Electroacoustics is using the analogous circuit to model the acoustical behavior including acoustic mass, acoustic resistance and acoustic compliance. The impedance type of analogy is the preferred analogy for acoustical circuits. The sound pressure is analogous to voltage in electrical circuits. The volume velocity is analogous to current.
Acoustic Resistance
An acoustic resistance is associated with dissipative losses that occur when there is viscous flow of air through a fine mesh screen or through a capillary tube. Fig.
3(a) is an example that illustrates a fine mesh screen with a volume velocity U flowing through it. The pressure difference across the screen is given byp= p1−p2, where p1 is the pressure on the side that U enters and p2 is the pressure on the side that U exits. The pressure difference is related to the volume velocity through the screen by
1 2 A
p= p − p = R U (22) where RA is the acoustic resistance of the screen.
The analogous circuit is shown in Fig. 3(b)
Acoustic Compliance
A short closed-end tube or the cavity with volume of air inside has acoustic input impedance that is modeled as compliance. Acoustic compliance is a parameter that is associated with any volume of air that is compressed by an applied force without an acceleration of its center gravity. In other words, compression without acceleration identifies an acoustic compliance.
According to Newton’s law, force is equaled to mass multiplied by acceleration.
When the acceleration of a volume of air is zero, an applied force causes the volume to be compressed without a displacement of its center of gravity.
Figure 4(a) is an example that illustrates a piston of area S is shown in one wall of the enclosure and an enclosed volume of air is distributed in the enclosure. We consider the piston to have zero mass and assume that it slides with zero friction.
When a force f is applied to the piston, it moves and compresses the air. Denote the displacement of the piston by x and its velocity by u. When the air is compressed, a restoring force is generated against the piston, which can be written f =k xM , where kM is the spring constant. The mechanical compliance is defined as the reciprocal of the spring constant. Thus we can write
1
M M
f x udt
C C
= =
∫
(23)The equation is one that involves mechanical variables and next be converted to one that involves the acoustic pressure p and the volume velocityU . By Eqs. (10), (11) and Eqs. (23) can be derived as where CA is the acoustic compliance of the air in the volume which is given by
2
A M
C =S C (25)
Integration in the time domain corresponds to a division jω for sinusoidal phasor variable. It follows from Eqs. 21 1
M A
p Udt Udt
S C C
=
∫
=∫
that the phasorpressure is related to the phasor volume velocity by
A
Acoustical impedanceZA, which varies inversely with jω is like a capacitor.
The analogous circuit is shown in Fig. 4(b). The figure shows one side of the capacitor connected to ground. This is because the pressure in a volume of air is measured with respect to zero pressure which is analogous to zero or ground voltage.
The acoustic compliance of the volume of air is given by the expression derived for the plane wave tube. It is
A short open-ended tube or a duct has impedance that can be modeled as an acoustic mass. All volume of air that is accelerated without being compressed acts an acoustic mass. In other words, acceleration without compression identifies an acoustic mass.
Fig. 5(a) is an example of the cylindrical tube of air having a length l and cross-section S. The mechanical mass of the air in the tube isMM =ρSl . If the air is moved with a velocityu, a force f is required that is given by f MM du
= dt.
The volume velocity of the air through the tube is U =Su and the pressure difference between the two ends isp p1 p2 f
= − = S . It follows from these relations that the pressure difference p can be related to the volume velocity U as follows:
1 2 2
A differentiation in the time domain corresponds to a multiplication by jω for Acoustical impedanceZ , which is proportional to jA ω is like an inductor.
The analogous circuit is shown in Fig. 5(b).
Radiation impedance of a baffled rigid piston
Radiation impedance can be easily explained by an example of the diaphragm vibration. When the diaphragm is vibrating, the medium reacts against the motion of the diaphragm. The phenomenon of this can be described as there is impedance between the diaphragm and the medium. The impedance is called the radiation impedance.
The detail of the theory of radiation impedance is clearly described by Beranek [7].
The analogous circuit of the radiation impedance for the piston mounted in an infinite baffle is shown in Fig. 6. The acoustical radiation impedance for a piston in an infinite baffle can be approximately over the whole frequency range by the analogous circuit. The parameters of the analogous values are given by
1 2
where ρ is the density of air, c is the sound speed in the air, a is the radius of the circular piston.
Radiation impedance on a piston in a tube
The flat circular piston in an infinite baffle that is analyzed in the preceding section is commonly used to model the diaphragm of a direct-radiator loudspeaker when the enclosure is installed in an wall or against a wall. If a loudspeaker is operated away from a wall, the acoustic impedance on its diaphragm changes. It is not possible to exactly model the acoustic radiation impedance of this case. An approximate model that is often used is the flat circular piston in a tube. The geometry for this model is shown in Fig. 7.
The analogous circuit for the piston in a long tube is the same from as that for the piston in an infinite baffle; only the element values are different. The analogous circuit is given in Fig. 6. The parameters of the analogous values are given by
1 the acoustic resistance elements in the acoustical system. They are modeled as an acoustic mass in series with the resistance. Fig. 8 is a simple diagram illustrates the geometry of a perforated sheet. If the holes in the sheet have centers that are spaced
more than one diameter apart and the radius a of the holes satisfies the The parameter µ is the kinematic coefficient of viscosity. For air at 20o C and 0.76m Hg,µ ; 1.56 10× −5m2/s.
A narrow slit is another acoustic element which can be modeled as an acoustic mass in series with the resistance. Fig. 9 is a simple diagram illustrates the geometry of such a slit. If the height t of the slit in meters satisfies the inequalityt<0.003 / f , the acoustic impedance of the slit is given by
3
Lumped Parameter Oscillator Model of a Duct
When sound propagates inside of a rigid duct, a resonance phenomenon occurs.
If the wavelength of the traveling wave is much larger than the diameter of the duct, the resonance properties will be governed by the pipe length and end conditions.
When the wavelength becomes equal to, or smaller than the diameter of the pipe, two
and three-dimensional standing waves can occur. By matching the impedance of the loudspeaker and the duct at the end conditions, the resonant points can be calculated.
First, consider a duct that is rigidly terminated at one end and is excited by a flat massless piston at the other. Fig. 10 shows a flat circular piston in one end of a circular tube of cross-sectionS, length l , and internal volumeV = lS .
Assuming that the piston is used to drive only low frequency content so that it produces a constant volume velocity, only plane waves will be produced. The pressure will take the form: The impedance seen by the source is given by
cos( ) ( ) sin( )
If the terminating impedance at Z = l is omitted so that the duct radiates into free air, it can be shown that approximation to the pressure at the end of the duct
isp l( ) ; 0. The impedance ( )
We seek an approximation of the form A 1 1
Z (0) A ( A )
From Fig. 11, matching the resonance frequency and the slop of the two curves can do this.
Thus the acoustic mass of the first resonant frequency can be lumped as 0.5 times acoustic mass of duct.
The acoustic compliance of the first resonant frequency can be lumped as 0.2 times acoustic compliance of duct.
The analogous circuit of the oscillator model of a duct is shown in Fig. 12.
Transmission Line Model of a Duct
In this section, the theory of electromagnetic about transmission line is discussed.
The transmission line can be represented by a simple equivalent circuit which is shown in Fig. 13.
where R, resistance per unit length, in Ω/ m; L, inductance per unit length, in H m/ ; G, conductance per unit length, in S m/ ; C, capacitance per unit length, in F m/ .
For solving the circuit diagram, the following ordinary differential equations for phasors ( )V x and ( )I x : These two equations are time-harmonic transmission-line equation.
The two equation (54) and (55) above can be combined to solve for V x and ( )( ) I x . Two equations can be derived as:
2 and imaginary parts, α andβ, are the attenuation constant and phase constant of the line.
The ratio of the voltage and the current at any x-direction for an infinite line is independent of x and is called the characteristic impedance of the line.
R j L R j L For the case of the Lossless line (R=0, G=0), there are three special significance listed below:
For acoustic system, phase velocity equals sound velocity.
Therefore According to Eqs. (66) and (67), analysis is similar with Eqs. (43) and (44).
From the three special significance, L and C can be solved by the characteristic impedance of the duct is c
S
ρ , and the phase velocity of the duct is c.
Thus, the two equations below can be combined to solve and the values of L and C can be got.
The detail of the transmission line can be found [8].
Two-Port Model of a Duct
Two port networks are another type of model which is widely used in analysis of the circuit diagram. They are used to model transistors, transformers, and other circuit blocks. In this section, the duct is modeled by a two-port network in which voltage is analogous to pressure and current is analogous to volume velocity.
The geometry of the duct is shown in Fig. 10. Two equations can be formulated as
From the above two equations, Eqs.46 and Eqs.47, the transmission matrix can be formulated as
For the convenience of the analysis of the circuit diagram, the impedance matrix is adopted. And the two equations can be defined as
1 11 1 12 2
p =z U +z U
2 21 1 22 2
p =z U +z U
Rewriting the two equations above in matrix form, we obtain P = ΖU
From [9], the conversion between the transmission and the impedance matrix can be easily solved.
After the transformation, the impedance matrix can be formulated as:
1 1
An equivalent circuit of a two-port model is shown in Fig. 14.
2. Simulation and experiment of results
In this section, the experiment results will be used to analyze miniature speaker and compared with the simulation results. Something that affects the performance of miniature speaker can be clearly realized by experiment results. A real example of mobile phone is presented and the structure of the mobile phone that causes some effects on the miniature speaker will be discussed. The theory of electroacoustics is used to simulate the conditions that the miniature speaker placed in mobile phone.