2. Theory and Method
2.3 Modeling acoustical systems
for acoustical circuits. The
2.3.1 Acoustic impedance
Acoustic resistance is associated with dissipative losses that occur when there is a viscous flow of air through a fine mesh screen or through a
The capillary tube. Fig. 4(a) illustrates a fine mesh screen with a volume velocity U flowing through it.
1 2
p= p −p , where p is the 1 pressure difference across the screen is given by
pressure on the side that U enters and p is the pr2 side that exits.
The pressure difference is related to
essure on the U the volume velocity through the screen by
1 2 A
p p p R U (25)
where A
= − =
R is the acoustic resistance of the screen. The circuit is shown in Fig. 4(b).
Theoretical formulas for acoustic resistance are generally not available. The values are usually determined by experiments. Table 1 gives the acoustic resistance of typical screens as a function of the area S of the screen, the number of wires in the screen, and the diameter of the wires.
2.3.2 Acoustic compliance
To illustrated an acoustic compliance, consider an enclosed volume of air as illustrated in Fig. 5(a). A piston of area is shown in one wall of the enclosure.
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 of gravity.
When a force
S
f is appli to the piston, it moves and compresses the air. Denote the piston displacement by
ed
x and its velocity by u . When the air is compressed, a restoring force is generated which can be written f =k xM , where k is the spring M constant. (This assumes that the displacement is not too large or the process cannot be modeled with linear equation.) The mechanical compliance is defined as the reciprocal of the spring constant. Thus we can write
1
M
M M
f =k x= x =
∫
udt (26)This equation involves the mech
C C
anical variables f and . We convert it to one that involves acoustic variables
u
This equation defines the acoustic compliance C of the air in the volume. It is A given by
2
A M
C =S C (28)
An integration in the time domain corresponds to a division by jω for phasor variable. It follows from Eqs. (27). That the phas re is rela to the phasor volume velocity by
or pressu ted
The impedance which varies inversely with
U jwC
jω is a capacitor. The analogous circuit is shown in Fig. 5(b). The
to gr
li onnect e ground
e acoustic compliance of the volume of air is given by the expression derived for the plane wave tube. It is
figure shows one side of the capacitor connected ound. This is because the pressure in a volume of air is measured with respect to zero pressure. One node of an acoustic comp ance always c s to th
node. Th
A 2
C c
V
= ρ (30)
2.3.3 Acoustic mass
Any volume of air that is accelerated without being compressed acts as an acoustic mass. Consider the cylindrical tube of air illustrated in Fig. 6 (a) having a length l and cross-section S. The mss of the air in the tube is MM =ρ0Sl. If
the air moved with u, the force required is given by f MM du
= dt. Th
velocity e
volume velocity of the air through the tube is U =Su and the pressure difference between the two ends is p p1 p2 f
A differentiation in the time domain corresponds to a m ltiplication by u jω for sinusoidal phasor variable. If follows from Eqs. (31) that the phasor pressure is related to the phasor volume velocity by p= j M Uω A . Thus the acoustic impedance of the mass is
A A
Z = p = j M (33)
An electrical impedance which is proportional to
U ω
jω
isfy
is an inductor. The analogous circuit is shown in Fig. 6(b). For a tube of air to act as a pure acoustic
e velocity.
. Oth
mass, each particle of air in e tube must move with the sam This is strictly true only if the frequency is low enough erwise, the motion of the air particles must be modeled by a wave equation. An often used criterion that the air in the tube act as a pure acoustic mass is that its length must sat
th
l≤λ8 , where λ is the wavelength.
Radiation imp nce
vibra otio
2.3.4 Radiation impedance of a baffled rigid piston
eda can be easily explained by an example of the diaphragm tion. When the diaphragm is vibrating, the medium reacts against the m n of the diaphragm. The phenomenon of this can be described as there is impedance
between the diaphragm and the medium. The is called the radiation impedance.
impedance
The detail of the theory of radiation impedance is clearly described by Bernek.
The analogous circuit of the radiation impedance for the piston mounted in an infinite baffle is shown in Fig. 7. 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
0 the circuit piston.
only used to model the diaphragm of a direct-radiator loudspeaker when the e
as that for the piston in an infinite b
circuit is given in Fig. 7. Th
a
2.3.5 Radiation impedance on a piston in a tube
The flat circuit piston in an infinite baffle that is analyzed in the preceding section is comm
nclosure is installed in a 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 circuit piston in a tube.
The analogous circuit for the piston in a long tube is the same from
affle; only the element values are different. The analogous e parameters of the analogous values are given by
1
2.3.6 Other acoustic elements A. Perforated sheets
Perforated sheets are often used as an acoustic resistance in application where an acoustic mass in series with the resistance is acceptable. Fig. 8 (a) illustrates the geometry. If the holes in the sheet have centers tat are spaced more than on diameter apart and the radius a of the holes satisfies the inequality 0.01 a 10
f < < f ,
where N is the number of holes. The parameters μ is the kinematic coefficient of viscosity. For air at 20 C° and 0.76 mHg, 1.56 10 m5 2
μ ≈ × − s . This parameter value app oximately as r T1.7
P0 , where perature and is the atmospheric pressure.
A tube having a very small diameter is another example of an acoustic element which exhibits both a resistance and a mass. If the tube radius in meters satisfies
T is the Kelvin tem P 0
a
0.002 a
the inequality
< f , the acoustic impedance is given by
0 corrections. The parameter η is the viscosity coefficient. For air,
5N S satisfies the inequality
f < < f used between the two equations.
A narrow slit also exhibits both acoustic resistance and mass. Fig. 8 (b) shows t of the slit in meters satisfies the inequality
the geometry of such a slit. If the heigh t 0.003
t< f , the acoustic impedance of the slit, neglecting end corrections
for the mass term, is given by
B. Vented-box system
s shown in Fig. 9.
SP with radius aP and length LP. The mechanism of low-frequency enhancement lies in the Helmholtz resonator comprised of the acoustic mass in the vent and the
nce c
istance. ss of
the port and acoustic compliance of the encl
=
The general diagram of a vented-box system i The system primarily consists of an enclosure of volume VAB and a port with a cross-sectional area
acoustic complia in the en losure. More precisely, the vent can be modeled as an acoustic mass and an acoustic res The acoustic resistance, acoustic ma
osure are given by [9]
0
The m pedance obtained using FEA mentioned above is changed into a lumped-parameter model.
0c ρ
echanical im
Therefore, the overall EMA analogous circuit of vented-box is shown in Fig. 1
odel of a duct
0.C. Transmission line m
The vent also can be modeled as a transmission line model. Consider a length of duct, the equivalent-circuit model for the transmission line is the T-circuit shown in Fig. 11. These equations of T-circuit are in transfer matrix form as follows [22], [40].