声学基础课件(许肖梅)fundamentals of acoustics 07-4共24页PPT资料
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声学基础课件(许肖梅)fundamentals of acoustics 07-9-26页精选文档
VE0 120u212p0c22
The instantaneous particle speed and acoustic pressure are functions of both position and time, and consequently the instantaneous energy density is not constant throughout the fluid.
3.5 Pressure and Intensity Level
• The average power produced by a person talking in an ordinary conversational tone is about 10-5 watt, or 100 ergs per sec.
• The range of power that can be produced by the voice varies from about 1 erg per sec for very weak speech to about 104ergs per sec for loud speech.
3.4.1 Sound Energy Density
The energy transported by acoustic waves through a fluid medium is of two forms:
• The kinetic energy of the moving particles ; • The potential energy of the compressed
The total acoustic energy of the volume element is:
The instantaneous particle speed and acoustic pressure are functions of both position and time, and consequently the instantaneous energy density is not constant throughout the fluid.
3.5 Pressure and Intensity Level
• The average power produced by a person talking in an ordinary conversational tone is about 10-5 watt, or 100 ergs per sec.
• The range of power that can be produced by the voice varies from about 1 erg per sec for very weak speech to about 104ergs per sec for loud speech.
3.4.1 Sound Energy Density
The energy transported by acoustic waves through a fluid medium is of two forms:
• The kinetic energy of the moving particles ; • The potential energy of the compressed
The total acoustic energy of the volume element is:
声学基础课件(许肖梅)fundamentals of acoustics 07-6-PPT课件
Transverse Waves on a String Let us consider a long, uniform string whose total mass is m , and length is L. What is important for the local behavior of each part is the linear mass density:
• A flexible string under tension provides the easiest example for visualizing how waves work and developing physical concepts and techniques for their study. • The vibrating string is interesting both for its own sake (as a source of sound on a guitar or violin) and as a model for the motion of other systems. • We study free motion of a string. The procedures we use will apply in our later study of other kinds of waves.
• Fig. A isolates an infinitesimal element of the string with equilibrium position x and equilibrium length dx. • When the string is at rest, the tensions at x and at x+dx are precisely equal in magnitude and opposite in direction, making zero total force.
FUNDAMENTALS OF ACOUSTICS(10) 声学基础(英文版教学课件)
• The simplest solutions to the wave
equation (3-4) are those that depend
on only one of the three spatial
coordinates.
2t2pc22p
(34)
We may as well call that one x. the equation (3-4)reduces to follow equation
the negative x direction
(x,t)Aej(tkx)
Now acoustic pressure
p 0 c 0 u j0 c 0A e j( t k x ) j0 c 0
Therefore the acoustic pressure legs the particle displacement by 900
uumcos(tkx)
0um 2co 2( stk)x
Average energy density
1
T
dt
T0
120um 2 1202m 2p0ce202
Acoustic intensity
I 1
T
pudt
T0
1
T
T 0
pm
cos(t
kx)
pm
0c0
cos(t
kx)dt
pm2
20c0
j(tx)
j(tx)
pA1e c0 A2e c0
pA 1 ei( t k)xA 2 ei( t k)x
• Where the wave number k is defined by
k 2 c0
A1,A2 are two arbitrary constants
《声学基础知识》课件
《声学基础知识》PPT课件
让我们一起探索声学的奥秘吧。从声学基础概述开始，深入了解声音的产生 机制、声音的特性和参数，以及声学波动的基本概念。
声学基础概述
声学是研究声音在空气、固体和液体中的传播和变化的学科。它涵盖了声音的起源、传播和感知等方面的内容。
声音的产生机制
声音的产生涉及物体振动，从声源传递到介质中形成声波。声波通过空气、固体或液体的震动传递，最终被我 们的耳朵接收。
声音的特性和参数
声音具有许多特性和参数，包括频率、振幅、声压级和声色。这些特性决定 了声音的音调、响度和音质。
声学波动的本概念
声学波动是指声音在空气、固体或液体介质中传播的过程。了解波动的基本概念可以帮助我们理解声音的行为 和传播规律。
声场的传播和测量
声场是声波在空间中的分布情况。了解声场的传播和测量方法有助于我们优 化声音的传递和改善声学环境。
声学信号的处理和分析
声学信号的处理和分析可以帮助我们理解和改善声音的质量。通过采用数字信号处理等技术，我们可以对声音 进行精确的控制和调整。
声学应用的案例研究
通过案例研究，我们可以了解声学在不同领域的应用，包括音乐演奏、建筑 设计、噪声控制等。这些案例可以帮助我们更好地理解声学的实际应用。
让我们一起探索声学的奥秘吧。从声学基础概述开始，深入了解声音的产生 机制、声音的特性和参数，以及声学波动的基本概念。
声学基础概述
声学是研究声音在空气、固体和液体中的传播和变化的学科。它涵盖了声音的起源、传播和感知等方面的内容。
声音的产生机制
声音的产生涉及物体振动，从声源传递到介质中形成声波。声波通过空气、固体或液体的震动传递，最终被我 们的耳朵接收。
声音的特性和参数
声音具有许多特性和参数，包括频率、振幅、声压级和声色。这些特性决定 了声音的音调、响度和音质。
声学波动的本概念
声学波动是指声音在空气、固体或液体介质中传播的过程。了解波动的基本概念可以帮助我们理解声音的行为 和传播规律。
声场的传播和测量
声场是声波在空间中的分布情况。了解声场的传播和测量方法有助于我们优 化声音的传递和改善声学环境。
声学信号的处理和分析
声学信号的处理和分析可以帮助我们理解和改善声音的质量。通过采用数字信号处理等技术，我们可以对声音 进行精确的控制和调整。
声学应用的案例研究
通过案例研究，我们可以了解声学在不同领域的应用，包括音乐演奏、建筑 设计、噪声控制等。这些案例可以帮助我们更好地理解声学的实际应用。
声学基础课件(许肖梅)fundamentals of acoustics 07-2-38页精选文档
Sound waves (in fluids) are longitudinal waves: the particles move in the direction of the wave motion. Propagation of sound waves involves the transfer of energy through space. While sound waves spread out in all directions from the source, they may be reflected and refracted, scattered and diffracted, interfered and absorbed. A medium is required for the propagation of sound wave, the speed of which depends on the density and temperature of the medium.
The Sound Wave Produces Sound Pressure Changes
The pressure changes produced by a sound wave are known as the sound pressure. Compared with atmospheric pressure (about 100 000 pascals(1Pa=1N/m2) they are very small (between 20 micropascals and 200 pascals) and are superimposed on it
• The acoustical engineer is interested in the fidelity of reproduction of sound • The conversion of mechanical and electrical energy into acoustical energy, • The design of acoustical transducers. • The architect is more interested in the absorption and isolation of sound in buildings, and in controlled reverberation and echo prevention in auditoriums. • The musician likes to know how to obtain rhythmic combinations of tones through vibrations of strings ,air columns.
FUNDAMENTALS OF ACOUSTICS(18) 声学基础(英文版教学课件)
• 3.When kb＜＜ 1
k b sin ＜＜1
2
sin(2 b sin)
R(
)
2 sin( b
sin
)
1
We can obtain a polar plot of pressure( in dB) versus angular displacements as shown in the following Fig., which is the radiation pattern or directivity of this particular arrangement of two point sources.
be the major lobe and the greater the number of side
The simple line array
• Consider a line of N simple sources with
adjacent elements spaced distance b apart, as shown in Fig.
can approximate by assuming all ri are parallel.
• In far field ,ri in amplitude can be replaced
2 sin( b
sin
)
1.When b s in m , b s in m (m 0 ,1 ,2 )
sin1m (m0,1,2,),
b
In these direction, We have two sound waves
of same magnitude and phase , hence R( ) 1
《声学基础》课件
声学与音乐学
声学研究为音乐学提供了 科学基础，有助于理解声 音在音乐中的产生、传播 和感知。
声学与医学
声学应用于医学领域，如 超声波成像、听力研究等， 为医学诊断与治疗提供了 重要工具。
结论
1 声音是什么？
声音是声波的感知，是人类与世界沟通的重要方式。
2 声学在生活中的应用
声学研究为我们提供了许多实用的应用，如语音识别、音乐欣赏、医学诊断等。
声波传播
1
声音的产生和传播方式
声音可以通过声源的振动产生，并在空气中以波的形式传播。了解声音传播的方 式对声学研究至关重要。
2
空气中声波传播的特性
空气中声波的传播速度、衰减和传播路径都受到温度、湿度和空气密度等因素的 影响。
3
物体表面反射和衍射
声波在物体表面上反射和衍射，这些现象会引起声音的反射、散射和聚焦。
《声学基础》PPT课件
# 声学基础 ## 概述 - 声波与声音的区别 - 声学基础概念 - 声学研究领域 ## 声波传播 - 声音的产生和传播方式 - 空气中声波传播的特性 - 物体表面反射和衍射 ## 声音特性 - 频率、波长及周期 - 振幅、声压和声强 - 速度和能量传播 ## 声学应用 - 声学与语音识别 - 声学与音乐学
3 声学的未来发展方向
随着科技的不断进步，声学研究将继续发展并为我们带来更多惊喜与可能。
声音特性
频率、波长及周期
声音的频率决定了它的音高； 波长和周期是描述声音波动特 征的声音的音量；声压和 声强是描述声音强度的指标。
速度和能量传播
声音传播速度的了解有助于研 究声音如何在空间中传递和传 播能量。
声学应用
声学与语音识别
声学在语音识别技术中发 挥着重要作用，帮助计算 机理解和转换人类的声音 信息。
声学基础课件(许肖梅)fundamentals of acoustics 07-18
Radiation of Sound (2)
The Radiation of Point Sources
• Two point sources with same phase • Consider two simple sources of equal
strengths, separated by a distance d, vibrating with the same frequency and same phase. • The pressure at a field point is the sum of the pressures produced by the individual sources.
In general , the larger the extent of the radiator (here we mean the spacing between the sources), the sharper will be the major lobe and the greater the number of side lobes.
• In far field ,ri in amplitude can be replaced with r .
r2 r1 b sin r3 r2 b sin r1 2b sin , rn r1 (N 1)b sin
b
d
p(r, )
A e j(tkr1 ) r
1 e j
e j2
The Directivity Function
R( ) pA ( )
pA (0)
Radiation of sound will be found equal to zero at certain angles from the maximum of the axial direction of the source, which is called the directivity acute angle.
The Radiation of Point Sources
• Two point sources with same phase • Consider two simple sources of equal
strengths, separated by a distance d, vibrating with the same frequency and same phase. • The pressure at a field point is the sum of the pressures produced by the individual sources.
In general , the larger the extent of the radiator (here we mean the spacing between the sources), the sharper will be the major lobe and the greater the number of side lobes.
• In far field ,ri in amplitude can be replaced with r .
r2 r1 b sin r3 r2 b sin r1 2b sin , rn r1 (N 1)b sin
b
d
p(r, )
A e j(tkr1 ) r
1 e j
e j2
The Directivity Function
R( ) pA ( )
pA (0)
Radiation of sound will be found equal to zero at certain angles from the maximum of the axial direction of the source, which is called the directivity acute angle.
声学基础课件(许肖梅)fundamentals of acoustics 07-11.ppt
Reflection, Refraction and Transmission of Plane Waves
When sound waves are traveling through a medium, they may be reflected or refracted, diffracted or scattered, interfered or absorbed. The transmission of sound involves the transfer of acoustic energy through the medium in which sound waves travel.
Transmission From One Fluid to Another: Normal Incidence
• As suggested in Fig. let the plane x=0 be the boundary between fluid I of characteristic acoustic impedance Z1 and fluid II of characteristic acoustic impedance Z2.
The first condition , continuity of pressure , means that there can be no net force on the (massless) plane separating the fluid.
In complex exponential form, the general solution can be written as
j (t x )
j (t x )
p A1e c0 A2e c0
When sound waves are traveling through a medium, they may be reflected or refracted, diffracted or scattered, interfered or absorbed. The transmission of sound involves the transfer of acoustic energy through the medium in which sound waves travel.
Transmission From One Fluid to Another: Normal Incidence
• As suggested in Fig. let the plane x=0 be the boundary between fluid I of characteristic acoustic impedance Z1 and fluid II of characteristic acoustic impedance Z2.
The first condition , continuity of pressure , means that there can be no net force on the (massless) plane separating the fluid.
In complex exponential form, the general solution can be written as
j (t x )
j (t x )
p A1e c0 A2e c0
FUNDAMENTALS OF ACOUSTICS(16) 声学基础(英文版教学课件)
FUNDAMENTALS OF ACOUSTICS (16)
Sound waves travel in the pipe
3. The nx, ny mode
• 1. (0,0) order mode • 2. (nx ny) order mode
For (0,0) order, the pressure is
n x 0n y 0
n x 0n y 0
f fnx ,ny
f fnx,ny
Corresponding high order waves can travel in z direction
• Corresponding high order waves can not
travel in z direction, these waves decay in z direction
x
c0
lx
jt
x
x
cos nxc0 lx
z
cos1, 00
Which means there are not high order
waves travel in the z direction (except nx 0,ny 0)
cos nx c0 nx 1
lx
2lx
so
fc
nxc0 2lx
kr k2 kz2
B=0
uz 0
z=0, z=l
ur 0
r=a
pp0cosm Jm ( c0 rr)cos c0 zzejt
z
(nz
l
)c0
r
(mn
a
)c0
m n is the solutions of
(
dJm dr
Sound waves travel in the pipe
3. The nx, ny mode
• 1. (0,0) order mode • 2. (nx ny) order mode
For (0,0) order, the pressure is
n x 0n y 0
n x 0n y 0
f fnx ,ny
f fnx,ny
Corresponding high order waves can travel in z direction
• Corresponding high order waves can not
travel in z direction, these waves decay in z direction
x
c0
lx
jt
x
x
cos nxc0 lx
z
cos1, 00
Which means there are not high order
waves travel in the z direction (except nx 0,ny 0)
cos nx c0 nx 1
lx
2lx
so
fc
nxc0 2lx
kr k2 kz2
B=0
uz 0
z=0, z=l
ur 0
r=a
pp0cosm Jm ( c0 rr)cos c0 zzejt
z
(nz
l
)c0
r
(mn
a
)c0
m n is the solutions of
(
dJm dr
FUNDAMENTALS OF ACOUSTICS(21) 声学基础(英文版教学课件)
n1
2
2
l1 dsin l22dsin
1
c
dsin22dscin
ln1
2
n1dsin
2
n1
2
n1dsin
c2
(n1) 2
n1dsin
c2
V
M0
p0
sin
n
d
sin
sin
d
sin
V(0)M0p0n
R
V V0
sin
n
nsin
dsin
dsin
Homework
• P279 6-3
6-4
FUNDAMENTALS OF ACOUSTICS (21)
RECEPTION OF ACOUSTIC WAVES
• A loudspeaker is an electronacoustic
transducer which converts electrical energy to acoustical energy. ( Transmitter )
V V0
0
The variation of intensity level with angle is the beam pattern
b,R2,V V220,, 0
R,M M F F0,， 0V V0,, 0
The receiver of the simple line array
• Consider a line of N simpler sources with
• A microphone is also an electroacoustic
transducer, but it converts acoustical energy to electrical energy. ( Receiver )
FUNDAMENTALS OF ACOUSTICS(14) 声学基础(英文版教学课件)
WaIS4r2I
Since
I
1 pm 2
2 0c0
A2
20c0r2
2 A 2 Wa 0c0
The average rate of energy flow through any spherical surface surrounding the origin is independent of the radius of the surface, a conclusion that is consistent with conservation of energy in a lossless medium
function
• N0(kr) is the zero order Neumann function
Introduce Hankel function
H 0 1(z)J0(z)jN 0(z)
H 0 2(z)J0(z)jN 0(z)
For the diverging waves
R(r)AH 0(2)(kr)
• For kr >>1, Za is predominantly resistive,
meaning that energy fed into such waves propagates away from the source, never to return.
• The larger the value of kr, the less curved are
I 1
T
RepReudt
T0
1 T
T 0
pmcos(t
kx)
pm
0c0
cos(tkx)dt
I
1
20c0
p2 m
Since
I
1 pm 2
2 0c0
A2
20c0r2
2 A 2 Wa 0c0
The average rate of energy flow through any spherical surface surrounding the origin is independent of the radius of the surface, a conclusion that is consistent with conservation of energy in a lossless medium
function
• N0(kr) is the zero order Neumann function
Introduce Hankel function
H 0 1(z)J0(z)jN 0(z)
H 0 2(z)J0(z)jN 0(z)
For the diverging waves
R(r)AH 0(2)(kr)
• For kr >>1, Za is predominantly resistive,
meaning that energy fed into such waves propagates away from the source, never to return.
• The larger the value of kr, the less curved are
I 1
T
RepReudt
T0
1 T
T 0
pmcos(t
kx)
pm
0c0
cos(tkx)dt
I
1
20c0
p2 m
声学基础课件(许肖梅)fundamentals of acoustics 07-6-文档资料
Standing Waves • Consider now a string of finite length L. Describing all motions of this string in terms of traveling waves remains possible in principle. • Because of repeated reflections between the two ends, that is usually not the most helpful description. • We find it more convenient to study standing waves.
l
General Solution of The Equation of Motion Equation (2-1) is a second-order, partial differential equation. Its complete solution contains two arbitrary functions. The most general solution is
Component of the tension at the two ends of the element is:
Fx (T sin 1 ) x
Fx dx (T sin 2 ) x dx
dFx (T sin 2 ) xdx (T sin 1 ) x
η is small, We get:
Transverse Motion - The Vibrating String
Vibrations of Extended Systems • In the previous chapter it was assumed that the mass moves as a rigid body so that it could be considered concentrated at a single point. • However, most vibrating bodies are not so simple. A loudspeaker has its mass distributed over its surface so that the cone does move as a unit .
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For a force oscillations, the general solution
x(t)A 0e tcos(0t)F Z m msin(t)
The solution of equation is the sum of two parts: • a transient term • a steady-state term
WM
1 T
T 0
Fm2 cos(t )costdt
Zm
1gFm2 2 Zm
cos
1 2
vm2 Rm
cos Rm Zm
vm Fm Z m
v x m
m
WM 122xm2Rm
Mechanical Resonance In the steady state, the displacement is equal to :
dA F dx Fv dt dt
Substituting the appropriate real expressions for force and speed
v dx Fm cos(t )
dt Zm
F Fm cost
dA Fm2 cos(t)cost
dt Zm
In most situations the average power is more significance than the instantaneous power.
At this frequency the mechanical impedance has its minimum value of Zm=Rm It is also the frequency of maximum speed amplitude.
Note that This frequency does not give the maximum displacement amplitude.
v(t) Fm cos(t)
Zm
The (angular) frequency of mechanical resonance is defined as that at which the mechanical reactance Xm vanishes, this is the frequency at which a driving force will supply maximum power to the oscillator. It was also found to be the frequency of free oscillation of a similar undamped oscillator.
has magnitude:
Zm
Rm2
(m
D
)2
and phase angle:
tg1 X m
Rm
tg
1
m
Rm
D
Energy Relation
The instantaneous power, supplied to the system in the steady state is equal the product of the instantaneous driving force and the resulting instantaneous speed.
x(t) F Z m mc o t s 2 F Z m msi n t ()
Speed gives:
v(t) Fm cos(t)
Zm
If the speed as given by above is plotted as a function of the frequency of a driving force , a curve is obtained as follow
The transient term is obtained by setting F equal to zero. The arbitrary constants are determined by applying the initial conditions to the total solution.
• When
0,
W MW M m ax
WM max
1 2
Fm2 Rm
For the case of a sinusoidal driving force f(t)=Fmcos(ωt) applied to the oscillator at some initial time, the solution of (1-3) is the sum of two parts –a transient term containing two arbitrary constants and a steady-state term which depends of F and ω but does not contain any arbitrary constants.
Zm is called the complex mechanical impedance; Rm is called the mechanical resistance; Xm is called the mechanical reactance
The mechanical impedance Zm nt time interval, The damping term makes this portion of the solution negligible. Leaving only the steady –state term whose angular frequency ω is that of the driving force
If the average power supplied to the system
as given by:
Wm
1 2
vm2 Rm
Therefore
vm Fm Zm
W m1 2F R m m 21m 1 R m D 22 1 2F R m m 21(m R m 0)21 ( 0m D 0)2
x(t)A 0e tcos(0t)F Z m msin(t)
The solution of equation is the sum of two parts: • a transient term • a steady-state term
WM
1 T
T 0
Fm2 cos(t )costdt
Zm
1gFm2 2 Zm
cos
1 2
vm2 Rm
cos Rm Zm
vm Fm Z m
v x m
m
WM 122xm2Rm
Mechanical Resonance In the steady state, the displacement is equal to :
dA F dx Fv dt dt
Substituting the appropriate real expressions for force and speed
v dx Fm cos(t )
dt Zm
F Fm cost
dA Fm2 cos(t)cost
dt Zm
In most situations the average power is more significance than the instantaneous power.
At this frequency the mechanical impedance has its minimum value of Zm=Rm It is also the frequency of maximum speed amplitude.
Note that This frequency does not give the maximum displacement amplitude.
v(t) Fm cos(t)
Zm
The (angular) frequency of mechanical resonance is defined as that at which the mechanical reactance Xm vanishes, this is the frequency at which a driving force will supply maximum power to the oscillator. It was also found to be the frequency of free oscillation of a similar undamped oscillator.
has magnitude:
Zm
Rm2
(m
D
)2
and phase angle:
tg1 X m
Rm
tg
1
m
Rm
D
Energy Relation
The instantaneous power, supplied to the system in the steady state is equal the product of the instantaneous driving force and the resulting instantaneous speed.
x(t) F Z m mc o t s 2 F Z m msi n t ()
Speed gives:
v(t) Fm cos(t)
Zm
If the speed as given by above is plotted as a function of the frequency of a driving force , a curve is obtained as follow
The transient term is obtained by setting F equal to zero. The arbitrary constants are determined by applying the initial conditions to the total solution.
• When
0,
W MW M m ax
WM max
1 2
Fm2 Rm
For the case of a sinusoidal driving force f(t)=Fmcos(ωt) applied to the oscillator at some initial time, the solution of (1-3) is the sum of two parts –a transient term containing two arbitrary constants and a steady-state term which depends of F and ω but does not contain any arbitrary constants.
Zm is called the complex mechanical impedance; Rm is called the mechanical resistance; Xm is called the mechanical reactance
The mechanical impedance Zm nt time interval, The damping term makes this portion of the solution negligible. Leaving only the steady –state term whose angular frequency ω is that of the driving force
If the average power supplied to the system
as given by:
Wm
1 2
vm2 Rm
Therefore
vm Fm Zm
W m1 2F R m m 21m 1 R m D 22 1 2F R m m 21(m R m 0)21 ( 0m D 0)2