expansions, (40) and
(61). These figures are similar to the results obtained by McAninch [4].
CONCLUSIONS
The acoustic field of a plane wave at grazing incident to a finite impedance has been
theoretically investigated. Exact numerical and asymptotic expansions are developed, which are very
similar to those found by Wenzel [6] for a point source and by McAninch [4] using the parabolic
approximation to the wave equation. When y is small, the incident wave is indeed canceled, but the
result is not zero due to the existence of a surface wave and the wave denoted as L. Near the ground,
the acoustic pressure decays as x -1/2 (assuming the surface wave is neglected). The asymptotic
expansion for large distance R shows that the acoustic pressure decays as R -1/2 when R---)_ and
when the receiver is not close to the surface only incident wave exists.
REFERENCES
[1] Dowling, A. P., and J. E. Ffowes Williams, Sound and Sources of Sound, (Ellis Horwood Ltd.,
Chichester, 1983), p. 83.
[2] Morse, P. M., Vibration and Sound, (American Institute of Physics, Woodbury, N.Y., 1983),
pp. 366-368.
[3] Rudnick, I., "The Propagation of an Acoustic Wave along a Boundary." J. Acoust. Soc. Am.,
19, 348-356, (1947).
[4] McAninch, G.L., "Propagation of Quasiplane Waves Along an Impedance Boundary." AIAA
26th Aerospace Sciences Meeting Paper 88-0179 (1974).
[5] Morse,P. M and H. Feshbach, Methods of Theoretical Physics, (McGraw-Hill,New York, 1953),
p. 441.
[6] Wenzel, A. R., "Propagation of Waves along an Impedance Boundary." J. Acoust. Soc. Am.,
55,956-963 (1974).
[7] Chien,C.F. and W.W. Soroka, "Sound Propagation Along an Impedance Plane." J. Sound and
Vibration, 43 ,(1), 9-20, (1975).
22
is
c2
m..._
v )t
Figure 1. Integration path for the integral 11in the complex A plane.
I
I
I F
I
I
I
i...._
v
k
Figure 2. The region F in the y plane.
23
iv
C'
I
Figure 3. The u-v plane, showing the integration contour C and the steepest descent path C'.
Figure 4.
v
..el
0.8
0.7
0.4
0.3 -
0.2 -
0.1 -
0.0
0.0
I I I
/
J
/"
/'
/"
/"
i I 1/ ///////
I
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
acoustic pressure
Comparison between a numerical result and the asymptotic result, (44), for the parameters,
k = 18.313 and "/= 6.935 + i 19.015.
24
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
I
I I | I I
0.2 0.4 0.6 0.8 1.0 12.
A]d]Pu'ruDz
1.4
2.0
1.8
1.6
1.4
1.2
1.0
0.0
0.6
0.4
0.2
0.0
0.0 0.2 0.4 0.6 0.8
MdPUTI_E
i
I !
1.0 1.2 1.4
2.0
1.8
1.6
1.4
1.2
1.0
O.O
0.6
0.4 :
0.2
0.0
0.0 0.2 0.4,
I ! I I
0.6 0.8 1.0 1.2 1.4
_,m_JILTDE
Figure 5. Amplitude of the acoustic pressure versus the receiver height for different parameters:
(a) f = 1000 Hz; y = 6.935 + i 19.015.
(b) f = 2000 Hz; _ = -5.031 + i 35.932.
(c) f = 4000 Hz; _ = 18.038 + i 37.002.
25


On the propagation of plane waves above an impedance surface

Abstract

Similar works

Full text

Available Versions

NASA Technical Reports Server