- 3 - A COUSTICS. k=ω /c the wavenumber in m - ¹ c = the speed of sound in the medium in ms - ¹

Transcription

1 - 3 - A COUSTICS WAVE MOTION IN A FLUID MEDIA 3. The Wave Equation The fundaental equation of acoustics is the Helholt Equation or the Wave Equation. This equation is a derivative of the uch ore general, but nonlinear Navier-Stokes equation for fluids. The Wave Equation can be derived fro the Navier-Stokes equation by assuing that the ediu is linear and retaining only first order ters linear in pressure. If the acoustic ediu is linear then waves of different frequencies will not interact with one another, i.e. superposition holds. The assuption of linearity of the ediu is generally accurate for audio acoustics, except in soe specific circustances, which will be discussed later. Assuing linearity in the transducer itself is not at all accurate, but fortunately for us we need only require the ediu to be linear. The ediu of air only deviates significantly fro this linear assuption at sound pressure levels (SPL) of about.nt/² or above 4dB (SPL). This is a substantial sound level that is seldo encountered in a free field and, hopefully, never at the receiver the ear. However, it can be encountered in sall spaces within the transducer itself. The Wave Equation can be substantially siplified by the use of coplex notation. When a coplex exponential is assued for the tie dependence i t Ψ () t = Ψ e ω (3..) and we insert this for into the Wave Equation, Ψ( t) Ψ() t = (3..) c t we get a second order partial differential equation where ² is the Laplacian operator. The new equation, which is the sipler Helholt Equation Ψ( ω) k Ψ( ω) = (3..3) k=ω /c the wavenuber in - ¹ c = the speed of sound in the ediu in s - ¹. See Morse, Methods of Theoretical Physics, Vibration and Sound or Theoretical Acoustics

2 ACOUSTICS - 43 Eq.(3..3) will for the basis of all our discussions in this chapter. For convenience we will usually drop any specific reference to either the tie dependence t or the frequency dependence ω as they will be assued. For linear solutions this is not a proble since an arbitrary solutions in frequency can be built up of a su of solutions at a particular frequency, as we saw in Chap.. Dropping these coplex dependencies greatly siplifies the writing, although we ust be careful to always reeber their continuous presence. The operator Ψ Ψ Ψ Ψ= + + (3..4) x y is the Laplacian, shown here in the failiar Rectangular coordinate syste. It is different in every coordinate syste. The Helholt Equation describes a scalar field in the scalar quantity Ψ, the velocity potential. It is called the velocity potential since the field velocity is its negative gradient. It has no physical significance but is a convenient quantity for us to use since either of the quantities of interest to us, pressure and particle velocity, can be derived fro it directly. The equations for the pressure p and the velocity v are p = iωρcψ v = Ψ (3..5) fro which it follows that iωρ v = p (3..6) ρ = the density of the ediu in kg /. The quantities c and ρ both vary with teperature, static pressure and huidity (aong other things), but these variations are usually negligible. (Although we will see that in roo acoustics these variations, however sall, have a profound effect.) The appearance of the Helholt Equation can be quite deceptive. Its solution in the general case would take an entire volue or ore to thoroughly investigate. We will investigate nuerous solutions of the Wave Equation, but in only a few different coordinate systes, and then with only a few specific boundary conditions. 3. The Helholt Equation in Rectangular Coordinates Solutions in Rectangular Coordinates are relatively easy copared to those in other coordinate systes. Unfortunately they are not a great deal of use to us in this text. Still, we can learn a significant aount about the general approach to the solution of the Helholt Equation in other coordinate systes by investigating the sipler solutions in Rectangular Coordinates.. See Morse, Vibration and Sound and Methods of Theoretical Physics.

3 44 - AUDIO TRANSDUCERS In Rectangular Coordinates the Helholt Equation is Ψ Ψ Ψ + + k Ψ = (3..7) x y Let us assue a solution that is the product of three separate solutions: Ψ ( x, y, ) = X( x) Y( y) Z( ) (3..8) We will find that, in general, the solutions to alost any proble starts with an assued solution which is then plugged back into the original equations in order to find out how well it fits. So assuing the general solution to be the product of three separate solutions in each of the three spatial coordinates is a reasonable place to start. Hindsight is also helpful. Using Eq.(3..8) in Eq.(3..7) results in X( x) Y( y) Z( ) k = X( x) x Y( y) y Z( ) which we can rewrite as (3..9) X( x) Y( y) Z( ) = k (3..) X( x) x Y( y) y Z( ) by siply rearranging the ters. For both sides of this equation to hold they ust both be equal to a coon constant. This last stateent is a powerful arguent, the validity of which is crucial to all of the following discussions. Its legitiacy is apparent by thinking about two functions, of different variables, varying independently, but yet still being equal. We will call this value the separation constant. Setting both sides of Eq.(3..) equal to the coon constant k x ² we will obtain x X( x) Y( y) Z( ) = k = k X( x) x Y( y) y Z( ) fro which it follows that (3..) X( x) = k x and X( x) x (3..) Y( y) Z( ) + = k k x Y( y) y Z( ) Following along with this sae process, once again, we find a coplete set of separated equations, each one conveniently in a single dependent variable, even though coupled together by a coon constant. This set is

4 ACOUSTICS - 45 k = kx + k y + k. X( x) k x X( x) = x Y( y) + ky( ) y y = y Z( ) k ( ) Z = (3..3) The wavenuber k is now in a for that is quite instructive, naely that it is actually a vector with eleents k x, k y, and k in the x, y and directions, respectively. This wavenuber vector concept, which we will call k-space, will becoe handy to us in the future. The three separated equations are, conveniently, all identical and they are known to have siple, well known solutions ikxx ikx x X ( x) = Axe + Bxe ik y y ik y y Y ( y) = A (3..4) ye + Bye ik ik Z( ) = Ae + Be where, as we has stated, we have ignored the tie factor. These equations are often written in the equivalent for of sines and cosines. Another useful for of this solution is to write iωt kir iωt+ kir Ψ ( r, t) = Ae + Be (3..5) where we have written all of the spatial coordinates siply as a vector r to the spatial point and the separation constants as a vector k in k-space. For Rectangular Coordinates this forulation and concept is hardly necessary, although instructive, for we can see that the general plane wave solution is ade up of wave coponents in the three orthogonal directions of the coordinate syste. This will hold true for any coordinate syste which is orthogonal. In Rectangular Coordinates it is hardly useful, but for ore coplex coordinate systes it will becoe very useful. An exaple of the utility of the k-space concept is in the study of the low frequency sound field of a sall rectangular roo which we will study in far greater detail in Chap.7. A sound wave propagating at a particular ode (l,, n) travels in a direction of k, given by l π π nπ k = i + j + k (3..6) Lx L y L where the roo has diension L x by L y by L. This iplies that a sound source which excites only a single ode in such a roo does not radiate sound in all directions, nor does the sound wave propagate freely about the roo. The sound

5 46 - AUDIO TRANSDUCERS wave is constrained to ove in precise fixed directions as given by the above equation. A loudspeaker exciting this ode would have propagating sound waves eitted only in the direction of k. There would be a sall direct field fro this source, but otherwise its directivity would be fixed in space by the vector k. The concept of source directivity, as we will coe to know it, does not exist in sall roos at low frequencies. 3.3 Cylindrical Coordinates 3 In Cylindrical Coordinates we will encounter soe significant coplications to our proble, naely, that the solutions are functions that are not as well known as siple coplex exponentials or sines and cosines. In the Cylindrical coordinate syste the Helholt Equation is Ψ Ψ Ψ r + (3.3.7) + k Ψ = r r r r ϕ Once again, we will assue that a solution exists which relies on only a single spatial variable Ψ (, r ϕ, ) = R() r Φ( ϕ) Z() (3.3.8) Following along lines, identical to the previous exaple, leads us to three separated equations in the three spatial variables, r, θ and d R( r) dr( r) r + r + dr dr d Φ( ϕ) + k Φ( ϕ) = ϕ dϕ ( r k k ) R( r) = d Z( ) + k ( ) Z = d We have already seen the solutions to the last two equations (3.3.9) ikϕϕ ikϕϕ Φ ( ϕ) = Ae + Be (3.3.) and ik ik Z( ) = Ae + Be (3.3.) In Eq.(3.3.) the separation constant k ϕ is not arbitrary, owing to the fact that the solution in φ ust be periodic. Therefore k ϕ ust be an integer, which we will call. The separation constant k reains copletely arbitrary and can take on any value within liits. As we will see, the k-space concept as a continuu is ost appropriate for this type of continuous wavenuber. ϕ 3. Morse, Methods of Theoretical Physics, and Theoretical Acoustics

6 ACOUSTICS - 47 Given the above restriction on the separation constants, the equation for R(r) in Eq.(3.3.9) becoes ( ) d R( r) dr( r) r + r + r k R( r) = (3.3.) dr dr This equation, known as Bessel s equation, and has been thoroughly studied 4. By assuing a solution as a power series in kr and inserting it into this equation we would find that a convergent series would result. This series has coe to be know as a Bessel Function. The solutions to Bessel s equation are then Rr ( ) = AJ (3.3.3) ( kr) J = are Bessel Functions of order n = n = n = n = 3 n = x Each value of exhibits a copletely different function, but one which is orthogonal to all the other functions in. The first five orders of this function are shown in Fig.3- as a function of the arguent kr (shown as x). Fro Eq.(3.3.) we can see that the equation is of second order in r and therefore ust have a second solution, which ust be orthogonal to the first. We can also see that the equation is singular at the origin, and as such, one of the solutions ust also be singular at the origin. The Bessel Functions are not singular anywhere. The techniques for finding this second solution are generally known 4. See Morse, Vibration and Sound. Figure 3- - Bessel Functions of order

7 48 - AUDIO TRANSDUCERS but these techniques are beyond the scope of this book. This second set of functions are known as Neuann functions and can be derived directly fro the Bessel Functions. The singularity of these functions at the origin generally liits their usefulness to probles that do not contain the origin. Soe of our probles are of this class, although any others are not. We ust consider each case separately and decide if we ust retain both solutions or not. Both of the solutions of Bessel s equation are a coplete orthogonal set on the interval r =, with a weighting function kr. The coplete solution to Eq.(3.3.) thus becoes R( r) = A J (3.3.4) ( k r) + B N ( k r) N = are Neuan functions of order. An alternative set of functions which we will have extensive use for are know as the Hankel Functions, and are denoted H (kr). They are defined as () H (3.3.5) ( k r) = J ( k r) i N ( kr) The superscript stands for of the second kind the first kind having a plus sign in the su. One set represents outgoing waves and the other incoing waves. We will usually consider only outgoing waves and drop the superscript. These functions have the characteristics of a traveling wave in Cylindrical Coordinates, like the coplex exponential functions in Rectangular Coordinates, the physical values being the real part. n = n = n = H(kr)... kr Figure 3- - Hankel Function agnitude for various orders

8 ACOUSTICS - 49 The agnitude of the Hankel Function waves is shown in Fig.3-. The real part of this function is identical to Fig.3-. The agnitudes of these waves all grow to infinity at the origin due to the presence of the Neuann functions. The higher the order of the function the faster they grow towards the origin, the singularity in the equation. As we will see the Bessel Functions will be useful for probles that require standing wave solutions and the Hankel Functions for probles that have propagating solutions. A useful fact is that the Bessel Functions are also orthogonal on a finite interval. An exaple is (one that we will have further use for shortly) when the proble dictates a two diensional (k =) solution and a boundary condition is given for R(r) at soe finite radius r = a a circle. If the boundary condition is Ra ( ) = (3.3.6) r then the characteristic functions for this proble (those functions which satisfy the boundary conditions) are r ψ n() r = J β, n (3.3.7) a β,n = the characteristic (eigen) values such that d r J (3.3.8) ( β, n ) = J( β, n) = dr a r = a The orthogonality integral for this set of Bessel Functions is π a n n ψ, n(, r ϕ) ψ, n (, r ϕ) rdrdϕ (3.3.9) π a J ( β, n ) n = n This integral represents the noraliation constants for the set of functions. 3.4 Spherical Coordinates 5 In Spherical Coordinates, the solutions will bear a striking reseblance to those found in the cylindrical case. In this coordinate syste the Helholt Equation is Ψ Ψ Ψ r θ k + sin( ) + Ψ = r r r r sin( θ) θ θ ϕ (3.4.3) 5. See Morse, Methods of Theoretical Physics

9 5 - AUDIO TRANSDUCERS Once again we will assue that a solution exists which relies on only a single spatial variable Ψ (, r θϕ, ) = R() r Θ( θ) Φ() φ (3.4.3) Inserting this assued solution into Eq.(3.4.3) yields the three separated equations d R dr ( ) r + r + k r n( n+ ) R = dr dr d dθ sinθ sin θ nn ( )sin θ dθ + + Θ= dθ (3.4.3) d Φ + Φ = dφ The equation for Φ(φ) is, by now, well known to us. We note that ust be an integer because of the periodic nature of the φ coordinate. For nearly all cases that we will encounter in this text will be ero, i.e. there will be axi-syetry about φ. The equation for R(r) looks new. However, with a siple change of variables this equation can be rewritten in a for whose solution we have already developed. By utiliing a new set of functions, which are based on Bessel Functions, we can obtain solutions to the radial equation. These new functions are obtained by ultiplying the Bessel Functions by kr -½ (and soe other constants) and using new orders of n+½. These new functions satisfy Eq.(3.4.3) for R(r). They are R( r) = = = h π J kr π H kr () n ( kr) n+ () n+ ( kr) in ( kr) ( kr) (3.4.33) where we have introduced a new sybol h n (kr) for these new functions. These functions are called the Spherical Hankel Functions of the second kind of order n. The spherical solutions are of sufficient iportance that they have been allocated their own nae, even though they are a siple derivative of the Bessel Functions. We will only be concerned with outgoing waves and so we will usually drop the superscript () which plays the sae role here as it did for the cylindrical case. Nuerical values and algoriths for these functions are readily available. 6,7 The equation inθ is new to us and warrants soe discussion. It is usually solved after transforing variables to µ = cos(θ ) giving us an equation in µ: n+ 6. see Morse and Ingard, Theoretical Acoustics 7. see Press, et. al. Nuerical Recipes

10 ACOUSTICS - 5 ( d T ( µ ) dt ( µ ) µ ) µ + ( ) ( ) (3.4.34) n n + ( T µ = dµ dµ µ ) where we have written these new function as T(µ). Solutions of this equation turn out to be describable by a finite series in with order n and degree. They are called the Associated Legendre Functions T n (µ) where d Pn ( µ ) T (3.4.35) n ( µ ) = ( µ ) dµ P n (u) = the Legendre Polynoials of order n. These functions are known by any different naes including Spherical Haronics, surface haronics and the Laplace Functions. For purposes of our iediate discussion we will restrict ourselves to the Legendre Polynoials where =. The first few of these functions are P ( µ ) =, P ( θ ) = P ( µ ) = µ, P ( θ ) = cosθ (3.4.36) P ( µ ) = (3µ ), P ( θ ) = 4 (3 cosθ + ) 3 P3 ( µ ) = (5µ 3µ ), P3 ( θ ) = 8 (5 cos3θ + 3 cosθ ) Fig.3-3 shows a plot of the Legendre Polynoials in the polar coordinate θ. They represent a onopole (n =) a dipole (n =) and higher order quadrapoles (n >). The Legendre Polynoials are an orthogonal set with n Pn ( µ ) P ( µ ) dµ = (3.4.37) n = + Now that we have coplete solutions for the individual coordinates in the spherical case, we can write the general solution as pkr (,, θ) = AP n n(cos θ) hn( kr) (3.4.38) where we have, as usual, dropped the tie exponential. n 3.5 A Spherical Exaple 8 The classic exaple proble in Spherical Coordinates is radiation fro a polar cap. This is a good approxiation to a loudspeaker in a box in free space even though it is not exactly correct. We will see later how to ake this proble ore realistic. 8. see Morse, Vibration and Sound or Theoretical Acoustics

11 5 - AUDIO TRANSDUCERS n= n= n= Figure Polar pattern for the angular odes of order n The boundary condition for this proble is that at the radius of the sphere, a, we ust have a given velocity v over a section of the sphere given by θ <3, a spherical cap. Thus we have v 3 θ 3 V ( θ ) = (3.5.39) otherwise We first assue a solution as a finite series of solutions of the type shown in Eq.(3.4.38), and then apply the boundary conditions to obtain our solution. We ust consider all possible solutions consistent with the separation constants. This will then deterine the values for the unknown coefficients that fit the given boundary conditions. Proceeding, let the pressure field be pr (, θ) = AP n n(cos θ) hn( kr) (3.5.4) n The boundary conditions can now be enforced by noting that (recall Eq.(3..6)) which results in pkr (, θ ) V ( θ ) = ik c r ρ r= a (3.5.4)

12 ACOUSTICS - 53 (3.5.4) Next we us the power of orthogonality by ultiplying both sides of this equation by P n (µ) and integrating fro to (3.5.43) Since the Legendre Polynoials are orthogonal we obtain (using Eq.(3.4.37), the orthogonality relationships for these polynoials) and finally n ' n n n AP(cos θ) h( ka) = iρcv( θ) ' = n n n n iρc V( θ ) P ( µ ) dµ A P ( µ ) P ( µ ) dµ h ( ka) ' iρc V( θ ) Pn( µ ) dµ = An hn( ka) n + (3.5.44) ( n ) An i ρ + = c V( ) Pn( ) d h ( ka) θ µ µ n (3.5.45) ( n + ) v = iρ c h ( ka) n Pn cos3 ( µ ) v (. Pn 87) Pn + (.87) = iρ c h n ( ka) We can now insert these values for A n back into Eq.(3.5.4) to get the final result iρcv hn ( kr) pr (, θ) = ( (.87) (.87) ) (cos ) (3.5.46) Pn Pn+ Pn θ hn ( ka) n Fig.3-4 shows the polar response for various values of ka. (This plot is noralied to the axial response.) As the frequency goes up (increasing ka) the polar response narrows until a value of about ka = 6, at which point it begins to take on a pattern which has a ore consistent directional response. This is the principle behind the concept of Constant Directivity (CD) which says that at sufficiently high frequencies the sound radiates directly fro the sources velocity profile, i.e. the angular variation of the radiated sound is the sae as the velocity distribution on the sphere. This is asyptotically true at high frequencies, but the next figure shows that this is not really true for ore practical frequencies where the wavelengths are not infinitesially sall. The frequency ust be fairly high for the polar response to even begin to approach CD at the angle of the wavefront. Shown in Fig.3-5 is the sae data as that shown in the figure above only in a different forat. We will use this new forat alost exclusively in later chapters, which is why we have introduced it. It is not the coon forat for presenting dµ ( )

13 54 - AUDIO TRANSDUCERS Figure Spherical radiation pattern for ka =,4,7& polar data, but this forat is able to present the entire frequency variation of the polar response in one plane. The ka or frequency value is along the horiontal axis and the angle the vertical axis. The contours are usually at 6 db intervals, although soeties they are at 3 or db intervals. When the plot shows a transparent portion (the grid shows through, then the data is above the axiu level usually db. In this figure, we can ore easily see CD occurring at about 3º, but not until we are substantially above ka=.. Note that the directivity first narrows until about ka=., and then it begins to widen asyptotically approaching 3º at about -3dB. If we exaine the pressure at the surface of the sphere r = a in Eq.(3.5.46) we can see that the only ters in k (the frequency variable) are the Hankel Functions. If we divide Eq.(3.5.46) by an expression for the velocity we will obtain an equation containing only those ters which deterine the frequency dependence of the radiation response. When noralied to ρ c, these ters are called the odal ipedances (ka) hn ( ka). (3.5.47) ( ka) = h n ( ka) The odal ipedances are shown in Fig These figures exhibit an iportant characteristic of sound radiation that we will find to be true of all radiation probles and that is ode cutoff. Note how each successive ode has a higher cutoff point below which it does not radiate sound to any appreciable degree.

14 ACOUSTICS Angle ka Figure Polar response plotted as a contour plot of ka value versus angle Each ipedance curve tends to peak at a ka value of n +, where n is the order of the ode. The ass loading also increases with order but tends to peak at n. At low frequencies only the lowest order ode can contribute any of significant sound to the radiation. No atter how uch we try to anipulate the velocity of the source, only that portion of the velocity that excites the lowest order ode will contribute to the sound radiation. We can see fro this that at low frequencies, there is nothing that we can do to affect the polar radiation pattern, except null out the ero ode and utilie only higher order odes. But in so doing we ust accept the extreely low radiation efficiency that will result. This characteristic is proof of soething that we already knew to be true, but perhaps had never known exactly why. Above about ka =, we can begin to affect the polar pattern to a liited degree. We will show a use for this effect in later chapters. These curves also show the degree to which the sound radiation diinishes at ever lower frequencies. Finally, the odal radiation ipedances are useful for calculating another iportant function. By siply integrating the angular ters over the surface of the spherical cap we can find a weighted su of the odal ipedance contributions to get the total radiation ipedance for the source 3 ( ) ( ka) = A P (.87) P (.87) P (cos θ)sin θ dθ ( ka) total n n n+ n n n (3.5.48)

15 56 - AUDIO TRANSDUCERS. (Iaginary) Ipedance (Real) n = n = n = n = 3 n = 4 n = 5 n = ka Figure Real and iaginary parts of the odal ipedances for n= 6 The results of this calculation are shown below. When ultiplied by the ipedance of the ediu ρ c and divided by the area this ipedance is the actual echanical ipedance seen by the diaphrag. There is an interesting trick that can be perfored with the spherical odal calculations. Since all odd odes have ero slope at µ = (θ =9 ), by excluding all the even odes we can force a boundary condition of ero velocity in a plane through the sphere perpendicular to the axis of the source. In other words, the sphere is now a heisphere placed against the wall. Fig.3-8 shows a coparison between the axial response for the free sphere and the heisphere ounted against a wall. The wall reflections cause a fairly large variation in the response as they alternately add and subtract fro the direct sound radiation due to phase delay effects. Fig.3-9 shows the polar ap for this exaple. Coparing this ap with the sae portion of Fig.3-5 shows that placing the source against the wall has a lessor effect on the polar response except at low ka. The axial response is affected to a greater degree because the reflections are all in the sae phase along this axis. If one ust put a source against a wall, it is better to do so with the source at an angle to the wall than noral to it. The ipedance of this source could be calculated just as we did previously, but only considering the odd odes in the calculation. This exercise is left to the reader. By now the reader should also be able to describe the syste (boundary conditions) that would result by taking only every fourth ode; or every sixth ode, etc. Another interesting exercise.

16 ACOUSTICS Magnitude (db) ka Figure Axial response coparison between free sphere and ounted sphere ,, Frequency Figure Showing the calculated radiation ipedance for a spherical cap

17 58 - AUDIO TRANSDUCERS Theta ka Figure Polar ap for heisphere against a wall 3.6 A Cylindrical Exaple We will now show an exaple of sound radiation calculations in Cylindrical Coordinates. This will give us an opportunity to discuss a new technique for sound radiation probles. Consider the source shown in the figure below. This is an infinite cylinder with a source placed in its shell. The cylinder has a radius a and the source is defined as 3 < θ < 3 (3.6.49) V ( θ, ) = V < b / There are two ways to think about this proble, and to a first approxiation the solutions are identical. First, we can think of this proble as a source placed on the shell of a cylinder such that it vibrates radially outwards. In reality it would be difficult to build such a source, at least not such that the velocity were unifor across the face. A b R.364 a 3 V(θ, ) Figure 3- - Geoetry for cylindrical radiation proble

18 ACOUSTICS - 59 bending source could be ade to approxiate this type of source, but it would likely have a diinished aplitude at the edges. Another way to think about this source is to consider a line source at the center of the cylinder such that at the walls, where the hole or aperture is, the velocity is approxiately unifor across the aperture. To a first order this is possible. To be exactly correct we would have to consider the effect of diffraction on the wavefront within the aperture and the effect that this diffraction would have on the velocity aplitude of the wave in the aperture. The diffraction in the aperture would act to alter the velocity distribution in the aperture so that it would not be unifor, as we have assued. It turns out that this odification of the aperture velocity takes place principally in a frequency range where the wavelength is approxiately the sae as the aperture diensions. Outside of this range this perturbation is negligible. Even in this range this effect is not a doinant one and ignoring it will only have a sall (second order) effect on our results. We will take the approxiate path and assue that the velocity is unifor in the aperture. The proble shown above is of great interest to designers of the current genre of high perforance sound reinforceent systes. These systes are being designed as large line arrays because of certain desirable features of these types of arrays. For the ost part the theory of these systes is based on solutions for an infinite line array, not for a finite one. We will see that the two solutions are really quite different. Since we have a know a solution for waves in the Cylindrical Coordinate syste we can define the solution in ters of this set of these cylindrical waves. Fro previous sections, we know that the coplete solution to an outgoing wave in this coordinate syste is ik r pr (, θ, ) cos( θ) B ( k) H ( kre ) dk = (3.6.5) We have ade two siplifications in this equation. First we have assued that the source will be syetric in θ and reduced the Fourier Series in θ to a single cosine function in (no sine function), and second, the exponential ter in k represents waves in both directions so long as we don t restrict the sign of k. We ust allow for all possible (syetric) angular solutions and since the solutions in this coordinate are a discrete set in the integer, this becoes an infinite su. The solutions in the direction have a separation constant k given by Eq.(3.6.5). The allowed values of k constitute a continuu in k-space and consequently the suation over all possible solutions in this coordinate becoes an integral in k. This integral is an exaple of the k-space foralis for sound radiation. The coefficients B (k ) consist of an indexed set of continuous functions which it is now our task to deterine. We have also excluded the second solution for the radial coordinate since we are interested only in waves that propagate outward fro the source. We ust also reeber that kr is not copletely arbitrary since

19 6 - AUDIO TRANSDUCERS ω k = = k r + k (3.6.5) c We are now in a position to apply the boundary conditions by setting Eq.(3.6.5) equal to the specified surface velocity which results in (3.6.5) (3.6.53) where the prie on the radial function eans that we ust take its derivative with-respect-to (wrt) its arguent (we ust not forget to take the derivative of the arguent wrt r which is where the kr coes fro). The functions f (θ ) and f () are siply the angular and vertical velocity functions of Eq.(3.6.49), respectively. Multiplying both sides of the above equation by cos(nθ ) e -ik and then integrating over θ fro θ =-π to π and =- to will yield (3.6.54) By the features of orthogonality (without which we are virtually helpless), and the Fourier Transfor in k-space, which in this case siply returns the original function (the double integral is a transfor and its inverse). Eq.(3.6.54) siplifies to fro which we can iediately deterine the coefficients B (k ) as where and pr (, θ, ) V( θ, ) = ik c r ρ r= a ' ik r r cos( θ) B ( k ) k H ( k a) e dk = ikρcv f( θ) f( ) π ik ikρcv f ( θ)cos( nθ) dθ f ( ) e d = π π π ' ik r r cos( θ)cos( nθ) dθ k H ( k a) B ( k ) e dk e d π ik ik ' r r ikρcv f ( θ)cos( nθ) dθ f ( ) e d = πb ( k ) k H ( k a) π kv B ( k ) i c A F( k ) = ρ ' kh r ( ka r ) π A = f( θ)cos( θ) dθ π (3.6.55) (3.6.56) (3.6.57)

20 ACOUSTICS - 6 ik Fk ( ) = f( e ) d (3.6.58) π We should iediately recognie these as a Fourier Series for the discreteθ coefficients and a Fourier Transfor for the continuous k coefficients. The last step in this calculation is to reinsert Eq.(3.6.56) into Eq.(3.6.5) Fk ( ) H( kr r ) e pr (, θ, ) = iρckv Acos( θ) dk ' kh r ( ka r ) Now using Eq.(3.6.5) we get the daunting equation (3.6.59) Fk ( ) H k k re pr (, θ, ) = iρckv Acos( θ) dk (3.6.6) ' k k H( k ka) Analytical solutions of this equation are not possible, but there are still ways that we can proceed. The ost direct way is to use the approxiate ethod of integration know as the ethod of stationary phase, which is siilar to other approxiate ethods of integration (steepest decent, saddle point integration, etc.). This subject as a whole is beyond the scope of what we are interested here, but we will suarie the pertinent results. The ethod of stationary phase states: for soe coplex integral I() where B itg( ) I( ) = f( ) e d ( + i) f( ) e t A (3.6.6) g ( ) = This ethod can be used on Eq.(3.6.6) if we consider only the far field, i.e. R. In this case we ust siplify the Hankel Function in the nuerator in order to get a for on which to apply the ethod. If we let = R sin ϕ with R large then ( ) itg( ) ik π tg ( ) ik e π r ir k k cosϕ ikr sinϕ i(+ ) π Fk ( ) 4 e e 3 ( ) 4 ' k k H( k k a) i(+ ) π 4 itg( ) π = e ( + i) f( ) e π r tg ( ) dk (3.6.6)

21 6 - AUDIO TRANSDUCERS where we have used t = R g( k ) = ( k k )cosϕ + k sinϕ x ϕ = vertical angle To find we set g' (k ) = off axis to get x d( k k cosϕ + k sin ϕ) = dk x k k sinϕ = k cosϕ k = ksinϕ = Using this result in leads to Fk ( sin ϕ) π cos ϕ Ik ( ) = = R ikr+ π k e ' ( k cos ϕ) H ( kasinϕ) π Fk ( sin ϕ) R cosϕ kh ka ( sinϕ ) ikr+ π 4 which after soe further siplifications yields ' e (3.6.63) ikr e iπ F( ksin ϕ) pr (, θϕ, ) = i ρcv (3.6.64) Ae cos( θ) ' kr cosϕ H ( kasinϕ ) This equation is only valid for large R. It has been shown that this result becoes exact as R. We are now in a position to find the values of the coefficients and look at soe results. Theθ coefficients are easily deterined π ( ) 6 sin π 6 A (3.6.65) = cos( θ) dθ = The F (k sin ϕ) coefficients are likewise straightforward to calculate since the Fourier Transfor of the Rect(kbsinϕ /) function is well known b / kbsinϕ ik sinϕ Fk ( sin ϕ) = Rect ( ) = f e d π b / (3.6.66) b sin( kbsin ϕ) = π kbsinϕ Finally the far field solution can be written as

22 -8 - ACOUSTICS - 63 ikr e pr (, θϕ, ) iρcbv kr kb sin( kbsin( ϕ)) = sin( ϕ) sin( π / 6) i cos( θ) cos( ϕ) H ( kacos( ϕ)) ' (3.6.67) Fig. 3- shows the polar pattern in the horiontal plane. The cylinder in this exaple is about two feet across (r =.3) and the source about eight inches across (c). The source is continuous about one eter in total height. The plots are noralied to the axial response which we will look at in ore detail later. In this figure the odal calculations are not likely to have sufficient content Angle ,, Frequency Figure 3- - Horiontal polar response for a line source on a cylinder after about 5 kh so the results in the higher frequencies ay not be correct. There were 4 odes in this calculation and the Hankel Functions for sall arguent, which go to infinity, tend to get unstable at higher ode nubers. Note that the horiontal response narrows to around ±3º reaining alost constant at about -6dB. Fig.3- shows the polar response in the vertical plane. Alost all of the energy is directed towards the central axis. We have plotted only to 6 since the results for larger angles fluctuate so rapidly that the plotting algorith fails. Note that at higher frequencies the results are suffering fro an instability in either the nuerical calculations of the function or the contour plotting routine (which uses

23 64 - AUDIO TRANSDUCERS 6 3 Angle -3-6,, Frequency Figure 3- - Vertical polar pattern for cylindrical radiator interpolative soothing) or both. The interesting thing to note here is that we ight expect this sae effect in an actual device, naely that the response has becoe hypersensitive to sall variations and exhibits an unstable condition like that shown in the plot. In a real situation the vertical polar response in this regie could fluctuate with sall variations in teperature, air currents, atospheric pressure or huidity. Unlike light, the acoustic wave speed can change by relatively large percentages in short distances. This can cause a sort of acoustic shiering siilar to the twinkling that we see when we look at the stars, only this effect occurs on a uch saller scale of wave travel. We will see these sae phenoena when we talk about roo acoustics in Chap.7. Finally Fig.3-3 shows the response of this array on axis for a constant acceleration array of drivers. The apparent constant directivity does not coe without a cost naely a continuously falling axial response. This response falloff ust be recovered with soe for of gain, but 4dB is a lot of loss to ake up. A calculation of the odal radiation ipedances can be obtained fro the previous exaple by setting the arguent of Hankel Function in the nuerator of Eq.(3.6.6) equal to ka, and perforing the integrations via other siplifications. However this would not be constructive in the general case since these ipedances would depend on the vertical arraignent of the source. An easy calculation is to calculate the odal ipedance for a unifor vertical velocity on the cylinder. In this case, the function F (k ) would becoe δ () and the integra-

24 ACOUSTICS - 65 Axial Response (db) ,, Frequency Figure Axial frequency response for cylindrical radiator tion would be trivial. The odal ipedances for this siplified exaple are shown in Fig.3-4 for the real and iaginary parts respectively of the first six odes. 3.7 The Cylindrical Near Field The characteristic of the above line source in the near field are very iportant owing to the fact that the near field extends a considerable distance into the sound field for this type of source. The solution of Eq.(3.6.64) cannot be directly analyed for the near field, but we can use another ethod. At this point we need to introduce another technique for sound field calculations that are based on the Green s Function. The Green s Function is the spatial analog to the ipulse response in systes theory. Once we know the pressure at the observation point r (the output response), due to a point source at point r (a spatial ipulse), the result for any source distribution is siply the integral over all of the point sources that ake up the desired source. This result is exactly analogous to calculating the output of a syste for an arbitrary input by using the systes ipulse response. This technique is very powerful, so long as the Green s Function is known (which is usually the difficult part). We will use such an approach here.

25 66 - AUDIO TRANSDUCERS Iaginary Ipedance Real ka n = n = n = n = 3 n = 4 n = 5 n = 6 Figure Modal radiation ipedances for cylindrical radiator The Green s Functions in infinite space (boundary at ) are Diensions Green s Function g(x x ) 3 π i ik x x k iπ H e ( k r ) r e ik r r r r Table 3.: Green s Functions Those for the single and three diensional spaces are quite siple, while the two diensional one is not so siple. Solutions of radiation probles involving boundaries which are not separable, and hence not analytically solvable, can be obtained by using the three diensional Green s Function along with Green s Theore to yield an integral equation over the surface of the radiating object. This technique can be applied to any shape boundary and is soeties referred

26 ACOUSTICS - 67 to as the Boundary Integral Method (BIM). (Although in general the Boundary Integral Method is ore general than applying to sound radiation probles.) These techniques are also very powerful, and the interested reader is encouraged to investigate the. They will not be discussed in this text because doing so would take us far away fro our intent, which is a thorough overview of the fundaental physics of transducers. Like the FEM, which we discussed in Chap., the BIM is best applied to specific cases where in depth analysis is warranted by application. BIM solutions are specific to the particular analysis being perfored and generaliations of results are difficult to ipossible to obtain, unlike the odal solutions that we have been studying here. Looking now at our particular proble we want to use the two diensional Green s Function to look at the near field of our radiating cylinder. When the source is independent of the axial coordinate then a siple integral results θ pr (, θ) = iρck v() θ g ( r r ) dθ θ k (3.7.68) where the Green s Function ust be one that satisfies the boundary conditions on the cylinder. This can be accoplished by using an expansion of the Hankel Function and adding a second expansion which represents the reflection fro the rigid boundary. This equation then has its gradient set equal to ero at the surface of the cylinder. We would eventually end up with exactly the sae equations that we obtained fro the odal expansion result for this case. On the other hand when we place ourselves in a plane of syetry (the r- plane) then the boundary condition for the Green s Function is siply twice the free space Green s Function when r is on the surface and r > r. There is one ore coplication to this proble and that is that the space into which the source is radiating is expanding, i.e. increasing with r at the rate of / r. We can still use the Green s Function approach so long as we recognie that the two diension axi-syetric proble ust have an additional / r ter in it to account for this coordinate expansion. This extra factor is obvious by considering that the far-field for any finite source ust fall as / r for large r. This is true in any coordinate syste. The Hankel Function for large arguent only falls as / r. Using the two diensional Green s Function, the equation for radiation in the r- plane fro an axi-syetric source which is finite in then becoes b R ( )cosϕ b H ( kr( )sin ϕ) pr (, ϕ) = iρck v( ) d (3.7.69) R ( ) = r r sinϕ + with the variables as shown in Fig.3-5. While this is not exactly the calculation that we wanted to do (it has a source that circunavigates the cylinder) it will give us a good indication of the near field

27 68 - AUDIO TRANSDUCERS a p (r, φ ) r R= r- b φ r' d Figure Geoetry for nearfield radiation calculation effect for the higher frequencies where the radiation fro the rear of the cylinder to the front becoes negligible. The results therefore becoe correct as the frequency increases. Eq.(3.7.69) is easily calculated nuerically. The calculation of the sound field on axis as a function of distance fro the source is shown in Fig.3-6 The solid lines at the botto of the graph show the -6 and - db/octave slopes indicating that the field falls off initially at -6 db/octave, but changes to - db/octave at a distance fro the source that oves out with higher frequency. The region where the falloff is slower is know as the near field. The near field extends out to approxiately r =. k a. a = height of the line array We can also see that within the near field the frequency response is highly irregular. Another way of viewing the near field is shown in Fig.3-7. This figure shows the response for a constant velocity source (not the usual situation; usually the source velocity falls and so this response would fall like that shown in the figure below). Here we can see that the frequency response for the cylindrical source is changing with the distance fro the source in a highly coplex anner. There is no hope of equaliing this response to flat at all locations. Generally speaking this characteristic changing frequency response with distance - is true of the near field of any source. It is for this reason that the near field is not usually a good place for a listener.

28 ACOUSTICS Axial Response (db) -6 - R =. R = 4. R = 8. R = 6. R = 3.,, Frequency Figure Axial frequency response at various distances fro the source H. H. 3 H. 4 H radius () Figure Nearfield axial pressure for a line array

29 7 - AUDIO TRANSDUCERS 3.8 Suary This chapter has shown the basic acoustical equations for sound radiation in a few coordinate systes. Several exaples were shown which typify the radiation characteristics of audio frequency transducers. In the next several chapters, we will look at ore specialied sound radiators and introduce soe new techniques for handling calculations in an efficient anner. Alost invariably these techniques will rely on the odal description of the sound field that we have studied in this chapter.