chapter 9 spherical waves, harmonics, and line arrays

Chapter 9Spherical Waves, Harmonics, and LineArrays

The wave equation, ∇2 p = p̈/c2, as derived in Sects. 7.1 and7.2, theoretically cov-ers all possible sound fields in idealized fluids, that is, gases and liquids. However,the task of computing specific sound fields requires solutions of the wave equa-tion for the particular boundary conditions in each case. In general, this task is bemathematically expensive, but there are helpful computer programs available, someof which are based on numerical methods like the finite-element method, FEM,or the boundary-element method, BEM. Yet, In practice, approximations are oftensufficient to understand the structure of a problem.

Closed solutions of the wave equation only exist for a limited number of specialcases. We already introduced the plane wave as one-dimensional solution in Carte-sian coordinates. A few further one-, two- and three-dimensional cases are solvable inclosed form, especially when symmetries allow simplified formulations using appro-priate coordinate systems as is the case for spherical or cylindrical coordinates.

In the current chapter, we discuss basic solutions of the wave equation in sphericalcoordinates. In the sameway that periodical time signals are decomposed intoFourierharmonics, spherical sound waves are decomposable into spherical harmonics.1

To start with the essential basics, we focus on the spherical harmonics of 0thand 1st order and the sound sources that emit them. This also makes sense fromthe engineering standpoint since 0th- and 1st-order sound sources are of significantpractical relevance, mainly for the following two reasons.

1. At low frequencies many sound emitters act approximately like sourcesof 0th- or 1st-order spherical waves.

2. According to Huygen’s principle, each point on a wavefront is the originof a spherical wave. Many sound fields are conceivable in a comparativelysimple way by employing this principle. With spherical sound waves, the

1 Spherical harmonics are eigen-functions of the wave equation in spherical coordinates.

© Springer-Verlag GmbH Germany, part of Springer Nature 2021N. Xiang and J. Blauert, Acoustics for Engineers,https://doi.org/10.1007/978-3-662-63342-7_9

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134 9 Spherical Waves, Harmonics, and Line Arrays

synthesis of sound radiators with arbitrary directional characteristics ispossible.

9.1 The Spherical Wave Equation

The wave equation allows for a one-dimensional, point-symmetric solution. This is asoundwave where all parameters only depend on the distance, r , from the origin. Thesolution does not depend on the direction of propagation, which is always radial anddirected either outward or toward the origin. This type of wave is called a sphericalwave of the 0th order, and a sound source that emits such a wave is called a sphericalsource of 0th order.

To derive the appropriate wave equation, it is helpful to transform the wave equa-tion from Cartesian coordinates, x, y, z, into spherical coordinates, φ, θ, r . Thisis accomplished with the following well known operator,2

� = ∇2 =[

∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2

]

= 1

r2

[∂

∂r

(r2

∂

∂r

)+ 1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+ 1

sin2 θ

∂2

∂ϕ2

]. (9.1)

Because the assumed sound field is point symmetric and only changes in the radialdirection, we state that

∂

∂θ= ∂

∂ϕ≡ 0 . (9.2)

This leads to the wave equation for the 0th-order spherical wave,

1

r2∂

∂r

(r2

∂

∂r

)p = ∂2 p

∂r2+ 2

r

∂ p

∂r= 1

c2∂2 p

∂t2. (9.3)

Note that this equation is identical to the wave equation for conical horns—whichwas derived in Sect. 8.2. The only difference is that x has been replaced by r . Thiscongruence is intuitively plausible when we think of the spherical wave as a soundfield composed of an infinite number of adjacent, very slim conical horns. Figure9.1illustrates this concept. When removing the “walls” between these conical horns, thesound field nevertheless remains the same because there is radial propagation only.

The solutions for the outward-progressing wave in the 0th-order spherical soundfield are

p→(r) = g →r

e−jβr , and (9.4)

v →(r) = g →

(1

� c r+ 1

jω � r2

)e−jβr . (9.5)

2 For a derivation of this expression see the solution to Problem9.6.

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9.1 The Spherical Wave Equation 135

Fig. 9.1 Spherical waves of 0th order as a composition of conical waves

Fig. 9.2 Sound sources for spherical waves (a) 0th-order source, also called breathing sphere(b) Example of a 1st-order source, also called rigid oscillating sphere (c) Example of a 2nd-ordersource. Note that there are 2n + 1 possible modes per order, n, with n = 0, 1, 2 . . . CompareFig. 9.5

The field impedance of the diverging wave is

Z f = � cj 2πr

λ

1 + j 2πrλ

= 11�c + 1

jω�r

. (9.6)

The region of r < λ/2π is called the near-field, the one of r > λ/2π the far-field,as discussed in Sect. 8.2.

Spherical sound fields of the 0th order are radiated by spherical sound sources of0th order, also called breathing spheres—shown in Fig. 9.2a.

The Co-vibrating Medium Mass

It is an interesting exercise to calculate which part of the near field medium massmoves back and forth without being compressed. Because this part does not transmitactive power, it is sometimes called the Watt-less-vibrating mass. The equivalentcircuit in Fig. 8.4 illustrates this situation. The diameter of the breathing sphere is r0.

In the far-field, the real term outweighs the imaginary one. As a result, there isno reactive power and no Watt-less vibrating mass. In the near-field the particlevelocity flows through the mass so that

∣∣∣∣ p

v

∣∣∣∣ = ω � r0, (9.7)

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and, therefore, by implementing Newton’s law, we get

∣∣∣∣ Fv∣∣∣∣ = ωm = ω � r0 A0 . (9.8)

By inserting the formula for the area of the sphere, A0 = 4π r20 , the co-vibratingmass is found to be

m co = 4π r30 � . (9.9)

This is three times the mass of the medium inside the sphere if the medium is thesame inside and outside.

Radiated Active Power and Source Strength

The radiated active power of a 0th-order spherical sound source is as follows—referto Sect. 8.4,

P = 1

2

radiation resistance, rrad︷︸︸︷A(r)Re{Z f(r)} | v(r)|2

= 1

2� c

(ωrc

)21 + (

ωrc

)2 4π r2 | v |2

= 1

2� c

(ωc

)24π

[1 + (

ωrc

)2] (4π r2| v |︸︷︷︸vol. velocity q

)2. (9.10)

In the near-field, we have 2π r/λ = ω r/c � 1, allowing us to write

P = 1

2� c

(ωc

)24π

(4π r2 | v |)2 = 1

2

�ω2

4π c| q

0|2 . (9.11)

With |v| ∼ 1/r2, the following also holds,

(4π r2 | v | )2 = | q

0|2 ≈ const , (9.12)

which means that in the near-field the volume velocity, q , is fairly independent ofthe distance, r , and converges to q

0. This primary volume velocity, q

0, is called the

source strength of spherical radiators.The active power transmitted, P , does not depend on the distance, r , given that

the medium is lossless. As a result of this and the fact that active power flows throughall spherical shells, we write

P = 4π r2 Re{ I } �= f (r) . (9.13)

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9.1 The Spherical Wave Equation 137

The term 4π in the denominator of (9.11) denotes the full spherical angle, �� = 4π.If a 0th-order spherical source with the source strength q

0radiates into a smaller

spherical angle, �1, that is only a section of the available volume, then the radi-ated power increases by a ratio of 4π/�1. Since this power is only radiated intothe smaller angle, the intensity, Re{ I }, in this section increases by (4π/�1)

2 or20 lg(4π/�1) dB.

This relationship is of practical relevance, for instance, for horn loudspeakers,further for all 0th-order spherical sound sources when placed in front of a wall or in acorner or edge of a room. The following level increases result from such placements.3

• Placement in front of a wall (hemisphere) =⇒ +6dB

• Placement in a room edge (quarter sphere) =⇒ +12dB

• Placement in a corner (1/8th sphere) =⇒ +18dB

Point Sources of the 0th Order (Monopoles)

In the 0th-order spherical sound field we have

p(r)

v(r)= 1

1� c + 1

jω � r

, or g →e−jβr

r= v(r)

1� c + 1

jω � r

, (9.14)

from which follows

g → = 4π v(r) r2

4π(

r� c + 1

jω �

) e+jβr . (9.15)

We now let the radius of the sphere go to zero while keeping g → constant.In this way we obtain

limr→0

[ 4π r2 v(r)] = q0, (9.16)

from which follows

limr→0

g → = jω � q0

4π. (9.17)

Finally, we arrive at the sound field of the point source of 0th-order, which is alsoknown as monopole,

p→(r) = jω � q0

e−jβr

4π r. (9.18)

Any 0th-order spherical sound source, that is, any breathing sphere, is representableby an equivalent monopole with the same source strength, q

0.

3 Note that loudspeakers in closed boxes become spherical radiators at low frequencies—refer toSect. 9.4. Adjustment of their-frequency response is thus possible by appropriate placement in thespace.

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9.2 Spherical Sound Sources of the First Order

A rigid sphere may oscillate according to the sketch in Fig. 9.2b, and a sound fieldcreated in this way is called a 1st-order spherical sound field. Such a sound fieldis no longer point-symmetric, which means that the shells around the sphere donot represent areas of equal phase. This may also be expressed as ∂/∂θ �= 0 and/or∂/∂ϕ �= 0.

Since the problem is axial-symmetric it is sufficient to deal with one sectionthrough the sphere. Here we consider a vertical section along the x–axis. The fol-lowing boundary condition is valid for the radial component on the surface of thesphere,

v(θ) = v(0) cos θ . (9.19)

Point Sources of the 1st Order (Dipoles)

The fields of two complementary monopoles with opposite phase are combinablefor creating a sound field like that of an oscillating sphere. This allow to derive thewave equation for 1st-order spherical sound fields in a relatively easily way.

Two point sources with equal strength but of opposite phase, that is, q1

= −q0

and q2

= +q0, are positioned a distance of 2 d apart, forming a so-called dipole.

Due to the linearity of the wave equation, the sound field of this arrangement is givenby superposition of the two individual sound fields, namely,

p→ = jω�

4πq0

(e−jβr2

r2− e−jβr1

r1

). (9.20)

Figure9.3a illustrates this situation. Since the two 0th-order point sources havezero radius, possible reflection or diffraction caused by their presence need not beconsidered.

Fig. 9.3 Derivation of the dipole sound field (a) Two monopoles of opposite phase at a distanceof 2 d (b) Equivalent situation with only one monopole

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9.2 Spherical Sound Sources of the First Order 139

The next step is to perform a limit-operation in such a way that 2 d goes to zero,while, by definition, the dipole strength,μ

d= 2 d q

0, is kept constant. This condition

prohibits the two monopoles from canceling each other, and we write

p→ = jω �

4πμ

dlim2d→0

[1

2 d

(e−jβr2

r2− e−jβr1

r1

) ]. (9.21)

Figure9.3b illustrates that the previous equation is interpretable as the result ofthe differentiation of a monopole sound field in the x–direction. By taking ∂x ≈∂r/ cos θ we get

∂

∂xf (x) = lim

x�

→ 0

[f (x + x�) − f (x)

x�

], where x� = 2 d . (9.22)

The resulting solutions of the wave equation for the dipole field are as follows—outward-progressing waves only,

p→(r, θ) = jω �μd

4πcos θ

∂

∂r

(e−jβr

r

)

= −jω �μd

(1

r+ jβ

)cos θ

e−jβr

4π r

= � cμd

(β2 − j

β

r

)cos θ

e−jβr

4π r, (9.23)

v →(r, θ) = μd

(β2 − 2

r2− j

2 β

r

)cos θ

e−jβr

4π r. (9.24)

The solution for v has again been derived via Euler’s equation. Note that thesound pressure possesses a 1/r2-component, which means that it has a near-field.From (9.23) and (9.24). the field impedance is, with β = 2π/λ = ω/c,

Z f = p→v →

= � c(β r)2 − jβ r

(β r)2 − 2 − j 2 β r= � c

(2πrλ

)2 − j 2πrλ(

2πrλ

)2 − 2 − j 2(2πrλ

) . (9.25)

The real part thereof is

Re{ Z f} = � c(β r)4[

2 + (β r)2]2 = � c

(2πrλ

)4[2 + (

2πrλ

)2]2 . (9.26)

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Fig. 9.4 Directional characteristic, Γ , of a dipole sound source

In the near-field, we thus have an approximate proportionality of Re{ Z f} ∝ ω4.Recall that for the monopole we found Re{ Z f} ∝ ω2 for the near-field, that is, aprofoundly less dependence on the frequency.

The dipole sound field shows the following directional characteristic for both thesound pressure and the particle velocity,

Γ = p→(r, θ)

p→(r, 0)= v →(r, θ)

v →(r, 0)= cos θ , (9.27)

as plotted in Fig. 9.4. Note that the plot only shows the vertical plane, but the direc-tional characteristics are axial-symmetric around the x-axis.

We see a figure-of-eight characteristic that complies with the boundary conditionsof the rigid oscillating sphere. It follows that sound field of oscillating spheres isrepresented by 1st-order spherical point sources (dipoles).

9.3 Spherical Harmonics

It is possible to consider any sound-field as being composed of a series of orthogonalspherical harmonics of different orders. These spherical harmonic waves are eigen-functions of the wave equation in spherical coordinates. Taking the spherical-waveequation in the Helmholtz form using (9.1) with a trial solution

p(r, θ,ϕ) = R(r)Θ(θ)�(ϕ) e jω t , (9.28)

yields

1

r2

[1

R

∂

∂r

(r2

∂R

∂r

)+ 1

Θ sin θ

∂

∂θ

(sin θ

∂Θ

∂θ

)

+ 1

� sin2 θ

∂2�

∂ϕ2

]+ β2 = 0. (9.29)

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9.3 Spherical Harmonics 141

We separate this equation4 in a radial, r , an elevation, θ, and an azimuth, ϕ, compo-nent, respectively,

1

r2∂

∂r

(r2

∂R

∂r

)+ β2 R − n(n + 1)

r2R = 0, (9.30)

1

sin θ

∂

∂θ

(sin θ

∂Θ

∂θ

)+

[n(n + 1) − m2

sin2 θ

]Θ = 0, (9.31)

and∂2�

∂ϕ2+ m2 � = 0. (9.32)

The solution of the radial-component equation (9.30) is expressible as

R(r) = R 1 jn(β r) + R 2 yn(β r), (9.33)

with R 1 and R 2 being constants.jn(ν) and yn(ν) are termed spherical Bessel functions, they read

jn(ν) = (−ν)n(1

ν

d

dν

)n sin ν

ν, (9.34)

yn(ν) = −(−ν)n(1

ν

d

dν

)n cos ν

ν, (9.35)

with n = 0, 1, 2, . . . being their order.The radial solution is also expressible as follow,

R(r) = R 3 h n(β r) + R 4 h∗n(β r), (9.36)

with R 3, R 4 being constants. h n(β r) is termed spherical Hankel function, writtenas

h n(β r) = jn(β r) + j yn(β r). (9.37)

The solutions of the elevation- and azimuth-component equations are

Θ(θ) = Θ̂Pmn (cos θ), (9.38)

�(ϕ) = �̂ e jm ϕ, (9.39)

where Θ̂ and �̂ are amplitudes, and Pmn (η) is the associated Legendre function

4 Done by adding a “magic-zero” quantity (m2 − m2)/(r2 sin2 θ) + [n(n + 1) − n(n + 1)]/r2 tothe right-hand side of (9.29).

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Pmn (η) = (−1)m(1 − η2)m/2 dm

dηmPn(η), (9.40)

with Pn(η) being the Legendre polynomial

Pn(η) = 1

2n n!dn

dηn(η2 − 1)n. (9.41)

For practical convenience, the solutions of the angular (elevation and azimuth)components in (9.38) and (9.39) are combined with specified amplitude values of

Ymn (θ,ϕ) =

√2n + 1

4π

(n − m)!(n + m)! P

mn (cos θ) e jm ϕ. (9.42)

The termYmn (θ,ϕ) is denominated spherical harmonics. The integern = 0, 1, 2, . . . ,

is called the order, and m = 0,±1,±2, . . . ,±n is named the degree (mode) of thespherical harmonics, respectively.

The spherical harmonics form a set of orthonormal basis functions.5 They areapplicable as expansion of any arbitrary function, g(θ,ϕ), on the surface of a sphere,namely,

g(θ,ϕ) =∞∑n=0

n∑m=−n

Amn Ym

n (θ,ϕ), (9.43)

with appropriate coefficients, Amn . The real-valued spherical harmonics

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

√2n+12π

(n−m)!(n+m)! P

mn (cos θ) cosm ϕ m > 0 ,

√2n+12π Pn(cos θ) m = 0 ,

√2n+12π

(n−m)!(n+m)! P

mn (cos θ) sin |m| ϕ m < 0 ,

(9.44)

up to the order of n = 3, are illustrated in Fig. 9.5 with Pn(·) being a Legendrepolynomial—see (9.41).

Combining the radial component in (9.33) with the above spherical harmonicsinto a single expression brings

p(r, θ,ϕ) =∞∑n=0

n∑m=−n

[Amn yn(β r) + Bm

n zn(β r)] Ymn (θ,ϕ) (9.45)

5 See the solution to Problem9.7.

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9.3 Spherical Harmonics 143

Fig. 9.5 Real-valued pherical harmonics in (9.44) up to the third order (n = 0, 1, 2, 3) as in (9.44),and for degrees, m, between −3 ≤ m ≤ 3. The lobes indicate positive and negative values bydifferent shading. For each given order, n, there are 2n + 1 degrees as listed in each row of the table

satisfies Helmholtz’ equation (9.29). This condition is convenient for solvingtraveling-wave problems with Am

n and Bmn being (constant) coefficients.

Similarly, combination of the radial component in (9.36) with (9.42)

p(r, θ,ϕ) =∞∑n=0

n∑m=−n

[Cmn hn(β r) + Dm

n h∗n(β r)] Ym

n (θ,ϕ), (9.46)

also satisfies the Helmholtz equation, which is convenient for solving standing waveproblems with Cm

n and Dmn being coefficients. The term containing Cm

n representsan outgoing wave component from the spherical coordinate origin, whereas the termcontaining Dm

n represents an incoming component, traveling toward the origin. Thevalues of the coefficients, Am

n , Bmn ,Cm

n , and Dmn , are specific for the actual problem

concerned and its boundary conditions.

9.4 Higher-Order Spherical Sound Sources

In the preceding section, we already introduced two spherical harmonics, namely,in form of the spherical waves of 0th and 1st order. Spherical waves of higher orderare radiated by spheres with surfaces that oscillate with velocities determined byhigher-order spherical functions. Figure9.2c depicts one possible 2nd-order spher-ical vibration, that is the one which is indexed (n = 2,m = 2)—compare as also inFig. 9.5.

For spherical sound emitters of nth order in the near-field, the resistive part of thefield impedance, Re{Z f}, increases with frequency as follows,

Re{Z f} ∝ ω2(n+1) . (9.47)

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Fig. 9.6 Schematic plots of the sound fields of breathing, (a), and oscillating sound sources, (b)

The increase of Re{ Z f}, with frequency opens the possibility of practical simplifi-cations. If the linear dimensions of the emitter are small compared to thewavelengths,that is, 2π r0 � λ, the radiation of higher-order spherical waves is negligible at lowfrequencies. Thus, amonopole source provides a good low-frequency approximationfor all breathing sound sources such as a loudspeaker mounted in a box (…cabinet).A dipole sound source serves well to approximate the behavior of oscillating sourcesat low frequencies, for instance, loudspeakers without a baffle—see Fig. 9.6a, b.

9.5 Line Arrays of Monopoles

Arrangements of several sources along a line in space are called line arrays (…lineararrays). They play a relevant role in practical applications. Because the sound fieldsof the individual sources interfere with each other, sharply bundled radiation areachieved with these arrays. Common applications of this principle are, for example,line arrays of loudspeakers.

The following discussion considers the directional characteristics of linear arrayscomposed of monopoles. Hereby we restrict our view to the sound field far awayfrom the array.

Line Array of Identical and Equidistant Monopoles

In an arrangement like the one depicted in Fig. 9.7, we refer to a position in space ata distance of r0 � 2h. The sum of the contributions of all monopoles of the array atthis reference point is

p→(r = r0, θ) = jω � q0

4π

n∑i=1

e−jβ [r0−(i−1)2d cos θ]

r0 − (i − 1)2d cos θ. (9.48)

We now neglect the differences in the magnitude of the individual contributionsbecause of r0 � 2h. This allows to consider the phase differences only.This approach leads to the following approximation,

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9.5 Line Arrays of Monopoles 145

Fig. 9.7 Linear array with monopole sources

p→(r, θ) ≈ jω � q0

4π

e−jβ r0

r0

n∑i=1

e+jβ (i−1) 2d cos θ . (9.49)

By substitutingβ d cos θwithb in the expression for the sum,weobtain an expressionwith a known series summation,

∑e+j(i−1)2b = 1 + e+j2b + e+j4b + · · · e+j2(n−1)b = 1 − e+j2nb

1 − e+j2b. (9.50)

Writing with an expansion using sin x = (e+jx − e−jx )/2 j, yields,

∑e+j(i−1)2b = e+jnb

e+jb

(e−jnb − e jnb

e− jb − e jb

)= e+j(n−1)b sin(nb)

sin(b). (9.51)

The term sin(nb)/ sin b determines the directional characteristic. We discuss itmore conveniently in the following paragraph, where a continuously loaded lineof monopoles is dealt with.

Continuously Loaded Line Array

First we perform a limit operation by letting the distance between the individualmonopoles, 2d, and, consequently, b, go to zero.With the length of line array, 2h, keptconstant,we then get n → ∞. To also keep the total source strength,n q

0= q ′(x) 2h,

constant, we normalize by n, with q ′(x) being a constant velocity load. The result ofthis operation,

lim2d→0

sin(nb)

n sin(b)= sin(nb)

nb= si(nb) , (9.52)

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Fig. 9.8 Directional characteristics, Γ , of a line array of the length 2h = 2λ

is a so-called sinc- or si-function ( ... sinus cardinalis).With 2h ≈ n 2d and, therefore,nb = βh cos θ, it follows that

Γ = si (βh cos θ) . (9.53)

This is the directional characteristic of the in-phase, continuously loaded line arraywith a constant source-strength load. This formula also loosely covers arrays with alimited number of monopoles.

As an example, Fig. 9.8 illustrates a line array with a length of two wavelengths,2h = 2λ, and βh = 2π. The upper panel illustrates the directional characteristic inCartesian coordinates. The lower panel shows the typical club-shaped form of thebeam in spherical coordinates—vertical section only.

9.6 Analogies to Fourier Transform in Signal Theory

In the preceding section, we assumed a continuous source-strength load, q ′(x),having the dimension [volume velocity/length]. We continued to presume that eachpoint on the line acts as a (differential) monopole—illustrated in Fig. 9.9.

After the following integration, we calculate the sound pressure at the observationpoint, r0 � 2h. The term in front of the integral is a constant term for a given r0.The complete expression is

p→(r, θ) = jω �

4π

e−jβ r0

r0︸︷︷︸const.

∫ +h

−hq ′(x) e+j(β cos θ)x dx . (9.54)

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9.6 Analogies to Fourier Transform in Signal Theory 147

Fig. 9.9 Sound fields of a line array with a continuous volume-velocity load

This expression is isomorphic to the well-known Fourier integral from signal theory,also termed Fourier transform, which we write as follows,

S(ω) =∫

s(t) e−jω tdt , or, symbolically, as (9.55)

s(t) ◦ • S(ω) , (9.56)

Here t corresponds to x and ω corresponds to −β cos θ. S(ω) is termed spectrum ofs(t) in frequency domain, ω.

Disregarding the constant factor, we find the following analogies between the timefunctions and the source-strength-velocity loads of line arrays.

Time function, s(t) Source-strength load, q ′(x)◦• ⇐⇒

◦•

Spectrum, | S(ω)| Directional characteristic, Γ (θ)

Table9.1 lists correspondences of examples that we have treated so far.6

At the end of this section. We discuss two additional directional characteristicsthat are relevant from an application point of view, and which are also obtainablefrom equivalent relationships in signal theory.

• A source-strength load with a Gaussian envelope leads to a Gaussiandirectional characteristic. This is a beam without side lobes—depicted inFig. 9.10a

• A source-strength load with a phase shift increasing linearly with position,

q ′2(x) = q ′

1(x) e−jβ cos(θ�) x (9.57)

6 For the definition of Γ see (9.27). In the table, the directional characteristics, Γ (θ), have beennormalized so that their maxima equal one. δ(z) is called Dirac impulse. It is a special mathemat-ical distribution that picks out the value of a function at the position of its argument as follows,∫ +∞−∞ y(z) δ(z − z0) dz = y(z0). The area under the Dirac impulse is

∫ ∞−∞ δ(z) dz = 1.

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Table 9.1 Some examples of the equivalence of time signal and frequency spectrum verses spatialdistribution of source-strength load and directional characteristics

Linear array with constant load Rectangular impulse

q ′ ={const for –h <x < h

0 others◦•

s(t) ={const for −τ < t < τ

0 others◦•

Γ = si (−h β cos θ) S(ω) = 2τ si (τ ω)

Monopole Dirac impulse

q ′(x) = q0δ(x)

◦•

s(t) = δ(t)◦•

Γ = 1 S(ω) = 1

Dipole Double Dirac impulse

q ′(x) = μd

ddx δ(x) = μ

dδ′(x)

◦•

s(t) = ddt δ(t) = δ′(t)◦

•Γ = cos θ S(ω) = ω

Fig. 9.10 Gaussian distribution of source strength and directional characteristic (a) Shaping thedirectional characteristic (b) Shifting the directional characteristic (schematic)

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NXian

Cross-Out

9.6 Analogies to Fourier Transform in Signal Theory 149

◦•

Γ2 = Γ1 [ β cos θ − β cos(θ�)] (9.58)

leads to a tilteddirectional characteristic—schematically shown inFig. 9.10b.The shifting theorem of Fourier transform is applied here.7

9.7 Directional Equivalence of Sources and Receivers

When a reversible transducer or transducer array is operated as sound emitter, itsdirectional characteristic is equivalent to its directional sensitivity characteristicwhenoperated as a receiver. We find this relationship by using the following two elements,

M... A transducer with arbitrary directional characteristics

X... An auxiliary point source with monopole characteristics

The point source is positioned far away from the transducer concerned. The proof isdone in two steps as follows. b

(1) The transducer is fed with an electric current, i 0. At the position of theauxiliary source we then have

| pX| = | T ip (ω,φ, θ, r)| | i 0| . (9.59)

(2) The auxiliary point source emits a volume velocity amounting to its sourcestrength, q0. At the position of the transducer, which is not present at thispoint, we get

| pM

| = ω �

4π r| q

0|. (9.60)

If the transducer is now introduced into the sound field, a voltage, ul, appears at itselectric output port according to

| u l| = | T pu (ω, r, θ,φ)| | pM

|. (9.61)

Since the sound field is linear and passive, reversibility according to (4.16) appliesas follows, ∣∣∣∣∣

i 0pX

∣∣∣∣∣q=0

=∣∣∣∣q0

u l

∣∣∣∣i=0

, (9.62)

and, hence, ∣∣∣∣ TipTpu

∣∣∣∣ =∣∣∣ ω�

4π r

∣∣∣ . (9.63)

7 An application of this algorithm, based on the directional equivalence of emitters and receivers—see next section—is the electronic steering of SONAR antennas.

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The striking attribute of this equation is that the transfer coefficient of the transmitterin emitting function increases with the frequency concerning the transfer coefficientfor receiver operation (sensitivity). This justifies a well known general rule in thefield, which is,

Transducers receive low frequencies better than they emit them!

A good example of this rule is a small reversible microphone.Finally, to show that the directional characteristic, Γ , for emitter operation of a

reversible emitter/receiver is identical to its characteristic for receiver operation, weset

| T ip (φ, θ)|| T pu (φ, θ)| = | T ip (0, 0)| Γip

| T pu (0, 0)| Γpu=

∣∣∣∣∣T ip (0, 0)

T pu (0, 0)

∣∣∣∣∣ , (9.64)

which results inΓ ip = Γ pu. (9.65)

Regarding the examples dealt with in this book, there are the following correspon-dences,

– Pressure receiver ⇐⇒ 0th-order spherical source– Pressure-gradient receiver ⇐⇒ 1st-order spherical source– Line microphone ⇐⇒ Line array with a constant

load and 90◦ shifteddirectional characteristic

9.8 Exercises

Spherical Waves and Line Arrays

Problem 9.1 In the context of the 0th order spherical wave equation,

(a) Show that

1

r2∂

∂r

(r2

∂

∂r

)p = ∂2 p

∂r2+ 2

r

∂ p

∂r= 1

r

∂2(rp)

∂r2= 1

c2∂2 p

∂t2. (9.66)

(b) Given the outward-progressingwave of the 0th-order spherical sound pres-sure

p→(r) = g →r

e−jβr , (9.67)

use the Euler’s equation to show that

v →(r) = g →

(1

� c r+ 1

jω � r2

)e−jβr . (9.68)

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9.8 Exercises 151

(c) Discuss the far-field behavior of the field resistance, R f = Re{Z f

}, of

a monopole source, and its field impedance, Z f , in the near-field as afunction of the angular frequency.

Problem 9.2 A sound source of 0th order in the free field—a breathing sphere—hasa radius of a, and the volume velocity on its surface is q(a).

(a) Determine the sound-pressure and velocity distribution in the space outsidethe sphere.

(b) Determine the source strength, q0, of a point source which generates the

same sound field.

Problem 9.3 Determine and compare the radiated power of spherical sound sourcesof

(a) 0th order and,

(b) 1st order.

(c) Find the dependency of the sound intensity on the distance.

Problem 9.4 An elementary sound source of 0th order is positioned at the originof an infinitely-long conical horn with an opening solid angle of �. The source ismounted in such a way that all of its volume velocity is released into the horn.

Compare the sound pressure, the velocity, the power, and the intensity in the hornwith those of an equivalent source in a free field.

Problem 9.5 Given a dipole with a dipole strength M ,

(a) Sketch the directional characteristics of this dipole.

(b) An additional 0th-order point source, q0, is brought to the same position

as that of the dipole. Find the magnitude and phase of q0so that in the

far-field the entire set-up exhibits the following directional characteristics,

Γ = 1 + cos θ

2. (9.69)

(c) Sketch this directional characteristic. What kind of directional character-istic does it represent?

Problem 9.6 Derive the spherical wave equation from the three-dimensional waveequation in Cartesian form.

Problem 9.7 The spherical wave equation (9.29) is equal to

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1

r2

[1

R

∂

∂r

(r2

∂R

∂r

)+ 1

Θ sin θ

∂

∂θ

(sin θ

∂Θ

∂θ

)

+ 1

� sin2 θ

∂2�

∂ϕ2

]+ β2 = 0. (9.70)

Add to the right-hand side of the spherical wave equation in (9.70) a zero-quantity

m2 − m2

r2 sin2 θ+ n(n + 1) − n(n + 1)

r2, (9.71)

with m and n being positive integers.

Show that Eqs. (9.30) and (9.32) is separable into a radial component, r,

1

r2∂

∂r

(r2

∂R

∂r

)+ β2 R − n(n + 1)

r2R = 0, (9.72)

and an elevation component, θ,

1

sin θ

∂

∂θ

(sin θ

∂Θ

∂θ

)+

[n(n + 1) − m2

sin2 θ

]Θ = 0, (9.73)

and an azimuth component, ϕ,

∂2�

∂ϕ2+ m2 � = 0. (9.74)

Problem 9.8 Given the spherical harmonics, Ymn (θ,ϕ), in (9.42) and the Legendre

functions in (9.40) and (9.41), determine the complex spherical harmonics of theorders n = 0, 1, 2, and the degrees of m = 0, 1.

Problem 9.9 Given a line array with a source-strength load [volume-velocity perlength] evenly distributed over the length, 2h, of the array, that is,

q ′(x) = q ′0e−x2/2h2 , −h ≤ x ≤ +h, (9.75)

(a) Determine and sketch the directional characteristics under the far-fieldassumption that the observation point is much farther away than the finitelength of the array.

(b) How does the directional characteristics change when the volume-velocityload of (a) is modified according to the following phase function,

φ = (β cos θ0) x . (9.76)

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chapter 9 spherical waves, harmonics, and line arrays

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