SOUND PROPAGATION AN IMPEDANCE BASED APPROACH
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
SOUND PROPAGATION AN IMPEDANCE BASED APPROACH Yang-Hann Kim Korea Advanced Institute of Science and Technology (KAIST), Republic of Korea
Copyright Ó 2010
John Wiley & Sons (Asia) Pte Ltd, 2 Clementi Loop, # 02-01, Singapore 129809
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For my children, Jue-Hee, Juelie and Jueseok, The very beginning of all hopes and the source of love And especially for my beloved wife, Eun-wha, In commemoration of our 30th wedding anniversary
Contents Preface
xi
Acknowledgments
xv
1
Vibration and Waves 1.1 Introduction/Study Objectives 1.2 From String Vibration to Wave 1.3 One-dimensional Wave Equation 1.4 Specific Impedance (Reflection and Transmission) 1.5 The Governing Equation of a String 1.6 Forced Response of a String: Driving Point Impedance 1.7 Wave Energy Propagation along a String 1.8 Chapter Summary 1.9 Essentials of Vibration and Waves 1.9.1 Single- and Two-degree of Freedom Vibration Systems 1.9.2 Fourier Series and Fourier Integral 1.9.3 Wave Phenomena of Bar, Beam, Membrane, and Plate Exercises
1 1 1 7 10 14 17 22 25 25 25 34 36 59
2
Acoustic Wave Equation and Its Basic Physical Measures 2.1 Introduction/Study Objectives 2.2 One-dimensional Acoustic Wave Equation 2.3 Acoustic Intensity and Energy 2.4 The Units of Sound 2.5 Analysis Methods of Linear Acoustic Wave Equation 2.6 Solutions of the Wave Equation 2.7 Chapter Summary 2.8 Essentials of Wave Equations and Basic Physical Measures 2.8.1 Three-dimensional Acoustic Wave Equation 2.8.2 Velocity Potential Function 2.8.3 Complex Intensity 2.8.4 Singular Sources Exercises
69 69 69 77 85 96 103 110 110 110 116 116 118 125
Contents
viii
3
4
5
Waves on a Flat Surface of Discontinuity 3.1 Introduction/Study Objectives 3.2 Normal Incidence on a Flat Surface of Discontinuity 3.3 The Mass Law (Reflection and Transmission due to a Limp Wall) 3.4 Transmission Loss at a Partition 3.5 Oblique Incidence (Snell’s Law) 3.6 Transmission and Reflection of an Infinite Plate 3.7 The Reflection and Transmission of a Finite Structure 3.8 Chapter Summary 3.9 Essentials of Sound Waves on a Flat Surface of Discontinuity 3.9.1 Locally Reacting Surface 3.9.2 Transmission Loss by a Partition 3.9.3 Transmission and Reflection in Layers 3.9.4 Snell’s Law When the Incidence Angle is Larger than the Critical Angle 3.9.5 Transmission Coefficient of a Finite Plate Exercises
129 129 129 134 140 144 149 153 156 156 156 159 159
Radiation, Scattering, and Diffraction 4.1 Introduction/Study Objectives 4.2 Radiation of a Breathing Sphere and a Trembling Sphere 4.3 Radiation from a Baffled Piston 4.4 Radiation from a Finite Vibrating Plate 4.5 Diffraction and Scattering 4.6 Chapter Summary 4.7 Essentials of Radiation, Scattering, and Diffraction 4.7.1 Definitions of Physical Quantities Representing Directivity 4.7.2 The Radiated Sound Field from an Infinitely Baffled Circular Piston 4.7.3 Sound Field at an Arbitrary Position Radiated by an Infinitely Baffled Circular Piston 4.7.4 Understanding Radiation, Scattering, and Diffraction Using the Kirchhoff–Helmholtz Integral Equation 4.7.5 Scattered Sound Field Using the Rayleigh Integral Equation 4.7.6 Theoretical Approach to Diffraction Phenomenon Exercises
177 177 178 188 196 201 213 214 214 217
Acoustics in a Closed Space 5.1 Introduction/Study Objectives 5.2 Acoustic Characteristics of a Closed Space 5.3 Theory for Acoustically Large Space (Sabine’s theory) 5.4 Direct and Reverberant Field 5.5 Analysis Methods for a Closed Space 5.6 Characteristics of Sound in a Small Space 5.7 Duct Acoustics 5.8 Chapter Summary
273 273 273 274 282 287 292 302 312
168 169 172
218 219 236 237 265
Contents
5.9 Essentials of Acoustics in a Closed Space 5.9.1 Methods for Measuring Absorption Coefficient 5.9.2 Various Reverberation Time Prediction Formulae 5.9.3 Sound Pressure Distribution in Closed 3D Space Using Mode Function 5.9.4 Analytic Solution of 1D Cavity Interior Field with Any Boundary Condition 5.9.5 Helmholtz Resonator Array Panels Exercises Index
ix
313 313 317 319 320 323 335 339
Preface Even before being born, we humans are said to be capable of listening to sound; after birth, we identify information conveyed by sound, enjoy sound, and create sound. Then how can we understand these things that seem so natural to us just as naturally as they are manifested? Though I have long contemplated the issue, this question has always remained in my mind. I have been lucky and privileged, however, to have solved this problem little by little, thanks to the invaluable help of my colleagues that I have worked with for over 30 years, and my students at KAIST with whom I have studied for more than 20 years. The moment when we realize that the advice, “Start from the very basics!” inevitably is the starting point of resolving all unanswered questions, we discover ways to present something new in our own form and in our uniquely beautiful expression and gestures. Applied to the field of acoustics, this concept is the very first direction of thought—that is, the first “coordinate axis”—reflected throughout this book. The second coordinate axis is non-dimensional time and space expressed in relative terms, as propagation—a physical value changing with time and space—needs units of time and space. The relative amounts of time and space (e.g. the ratio of frequency to resonant frequency, the ratio of an object’s size to wave) are the basic axes for measuring and appreciating the physical values covered in this book. In short, it is emphasized that all phenomena should be observed in these relative units of time and space. For the third coordinate axis, I considered major physical values to provide an easily comprehensible but extensive analysis of an acoustic phenomenon (i.e. sound). In this regard, impedance is a prime example of such physical values. In this book, five grand “mountains” constitute a mountain range that embodies the tenet, “Even the most complicated thing is made up of very basic elements.” The vibration of a simple string, exemplified in Chapter 1, is the very first mountain representing this concept. In Chapter 2, the equation governing the propagation of sound waves is explained using the onedimensional phenomenon of their transmission through pipes; the propagation of sound waves on a discontinuous surface, covered in Chapter 3, also has its basis in this one-dimensional phenomenon. In other words, by fully understanding what occurs in the case of a string, one can build on this and understand the much more complicated phenomenon of sound wave propagation. The fourth and fifth mountains (Chapters 4 and 5) provide an opportunity to comprehensively understand and get a sense of the phenomenon of sound, where all the concepts grasped in the previous chapters present themselves in a complex and intermingled manner.
xii
Preface
In Chapter 1, vibration of strings is used to specifically illustrate how this physical value, impedance, affects the propagation of sound waves. In Chapter 2, the significance and units of physical values other than impedance—sound pressure, speed, energy, power, and intensity— are explained, and it is emphasized that these physical values are the very basic concepts for understanding the propagation of sound waves, or sound. In Chapter 3, the basic concepts mentioned above are more actively utilized to explain the reflection and transmission of sound waves on a discontinuous surface. In Chapter 4, the phenomena of radiation, scattering, and diffraction are explained in a unified concept: change in sound waves depending on the spatial distribution of a discontinuous surface, or the pattern by which an impedance mismatch is distributed in space. Lastly, in Chapter 5, change in sound—depending on the size and form of the space propagating it—is explained. Here, that which occurs in a given space is also understood under the assumption that the walls constituting the space are an impedance mismatch distributed in that region. In other words, impedance and other major physical values are the third coordinate axis of this book. I sincerely ask the readers to review the three coordinate axes—and subscribe to the philosophy of knowledge inquiry that one should “start from understanding the very basics” —that is, the concepts of relative time and space, impedance, and other major physical values, in a comprehensive manner when reading this book. Through this three-dimensional understanding, I hope the readers can understand a wide variety of basic physical phenomena with regard to sound in an easier and more extensive manner. If one can create something related to sound on the basis of this understanding, then we will bear witness to a new achievement that humanity has not yet seen in its long history.
This may seem too much of an imagination and expectation about this book, which still needs lots of improvement. However, if we can fully understand the basic concepts, and if this understanding leads us to intuitive perception and judgment on sound without further analysis, wouldn’t it be hard to say the suggestion to recognize and understand sound based on the three coordinate axes and the expectations on this are still too groundless? Another characteristic of this book is that it includes numerous interpretative/analytic approaches in the appendices to elucidate basic concepts in a more consistent and in-depth
Preface
xiii
manner. The main text of this book is designed to help readers understand the nature of sound and its physical significances. I believe it will be a good bit of reading material for undergraduate or master’s students or the general public interested in acoustics. The appendices contain more detailed theoretical background and derivation processes. I expect they serve as a solid reference for master’s and doctoral students in need of more in-depth mathematical and theoretical background. Yang-Hann Kim KAIST
Acknowledgments This book describes what I have learned while teaching students at KAIST on what I had been inspired by the teachers at MIT who opened my eyes to the world of vibration engineering and acoustics. I have been fortunate to have had tremendous academic influence from Professor J. K. Vandiver, who had a major impact on vibration engineering studies and experimental approaches; Professor S. H. Crandall, who taught me random vibration and acoustics; Professor R. H. Lyon, from whom I learned structural acoustics; and Professor U. Ingard, who considerably influenced the theories of acoustics. Professor P. E. Doak of the Institute of Sound and Vibration Research (ISVR) has told me many intriguing stories about great scholars of contemporary acoustics and historical milestones that they achieved, and he has always provided sincere encouragement for me and my somewhat bold approach to acoustics. Professor J. K. Hammond (ISVR) who has motivated me to transform general acoustics problems to signal processing issues must be well acknowledged. I still cherish these wonderful memories with my two great friends. I also feel very privileged and honored to publish this book as part of an ongoing effort to “transform” voluminous books on physical acoustics— written by Professors Morse, Feshbach, and Ingard and commonly seen on the bookshelves— into a single volume shedding light on the very basic concepts needed for engineering applications. I would like to also thank to Professors J. S. Bolton of Purdue University, Karl Grosh of University of Michigan, and Hideki Tachibana of Tokyo University for their kind provision of materials that make this book more informative. Their long-time friendships are the essentials of this book. Finally, my sincere thanks must go to Professor Thomas Rossing of Stanford University who encouraged me to write the book that can convey fundamentals of acoustics in terms of impedance. My deepest appreciation goes to Professors Chong-Won Lee, Yun-Sik Park, Kwang-Joon Kim, Jeong-Guon Ih, and Young-Jin Park of the Center for Noise and Vibration Control (NOVIC) at the KAIST Department of Mechanical Engineering for their valuable advice during the completion of this book. I also extend my special thanks to Professor Duck-Joo Lee of the KAIST Department of Aerospace Engineering. Since becoming a professor at KAIST, I have been lucky and blessed enough to teach and study in such an excellent environment, and all this would not have been possible if it were not for the help of these distinguished colleagues. I would also like to convey my respect, devotion, and thanks to all the professors at the Department of Mechanical Engineering. I extend my sincere gratitude to Drs. Hee-Seon Seo of the Agency for Defense Development (ADD) and Jung-Woo Choi of Samsung SAIT who, during their doctoral years, provided
xvi
Acknowledgments
tremendous support for the editing and proofreading of this book with sincere interest and passion. After all, this book would have never come to fruition without the unrelenting encouragement of KAIST students and graduates that helped me overcome my hesitations several times and eventually gave me the courage to write this book. I especially thank Dr. Kyoung-Uk Nam for his theoretical contributions of deriving interpretative methods needed for a comprehensive understanding of the material discussed in Chapter 4—radiation, scattering, and diffraction—and obtaining results in a physically significant manner. Indeed, it would be fair to say that the bulk of the appendix for Chapter 4 was contributed by Dr. Nam. Whenever struggling to find a way to teach my class in an easy-to-understand but extensive manner, I have always been fortunate and honored to learn a fresh and new approach from my students. Last but not least, I sincerely thank Ms. Young-Hwa Choi for her splendid efforts for the publication of this book. Yang-Hann Kim
1 Vibration and Waves 1.1
Introduction/Study Objectives
Vibration can be considered as a special form of a wave (wave propagations, Figure 1.1). We shall begin with how vibration can be envisaged as a kind of wave, mathematically and physically, and then move to the study of acoustic waves.
1.2
From String Vibration to Wave
There are many different types of waves, including acoustic waves propagating in general three-dimensional space. To understand how a wave propagates in space, let us start with the simplest case. Figure 1.2 shows how two sinusoidal vibrations, whose frequencies are f2 and f3, are actually composed of two different vibrations, that is, modes. This can be mathematically expressed as yðx; tÞ ¼ A2 sin
2px 3px sin 2pf2 t þ A3 sin sinð2pf3 t þ FÞ; L L
ð1:1Þ
where F represents the phase difference between the second and third modes that are participating in the vibration. In other words, the phase difference determines how much the third mode vibrates in advance of the second mode. This is the phase difference in time. However, there is also a phase difference in space, as demonstrated by Figure 1.3. By carefully examining the motion between the nodal points, for example, we can see that there is a phase difference of p. The motion of the second mode at x ¼ L=4 can be written as A2 sin 2pf2 t, where A2 is the maximum amplitude of the second mode. On the other hand, the motion at x ¼ 3L=4 can be expressed as A2 sin 2pf2 t or as sinð2p f2 t þ pÞ. Note that there is always a phase difference of p between nodal points. It is also interesting to change
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
2
Sound Propagation
Figure 1.1 The first, second, and third modes of a string (demonstration by C.-S. Park and S.-H. Lee, 2005, at KAIST) y y
A2
(Second mode)
A2 + A3
x
x
=
L (An example of a string’s vibration)
+ y
A3
(Third mode)
x
Figure 1.2 Vibration of a string fixed at both ends (this demonstrates that the vibration can be expressed as the sum of two modes: the second and third modes of the string)
the first term of Equation 1.1 to an additive form from its multiplication form. It can then be rewritten as 2px sin 2pf2 t L : 1 2px 2px ¼ A2 cos 2p f2 t cos þ 2p f2 t 2 L L
A2 sin
ð1:2Þ
Rearranging this equation in terms of x gives A2 sin
2px 1 2p L 2p L sinð2pf2 tÞ ¼ A2 cos x xþ t cos t L 2 L ð1=f2 Þ L ð1=f2 Þ
ð1:3Þ
3
Vibration and Waves
Figure 1.3
How the second and third modes create the vibration
where L=ð1=f2 Þ indicates a velocity that travels along the string. In fact, Equation 1.3 essentially means that there are two waves propagating along the string in opposite directions with a velocity of L=T2 ðT2 ¼ 1=f2 Þ. In other words, the second mode can be regarded as a wave created by two waves that propagate in opposite directions. We often call this kind of wave a standing wave, simply because it has a fixed phase difference of p between the nodal points. Similarly, the first or even nth mode can be interpreted in the same manner as for the second and third mode cases. Therefore, the string vibration can generally be written as yðx; tÞ ¼
1 X n¼1
An sin
np x sin ð2pfn t þ Fn Þ; L
ð1:4Þ
4
Sound Propagation
where Fn is the phase of the nth mode.1 If we rewrite Equation 1.4 with respect to x, then yðx; tÞ ¼
1 X 1 n¼1
2
An
np 2L Fn np 2L Fn cos x t cos xþ tþ : ð1:5Þ L nTn ðnp=LÞ L nTn ðnp=LÞ
This equation states that each mode propagates with a velocity of 2L=nTn in opposite directions.2 If we investigate Equation 1.5 from another perspective, we obtain a more general mathematical expression for a wave. This equation essentially states the following: “There are two cosine waves propagating in the positive ( þ ) and negative () directions with respect to space, x.” This certainly suggests that the general wave form, which is not simply a cosine wave, can be mathematically expressed as yðx; tÞ ¼ gðxctÞ þ hðx þ ctÞ;
ð1:6Þ
where g() and h() generally denote a wave form. Note that a wave g or h essentially depicts a wave form in arbitrary space and time. These also propagate in space and time with the relation x þ ct or xct. To understand this concept in greater depth, refer to the wave depicted in Figure 1.4. This graph demonstrates how the function g moves along the axis x with time. We can see that the function g moves in space by a distance ct. Notably, we are now seeing the wave, expressed by the function g, in space; in other words, with respect to the x coordinate, we can now see how it changes in time with respect to space. g(x), t = 0 g(x −cτ), t = τ
ct
x
Figure 1.4 The wave propagates in the positive ( þ ) x direction; g expresses the shape of the wave, c the wave propagation speed, and t and x are the time and coordinate
If we rewrite the function or wave g with regard to time, then we obtain x gðxctÞ ¼ g c t : c
ð1:7Þ
Equation 1.7 states that the right-going wave in space can be seen as the wave propagates in time. Figure 1.5 essentially illustrates that what we can see in space is related to what we observe in time; this graph is typically referred to as a wave diagram. 1
This phase has to be determined by the relation between the excitation force and vibration amplitude response (see Section 1.9.1). 2 Since Tn ¼ T1 =n, 2L=nTn ¼ 2L=T1 . In other words, the propagation speed of the wave that creates every mode has to be the same.
5
Vibration and Waves
y(x,t)
t=0
x
t = Δt
t Wave line Wave diagram y(x,t) t = t1
t=0
x1
x2
x
y(x,t) x = x1
x = x2
t1
t2
t
Figure 1.5 Wave diagram: waves can be observed at the x coordinate (space) and t axis (time), where Dt denotes infinitesimal time, x1;2 and t1;2 indicates arbitrary position and time, and y is the wave amplitude
The sine wave is a special wave that can be expressed by Equation 1.7. The sine wave, propagating to the right, is expressed yðx; tÞ ¼ Y sin ðkðxctÞ þ FÞ
ð1:8Þ
where k converts the units of the independent variable of the sine function to radians; x and ct are in units of length; Y represents amplitude and F is an arbitrary phase depending on the initial value of the amplitude of the sine wave. We rewrite Equation 1.8 as yðx; tÞ ¼ Y sin ððkxkctÞ þ FÞ ¼ Y sin ðkxot þ FÞ;
ð1:9Þ
where kc ¼ o. This is simply because the independent variable of the sine function must be measured in radians and therefore o is measured in rad/sec. This rather direct consequence has a very significant physical implication. It relates the variable that expresses the change of space (x), k, with that related to time (t), o. That is, k¼
o : c
ð1:10Þ
6
Sound Propagation
However, Equation 1.10 does not allow us to interpret any significant physical meaning. This is simply because the radian frequency does not coincide with our perceptual sensation. It is instead preferable to write Equation 1.10 in terms of frequency (cycles/sec, Hz), or period (sec), that is k¼
o 2pf 1 ¼ ¼ 2p : c c cT
ð1:11Þ
Note that the distance across which a wave travels for a period T with a propagation speed c will be a wavelength (l) (see Figure 1.6). Therefore, we can rewrite Equation 1.11 as k¼
2p ; l
ð1:12Þ
where k represents the number of waves per unit length (1=l). We call this the wave number or a propagation constant. c 0 x T 4
x
T 2
x
3T 4
x
T
t
xT = λ
x Wave line Wave line x = ct
Figure 1.6 Waves can be seen for one period: T is period (sec), c is propagation speed (m/sec), and x and t represent the space and time axis, respectively
Figure 1.6 illustrates Equations 1.10, 1.11 and 1.12. We can also obtain an additional relation from Equations 1.10 and 1.12. That is, c l¼ : ð1:13Þ f This states that the variables which express space (l) and time (f ) are not independent of each other; that is, they have a very direct relation. We normally call this a “dispersion relation,” and
7
Vibration and Waves
Equation 1.10 is one such case. For example, we can determine the wavelength if we know the speed of sound and frequency. For example, if the speed of propagation is 340 m/sec and its frequencyis 1 kHz,then thedispersion relationstates that itscorrespondingwavelengthis 34 cm.3 Waves having a sine or cosine form are considered to be special types and not general. The dispersion relation is limited to these special cases. However, we also know that any wave can be obtained or constructed from a Fourier series or Fourier integral (Section 1.9.2) by superposing sine and cosine waves. Understanding what we have anticipated from harmonic waves is the key process in wave physics and engineering. Any steady-state waves can be regarded as a superposition of harmonic waves; understanding single harmonic waves is therefore the starting point as well as the source to which we must always return. If we recall that sines and cosines can be compactly expressed by using a complex function, then Equation 1.9 can be rewritten as yðx; tÞ ¼ ImfYe jðkxot þ FÞ g ¼ ImfYe jðkxotÞ g:
ð1:14Þ
where Y is the complex amplitude and Im indicates the operator, which takes only an imaginary part of the complex function. For the sake of simplicity, Equation 1.14 will be written as yðx; tÞ ¼ Ye jðkxotÞ :
ð1:15Þ
Note that this equation expresses a harmonic wave propagating in the right direction.4 As noted previously, we can also express Equation 1.15 with respect to time instead of space, that is yðx; tÞ ¼ YejðotkxÞ :
ð1:16Þ
To summarize, we can regard vibration as a special wave that has a fixed phase p. We also learned that any one-dimensional wave can be expressed using Equation 1.6. Harmonic waves, which fall under the realm of Equation 1.6, can be described by Equation 1.15 or 1.16. We obtained the dispersion relation from these harmonic wave expressions, linking time and space using Equation 1.6.
1.3
One-dimensional Wave Equation
We have determined that any one-dimensional wave can be expressed as yðx; tÞ ¼ gðxctÞ þ hðx þ ctÞ:
3
ð1:17Þ
The essential length scale for waves is not what is normally used for measuring length, area, and volume (i.e., m, m2, m ), but what can be compared with the wavelength in which we are interested. 4 If we express the wave propagation by a complex function, then we have many mathematical advantages. For example, it does not change its form as we calculate the derivative or integral of it (albeit the sine and cosine function changes its form). If we are interested in the response of a sine or cosine form excitation, then we can just consider the imaginary or real part provided that we are handling a linear system. 3
8
Sound Propagation
We have also examined its fundamental physical meanings, especially the dispersion relation. Let us look at Equation 1.17 in a rather different way to Section 1.2. Mathematical analysis of Equation 1.17 allows us to explore the physical and engineering derivations that will enhance our understanding of wave propagation. Our first attempt based on this philosophy is to examine how Equation 1.17 behaves if we view it with respect to small changes in time and space. In other words, we would like to determine the derivative of Equation 1.17 with regard to time and space and thereby examine its underlying physical meaning.5 First, let us see how Equation 1.17 behaves in the case of a small spatial change: @y ¼ g0 þ h 0 ; @x
ð1:18Þ
where 0 denotes the derivative of each function with respect to its argument (e.g., gðzÞ0 ¼ dg=dz). Its time rate of change is expressed as @y ¼ cg0 þ ch0 @t
ð1:19Þ
@g @g ¼ cg0 ¼ c ; @t @x
ð1:20Þ
@h @h ¼ ch0 ¼ c : @t @x
ð1:21Þ
which leads to
Equation 1.20 essentially states that the rate of change of a wave g is related to its change in space by the factor c. On the other hand, the rate of change of wave h which is propagating in the negative direction is related to its change in space by the factor c. In other words, the g wave moves upward if it has a negative slope with respect to space and moves downward if it has a positive slope. Figure 1.7 illustrates the associated kinematics of the right-going and left-going wave. If we differentiate Equations 1.20 and 1.21, we obtain 2 @2g 2@ g ¼ c ; @t2 @x2
ð1:22Þ
2 @2h 2@ h ¼ c : @t2 @x2
5
We often look at the rate of change of physical variables such as wave amplitude or pressure with respect to time or space, in order to understand its underlying physics. This is a very popular means that is often used in continuum mechanics, in which we can assume that what happens in the small element will eventually governs general behavior. The opposite concept is to integrate the physical parameters in which we are interested. For example, see Hamilton’s Principle (Crandall, S.H., Karnopp, D.C., Kurtz, E.F. Jr, and Pridmore-Brown, D.C. (1968) Dynamics of Mechanical and Electro Mechanical System, McGraw-Hill) or Kirchhoff–Helmholtz integral equation in Chapter 2.
9
Vibration and Waves Right-going wave, g(x –ct) Wave with positive slope
Wave with negative slope
x
x =0
Figure 1.7 Understanding waves from the perspective of wave kinematics (a wave that has a positive slope or negative slope has a negative or positive rate of change, i.e., velocity)
This indicates that any one-dimensional wave ðyðx; tÞÞ which has left-going and right-going waves with respect to the selected coordinates satisfies the partial differential equation: @2y @2y ¼ c2 2 ; 2 @t @x
ð1:23Þ
emphasizing that the acceleration is an expression of result of motion. Equation 1.23 can then be rewritten as @2y 1 @2y ¼ 2 2: 2 @x c @t
ð1:24Þ
The above equation is normally used for one-dimensional waves. This wave equation essentially expresses all possible waves (right- or left-going). This equation can also express the relation between curvature and acceleration. A simple extension of Equation 1.24 yields a general three-dimensional wave equation. If we denote the amplitude of the three-dimensional wave as xðx; y; z; tÞ to avoid any ambiguity that could arise by using y as a displacement, where y denotes one of the rectangular coordinates, then a three-dimensional version of Equation 1.24 can be written as r2 x ¼
1 @2x : c2 @t2
ð1:25Þ
Any wave can be mathematically regarded as one of the solutions that satisfies this equation and the boundary condition, in steady state. The boundary condition can generally be written as ax þ b
@x ¼ c; @x
ð1:26Þ
where c expresses the general force acting on the boundary. a and b are coefficients that are proportional to force and spatial change of force, respectively.6 There are normally two types of boundary conditions: passive and active. The former expresses the case which essentially does not exert force or pressure and the latter is related to that which applies forces to the boundary (Figure 1.8). 6
Equation 1.26 is often referred to as the Robin boundary condition. If b is 0, then it is referred to as the Dirichlet @x boundary condition; if a is 0, then it becomes Neumann boundary condition. If x and @x are given independently, then we refer to is as the Cauchy boundary condition (Morse, P.M. and Feshbach, H. (1953) Methods of Theoretical Physics, McGraw-Hill, New York, pp. 678–679).
10
Sound Propagation
z=∞
m s
rd
Example of passive boundary condition
Ae–jω t Example of active boundary condition
Figure 1.8
Examples of passive and active boundary conditions
To summarize, the general wave that can be expressed by Equation 1.17 satisfies the governing equation, Equation 1.24. This means that every possible wave that we can imagine, that is, g and h, are the solutions of the governing equation. The boundary condition determines the specific waves among every possible wave. We can therefore conclude that the waves that can be seen physically satisfy the governing equation and the boundary conditions. This is one reason why we are interested in solving the governing equation that satisfies the boundary condition, that is, it shows what we cannot see.
1.4
Specific Impedance (Reflection and Transmission)
We have studied the wave equation that governs every possible wave in three-dimensional space which has arbitrary boundaries. To explore how waves propagate and behave in relation to their medium and boundaries, we first examine the one-dimensional wave equation. This will allow us to understand more complicated general waves in a compact and simple form. Waves traveling along a string are representative of the many possible one-dimensional waves. Let us first examine waves propagating along two different strings, as illustrated in Figure 1.9. As we can see in this figure, a wave propagates in the right-hand direction. Let us call this an incident wave, g1, which propagates in the positive x direction. Observing this wave with respect to time, it can be seen as a propagating wave in the positive time axis. Assume there are an incident wave g1 and a reflected wave h1 along a thinner string (see Figure 1.9). Assume also that there is only a transmitted wave g2 along the thicker string (#2
11
Vibration and Waves
Figure 1.9 Waves in two strings of different thickness (g1 is an incident wave, h1 represents a reflected wave, and g2 is a transmitted wave)
string in Figure 1.9). Note that this assumption includes every possible case that can physically occur in the system under consideration. At this point, we wish to determine the relation between the incident wave g1, the reflected wave h1 and the transmitted wave g2. The relation has to be determined from the mathematical realization of what is physically observed at the junction (x ¼ 0), where there is a discontinuity in the string thickness. Let’s envisage what really happens at this discontinuity, and then express it mathematically. First, we can easily see that the velocities in the y direction (uy) of the thin string and thick string have to be identical.7 In addition, the resultant force in the y direction ( fy) has to be balanced according to Newton’s second Law. These two requirements at the discontinuity are expressed mathematically as uy x¼0 ¼ uy x¼0 þ ;
ð1:27Þ
fy x¼0 þ fy x¼0 þ ¼ 0:
ð1:28Þ
As illustrated in Figure 1.9, we denote the waves on the negative x axis region, #1 string, as y1 and express the wave that propagates in the positive x axis as y2. Describing these waves with regard to time, they can be written as x x þ h1 t þ ; ð1:29Þ y1 ¼ g1 t c1 c1 y 2 ¼ g2
x t : c2
ð1:30Þ
The velocity in the y direction at x ¼ 0 can be written as uy x¼0 ¼
7
@y1 i ¼ g01 x¼0 þ h01 x¼0 : @t x¼0
ð1:31Þ
We can assume that the velocity in the x direction is much smaller than that in the y direction. This assumption is valid when the string’s displacement is small enough to be linearized (see Section 1.5).
12
Sound Propagation
At x ¼ 0 þ , it is uy x¼0 þ ¼
@y2 @t
x¼0 þ
¼ g02 x¼0 þ :
ð1:32Þ
We therefore obtain the following equality since the velocity must be continuous: g01 x¼0 þ h01 x¼0 ¼ g02 x¼0 þ :
ð1:33Þ
Note also that the forces in the y direction (fy) are related to the tension along the string TL and the slope (Figure 1.10) as fy x¼0 ¼ TL
@y ; @x
fy x¼0 þ ¼ TL
@y : @x
ð1:34Þ
y TLsinθ ≈ TL
TL
∂y ∂x
String #2 θ 0
x
dy dx
String #1 TL
TLsinθ ≈ TL
∂y ∂x
Figure 1.10 Forces acting on the end of the string where TL is tension, fy describes the force in the y direction, y indicates the amplitude of the string, and x denotes the coordinate)
Using Equations 1.29, 1.30 and 1.34, we can rewrite Equation 1.28 as TL 0 TL TL g1 x¼0 h0 x¼0 ¼ g02 x¼0 þ : c1 c1 c2
ð1:35Þ
It should be noted that Equations 1.33 and 1.35 are composed of the first derivatives of the functions g1, h1, and g2 which describe the waves that can exist in the strings. Without loss of generality, we can postulate that the string’s wave amplitude at x ¼ 0, t ¼ 0 is zero. We can therefore write Equations 1.33 and 1.35 as g01 ð0Þ þ h01 ð0Þ ¼ g02 ð0Þ;
ð1:36Þ
TL 0 TL TL g ð0Þ h01 ð0Þ ¼ g02 ð0Þ: c1 1 c1 c2
ð1:37Þ
Equation 1.37 essentially states that the rate of change of each wave amplitude is scaled by the ratio between the tension (TL ) and propagation speeds (c1 ; c2 ). If we use Equation 1.34 to
13
Vibration and Waves
examine the ratio between the forces acting on the waves (g1 ; h1 , and g2 ) and the corresponding velocities (their magnitudes), then it can be easily found that this is in fact the ratio of the string’s force in the y direction (fy) and the associated velocity (uy), i.e. fy T L ¼ : uy c
ð1:38Þ
The force with respect to velocity, or the force that can generate the unit velocity, is generally defined as the impedance. We normally express this using the complex function Z, which allows us to express any possible phase difference between the force and velocity. Therefore, Equation 1.37 can be rewritten as Z1 g01 ð0ÞZ1 h01 ð0Þ ¼ Z2 g02 ð0Þ
ð1:39Þ
where Z1 and Z2 are equal to TL =c1 and TL =c2 , respectively. Z1 and Z2 describe the propagation characteristics of the string or the impedance or characteristic impedance of the string. We can express Equations 1.36 and 1.39 in terms of the ratio of the amplitude of the reflected wave with respect to the unit incident wave, that is, the reflection ratio (h1/g1). Using a similar comparison for the transmitted wave with regard to the incident wave, that is, the transmission ratio (g2/g1), we obtain the following: h1 ð0Þ Z1 Z2 ¼ ; g1 ð0Þ Z1 þ Z2
ð1:40Þ
g2 ð0Þ 2Z1 ¼ : g1 ð0Þ Z1 þ Z2
ð1:41Þ
Equations 1.40 and 1.41 essentially demonstrate that the ratio of the reflected wave and transmitted wave to the incident wave depends entirely on the string’s impedance, ðTL =cÞ. We may examine these equations by considering extreme cases, for example, when Z1 Z2 or Z1 Z2 . Figure 1.11 exhibits how the waves on a string propagate when they meet a change of impedance or, in this case, a change of thickness of string. We can see that the transmitted and reflected waves completely depend on the impedance mismatch, that is, mass or thickness discontinuity in the present cases. The first case shows that the reflected wave is completely reversed. In other words, there is a phase difference of p. This can be readily seen from Equation 1.41 when the impedance of # 1 string is considerably smaller than that of # 2 string; in other words, Z2 is much larger than Z1 . In this case, Equation 1.40 approaches 1, which means that the reflected wave amplitude is the same as that of the incident wave but reversed in phase. As Z2 becomes far greater than Z1 , the transmitted wave will be very small and Equation 1.41 will approach 0. However, from the sixth case in Figure 1.11 where there is a perfect impedance match, the phases of the reflected waves are not reversed. If Z1 becomes substantially larger than Z2 (Z1 Z2 ), then Equation 1.40 predicts that the incident wave will be totally reflected without any change in phase. Equation 1.41 demonstrates that the amplitude of the transmitted wave is twice that of the incident wave; this is extremely unlikely in reality, however. This rather surprising paradox can be explained by examining how
14
Sound Propagation
Figure 1.11 Incident, reflected, and transmitted waves on a string; note the phase changes of the reflected and transmitted waves compared to the incident wave. The thin line has impedance Z1 and the thick line has impedance Z2
much power is transmitted. We can readily see that the power becomes zero because the transmitted velocity is zero, even although the transmitted wave amplitude is twice that of the incident wave.8 To summarize, the reflected and transmitted waves satisfy the governing equation. The amount/degree that is reflected and transmitted is completely determined by the impedances of the strings. This is the most important part of this chapter. Recall that the governing equation essentially and completely expresses everything associated with wave propagation. The solutions, which represent specific physical phenomena of a wave, are determined by the boundary and initial conditions. For the waves on a string that has two different thicknesses, the impedance mismatch is related to the boundary condition and therefore determines the ratio of the reflected and transmitted wave. Next, we will study in greater detail the physical concepts associated with impedance.
1.5
The Governing Equation of a String
We have seen that the transmission and reflection are physical and mathematical consequences of impedance mismatch. In other words, the impedance determines the waves on the string. We have also discovered that the string tension and speed of propagation determine the impedance 8 The transmitted velocity can be easily obtained by multiplying the numerator of Equation 1.41 byZ2 and dividing the denominator by Z1 ; note that the impedance is the ratio of force and velocity. In fact, we can also obtain the velocity transmission and reflection ratios by using similar equations to Equations 1.40 and 1.41.
15
Vibration and Waves
of the string. The next problem to resolve is what determines or governs the wave propagation speed on the string. To answer this, let us examine an infinitesimal element of string (Figure 1.12). For convenience, x represents the length scale along the string and y the perpendicular direction as illustrated in Figure 1.12. Applying Newton’s second law to the sselected element, and expressing the law according to the coordinates and sign convention defined in Figure 1.12, we obtain the following relations. First, Newton’s second law in the x direction can be written: TL cos y þ ðTL þ dTL Þcosðy þ dyÞ ¼ rL ds
y
y
TL + dTL
ρLds
@2x : @t2 ∂ 2y ∂ t2
θ + dθ dy
2 ρLds ∂ x
=
∂ t2
ds
θ
ð1:42Þ
TL 0
0
x
dx
x
Figure 1.12 Newton’s second law on an infinitesimal element of a string (notation as for Figure 1.10)
The force and motion in the y direction can be written: TL sin y þ ðTL þ dTL Þsin ðy þ dyÞ ¼ rL ds
@2y ; @t2
ð1:43Þ
where y expresses the slope of the string with respect to the x axis at an arbitrary position of x: tan y ¼
@y : @x
ð1:44Þ
The change of this slope with regard to a small change in x (dx) can be written y þ dy ffi
@y @2y þ 2 dx @x @x
ð1:45Þ
using a Taylor expansion. Assuming that the displacement of the string is small enough to be linearized, then sin y ffi y;
ð1:46Þ
cos y ffi 1: Equations 1.42 and 1.43 thus become TL þ ðTL þ dTL Þ ¼ rL ds
@2x ; @t2
ð1:47Þ
16
Sound Propagation
@y @y @2y @2y þ ðTL þ dTL Þ þ 2 dx ¼ rL ds 2 : TL @x @x @x @t The small ds can be rewritten as qffiffiffi ds ¼ ðdxÞ2 þ ðdyÞ2 ¼ dx
sffiffiffiffi ! @y 2 1 @y 2 ffi dx 1 þ ; 1þ @x 2 @x
ð1:48Þ
ð1:49Þ
@y using Pythagoras’ Theorem. In Equation 1.49, the slope of the string @x can be assumed to be small. Its square can therefore be neglected compared to other variables. Therefore, we can approximate
ds ffi dx:
ð1:50Þ
Note that the small change of tension dTL can be expressed by a first-order approximation as @TL dx: @x
ð1:51Þ
@TL @2x ffi rL 2 : @t @x
ð1:52Þ
dTL ffi Therefore, Equation 1.47 can be rewritten as
This is a first-order approximation of Equation 1.47, or a linearized equation of Equation 1.47. However, it is must be noted that the tension TL does not exhibit a significant change in the x direction, because we assumed a small displacement. Equation 1.52 allows us to neglect the motion in the x direction. Therefore, we can conclude that Equation 1.48 essentially governs the wave propagation of the string under study, and we can easily write Equation 1.48 as TL
@2y @2y ffi rL 2 : 2 @x @t
ð1:53Þ
This is a linearized version of Equation 1.48. Rearranging Equation 1.53 results in @ 2 y rL @ 2 y ¼ : @x2 TL @t2
ð1:54Þ
We see that this has the same form as Equation 1.24; a typical form of a wave equation. Equation 1.54 can be summarized as @2y 1 @2y ¼ 2 2; 2 @x cs @t c2s ¼
TL ; rL
ð1:55Þ
ð1:56Þ
17
Vibration and Waves
where cs represents the propagation speed of the string. Note that Equation 1.55 is exactly the same as Equation 1.24. Recall that Equation 1.24 was obtained by first imagining all possible waves, and then attempting to find a complete and compact mathematical form that can express these in space and time. Such a form turns out to be x ct. It is then necessary to determine the relations between these mathematical functions. We first selected the solutions and then identified what governs these solutions, that is, the governing equation. However, Equation 1.55 originated from an examination of the small element of a string: forces and motion relation. These two approaches must lead us to the same conclusion. However, we must recall that the final results stem from very different approaches.9 Note that a general boundary value problem entails nothing more than finding solutions to satisfy Equation 1.55 and the boundary conditions. Recall that the impedance of the string Z is Z¼
TL : cs
ð1:57Þ
Using Equation 1.56, we can rewrite Equation 1.57 as ð1:58Þ Z ¼ rL cs ; pffiffiffi or it can be expressed as Z ¼ TL rL . This means that the impedance of the string becomes larger as the string’s density per unit length increases (Figure 1.11). It is stressed here that impedance has two different implications. First, as noted by Equation 1.38, the impedance is a measure of how effectively the force can generate velocity (response), that is, the input and output relation between force and velocity. We also see that the impedance is determined by the string’s own characteristics. Note that the right-hand side of Equation 1.57 is composed of the string’s density and the tension along the string. Because it represents the characteristics of the medium (in this case, the string), we call this the characteristic impedance. The following addresses how these impedances affect the wave propagation on a string. To summarize, we have seen that the wave propagation along a string is determined by the impedance of a string. If there is an impedance mismatch, then waves are reflected and transmitted, depending on the degree of impedance mismatch. The characteristic impedance of a string is related to the tension along the string, the mass per unit length, and the propagation speed. We also learned that the speed of propagation becomes faster as the tension increases, and tends to slow down as the mass of string becomes larger.
1.6
Forced Response of a String: Driving Point Impedance
We have seen that the characteristic impedance essentially governs the propagation phenomena on a string. For example, it determines the degree to which the wave is transmitted and is reflected due to impedance mismatch. Another aspect concerning impedance is also possible. First, recall that the impedance expresses the ratio between the force (or pressure (input)) and velocity (output) (see
9
These two approaches can be applicable in any system of interest.
18
Sound Propagation
Equation 1.38). This implies that we can use impedance as a representative measure to indicate how well velocity responds to external excitation (force).10 We first investigate what happens if we harmonically excite one end of a semi-infinite string, as illustrated in Figure 1.13. For mathematical convenience, we begin by expressing the waves in Figure 1.13 using a complex function: yðx; tÞ ¼ gðxcs tÞ:
ð1:59Þ
Figure 1.13 Wave propagation by harmonically exciting one end of a semi-infinite string (T is period, cs is propagation speed, l is the wavelength, f is the frequency in Hz (cycles/sec), and o is the radian frequency in rad/sec)
If we denote the response of the string as Yejot due to the excitation (Fejot ) at x ¼ 0, then the boundary condition at x ¼ 0 can be written as yð0; tÞ ¼ gðcs tÞ ¼ Yejot : 10
We often refer to this kind of impedance as mechanical impedance.
ð1:60Þ
19
Vibration and Waves
Notably, the second expression in Equation 1.60 indicates that the solution is a function of cs t. We can therefore rewrite Equation 1.60 as gðcs tÞ ¼ Yejkðcs tÞ ;
ð1:61Þ
where we use the dispersion relation k ¼ o=cs . k is the wave number, relating the frequency and speed of propagation (Figure 1.13 illustrates this dispersion relationship and the physical meanings). If we rearrange Equation 1.61 using an independent variable a, then we obtain gðaÞ ¼ Ye þ jka
ð1:62Þ
We can therefore substitute a by xcs t, which gives us gðxcs tÞ ¼ Ye þ jkðxcs tÞ ¼ YejðotkxÞ :
ð1:63Þ
This equation expresses a wave that propagates in the right-hand side with wave number k and frequency o (Figure 1.13). The impedance at x ¼ 0 is the ratio between the force ðFejot Þ at x ¼ 0 and the velocity uy ð0; tÞ. The velocity can be expressed using Equation 1.60: @y uy ð0; tÞ ¼ ¼ jo Yejot : ð1:64Þ @t x¼0 We also know that the force at the end of the string is related to the tension and the slope of the string (Figure 1.10): @y fy ð0; tÞ ¼ Fejot ¼ TL : ð1:65Þ @x x¼0 From Equations 1.64 and 1.65, we can rewrite the impedance at the end (Zm0 ) as Zm0 ¼
fy ð0; tÞ ¼ rL cs ; uy ð0; tÞ
ð1:66Þ
where the subscript m denotes the mechanical impedance and 0 is used to expresses the end of the string (x ¼ 0). Equation 1.66 is an indicator of how the string actually moves or responds to the excitation. Note that the impedance is a complex function; this means that it has magnitude and phase. In this respect, Equation 1.66 predicts that the string moves, in terms of velocity, with the same phase as the harmonic excitation force. In other words, it moves in the direction of the force. As we have seen in Figure 1.13, the only possible wave in this situation is one which propagates in the positive x direction. We can therefore conclude that the wave can only propagate in the positive x direction if and only if the impedance at the end of the string is real and has a value of rL cs . No reflected wave is then possible. This highlights the importance of the driving point impedance, which we now define as the impedance at a specific point where force is applied. This impedance describes how the wave propagates along the string. Surprisingly, the characteristics of the driving point impedance determine the spatial phenomenon of wave propagation, that is, the ways in which waves propagate in space. This is what is physically meant by the driving point impedance. Another extreme case that can demonstrate how the driving point impedance reflects the wave propagation along a string is a string that has finite length L. Consider again a string where
20
Sound Propagation
one end (x ¼ 0) is harmonically excited and the other end (x ¼ L) is fixed, not allowing any motion. The boundary condition at x ¼ L requires that the displacement yðx; tÞ always be 0. We find that the solution that satisfies the governing wave equation and this boundary condition can be written as yðx; tÞ ¼ Y sin kðLxÞejot ;
ð1:67Þ
where Y is an arbitrary complex amplitude. The reason why we use Lx instead of xL is simply because we want a sine wave with respect to the fixed end ðx ¼ LÞ. If we calculate the velocity using Equation 1.67 at x ¼ 0, then we have @y uy ð0; tÞ ¼ ¼ jo Y sin kL ejot : ð1:68Þ @t x¼0 On the other hand, the force at x ¼ 0 is @y fy ð0; tÞ ¼ TL @x
¼ TL kY cos kL ejot :
ð1:69Þ
x¼0
Equations 1.68 and 1.69 eventually give us the impedance (specifically, the driving point impedance Zm0 ) at x ¼ 0. That is, Zm0 ¼
fy ð0; tÞ TL ¼ j cot kL ¼ jrL cs cot kL: uy ð0; tÞ cs
ð1:70Þ
Equation 1.70 shows, in contrast to the driving point impedance of a semi-infinite string, that the impedance only has an imaginary part. In other words, there is a phase difference of 90 between the force and velocity at the driving point. It is more likely that there is a reflected wave due to the presence of a boundary at x ¼ L.11 Figure 1.14 depicts the driving point impedance at x ¼ 0. First, it must be stressed that the impedance is described in terms of kL which is a measure of the lengthscale L with regard to the wavelength l. When kL is very small or, in other words, when the wavelength is large compared to the length of the string, then Equation 1.70 reduces to Zm0 ffi jrL cs
1 : kL
ð1:71Þ
Rearranging this equation, we obtain Zm0 ffi j
TL : oL
ð1:72Þ
This conveys that the driving point impedance, in this case, is mainly dominated by the tension along the string and has a significant value (see Figure 1.14). This means that when we excite the end point with a very low frequency, the tension dominates the impedance. This agrees with intuition.
11
The phase difference between force and velocity are addressed in Section 1.9.1.
Vibration and Waves
Figure 1.14 string)
21
The driving point impedance of a finite string (k is wave number and L is the length of the
In addition, if kL becomes larger and approaches p=2, as in the case where the wavelength is 4 times longer than the length of the string, then the driving point impedance will be 0 and there will be resonance. When kL passes the value of p=2 (90 ), then the impedance will have a negative imaginary part. This means that the force and velocity have p=2 (90 ) phase difference. If kL continues to become larger and approaches p, then the impedance tends to minus infinity. In this case, we have zero velocity or, in other words, anti-resonance. To summarize, the wave propagation is dominated by kL; that is, the ratio of the length to the wavelength. We can readily anticipate that the driving point impedance for a more general case, such as a semi-infinite or finite length string that is completely fixed at one end, would be somewhere between these two extreme cases. The impedance will generally have both real and imaginary parts. The most important observation is that the driving point impedance, as its name implies, represents how much force is required to obtain unit velocity, or how much velocity will be generated by a unit force at the point of interest. However, it is dependent upon the characteristics of the field or medium. For example, Equation 1.70 depends on kL and certainly expresses how the wave propagates from the point to the field. Section 1.9.3 describes the driving point impedance of rod, membrane, beam, and plate. Table 1.1 summarizes these driving point impedances. To summarize, we have studied the wave propagation along two distinctly different strings, a semi-infinite string and a finite string, using driving point impedance. The main point of this section is the realization that the point impedance possesses information on how the waves propagate along the string. This introduces the idea of the possible role of impedance in understanding acoustic wave propagation. Finding solutions which satisfy the governing acoustic wave equation and boundary condition is therefore not the only approach to understanding acoustic waves. It should also be clear at this point that wave reflection is not likely to occur if the driving point impedance has the same value as the characteristic impedance of the string.
22
Sound Propagation
Table 1.1
Driving point impedances Infinite
String Rod (longitudinal wave)
Finite
Zm0 ¼ rL cs pffiffi Zm0 ¼ YSrL
Bar or beam ZM m0 ¼ (flexural wave)
Zm0 ¼ jrL cs cot kL pffiffi Zm0 ¼ j YSrL cot kL
ð1jÞ YSw2 2 vb
ð1 þ jÞ YSw2 o2 2 v3b pffiffi ¼ rA cm ð¼ Tm rA Þ
ZFm0 ¼ Membrane Plate
Zm0
4 ZFm0 ¼ pffi ðrd 2 Þcp 3
YSw2 o sinh kb L vb cos kb L
ZFm0 ¼ j
YSw2 o2 sin kb L þ cosh kb L cos kb L þ sinh kb L v3b
sffiffi ¼ 4d 2
ZM m0 ¼ j
Yr 3ð1l2p Þ
!
Nomenclature: rL : mass per unit length of string, rod; rA : mass per unit area of membrane; r: mass per qffi pffi volume of plate; lp : Poisson’s ratio; cs : speed of propagation of string; cb ¼ Y=r; cp ¼ Y=rð1l2p Þ; o: angular frequency; k: wavenumber; L: length of string, rod, and bar; Y: Young’s modulus; S: crosssectional area of rod and beam; w: radius of gyration of beam and plate; d: thickness of plate; pffiffi Tm : membrane tension (N/m); vb : propagation speed of bar (¼ ocb w, depending on frequency); pffiffi vp : propagation speed of plate (¼ ocp w, depending on frequency); ZM m0 : driving point impedance by bending moment of beam; ZFm0 : driving point impedance by shear force of beam and plate.
In other cases, reflected waves will be obtained12 regardless of whether there is an imaginary part or real part, as long as the driving point impedance has the same value as the characteristic impedance. For a finite string case, kL (which describes the ratio of the length of the string to the wavelength of interest) governs the waves in the string. We can envisage that the driving point impedance for a more general case will have a value between those which we obtained from two extreme cases, that is, it will have a real and imaginary part.
1.7
Wave Energy Propagation along a String
We have seen that the driving point impedance and characteristic impedance are useful measures or physical concepts to understand the waves on a string. We are now ready to explore other physical variables associated with waves along a string. In the following, we study the waves on a string in terms of energy. First, let’s determine how much energy can be stored in an infinitesimal element of string as depicted by Figure 1.15. If the energy that is dissipated by the motion of the element is small compared to other energies, then we can postulate that the string mainly has kinetic and potential energy (EK and EP, respectively). 12 Note that even if we have a driving point impedance that only has a real part, it does not mean that we do not have a reflected wave. When we have impedance mismatch, we generally have reflected waves. The type of impedance mismatch determines the magnitude and phase difference.
23
Vibration and Waves
y
at f = Δt ds dy
0
Figure 1.15
x
at t = 0
dx
The change of an infinitesimal element of a string in infinitesimal time
The kinetic and potential energy in the infinitesimal element of the string (dEK and dEP, where d expresses a small element) can be written 2 1 @y dEK ¼ rL dx ; ð1:73Þ 2 @t dEP ¼ TL
qffiffiffi ðdxÞ2 þ ðdyÞ2 dx :
2 1 @y ffi TL dx 2 @x
ð1:74Þ
Note that we neglected the potential energy due to the elevation of the string in the y direction. This is possible because it can be assumed to be much smaller than the potential energy due to the elongation of the string, in the case of a small motion. Using Equation 1.73 the total energy of the string can be written ( 2 ) 1 @y 2 @y dE ¼ rL þ TL dx: 2 @t @x
ð1:75Þ
If we introduce the concept of energy density as e¼
dE ; dx
ð1:76Þ
then the total energy in the string can be written as ð
1 E ¼ edx ¼ rL c2s 2
ð (
@y @x
2
1 @y þ cs @t
2 ) dx:
ð1:77Þ
Equation 1.77 can be easily obtained from Equations 1.75 and 1.76 and by considering the relation between the tension TL, mass per unit length rL , and the propagation speed of a wave on
24
Sound Propagation
the string cs . Equation 1.77 demonstrates that the greater the slope along the string (with regard to x) and the faster the speed of wave propagation, the more energy we have. We then determine how this energy propagates along a string and what the speed of energy propagation is. We have already found that the wave along a string propagates at speed cs, that is, the phase speed. Consider that we raise one end of the string (see Figure 1.16). We can now study how energy and the wave propagate.13 This postulation is quite rational because we are doing work on the string which will be converted to kinetic and potential energy according to the principle of work done and energy. TL
TL
uy cE
uyτ
TL
cEτ
Figure 1.16 Energy propagates along a string by raising one end (TL is tension along the string, cE energy propagation speed, t time, and uy lifting velocity at the end)
Assume that we vertically raise the end of the string with a speed of uy. The work done is u u TL cEy uy t; where TL cEy is the force in the y direction and uy t is the distance across which the force is applied. We postulate that the energy propagates with a speed of cE. The kinetic energy can be approximated as 12 rL ðcE tÞu2y . We make the reasonable assumptions that uy is small compared to cE when we calculate the length of the string that is raised and that the raised string moves with an average speed of uy. 2 u The potential energy is 12 TL cEy cE t; this can be readily obtained by the work done due to the elongation of string. Alternatively, we can obtain these results by simply using Equation 1.75. These lead to the equation: 2 uy 1 1 uy TL uy t ¼ rL ðcE tÞu2y þ TL cE t; ð1:78Þ cE 2 2 cE which gives us c2E ¼
TL : rL
ð1:79Þ
We can therefore conclude that the speed of energy propagation is identical to the phase velocity of a string.14
13
Morse, P.M. and Ingard, U. (1953) Theoretical Acoustics, Princeton, New Jersey, pp. 102. In many cases, the energy propagates with the speed of phase velocity as for the string. We refer to this type of wave as non-dispersive and the corresponding medium is a non-dispersive medium. If the phase and energy velocity of a wave are not equal, then we often regard it as dispersive wave. Typical dispersive waves are bending and water waves. 14
Vibration and Waves
1.8
25
Chapter Summary
Wehavestudiedwavepropagationalongapieceofstring,whichisatypicalone-dimensionalwave.By simply extending what we have learned from string wave propagation, we will be able to understand moregeneral three-dimensional acousticwaves.Webeganwiththevibrationofastring,whichisquite familiar to many people, and demonstrated that this is a special form of wave propagation. We extended this observation to introduce a general mathematical form of wave propagation. Harmonic waves were selected in particular to look at the specific relation between wave number and frequency. From this relation, we envisaged how the space in which we are interested is linked to time. We know that a wave is an expression of a space–time relation. We also found that this general mathematical form satisfies the governing equation, and is therefore a solution of the equation. A harmonic wave solution gives us the dispersion relation, which determines the relation between wave number and frequency and is determined by the characteristics of the medium. We subsequently examined how waves are transmitted and reflected when they meet a discontinuity (created by employing two strings of different thicknesses). We learned that the waysinwhichwavesarereflectedandtransmittedarecompletelydeterminedbythecharacteristic impedances of two strings, which create an impedance mismatch between the strings. The driving point impedance is another measure that represents how the waves on a string propagate. It is understood that while the driving point impedance is measured at one point, it expresses what is happening in the wave field. This highlights the concept of the driving point impedance. From a mathematical point of view, it is sufficient to find solutions that satisfy the governing wave equation and the boundary and initial conditions. We can then examine what the solutions physically imply. This chapter acknowledges the rationale provided by the mathematical procedure while also attempting to elucidate the fundamentals associated with wave propagation, that is, the basic physical parameters. We have used the wave along a string as an example. This is a special case of one-dimensional waves, but one that can represent every one-dimensional wave. Examples include the waves in a bar and bending waves, as described in Section 1.9.3. We also see that the waves in a membrane can be considered as an extended one-dimensional case, as described in Section 1.9.3.
1.9 1.9.1
Essentials of Vibration and Waves Single- and Two-degree of Freedom Vibration Systems
We first study a single-degree-of-freedom vibration, which is the simplest form of vibration. Understanding the simplest case creates a firm foundation for understanding more complicated vibration. Its importance is confirmed if we consider that multi-degree-of-freedom system vibrationisoftenunderstoodasthesuperpositionofasingle-degree-of-freedomvibrationsystem. Figure 1.17 is a symbolic illustration of single-degree-of-freedom vibration.15 Let’s examine the relations between the displacement of mass (xðtÞ) and other physical values.16 15 It is important to represent a physical phenomenon with figures and symbols. Symbolic representations, in this regard, must have the ability to encapsulate physical features as much as possible, while those that can distort the understanding of physical phenomena must be excluded. The process of symbolization requires meticulous attention. 16 The process of finding a rule that dominates the relations among physical values of interest is diverse (deductive, inductive, experimental, and theoretical methods) and has a long history.
26
Sound Propagation
Figure 1.17 Single-degree-of-freedom (SDOF) vibration system
First, the positive direction of displacement is set by a coordinate system in Figure 1.17. The displacement of the vibration system is measured by using this coordinate. In many cases, it is helpful to consider a number of physically plausible situations in order to analyze specific behavior of the selected physical system. When the mass m is moving in the selected coordinate system (Figure 1.17), the linear spring s and the viscous damper rd are likely to apply their force toward the opposite direction of the mass motion, that is, the negative direction of the coordinate system (Figure 1.18). When Newton’s law of motion is applied, the second-order inhomogeneous differential equations with constant coefficients are m€ x þ rd x_ þ sx ¼ f ;
ð1:80Þ
where x ¼ xðtÞ, f ¼ f ðtÞ, and t is time. The solution of Equation 1.80 can be found in several ways.17
Figure 1.18
Illustrative adaptation of free-body diagram to single-degree-of-freedom vibration system
If the reader is interested in the steady-state response, Fourier transform is a method of finding the response of the vibration system of Equation 1.80 (see Section 1.9.2). Fourier transform allows us to observe the response of the vibration system with regard to selected frequencies of interest. Using this, Equation 1.80 can be expressed as ðo2 mjord þ sÞXðjoÞ ¼ FðjoÞ:
ð1:81Þ
17 Time and frequency domain analysis can be considered as representative examples. The choice of method depends on the given physical circumstances. For instance, it may be appropriate to use a Laplace transform when the given physical circumstance or condition is either initial velocity or displacement.
27
Vibration and Waves
pffi Ð1 Ð1 Here, XðjoÞ ¼ 1 x ðtÞejot dt, FðjoÞ ¼ 1 f ðtÞejot dt, and j denotes 1. X and F represent the Fourier transform of the displacement xðtÞ and external force, respectively. XðjoÞ is a response that occurs when exciting the vibration system with a selected frequency o, and this means that the displacement of the vibration system is affected by the amount specified by parentheses in Equation 1.81. Because X and F are complex numbers, they can be represented in the complex plane as shown in Figure 1.19. jXj expresses the magnitude of the displacement ðXÞ and y represents the phase difference between RefXg and ImfXg: X ¼ jXje jy :
Im X 0
φ
F
Figure 1.19
Re: Real axis Im: Imaginary axis
X
θ
ð1:82Þ
Re F
Complex displacement X and complex excitation F expressed on a complex plane
Figure 1.19 indicates that both the complex displacement and force have their own magnitudes and phase differences. Equation 1.81 indicates that the complex value of right and left sides must be identical, and this is expressed on a complex plane in Figure 1.20. To understand the physical meaning of the complex displacement x, it is assumed that the displacement x(t) has a frequency confined to a certain narrow frequency bandwidth, Do. In this case, Equation 1.86 can be written as xðtÞ ¼
1 Xð joÞejot Do: 2p
ð1:83Þ
Figure 1.20 Equilibrium relationships within vibration system expressed on a complex plane (F and X represent complex force and displacement, respectively, while m; rd ; s represent mass, viscous damping coefficient, and linear spring constant, respectively)
28
Sound Propagation
Equation 1.83 can be expressed in a form so that it is easier to understand its physical aspects on the basis of the relationship of o ¼ 2pf as xðtÞ ¼ Xð j2pf Þej2pft Df :
ð1:84Þ
Here, Xðj2pf Þ is called the spectral density. The right-hand side of Equation 1.84 comprises the complex displacement and values that change periodically with respect to time, and can be expressed as xðtÞ ¼ XS ejot ;
ð1:85Þ
where XS denotes the spectrum or the displacement within the width of frequency. Here, the subscript s represents a steady state. In Figure 1.20, x is equivalent to XS in its steady state. Meanwhile, we can return to x(t) from an inverse Fourier transform of XðjoÞ: ð 1 1 xðtÞ ¼ Xð joÞejot do: ð1:86Þ 2p 1 The following relationship between Fourier transforms of displacement and velocity can be obtained as ð dx 1 1 uðtÞ ¼ ¼ joXð joÞejot do; ð1:87Þ dt 2p 1 Ð1
Uð joÞ ¼ joXð joÞ;
ð1:88Þ
where UðjoÞ ¼ 1 u ðtÞ e dt. The relationship between velocity and force can be found by substituting Equation 1.88 into Equation 1.81: s Uð joÞ ¼ Fð joÞ: ð1:89Þ jom þ rd jo jot
The relationship between velocity and force is expressed on a complex plane in Figure 1.21 in the same manner as Equation 1.81 is illustrated in Figure 1.20.
Figure 1.21
Relationship between velocity and force of vibration system on a complex plane
29
Vibration and Waves
As we defined the mechanical impedance in Section 1.4, mechanical impedance (Zm ) of the single-degree-of-freedom system is expressed as Zm ð joÞ ¼
Fð joÞ s ¼ rd þ j om þ : Uð joÞ o
ð1:90Þ
Equation 1.90 can be rewritten as Zm ð joÞ ¼ rd j
m 2 ðo o2n Þ; o
ð1:91Þ
pffiffi where on ¼ ð s=mÞ is the natural frequency of a single-degree-of-freedom system. Now let us consider three limit cases: (i) o ¼ on , (ii) o on , and (iii) o on . Firstly, in the case of o ¼ on , mechanical impedance becomes a real value equivalent to the viscous damping coefficient rd, that is, Zm ð joÞ ¼ rd :
ð1:92Þ
Therefore, the magnitude and phase are obtained as jZm ð joÞj ¼ rd ÊZm ð joÞ ¼ 0
:
ð1:93Þ
Secondly, when o on , the mechanical impedance is Zm ð joÞ rd jom:
ð1:94Þ
The magnitude and phase are then obtained as qffiffiffi r2d þ ðomÞ2 : om ÊZm ð joÞ ¼ tan1 rd jZm ð joÞj ¼
ð1:95Þ
In other words, when the frequency becomes larger, (i) magnitude tends to increase linearly and is dominated by mass m and (ii) phase approaches p=2. Thirdly, when o on , the mechanical impedance is s ð1:96Þ Zm ð joÞ rd þ j : o The magnitude and phase are obtained as
rffiffiffi
s 2 r2d þ o : s 1 ÊZm ð joÞ ¼ tan rd o jZm ð joÞj ¼
ð1:97Þ
30
Sound Propagation
In this case, magnitude tends to decrease and is dominated by spring constant s. Phase is converging to p=2 when the frequency becomes smaller (o on ). Figure 1.22 illustrates the magnitude and phase of the mechanical impedance (Zm ¼ F=U) with respect to frequency. It is interesting to note that the phase difference between the force and velocity is p=2f, where f is the phase difference of the force to displacement. When there is no phase difference, in other words, the force is completely used to drive velocity (Figures 1.20 and 1.21) and the magnitude is at a maximum (Figure 1.22). When the excitation frequency becomes larger, then its phase difference approaches p=2; see Figure 1.21. It is therefore quite useful to use Figures 1.21 and 1.22 together to explore the associated physics of the complex mechanical impedance. Zm ( jω)
rd Zm ( jω) π 2 –π 2
ω n (= √s/m)
ω (2π f )
ω (2π f )
Figure 1.22 Magnitude and phase of complex impedance (Z ¼ F=U) with respect to angular frequency (o)
To study more general cases, let us consider a two-degrees-of-freedom (2-DOF) system (Figure 1.23) and obtain its driving point impedance when a force is applied at each mass (m1 and m2). In Figure 1.23, si is the linear spring constant of the ith spring and rdi is the viscous
Figure 1.23 Two-degree-of-freedom (2DOF) vibration system. m1 and m2 indicate each vibrating mass, s1 and s2 are linear spring components, rd and rd2 represent viscous damping coefficients, x1 ðtÞ is displacement of mass 1, x2 ðtÞ is displacement of mass 2, and f1 ðtÞ and f2 ðtÞ are forces acting on the masses
31
Vibration and Waves
damping coefficient of ith damper. To obtain the equation of motion, we graphically express Newton’s second law for each mass (see Figure 1.24).
Figure 1.24 Graphical expression of Newton’s second law for each mass (m1 and m2 ) when only the force f1 ðtÞ excites mass 1 (m1 ). s1 and s2 are linear spring component, rd and rd2 represent each viscous damping coefficient, x_ 1 and x_ 2 are the rate change of each displacement (x1 and x2 ) with respect to time, that is, velocity
Firstly, suppose that there is only a driving force (f1 ðtÞ ¼ F1 ejot ) acting at m1. Then the force acting on each object can be depicted as Figure 1.24. The equation of motion can be easily written as f1 ðrd2 þ rd1 Þx_ 1 þ rd2 x_ 2 ðs1 þ s2 Þx1 þ s2 x2 ¼ m1 x €1 þ rd2 x_ 1 rd2 x_ 2 þ s2 x1 s2 x2 ¼ m2 x €2
;
ð1:98Þ
Where x(t) is the displacement with respect to time. We can rewrite the equation of motion in a matrix form as ! " # ! " # ! " # ! x €1 rd1 þ rd2 rd2 x_ 1 s1 þ s2 s2 x1 f1 m1 0 : þ þ ¼ x €2 rd2 rd2 s2 s2 x2 0 x_ 2 0 m2 ð1:99Þ If we assume that force is harmonic in time, then we can attempt the solution x1 ðtÞ x2 ðtÞ
! ¼
X1 X2
! ejot :
ð1:100Þ
32
Sound Propagation
Equations 1.99 and 1.100 lead us to (
" o
2
m1
0
0
m2
#
" jo
rd1 þ rd2
rd2
rd2
rd2
#
" þ
s1 þ s2
s2
s2
s2
#)
X1
!
X2
F1
¼
0
! : ð1:101Þ
This equation can be rewritten as "
m1 o2 joðrd1 þ rd2 Þ þ ðs1 þ s2 Þ
jord2 s2
jord2 s2
m2 o2 jord2 þ s2
#
X1 X2
! ¼
F1 0
! :
ð1:102Þ
We can obtain the displacement of each mass by matrix inversion, that is " # ! ! F1 X1 m2 o2 jord2 þ s2 jord2 þ s2 1 ; ¼ det A X2 0 jord2 þ s2 m1 o2 joðrd1 þ rd2 Þ þ ðs1 þ s2 Þ ð1:103Þ where A is the 2 2 matrix of Equation 1.102. By multiplying jo by displacementX1 , the velocity of m1 can be derived from Equation 1.103 as U1 ¼ jo
ðm2 o2 jord2 þ s2 Þ F1 : det A
ð1:104Þ
We can also find a driving point impedance at mass m1, that is Zm;1 ¼
F1 det A : ¼ U1 ½o2 rd2 þ joðm2 o2 s2 Þ
ð1:105Þ
This can be rewritten as Zm;1
¼
fm1 m2 o4 þ ðm1 s2 ðs1 þ s2 Þm2 rd1 rd2 Þo2 þ s1 s2 g ½o2 rd2 þ joðm2 o2 s2 Þ fðrd m1 þ ðrd1 þ rd2 Þm2 Þo3 ðs1 rd þ s2 rd1 Þog : þj ½o2 rd2 þ joðm2 o2 s2 Þ
ð1:106Þ
If there is no damping (rd1 ¼ rd2 ¼ 0), then the driving point impedance can be expressed as s2 2 4 2 2 2 m o o þ o þ þ ðo o Þ o 1 n;1 n;2 n;2 n;1 m1 Zm;1 ; ð1:107Þ ¼ j rd1 ¼rd2 ¼0 oðo2 o2n;2 Þ where on;1 and on;2 are the natural frequencies of m1 and m2, respectively. From Equation 1.107, the driving point impedance of mass 1 (Zm;1 jrd1 ¼rd2 ¼0 ) approaches infinity as frequency approaches the natural frequency of mass 2 (on;2 ). This means that mass 1(m1 ) does not move, although force is applied to mass 1. The other limit case occurs when the frequency is
33
Vibration and Waves
zero (in a static condition). Equation 1.107 tells us that the impedance also becomes infinite. The driving point impedance increases with frequency because the order of angular frequency in the numerator is higher than that of the denominator. We now investigate the case where the driving force (f2 ðtÞ ¼ F2 ejot ) is acting at m2. The equation of motion can be written as " # ! " # ! " # ! ! m1 0 x €1 rd1 þ rd2 rd2 x_ 1 s1 þ s2 s2 x1 0 þ þ ¼ : x €2 rd2 rd2 s2 s2 x2 x_ 2 0 m2 f2 ð1:108Þ Similarly, we obtain the velocity of m2 as U2 ¼ jo
ðm1 o2 joðrd1 þ rd2 Þ þ s1 þ s2 Þ F2 : det A
ð1:109Þ
The driving point impedance of m2 can be written as Zm;2 ¼
F2 det A : ¼ U2 ½o2 ðrd1 þ rd2 Þjoðm1 o2 þ s1 þ s2 Þ
ð1:110Þ
Equation 1.110 can be also written as Zm;2 ¼
fm1 m2 o4 þ ðm1 s2 ðs1 þ s2 Þm2 rd1 rd2 Þo2 þ s1 s2 g ½o2 ðrd1 þ rd2 Þjoðm1 o2 þ s1 þ s2 Þ fðrd m1 þ ðrd1 þ rd2 Þm2 Þo3 ðs1 rd þ s2 rd1 Þog þj : ½o2 ðrd1 þ rd2 Þjoðm1 o2 þ s1 þ s2 Þ
ð1:111Þ
As we discussed earlier, if we have no damping then the driving point impedance of mass 2 is s2 2 4 2 2 2 m o o þ o þ þ ðo o Þ o 1 n;1 n;2 n;2 n;1 m1 ¼ j : ð1:112Þ Zm;2 m1 2 rd1 ¼rd2 ¼0 o ðo o2n;1 Þo2n;2 m2 Note that the driving point impedance becomes Zm;2 ðo ¼ on;1 Þ
rd1 ¼rd2 ¼0
on;1 s2 ¼ j 2 ¼ jm2 on;2
rffiffi s1 m1
ð1:113Þ
when the excitation frequency reaches the natural frequency of m1 , on;1 . It reaches its maximum value, as expected. The driving point impedance also increases with frequency. However, the increasing rate is lower than the previous case. Note also that the driving point impedance at m2 has maximum value but is finite. It is strongly related to the natural frequency of m1 when the force is exciting the system with the natural frequency of mass 1 (o1 ).
34
1.9.2
Sound Propagation
Fourier Series and Fourier Integral
A Fourier series is defined as xðtÞ ¼ b0 þ
N X
ðan sinon t þ bn coson tÞ;
ð1:114Þ
n¼1
where N represents integers including infinity. When a Fourier series is expressed in terms of frequency on ¼ no1 and, when each frequency on is distributed at even frequency separation against the basic frequencyðo1 Þ, Equation 1.114 can be written as xðtÞ ¼ b0 þ
N X
ðan sin no1 t þ bn cos no1 tÞ:
ð1:115Þ
n¼1
This equation indicates that a certain x(t) or a time signal can always be expressed using sine and cosine functions (see Figure 1.25). Note here that the sine and cosine functions used in the Fourier series have an especially periodic characteristic and convenient orthogonality (or orthogonal property). Using this characteristic, an and bn the contribution of sine and cosine of a certain frequency on ðno1 Þ, can be calculated as ð 2 T an ¼ xðtÞsin on t dt; ð1:116Þ T 0
Fourier transform
X (ω)
x(t ) t
ω x(t )
X (ω) a1
a1 sin ω1t
t
x(t )
+
X (ω)
a 2 sin 2ω1t
ω1
+
ω
a2
t x(t )
+
X (ω)
a3 sin3ω1t t
ω
2ω1
+
a3 3ω1
Figure 1.25
Meaning of Fourier series
ω
35
Vibration and Waves
2 bn ¼ T
ðT xðtÞ cos on t dt:
ð1:117Þ
0
Here, T is the period and is equal to 2p=on . This means that each frequency o1 ; o2 ; . . . ; on has an orthogonal coordinate and the level of contribution for each of those coordinates is measured (see Figure 1.25). To express Equation 1.114 conveniently using complex functions, we can write Equation 1.114 as xðtÞ ¼
N X ðjan sin on t þ bn cos on tÞ n¼0
N X ðjan sin no1 t þ bn cos no1 tÞ: ¼
ð1:118Þ
n¼0
The following equation can be obtained using Euler’s relation between trigonometric function and exponential function, which is xðtÞ ¼
N X
Cn e jno1 t :
ð1:119Þ
n¼0
Here, Cn is a complex number and is related to an ; bn by RefCn g ¼ bn ;
ð1:120Þ
ImfCn g ¼ an :
ð1:121Þ
The Fourier transform is defined as xðtÞ ¼
1 2p
ð1
Xð joÞ ejot do;
1
Xð joÞ ¼
ð1 1
xðtÞe jot dt:
ð1:122Þ
ð1:123Þ
XðjoÞdo represents the property of xðtÞ that has frequency o. In other words, it denotes the spectrum of frequency o. In general, XðjoÞ is called a spectral density function. That is, it represents the displacement per unit frequency.18 Equations 1.119 and 1.122 have in fact a direct relationship. When an infinite sum (instead of finite N) is applied in Equation 1.119 and when Cn is the displacement spectrum of each frequency, or the amplitude of the displacement of each frequency, the two equations are equivalent.
18
In many cases, a spectral density function refers to the energy per unit frequency. In this case, the correlation functions related to the square of the displacement become a Fourier transform pair, unlike Equations 1.91 and 1.92.
36
1.9.3
Sound Propagation
Wave Phenomena of Bar, Beam, Membrane, and Plate
1.9.3.1 Driving Point Impedance of Thin Rods (Longitudinal Waves) The propagation of one-dimensional waves was examined by investigating the waves of strings (ropes and cords). Wave motions in bars and beams are also examples of one-dimensional waves that have quite similar propagation characteristics but also interesting differences to the waves of strings. We examine the characteristics of the two wave motions and, in particular, how the characteristic impedance of bars and beams is different from that of strings and the effect of these differences on wave propagation. Consider a situation where a bar (Figure 1.26) vibrates to the longitudinal direction due to an external force. The coordinate of the length direction is denoted x and the displacement at x is denoted x. We can then assume that a small element dx induces displacement change, as shown in the figure. The strain e can be expressed as the rate of change against the unit distance of displacement as e¼
@x : @x
ξ+
ξ
∂ξ dx ∂x
dx
x
Figure 1.26
ð1:124Þ
Bar displacement, its rate of change, and its coordinate system
This is known as the strain–displacement relation. Because of the force applied to the right and left sides, the small element is accelerated as shown in Figure 1.27. Newton’s law allows us to write the relation between force and acceleration as
@f @2x ¼ rL 2 : @x @t
ð1:125Þ
ρ L dx f+
f
dx
Figure 1.27
∂ 2ξ ∂t 2
∂f dx = ∂x dx
Relationship between force and motion applied to small sections of a bar
With two equations (Equations 1.124 and 1.125), there are three unknowns (x, e and f ). Another equation is therefore required to completely characterize the motion of the bar. This is
37
Vibration and Waves
the constitutive relation that represents the relation between the rate of change of a medium and force, that is f @x ¼ Ye ¼ Y : ð1:126Þ S @x Here, S denotes cross-sections of a bar and Y represents Young’s modulus. When this equation is differentiated with respect to x, we obtain @f @2x ¼ YS 2 : @x @x
ð1:127Þ
When this is substituted into Equation 1.125, we arrive at
cb ¼
@2x 1 @2x ¼ 2 2; 2 @x cb @t
ð1:128Þ
pffiffiffi pffi YS=rL ¼ Y=r:
ð1:129Þ
Here, r is the mass per unit volume and rL is the mass per unit length. This equation is the same as the wave motion equation for string. It is noticeable that only the propagation velocity of the pffiffi string is different, which is cs ¼ TL =rL . Physical phenomena such as reflection and damping by the boundary condition will therefore occur in the same manner as in the case of strings. The behavior of longitudinal waves is dependent on boundary conditions. Reflection and damping by the boundary condition will occur in the same manner as in the case of string F at x ¼ 0 as depicted in Figure 1.28. Because the rod has infinite length, the solution is only composed of a right-going wave. We can predict that the solution would be expressed similarly with the case of string, because the driving point impedance is defined as Equation 1.131. The solution can therefore be written as xðx; tÞ ¼ AejðotkxÞ ;
Figure 1.28
ð1:130Þ
A semi-infinite rod under longitudinal excitation force fð0; tÞ ¼ Fejot at x ¼ 0
and the driving point impedance is defined as Zm;0 ¼
f ð0; tÞ : uð0; tÞ
ð1:131Þ
The velocity can be obtained by taking the time derivative of the displacement as uðx; tÞ ¼
@xðx; tÞ ¼ joAejðotkxÞ : @t
ð1:132Þ
38
Sound Propagation
Therefore, the velocity at the driving point is @xðx; tÞ uð0; tÞ ¼ @t
¼ joAejot ;
ð1:133Þ
x¼0
and the force at the driving point is f ð0; tÞ ¼ Fejot :
ð1:134Þ
Finally, the driving point impedance can be written as Zm;0 ¼
f ð0; tÞ F ¼ : uð0; tÞ joA
ð1:135Þ
This equation highlights that we need to know the relation between magnitude of excitation force F and magnitude of displacement A to obtain the driving point impedance. This relation can be explained by the constitutive equation between force and displacement. In the case of longitudinal waves, the relation between stress (s ¼ F=S, S is a cross-section area of the rod) and strain (e ¼ @x=@x) can be written as s ¼ Ye ¼ Y
@x : @x
ð1:136Þ
The minus sign means strain by compression. Equation 1.136 can be rewritten as f ðx; tÞ @xðx; tÞ ¼ Y : S @x
ð1:137Þ
@xðx; tÞ ¼ jkAejðotkxÞ ; @x
ð1:138Þ
The strain can also be rewritten as
and the complex magnitude of force at the driving point (x ¼ 0) is then obtained as F ¼ jkYSAejot :
ð1:139Þ
Finally, the driving point impedance of the semi-length of rod is Zm;0 ¼
pffiffiffi jkYSA YS YS ¼ ¼ YSrL : ¼ pffi joA cb YS=rL
ð1:140Þ
The driving point impedance of the semi-infinite rod has only a real value, thus the force and velocity have the same phase. We next consider the case where one end of the rod is fixed, as depicted in Figure 1.29. In this case, the solution is composed of not only a right-going wave but also a reflected wave at the fixed boundary. Therefore, the solution can be written as xðx; tÞ ¼ AejðotkxÞ þ Bejðot þ kxÞ :
ð1:141Þ
39
Vibration and Waves
Figure 1.29 A rod of length L clamped at the right end and excited by longitudinal excitation force fð0; tÞ ¼ Fejot at x ¼ 0
Because there are two unknowns we need one more condition to obtain the solution, which is the condition that one end is fixed. It can be written as xðL; tÞ ¼ 0:
ð1:142Þ
Therefore, the solution should satisfy the following equality, xðL; tÞ ¼ ðAe jkL þ BejkL Þejot ¼ 0:
ð1:143Þ
From Equation 1.143, the relation between A and B is B ¼ Ae j2kL :
ð1:144Þ
Using this relation, the displacement at x can be obtained as xðx; tÞ ¼ Aejot fe jkx ejkðx2LÞ g: The velocity and force at the driving point is defined
@xðx; tÞ ¼ joAejot 1e jk2L ; uð0; tÞ ¼ @t
ð1:145Þ
ð1:146Þ
x¼0
and the force is fð0; tÞ ¼ jkYSAejot f1 þ ejk2L g: The driving point impedance is found to be pffiffi Zm;0 ¼ j YSrL cot kL
ð1:147Þ
ð1:148Þ
The driving point impedance has only an imaginary part. This means that the rod is oscillating with a phase difference of 90 between the force and velocity at the driving point. This result is the same as for that of transverse vibration of a string, with the exception of the direction of vibration and wave speed. 1.9.3.2 Driving Point Impedance of Thin Bars (Flexural Wave) The governing equation of flexural waves can be found from the relationship between the force (shear force and moment) applied to small elements of a bar and displacement. Consider a small
40
Sound Propagation
element of a bar as shown in Figure 1.30. Suppose that we have a shear force (Fy ) at the left end of the element upward, and the associated shear force on the right end of the element acts downward. By taking moments about x (positive for counter-clockwise) of the element as shown in Figure 1.30, we have MðxÞMðx þ dxÞFy ðx þ dxÞ dx ¼ 0;
ð1:149Þ
Figure 1.30 A small element of a bar of length dx for transverse vibration. Bending moments and shear forces are present at the both ends of the element by the transverse displacements
since we can decide that the product of rotational moment of inertia and angular acceleration is small enough to be neglected. For small dx,Mðx þ dxÞ and Fy ðx þ dxÞ can be rewritten using Taylor’s expansion with respect to the moment and shear force at x. Furthermore, using the relation between moment and shear force19 under the assumption of small transverse displacement, we can write the shear force as 3 @M 2 @ y ¼ YSw Fy ¼ : ð1:150Þ @x @x3 Here, second-order terms in dx have been neglected and w is the radius of gyration, defined w2 ¼ I=S where ð ð1:151Þ I ¼ r2 dS: S
Net upward force dFy of the element can be given by 4 @Fy 2 @ y dx: dFy ¼ Fy ðxÞFy ðx þ dxÞ ¼ dx ¼ YSw @x4 @x
ð1:152Þ
We now obtain the force for the transverse vibration, and this force will yield the mass (rSdx) of the small element and acceleration (@ 2 y=@t2 ). This gives us the equation of motion for the transverse vibration of a bar as @4y 1 @2y ¼ ; @x4 c2b w2 @t2
19
Morse, P.M. and Ingard, U. (1953) Theoretical Acoustics, Princeton, New Jersey, pp. 102.
ð1:153Þ
41
Vibration and Waves
where cb ¼ is given by
pffiffiffi Y=r. The propagation velocity vb changes in accordance with the frequency and vb ¼
pffiffi ocb w:
ð1:154Þ
This indicates that the bending wave is dispersive. From a mathematical point of view, this happens because there is a fourth order of differentiation term, unlike wave equations for strings or bars. The wave number is proportional to the power of 1/2, that is rffi o o kb ¼ ¼ : ð1:155Þ vb cb w Note that this dispersion relation is different from the strings; it is proportional to frequency k ¼ o=cs . Figure 1.31 represents this dispersion relation. It is interesting that each frequency has a different propagation speed; the shape of the wave therefore changes with time.
Figure 1.31 Beam’s dispersion curve; wave motions with different frequencies have different propagation (vb ) and energy propagation speed (cg ). Energy propagation velocity is the gradient of a tangent of a dispersion curve at the corresponding frequency
In the case of beams, high-frequency waves propagate faster than low-frequency waves.20 Meanwhile, the propagation speed of wave motion energy ðcg Þ satisfies the relationship cg ¼ do=dkb . Intuitively, this describes a situation in which two sinusoidal waves with two very minor frequency differences are being propagated. It can be imagined that this wave motion will propagate with a wave packet that has a cycle based on the difference between the frequencies of the two wave motions. The propagation speed of this wave packet will be Do=Dkb , different to the phase propagation speed. It can be deduced that the correlation with the propagation speed of 20
Low-frequency water waves propagate faster than high-frequency waves, in direct contrast to that for beams. As such, it can be assumed that the more motions vary per unit length, the faster propagation velocity becomes. In the case of beams, high-frequency wave motions can deliver more motions (the energy stored on short flexural waves will be larger than the energy stored on long flexural waves).
42
Sound Propagation
energy can be satisfied when the frequency difference becomes extremely small. It should be mentioned again that when a number of waves propagate, it is reasonable to express the representative propagation speed of the entire wave packet as a physical value, that is, the velocity of the wave group. Energy propagates at the propagation speed of the group. We now obtain driving point impedances of a bar for flexural waves. The mechanical impedance (or characteristic impedance) for flexural waves can be defined in two ways with respect to excitation force, in terms of either moment or shear force. Those are bending moment angular velocity of section
ð1:156aÞ
shear force : transverse velocity of section
ð1:156bÞ
ZM ¼ and ZF ¼
The superscripts M and F represent moment and shear force, respectively. First, let us consider an infinite bar as shown in Figure 1.32. Suppose that there is a transversal displacement (y) which is generated by a bending moment and propagates in the x direction. A solution of the governing equation (Equation 1.153) can be attempted in the form yðx; tÞ ¼ fAe jkb x þ Bejkb x þ Cekb x þ Dekb x gejot :
ð1:157Þ
Figure 1.32 Transverse vibration of an infinite bar under a moment at x ¼ 0; y is the direction of displacement
If we assume that the bar is semi-infinite (x 0) and there is harmonic excitation of bending moment at x ¼ 0, then the solution can be expressed in terms of a propagating wave and evanescent wave in the positive x direction: yðx; tÞ ¼ fAe jkb x þ Cekb x gejot :
ð1:158Þ
In this case, the bending moment in a beam is given as M ¼ YSw2
@2y : @x2
ð1:159Þ
The second derivative of the transversal displacement (y) is expressed as
@2y ¼ kb2 Ae jkb x þ Ce kb x ejot : 2 @x
ð1:160Þ
43
Vibration and Waves
The bending moment at x ¼ 0 (driving point) is given as mð0; tÞ ¼ YSw2 kb2 fA þ Cgejot :
ð1:161Þ
_ is the rate of change of angular displacement (y) with In addition, the angular velocity (y) respect to time (see Figure 1.33). Since the transverse displacement is assumed to be small, the angle between arc length (l) and dx becomes y. Therefore, we can express it as y
@y ; @x
ð1:162Þ
Figure 1.33 The relationship between angular displacement (y) and transverse displacement (y) for a small vibration at an arbitrary cross-section of a beam; l is arc length corresponding to the angle y
because tan y y when y is small. Therefore, the angular velocity can be expressed as _ tÞ ¼ jo @y : yðx; @x
ð1:163Þ
_ tÞ ¼ jokb f jAe jkb x þ Ce kb x gejot ; yðx;
ð1:164Þ
This gives us
and the driving point impedance of moment is finally obtained as M ¼ j Zm;0
YSw2 kb A þ C : jA þ C o
ð1:165Þ
44
Sound Propagation
Because the bending moment acts at the driving point (x ¼ 0), there is no shear force at the left end of the bar. We can therefore obtain the relation between A and C. The shear force is related to the third derivative of the displacement y with respect to x, that is, @M @3y ¼ YSw2 3 : Fs ¼ ð1:166Þ @x @x The boundary condition (no shear force) leads us to the relation between A and C: @ 3 y ¼0 ð1:167Þ @x3 x¼0
@ 3 y @x3
¼ kb3 fjA þ Cgejot ¼ 0
ð1:168Þ
x¼0
C ¼ jA:
ð1:169Þ
If we substitute Equation 1.169 into Equation 1.165, we obtain the driving point impedance as ZM m;0 ¼ j
YSw2 kb A þ C YSw2 kb 1 þ j ¼ j o o jA þ C 2j
YSw2 : ¼ ð1jÞ 2vb
ð1:170Þ
From Equation 1.170, the driving point impedance of an infinite bar has a complex value. The magnitude of the real part is equivalent to that of the imaginary part. Next, let us consider the finite beam which has the fixed end at x ¼ L as shown in Figure 1.34. The excitation point is assumed to be located at x ¼ 0. In this case, there might be left-going waves and reflected waves at x ¼ L in the solution: yðx; tÞ ¼ fAe jkb x þ Bejkb x þ Cekb x þ Dekb x gejot :
Figure 1.34
ð1:171Þ
Transverse vibration of a finite bar of length L under a moment at x ¼ 0
45
Vibration and Waves
The boundary conditions require that the slope of the displacement must be zero at x ¼ L and that the shear force at x ¼ 0 should be zero. There are mathematically expressed as @y ¼0 ð1:172Þ yjx¼L ¼ 0 and @x x¼L
@ 3 y @x3
¼ 0:
ð1:173Þ
x¼0
By substituting Equations 1.171 and 1.172 into Equation 1.173, we obtain the following relations: B¼A
and
C ¼ D:
ð1:174Þ
Using these relations, we derive the expression as yðx ¼ L; tÞ ¼ fAðe jkb L þ ejkb L Þ þ Cðe jkb L ekb L Þgejot ¼ 0:
ð1:175Þ
Equation 1.175 can be rewritten using Euler’s relation between trigonometric and exponential function as yðL; tÞ ¼ 2fA cos kb L þ C sinh kb Lgejot ¼ 0:
ð1:176Þ
We therefore obtain the relation between A and C as C¼
cos kb L A: sinh kb L
ð1:177Þ
Finally, using the relations Equations 1.174 and 1.175, the displacement of the finite beam which has the fixed end at x ¼ L is obtained as cos kb L yðx; tÞ ¼ 2A cos kb x þ sinh kb x ejot : ð1:178Þ sin h kb L The angular velocity at the driving point (x ¼ 0) can be obtained from Equation 1.163, and the moment at the driving point is given by Equation 1.159. These are _ tÞ ¼ 2jokb A cos kb L ejot yð0; sin h kb L
ð1:179Þ
mð0; tÞ ¼ 2k2 YSw2 Aejot :
ð1:180Þ
and
From Equations 1.179 and 1.180, the driving point impedance of the finite bar is M ¼ j Zm;0
YSw2 o sin h kb L : vb cos kb L
ð1:181Þ
46
Sound Propagation
The driving point impedance of a finite bar under bending moment is dependent on the frequency, and also has a purely imaginary value. In other words, it means the force and the velocity always have 90 phase difference. We now derive the driving point impedance for the case where we apply shear force at the left end. At first, as shown in Figure 1.35, let us consider the wave in the infinite beam that is excited by the shear force at x ¼ 0 (fs ð0; tÞ ¼ Fs ejot ). In this case, the governing equation and the solutions can be readily obtained as the case of the bending moment excitation (see Equation 1.158); the only difference is the boundary condition. The relation between the shear force and displacement is given as Fs ¼
@M @3y ¼ YSw2 3 ; @x @x
ð1:182Þ
Figure 1.35 Transverse vibration of an infinite bar under a shear force at x ¼ 0
and the third derivative of the displacement (y) with respect to x is obtained as
@3y ¼ kb3 jAe jkb x þ Cekb x ejot : 3 @x
ð1:183Þ
Finally, we derive the force at the driving point as fs ð0; tÞ ¼ YSw2 kb3 fjA þ Cgejot :
ð1:184Þ
The velocity is obtained by differentiating yðx; tÞ with respect to time as uðx; tÞ ¼ jofAe jkb x þ Cekb x gejot
ð1:185Þ
and the driving point impedance is found to be F Zm;0 ¼j
YSw2 kb3 jA þ C : o AþC
ð1:186Þ
To find the relation between A and C, we use the boundary condition that the moment is zero at the driving point (x ¼ 0): @ 2 y ¼ kb2 fA þ Cgejot ¼ 0: ð1:187Þ @x2 x¼0
47
Vibration and Waves
From Equation 1.187, we obtain the relation C ¼ A:
ð1:188Þ
Using this relation, Equation 1.186 can be rewritten as ZFm;0 ¼ j
YSw2 kb3 jA þ C YSw2 kb3 ð1jÞ ¼j o o AþC 2
ð1:189Þ YSw2 kb3 : ¼ ð1 þ jÞ 2o pffi pffi By using the relations k ¼ o=v ¼ o=cb w and cb ¼ Y=r, Equation 1.189 can be rewritten as F ¼ Zm;0
ð1 þ jÞ YSw2 o2 : 2 v3b
ð1:190Þ
In this case, the driving point impedance has a complex value; the value of the real part is equivalent to that of the imaginary part. Next, let us consider the finite bar which has a fixed end at x ¼ L. The shear force is excited at x ¼ 0. In this case, the solution must have the same form as Equation 1.171, which is yðx; tÞ ¼ fAe jkb ðxLÞ þ Bejkb ðxLÞ þ Ce kb ðxLÞ þ Dekb ðxLÞ gejot :
ð1:191Þ
The boundary conditions are given as the displacement and the slope are zero at x ¼ L and the bending moment is zero at x ¼ 0. They are, yjx¼L ¼ 0
ð1:192Þ
@y ¼ 0: @x x¼L
ð1:193Þ
By substituting Equation 1.193 into Equation 1.191, we obtain the results B¼A
and
C ¼ D:
ð1:194Þ
By substituting Equation 1.191 into Equation 1.192 and taking into account Equation 1.194, we have yjx¼L ¼ fA þ Cgejot ¼ 0:
ð1:195Þ
The relation between A and C can then be written as C ¼ A:
ð1:196Þ
Finally, using the relations Equations 1.194 and 1.196, the displacement of the finite beam which has the fixed end at x ¼ L is obtained as yðx; tÞ ¼ 2Afcos½kb ðxLÞsinh½kb ðxLÞgejot :
ð1:197Þ
48
Sound Propagation
The velocity at the driving point (x ¼ 0) is found to be uð0; tÞ ¼ j 2oAfcos kb L þ sinh kb Lgejot :
ð1:198Þ
The shear force at the driving point is given by Equation 1.182 so we can therefore write fs ð0; tÞ ¼ YSw2
@ 3 y jot e ¼ 2YSw2 kb3 A½sin kb L þ cosh kb Lejot : @x3
ð1:199Þ
From Equations 1.198 and 1.199, the driving point impedance of the finite bar is given as F ¼ j Zm;0
YSw2 o2 sin kb L þ cosh kb L : cos kb L þ sinh kb L v3b
ð1:200Þ
The driving point impedance of a finite bar under shear force is dependent on the frequency, and also has a purely imaginary value. In other words, it means the force and the velocity always have 90 phase difference in vibration. 1.9.3.3 Driving Point Impedance of Membranes Waves in membranes, just as those in strings, can be considered as a dynamic phenomenon between a force of restoration by tension and mass per unit area. It can be also seen as repetitive exchange between kinetic and elastic energy. The only difference is that while strings are onedimensional waves, membranes are two-dimensional. The governing equation of membranes can be derived in the same way as for string. For simplicity, a membrane can be assumed to be thin and stretched uniformly in all directions. In addition, let us also assume that the membrane vibrates transversely with small displacement. Suppose that we have a small element of a membrane as shown in Figure 1.36. The force acting on the element (on the area dx dy) is the sum of the transverse forces acting on the Tmdx (x,y+dy)
(x+dx,y+dy)
y dy
Tmdy 0
x
Tmdy
dx (x+dx,y)
(x,y) Tmdx
Figure 1.36
Forces acting on an element of a membrane and coordinate setup
49
Vibration and Waves
edges parallel to the x and y axes. The net vertical force arising from the pair of opposing tensions is @x @x @2x Tm dy ¼ Tm 2 dxdy; ð1:201Þ @x x þ dx @x x @x where Tm is the membrane tension in units of N/m. Furthermore, the tension Tm dx is obtained similarly as Tm dx ¼ Tm
@2x dxdy: @y2
ð1:202Þ
By Newton’s second law, the force induces the mass of membrane element (rA dxdy) to accelerate. The wave equation for membranes can then be derived as @2x @2x 1 @2x þ 2¼ 2 2: 2 @x @y cm @t
ð1:203Þ
Thispresults in differentiation terms for two directions (x, y) in space. Here, we have ffiffiffi cm ¼ Tm =rA , Tm is the membrane tension per unit length, and rA is the mass per unit area (kg/m2). The wave propagation velocity cm is expressed similarly with strings. Consider an infinitely large membrane in the xy plane. The membrane is assumed to be driven at the origin with harmonic force perpendicular to the surface of membrane with magnitude F (fð0; tÞ ¼ Fejot ), as shown in Figure 1.37. The possible waves that satisfy Equation 1.203 are only what propagate away from the origin. The solution will be similar to that of the infinite string: xðx; y; tÞ ¼ Ae jðkx þ kyÞ ejot :
ð1:204Þ
∞ ξ(x,y,t) y
∞
0
x
∞
f(0,0,t) = Fe –jωt
∞ Figure 1.37 Transverse vibration of an infinite membrane excited harmonically at x ¼ 0. x is transversal displacement with respect to the plane on which the membrane is located
50
Sound Propagation
The driving point impedance at its origin is fð0; 0; tÞ : uð0; 0; tÞ
Zm;0 ¼
ð1:205Þ
At an arbitrary point, the velocity can be written as uðx; y; tÞ ¼
@xðx; y; tÞ ¼ joAe jðkx þ kyÞ ejot : @t
ð1:206Þ
The velocity at the driving point is then @xðx; y; tÞ uð0; 0; tÞ ¼ ¼ joAejot ; @t x¼0
ð1:207Þ
and the force at the driving point is given as f ð0; 0; tÞ ¼ Fejot :
ð1:208Þ
Therefore, the driving point impedance is Zm;0 ¼
fð0; ; 0tÞ F ¼ : uð0; 0; tÞ joA
ð1:209Þ
The relation between the driving force magnitude F and the displacement magnitude A should be found in order to determine the driving point impedance. Recall that the force acting in the y direction of the string is f ðx; tÞ ¼ Tm
@y : @x
ð1:210Þ
Equation 1.210 can be extended to a membrane case. Figure 1.38 shows the transverse membrane displacement at the origin when the force is applied to the perpendicular direction with respect to the xy plane. Because the tension can be assumed to be constant, the displacement induced by a point force has a symmetric cone shape. Let Tm be the constant membrane tension for a cone-shaped membrane, dx the displacement change and dr the radius of the infinitesimal cone. The force acting at the origin is then 2p ð
f ð0; 0; tÞ ¼
0
Tm @x dy: 2p @r
ð1:211Þ
In Equation 1.211, Tm =2p is a tension that is acting in a radial direction (y) (Figure 1.38). Equation 1.211 can be rewritten as f ð0; 0; tÞ ¼ Tm
@x : @r
ð1:212Þ
51
Vibration and Waves
f (0,0,t) = Fe –jωt
y dξ θ
dr
x
Tm
Figure 1.38
Small element extracted form an infinite membrane around the driving point
For simplicity, let us express the displacement using polar coordinates. Equation 1.204 can then be rewritten as xðr; f; tÞ ¼ Ae jkr ejot :
ð1:213Þ
The rate of change of the displacement with respect to r (@x=@r) of Equation 1.213 is @xðr; f; tÞ ¼ jkAe jkr ejot : @r
ð1:214Þ
The force at the driving point (x ¼ 0) is f ð0; 0; tÞ ¼ jkAejot :
ð1:215Þ
Finally, the driving point impedance of the infinitely large membrane is Zm;0 ¼
jkTm A pffi ¼ Tm rA : joA
ð1:216Þ
This result is the same as that for infinitely long string. Let us investigate the vibration of a membrane which has finite dimensions. We continue to visualize this using driving point impedance. The membrane has a rectangular shape and is fixed at x ¼ 0, x ¼ Lx , y ¼ 0, y ¼ Ly as shown in Figure 1.39. This membrane must satisfy Equation 1.203 and the boundary conditions, which are xð0; y; tÞ ¼ xðLx ; y; tÞ ¼ xðx; 0; tÞ ¼ xðx; Ly ; tÞ ¼ 0:
ð1:217Þ
A solution of the fom xðx; y; tÞ ¼ Csinðkx xÞsinðky yÞejot
ð1:218Þ
52
Sound Propagation y
Ly
Lx
Figure 1.39 fixed rim
x
A rectangular membrane of length Lx and Ly with respect to x and y directions, with
can be attempted, where C is complex displacement amplitude and kx and ky are wavenumbers in the x and y directions. Equation 1.218 must satisfy the boundary condition (Equation 1.217), therefore mp ; Lx np ky ¼ ; Ly
kx ¼
m ¼ 1; 2; 3; . . . n ¼ 1; 2; 3; . . .
ð1:219Þ
Equation 1.218 can be rewritten as xðx; y; tÞ ¼
X
Cmn sin
m;n
mp np x sin y ejot Lx Ly
ð1:220Þ
where Cmn is modal coefficient. Figure 1.40 shows mode shapes of rectangular membranes (Equation 1.220). Figure 1.40 and Equation 1.220 essentially imply that the location of driving force determines how each mode participates in its response. Figure 1.41 depicts natural frequencies and mode shapes of a circular membrane. It is interesting that its natural frequencies are not integer multiples of fundamentals, that is, not harmonic. Figure 1.42 depicts the natural frequencies and mode shapes of a jing (a gong).21 1.9.3.4 Driving Point Impedance of Plates The relation between the bending wave of a bar and plate is similar to that which has been envisaged in the vibration of string and membrane. Assuming that the vibration amplitude x is 21
The ggaenggwari (a small gong) is a popular Korean traditional folk instrument. It does not have soft sounds like a jing (a gong), but has unique sounds since the natural frequencies of the ggaenggwari are not integer multiples of the fundamental natural frequency, meaning it is not harmonic frequency. While the jing is also a round-shaped instrument, its center is thick and bulging from the edge.
53
Vibration and Waves
Figure 1.40 Schematic diagrams of four typical normal modes of a rectangular membrane with fixed rim and displacement graphs corresponding to each mode
small, the energy loss, the rotary inertia, and the transverse shear flexibility are negligible for the sake of simplicity. The governing equation of a thin plate is then22 r4 x ¼ b4
@2x ; @t2
ð1:221Þ
3 where b4 ¼ rð1l2p Þ=Yw2 , r represents the pffiffi density of the plate (kg/m ), lp is Poisson’s ratio, w is the radius of gyration (where w ¼ d= 12 and d is the thickness of plate) and Y is Young’s modulus. Note that Equation 1.221 has a fourth-order derivative with respect to space. To start with the simplest case, let us look at the vibration of an infinite plate which has been excited at its center. The waves will propagate from the center and will be symmetric. It is therefore convenient to change Equation 1.221 in terms of radial coordinate. This
22
Cremer, L., and Heckl, M. (1973) Structure-borne Sound-structural Vibrations and Sound Radiation at Audio Frequencies, Springer-Verlag, pp. 130–136.
54
Sound Propagation
Figure 1.41 Vibration pffi mode and natural frequency examples of round-shaped membrane with fixed boundary (cm =a ¼ Tm =rA a2 ¼ 300, where a is the radius of the membrane): (a) 115 Hz; (b) 176 Hz; (c) 183 Hz; and (d) 246 Hz
gives us
rð1l2p Þ @ 2 x 1d d 1 d dx r r ¼ : r dr dr r dr dr w2 Y @t2
ð1:222Þ
If the plate undergoes simple harmonic excitation, that is, xðr; tÞ ¼ CðrÞejot ; then Equation 1.222 can be rewritten as 1d d 1 d dC r r kp4 C ¼ 0; r dr dr r dr dr
ð1:223Þ
ð1:224Þ
where kp4 ¼ b4 o2 . Equation 1.224 is another form of r4 Ckp4 C ¼ 0, whose solution is known as CðrÞ ¼ AJ0 ðkp rÞ þ BY0 ðkp rÞ þ CI0 ðkp rÞ þ DK0 ðkp rÞ;
ð1:225Þ
where J0 ðkp rÞ and Y0 ðkp rÞ are the first and second type Bessel functions of 0th order and I0 ðkp rÞ and K0 ðkp rÞ are the first and second type modified Bessel function of 0th order. The coefficients of Bessel functions in Equation 1.225 are determined by the boundary
Vibration and Waves
55
Figure 1.42 Vibration mode and natural frequency of the jing. (Adapted from Kwon, H.-S., Kim, Y.-H. and Rim, R., “Acoustical characteristics of the jing: an experimental observation using planar acoustic holography,” The Journal of the Acoustical Society of Korea, 16, no. 2E, 3–13, 1997, ASK.)
conditions. As we observed in the bending waves of a bar, the vibration of a plate may be expressed as the sum of the incoming wave and the outgoing wave from the excitation point, which have both the propagating and evanescent components. This physical situation, however, cannot be shown directly in the solution. Suppose first that the shear force fs ð0; tÞ ¼ Fs ejot is acting at the origin of the infinite plate as shown in Figure 1.43. In this case, the boundary conditions can be assumed to be23 1. The vibration amplitude at the excitation point (r ¼ 0) has a finite value. 2. The vibration amplitude far from the excitation point (r ! 1) is also finite. 3. There is no incoming wave since the infinite plate has no reflection. In order to apply the above boundary conditions to the solution, we need to investigate Bessel functions which express the solution. For example, we have to know how Bessel functions 23
Thomas, D.A. (1958) Characteristic impedances for flexure waves in thin plates. Journal of Acoustic Society of America, 30 (3), 220–221.
56
Sound Propagation
∞ ξ (x,y,t) y
∞
0
x
∞
fg(0,t) = Fge jωt
∞ Figure 1.43 Transverse vibration of an infinite thin plate excited harmonically at x ¼ 0. x is transversal displacement with respect to the plane on which the membrane is located
behave when r is equal to zero (the first boundary condition) and when r tends to infinity (the second boundary condition). Figure 1.44 shows Bessel functions with respect to kp r. According to Figure 1.44, J0 ðkp rÞ and I0 ðkp rÞ satisfy the first boundary condition. However, I0 ðkp rÞ increases as r increases and it does not meet the second boundary condition. Therefore, the coefficient C should be zero. Although Y0 ðkp rÞ and K0 ðkp rÞ approach þ 1 and 1, respectively, the sum may have the finite value. When r approaches zero, Equation 1.225 can be written as lim CðrÞ ¼ lim
r!0
r!0
2 AþB lnðkp rÞD lnðkp rÞ p
ð1:226Þ
where Y0 ðkp rÞ and K0 ðkp rÞ can be approximated as follows: kp r 2 ln þg ; r!0 p 2 kp r lim K0 ðkp rÞ ð1Þ ln þg ; r!0 2 lim Y0 ðkp rÞ
ð1:227Þ
where g is Euler’s constant (¼0:5772156 . . .). B and D are related by D ¼ 2B=p so that Equation 1.226 satisfies the first boundary condition. Now consider the third boundary condition. When r approaches infinity, CðrÞ can be approximated as rffi lim CðrÞ A
r!1
rffi rffiffiffi 2 p 2 p 2B p kp r cos kp r þ B sin kp r þ e : pkr 4 pkr 4 p 2kp r
ð1:228Þ
57
Vibration and Waves J0
Y0
2
2
1
1
0
0
–1
–1
–2
–2 0
5 Kp r
10
0
5 Kp r
I0
10
K0
3000
5 5
20
2000
15
4
10
3
3
5
1000
0 0
4
2 1
2 1
2
3
4
0 0
5
Kp r
0.5
0
1
1.5
2
Kp r
1 0
0
5 Kp r
10
0
5
10
Kp r
Figure 1.44 Bessel functions with respect to kp r. J0 ðkp rÞ and Y0 ðkp rÞ are the first and second kind Bessel functions of 0th order. I0 ðkp rÞ and K0 ðkp rÞ are the first and second kind modified Bessel function of 0th order
Euler’s relation, which explains the relation between the trigonometric and exponential function, enables us to express Equation 1.228 as the sum of the incoming and outgoing wave: sffiffiffi sffiffi o 1 n 2p kp r j ðkp rp4Þ j ðkp rp4Þ lim CðrÞ ðAjBÞe þ ðA þ jBÞe þB e ; ð1:229Þ r!1 2pkp r 2kp r where the second term within the braces represents the incoming wave. In order to satisfy the third boundary condition, the relation A ¼ jB should be obtained. Therefore, the desired solution is 2 xðr; tÞ ¼ A J0 ðkp rÞ þ jY0 ðkp rÞ þ j K0 ðkp rÞ ejot : ð1:230Þ p Now consider the driving point impedance by the bending moment (ZM m;0 ) and by the shear force (ZFm;0 ) of an infinitely thin plate. Note that the driving point impedance by moment is
58
Sound Propagation
infinite because the plate does not bend at the center. This motivates us to look at the wave propagation characteristics of a thin plate using the driving point impedance due to shear force (ZFm;0 ). The time derivative of Equation 1.230 enables us to obtain the velocity: 2 uðr; tÞ ¼ joA J0 ðkp rÞ þ jY0 ðkp rÞ þ j K0 ðkp rÞ ejot : p
ð1:231Þ
The velocity at the excitation point can be written uð0; tÞ ¼ joAejot :
ð1:232Þ
When the shear force fs ðr; tÞ is acting on the infinitesimal symmetric element of the twodimensional space, fs ðr; tÞ can be written as fs ðr; tÞ ¼
Yw3 @ ðr2 xÞ: ð1l2p Þ @r
ð1:233Þ
The force at the origin can be obtained from the third-order derivative of x with respect to r. This is expressed as the sum of Bessel functions of zeroth and first order by using the consecutive relation between Bessel functions of different order: 1 2K1 ðkp rÞ jJ1 ðkp rÞ Y1 ðkp rÞ B C jot p C ¼ jA8drv2p ejot ; e fs ð0; tÞ ¼ drcp B @ 2K0 ðkp rÞ A J0 ðkp rÞ þ j Y0 ðkp rÞ þ p 0
ð1:234Þ
where vp ¼ o=kp . The right-hand side of Equation 1.234 can be obtained by using the relation between Y1 ðkp rÞ and K1 ðkp rÞ as r approaches zero. The driving point impedance can be written as
Zm;0 ð0; tÞ ¼
jA8drv2p ejot jAoejot
sffiffiffi ¼ 4d
2
Yr ; 3ð1l2p Þ
ð1:235Þ
which is always real regardless of frequency. To better understand this, we change the form of Equation 1.235. Firstly, the relation between kp and o can be written as kp4 ¼ b4 o2 ¼
o2 ; w2 c2p
ð1:236Þ
59
Vibration and Waves
where c2p ¼ Y=rð1l2p Þ. Substituting Equation 1.236 into Equation 1.234 gives Zm;0 ð0; tÞ ¼
8drv2p 8dro 4 ¼ sffi ¼ 8drwcp ¼ pffi ðrd 2 Þcp ; 2 o 3 o
ð1:237Þ
w2 c2p where rd2 can be interpreted as density per unit length (kg/m) and v is the velocity. Equation 1.237 demonstrates that the driving point impedance of the thin plate has a similar form to that of an infinite string. Again, the driving point impedance allows us to see how waves will propagate in the selected medium, depending on boundary conditions. The driving point impedance of a finite plate, therefore, will be quite different from that of the infinite case (Equation 1.237). It will depend strongly on the possible modes that are allowed. As explained for membranes, the displacement can be regarded as the weighted sum of eigenmodes (or natural modes). The displacement of the plate with fixed rim is determined by its shape and boundary condition. If the driving force acts on the nodal line, the corresponding mode will not be driven. Furthermore, although the location of the driving point is the same, the driving point impedance will vary with respect to its mode shape depending on the frequency of excitation. For example, if we have a circular thin plate with fixed rim, we can obtain the mode shape of the plate as ! cosðmfÞ pbmn r Jm ðpbmn Þ pbmn r ; ð1:238Þ xmn ðr; fÞ ¼ Im Jm a Im ðpbmn Þ a sinðmfÞ where bmn is a constant determined with respect to m and n. In particular, if m ¼ 0 we have to use cosðmfÞ. In case of m > 0, sinðmfÞ should be used to obtain the mode shape of the circular plate with fixed rim. Figure 1.45 depicts mode shapes for some values of m and n. In addition, a rectangular plate also vibrates depending on the boundary conditions. Figure 1.46 depicts experimental results for a rectangular steel plate with fixed rim.
Exercises 1. Wave propagation can be expressed as follows: expression with regard to time : expression with regard to space :
yðx; tÞ ¼ gðtx=cÞ yðx; tÞ ¼ gðxctÞ
Suppose that we have the following wave at t ¼ 0:0. Draw the shape of the wave at x ¼ 0:1, 0.2, 0.3, 0.4, 0.5 (Figure 1.47). c is the speed of propagation. 2. Suppose that we have waves that obey the dispersion relation as depicted in Figure 1.48 Assume also that we have a harmonic wave. (a) What does the wave look like at t ¼ 1 for o ¼ 1, 4 if its initial amplitude is 1.0? (b) If two waves have the same amplitude of 1.0 but different radian frequencies (o ¼ 1:0, 1.2) and propagate together, what will the group velocity be and its physical meaning? Hint: group velocity can be written as
do . dk
60
Figure 1.45 fixed rim
Sound Propagation
Schematic diagrams of four typical normal modes (m, n) of a circular thin plate with
Figure 1.46 Normal modes of a flat thin steel plate. (Scatter sand on a steel plate with fixed boundary and excite it at natural frequency; the spot where sand is concentrated represents the nodal line of each mode)
61
Vibration and Waves
y c = 0.2
0.5
1
0
Figure 1.47
x
Shape of the wave at t ¼ 0
k k= ω
0
1
2
3
4
ω
Figure 1.48 Dispersion relation
3. Describe how yðx; tÞ ¼ 1 ejðotkxÞ expresses a harmonic wave, using a graph or any kind of illustration. (Assume that k ¼ 1 and o ¼ 1 for convenience). 4. We have a semi-infinite string attached to a mass (m), spring (s), and dash pot (rd ) system, as illustrated in Figure 1.49. Assume a harmonic wave Aejðot þ kxÞ approaches from the left side of the system. (a) How much reflected wave will there be? (b) If m; s; rd ! 0; 1, what would happen to the magnitude of the reflected wave? (c) What would the effect of frequency be on the reflection?
Ae–j(ω t+ks)
∞
m
s
Figure 1.49
rd
A semi-infinite string attached to a mass (m), spring (s), and dashpot (rd)
62
Sound Propagation
5. A piece of string has fixed ends and a length L, as illustrated in Figure 1.50. Knowing that a Green’s function is one way to analytically predict the response of any linear system that can be described by a linear partial or ordinary differential equation, we attempt to find an appropriate Green’s function for the given system. Using this hint, first determine the Green’s function Gðxjx0 Þ and then answer the following questions. The Green’s function is simply the solution that predicts the response at x due to the unit impulse excitation at x ¼ x0 . L
x = x0
f (x,f ) = δ (x–x0)e –jωt
Figure 1.50
A string fixed at each end with a point excitation at x ¼ x0
(a) Assume there is a distributed excitation between x ¼ x1 and x ¼ x2 , as illustrated in Figure 1.51. Determine the response at an arbitrary position of the string by using your Green’s function. This problem highlights the concept of “principle of superposition,” which is always valid for a linear system. (b) We also want to compare this result with that obtained by using eigenfunctions of the string. L
x = x2
x = x1
f (x,t) = e –jωt , if x1 ≤ x ≤ x2 =0
Figure 1.51
, otherwise
A string fixed at each end with a distributed excitation at x1 < x < x2
Hint: When there is a unit impulse excitation at x ¼ x0 , then the response of the string follows the following one-dimensional inhomogeneous partial differential equation: @2y 1 @2y 1 ¼ dðxx0 Þejot : @x2 c2 @t2 TL
ð1:239Þ
63
Vibration and Waves
Noting that the excitation is harmonic, the response can be assumed to be yðx; tÞ ¼ YðxÞejot :
ð1:240Þ
Y is what is being sought in this problem, that is, Green’s function. We can therefore write YðxÞ ¼ Gðxjx0 Þ:
ð1:241Þ
Substituting Equations 1.240 and 1.241 into Equation 1.239 leads to @2G 1 þ k2 G ¼ dðxx0 Þ: 2 @x TL
ð1:242Þ
The solution of Equation 1.242 can be assumed to be Gðxjx0 Þ ¼ Ae jkjxx0 j :
ð1:243Þ
(This approach can be justified by simply recognizing that the directions of propagation will be opposite to each other when x > x0 or when x < x0 ). If we integrate Equation 1.242 at the vicinity of x ¼ x0 , ð x¼x þ 0
x¼x 0
@2G 1 2 þ k G dx ¼ : @x2 TL
ð1:244Þ
This can be rewritten as @G @x
#
@G @x þ
x¼x0
# ¼ x¼x 0
1 : TL
ð1:245Þ
Inserting Equation 1.243 into Equation 1.245 gives 1 : 2jkTL
ð1:246Þ
1 ejkjxx0 j : 2jkTL
ð1:247Þ
A¼ We therefore have Gðxjx0 Þ ¼
This is the Green’s function for an infinite string, because we have not yet considered the boundary condition. If the length of the string is L, then Gðxjx0 Þ has to satisfy the boundary condition Gðxjx0 Þ ¼ 0
at x ¼ 0; L:
ð1:248Þ
64
Sound Propagation
Note that there are two possible forms that belong to Equation 1.247: a sine function and a cosine. It is not possible to satisfy the given boundary condition by using the cosine function, and we can therefore write the solutions that satisfy both Equations 1.247 and 1.248 as ( Gðxjx0 Þ ¼
A sin kðLx0 Þ sin kx; x < x0 A sin kðLxÞ sin kx0 ; x > x0
:
ð1:249Þ
If we substitute Equation 1.249 into Equation 1.242 and integrate in the vicinity of x ¼ x0 , then we obtain the following: A¼
1 : TL ksin kL
ð1:250Þ
Equations 1.249 and 1.250 provide the Green’s function that predicts the response at any position along the string due to the impulse excitation at x ¼ x0 . 6. Consider the two different bars depicted in Figure 1.52. They have the same cross-sectional area and material properties, but different boundary conditions. (a) Which case results in faster wave propagation? (b) Explain the reasons. Hint: The speed of propagation is related to how easily the wave can move, as noted for the string case.
Rigid wall
Air
A stick
A stick
Rigid wall
Air
I
II
Figure 1.52
Two identical bars with different boundary conditions
7. For a bar that has the impedance boundary condition described by Figure 1.53, find the solution of free vibration and then the forced response due to the excitation on x ¼ 0. What would be the driving point impedance? 8. Determine the response at any point along the bar (Figure 1.54) due to a given excitation. Compare the result with that obtained for the above problem (Problem 7). Comparing both cases by using the driving point impedance will also be valuable. 9. A membrane is fixed at the rims of four sides, as depicted in Figure1.55. (a) Derive the response at ðx; yÞ due to a unit impulse at A Lx ; Ly . 2
2
65
Vibration and Waves
y
F0e –jω t
ζ (x,t) x
x=0 Zmc
Figure 1.53
x=L ZmL
A bar with impedance boundary condition and harmonic excitation at x ¼ 0
y
F0e –jω t
ζ (x,t)
Figure 1.54
x=L ZmL
x
x=0
A bar with impedance boundary condition and harmonic excitation at x ¼ 0
y
Ly C
3Ly 4
B A
Ly 4 0
Lx
3Lx
4
4
Figure 1.55
Lx
A membrane fixed at four edges
x
66
Sound Propagation
(b) What would be the response if impulsive force is applied at B
7Lx 3Ly ; 8 4
?
(c) If we have a unit impulse/unit area over the region C, then what would be the response at an arbitrary position on the membrane? 10. Determine the total energy when a circular membrane, which has a radius a and mass per unit area rA , vibrates with its fundamental mode. Assume that the amplitude at the center is A and the radian frequency is o. o 11. Determine the frequency or frequencies ( f ¼ 2p ) that makes/make the driving point impedance zero for the four cases depicted in Figure 1.56. Assume that the mass per unit length of the string is rL and the phase velocity is cs .
L L 2
TL : Tension of the string TR : Mass rd : Viscous friction coefficient s : Spring constant
F0e–jω t
L 2
–jω t
F0e L m
m s
L 2
s F0e–jω t L
m rd
m L 2
Figure 1.56
rd F0e–jω t
Strings with various boundary conditions
12. A longitudinal wave is excited by a harmonic force at the end of a bar of length L, as depicted in Figure 1.57. The impedance of the other end is ZmL and the cross-sectional area of the bar is A0 . Assume that F0 is real. (a) What will be the intensity, which is the velocity multiplied by the force at the point of interest, when ZmL ¼ rL c (rL is mass per unit length of the bar and c is phase velocity) at an arbitrary position along the bar? What will be the direction and magnitude of intensity? (b) When ZmL ¼ 1, how does the intensity change?
67
Vibration and Waves
y
F0e –jω t
ζ (x,t)
Figure 1.57
x=L ZmL
x
x=0
A bar with impedance boundary condition and harmonic excitation at x ¼ 0
13. Suppose that we have a semi-infinite string that has a spring, mass, and viscous damping system whose undamped natural frequency f0 is 10 Hz, as depicted in Figure 1.58. What will be the phase difference between the reflected wave and incident wave for the following three cases? (a) 0.1 Hz (b) 1 kHz (c) 10 Hz Incident wave m Reflected wave
Figure 1.58
s
rd
A semi-infinite string attached to a mass (m), spring (s), and dashpot (rd) system
14. We have studied the driving point impedance of a string that is fixed at x ¼ L. We can simply reapply the procedures that were used thus far to the more general case depicted in Figure 1.59 where m denotes the mass and S indicates the spring constant. Express the result using a graph which shows the driving point impedance with respect to kL and
L x F0e –jω t
m s
Figure 1.59
A finite string fixed at x ¼ L
68
Sound Propagation L
L
x m
m F0e –jω t
s
Figure 1.60
s
A finite string fixed at both ends
frequency f. Discuss the behavior of the impedance when the wave number tends to infinity or kL tends to infinity or to zero. It is also interesting to examine its behavior when the frequency of interest is much larger or smaller than the resonance frequency of the massspring-viscous damping system at the end of the string. You may also wish to examine the somewhat different case depicted in Figure 1.60.
2 Acoustic Wave Equation and Its Basic Physical Measures 2.1
Introduction/Study Objectives
In Chapter 1, it was explained that the governing equation is the total expression of every possible wave. A one-dimensional wave is the simplest wave, which essentially has all the mathematical and physical characteristics of general waves. The waves along a string were extensively studied, as this case represents the fundamental characteristics of every onedimensional wave. The waves along a string propagate along its length, but the string itself moves perpendicular to the propagation direction. It therefore forms a transverse wave. If the particle of a medium moves in the direction of propagation, we refer to it as a longitudinal wave. The waves in air, water, or any compressible medium are longitudinal waves, which are often referred to as acoustic waves. This chapter explores the underlying physics and sensible physical measures that are related to acoustic waves, including pressure, velocity, intensity, and energy. Impedance plays a central role with regard to its effect on these measures. The final objective of this chapter is to determine rational means of finding the solutions of acoustic wave equations.
2.2
One-dimensional Acoustic Wave Equation
The simplest case is illustrated in Figure 2.1. The end of a pipe or duct which is filled with a homogeneous compressible fluid (air, water, etc.) is excited with a radian frequency (o). If the pipe is semi-infinitely long, then the pressure in the pipe ðpðx; tÞÞ can be mathematically written as pðx; tÞ ¼ P0 cosðkxot þ fÞ
ð2:1Þ
where P0 is the pressure magnitude and f is an initial phase. Note that the driving point impedance is somewhat similar to Equation 1.66.1 1 There will be no reflected wave; the wave will therefore propagate in the positive x direction and the driving point impedance will be equivalent to that of the medium.
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
Sound Propagation
70
cosω t
This depicts the waves that can be generated when we excite one end of the pipe, harmonically. Δx pS
p+
∂p dx S ∂x
x
0
=
ρ
∂u ∂u +u ∂t ∂x
x+Δx
x 0
S: cross section area (m2) u: fluid particle velocity in x – direction (m/sec) 3
ρ: density of fluid (kg/m )
Figure 2.1 Relation between forces and motion of an infinitesimal fluid element in a pipe (expressing momentum balance: the left-hand side shows the forces and the right exhibits the change of momentum)
If the pipe is of finite length L, then the waves in the pipe can be expressed by Equation 1.67. Recall that the displacement of the string, in this case, can be regarded as the real part of Equation 1.67, that is yðx; tÞ ¼ A sin kðLxÞcos ot:
ð2:2Þ
However, the possible acoustic pressure in the pipe can be written as pðx; tÞ ¼ P0 cos kðLxÞcos ot:
ð2:3Þ
Equations 2.2 and 2.3 are different simply because the displacement of the string has to be 0 at the end, but the acoustic pressure is maximal at x ¼ L. We can also predict that the driving point impedance is governed by kL (wave number k multiplied by its length L), as given in Equation 1.70. To understand what is happening in the pipe, we have to understand how pressures and velocities of the fluid particles behave and are associated with each other.2 This motivates us to look at an infinitesimal element of the volume of the fluid in the pipe; specifically, we will investigate the relation between force and motion.
2
The force and displacement relation is expressed in Equation 1.34, which states that the force acting perpendicular to the normal of the cross-section is the product of the tension and the rate of the displacement with respect to space. Similarly, it is important to define the relationship between the force per unit area of the fluid and the response of the velocity of the fluid.
Acoustic Wave Equation and Its Basic Physical Measures
71
As illustrated in Figure 2.1, the forces acting on the fluid between x and x þ Dx and its motion will follow the conservation of momentum principle. That is, Sum of the forces acting on the fluid ¼ momentum change
ð2:4Þ
We can mathematically express this equality as ðpSÞx ðpSÞx þ Dx ¼ rS
du Dx dt
ð2:5Þ
where it has already been assumed that the viscous force, which likely exists in the fluid, is small enough (relative to the force induced by pressure) to be neglected. The rate of change of the velocity ðdu=dtÞ can be expressed by du @u @u @x ¼ þ dt @t @x @t
ð2:6Þ
where u is a function of position (x) and time (t) and the velocity is the time rate change of the displacement. Therefore, we can rewrite Equation 2.6 as du @u @u ¼ þu : dt @t @x
ð2:7Þ
If the cross-section between x and x þ Dx is maintained constant and Dx becomes small ðDx ! 0Þ, then Equation 2.5 can be expressed as3 @p @u @u Du ¼r þu ð2:8Þ ¼r @x @t @x Dt where p ¼ p0 þ p0
ð2:9Þ
r ¼ r0 þ r0
ð2:10Þ
D @ @ ¼ þu : Dt @t @x
ð2:11Þ
Note that the pressure ðpÞ is composed of the static pressure (p0 ) and the acoustic pressure (p0 ), which is induced by the small fluctuation of fluid particles. The density also has two components: the static density (r0 ) and the small fluctuating density (r0 ).
Note that we used @x=@t ¼ u in Equation 2.7. This is the Lagrangian description, which describes the motion of a mass of fluid at Dx. The other method to describe the momentum change through fixed infinitesimal control volume is by using the Euler description (Section 2.8.1). Note also that a more precise momentum balance can be expressed as @p ¼ Dru @x Dt . 3
Sound Propagation
72
Equation 2.11 is the total derivative, and is often called the material derivative. The first term expresses the rate of change with respect to time, and the second term can be obtained by examining the change with respect to space as we move with the velocity u.4 As can be anticipated, the second term is generally smaller than the first. If the static pressure (p0 ) and density (r0 ) do not vary significantly in space and time, then Equation 2.8 becomes
@p0 @u ¼ r0 @x @t
ð2:12Þ
where p0 is acoustic pressure and is directly related to acoustic wave propagation. As already implied in Equation 2.9, the acoustic pressure is considerably smaller than the static pressure.5 Equation 2.12 essentially means that a small pressure change across a small distance (@x) causes the fluid of mass/unit volume r0 to move with the acceleration of @u=@t. This equation is generally referred to as a linearized Euler equation. Equation 2.8, on the other hand, is an Euler equation. Equations 2.8 and 2.12 describe three physical parameters, pressure, fluid density, and fluid particle velocity. In other words, they express the relations between these three basic variables. In order to completely characterize the relations, two more equations are needed. The relation between density and fluid particle velocity can be obtained by using the conservation of mass. Figure 2.2 shows how much fluid enters the cross-section at x and how much exits through the surface at x þ Dx. If we apply the principle of conservation of mass to the fluid volume between x and x þ Dx, the following equality can be written. the rate of mass increase in the infinitesimal element ¼ the decrease of mass resulting from the fluid that is entering and exiting through the surface at x and x þ Dx
∂ (ρ S Δx ) ∂t
( ρ u S )x x
0
x
ð2:13aÞ
( ρ u S )x+Δx
x + Δx
S : cross section area (m2) u : fluid particle velocity in x – direction (m/sec) ρ : density of fluid (kg/m3)
Figure 2.2 Conservation of mass in an infinitesimal element of fluid (increasing mass of the infinitesimal volume results from a net decrease of the mass through the surfaces of the volume)
4 5
We assume that the effect of mass transport is negligible. We also refer to the acoustic pressure as “access pressure” or “sound pressure”.
Acoustic Wave Equation and Its Basic Physical Measures
73
Expressing this equality mathematically leads to @ ðrSDxÞ ¼ ðruSÞx ðruSÞx þ Dx @t
ð2:13bÞ
as illustrated in Figure 2.2. As assumed before, if the area of the cross-section (S) remains constant, then Equation 2.13 can be rewritten as @r @ ¼ ðruÞ: @t @x
ð2:14Þ
We can linearize this equation by substituting Equation 2.10 into Equation 2.14. Equation 2.14 then becomes @r0 @u ¼ r0 : @t @x
ð2:15Þ
Equations 2.12 and 2.15 express the relation between the sound pressure and fluid particle velocity, as well as the relation with the fluctuating density and fluid particle velocity, respectively. One more equation is therefore needed to completely describe the relations of the three acoustic variables: acoustic pressure, fluctuating density, and fluid particle velocity. The other equation must describe how acoustic pressure is related to the fluctuating density.6 Recall that a pressure change will induce a change in density as well as other thermodynamic variables, such as entropy. This leads us to postulate that the acoustic pressure is a function of density and entropy, that is, p ¼ pðr; sÞ
ð2:16Þ
where s denotes entropy. We can then write the change of pressure, or fluctuating pressure, dp or p0 , by modifying Equation 2.16 as follows: @p @p dp ¼ dr þ ds: @r s @s r
ð2:17Þ
This equation simply states that a pressure change causes a density change (dr) and entropy variation (ds). It is noticeable that the fluid obeys the law of isentropic processes when it oscillates within the range of the audible frequency: 20 Hz–20 kHz.7 The second term on the right-hand side of Equation 2.17 is therefore negligible. This implies that the small change of sound pressure with regard to the infinitesimal change of density can be assumed to have certain proportionality. (An alternative way to deduce the same relation can be found in Section 2.8.1.3.) Note that the second relation of Equation 2.18 is mostly found experimentally. This 6 For the string, Equation 1.56 shows that the square of the propagation velocity is proportional to the tension per unit length, and inversely proportional to the string’s density per unit length. On the other hand, for the compressible fluid the three acoustic variables constitute the relationship. 7 This is possible if the period of oscillation by the fluid particle is much smaller than the time required to dissipate or transfer the heat energy within the wavelength of interest.
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74
reduces Equation 2.17 to the form
p0 B ¼ ¼ c2 r0 r0
ð2:18Þ
where B is the bulk modulus that expresses the pressure required for a unit volume change and c is the speed of sound. We may obtain Equation 2.18 by introducing a gas dynamics model. This equation is an equation of state; for example, Equation 1.56 is an equation of state for the string. Tables 2.1 and 2.2 summarize the speed of sound in accordance with the state of gas. An alternative method of deducing Equations 2.17 and 2.18 can be found in Section 2.8.1.3. Table 2.1
Dependency of speed of sound on temperature
Temperature ( C) 100 95 90 85 80 75 70 65 60 55 50 45 40
Speed of sound (m/s)
Temperature ( C)
Speed of sound (m/s)
Temperature ( C)
Speed of sound (m/s)
263.5 267.3 271.1 274.8 278.5 282.1 285.7 289.2 292.7 296.1 299.5 302.9 306.2
35 30 25 20 15 10 5 0 5 10 15 20 25
309.5 312.7 315.9 319.1 322.3 325.3 328.4 331.5 334.5 337.5 340.4 343.4 346.3
30 35 40 45 50 55 60 65 70 75 80
349.1 352.0 354.8 357.6 360.4 363.2 365.9 368.6 371.3 374.0 376.7
Note that Equation 2.18 expresses how the access pressure or acoustic pressure communicates with the fluctuating density. Equations 2.12 and 2.15 completely express the laws that govern the waves in which we are interested. Therefore, we can summarize the relations as @p0 @u ¼ r0 @x @t
ð2:12Þ
@r0 @u ¼ r0 @x @t
ð2:15Þ
p0 ¼ c2 : r0
ð2:18Þ
Figure 2.3 demonstrates how these equations and physical variables are related. If we eliminate r0 and u from Equations 2.12, 2.15 and 2.18, then we obtain @ 2 p0 1 @ 2 p0 ¼ 2 2 : 2 @x c @t
ð2:19Þ
This is a linearized acoustic wave equation.8 8
If we eliminate p0 and u or change to p0 and r0 , then we can obtain the equation for r0 and u, respectively.
Acoustic Wave Equation and Its Basic Physical Measures
75
Table 2.2 The dependency of the speed of sound on relative humidity and on frequency 0%
30%
60%
100%
Relative humidity/Frequency
Decay rate (%)
Speed of sound (m/s)
Decay rate (%)
Speed of sound (m/s)
Decay rate (%)
Speed of sound (m/s)
Decay rate (%)
Speed of sound (m/s)
20 40 50 63 100 200 400 630 800 1250 2000 4000 6300 10 000 12 500 18 000 20 000
0.51 1.07 1.26 1.43 1.67 1.84 1.96 2.11 2.27 2.82 4.14 8.84 14.89 26.28 35.81 52.15 75.37
343.477 343.514 343.525 343.536 343.550 343.559 343.561 343.562 343.562 343.562 343.562 343.564 343.565 343.566 343.566 343.567 343.567
0.03 0.11 0.17 0.25 0.50 1.01 1.59 2.24 2.85 5.09 10.93 38.89 90.61 204.98 294.08 422.51 563.66
343.807 343.808 343.810 343.810 343.814 343.821 343.826 343.827 343.828 343.828 343.829 343.831 343.836 343.846 343.854 343.865 343.877
0.02 0.06 0.09 0.15 0.34 0.99 1.94 2.57 2.94 4.01 6.55 18.73 42.51 101.84 155.67 247.78 373.78
344.182 344.183 344.183 344.184 344.185 344.190 344.197 344.200 344.201 344.202 344.203 344.204 344.204 344.206 344.208 344.211 344.215
0.01 0.04 0.06 0.09 0.22 0.77 2.02 3.05 3.57 4.59 6.29 13.58 27.72 63.49 96.63 154.90 237.93
344.685 344.685 344.685 344.685 344.686 344.689 344.695 344.699 344.701 344.704 344.705 344.706 344.706 344.706 344.707 344.708 344.709
(Adapted from CRC Handbook of Chemistry and Physics, 79th edition, 1998, pp. 14–38, CRC Press, 1998. With kind permission of Taylor & Francis Group LLC-Books.)
Figure 2.3 Pictorial relation between three variables that govern acoustic wave propagation (p0 and r0 express the mean pressure and static density, respectively; p0 and r0 denote acoustic pressure and fluctuating density, respectively; c denotes the speed of propagation, and u is the velocity of the fluctuating medium)
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76
Based on what we have studied so far, two conclusions can be drawn. The first is that there is an analogy between the wave propagation along a string and acoustic waves, that is, the waves in compressible fluid. Note that Equation 2.19 is essentially the same as that given by Equations 1.24 and 1.55 which satisfy Equation 1.17 and represent a general one-dimensional wave. This implies that the theory outlined in Chapter 1 (e.g. the relation between wave number and frequency, how the wavelength is associated with the period, characteristic impedance, driving point impedance, and the role of the boundary condition, etc.) can be directly applicable to acoustic waves. The only difference between the waves along a string and acoustic waves lies in whether the directions of wave propagation and velocity fluctuation of medium are collinear or perpendicular. Note that the propagation direction of the waves along a string is perpendicular to that of the motion of the string. Conversely, the acoustic wave propagates in the direction of the fluid particle’s velocity.9 The former is called a transverse wave, while the latter is considered a longitudinal wave.10 This means that the acoustic wave, at least a one-dimensional case, is analogous to that a wave on a string. It therefore appears that a unified concept or theory is very possible. We can also optimistically consider that any three-dimensional wave can be decomposed into a one-dimensional wave, at least conceptually, implying that we can extend what was studied in Chapter 1 to three-dimensional acoustic waves.11 The second conclusion involves the relations between three acoustic variables, which are illustrated in Figure 2.3. Euler’s equation, Equation 2.12, states that the spatial pressure change makes the fluid particle move. Equation 2.15 shows that a higher compression rate in time makes a steeper negative velocity gradient in space. Lastly, Equation 2.18 indicates that the speed of propagation depends on the characteristics of the medium. If we have a smaller density change for a unit pressure change, then we will obtain faster propagation speed. For example, compare the speed of propagation of water and air (see Tables 2.1 and 2.2). We now extend Equations 2.12, 2.15 and 2.19 to a three-dimensional case. First, Euler’s equation can be written as @p0 @u ¼ r0 @x @t @p0 @v ¼ r0 @y @t
ð2:20Þ
@p0 @w ¼ r0 @t @z
where we use ðx; y; zÞ coordinate for convenience. Each equation simply states that the change
9
The linearized Euler equation (2.12) essentially states that the pressure difference induces time rate of velocity u in the x direction. 10 Depending on whether the direction of wave propagation and that of the medium’s oscillation are perpendicular or collinear, the wave can be transverse or longitudinal, respectively. The bending wave essentially makes the change of angle measured from the mid plane (Section 1.9.3). 11 The principle of superposition holds for the waves that are governed by the linearized acoustic wave equation. Therefore, any combination of the solutions with respect to each component of the coordinate – for example, ðx; y; zÞ rectangular coordinate, ðr; y; zÞ cylindrical, and ðr; y; fÞ spherical coordinates – also satisfies the governing equation. Simply extending this concept leads us to use the Fourier series. This means that we attempt to look at what we are interested in, in terms of the orthogonal series.
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77
of pressure with respect to the ðx; y; zÞ direction, that is, the pressure gradient, makes the fluid particle move. u, v, and w denote the velocity with respect to the ðx; y; zÞ coordinate system. We may use a vector notation to express Equation 2.20, which will yield a more compact form. This gives rp0 ¼ r0
@~ u : @t
ð2:21Þ
This is the linearized Euler equation, where ~ u ¼ ðu; v; wÞ:
ð2:22Þ
Similarly, Equation 2.15 can also be extended to the three-dimensional form, that is, @r0 u ¼ r0 r ~ @t
ð2:23Þ
(see Section 2.8.1 for details). The right-hand term of Equation 2.23 represents the net mass flow into the unit volume in space. Section 2.8.1 provides the detailed derivation and theoretical background. If we eliminate r0 and ~ u using Equations 2.21, 2.23 and 2.18, then r2 p0 ¼
1 @ 2 p0 c2 @t2
ð2:24Þ
is obtained, which is a three-dimensional form of a wave equation.12 To summarize, three physical variables (acoustic pressure, fluid particle velocity, and fluctuating density) govern acoustic waves in a compressible fluid. These variables are not independent. The relations between them are expressed by laws or equations, that is, the linearized Euler equation, conservation of mass, and the state equation of gas. It is also clear that the acoustic waves for a compressible fluid generally follow the theory already determined for one-dimensional wave propagation along a string. The physical concept and all details, including driving point impedance, can be used to explore acoustic phenomena that are possible in a compressible fluid. Lastly, we envisage that three-dimensional acoustic waves can also be understood using theory developed to explain one-dimensional acoustic waves.
2.3
Acoustic Intensity and Energy
We acknowledged that the acoustic pressure (p0 ), density (r0 ), and fluid particle velocity (u for one dimension, or~ u for the three-dimensional case) are major physical variables that determine everything related to acoustic wave propagation. We also recall that the characteristic impedance and driving point impedance measure how the waves behave in the presence of 12
We can also express the governing equation by using a potential function, that is, the velocity potential. This is possible because the viscous force is negligible compared to others. This means that r ~ u ¼ 0; in other words, the angular deformation can be neglected. Note that rF ¼ ~ u always satisfies r ~ u ¼ 0, where F is the velocity potential (Section 2.8.2).
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a discontinuity or boundary, as described in Chapter 1. Power and energy are also major descriptors of wave propagation along a string. Understanding that the waves in a compressible fluid also obey what we have learned in the case of one-dimensional string waves, we explore what is physically meant by the energy and intensity of acoustic waves. For simplicity, we consider a one-dimensional case (Figure 2.4). We denote acoustic pressure (p0 ) as p, and fluctuating density (r0 ) as r.
− Δl
p0 + p
p0 l
p
εp =
1 − Δl .p. 2 l
− Δl / l
Figure 2.4 Volume change and energy for a one-dimensional element (ep is potential energy density and p is p0 for convenience)
As illustrated in Figure 2.4, there will be a volume change of Dl S because of the pressure difference along the element. The length of the element will be shortened by Dl due to the small pressure change p. The energy stored in the unit volume, potential or elastic energy, can then be written as 1 Dl ep ¼ p 2 l
ð2:25Þ
where Dl has to obey the conservation of mass. We therefore have r0 lS ¼ ðr0 þ rÞðl þ DlÞS:
ð2:26Þ
r0 l ¼ r0 l þ rl þ r0 Dl þ rDl:
ð2:27Þ
Rearranging this, we obtain
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79
Note that the last term on the right-hand side is much smaller than the others. Equation 2.27 therefore reduces to
Dl r ¼ : l r0
ð2:28Þ
This equation simply states that the change of density is proportional to the decrease of volume, Dl S, or the length reduction (Dl). Substituting Equation 2.28 into Equation 2.25 then gives 1 r ep ¼ p : 2 r0
ð2:29Þ
Using the state equation, Equation 2.18, and changing r to p, then gives ep ¼
1 p2 2 r0 c 2
ð2:30Þ
where ep denotes the acoustic potential energy. The kinetic energy per unit volume can be written as 1 ek ¼ r0 u2 : 2
ð2:31Þ
If we assume that the dissipated energy in the fluid is much less than the potential energy or kinetic energy, then the total energy has to be written:13 et ¼ ep þ e k ¼
1 p2 1 þ r0 u 2 : 2 2 r0 c 2
ð2:32Þ
Note that the potential and kinetic energy are identical if the wave of interest is a plane wave in an infinite domain;14 in other words, the wave propagates without any reflection. This can be readily understood by recalling the impedance of a plane wave, which is r0 c. The next question then is how the acoustic energy changes with time. We can see that the energy per unit volume has to be balanced by the net power flow through the surfaces that enclose the volume of interest, as illustrated in Figure 2.5. This observation can be written conceptually as the rate of increase of energy ¼ the power entering through the surface at ðx ¼ xÞ the power exiting through the surface at ðx ¼ x þ DxÞ
13
ð2:33Þ
The wave energy along the string (1.75) is the same as what we have in a compressible fluid. The energy dissipation mechanism would be different in each case, but the global behavior is identical. 14 We have to measure both velocity and pressure to derive sound energy. The sound pressure can be easily measured using a microphone; however, measuring fluid particle velocity is relatively hard. The linear Euler equation certainly implies that the velocity can be measured by using two pressure sensors. Equation 2.12 states that the velocity can be measured by integrating the signals from the microphones that are separated by a small distance apart.
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∂εt Δx ∂t
pu
(pu)x
pu + x + Δx
x
∂(pu) dx ∂x
(pu)x+Δx
Cross sectional area : S
Figure 2.5 Relation between energy ðet Þ and one-dimensional intensity ðpuÞ (energy in the volume SDx and the intensity through the surface at x and x þ Dx must be balanced)
This can be translated into a mathematical expression as follows: @et SDx ¼ ðp u SÞx ðp u SÞx þ Dx @t @pu ¼ SDx: @x
ð2:34Þ
Equation 2.34 can then be reduced to @et @I þ ¼0 @t @x
ð2:35Þ
where I ¼ pu, which we call “acoustic intensity” or “sound intensity”. Acoustic intensity is the acoustic power per unit area.15 If we simply extend Equation 2.35 to a three-dimensional case, then @et þ r ~ I ¼ 0: @t
ð2:36Þ
The second term on the left-hand side of Equation 2.36 expresses net outflow power through the fluid surfaces (Figure 2.5). Two major points must be noted in relation to the expression of the intensity. First, the intensity is a vector which has direction.16 Second, the intensity is a product of two different physical quantities. Recall that the impedance is also composed of two physical quantities, pressure and velocity, but in a division form; it can therefore be regarded as having an input and output relation. Note, however, that the intensity expresses how much power is transported in which direction. When we have two physical variables, the phase difference between them has significant meaning; for example, the phase relationship between force (pressure) and velocity of impedance. As explained in detail in Section 1.9.1, the phase between the force and velocity expresses how well the force generates the velocity (response). In this regard, the intensity can be classified as two different categories: active intensity and reactive intensity. The phase of the former is in-phase and that of the latter has 90 (p=2) phase difference.
15
The power on the cross section S is puS. Strictly speaking, the intensity in one dimension is also indicated in Figure 2.5. The velocity u is in the positive direction of x. 16
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81
To understand the meaning of the intensities in physical terms, we look again at the simplest case: the intensity of waves propagating in a one-dimensional duct. Figure 2.6 depicts the waves in an infinite-length duct and Figure 2.7 shows the waves for a finite-length L duct.17 When the waves propagate in an infinite duct, where no reflection is possible, the pressure and velocity have the same phase and the frequency of the intensity is double that of the frequency of the pressure and velocity. It can also be observed that the average intensity with respect to time hpuiavg is constant, as can be seen in Figure 2.6. The instantaneous intensity changes, however, with regard to the position along the duct; note also that a similar result was obtained for the string (Figure 1.6). This is a simple consequence that we already observed in Section 1.6, and is related to the driving point impedance. This driving point impedance is matched precisely with the characteristic impedance of the medium (r0 c), and therefore the generated
Figure 2.6 The acoustic pressure and intensity in an infinite duct. Note that the pressure and velocity are in phase with each other. Also, the active intensity hpuiavg (or average intensity with respect to time) is constant
17
The pressure and velocity of both the infinite as well as the finite duct satisfy the governing equation and the linearized Euler equation. The waves in the finite duct also meet the boundary condition at x ¼ L.
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Sound Propagation
Figure 2.7 The acoustic pressure and intensity in a duct of finite length of L. Note that the phase difference between the pressure and velocity is 90 (p=2)
waves only propagate in the right-going direction. The fact that the waves all propagate to the right implies that only active intensity is allowed. Therefore, the excitation effectively supplies energy to the system, which means that the pressure and velocity are in phase. Conversely, if we have the same excitation at one end, the duct has a finite length of L, and a rigid boundary condition exists at the other end x ¼ L, then the phase difference between the pressure and velocity will be 90 (p=2) as depicted in Figure 2.7. Therefore, it is not possible to effectively put energy into the system (Section 1.9.1). The instantaneous acoustic intensity in this case is completely different from what we have for the infinite duct case. The mean intensity is zero, independent of the position of the duct. Also, we see that the intensity is always zero at the nodal point of the duct x ¼ ðn þ 1ÞL=4, but it oscillates between these points where the energy vibrates and does not propagate anywhere. This is a typical characteristic of a finite duct. We can envisage that a more general case will exhibit behavior which is between those two extreme cases depicted in Figures 2.6 and 2.7. In general, the intensity may be both active and reactive. We have studied fundamental physics and its implications by examining two typical but extreme cases. We now need to explore more specific characteristics of the sound intensity, such as how to calculate and measure the intensity.
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The mathematical definition of intensity can be written as ~ I ¼ p~ u:
ð2:37Þ
The one-dimensional expression is simply I ¼ pu: The velocity can be obtained from the Euler equation (2.12): ð 1 @p dt: u¼ r0 @x
ð2:38Þ
ð2:39Þ
To obtain the derivative with respect to space, we may use two microphones. This means that we approximate the derivative as @p p1 p2 : Dx @x
ð2:40Þ
This equation allows us to measure the rate of change of pressure in space. The pressure (p) at the position of the measurement can be approximately obtained as: p
p1 þ p 2 2
ð2:41Þ
where the pressure fluctuates in time and is therefore a dynamic quantity. This means that the microphones that are used to implement Equations 2.40 and 2.41 have to be dynamically identical. In particular, the phase difference has to be treated carefully; this requires accurate calibration for magnitude and phase. We now look at the intensity measurement and calculation by considering a plane wave with a radian frequency o. The pressure can then be written as pðx; tÞ ¼ PðxÞejðot þ fp ðxÞÞ
ð2:42Þ
where PðxÞ denotes the pressure magnitude which has a real value and fp ðxÞ represents the possible phase change in space. This is a more general expression than what we had before, and is devised specially to look at the physics associated with the phase. To obtain the velocity using the linearized Euler equation a pressure gradient is needed, that is dfp @p dP ¼ j P ejðot þ fp Þ : @x dx dx
ð2:43Þ
Equations 2.39 and 2.43 then allow us to obtain the particle velocity: dfp 1 dP jðot þ fp Þ u¼ þj P : e or0 dx dx
ð2:44Þ
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The first term in Equation 2.44 has the same phase as that of the pressure of Equation 2.42; however, the second has a 90 phase difference. These two different velocities obviously result in very physically distinct intensities. The intensity that is generated by the real part of pressure (2.42) and the corresponding velocity (2.44), which is in phase with the real part of the pressure, can be obtained as P dfp Ia ðx; tÞ ¼ Pcosðot þ fp Þ cosðot þ fp Þ or0 dx ð2:45Þ df 1 p cos2 ðot þ fp Þ: ¼ P2 or0 dx This is normally referred to as the “active component of sound intensity”. The time average of this intensity is often referred to as a mean intensity, or an active intensity, and can be written: dfp 1 P2 : ð2:46Þ Iavg ðxÞ ¼ dx 2or0 This intensity can effectively supply power to space, because the velocity and pressure have the same phase. On the other hand, the multiplication of the real part of the pressure and the imaginary part of the velocity that has 90 phase difference (with respect to the real part of the pressure (2.42)) will generate the following intensity: 1 dP sin ðot þ fp Þ Ire ðx; tÞ ¼ Pcos ðot þ fp Þ or0 dx ð2:47Þ 1 dP2 ¼ sin 2ðot þ fp Þ: 4or0 dx We refer to this intensity as the “reactive component of sound intensity”. The time average of this intensity is 0 and, therefore, there is no net energy transport; it only oscillates. We now look at the direction of intensity. From Equations 2.45 and 2.47, we see that the active intensity is proportional to the change of phase with space and the reactive intensity is linearly related to the rate of change of pressure squared with space. From these observations, we can conclude that the direction of the active intensity is perpendicular to the wave front where the phase is constant. In addition, the direction of reactive intensity has to be perpendicular to the surface over which the mean square pressure is constant.18 In order to clearly distinguish the intensities at a specific time from the mean intensity (2.46), Equations 2.45 and 2.47, referred to as the instantaneous active intensity and instantaneous reactive intensity, respectively, are employed. Normally, when we say active intensity, we are 18
Depending on the type of exponential function (exp(–jot) or exp(jot)), the intensity vectors direction may change. Note that the direction of reactive intensity does not matter, but the active intensity direction is in the opposite direction.
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85
referring to a time average of the instantaneous active intensity, that is, Equation 2.46. For the reactive intensity case, we describe its amplitude Ir ðxÞ ¼
1 dP2 4or0 dx
ð2:48Þ
as reactive. Note that the instantaneous intensity expressed by Equation 2.38 is composed of two components: the instantaneous active intensity (2.45) and instantaneous reactive intensity (2.47). We can therefore write them as: Iðx; tÞ ¼ Iavg ðxÞ½1 þ cos2ðot þ fp Þ þ Ir ðxÞsin 2ðot þ fp Þ:
ð2:49Þ
Using a complex function, Equation 2.49 can be expressed in simpler form, that is, h n io Iðx; tÞ ¼ Re CðxÞ 1 þ e2jðot þ fp Þ
ð2:50Þ
where CðxÞ ¼ Iavg ðxÞ þ jIr ðxÞ. This is often referred to as complex intensity. The real part of the complex intensity is the active intensity 2.46 and its imaginary part is the amplitude of the reactive intensity (2.48). (Section 2.8.3 provides details associated with the derivation of intensities.) To summarize, the acoustic energy and intensity represent how acoustic waves propagate in space and time. The acoustic energy is composed of two components: acoustic kinetic energy and acoustic potential energy. The former is due to the motion of the fluid and the latter is induced by the compression and expansion of the medium. The rate of propagation of these energies with regard to space and time is known as the intensity. In other words, the intensity is the energy per unit area in time and space. The intensity is the product of pressure and velocity; it therefore has both magnitude and direction. Because it is a product of two physical quantities, the phase relationship is an important characteristic. Note that the impedance is also composed of pressure and velocity but is a division of two physical quantities.
2.4
The Units of Sound
We have studied how the physical quantities, such as pressure, velocity, intensity, energy, and impedance, are related to the propagation characteristics of acoustic waves. We have also seen that a one-dimensional acoustic system such as a duct can express the associated fundamentals. In considering these physical quantities, it is necessary to know how to communicate them. The units of the physical quantities are the basis of such means. In fact, units are a currency that form the basis of exchanging these physical quantities. Note also that sound exists before the governing equation was formulated; the measures and units therefore form the essential tenets of acoustics. The units that are relevant to sound can be classified into two groups: absolute units and subjective units. The former express the acoustic expression of physical dimensions while the latter take into account the subjective recognition of sound.
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Let us first start with the absolute units. The force per unit area, which is pressure, has the units: the unit of pressure ¼ Pascal ðPaÞ ¼ N=m2 : Velocity, which is defined as the time rate change of displacement, has units: the unit of velocity ¼ m=sec: Intensity, which is the power per unit area, has units: the unit of intensity ¼ Pa m=sec ¼ watt=m2 : Finally, the unit of energy is defined: the unit of energy ¼ joule ¼ watt sec: To understand subjective units, we need to understand how we hear which means that we need to study our hearing system. Figure 2.8(a) depicts the human hearing system. Sound arrives at the ear and vibrates the ear drum (Figure 2.8(a), external ear, Figure 2.8(b)). This vibration is amplified by three ossicles before arriving at the cochlea (middle ear, Figure 2.8(b)). The cochlea has a membrane referred to as the basilar membrane in its middle region, which is filled with fluid (Figure 2.8(c)). Hair cells are planted on the membrane (Figure 2.8(d)) and the motion of these cells generates signals that are transported to the brain. The brain processes the signals, feeding them to our cognition system. We recognize the sound and the information that it carries.19 An interesting characteristic of the hair cells is that they behave like a spatially distributed frequency band-pass filter, as illustrated in Figure 2.8(e) and (f).20 It is well known that humans do not hear the frequency of sound in absolute scale, but rather relatively.21 Due to this reason, we normally use relative units for the frequencies. The octave band is a typical relative scale (Figure 2.9). The band between the reference frequency (f1 ) and the frequency that is twice that (f2 ) is referred to as an “octave”. The 1/3 octave band, which is popular in noise control engineering, has a frequency of 21/3 f1 ; the reference frequency has the frequency of upper band f2 . The center frequency (f0 ) of each band is at the geometrical center of the band.
19 The acoustic signal is expected to be effectively transmitted to the fluid of the cochlea. As noted earlier, there is an unavoidable impedance mismatch when sound propagates from air to fluid. To achieve best possible impedance matching our ear attempts, in various ways, to somehow release impedance mismatch. One source is the area ratio between the eardrum size and that of the oval window (about 17:1). An additional amplification is due to the lever effect of the middle ear ossicle. Other amplification factors vary with the frequency. For example, the ear canal has maximum amplification of 2.5 kHz. The shoulder, head, and pinna also contribute the amplification. The total of these amplifications is around 30 dB. 20 Due to the physical arrangement of the hair cell, if a single part is defect then hearing loss corresponding to that frequency band occurs. Because of the characteristics of the hair cell, we often regard it as a spatially distributed spectrum analyzer. 21 The cognition of frequency or pitch varies with respect to individuals.
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Figure 2.8 The structure of the ear and its frequency band characteristics. (a) The structure of a human ear. (Redrawn with permission from D. Purves et al., Neuroscience, 3rd edition, 2004, pp. 288 (Figure 12.3), Sinauer Associates, Inc., Massachusetts, USA. 2004 Sinauer Associates, Inc.) (b) External, middle, and inner ear. (c) Basilar membrane and Corti organ. (d) The cross-section of the cochlea shows the sensory cells (located in the organ of Corti) surrounded by the cochlear fluids. (e) Space-frequency map: moving along the cochlea, different locations are preferentially excited by different input acoustic frequencies. (f) Tonotopic organization. (Figure 2.8(b–f) drawings by Stephan Blatrix, from “Promenade around the cochlea,” EDU website: http://www.cochlea.org by Re´my Pujol et al., INSERM and University Montpellier.)
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Figure 2.8
(Continued)
Acoustic Wave Equation and Its Basic Physical Measures
Figure 2.8
89
(Continued)
According to this definition, the center frequency can be any frequency. However, for convenience, we use the standard center frequencies (Table 2.3). The frequency bandwidth of the octave band is about 70% and that of the 1/3 octave band is about 23% of the corresponding band (see Figure 2.9). Therefore, the higher the center frequency, the wider the frequency band. As we can see in the standard frequency band, humans can hear sound at a broad range of frequencies. The audible frequency range is between 20 Hz and 20 kHz. There are two important reasons why we scale or measure the frequency of our interest by an octave or 1/3 of an octave: (i) we hear frequency in a relative scale and (ii) we cannot express our audible frequency range by using a linear frequency scale.
Octave scale
f 0 = 21/2 f1 f 2 = 2f1
f1
f0
f 2 = 2f
1
1/3 octave scale f1
f0
f 0 = 21/6 f1 1 3
f 2 = 2 f1
Figure 2.9
Δ f = f 2 − f1 = 0.23 f 0
f 0 = 2 2n f 1 1
f0
f 2 = 21/3 f1
1
1/n octave scale f1
Δf = f 2 − f 1 = 2 −1/2 f 0 = 0.7f
f 2 = 2 n f1
1 n
f 2 = 2 f1
Octave, 1/3 octave, and 1/n octave
0
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Table 2.3
The center frequency of octave and 1/3 octave Band limits
Octave band center frequency 31.5
63
125
250
500
1000
2000
4000
8000
16 000
One-third octave band, center frequency 25 31.5 40 50 63 80 100 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000 6300 8000 10 000 12 500 16 000 20 000
Lower 22 28 35 44 57 71 88 113 141 176 225 283 353 440 565 707 880 1130 1414 1760 2250 2825 3530 4400 5650 7070 8800 11 300 14 140 17 600
Upper 28 35 44 57 71 88 113 141 176 225 283 353 440 565 707 880 1130 1414 1760 2250 2825 3530 4400 5650 7070 8800 11 300 14 140 17 600 22 500
For the amplitude of the sound pressure, we use the sound pressure level (SPL or Lp ). It is defined as ! p2avg ð2:51Þ SPL ¼ Lp ¼ 10 log10 2 pref and is measured in units of decibels (dB); pref is the reference pressure, pavg is the average pressure, and log10 is a log function that has a base of 10. pref is 20 mPa (20 106 N/m2). The reference pressure is the smallest sound that a human being can hear. The first reason why we use a log scale and not use a linear scale is that we can hear sound in a range from small amplitude to extremely large amplitude (Figure 2.10). From Figure 2.10, we can see that the human can hear from about 0 dB to somewhere in the range of 130–140 dB.
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Figure 2.10 Equal-loudness contour: each line shows the SPL with respect to the frequency which corresponds to a loudness (phon) of 1 kHz pure sound. (Reproduced from ISO 226 (2003), “Normal equalloudness-level contours,” International Standards Organization, Geneva.)
The second reason is that we recognize the level of sound relatively and not in an absolute scale, similar to frequency recognition. Equation 2.51 essentially expresses how large the sound is compared to the reference sound pressure level, and is therefore a good representation of the human hearing system. Table 2.4 collects some typical samples of sound levels that we can encounter, providing some practical references of the sound pressure level (SPL). In order to calculate SPL, we write ð 1 T 2 p ðtÞdt ð2:52Þ p2avg ¼ T 0 where T denotes the measurement time. Equation 2.52 expresses pðtÞ as the sum of every frequency component, equivalent to ð 1 T XX 2 pavg ¼ ð2:53Þ Re Pm ejom t Re Pn ejon t dt: T 0 n m
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Daily life noise level in SPL
Table 2.4
SPL (dB ref. 20 mPa)
Example
120 110 100 90 80 70 60 50 40 30 20 10 0
When a jet takes off (60 m) Construction site Loud shouting (1.5 m) Truck noise (15 m) Busy street Moving car interior Normal conversation (1 m) Office Living room Bedroom (night) Broadcasting studio Leaves oscillating by a breeze Minimum audible sound pressure level
(Rossing, Thomas D.; The Science of Sound, 2nd edition. Copyright 1990, pg. 86. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, New Jersey.)
We then use the well-known relation 1 Re Pejot ¼ ðPejot þ P* e jot Þ 2
ð2:54Þ
where denotes the complex conjugate. If we rearrange Equation 2.53 using Equation 2.54, then we obtain
p2avg ¼
1 XX 4T n m
ðT h 0
Pm Pn ejðom þ on Þt þ Pm P*n ejðom on Þt
i þ P*m Pn e jðom on Þt þ P*m P*n ejðom þ on Þt dt ¼
1 XX 4T n m
1 X ffi 4T m ¼
X1 m
2
ðT
ðT h n o n oi 2Re Pm Pn ejðom þ on Þt þ 2Re Pm P*n ejðom on Þt dt 0
n o 2Re jPm j2 dt
0
jPm j2 :
ð2:55Þ
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93
Equation 2.55 states that the mean square average of sound pressure is the sum of the mean square average of each frequency’s sound pressure.22 This is only valid if and only if Equation 2.55 has a maximum when n ¼ m. Note that when n 6¼ m, the slowly fluctuating terms with frequency ðom on Þ are much greater than those with a frequency of ðom þ on Þ. Figure 2.11 illustrates the relation between the sound pressure level and the mean square pressure.
P1
p
P2
B
Fourier
0.00
0.01
p2avg
| P2 |
0.0014
0.02
0.03
0.04
Pm
2
Linear
2
SPL
| P1 |
0.0008
2
2
0.0006
SPL (dB ref. 20μPa)
0.0012 0.0010
| P3 |
2
2
0.0004 0.0002
ω1 ω2
0.0000 0
P3
transform
t
70 60
ω1
dB
ω2
ω3
50 40 30 20 10
ω3
500 1000 1500 2000 2500 3000 3500 4000 Frequency (Hz)
ω
0 500 1000 1500 2000 2500 3000 3500 4000
ω
Frequency (Hz)
SPL (dB ref. 20μPa)
SPL
70
ω1
60
ω2
B
ω3
50 40 30 20 10 0 63
Figure 2.11
125 250 500 1000 2000 4000 8000 1/3 octave center frequency (Hz)
Total mean square pressure and the mean square pressure of each frequency band
Equations 2.51 and 2.55 indicate that it is important to carefully sum the sound levels or the mean square values of different sounds. Let us begin with two sound pressures that have different frequencies, f1 and f2 . According to Equation 2.51, the sound pressure level of each individual tone can then be written as p21;avg tone 1 : SPL ¼ Lp1 ¼ 10 log10 2 ð2:56Þ pref tone 2 : SPL ¼ Lp2 ¼ 10 log10 22
p22;avg p2ref
:
ð2:57Þ
This indicates that mean square is basically nonlinear but, for this case, the total mean square value is the sum of the mean square value of each individual frequency band and is therefore linear.
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If these two tones occur at the same time, the SPL can be written SPL1 þ 2 ¼ Lp1 þ 2 ¼ 10 log10
p21; avg þ p22; avg
¼ 10 log10 ð10
p2ref Lp1 =10
ð2:58Þ þ 10
Lp2 =10
Þ:
For example, if each tone has an SPL of 80 dB, that is, Lp1 ¼ 80 dB and Lp2 ¼ 80 dB, then the sum of these two must be Lp1 þ 2 ¼ 10 log10 ð108 þ 108 Þ. This simply means that the SPL increases by 3 dB.23 This method is valid if we add any sound pressure that has the same level. It also implies that the biggest sound level dominates the total SPL when there are many sounds together. For example, if we have two sounds of SPL 75 dB and 80 dB, the resulting SPL of the sounds is 81.2 dB. If we generalize this result to N different pure tone cases, the SPL is defined
SPL1 þ 2 þ þ N ¼ Lp1 þ 2 þ þ N ¼ 10 log10 10Lp1 =10 þ þ 10Lp2 =10 : ð2:59Þ The SPL can be obtained from Equation 2.59, but it is not yet clear how humans perceive sound. Awell-known fact is that we acknowledge sound depending on its frequency content. As illustrated in Figure 2.10, we interpret equal loudness for sounds with different sound pressures. A primary conclusion based on these characteristics of hearing perception is that there is a scaling factor with respect to frequency. This motivates us to introduce a weighting curve to the SPL (Figure 2.12). The A-weighting is the most widely used weighting curve.
0
Level weighting (dB)
-5
C
-10 -15
B
-20 -25 -30 -35 -40
A
-45 -50 31.5 63 125 250 500 1k 2k 4k 8k 16k One-third octave band center frequency (Hz)
Figure 2.12 Various weighting curves. A-weighting: 40 phon curve (SPL < 55 dB); B-weighting: 70 phon curve (SPL ¼ 55–85 dB); C-weighting: 100 phon curve (SPL > 85 dB)
23
The minimum difference in SPL which we can hear is 2 dB.
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The weighting curve is based on our frequency-dependent sensation. The human ear is normally very sensitive to a frequency of 1–3 kHz in the case of a pure tone, and less sensitive to frequencies lower than this range (Figure 2.10). Recall that the weighting curves are all based on pure tones. For more general sound or noise, the use of these curves is somewhat different from reality. It is therefore reasonable to use the curves as a guideline to considering human perception with regard to frequency. To summarize, the most commonly used unit to measure the amplitude of sound is sound pressure level (SPL, Lp ) which is defined as the ratio of the mean square pressure to the mean square reference pressure. The acoustic intensity expresses the power transmitted through the unit surface, and its reference intensity (I0 ) is defined as 1012 W=m2 ¼ 1012 N=ðm sÞ. Table 2.5 summarizes the measurement units and its criteria. SPL is the sum of the mean square pressure of individual frequency components of sound. We also see that weighting is necessary to consider the frequency perception of sound. The basic frequency scale has to be an octave or 1/3 octave to take into account human perception. For the same reason, a logarithmic scale is also necessary to measure the SPL.
Table 2.5 Measurement standards Physical measure
Definition
Measurement reference (SI)
ANSI SI.8-1989. (1989) American National Standard Reference Quantities for Acoustical Levels, American National Standards Institute, New York. Sound pressure level, dB (gases) Lp ¼ 20log10(P/P0) P0 ¼ 20 mPa ¼ 2 10-5 N/m2 Sound pressure level, dB (other Lp ¼ 20log10(P/P0) P0 ¼ 1 mPa ¼ 106 N/m2 than gases) Sound power level, dB LW ¼ 10log10(W/W0) W0 ¼ 1pW ¼ 1012 N m/s W0 ¼ 1pW ¼ 1012 N m/s Sound power level, bel LW ¼ log10(W/W0) bel Sound intensity level, dB LI ¼ 10 log10(I/I0) I0 ¼ 1pW/m2 ¼ 1012 N/m s F0 ¼ 1 mN ¼ 106 N Vibratory force level, dB LFy ¼ 20 log10(F/F0) Frequency level, dB N ¼ log10(f/f0) f0 ¼ 1 Hz E0 ¼ (20 mPa)2 s ¼ (2 105 Pa)2 s Sound exposure level, dB LE ¼ 10 log10(E/E0) According to ISO or suggested by the author Sound energy level given in ISO Le ¼ 10 log10(e/e0) 1683;1983 Sound energy density level given LD ¼ 10 log10(D/D0) in ISO 1683;1983 Vibration acceleration level La ¼ 20 log10(a/a0) Vibration acceleration level in La ¼ 20 log10(a/a0) ISO 1683;1983 Vibration velocity level Ln ¼ 20 log10(n/n0) Vibration velocity level in ISO Ln ¼ 20 log10(n/n0) 1683; 1983 Vibration displacement level Ld ¼ 20 log10(d/d0)
e0 ¼ 1pJ ¼ 1012 N m D0 ¼ 1pJ/m3 ¼ 1012 N/m3 a0 ¼ 10 mm/s2 ¼ 105 m/s2 a0 ¼ 1 mm/s2 ¼ 106 m/s2 n0 ¼ 10 nm/s ¼ 108 m/s n0 ¼ 1 nm/s ¼ 109 m/s d0 ¼ 10 pm ¼ 1011 m
(Reprinted from L.L. Beranek and I.L. Ver (eds.), Noise and Vibration Control Engineering: Principles and Applications, 2nd edition, 1992, with the permission of John Wiley & Sons, Inc., New York.)
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2.5
Analysis Methods of Linear Acoustic Wave Equation
We have studied the physical variables – pressure, density, and fluid particle velocity – which determine acoustic wave propagation as well as their relations (Section 2.2). The physical implications of intensity and energy on acoustic waves have been also studied in detail (Section 2.3). We have subsequently introduced how these physical quantities, especially sound pressure and intensity, excite our hearing system. What we hear is the output of our hearing system from the input of physical acoustic variables, sound pressure and intensity. The major attributes of our hearing system are how we hear frequency and how we acknowledge sound pressure level. The former is described by the octave band frequency unit while the latter is expressed in units of dB (Section 2.4). This section addresses how we mathematically predict or describe sound in space and time. We do not want to imply that a mathematical approach is the best way to understand acoustic wave propagation, but want to demonstrate that a mathematical approach provides us with a logical way to understand the characteristics of sound propagation. A mathematical approach does not always mean that we will eventually arrive at an exact solution, but is a guide to understanding the fundamentals of all forms of acoustic wave propagation. The acoustic wave equation that we learned in Section 2.2 represents the characteristics of acoustic wave propagation. However, it only expresses that which is related to the medium. It can be applied to every possible wave that satisfies the governing equation. For example, for waves that propagate in a one-dimensional duct, Equations 2.1 and 2.3 satisfy the governing equation. The next point to consider is what makes the solution unique: the answer is the boundary and initial conditions. We conceive a method to express the sound source. In fact, the governing equation previously studied satisfies waves in time and space without regard to acoustic sources, that is, excitations. The equation is, therefore, a homogeneous governing equation. The inhomogeneous terms that express acoustic sources or excitations include anything that can make acoustic waves; however, for convenience, we consider only pressure and velocity sources. The former creates waves by changing pressure; the latter is related to volume change. Let us begin with a situation where we have a volume source in one-dimensional infinite space. The volume velocity source makes the mass change by the velocity excitation. We can express this mathematically by modifying Equations 2.13 and 2.14, that is @r @u ¼ r0 þ r0 q; @t @x
ð2:60Þ
where q ¼ qðx; tÞ is the volume velocity at x. Therefore, the last term of the right-hand side of Equation 2.60 expresses the time rate of mass supply to the fluid. In summary, Equation 2.60 states that the time rate of mass change per unit volume is balanced by the net supply of mass and net mass flux through the surface of the fluid. Substituting Equations 2.18 and 2.12 into this new mass law equation 2.60, we obtain the governing equation that includes the acoustic source: @2p 1 @2p @q ¼ r0 : @x2 c2 @t2 @t
ð2:61Þ
The right-hand side of Equation 2.61 is the multiplication of fluid density and volume acceleration, and therefore expresses the inertia force of the source.
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97
We first attempt a harmonic solution, as follows: pðx; tÞ ¼ PðxÞejot :
ð2:62Þ
Equation 2.61 can then be written as d2P þ k2 P ¼ fðxÞ; dx2
ð2:63Þ
where fðxÞ represents the right-hand side of Equation 2.61 in an abbreviated form. In fact, we assume that it can be expressed by f ðxÞejot. Equation 2.61 is a typical inhomogeneous governing equation. Note that Equation 2.63 is strictly only valid where the sound source exists; otherwise a homogeneous equation is valid. For example, if there is a point source at x0 , then Equation 2.63 can be rewritten as d 2P þ k2 P ¼ fðx0 Þdðxx0 Þ; dx2 where dðxx0 Þ is a Dirac delta function, that is, ð1 dðxx0 Þ ¼ 1;
ð2:65Þ
1
dðxx0 Þ ¼ 0;
ð2:64Þ
x 6¼ x0 :
If the source exists only in the region L0 , then we can write the governing equation as ð d 2P 2 þ k P ¼ fðx0 Þdðxx0 Þdx0 : ð2:66Þ dx2 L0 Expanding this equation to a three-dimensional case yields ð 2 2 r 0 Þdð~ r~ r 0 ÞdVð~ r 0 Þ; r P þ k P ¼ fð~
ð2:67Þ
V0
where~ r 0 and V0 express the source position and the volume where the source is, respectively. To summarize, if there is a sound source or sources, then the mechanism that we investigated in Section 2.2 is no longer valid at the source position and the governing equation becomes Equation 2.67, that is, the inhomogeneous wave equation.24 We now look at how to mathematically express the boundary condition. If a space is closed by a boundary, then the solutions have to satisfy not only the wave equation but also the boundary condition. We first study the one-dimensional case. As already expressed (1.26), the boundary condition can generally be shown as a0;L P þ b0;L U ¼ g0;L ;
x ¼ 0; L;
ð2:68Þ
where the subscript 0 and L represent boundary value at x ¼ 0; L. To understand the boundary conditions that are expressed by Equation 2.68, let us investigate several typical cases. First, 24
It is possible to express any sound source by pressure source, volume source, or their combination.
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when b ¼ 0, the condition takes the form g P¼ : a
ð2:69Þ
This type of boundary condition, which describes the pressure on the boundary, is generally known as the Dirichlet boundary condition. On the other hand, if a ¼ 0, then the equation becomes g U¼ : b
ð2:70Þ
Equation 2.70 essentially describes the situation where the boundary moves with the prescribed velocity. If we recollect that the velocity is related to the change of pressure with regard to space (according to Euler’s equation), then Equation 2.70 describes the boundary condition in terms of pressure derivatives on the boundary. This type of boundary condition is called a Neumann boundary condition. If g ¼ 0, then Equation 2.68 reduces to a0;L P þ b 0;L U ¼ 0
ð2:71Þ
P b ¼ : U a
ð2:72Þ
and
The impedance, which is the ratio between pressure and velocity, is described on the boundary. We refer to the condition where the right-hand side is 0 (as in Equation 2.71) as a homogeneous boundary condition. If the equation follows the form of Equation 2.69 or Equation 2.70, then we refer to it as an inhomogeneous boundary condition. The latter implies that there is an active element on the boundary while the former means the boundary behaves rather passively. More generally, the three-dimensional case of Equation 2.71 can be written as aP þ bU ¼ 0;
on
S0 ;
ð2:73Þ
where S0 is the surface that encloses the space of interest, as depicted in Figure 2.13 where U is the particle velocity that is normal to the surface. Equation 2.68 is the more generally acceptable form in practice, but it must be noted that we can always change the inhomogeneous boundary condition in terms of the homogeneous boundary condition and the inhomogeneous governing equation.25 Therefore, we will only consider the problem that is governed by the inhomogeneous governing equation and homogeneous boundary condition. That is, ð d 2P 2 þ k P ¼ fðx0 Þdðxx0 Þdx0 ð2:66Þ dx2 L0 25
Meirovitch, L. (1967) Analytical Methods in Vibrations. Macmillan Publishing, New York, pp. 301–305.
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99
αP+βU = γ
S0
V0
∇2P + k2P = 0
r
r0 0
Figure 2.13 General boundary value problem (P is complex amplitude, U is complex velocity, k is the wave number, and S0 expresses the boundary; ~ r and ~ r 0 indicate the observation position and boundary, respectively)
and a0;L P þ b 0;L U ¼ 0:
ð2:71Þ
One very well-known method for obtaining the solutions which satisfy Equations 2.66 and 2.71 uses eigenfunctions to express the solution. This means, specifically, that we first try to find the function Cn which satisfies d 2 Cn þ k2n Cn ¼ 0 dx2
ð2:74Þ
and also satisfies the boundary condition of Equation 2.71; that is a0;L Cn þ
b0;L dCn ¼ 0: jr0 o dx
ð2:75Þ
The function which satisfies Equations 2.74 and 2.75 is the eigenfunction or eigenmode, and the constant kn is the eigenvalue. To shed more light on this problem, we consider the special case when a ¼ 0. In this case, we have a rigid-wall boundary condition and the eigenmode can be found, intuitively, as: Cn ðxÞ ¼ cos
np x: L
ð2:76Þ
If b ¼ 0, which is the case for the pressure release boundary condition, then the solution has to take the form Cn ðxÞ ¼ sin
np x: L
ð2:77Þ
We generally call this method, which attempts to obtain the solution by superimposing the eigenfunctions, a modal analysis. The advantage of this method is that a linear combination of the eigenmodes also satisfies the given boundary condition. For example, for the
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one-dimensional case studied previously, the pressure can be written as PðxÞ ¼
1 X
an Cn ðxÞ:
ð2:78Þ
n¼0
It is obvious that Equation 2.78 always satisfies the boundary condition of Equation 2.75. The main obstacle in finding the solution by a linear combination of eigenfunctions is finding each mode’s contribution, or weighting, on the solution. In other words, we must attempt to find an (2.78) that satisfies Equation 2.66. For example, if we have one source at a point where Equation 2.64 is the governing equation, then we can attempt to construct the solution as given by Equation 2.78. The coefficients can be found by using the property of the eigenfunctions (orthogonality condition), that is, ð 1 L Cm ðxÞC*n ðxÞdx ¼ Lm dmn ð2:79Þ L 0 and
( dmn ¼ 1 L
ðL
1
if
m¼n
0
if
m 6¼ n
;
jCn ðxÞj2 dx ¼ Ln :
ð2:80Þ
0
Using Equations 2.66, 2.74, 2.78, 2.79 and 2.80, we can obtain the weighting as an ¼
1 Ln Lðk2 k2n Þ
ðL 0
fðxÞ C*n ðxÞ dx;
ð2:81Þ
where * denotes the complex conjugate. This approach entails first finding all possible solutions that satisfy the homogeneous wave equation. We then try to construct the solution for an inhomogeneous wave equation, which has a source excitation on the right-hand side of the equation, by a linear combination of the eigenfunctions. This is valid because the equation is linear and therefore the principle of superposition holds. In other words, it is akin to spreading a net over a body of water surrounded by a boundary and catching the fish therein, that is, the coefficients. Alternatively, we can try to obtain the solution that satisfies the boundary condition by introducing Green’s function. As noted earlier, the sound field induced by the source fðxÞ can be expressed by Equation 2.63. If we denote the sound pressure due to a unit point source at x ¼ x0 as Gðxjx0 Þ, then G has to satisfy the equation:26 d 2G þ k2 G ¼ dðxx0 Þ: dx2
26
The solution that satisfies Equation 2.82 is Green’s function.
ð2:82Þ
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101
Multiplying G by Equation 2.63 and P by Equation 2.82, subtracting the former from the latter and finally integrating with respect to x lead us to ðL 2 ðL d G d 2P P 2 G 2 dx ¼ ðG fðxÞPdðxx0 ÞÞ dx: ð2:83Þ dx dx 0 0 Then, integration by parts yields: L L ð L dG dP P G ¼ G fðxÞdxPðx0 Þ: dx 0 dx 0 0 Changing the variable x for x0 reduces Equation 2.84 to the form27 0 0 ðL dG dP PðxÞ ¼ G fðxÞdx0 þ P G : dx0 L dx0 L 0
ð2:84Þ
ð2:85Þ
We now investigate how to apply Equation 2.85 when we have a unit amplitude sound source at x ¼ x0 , as illustrated in Figure 2.14. This specific case reduces Equation 2.85 to dG 0 dP 0 PðxÞ ¼ Gðxjx0 Þ þ P G : ð2:86Þ dx0 L dx0 L x = x0 e− jωt
Z0
ZL x=L
x 0 a. 1 dimensional case
V0 S0 r r0 0 b. 3 dimensional case (V: volume, S0: surface which surrounds the V)
Figure 2.14
One-dimensional and three-dimensional boundary value problems
The integral sequence is reversed in [0, L] to make the sign of P0 @G @x positive for consistency with the 3-D case. In this case, the direction of the surface normal vector ~ n is inwards. 27
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If the velocity at x ¼ 0; L is 0 (rigid-wall boundary condition), or the pressure is 0 (pressure release boundary condition), then Equation 2.86 becomes 0 dG ð2:87Þ PðxÞ ¼ Gðxjx0 Þ þ P dx0 L or PðxÞ ¼ Gðxjx0 ÞG
dP dx0
0 ;
ð2:88Þ
L
respectively. In other words, the pressure at x is the sum of two pressure components. One is due to the sound source of unit magnitude with radian frequency o at x0 , and the reflected pressure due to the boundary. Note also that the pressure at the right-hand side is induced by the pressure at the boundary. Equation 2.86 essentially states that the sound pressure in the space of interest is determined by the pressure at the boundary.28 Interestingly, Green’s function G can be anything that satisfies Equation 2.82. That is, Equation 2.86 is always valid for any function that does satisfy Equation 2.82 (see further details in Section 4.7.4). We can therefore select G to make Equation 2.86 simpler; that is, choosing G that satisfies the Dirichlet boundary condition or Neuman boundary condition. Equation 2.86 states that the sound pressure at x consists of two components: one is a direct effect from the sound source and the other is due to the reflection from the boundary. In comparison to the modal analysis method (discussed earlier in this section), this approach is akin to catching fish from the boundary using a fishing rod, that is, Green’s function. Equation 2.86 is, in fact, an integral form and we have to assign or know Green’s function in advance. Expanding Equation 2.86 to a three-dimensional form yields the integral equation ð ð Pð~ rÞ ¼ G fð~ r 0 ÞdV þ ðPr0 G Gr0 PÞ ~ n 0 dS: ð2:89Þ V0
S0
If we do not have the sound source fð~ r 0 Þ in the integral volume V0 (Figure 2.14(b)), then Equation 2.89 becomes ð ð2:90Þ Pð~ rÞ ¼ ðPr0 G Gr0 PÞ dS: S0
(See Section 4.7.4 for the detailed derivations.) Equations 2.89 and 2.90 are referred to as Kirchhoff–Helmholtz integral equations. As noted with respect to the one-dimensional case, Green’s function G can be anything that satisfies Equation 2.82; we can use any type of fishing rod as long as it is a fishing rod. Note that Equation 2.90 is essentially the basis of acoustic holography and the boundary element method, which are quite popular in acoustics. The concept of acoustic holography is to start by constructing the integral surface (S0 ) by using any orthogonal coordinate that conveniently expresses what we wish to express. The pressure on the surface is then measured 28
This type of equation is an integral equation for the one-dimensional case.
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103
at discrete points. The last step is required to predict the sound pressure at any point (~ r ¼ ðx; y; zÞ: for the rectangular coordinate) by using the predictor G, that is, Green’s function. Acoustic holography can therefore provide information that cannot be measured or is not otherwise available on any surface of interest. On the other hand, the boundary element method predicts acoustic information (e.g. sound pressure, particle velocity, intensity, etc.) at a position or on a surface. This information cannot be measured by constructing the integral surface as an arbitrary shape and dividing the surface into small elements. These two methods use the prediction property of Green’s function. To summarize, the mathematical methods that can analytically predict acoustic wave propagation in time and space may employ either eigenfunctions or Green’s function. The former attempts to describe waves that can be generated by a prescribed acoustic source or sources by seeking the contribution from each eigenfunction; here we are catching fish using an eigenfunction net. The latter attempts to describe the sound field by using Green’s function that transforms the information of sound pressure and velocity on the boundary to the point or field of interest.29 In fact, this approach distributes a function that satisfies the governing equation on the boundary. It can be a singular function or a Green’s function. We then try to find the contribution of each function which satisfies the boundary condition; we have many fishing rods on the boundary to catch the fish.
2.6
Solutions of the Wave Equation
We have seen that the solutions which satisfy the acoustic wave equation can be obtained using two different approaches: the modal analysis method or Green’s function method. The boundary condition and the source determine the solution that expresses the waves in the area of interest and time. However, this statement is too general to grasp the idea of how an acoustic wave would behave in space and time. To be more specific and realistic, we again have to start with the simplest case. We therefore study a representative acoustic wave that satisfies the governing equation. As attempted frequently, we will start with a one-dimensional, planar acoustic wave at position x and time t, pðx; tÞ. This can be written as pðx; tÞ ¼ PðxÞejot ¼ AejðotkxÞ :
ð2:91Þ
A wave in a certain direction in space can be expressed as ~
pð~ r; tÞ ¼ Pð~ rÞejot ¼ Aejðotk ~rÞ ;
ð2:92Þ
where A is a complex amplitude. Equation 2.91 obviously satisfies Equations 2.19 as Equation 2.92 satisfies Equation 2.24. The plane wave 2.92, as the name implies, essentially has all the same physical properties (i.e., pressure and velocity) at the plane perpendicular to ~ k at ~ r (Figure 2.15). Note that its impedance at any position and time is30 Zp ¼ r0 c:
ð2:93Þ
29 The ideal technique is to find superposition of the solutions that satisfy the boundary condition and governing equation. 30 This can be derived by obtaining velocity from the 1-D Euler equation then finding impedance.
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104 Planes of constant phase
k
r
p ( r , t ) = Ae − j ( ω t − k ⋅r )
0
Figure 2.15
A plane wave (~ k is normal to the planes of constant phase)
This is precisely the same as the characteristic impedance of the medium. This means that the plane wave in an unbounded fluid propagates in the wave number vector direction, independent of the position, frequency, wave number, and wavelength. Intensity which expresses the power through the unit area, specifically the average intensity (active intensity), can be expressed as follows (see details in Section 2.8.3): 1 1 Iavg ¼ Re PU* ¼ jPj2 ; 2 2r0 c
ð2:94Þ
where U is the velocity in the direction of propagation and U ¼ P=r0 c. Therefore, the intensity (2.94) can be written as 1 Iavg ¼ r0 cjUj2 : 2
ð2:95Þ
The governing equation that can express this type of wave can also be written in terms of the spherical coordinate. Let us therefore express Equation 2.24 in terms of the spherical coordinate. We assume that the pressure is independent of the polar and azimuth angles and only depends on the distance from the origin (r). Equation 2.24 then becomes @2 1 @2 ðrpÞ ¼ ðrpÞ: @r2 c2 @t2
ð2:96Þ
Its solution will be ~
rp ¼ Aejðotk ~rÞ ;
ð2:97Þ
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105
where A is a complex amplitude. Equation 2.97 can be rewritten as A jðot~k ~rÞ ð2:98Þ e r which tends to infinity at r ¼ 0 and is inversely proportional to the radius r. To assess the velocity, consider Euler’s equation in the spherical coordinate: pðr; tÞ ¼
@p @ur ¼ r0 ; @r @t
ð2:99Þ
where ur is the velocity in the radial direction. Equations 2.98 and 2.99 allow us to calculate the velocity in the radial direction, that is A 1 j ~ ejðotk ~rÞ : ur ¼ 1þ ð2:100Þ r r0 c kr Therefore, the impedance at r can be written as ( ) p ðkrÞ2 kr : Zr ¼ ¼ r0 c j ur 1 þ ðkrÞ2 1 þ ðkrÞ2
ð2:101Þ
Figure 2.16 illustrates the implication of the impedance in terms of the dimensionless distance kr. If the distance from the origin is large compared to the wavelength of interest ðkr 1Þ, then the impedance approaches the plane wave impedance. This means that the wave propagates like that seen on an infinite string. On the other hand, if the position is very close to the origin with respect to wavelength ðkr 1Þ, then the pressure and velocity has a 90 phase difference. This means that the fluid particle is accelerated by the acoustic pressure. Interestingly, kr ¼ 1 makes the real and imaginary part even. We can therefore argue that kr is a measure of how to share the real and imaginary parts. Intensity is the product of Equations 2.98 and 2.100. The active intensity, where the pressure and velocity have no phase difference, is inversely proportional to the square of the distance but independent of kr. The reactive intensity on the other hand, where the pressure and velocity have a 90 phase difference, is inversely proportional to the square of the distance and also inversely proportional to kr. Therefore, the reactive intensity tends to decrease rapidly as we move away from the origin. This means that, in the vicinity of the origin, the reactive intensity dominates the acoustic behavior in such a way that the waves do not propagate well. We call this region the “near field” and this effect the “near field effect”. The “far field”, therefore, indicates a position that is relatively far from the origin with respect to the wavelength of interest. Note also that the distance is measured in a relative scale with regard to wavelength kr.31 The monopole sound source is defined by Equations 2.98 and 2.100 and has a singularity at r ¼ 0. This simple solution satisfies the linear wave equation. This implies that superposition of this type of solution also satisfies the governing wave equation. We can therefore attempt to construct any type of wave by using the monopole. This concept is illustrated in Figure 2.17, a graphical expression of Huygen’s principle. If the two monopoles are close together with the
31
Note that the information measured at the far field does not contain the information of the near field. It is therefore not possible to predict the sound source based on what is measured in the far field.
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Sound Propagation
Figure 2.16 Monopole radiation. (a) The monopole’s radiation impedance where k is wave number, l indicates the Note is noteworthy that it behaves as a plane wave, as the observation position is far from the origin. (b) Pressure and particle velocity in near field (kr is small), magnitude (left) and phase (right) of pressure (top) and particle velocity (bottom); arrows indicate intensity. (c) As for (b) for far field case
opposite phase, then a dipole is formed. Having 4 monopoles but alternating opposite phases makes a quadrupole. Section 2.8.4 explains the various singular sound sources (Figure 2.18). If we use the monopole and dipole as the Green’s functions of Equation 2.90, then the integral equation describes how the resulting sound pressure at the position of interest is caused by monopole propagation of the velocity on the surface and dipole propagation of the surface pressure. To summarize, the plane wave and monopole sound source satisfy the governing wave equation and represent the simplest but most representative sound sources. The near field and far field are measured by the distance compared to the wavelength of interest. In the near field, the monopole source does not effectively drive the sound field, but in the far field, the monopole behaves as if it were a planar wave. The impedance of plane wave is exactly the same as the characteristic impedance of medium. The impedance of monopole depends on kr, which
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107
Figure 2.17 Huygen’s principle. The wave front constructed by many monopole sound sources: (a) graphical illustration and (b) shallow ripple tank
Figure 2.18 Dipole and quadrupole distributions and their characteristics where ðr; y; fÞ indicates an arbitrary point in spherical coordinate, k is wave number, D represents the dipole-moment amplitude vector, and Q represents the amplitude of quadrupole: (a) pressure of the spatial pattern of dipole sound; (b) impedence of a dipole at r; (c) magnitude (left) and phase (right) of particle velocity of a dipole in near field (top) and far field (bottom) (arrows indicate intensity); (d) pressure of a quadrupole pattern in space; (e) impedance of a quadrupole at r and (f) magnitude (left) and phase (right) of particle velocity of a quadrupole in near field (top) and far field (bottom) (Section 2.8.4)
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Figure 2.18
(Continued )
measures the distance from the origin with regard to the wavelength of interest. In the far field, the impedance approaches that of a plane wave. In the near field, the imaginary component is no longer negligible. Again, we found that impedance describes the underlying physics well. Note also that we can use monopole sources to describe the sound field: this is the foundation of the Kirchhoff–Helmholtz equation and Huygen’s principle: “Any complicated phenomena can be expressed by the superposition of simple and fundamental elements.”
Acoustic Wave Equation and Its Basic Physical Measures
Figure 2.18
(Continued )
109
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2.7
Chapter Summary
We have attempted to understand how acoustic waves are generated and propagated in a compressible fluid by examining what takes place in a duct or pipe. Generally, the behavior of three-dimensional acoustic waves can be envisaged from the observations and understanding gained from a one-dimensional acoustic wave. Newton’s second law describes how the forces acting on a compressible fluid create the motion of the fluid. Conservation of mass and the state equation of fluid, together with Newton’s law, provide three relations between mass per unit volume (density), fluid velocity, and force per unit area (pressure). We have also learned that acoustic intensity, which is power transmission through a unit area in space, is another representative measure in addition to the three basic physical parameters of density, velocity, and pressure. The intensity is a vector, which expresses not only the power per unit area but also the direction of power propagation. It was also stressed that acoustic waves are particularly meaningful to our hearing system. We therefore need to establish a way to measure the associated acoustic variables in accordance with human perception. As a result, the primary measure of frequency is the octave or 1/3 octave band and the dB scale is employed to indicate the magnitude of pressure. The A, B, C, and other weightings essentially reflect the dependency of human perception on frequency and magnitude. Knowing that acoustic waves are governed by the acoustic wave equation and boundary and initial conditions, and recognizing that this can be considered as a typical mathematical problem related to linear partial differential equation, we have investigated possible solution methods that predict how sound waves propagate in space and time. The modal analysis method and Green’s function method were extensively studied. The boundary condition determines the solution of an acoustic wave equation. Alternatively, we can also say that the impedance distribution in space certainly determines all possible sound waves as studied in Chapter 1.
2.8 2.8.1
Essentials of Wave Equations and Basic Physical Measures Three-dimensional Acoustic Wave Equation
2.8.1.1 Conservation of Mass Consider the conservation of mass in a small volume located in an arbitrary three-dimensional space. According to the conservation of mass, a net increase of mass per unit time is equal to a net mass flux per unit time, which enters and exits through the surface of small volume. For convenience, mass per unit time that enters and exits through dy dz faces, namely two faces orthogonal to the x coordinate, can be written as (Figure 2.19): @ru @ru ru dy dz ru þ dx dy dz ¼ dV: @x @x
ð2:102Þ
In the same manner, increased mass that enters and exits through dz dx; dx dy faces in the @rw volumes dx dy dz are equal to @rv @y dV and @z dV, respectively.
Acoustic Wave Equation and Its Basic Physical Measures ∂pw dz dxdy pw+ ⎝ ∂z
⎝
dy
pv+
∂pv dy dxdz ⎝ ∂y
⎝ pu dydz
z
⎝
dz
pv dxdz y
pu+
∂pu dx dydz ⎝ ∂x ⎝
⎝
⎝
⎝
⎝
⎝
111
dx pw dxdy
x
Figure 2.19 Conservation of mass for a small volume located in three-dimensional space
In addition, increased mass in the volumes per unit time dx dy dz due to fluid compression can be written as: @ @ @r ðr dx dy dzÞ ¼ ðrdVÞ ¼ dV : ð2:103Þ @t @t @t By considering Equations 2.102 and 2.103 and the mass increase rates through the surfaces in the y, z directions, we can write @r @ru @rv @rw dV ¼ þ þ dV ¼ r r~ u dV; ð2:104Þ @t @x @y @z where ~ u ¼ ðu; v; wÞ. By applying the conservation of mass, we obtain @r ¼ r r~ u: @t
ð2:105Þ
The assumption employed here is that the fluid is a continuum and compressible. 2.8.1.2 Conservation of Momentum As shown in Figure 2.20, Newton’s second law dictates that the force applied to a small volume and the change of momentum should be balanced. We represent this law with selected coordinate systems and physical measures including velocity and mass. First, the balance between the change of momentum and force can be specified as Change of momentum through the surface of small volume þ Change of momentum within the body of small volume ¼ Sum of forces ðsurface or bodyÞ applied to small volume
ð2:106Þ
Here, the overall change of momentum can be written as:32 32 The terms in Equation 2.107 involved with changes of momentum through the surface of small volume are expressed in a similar form (dx dy dz ¼ dV) to the changes in momentum within a small volume; the area integral can be converted into a volume integral based on Gauss’ theorem.
Sound Propagation
112
p+
∂p dz dxdy ⎝ ∂z ⎝
⎝
p dydz
⎝
dz
p dxdz
y
p+
∂p dy dydz ∂x ⎝ ⎝
dy
z
∂ (ρwdV) ∂t
∂p dy dxdz p+ ⎝ ∂y ⎝ ⎝
⎝
⎝
⎝
dy
=
∂ (ρvdV) ∂t
dz
dx
∂ (ρvdV) ∂y
dx
p dxdy
x
Figure 2.20 space
Relationship between force and motion for a small volume located in three-dimensional
@ @ ður~ uÞdx dy dz þ ðvr~ uÞdy dz dx @x @y @ @ þ ðwr~ uÞdz dx dy þ ðr~ udVÞ @z @t
ð2:107Þ
There are two types of force that act on the surface of a small volume. One is the surface normal force that works normal to the surface and the other is shear force, which is a surface tangential force that works parallel to the surface. Of these, the forces that work parallel to the surface bring about an angular displacement change of a small amount of fluid in a hexahedronal shape (See Figure 2.20). The force that works normal to the plane, or pressure, causes a small amount of fluid to have pure dilation and compression. If the shear force by the viscosity of fluid is negligibly small compared to the fluid pressure, the net force applied to surface dy dz in Figure 2.20 can be written as:33 @p @p p dy dz p þ dx dy dz ¼ dx dy dz: ð2:108Þ @x @x When the force applied to the surface in Equation 2.108 is applied to dz dx; dx dy faces in the same manner, the net force that is applied to a small volume dx dy dz can be expressed as @p @p @p ~ ex þ ~ ey þ ~ e z dx dy dz ¼ rpdV; ð2:109Þ @x @y @z where ~ e x ;~ e y ;~ e z represent unit vectors in the x; y; z directions, respectively. Another force that is not considered here is the body force, which is the gravitational acceleration applied to an entire fluid. Considering this, the force applied to a small volume is The kinematic viscosity of water and air at normal temperature is 1.45 105 m2/s and 1.14 106 m2/s, respectively. 33
Acoustic Wave Equation and Its Basic Physical Measures
113
written as: r dVg~ e z rpdV
ð2:110Þ
where g is the gravitational acceleration of Earth. Here, the balance of momentum without considering gravity can be written as:34 @ @ @ @ ður~ uÞ þ ðvr~ uÞ þ ðwr~ uÞ þ ðr~ uÞ ¼ rp: ð2:111Þ @x @y @z @t The terms in the parentheses on the left side of Equation 2.111 can be rewritten as: p~ u
@u @ðr~ uÞ @v @ðr~ uÞ @w @ðr~ uÞ þu þ r~ u þv þ r~ u þw @x @x @y @y @z @z :
ð2:112Þ
¼ r~ uðr ~ uÞ þ ð~ u rÞr~ u The final term of the left side in Equation 2.111 can be rearranged to yield @ @~ u @r ðr~ uÞ ¼ r þ~ u @t @t @t
ð2:113Þ
And, by using the conservation of mass, @ @~ u ðr~ uÞ ¼ r ð~ u rÞr~ u @t @t
ð2:114Þ
is finally obtained. The left side in Equation 2.111 can be rewritten r~ uðr ~ uÞ þ ð~ u rÞr~ uþr ¼r
@~ u þ rð~ u rÞ~ u @t
¼r
D~ u Dt
@~ u ð~ u rÞr~ u @t ð2:115Þ
D where Dt represents total derivative or material derivative. Consequently, the balance between the force and momentum can be defined as
r
D~ u ¼ rp: Dt
ð2:116Þ
This equation is called the three-dimensional Euler equation.
34 As it can be assumed that the body force due to the density change in a small volume is very small compared to the force by sound pressure, the former is not usually included in the momentum balance when deriving governing equations in acoustics.
Sound Propagation
114
2.8.1.3 Equation of State Equations 2.105 and 2.116 represent the relationship between the density of the fluid ðrÞ and the velocity of the fluid particles ð~ uÞ, and the relationship between fluid particle velocity and pressure. Therefore, another equation is needed to identify the relationship between the three physical variables of density, fluid particle velocity, and pressure. This is the equation of state that determines the relationship between fluid pressure and density. As sound pressure in a fluid is generally thought to be governed by density and entropy, it can be expressed as p ¼ pðr; sÞ;
ð2:117Þ
where s is entropy. A small change of pressure can occur due to changes in density and entropy (or dr; ds), respectively. This can be expressed as @p @p dp ¼ dr þ ds: ð2:118Þ @r s @s r For sounds smaller than 109 Hz, it can be assumed that during compression and expansion of a fluid it undergoes an isentropic reversible process. Equation 2.118 can then be written as @p 0 p0 ¼ r; ð2:119Þ @r s0 where p0 and r0 represent small changes in pressure and density respectively. Moreover, @p ¼ c2 ; ð2:120Þ @r s0 where c is speed of sound. In other words, the square of the speed of sound is equal to the pressure change with respect to the density change. Alternatively, we can obtain Equation 2.120 by considering wave front propagation in a duct. Suppose that we make a disturbance which induces a small volume change in the onedimensional duct as illustrated in Figure 2.21. This disturbance, which is assumed to compress the fluid by a small volume (dV), will introduce an infinitesimal increase in pressure and density (Figure 2.21). As a result of this increase, we can postulate that a wave front will propagate with the speed of sound c (Figure 2.21). We now want to find the relation between the speed of sound propagation c and other physical variables, such as pressure and density. dV = S . dv dv
Wave front ∂P p+ dx ∂x ∂ρ dx p+ ∂x
0
x
p, ρ
c
Figure 2.21 S is cross-sectional area of the duct, x is the coordinate that measures the distance from the disturbance, and v is the disturbance velocity
Acoustic Wave Equation and Its Basic Physical Measures
115
Conservation of mass implies the identity: ðr þ drÞðcdvÞS ¼ rcS:
ð2:121Þ
The left-hand side simply represents the amount of mass change per unit time due to the disturbance. The right-hand side is the mass flux of the fluid at rest. These two have to be balanced, and can be written as rdv ¼ cdr:
ð2:122Þ
We next apply Newton’s second law to the fluid of interest. The force difference will be pSðp þ ð@p=@xÞdxÞS, and the corresponding momentum change under consideration is fðcSSdvÞcSgrc which neglects higher order terms induced by dr. We can therefore write dp ¼ rcdv:
ð2:123Þ
Equations 2.122 and 2.123 lead to the following relation which describes the relation between the speed of propagation and other physical variables of fluid: c2 ¼
dp : dr
ð2:124Þ
Equation 2.124 states that the square of the speed of sound depends on the rate of compression with respect to density, that is, the amount of pressure requires to generate a unit change in density. Note, however, that the change in pressure and density of the fluid also depends on temperature or entropy. Therefore, Equation 2.124 has to be rewritten as c2 ¼
@p @p þ : @r @s
ð2:125Þ
For isentropic process, Equation 2.125 can be written as @p 2 : c ¼ @r sc
ð2:126Þ
Equation 2.126 simply implies that the speed of sound can be predicted if we know the relation between pressure and density. For example, if the fluid can be assumed to be an ideal gas in isentropic process, then we can obtain the relations between pressure and density (the ideal gas law and the isentropic relation) as p ¼ nRT; ð2:127Þ r @p p ¼g or @r r
p ¼ constant rg
ð2:128Þ
Sound Propagation
116
where n is the number of moles defined as mass M (kg) per unit molar mass (kg=mol), R is the universal gas constant (¼ 8.314 J/(K mol) in standard air), T is the absolute temperature (K) and g is the heat capacity ratio which is defined as the ratio of the specific heat capacity under constant pressure to the specific heat capacity under constant volume. Consequently, we can predict the speed of sound for an ideal gas as pffiffiffiffiffiffiffiffiffiffiffi c ¼ gnRT ð2:129Þ under isentropic (i.e., no change in entropy) conditions.
2.8.2
Velocity Potential Function
When a viscous force is negligibly small compared to other forces, the angular deformation or velocity of the fluid (Figure 2.20) will satisfy the relation r ~ u ¼ 0:
ð2:130Þ
~ u ¼ rF;
ð2:131Þ
The velocity can then be expressed as
because r r ¼ 0. In other words, when sound waves propagate towards a compressible fluid, we can consider a certain function which expresses the velocity vectors of the fluid particles. This function is called the velocity potential function. Substituting Equation 2.131 into the linear Euler equation yields: p ¼ r0
@F : @t
ð2:132Þ
This means that sound pressure can be obtained from the potential function. In addition, when Equation 2.132 is substituted into a linear sound wave equation, we obtain @2F 1 @2F ¼ : @x2 c2 @t2
ð2:133Þ
From the above equation, we can conclude that the velocity potential satisfies the wave equation just as sound pressure does. Note also that the velocity and sound pressure can be directly obtained by differentiating the velocity potential function with respect to space or time, as indicated in Equations 2.131 and 2.132. As also indicated by Equation 2.131, the velocity along an equipotential line (the line along which the potential is constant) is zero.
2.8.3
Complex Intensity
We examine basic concepts underlying complex intensity by using one-dimensional sound waves of single frequency. Instant intensity is represented by multiplying the sound pressure pðx; tÞ ¼ Pcosðot þ fp ðxÞÞ
ð2:134Þ
Acoustic Wave Equation and Its Basic Physical Measures
117
by the particle velocity, uðx; tÞ ¼ Ucosðot þ fu ðxÞÞ;
ð2:135Þ
where P and U are real numbers and fp ðxÞ and fu ðxÞ represent the phase of the sound pressure and of the particle velocity, respectively. Therefore, the instant intensity is expressed as Iðx; tÞ ¼ PUcosðot þ fp Þcosðot þ fu Þ;
ð2:136Þ
and can also be written h i 1 Iðx; tÞ ¼ PU cosð2ot þ fp þ fu Þ þ cosðfp fu Þ 2 h i 1 ¼ PU cosð2ðot þ fp Þ ðfp fu ÞÞ þ cosðfp fu Þ 2 h 1 ¼ PU ðfp fu Þþ sinð2ðot þ fp ÞÞ sinðfp fu Þ þ cosðfp fu Þ: 2
ð2:137Þ
Let us consider the result of Equation 2.137 in conjunction with Equation 2.50. First, we define complex intensity as CðxÞ ¼ Iavg ðxÞ þ jIr ðxÞ 1 ¼ PUejðfp fu Þ : 2
ð2:138Þ
The magnitude of complex intensity is then 1 jCðxÞj ¼ PU: 2
ð2:139Þ
The average intensity or active intensity is found to be 1 Iavg ðxÞ ¼ PU cosðfp fu Þ 2
ð2:140Þ
and the reactive intensity is expressed 1 Ir ðxÞ ¼ PU sinðfp fu Þ: 2
ð2:141Þ
Note also that the phase difference between Equations 2.140 and 2.141 is 90 (or 90 ). Furthermore, when the phase of sound pressure is identical to that of particle velocity (for instance, plane waves), only active intensity exists since reactive intensity becomes 0. On the other hand, when the phase difference between sound pressure and particle velocity is 90 (for instance, an interior sound field in a rigid boundary), the active intensity becomes 0 while only reactive intensity exists.
Sound Propagation
118
We now express the intensity using complex magnitudes and relate them to Equations 2.138, 2.140 and 2.141. The sound pressure and particle velocity can be written as pðx; tÞ ¼ PðxÞejot
ð2:142Þ
uðx; tÞ ¼ UðxÞejot ;
ð2:143Þ
where P and U represent the complex magnitude of the sound pressure and particle velocity, respectively. The complex intensity defined by Equation 2.138 can be written as 1 C ¼ PU * ; 2
ð2:144Þ
and the active or reactive intensity can be readily written as
2.8.4
1 Iavg ðxÞ ¼ Re PU * ; 2
ð2:145Þ
1 Ir ðxÞ ¼ Im PU * : 2
ð2:146Þ
Singular Sources
2.8.4.1 Monopole Sources The sound pressure of a monopole source can be expressed as follows: pðrÞ ¼ S
e jkr ; r
ð2:147Þ
where the complex variable S is the monopole amplitude. Here, the amplitude of the sound source is the inertial force of fluid mass that the sound source exerts per unit time and unit solid angle. In other words, it can be expressed as S¼
1 r ðjoqÞ; 4p 0
ð2:148Þ
where q represents the volume velocity which is radiated by a monopole sound source per unit time. From Euler’s equation, the velocity for the r direction can be written as 1 @P S e jkr 1 1þj Ur ðrÞ ¼ ¼ : jkr0 c @r r0 c r kr
ð2:149Þ
The velocity in near-field and far-field sound fields are approximately obtained as Near field ðkr 1Þ :
S e jkr 1 j Ur ðrÞ r0 c r kr
ð2:150Þ
Acoustic Wave Equation and Its Basic Physical Measures
Far field ðkr 1Þ :
Ur ðrÞ
119
S e jkr r0 c r
ð2:151Þ
From Equations 2.147 and 2.149, the impedance can also be obtained as Zr ðrÞ ¼
PðrÞ 1 : ¼ r0 c Ur ðrÞ 1 þ j kr1
ð2:152Þ
The impedance in the near-field and far-field sound fields is defined: Zr ðrÞ jr0 cðkrÞ
Near-field : Far-field :
Zr ðrÞ r0 c
ð2:153Þ ð2:154Þ
Equation 2.154 implies that it becomes a plane wave in a far field. The average intensity in the r direction can be calculated from Equations 2.147 and 2.149 as follows: o 1 n 1 jSj2 : Iavg ðrÞ ¼ Re PðrÞUr ðrÞ* ¼ 2 2r0 c r2
ð2:155Þ
For a monopole sound source, the average acoustic power can be obtained by integrating Equation 2.155 over an area with a radius r, that is ð p ð 2p 2pjSj2 : ð2:156Þ Pavg ¼ Iavg ðrÞr2 sin y df dy ¼ r0 c 0 0
2.8.4.2 Dipole Source A dipole sound source can be constructed by two monopole sound sources that are closely located with a 180 phase difference. It can be mathematically written as: jkr1 e e jkr2 Pðr; y; fÞ ¼ lim S ; ð2:157Þ D!0 r1 r2 where r21 ¼ x2 þ y2 þ ðzD=2Þ2 ;
ð2:158Þ
r22 ¼ x2 þ y2 þ ðz þ D=2Þ2 ;
ð2:159Þ
SD ¼ D:
ð2:160Þ
Sound Propagation
120
Here, ðx; y; zÞ is a point midway between two sound sources and D represents the dipolemoment amplitude vector.35 When Equation 2.157 is expanded with a Taylor series, and higher order terms are assumed to be negligible, Equation 2.157 can be written as: jkr1 e ejkr2 d ejkr1 ejkr2 : Pðr; y; fÞ ¼ lim S þ D D!0 r1 r2 D¼0 dD r1 r2 D¼0
ð2:161Þ
Equations 2.158 and 2.159 can be rewritten as lim r1 ¼ r;
ð2:162Þ
lim r2 ¼ r:
ð2:163Þ
D!0
and D!0
Then the first term of Equation 2.161 can therefore be neglected. The second term can be rewritten as Pðr; y; fÞ ¼ lim SD D!0
d ejkr1 dr1 d ejkr2 dr2 : dr1 r1 D¼0 dD D¼0 dr2 r2 D¼0 dD D¼0
ð2:164Þ
Note that d ejkr jkrejkr ejkr ¼ : r2 dr r D¼0
ð2:165Þ
From Equations 2.158 and 2.159, we have dr1 1z 1 ¼ cos y; ¼ 2r 2 dD D¼0
ð2:166Þ
dr2 1z 1 ¼ cos y: ¼ dD D¼0 2 r 2
ð2:167Þ
and
When a dipole sound source is expressed using two sound sources with 180 phase difference, their sound fields change if the locations of the two sound sources change. Representing the directions of the two sound sources with vectors is therefore a more accurate means of representation. It is desirable to represent them as ~ D; ~ D; however, in this chapter, it is assumed that the two sound sources are located on the z axis for convenience. 35
Acoustic Wave Equation and Its Basic Physical Measures
121
By substituting Equations 2.160, 2.165, 2.166 and 2.167 into Equation 2.164 and rearranging them, the sound pressure by the dipole sound source can be written ejkr z 1 1þj Pðr; y; fÞ ¼ jkD r r kr ð2:168Þ ejkr 1 ¼ jkD cos y 1 þ j : r kr In the near field and far field, the pressure can be estimated as Near field : Far field :
Pðr; y; fÞ D
ejkr cos y r2
Pðr; y; fÞ jkD
ejkr cos y: r
ð2:169Þ
ð2:170Þ
From Euler’s equation, the velocity of direction r (Figure 2.22) can be written as ( 2 ) kD ejkr 1 1 Ur ðr; y; fÞ ¼ j cos y 1 þ 2j 2 ð2:171Þ r0 c r kr kr z θ
r y
φ
x = r sinθ cosφ y = r sinθ sinφ z = r cosθ
x
Figure 2.22
Spherical coordinate system
and the near field and far field can be written as: Near field :
Far field distance :
Ur ðr; y; fÞ j
2D ejkr cos y kr0 c r3
Ur ðr; y; fÞ j
kD ejkr cos y: r0 c r
ð2:172Þ
ð2:173Þ
Impedance in the r direction can be obtained from Equations 2.168 and 2.171 as
1 þ j kr1 Zr ðr; y; fÞ ¼ r0 c ð2:174Þ 2 1 þ 2j kr1 2 kr1
Sound Propagation
122
and the near field and far field can be estimated: Near field :
Far field :
j Zr ðr; y; fÞ r0 cðkrÞ 2 Zr ðr; y; fÞ r0 c:
ð2:175Þ
ð2:176Þ
The average intensity in the r direction can be written (see Equation 2.155): Iavg ðr; y; fÞ ¼
k2 jDj2 1 cos2 y: 2r0 c r2
ð2:177Þ
As examined in Equation 2.156, the power obtained by the area integral of the intensity is: Pavg ¼
2pk2 jDj2 : 3r0 c
ð2:178Þ
Radiation characteristics of monopole and dipole sound sources can be easily understood by comparing Equation 2.155 with 2.177 or Equation 2.156 with 2.178. 2.8.4.3 Quadrupole Source Quadrupole sources can be made by having very closely located two dipole sound sources with opposite phases. It can be mathematically expressed: jkr1 e 1 Pðr; y; fÞ ¼ lim ðjkDÞ cos f1 1 þ j D!0 kr1 r1
ð2:179Þ
ejkr2 1 cos f2 1 þ j kr2 r2 where r21 ¼ ðxD=2Þ2 þ y2 þ z2
ð2:180Þ
r22 ¼ ðx þ D=2Þ2 þ y2 þ z2
ð2:181Þ
cosy1 ¼
z r1
ð2:182Þ
Acoustic Wave Equation and Its Basic Physical Measures
cos y2 ¼
z r2
DD ¼ Q
123
ð2:183Þ
ð2:184Þ
and Q represents amplitude of the quadrupole.36 If Equations 2.182 and 2.183 are substituted into Equation 2.179 and rearranged with respect to r1 and r2 , we obtain 1 1 jkr1 jkr2 : ð2:185Þ Pðr; y; fÞ ¼ lim Dz 3 ð1jkr1 Þe 3 ð1jkr2 Þe D!0 r1 r2 After expanding the Taylor series as in dipole sources and rearranging, we can arrive at dr1 d 1 jkr1 Pðr; y; fÞ ¼ lim DzD ð1jkr Þe 1 dD D!0 dr1 r31 D¼0 D¼0 d 1 dr2 ð1jkr2 Þejkr2 : ð2:186Þ dr2 r32 D¼0 dD D¼0 A similar procedure that is taken for the dipole sound sources can also be applied, that is, d 1 1 k k2 ð1jkrÞejkr ¼ ejkr 3 4 þ 3j 3 þ 2 ; r dr r3 r r D¼0
ð2:187Þ
dr1 1x 1 ¼ sinycosf; ¼ 2r 2 dD D¼0
ð2:188Þ
dr2 1x 1 ¼ ¼ sinycosf: dD D¼0 2 r 2
ð2:189Þ
By substituting Equations 2.184, 2.187, 2.188 and 2.189 into Equation 2.186 quadruple sound pressure can be written as ( 2 ) jkr 1 1 2 e xz 3 1 þ 3j Pðr; y; fÞ ¼ k Q ; ð2:190Þ kr kr r r2 36
The amplitude of a quadrupole can be defined in two forms. One is a longitudinal quadrupole, in which the dipole moment amplitude vector and the vector connecting two dipole sources are parallel. The other is a lateral quadrupole, in which the vector directions are perpendicular to one another. The quadrupole velocity (Q) is defined as the average of the two cases.
Sound Propagation
124
or as
( 2 ) ejkr 1 1 sin 2y cos f 1 þ 3j Pðr; y; fÞ ¼ k Q 3 : r kr kr 2
ð2:191Þ
From Equation 2.191 we can see that sound pressures in the near field and far field are approximately: 3k2 Q ejkr 1 2 sin 2y cosf ð2:192Þ Near field : Pðr; y; fÞ 2 r kr Far field :
Pðr; y; fÞ
k2 Q ejkr sin 2y cosf: 2 r
ð2:193Þ
The velocity of the r direction can be written as: k2 Q ejkr sin 2y r0 c r ( 2 3 ) 1 1 1 9j cosf 1 þ 4j : 9 kr kr kr
Ur ðr; y; fÞ ¼
The near field and far field velocity can then be written as: 9k2 Q ejkr 1 3 sin 2y cosf Near field : Ur ðr; y; fÞ j 2r0 c r kr
ð2:194Þ
ð2:195Þ
k2 Q ejkr sin 2y cos f: 2r0 c r
ð2:196Þ
2 1 1 3 1 þ 3j kr kr Zr ðr; y; fÞ ¼ r0 c 2 3 ; 1 1 1 1 þ 4j 9j 9 kr kr kr
ð2:197Þ
Far field :
Ur ðr; y; fÞ
Impedance in the r direction can be written
and the near-field and far-field behavior is Near field :
Far field :
j Zr ðr; y; fÞ r0 cðkrÞ 3 Zr ðr; y; fÞ r0 c:
ð2:198Þ
ð2:199Þ
Acoustic Wave Equation and Its Basic Physical Measures
125
Intensity in the r direction is Iavg ðr; y; fÞ ¼
k4 jQj2 1 2 sin 2y cos2 f: 2r0 c r2
ð2:200Þ
Power in the r direction is Pavg ¼
k4 jQj2 : 15r0 c
ð2:201Þ
Note that in the far field, impedance in the r direction is the same as a plane wave regardless of the type of sound source (see Equations 2.154, 2.176, and 2.199). One of the notable physical phenomena is that the entire power radiating in the r direction is independent of wave number k in the case of a monopole sound source. It is proportional to k2 and k4 in the case of dipole and quadruple sound sources, respectively.
Exercises 1. Newton assumed that a gas would follow an isothermal process, and calculated the speed of sound. His results, however, differed from the measured values. It was found that we can obtain more acceptable results if we assume that the gas follows an isentropic process. We want to calculate the speed of sound in terms of p0 and r0 , when the gas is assumed to follow an isothermal or isentropic process at 20 C. The specific heat conductivity is 1.402 at 20 C, the atmospheric pressure (p0 ) is 101.325 kPa, and density r0 is 1.12 kg/m2. It is recommended that you use the ideal gas state equation and apply the relation c2 ¼ @p=@r. 2. Mean intensity is obtained by averaging the product of pressure and velocity with respect to time. Prove that the mean acoustic intensity is expressed 1 Iavg ¼ Re PU* ; 2 where p ¼ RefPejot g and u ¼ RefUejot g. P and U are complex and implies complex conjugate (see Section 2.8.3). 3. Derive the governing wave equation for an inhomogeneous medium, provided that there is no mean flow. The effect of gravity (the body force) is assumed to be small compared to other forces. We also assume that the atmospheric pressure p0 is constant with respect to time and space, but the density r0 and the speed of sound c can vary with regard to space. (a) Prove that the linearized governing equation can be written as @p0 þ r0 c2 r ~ u ¼ 0; @t
r0
@~ u ¼ rp0 @t
ðp ¼ p0 þ p0 Þ
(b) From the two equations of (a), prove that the wave equation can be obtained in the form 1 1 @ 2 p0 rp0 2 2 ¼ 0: r0 r r0 c @t
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4. Suppose that the SPL was 90 dB when one of the machines ceased to operate; it was 95 dB when it was working. What is the SPL of the machine? 5. Table 2.6 lists the SPL in the octave band measured in an anechoic chamber. The SPL was measured at a distance of 1 m from the motor. (a) What is the bandwidth of each octave band and its lower and higher frequency? (b) What would the total SPL of the motor be? Table 2.6 SPL in the octave band Center frequency of the octave band (Hz) SPL (dB)
63
125
250
500
1000
2000
4000
8000
65
70
72
78
84
86
75
71
6. Suppose that we have a constant OBL (Octave Band Level), as illustrated in Figure 2.23. We want an A-weighted SPL in the graph, and want to estimate the total A-weighted SPL (see Figure 2.12 for reference). OBL (dB) 70
f (Hz) 16 31.5
63
8000
Figure 2.23 Octave band level
7. The noise level measured at a factory when not in operation, known as the “ambient noise level”, is 60 dB. We also know that the SPL of the factory is 63 dB when only one machine is in operation. What would be the SPL when an identical machine is added? 8. The campus police decided to put a speaker on top of the patrol car to announce the new rule which does not permit a car to exceed a speed of 20 km/hour. To obtain the necessary equipment, PC Smith visited a local audio shop where he was informed that a new speaker which radiates sound omni-directionally could be bought. PC Smith was so excited that he bought two speakers and installed them himself. He then tested his new set-up, and found that the radiated sound is not omni-directional but rather very directional. How could you explain this? 9. A student was disturbed by noise coming from the nearby soccer field. The campus police agreed that the loud speaker was too loud, and decided to estimate the sound level with the help of two students. One student came up with data indicating that it was higher than 70 dB, whereas the other student said it was lower than 70 dB if the subjective effects were considered. Repeat the students’ calculation for the campus police using the measured data listed in Tables 2.7 and 2.8.
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127
Table 2.7 1/3 octave data Frequency (Hz) 50 63 80 100 125 160 200 250 315
SPL (dB)
Frequency (Hz)
52 59 63 68 70 71 69 67 62
SPL (dB)
Frequency (Hz)
SPL (dB)
400 500 630 800 1000 1250 1600 2000 2500
58 57 55 53 52 54 50 57 53
3150 4000 5000 6300 8000 10 000 12 500 16 000 20 000
55 55 58 58 53 52 60 61 58
Table 2.8 A weighting Frequency (Hz)
A weighting (dB)
Frequency (Hz)
A weighting (dB)
Frequency (Hz)
A weighting (dB)
50 63 80 100 125 160 200 250 315
30.2 26.2 22.5 19.1 16.1 13.4 10.9 8.6 6.6
400 500 630 800 1000 1250 1600 2000 2500
4.8 3.2 1.9 0.8 0 þ 0.6 þ 1.0 þ 1.2 þ 1.3
3150 4000 5000 6300 8000 10 000 12 500 16 000 20 000
þ 1.2 þ 1.0 þ 0.5 0.1 1.1 2.5 4.3 6.6 9.3
10. A student bought two pairs of speaker units. One pair was for low frequency and the other pair for high frequency sound. He was told that the size of the woofer has a significant effect on sound radiation. The size of the woofer he selected was V (m3). However, the student was not quite satisfied with this speaker, especially at low frequency. What would be a rational way to improve the low frequency sound? Hint: Consider the woofer’s enclosure. 11. Examine the typical ducts that are excited by a velocity source uðtÞ in Figure 2.24. The walls are assumed to be rigid, and the wavelength of interest is much longer than the crosssectional dimension of the ducts. x=0
x=0
Infinite duct u(t) = Asin(ω t)
Figure 2.24
∞
x=L
Finite duct
Rigid surface
u(t) = Asin(ω t)
Left: an infinite duct and right: a finite duct
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(a) What would be the mean intensity and energy, with respect to time, at an arbitrary position x? What are the associated physical meanings? (b) What will the average acoustic energy from x ¼ 0 to x ¼ L be, and its physical implications? 12. We can express the momentum balance of an infinitesimal element of fluid as ðpSÞx ðpSÞx þ Dx ¼ S
@ru Dx: @x
Prove that we can have the same form of Euler equation if we linearize the equation.
3 Waves on a Flat Surface of Discontinuity 3.1
Introduction/Study Objectives
We wish to understand issues such as: what if we have a distributed impedance mismatch in space? How does this change the propagation characteristics of waves in space? We learned in Chapter 1 that wave propagation depends on the change of characteristic impedance of the medium and the boundary conditions of a string, which can also be expressed in terms of impedance. Note that the waves of string are one-dimensional. What if the waves propagate in three-dimensional space? Do we observe the same transmission and reflection as in the string case? We can easily anticipate that, in the three-dimensional space, the impedance mismatch in space makes the wave propagation rather complicated compared to that for the string. The transmission and reflection will occur in three dimensions. However, we first investigate how a plane wave in two-dimensional space is transmitted and reflected in the presence of impedance mismatch. The flat surface of a discontinuity in space, that is, a wall, creates an impedance mismatch in space. We will study how this mismatch transmits and reflects waves. This chapter begins with the simplest wall, which is modeled as a limp wall. A limp wall is defined as one which has only mass. A more general wall creating an impedance mismatch is then introduced.
3.2
Normal Incidence on a Flat Surface of Discontinuity
As illustrated in Figure 3.1, suppose that we have a flat surface of discontinuity that separates two different media. Let us also assume that a wave propagates in the direction perpendicular to the flat surface. We usually call this type of incident wave to the surface “normal incidence” or “perpendicular incidence”. In this case, a water–air interface, ocean sediment, and ground surface can be thought of as boundary conditions that belong to this physical situation.
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
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Incident wave
pi = Pi e−jω (t−x/c0)
Transmitted wave
pt = Pt e−jω (t−x/c1) Reflected wave
pr = Pr e–jω (t+x/c0) Medium 0 Z0 = ρ0c0
Medium 1
Z1 = ρ1c1 x
Figure 3.1 The reflected and transmitted wave for a normal incident wave. (The subscripts i; r; t denote the incident, reflected, and transmitted wave, respectively. p expresses the sound pressure with regard to time and space and P denotes the complex pressure amplitude)
Our objective is to determine how much pressure or energy of the incident wave is reflected and transmitted due to the flat surface of discontinuity, in other words, impedance mismatch in space. What causes the incident wave to be reflected and transmitted? It is quite obvious that the flat surface of discontinuity is responsible for these phenomena. This leads us to mathematically express what happens to fluid particles on the flat surface. The first point that we realize is that the pressure must be continuous on the surface ðx ¼ 0Þ; otherwise the surface will move according to Newton’s second law. In addition, the velocity of a fluid particle at the surface must also be continuous. In other words, the resultant velocity due to the incident and reflected waves at x ¼ 0 must be the same as that of the transmitted wave at x ¼ 0. This is a simple consequence of assuming that the medium is a continuum; velocity is therefore continuous. These two requirements can be mathematically expressed as follows. First, the pressure continuity at x ¼ 0 can be written as Pi þ Pr ¼ Pt :
ð3:1Þ
The velocity continuity is expressed as Ui Ur ¼ Ut
ð3:2Þ
where the subscripts i, r, and t represent the incident, reflected, and transmitted wave, respectively. P and U are the complex amplitude of pressure and velocity. The incident wave ðpi Þ, reflected waveðpr Þ, and transmitted wave ðpt Þ can therefore be written as pi ðx; tÞ ¼ Pi ejðotk0 xÞ
ð3:3Þ
pr ðx; tÞ ¼ Pr ejðotk0 xÞ
ð3:4Þ
pt ðx; tÞ ¼ Pt ejðotk0 xÞ
ð3:5Þ
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131
where k0 and k1 are defined k0 ¼
o ; c0
ð3:6Þ
k1 ¼
o ; c1
ð3:7Þ
where c0 and c1 are the speed of sound in medium 0 and 1, respectively. Note that the signs of the reflected wave pressure and velocity are different in Equations 3.1 and 3.2. This is because the pressure, which is always normal to the surface by definition, is a scalar quantity and does not depend on the propagation direction. On the other hand, velocity is a vector and therefore must take into account the propagation direction. For a plane wave, we can rewrite Equation 3.2 as Pi P r Pt ¼ ; Z0 Z0 Z1
ð3:8Þ
in which we use the relation Z ¼ P=U where Z is the characteristic impedance of the medium. The ratio of Pr to Pi , that is, the reflection coefficient R, can be obtained from Equations 3.1 and 3.8: Pr Z1 Z0 ¼R¼ : Pi Z1 þ Z0
ð3:9Þ
Note that the reflection coefficient depends totally on the characteristic impedance of the mediaðZ1 ; Z0 Þ. The transmission coefficient, which is the ratio of the transmitted wave with respect to the incident, can also be obtained from Equations 3.1 and 3.8: Pt 2Z1 ¼s¼ : Pi Z1 þ Z0
ð3:10Þ
Equations 3.9 and 3.10 are essentially the same as Equations 1.40 and 1.41. This certainly indicates that the transmission and reflection depend solely on the impedance at the discontinuity, regardless of whether the medium is string or fluid. This also implies that what we have learned in Chapter 1 is equally applicable to the concepts introduced in this chapter. The best method of exploring what Equations 3.9 and 3.10 physically convey is to examine their extreme behaviors. For example, let us investigate how a wave is reflected and transmitted when Z0 is extremely smaller than Z1 . This is the case when a sound wave generated in air meets water. In other words, “Can a fish hear what a fisherman says?” The first point to note from Equation 3.10 is that the transmitted sound pressure is twice as large as that of the incident sound pressure. This contradicts what we would normally expect. The opposite extreme case is when Z0 is much larger than Z1 . In this case, Equation 3.10 predicts that there will be negligible transmission. This means that even if the fish are holding a rock concert, it will not be possible to hear the sound in air. Note also that Equation 3.9 predicts that the amplitude of the reflected wave has a negative sign.1
1
As we learned from the waves on a string in Chapter 1 (e.g., the thin and thick strings in Figure 1.11), the phase is completely opposite to the incident wave.
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What we observe in considering the extreme behaviors of Equations 3.9 and 3.10 appears to contradict common sense. However, these behaviors essentially stem from not distinguishing clearly between sound pressure and power, which is necessary to excite our hearing system. In other words, we cannot say that we would hear twice as loud even if the amplitude of transmitted sound was twice as large as the incident sound. To see how much power is essentially transmitted, we must determine the velocity reflection and transmission coefficient.2 These can be obtained by using the impedance relation of a plane wave, Z ¼ P=U, from Equations 3.9 and 3.10. These are velocity reflection coefficient : velocity transmission coefficient :
Ur Z1 Z0 ¼ Ui Z1 þ Z0 Ut 2Z0 ¼ : Ui Z1 þ Z0
ð3:11Þ
ð3:12Þ
These velocity coefficients clarify the apparent discrepancy with common sense. For example, when Z0 is much smaller than Z1 , the transmitted pressure is twice as large as the incident pressure. As we can see from Equation 3.12, the transmitted velocity is negligible and the transmitted power is therefore trivial. Figure 3.2 summarizes pressure reflection and transmission coefficient (Equations 3.9 and 3.10). The velocity reflection and transmission (Equations 3.11 and 3.12) are illustrated in Figure 3.3.
Figure 3.2 Pressure reflection and transmission coefficient, where R ¼ ðZ1 Z0 Þ=ðZ1 þ Z0 Þ and s ¼ ð2Z1 Þ=ðZ1 þ Z0 Þ
The power reflection/transmission coefficients are defined as the ratio between the reflected/ transmitted power and the power of the incident wave. Equations 3.9–3.12 lead us to obtain 2
We can derive the reflected and transmitted velocities by using Equations 3.3–3.5, and the impedance of plane wave.
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133
Figure 3.3 Velocity reflection and transmission coefficient, where Ur =Ui ¼ ðZ1 Z0 Þ=ðZ1 þ Z0 Þ and Ut =Ui ¼ ð2Z0 Þ=ðZ1 þ Z0 Þ
power reflection coefficient :
power transmission coefficient :
* 1 2 P r Ur * 1 2 P i Ui * 1 2 Pt Ut * 1 2 P i Ui
¼
jZ1 Z0 j2
ð3:13Þ
jZ 1 þ Z 0 j2
¼
4Z1 Z*0 jZ 1 þ Z 0 j2
:
ð3:14Þ
Note that the power per unit area (intensity) is defined 12 PU* (Equation 2.144). If the characteristic impedances of two media only have a real part (e.g., water or air), then the power reflection and transmission coefficients can be written as PR ¼ Pt ¼
ðZ1 Z0 Þ2 ðZ1 þ Z0 Þ2 4Z1 Z0
ðZ1 þ Z0 Þ2
:
ð3:15Þ ð3:16Þ
This can be derived from Equations 3.13 and 3.14. The sum of transmitted and reflected power at the flat surface, that is, the incident power, is derived by adding Equations 3.15 and 3.16.3 Figure 3.4 illustrates how these two power coefficients behave with varying characteristic impedances of the media. To summarize, when the waves meet a flat surface or discontinuity they are reflected and transmitted. The degree to which they are reflected and transmitted is completely determined 3 The difference between Equations 3.13 and 3.15 must be recognized. The power reflection coefficient (Equation 3.13) includes the reflected wave’s direction. The – sign means that the direction of the reflected wave is opposite to that of the incident wave. On the other hand, the power reflection coefficient of Equation 3.15 only shows the magnitude.
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Figure 3.4 Change of the power reflection and transmission coefficients with regard to variation of the characteristic impedances of the media, where PR ¼ ðZ1 Z0 Þ2 =ðZ1 þ Z0 Þ2 and Pt ¼ ð4Z1 Z0 Þ= ðZ1 þ Z0 Þ2
by the characteristic impedances of the two media. The theory corresponds to that which we observed for waves on string with two different thicknesses. The representative measures that express the waves induced by a flat surface of discontinuity are the reflection/transmission coefficients.
3.3
The Mass Law (Reflection and Transmission due to a Limp Wall)
The second simplest case, but one that has significant implications in practice, is a limp wall. A limp wall is a wall that only has mass. In other words, the mass effect is dominant compared to the stiffness or internal damping. To understand what physically occurs when incident waves meet the limp wall, let us envisage how fluid particles at the front side of the wall behave. When an incident wave with a certain frequency arrives at the limp wall, the particles will vibrate with the frequency of the incident wave. The wall also has to vibrate with the same frequency and magnitude of the fluid particles in order to avoid violating the continuum assumption. The vibration of the limp wall oscillates the fluid particles on the right-hand side of the wall. Our hypothesis is based on several assumptions that can approximate the underlying physics. One is obviously related to linear acoustics. In other words, we assume that the acoustic pressure, density, and velocity are small enough to be linearized. The second is that the fluid particles only oscillate the mass particles that they are in contact with, and the mass particles vibrate the fluid particles that they contact. This rather physically simplified postulate is often referred to as “locally reacting.”4 This indeed makes sense for the limp wall case, because there is no mechanism that can transfer 4
Section 3.9.1 addresses the topic of locally reacting wall.
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135
the motion or force at one position to the perpendicular direction of propagation. It is essentially a one-dimensional problem. We are now ready to express what we have observed physically with rational assumptions that approximate what is happening on both sides of the limp wall. To this end, we seek a suitable expression in a compact form that precludes any possibility to induce faulty information; that is, a universally acceptable expression. The most popular and well-developed approach is to express the phenomena in mathematical form. With this approach we can explore what actually occurs in practice on the basis of the underlying mathematical tools. Figure 3.5 depicts waves on a limp wall. Let us first look at the forces acting on the unit surface area of a limp wall. Newton’s second law states that an unbalanced force, which is ðPi þ Pr ÞPt , will produce motion of the limp wall. This is defined to be the mass/unit area of the limp wall multiplied by the acceleration. This will be mo2 Y, where Y is the wall displacement magnitude. We assume that the wall harmonically oscillates. This is summarized by the following force balance equation on x ¼ 0, that is ðPi þ Pr ÞPt ¼ mo2 Y;
ð3:17Þ
2
where m is mass per unit area (kg/m ). The fluid particle on the left and right side of the wall has to be the same to satisfy the velocity continuity. The velocity of the fluid particle on the lefthand side of the wall should correspond to the vibration velocity of the wall, that is, Pi Pr ¼ joY: Z0 Z0
ð3:18Þ
Incident wave
pi = Pi e−jω (t−x/c0)
Transmitted wave
pt = Pt e−jω (t−x/c0) Reflected wave
pr = Pr e−jω (t+x/c0) Medium 0 Z0 = ρ0c0
y = Ye−jω t (displacement) Medium 0
Z0 = ρ0c0
x Limp wall (m: mass/unit area)
Figure 3.5 Reflection and transmission due to the presence of a limp wall
The velocity of the fluid particles on the right-hand side of the wall is also equal to the vibration velocity of the wall, that is, Pt ¼ joY: ð3:19Þ Z0 We have assumed that the waves are planar and that the wall is locally reacting. Therefore, the problem is simply one-dimensional and requires only one spatial variable to describe it.
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We now have three equations (Equations 3.17–3.19) and four unknowns (Pi ; Pr ; Pt ; Y).We can therefore only obtain the ratios. We are interested in how much the waves are reflected and transmitted relative to the incident wave amplitude. These ratios can be obtained from Equations 3.17–3.19, which are Pr jom ¼R¼ Pi jom þ 2Z0
ð3:20Þ
Pt 2Z0 ¼s¼ ; Pi jom þ 2Z0
ð3:21Þ
where R and t are the reflection and transmission coefficients, respectively. The power transmission coefficient, which expresses how much power is transmitted relative to the incident power, can be written as 2 * 1 P t 2 Pt Ut : ¼ ð3:22Þ P * 1 i 2 P i Ui Substituting Equation 3.22 into Equation 3.21 gives5 jsj2 ¼
ð2Z0 Þ2 ðomÞ2 þ ð2Z0 Þ2
:
ð3:23Þ
If we want to use this limp wall to block the transmission of noise from one place to another, then the transmission loss has to be as small as possible. This necessitates a measure that determines how the wall is designed in terms of obtaining the lowest possible transmission. Equation 3.23 again indicates that the coefficient is totally determined by the impedances. This is the transmission loss (TL or RTL ), which is defined as 1 ð3:24Þ RTL ¼ 10 log10 2 : j tj Note that the log scale is used to consider SPL (sound pressure level) in practice. From Equations 3.23 and 3.24, the limp wall transmission is " # om 2 RTL ¼ 10 log10 1 þ : ð3:25Þ 2Z0 This equation essentially states that the transmission loss is strongly dependent on the frequency ðoÞ, mass per unit area ðmÞ, and the characteristic impedance of the medium ðZ0 Þ om expresses the gain of that exists on both the left and right-hand sides of the wall. The term 2Z 0 transmission loss and how the motion of the limp wall affects the fluid particles with impedance om Z0 on both sides of the wall. If 2Z is much greater than 1, then Equation 3.25 becomes 0 om dB: ð3:26Þ RTL ffi 20 log10 2Z0 5
The acoustic power per unit area is the product of sound pressure and velocity. Since we have been studying plane waves, that is, Ut ¼ Pt =Z0 , Ui ¼ Pi =Z0 , we can obtain Equation 3.22.
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137
Equation 3.26 predicts that the transmission loss increases to 6 dB as we double the frequency; that is, 1 octave increase of frequency and mass per unit area is doubled. This is referred to as the mass law, a well-known and commonly used law in partition design. We are now ready to explore what Equation 3.26 implies in practice. First, it is noted that the transmission loss expresses how much power is blocked by the wall. Second, the transmission loss is a measure that is related to human beings, and therefore uses dB scale. For noise control from an engineer’s point of view, the objectives are to obtain maximum transmission loss under the given circumstances. The next question is whether there is a rational way to design the wall to obtain maximum possible transmission loss. To establish certain design guidelines, let us look at the frequency that makes the transmission loss zero. We refer to this frequency as “blocked frequency” ðfb Þ (see Figure 3.6). This can be obtained from Equation 3.26, because the blocked frequency must satisfy 2pfb m ¼ 1: 2Z0
ð3:27Þ
Therefore, fb ¼
Z0 : pm
ð3:28Þ
Figure 3.6 Mass law: the graph shows that RTL increases by 6 dB when the frequency doubles (1 octave). The mass law is applicable from the blocked frequency
As illustrated in Figure 3.6, we can utilize the mass law to estimate how much transmission loss we would obtain. This allows us to design the wall or partition. For example, if Z0 is 415 kg/(m2 s) (the characteristic impedance of air at 20 C), then Equation 3.28 predicts
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the blocked frequency, that is, fb ¼
132 : m
ð3:29Þ
This rather simple equation leads us to predict the transmission we can obtain using our chosen material. For example, if we use a material whose mass per unit area is 10 kg/m2, then Equation 3.29 gives us a blocked frequency of 13.2 Hz. The mass law predicts that the noise will be reduced by 6 dB for a 1 octave increase of frequency; we can simply estimate how much noise reduction level can be achieved for the relevant frequency of noise. If the noise we want to block has a frequency of 800 Hz, then we may count 7 octaves (e.g., 13.2 ! 26 ! 52 ! 104 ! 208 ! 416 ! 832) and therefore a transmission loss of 36 dB is estimated. Table 3.1 summarizes the properties of some well-known sound insulation materials according to their densities. Table 3.1
Sound insulation materials and their density Mass per unit volume (g/cm3)
Insulation material Asbestos concrete Asphalt Oak Fiber mats, including matrix and air stiffness Gypsum board Glass Pressed-wood panels Plaster Cork Light concrete Plexiglas Porous concrete Dry sand Dense concrete Plywood Brick
2.0 1.8–2.3 0.7–1.0 0.08–0.3 1.2 2.5 0.6–0.7 1.7 0.12–0.25 1.3 1.15 0.6 1.5 2.3 0.6 1.9–2.2
(Structure-Borne Sound: Structural Vibrations and Sound Radiation at Audio Frequencies, 2nd edition, 1988, pp. 242 (Table 4.5), L. Cremer and M. Heckl, Springer-Verlag New York, Inc.: With kind permission of Springer Science þ Business Media.)
We should also note that the sound on the left and right-hand sides of the wall is composed of two different components: incidence and reflection on the left, and transmission on the right. The pressure on the left side of the wall, Pi þ Pr , is found to obey Pi þ Pr ¼ 2Pi Pt :
ð3:30Þ
This can be obtained by using Equations 3.17–3.19. Using Equation 3.19, we can rewrite Equation 3.30 as Pi þ Pr ¼ 2Pi þ joYZ0 :
ð3:31Þ
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139
This equation essentially provides a new way to look at the problem of a flat surface of discontinuity. The pressure on the left (Equation 3.31), which is the sum of the incident and reflected waves, is composed of two different pressures. The former ð2Pi Þ is the pressure when an incident wave hits a rigid wall. This is called “blocked pressure”. The latter is what can be generated by the motion of the wall ðjoYZ0 Þ. Note that the latter, which is the magnitude of the pressures induced by the motion of the wall, has the same magnitude but different sign of the transmitted wave. The magnitude of the transmitted wave ðPt Þ can be written as Pt ¼ joYZ0
ð3:32Þ
by rearranging Equation 3.19. In summary, this observation leads us to conclude that the incident, reflected, and transmission phenomena can be seen as the superposition of the blocked pressure and the radiation pressure induced by the wall’s motion (Figure 3.7).6 This is based on the assumption that the wall has only mass, which is very unlikely to exist in practice. However, this leads us to understand how the wall achieves transmission loss. We may therefore predict that the wall would generally follow what was observed for the case of external excitation, that is, the incident wave has a frequency that is higher than its natural frequency. This also helps us understand that the assumption is not merely to avoid complexity: it is a tenet that provides information on the actual phenomena when the assumption is reasonably valid in practice. Rigid wall Incident wave Reflected wave
Transmitted wave =
Blocked pressure (Pb = 2Pi)
Radiation pressure + (Prad)
Ye−jω t
Figure 3.7 The principle of superposition allows us to regard the incidence, reflection, and transmission phenomena as the sum of the blocked and radiation pressure (blocked pressurePb ¼ 2Pi and radiation pressure Prad ¼ joYZ0 )
To summarize, the mass law governs limp wall behavior. The blocked frequency provides a stepping stone to practically apply the mass law. The blocked frequency, in other words, is the fundamental frequency of the consecutive octave frequencies. The mass law essentially dictates a 6 dB/octave increase of the transmission loss. We also learned that the incidence reflection, and transmission phenomena can also be envisaged as the sum of the blocked pressure and the radiation pressure due to the motion of the wall. This essentially transforms the flat surface of discontinuity problem to a blocked pressure on a rigid wall and radiation pressure problem. The radiated pressure field depends on the wall velocity and characteristic impedance of medium. This provides the concept of radiation impedance (Chapter 4). 6
This concept of superposition is applicable to any linear system. Transmission minimization problem, therefore, is equivalent to the minimization of radiation.
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3.4
Transmission Loss at a Partition
We have investigated what would happen if waves impinge on a surface of discontinuity (impedance discontinuity). We have studied this problem by starting with the simplest case. The waves are considered to be plane waves, recognizing that any wave front in space can be expressed as the sum of plane waves. The surface of discontinuity is assumed to be the interface between two media, as well as a limp wall. From these examples, we understand that reflection and transmission entirely depends on the impedances, as we already anticipated by studying waves on two different strings (in Chapter 1). These findings assure us that waves on a string and in fluids obey the same laws or principles. The only difference is the way that we look at the problem; for example, the reflection and transmission obey impedances completely. Therefore, impedance is what governs or represents the transmission and reflection, not the other physical parameters or geometry. We now extend our understanding to more general cases. Figure 3.8 illustrates a partition that represents a more general flat surface of discontinuity. To determine how much transmission will occur, we have to apply the same laws that we used for the limp wall case: the velocity continuity and the force balance between the wall and forces acting on the wall of unit area (see Section 3.9.2). The transmission coefficient t can be readily obtained as t¼
2Z0 : jðoms=oÞ þ ð2Z0 þ rd Þ
ð3:33Þ
The first point to note about Equation 3.33 is that the denominator has imaginary and real parts. The imaginary part is related to mass m and the linear spring constant s. The mass contribution ðomÞ and the spring contribution ðs=oÞ have a 180 phase difference. This is because the former is the product of acceleration and mass, and the latter is the product of the spring constant and displacement (see Section 1.9.1). The second term in the denominator, m
Incident wave
s
pi = Pi e−jω (t−x/c0) rd
Transmitted wave
pt = Pt e−jω (t−x/c0)
Reflected wave
x
pr = Pr e−jω (t+x/c0) s Medium 0 Z0 = ρ0c0
Medium 0
Z0 = ρ0c0 rd
Figure 3.8 The reflection and transmission due to the partition (rd is linear damping coefficient and s is linear spring constant)
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141
2Z0 þ rd , has two components. One expresses the radiation at both sides of the wall, and the second term is what is lost by the linear damping ðrd Þ. The phases of the inertia and spring force term differ with jo and j=o, respectively, with regard to the damping force term. The numerator of 2Z0 essentially implies that the wall is radiating sound from the wall to both the left and right directions. The greater the loss of energy, the larger is the transmission coefficient (Figure 3.9). Rigid wall
Incident wave Reflected wave
Z0
Figure 3.9 joYZ0 Þ
Transmitted wave
Blocked pressure (Pb = 2Pi)
Radiation pressure
=
+
Z0
(Prad)
Z0
Z0
Ye−jωt
Z0
Transmission and reflection, utilizing the concept of superposition ðPb ¼ 2Pi ; Prad ¼
If we express Equation 3.33 using dimensionless parameters that can be obtained by dividing every term by Z0, then we have t¼
j
2
om s Z0 oZ0
þ 2þ
rd Z0
:
ð3:34Þ
The transmission coefficient, which is the amplitude ratio of the transmitted wave to the incident, determines how effectively the wall vibrates to move the fluid particles on it. The numerator (2) of Equation 3.34 essentially indicates that the motion occurs in both directions. This leads us to rewrite Equation 3.34 as t¼
j
om 2Z0
1
s 2oZ þ 1þ 0
rd 2Z0
:
ð3:35Þ
We may also modify Equation 3.33 to understand the underlying physics differently: t¼
2Z0 : jðom þ jrd s=oÞ þ 2Z0
ð3:36Þ
This expression can be rewritten as t¼
Zf ; Zp þ Zf
ð3:37Þ
142
Sound Propagation
where Zp ¼ jðom þ jrd s=oÞ is the partition impedance and Zf ¼ 2Z0 is the fluid loading impedance in both directions. The transmission loss is expressed similarly to Equation 3.21. Figure 3.10 shows typical normal incidence absorption coefficients and the real and imaginary parts of the impedance of sound absorbing materials.
Figure 3.10 (a) Normal incidence absorption coefficients. (b) Real and (c) imaginary surface normal impedances of some typical sound absorbing materials (thinsulate, foam, and fiberglass). (Data provided by Taewook Yoo and J. Stuart Bolton, Ray W. Herrick Laboratories, Purdue University.)
Waves on a Flat Surface of Discontinuity
143
We can obtain the transmission loss in a similar manner to that described in Section 3.3, that is7 RTL ¼ 10 log10
1 j tj 2
i 2 1 h ¼ 10 log10 om 1o20 =o2 =Z20 þ ð2 þ rd =Z0 Þ2 ; 4
ð3:38Þ
pffiffiffiffiffiffiffiffi where o0 ¼ s=m, which is the resonant frequency of the partition in vacuum. Note that Equation 3.38 assumes that the direction of the incident wave is normal to the wall, that is, normal incidence. This is analogous to waves propagating along a string which is attached to a mass-spring-dash pot system, as illustrated in Figure 3.11. Incident wave
Transmitted wave
m Reflected wave
s
rd
Figure 3.11 The incident, reflection, and transmission waves on strings that are attached to a single degree of freedom vibration system (m is mass, s is linear spring constant, and rd is viscous damping coefficient)
To understand what Equation 3.38 implies physically and practically, we look at the equation in various ways. First, if the frequency of the impinging wave is much higher than the natural frequency of the partition ðo o0 Þ, then Equation 3.38 becomes8 om RTL ffi 20 log10 : ð3:39Þ 2Z0 This result is essentially the same as Equation 3.26. In other words, the transmission loss follows the mass law when the excitation frequency is much higher than the natural frequency (Figure 3.12). When the frequency of interest is far less than the natural frequency ðo o0 Þ, then Equation 3.38 becomes s RTL ffi 20 log10 : ð3:40Þ 2oZ0 This means that the transmission loss mostly depends on the linear spring constant. If we increase s twofold, then the transmission loss increases by 6 dB (6 dB/octave) (Figure 3.12). 7 8
Equation 3.25, which is one of the results of Section 3.3, is the special case of Equation 3.38 when s ¼ rd ¼ 0. The effect of the damping, which is in the denominator of Equation 3.35, is generally much less than the mass effect.
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144
Figure 3.12 Transmission loss at a partition where o0 ¼ the partition
pffiffiffiffiffiffiffiffi s=m is the undamped resonant frequency of
Finally, we look at what the wall experiences when it resonates ðo ¼ o0 Þ. The transmission loss at the resonance turns out to be rd : ð3:41Þ RTL ffi 20 log10 1 þ 2Z0 In other words, the transmission loss is entirely dominated by the damping coefficient. This implies that we have to increase the damping to make the transmission loss larger (as illustrated in Figure 3.12) which is obviously not a practical solution because of the cost. To summarize, the transmission characteristics of a partition are dependent upon the frequency of the incident wave relative to the resonant frequency. If the frequency is much higher than the resonant frequency, it follows the mass law. For the case of lower incident frequency than resonant frequency, the stiffness of the wall controls the transmission. In the region of resonant frequency, the transmission loss depends upon the degree of damping, which is obviously the most expensive case with regard to transmission control. We can therefore see that what has been studied in Section 3.3 can be applied to many practical cases which follow the mass law. Similar approaches can be considered if we have more than one partition or we have two different media in the incident and transmitted space. These are briefly addressed in Section 3.9.3.
3.5
Oblique Incidence (Snell’s Law)
If incident waves impinge on a flat surface of discontinuity with an arbitrary angle other than p=2, the fluid particles on the wall will oscillate in directions both parallel and perpendicular to the wall (see Figure 3.13).
Waves on a Flat Surface of Discontinuity
145
pi Transmitted wave B
A
t
A B
λi
Reflected wave
Δy y θr
θt
θi A'
x
A"
λt
λr B"
B'
Incident wave Medium 0
Medium1
Δy sin θi = λi Δy sin θr = λr Δy sin θt= λt
Figure 3.13 Reflection and transmission for oblique incidence (yi ; yr ; yt : incidence, reflection, and transmission angle; li ; lr ; lt : incidence, reflection, and transmission wavelength)
Suppose that an incident wave having pressure pi ð~ r; tÞ reaches a flat surface of discontinuity at x ¼ 0, where mediums 0 and 1 intersect. If the incident wave reached (B) after a period T, arrives the surface (A) at x ¼ 0, then the following geometrical relations must hold: Dy sin yi ¼ li ;
Dy sin yr ¼ lr ;
Dy sin yt ¼ lt ;
ð3:42Þ
where yi ; yr and yt are the incident, reflected, and transmitted angles, respectively. li ; lr , and lt are the corresponding wavelengths. Equation 3.42 can then be written as li lr lt ¼ ¼ : sin yi sin yr sin yt
ð3:43Þ
The relation between wave number and wavelength ðk ¼ 2p=lÞ transforms Equation 3.43 to ki sin yi ¼ kr sin yr ¼ kt sin yt :
ð3:44Þ
If the medium is non-dispersive, the dispersion relation must be k ¼ o=c. Equation 3.44 can then be rewritten as sin yi sin yr sin yt ¼ ¼ : c0 c0 c1
ð3:45Þ
Note that we have assumed that the waves propagating in mediums 0 and 1 satisfy the linear acoustic wave equation.9 9
Because we are discussing linear waves, the wave with frequency o in medium 0 has to propagate with the same frequency in medium 1.
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146
We can then deduce from Equation 3.45 that yi ¼ yr (the incident and reflection angles are equal). Hence, Equation 3.45 can be rewritten as sin yi sin yt ¼ : c0 c1
ð3:46Þ
To emphasize the characteristics of the media, let us denote yi and yt as y0 and y1 . Equation 3.46 can then be written as sin y0 sin y1 ¼ : c0 c1
ð3:47Þ
Similarly, we can write the relations as k0 sin y0 ¼ k1 sin y1 , ki ¼ kr ¼ k0 , kt ¼ k1 . This is what we refer to as Snell’s law. This law simply expresses that the wave number in the y direction at x ¼ 0 must be continuous in both media. This wave number ðk0 sin y0 ; k1 sin y1 Þ is often called the trace wave number. Figures 3.14 and 3.15 depict the implications of these wave number relations. Note that we can obtain these laws by simply considering the geometric relations of the waves on the flat surface of discontinuity. These relations can also be obtained by considering what we already observed in Section 3.2, that is, the pressure and velocity continuity condition on the flat surface of discontinuity that has impedance mismatch. In other words, the pressure and velocity made by the incident, reflected, and transmitted waves have to be continuous on the surface. We denote the incident, reflected, and transmitted waves as ~
pi ð~ r; tÞ ¼ Pi ejðotk i ~rÞ ~
ð3:48Þ
pr ð~ r; tÞ ¼ Pr ejðotk r ~rÞ ~
pt ð~ r; tÞ ¼ Pt ejðotk t ~rÞ ; ky k0 k1
θ1
θ0 θ0
θ crit
kx
sin θ 0 sin θ1 = c0 c1
k0 sin θ 0 = k1 sin θ1
Figure 3.14 Snell’s law expressed in the wave number domain ðk0 > k1 ; c0 < c1 Þ. ycrit denotes the critical angle
Waves on a Flat Surface of Discontinuity
147 ky k1 k0
θ0
k0 θcrit
θ0
θ1
θ = 90º
kx
sin θ 0 sin θ1 = c0 c1 k 0 sin θ 0 = k1 sin θ1
Figure 3.15
Snell’s law in the wave number domain ðk0 < k1 ; c0 > c1 Þ
where ~ k i ¼ ki cos yi~ e x þ ki sin yi~ ey ~ k r ¼ kr cos yr~ e x þ kr sin yr~ ey
ð3:49Þ
~ k t ¼ kt cos yt~ e x þ kt sin yt~ ey: ~ e x ;~ e y are the unit vectors in the x; y directions, respectively. It is important to realize that the wave number vectors in Equation 3.49 essentially express both the direction of the propagation of the waves and their wave numbers. ~ r ¼ ðx; yÞ is the position vector. From Equations 3.48 and 3.49, the pressure continuity at x ¼ 0 can be written as Pi ejki sin yi y þ Pr ejkr sin yr y ¼ Pt ejkt sin yt y :
ð3:50Þ
Equation 3.50 can be simply written as Pi þ Pr ¼ Pt ;
ð3:51Þ
which is exactly the same as what was derived for the case of normal incidence (Sections 3.2 and 3.3). The velocity continuity has to be satisfied by the fluid particles on the surface of discontinuity, that is, the partition. Because we assumed that the fluid is inviscid, the continuity is only related to the velocity in the x direction on the surface. In other words, the fluids on the surface have to move with the same magnitude and phase in the x direction, but can freely move in the y direction because there is no viscosity. This continuity requirement can be written10 f~ u i ð~ r; tÞ þ~ u r ð~ r; tÞg ~ ex ¼ ~ u t ð~ r; tÞ ~ e x ; x ¼ 0;
10
The velocity continuity in the y direction must also be satisfied. It is, in fact, satisfied by Snell’s law.
ð3:52Þ
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148
where11
The term r
0
1 ~ Pi;r;t B k i;r;t C jðot~k i;r;t ~rÞ Ae ~ r; tÞ ¼ : u i;r;t ð~ @ r0;0;1 c0;0;1 ~ k i;r;t Pi;r;t 0;0;1 c0;0;1
ð3:53Þ
in Equation 3.53 confirms that the velocity of a plane wave can be obtained by
the pressure divided by the characteristic impedance of the medium. On the other hand,
~ k i;r;t
j~k i;r;t j implies that the directions of velocity and of propagation are the same.12 Equations 3.49, 3.52 and 3.53 yield Pi ki cos yi jki sin yi y Pr kr cos yr jkr sin yr y e e r0 c 0 k i r0 c0 kr ð3:54Þ Pt kt cos yt jkt sin yt y ¼ e : r1 c 1 k t The exponents of Equation 3.54 must be identical. This again gives us Snell’s law, that is, Equation 3.44. Equation 3.54 can be rewritten as Pi Pr Pt cos y0 cos y0 ¼ cos y1 ; r0 c 0 r0 c 0 r1 c1
ð3:55Þ
where we adapted Equation 3.47 to simply emphasize the medium: cos yi ¼ cos yr ¼ cos y0 ;cos yt ¼ cos y1 . Equation 3.55 can also be rewritten as Pi Pr Pt ¼ : r0 c0 =cos y0 r0 c0 =cos y0 r1 c1 =cos y1
ð3:56Þ
If we define the oblique wave impedance as r0;1 c0;1 ; cos y0;1
ð3:57Þ
Pi Pr Pt ¼ : Z0 Z0 Z1
ð3:58Þ
Z0;1 ¼ we can rewrite Equation 3.56 as
The result is somewhat surprising. The relation between the incident, reflected, and transmitted waves is identical to that derived in Section 3.2. This means that even if a twodimensional wave approaches the surface with an incident angle ðyi ¼ y0 Þ, the reflected and
11 We can obtain the magnitude of the incident, reflected, and transmitted wave and their corresponding velocities using the linearized Euler equation (Equation 2.21). 12 The ejot term expresses the oscillation with respect to time.
Waves on a Flat Surface of Discontinuity
149
transmitted wave amplitudes follow Equations 3.50 and 3.51. The only difference is the impedance, which has to be oblique incidence (Equation 3.57). The power coefficients for the reflected and transmitted waves are also the same as those obtained in Section 3.3. We can therefore conclude that the degree to which the wave is reflected and transmitted is determined by the same approach as for the normal incident wave. It is important to note that our findings are based on the assumption that the continuity on the surface is independent of y. Equations 3.50 and 3.54 are based on this assumption. A surface of discontinuity that follows this assumption is called a “locally reacting surface”. The material covered in Sections 3.3 and 3.4 is also based on this assumption.13 Before we proceed to the next section, consider a special case as depicted in Figure 3.14. This is the case where we have a smaller propagation speed for medium 0 than that for medium 1. In other words, the wave number of medium 0 is larger than that of medium 1. In this case, if the incident angle is the same as the critical angle ðycrit Þ, then we have a transmission angle of 90 . If the angle of incidence is larger than the critical angle, then the transmission angle has to be somewhat larger than 90 which is not physically allowable. This situation can be modeled by writing the reflected angle as p yt ¼ jb: 2
ð3:59Þ
By substituting Equation 3.59 into Equation 3.49 and then substituting the result into Equation 3.48 we can obtain the reflected wave that is exponentially decaying in the x direction. We call this an “evanescent wave”. In contrast to this case, if we consider that the speed of propagation of medium 0 is larger than that of medium 1 (Figure 3.15), then the transmitted angle cannot be larger than any critical angle. To summarize, when waves impinge on a wall with an oblique angle and are reflected and transmitted we can decompose the waves into those which are perpendicular and parallel to the wall. This is also possible because the waves satisfy the linear acoustic wave equation. In other words, the principle of superposition allows us to decompose oblique waves into two parts. We found that how much is transmitted and how much is reflected is again determined by the wall impedance, as we saw for the case of waves that are normal to the wall. The only difference is the impedance, which reflects the oblique angle of incidence, reflection, and transmission. That is the oblique wave impedance. We also observed that it is possible to have an exponentially decaying transmission (evanescent waves) if the angle of incidence is larger than the critical angle of incidence. The critical angles of incidence and transmission are determined by the characteristics of the media. The main point of this section is that transmission and reflection are determined by the oblique impedance and other physical measures; for example, the transmission loss follows exactly the same function as that of the impedance.
3.6
Transmission and Reflection of an Infinite Plate
Suppose that the surface of the discontinuity does not locally react to the waves, that is, incident, reflected, and transmitted waves as illustrated in Figure 3.16. For simplicity, let us assume that we have plane waves. The incident, reflected, and transmitted waves can be written 13
The assumptions are often regarded as what restrict the physical situation that can be applied. However, we could also regard the assumption as a vehicle that guides us to explore unknown worlds.
Sound Propagation
150 y
Pr
Pt
λi
θr θi
θt x λt
λr Pi ρ0c0
Figure 3.16 identical
η =Ae−j(ωt−kby) ρ0c0
Incident, reflected, and transmitted waves on the plate. The media are assumed to be
as for Equation 3.48, that is, ~
pi;r;t ð~ r; tÞ ¼ Pi;r;t ejðotk i;r;t ~rÞ :
ð3:60Þ
The wave number vectors can also be expressed as for Equation 3.49. The displacement of the surface of discontinuity ðgÞ, a plate, only propagates in the y direction. We can therefore write the displacement as gðy; tÞ ¼ Aejðotkb yÞ ;
ð3:61Þ
where A is the amplitude of the displacement. We can assume that the thickness of the plate is much smaller than the wavelength of interest. The response of the plate to the pressures can then be written as m
@2g @4g þ B 4 ¼ pi þ pr pt x¼0 ; 2 @t @y
ð3:62Þ
where we assumed that the plate is thin enough to neglect the shear effect.14B is the bending rigidity (or flexural rigidity) of the plate and can be written in the form:15 B¼
Yd 3 ; 12 1l2p
ð3:63Þ
where lp is the Poisson ratio,16d is the thickness of the plate, and Y is Young’s modulus. 14
Thin plate means that the thickness is small enough to neglect the shear effect across the cross-section. Therefore, the bending dominates the overall deformation. 15 Graff, K.F. (1975) Wave Motion in Elastic Solids. Ohio State University, pp. 229–244. 16 Note that it is not wavelength.
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151
The characteristic equation that describes how the bending waves generally behave can be obtained by substituting Equation 3.61 into the homogeneous form of Equation 3.62. The excitation term (the right-hand side term of Equation 3.62) is zero. This yields mo2 þ Bkb4 ¼ 0:
ð3:64Þ
Equation 3.64 gives us the relation between the bending wave number ðkb Þ and radian frequency ðoÞ, that is m kb4 ¼ o2 : ð3:65Þ B If we use the dispersion relation, then the speed of propagation of the bending wave ðcb Þ can be obtained as 2 14 o Bo : ð3:66Þ cb ¼ ¼ kb m Equation 3.66 simply states that the speed of propagation depends on frequency. Therefore, the bending wave is dispersive. Note that a wave of higher frequency propagates faster than a wave of lower frequency (see Figure 3.17). This characteristic causes the shape of the wave to change as it moves in space. If we have a wave that has many frequencies such as a triangularshaped wave, then the triangular shape cannot be preserved as it travels because the wave of each frequency component travels at different speeds.
Figure 3.17
Dispersive wave propagation in an infinitely thin plate
If we express the velocity continuity mathematically, considering that the velocity of the surface at the discontinuity ðx ¼ 0Þ has to be exactly the same as the velocity of the fluid
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152
particle on the surface, we arrive at Pi ejki sin yi y Pr ejkr sin yr y r0 c0 =cos yi r0 c0 =cos yr ¼ joAejkb y
ð3:67Þ
Pt ¼ ejkt sin yt y : r0 c0 =cos yt
Equation 3.67 has to be valid for all y and, therefore, the exponents have to be identical. This leads us to write yi ¼ yr ¼ yt ð y0 Þ ki ¼ kr ¼ kt ¼
o c0
ð3:68Þ
k0 sin y0 ¼ kt : Equation 3.67 can therefore be rewritten as Pi Pr ¼ joA; Z0 Z0
ð3:69Þ
Pt ¼ joA; Z0
ð3:70Þ
where Z0 ¼ r0 c0 =cos y0 , which is the oblique impedance of the incident wave. Equations 3.60, 3.61, 3.62, 3.69, and 3.70 give us the transmission loss of an infinite plate, that is 2 s¼ : ð3:71Þ kb4 B j om Z0 oZ0 þ 2 Note that Equation 3.71 has essentially the same form as Equations 3.21, 3.34 and 3.35. For example, Equation 3.71 is a special case of Equation 3.37 when the partition impedance ðZp Þ is
kb4 B Zp ¼ j om : o
ð3:72Þ
The transmission loss of an infinite plate follows the behavior described in Section 3.3. For example, the mass law works for frequencies higher than the natural frequency. Note also that the radiation tends to be maximum; in other words, the transmission loss is minimal as the frequency approaches a level that satisfies Equation 3.65. The magnitude of the transmitted wave ðPt Þ can be obtained from Equations 3.68 and 3.70, that is, Pt ¼
jor0 c0 A: cos yt
ð3:73Þ
Waves on a Flat Surface of Discontinuity
Recalling that cos yt ¼ Equation 3.73 as
153
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1sin2 yt and using Equation 3.68 once more, we can rewrite A Pt ¼ jor0 c0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 1
kb k0
2
ð3:74Þ
Equation 3.60 can therefore be written as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A jðotkb yÞ jk0 1ðkb =k0 Þ2 x pt ð~ e r; tÞ ¼ jor0 c0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e : 1
kb k0
2
By setting k0 ¼ o=c0 and kb ¼ o=cb , Equation 3.75 can be rewritten as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A jðotkb yÞ jk0 1ðc0 =cb Þ2 x e pt ð~ r; tÞ ¼ jor0 c0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e : 1
c0 cb
2
ð3:75Þ
ð3:76Þ
The key feature of Equations 3.75 and 3.76 is the square root of the propagation constant or the wave number in the x direction. If the bending wave’s propagation speed ðcb Þ is greater than the speed of sound (supersonic), then the transmitted wave propagates in the x direction. However, if the propagation speed of the bending wave is smaller than the speed of sound (subsonic), then we have an exponentially decaying wave (an evanescent wave) in the x direction. This means that there are only waves in the vicinity of the plate and there is no transmission (Figure 3.18).17 To summarize, if the surface of a discontinuity does not locally react to an incident wave then the transmission occurs very differently from what we saw in Section 3.5. It is found that the trace wave number plays a key role in determining the transmission characteristics. Subsonic and supersonic behavior of a thin plate introduces a fundamental concept that is useful for exploring the transmission of non-locally reacting surfaces. Note that the transmission loss essentially follows what we have observed for the case of a locally reacting surface of discontinuity. The only difference comes from the effect of bending stiffness, which is the second term in the parenthesis of the denominator of Equation 3.71.
3.7
The Reflection and Transmission of a Finite Structure
We have studied the reflection and transmission of the discontinuity of an infinite flat surface. In other words, we studied a very ideal case that may or may not provide any realistic or practical value. The main reason for considering these rather unrealistic cases is because they are relatively easy to tackle mathematically and therefore they are easy to understand physically. The other reason is that the basic characteristics are often preserved for more practical cases, or at least they provide a guideline to understand what would happen in practical (more realistic) cases.
17
For a finite case, the wave propagating along the y direction is the sum of y-eigen modes. See Section 3.9.5.
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154
Figure 3.18 The propagation characteristics of bending waves in subsonic, critical speed, and supersonic range
For example, the mass law can be applied to any finite partition or wall. Because it is a law for unit area, it can therefore be applied to a finite structure as illustrated in Figure 3.19. The transmission loss of a partition that can be modeled as a mass-spring-dash pot system is also related to unit area. It can therefore also be applied to a finite partition case (Figure 3.19). As we can see in Figure 3.19, the transmission depends entirely on the motion of the partition which induces radiation (see Chapter 4). The difference between an infinite flat surface of a discontinuity and a finite structure lies only in the vicinity of edges; that is, the edge effect (see Section 4.4). The radiation pattern at these regions will be quite different. However, these effects tend to be smaller as the wavelength becomes larger. This edge effect makes the transmission with regard to unit area of the partition larger than that of an infinite flat surface of discontinuity. We can therefore express it as sf > s¥ ;
ð3:77Þ
where sf and s¥ express the transmission coefficient of finite and infinite flat surfaces of discontinuity, respectively. The transmission loss will therefore obey the relation: RTL;f < RTL;¥ :
ð3:78Þ
Waves on a Flat Surface of Discontinuity
155
pi pt pr
2 pi =
+
Blocked pressure
Radiation pressure
Limp wall
pi pt pr
2 pi =
+
Radiation pressure
Blocked pressure
Partition
pi pt pr
2 pi =
Blocked pressure
+
Radiation pressure
Finite plate
Figure 3.19 Blocked pressure and radiation of a finite flat surface of discontinuity
Equation 3.78 is very useful for practical applications. Because RTL;¥ overestimates the transmission loss, it can be safely used as a design value. In fact, however, we can readily estimate the transmission loss of any finite structure. Figure 3.19 essentially illustrates the method to calculate the transmission coefficient or loss. Transmission in the result of vibrating structure, that is, the radiation of a structure transmits the incident sound to the other side. The radiated sound can be obtained by using a Rayleigh integral equation (Chapter 4). To summarize, the radiation due to the oscillating motion of a finite partition determines its transmission loss. The transmission coefficient, which is defined as the ratio of the transmitted power to the incident power with regard to unit area, of a finite partition is therefore larger than that of an infinite case. This is because of the edge effects of the finite structure. We can therefore use the transmission loss of the infinite case as a design value, because it predicts larger transmission loss than in the case of the finite partition. We also found that the transmission loss can be more easily understood if we better understand radiation, which will be introduced in Chapter 4.
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156
3.8
Chapter Summary
We have examined how acoustic waves propagate in space and time with regard to the characteristic impedance of the medium and the driving point impedance. When we have two media that create a surface of discontinuity, then their characteristic impedances determine the size of the reflected and transmitted wave with respect to unit incident wave amplitude. This is analogous to what we learned in Chapter 1 regarding the transmission and reflection of waves on a string. From this analogy, we learned that the laws are essentially the same but the expression of the laws is different, depending on the medium under study. The limp wall example highlights that the mass law is the simplest case and is a fundamental principle that can be used in practice. We also found that the reflection and transmission can be regarded as radiation due to the motion of the structure or partition. Note that resonance is a unique and important phenomenon of the motion of the wall. The frequency therefore has to be seen with regard to its relative scale compared to the resonant frequency of the partition. The resonant frequency, often called the Helmholtz frequency, is a flag that directs the region to consider, for example, the mass controlled, stiffness controlled, or damping controlled region. The mass law is highly applicable in regions of frequency which are higher than the resonant frequency. If we have oblique incidence, the reflection and transmission are characterized by the surface of discontinuity. Transmission, which occurs through the surface, is referred to as refraction. If the surface is non-locally reacting to the waves, then the transmission depends on whether the trace wave number is larger than the propagation wave number. Evanescent waves are also possible if the waves are subsonic, which means that the propagation speed of the bending wave is smaller than that of the speed of sound. Transmission through a finite partition is due to the radiation of the partition. The radiation pattern at the edges of the partition makes the transmission different from that of an infinite partition per unit area. The transmission loss of a finite partition per unit area is smaller than that of an infinite partition per unit area. Therefore, the transmission loss of an infinite partition per unit area can be safely used as a design guideline. Finally, every transmission can be expressed in terms of partition and fluid loading impedance. Depending on the type of discontinuity, the two impedances are changed but keep their general form.
3.9 3.9.1
Essentials of Sound Waves on a Flat Surface of Discontinuity Locally Reacting Surface18
When reactions (that is, velocity, displacement, acceleration, etc.) at a certain point on the surface are related to the sound pressure of only the point and are not affected by sound pressures of other locations, the surface is called a locally reacting surface. The surface velocity is represented by Uð~ r 00 Þ. Here,~ r 00 denotes a random location on the surface (S). In general, this velocity can be written as ð 0 0 r 0 ÞdS; r 0 j~ r 0 Pð~ ð3:79Þ U~ r0 ¼ A ~ S
18
Kim, Y.-K. (2000) Locally reacting surface. Technical notes on Acoustics and Vibration Laboratory, NCL TN 2000–2, http://soundmasters.kaist.ac.kr/data%20bank/technical%20note/tnote2.pdf.
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157
0 measured at~ r 00 by the unit sound pressure applied to the where A ~ r 0 j~ r 0 representsthe velocity 0 r 0 j~ r 0 is the admittance or surface admittance (impedance inverse). point~ r 0 . In other words, A ~ In the case of a local reaction, Equation 3.79 can be simplified as 0 0 0 r0 P ~ r0 : ð3:80Þ U~ r0 ¼ A ~ The following provides some cases in which Equation 3.79 can be simplified to Equation 3.80. When a motion at one point of the surface is not related to motions in other areas, the admittance can be written as 0 0 A~ r 0 j~ r 0 ¼ Að~ r 0 ~ r0 : r 0 Þd ~ ð3:81Þ Equation 3.79 can then be expressed as ð 0 0 0 0 r 0 j~ r 0 Pð~ r0 P ~ r 0 ÞdS ¼ A ~ r0 ; U~ r0 ¼ A ~
ð3:82Þ
S
implying that this case is an example of a locally reacting surface. When the admittance is governed bythelocation of the reaction and by the distance from the location of applied sound pressure, U ~ r 00 can be expressed from Equation 3.79 as ð 0 r 0 ÞdS: U~ r0 ¼ A ~ r 0 ~ r 00 Pð~ ð3:83Þ S
The spatial Fourier transform of Equation 3.83 gives Uðkx ; ky Þ ¼ Aðkx ; ky ÞPðkx ; ky Þ:
ð3:84Þ
If the surface impedance is independent of the incident angle, then Aðkx ; ky Þ is independent of kx ; ky . This means that Aðkx ; ky Þ ¼ A0 :
ð3:85Þ
Uðkx ; ky Þ ¼ A0 Pðkx ; ky Þ
ð3:86Þ
Equation 3.84 can be rewritten as
to meet the local reaction condition. Note also that a local reaction can be achieved even when the propagation velocity of a surface wave is slow. The surface impedance of such a case (see Figure 3.20) can be written as P r c1 r 1 c1 ffi; ¼ 1 ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 U cos yt 1 kk01 sin2 yi
ð3:87Þ
making the assumption that there is a wave that propagates from the surface in the y direction and by using continuous conditions of velocity and sound pressure.
Sound Propagation
158 y Pt Pr
θr θi
θt x
Pi ρ0c0
Figure 3.20
ρ1c1
When waves exist on the boundary of two media
If the propagation velocity ðc1 Þ is much slower than the sound in the air denominator of Equation 3.87 tends to 1. In other words, 2 k0 sin2 yi 1: k1
c1 c0
1 , then the
ð3:88Þ
The surface impedance is therefore independent of the incident angle. The locally reacting case also occurs when the motion of the surface is rapidly attenuated. While the wave number of the air can be written as k0 ¼ co0 , the wave number of the surface can generally be written as o alsw k1 ¼ bja ¼ 1j : ð3:89Þ csw 2p The imaginary part of k1 represents the attenuation and csw and lsw denote the propagation velocity and wavelength, respectively. The magnitude ratio of k0 and k1 can be written as 2 2 k0 1 ¼ csw ð3:90Þ lsw 2 : k c0 1 1 þ a 2p If we have fairly large attenuation (a 1), then the magnitude ratio is smaller than 1. When 2 2 k0 k1 is much smaller than 1, kk01 sin2 yi is also much smaller than 1; the magnitude ratio therefore becomes independent of the incident angle and satisfies the local reaction condition. The final case which we will consider is that of normal reaction, that is, yt ¼ 0 in Figure 3.20. We can write P r c1 ¼ 1 ¼ r 1 c1 : ð3:91Þ U cosyt The surface impedance is obviously independent of the incident angle and therefore satisfies the local reaction condition.
Waves on a Flat Surface of Discontinuity
3.9.2
159
Transmission Loss by a Partition
As illustrated in Figure 3.8, the relation between pressure and the motion (displacement) of a partition can be written as ð3:92Þ Pi þ Pr Pt ¼ mo2 jord þ s Y; where Y is the amplitude of partition displacement. The velocity continuity must be satisfied on the left side of the partition, which can be written as Pi Pr ¼ joY: Z0 Z0
ð3:93Þ
The right-hand side of Equation 3.93 expresses the net velocity of fluid particles and the lefthand side expresses the amplitude of the partition velocity. Similarly, we can write the velocity continuity on the right side, that is, Pt ¼ joY: Z0
ð3:94Þ
From these three equations, Equation 3.33 for the transmission coefficient ðt ¼ Pt =Pi Þ is obtained. However, it is noticeable that sound pressure on the left side of the partition Pleft can be written as Pleft ¼ Pi þ Pr ¼ 2Pi þ joYZ0 ;
ð3:95Þ
and sound pressure on the right is Pright ¼ Pt ¼ joYZ0 :
ð3:96Þ
From Equation 3.95, we can physically interpret the sound pressure on the left side of the partition as the linear combination of blocked pressure and acoustic radiation pressure, as in the case of a limp wall (see Figure 3.7).
3.9.3
Transmission and Reflection in Layers
If we have two flat surfaces of discontinuity as illustrated in Figure 3.21, the transmission and reflection occur in three different regions. The boundary conditions at the discontinuities determine how much is reflected and transmitted. The pressure continuity at x ¼ 0 requires that Pi þ Pr ¼ A þ B:
ð3:97Þ
The continuity of the velocity at x ¼ 0 can be written as Pi Pr A B ¼ : r1 c 1 r1 c 1 r 2 c 2 r 2 c 2
ð3:98Þ
Similarly, we can write the continuity conditions of sound pressure and velocity at x ¼ L as Aejk2 L þ Bejk2 L ¼ Pt ejk3 L
ð3:99Þ
Sound Propagation
160 ρ1c1
ρ2c2
y
ρ3c3
B
Pr
x Pi
Pt A x=L
x=0
Figure 3.21
Normal incidence at three different media
Aejk2 L Bejk2 L Pt ejk3 L ¼ : r2 c 2 r2 c 2 r3 c 3
ð3:100Þ
Our objective is to obtain the ratio of the transmitted waves to the incident wave from the above four expressions. From Equations 3.97 and 3.98, we have Pi þ Pr r2 c2 A þ B ¼ Pi Pr r1 c1 AB
ð3:101Þ
which can be rewritten as Pr Pi 1 PPri
1þ
¼
r2 c2 1 þ AB : r1 c1 1 AB
ð3:102Þ
2 Let us denote PPri ¼ R, AB ¼ a, and rr2 cc21 ¼ Z Z1 ¼ m1 . 1 Equation 3.102 can then be written as
1 þ R Z2 1 þ a 1þa ¼ ¼ m1 : 1R Z1 1a 1a
ð3:103Þ
þa m1 11a 1 m1 ð1 þ aÞð1aÞ ¼ : 1þa m1 1a þ 1 m1 ð1 þ aÞ þ ð1aÞ
ð3:104Þ
That is, R¼
By eliminating Pt ejk3 L from Equations 3.99 and 3.100, we obtain Aejk2 L þ Bejk2 L ¼
r3 c3 jk2 L Ae Bejk2 L ; r2 c 2
ð3:105Þ
which can also be expressed as Aejk2 L þ Bejk2 L r3 c3 ¼ : Aejk2 L Bejk2 L r2 c 2
ð3:106Þ
Waves on a Flat Surface of Discontinuity
161
If we divide the denominator and the numerator by Aejk2 L, then we obtain 1 þ AB e2jk2 L r3 c3 Z3 ¼ ¼ ¼ m2 r2 c2 Z2 1 AB e2jk2 L
ð3:107Þ
1 þ ae2jk2 L ¼ m2 : 1ae2jk2 L
ð3:108Þ
1 þ ae2jk2 L ¼ m2 1ae2jk2 L
ð3:109Þ
which can be rewritten as
The identity
is also true. It can therefore be written as aðm2 þ 1Þe2jk2 L ¼ m2 1
ð3:110Þ
where a can be expressed as a¼
m2 1 2jk2 L e : m2 þ 1
ð3:111Þ
From Equation 3.111, we have 1þa ¼
m2 þ 1 þ ðm2 1Þe2jk2 L m2 þ 1
ð3:112Þ
1a ¼
m2 þ 1ðm2 1Þe2jk2 L : m2 þ 1
ð3:113Þ
Substituting Equations 3.112 and 3.113 into Equation 3.104 yields R¼
¼
m1 ð1 þ aÞð1aÞ m1 ð1 þ aÞ þ ð1aÞ 2 1Þe m1 m2 þ 1 þmðm 2 þ1 2 1Þe m1 m2 þ 1 þmðm 2 þ1
2jk2 L
2jk2 L
2 1Þe m2 þ 1ðm m2 þ 1
þ
ð3:114Þ
2jk2 L
m2 þ 1ðm2 1Þe2jk2 L m2 þ 1
:
By multiplying the denominator and numerator of Equation 3.114 by ejk2 L, we have R¼
ðm1 m2 m1 þ m2 1Þejk2 L þ ðm1 m2 þ m1 m2 1Þejk2 L : ðm1 m2 m1 m2 þ 1Þejk2 L þ ðm1 m2 þ m1 þ m2 þ 1Þejk2 L
By using cos k2 L ¼ e
jk2 L
þ ejk2 L 2
R¼
; sin k2 L ¼ e
jk2 L
ejk2 L 2j
ð3:115Þ
again, we obtain
ðm1 m2 1Þcos k2 L þ jðm2 m1 Þsin k2 L : ðm1 m2 þ 1Þ cos k2 Ljðm2 þ m1 Þsin k2 L
ð3:116Þ
Sound Propagation
162
Dividing the denominator and numerator by m1 m2 yields 1 m11m2 cos k2 Lj m12 m11 sin k2 L R¼ : 1 þ m11m2 cos k2 Lj m12 þ m11 sin k2 L
ð3:117Þ
Z3 Z3 2 Because m1 ¼ Z Z1 ; m2 ¼ Z2 ; m1 m2 ¼ Z1 , the reflection coefficient R is ultimately expressed as
Z2 Z1 1 1 Z Z3 cos k2 Lj Z3 Z2 sin k2 L : R¼ Z2 Z1 1 1þ Z Z3 cos k2 Lj Z3 Z2 sin k2 L
ð3:118Þ
The intensity transmission coefficient, or power transmission coefficient, is defined as jsj2 ¼ 1jRj2 :
ð3:119Þ
By using Equation 3.118, 1 Z1 cos k Lj Z2 Z1 sin k L 2 2 2 Z3 Z3 Z2 jsj2 ¼ 1 Z2 Z1 1 1þ Z cos k sin k Lj þ L 2 2 Z3 Z3 Z2
¼
1þ
¼
Z21 Z23
Z1 Z3
2
2 2 2 Z1 Z1 Z2 Z1 2 2 2 2 cos2 k2 L þ Z þ sin k L 1 cos k L 2 2 Z3 Z2 Z3 Z3 Z2 sin k2 L 2 2 Z2 Z1 2 1 2 1þ Z Z3 cos k2 L þ Z3 þ Z2 sin k2 L
2 2 1 4Z Z3 cos k2 L þ sin k2 L 2 : Z Z2 2 1 2 þ 1 cos2 k2 L þ Z22 þ Z12 sin2 k2 L þ 2 Z Z3 cos k2 L þ sin k2 L 3
2
ð3:120Þ
This can be rewritten as jsj2 ¼
2þ
Z3 Z1
þ
Z1 Z3
4 cos2
k2 L þ
Z22 Z1 Z3
þ
Z1 Z3 Z22
: sin2 k2 L
ð3:121Þ
If Z1 ¼ Z3 , then we have jsj2 ¼ 2ð1 þ cos2 k2 LÞ þ
4 2 Z2 Z1
þ
2 Z1 Z2
: sin2 k2 L
ð3:122Þ
Waves on a Flat Surface of Discontinuity
163
This has the most practical significance. The transmission coefficient is shown in Figure 3.22. When k2 L is smaller than 1 or when the thickness of the layer compared to the wave field is small, Equation 3.122 can be expressed as jsj2 ffi 4þ
2 Z2 Z1
4 : 2 2 1 þ Z LÞ ðk 2 Z2
ð3:123Þ
Equation 3.121 predicts the level of transmission when three layers have different values of impedance. For example, it allows us to estimate the amount of sound reduction from water to a structure of impedance Z2 .
Figure 3.22
Transmission loss RTL (RTL ¼ 10 log10 ðjtj2 Þ) with respect to Z2 =Z1 and k2 L
Sound Propagation
164
If waves are obliquely incident to the surfaces, there will be countless reflections and transmissions as illustrated in Figure 3.23. z (i)
(ii)
(iii)
ρ , c : medium3 3 3
d T32 T32R21 ρ2 , c 2 : medium 2
x T32T21
T32T21R23T21
ρ1, c1 : medium1
Figure 3.23 Oblique incidence at three different media (when a number of reflections and transmission properties are considered)
In Figure 3.23, case (i) describes the reflected wave due to the incident wave generated in medium 3, and case (ii) represents the transmitted wave that is reflected due to the impedance mismatch between medium 2 and 1. This repetition is presumed to take place indefinitely. Let (i) be R32 , (ii) be T32 R21 T23 e2jk2z d , and (iii) be T32 R21 R23 R21 T23 e4jk2z d . Here, Tij and Rij denote the reflection coefficient and transmission coefficient of a wave passing through the ith medium to the ij boundary, respectively. For example, T32 means that the amplitude decreases as the waves pass through the boundary between medium 2 and medium 3, while R32 is a coefficient that represents the ratio of amplitude reduction as they are reflected at the boundary. e4jk2z d denotes phase differences that occur over the distance of two round trips in the z direction due to incidence and reflection, which is Rij ¼
Zj Zi Zj þ Zi
ð3:124Þ
where Zi is the oblique incidence wave impedance at the ith medium. When all these properties are added together, the properties of the reflection waves can be expressed as R ¼ R32 þ T32 R21 T23 e2jk2z d þ T32 R21 R23 R21 T23 e4jk2z d þ T32 R21 ðR23 R21 Þ2 T23 e6jk2z d þ ¼ R32 þ T32 R21 T23 e2jk2z d
¥ X
R23 R21 e2jk2z
d n
ð3:125Þ :
n¼0
By using the sum of finite geometric progression and the relationship of 1 þ R ¼ T, the same result as that for reflection coefficient above is obtained as R¼
R32 þ R21 e2jk2z d 1 þ R32 R21 e2jk2z d
ð3:126Þ
Waves on a Flat Surface of Discontinuity
165
where R32 þ R23 ¼ 0.The transmission coefficient can be similarly found: T¼
T32 T21 ejk2x d 1R32 R21 e2jk2x d
4Z1 Z2 1 ¼ : ðZ1 þ Z2 ÞðZ2 þ Z3 Þ ejk2z d þ R32 R21 ejk2z d
ð3:127Þ
Let us examine Equation 3.127 by looking at a few special cases. First, consider that the wavelength of interest is 1/2 that of the thickness. In other words, k2z d ¼ np; n ¼ 1; 2; . . .
ð3:128Þ
The thickness has to be d¼
nl2 ; 2
ð3:129Þ
the input impedance of medium 2 at z ¼ d is Zin ¼
Z1 cos k2z djZ2 sin k2z d Z2 ¼ Z1 ; Z2 cos k2z djZ1 sin k2z d
ð3:130Þ
and the reflection coefficient is written as R¼
Zin Z3 Z1 Z3 ¼ ¼ R31 : Zin þ Z3 Z1 þ Z3
ð3:131Þ
Equations 3.130 and 3.131 state that the layer with half-wave thickness (medium 2) is acoustically transparent because the reflection coefficient ðRÞ is not related to the wave transmission through medium 2. Therefore, if media 1 and 3 have the same characteristic impedance, the reflection coefficient becomes 0 and perfect transmission occurs. We can also see that the layer (medium 2) functions as a spatial filter for a certain frequency or a specific direction of wave propagation. If a wavelength satisfies Equation 3.129, then it achieves perfect incidence. If not, a certain amount of reflection occurs. When the transmission layer is 1/4 of a wavelength, the input impedance is expressed as Z22 ; Z1
ð3:132Þ
Z22 Z1 Z3 : Z22 þ Z1 Z3
ð3:133Þ
Zin ¼ and the reflection coefficient is R¼
When Z22 Z1 Z3 ¼ 0, the reflection coefficient is 0; perfect transmission can therefore take place. In other words, if there is a transmission layer that has a thickness of 1/4 of a wavelength between two media and the impedance of the layer is the geometric mean of two
Sound Propagation
166
media (medium 1 and medium 3), then the reflection coefficient becomes zero and we have impedance matching. θ n+1
z
n+ 1
Zin(n) dn dn–1
Zin(n–1)
n
Zin(n–2)
n− 1
Zin(2) d2
Zin(1)
2 x
1 θ1
Figure 3.24 Oblique incidence from several different media (when a number of reflections and transmission properties are considered)
Let us consider the more general situation where we have n 1 layers, as illustrated in Figure 3.24. To find the reflection coefficient, the input impedances at all boundaries must be prescribed. The input impedances at all boundaries can be expressed as ð1Þ
ð2Þ
Zin ¼
Zin jZ2 tan k2z d2 ð1Þ
Z2 jZin tan k2z d2
Z2
ð2Þ
ð3Þ
Zin ¼
Zin jZ3 tan k3z d3 ð2Þ
Z3 jZin tan k3z d3
Z3
ð3:134Þ
.. . ðnÞ
ðn1Þ
Zin ¼ ð1Þ
Zin
jZn tan knz dn ðn1Þ
Zn jZin
tan knz dn
Zn ;
ðnÞ
where Zin ¼ Z1 , Zin can be determined from Equation 3.134. The reflection coefficient can also be obtained from Equation 3.134, that is ðnÞ
R¼
Zin Zn þ 1 ðnÞ
Zin þ Zn þ 1
:
ð3:135Þ
We now determine the transmission coefficient when there are n 1 layers between two media, or we have n þ 1 different media. The sound pressure at each layer can be expressed
Waves on a Flat Surface of Discontinuity
167
Pn þ 1 ¼ An þ 1 ejkn þ 1;z ðzzn Þ þ Bn þ 1 ejkn þ 1;z ðzzn Þ .. . Pi
¼ Ai ejkiz ðzzi1 Þ þ Bi ejkiz ðzzi1 Þ .. .
P1
¼ A1 ejkiz z ;
ð3:136Þ
provided that the coordinates of the upper boundary of the ith media are zi ðz0 ¼ 0Þ. Let us consider a boundary condition in which the pressure and impedance are continuous at z ¼ zi , then ½Pi ¼ Pi þ 1 z¼zi h i jori Pi ð@Pi =@zÞ1 ¼ jori þ 1 Pi þ 1 ð@Pi þ 1 =@zÞ1
z¼zi
ðiÞ
¼ Zin
ð3:137Þ
If we substitute Pi and Pi þ 1 from Equation 3.136 into Equation 3.137 and denote zi þ 1 zi ¼ di þ 1 and kiz di ¼ fi , we have Ai ejfi þ Bi ejfi ¼ Ai þ 1 þ Bi þ 1 :
ð3:138Þ
ðiÞ ðiÞ Bi 2jfi Zin Zi Bi þ 1 Zin Zi þ 1 ¼ ðiÞ ; : e ¼ ðiÞ Ai Ai þ 1 Zin þ Zi Zin þ Zi þ 1
ð3:139Þ
Zi
Ai ejfi þ Bi ejfi Ai þ 1 þ Bi þ 1 ðiÞ ¼ Zin ¼ Zi þ 1 Ai ejfi Bi ejfi Ai þ 1 Bi þ 1
Equation 3.138 can be expressed as
Substituting Equation 3.138 into the above equation, we have ðiÞ
Ai Z þ Zi jfi ¼ ðiÞin e : Ai þ 1 Z þ Zi þ 1 in
ð3:140Þ
By multiplying the ratios of the amplitude of incident to that of transmitted wave for all layers, the ratio for A1 to An þ 1 is ðiÞ
n A1 Z þ Zi jfi ¼ P ðiÞin e : An þ 1 i¼1 Z þ Zi þ 1
ð3:141Þ
in
Consider a single impedance layer. As illustrated in Figure 3.25, suppose that a wave travels from seawater through the sedimentary layer into the bottom layer. Assuming that the density of seawater is 1023 kg/m3, velocity of sound is 1460 m/s, density of the sedimentary layer is
Sound Propagation
168
Reflection coefficient on the incident angle (90-θ) 0
Seawater abs(R) (dB)
Sedimentary layer
-2 -4 -6 -8 -10
Bottom water
-12 0
10
20
30
40 50 90-θ
60
70
80
90
Figure 3.25 Model of plane wave that enters seawater through sedimentary layer to bottom layer and reflection coefficient based on incident angle
1780 kg/m3, the corresponding sound velocity is 1531.3 m/s, the density of the bottom layer is 1920 kg/m3, and its allowable sound velocity 1656.6 m/s. For simplicity, there is no sound absorption. The reflection coefficient with regard to incident angle is illustrated in Figure 3.25.
3.9.4
Snell’s Law When the Incidence Angle is Larger than the Critical Angle
Snell’s law, which is expressed as Equation 3.47, is illustrated in Figures 3.14 and 3.15. Let’s look at a few interesting cases, for example, when the incident angle ðy0 Þ is larger than ycrit (Figure 3.14). Snell’s law can always be expressed as k0 sin y0 ¼ k1 sin y1 :
ð3:142Þ
To satisfy the above relationship, one possible way is to allow y0 to be expressed as p y1 ¼ jb: 2
ð3:143Þ
The wave vector of transmitted waves (~ k t ) can then be rewritten as ~ k t ¼ k1 cos y1~ e x þ k1 sin y1~ ey p p ¼ k1 cos cos jb þ sin sin jb ~ ex 2 2 p p þ sin cos jbcos sin jb ~ ey 2 2 ¼ k1 ðj sinh bÞ~ e x þ k1 cosh b~ ey:
ð3:144Þ
Waves on a Flat Surface of Discontinuity
169
The transmitted wave is therefore pt ¼ Pt ej ðotk t ~r Þ : ~
ð3:145Þ
Equation 3.145 can be rewritten as pt ¼ Pt ejot ejk1 ðj sinh bÞx ejk1
cosh by
ð3:146Þ
¼ Pt ej ðotk1 cosh byÞ ek1 sinh bx
using Equation 3.143. This states that transmitted waves decrease exponentially along the x direction and propagate to the boundary surface, that is, in the y direction.
3.9.5
Transmission Coefficient of a Finite Plate19
We examine reflection and transmission of a finite, rectangular plate lying on an infinite baffle, as depicted in Figure 3.26. The incident sound field P1 ðz < 0 Þ, which consists of incident and reflected sound waves, is shown in Figure 3.27. The sound field radiated by a plate P2 ðz < 0 þ Þ can be represented by the Kirchhoff–Helmholtz integral equation (see Section 4.7.4) as ð P1 ð~ r Þ ¼ 2Pi 2 Gð~ rj~ r 0 Þr0 Pð~ r 0 Þ ~ n 0 dS; ðz < 0 Þ
ð3:147aÞ
S0
ð P2 ð~ rÞ ¼ þ 2
Gð~ rj~ r 0 Þr0 Pð~ r 0 Þ ~ n 0 dS; ðz > 0 þ Þ;
ð3:147bÞ
S0
y
x
φ
ρ → r
ρ → r
θ
0
Ly
Lxx z
infinite rigid Infinite rigidbaffle baffle
Figure 3.26 19
Coordinate system of a rectangular plate on a rigid baffle
Lee, J.-H. (2003) A study on prediction of sound isolation performance of finite laminate. PhD Thesis, Korea Advanced Institute of Science and Technology, Department of Mechanical Engineering.
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170
Infinite rigid baffle
Pr Pt
h
θr
θt
θi
z
Pi
ρ0c0
Figure 3.27
ρ0c0
Transmission and reflection by finite barriers
where P1 is composed of a blocked sound wave which is induced by movement of the plate at z ¼ 0 and propagates in the negative z direction. On the other hand, P2 represents radiation pressure due to the sound wave propagating in the positive z direction. Gð~ rj~ r 0 Þ represents Green’s function, r0 is the gradient of ~ r0, ~ r 0 is a surface location vector of the plate, and dS denotes the small area on the plate surface. The subscript 0 of the integral section S0 represents the integral of~ r 0 . Assuming that the incident sound pressure Pi is a harmonic function, we can express it as Pi ¼ exp j kx x þ ky y þ kz z
ð3:148Þ
where k ¼ o=c, kx ¼ k sin y cos f, ky ¼ k sin y sin f, and kz ¼ k cos y. Meanwhile, the pressure applied to a plate and the plate’s response can be expressed as
Br4 rs ho2 Wð~ r 0 Þ ¼ ½P1 ð~ r ÞP2 ð~ r Þ z¼0
ð3:149Þ
where B is the plate’s flexural stiffness. rs and h represent the density and thickness of each plate, respectively. ~ r denotes ðx; y; zÞ and ~ r 0 represents ðx; y; 0Þ. Equation 3.149 is valid at L2x x L2x , L2z z L2z . In Equation 3.149, if the media at both sides of the plate have the same characteristic impedance, then the pressure acting on the plate can be expressed in terms of the blocked pressure and radiation sound pressure, that is ½P1 ð~ r ÞP2 ð~ r Þ z¼0 ¼ 2Pi Pt ;
ð3:150Þ
Waves on a Flat Surface of Discontinuity
171
where Pi is the same as for Equation 3.148. Pt can be written as ð r Þ ¼ 2r0 o2 Pt ð~
Gð~ rj~ r 0 ÞWð~ r 0 Þ dS
ð3:151Þ
S0
to satisfy the continuity of the particle velocity on the surface of the plate. Note also that Equation 3.149 can be expressed by representing the velocity of the plate as a mode combination, that is
X 4 Bkmn rs ho2 Wmn þ Cmnqr Wqr ¼ 2Pi;mn :
ð3:152Þ
q;r
Here, we have Wð~ r0Þ ¼
X
ð Wmn Cmn ð~ r 0 Þ; S0
m;n
Wmn
1 ¼ Lmn
Pi;mn ¼
Cmnqr ¼
2r0 o2 Lmn
1 Lmn
C2mn ð~ r 0 ÞdS ¼ Lmn ;
ð3:153a; bÞ
ð Wð~ r 0 ÞCmn ð~ r 0 ÞdS;
ð3:153cÞ
Pi ð~ r 0 ÞCmn ð~ r 0 ÞdS;
ð3:153dÞ
S0
ð S0
ð ð Gð~ rj~ r 0 ÞCmn ð~ r 0 ÞCqr ð~ r ÞdS dS:
ð3:153eÞ
S S0
In the above expressions Cmnqr are coefficients, representing structure–fluid interactions. That is, the amount radiating as sound pressures of ðm; nÞ mode and ðq; rÞ mode of a plate excited by an incident sound field is being considered. The main effect of this modal coupling is the change in resonance frequencies of the plate. However, it is known that the modal coupling can be negligible in case of higher modal density, even if there is attenuation of vibration amplitude due to fluid loading increase. If the modal coupling is neglected, the mode coefficient the of vertical velocity of a plate can be obtained as Umn ¼
1 j2oPi;mn h i ; rs h oe 2 o2 þ jZe o2 mn mn mn
ð3:154Þ
where oemn and Zemn represent the fluid loaded resonant frequency and effective loss factor, respectively, and are given as e 2 rc r c ymn ð3:155Þ omn ¼ o2mn 2 0 owmn ; Zemn ¼ Z þ 2 0 o 2 ; omn rs h rs h where ymn and wmn denote the real part and the imaginary part of impedance, respectively, while omn means the in-vacuo resonant frequency of a plate. Recall that the acoustic power radiating from a plate can be written as
Sound Propagation
172
1 Pt ¼ Re 2
ð
P2 U dS *
s
( ) XX 1 * ¼ Re Cmnqr Uqr Umn : 2 m;n q;r
ð3:156Þ
The radiation impedance Cmnqr term in Equation 3.156, which represents interaction between fluid and structure of different modes (as also used in Equation 3.152) is the important factor. Each interaction between plate modes is known to be very small. When it is ignored, the radiated acoustic power is 2 2 1 X r0 c 0 ymn Y2mn Pi;mn ; ð3:157Þ Pt ¼ 2 2 m;n ðrs hÞ where Ymn is the vibration admittance of the plate mode and is defined hn o i1 2 oemn o2 þ jZemn o2mn : Ymn ¼ j2o
ð3:158Þ
Meanwhile, the acoustic power entering the plate is Pi ¼
Lx Ly cos y : 2r0 c0
ð3:159Þ
2 The average of the mode coefficient of the incident sound pressure (Pi;mn , Equation 3.157) will be ð ð 2 1 2p p=2 4pc2 Pi;mn sin y dy df ¼ 2 0 smn : ð3:160Þ p 0 0 o This is proportional to the plate’s mode acoustic radiation efficiency smn, which is the real part of radiation impedance. The acoustic transmission coefficient t can be found to be pc20 2r0 c0 2 X 2 2 2 s Y : ð3:161Þ tðoÞ ¼ Lx Ly rs ho m;n mn mn Note also that the average transmission coefficient with respect to frequency band of interest ðol < o < ou Þ can be written as ð 1 pc20 ou X 2r0 c0 2 2 2 2 t ¼ smn Ymn do; ð3:162Þ Do Lx Ly ol m;n rs ho where Do denotes the bandwidth.
Exercises 1. The waves depicted in Figure 3.28 are all normal to the surface of discontinuity. (a) What are the reflection and transmission coefficients? (b) What are the reflection and transmission coefficients of intensity? You may first obtain the velocity reflection and transmission coefficient.
Waves on a Flat Surface of Discontinuity
173
pi pt pr
ρ1c1
Figure 3.28
ρc r= 1 1 ρ2c2
ρ2c2
Normal incidence at two different media
(c) If r 1 or r 1, then how do the pressure and intensity reflection and transmission coefficient behave? 2. We decided to use a curtain to block noise coming from outside the lecture hall. To select the curtain, we have to know the mass per unit area of the curtain. Assuming the waves are normally incident to the curtain, you are asked to estimate the minimum mass per unit area to obtain 10 dB TL (transmission loss) for the frequencies that are given. Hint: The curtain can be assumed to be a limp wall. Therefore, the mass law can be applied. The characteristic impedance of air ðr0 cÞ is 415 kg/m2sec. (a) 100 Hz (b) 500 Hz (c) 1 kHz. 3. We would like to measure the specific impedance of a medium. We devised a plane wave to be incident to the surface with an angle of incidence yi . To measure the specific impedance, we took the following measurements. Measurement 1: We measured the pressure immediately at the reflected surface, which was B cosðotfÞ where f is the initial phase. Measurement 2: The acoustic pressure at the same position was A cos ot when the surface was removed. What will the specific acoustic impedance be? This will be expressed in terms of c, r, A, B, yi , f (Ingard, U., and Bolt, R.H. (1951) A free field method of measuring the absorption coefficient of acoustic materials. Journal of Acoustic Society of America, 23 (5), pp. 509–516). 4. We conducted two experiments on a surface that is comprised of two media whose characteristic impedances areðrI ; cI Þ and ðrII ; cII Þ, respectively. Experiment 1: We sent a plane wave with an angle of yI , which is less than the critical angle. The measured transmission angle was yII . Experiment 2: We then sent a plane wave with an incident angle yII from the second medium to the first. The power reflection and transmission coefficients obtained from the two experiments are equal. Prove this intuitive prediction.
Sound Propagation
174
5. As illustrated in Figure 3.29, Snell’s law determines the transmitted angle, or refracted angle, when we have an incident plane wave at frequency, f (Hz). However, one may argue that the transmitted angle must always be zero if the fluid can be assumed to be inviscid. Because the fluid has negligible viscosity in the y direction on the surface, the fluid particles’ motion in the y direction may not affect the motion of fluid particles in the x direction. Thus, the angle of transmission is always zero. What is the origin of this paradox? y
Φr
Φt
Φi
Medium 1
x
Medium 2
Figure 3.29 Oblique incidence at two different media
6. If two fluids or materials make a flat surface of discontinuity as illustrated Figure 3.30, then there is a possibility to have a region where no waves are transmitted. What is the condition that would make this possible? You may use the propagation vectors to explain this phenomenon. Medium 2
c2
Medium 1
c1
Figure 3.30 Two media that have a flat surface of discontinuity
7. We are asked to make a partition that can effectively reduce the noise from a machine that emits noise primarily at 1 kHz. The available materials are bricks and a steel plate. What would be the transmission loss that can be obtained using these materials? The mass of a brick is 5 kg and its dimensions are 40 cm, 30 cm, and 15 cm. The mass per unit area of the steel plate is 10 kg/m2. 8. One intrepid student attempts to heat or cool the plate to control the transmission loss of the steel plate. Do you think this is possible? What principles would support this idea? 9. A construction company is planning to build a hotel that is located about 10 km from the airport. The airport noise is the primary concern in the design of the walls of the hotel. One engineer measured the noise from the airport at the location of the hotel. Table 3.2 summarizes the engineer’s results from a 1/3 octave analysis. We would like to find the minimal thickness of the wall of the hotel to obtain a sound pressure level inside of less than 40 dB.
Waves on a Flat Surface of Discontinuity
175
Table 3.2 Noise levels from the airport at the future location of the hotel Center frequency (Hz) Sound pressure level (dB)
500 80
630 85
800 90
1000 95
1250 90
1600 85
2000 80
Hint: Assume that the effect of the windows is negligible, compared with that coming through the walls. The wall is made of concrete, which has a density of 2600 kg/m3. 10. One student was extremely impressed by the mass law due to its simplicity and practicality. He found a machine that generates noise with the spectrum depicted in Figure 3.31. (a) What will the total SPL be? (b) He attempted to use a curtain that has 10 kg per unit area. How many folds of a curtain should he use to obtain an SPL of less than 70 dB? SPI. 100 dB 80 dB
500Hz
Figure 3.31
750Hz
1000Hz
f
Spectrum of the machine noise
11. As illustrated in Figure 3.32, we installed a wall to reduce the noise from the machine. The result was not satisfactory, however. The sound level is 80 dB, which is not likely to be acceptable for an office. We would like to reduce it to 50 dB. The available material that can be used for this is a sheet of plywood of thickness 2 mm. The mass per unit area is 10 kg/m2. Assume that the machine noise is mainly at 100 Hz and the resonant frequency of the existing wall is 5 Hz. Assuming that adding plywood will not change
Figure 3.32
A wall to reduce the noise from the machine to the office room
Sound Propagation
176
the resonant frequency of the wall, you are asked to design the partition. The machine’s noise is 90 dB. The density of air is r ¼ 1:25 kg=m3 and the speed of sound is c ¼ 320 m=sec. 12. We would like to construct a partition between the machine room and our office, as depicted in Figure 3.33. The noise of the machine room is measured as 100 dB and its center frequency is 500 Hz. The two rooms can be assumed to be large enough to be regarded as having a diffuse sound field. This means that we have a uniform sound field at any position of the rooms. The available materials are as follows: #1 partition : f1 ¼ 25 Hz; #2 partition : f2 ¼ 50 Hz; #3 partition : f3 ¼ 50 Hz;
r1 ¼ 7700 kg=m3 ; r2 ¼ 6000 kg=m3 ; r3 ¼ 5000 kg=m3 ;
Machine room
Y1 ¼ 210 103 Pa Y2 ¼ 150 103 Pa Y3 ¼ 100 103 Pa
Office l = 3m
500 Hz
d = 0.01m
Figure 3.33
A partition to reduce the noise from the machine to the office room
where fi are the resonant frequencies of the partitions, ri are their densities, and Yi are Young’s moduli. You are asked to determine the maximum possible transmission loss by using the partitions that are available. Hint: Use the following equalities and assume that Poisson’s ratio ðlp Þ is 0.3. Yd 3 12 1l2p vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u Y ðlongitudinal wave speedÞ cl ¼ u t r 1l2p ðbending rigidityÞ B ¼
c2 ðcoincident frequencyÞ fc ¼ 0 2p
rffiffiffiffi m ; m ¼ rd B
13. If the media on both sides of a limp wall are different, what are the power transmission and the transmission coefficient? From this result, decide if the sound made by the crew in a submarine can propagate to the water or if the crew can hear sound from outside.
4 Radiation, Scattering, and Diffraction 4.1
Introduction/Study Objectives
We have studied how sound waves are reflected and transmitted when they meet a surface of discontinuity in space. We also noted that the discontinuity is expressed by its impedance. The impedance is a function of two spatial variables (e.g., ðx; yÞ for Cartesian coordinate), which means that it is two-dimensional in Chapter 3. The reflected and transmitted waves are therefore also two-dimensional. If we have a certain discontinuity that is a function of three spatial variables (e.g., (x, y, z) for Cartesian coordinate), then the waves will have more complicated patterns than those in two- or one-dimensional cases. Figure 4.1 depicts some representative cases. Depending on the type of discontinuities or the type of waves, scattering, diffraction, or radiation may occur. A mathematical approach to analyze the phenomena can be one of the following two methods: find the solutions that satisfy the linear acoustic wave equation (2.24) and boundary conditions, or use the integral equation (2.90). These two approaches enable us to understand radiation, scattering, and diffraction.1 Scattering is a phenomenon that describes the reflection of waves due to the presence of discontinuities in space. As its name implies, “scattering” typically describes waves scattered by the discontinuities in space. Radiation, on the other hand, describes waves generated by the vibration of structures or by fluid fluctuation (see Figure 4.1(a)). In this respect, scattering can also be a form of radiation. In particular, scattering occurs when incident waves meet scatterers (e.g., discontinuities) which radiate the scattered sound field. We also refer to “diffraction” as illustrated in Figure 4.1(b, d), which exhibits rather different types of waves. The latter are waves which are curved around corners or edges. Diffraction implies that we cannot see the sound source but can hear the sound. In addition, diffraction introduces the concept of a shadow region where we cannot hear the sound. 1
Scattering and diffraction are physical phenomena, which represent waves deflected by characteristics of the discontinuity. Both can therefore be expressed by solutions of the wave equation which satisfy boundary conditions.
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
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Sound Propagation
Figure 4.1 (a) Radiation, (b) scattering, and (c) diffraction: waves are visualized using a ripple tank. Incidence waves for (b)–(d) are plane waves coming from the left. The depth of the water has to be sufficiently smaller than 1/8th the wavelength to create a non-dispersive wave (kh has to be smaller than roughly 0.5, where k is wave number, and h is water depth)
To understand these rather complicated phenomena, we need to understand sound fields induced by the basic unit sources: the radiation of a breathing sphere and a trembling sphere. These are, in fact, quite similar to what we have learned with regard to a monopole and dipole.
4.2
Radiation of a Breathing Sphere and a Trembling Sphere
We first study the radiation due to the vibration of a sphere of radius a (Figure 4.2). The sphere is assumed to vibrate omni-directionally with equal magnitude; we refer to this as a “breathing sphere.” Let us attempt to use the velocity potential to describe the sound radiation from the sphere. The velocity potential can directly provide the velocity and pressure by differentiation with respect to space (~ u ¼ rU) and time (2.132), respectively. This is the advantage of using the velocity potential. The velocity potential (U) (Section 2.8.2), which can effectively describe the acoustic wave field induced by the breathing sphere, must satisfy the following linear differential equation (a linear acoustic wave equation) in polar coordinates, @2 1 @2 ðrUÞ ¼ ðrUÞ: c2 @t2 @r2
ð4:1Þ
179
Radiation, Scattering, and Diffraction
U0e–jω t
r
r=a
0
Figure 4.2 A breathing sphere and its radiation pattern: a is the radius, r indicates the radial distance, and U0 denotes the velocity magnitude
Equation 4.1 assumes that the velocity potential does not depend on the angle of the polar coordinate because of its symmetry characteristics. The solution of Equation 4.1 can be written, as confirmed in Chapter 2, in the form U¼
A jðotkrÞ e r
ð4:2Þ
where k is the wave number in the r direction. If the surface of the sphere harmonically vibrates, then the velocity on the surface (ur ða; tÞ) can be written as ur ða; tÞ ¼ U0 ejot :
ð4:3Þ
The relation between the two constants A and U0 in Equations 4.2 and 4.3 can be obtained by recalling that the rate of change of the velocity potential with regard to r has to be the velocity at the sphere (4.3). That is, @U 1 jk jðotkrÞ ur ðr; tÞ ¼ ¼ A 2 e : @r r r
ð4:4Þ
The velocity at r ¼ a will then be 1 jk jðotkaÞ ur ða; tÞ ¼ A 2 e : a a
ð4:5Þ
From Equations 4.3 and 4.5, we can obtain A ¼ U0
a2 jka e : 1jka
ð4:6Þ
180
Sound Propagation
The velocity potential, the velocity, and the acoustic pressure can therefore be written as a2 1 jðotkðraÞÞ e ; 1jka r
ð4:7Þ
1jkr a2 jðotkðraÞÞ e ; 1jka r
ð4:8Þ
Uðr; tÞ ¼ U0 ur ðr; tÞ ¼ U0
pðr; tÞ ¼ r0 cU0
jka a 2 jðotkðraÞÞ e : 1jka r
ð4:9Þ
Note that Equations 4.7, 4.8, and 4.9 are only valid if r is larger than a. We now have expressions for the velocity potential, radial velocity, and acoustic pressure that satisfy the governing differential equation (4.1) and boundary condition (4.3). Note that the velocity and pressure depend on the relative scales such as the ratio between the radius of the sphere (a) and the observation position (r): the sphere’s radius with respect to the wavelength (ka) and the observation position with regard to the wavelength (kr). In other words, the absolute length scale does not have any significance. For example, the location where we measure the sound pressure should be determined with respect to how far we are in terms of the wavelength and the sphere’s diameter. The significance of the relative scale becomes more important if we look at the acoustic impedance of the sphere (Zr ). This can be obtained from Equations 4.8 and 4.9, that is, ( Zr ¼ r0 c
ðkrÞ2 1 þ ðkrÞ2
j
kr
)
1 þ ðkrÞ2
9 > 1 > > = 1 kr ¼ r0 c j 2 2 : > 1 1 > > > > > ; :1 þ 1þ kr kr 8 > > > <
ð4:10Þ
Figure 4.3(a) illustrates Zr with respect to kr, which scales the observation position with respect to the wavelength. From this graph, we can explore many interesting radiation characteristics of the breathing sphere. For example, the real part of the specific acoustic impedance tends to 1 and its imaginary part is negligible when kr is much larger than 1. This means that we are observing the waves at a distance far from the origin compared to the wavelength of interest. In fact, we often say that we are seeing the waves in a far field. In this case, the radiation from the breathing sphere resembles a one-dimensional acoustic wave (plane wave), because the acoustic impedance is 1. We call this distance, or field, the “far field,” as noted in Chapter 2. In the far field, we observed that the acoustic impedance approaches the characteristic impedance of the medium, and is purely resistive. This means that the pressure and velocity are in phase with each other. On the other hand, as we approach the near field, the real part of the impedance becomes smaller and the contribution of the imaginary part tends to be larger. In particular, when kr ¼ 1 (i.e., the wavelength is 2p times the wavelength of interest), then the
181
Radiation, Scattering, and Diffraction
(kr )2 , 2 1 + (kr )
Zr Zr = a , ρ0c ρ0c 1
0.5
(ka)2 2 1 + (ka)
kr ka , 2 2 1 + (kr ) 1 + (ka)
1
kr, ka
(a)
Πavg 1
Πavg =
(ka) 2 1 + (ka) 2
6 dB/octave
ka
(b)
Figure 4.3 Impedances of the breathing sphere: (a) the acoustic impedance and the radiation impedance, and (b) the radiation power. (kr and ka dominate the characteristics of the impedances; as they become larger, the wave behaves as if it is planar)
contribution of the real part and the imaginary part to the radiation is the same. In the region where kr < 1, the imaginary part (i.e., the reactive part of the impedance) dominates the radiation characteristics. Therefore, the radiation is not likely to be effective. We generally call this region the near field. It is also interesting to study the radiation impedance of a breathing sphere at r ¼ a. This impedance exhibits how well the sphere radiates sound from its surface. This can be obtained from Equation 4.10, that is, ( Zr¼a ¼ r0 c
ðkaÞ2
1 þ ðkaÞ2 8 > > > <
j
ka
)
1 þ ðkaÞ2
9 > 1 > > = 1 ka ¼ r0 c 2 j 2 : > 1 1 > > > > > ; :1 þ 1þ ka ka
ð4:11Þ
182
Sound Propagation
This impedance is analogous to the driving point impedance of the string.2 When ka becomes larger, the radiation from the sphere tends to be similar to that of a plane wave. Note that ka determines the size of the sphere with regard to the wavelength. As ka becomes larger, the size of the sphere will therefore be bigger compared to the wavelength. This implies that the size of the radiator has to be larger as the frequency of radiation becomes lower, thus illustrating why a sub woofer is typically much larger than a mid-frequency speaker. Figure 4.3(a) illustrates how well the sphere radiates sound with respect to ka, which is the fundamental scale of the radiator. The phenomenon that we observe when ka becomes larger can also be envisaged as many infinite strings attached to a sphere. In this case, we only have an outgoing wave propagating from the surface of the sphere.3 On the other hand, in the near field, the sphere has many strings that have finite length and an impedance boundary condition. We can also define the mechanical impedance (Zm ) by multiplying Equation 4.11 by the surface area (4pa2 ). We then obtain ( ) ðkaÞ2 ka 2 j Zm ¼ 4pa r0 c 1 þ ðkaÞ2 1 þ ðkaÞ2 9 > 1 > > = 1 ka 2 ¼ 4pa r0 c 2 j 2 : > 1 1 > > > > > ; :1 þ 1þ ka ka 8 > > > <
ð4:12Þ
This again allows us to look at the radiation characteristics of the breathing sphere. For example, if ka is much larger than 1, then the mechanical impedance will be written as Zm ¼ 4pa2 r0 c:
ð4:13Þ
When ka is much smaller than 1, the mechanical impedance will be Zm ¼ jk4pa3 r0 c:
ð4:14Þ
Equation 4.14 can be rewritten as Zm ¼ 3 jomd
ð4:15Þ
by introducing a new term, md ¼ ð4=3Þpa3 r0 , which is the mass of the medium occupied by the sphere. Equation 4.15 shows that the pressure simply accelerates the fluid particles when ka is much smaller than 1.4 The imaginary part of the mechanical impedance (4.12) using md is given by md ImfZm g ¼ 3o : ð4:16Þ 1 þ ðkaÞ2
2
This radiation impedance is the input (force or pressure) divided by the output (velocity), which shares the concept of mechanical impedance (Chapter 2). 3 We can also see why the low frequency speaker is louder than the high frequency speaker from this observation. 4 If there is a phase difference of jo in the impedance, then it means that the fluid particle behaves as if it is a mass.
183
Radiation, Scattering, and Diffraction
Equation 4.16 implies that the accession to inertia depends on both frequency and ka. Note that the inertia tends to be smaller as the sphere size increases relative to the wavelength of interest. This means that we have a more efficient radiator. If we calculate the mean intensity (Iavg ) by using Equations 4.8 and 4.9, then we obtain 1 Iavg ¼ RefPU* g 2 a2 ðkaÞ2 a2 1 1 ¼ r0 cU20 ¼ r0 cU20 2 2 r 1 þ ðkaÞ 2 r
; 1 2 1 1þ ka
ð4:17Þ
where P and U* are the sound pressure magnitude (4.9) and the complex conjugate of the velocity magnitude (4.8). If we want to consider the radiation power, it can be obtained from Equation 4.17 by multiplying the area of interest by 1 ðkaÞ2 1 ¼ r0 c4pa2 U20 Iavg 4pr2 ¼ r0 c4pa2 U20 2 2 2 1 þ ðkaÞ
1 2 : 1 1þ ka
ð4:18Þ
Equation 4.17 expresses the intensity at r, and Equation 4.18 denotes the radiation power on the surface of the radius r. Note that the term ð1=2Þr0 c4pa2 U20 in Equation 4.18 is the radiation power in the case of a plane wave that is radiating from the surface of the area 4pa2 . In other words, the difference in radiation power (4.18) between the breathing sphere and the plane wave is the last term of Equation 4.18, i.e. the function of ka. Consider the normalized radiation power ðPavg Þ of a circular plate of radius a with a velocity of U0 and frequency o. This is expressed as Pavg ¼
Iavg 4pr2 1 r c4pa2 U20 2 0
¼
ðkaÞ2 1 þ ðkaÞ
2
¼
1 2 ; 1 1þ ka
ð4:19Þ
which highlights that the radiation power becomes larger as we increase ka (Figure 4.3(b)). It indicates that we can increase the radiation power by 6 dB by doubling the size or dimension of radiator when the wavelength is large relative to the size of the radiator (a). If the wavelength is much smaller than a, then the average power asymptotically approaches that of plane waves, which is the maximum possible radiation. A second type of basic radiation unit is a trembling sphere (Figure 4.4). Assuming that the sphere vibrates in the z direction with a velocity of Uo , the acoustic waves generated by this trembling sphere would satisfy the acoustic wave equation in the spherical coordinate, that is, 1 @2 1 @ @U 1 @2U 1 @2U ðrUÞ þ ¼ : siny þ 2 2 2 2 r @r r sin y @y @y r sin y @f2 c2 @t2
ð4:20Þ
184
Sound Propagation z Uc
Ur
0 r ϕ
y
x
Figure 4.4 The trembling sphere. (The direction of vibration is z, the velocity magnitude is Uc , and Ur is the velocity in the r direction. Other symbols represent coordinates)
It is obvious that the waves are independent of the f coordinate. In this case, the boundary condition on the surface of the trembling sphere (r ¼ a) can be written as Ur¼a ¼ Uc cos y ¼
@U @r
;
ð4:21Þ
r¼a
where Ur¼a denotes the velocity of the trembling sphere in the r direction. The tangential velocity components in the r direction are not considered, because we assume that the fluid is inviscid. In this case, the solution satisfying Equation 4.20 is found to be5 Uðr; y; fÞ ¼ A cos y
@ e jkr : @r r
ð4:22Þ
The pressure (p) can be calculated from Equation 4.22 as @ ðUejot Þ @t @ e jkr jot ¼ Ar0 cos yjo : e @r r
p ¼ r0
5
ð4:23Þ
It is known that we can obtain the solution of the trembling sphere by either applying a coordinate transformation or using a separation of variables in the governing equation. However, we can try to use the solution (Equation 4.22) that satisfies the governing equation and boundary condition. Intuitively, the solution would be of dipole type because the sphere trembles in one direction.
185
Radiation, Scattering, and Diffraction
The velocity (Ur ) can be obtained by taking the derivative of the potential function with respect to r as below: @U @ 2 e jkr : ð4:24Þ Ur ¼ ¼ A cos y 2 @r r @r Equations 4.21 and 4.24 lead us to write @ 2 e jkr ; Uc cos y ¼ A cos y 2 @r r r¼a
ð4:25Þ
and we can obtain A as A¼
U c a3 2ðkaÞ2 2jka
ejka :
ð4:26Þ
The pressure magnitude (p) can therefore be obtained by using Equations 4.23 and 4.26, that is, 1 jk 1 jkðraÞ p ¼ jkr0 cUc a3 cos y : ð4:27Þ e 2ðkaÞ2 2jka r r2 The velocity (Ur ) can be calculated using Equations 4.24 and 4.26. This gives us U c a3 2 2jk k2 jkðraÞ Ur ¼ cos y 3 2 e : ð4:28Þ r r r 2ðkaÞ2 2jka The acoustic impedance (Zr ) can be obtained from Equations 4.27 and 4.28 as ( Z r ¼ r0 c
ðkrÞ4
2kr þ ðkrÞ3
)
j 4 þ ðkrÞ4 4 þ ðkrÞ4 8 3 9 > 1 1 > > > > > þ2 = < 1 kr kr ¼ r0 c 4 j 4 ; > 1 1 > > > > > ; :1 þ 4 1þ4 kr kr
ð4:29Þ
which states that the impedance is independent of y although the pressure and velocity magnitude depend on cos y. This is a rather surprising result. Note that the maximum pressure and velocity occur in the direction in which the sphere vibrates (y ¼ 0 ) (Figure 4.5). On the other hand, the pressure and velocity are both zero along the line that is perpendicular to the direction of vibration, which is quite different from the case of the breathing sphere. However, as we can see from Figure 4.6, the sound waves in the far field are somewhat similar to those in the case of the breathing sphere; the waves become plane waves in the far field. In relation to our investigation in the breathing sphere case, let us look at the radiation characteristics of the trembling sphere, in particular, the radiation impedance. From
186
Sound Propagation 0
1 330
30 0.8 0.6 60
300 0.4 0.2
90
270
120
240
150
210 180 D(θ ) = cos2 θ
Figure 4.5 The directivity factor of a trembling sphere; the circumferential angle is expressed in degrees and the radial distance is a non-dimensional arbitrary unit
Equation 4.29, we obtain the impedance as ( ) ðkaÞ4 2ðkaÞ þ ðkaÞ3 Zr¼a ¼ r0 c j 4 þ ðkaÞ4 4 þ ðkaÞ4 8 3 9 > 1 1 > > > > > þ2 < = 1 ka ka ¼ r0 c 4 j 4 : > > 1 > > >1 þ 4 1 > : ; 1þ4 ka ka
ð4:30Þ
Equations 4.11 and 4.30 show that the radiation of the breathing sphere and trembling sphere depend on ðkaÞ2 and ðkaÞ4 , respectively. If ka is small (i.e., if the radius of the sphere is small compared to the wavelength of interest), then the reactive term which is the imaginary part of the impedance dominates the radiation characteristics. On the other hand, when ka is large, then the resistive term governs the impedance and the trembling sphere effectively radiates sound waves. These observations make it possible to understand how the sound waves of percussion instruments are radiated and why we hear lower frequency sounds better than higher frequency sounds as we move away from the instruments. We can also obtain the expression for the normalized radiation power ðPavg Þ as Pavg ¼
ðkaÞ4 4 þ ðkaÞ
4
cos2 y ¼
1
2 cos y: 1 4 1þ4 ka
ð4:31Þ
187
Radiation, Scattering, and Diffraction Zr
ρ0 c
,
Z r =a ρ0c
1
(kr ) 4 , 4 + (kr ) 4
2 2
2kr + (kr )3 , 4 + (kr ) 4
(ka ) 4 4 + (ka ) 4
2ka + (ka ) 3 4 + (ka ) 4
kr, ka
2 (a)
Π avg 1
( ka ) 4 4 + ( ka ) 4
Π avg =
12 dB/octave
ka
(b)
Figure 4.6 Impedances of the trembling sphere: (a) the acoustic impedance and the radiation impedance; and (b) the radiation power. Note that the radiation impedance and the radiation power are proportional to ðkrÞ4 and ðkaÞ4 , respectively. The corresponding characteristics of the breathing sphere are proportional to ðkrÞ2 and ðkaÞ2 . The radiation of the trembling sphere therefore depends on the viewing location and the size of the diameter relative to the wavelength of interest
Figure 4.6(b) illustrates the behavior of the radiation power. From this figure, we can readily see that the radiation impedance rapidly decreases as ka becomes small; it decreases by 12 dB/octave.6 It should be noted that the pressure and velocity depend on y. When y ¼ 90 or 270 , the radiation pressure is zero. On the other hand, the radiated pressures reach their maxima when y ¼ 0 or 180 . To effectively express this angle dependency of the radiation, we define the directivity factor as D¼
I Isphere
;
ð4:32Þ
where Isphere is the intensity radiated from the breathing sphere with radiation power equivalent to that of the radiator of interest. I denotes the intensity of the radiator whose directivity we wish 6
It is 6 dB/octave in the case of the breathing sphere.
188
Sound Propagation
to characterize.7 Figure 4.5 illustrates the directivity factor of the trembling sphere. As indicated by Equations 4.31 and 4.32, the radiated sound from a trembling sphere (e.g., sound radiation from a volume source such as a conventional audio speaker or percussion musical instrument) tends to have more directional dependency as the wavelength becomes smaller than its radiator size. This is one reason why we normally need two speakers separated by a significant distance, effectively generating a stereo sound. To summarize, we have studied the basic units of radiators, that is, the breathing sphere and the trembling sphere. The characteristics of the radiators are expressed in terms of sound pressure, velocity, intensity, acoustic impedance, radiation impedance, and radiation power. For the breathing sphere, the sound pressure was found to be inversely proportional to the observation distance from the origin, and the power is inversely proportional to the square of the distance. The expression ka, which describes the size of the wavelength of interest relative to the radiator size, determines the radiation characteristics. As ka becomes larger, the radiated sound becomes planar. Note that for relatively small values of ka, the radiated power increases by 6 dB when ka doubles in size. For a trembling sphere, the radiated sound depends strongly on the axis of propagation; in other words, it has very strong directivity. For relatively small values of ka, the radiated power increases by 12 dB when ka doubles in size. In this case, it is more rapid than in the breathing sphere.8 We anticipate that these two radiators can be used to construct any type of radiator by the principle of superposition. In other words, we can draw any radiated sound field in space using these two basic paint brushes. Their characteristics are understood by using their impedances.
4.3
Radiation from a Baffled Piston
We have seen that the vibration of fluid particles induces sound propagation in space. The propagation strongly depends on the characteristic impedance of the medium. However, radiation is not only governed by the medium impedance but also by the relation between the fluid particle velocity and the pressure on the surface of the radiator; this is defined to be the radiation impedance. Section 4.2 introduces two basic radiators that feature all the necessary physics to understand radiation. There are basically two ways to make sound. As the Kirchhoff–Helmholtz integral equation (2.90) states, sound can be generated by the fluctuation of fluid particles which are attached to the surface of a vibrating structure, or, if there is a certain pressure fluctuation, then sound propagation will be induced from the pressure changes. The former is generally called “velocity source or velocity sound source” and the latter is called “pressure source or pressure sound source.”9
7
Section 4.7.1 summarizes all kinds of coefficients and indices that express the directivity of the sound source. The radiation power decreases more rapidly compared to that of the breathing sphere as ka becomes smaller. The breathing sphere is therefore a better radiator than the trembling sphere for a given velocity U. 9 These two methods of sound generation are usually used in musical instruments. For example, a flute uses the pressure fluctuation. On the other hand, the violin and cello use the vibration of the structure that is induced by the excitation of the bow on the string. 8
189
Radiation, Scattering, and Diffraction
Figure 4.7 illustrates the boundary condition for the integral equation (2.90). We can rewrite the integral equation that emphasizes the individual contribution of pressure and velocity sources as ð Pð~ rÞ ¼ ðPð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r 0 ÞÞ ~ n 0 dS Sp
ð4:33Þ
ð
þ
ðPð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r 0 ÞÞ ~ n 0 dS: Su
S∞ ∇2P + k2 P = 0 Sp
Su
U(r0)
P(r0)
S∞
r0 r O
Figure 4.7 Volume integral of the Kirchhoff–Helmholtz integral equation. (Sp and Su express the boundary surface for the pressure P and velocity U, S1 denotes a surface that is infinitely far from the origin, and ~ r and ~ r 0 indicate the observation position and the boundary position vector, respectively)
Equation 4.33 conveys that the sound pressure at an arbitrary location is what is propagated from the location where it is produced by the pressure and velocity sources. To understand how we can use this equation, consider the sound radiation from a baffled piston as illustrated in Figure 4.8. Note that we only have the second integral of Equation 4.33 in this case. If we apply Equation 4.33 to this specific case by employing the surface of the integral ðS0 Þ (Figure 4.8), then we obtain10 ð jkr0 c e jkR Pð~ rÞ ¼ Un dS0 : ð4:34Þ 2p S0 þ R The surface integral S0 is composed of S0þ and S 0 . The velocity distributions on each surface are equal in magnitude but different in direction. The boundary condition of the baffle, which requires the velocities normal to the surface to be 0, is therefore automatically satisfied. Green’s functions which are employed satisfy the aforementioned boundary conditions and the linear acoustic wave equation. 10
190
Sound Propagation
Un = 0
Un = 0 R = r – ro
R
=
Un
Un r
r ro
ro S0 – S0 +
O
O
Figure 4.8 Surface integral to calculate the sound radiation from a baffled piston and nomenclature of the coordinate
@G @G on S0 þ and on S0 (Figure 4.8) have @n @n opposite signs, but the pressure on S0 þ and S0 is equal. Therefore, the first term of the second integral in Equation 4.33 disappears, and we only have the second term. Using free-space Green’s function in Equation 4.33, we can derive Equation 4.34 (see Section 4.7.4.4). Note that Equation 4.34 states that the radiated sound pressure is induced by the vibrating surface velocity. The waves generated by the vibrating surface propagate like a monopole in space. The magnitude of vibration is simply a scale factor of the sound pressure that is propagated from the surface to the observation position of interest. It can be demonstrated by dividing the integral surface into small elements. We can then see that the velocity of each element (Un ) causes fluid particles to vibrate and that the disturbances then propagate to the observation position (~ r) as the monopole propagator. Equation 4.34 can also be regarded as an expression of Huygen’s principle in an integral form. Note that the denominator of the integrand takes the absolute distance. This is because the propagator, which is monopole, depends only on the distance. This is one of the characteristics of a monopole source. We start with the case where Un is constant. In other words, the piston is a rigid vibrator (Figure 4.9). By using Equation 4.34, the pressure at an arbitrary position (z) can be expressed by pffiffiffiffiffiffiffiffiffiffi2 ð 2 jkr0 c a e jk z þ z pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Un 2pzdz: PðzÞ ¼ ð4:35Þ 2p 0 z2 þ z2 This result can be obtained by realizing that
This equation can be simplified as PðzÞ ¼ r0 cUn ðeikz e jkR0 Þ;
ð4:36Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi where R0 ¼ z2 þ a2 is the distance from the edge of a circular piston to z (detailed derivation is given in Section 4.7.2). This result has significant physical implications. There are two distinct contributions in the z direction. The first ðeikz Þ is the pressure coming from the center of the piston, and the second term ðe jkR0 Þ is from the piston’s rim. The coefficient (r0 cUn ) is the pressure of the plane wave that is generated by the piston’s motion. Note that these two pressure waves interfere with each other. The interference sometimes mutually cancels the waves or reinforces them. In the far field (i.e., when z is large compared to the wavelength of interest), the
191
Radiation, Scattering, and Diffraction
Une–jω t
a
z
Un ζ
R=
z2 + ζ 2 z
a Ra =
z2 + a2
dS = 2πζdζ
Figure 4.9
The radiated sound field from an infinitely baffled circular piston
cancellation is more likely to dominate the result. If the wavelength of interest is fairly large compared to the piston’s diameter (a=l 1), then perfect cancellation occurs independently of the observation position. If z is significantly large compared to the radius of the piston, then Equation 4.36 becomes11 PðzÞ ¼ r0 cUn
jka a jkz e : 2 z
ð4:37Þ
This essentially states that the far field radiation of the baffled piston is similar to monopole radiation. This is a general observation which can be made in most cases. The near field effect ceases rapidly and only monopole behavior persists in the far field (Section 4.7.2). Note that Equation 4.37 describes the sound pressure along the center line of the piston (z axis). Figure 4.10 depicts the coordinates and nomenclature that we use to predict the radiation from the vibrating piston to an arbitrary position, including the z axis. The sound pressure (Pðr; yÞ) on the xz plane is given as12 Pðr; yÞ ¼ jkr0 cUn
e jkr 2 J1 ðka sin yÞ a ; r ka sin y
ð4:38Þ
where J1 is a Bessel function of the first kind. It is interesting to look at the sound pressure at y ¼ 0 , that is, Pðr; 0Þ ¼ jr0 c
11 12
For a detailed derivation, see Section 4.7.2. Section 4.7.3 includes the detailed derivation.
ka2 jkr Un e : r 2
ð4:39Þ
192
Sound Propagation y
x
R
ζ
R0
dS r
φ θ
z a
Figure 4.10 Coordinate set-up and variables for obtaining radiated sound field on xz plane from an infinitely baffled circular piston
This is identical to Equation 4.37, meaning that Equation 4.38 is a general expression of the radiation sound pressure on the xz plane. Note that Equations 4.38 and 4.39 do not exhibit any dependency with respect to f (Figure 4.10). This is a simple consequence of the symmetry property of radiation: we assumed that the velocity distribution on the piston is uniform (Un ). The sound pressure on the yz plane, therefore, will be the same as that on the xz plane because the piston’s vibratory motion is symmetrical with respect to the x and y axes. Using Equations 4.38 and 4.39, we can obtain an expression which provides further significant physical insight, that is, Pðr; yÞ ¼ Pðr; 0Þ 2
J1 ðka sin yÞ : ka sin y
ð4:40Þ
Also, note that the mean intensity is Iavg ¼
jPj2 : 2r0 c
ð4:41Þ
Therefore, the intensity ratio between the time-averaged intensity at an arbitrary position with respect to that of the axis attached to the center of the piston can be expressed as Iavg ðr; yÞ J1 ðka sin yÞ 2 ¼ 2 ð4:42Þ Iavg ðr; 0Þ ka sin y by using Equations 4.40 and 4.41. Figure 4.11 depicts Equation 4.42 in terms of ka. Equation 4.42 is often called the directivity or directivity index. Other names such as spreading and spreading index are also widely used.
Radiation, Scattering, and Diffraction
193
From Figure 4.11, we can see that a key factor governing the radiation characteristics is ka. This basically measures how large the radiator is compared to the wavelength of interest. Equation 4.42 states that the angular dependency of the radiation significantly decreases as the radiator size decreases relative to the wavelength, or when we have lower frequency radiation. On the other hand, when ka becomes larger, or the frequency becomes higher, the radiation strongly depends on the angle. In other words, the directivity tends to be angle dependent. As illustrated in Figure 4.11, there is a silent region. In other words, there are some angles where we cannot hear sound coming from the piston.
Figure 4.11
The directivity or directivity index of a baffled circular piston
We now examine the radiation impedance of the circular baffled piston, which shows the relation between pressure and particle velocity on the surface of the piston. As we have
194
Sound Propagation
observed previously, the driving point impedance of a string (Chapter 1) is similar in concept to that of radiation impedance. Note that the radiation impedance of the plate varies with the position of the plate. The velocity is constant (Un ) on the surface, but the resulting pressure is not uniform. We therefore need to define the radiation impedance of this case. One possible definition can be written as Zr ¼
Pavg F=S ¼ ; Un Un
ð4:43Þ
where F is force acting on the source surface (S), and Pavg is the average pressure on the surface. To obtain F, we need to calculate the pressure on the surface of the piston (PS ðz0 ; f0 Þ). This can be regarded as the sum of the pressure induced by the other area over the entire piston surface, that is, jkr0 cUn PS ðz ; f Þ ¼ 2p 0
0
ð
e jkR jkr0 cUn dS ¼ R 2p S
ð 2p ð a 0
e jkR zdzdf; 0 R
ð4:44Þ
where Figure 4.12(a) illustrates the variables for the integration. The force acting on the piston surface can be given by ð
PS dS0
F ¼ S
ð 2p ð a
¼
0
PS ðz0 ; f0 Þz0 dz0 df0
jkr0 cUn ¼ 2p
ð 2p ð a ð 2p ð a 0
0
0
y dS ζ
ð4:45Þ
0
e jkR zdzdf z0 dz0 df0 : 0 R
y R φ ζ′ φ′
(ζ ′, φ ′ )
R x
θ ζ′
φ′
(ζ ′, φ ′) x
a
(a)
Figure 4.12 and y
(b)
The variables on the surface of the disk for integrating with respect to (a) z and f, and (b) R
195
Radiation, Scattering, and Diffraction
The integration of Equation 4.45 can be written as: ð 2p ð a ð 2p ð a 0
e jkR zdzdf z0 dz0 df0 0 0 0 R # ð 2p ð a "ð p=2 ð 2z0 cos y jkR e RdRdy z0 dz0 df0 ¼2 R 0 0 p=2 0 ¼
2p jk
ð 2p ð a 0
ð4:46Þ
½1J0 ð2kz0 ÞjH0 ð2kz0 Þz0 dz0 df0 ;
0
where J0 ð2kzÞ is the first kind of Bessel function of zero order, and H0 ð2kzÞ is the zero-order Struve function. Figure 4.12(b) illustrates the variables of the disk used to calculate the integration. The force acting on the piston surface can then be given by ð ð jkr0 cUn 2p 2p a ½1J0 ð2kz0 ÞjH0 ð2kz0 Þz0 dz0 df0 F ¼ 2p jk 0 0 ð 2p ð a ¼ r0 cUn ½1J0 ð2kz0 ÞjH0 ð2kz0 Þz0 dz0 df0 0
ð4:47Þ
0
J1 ð2kaÞ H1 ð2kaÞ j ; ¼ r0 cUn pa2 1 ka ka where J1 ð2kaÞ is the first kind of Bessel function of the first order, and H1 ð2kaÞ is the Struve function of the first order. Pavg is then defined as J1 ð2kaÞ H1 ð2kaÞ j ; Pavg ¼ r0 cUn 1 ka ka
ð4:48Þ
and the radiation impedance of a baffled circular piston with radius a can be given by J1 ð2kaÞ H1 ð2kaÞ Zr ¼ r0 c 1 j : ka ka
ð4:49Þ
Figure 4.13 shows the normalized radiation impedance ðZr=r0 cÞ with respect to ka. As ka becomes large (i.e., radius becomes larger than wavelength), the resistance term approaches the medium impedance (r0 c). On the other hand, the reactance is quickly diminished. It shows that the baffled piston radiates very effectively, as if generating a plane wave, when the wavelength of interest is larger than the radius a. To summarize this section, we have studied the radiation characteristics of the sound generated by the vibration of a baffled piston. These are essentially governed by the size of the radiator and the distance from which the sound pressure is observed compared to the
196
Sound Propagation Zr ρ0c 1–
J1 (2ka) ka
1
0.7
H1 (2ka) ka
1.32
ka
Figure 4.13 Radiation impedance, which is normalized to r0 c, of the circular baffled piston with respect to ka. The solid line represents the resistance term, and the dashed line the reactance
wavelength. The size and distance of the observation position have to be scaled by the wavelength, and the absolute scale does not possess any significant meaning with respect to the radiation characteristics. We also found that the directivity index is a good measure to examine the angle dependency of the radiation. The radiation tends to be of monopole type when we have a smaller radiator compared to the wavelength. On the other hand, as the radiator increases in size compared to the wavelength, there will be more interference. As a result, there will be a very rapid angle fluctuation in the directivity index. If the size of a baffle is finite, we cannot expect similar results. However, if for example we produce a wave using a speaker unit that is installed in a box, then radiation characteristics as described in this section may or may not be observed depending on the size of the box compared to the wavelength. Overall, examination of the characteristics of a baffled piston will provide basic understanding of sound radiation. It is obvious that if we want to determine precisely how sound is radiated from an arbitrarily vibrating body or surface, it is necessary to use numerical methods (addressed in Section 4.7) or conduct experiments.
4.4
Radiation from a Finite Vibrating Plate
It may be possible to apply what we learned regarding the radiation from a baffled piston to the radiation from a vibrating plate. This idea stems from the realization that the vibrating plate can be modeled as numerous vibrating pistons, as illustrated in Figure 4.14.13
13 We can assume that the plate vibrates with sufficiently small amplitude so that the radiated pressure obeys the linear acoustic wave equation. Therefore, the principle of superposition holds. Figure 4.14, in fact, shows how we can apply the principle to predict the radiated sound field.
Radiation, Scattering, and Diffraction
Figure 4.14
197
(a) Radiation from a finite plate and (b) its possible modeling
Figure 4.14(a) depicts how the vibration of a plate can be considered to be made by the n modes of the plate. The radiation due to the plate vibration can therefore be considered to be composed of the radiation of n modes of vibration. The radiation due to each mode of vibration can be superimposed by the vibration of many pistons, as illustrated in Figure 4.14(b). We can assume that each mode of vibration can have equivalent pistons.14 The piston’s amplitude of vibration has to have the same volume velocity as the corresponding mode of vibration. Equation 4.34, the Rayleigh integral equation, can be applied to each equivalent piston. The superposition concept can be used, as illustrated in Figure 4.14. The radiated pressure field from the vibrating plate can be readily obtained. To understand the radiation from the vibrating plate, we look at basic radiations patterns such as those illustrated in Figure 4.15. The radiated fields are obtained using the Rayleigh integral equation. The radiator’s typical dimensions are larger than the wavelengths that are generated; the radiation efficiencies are therefore fairly good for all three cases. It should be noted that the sound pressures at the distance from the radiators (zl ) become negligible when the distance is long compared to the wavelength, except for case (a). This is because the far field sound propagation tends to a plane wave, resulting in 14
This concept essentially motivates us to establish the boundary element method (BEM).
198
Sound Propagation
a perfect cancellation for the cases of (b) and (c) in Figure 4.15. Note that the observation positions of these cases are exactly on the axis of symmetry, therefore the waves have a phase difference of 180 . This kind of cancellation becomes more and more significant when we have higher order modes.
Figure 4.15 Basic examples of plate vibration and radiation (A indicates the observation position, and can be expressed as (0, 0, zl ) in the Cartesian coordinate)
To look more specifically at the radiation from a plate, we describe the waves in space using rectangular coordinates for convenience. Let us also assume that the wave is harmonic in space and time, without loss of generality. The acoustic wave can now be written as ~
pðx; y; z; tÞ ¼ Pe j k ~r ejot ¼ Pe jðkx x þ ky y þ kz zÞ ejot :
ð4:50Þ
Equation 4.50 must satisfy the linear acoustic wave equation; the following equality must therefore hold, that is, kx2 þ ky2 þ kz2 ¼ k2 ;
ð4:51Þ
where k ¼ o=c, which is a dispersion relation that relates spatial and time domain fluctuation. If we have a vibrating body only on the ðx; yÞ plane which vibrates as illustrated in Figures 4.14 and 4.15, then the wave numbers in the x and y directions, (kx and ky ) can be written as mp ð4:52Þ kx ¼ Lx
199
Radiation, Scattering, and Diffraction
ky ¼
np ; Ly
ð4:53Þ
where Lx and Ly are the length of the plate in the x and y directions, respectively, and m and n are integers. These relationships can also be obtained from the boundary conditions of the plates. Figure 4.16(a) illustrates what is implied by these expressions. Figure 4.16(a), Equations 4.51, 4.52, and 4.53 tell us that the wave number in the z direction (kz ), which describes how the wave in the z direction propagates, could be real or imaginary. This depends on whether the free space wave number (k) is smaller or greater than the wave number in the ðx; yÞ plane, which is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kx2 þ ky2 . The former case can be written in terms of the inequality between the wave numbers, that is kx2 þ ky2 < k2 :
ð4:54Þ
Figure 4.16 Radiation from a finite plate: (a) corner mode, edge modes, and radiation circle in wave number domain and (b) edge (corner) mode dominates the radiation in the left- (right-) hand image
In this case, ðkx ; ky Þ is located inside a radiation circle whose radius is the free space wave number k and its equation is k2 ¼ kx2 þ ky2 þ kz2 . The wave number in the z direction is always positive and real, as indicated by Equations 4.51 and 4.54. This means that a wave in the z direction continuously changes its phase as it propagates. In other words, the propagator
200
Sound Propagation
can be mathematically written as e jkz z . On the other hand, the latter case satisfies the inequality: kx2 þ ky2 > k2
ð4:55Þ
In this case, kz has to be imaginary and the wave therefore decays exponentially. Note that ðkx ; ky Þ is located at the outside of the radiation circle (Figure 4.16(a)).15 This figure also illustrates that the wave is less likely to propagate in the z direction with a larger kx or ky, in other words, as the wavelength becomes increasingly smaller. This is simply because the wave numbers in the x and y directions approach the region far away from the radiation circle as the wavelength becomes smaller. Therefore, propagation in the z direction is not highly plausible. To summarize, the development of either exponentially decaying propagation (evanescent wave) in the z direction or a continuous phase changing wave is determined by the location ðkx ; ky Þ in reference to the radiation circle. Figure 4.16 also illustrates the radiation characteristics of the plate when it has higher order modes in either the x or y directions. If kx ¼ 0 but ky is getting larger, then the vibration mode tends to have more oscillation in the y direction as illustrated in Figure 4.16(b). This kind of mode is often referred to as edge mode. We can easily see that the radiation due to this edge mode will be far less effective than that of the baffled piston because of possible wave cancellation, as illustrated in Figure 4.15. Corner modes occur when the vibrations and radiations with respect to x and y are equally likely to participate.16 The radiation impedance of a vibrating structure that has a finite area (for example, baffled piston or plate, etc.) should be considered by averaging over the area as explained in Section 4.3. In particular, a vibrating plate that has corner and edge modes must not only consider average sound pressure on the plate but also averaged velocity on the area. For instance, the normal velocity distribution on a baffled plate can be expressed as ( 0 x0 Lx Un ðx0 ; y0 Þ ¼ Umn sinðkx x0 Þsinðky y0 Þ; ð4:56Þ 0 y 0 Ly where Umn is the velocity magnitude on the plate, the subscript 0 indicates source coordinate, and the coordinate set-up is the same as for the rectangular membrane in Section 1.9.3 (see Figure 1.39). Sound pressure at an arbitrary observation point in the positive z direction can be determined by Rayleigh’s integral equation of Equation 4.34. Sound pressure by the vibration (4.56) can therefore be written as jkr0 cUmn Pðx; y; zÞ ¼ 2p
Lðy Lðx
sinðkx x0 Þsinðky y0 Þ
e jkR dx0 dy0 ; R
ð4:57Þ
0 0
where R represents the distance between an observation point and an infinitesimal area (dx0 dy0 ) on the plate. The radiation impedance of a vibrating baffled plate can be obtained 15 The radiation circle is normally defined in this context. For example, the evanescent wave and propagating wave are decided by the region in which the wave number is located. 16 These characteristics may or may not be good depending on how we want to use them. For example, if we want to use these characteristics to reduce the sound or noise, then we would like to have more corner or edge modes. On the other hand, to create a good radiator, we want to avoid these modes.
Radiation, Scattering, and Diffraction
201
according to the definition of Equation 4.49 in Section 4.3. Note, however, that the velocity in the denominator also has to be averaged over the area of the plate because the velocity at a point on the plate cannot represent the entire velocity distribution. We can also extend the concept of studying sound propagation by using wave numbers into a general three-dimensional case. Acoustic holography is essentially based on this concept. In acoustic holography, the spatial distribution of a sound wave is expressed in terms of the wave number domain of interest, and then propagated to a plane that is not measured by using a propagator. For example, we transform the sound wave field in the ðx; yÞ plane to the ðkx ; ky Þ plane, and then use the kz propagator. Figures 4.17 and 4.18 illustrate the basic procedures of acoustic holography.
Figure 4.17
Conceptual diagram of acoustic holography in rectangular coordinates17
To summarize, the radiation from a plate vibration depends very strongly on the vibration characteristics of the plate. It is understood that the radiation from a plate is effective when the size of the plate is much greater than the wavelength that we wish to generate. On the other hand, if the wavelength is much smaller than the size of the plate, then an edge or corner mode will arise. The associated radiation efficiency tends to be smaller as there are more modes of vibration. This result can be visualized if we consider the possible motions of fluid particles on a plate.
4.5
Diffraction and Scattering
Figure 4.19 illustrates typical diffraction and scattering phenomena. We can clearly see from this figure that the diffraction tends to be stronger as the wavelength becomes larger. In other words, 17
Kim, Y.-H. (2004) Can we hear the shape of a noise source? ICA 2004, plenary lecture, Kyoto International Conference Hall, 4–9 April 2004. http://soundmasters.kaist.ac.kr.
202
Sound Propagation
Figure 4.18
Illustration of acoustic holography and its fundamental procedure
we have more diffraction at the back of the wall as the wavelength increases. Strictly speaking, as the wall’s height decreases compared to the wavelength there will be more diffraction. Diffraction is generally used to describe the physical circumstances under which we can hear sound but cannot see the sound source. As illustrated in Figure 4.19, these circumstances
Figure 4.19 wavelength)
Diffraction around a straight barrier (note that the diffraction strongly depends on the
203
Radiation, Scattering, and Diffraction
typically occur when a sound wave meets a discontinuity that is open in half space. The sound waves arriving at the lower part of the barrier are more likely to be reflected whereas, at the upper part of the barrier, the waves are mostly scattered. The reflected and incident waves are mostly dominant in the left part of the barrier, and the scattered waves dominate the remainder of the region. We often call this phenomenon diffraction. However, we also refer to such waves as “scattering.” Scattering describes waves that are induced due to an abrupt impedance change in space, especially when the waves spread out in space. Notably, reflection and scattering are due to the presence of an impedance discontinuity, and the type of acoustic wave will dictate whether reflection or scattering occurs. Section 4.7.4 mathematically describes how diffraction depends on the wavelength, frequency and observation position with respect to wavelength. As we noted in the earlier part of this chapter, radiation, diffraction and scattering can be intuitively understood in a unified way. All of these phenomena are consequences of the impedance mismatch in space and frequency. Depending on the type of impedance mismatch, the sound waves that can occur are diffraction, scattering, refraction or reflection. Note that when an incident wave is oblique on the flat surface of a discontinuity (Chapter 3), the transmitted wave normally has a different angle of transmission; we call this phenomenon “refraction.” Diffraction can also be regarded as a phenomenon that can occur when an incident wave meets an impedance mismatch in space. The impedance mismatch in space produces not only reflection but also scattering at a relatively sharp edge or edges, as illustrated in Figure 4.19. In other words, we can regard the impedance mismatch as a predictor of how much sound pressure will be propagated because it always creates pressure and velocity fluctuation. Thus, understanding of the diffraction problem is possible by finding the solution in a half space for the case of a piston excitation on an infinite baffle or finite baffle (Section 4.7.4 mathematically addresses the diffraction and radiation problems). We begin by looking at a scattering problem in terms of a unified concept that regards diffraction, scattering and other phenomenon as spatial- and frequency-dependent impedance mismatch problems. Beginning with the simplest case which considers all the necessary fundamentals, suppose that we have a plane wave impinging on an arbitrary scatterer, as illustrated in Figure 4.20. The complex amplitude in space of the incident wave is denoted Pi and the complex amplitude of the scattering wave is denoted Psc . The total sound pressure can then be expressed as Pt ¼ Pi þ Psc :
ð4:58Þ
The total pressure has to satisfy the following boundary condition, that is, n ¼ 0: rPt ~
ð4:59Þ
We assumed that the boundary is acoustically rigid and ~ n is a unit normal vector on the k. The surface (Figure 4.20). Let Pi propagate in the direction of the wave number vector ~ incident wave at the position ~ r can then be written as ~
Pi ¼ Be jk ~r ;
ð4:60Þ
204
Sound Propagation a Psc Pi = Be
jk ⋅ r
S0
n (a)
Psc
n y
Pi
Af Ab
x a
S0
Pi = Be jk ⋅r = Be
jk x x
(b)
Figure 4.20 Scattering by (a) arbitrary scatterer and by (b) a sphere. (Pi , Psc represent the complex amplitude of incident and scattered waves; S0 and~ n represent the surface of the scatterer and a unit vector normal to the surface; and a is the characteristic length of the scatterer)
where B is the complex amplitude of the incident wave (Figure 4.20). Equations 4.58, 4.59, and 4.60 lead us to a relation between the scattered wave and the incident wave, that is, ~ rPsc ~ n ¼ jB ~ k ~ n e jk ~r :
ð4:61Þ
Equation 4.61 essentially states that fluid particles on the scatterer have to move as if they are attached to it.18 If we rewrite Equation 4.61 in terms of the velocity of the scatterer by using the linearized Euler equation, then we obtain ~ sc ; rPsc ¼ r0 ð joÞU
ð4:62Þ
where we assume that the scattering velocity(~ u sc ) is harmonic in time, that is, ~sc ejot ; ~ u sc ¼ U
ð4:63Þ
~sc is the magnitude of the scattering velocity in vector form. where U By starting with the simplest case, we consider the scattering of a rigid sphere to explore what is meant by Equations 4.61 and 4.62. Rewriting Equation 4.61 using the coordinate of 18
Note that the pressure gradient induces the motion of fluid particles as described by the linearized Euler equation.
205
Radiation, Scattering, and Diffraction
Figure 4.20(b), we obtain @Psc ~ ¼ jkB cos ye jk ~r ; @n
ð4:64Þ
where y represents the angle between the propagation vector ~ k and the normal vector ~ n. ~sc ~ Equations 4.62 and 4.64 allows us to find the velocity in the normal direction, which is U n, that is, 1 ~ ~sc ~ U n ¼ jkB cos y ; ð4:65Þ e j k ~r jkr0 c on S0 where S0 is the surface of the sphere (Figure 4.20(b)). Since the radius of the sphere is a, we can rewrite Equation 4.65 as B ~sc ~ U cos ye jka cos y : n¼ ð4:66Þ r0 c The normal velocity on the rigid sphere is a function of y as shown in Figure 4.21, which shows that the real part of the directional component is cos ye jkacos y with respect to ka.
~sc ~ Figure 4.21 Directional component (cos yejka cos y ) of the normal velocity (U n) on the rigid sphere with respect to ka
206
Sound Propagation
If the size of the scatterer is substantially smaller than the wavelength of interest (i.e., if ka 1), then Equation 4.66 can be approximately written as follows by expanding it in a Taylor series and taking only the first term: B ~sc ~ nffi U cos yð1 þ jka cos yÞ: r0 c
ð4:67Þ
This means that the scattering field is essentially induced by the radiation of the sphere, which vibrates with a velocity as described by Equation 4.67. Note that Equation 4.67 is composed of two distinctly different terms. The scattered field results from two different types of vibration. The first term is mainly governed by cos y. Therefore, the vibration is an asymmetric velocity distribution with respect to the angles of 90 and 270 . When y ¼ 0 , the angle between the wave number vector and the surface normal of the sphere is 0 (indicated by the location of Ab on Figure 4.20(b)). The first term of Equation 4.67 is defined by B=r0 c and the second by jkaB=r0 c. On the other hand, when y ¼ 180 (which corresponds to the position Af ; see Figure 4.20(b)), the first term of Equation 4.67 will be B=r0 c and the second becomes jkaB=r0 c. The first term states that we have a velocity of B=r0 c at Af in the direction of the surface normal. Velocity has the same magnitude but exactly the opposite direction, that is, the negative surface normal direction at Ab . This simply means that the sphere oscillates in the same manner as a trembling sphere. The second term, on the other hand, has j and j phase difference at Ab and Af . This means that the sphere vibrates as if it were a breathing sphere. Lastly, it is also valuable to note that the second term becomes smaller as ka decreases. In other words, if we have a sphere with diameter smaller than the wavelength of interest, then the second term contributes less than the first term. This means, rather surprisingly, that the trembling sphere effect become stronger as the size of the scatterer becomes smaller. We can generalize this relation by using Equation 4.61 and the linearized Euler equation, that is, ~ u sc ~ n¼
B j~k ~r 0 ~ n ejot ; e k ~ e r0 c
ð4:68Þ
where~ ek ¼ ~ k=k is the unit propagation vector and~ r 0 is the position vector, which indicates the surface of the scatterer. If the wavelength is much larger than the characteristic length of the scatterer, then we can linearize Equation 4.68 such that it becomes
B B ~ ~ ~ ðk ~ r 0 Þ~ u sc ~ nffi nþj e k ~ n ejot : ð4:69Þ e k ~ r0 c r0 c This is a general form of Equation 4.67. The radiated sound field due to this velocity distribution can be readily obtained by using the Rayleigh integral equation. Section 4.7.5 presents details of the associated integration. We attempted to understand Equation 4.67 by examining the contribution of its two terms. Similarly, Equation 4.69 can be conceptually envisaged. The first term is real and the second is imaginary. The vector expressions describe how the scatterer, in general, vibrates. We have seen that scattering is mainly dominated by the ratio of the size of the scatterer to the wavelength of interest (ka) as well as the scattering direction, that is, the angle y in Equation 4.67. The first type dependency is in line with our expectations because the scattering
207
Radiation, Scattering, and Diffraction
is induced by the radiation and the radiation strongly depends on the scale factor ka, as we already investigated using breathing and trembling spheres. However, the angle dependency has not been well investigated, and is presumably a characteristic of scattering. Figure 4.22 exhibits two fundamental scatterers which can demonstrate the scatterers’ geometrical effect on the radiation. The first is a two-dimensional case and the second is its extension to three dimensions.
y
y
z x
r
ψ φ
R y0 b
r0 O
r
r0
θ
(a)
φ0 0 O
θ
r
R r z
(b)
Figure 4.22 (a) Two-dimensional and (b) three-dimensional rectangular slit with corresponding nomenclature
The radiated sound pressure for the two-dimensional slit can be found as rffiffiffiffiffi pffiffiffi k jkr sinððkb=2Þsin yÞ Pðr; yÞ ¼ 2ð1jÞr0 cU0 e pr ðkb=2Þsin y
ð4:70Þ
(see Section 4.7.4.5 for details). This result says that the scattered sound propagates with jkr respect to its radial direction (Figure 4.22) term of pffiffiffiffiffiffiffias if it is a plane wave (the e Equation 4.70). However, it decays with 1=r, which is a typical characteristic of a twodimensional free field. Note also that the real and imaginary parts contribute equally. The most distinctive characteristics, however, are described in the last term of Equation 4.70, which depends mainly on the size of the slit b relative to the wavelength l, which is kb, and the angle y. The last term is a sinc function. It implies that the scattered field becomes more directional for smaller kb and y, as intuitively seen. These phenomena are maintained even for a rectangular slit case. The scattered field of the rectangular slit (Figure 4.22) can be obtained as Pðr; y; fÞ ¼ j
kr0 c e jkr sinððka=2Þsin ycos fÞ abU0 2p r ðka=2Þsin y cos f
sinððkb=2Þsin y cos fÞ :
ðkb=2Þsin ycos f
ð4:71Þ
208
Sound Propagation
(See Section 4.7.4.5 for the detailed derivation.) U0 is the velocity in the z direction (Figure 4.22). This result is quite similar to Equation 4.70 which describes how the scattered field decays as a monopole as r increases, but is proportional to the magnitude of the velocity flux through the slit (abU0 ). The directivity that is expressed in the last term is a function of the relative geometry of the slit to the wavelength, that is, ka and kb. We can also consider that Equations 4.70 and 4.71 describing a diffraction field due to the slits or scatterers. It can also be argued that the diffractions in these cases strongly depend on the non-dimensional scale factors (ka and kb) and observation angle y. The next step is to investigate diffractions that are induced by more general cases. Any general diffraction problem can be tackled using numerical analysis: for example, Boundary Element Method. However, it will be interesting to investigate the general diffraction problem by starting with a very simple case (Figure 4.23) then expanding to more general case as illustrated in Figure 4.24.
y
y
r
θ
O
Φ
r
θ
O
z
Φ
z
h
−∞ (a)
(b)
Figure 4.23 Two-dimensional diffraction problem for a plane wave source (it is assumed that the wavelength is much larger than the thickness of the wall, and the wall is acoustically rigid): (a) semiinfinite and (b) finite barrier case
To understand the diffraction phenomenon, we begin by studying a typical example: diffraction phenomenon by a sound barrier as shown in Figure 4.23, for example. The closed form solution that describes diffraction due to a semi-infinite barrier (Figure 4.23(a)), the simplest case, can be obtained (see Section 4.7.6.3): ( 0; y
F where Dðy; FÞ ¼
1 1 : cos ðy=2 þ F=2Þ sin ðy=2F=2Þ
ð4:73Þ
Equation 4.72 states that the diffraction depends upon the angles y and F. When y < F, we cannot see the sound source. In other words, we are in the shadow zone and the diffraction is
Radiation, Scattering, and Diffraction
209
Figure 4.24 Diffraction of barriers when we have a monopole source, obtained using FDTD (Finite Difference Time Domain): (a) by a straight barrier and (b) by a curved barrier with respect to time (S denotes a monopole source on the ground). (Photographs courtesy of H. Tachiban, University of Tokyo.)
dominated by the scattered field induced at the edge of the wall. On the other hand, when y > F, we can see the sound source on the positive z axis (Figure 4.23). The diffraction field is composed of two parts: the scattered field from the edge of the wall and the direct sound field. In this regard, we can say that the diffraction is the result of the edge scattering. In other words, we hear sound coming from the sound sources. This means that the edge condition strongly affects the diffraction in the shadow zone. Section 4.7.6 describes the diffraction of many different barriers and sound sources. The analytic method to obtain the solution is also summarized.19 The analytic solution provides many of the elements of the physics associated with the diffraction of a barrier. It also serves as a source for understanding diffraction. However, all this information cannot be taken into account when designing the barrier. In this respect,
19
Most analytic methods have been contributed by K.-U. Nam (Hyundai Automobile Co.).
210
Sound Propagation
Figure 4.25 The sound barrier and associated nomenclature, where NF is the Fresnel number, S represents the source position and R is the receiver’s location. We also assume that the wall thickness is small relative to the wavelength, and acoustically hard
Figures 4.25 and 4.26 effectively provide the practical parameters that are associated with the diffraction of a barrier. In this respect, we first consider the simplest two-dimensional diffraction problem, illustrated in Figure 4.25. First, let us investigate the possible geometrical scales that have to be considered in the diffraction. These are the direct distance d between the source S and the receiver R, unless there is a barrier, and the distance (A þ B), which is the
Figure 4.26
The sound attenuation due to a sound barrier (NF is the Fresnel number)
211
Radiation, Scattering, and Diffraction
shortest distance that the sound from the source can travel to the receiver. It would also be rational to scale these geometrical distances with respect to the wavelength of interest. This quickly becomes clear if we imagine what the wave would look like when we have a wavelength that is much larger than the distance A þ B. As the wavelength becomes increasingly larger, the barrier’s height would lose its presence in terms of diffraction. On the other hand, if the wavelength is much smaller than A þ B, then the listener would perceive that the sound comes from the edge of the barrier. Considering a triangle of sides A, B, and d leads us to deduce the relative distance between A þ B and d with regard to the wavelength. This parameter, which is needed to view the diffraction problem, is the Fresnel number (NF ) and is defined NF ¼
A þ Bd : l=2
ð4:74Þ
The transmission loss of a barrier ðTLbr Þ is generally expressed as20 pffiffiffiffiffiffiffiffiffiffiffi 2pNF pffiffiffiffiffiffiffiffiffiffiffi þ 5 dB TLbr ¼ 20 log10 tanh 2pNF
ðNF > 0Þ:
ð4:75Þ
Figure 4.26 depicts Equation 4.75 in terms of a log-log diagram. This provides a very practical means to design a barrier. For example, once we know the desired sound attenuation by using a barrier, then the graph provides us with the corresponding Fresnel number (NF ). We can then estimate the required height of the barrier from the Fresnel number. We could also regard the Fresnel number as a means to represent the effect of the barrier or, more generally, a spatially distributed impedance discontinuity, on diffraction. When sound meets this discontinuity, it is reflected and scattered, depending on the relative size compared to that of the geometry of impedance mismatch. Fresnel number determines or represents a complicated sound wave propagation associated diffraction; it is surprising that one number can express everything about the diffraction. A similar phenomenon to that found with diffraction can also be found for refraction. Refraction is generally due to a change in the media’s characteristic impedance. For example, as illustrated in Figure 4.27, we can observe the refraction of sound due to the inhomogeneous characteristics of the media because of varying temperature. If we have a gradual increase in temperature from the ground surface which, for example, could occur during the night, then the sound will propagate toward the ground surface (Figure 4.27). Note also that there is a region where we cannot hear the sound, which is often called the “shadow zone.” 21 This is due to a gradual increase or decrease of the characteristic impedance from the ground surface.22 To understand this in more depth, consider the case of multi-layer The total transmission loss ðTLb Þ can be written as TLb ¼ TL þ TLbr , where TL is the transmission loss when there is no sound barrier. 21 We often refer to this region as the “acoustical shadow zone.” 22 The characteristic impedance increases as we rise from ground level for Figure 4.27(a), and decreases for the case of Figure 4.27(b). However, the speed of sound propagation decreases for the former, and increases for the latter. 20
212
Sound Propagation
Figure 4.27 Refraction due to the media’s impedance change (T is temperature and H is height) (a) during the day and (b) during the night
media as illustrated in Figure 4.28. As the propagation vectors depict, the propagation becomes stiffer in the x direction as we have greater propagation speed in the y direction. In reality, within the sea and atmosphere, it is possible to have more complicated media characteristics. As a result, some sound can propagate distances of more than 10 km or 10 000 km. Impedance change in space brings refraction.
Figure 4.28
Sound propagation in a medium where the characteristic impedance changes smoothly
Radiation, Scattering, and Diffraction
213
To summarize this section, the scattered field due to the presence of a sphere in an incident sound field is composed of the part that is induced by a breathing sphere and that made by a trembling sphere. The magnitude is proportional to the magnitude of the incident wave (B). The contribution of the breathing sphere becomes smaller as the wavelength becomes shorter than the size of the scatterer or if ka becomes smaller, when its effect on the scattered field is linearly reduced.23 For scatterers that have general geometries and impedance distributions in space, however, similar scattered fields are produced (as described in Section 4.7.4). Diffraction occurs when the scatterer is large compared to the wavelength of interest. In this case, waves are reflected and scattered as they experience continuous distribution of impedance mismatch in space. It is interesting to note that diffraction in a shadow zone is dominated by the scattering at the edge of barrier: we hear sound from the distributed sound sources on the edge. Refraction is a phenomena induced by impedance changes in space, in particular when there is a change in impedance perpendicular to the direction of sound propagation.
4.6
Chapter Summary
It is plausible that radiation, scattering, diffraction, and refraction can be understood in a unified concept. Depending on the type of impedance mismatch and the wavelength of interest, the radiated sound will be scattered, diffracted, or refracted in space. We started to explore the nature of radiation, scattering, diffraction, and refraction, as well as their implicit relationship by looking at the radiation of the breathing sphere and the trembling sphere. These two radiators are basic units that can create any radiator or any radiation field by their linear combinations. The radiations are mainly dominated by the relative size (ka) of the radiator compared to the wavelength, and the radiated sound field is mainly governed by the relative distance (kr) from the radiator compared to the wavelength of interest. If ka is very small, the radiation power of the breathing sphere increases by 6 dB as ka doubles. For the case of the trembling sphere, such an increase in ka would lead to an increase in radiation power of 12 dB. The scattered sound field was investigated in the cases where a wave is scattered by a rigid sphere and passes through a two-dimensional slit. In these cases, the scattered field strongly depends on the angle of the incident wave, the boundary condition of a scatterer, and the ratio of the wavelength of interest relative to the size of a scatterer (ka or kb). Diffraction was also studied using a typical two-dimensional example. As we saw for the scattering of a slit, diffraction also depends on the angle of the incident wave and the angle from the edge. In the region where we can see the source, the observed sound field is the sum of the incident and the scattered sound fields. However, in the region where the sound source is invisible, the diffraction from the edge of the barrier is dominant. The Fresnel number is widely accepted as a practical means to design a barrier. This parameter requires the difference between the distance that the wave traveled passing the edge of the barrier and the straight
23
The second term in Equation 4.67 has –j. This means that the radiation due to the trembling sphere is due to the acceleration of the fluid particle on the surface.
214
Sound Propagation
distance between the receiver and the source, normalized by the half wavelength. The number, which is compact and less complex compared to how diffraction wave behave in space, is a guideline in the design of a sound barrier.
4.7 4.7.1
Essentials of Radiation, Scattering, and Diffraction Definitions of Physical Quantities Representing Directivity
4.7.1.1 Acoustical Axes In the far field, the directivity of sound source does not change with respect to distance (r) from the source. The amplitude of sound pressure can be expressed as Pðr; y; fÞ ¼ Pax ðrÞHðy; fÞ;
ð4:76Þ
where y and f are the angular directions in space. Pax ðrÞ is called the on-axis pressure, and its direction is called the acoustical axes (Figure 4.29). In general, this indicates the direction in which the sound source emits maximum power, while the on-axis pressure is the maximum sound pressure radiated by the sound source a distance r. Hðy; fÞ is called the directional factor (see Section 4.7.1.3).
Figure 4.29 Acoustical axes
4.7.1.2 Directivity Factor D The directivity factor of a sound source (Figure 4.30) is given by the intensity ratio of a given sound source and an omni-directional sound source which emits the same acoustic power as the source of interest, that is, D¼
jPj2 jPS j
2
¼
I ; IS
ð4:77Þ
where PS and IS are the pressure amplitude and the intensity radiated by the omni-directional sound source, respectively. The directivity factor on the acoustical axes therefore has the
215
Radiation, Scattering, and Diffraction
Figure 4.30 Directivity factor, D. The sound field radiated by an infinitely baffled circular piston with ka ¼ 10 is axially symmetric to acoustical axes
minimum value of 1 in the case of spherical sources. The direction ðy; fÞ is frequently omitted and the directivity factor on the acoustical axes is used to represent the directivity of a sound source. 4.7.1.3 Directional Factor H The directional factor (Figure 4.31) is normalized by Pax ðrÞ, the on-axis pressure, and always has a maximum value of 1. The maximum value of the directivity factor is determined by the type of sound source. The directional factor Hðy; fÞ is given by Hðy; fÞ ¼
Figure 4.31
jPðr; y; fÞj : jPax ðrÞj
The directional factor, H
ð4:78Þ
216
Sound Propagation
Alternatively, the directivity of a sound source can be expressed in terms of the directivity factor D, that is, D¼ð
4pH 2 ðy; fÞ H 2 ðy; fÞdO
:
ð4:79Þ
4p
4.7.1.4 Directivity Index DI The directivity index (Figure 4.32) represents the directivity factor ðDÞ in log scale, that is, DI ¼ 10 log D:
Figure 4.32
ð4:80Þ
The directivity index, DI
Assuming that the power radiated by an omni-directional sound source is the same as the total power radiated by the sound source with respect to distance r, the directivity index is estimated by the spatial average of the power at the measurement surface, that is, I ð4:81Þ DI ¼ 10 log ; IS where I S denotes the spatially averaged intensity at the surface, Sð¼ 4pr2 Þ. 4.7.1.5 Beam Pattern b The beam pattern (Figure 4.33) represents the directional factor Hðy; fÞ in log scale, and is defined as the ratio of relative intensity to that of intensity on the acoustical axes, that is, bðy; fÞ ¼ 10 log
Iðr; y; fÞ ¼ 20 log Hðy; fÞ: Iax ðrÞ
ð4:82Þ
217
Radiation, Scattering, and Diffraction
b
90 0
120
60
-10 -20
30
150
-30 -40 180
0
Figure 4.33
4.7.2
The beam pattem, b
The Radiated Sound Field from an Infinitely Baffled Circular Piston
The sound pressure, radiated by an infinitely balled circular piston that has a radius a and vibrates with a velocity U0, can be written as ð a ikpffiffiffiffiffiffiffiffiffiffi 2 2 jk e z þz PðzÞ ¼ r0 c pffiffiffiffiffiffiffiffiffiffiffiffiffiffi U0 2pzdz: ð4:83Þ 2p 0 z2 þ z2 Figure 4.9 depicts the associated notations and their definitions. PðzÞ is the sound pressure along the line of symmetry of the circular piston. If we define the variables z2 þ z2 ¼ a2 and 2zdz ¼ 2ada for a fixed z, Equation 4.83 can be rewritten as ð R0 jk PðzÞ ¼ r0 cU0 2p e jka da 2p ð4:84Þ z ¼ r0 cU0 fe jkz e jkR0 g;
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi where R0 ¼ z2 þ a2 is the distance from the edge of a circular piston to z. Let us examine what happens when z becomes very large. It is obvious that the difference between Ra and z is negligible when the observation point becomes farther from the origin. In this case, the pressure at z is expressed as PðzÞ ¼ r0 cU0 fe jkðRM þ dÞ e jkðRM dÞ g ¼ r0 cU0 e jkRM 2j sin kd;
ð4:85Þ
where RM and d represent long and short distances. These are defined as RM ¼
d¼
Ra þ z ; 2
ð4:86Þ
Ra z : 2
ð4:87Þ
218
Sound Propagation
Equations 4.86 and 4.87 can be approximated as 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi z2 þ a2 z 2 ! rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 1 ¼ z 1þ 1 2 z 1 1 a 2 þ 1 ¼ z 1þ 2 2 z
d ¼
ffi
ð4:88Þ
1 a2 4z
and RM ffi z:
ð4:89Þ
Pressure PðzÞ is then defined by PðzÞ ¼ r0 cU0
jka a jka e : 2 z
ð4:90Þ
In conclusion, the sound pressure in the far field is inversely proportional to the distance from the source, as for the sound field generated by a monopole source.
4.7.3
Sound Field at an Arbitrary Position Radiated by an Infinitely Baffled Circular Piston
From Figure 4.10, R can be defined z z2 R2 ¼ r2 þ z2 2rz sin ycos f ¼ r2 12 sin ycos f þ 2 r r
ð4:91Þ
which can be written as 1 z z2 2 R ¼ r 12 sin y cos f þ 2 : r r By using the binomial expansion, this can be approximated as z z2 1 1 2 2 sin y cos f : R ffi r 1 sin ycos f þ 2 r r 2 2
ð4:92Þ
ð4:93Þ
Therefore, kR can be expressed as kR ffi krkz sin y cos f þ
kz2 ð1sin2 f cos2 yÞ: 2r
ð4:94Þ
219
Radiation, Scattering, and Diffraction
r Consider the case of ka , that is, the observation point is at a great distance and the a wavelength is very long compared to the radius of the circular piston. In this case, the third term in the right-hand side of Equation 4.94 is negligible compared to the other terms. We therefore have e jkR e jkr jkz sin y cos f : ffi e R r
ð4:95Þ
By using the Rayleigh integral equation, the pressure at ~ R 0 (Figure 4.10) is expressed as ð ð a 2p e jkr jkr0 c ejkz sin y cos f zdzdf U0 Pðr; yÞ ¼ r 2p 0 0 " ð # ð4:96Þ ð a p=2 jkr0 c e jkr ¼ U0 zdz 4 cosðkz sin ycos fÞdf : r 2p 0 0 According to the characteristics of the Bessel function, which is 2 J0 ðzÞ ¼ p
ðp 2
cosðz cos aÞda;
ð4:97Þ
0
the sound pressure (4.96) can be rewritten as jkr0 c e jkr Pðr; yÞ ¼ U0 r 2p e jkr ¼ jkr0 cU0 r ¼ jkr0 cU0
ða
p 4J0 ðkzsin yÞzdz 0 2
ða
J0 ðkzsin yÞzdz
ð4:98Þ
0
e jkr 2 J0 ðka sin yÞ : a ka sin y r
The sound pressure (4.98) on the z axis is the same as that of Equation 4.90 because when y approaches 0,
4.7.4
J0 ðka sin yÞ ka sin y
tends to 1/2.
Understanding Radiation, Scattering, and Diffraction Using the Kirchhoff–Helmholtz Integral Equation24
Radiation, diffraction, and scattering can be understood as various forms of sound fields that are produced by spatially distributed impedance mismatch. Depending on the impedance distribution in space, the resulting sound field can vary widely. It is well known that the sound fields that are induced by the spatial impedance distribution can be predicted by various numerical or theoretical approaches. In this chapter, analytical solutions of particular significance, which allow us to understand associated fundamental physical meanings, are introduced. We first define and establish notation and summarize basic equations. 24
Technical note written by Kyoung-Uk Nam.
220
Sound Propagation
4.7.4.1 Symbols and Basic Equations . .
.
. . . .
Expression of a time harmonic function: ejot . Position of prediction: ~ r - in the orthogonal coordinate systems, - one-dimensional coordinate system: ðxÞ, - two-dimensional coordinate system ðx; yÞ or ðr; yÞ for polar coordinate system, - three-dimensional coordinate system ðx; y; zÞ for Cartesian coordinate, ðr; y; zÞ for cylindrical coordinates, or ðr; y; fÞ for spherical coordinate system, Position of sound source: ~ r0 - in the orthogonal coordinate systems, - one-dimensional coordinates: ðx0 Þ, - two-dimensional coordinates: ðx0 ; y0 Þ; ðr0 ; y0 Þ, - three-dimensional coordinates: ðx0 ; y0 ; z0 Þ; ðr0 ; y0 ; z0 Þ; ðr0 ; y0 ; f0 Þ, The distance between the sound source and prediction point: R, Complex amplitude of sound pressure: Pð~ rÞ, Complex amplitude of particle velocity: Un ð~ rÞ, n denotes the direction, Relationship between sound pressure and velocity: linearized Euler equation rp ¼ r0
@~ u @t
ð4:99Þ
Green’s function: Gð~ rj~ r0Þ .
Green’s law:
þ
~ F ~ n dl ¼
C
.
ð
r~ F dS
ð4:100Þ
S
(Line integral calculated for counter-clockwise direction. C represents a contour formed by a certain line, and S denotes the surface. In addition, ~ n represents the unit normal vector pointing toward outside the domain wrapped by the contour.) Gauss’s theorem: ð ð ~ F ~ n dS ¼ r ~ F dV ð4:101Þ S
V
(V is volume of the region of interest.) 4.7.4.2 One-dimensional Kirchhoff–Helmholtz Integral Equation In order to understand the Kirchhoff–Helmholtz integral equation, we shall first investigate it for a one-dimensional case. The one-dimensional acoustic governing equation can be expressed as a Helmholtz integral equation in the frequency domain, that is, d 2 PðxÞ þ k2 PðxÞ ¼ 0; dx2
ð4:102Þ
221
Radiation, Scattering, and Diffraction
where P is the complex sound pressure and k represents the wave number. x denotes a position in the defined domain a x b (See Figure 4.34).
x0
x=0
Figure 4.34
x=a
x
x=b
One-dimensional coordinate system with free field boundary conditions
Consider Green’s function G which satisfies d 2 Gðxjx0 Þ þ k2 Gðxjx0 Þ ¼ dðxx0 Þ; dx2
ð4:103Þ
where dðxx0 Þ is a one-dimensional delta function. By multiplying Equation 4.102 by G and Equation 4.103 by P and subtracting the latter from the former, we arrive at 2 2 d d 2 2 G þ k PP þ k ¼ P dðxx0 Þ: ð4:104Þ dx2 dx2 The left-hand side of Equation 4.104 is reduced to d dG dP P G ¼ P dðxx0 Þ: dx dx dx
ð4:105Þ
Integrating Equation 4.105 for x over the domain between a and b, we have ðb P dðxx0 Þdx a
or
d dGðxjx0 Þ dP ¼ P Gðxjx0 Þ dx dx dx a dx ðb
dGðxjx0 Þ dP x¼a Pðx0 Þ ¼ P Gðxjx0 Þ : dx dx x¼b
ð4:106Þ
ð4:107Þ
The reciprocity of the Green function, Gðxjx0 Þ ¼ Gðx0 jxÞ, is applied to Equation 4.107 and x is replaced with x0 , we obtain dGðxjx0 Þ dPðx0 Þ x0 ¼a PðxÞ ¼ Pðx0 Þ Gðxjx0 Þ : ð4:108Þ dx0 dx0 x0 ¼b This is the one-dimensional Kirchhoff-Helmholtz integral equation. Equation 4.108 means that if the sound pressure amplitude and the velocity amplitude at x0 ¼ a; b are given, then sound pressure at any position x can also be obtained. In other words, the velocity and sound
222
Sound Propagation
pressure at a and b propagate to x using Green’s function as the propagator. Note that any function that satisfies Equation 4.103 can be used as a Green’s function. The most commonly selected is a free field Green’s function, that is GF ðxjx0 Þ ¼
e jkR ; j2k
R ¼ jxx0 j;
ð4:109Þ
where the subscript F denotes a free field. This Green’s function can be obtained from Equation 4.103. Because it is a one-dimensional free field, the solution of Equation 4.103 can be written as ( Ae jkðxx0 Þ ; x > x0 GF ðxjx0 Þ ¼ ð4:110Þ Bejkðxx0 Þ ; x < x0 : Since the sound pressure must be the same at x ¼ x0 , A and B are the same. If Equation 4.103 is integrated from x ¼ x0 e to x ¼ x0 þ e, we obtain dGF x¼x0 þ e ¼ 1: ð4:111Þ dx x¼x0 e This means that, while the sound pressure is continuous, the rate change of spatial distribution of pressure is not continuous and the difference is 1. Substituting Equation 4.110 into Equation 4.111, we have Ajk þ Ajk ¼ 1:
ð4:112Þ
Hence, Equation 4.110 can be rewritten over the entire domain as GF ðxjx0 Þ ¼
e jkR ; j2k
ð4:113Þ
R ¼ jxx0 j: Note also that the Green’s function which satisfies the Neumann boundary condition (see Figure 4.35) in a semi-infinite space must have zero velocity at x ¼ 0. The function can be obtained by superimposing the free field Green’s function (4.109), which can be expressed as GN ðxjx0 Þ ¼ GF ðxjx0 Þ þ GF ðxjx0 Þ ð4:114Þ
0
¼
e jkR e jkR ; j2k j2k
+ x = – x0
x0 x=0
+
x
x = x0
Figure 4.35 Coordinate system to induce one-dimensional Green’s function which satisfies Neumann boundary condition
223
Radiation, Scattering, and Diffraction
where R ¼ jxx0 j, R0 ¼ jx þ x0 j and the subscript N denotes Neumann boundary condition. When x0 ! 0, Equation 4.114 becomes GN ðxjx0 Þ ¼ lim GN ðxjx0 Þ ¼ N In this case, dG dx0
x0 ! 0
i x0 ¼0
e jkx : jk
ð4:115Þ
¼ 0.
In Equation 4.108, the boundary value at x ¼ 1 is negligible. By using the linear Euler equation and Equation 4.115, we have PðxÞ ¼ r0 ce jkx Uð0Þ;
ð4:116Þ
which is a one-dimensional Rayleigh integral equation. We now consider a Green’s function that satisfies the Dirichlet boundary condition in a semiinfinite space. A Green’s function whose value becomes 0 at x ¼ 0 can be written as GD ðxjx0 Þ ¼ GF ðxjx0 ÞGF ðxjx0 Þ ¼
e jkR e jkR þ j2k j2k
ð4:117Þ
0
where R ¼ jxx0 j, R0 ¼ jx þ x0 j, and the subscript D denotes the Dirichlet boundary condition (see Figure 4.36). If x0 ! 0 in Equation 4.117, dGD ðxjx0 Þ dGD ðxjx0 Þ ¼ lim ¼ e jkx : ð4:118Þ x0 ! 0 dx0 dx 0 x0 ¼0 x
–
x = –x0
x=0
x0
+
x = x0
Figure 4.36 Coordinate system to induce one-dimensional Green’s function which satisfies Dirichlet boundary condition
Because GD ¼ 0 at x ¼ 0, and using Equation 4.118, Equation 4.108 can be reduced to PðxÞ ¼ e jkx Pð0Þ:
ð4:119Þ
We assumed that the boundary value at x ¼ 1 is negligible. We now expand the theory to two-dimensional cases. 4.7.4.3 Two-dimensional Kirchhoff–Helmholtz Integral Equation In the two-dimensional case, the governing equation can be written as r2 Pð~ rÞ þ k2 Pð~ rÞ ¼ 0;
ð4:120Þ
224
Sound Propagation
where ~ r represents all points other than ~ r 0 (see Figure 4.37). For ~ r ¼~ r 0, it satisfies r2 Gð~ rj~ r 0 Þ þ k2 Gð~ rj~ r 0 Þ ¼ dð~ r~ r 0 Þ:
ð4:121Þ
r
R
n
r0
0
Figure 4.37
Coordinate system used for a two-dimensional Kirchhoff–Helmholtz integral equation
From Equations 4.120 and 4.121, we have Gðr2 þ k2 ÞPðxÞPðr2 þ k2 ÞG ¼ Pð~ rÞdð~ r~ r 0 Þ:
ð4:122Þ
Rewriting the left-hand side, we have r ðGrPPrGÞ ¼ Pð~ rÞdð~ r~ r 0 Þ:
ð4:123Þ
Integrating over the surface of interest S and applying Green’s law in Equation 4.121 in Section 4.7.4.1, ð Pð~ rÞdð~ r~ r 0 ÞdS S ð4:124Þ ð ¼ ½Gð~ rj~ r 0 ÞrPð~ rÞPð~ rÞrGð~ rj~ r 0 Þ dS S
or þ Pð~ r0Þ ¼
½Pð~ rÞrGð~ rj~ r 0 ÞGð~ rj~ r 0 ÞrPð~ rÞ ~ ndL;
ð4:125Þ
L
where ~ n is a unit normal vector from the boundary to the region of interest, as depicted in Figure 4.37. By using the reciprocity of Green’s function, Gð~ rj~ r 0 Þ ¼ Gð~ r 0 j~ rÞ. By replacing~ r with ~ r 0 , we obtain the two-dimensional Kirchhoff–Helmholtz integral equation as þ Pð~ rÞ ¼
½Pð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r 0 Þ ~ ndL; L0
where
Þ L0
denotes the line integral.
ð4:126Þ
225
Radiation, Scattering, and Diffraction
The two-dimensional free field Green’s function is expressed by a Hankel function of the first kind as j ð1Þ GF ð~ rj~ r 0 Þ ¼ H0 ðkRÞ; 4
R ¼ j~ r~ r 0 j:
ð4:127Þ
In the far field ðkR ! 1Þ, it becomes approximately 1 e jkR GF ð~ rj~ r 0 Þ ffi pffiffiffi ð1 þ jÞ pffiffiffiffiffiffiffiffiffi : 2 pkR
ð4:128Þ
First, the Green’s function that satisfies Neumann boundary condition is similar to the onedimensional case (see Figure 4.38), that is GN ðx; yjx0 ; y0 Þ ¼ GF ðx; yjx0 ; y0 Þ þ GF ðx; yjx0 ; y0 Þ ¼
j ð1Þ j ð1Þ H0 ðkRÞ þ H0 ðkR0 Þ 4 4
r0 (–x0, y0)
r0 (x0, y0)
+
+
ð4:129Þ
R
R′
r (x, y)
y x
Figure 4.38 Coordinate system used for a two-dimensional Green’s function which satisfies the Neumann boundary condition
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where R ¼ ðxx0 Þ2 þ ðyy0 Þ2 and R0 ¼ ðx þ x0 Þ2 þ ðyy0 Þ2 . As x0 ! 0, Equation 4.129 can be rewritten as GN ðx; yjx0 ; y0 Þ ¼ lim GN ðx; yjx0 ; y0 Þ x0 ! 0
¼ where R ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ ðyy0 Þ2 . Note that
dGN dx0
j ð1Þ H ðkRÞ; 2 0
ð4:130Þ
¼ 0, and the boundary value at infinity can be
neglected. Using Equations 4.125, 4.130, and 4.120 from Section 4.7.4.1, Equation 4.126 becomes ð kr c 1 ð1Þ Pðx; yÞ ¼ 0 H ðkRÞUx ð0; y0 Þdy0 : ð4:131Þ 2 1 0
226
Sound Propagation
This equation is a two-dimensional Rayleigh integral equation. By using Equation 4.128, Equation 4.131 can be approximated in the far field as rffiffiffi ð pffiffiffi k 1 e jkR pffiffiffi Ux ð0; y0 Þdy0 : Pðx; yÞ ffi 2ð1jÞr0 c ð4:132Þ p 1 R We now find the Green’s function that satisfies the Dirichlet boundary condition in a semiinfinite space (see Figure 4.39). By using a two-dimensional Green’s function as used in the one-dimensional case, it can be written as GD ðx; yjx0 ; y0 Þ ¼ GF ðx; yjx0 ; y0 ÞGF ðx; yjx0 ; y0 Þ ¼
j ð1Þ j ð1Þ H0 ðkRÞ H0 ðkR0 Þ; 4 4
r0 (–x0, y0)
ð4:133Þ
r0 (x0, y0)
–
+
R
R′
r (x, y)
y x
Figure 4.39 Coordinate system used for a two-dimensional Green’s function which satisfies the Dirichlet boundary condition
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 0 where R ¼ ðxx0 Þ þ ðyy0 Þ and R ¼ ðx þ x0 Þ2 þ ðyy0 Þ2 . As x0 ! 0, @GD ðx; yjx0 ; y0 Þ @GD ðx; yjx0 ; y0 Þ ¼ @n @x x0 ¼0
0
x0 ¼0
ð4:134Þ
@GD ðx; yjx0 ; y0 Þ j kx ð1Þ ¼ lim ¼ H ðkRÞ; x0 ! 0 @x0 2 R 1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where R ¼ x2 þ ðyy0 Þ2 . Because GD ¼ 0, Equation 4.126 can be rewritten as (by using Equation 4.134) j Pðx; yÞ ¼ kx 2
ð1
ð1Þ
H1 ðkRÞ Pð0; y0 Þdy0 : R 1
ð4:135Þ
Again, it is assumed that the boundary value at infinity is negligible, just as in the onedimensional case. We now consider the three-dimensional case.
227
Radiation, Scattering, and Diffraction
4.7.4.4 Three-dimensional Kirchhoff–Helmholtz Integral Equation The sound pressure at an arbitrary position ~ r ð~ r 6¼ ~ r 0 Þ must satisfy r2 Pð~ rÞ þ k2 Pð~ rÞ ¼ 0:
ð4:136Þ
At r ¼ ~ r 0 , the Greens’ function must satisfy the Helmholtz equation, that is, r2 Gð~ rj~ r 0 Þ þ k2 Gð~ rj~ r 0 Þ ¼ dð~ r~ r 0 Þ:
ð4:137Þ
By multiplying Equation 4.136 by G and Equation 4.137 by P, and by subtracting the latter from the former, we have Gðr2 þ k2 ÞPPðr2 þ k2 ÞG ¼ Pð~ rÞdð~ r~ r 0 Þ:
ð4:138Þ
Rewriting the left-hand side of Equation 4.138, we obtain Pð~ rÞdð~ r~ r 0 Þ ¼ r ðGrPPrGÞ: Integrating over the volume of interest V yields ð Pð~ rÞdð~ r~ r 0 ÞdV V ð rj~ r 0 ÞrPð~ rÞPð~ rÞrGð~ rj~ r 0 ÞdV: ¼ r ½Gð~
ð4:139Þ
ð4:140Þ
V
Note that the volume excludes all sound sources. In other words, it is a source free volume. By using Gauss’s theorem to make the right-hand side of Equation 4.140 a surface-integral form, it can be rewritten as ð Pð~ r 0 Þ ¼ ½Pð~ rÞrGð~ rj~ r 0 ÞGð~ rj~ r 0 ÞrPð~ rÞ ~ ndS; ð4:141Þ S
where ~ n is a unit normal vector from the boundary to the region of interest. By using r 0 j~ rÞ. By replacing~ r by~ r 0, the three-dimensional Green’s function’s reciprocity, Gð~ rj~ r 0 Þ ¼ Gð~ Kirchhoff–Helmholtz integral equation can be obtained as ð Pð~ rÞ ¼ ½Pð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r 0 Þ ~ n 0 dS: ð4:142Þ S0
The three-dimensional free field Green’s function is rj~ r0Þ ¼ GF ð~
e jkR ; 4pR
R ¼ j~ r~ r 0 j:
ð4:143Þ
The Green’s function (Figure 4.40) that satisfies the Neumann boundary condition in a semiinfinite space is GN ðx; y; zjx0 ; y0 ; z0 Þ ¼ GF ðx; y; zjx0 ; y0 ; z0 Þ þ GF ðx; y; zjx0 ; y0 ; z0 Þ 0
e jkR e jkR ¼ þ ; 4pR 4pR0
ð4:144Þ
228
Sound Propagation z P(x, y, z)
P0(x0, y0, z0) +
R r
y
r0
z0
R′ x
z0
r0
+ P′0(x0, y0, –z0)
Figure 4.40 Coordinate system used for a three-dimensional Green’s function which satisfies the Neumann boundary condition
where R¼ and
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxx0 Þ2 þ ðyy0 Þ2 þ ðzz0 Þ2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R ¼ ðxx0 Þ2 þ ðyy0 Þ2 þ ðz þ z0 Þ2 : 0
As z0 ! 0, Equation 4.144 becomes GN ðx; y; zjx0 ; y0 ; z0 Þ ¼ lim GF ðx; y; zjx0 ; y0 ; z0 Þ ¼ z0 ! 0
where R ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxx0 Þ2 þ ðyy0 Þ2 þ z2 . In this case,
dGN dz0
e jkR ; 2pR
ð4:145Þ
¼ 0 at the boundary. Using
Equation 4.136 (Section 4.7.4.1) and Equation 4.145, we then have jkr0 c Pðx; y; zÞ ¼ 2p
ð1 ð1 1
e jkR Uz ðx0 ; y0 ; 0Þdx0 dy0 ; 1 R
ð4:146Þ
which is the Rayleigh integral equation in terms of Cartesian coordinates. We now find the Green’s function that satisfies the Dirichlet boundary condition. By using the free field Green’s function as for the cases of one and two dimensions, it can be written (see Figure 4.41): GD ðx; y; zjx0 ; y0 ; z0 Þ ¼ GF ðx; y; zjx0 ; y0 ; z0 ÞGF ðx; y; zjx0 ; y0 ; z0 Þ ¼
jkR
jkR0
e e 4pR 4pR0
ð4:147Þ
229
Radiation, Scattering, and Diffraction z P(x, y, z)
P0(x0, y0, z0) +
R r
y
r0
z0
R′ x
z0
r0
– P′0(x0, y0, –z0)
Figure 4.41 Coordinate system used for a three-dimensional Green’s function which satisfies the Dirichlet boundary condition
where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R ¼ ðxx0 Þ2 þ ðyy0 Þ2 þ ðzz0 Þ2 ; and R0 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxx0 Þ2 þ ðyy0 Þ2 þ ðz þ z0 Þ2 :
As z0 ! 0, @GD ðx; y; zjx0 ; y0 ; z0 Þ @GD ðx; y; zjx0 ; y0 ; z0 Þ ¼ @n0 @z0 z0 ¼0 z0 ¼0 ¼ lim
z0 ! 0
¼ where R ¼
@GD ðx; y; zjx0 ; y0 ; z0 Þ @z0
ð4:148Þ
1 z ð1jkRÞe jkR ; 2p R3
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxx0 Þ2 þ ðyy0 Þ2 þ z2 . Since GD ¼ 0 at the boundary, by using Equation 4.148
we can predict the pressure at any position in the source free volume by Pðx; y; zÞ ¼
z 2p
ð1 ð1 1
ð1jkRÞ jkR e Pðx0 ; y0 ; 0Þdx0 dy0 : R3 1
ð4:149Þ
4.7.4.5 Diffraction by a Slit We have obtained integral functions that can be used to predict sound fields in all possible dimensions. This allows us to explore how sound propagates and changes under various spatially distributed impedance boundary conditions or impedance mismatch.
230
Sound Propagation
We first consider diffraction when there is a slit in an infinite baffle, as shown in Figure 4.42. Assume that the baffle is rigid and the incident wave is a plane wave that arrives at the plate surface (baffle) in the normal direction. Assume that the plane wave’s velocity at the slit is uniform over the surface, which is U0 .
Incident plane wave
y x
z slit
Figure 4.42
Infinite baffle with a slit
Two-dimensional Slit If the width of the slit is b and it is placed from 1 to 1 in the x direction, it becomes a twodimensional diffraction problem as illustrated in Figure 4.43. In this case, applying the Rayleigh integral equation (Equation 4.132 of Section 4.7.4.3) to the far field, we can write pffiffiffi Pðr; yÞ ffi 2ð1jÞr0 cU0
rffiffiffi ð k b=2 e jkR pffiffiffi dy0 ; p b=2 R
ð4:150Þ
where U0 is the impinging velocity at the slit’s entrance. Note that R(4.150) in the far field can be written using a Taylor series expansion as R¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 þ y20 2ry0 cosðp=2yÞ
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y 2 y 0 0 ¼ r 1þ 2 sin y ffi ry0 sin y: r r
ð4:151Þ
231
Radiation, Scattering, and Diffraction y z r R r y0 b
r0
θ
O
Figure 4.43
Coordinate system of two-dimensional slit
If Equation 4.151 is substituted into Equation 4.150, the sound pressure at the far field is expressed as rffiffiffi ð b=2 pffiffiffi k e jkðry0 sin yÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dy0 Pðr; yÞ ffi 2ð1jÞr0 cU0 p b=2 ry0 sin y pffiffiffi ffi 2ð1jÞr0 cU0
rffiffiffiffiffi ð k jkr b=2 jky0 sin y e e dy0 pr b=2
rffiffiffiffiffi k jkr ejkðb=2Þsin y e jkðb=2Þsin y e pr jk sin y rffiffiffiffiffi pffiffiffi k jkr 2jsinððkb=2Þsin yÞ ffi 2ð1jÞr0 cU0 e pr jk sin y rffiffiffiffiffi pffiffiffi k jkr sinððkb=2Þsin yÞ ffi 2ð1jÞr0 cU0 e : pr ðkb=2Þsin y pffiffiffi ffi 2ð1jÞr0 cU0
ð4:152Þ
Equation 4.152 states that directivity pattern of jPj2 can be expressed as sinððkb=2Þsin yÞ 2 ðkb=2Þsin y :
ð4:153Þ
This is a sinc function with regard to kb and sin y. The smaller the slit’s height ðbÞ relative to the wave, the stronger is the directivity of the field that is formed. The greater the value of kb or y, less diffraction by the slit is expected. This can also be understood as a typical scattering phenomenon by a two-dimensional slit.
232
Sound Propagation
Diffraction by a Round Slit By applying the Rayleigh integral equation to the circular slit, as shown in Figure 4.44, the sound pressure can be expressed as jkr0 c Pðr; y; fÞ ffi U0 2p
ð 1 ð 2p 0
0
e jkR r0 df0 dr0 : R
ð4:154Þ
y x ψ r0
φ
φ0 r0
O
r θ
a
Figure 4.44
r
R
z
Coordinate system of round slit
The far field approximation of R (4.154) is R¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 þ r20 2rr0 cos c
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r 2 r 0 0 ¼ r 1þ 2 cos c ffi rr0 cos c: r r
ð4:155Þ
In Equation 4.155, cos c can be obtained by the inner products of unit vectors in the~ r and~ r0 directions (~ e r and ~ e r0 ). Note also that, due to symmetry, Pðr; y; fÞ does not change its value relative to f. Therefore, when f ¼ 0, cos c is written as cos c ¼ ~ e r ~ e r0 ¼ ðsin y~ e x þ cos y~ e z Þ ðcos f0~ e x þ sin f0~ ezÞ ¼ sin y cos f0 :
ð4:156Þ
233
Radiation, Scattering, and Diffraction
Equation 4.155 can therefore be rewritten as R ffi r r0 sin y cos f0 :
ð4:157Þ
Substituting Equation 4.157 into Equation 4.154, we have ð 1 ð 2p
e jkðrrsin y cos f0 Þ r0 df0 dr0 0 0 rrsin y cos f0 ð ð 2p jkr0 c e jkr a jkr0 sin y cos f0 ffi r0 e df0 dr0 : U0 r 0 2p 0
Pðr; y; fÞ ffi
jkr0 c U0 2p
ð4:158Þ
Note that the imaginary part is an odd function with respect to f0 ; its integral is therefore 0. The integral of Equation 4.158 is reduced to ð 2p e
jkr0 sin y cos f0
df0 ¼
ð 2p
0
cosðkr0 sin y cos f0 Þdf0
0
¼2
ðp
ð4:159Þ cosðkr0 sin y cos f0 Þdf0 :
0
Noting that one of the characteristics of a Bessel function is ð1Þn J2n ðuÞ ¼ p
ðp
cos 2nf0 cosðucos f0 Þdf0 ;
ð4:160Þ
0
Equation 4.159 can be rewritten as ð 2p
ejkr0 sin y cos f0 df0 ¼ 2pJ0 ðkr0 sin yÞ:
ð4:161Þ
0
Equation 4.158 then becomes Pðr; y; fÞ ffi jkr0 cU0
e jkr r
ða
r0 J0 ðkr0 sin yÞdr0 :
ð4:162Þ
0
By changing variable using u ¼ kr0 sin y, Equation 4.162 can be rewritten as ða 0
r0 J0 ðkr0 sin yÞdr0 ¼
1 ðk sin yÞ2
ð ka sin y
uJ0 ðuÞdu:
ð4:163Þ
0
Using another characteristic of the Bessel function, which is u Jn0 ðuÞ ¼ u Jn1 ðuÞn Jn ðuÞ;
ð4:164Þ
234
Sound Propagation
Equation 4.163 can be simplified by integration by parts as 1 ðk sin yÞ2
ð ka sin y
uJ0 ðuÞdu
0
¼ ¼ ¼ ¼
ð ka sin y
1 ðk sin yÞ2
0
u J10 ðuÞdu þ
1 ðk sin yÞ2 ka sin y ðk sin yÞ2
sin y u J1 ðuÞjka 0
ð ka sin y
u J1 ðuÞdu
0
ð ka sin y
u J1 ðuÞdu þ
0
ð ka sin y
u J1 ðuÞdu
ð4:165Þ
0
J1 ðka sin yÞ
a2 J1 ðka sin yÞ : ka sin y
If this is substituted into Equation 4.162, the sound pressure in the far field is ultimately expressed as Pðr; y; fÞ ffi jkr0 cU0
e jkr 2J1 ðka sin yÞ : ka sin y r
From Equation 4.166, the directivity pattern of jPj2 is 2J1 ðka sin yÞ 2 ka siny :
ð4:166Þ
ð4:167Þ
This is a function of ka and sin y, and behaviors similarly to that predicted for the twodimensional slit. Diffraction by a Rectangular Slit By applying the Rayleigh integral equation (Figure 4.45 shows the corresponding coordinate), sound pressure is generally expressed as Pðr; y; fÞ ¼
jkr0 c U0 2p
ð b=2 ð a=2 b=2
e jkR dx0 dy0 : a=2 R
ð4:168Þ
In Equation 4.168, R in far field is expressed as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 þ y20 2rr0 cos c rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r 2 r 0 0 ¼ r 1þ 2 cos c ffi rr0 cos c: r r
R¼
ð4:169Þ
235
Radiation, Scattering, and Diffraction
y x ψ
φ
φ0 r0
Figure 4.45
r
R θ
O
r z
Coordinate system of rectangular slit
In Equation 4.169, as noted before, we can obtain cos c by the inner products of the unit factors in the ~ r and ~ r 0 directions (~ e r and ~ e r0 ): cos c ¼ ~ e r ~ e r0 ¼ ðsin y cos f~ e x þ sin ysin f~ e y þ cos y~ ezÞ ¼
x0 y0 ~ ex þ ~ ey r0 r0
ð4:170Þ
x0 y0 sin y cos f þ sin ysin f: r0 r0
Equation 4.169 can therefore be rewritten as R ffi rx0 sin y cos fy0 sin ysin f:
ð4:171Þ
When Equation 4.171 is substituted into Equation 4.168, Pðr; y; fÞ can be written approximately as Pðr; y; fÞ jkr0 c ffi U0 2p ffi
jkr0 c U0 2p
ð b=2 ð a=2 b=2
ð b=2 b=2
e jkðrx0 sin y cos fy0 sin y sin fÞ dx0 dy0 a=2 rx0 sin y cos f y0 sin y sin f "ð #
ejky0 sin y sin f
a=2
a=2
ejkx0 sin y cos f dx0 dy0 :
ð4:172Þ
236
Sound Propagation
The first integral of Equation 4.172 can be written as ð a=2 e jka sin y cos f=2 ejka sin y cos f=2 ejkx0 sin y cos f dx0 ¼ jk sin ycos f a=2
ð4:173Þ
sinððka=2Þsin y cos fÞ : ¼a ðka=2Þsin y cos f Substituting Equation 4.173 into Equation 4.172 and solving the second integral provides the sound pressure level in the far field as Pðr; y; fÞ ¼
jkr0 c e jkr sinððka=2Þsin y cos fÞsinððkb=2Þsin ycos fÞ abU0 : r 2p ðka=2Þsin ycos fðkb=2Þsin y cos f
ð4:174Þ
Therefore, the directivity pattern of jPj2 can be expressed as sinððka=2Þsin y cos fÞsinððkb=2Þsin y cos fÞ 2 ; ðka=2Þsin y cos fðkb=2Þsin ycos f
ð4:175Þ
where the maximum value is 1. Again note that the directivity strongly depends on the normalized scale factors (ka, kb), which describe the size of the rectangular slit relative to the wavelength of interest.
4.7.5
Scattered Sound Field Using the Rayleigh Integral Equation
From Section 4.5, the scattered field is defined as jkr0 c Psc ¼ 2p
e jkR B B ~ ~ n þj e k ~ n dS: ðk ~ r a Þ~ e k ~ r0 c r0 c S0 R
ð
ð4:176Þ
The first term can be rewritten as ð jkR jkr0 c e B ~ e k ~ n dS 2p S0 R r0 c ð jkR ð jkB e jkB e jkR ~ ~ n dS ¼ r e k ~ e k dV; ¼ 2p S0 R 2p V0 R
The first term ¼
ð4:177Þ
and the second term can be rearranged as ð jkr0 c B e jkR ~ ðk ~ r 0 Þð~ j e k ~ nÞ dS 2p S0 r0 c R ð jkR k2 B e ¼ ð~ e k ~ r 0 Þð~ e k ~ nÞ dS: 2p S0 R
The second term ¼
ð4:178Þ
237
Radiation, Scattering, and Diffraction
In a far field, The second term ¼
k2 B e jkR 2p R
ð ð~ e k ~ r 0 Þð~ e k ~ nÞ dS:
ð4:179Þ
S0
If Gauss’s theorem is applied, then The second term ¼
k2 B e jkR 2p R
ð r ð~ e k ~ r 0 Þ~ e k dV V0
k2 B e jkR k2 B V0 e jkR ¼ V0 ¼ 2p 2pR R
ð4:180Þ
where V0 is the volume of the scatterer. Therefore, Equation 4.176 can be rewritten as Psc ¼
jkB 2p
ð r V0
e jkR k2 B ~ V0 e jkR : e k dV þ 2pR R
ð4:181Þ
This means that the scattered sound field can be divided into two different fields: that constructed by a monopole sound source (proportional to the volume of the scatterer, second term) and that due to a dipole source contribution (first term).
4.7.6
Theoretical Approach to Diffraction Phenomenon
We have studied scattering and diffractions by using the very simple but highly representative general cases of scattering and diffraction due to a two-dimensional slit, circular and rectangular openings. Very common diffraction problems occur when we have a sound barrier. We next introduce a theoretical approach which can handle these typical diffraction problems. We begin with fundamental mathematics which will allow us to tackle the problem theoretically. 4.7.6.1 Definition of Relevant Complex Variable Theories Conventional Fourier Transform A conventional Fourier transform pair can be defined as ð1 f ðxÞe jsx dx; FðsÞ ¼ 1
and f ðxÞ ¼
1 2p
where s is a real number and FðsÞ exists if
ð1
FðsÞejsx ds;
1
ð1 1
j f ðxÞjdx is finite.
ð4:182Þ
ð4:183Þ
238
Sound Propagation
Generalized Fourier Transform When s is a complex number (i.e., s ¼ s1 þ js2 ), the following two theorems are available. Theorem 4.1 If j f ðxÞj < Aeax when x ! 1 (where A, a are constants), there exists F þ ðsÞ such that, ð1 F þ ðsÞ ¼ f ðxÞe jsx dx; ð4:184Þ 0
which is analytic in an area where s2 > a. Theorem 4.2 If j f ðxÞj < Bebx when x ! 1 (where B, b are constants), there exists F ðsÞ such that, ð0 F ðsÞ ¼ f ðxÞe jsx dx; ð4:185Þ 1
which is analytic in an area where s2 > b. Given Theorems 1 and 2, if j f ðxÞj < Aeax as x ! 1, j f ðxÞj < Bebx as x ! 1, and a < b, there exists FðsÞ such that ð1 FðsÞ ¼ f ðxÞe jsx dx; ð4:186Þ 1
which is analytic in an area where a < s2 < b. In this case, the inverse Fourier transform can be written as ð 1 f ðxÞ ¼ FðsÞejsx ds; 2p p
ð4:187Þ
where the integral path p (Figure 4.46) can be any path in the strip of a < s2 < b, where s1 ranges from 1 to 1. s2
Region that F,(s) is analytic
β Integration path p s1 α
Strip that F(s) is analytic
Region that F,(s) is analytic
Figure 4.46 Analytic area of generalized Fourier transform
239
Radiation, Scattering, and Diffraction
Relevant Theories and Theorems Theorem 4.3
Abelian theorem
(a) If j f ðxÞj < Aeax when x ! 1, if f ðxÞ is infinitely differentiable for x > 0, and if f ðxÞ ¼ Oðxl Þ (l < 1) when x > 0 and x ! 0 þ in an area where s2 > a, then ð 1 1 F þ ðsÞ ¼ O xl e jsx dx ¼ O l þ 1 as jsj ! 1: ð4:188Þ s 1 (b) If j f ðxÞj < Bebx when x ! 1, if f ðxÞ is infinitely differentiable for x < 0, and if f ðxÞ ¼ O½ðxÞm (m > 1) when x ! 0 in an area where s2 < b, then F ðsÞ ¼ O
ð 0
m jsx
1
ðxÞ e
dx
¼O
1
smþ1
as jsj ! 1:
ð4:189Þ
Theorem 4.4 If RðsÞ is analytic in an area where a < s2 < b and if RðsÞ ¼ Oðsd Þ in an area where d > 0 when jsj ! 1, then RðsÞ ¼ R þ ðsÞ þ R ðsÞ;
ð4:190Þ
where there exist R þ ðsÞ and R ðsÞ such that R þ ðsÞ ¼
ð ja1 þ 1
1 2pj
R ðsÞ ¼
ja1 1
1 2pj
RðzÞ dz zs
ð jb1 þ 1 jb1 1
RðzÞ dz zs ;
ð4:191Þ
ð4:192Þ
a < a1 < s 2 < b 1 < b where R þ ðsÞ is analytic when s2 > a1 and R ðsÞ when s2 < b1 . Figure 4.47 shows integration paths and analytic areas. Theorem 4.5 Liouville’s theorem If FðsÞ is analytic in all s and jFðsÞj < M, for some constant M > 0, then FðsÞ is a constant function. Similarly, if FðsÞ is analytic for all s and FðsÞ ! 0 when jsj ! 1, then FðsÞ ¼ 0. Asymptotic Method: Stationary Phase We define IðxÞ ¼
ðb a
f ðtÞe jxcðtÞ dt:
ð4:193Þ
240
Sound Propagation
s2
Region that R+ (s) is analytic
β Integration path of R– (s) s1 Integration path of R+ (s) α
Region that R– (s) is analytic
Strip that R(s) is analytic
Figure 4.47
Integration paths and analytic areas
As x ! 1, it is true that the value at the stationary phase point ðt ¼ aÞ contributes most to the integral if c0 ðtÞ 6¼ 0 at all t when t ¼ aða < a < bÞ. Therefore, we can write IðxÞ ffi
ða þ e
f ðtÞe jxcðtÞ dt ffi
ae
ffi f ðaÞe jxcðaÞ
ða þ e
c00 ðaÞ 2 f ðtÞe jx cðaÞ þ 2 ðtaÞ dt
ae
ðe e
xc00 ðaÞ j 2 t2
e
dt ffi 2f ðaÞ e jxcðaÞ
ð4:194Þ
ð1 e
xc00 ðaÞ j 2 t2
dt;
0
where ð1 0
ð1 0
pffiffiffi p jp=4 e dt ¼ e ; 2 jt2
ejt dt ¼ 2
ð4:195Þ
pffiffiffi p jp=4 e : 2
ð4:196Þ
Therefore, Equation 4.194 can be summarized as 8 2p > jxcðaÞ > f ðaÞe e jp=4 ; > > < xc00 ðaÞ IðxÞ ffi > > 2p > jxcðaÞ > e jp=4 ; : f ðaÞe xc00 ðaÞ
c00 ðaÞ > 0 ð4:197Þ c00 ðaÞ < 0:
241
Radiation, Scattering, and Diffraction
4.7.6.2 Sound Pressure Above Barrier Based on the theorems outlined above, we now solve diffraction problems for a barrier. Inhomogeneous Kirchhoff–Helmholtz Integral Equation The inhomogeneous Helmholtz equation that determines Green’s function can be written as r2 P þ k2 P ¼ fð~ rÞ;
ð4:198Þ
r2 Gð~ rj~ r 0 Þ þ k2 Gð~ rj~ r 0 Þ ¼ dð~ r~ r 0 Þ:
ð4:199Þ
The Kirchhoff–Helmholtz integral equation can also be expressed as (Figure 4.48) ð Pð~ rÞ ¼ fð~ r 0 ÞGð~ rj~ r 0 ÞdV V
ð4:200Þ
ð
þ
ðPð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r 0 ÞÞ ~ n 0 dS: S0
Incident wave
y x O
r
z
r0
Barrier
Figure 4.48
Coordinate system of barrier problem
Sound Pressure at z < 0 If we employ the Green’s function that satisfies the Neumann boundary condition G<0 N , then Equation 4.200 can be rewritten as ð Pðx; y; zÞ ¼ fðx0 ; y0 ; z0 ÞG<0 N ðx; y; zjx0 ; y0 ; z0 Þdx0 dy0 dz0 V0
ð S0
ð4:201Þ
@P G<0 N ðx; y; zjx0 ; y0 ; 0Þ @z
dx0 dy0 : z¼0
242
Sound Propagation
We assumed that noise source exists for z < 0. Using the image source method, G<0 N can be expressed as G<0 N ðx; y; zjx0 ; y0 ; z0 Þ ¼ GF ðx; y; zjx0 ; y0 ; z0 Þ þ GF ðx; y; zjx0 ; y0 ; z0 Þ:
ð4:202Þ
where it can be written as e jkR GF ðx; y; zjx0 ; y0 ; z0 Þ ¼ 2pR qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 R ¼ ðxx0 Þ þ ðyy0 Þ2 þ ðzz0 Þ2
ð4:203Þ
which is the free field Green’s function. Therefore, Equation 4.201 can be rewritten as ð fðx0 ; y0 ; z0 Þ½GF ðx; y; zjx0 ; y0 ; z0 Þ Pðx; y; zÞ ¼ V0
þ GF ðx; y; zjx0 ; y0 ; z0 Þdx0 dy0 dz0 Ð @P 2 S0 GF ðx; y; zjx0 ; y0 ; 0Þ dx0 dy0 : @z z¼0
ð4:204Þ
Sound Pressure at z > 0 Using the Green’s function that satisfies the Neumann boundary condition G>0 N , Equation 4.200 can be written as ð @P >0 Pðx; y; zÞ ¼ GN ðx; y; zjx0 ; y0 ; 0Þ dx0 dy0 ; ð4:205Þ @z z¼0 S0 assuming that there is no noise source for z > 0. Using the image source method as before, G>0 N can be written as G>0 N ðx; y; zjx0 ; y0 ; z0 Þ ¼ GF ðx; y; zjx0 ; y0 ; z0 Þ þ GF ðx; y; zjx0 ; y0 ; z0 Þ: Therefore, Equation 4.205 can be summarized as ð @P Pðx; y; zÞ ¼ 2 GF ðx; y; zjx0 ; y0 ; 0Þ dx0 dy0 : @z z¼0 S0
ð4:206Þ
ð4:207Þ
Pressure Continuity Conditions at z ¼ 0 The sound pressure has to be continuous at y > 0 in a plane where z ¼ 0; this can be written as: Pðx; y; 0 Þ ¼ Pðx; y; 0 þ Þ;
y > 0:
ð4:208Þ
243
Radiation, Scattering, and Diffraction
Using Equations 4.204 and 4.207, we can rewrite Equation 4.208 as ð @P 4 GF ðx; y; zjx0 ; y0 ; 0Þ dx0 dy0 @z z¼0 S0 ð ¼ fðx0 ; y0 ; z0 Þ½GF ðx; y; zjx0 ; y0 ; z0 Þ
ð4:209Þ
V0
þ GF ðx; y; zjx0 ; y0 ; z0 Þdx0 dy0 dz0 : Note that GF ðx; y; zjx0 ; y0 ; z0 Þ ¼ GF ðx; y; zjx0 ; y0 ; z0 Þ
ð4:210Þ
in Equation 4.203. Equations 4.207, 4.209, and 4.210 allow us to write the following equality: ð @P Pðx; y; zÞ ¼ 2 GF ðx; y; 0jx0 ; y0 ; 0Þ dx0 dy0 @z z¼0 S0 ð4:211Þ ð ¼
fðx0 ; y0 ; z0 ÞGF ðx; y; 0jx0 ; y0 ; z0 Þdx0 dy0 dz0 : V0
4.7.6.3 Solution for 2D Plane Wave Incidence Problem Definition and Fourier Transform If a 2D plane wave is incident, then the incident wave can be expressed as (Figure 4.49) Pin ðy; zÞ ¼ e jky þ jkz z ¼ e jðk sin FÞy þ jðk cos FÞz :
ð4:212Þ
( y, z) or (r, θ ) y O
r θ
z
Φ
Pin (y, z) = e j(k sinΦ )y + f(k cosΦ )z
Figure 4.49
Coordinate system of 2D barrier problem (plane wave incidence)
The observation location is ðy; zÞ, but the above can also be expressed in cylindrical coordinates ðr; yÞ. The following relation must hold: y ¼ r sin y; and the area of interest is for z > 0.
z ¼ rcos y;
ð4:213Þ
244
Sound Propagation
Because there is no noise source for z > 0, the governing equation can be expressed as @2P @2P þ 2 þ k2 P ¼ 0: @y2 @z
ð4:214Þ
The boundary condition at z ¼ 0 can be expressed as Pðy; 0Þ ¼ e jk sin Fy ; @P @z
¼ 0;
y<0
ð4:215Þ
y < 0:
ð4:216Þ
z¼0
This is a typical mixed boundary condition. If it is assumed that wave number k used in Equation 4.214 has a small dissipation (e.g., k ¼ k þ je), and we follow the definition of the generalized Fourier transform (Section 4.7.6.1), then the sound pressure at z > 0 can be written as ð1 Pðs; zÞ ¼ Pðy; zÞe jsy dy ð4:217Þ 1
If y is an infinitely large number, then jPðy; zÞj < Aeey and jPðy; zÞj > Beey . As we discussed in Section 4.7.6.1, then Pðs; zÞ is analytic in an area where e < s2 < e. If we transform Equation 4.214 accordingly, then it can be written as d2P 2 c Pðs; zÞ ¼ 0 dz2
ð4:218Þ
where c ¼ ðs2 k2 Þ1=2 ¼ ðs þ kÞ1=2 ðskÞ1=2 . In Equation 4.218, c is a multi-valued function. For c, the branch cut (Figure 4.50) can be used, and cð0Þ ¼ jk has to hold. In this case, Imfcg < 0 for all s, and Refcg > 0 when e < s2 < e.
s2
ε
Branch cut
–k X
–ε
0 ∼ 2π Branch cut X k
s1
π ∼ 3π Strip that R(s) is analytic
Figure 4.50
Branch point and branch cut
When such a branch is used, the solution of Equation 4.218 can be written as Pðs; zÞ ¼ AðsÞecz
ð4:219Þ
245
Radiation, Scattering, and Diffraction
because ecz diverges when z ! 1. Also, the boundary conditions of Equations 4.215 and 4.216 can be expressed as ð1 j ; ð4:220Þ P þ ðs; 0Þ ¼ e jk sin Fy þ jsy dy ¼ s þ ksin F 0 P0 ðs; 0Þ
¼
ð0 1
0e jsy dy ¼ 0:
ð4:221Þ
Using Equation 4.219, we can write Pðs; 0Þ ¼ P þ ðs; 0Þ þ P ðs; 0Þ ¼
j þ P ðs; 0Þ ¼ AðsÞ: s þ k sin F
ð4:222Þ
Equation 4.222 can also be written as P0 ðs; 0Þ ¼ P0þ ðs; 0Þ þ P0 ðs; 0Þ ¼ P0þ ðs; 0Þ ¼ cAðsÞ:
ð4:223Þ
Equations 4.222 and 4.223 are then summarized as P0þ ðs; 0Þ þ cP ðs; 0Þ ¼
jc : s þ k sin F
ð4:224Þ
Equation 4.224 has two unknowns: P0þ ðs; 0Þ and P ðs; 0Þ. To derive the two unknowns from one equation, we may try to use the Wiener–Hopf technique which is based on the characteristics of an analytic function. Wiener–Hopf Technique Equation 4.224 can be rewritten as ðs þ kÞ1=2 P0þ ðs; 0Þ þ ðskÞ1=2 P ðs; 0Þ ¼ RðsÞ:
ð4:225Þ
We can also write: RðsÞ ¼ j
ðskÞ1=2 : s þ k sin F
ð4:226Þ
Given Theorem 4.4, RðsÞ can be written as RðsÞ ¼ R þ ðsÞ þ R ðsÞ;
ð4:227Þ
where R þ ðsÞ ¼
1 2pj
ð je þ 1
1 R ðsÞ ¼ 2pj
je1
RðzÞ dz; zs
ð je þ 1 je1
RðzÞ dz: zs
ð4:228Þ
ð4:229Þ
246
Sound Propagation
R þ ðsÞ is analytic at s2 > e, and R ðsÞ is analytic at s2 > e. A contour integration on a half plane in the lower portion of Figure 4.51 allows us to write R þ ðsÞ as R þ ðsÞ ¼ j
ðk sin FkÞ1=2 ðk þ k sin FÞ1=2 ¼ : ðk sin FsÞ ðs þ k sin FÞ
ð4:230Þ
ς2
ε x – k sinΦ
Figure 4.51
x
s
x k
ς1
–ε
Integration contour for evaluating R þ ðsÞ
We do not evaluate R ðsÞ separately, because the value is not necessary. Given Equation 4.227, Equation 4.226 can be rewritten as ðs þ kÞ1=2 P0þ ðs; 0ÞR þ ðsÞ ¼ ðskÞ1=2 P ðs; 0Þ þ R ðsÞ: Define the function EðsÞ to utilize the characteristics of an analytic function: 8 < E þ ðsÞ ¼ ðs þ kÞ1=2 P0þ ðs; 0ÞR þ ðsÞ; s2 > e EðsÞ ¼ : E ðsÞ ¼ ðskÞ1=2 P ðs; 0ÞR ðsÞ; s2 < e:
ð4:231Þ
ð4:232Þ
Since ðs þ kÞ1=2 , P0þ ðs; 0Þ and R þ ðsÞ are analytic in an area where s2 > e, E þ ðsÞ is analytic for s2 > e. ðskÞ1=2 , P ðs; 0Þ and R ðsÞ are analytic in an area where s2 < e. Therefore, E ðsÞ is analytic for s2 > e. In the vicinity of s2 ¼ 0, E þ ðsÞ ¼ E ðsÞ from Equation 4.231. Therefore, EðsÞ is analytic everywhere. We now examine what happens when jsj ! 1. The sound pressure has to be finite around y ¼ 0, because there is no noise source. Therefore, if y is a positive number very close to 0, @p then ¼ Oðyd Þ. Given Theorem 4.3 it can therefore be written as @z z¼0 P0þ ðs; 0Þ ¼ Oðsd1 Þ as jsj ! 1: ð4:233Þ Also, Equation 4.230 allows us to write R þ ðsÞ as R þ ðsÞ ¼ Oðs1 Þ as jsj ! 1:
ð4:234Þ
247
Radiation, Scattering, and Diffraction
E þ ðsÞ in Equation 4.232 can therefore be written as E þ ðsÞ ¼ Oðs1=2 ÞOðsd1 ÞOðs1 Þ ¼ 0 as jsj ! 1:
ð4:235Þ
For the same reason that we considered Equation 4.233, it can be written as P ðs; 0Þ ¼ Oðs1 Þ as jsj ! 1:
ð4:236Þ
Also, Equations 4.227 and 4.234 allow us to write R ðsÞ ¼ Oðs1=2 ÞOðs1 Þ ¼ Oðs1=2 Þ as jsj ! 1:
ð4:237Þ
Therefore, E ðsÞ in Equation 4.232 can be expressed as E ðsÞ ¼ Oðs1=2 ÞOðs1 Þ þ Oðsd1 Þ ¼ 0 as jsj ! 1:
ð4:238Þ
Since EðsÞ is analytic according to the results that we have, and from Equations 4.235 and 4.238, EðsÞ ¼ 0 when jsj ! 1. Therefore, we can finally write EðsÞ ¼ 0;
ð4:239Þ
which is based on Theorem 4.5. Evaluating P0þ ðs; 0Þ from Equation 4.232, and solving AðsÞ from Equation 4.223 and substituting it into Equation 4.219, the solution in the wave number area can ultimately be found as Pðs; zÞ ¼
ðk þ k sin FÞ1=2 ðs þ k sin FÞðskÞ
Equation 4.240 can be written as
ð
1=2
ecz :
fðsÞeczjsy ds:
Pðy; zÞ ¼
ð4:240Þ
ð4:241Þ
p
By using an inverse Fourier transform, the sound pressure in a space area can be obtained as fðsÞ ¼
1 ðk þ k sin FÞ1=2 : 2p ðs þ k sin FÞðskÞ1=2
ð4:242Þ
Figure 4.52 shows the integration path that is used for the inverse Fourier transform. The eczjsy term in Equation 4.241 is difficult to handle, so we change its integration path in order to make it easier to deal with. We first examine what happens when jsj ! 1. In this section, we want to look at eczjsy with s ¼ Re jf (Figure 4.53) and R ! 1. If 0 < f < p, and when R ! 1, then ðs þ kÞ1=2 ½Re jðf þ 2pÞ 1=2 e jðf=2 þ pÞ
ð4:243Þ
ðskÞ1=2 ½Re jf 1=2 ¼ R1=2 e jf=2 :
ð4:244Þ
and
248
Sound Propagation s2
ε –k x
Figure 4.52
x
x k
–ε
s1
Integration path of inverse Fourier transform
s2
–k x π ∼ 3π
x
R
0 ∼ 2π
φ
x k
–ε
s1
– k sin Φ
Figure 4.53 Complex coordinate for observing movements at jsj ! 1
The real part of the exponential term eczjsy can therefore be expressed as Refczjsyg ¼ Refðs þ kÞ1=2 ðskÞ1=2 zjsyg ¼ RefR jfz jR jfy g
ð4:245Þ
¼ Rðcos fz þ sin fzÞ: If we substitute Equation 4.213 into the equation above, then Refczjsyg ¼ Rrðcos f cos y þ sin f sin yÞ ¼ RrcosðfyÞ:
ð4:246Þ
Therefore, considering p=2 < y < p=2, it can be written as Refczjsyg < 0;
p=2 þ y < f < p:
ð4:247Þ
As shown in Figure 4.54, eczjsy ¼ 0 on path p1 when R ! 1. If p < f < 2p, by the same reasoning, Refczjsyg < 0;
3p=2y < f < 2p:
On path p2 in Figure 4.54, therefore, it becomes eczjsy when R ! 1.
ð4:248Þ
249
Radiation, Scattering, and Diffraction s2
R Path
θ
P1 x –k – k sin Φ x x
R
θ Path
Figure 4.54
This therefore leads us to
ð
s1
k
P2
Paths with no contribution to integration
fðsÞeczjsy ds ¼ 0 as R ! 1
ð4:249Þ
fðsÞeczjsy ds ¼ 0 as R ! 1:
ð4:250Þ
p1
and
ð p2
We now consider p3, a path that is associated with t: s ¼ k sinðy þ jtÞ;
i:e:;
s1 ¼ k sin y cosh t;
s2 ¼ k cos y sinh t:
Based on the relation of a hyperbolic function, it can be written as s 2 s 2 1 2 ¼ 1: k sin y k cos y
ð4:251Þ
ð4:252Þ
Hence, the path p3 (Figure 4.55) represents the path of a hyperbolic function. In the path p3 it can be written as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi czjsy ¼ s2 k2 zjsy ¼ jkr cosðy þ jtÞcos y þ jkrsinðy þ jtÞsin y ¼ jkrðcos y cosh t cos yj sin y sinh tcos y þ sin y cosh t sin y þ j cos ysinh t sin yÞ ¼ jkr cosh t:
ð4:253Þ
250
Sound Propagation s2 t = −∞ θ
p3
Path −k sinθ
x −k x
s1
k
N
−k sinΦ θ t=∞
Figure 4.55 Path of hyperbolic function
The integral over path p3 in Equation 4.241 can therefore be expressed as ð p3
fðsÞeczjsy ds ¼
ð1 1
f ½k sinðy þ jtÞe jkr cosh t ½jt cosðy þ jtÞdt:
ð4:254Þ
We next take two contour integrations in order to transform the integral over the path p in Equation 4.241 into that over path p3 in Equation 4.254. If the integration is performed over a path in Figure 4.56, we can write ð
ð
ð
þ p
ð
þ
þ
p2
v1
¼ 2pjResI ;
ð4:255Þ
v2
s2
−k sinΦ −k
Path
k x s1
x
Path v2 Path v1
p
R
θ Path
Figure 4.56 Path I
p2
251
Radiation, Scattering, and Diffraction
where ResI represents the residue of the pole surrounded by Contour I. Since Equation 4.250 holds when R ! 1, Equation 4.255 can be rewritten as ð
ð
ð
¼ p
v1
2pjResI :
ð4:256Þ
v2
If integration is performed on the path in Figure 4.57, we can write the second integration as ð ð ð ð þ þ þ ¼ 2pjResII ð4:257Þ v1
v2
p1
p2
s2
θ
Path p1
p3
Path
−k sinθ x
−k Path Path
k
x
X
v2
s1
−k sinΦ θ
v1
Figure 4.57 Path II
where ResII represents the residue of the poles (Figure 4.57). Because Equation 4.249 holds when R ! 1, Equation 4.257 can be rewritten as ð ð ð ¼ þ 2pjResII : ð4:258Þ v1
v2
p2
From Equations 4.256 and 4.258, the integral of path p in Equation 4.241 can be expressed (by changing the integration path) as ð ð Pðy; zÞ ¼ ¼ þ 2pjðResII ResI Þ: ð4:259Þ p
p2
The term in Equation 4.259, 2pjResI , can be written as 2pjResI ¼ 2pj
ffi 1 ðk þ ksin FÞ1=2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2 sin Fk2 z þ jksin Fy e 2p ðkksin FÞ1=2
¼ e jksin Fy þ jkcos Fz
ð4:260Þ
252
Sound Propagation
and 2pjResII can be obtained as ( 2pjRes ¼ II
e jksin Fy þ jkcos Fz ;
y
0;
y > F:
ð4:261Þ
Therefore, Equation 4.259 can be rewritten as ( ð e jksin Fy þ jkcos Fz ; y < F Pðy; zÞ ¼ fðsÞ þ p2 0; y > F:
ð4:262Þ
In Equation 4.254, ðcosh tÞ0 ¼ sinh t ¼ 0 as t ¼ 0:
ð4:263Þ
Therefore, in a far field where kr ! 1, the area around t ¼ 0 provides the largest contribution to the integration. From this, using the stationary phase method, the integral term of Equation 4.262 can be written as ð1 ð fðsÞeczjsy ds ¼ f ½ksinðy þ jtÞe jkr cosh t ½jkcosðy þ jtÞdt 1
p2
f ðksin yÞðjkcos yÞ
ðe
e jkrð1 þ t
2
=2Þ
dt
e
1 ðk þ k sin FÞ1=2 ðjk cos yÞ e jkr ¼ 2p ðk sin Fk sin yÞðk sin ykÞ1=2
rffiffiffiffiffiffi 2p jp4 e kr
ð4:264Þ
e jkr þ jp=4 ¼ pffiffiffiffiffiffiffiffiffiffi Dðy; FÞ: 8pkr Here, Dðy; FÞ can be approximately expressed as Dðy; FÞ ffi 2
ð1 þ sin FÞ1=2 ð1sin yÞ1=2 : sin Fsin y
ð4:265Þ
Dðy; FÞ of Equation 4.265 also can be rewritten as Dðy; FÞ ¼
1 1 cosðy=2 þ F=2Þ sinðy=2F=2Þ
ð4:266Þ
by using the relations: ð1 þ sin FÞð1sin yÞ ¼ ½cosðy=2 þ F=2Þsinðy=2F=2Þ2
ð4:267Þ
sin Fsin y ¼ 2cosðy=2F=2Þsinðy=2F=2Þ:
ð4:268Þ
and
253
Radiation, Scattering, and Diffraction
In summary, Equation 4.262 can be written approximately as ( 0; e jkr þ jp4 Pðy; zÞ ¼ pffiffiffiffiffiffiffiffiffiffi Dðy; FÞ þ 8pkr e jðksin FyÞ þ jðkcos FzÞ ;
y F:
ð4:269Þ
Equation 4.269 is a weak solution because Dðy; FÞ diverges around y ¼ F. This is because, when we derive Equation 4.264, we did not rigorously obtain fðsÞ. In fact, fðsÞ does not well behave when the pole of f ðsÞ comes closer to the integral path as y tends to F. e jkr þ jp=4 In Equation 4.269, the first term represents a diffraction field. Note that pffiffiffiffiffiffiffiffiffiffi is a far field 8pkr solution of a 2D monopole source. This simply means that the diffraction can be seen as the radiated field due to a noise source on the edge of the barrier and whose direction is defined by Dðy; FÞ. The second term of Equation 4.269 shows that the diffraction field appears without a direct field in the shadow region, but the direct field appears with the diffraction field elsewhere. 4.7.6.4 Solution for 2D Monopole Source If we have a monopole source, then the solution becomes more complicated than Equation 4.229 of Section 4.7.6.3. Let us assume that the location of the observer is ðy; zÞ and that of the noise source ðb; aÞ. We also introduce a cylindrical coordinate to express the shape of diffraction. The following relations (Figure 4.58) must hold: z ¼ rcos y;
y ¼ sin y
a ¼ rs cos F;
b ¼ rs sin F:
ð4:270Þ ð4:271Þ
( y,z) or ( r,θ) y r O
b
r3
Φ
θ
z
r4
Ψ α
Figure 4.58
Coordinate system of 2D barrier problem (for 2D monopole source)
These can also be written as z þ a ¼ rd cos c;
y þ b ¼ rd sin c:
ð4:272Þ
254
Sound Propagation
Next, we try to express the diffraction field due to the monopole source at ðb; aÞ by superimposing plane waves that are already evaluated. If the sound pressure by an incident wave at the barrier is Pin ðy; 0Þ ¼ e jky y ;
ð4:273Þ
then Equation 4.262 of Section 4.7.6.3 can be written as ð Pðy; zÞ ¼
1 ðk þ ky Þ1=2 eczjsy ds 1=2 p2 2p ðs þ ky ÞðskÞ ( 0; ky > ksin y þ ejkr þ jp4 ; ky < ksin y
b ¼ ðk2y k2 Þ1=2 ¼ ðky þ kÞ1=2 ðky kÞ1=2 ;
ð4:274aÞ
ð4:274bÞ
where b is a multi-valued function, and the branch cut and the branch used are the same as for c in Section 4.7.6.3. If the sound pressure at the boundary by an incident wave is Pin ðy; 0Þ ¼ ejky y ; then Equation 4.274 can be rewritten as ð 1 ðkky Þ1=2 Pðy; zÞ ¼ eczjsy ds 1=2 p2 2p ðsky ÞðskÞ ( 0; ky > ksin y þ ebzjky y ; ky < ksin y: Note that the incident wave can be expressed by Fourier transform pair, that is ð1 Pin ðy; 0Þeky y dy Pin ðky ; 0Þ ¼ 1
Pin ðy; 0Þ ¼
ð1 1
Pin ðky ; 0Þejky y dky :
ð4:275Þ
ð4:276Þ
ð4:277Þ
ð4:278Þ
The principle of superposition leads us to express the sound pressure as Pðy; zÞ ¼ uðy; zÞ þ vðy; zÞ;
ð4:279Þ
where 1 uðy; zÞ ¼ 2p
(ð
ð p0
Pin ðky ; 0Þ
) 1 ðkky Þ1=2 eczjsy ds dky 1=2 p3 2p ðsky ÞðskÞ
ð4:280Þ
255
Radiation, Scattering, and Diffraction
and 1 vðy; zÞ ¼ 2p
ð p00
Pin ðky ; 0Þebzjky y dky :
ð4:281Þ
Figure 4.59 shows integral paths p0 and p00 in the ky complex plane. We now express the incident wave in the wave number area. The incident wave must satisfy the governing equation, that is, @ 2 Pin @ 2 Pin þ þ k2 Pin ðy; zÞ ¼ dðy þ bÞdðz þ aÞ: @y2 @z2
ð4:282Þ
s2
Path −k X
p′ ε
X
−ε Path p′ k
s1
−k sinθ
Figure 4.59
Integral paths p0 and p00
Also, when the Fourier transform is defined as ð1 Pin ðky ; zÞ ¼ Pin ðy; zÞeky y dy; 1
ð4:283Þ
Equation 4.282 can be rewritten as @ 2 Pin b 2 Pin ðy; zÞ ¼ ejky b dðz þ aÞ: @z2
ð4:284Þ
The real term of b is a positive number similar to c on the integral path that is used for inverse transformation. The solution of Equation 4.284 can therefore be expressed as 8 < Aðky Þebðz þ aÞ ; z < a Pin ðky ; zÞ ¼ ð4:285Þ : Bðk Þebðz þ aÞ ; z > a: y It has to be continuous at z ¼ a, therefore Aðky Þ ¼ Bðky Þ holds. Also, if we integrate Equation 4.284 around z ¼ a, then þ @Pin z¼a ¼ ejky b : ð4:286Þ @z z¼a If we substitute Equation 4.285, then we obtain bAðky ÞbAðky Þ ¼ ejky b :
ð4:287Þ
256
Sound Propagation
We therefore have Aðky Þ ¼
ejky b : 2b
ð4:288Þ
Finally, the sound pressure of the wave number area at z ¼ 0 is obtained as Pin ðky ; 0Þ ¼
ebajky b : 2b
ð4:289Þ
Substituing Equation 4.289 into Equation 4.280, we obtain ð 1 ebajky b uðy; zÞ ¼ 2p p0 2ðky þ kÞ1=2 ðky kÞ1=2 (ð ) 1 ðkky Þ1=2 czjsy e ds dky 1=2 p0 2p ðsky ÞðskÞ (ð ) ð 1 eczjsy 1 ðkky Þ1=2 ebajky b dky ds ¼ 2p p3 ðskÞ1=2 p0 4p ðky þ kÞ1=2 ðky kÞ1=2 ðsky Þ ð 1 eczjsy ¼ Iðs; zÞds 2p p3 ðskÞ1=2 where
ð Iðs; zÞ ¼
p0
j ebajky b dky : 4p ðky þ kÞ1=2 ðky sÞ
ð4:290Þ
ð4:291Þ
Defining the integral path p03 as ky ¼ k sinðFjt0 Þ
ð4:292Þ
and defining the integral path as depicted in Figure 4.60 allows us to define Iðs; zÞ(4.291) as ð j ebajky b Iðs; zÞ ¼ dky þ 2pjðResII ResI Þ: ð4:293Þ 1=2 p3 4p ðky þ kÞ ðky sÞ A similar derivation to that was used to obtain Equation 4.259 (Section 4.7.6.3) has been used. If y > F, then the residues of Equation 4.293 can be written as ! j ecajsb 1 ecajsb II 2p jRes ¼ 2pj ð4:294Þ ¼ 4p ðs þ kÞ1=2 2 ðs þ kÞ1=2 8 > < 0; I 2p jRes ¼ 1 ecajsb > ; :2 ðs þ kÞ1=2
Imfsg > 0 Imfsg < 0:
ð4:295Þ
257
Radiation, Scattering, and Diffraction ky2 ky2
s
s
x
k x
−k x
Φ
x ky1
− k sinθ
− k sinθ
−k k
Φ
x k
ky1
− k sinΦ Φ
Figure 4.60 Paths I and II
uðy; zÞ is therefore defined uðy; zÞ ¼ u1 ðy; zÞ þ u2 ðy; zÞ;
ð4:296Þ
where 1 u1 ðy; zÞ ¼ 2p
ð p3
eczjsy
(ð
ðskÞ1=2
) j ebajky b dky ds 1=2 p0 3 4p ðky þ kÞ ðky sÞ
ð4:297Þ
and 1 u2 ðy; zÞ ¼ 2p
ð Imfsg>0 on
ecðz þ aÞjsðy þ bÞ ds: 2c p3
ð4:298Þ
If Equation 4.289 is substituted into Equation 4.281, then vðy; zÞ can be written as ð 1 ecðz þ aÞjsðy þ bÞ vðy; zÞ ¼ ds ð4:299Þ 2p p 2c And Pðy; zÞ can be rewritten as Pðy; zÞ ¼ u1 ðy; zÞ þ u2 ðy; zÞ þ vðy; zÞ:
ð4:300Þ
We now consider u2 ðy; zÞ þ vðy; zÞ in Equation 4.300 in some detail. u2 ðy; zÞ and vðy; zÞ have the same integral functions, but their integral paths are Imfsg > 0 on p3 and p00 , respectively (see Figure 4.61). y > f when y > F, therefore the integral along the path p1 in Figure 4.61 is zero. u2 ðy; zÞ þ vðy; zÞ can therefore be rewritten as ð 1 ecðz þ aÞjsðy þ bÞ ds: ð4:301Þ u2 ðy; zÞ þ vðy; zÞ ¼ 2p p 2c
258
Sound Propagation s2
Ψ Im(s) > 0 on path p3
θ
Path p1
−k
x
Path
p″
x
Path
p
k
s1
−k sinθ
Figure 4.61 Path for calculation of u2 ðy; zÞ þ vðy; zÞ (when y > F)
As we can expect from Equation 4.289, it represents the sound field when a monopole source is at ðb; aÞ in a free field. It can therefore be written as j ð1Þ u2 ðy; zÞ þ vðy; zÞ ¼ H0 ðkrd Þ: 4
ð4:302Þ
If we first take a look at the situation when y < F, the residues in Equation 4.293 can be written as 2p jResII ¼ 0;
ð4:303Þ
8 0; Imfsg > 0 > < I 2pjRes ¼ 1 ecajsb > ; Imfsg < 0: : 2 ðs þ kÞ1=2
ð4:304Þ
In Equation 4.296, u2 ðy; zÞ can therefore be written as ð 1 ecðz þ aÞjsðy þ bÞ u2 ðy; zÞ ¼ ds: 2p Imfsg<0 on p3 2c
ð4:305Þ
Note that that u2 ðy; zÞ and vðy; zÞ of Equation 4.300 have the same integral but their integral paths are Imfsg < 0 on p3 and p00 , respectively (see Figure 4.62). y < f when y < F, therefore the integral for path p2 in Figure 4.62 is zero. Because there is no pole in the contour of Figure 4.62, u2 ðy; zÞ þ vðy; zÞ is u2 ðy; zÞ þ vðy; zÞ ¼ 0: These allow us to write Equation 4.300 as 8 y F: d 4 0
ð4:306Þ
ð4:307Þ
259
Radiation, Scattering, and Diffraction s2
−k sinθ Path p″ −k
x s1
k
x Ψ
Im(s) < 0 on path −p3
θ Path
Figure 4.62
p2
Path for calculation of u2 ðy; zÞ þ vðy; zÞ ¼ 0 (when y < F)
We now calculate u1 ðy; zÞ of Equation 4.297. Using the stationary phase method when krs ! 1, the integral of ky in Equation 4.297 can be approximately expressed as ð j ebajky b e jkrs þ jp=4 ðk þ k sin FÞ1=2 pffiffiffiffiffiffiffiffiffiffiffiffi dk ffi : ð4:308Þ y 1=2 s þ k sin F 8pkrs p0 3 4p ðky þ kÞ u1 ðy; zÞ is therefore expressed as e jkrs þ jp=4 u1 ðy; zÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi 8pkrs
ð
1 ðk þ k sin FÞ1=2 eczjsy ds: 1=2 2p p3 ðs þ k sin FÞðskÞ
ð4:309Þ
Hence, if we use Equation 4.264 of Section 4.7.6.3, we can write e jkrs þ jp=4 e jkr þ jp=4 u1 ðy; zÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi Dðy; FÞ: 8pkrs 8pkr From Equation 4.307, the sound pressure can finally be written as 8 y < 0; jkrs þ jp=4 jkr þ jp=4 e e Pðy; zÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi Dðy; FÞ þ j > 8pkrs 8pkr : Hð1Þ ðkrd Þ; y > F: 4 0
ð4:310Þ
ð4:311Þ
e jkrs þ jp=4 Equation 4.311 is very similar to that of a plane wave. The difference is the term pffiffiffiffiffiffiffiffiffiffiffiffi , 8pkrs which acts as the source strength of the diffraction field at the edge of the barrier. We next consider the case when a monopole source is located at any point on the surface. Figure 4.63 depicts the geometrical relation and coordinate axis for defining the rigid ground problem. In this case, the sound field at y > 0 can be regarded as the same as that of the case which is illustrated in Figure 4.64. In Figure 4.63, the plane at z ¼ 0 can be considered as the boundary for the region of z > 0. The boundary condition can be expressed as Pðy; 0Þ ¼ PF ðy; 0Þ;
y>0
ð4:312Þ
260
Sound Propagation y θ
O z
r1 Φ h
+
Figure 4.63
r
Coordinate system of barrier problem with rigid body surface
y r θ
O + +
z
r1 Φ h r′ Φ′ h
r′
−θ′
Figure 4.64
Expression of rigid body surface using image source
vðy; 0Þ ¼ 0;
2h < y < 0
Pðy; 0Þ ¼ PF ðy; 0Þ;
y < 2h;
ð4:313Þ ð4:314Þ
where PF ðy; zÞ is sound pressure in a free field. On the other hand, for the case of Figure 4.65(a) the boundary condition can be written as P1 ðy; 0Þ ¼ PF ðy; 0Þ; v1 ðy; 0Þ ¼ 0;
y>0
ð4:315Þ
2h < y < 0
ð4:316Þ
P1 ðy; 0Þ PF ðy; 0Þ;
y < 2h:
ð4:317Þ
If we denote the sound field at z > 0 as P1 ðx; yÞ then the height of the barrier h is sufficient. Similarly, for the case of Figure 4.65(b), the boundary condition can be expressed as P2 ðy; 0Þ ¼ 0; v2 ðy; 0Þ ¼ 0;
y>0
ð4:318Þ
2h < y < 0
ð4:319Þ
P2 ðy; 0Þ PF ðy; 0Þ;
y < 2h:
ð4:320Þ
If we denote the sound field at z > 0 to be P2 ðx; yÞ then the height of the barrier h is high enough. P1 ðy; zÞ þ P2 ðy; zÞ therefore satisfies the boundary condition and governing equation
261
Radiation, Scattering, and Diffraction y O z
r′2 + +
θ
∞
r
y
Φ h r′1 Φ′ h
O z +
h
− θ′
r′2
−∞ (a)
Figure 4.65
r′
Φ′ r′1 Φ h
(b)
Coordinate system using two semi-infinite barriers
of Figure 4.64 at z > 0. This is the solution that satisfies the condition illustrated by Figure 4.64. Thus, the solution for the case where the surface is rigid can be written as Pd ðy; zÞ ¼
jkr þ jp=4 0 e jkrs þ jp=4 e jkr s þ jp=4 e pffiffiffiffiffiffiffiffiffiffiffiffi Dðy; FÞ þ pffiffiffiffiffiffiffiffi Dðy; F0 Þ pffiffiffiffiffiffiffiffiffiffi 8pkrs 8pk 8pkr jkrs þ jp=4 jkr0 þ jp=4 0 e e jkr s þ jp=4 e 0 0 þ pffiffiffiffiffiffiffiffiffiffiffiffi Dðy ; FÞ þ pffiffiffiffiffiffiffiffi Dðy; F Þ pffiffiffiffiffiffiffiffiffiffiffi 8pkrs 8pk 8pkr0
ð4:321Þ
if we only express the diffracted sound. 4.7.6.5 Solution for 3D Monopole Source Assume that the observer is located at the orthogonal coordinate of ðx; y; zÞ and the cylindrical coordinate of ðx; y; yÞ (Figure 4.66). We also assume that the location of the noise source is ðc; b; aÞ, or ðc; rs ; FÞ in terms of the cylindrical coordinate. The distance between the noise source and the observer is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 rd ¼ ðx þ cÞ2 þ ðy þ bÞ2 þ ðz þ aÞ2 : ð4:322Þ If we do not consider the x axis, then the distance is rd ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðy þ bÞ2 þ ðz þ aÞ2 :
ð4:323Þ
The shortest distance that connects the noise source, the edge of the barrier, and the observer is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L ¼ ðx þ cÞ2 þ ðrs þ rÞ2 : ð4:324Þ
262
Sound Propagation
r θ
y x r2
O z
Φ
Figure 4.66 Coordinate system of a 3D barrier problem
As previously done, we attempt to evaluate the solution in the wave number domain. If we have a monopole source, then the governing equation of the incident wave can be written as @ 2 Pin @ 2 Pin @ 2 Pin þ þ þ k2 Pin ðx; y; zÞ @x2 @y2 @z2
ð4:325Þ
¼ dðx þ cÞdðy þ bÞdðz þ aÞ: Defining the Fourier transform with respect to x as ð1 Pin ðky ; y; zÞ ¼ Pin ðx; y; zÞe jkx x dkx 1
ð4:326Þ
and performing the Fourier transform of Equation 4.325, we obtain @ 2 Pin @ 2 Pin þ þ a2 Pin ðkx ; y; zÞ ¼ Gðkx Þdðy þ bÞdðz þ aÞ @y2 @z2 where Gðkx Þ ¼ e jkx x ;
a¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2 k2x :
ð4:327Þ
ð4:328Þ
k in Equation 4.282 of Section 4.7.6.4 is replaced by a and the source term is multiplied by Gðkx Þ in Equation 4.327. The boundary condition of Section 4.7.6.4 can also be applied to Pin ðkx ; y; zÞ. By using the results of Section 4.7.6.4, the solution of the wave number domain with a 3D monopole source can be expressed as e jars þ jp=4 e jar þ jp=4 Pðkx ; y; zÞ ¼ Gðkx Þ pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi Dðy; FÞ 8pars 8par 8 0; y F: 4 0
ð4:329Þ
263
Radiation, Scattering, and Diffraction
The solution in terms of space can be found by performing an inverse Fourier transform, that is ð 1 1 Pðx; y; zÞ ¼ Pðkx ; y; zÞejkx x dkx : ð4:330Þ 2p 1 First, the diffraction field can be written as ð1
e jars þ jp=4 e jar þ jp=4 Gðkx Þ pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi Dðy; FÞejkx x dkx 8pars 8par 1 ð 1 jaðrs þ rÞjkx ðx þ cÞ jDðy; FÞ e dkx ¼ pffiffiffiffiffiffi 2 16p rs r 1 a
Pd ðx; y; zÞ ¼
1 2p
ð4:331Þ
where a0 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2x k2 ¼ j k2 k2x ¼ ja:
ð4:332Þ
Therefore, Equation 4.331 can be rewritten as jDðy; FÞ Pd ðx; y; zÞ ¼ pffiffiffiffiffiffi 16p2 rs r
ð1
0
eja ðrs þ rÞjkx ðx þ cÞ dkx : a0 1
ð4:333Þ
This equation is similar to the integral derived in Sections 4.4 and 4.5. We can apply a similar processes using Equation 4.324; the diffraction field is then obtained as Pd ðx; y; zÞ ¼
1 e jkL þ jp=4 pffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Dðy; FÞ: 4p rs r 8pkL
ð4:334Þ
The geometrical solution can also be obtained by following a similar processes, and it becomes 0 when y < F and has a free field solution when y > F. Ultimately, the sound pressure is expressed as 8 y F: 4prd 0
ð4:335Þ
If a sound source is on the rigid surface, then similar results can be obtained by executing similar processes as those described in Section 4.6. 4.7.6.6 Experiments on Diffraction In the previous section, we studied diffraction by looking at analytic solutions. Figures 4.67–4.70, obtained experimentally, allow us to understand diffraction. We can match the analytic solutions to these experimental results.
264
Sound Propagation
Figure 4.67 Diffraction by a wall; the arrow is the direction of incident waves (note that the thickness of the walls are different)
Figure 4.68 Diffraction and scattering by finite size over changing time; the arrow indicates the direction of incident waves (note that the length of the bar is 90 mm, and the width is 13.5 mm)
Figure 4.69 Diffraction by different scatterers on the edge of barrier; the arrow indicates the direction of incident waves
265
Radiation, Scattering, and Diffraction
Figure 4.70
Diffraction due to different scatterers at the edge of barrier
Exercises 1. The acoustic power induced by the radiation of a breathing sphere is 1 ðkaÞ2 ð4pa2 ÞjU0 j2 : Iavg 4pr2 ¼ r0 c 2 1 þ ðkaÞ2 Prove that Iavg 4pr2 ¼
1 jPj2 4pr2 2 r0 c
for a breathing sphere. 2. The governing equation in the frequency domain for the sound field radiated by the point source at ~ r ¼~ r s with amplitude (S) can be expressed by the inhomogeneous Helmholtz equation, that is, r~ r s Þ: r2 P þ k2 P ¼ 4pSdð~ (a) Express the solution to the above equation in terms of S. Hint: Compare the solution to the free field Green’s function (b) Express qs in the term of S, where qs is the volume velocity of the point source. Hint: Use the result of (a) and the linearized Euler equation at the distance e, close to the observation point (~ r). (c) Rewrite the governing equation mentioned above using the volume velocity, qs . Hint: The governing equation expressed by the velocity potential (U) is given by r2 U þ k2 U ¼ qs dð~ r~ r s Þ:
266
Sound Propagation
The velocity potential is related to the particle velocity by uð~ rÞ ¼ rUð~ rÞ, and the rÞ by using the Euler equation. complex amplitude of the pressure by Pð~ rÞ ¼ jor0 Uð~ In this case, qs is often referred to as source strength. (d) The term 4p appears in the right-hand side of the inhomogeneous Helmholtz equation. Explain why it appears and what its physical meaning is. 3. Consider a region of interest with no sound source which is bounded by an arbitrary boundary (Figure 4.71). Then, the pressure at an arbitrary position is given by ð cð~ rÞPðrÞ ¼ ðPð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r 0 ÞÞ ~ n 0 dS: S0
Figure 4.71
Illustration for Exercise 3
G denotes the free field Green’s function: Gð~ rj~ r0Þ ¼ where R ¼ j~ r~ r 0 j.
e jkR ; 4pR
Radiation, Scattering, and Diffraction
267
(a) Show that cð~ rÞ ¼ 1 at the observation point~ r in the region (Oo ) outside the boundary surface. Hint: Derive the integral equation by surrounding the point by the sphere which has the radius e and is connected to So (Figure 4.71(b)) (b) Show that cð~ rÞ ¼ 1=2 at the observation point ~ r on the boundary surface, So . Hint: Derive the integral equation by surrounding the point on the boundary surface by the semi-sphere of radius e. Note that the perimeter is the half of that of (a) (Figure 4.71(c)). (c) Show that cð~ rÞ ¼ 0 at the observation point ~ r in the region (Oi ) inside the boundary surface by using the results of (a) and (b). 4. Velocity distribution of an arbitrary boundary surface is propagated by monopole radiation process, and pressure distribution of that surface is propagated by dipole radiation process. Each process creates a sound field at an arbitrary position by superposition (integral in mathematical sense). Analyze each term of the integral equation as follows: (a) Indicate terms representing velocity distribution and pressure distribution on the boundary surface. (b) Indicate terms representing monopole radiation process and dipole radiation process (that is, the Kernel of integral equation), and explain physical meaning of them. Hint: Use (i) Euler’s equation and (ii) free space Green’s function. 5. Obtain the sound field radiated by a breathing sphere of radius (a) and surface velocity (U0 ) using the integral equation and assuming the free field boundary condition. Compare your solution to that obtained by using the assumed solution of a spherical wave and boundary condition at the surface in the case of the breathing sphere. Are they the same? Compare them by checking their limit cases. 6. Use the Rayleigh integral to derive the sound field due to an infinitely baffled vibrating piston in a far field region. (a) Write the Rayleigh integral equation and explain its meaning. (b) Express R by using r; u; c, and prove that R ru cos c in the far field region (where u is the surface velocity). (c) Prove that cos c ¼ sin y cos f. Hint: Note that cosc ¼ ~ e r ~ eu. e jkR e jkr ejku sin y cos f ¼ from the results of (b) and (c). R r (e) Integrate the result of (d) with respect to the surface, and derive the far field sound pressure distribution (Figure 4.72). ðp ð 2 2 1 p cosðz cos yÞdy ¼ cosðz sin yÞdy Hint: Use J0 ðzÞ ¼ p 0 p 0 (d) Show that
268
Sound Propagation
Figure 4.72
An infinitely baffled vibrating piston
7. Explain the following expressions for the case where a monopole is laid on an infinite surface which has boundary conditions as follows. (Note: this provides a basic foundation of the image method. This method tries to distribute the well-known solutions – for example, monopole or dipole solution – and attempts to find the magnitude and phase that satisfy the given boundary condition.) (a) If the surface is rigid, that is, its impedance is infinite, the monopole and the boundary condition can express the sound field generated by the monopole source whose strength is doubled. (b) If the surface can be regarded as pressure released, the impedance is zero, and the monopole and boundary conditions can express the sound field generated by the dipole source. 8. Derive the acoustic reciprocity principle by the following procedures. (a) The sound pressure P1 generated by a volume source having amplitude S1 at~ r 1 can be expressed by using the Helmholtz equation: rÞ þ k2 P1 ð~ rÞ ¼ 4pS1 dð~ r~ r 1 Þ: r2 P1 ð~
ð4:336Þ
r2. Hint: Write the Helmholtz equation on P2 generated by the volume of amplitude S2 at ~ (b) Multiply Equation 4.335 by P2 ð~ rÞand the solution of exercise 8(a) byP1 ð~ rÞ, and subtract the latter from the former. (c) Take the volume integral of both sides of the solution to exercise 8(b), and use the Gauss theorem to convert the volume integral of the pressure term to surface integral at the boundary. (d) Derive the following equation using the fact that boundary conditions of the two cases are the same: P1 ð~ r 2 Þ P2 ð~ r1Þ ¼ : S1 S2 Provide some physical interpretation of Equation 4.337.
ð4:337Þ
269
Radiation, Scattering, and Diffraction
(e) You want to select the best position for your new speaker which provides the best sound at your listening position. However, it is not easy to move the speaker several times. Instead, you can set your speaker at a fixed place and move your microphone several times to find the best location. Why does this method work? 9. Compare the radiation impedances on the z axis for the two cases depicted in Figure 4.73 and discuss the results (assume the surface areas of the two cases are identical).
d a1
e−jωt
z
Case I: vibrating annular
a2
e−jωt
z
Case II: vibrating circular disk
Figure 4.73
Radiation impedance for two separate cases
10. Consider a sound source of a single frequency f at the position ðx; y; zÞ ¼ ð0; 0; 0Þ in free field. A student wants to determine the acoustic pressure distribution in the entire space. A technicality implies that he can only measure the acoustic pressure at the infinite plane on x ¼ h. The student therefore attempts to predict the sound pressure at positions other than x ¼ h by using the famous Kirchhoff–Helmholtz integral equation. He also found that the following are true. (a) The sound pressure can be predicted if we know the sound pressure on the surface of the infinite semi-sphere whose center is at ðx; y; zÞ ¼ ðh; 0; 0Þ. Furthermore, on the infinite sphere other than x ¼ h, the sound pressure will be very small and therefore negligible. (b) If we set Green’s function to be 0 at x ¼ h, the sound pressure can be predicted even although we do not measure pressure on the surface on x ¼ h (Figure 4.74). (i) Suggest the Green’s function that can be used by the student. Hint: Use the free space Green’s function
e jkR ; 4pR
R ¼ j~ r~ r 0 j.
(ii) Verify the student’s Green’s function. When we have a source that radiates a plane wave, pðx; tÞ ¼ Aejðkx xotÞ .
270
Sound Propagation y r→0 →
r
x
x=h
Figure 4.74
Sound source and its sound field
11. An plane wave of 80 dB SPL is incident at a rigid sphere which has a radius aðka 1Þ in free field, as depicted in Figure 4.75. When the plane wave is scattered by the sphere, the SPL at the point rðkr 1Þ which is far away from the center of sphere is measured by 86 dB. Determine the SPL at the position 2r (assume 10 log102 ¼ 3, 10 log103 ¼ 5).
Rigid sphere r
2r
SPL : 86dB
?
a Plane wave SPL : 80dB
Figure 4.75
Plane wave incident at a rigid sphere
12. Obtain the radiation impedance of the baffled piston depicted in Figure 4.76 on the z axis, and discuss its physical meaning.
y x dS ζ
a
Figure 4.76
R = ζ 2 + z2 z
An infinitely baffled vibrating piston
271
Radiation, Scattering, and Diffraction dB(A)
70
B A
60
h 50
d
4m
200
200
100
500
700
850
1.5m
8m
Figure 4.77
8m
Frequency (Hz)
Left: a wall to block the noise and right: the spectrum of the noise
13. A student is continually woken early in the morning by noise from his neighbor. He decides to build a wall to block the noise. The frequency characteristics of the noise are depicted in Figure 4.77. (a) The student thinks that reducing the total SPL by 30 dB would be enough to avoid being woken early. Based on this, determine the height of the wall. (b) Do you think the judgment of the student to be appropriate? Suggest a more effective method for this noise control problem.
5 Acoustics in a Closed Space 5.1
Introduction/Study Objectives
How sound propagates depends on the characteristics of the medium and what it meets in space and time, for example, walls or furniture. The propagation of sound is induced by the motion of fluid particles and pressure. The relationship between the pressures and velocity is therefore what determines the propagation. This relationship indicates the characteristic impedance when it applies to the medium or describes the impedance of the objects located in the space. Depending on the distribution of the impedance,1 the sound propagation differs significantly. We found that scattering, diffraction, and refraction constitute the factors that are induced by different types of impedance distribution in space. This impedance distribution can also be considered to create a closed or partially closed space. In other words, the walls that bound a space can be thought of as widely spatially distributed impedance. Therefore, sound propagation will be determined by the overall volume of the space and the wall impedances which characterize the space. The volume of space has to be considered with regard to the wavelength of interest. If it is fairly large, the waves would behave as if in a large space, and would reach all possible places. If this volume is small compared to the wavelength, then the wave would appear to be everywhere in the space instantly. Therefore, two distinct spaces are examined: an “acoustically large space,” and an “acoustically small space.” In addition, the measures that represent these two acoustic spaces are also introduced.
5.2
Acoustic Characteristics of a Closed Space
For a mathematical point of view, sound is nothing but a solution that satisfies the acoustic wave equation and boundary conditions. It is therefore possible to explore how the sound propagates and what sort of physical implications it makes by finding one of the many possible solutions that satisfy the governing equation and boundary conditions. However, it is usually 1 By spatial distribution, we mean the geometrical properties of the space and how they respond to the incident waves of certain frequency components.
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
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not plausible to express the sound that is likely to propagate in a space of interest mathematically. In particular, if the space of interest is large compared to the wavelength of interest, and the walls that enclose the space are complex in geometry as well as in terms of impedance distribution, then determining these types of mathematical solutions becomes highly complex. On the other hand, if the space is small compared to the wavelength, it may become possible to regard the space as one small physical element that can be easily modeled. Determining a solution that is valid in every position in the space is therefore also unnecessary. These arguments lead to the conclusion that the volume of a space of interest determines the major acoustical characteristics of sound propagation in the space. This immediately motivates a search for a measure that can define the size of the space. Intuitively, such a measure has to be scaled with pffiffiffiffi respect to the wavelength of interest. For example, if the representative length of the space p ( 3ffiffiffiVffi , where V is the volume of the space) is much smaller than the wavelength of interest (l 3 V ), then the fluid particles in the space can be regarded as if they are all moving with the same phase. On the other hand, if the characteristic length of the space is much larger than the wavelength, then the acoustic wave travels in the space as a ray. From these physical explanations, it is rational to classify the space into an “acoustically small space,” and an “acoustically large space.” It is, of course, natural to have very different means of understanding and analyzing how acoustic waves behave in the space. For an acoustically large space, Sabine2 found that the reverberation period represents the acoustic characteristics of the space well. This leads to a search for methods appropriate for a space that could be regarded as something in between. It may be possible to use a modal analysis method or to use a method involving Green’s function.
5.3
Theory for Acoustically Large Space (Sabine’s theory)
Figure 5.1 illustrates the propagation pattern of acoustic waves in two different types of space. This figure shows that acoustic waves tend to have very complicated patterns in space as time advances. This phenomenon pffiffiffiffi is, of course, possible when the space is relatively large compared to the wavelength: l 3 V . The figure also shows that the spatial distribution of the acoustic waves is not hugely dependent upon the location of the space. In other words, if the pressure is measured at any position in the space, it would be almost identical to the mean value. This phenomenon would be more likely if more randomly distributed wall impedance exists. Therefore, the sound of any frequency cannot dominate the sound propagation pattern in space: the sound of any frequency is equally likely to contribute in the space. In other words, the sound is well mixed in the space and is independent of position. This observation can also be made by comparing sounds in two different rooms, as illustrated in Figure 5.1. A room with a more complex shape compared to a simple rectangular room mixes the sound in a more random manner (Figure 5.1). A diffuse field implies a space in which the sound is likely to be equally distributed irrespective of the position. A somewhat loosely defined concept of a diffuse field is to describe a space that has uniformly averaged acoustic energy independent of the position. 2
Wallace Clement Sabine (1868–1919) is often referred to as the father of architectural acoustics. He established the basic theory and measures that characterize the sound propagation in a closed space: room acoustics.
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275
Figure 5.1 Illustration of sound propagation patterns in rooms. (These are the cases in which the size of room is acoustically large; in other words, the typical dimension of the room is much larger than the typical wavelength)
We first define acoustic energy density as 1 1 p2 eð~ r; tÞ ¼ r0 u2 þ 2 2 r0 c 2
ð5:1Þ
where eð~ r; tÞ is the energy density,~ r is the position vector, and t indicates the time. The first term on the left-hand side of the equation expresses the acoustic kinetic energy per unit volume, while the second term is potential energy induced by the expansion and contraction of the unit volume of the medium.3 3
As discussed in Chapter 2, we assume that the dissipated energy due to the viscosity of fluid is relatively small compared to potential and kinetic energy (Equation 5.1).
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Here, the sound energy at an arbitrary location is not expected to be perfectly uniform, even if the energy has been well mixed. The energy would depend on the position. However, if considering an averaged sound energy density ðeÞ with respect to a certain time (DT) and a small volume ðVÞ, expressed as ð ð 1 t þ DT 1 eð~ eð~ r; t0 ÞdVdt0 ; ð5:2Þ r; tÞ ¼ DT t DV DV then this averaged energy density is relatively independent of the position~ r and time. This is a very common measure that describes the sound field of a room. If a diffuse field is expressed using this measure, then the sound field would satisfy the equality eð~ r; tÞ ¼ eðtÞ.4 In other words, the sound field fluctuates in time but is independent of the location of the room: the same magnitude of sound is heard irrespective of where the listener sits in the room. This conjecture becomes even more acceptable considering the patterns shown in Figure 5.1. As time passes, the sound field in the rooms becomes more complex. Therefore, the magnitude of sound pressure at any position can be regarded as approximately uniform, averaged with respect to the cube of the typical wavelength. It is also clear that the sound field before the sound wave is reflected from the walls (for example, before t ¼ t0 ) is quite different compared to after the wave has been reflected as sound from the walls (for example, t ¼ 4t0 ) (Figure 5.1). The former is typically referred to as a “direct sound field” and the latter a “reverberant sound field,” or a “reflected sound field.”5 The sound energy of a reverberant field can be determined using an equation that expresses the conservation of sound energy (Equation 2.36): de ! þ r I ¼ 0: dt
ð5:3Þ
Equation 5.3 essentially states that the rate of change of the sound energy in the unit volume, which is in fact the sound energy density per unit time (unless there are sound sources), has to be balanced by the net sound flux through the surface of the unit volume. Equation 5.3 is then applied to the closed volume V, as illustrated in Figure 5.2. With the assumption that the volume does not include any sound source bounded by the surface of the room as well as by the sound source, if this equation is integrated with regard to the volume we have ð Y Y d edV ¼ out : ð5:4Þ in dt V Equation 5.4 essentially states that the energy in the volume Q must be balanced by the energy flux through the surfaces that create the volume. Here, denotes the power, which is Q denotes the power that comes into the the rate of energy change per unit time. Additionally, in Q volume and out represents the power leaving the volume through the boundary as shown in Figure 5.2. 4
There is a definition that takes into account the directivity of energy. For example, a diffuse sound field is defined as the sound in a region where the intensity is the same in all directions and at every point (ASTM C634-09. (2009) Standard Terminology Relating to Building and Environmental Acoustics, American Society for Testing and Materials (ASTM) International, West Conshohocken, Pennsylvania). 5 The expression ‘reverberant sound field’ is often used to describe a sound field that has very rich frequency and sound fluctuations, which is a rather subjective description.
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277
out
V
in
Figure 5.2 The energy balance by the power through the boundary (V is volume, Pin is the incoming power, and Pout is the outgoing power)
As noted before, it is possible to regard the sound in a closed space as being composed by two sound fields: the first is direct and the second is reverberant. It is therefore possible to rewrite Equation 5.4 as:6 ð Y ð Y Y d edirect dV þ erev dV ¼ þ : ð5:5Þ in;direct out;direct out;rev dt V V QThe right-hand side of Equation 5.5 states that the acoustic power lossQthrough the walls ( out ) occurs in two different Q ways: the loss induced by the direct sound ( out;direct ) andQ that dueQto the reverberant sounds ( out;rev ). It is also quite rational to denote the input power ( in ) as in;direct , simply because the input power only comes from the direct sound field. The lefthand side of Equation 5.5 also states that the sound energy in a selected space can be regarded as the sum of direct and reverberant sound fields. How the sound behaves under a reverberant field that is induced by the reflections of all the direct sounds is also important. If Equation 5.5 is applied when only a reverberant sound field exists which, in fact, implies that it would be possible to eliminate the factors related to the direct sounds from Equation 5.5, the energy conservation equation for the reverberant sound is: ð Y d erev dV ¼ out;rev : ð5:6Þ dt V By conducting many clever experiments, Sabine found that the reverberant sound field created by the reflection from the walls can be regarded as a diffuse sound field.7 Equation 5.6 6 This is based on the assumption that the reverberant sound field is so random in space that there is very little chance of the interaction between the direct and reverberant sounds inducing energy or power. Note that the sum of the power or energy cannot be the sum of the power or energy of individual sound components. 7 Sabine’s result might be considered physically unreasonable. For example, when t ¼ 2t0 in Figure 5.1, the reflected acoustic energy is heavily distributed in the vicinity of the walls. Note, however, that what we have been handling is an average acoustic energy with regard to time and space. If acoustic energy propagates very rapidly in space and time, then it would be distributed quickly. Sabine’s theory can be applied to this type of case.
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278
can therefore be rewritten using the definition of the diffuse sound field. This implies that erev is constant with regard to space, therefore Equation 5.6 can be written as V
Y derev ¼ out;rev : dt
ð5:7Þ
Here, erev is the energy density that is constant for the diffuse field, independent of the position of the selected space. QSabine also noted that the rate of energy reduction per unit time, the power dissipation ( out;rev ), is proportional to the average energy density of a diffuse field. This gives Y / Verev : ð5:8Þ out;rev To convert Equation 5.8 into a formula, a coefficient that has a time scale must be used. Here, time scale is denoted as t. Equations 5.7 and 5.8 then lead to derev derev ¼ : dt t
ð5:9Þ
It is straightforward to obtain the solution of Equation 5.9, as erev ðtÞ ¼ e0 et=t :
ð5:10Þ
Equation 5.10 predicts that the sound energy in a closed space would decay exponentially irrespective of the specific location in the space. This prediction, in fact, is consistent with the suggestion that the reverberant field can be regarded as a diffuse field. The concept of energy decay as expressed by 1=t or the characteristic decay time (t) is very strongly related to the walls that form a closed space as well as the items located in the space (e.g., the furniture), as these items act as sound absorbing elements. They can be regarded as an open window that dissipates sound energy from the closed space to outside. This introduces the concept of the “area of an open window” shown as t/
1 ; As
ð5:11Þ
where As is the area of the open window.8 Equation 5.11 essentially states that a larger area of open window leads to quicker sound energy decay, which agrees with common sense. Recalling that the closed space is created by the walls, the factors related to the walls shall typify the characteristics of the space. The next factor that must affect the sound characteristics of the space must be its volume. The characteristic decay time suitably expresses how the sound of the closed space would decay and how it is related to the walls, as an open area of a window. If a relationship is found between the volume and the characteristic decay time, this would imply that the characteristic decay 8 It is obvious that the energy loss from a room is directly determined by the window size. To confirm this observation, Sabine conducted a well-designed series of experiments to put materials in the room which dissipate sound energy as if the room had an equivalent window. The concept of equivalent open area window comes from this realization.
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279
time represents the sound property of the closed space. This is a rather surprising result, as the sound propagation in a closed space is very complicated (Figure 5.1). We can envisage whether the characteristic decay time would be proportional to the volume or inversely proportional to the volume. Intuitively, it is natural to postulate that a greater size would lead to a longer time required to dissipate the acoustic energy in the room. Equation 5.11 could then be rewritten in the proportional form: t/
V : As
ð5:12Þ
Sabine successfully found a coefficient that can convert the proportional form of Equation 5.12 into the following equality: t¼
4V : c As
ð5:13Þ
In fact, this coefficient is based on many well-designed experiments. In Equation 5.13, c represents the speed of sound. This equation essentially states that the sound in the room (strictly speaking, the sound in a diffuse field) can be represented by only one parameter: the characteristic decay time t. It simply holds that, regardless of what the room has inside or the type of window or sound-absorbing material on the walls, Equation 5.13 can represent the ultimate characteristics of the sound in the room. In other words, Equation 5.13 is the essential measure of the sound property of the room. It should be noted, however, that Equation 5.13 has to be modified to accommodate the scale that is used in practice: the dB scale in magnitude. Generally, for this purpose, T60 (the reverberation time or the reverberation period) is defined as the time required to reduce the sound by 60 dB.9 Applying this definition to Equation 5.10 yields e0 ¼ e0 eT60 =t : 106
ð5:14Þ
Rearranging Equation 5.14 provides T60 ¼ tln 106 :
ð5:15Þ
Therefore, Equations 5.13 and 5.15 result in T60 ¼
55:3 V V ¼ 0:161 ; c As As
ð5:16Þ
where the area of the open window As can be rewritten as As ¼
N X
an A n ;
ð5:17Þ
n¼1
where N is the number of elements that comprise the room of interest, an is the absorption coefficient (which is the ratio of the absorbed sound power to the incident sound power), and n is an index that represents each material. This equation, in fact, was conceptually implied when 9
We might obtain some insight into the reverberation period by considering the time from the moment we hear some sound (which is initially larger than the minimum audible sound of 60 dB) until we cannot hear it any longer.
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280
Equation 5.11 was derived. In other words, As acts as the sound-absorbing element. This value can be obtained by multiplying the sound-absorption coefficient and the corresponding area (An ).10 Using Equations 5.16 and 5.17, an expression that relates the reverberation period to the open window area and the volume of the closed space is found to be T60 ¼ 0:161
V N P
:
ð5:18Þ
a n An
n¼1
Sabine claims that the reverberation period represents the characteristics of the room of interest. It is irrelevant whether the room or closed space is a lecture hall, concert hall, or living room. Additionally, it does not matter whether the closed space has certain furniture or other items. This equation assumes that as long as it has the same volume and an open window area (total sound absorption), then the room has equal sound quality as measured by the reverberation period or time. In other words, the characteristics of a room are not a direct function of wall impedance or the contents of the room. However, we may note that the open window area is a measure that represents an averaged total impedance of the room. We could argue that this is too simple to be accepted in general, considering the complex nature of sound propagation in a room. However, this simple conclusion has great value in practice. In fact, the measure of the reverberation period is the first measure considered in most architectural acoustic designs. There are many derivatives that modify Equation 5.18 to take into account some of the details through which sound waves actually behave in a room (see Section 5.9.2). Sabine likely found that the reverberation period expresses the fingerprint of sound, which certainly expresses how sound propagates and is reflected in a complex manner which depends on the geometry. It also accounts for the sound-absorbing characteristics of the wall as a type of law that governs the time and space of a closed space. Table 5.1 shows the reverberation periods of several famous concert halls. Figure 5.3 shows photographs of these locations. Equation 5.18 also shows that there are two techniques for lengthening or shortening reverberation periods. For example, a greater volume or a smaller open window area can be set for a longer reverberation period. To summarize, Sabine’s theory certainly determines (at least for the first-order approximation) the sound quality of a closed room which has dimensions much larger than the wavelength of sound in the room, that is, an acoustically large space. The propagation of sound waves in an acoustically large space is mainly controlled by how sound is reflected from the wall and the extent to which it radiates from the room or is dissipated. Detailed patterns of sound in a room are highly complex, but Sabine found that the reverberation period can determine the property of an acoustically large space. The reverberation period is not only a measure of an acoustically large space but also a means to determine a number of other details that are associated with architectural acoustics. The reverberation period is the term used to denote the room acoustics. 10
We use such the relation between the reverberation time and the absorption coefficient to measure the absorption coefficient with accuracy. By definition, the measurement of absorption coefficient of a material requires the incident and the absorbed sound power. In the case of the normal incidence and reflection, we can measure the absorption coefficient by using the standing wave apparatus (see Section 5.9.1). On the other hand, when the material is set in the room in which the wave generates and scattered in an arbitrary direction, we measure the absorption coefficient in a reverberation room (see Section 5.9.1).
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281
Table 5.1 Reverberation time of famous concert halls (RToc and RTunoc are reverberation time when occupied and unoccupied) City
Hall name
Vienna Birmingham Berlin Tokyo Amsterdam Christchurch Zurich Boston Dallas Cardiff
Grosser Musikvereinssaal Birmingham Symphony Hall Berlin Philharmonie Suntory Hall Concertgebouw Town Hall Grosser Tonhallesaal Boston Symphony Hall Meyerson Concert Hall St David’s Hall
Volume (m3)
RToc
RTunoc
15 000 25 000 21 000 21 000 18 780 20 500 11 400 18 750 23 900 22 000
2.00 1.85 1.90 1.80 2.00 1.90 2.05 1.90 2.80 1.96
3.05 2.52 2.20 2.00 2.59 2.34 3.18 2.52 2.90 2.15
(Concert Halls and Opera Houses: Music, Acoustics, and Architecture, 2nd edition, 2004, pp. 50–489, L. Beranek, Springer-Verlag New York, Inc.: With kind permission of Springer Science þ Business Media.)
Figure 5.3 Several famous concert halls: (a) Vienna Grosser Musikvereinssaal (Concert Halls and Opera Houses: Music, Acoustics, and Architecture, 2nd edition, 2004, pp.174, ‘‘Vienna Grosser Musikvereinssaal,’’ L. Beranek, Springer-Verlag New York, Inc.: With kind permission of Springer Science þ Business Media); (b) Berlin Philharmonie (Concert Halls and Opera Houses: Music, Acoustics, and Architecture, 2nd edition, 2004, pp. 298, “Berlin Philharmonie,” L. Beranek, Springer-Verlag New York, Inc.: With kind permission of Springer Science þ Business Media.); (c) Tokyo Suntory Hall (Concert Halls and Opera Houses: Music, Acoustics, and Architecture, 2nd edition, 2004, pp.408, ‘‘Tokyo Suntory Hall,’’ L. Beranek, Springer-Verlag New York, Inc.: With kind permission of Springer Science þ Business Media.); and (d) Boston Symphony Hall (Concert Halls and Opera Houses: Music, Acoustics, and Architecture, 2nd edition, 2004, pp.48, ‘‘Boston Symphony Hall,’’ L. Beranek, SpringerVerlag New York, Inc.: With kind permission of Springer Science þ Business Media.)
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P(r¯,t) r1
r2
r¯
0
Figure 5.4 Sound propagation in an open space (direct sound field)
5.4
Direct and Reverberant Field
A direct, reverberant, and diffuse sound field was investigated in Section 5.3. These three different types of sound can exist together in reality. However, this distinction allows us to understand a sound field easier. As defined earlier, a direct sound field refers to a field that does not have any reflected sound waves. For simplicity, a sound field created by a monopole, or a breathing sphere located at ~ r as illustrated in Figure 5.4, is investigated. It emits sound at time t. Its wave front propagates to r1 or r2, depending on the time. If there is no reflection, as can be anticipated in the direct field, then the total sound power through the surface at r1 or r2 has to be conserved provided that there are no energy loss in the medium. The following equality can therefore be written: Ir1 4pr21 ¼ Ir2 4pr22 ;
ð5:19Þ
where Ir1 ; Ir2 denote the magnitude of intensity at r1 or r2. If r2 ¼ 2r1 , then the intensity ratio is Ir2 1 ¼ : Ir1 4
ð5:20Þ
The following relationship becomes apparent: 10 log10
Ir2 1 ¼ 10 log10 ¼ 6 dB: Ir1 4
ð5:21Þ
This indicates that the sound intensity will be reduced by 6 dB.11 On the other hand, for a diffuse field, the acoustic properties are independent with respect to the location. The solution of Equation 5.21 is therefore 0 dB. As discussed in Section 5.3, if a reverberant field satisfies the definition of a diffuse field, as Sabine assumed, the sound energy must be uniform in space, that is, independent of the position. At this point, a sound field in a closed space can be considered using three basic concepts (direct, reverberant, and diffuse field) that describe what can exist in reality. In fact, the sound 11 The ratio of magnitude of intensity is inversely proportional to the square of distance. The phenomenon where sound power is decreased by 6 dB when the distance from source to measurement is doubled is normally used to evaluate an anechoic chamber (see Equation 4.17).
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283
field as normally experienced is neither a direct field nor a diffuse field. Depending on the relative position between the listener, sources, and walls, what is heard will differ greatly. The sound that we hear is generally the sum of the direct and the reverberant sound. As Equation 5.21 indicates, the magnitude of the direct sound depends on its location of origin. What the listener hears therefore depends, of course, on their location with regard to the walls and sources. This implies that the manner in which the reverberant sound and direct sound affects listener sensation depends on the listeners’ relative position with regard to the walls as well as to the relative location of the source. It is clear that the direct sound would be dominant if a listener is close to the source; however, reverberant sound would be more likely to dominate when the listener is further away from the sources and close to the wall or walls.12 As Figure 5.5 illustrates, the sound pressure decays 6 dB as the distance from the sound source in the sound field near the source is doubled. In the field close to the source (where distance with respect to the wavelength of interest is short enough to be regarded as a near field), the sound field is virtually dominated by the direct sound from the source. However, as the listener moves away from the source, the contribution by the reverberant sound becomes stronger and the sound field does not monotonically depend on the distance from the source. It is clear that the manner in which reflected sound is mixed with the direct sound determines how the sound pressure is distributed in the space. It is also very straightforward to predict that a sound field very close to the walls will be very much dominated by reflected sound waves; this approaches a diffuse sound, which implies that the sound pressure is constant irrespective of the distance from the source. Sound power level (dB re 10-12.Nm)
-6dB / Double distance
Free field
Diffuse field
Reverberant field Distance
Figure 5.5 Spatial variation of sound field with respect to the distance from source
From these observations, it is necessary to derive a certain measure or scale that can determine the degree of participation of the direct and reverberant fields, or the direct and reflected sound waves in a room. This measure will provide a distance through which the degree to which the direct field dominates the reverberant field can be determined. To obtain this representative measure, the relationship between the acoustic energy of the direct and the reverberant sound must be analyzed. The sources are assumed to emit sound 12
This can be replicated using water in a bath. Place one palm on the water surface and excite the water; watch the water wave at the palm and the edge and corner of the bath.
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284
energy in such a way as to reach a steady state quickly. Equation 5.5 can then be rewritten, for this special case, as 0 ¼ Pin;direct ðPout;direct þ Pout;rev Þ:
ð5:22Þ
Equation 5.22 states that the rate change of energy with regard to time (i.e., the power) in the room is zero, as the sound in the room is assumed to reach a steady state. This implies that the sound power generated by the sound sources is balanced by the sound power reflected due to the direct sound and due to what is induced by the reverberant sound on the surface that we select. This distinction is intentionally made to obtain a measure that determines the distance that expresses the degree of participation by the direct and reverberant sound. How much is reflected is directly related to the absorption coefficient of the walls, or how much energy is absorbed by the walls. The average absorption coefficient of the walls is denoted a ¼ As =At :
ð5:23Þ
The introduction of this average absorption coefficient is necessary because one measure that relates the direct and reverberant field is examined here rather than the details of each wall. Here, At expresses the total area of the closed space, and As is the equivalent area of an open window. Note also that Pout;direct , which is the power reflected from the walls by the incident sound power (Pin;direct ), are related as Pout;direct ¼ a Pin;direct :
ð5:24Þ
Equations 5.22 and 5.24, the time rate change of the reverberant sound energy, are related to the direct sound power, that is: aÞPin;direct : Pout;rev ¼ ð1
ð5:25Þ
This study has only considered the power relationship between the direct and reverberant sound so far. Nothing about the location of the listener has been considered. The extent to which the listener’s position depends on this sensation should be examined. Let r denote the distance between the source and the listener. The sound power passing through the surface of a sphere with a radius of r has to be identical to what the sound source generates. This physical balance can be mathematically written as Pin;direct ¼ Ir 4pr2 ;
ð5:26Þ
where Ir expresses the intensity at distance r from the source. If a monopole source and the radius are widely spaced relative to the wavelength of interest, the acoustic intensity and power can be written as: Ir ¼
p2avg jPj2 ¼ ; 2r0 c r0 c
Pin;direct ¼ 4pr2
p2avg : r0 c
ð5:27Þ
ð5:28Þ
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285
These are obtained because the acoustic impedance in the far field can be assumed to be r0 c.13 Similarly, the acoustic energy density of the direct sound can be obtained as (see Equation 5.1): 1 1 p2avg p2avg ¼ : edirect ¼ r0 u2avg þ 2 2 r0 c r0 c
ð5:29Þ
Equations 5.28 and 5.29 provide the following relationship between the sound energy that is experienced directly from the source and the corresponding sound power, that is, edirect ¼
Pin;direct : 4pr2 c
ð5:30Þ
A similar relationship can be obtained for the reverberant sound. The reverberant sound energy can be regarded to be distributed in a closed space, which can be envisaged as the space surrounded by the surfaces of discontinuities that have various wall impedances (see Section 5.3). From Equation 5.7 to Equation 5.9, the rate of energy decay per unit time (i. e., the power loss) is related to the energy density of the reverberant sound field as Pout;rev ¼
Verev : t
ð5:31Þ
Substituting Equation 5.13, which shows the characteristic decay time with the other properties of the closed space (t ¼ 4V=cAs ), into Equation 5.25 gives us the manner in which the reverberant energy density is related to the direct input sound power, that is, erev ¼
4 ð1aÞPin;direct : cAs
ð5:32Þ
The total energy density comprising the direct and reverberant sound can therefore be written as eðrÞ ¼ edirect þ erev Pin;direct 2 16p 1þr ¼ ð1aÞ 4pr2 c As ( 2 ) r ¼ edirect 1 þ : r0
ð5:33Þ
Equations 5.30 and 5.32 are used to formulate Equation 5.33. Equation 5.33 essentially expresses the composition of the sound energy that is heard, relative to how much direct or reverberant sound energy is experienced. Equation 5.33 shows that the effect of direct sound diminishes as the listener moves away from the source, and vice versa (see Figure 5.6). The new parameter used in Equation 5.33 is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi As : ð5:34Þ r0 ¼ 16pð1aÞ 13
See Section 2.8.4.1 and Equations 5.19 and 5.27.
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Figure 5.6 Total energy density (the sum of direct and reverberant sound energy density; r0 , an intersection point, is the radius of reverberation)
This expresses the radius at which it is likely that the direct and reverberant sound participate equally. Therefore, if r is greater than r0 , then the effect of the direct sound is greater than that of the reverberant sound. For this reason, this radius is often referred to as the radius of reverberation. The property or the physical meaning of the radius of reverberation can be seen in a different manner if the open window area (As ) is expressed in terms of the spatially averaged soundabsorption coefficient. This is given as sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a At r0 ¼ : 16p ð1aÞ
ð5:35Þ
This implies that the radius of reverberation becomes longer as the sound-absorbing area increases if the sound-absorbing power is stronger than what is reflected. What is essentially meant by the radius of reverberation is that it measures the effective radius in which the direct sound is more likely to dominate. The radius will therefore increase as the sound-absorbing area becomes larger and the sound-absorption coefficient increases, as shown by Equation 5.35. Note again that the radius of reverberation depends strongly on the wall impedances that form the space of interest. However, the details of impedance distribution do not determine the radius of reverberation; this is only determined by the mean value of the absorption coefficient. This is simply because the radius is the measure of average room characteristics. Note also that the radius of reverberation is indirectly related to wall impedance and furniture. To summarize, the sound field that can be constructed in a space pffiffiffiffi with a much larger dimension than a typical wavelength of interest (for example, if l ¼ 3 V ) can be categorized as a direct, diffuse, or reverberant sound field, depending on the spatial characteristics of the sound. The direct sound field, as its name implies, is a sound field that does not have any reflection. The diffuse and reverberant sound fields, on the other hand, are fields that are
Acoustics in a Closed Space
287
characterized by reflections. The diffuse sound field is an ideal field or is an example of an extreme concept that defines a sound field with a uniform sound distribution in space. In other words, the sound is equally likely to be distributed in space. Therefore, an identical sound pressure and intensity distribution is given irrespective of the location in the space. This implies that the reflection from the walls enclosing the space has to be spatially random. In other words, the specular angle of reflection is equally likely to be distributed. The conditions of the reverberation sound field lie between these cases. The sound field has some locations at which the direct sound field is more likely to be dominated by the reverberant sound field, or vice versa, depending on where the listener is in the space. The sound pressure or intensity distribution depends very strongly on the location of the space. This is what we normally observe in practice. The measure that highlights how the direct sound dominates the reverberant sound field at the position selected is the radius of reverberation. A typical example of a direct sound field is what is experienced in an anechoic room. The reverberation chamber that is normally used to measure the transmission loss or measure the sound-absorption coefficient of a panel is a space designed to create a diffuse or very good reverberation sound field.
5.5
Analysis Methods for a Closed Space
The acoustic characteristics of a closed space have been explained in terms of the reverberation period and the radius of reverberation. Another way to understand sound wave propagation in a closed space is to use mathematical approaches. These can be regarded as integrated expressions of sound waves that are possible in the space enclosed by the walls of impedances. The next step is to examine the details of sound wave propagation in the space. Mathematical tools allow us to meet this objective. Sound waves in a closed space can be regarded as the solutions that satisfy the boundary conditions of the closed space and the governing equation. It is well known that there are two distinct approaches to acquire these solutions. The first is to obtain the solutions in the time domain, and the second is to acquire them in the frequency domain.14 The latter expresses the sound wave distribution with regard to its possible spatial distribution for each frequency. In other words, it describes the sound waves in terms of the superposition of mode shapes. These approaches can be implemented by the following three methods. The first regards the sound field of interest as the superposition of natural or normal modes that satisfy the boundary condition and the governing equation. The governing equation, in this case, has to be the linear acoustic wave equation. Note, however, that it is not always straightforward to obtain the normal modes (mode shape) that exactly satisfy the governing equation as well as the boundary condition if the boundary condition is neither a Neumann nor a Dirichlet boundary condition (i.e., a homogeneous boundary condition). However, even if the boundary condition has a more general form (i.e., an inhomogeneous boundary condition, e.g., the Robin boundary condition which has non-zero velocity and pressure), it is acceptable to use the normal modes which satisfy the homogeneous boundary condition. This is the reason why the inhomogeneous boundary condition can be expressed as the homogenous boundary condition and the 14
This classification is also relevant to general linear acoustics. For example, Figure 5.1 is obtained in time domain by using computer simulation.
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inhomogeneous governing equation. Having the normal modes, the extent to which each mode contributes to the sound field is then determined. This provides the value of the coefficients of the normal modes. This method can be observed by extending what is known about the vibration of a string. The second method describes the sound field using singular functions that satisfy the governing equation. A typical example of these types of singular functions is a solution that describes a sound field induced by an acoustic monopole source at an arbitrary location. This method assumes that any sound field can be constructed by distributing acoustic monopole sources. Because the solution of a monopole source in the field is known, the principle of superposition supplies a solution that completely describes the sound field. The Kirchhoff–Helmholtz integral equation essentially describes this approach. The boundary element method (BEM) is a type of discrete form of this integral equation. The latter method describes the sound field using acoustic rays, and is often referred to as ray acoustics. It assumes that the wavelength of interest is very much smaller than the characteristic length of the surface of reflection. Acoustic waves, in fact, are assumed to be propagated as for light. The specular reflection dominates how the waves are reflected on the surface. Figure 5.7 illustrates the sound field that can be anticipated if the sound is assumed to propagate as a ray.
Figure 5.7
Conceptual example of ray acoustics
This method cannot be applied if the walls are no longer considered as locally reacting surfaces,15 or if the acoustic wavelength fails to meet the basic assumption of a locally reacting surface. Here, we examine the first method for predicting sound fields generated in a closed space.16 As discussed above, a sound field that falls into a given frequency within the closed space can be expressed (according to the principle of linear superposition) by superposition of unique modes that meet the boundary condition and the governing equation, as 15 16
See Section 3.9.1. The second method which uses Green’s function is summarized in Section 5.9.4.
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289
Pð~ rÞ ¼
1 X
almn wlmn ð~ rÞ;
ð5:36Þ
l;m;n¼0
where subscripts l; m; n refer to the respective orders of modes that correspond to individual coordinate directions of the Cartesian coordinate system. For instance, if the space of interest can be expressed by the coordinate system (x; y; z), wlmn denotes the lth mode in the x direction, the mth mode in the y direction and the nth mode in the z direction, and~ r represents (x; y; z). Let us consider a cube-shaped space in which sound can potentially be generated. Under the rigid wall boundary condition, the patterns of possible sound pressures would be cosine functions, which can be written lpx mpy npz wlmn ðx; y; zÞ ¼ cos cos cos ; ð5:37Þ Lx Ly Lz where Lx ; Ly ; Lz represent the lengths in each direction. In the case of relatively simple single dimension (i.e., a square tube with of length L), this can be expressed as lpx ; L
ð5:38Þ
al wl ðxÞ:
ð5:39Þ
wl ðxÞ ¼ cos PðxÞ ¼
1 X l¼0
In summary, mode function wl is an ordinary type of function that satisfies the sound field of a one-dimensional tube, which has to respond in the form of Equation 5.39 despite any excitations. What is important here is which of the several items in Equation 5.39 would contribute the most dominant response. almn in Equation 5.36 or al in Equation 5.39 express how each unique mode contributes to the sound field. A constant that represents the level of contribution that each unique mode makes to the entire sound field is called “modal coefficient.” Here, al needs to become greater as the excitation power of the sound source increases; it is also determined by the location of the sound source, the excited frequency and other factors. Mathematically speaking, Equation 2.67 becomes the governing equation if it is inhomogeneous. This phenomenon can therefore be considered a problem of obtaining the coefficients of Equation 5.39 that can satisfy the excitation term on the right-hand side of the governing equation. To look at the behavior of modal coefficients in detail, let us observe sound fields that are radiated from a monopole sound source placed in a three-dimensional space. As stated in Equation 5.36, sound fields in a three-dimensional space are expressed as the sum of unique modes. Therefore, if the excitation is generated using a monopole sound source at the location of~ r 0 , the sizes of modal coefficients and sound pressure can be described as (for more details on the derivation, see Section 5.9.3): almn
Pð~ rÞ ¼
¼
r0Þ 4pS w*lmn ð~ 2 Þ 2 V Llmn ðk klmn
¼
4pShlmn ðkÞw*lmn ð~ r 0 Þ;
1 wlmn ð~ rÞw*lmn ð~ r0Þ 4pS X ; 2 2 V l;m;n¼0 Llmn ðk klmn Þ
ð5:40Þ
ð5:41Þ
290
Sound Propagation
where S denotes monopole amplitude. The physical meaning of Equation 5.40 is that the contributions of each mode are proportional to monopole amplitudes, the magnitude of r 0 Þ), and what is governed by the particular frequencies modes at sound source locations (w*lmn ð~ which the space has (hlmn ðkÞ). For instance, Figure 5.8 depicts some individual modes contributing to the entire sound field, with each extent in a cubic room described by a given volume.
Figure 5.8 Sound pressure generated in a cubic space (0.8 0.6 0.1 m3); contributions by individual unique modes
Now consider hlmn ðkÞ, a function that represents the frequency characteristics of a space. If the walls of a cubic room have the rigid body boundary condition, k2lmn has a real-number value 2 ) and is expressed as (klmn 2 2 2 lp mp np 2 2 2 2 þ þ : ð5:42Þ klmn ¼ ðkx Þ þ ðky Þ þ ðkz Þ ¼ Lx Ly Lz If the excitation frequency (f ¼ kc=2p) of Equation 5.40 is the same as or similar to c flmn ¼ klmn ; ð5:43Þ 2p
Acoustics in a Closed Space
291
then the particular mode contribution (almn ) will be infinite or significantly amplified.17 This frequency characteristic function, like the transfer function of a 1-DOF vibratory system, therefore serves to adjust the extent of amplification for each mode depending on excitation frequency. In fact, the number of modes that are participating in the response tends to increase exponentially with frequency, as illustrated in Figure 5.9.
Figure 5.9 Number of modes and modal density in a closed rectangular space (L ¼ 5.8 m; W ¼ 4.5 m; H ¼ 2.5 m) with the walls being rigid bodies
Figure 5.9 shows the number of modes contributing to all frequency ranges below the excitation frequency within a cubic room and modal density (i.e., the number of modes taking part per unit frequency). It implies that the total number of participating modes and modal density increase dramatically as the frequency increases. In other words, a larger number of modes are needed to express sound fields as the frequency becomes higher; spatially, sound pressure or acoustic energy becomes flat depending on the space. As explained in Section 5.2, with numerous modes being mixed, the acoustic characteristics of the space turn into a diffuse field. To summarize, the sound field in a reverberant space (i.e., a space surrounded by the walls) is determined by the location of the sound source, radiation characteristics of sound source and frequency characteristics of the walls. The sound field can be its pressure, velocity, energy, or intensity distribution in the space that is surrounded by the walls. The walls can be regarded as spatially distributed impedance. We can therefore envisage that the characteristics of sound must depend on how the walls react to the sound and the size of the space that are made by the walls. Discovered by Sabine, reverberation time comprehensively expresses the characteristics of a closed space. This is a very comprehensive measure and does not reveal the detailed properties 17
The frequency in Equation 5.43 is referred to as undamped natural frequency.
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292
of sound distribution. Sound field analysis on a closed space is one way to reveal the details. This is a classic problem in finding solutions that meet the given governing equation and the boundary condition of the closed space. Representative techniques for this purpose include the ray method, the mode analysis method, the boundary element method, and the finite element method. The mode analysis method shows that the participated number of modes in a room normally exceeds several thousand.
5.6
Characteristics of Sound in a Small Space
An acoustically small space is one whose representative length or size is small relative to wavelength. Sound waves in this space have unique characteristics; the dynamic characteristics of a medium moving in the space (e.g., pressure and particle velocity) can be considered relatively constant. In other words, the entire medium is not dominated by propagation properties of sound waves but moves in the same phase. Thus, an acoustically small space can generally be regarded as a vibratory system. A prime example of this is the Helmholtz resonator. Figure 5.10(a) shows a simple resonator, which consists of body and neck.18 If the wavelength is considerably longer than these two parts, it can be easily imagined, as explained above, that the movements of fluid in the neck or the body will have almost identical phases.19 From Figure 5.10(a), the pressure change (pin ) per unit time will induce the volume change in the cavity of the resonator. If the pressure changes and volume are small enough to be linearized, the relation can be expressed as dp dV / ; dt dt
ð5:44Þ
where p is pin (Figure 5.10(a)), and V represents volume of the cavity. In order to convert this relation into an equation, a certain proportional constant is needed. Using acoustic compliance CA (which represents the volume change induced by unit sound pressure) as a proportional constant, Equation 5.44 can be rewritten as CA
dp dV ¼ : dt dt
ð5:45Þ
When we have a large CA (e.g., a larger cavity case with respect to the same neck as shown in Figure 5.10(c)), the resonator undergoes a massive volume change. Equation 5.45 only highlights the correlation between pressure and volume and so needs to be rewritten using variables that express the fluid motion (i.e., force, displacement, and velocity, etc.) in order to observe how the fluid at the neck and cavity behave. First of all, the volume change with respect to time in the cavity can be written as20 18 It is reasonable to design the body and the neck to occupy as small a space as possible unless their characteristics are changed as illustrated in Figure 5.10(b). 19 This assumption is not valid at the frequencies where the contribution of higher modes is not neglected. Generally, the contribution of higher modes appears at higher frequencies than the resonance frequency, which will be considered in the latter part of Section 5.5. 20 It is assumed that the velocity of fluid is uniform in the neck, because the length and area of the neck is much smaller than the wavelength.
Acoustics in a Closed Space
293
A
m
Pout
V
Pin
Equivalent vibratory system
(a)
(b)
Pout
u
Pout
Pin
u
ΔV
Pin
ΔV
CA is relatively small.
CA is relatively large.
(c)
Figure 5.10 (a) Shape of simple resonator and equivalent vibratory system; (b) various types of resonator and conceptual samples; and (c) meaning of acoustic compliance
dV ¼ AuðtÞ; dt
ð5:46Þ
where uðtÞ is the velocity of fluid at the neck as depicted in Figure 5.10(c), and A is the crosssectional area of the neck (see Figure 5.10(a)). We can then rewrite Equation 5.44 as21 dp 1 ¼ AuðtÞ: dt CA
ð5:47Þ
Now consider the fluid motion at the neck induced by sound pressure. As shown in Figure 5.10(c), the balance between sound pressure acting on the fluid at the neck and the momentum of the fluid can be formulated as ðpout pin ÞA ¼ r0 Al
du ; dt
ð5:48Þ
where l is the length of the neck (or effective length of the neck, to be more precise) in Figure 5.10(a). Equation 5.48 can be rewritten as pout pin ¼ r0 l
21
du : dt
ð5:49Þ
In this case, precisely speaking, p is the sound pressure inside the neck of the resonator (see Figure 5.10(c)).
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294
Substituting Equation 5.49 into Equation 5.47, we can obtain CA d 2 p : pout ¼ pin þ r0 l A dt2 As noted before, pin ¼ p; Equation 5.50 can therefore be rewritten as rl d 2p p þ 0 CA 2 ¼ pout : A dt
ð5:50Þ
ð5:51Þ
Equation 5.51 essentially describes how sound pressure just inside the neck (p) reacts to the excitation sound pressure (pout ) just outside the neck. It is a typical second-order ordinary differential equation, which can be found in the well-known single degree of vibration system. From Equation 5.51, the resonance radial frequency (on ) can be obtained as o2n ¼
1 : r0 l CA A
ð5:52Þ
To better understand the physical meaning of Equation 5.52 and utilize it in practice, it is essential to express acoustic compliance in terms of three representative physical quantities in acoustics. Those are sound pressure, velocity and density (Chapter 2). The law of conservation of mass on the fluid in the cavity requires us to write r0 V ¼ ðr0 þ DrÞðV þ DVÞ:
ð5:53Þ
Neglecting higher-order terms implies Equation 5.53 becomes r0 DV þ VDr ¼ 0:
ð5:54Þ
dV V dr ¼ : dt r0 dt
ð5:55Þ
We then have
Note that the compliance was introduced to represent the proportionality of Equation 5.44. The result is Equation 5.45. We therefore need to convert Equation 5.55 to the form of Equation 5.45. This can be done by introducing the state equation, that is, dp=drs ¼ c2 . Equation 5.55 can then be written as dV V dp ¼ : dt r0 c2 dt
ð5:56Þ
Comparing this with Equation 5.45 leads us to express acoustic compliance as CA ¼
V : r0 c 2
ð5:57Þ
If we regard Equation 5.52 as the equivalent one-dimensional vibratory system in Figure 5.10(a), we can obtain mA ¼
r0 l ; A
ð5:58Þ
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295
where mA is an acoustic inertance. Equation 5.52 can be written as 1 mA CA
ð5:59Þ
rffiffiffiffiffi A : on ¼ c lV
ð5:60Þ
o2n ¼ or
In other words, the resonance frequency increases as the area of the neck becomes larger, but falls as the volume of the cavity becomes larger. Equation 5.59 shows that the resonance frequency of a resonator is governed by the mass of the system and the spring constant (i.e., inversely proportional to the acoustic compliance) as in the case of a 1-DOF vibratory system. In addition, the mass (acoustic inertance) and the spring constant (acoustic compliance) are entirely determined by the size of the resonator’s neck and the size of the cavity, respectively. This is because the wavelength associated with the resonance frequency is very long relative to the size of the resonator. This causes the entire fluid at the neck to move in the same phase22 and the volume in the cavity to sustain the entire fluid at the neck as a kind of spring element. Strictly speaking, what matters here is what amount of fluid particles around the neck of the resonator move 180 out of phase to external pressure excitation.23 If a diameter of the neck is considerably smaller than the wavelength, the effective length (including end correction) of the neck can be expressed depending on whether it has a flange or not: l 0 ¼ l þ 0:85a
ðwith a flangeÞ
l 0 ¼ l þ 0:6a
ðwithout a flangeÞ;
ð5:61Þ
where l represents the length of the neck and a the radius of its cross-section.When we make use of a resonator in practice, the required resonance frequency is usually determined first and a resonator with this frequency is designed. The performance of the resonator is then verified through experiments. To design the resonance frequency of a resonator precisely, the end correction factor should be taken into account. The end correction factor is associated with the geometrical shape of a resonator (for example, flange at the neck in Equation 5.61). In addition, neck and cavity are also basic components that consist of the geometrical shape of a resonator. We therefore observe the characteristics of a resonator by separating the body, neck and other effects (for example, flange, and radiation effect, etc.). In particular, impedance of a resonator, which can have significant physical meaning, can also be expressed as ZHR ¼ Zr þ Zneck þ Zcavity ;
ð5:62Þ
where Zr represents radiation impedance, and Zneck and Zcavity are impedances for the neck and the cavity, respectively. In particular, the reactance (imaginary part) of the impedance 22
We are interested in the fluid particles around the neck as well as the fluid particles which participate in the movement in the neck. 23 This is referred to as “end correction” or “end correction factor”, and is a function of cross-sectional area or the crosssectional area multiplied by the wave number.
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mainly determines resonance frequency. The reactive impedance of Equation 5.62 can be written as ImfZHR g ¼ r0 o
8a pa2 : þ r0 ol 0 þ r0 c2 joV 3p
ð5:63Þ
The first term (Zr ) is the reactive radiation impedance which can be obtained by open end correction. The radiation impedance is obtained from the impedance of a circular piston that has a radius a (Equation 4.49) by low frequency approximation. In particular, the third term in Equation 5.63 (the reactive impedance of cavity Zcavity ) can be derived under the assumption that the pressure in the cavity is maintained uniformly. However, in reality, the abrupt change between neck and cavity and wave motion in the cavity does exist. To explore how Equation 5.63 is obtained, a more rigorous theoretical approach is required. We first look at the Green’s function (see Section 4.7.4 for details) to analyze sound field in rÞ of the system exist so that it is possible the cavity. We can also assume that eigenfunctions wn ð~ rÞ, which satisfies homogeneous boundary to expand Green’s function in a series of wn ð~ conditions. Then we can write Gð~ rj~ r0Þ ¼
Xw n ð~ r 0 Þ wn ð~ rÞ ; 2 2 kn k n
ð5:64Þ
n ð~ where wn ð~ rÞ are the eigenfunctions of the system, w rÞ is the orthonormal set of eigenfunctions, and kn is the eigenfrequency of the system. These eigenfunctions can be obtained by selecting the appropriate coordinate system and by choosing the characteristic functions representing the acoustic field in terms of each coordinate system. Based on these fundamental understandings, the Green’s solution for a resonator (which has arbitrary a neck) can be readily obtained. A resonator which we want to observe is shown in Figure 5.11, which has cylindrical neck and body. The sound field in the cavity is also governed by the Kirchhoff–Helmholtz integral equation. The boundary conditions expressed by the velocity potential (f)24 can be given by S1 :
@f ¼ u hðr; yÞ; @z z¼0
ð5:65aÞ
@f S2 : ¼ 0; @r r¼R
ð5:65bÞ
@f ¼ 0; S3 : @r z¼L
ð5:65cÞ
where h is the Heaviside step function. By applying these boundary conditions to the Kirchhoff–Helmholtz integral equation, we can obtain the velocity potential in the cavity as
24
See Section 2.8.2.
Acoustics in a Closed Space
297
Figure 5.11 A simple resonator of interest which has a round neck (diameter a) and body (radius R). The neck is located at an arbitrary position which is offset (e) to the center of body, region S1. The wall is assumed to be rigid, and harmonically excited at the neck by a circular piston
ZZ fð~ r0Þ ¼
Gð~ rj~ r 0 Þ u hðr0 ; y0 ÞdS;
ð5:66Þ
region S1
where G is Green’s function, u is the particle velocity of the source, and h is the Heaviside step function. The solution to Equation 5.66 can be given by25 fð~ r0Þ
¼
P 2uem ep a2 ðkrmn þ mmn Þ2 fJ1 ðkrmn aÞ þ mmn gJm ðkrmn eÞ Lðkrmn a þ 2mmn Þfðkrmn þ mmn Þ2 R2 m2 gJm2 ðkrmn RÞ 1 Jm ðkrmn rÞcosðmyÞcosðkzp zÞ; 2 kmnp k2
ð5:67Þ
where em , ep is 1 when m; p ¼ 0 or 2 when m; p 6¼ 0, and mmn is 1 when m; n ¼ 0 or 0 otherwise. The subscripts m; n; p indicate the mode with respect to circumferential, radial, and axial, respectively. kzp and krmn represent axial and radial wave number, respectively. The cavity impedance (Zcavity ) can be obtained by using the force (F) acting on the piston, that is, ZZ prdrdy F ¼ source
¼ r0 o pa2
X
4em ep a2 ðkrmn þ mmn Þ2
m; n; p Lðkrmn a þ 2mmn Þ
2
fðkrmn þ mmn Þ2 R2 m2 g
ð5:68Þ
fJ1 ðkrmn aÞ þ mmn g2 Jm2 ðkrmn eÞ : 2 k 2 Þ Jm2 ðkrmn RÞðkmnp
25 Kim, Y.-H., and Kang, S.-W. (1993) Green’s solution of the acoustic wave equation for a circular expansion chamber with arbitrary locations of inlet, outlet pot, and termination impedance. Journal of Acoustic Society of America, 94(1), pp. 473–490.
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The cavity impedance is then obtained by Zcavity ¼
F pa2 u
¼ jr0 co
X
4em ep a2 ðkrmn þ mmn Þ2
2 2 2 2 m; n; p Lðkrmn a þ 2mmn Þ fðkrmn þ mmn Þ R m g fJ1 ðkrmn aÞ þ mmn g2 Jm2 ðkrmn eÞ : 2 k 2 Þ Jm2 ðkrmn RÞðkmnp
ð5:69Þ
The limit case (a lumped parameter system) which has m ¼ n ¼ p ¼ 0 determines the validity of Equation 5.69. In other words, Equation 5.69 gives the same result with the impedance of a one-dimensional vibratory system, which is jr0 cpa2 =kV. This result is the same as the third term in Equation 5.63. By comparing Equation 5.63, we can see the contribution from higher-order terms as well as plane wave components (when m ¼ n ¼ p ¼ 0) in Equation 5.69. We can then rewrite Equation 5.63 as ImfZHR g ¼ r0 o
X 8a þ r0 ol 0 þ r0 o 3p m; n; p
ð5:70Þ
Equation 5.70 considers wave motion and higher-order terms in ZHR compared to the classical approach, and we can also express the cavity impedance as P Zcavity ¼ Zlumped þ jr0 ck
k
¼ Zlumped þ Zaddition ;
ð5:71Þ
where Zaddition indicates additional impedance to the impedance the of lumped model. The additional impedance represents the effect of higher-order terms. In addition, suppose that we define the additional impedance as end correction (di ). We can then understand how the end correction varies with respect to the cavity shape or size and so on. The end correction can then be written as P Zaddition ¼ jr0 ck
k
¼ jr0 ckdi :
ð5:72Þ
We can therefore write the end correction as di ¼
X pa2 p¼1
V
2 þ 2 kzp k2
X m;n is not zero simultaneously
em ep 2 Lðkrmn R2 m2 Þ
1 2 k2 kmnp
ð5:73Þ
J12 ðkrmn aÞJm2 ðkrmn eÞ : Jm2 ðkrmn RÞ
Acoustics in a Closed Space
299
The first term in Equation 5.73 is the effect of a plane wave that propagates in the axial direction of the resonator. The second term shows the effect of higher-order terms. As shown in Equation 5.73, the end correction (di ) depends on the characteristic length of a resonator (diameter and length of cavity), the location of the neck, and the ratio of the diameter between neck and cavity. Suppose that we are interested in a low frequency range relative to kzp and kmnp . Equation 5.73 can then be rewritten as di ¼
X pa2 p¼1
V
1 þ 2 kzp
X m;n is not zero simultaneously
em ep 2 Lðkrmn R2 m2 Þ
1 2 kmnp
J12 ðkrmn aÞJm2 ðkrmn eÞ : Jm2 ðkrmn RÞ
ð5:74Þ
If we have the same diameter of the neck with the cavity as a particular case (often referred to as l=4 resonator), the second term in Equation 5.74 is canceled out because J1 ðkrmn aÞ tends to zero as a tends to R. Equation 5.74 can then be obtained as di ¼
1 X 2L 2L p2 L ¼ : ¼ p2 p2 p2 6 3 p¼1
Therefore, resonance frequency of the l=4 resonator can be written as c : fn jl ¼ 4 4ðL þ L=3Þ
ð5:75Þ
ð5:76Þ
We can see various phenomena on the end correction (di ) with respect to geometrical dimensions of a resonator using Equation 5.73. This may provide rather precise resonance frequency estimation when designing a resonator of interest. From these observations, we can conclude that the geometry (the shape, location, and size of the neck and cavity) mainly affects the performance of a resonator. In addition, the shape of the neck is one of the main attributes which changes the absorption characteristics of resonator impedance. By changing the shape of the neck, we can therefore improve the absorption performance of a resonator. The necks can be any shape depending on practical requirements other than acoustical. The shape is not very important if its spatial variation is considerably smaller than the wavelength of interest, such as the case of the neck of the Helmholtz resonator. We therefore consider a hornshaped neck. The horn causes the impedance of propagating sound from a small source to gradually change to that of the impedance at the end of the horn, which lets the sound radiate well. A resonator with such a horn shape is illustrated in Figure 5.12. Suppose that we have a plane wave propagating in the neck, then the wave is governed by Webster’s horn equation (see Section 5.7 for details). This can be written as 1 d dp B þ k2 p ¼ 0; B dx dx
ð5:77Þ
Sound Propagation
300
Figure 5.12 A schematic model of a round resonator that has a gradually changing neck. L is the axial length of the cavity, l is the length of the neck, R is the radius of the cavity, ri is the inlet radius of the neck, and ro is the outlet radius of the neck
where B ¼ pðmx þ ri Þ2 , and m ð¼ ðr0 ri Þ=lÞ is the slope of the neck. ri , r0 , and l are depicted in Figure 5.12. The solution to Equation 5.77 is then p¼
n o mx þ ri mx þ ri m a1 e jk m þ a2 ejk m ; kðmx þ ri Þ
ð5:78Þ
where a1 and a2 are the magnitude of the incident and reflected wave, respectively. Particle velocity can be obtained by linearized Euler’s equation, that is,
n o mx þ r mx þ r j m2 jk m i jk m i a e þ a e u ¼ 1 2 r0 ck kðmx þ ri Þ2 n o mx þ r mx þ r m jk m i jk m i : a1 ðjkÞe þ a2 ðjkÞe þ kðmx þ ri Þ
ð5:79Þ
Then the impedance (Z0 ) at x ¼ l can be written as ro e j2k m þ aa21 p : Zo ¼ ¼ r0 c kro ro u x¼l ðkro jmÞe j2k m ðkro þ jmÞ aa2
ð5:80Þ
1
Writing Equation 5.80 with respect to a2 =a1 yields a2 ðkro jmÞZ0 r0 c kro j2kro ¼ e m: a1 ðkro þ jmÞZ0 þ r0 c kro
ð5:81Þ
In addition, the impedance (Zi ) at x ¼ 0 also can be obtained as ri e j2km þ aa21 p ¼ r0 c kri Zi ¼ ri u x¼0 ðkri jmÞe j2km ðkri þ jmÞ a2
a1
ð5:82Þ
Acoustics in a Closed Space
301
using Equations 5.78 and 5.79. By substituting Equation 5.81 into Equation 5.82, we can rewrite the impedance at the inlet of the neck (Zi ) as Zi ¼ r0 c kri "
kro Zo ðmZo jr0 c kro Þtan kl #: fr0 c kro kri jmkðro ri ÞZo g
þ r0 c kro m þ jðkro kr þ m2 ÞZo tan kl
ð5:83Þ
If we assume that the wavelength of interest is much larger than the length of the neck, tan kl tends to kl. Equation 5.83 can then be simplified as Zi ¼ r0 c
r2i Zo þ jr0 ckri ro l : r0 cr2o þ jkri ro lZo
ð5:84Þ
We now examine the impedance at x ¼ l (Zo ). If fluid around the neck is moved about d, the pressure change in the cavity can be expressed as p¼
r0 c2 Bx¼l d r0 cpr2o u ¼ : V jkV
ð5:85Þ
The impedance without regard to energy dissipation (resistance) at x ¼ l can be written as p pr2 Zo ¼ ¼ r0 c o : ð5:86Þ jkV u x¼l Substituting Equation 5.86 into Equation 5.84, the impedance at the inlet of the neck (Zi ) can be rewritten as pr2
Zi ¼ r0 c
r2i jkVo þ jkri ro l r2o þ ri ro l
pr2o V
;
ð5:87Þ
where l (or l 0, to be more precise) denotes neck length that generally includes end correction. Therefore, the resonance frequency that sets reactance to zero can be obtained as rffiffiffiffiffiffiffiffiffiffi c pri ro : ð5:88Þ fn ¼ lV 2p If ro ¼ ri , which is a regular cylinder neck, Equation 5.88 shows that the resonance frequency is the same as Equation 5.60. A very common misunderstanding of a resonator is that it reduces sound by absorption. In reality, an abrupt impedance mismatch takes place at the resonance frequency of a resonator when installed on a noise transmission path (e.g., automotive engine suction/exhaustion units). This impedance mismatch reflects incident waves, and transmitted noise is finally reduced. In other words, it acts like an invisible wall. On the other hand, the amount of sound absorbed by a resonator is governed by its dissipation properties.26 The energy dissipation occurs primarily 26 If we look at the position of the resonator, the radiated sound power through the entrance of the resonator is equivalent to the sum of attenuated sound power of the resonator itself and reradiated sound power through the entrance of the resonator.
Sound Propagation
302
around the neck of the resonator, which is induced by friction between the fluid moving around the resonator’s neck and the confronting surface of the neck. The amount of dissipated energy, however, is generally much smaller than what is reflected by an impedance mismatch. To summarize, a sound field formed in a space of relatively shorter characteristic length than a wavelength has almost evenly distributed acoustic properties, because the space is too small to have propagation properties. A case in point is a resonator, which generates massive impedance mismatch at a resonance frequency. Such an impedance mismatch returns incident waves, if the resonator is used as noise reduction. On the other hand, it can also serve as an effective radiator for radiation. Attached to an acoustic transmission path to reduce noise levels, a resonator serves as a type of notched band stop filter that has a band with resonance frequency as its center frequency.27 On the other hand, if the resonator is excited in an open space, a resonator acts like a sort of narrow band pass filter that has a band with the resonance frequency as its center frequency. In the former, the resonator serves as a means to reduce noise, while it functions as a sound amplifier or a musical instrument in the latter case.
5.7
Duct Acoustics
A duct is a space where the length of one direction is significantly greater than the crosssectional direction. Because of such geometric characteristics, the sound propagation within a duct can be primarily expressed with respect to a single direction or coordinate.28 For better understanding, let us consider the sound at a certain frequency in a duct (Figure 5.13) which is infinitely long in the z direction and is enclosed by rigid walls. This wave will take the easiest way to propagate in the compressive fluid, which constitutes the medium within the duct.29 In other words, as depicted in Figure 5.13, any propagating sound inside a duct travels in the direction of acoustically longest length.30 On the other hand, there may be innumerable “standing waves” in the cross-sectional direction to meet the rigid boundary condition.
Figure 5.13 27
Three directions in which compressive fluids can move within an infinite duct
To improve the performance of a resonator, a Helmholtz resonator array is proposed and utilized in practice (see Section 5.9.5 for details). 28 As an exemplary case of 2D wave fields, we can consider the wave phenomenon in a plate or a membrane. In the case of double-layered partition including air or absorption materials, the wave phenomena are also 2D when the length of the cross-sectional area is much smaller than the wavelength in the propagating direction. 29 We can observe similar phenomena in solid materials. In the case of solid materials, unlike compressible fluid, we need to consider not only simple expansion and compression but also shear deformation and bending mode as basic modes. 30 The terminology “wave guide” implies the wave phenomena are dominantly governed with respect to one direction or a single coordinate.
Acoustics in a Closed Space
303
For an infinite duct, this qualitative idea can be expressed as a mathematical formula: pðx; y; z; tÞ ¼
1 X
Pmn ðx; yÞejðotkz zÞ ;
ð5:89Þ
m;n¼0
where Pmn ðx; yÞ represents the ðm; nÞth mode in the sectional direction, or the sound pressure distribution of a standing waves in the cross-sectional direction. kz is the wave number of waves that propagate in the z direction; it is also a function of ðm; nÞ. Equation 5.89 tells us that a sound field within a duct can be expressed as the acoustic waves in the cross-section and propagating waves in the z direction. In the case of a square duct as in Figure 5.13, its modes in the crosssection will take the cosine function. Equation 5.89 can therefore be rewritten as pðx; y; z; tÞ ¼
1 X
amn cosðkx xÞcosðky yÞejðotkz zÞ ;
ð5:90Þ
m;n¼0
where the wave numbers in the x and y directions are given as mp np kx ¼ ; ky ¼ : Lx Ly
ð5:91Þ
As Equation 5.90 has to meet the linear wave equation, there should be the following relationship among wave number components: k 2 ¼ kx 2 þ ky 2 þ kz 2 :
ð5:92Þ
Equation 5.92 can be expressed with respect to kz , the wave number (or propagation constant) in the z-direction in which propagation is concentrated, as kz2 ¼ k2 ðkx2 þ ky2 Þ:
ð5:93Þ
In terms of wavelength instead of wave number, Equation 5.93 can be rewritten as ( ) l2 ml 2 nl 2 ¼ 1 þ Lx 2 Ly 2 lz 2 ð5:94Þ ( 2 ) l 2 l þ : ¼ 1 lx ly Equations 5.93 and 5.94 delineate the dispersion properties of a sound wave being propagated within a duct. According to the propagation constant of the propagating direction (kz in this case), the properties of the propagating wave are closely related to the frequency of the wave and the propagation properties in the sectional direction (which are tightly bound to each other). It can also be observed that kz , the propagation constant in the z direction, can be a real or imaginary number. If it is a real number, it is propagated in the positive z direction. If it is an imaginary number, the magnitude of sound waves attenuates exponentially as it progresses toward the propagation direction (see Equation 5.89). This wave is referred to as an “evanescent wave.” Equation 5.94 indicates that this phenomenon will occur if wavelengths that propagate in the cross-sectional directions (ð1=lx 2 þ 1=ly 2 Þ1=2 in the case of a rectangular section) are shorter than the wavelength of the frequency (l). In the wave number domain, only those modes whose wave numbers in the cross-sectional direction are lower than k ¼ o=c can propagate
Sound Propagation
304
without being attenuated, that is kxy
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ kx 2 þ ky 2 :
ð5:95Þ
In other words, the duct serves as a sort of low pass filter with the cut-off wave number of k. As illustrated conceptually in Figure 5.14, if the cross-sectional area of a duct changes dramatically, this also significantly alters the way that a wave is propagated. First of all, if the sectional area increases sharply, the distribution of modes in the sectional direction becomes very dense at the wave number domain. For this reason, even those waves with the same mode number ðm; nÞ may be propagated easily in a tube with large sectional area while they are attenuated exponentially in a tube with small sectional area (i.e., wave blocking). Modes being attenuated exponentially in a larger tube, on the other hand, may become propagated efficiently once the sectional area becomes smaller (wave tunneling).
Ky Ky Kx Kx
Ky Ky Kx Kx
Ky
Ky Ky
Kx
Kx Kx
Figure 5.14
Propagation of waves in a wave guide with square section
We now examine Equation 5.94 for a special case: the length in each sectional direction being shorter than half a wavelength ðl=2Þ. In this case, all modes in the sectional direction, excluding one where ðm; nÞ ¼ ð0; 0Þ, will continue to be attenuated exponentially while being propagated. The only mode that is propagated without attenuation, ð0; 0Þ, is a plane wave whose sound pressure remains constant in the sectional direction and whose wave number in the z direction is k. Using Equation 5.90, the wave in this case can be expressed as pðz; tÞ ¼ a00 ejðotkzÞ :
ð5:96Þ
Acoustics in a Closed Space
305 S2 S1 Pr Pi
Psi
S1
Psr
Pt
z
z=L (a)
Psi Pr Pi
Psr
Pt
z=0
z=L (b)
Figure 5.15 Reflection and transmission of waves in simple divergent tube (pi is an incident wave; pr a reflected wave; pst and psr waves transmitted into and reflected by a silencer; and pt a transmitted wave)
This implies that, if the characteristic length of a section is considerably smaller than the wavelength, the wave of a duct may be considered a one-dimensional problem. Even in the absence of higher-order modes, massive changes takes place in the propagation of waves when the section experiences dramatic change. For instance, let us examine the properties of an expansion chamber-based silencer, which is one of the most representative silencers (see Figure 5.15). First, the waves in a small tube before expansion can be divided into incident wave (pi ) and one being reflected by impedance changes on the boundary (pr ), as described below: pi ðz; tÞ ¼ Pi ejðotkzÞ ;
pr ðz; tÞ ¼ Pr ejðot þ kzÞ :
ð5:97Þ
Waves at 0 z L, an interval where the section is expanded, can also be divided into those being propagated in the positive and negative directions in the z direction. Those are psi ðz; tÞ ¼ Psi ejðotkzÞ ;
psr ðz; tÞ ¼ Psr ejðot þ kzÞ :
ð5:98Þ
Meanwhile, in the area where L < z, only those waves being transmitted via a silencer and propagated in a single direction are present. This can be expressed as pt ðz; tÞ ¼ Pt ejðotkzÞ :
ð5:99Þ
31
These waves should meet the continuity condition (i.e., sound pressure and velocity need to be continuous) at the planes whose sections are expanded (z ¼ 0) and contracted (z ¼ L), respectively.
31
These conditions were physically applied in case of the reflection and transmission of waves in two strings with different thicknesses in Chapter 1 and in a flat surface of discontinuity in Chapter 3.
Sound Propagation
306
First of all, the pressure continuity condition at z ¼ 0 can be expressed as Pi þ Pr ¼ Psi þ Psr :
ð5:100Þ
Given that the impedance of a one-dimensional wave which is transmitted in a single direction is r0 c, the amounts of mass flowing in and out by unit time at z ¼ 0 should be identical. On this basis, the velocity continuity condition is obtained as Pi Pr Psi Psr r0 S1 ¼ r 0 S2 ; ð5:101Þ r0 c r0 c r0 c r 0 c where S1 and S2 refer to the cross-sectional areas of the two tubes before and after expansion. The continuity condition at z ¼ L can also be written as Psi ejkL þ Psr ejkL ¼ Pt ejkL ; Psi jkL Psr jkL Pt jkL ¼ r 0 S1 r0 S 2 e e e : r0 c r0 c r0 c
ð5:102Þ ð5:103Þ
On this basis, the magnitude ratio of transmitted waves against incident waves is derived: t¼
Pt ejkL ¼ : Pi cos kL j S1 þ S2 sin kL S1 2 S2
ð5:104Þ
Transmission loss ðTLÞ, which indicates the power of incident waves being lost while passing through a silencer, are therefore expressed as 1 TL ¼ 10 log10 2 jtj ( ) ð5:105Þ 1 S1 S2 2 2 ¼ 10 log10 1 þ sin kL : 4 S2 S1 An interesting fact here is that the amounts of transmission and reflection are related to the sectional area and frequency of the two tubes. In Equation 5.105, transmission loss reaches its peak when sinkL has the highest value of 1; transmission losses becomes zero, which is the minimum value, when sin kL is zero. In the former case, kL ¼ ðm þ 1=2Þp ðm ¼ 0; 1; 2; Þ; in the latter, kL ¼ mp. In other words, the first case occurs when the divergent tube is ð2m þ 1Þ=4 longer than the wavelength, while the second case occurs when the divergent tube is an integer multiple of the half wavelength. To explain the first case (i.e., incidence of the 1/4 wavelength on the divergent tube) more specifically, the incident sound wave will make regular wave movements: with minimum sound pressure at z ¼ L, being reflected based on the boundary condition of z ¼ L, and then reaching its maximum sound pressure at z ¼ 0 (see Figure 5.16(a)). In contrast, if a wave of wavelength twice the length of the divergent tube reaches the divergent tube, its sound pressure will be highest at z ¼ L. This wave will be successfully transmitted via a tube linked to z ¼ L (see Figure 5.16(b)). Equation 5.105 demonstrates greater transmission loss as the sectional area ratio (S2 =S1 ) increases; if the section of a simple divergent tube remains unchanged, its transmission loss will be 0. Figure 5.17 shows the transmission loss of a simple divergent tube as a function of sectional area ratio and frequency.
Acoustics in a Closed Space
307
Figure 5.16 Comparison of transmission losses in simple divergent tube by wavelength (a) maximum transmission loss (L ¼ l/4); and (b) minimum transmission loss (L ¼ l/2)
Figure 5.17
Transmission loss by frequency and cross-sectional area ratio of simple divergent tube
A similar phenomenon occurs in a pipe of shape is illustrated in Figure 5.18 (otherwise referred to as “1/4 wavelength tube”). In this case, however, the length of the tube ðlÞ needs to be understood as the length of an effective tube ðl 0 Þ, as described in Equation 5.60 Based on such observations, a basic idea can be formulated for the design of silencers.32 Using an expansion chamber-based silencer, certain frequency elements in the noise of your choice can be reduced dramatically by adjusting the length of the expansion chamber. In the presence of multiple frequencies, several expansion chambers on a repetitive basis can be investigated. 32
In case of the silencer utilized in internal combustion engines or air blowers, back pressure should be considered. The ideal silencer might reduce the noise without any other effects due to the silencer. However, in general, the silencer with the expansion chamber induces a large amount of back pressure because it uses the impedance mismatch due to the abrupt change of cross-sectional area of the duct. These types of silencer are inexpensive and durable.
Sound Propagation
308
Figure 5.18
Impedance mismatch due to duct resonance in 1/4 wavelength tube
A silencer that reduces noise using an impedance mismatch generated by a sharp change in shape is referred to as a reactive silencer or reactive muffler. One that tries to reduce noise using a perforated tube or sound-absorbing material is referred to as a dissipative silencer. In general, a dissipative silencer is known to be more effective for controlling high-frequency noise and absorbing noise better at relatively wider bandwidths. As a special case of a duct, acoustic horns have been considered frequently. Webster’s horn equation governs fluid particles in an acoustic horn, which is a kind of duct with a slowly varying cross-section (for example, see Figure 5.19). The acoustic horn can reduced reflection waves at the right end (exit of the horn) by slowly changing cross-section (or impedance). This enables sound radiated from the horn to be more effective than the constant duct. Webster’s horn equation can be derived by using the conservation of momentum, the conservation of mass, and the state equation for an infinitesimal fluid element (Figure 5.19), which can be formulated along the same line as the linearized wave equation in Section 2.2.
Δx ( pS)x 0
( pS)x+Δx
x
x
x + Δx ΔS ρ
0
x
x
∂u ∂t
+u
∂u ∂x
x + Δx
Figure 5.19 The relation between forces and the motion of an infinitesimal fluid element in an acoustic horn, which expresses momentum balance: forces (left-hand side) and momentum change (righthand side). S denotes the cross-sectional area of the horn, u is the particle velocity in the x direction, r is volume density of the fluid, and DS is the ratio of the projected area of the area at x þ Dx (Sx þ Dx ) to the area at x (Sx )
Acoustics in a Closed Space
309
As depicted in Figure 5.19, the forces acting on the fluid between x and x þ Dx and its motion will obey the conservation of momentum, that is ðpSÞx ðpSÞx þ Dx þ ðpDSÞx þ Dx ¼ rS 2
du Dx; dt
ð5:106Þ
where S represents the cross-sectional area of the horn, u is the particle velocity in the x direction, r is volume density of the fluid, and DS is the projected area of the area at x þ Dx (Sx þ Dx ) to the area at x (Sx ). The third term in the left-hand side of Equation 5.106 represents the average forces acting through x and x þ Dx on the projected area, DS. The pressure and area at x þ Dx in Equation 5.106 can be expanded using a Taylor series. By neglecting higher-order terms and using the assumptions in Section 2.2, we can eventually simplify Equation 5.106 as
@p @u ¼ r0 ; @x @t
ð5:107Þ
which is the same result as for a constant duct case in Section 2.2. p is the sound pressure (or access pressure), which is also expressed as p0 in Section 2.2. r0 is the static volume density. Δx
Throat (ρuS)x 0
Mouth
∂ (ρSΔ x) ∂t
x
x
(ρuS)x+Δx
x + Δx
Figure 5.20 Conservation of mass in an infinitesimal fluid element in an acoustic horn. S denotes the cross-sectional area of the horn, u is the particle velocity in the x direction, and r is volume density of the fluid
Secondly, let us obtain the relation between the volume density and fluid velocity using the conservation of mass. Figure 5.20 shows how much fluid enters the cross-section at x and how much exits through the surface at x þ Dx. The conservation of mass can be expressed as @ ðrSDxÞ ¼ ðruSÞx ðruSÞx þ Dx : @t
ð5:108Þ
As Dx ! 0, Equation 5.108 becomes S
@r @ ¼ ðruSÞ: @t @x
ð5:109Þ
By differentiating the right-hand side of Equation 5.109 with respect to ru and S, Equation 5.109 yields @r @u 1 @S ¼ r0 r u ; @t @x 0 S @x
ð5:110Þ
which is different from the case of a constant cross-sectional duct of Equation 2.15. The last term in Equation 5.110 describes the effect of gradual change of cross-section of the horn. As a limiting case, when we have a constant cross-section we can see that Equation 5.110 reduces to
Sound Propagation
310
Equation 2.15. With Equations 5.107 and 5.110, the state equation (Equation 2.18) can, finally, provide us with Webster’s horn equation as explained in Section 2.2, that is, @ 2 p 1 @S @p 1 @ 2 p þ ¼ 0: @x2 S @x @x c2 @t2
ð5:111Þ
By assuming harmonic oscillation, we can obtain Equation 5.77 which expresses the waves in the neck of the resonator with the tapered neck as shown in Figure 5.12. We now obtain a solution to Equation 5.111 for an exponential acoustic horn as a simple and widely used case. The exponential acoustic horn has a gradual change of cross-section exponentially, for instance, S ¼ S0 eax ;
ð5:112Þ
where S0 is the area at the left end of the horn (often referred to as throat, see Figure 5.20), and a is a flare constant which expresses exponential increase as x becomes larger. Webster’s horn equation (Equation 5.111) can then be written as @2p @p 1 @ 2 p þ a ¼ 0: @x2 @x c2 @t2
ð5:113Þ
We assume the solution to Equation 5.113 to be of the form pðx; tÞ ¼ PejðotkxÞ
ð5:114Þ
as for Chapters 1 and 2. Substituting into Equation 5.113, we obtain a relation between wave numbers as k2 þ jak þ kc2 ¼ 0;
ð5:115Þ
where kc ¼ o=c, which is also the wave number for a constant cross-sectional duct. Possible solutions to k in Equation 5.114 are therefore obtained as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ja a2 þ 4kc2 : ð5:116Þ k¼ 2 By rearranging Equation 5.116, we can obtain sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a a þj : k ¼ kc 1 2kc 2
ð5:117Þ
By substituting Equation 5.117 into Equation 5.114, the solution to Equation 5.113 can be given by 2 8 0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 93 2 = < a a Ax 5: ð5:118Þ pðx; tÞ ¼ Pe2x exp4j ot @kc 1 : ; 2kc We can see the right-going waves are amplified as the waves propagate to the mouth (see Figure 5.20). In addition, we can also obtain the phase velocity (cf ) for the horn as c ð5:119Þ cf ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ac 2ffi : 1 2o
Acoustics in a Closed Space
311
Equation 5.119 tells us that the phase velocity varies with frequency. There is a certain frequency which causes the phase velocity to be infinite, referred to as the cut-off frequency in the case of wave guides, that is fcutoff ¼
ac : 4p
ð5:120Þ
Figure 5.21 shows the phase velocity with respect to frequency, which depicts how the phase velocity increases abruptly around the cut-off frequency and approaches the speed of sound in the air as frequency increases. Cφ
C
fcutoff
f
Figure 5.21 Phase velocity (cf ) and cutoff frequency (fcutoff ) for an exponential acoustic horn, where c is the speed of sound in air (343 m/s at 20 C)
The driving point impedance of this exponential acoustic horn allows us to understand the phenomenon in the horn. Suppose that we have a velocity source at the throat (x ¼ 0) with u0 ejot ; we can easily obtain the pressure from the principle of momentum (Equation 5.107) as or0 p¼ u0 : ð5:121Þ k By inserting the positive values of Equation 5.117 and Equation 5.120, we can obtain the radiation impedance of the exponential horn as r0 c Zr ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
fcutoff f
2
þj
fcutoff f
:
ð5:122Þ
When we have determined the cut-off frequency, the radiation impedance has a purely imaginary value (jr0 c). This means that the waves in the horn cannot propagate well. Figure 5.22 depicts the radiation impedance in the horn with respect to frequency. As frequency increases, the resistance term approaches the characteristic impedance of the medium (r0 c) but the reactance term tends to 0. Note that the resistance term is 0 below the cut-off frequency. This means that the frequency cannot propagate in the horn at all if below the cut-off frequency. In
Sound Propagation
312 1.2 1 Re
0.8 Zr ρ0C
{ { Zr
ρ0C
0.6 0.4 Im 0.2 0
0
1
2
4
6
{ {
8
Zr
ρ0C
10
fcutoff f
Figure 5.22 Driving point impedance of an exponential acoustic horn. The solid line denotes the real part of the impedance (resistance), and the dashed line represents the imaginary part of the impedance (reactance)
addition, the impedance is independent of the position x, and this implies that the impedance does not change as the waves propagate through the exponential horn. To summarize, we have examined the physical properties of a sound wave propagating inside a duct using the wave number of the propagation direction. During wave propagation within a duct, some elements decrease exponentially while being propagated in accordance with the distribution of sound pressure in the sectional direction. Other elements, on the other hand, are propagated without experiencing any reduction. A change in the duct’s shape results in an impedance mismatch and this, in turn, reflects sound waves and causes transmission loss. These properties can be widely utilized in designing silencers on the basis of an impedance mismatch.
5.8
Chapter Summary
The characteristics of sound generated in a relatively large space compared to the wavelength differ significantly from those of sound created in a smaller space. The sound generated in the former case can be classified into: direct sound field (propagated directly from the sound source) and reverberant sound field (reflected from the wall). A reverberant sound field generally has the properties of a diffuse field, so its sound propagation characteristics are very complicated. The physical quantity of reverberation time, suggested by Sabine, represents the properties of a larger space in a more simplified way, and this provides the groundwork for architectural acoustic design. In contrast to a larger space, the resonator properties of a small space are more similar to those of a 1-DOF vibratory system than its propagation properties, and this phenomenon can be utilized in controlling a variety of noises. If, as in the case of a duct, the characteristic length of its section is shorter than a wavelength and the length of its propagation direction is considerably longer than that of a wavelength (i.e., if acoustically small and large spaces are mixed together), unique phenomena such as wave blocking and tunneling can be observed.
Acoustics in a Closed Space
5.9 5.9.1
313
Essentials of Acoustics in a Closed Space Methods for Measuring Absorption Coefficient
5.9.1.1 Using an Impedance Tube We assume that, as in Figure 5.23, a specimen is established on one edge of a slick rigid body straight tube (impedance tube) with a speaker on the opposite side and a rigid wall built on the rear of the speaker. When a sine wave is generated from the speaker, sound pressure in the tube makes a standing wave by superposition of the incident wave (Pi ) from the speakers and the reflected wave (Pr ) from the surface of the specimen. In other words, the sound pressure (P) measured at x is given by PðxÞ ¼ Pi ðxÞ þ Pr ðxÞ;
ð5:123Þ
Pi ðxÞ ¼ P0 ejkx ;
ð5:124Þ
Pr ðxÞ ¼ R P0 ejkx ;
ð5:125Þ
where k ¼ oc ¼ 2pf c is a wave number, R ¼ Pr ð0Þ=Pi ð0Þ a reflection coefficient defined by the aspect ratio of a reflected wave compared to an incident wave, and P0 the amplitude of the incident wave.
Figure 5.23 Impedance tube with a specimen established on one edge
The particle velocity of the incident and reflected waves, Ui ; Ur , are UðxÞ ¼ Ui ðxÞ þ Ur ðxÞ; Ui ðxÞ ¼
Pi ðxÞ ; Z0
Ur ðxÞ ¼
Pr ðxÞ ; Z0
ð5:126Þ ð5:127Þ
where Z0 ¼ r0 c is the characteristic impedance of the air. In Figure 5.23, the impedance at any point x in the tube is the ratio of the sound pressure and particle velocity measured. Hence, when Equations 5.123–5.127 are used, the result is given by ZðxÞ ¼
PðxÞ Pi ðxÞ þ Pr ðxÞ : ¼ Z0 UðxÞ Pi ðxÞPr ðxÞ
ð5:128Þ
Note that since the impedance of the incident wave has a minus value, a minus sign () is added so that it can be transformed into a positive resistance value. Using the definition of a reflection coefficient in Equation 5.128, the relation between impedance Z and reflection
Sound Propagation
314
coefficient R at x ¼ 0 can be evaluated as Zð0Þ ¼ Z0
Pi ð0Þ þ Pr ð0Þ 1þR ; ¼ Z0 Pi ð0ÞPr ð0Þ 1R
R¼
Zð0Þ=Z0 1 : Zð0Þ=Z0 þ 1
ð5:129Þ
ð5:130Þ
Also, the sound absorption coefficient of the specimen (a) is the ratio of absorbed energy to the energy of the incident wave (which equals the energy of the incident wave less the energy of the reflected wave), and can therefore be expressed as: a¼
jPi j2 jPr j2 jPi j2
¼ 1jRj2 :
ð5:131Þ
The material’s absorption coefficient can therefore be evaluated by measuring the reflection coefficient (or impedance) on the material surface, given Equations 5.130 and 5.131. Using Standing Wave Ratio33 In Equation 5.123, the maximum value of sound pressure level occurs when Pi and Pr are in phase, that is, jPmax j ¼ jP0 jð1 þ jRjÞ:
ð5:132Þ
When the two sound waves are out of phase with each other, the minimum value of sound pressure level can be obtained as jPmin j ¼ jP0 jð1jRjÞ:
ð5:133Þ
When the standing wave ratio s is defined as s¼
jPmax j ; jPmin j
ð5:134Þ
the relation between the standing wave ratio (Equation 5.134) and reflection coefficient, obtained from Equations 5.132 and 5.133, can be expressed as s¼
1 þ jRj ; 1jRj
ð5:135Þ
s1 : sþ1
ð5:136Þ
jR j ¼
The standing wave ratio (Equation 5.134) can therefore be evaluated by measuring the maximum and minimum sound pressure levels within the tube at the excitation frequency. The reflection coefficient (Equation 5.135) can therefore be found, and the absorption coefficient can then be determined from Equation 5.131.
33
ISO 10534-1. (1996) Acoustics-Determination of Sound Absorption Coefficient and Impedance in Impedance Tube, Part 1: Method Using Standing Wave Ratio. International Standards Organization.
Acoustics in a Closed Space
315 xmin, 2 xmin, 1
P(x)
P(xmin, 2)
P(xmin, 1)
P(xmax, 1)
Figure 5.24
Standing wave pattern within an impedance tube
It should be noted here that the first point of maximum sound pressure should be chosen between the first two minimum sound pressure points. Due to viscosity and heat loss, however, the incident and reflected waves within the tube generally attenuate in size while they proceed; the maximum and minimum values of sound pressure, as in Figure 5.24, change along the x axis. When k, the real wave number in Equation 5.123, is replaced by k, the complex wave number where the attenuation term is included, the equation describing the standing wave ratio and reflection coefficient is therefore k ¼ k0 þ jk0 ;
ð5:137Þ
Pðxmax;n Þ ¼ jP0 j ek0 xmax;n þ jRjek0 xmax;n ;
ð5:138Þ
Pðxmin;n Þ ¼ jP0 j ek0 xmin;n jRjek0 xmin;n ;
ð5:139Þ
Pðxmax;n Þ ek0 xmax;n þ jRjek0 xmax;n ¼ k0 x sn ¼ ; 0 e min;n jRjek xmin;n Pðxmin;n Þ 0
jR j ¼
ð5:140Þ
0
sn ek xmin;n ek xmax;n ; 0 0 sn ek xmin;n þ ek xmax;n
ð5:141Þ
where k0 ¼ 2p=l and l ¼ c=f is wavelength. Also, with respect to the points of the minimum sound pressure level, the subscripts (n ¼ 1, 2, 3, . . .) are assigned from the left to the right started from the closest point of minimum sound pressure level to the specimen. On the other hand, for maximum sound pressure level points, the subscripts are assigned from the first point of maximum sound pressure level on the right of the first point of minimum sound pressure level. When the same experiment is performed on an empty tube with a rigid body edge (jRj ¼ 1) to measure maximum and minimum sound pressure levels, the attenuation term of the impedance tube k0 can be evaluated: 0 Pðxmin;n þ 1 Þ Pðxmin;n Þ kl ¼ 2 sinh : Pðxmax;n Þ 4
ð5:142Þ
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316
Δx
x2 x1
x
Figure 5.25 Method for measuring absorption coefficient using transfer function: location and interval of microphones
Using a Transfer Function34 When sound pressure is measured at any two points (x ¼ x1 , x ¼ x2 ) as in Figure 5.25, the sound pressure at each point according to Equation 5.123 is expressed as P1 ¼ P0 ejkx1 þ R P0 ejkx1 ;
ð5:143Þ
P2 ¼ P0 ejkx2 þ R P0 ejkx2 :
ð5:144Þ
H12 , the transfer function of sound pressure measured at two points, is defined as H12 ¼
P2 ejkx2 þ Rejkx2 ¼ : P1 ejkx1 þ Rejkx1
ð5:145Þ
The following relation between the reflection coefficient and the transfer function, H12 , at x ¼ 0 can be obtained as R¼
H12 ejkDx j2kx1 e ; ejkDx H12
ð5:146Þ
where Dx ¼ x1 x2 . 5.9.1.2 Using Reverberation Room35 From the given Sabine theory, the equivalent absorption area A within a reverberation room with a very large space is expressed as A¼
55:3V ; c0 T60
ð5:147Þ
where V represents the volume (m3) of an empty reverberation room and T60 the time for the sound pressure to decrease by 60 dB, that is, reverberation time (s). By measuring the 34
ISO 10534-2. (1998) Acoustics-Determination of Sound Absorption Coefficient and Impedance in Impedance Tube, Part 2: Transfer-Function Method. International Standards Organization. 35 ISO 354. (2003) Acoustic-Measurement of Sound Absorption in a Reverberation Room. International Standards Organization.
Acoustics in a Closed Space
317
reverberation time before and after establishing a specimen in the reverberation room, the absorption coefficient of the specimen as can therefore be calculated from the equivalent absorption area of Equation 5.147, that is, A2 A1 V 1 1 ; ð5:148Þ as ¼ ¼ 55:3 c0 S T2 T1 S where S is the area of the specimen (m2) and the subscripts 1 and 2 are the reverberation time before and after establishment of the specimen, respectively. Reverberation time is usually measured by the 1/3 octave band center frequency (generally between 100 Hz and 5000 Hz). Also, as the sound field in a reverberation room is not a perfect diffuse field, reverberation time is measured repeatedly with differing sound sources and measurement locations within the reverberation room. The average value is used as the reverberation time of Equation 5.148.
5.9.2
Various Reverberation Time Prediction Formulae
In outdoor space, the moment when sound which has been turned off or blocked is no longer be heard can easily be determined from the distance between the source and the receiving point. For indoor space, however, this is not the case. The sound in indoor space does not disappear simultaneously with the blockage of the sound source but rather, due to reflection from the wall surface, gradually diminishes over a certain period of time. This phenomenon – sound being heard continuously even after removal of the sound source – is reverberation; the time characterizing this phenomenon is called reverberation time. Reverberation time is one of the major objective indicators that numerically express the acoustic characteristics of a space. It is defined by the amount of time (T60 ) for a sound pressure level to decrease by 60 dB from the normal state of a sound field. When Sabine first suggested the applied frequency, 500 Hz was preferred from experiments. The octave band or 1/3 octave band center frequencies in the audio frequency are now widely used. When this phenomenon of reverberation takes place, a curve can be drawn that represents the energy of sound gradually diminishing over time from the steady state. This curve is called a sound decay curve; reverberation time is calculated from this curve. Reverberation time in real space can be evaluated by drawing a sound decay curve through the experiment. 5.9.2.1 W. C. Sabine36 T60 ¼
55:3 V ; c At a
ð5:149Þ
where At is the total surface area of the space, V isPthe volume, and c is the sound speed. The number 55.3 ¼ 24 ln 10 (in SI units), a ¼ ð1=At Þ n An an is the averaged absorption coefficient of the space, and An and an are the area and absorption coefficient of the nth boundary surface. Sabine’s reverberation formula is the most general and can be used when the absorption coefficient is small. 36
Kuttruff, A. (1973) Room Acoustics. Applied Science Publishers LDT, pp. 96–123.
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318
5.9.2.2 C. F. Eyring (1930)37 T60 ¼
55:3 V : c At lnð1aÞ
ð5:150Þ
Eyring’s reverberation formula can also be used when the absorption coefficient is significantly large. Sabine’s formula is the same as the approximation of Eyring’s when the absorption coefficient is considerably small. Equations 5.149 and 5.150 have the disadvantage that even when absorption coefficients of walls vary, every wall is erroneously calculated as having the same average absorption coefficient. To address this shortcoming, Millington’s formula was suggested. 5.9.2.3 G. Millington (1932)38 T60 ¼
55:3 V P : c An lnð1an Þ
ð5:151Þ
n
Millington’s reverberation formula overcomes the weakness of Equations 5.149 and 5.150 noted above, and is appropriate when absorption is not diffused evenly in the space. However, if even the smallest area of a wall consists of completely absorbent material with an absorption coefficient of 1, this will result in miscalculation of the reverberation time as 0. 5.9.2.4 D. Fitzroy (1959)39 T60
55:3 V Ax Ay Az þ þ ; ¼ 2 c At ln ð1ax Þ ln ð1ay Þ ln ð1az Þ
ð5:152Þ
where Ax ; Ay ; Az are the total areas of the boundary surfaces normal to the x; y; z axes, respectively, and ax ; ay ; az represent the average absorption coefficients of the boundary surfaces normal to the x y; z axes. Fitzroy’s reverberation formula is an experimental formula suggested to overcome the weakness of Sabine’s and Eyring’s formulae. When ax ; ay and az are equal, it becomes the same formula as Eyring’s. As the wall surfaces are classified in only three directions, however, it cannot reflect subtle changes in the wall surface absorption coefficients (unlike for Equation 5.152). It is therefore utilized in processes such as designing a concert hall, that is, a relatively large space. Of Sabine’s, Eyring’s, and Fitzroy’s formulae, the latter makes the most accurate predictions in relatively large structures. As it is difficult to determine the exact absorption coefficient prior to construction, however, it has been reported that Sabine’s formula is adequate in actual application.40 Unlike these, Millington’s formula is the right choice for places such as vehicle interiors (which are small in size) and where wall surface absorption coefficients change significantly. 37
Eyring, C. F. (1930) Reverberation time in dead rooms. Journal of Acoustical Society of America, 1(2), 217–241. Millington, G. (1932) A modified formula for reverberation. Journal of Acoustical Society of America, 4(1), 69–82. 39 Fitzroy, D. (1959) Reverberation formula which seems to be more accurate with non-uniform distribution of absorption. Journal of Acoustical Society of America, 31(7), 893–897. 40 Beranek, L. L. (1992) Concert hall acoustics: 1992. Journal of Acoustical Society of America, 92(1), 1–39. 38
Acoustics in a Closed Space
319 z
r¯
r¯0 y
x
Figure 5.26 Locations of sound source (~ r 0 ) and sound reception (~ r) in a closed space (note that the size of the space is Lx > Ly > Lz )
5.9.3
Sound Pressure Distribution in Closed 3D Space Using Mode Function
As shown in Figure 5.26, we assume that a monopole source with radial frequency o and the size of sound source S excites the interior field at the location of~ r 0 within a closed space. Sound pressure in a closed 3D space can be expressed using an inhomogeneous Helmholtz equation, that is r2 Pð~ rÞ þ k2 Pð~ rÞ ¼ 4pSdð~ r~ r 0 Þ:
ð5:153Þ
The boundary of the room can be expressed with acoustic impedance Z, and can have different values in different locations and is expressed mathematically as r c rPð~ r s Þ ~ n out ¼ jk 0 Pð~ r s Þ; ð5:154Þ Z where~ r s represents the location of the boundary and~ n out the unit vector in an outward direction normal to the wall surface. We now define Wlmn , a mode function that satisfies the current boundary condition of the space. The mode function Wlmn should satisfy the following governing equation, that is, r2 Wlmn þ k2lmn Wlmn ¼ 0: It also has to satisfy the boundary condition, which is r c rWlmn ð~ r s Þ ~ n out ¼ jk 0 Wlmn ð~ r s Þ: Z Also, orthogonality among different modes holds as ð 1 Wlmn W*l 0 m0 n0 dV ¼ Llmn dll 0 dmm0 dnn0 V V
ð5:155Þ
ð5:156Þ
ð5:157Þ
where Llmn represents the normalizing constant of the mode and d the Kronecker delta function.
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320
When expressed with a linear combination of the mode function Wlmn , the sound pressure distribution within the space can be written as X Pð~ rÞ ¼ almn Wlmn ð~ rÞ: ð5:158Þ l;m;n
Therefore, evaluating the sound pressure distribution within a closed 3D space using a mode function translates into a problem of solving the modal coefficient almn . When the sound pressure of Equation 5.153 is expressed as the sum of mode functions, the result is X almn ðr2 Wlmn þ k2 Wlmn Þ ¼ 4pSdð~ r~ r 0 Þ: ð5:159Þ l;m;n
Using the governing equation of the mode function (Equation 5.155), the first term of Equation 5.159 can be expressed as X almn ðk2lmn Wlmn þ k2 Wlmn Þ ¼ 4pSdð~ r~ r 0 Þ: ð5:160Þ l;m;n
rÞ and integrating over the entire space By multiplying both sides by the mode function W*l 0 m0 n0 ð~ V, we obtain: ð X almn ðk2lmn Wlmn þ k2 Wlmn ÞW*l 0 m0 n0 dV V l;m;n
ð r~ r 0 ÞdV: ¼ 4pS W*l 0 m0 n0 dð~
ð5:161Þ
V
r 0 Þ. Due to orthogonality of the The right-hand side of Equation 5.161 becomes 4pSW*l 0 m0 n0 ð~ mode, the integral of the left side of Equation 5.161 has a value only when l 0 m0 n0 ¼ lmn. Hence, it can be simplified as VLl 0 m0 n0 al 0 m0 n0 ðk2l 0 m0 n0 þ k2 Þ ¼ 4pSW*l 0 m0 n0 :
ð5:162Þ
Therefore, the sound pressure distribution in any place in closed 3D space with its boundary condition having any value is eventually obtained as Pð~ rÞ ¼
5.9.4
4pS X W*lmn ð~ r0Þ rÞ: Wlmn ð~ 2 V l;m;n Llmn ðk k2lmn Þ
ð5:163Þ
Analytic Solution of 1D Cavity Interior Field with Any Boundary Condition
5.9.4.1 Exact Solution by Green’s Function Method Figure 5.27 shows a 1D cavity (e.g., duct system). A 1D cavity means that an acoustic wave in only one direction in the area of interest is considered in the frequency region under the occurrence of the first mode in the other two directions. Assuming that the direction of interest
Acoustics in a Closed Space
321 L x0 x
→
→
n
n G1
Z0
Figure 5.27
G2
ZL
An example of a 1D cavity
is x and that there is acoustic excitation at x ¼ x0 , the following governing equation (or Helmholtz equation) holds: r2 G þ k2 G ¼ dðxx0 Þ:
ð5:164Þ
In Equation 5.164, G represents Green’s function and k a wave number. The solution of the governing equation, expressed as in Equation 5.164, is determined uniquely by the boundary condition of the cavity boundary (or the wall surface). By introducing acoustic impedance Z, Equation 5.164 can be represented as r c rG1 ~ n out ¼ jk 0 G1 ; x ¼ 0; ð5:165Þ Z0 rG2 ~ n out ¼ jk
r0 c G2 ; ZL
x ¼ L;
ð5:166Þ
where the subscripts of the Green’s function G represent Green’s function on the left and right of the sound source location x ¼ x0 (Figure 5.27). The subscripts of acoustic impedance Z (0 and L) are used to distinguish the acoustic impedance at x ¼ 0 and x ¼ L, respectively. The general solution of Equation 5.164 can be expressed as G1 ¼ A cos kx þ B sin kx;
ð5:167Þ
G2 ¼ C cos kx þ D sin kx:
ð5:168Þ
The four coefficients in Equations 5.167 and 5.168 (A; B; C; D) can be determined by using two boundary conditions (Equations 5.165 and 5.166) and the following continuity condition of the Green’s function at x ¼ x0 : G1 jx0 ¼ G2 jx0 ;
ð5:169Þ
@G2 @G1 ¼ 1: @x x þ @x x
ð5:170Þ
0
0
5.9.4.2 Solution of Cavity Interior Field using Free-space Green’s Function Sound pressure Pð~ rÞ at any one point within the space ~ r can be expressed (using the Kirchhoff–Helmholtz integral equation) as a function of the sound pressure on the wall
Sound Propagation
322
surface surrounding the space, Pð~ r s Þ, and the normal sound pressure gradient, @n@ 2 Pð~ r s Þ: ð f ð~ r 0 ÞGð~ rj~ r 0 ÞdV
Pð~ rÞ ¼ V0
ð5:171Þ
ð ½Pð~ r 0 Þr0 Gð~ rj~ r 0 ÞGð~ rj~ r 0 Þr0 Pð~ r s Þ ~ n 0 dS;
þ S0
where f ð~ r 0 Þ represents the magnitude of the sound source excited at the point of~ r 0 , and G is Green’s function in the space of interest. Note that in deriving Equation 5.171, Green’s function G only needs to be the solution of an inhomogeneous wave equation as expressed in Equation 5.164; it does not have to satisfy the boundary condition on the space of interest. As seen in Equation 5.171, the Kirchhoff–Helmholtz integral equation is merely a simultaneous expression of wave equation and boundary condition. This is because, in order to calculate sound pressure inside the cavity, it is necessary to know the sound pressure and sound pressure gradient on the boundary. In other words, to calculate the desired value of sound pressure Pð~ rÞ requires knowledge of the sound pressure on the wall surface Pð~ r 0 Þ as prior information. As this section provides exercises on a 1D cavity, the Green’s function in Equation 5.171 uses the following 1D free field Green’s function: G¼
j jkjxx0 j e : 2k
ð5:172Þ
When Equation 5.172 and the second boundary condition (Equation 5.166) are substituted into Equation 5.171 and summarized (where f ð~ r 0 Þ ¼ dðxx0 Þ), the result is expressed as functions of sound pressures on the boundaries, P0 and PL , in the following way: PðxÞ ¼
j jkjxx0 j ð1b L ÞPL jkðLxÞ ð1b 0 ÞP0 jkx e e ; þ þ e 2 2 2k
ð5:173Þ
P0 ¼
j ð1 þ b L ÞejkðLx0 Þ þ ð1b L ÞejkðLx0 Þ ; k ð1 þ b 0 Þð1 þ b L ÞejkL ð1b 0 Þð1b L ÞejkL
ð5:174Þ
PL ¼
j ð1 þ b 0 Þejkx0 þ ð1b 0 Þejkx0 ; k ð1 þ b 0 Þð1 þ b L ÞejkL ð1b 0 Þð1b L ÞejkL
ð5:175Þ
where b 0 ¼ r0 c=Z0 , b L ¼ r0 c=ZL and P0 and PL represent sound pressures at x ¼ 0 and x ¼ L, respectively. If b 0 and b L become 1, then only the impact by the sound source on sound pressure exists at any place. This is because the impact of a reflected wave is non-existent, as impedance mismatch on the boundary does not occur. Meanwhile, when b 0 and b L become 0, this constitutes a rigid body wall boundary condition where impedance on the boundary is infinite. It can therefore be determined that P0 and PL have maximum values when k ¼ np=L (n ¼ 0; 1; 2; . . .) and resonance occurs. Equation 5.173 can be considered the result of the basic concept of the boundary element method applied to the 1D cavity.
Acoustics in a Closed Space
5.9.5
323
Helmholtz Resonator Array Panels41
To reduce acoustic energy, many numbers of Helmholtz resonators (for example, see Figure 5.28) can be used. In this case, interaction between each resonator may cause different phenomena at the inlet of the resonator from its case.
Figure 5.28 Examples of resonator array panels: (a) a panel composed of identical resonator units and (b) a panel composed of different resonator units
Even although the mathematical formulation for a resonator in an array panel can be expressed similarly, radiation impedance is used differently due to the interactions between resonators. The radiation impedance can be obtained by regarding the inlet of a resonator as a rigid disk; Figure 5.28 shows a typical array of panels. The motion of a disk that has a mass velocity u1 (analogous to the radiation impedance in Figure 5.29(c)) must be affected by reflection (pressure on the disk) from neighboring disks, which have mass velocity u2 and u3 . This indicates that the radiation impedance must consider the effect from resonators around the resonator of interest. Figure 5.29(b) is an equivalent model that can mimic the corresponding acoustic characteristics. 5.9.5.1 Performance Analysis of Helmholtz Resonator Array Panels Consider an infinite plane which is composed of various resonators as depicted in Figure 5.30. The surface of the plate has distributed necks of the resonators, and has periodic square area (Lx Ly ) as illustrated in Figure 5.30(b). We refer to the periodic square area as a unit cell of a resonator array panel. Various types of resonators (more than one) can be placed in this unit cell, and the shape of the neck is assumed to be a circle for simplicity. In addition, the diameter of the neck is assumed to be small compared to the shortest wavelength of interest. The fluid particles in the neck can therefore be regarded as moving with same phase. The other surface on the 41
Kim, S.-R., (2006) Design method of Helmholtz resonator array panels for low-frequency sound absorption. PhD thesis, Korea Advanced Institute of Science and Technology.
Sound Propagation
324
2 pin
pin u
Kr
Rr Mr
u M
R
K (a)
(b)
u
u2
u1
u3
Kr
u1 u3
Rr
Mr
u1
u2 (c)
Figure 5.29 One-dimensional vibratory analogous models that reflect acoustic phenomena at the inlet of a resonator in a resonator array panel: (a) an incident plane wave and particle velocity at the inlet of resonators; (b) 1D vibratory analogous model; (c) 1D vibratory analogous model at the inlet of a resonator with particle velocity u1. (pin indicates sound pressure of an incident wave, u represents mass velocity at the neck, and the subscript of u is the index for each resonator in an array panel. M; K and R indicate mass, spring constant, and damping coefficient of an analogous model for a resonator, and Mr ; Kr and Rr are mass, spring constant, and damping coefficient of an analogous model for radiation impedance, respectively)
plane, except for the inlet of the neck, is also assumed to be rigid. We use a plane wave as an incident wave, which has a complex amplitude Pin , a polar angle yin , and an azimuthal angle fin as shown in Figure 5.30, that is, pin ¼ Pin ejðkx x þ ky ykz zÞ ;
ð5:176Þ
where kx ¼ k sin yin cos fin , ky ¼ k sin yin sin fin , and kz ¼ k cos yin . The time harmonic function is ejot . As explained in Section 5.6, the Helmholtz resonator reduces sound by using impedance mismatch and also sound absorption around the neck. In general, for a resonator, sound is dominantly reduced by impedance mismatch. In case of a resonator array panel, however, the absorbed energy cannot be considered negligible. The quantity of energy absorbed by a resonator array panel should therefore be determined to accurately estimating the sound
Acoustics in a Closed Space
325
Plane wave pin = Pine-f(kxx+kyy-kzz)θ in
z
y ∞
∞
x φm
Ly
∞
∞
Lx (a)
y
Lx
i″
yc,i = yc,i + n″Ly
yc,i yc,i
Ly
i′ i x
xc,i = xc,i + m″Lx
xc,i , xc,i
m″, n″=0, ±1, ±2, ... (b)
Figure 5.30 Theoretical model for Helmholtz resonator array panel: (a) an incident plane wave into the surface of a rigid infinite plane and (b) Cartesian coordinate located at the origin. (kx ¼ k sin yin cos fin , ky ¼ k sin yin sin fin and kz ¼ k cos yin , where k is wave number. yin and fin represent incident angle. This model is composed of many periodic unit square cells, of area Lx Ly , and the number of resonators is N)
reduction performance of the resonator array panel. The absorbed energy (Pa ) for a single Helmholtz resonator with cross-sectional area of the neck A is given by
1 Pa ¼ Re Zsurf juj2 A; 2
ð5:177Þ
where Zsurf is surface impedance, which relates total pressure42 and particle velocity at the inlet point of the neck, and u is the particle velocity at the inlet of the neck. By using the relation 42
For an incident wave (pin ) and radiation impedance (Zr ), the total pressure (p) at the inlet of the neck is expressed as p ¼ 2pin þ Zr u, where u is the particle velocity of the fluid particle at the neck. This enables us to express the incident wave using Zsurf and Zr by the relation p ¼ Zsurf u, that is, ðZsurf þ Zrad Þu ¼ 2pin . The minus sign in the right-hand side is a result of positive direction of motion being outwards from the inlet of the neck.
Sound Propagation
326
between the incident wave and the impedance (see Section 3.2), Equation 5.177 can be rewritten as 4ReðZsurf Þ jpin j2 ; Pa ¼ A 2 Iin where Iin ¼ Zr þ Zsurf 2r0 c
ð5:178Þ
where Iin represents the intensity of the incident wave. Equation 5.178 tells us that the absorbed energy of the resonator depends on both the radiation impedance and the surface impedance. How well we estimate the performance of the resonator array panel depends, therefore, on how well we measure the impedance at the inlet of the neck. To this end, the following three analytic methods are available. 5.9.5.2 Equivalent Surface Impedance Method This method has been widely used for analyzing the characteristics of an absorbing system with a perforated surface.43,44 This method regards a perforated surface as a homogeneous surface with an equivalent surface impedance, and assumes that a spatially periodic unit cell is smaller than the wavelength. The equivalent impedance is defined as Zavg surf ¼
hpi ; hun i
ð5:179Þ
where h i denotes spatial average on a unit cell area, and p and un represent sound pressure and normal particle velocity on the surface of the unit cell, respectively. The positive direction of un in Equation 5.179 is the negative z direction in Figure 5.30(a). If we assume that the panel is a locally reacting homogeneous surface, the absorption coefficient a can be given by a¼
4ReðZavg surf Þ r0 c cos yin 2
avg fr0 c þ ReðZavg surf Þ cos yin g þ fImðZsurf Þ cos yin g
2
ð5:180Þ
from the conventional formula for the absorption coefficient of a homogeneous surface. In addition, the absorption coefficient for random incidence (ast ) in a diffuse field can be obtained by inserting Equation 5.180 into the following formula:45 ð ð 1 2p p=2 ast ¼ a cosyin sinyin dyin dfin : ð5:181Þ p 0 0 The spatially averaged surface impedance of a resonator array panel composed of many identical resonators (Figure 5.31) is obtained as Zavg surf ¼ 43
ZHR A with e ¼ : Lx Ly e
ð5:182Þ
Bolt, R. H. (1947) On the design of perforated facings for acoustic materials. Journal of Acoustical Society of America, 19, 917–921. 44 Ingard, U., and Bolt, R. H. (1951) Absorption characteristics of acoustic material with perforated facings. Journal of Acoustical Society of America 23, 533–540. 45 Beranek, L. L., and Ver, I. L. (1992) Noise and Vibration Control Engineering, Principle and Application. John Wiley & Sons, Inc., New York.
Acoustics in a Closed Space
327
Plane wave
pin
Pin e
z
j ( k x x k y y kz z)
A
e
y
Ly
e
x
Lx
Figure 5.31 Theoretical model for Helmholtz resonator array panel composed of many identical resonators. (Pin is a complex amplitude of the incident wave, A is the cross-sectional area of the neck, the area of the cell is Lx Ly , and the number of resonators is only one in a unit cell)
As shown in Equation 5.182, the averaged surface impedance is equivalent to the resonator impedance46 (ZHR ) divided by surface porosity (e), which is the ratio between the area of the neck and the area of a unit cell. On the other hand, the spatially averaged surface impedance for a resonator array panel that has various kinds of resonators can be obtained from47 1 Zavg surf
¼
N X ei Ai with ei ¼ ; Z L HR;i x Ly i¼1
ð5:183Þ
where N is the number of resonators in a unit cell, and subscript i denotes each resonator. 5.9.5.3 Fourier–Rayleigh Method The sound field with the surface shown in Figure 5.30 as a boundary can be obtained by using the Fourier–Rayleigh method, because the boundary is periodic. This method is based on Rayleigh’s theory48 which allows us to expand the reflected waves in terms of Fourier series. It has been applied to absorption performance analysis of a periodic absorbing strip or absorption materials that have a perforated surface. Note that the method does not assume that the unit cell is much smaller than the wavelength. Firstly, let us consider the case when we have an incident plane wave to the surface which is composed of different impedances periodically as shown in Figure 5.30. Total pressure (p) on the surface can be regarded as the sum of the incident wave (pin ) and scattered wave (psc ), that is, pðx; y; zÞ ¼ pin ðx; y; zÞ þ psc ðx; y; zÞ:
ð5:184Þ
46 The resonator impedance (ZHR ) includes a mutual interaction effect (i.e., end correction) between the resonator and external sound field, which includes a mutual effect between resonators as discussed in Section 5.5. 47 Zwikker, C., Kosten, C. W. (1949) Sound Absorbing Materials. Elsevier, New York, pp. 127–163. 48 Strutt Baron Rayleigh, J. W. (1907) On the dynamical theory of grating. Proceedings of the Royal Society of London, A, 79, 399–416.
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328
The scattered wave can then be expressed by using the Fourier series expansion as psc ðx; y; zÞ ¼
1 1 X X
Amn ejðxm x þ Zn y þ gmn zÞ ;
ð5:185Þ
m¼1 n¼1
where xm ¼ kx þ 2mp=Lx and Zn ¼ ky þ 2np=Ly . Equation 5.185 has to satisfy the Helmholtz equation; the following wave number relation must therefore hold: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5:186Þ gmn ¼ j k2 xm 2 Zn 2 : Moreover, the acoustic energy passing through the surface (Lx Ly ) can be written as ð Lx ð Ly 1 P¼ Reðp u*n Þdxdy; ð5:187Þ 0 0 2 where * denotes complex conjugate. Using Euler’s equation, incident acoustic energy (Pin ) and reflected acoustic energy (Pref ) can be obtained as Pin ¼
Lx Ly jPin j2 cos yin 2r0 c
ð5:188Þ
and Pref ¼ P
1 1 X 1 X Reðgmn Þ 1 X Reðgmn Þ jAmn j2 ¼ jAmn j2 ; 2r0 c m¼1 n¼1 k 2r0 c Reðg Þ k
ð5:189Þ
mn
where Re(gmn) implies summation with respect to every m and n that have real-value gmn . In addition, from the definition implies that the absorption coefficient at the boundary can be written as X g Amn 2 Pin Pref 1 mn : a¼ ¼ 1 ð5:190Þ cos yin Reðg Þ k Pin Pin mn
On the other hand, acoustic admittance Bðx; yÞ, which is periodic, can be expanded using the Fourier series as 1 1 2np X X j 2mp x þ y Lx Ly bmn e ; ð5:191Þ Bðx; yÞ ¼ m¼1 n¼1
and bmn
1 ¼ Lx Ly
ð Lx ð Ly 0
Bðx; yÞe
j
2mp 2np Lx x þ Ly y
dxdy:
ð5:192Þ
0
By considering Equations 5.184 and 5.191 at the boundary of the xy plane, we obtain 1 @p Bðx; yÞpðx; y; 0Þ ¼ : ð5:193Þ jk @z z¼0
Acoustics in a Closed Space
329
By multiplying both sides by ejð2mpx=Lx þ 2npy=Ly Þ, and then integrating over the area of a unit cell (Lx Ly ), we can obtain infinite simultaneous equations for the Fourier coefficients Amn of the scattered wave psc as g 0 0 Amn bm0 m n0 n þ dm0 m dn0 n m n ¼ Pin ðdm0 0 dn0 0 cos ybm0 n0 Þ k m¼1 n¼1 1 1 X X
for m0 ; n0 ¼ 0; 1; 2; . . .. dmn is the Kronecker delta function, defined as ( 1 if m ¼ n dmn ¼ 0 if m 6¼ n:
ð5:194Þ
ð5:195Þ
The admittance at the boundary Bðx; yÞ is zero on the surface except for at the inlet of resonators where admittance is r0 c=Zsurf ;i ; this can be written as ( r0 c=Zsurf ;i part of the ith hole Bðx; yÞ ¼ ð5:196Þ 0 another part of the surface: Once the surface impedance is determined at the inlet of the neck, we can obtain the solution to Equation 5.194. By substituting the solution into Equation 5.190, the absorption coefficient for a resonator array panel can be obtained. The absorption coefficient, moreover, enables us find the absorption coefficient for random incidence by using Equation 5.181. Note that the surface impedance used in Equation 5.196 is not the same as the resonator impedance (ZHR;i ) used in Equation 5.183. In other words, Zsurf ;i does not include the end correction factor. This is because the mutual effect between resonator and external sound field is already included when we calculate sound field. Furthermore, the Fourier coefficient bmn of the admittance Bðx; yÞ is obtained as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2J1 ai ð2mp=Lx Þ2 þ ð2np=Ly Þ2 N 2np X j 2mp r0 c pai 2 Lx xc;i þ Ly yh;i qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e bmn ¼ ð5:197Þ Zsurf ; i Lx Ly i¼1 ai ð2mp=Lx Þ2 þ ð2np=Ly Þ2 by substituting Equation 5.196 into Equation 5.192. Here, N is the number of resonators in a unit cell, the subscript i denotes an index for each resonator, ai is the radius of the neck, xc;i and yc;i represent the x and y values from the center of the inlet area of the resonator, and J1 is the Bessel function of first kind of order 1. Note that the second term in the right-hand side of Equation 5.190 is the sum of all the reflected waves for Reðgmn Þ > 0, that is, the summation of the reflected waves that propagate along the z direction. As the reflected waves increase, the absorption coefficient becomes lower. To propagate along the z direction, the reflected waves satisfy Reðgmn Þ > 0, which is xm 2 þ Zn 2 < k2 . Therefore, ðsin yin cos fin þ ml=Lx Þ2 þ ðsin yin sin fin þ nl=Ly Þ2 < 1
ð5:198Þ
by using the relation between wave numbers and incident angle. Here, l is the wavelength (2p=k). Equation 5.198 shows us that whether the reflected waves propagate or not depends on
Sound Propagation
330
the incident angle, even although the size of a unit cell and the frequency of the incident waves are determined. When we have the condition maxðLx ; Ly Þ < 0:5; l
ð5:199Þ
all of the reflected waves except those with ðm; nÞ ¼ ð0; 0Þ (i.e., geometrical reflection) cannot propagate from the surface. In this case, the absorption coefficient is given by 2 A00 a ¼ 1 : ð5:200Þ Pin To examine A00 in detail, we express Equation 5.194 in a matrix form, that is Hd ¼ q; where
2
6 6 H ¼ fhmn g ¼ 6 6 4
b00 þ cosyin
b10
b10
..
. ..
.. .
9 8 cosyin b00 > > > > > > = < b 10 q ¼ fqi g ¼ : > > > > > > .. ; : .
.
3 7 7 7; 7 5
ð5:201Þ
9 8 A00 =Pin > > > > > > = < A =P 10 in ; d ¼ fdi g ¼ > > > > > > .. ; : .
The solution to Equation 5.201 can be obtained as
and ð5:202Þ
X
Hmn qm adj H m d¼H q¼ ; q ) dn ¼ X det H hm1 Hm1 1
ð5:203Þ
m
where adj H represents the adjoint matrix, and Hmn is the cofactor of H. If we have h11 ¼ b00 þ cos yin , q1 ¼ cos yin b00 , and m > 1 in Equation 5.202, then we can obtain hm1 ¼ qm . This enables us to obtain the first term of vector d in Equation 5.203 as X Hm1 qm A00 m ¼ d1 ¼ X Pin hm1 Hm1 m
X Hm1 qm m>1 X ¼ : H11 cos yin þ H11 b00 qm Hm1 H11 cos yin H11 b00 þ
ð5:204Þ
m>1
By defining effective admittance beff , Equation 5.204 can be rewritten as X A00 cos yin beff ¼ where beff ¼ b00 hm1 Hm1 =H11 : Pin cos yin beff m>1
ð5:205Þ
Acoustics in a Closed Space
331
Here, we can recognize that b00 is the spatially averaged admittance from Equation 5.192. By substituting Equation 5.205 into Equation 5.200, absorption coefficient where Equation 5.198 is satisfied can be obtained as cos yin beff 2 ; a ¼ 1 cos yin þ beff
ð5:206Þ
which is similar to the absorption coefficient obtained using an equivalent surface impedance method in Section 5.9.5.2. 5.9.5.4 Radiation Impedance Method In Figure 5.30, we can also regard the plane wave incidence as sound radiated from a monopole source in the far field. Using Green’s function, we can then obtain the sound field on the xy plane which satisfies a semi-infinite rigid wall boundary condition by the Kirchhoff–Helmholtz integral equation. The integral equation can be written as ZZ pðx; y; 0Þ ¼ 2pin ðx; y; 0Þjkr0 c un ðx0 ; y0 ; 0ÞGN ðx; y; 0jx0 ; y0 ; 0Þ dAðx0 ; y0 Þ; ð5:207Þ where GN ¼
ejkR with R ¼ 2pR
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxx0 Þ2 þ ðyy0 Þ2 þ ðzz0 Þ2 ;
ð5:208Þ
and un is a surface normal velocity which has positive value in the direction from the exterior region to the boundary surface. In other words, we regard the negative z direction in Figure 5.30(a) as positive. The first term in the right-hand side of Equation 5.207 is the pressure induced by a source, and second term denotes the sound field generated by the boundary condition. By recalling the assumption that the velocity on the surface (except at the inlet) of the neck is 0 and the velocity (i.e., the inlet velocity at the neck) within the area of the neck is constant, the second term can be expressed as a summation of the integration of the neck area, that is, ZZ 1 X pðx; y; 0Þ ¼ 2pin ðx; y; 0Þ þ jkr0 c ui GN ðx; y; 0jx0 ; y0 ; 0ÞdAi ; ð5:209Þ i¼1
Ai
where the subscript i represents the index on the inlet of each resonator, ui is the inlet velocity at the neck, and Ai is the inlet area of the neck. The inlet velocity ui is defined positive when directed along the positive z direction, that is, ui ¼ un jorifice i . By integrating Equation 5.209 with respect to the inlet area (Ai0 ) for each i0 resonator, and dividing them by the area (Ai0 ), the first terms in both the left- and right-hand side become the sound pressure at ðxc;i0 ; yc;i0 ; 0Þ which is the center point for each i0 resonator. This can be written as ZZ 1 pdAi0 ; ð5:210Þ pi 0 Ai 0 Si 0
Sound Propagation
332
and pin;i0
1 Ai 0
ZZ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2J1 ai0 kx2 þ ky2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Pin ejðkx xc;i0 þ ky yc;i0 Þ Pin ejðkx xc;i0 þ ky yc;i0 Þ : pin dAi0 ¼ 2 2 Ai0 ai 0 k x þ k y ð5:211Þ
Here, pi0 is the first term on the left-hand side of Equation 5.209 and pin;i0 is the first term on the right-hand side of Equation 5.209, obtained by integrating and dividing by Ai0 . In addition, by using the mutual radiation impedance between two resonators49 which is defined as Zr;i0 i ¼
jkr0 c Ai 0
ZZ ZZ GN dAi0 dAi ; Ai 0
ð5:212Þ
Ai
1 P the second term on the right-hand side of Equation 5.209 can be simplified to ui Zr;i0 i . i¼1 These results enable us to simplify Equation 5.209 to
pi0 ¼ 2pin;i0 þ
1 X
ui Zr;i0 i ;
ð5:213Þ
i¼1
where Zr;i0 i is the mutual radiation impedance between two resonators. The boundary condition on the inlet of the resonator can be expressed as pi ¼ Zsurf ;i ui on the orifice i
ð5:214Þ
using the surface impedance (Zsurf ;i ). By substituting Equation 5.214 into Equation 5.213 and rearranging, we can obtain infinite numbers of simultaneous equations as 1 X
ðZr;i0 i þ di0 i Zsurf ;i Þui ¼ 2pin;i0 for i0 ¼ 1; . . . ; 1; 0; 1; . . . ; 1;
ð5:215Þ
i¼1
where di0 i is the Kronecker delta function (defined in Equation 5.195). Consider two identical resonators, i and i0 , which are located at the same position in each unit cell (see Figure 5.30(b)). These two inlets of the resonator not only have the same radius but also the same impedance. Because the boundary is periodic and infinite, the acoustic characteristic for each inlet of the resonators should be the same except acoustic excitation by the incident plane waves. In other words, the inlet velocity ratio between two resonators is the same as the pressure ratio between two incident plane waves corresponding to each resonator, that is, pin;i ui ¼ : ð5:216Þ ui00 pin;i00
49
The mutual radiation impedance (Zr;i0 i ) physically means the ratio of the averaged pressure acting on the inlet area of resonator i0 generated by the motion of the inlet area of resonator i and the inlet velocity of resonator i0.
Acoustics in a Closed Space
333
By substituting Equation 5.216 into Equation 5.215, we can obtain finite numbers of simultaneous equations as N X r;i0 i þ di0 i Zsurf ;i Þui ¼ 2pin;i0 for i0 ¼ 1; 2; 3; . . . ; N; ðZ
ð5:217Þ
i¼1
where N is the number of resonators in a unit cell. The first term on the left-hand side of Equation 5.217 is defined as 1 X
r;i0 i ¼ Z
m0 ¼1 i00 ¼i þ m0 N
Zsurf ;i0 i00
pin;i00 : pin;i
ð5:218Þ
For a circular inlet of a resonator, Equation 5.218 can be given by 1 1 2 X X jk 2J1 ðai0 bmn Þ 2J1 ðai bmn Þ r;i0 i ¼ r0 c pai qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z ai0 bmn ai bmn Lx Ly m¼1 n¼1 b2mn k2
ð5:219Þ
ejfð2pm=Lx kx Þðxc;i0 xc;i Þ þ ð2pn=Ly ky Þðyc;i0 yc;i Þg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where bmn ¼ ð2pm=Lx kx Þ2 þ ð2pn=Ly ky Þ2 . Furthermore, by inserting the boundary condition (Equation 5.214) into Equation 5.188, the absorbed energy by a unit cell can be obtained as Pa ¼
N 1X ReðZsurf ;i Þjui j2 Ai : 2 i¼1
ð5:220Þ
In addition, the absorption coefficient is the ratio of the absorbed energy to the incident energy. The absorption coefficient with a circular inlet area can therefore be obtained as N P
a¼
i¼1
ReðZsurf ;i Þjui j2 Ai =2
ðLx Ly =2r0 cÞjPin j2 cos yin
¼
N X i¼1
pa2i ReðZsurf ;i Þ ui 2 r0 c Lx Ly cos yin Pin
ð5:221Þ
by using Equations 5.188 and 5.220. Once the surface impedances for each inlet of resonators are determined, we can obtain solutions to Equation 5.217. By putting the solution into Equation 5.221, we can then calculate the absorption coefficient for a resonator array panel. This method does not have a restriction on the size of a unit cell as the Fourier–Rayleigh method. On the other hand, the radiation impedance method has the same number of simultaneous equations as the number of resonators in a unit cell, regardless of the size of a unit cell or the distribution of resonators. The absorption coefficient for random incidence can also be obtained by substituting the absorption coefficient from Equation 5.221 into Equation 5.181.
Sound Propagation
334
A simple case has a resonator in a unit cell (N ¼ 1) and is useful to investigate the physical meaning of the results. The number of simultaneous equations is reduced to only 1, which simply gives the inlet velocity. By inserting the inlet velocity into Equation 5.221, we can obtain the absorption coefficient when N ¼ 1 as a¼
pa21 r0 c 4ReðZsurf ;1 Þ ; Lx Ly cos yin Z r;11 þ Zsurf ;1 2
ð5:222Þ
r;11 is net radiation impedance50 at the inlet of the resonator and is expressed as where Z Zr;1
1 1 X pa21 1 pa21 X jk 2J1 ða1 bmn Þ 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Z11 ¼ r0 c þ r0 c ; ð5:223Þ a1 bmn Lx Ly cos yin Lx Ly m¼1 n¼1 b2 k2 m;n6¼0;0
mn
from Equation 5.219. If all bmn are larger than k except ðm; nÞ ¼ ð0; 0Þ, in other words, in the case of ðml=Lx sin yin cos fin Þ2 þ ðnl=Ly sin yin sin fin Þ2 > 1 for ðm; nÞ 6¼ ð0; 0Þ;
ð5:224Þ
the summation in the second term on the right-hand side of Equation 5.223 becomes purely imaginary, and this is directly related to exterior end correction factor. If we denote the exterior end correction factor as dex;1 , the exterior end correction factor is the same as the imaginary part of a radiation impedance divided by wavelength. The absorption coefficient can therefore be rewritten as 4ReðZr;1 Þr0 c cosyin =e1 pa21 ; a¼ 2 ; where e1 ¼ r c þ ðjkdex;1 þ Zr;1 Þ cosyin =e1 Lx Ly 0
ð5:225Þ
which is similar to the absorption coefficient obtained by using the equivalent surface impedance method (Section 5.9.5.2). To express the absorption coefficient as Equation 5.225, the condition in Equation 5.224 should be satisfied. Even although the range of a unit cell size (Lx ; Ly ) can be vary with incidence angle, the condition of Equation 5.224 is always satisfied when we have maxðLx ; Ly Þ < 0:5; l
ð5:226Þ
regardless of angle of incidence. The condition for a unit cell size is the same as the results from the Fourier–Rayleigh method (see Section 5.9.5.3).
50
Prichard, R. L. (1960) Mutual acoustic impedance between radiators in an infinite rigid plane. Journal of Acoustical Society of America, 32, 730–737.
Acoustics in a Closed Space
335
Exercises 1. Consider a room that is 6 m long, 5 m wide and 3 m high and whose walls, floor and ceiling have the same sound absorption coefficient (a ¼ 0:01). When all machines are in use, the sound pressure level (SPL) within the room amounts to 90 dB. A carpet is to be placed on the floor to improve the working environment. If the noise level within the room is to be lowered to 80 dB, what should the carpet’s sound absorption coefficient (ac ) be? 2. Jenny tells Tom that a known sound source can be used to measure the reverberation time of a room with the extent of reverberation illustrated in Figure 5.32. According to her, the room’s reverberation time can be calculated by turning on the sound source within the room and observing sound pressure in the normal state after a while. (a) How will Jenny calculate reverberation time? What would be the reverberation time of this room found by Jenny? In this case, the sound source that she uses is one whose sound pressure level is observed to be 45 dB at a 5 m distance from the free field; the sound pressure that Jenny has measured within the room is 64 dB. (b) The owner of this room wants its reverberation time to be 1 sec. Jenny wants to reach the reverberation time that the owner hopes for by placing a carpet on the floor. For this purpose, what should the sound absorption coefficient of the carpet be? (Assume that the original walls are covered with identical sound absorption materials.)
Figure 5.32
The geometry of a room
3. After learning about room modes during class, Jenny wants to observe this phenomenon. For this experiment, she decides to generate an excitation using speakers within the room depicted in Figure 5.32 which has rigid walls. In an endeavor to excite as many modes as possible, which location would have been chosen by Jenny? How would she explain the reason for her choice to Tom? 4. When designing a cubic room as depicted in Figure 5.33, the walls need to be processed with the right sound absorbing materials and other wall materials so that its reverberation time can be set as required. Suppose this room is sized as follows: H ¼ 3 m, L ¼ 20 m, and W ¼ 12 m; three types of material can be applied to the ceiling and four walls (with the floor excluded), and their respective sound absorption coefficients are: a1 ¼ 0.10, a2 ¼ 0.15, and a3 ¼ 0.20.
Sound Propagation
336
Figure 5.33
The geometry of a room
5. The floor needs to undergo total reflection and the same material should be used in each pair of walls which face each other. Note that all three types of material should be applied. (a) What would be the maximum reverberation time for this room and the arrangement of sound-absorbing materials to ensure maximum reverberation time? (b) What would be minimum reverberation time and the arrangement of sound absorbing materials to ensure minimum reverberation time? Wall A is the HL plane (Plane 2), Wall B the HW plane (Plane 2) and the ceiling the LW plane. 6. Suppose a sound source generating a frequency of 1000 Hz and a sound wave of 1 mW in a free space is located in the middle of an infinitely long square duct of cross-section 100 200 mm. (a) How many propagating sound wave modes are there? (b) Design the length of the duct in the sectional direction for propagating plane waves only (the width and height of the section should be identical). 7. An infinitely long tube with a cross-sectional area of A is linked to another infinitely long tube with a cross-sectional area of 10A. If the noise of 95 dB enters via the left-hand tube (Figure 5.34), what would be the sound pressure level (SPL) of the noise passing through the right-hand tube?
Figure 5.34 Simple expansion chamber
8. The frequency characteristics depicted in Figure 5.35 have been obtained from an air handling unit at a hotel. State the most suitable method for reducing this noise and explain the reasons. If three peak elements are to be reduced by 10 dB using a simple divergent tube, derive the area ratio and length.
Acoustics in a Closed Space
Figure 5.35
337
Sound pressure level of a hotel air handling unit
9. Suppose there is a resonator with the noise spectrum as depicted in Figure 5.36 and that we seek to apply one divergent tube and one Helmholtz resonator to reduce the resonator’s noise. Design a silencer that can reduce noise levels to the maximum. (It is assumed that the section of the divergent tube is very small so plane waves are propagated at the frequency of interest and that the resonator is so small in size that it can be approximated by a 1-DOF vibratory system. In the case of the resonator, it is assumed that its maximum transmission loss at the frequency area cannot exceed 50 dB due to limits in its crosssectional area and that the sound absorption effects of the resonator are so minimal that they are negligible.)
Figure 5.36
Sound pressure level of a noise source
10. Consider a silencer consisting of a divergent tube and a resonator. If its frequencies decreased by 10 dB in the noise spectrum depicted in Figure 5.36, what would be the entire noise level (i.e., sum of noise elements in individual frequencies)? (Note that log2 ¼ 0.30, log3 ¼ 0.48, and log5 ¼ 0.70.)
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338
Figure 5.37
Four types of Helmholtz resonator
11. If A-weighting is applied to the noise spectrum discussed in exercise 8, will the resultant noise level be higher or lower than obtained previously? 12. List the four types of Helmholtz resonator depicted in Figure 5.37 in ascending order of resonance frequency. (The diameter of the neck, D, is identical for all of the four resonators, and L ¼ 2D.)
Index absorption coefficient, 142, 279, 313 acoustic compliance, 294 acoustic energy density, 275, 285 acoustic holography, 201 acoustic potential energy, 79 acoustic pressure, 70–71 acoustic wave equation, 74, 96 acoustically large space, 273–274, 280 acoustically small space, 273–274, 292 admittance, 157 acoustic admittance, 328 amplitude of the quadrupole, 123 area of an open window, 278, 284 audible frequency range, 89
critical angle, 149 cut-off frequency, 311 decibel, 90 density, 71 diffraction, 177 diffuse field, 274, 278, 283 dipole, 119 dipole-moment amplitude vector, 120 direct field, 281 directional factor, 215 directivity factor, 214 directivity index, 192, 216 dispersion relation, 6, 19, 41 duct acoustics, 302
baffled piston, 188 basic frequency scale, 95 beam pattern, 216 blocked frequency, 137, 139 blocked pressure, 139 boundary condition, 9–10 Dirichlet boundary condition, 98, 223, 287 Neumann boundary condition, 98, 222, 287 pressure release boundary condition, 99, 102 rigid-wall boundary condition, 99, 102 boundary element method, 208 boundary value problem, 99 breathing sphere, 178
edge effect, 154 effective length, 293, 295, see also end correction factor eigenfunction, 99 end correction factor, 295 equal-loudness contour, 91 equation of state, 114 equivalent absorption area, 316 Euler equation, 72, 113 evanescent wave, 149, 153, 200 expansion chamber-based silencer, 305, 307, 336
characteristic decay time, 278–279, 285 conservation of mass, 72, 110 conservation of momentum, 111
far field, 105, 180 flexural wave, 42 Fourier integral, 34 Fourier series, 34
Sound Propagation: An Impedance Based Approach Yang-Hann Kim © 2010 John Wiley & Sons (Asia) Pte Ltd. ISBN: 978-0-470-82583-9
340
Fourier-Rayleigh method, 327 Fresnel number, 211 Green’s function, 62, 100, 296, 320 harmonic wave, 7 Helmholtz frequency, 156 Helmholtz resonator, 292, 323–324 Huygens’s principle, 107 impedance, 10, 13 acoustic impedance, 180 cavity impedance, 297 characteristic impedance, 13, 17, 129 driving point impedance, 17, 22 fluid loading impedance, 142 mechanical impedance, 19, 42 partition impedance, 142, 152 radiation impedance, 106, 139, 181, 193 reactive radiation impedance, 296 specific impedance, 10 surface impedance, 325 impedance mismatch, 13, 129, 203 impedance tube, 313 incident wave, 10, 129 inhomogeneous Helmholtz equation, 319 inhomogeneous wave equation, 62 intensity acoustic intensity, 80, 284 active intensity, 80, 84, 117 complex intensity, 85, 116 instantaneous active intensity, 84 instantaneous intensity, 116 instantaneous reactive intensity, 84 mean active intensity, 125 reactive intensity, 84, 117 reference intensity, 95 intensity transmission coefficient, see power transmission coefficient kinetic energy per unit volume, 79 Kirchhoff-Helmholtz integral equation, 220, 223, 227 limp wall, 129, 134 locally reacting, 134 locally reacting surface, 149, 156, 288 longitudinal wave, 37–38, 76
Index
mass law, 134, 143 mean intensity, 183 measurement standards, 95 modal coefficient, 289 modal density, 291 mode function, 289, 319 monopole, 118 near field, 105, 180 normal incidence, 129, 160 normalized radiation power, 183 oblique impedance, 152 oblique incidence, 144 octave, 90 one-third octave (1/3 octave), 90 particle velocity, 73, 83, 118 perpendicular incidence, see normal incidence phase speed, see speed of propagation power reflection coefficient, 133 power transmission coefficient, 133, 136, 162 pressure source, 188 propagation constant, see wave number quadrupole, 122 radiation, 177 radiation circle, 199–200 radiation impedance method, 331 radiation power, 183 radiation pressure, 139, 170 radius of reverberation, 286 reference pressure, 90 reflected sound field, 276 reflected wave, 10, 130 reflection coefficient, 131 reflection in layers, 159 reflection of an finite plate, 153 reflection of an infinite plate, 149 refraction, 203 resonance radial frequency, 294 reverberant sound field, 276–277, 281 reverberation formula, 318 reverberation period, 274, 280 T60, 279–280, 317 reverberation room, 316–317
341
Index
Sabine’s theory, 274, 277, 280 scattering, 177 semi-infinite barrier, 208 shadow zone, 211 silencer, 305, 308 dissipative silencer, 308 reactive muffler, 308 simple divergent tube, 305–307 slit, 230 rectangular slit, 207, 234 round slit, 232 Snell’s law, 144, 168 sound decay curve, 317 sound insulation material, 138 sound pressure level, 90 speed of energy propagation, 24 speed of propagation, 4, 7, 24 speed of sound, 7, 74, 114 spreading index, 192 standing wave, 3 standing wave ratio, 314 static pressure, 71 structure of a ear, 87–89 total energy, 79 transfer function, 316 transmission coefficient, 131
transmission in layers, 159 transmission loss, 136, 143, 306 transmission of an finite plate, 153 transmission of an infinite plate, 149 transmitted wave, 10, 130 transverse wave, 76 trembling sphere, 178 velocity potential, 116, 178 velocity reflection coefficient, 132 velocity source, 188 velocity transmission coefficient, 132 vibrating plate, 196 vibration of a string, 1 wave blocking, 304, 312 wave diagram, 4 wave equation, 7, 77 wave equation of a membrane, 49 wave equation of a string, 16 wave number, 6, 41 wave propagation speed, see speed of propagation wave tunneling, 304, 312 wavelength, 6 Webster’s horn equation, 299, 310 Wiener-Hopf technique, 245
