Acoustics

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

3.10. Transducer impedances

Let us look a little closer at the impedances at the terminals of electromechanical transducers. It has become popular over the years for electrical circuit specialists to express the equations for their circuits in matrix form. The matrix notation is a condensed manner of writing systems of linear equations [10,11]. We shall express the properties of transducers in matrix form for those who are familiar with this concept. An explanation of the various mathematical operations to be performed with matrices is beyond the scope of this book. The student not familiar with matrix theory is advised to deal directly with the simultaneous equations from which the matrix is derived. A knowledge of matrix theory is not necessary, however, for an understanding of any material in this text.

Transmission matrix for an electrical two-port network

As we shall see, transmission matrices are particularly useful because an overall transmission matrix M can be easily obtained by multiplying together the individual transmission matrices for each circuit element. The general two-port network shown in Fig. 3.44 can be represented by the following matrix equation:

$[\begin{matrix} {\tilde{e}}_{i n} \\ {\tilde{i}}_{i n} \end{matrix}] = A \cdot [\begin{matrix} {\tilde{e}}_{o u t} \\ {\tilde{i}}_{o u t} \end{matrix}] = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}_{o u t} \\ {\tilde{i}}_{o u t} \end{matrix}],$

(3.46)

where the transmission parameters are given by

Figure 3.44 Electrical two-port network.

$a_{11} = \frac{{\tilde{e}}_{i n}}{{\tilde{e}}_{o u t}} | {\tilde{i}}_{o u t} = 0,$

(3.47)

$a_{12} = \frac{{\tilde{e}}_{i n}}{{\tilde{i}}_{o u t}} | {\tilde{e}}_{o u t} = 0,$

(3.48)

$a_{21} = \frac{{\tilde{i}}_{i n}}{{\tilde{e}}_{o u t}} | {\tilde{i}}_{o u t} = 0,$

(3.49)

$a_{22} = \frac{{\tilde{i}}_{i n}}{{\tilde{i}}_{o u t}} | {\tilde{e}}_{o u t} = 0,$

(3.50)

In other words:

a ₁₁ is ratio of applied input voltage to output voltage measured with the output terminals open-circuited.
a ₁₂ is ratio of applied input voltage to output current measured with the output terminals short-circuited.
a ₂₁ is ratio of applied input current to output voltage measured with the output terminals open-circuited.
a ₂₂ is ratio of applied input current to output current measured with the output terminals short-circuited.

Transmission matrix for an electromagnetic-mechanical transducer

Let us determine the transmission matrix for the electromagnetic-mechanical transducer of Fig. 3.45. In that circuit Z _E is the electrical impedance measured with the mechanical terminals “blocked,” that is,

$\tilde{u}$

= 0; Z _M is the mechanical impedance of the mechanical elements in the transducer measured with the electrical circuit “open-circuited”; and Z _L is the mechanical impedance of the acoustic load on the diaphragm. The quantity Bl is the product of the flux density times the effective length of the wire cutting the lines of force perpendicularly. The individual transmission matrices for each element can be written from inspection

Figure 3.45 Analogous circuit for an electromagnetic-mechanical transducer. The mechanical side is of the impedance type.

$[\begin{matrix} {\tilde{e}}_{0} \\ {\tilde{i}}_{0} \end{matrix}] = [\begin{matrix} 1 & Z_{E} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = M_{0} \cdot [\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}],$

(3.51)

$[\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = [\begin{matrix} 0 & B l \\ 1 / B l & 0 \end{matrix}] \cdot [\begin{matrix} \tilde{f} \\ \tilde{u} \end{matrix}] = M_{1} \cdot [\begin{matrix} \tilde{f} \\ \tilde{u} \end{matrix}],$

(3.52)

$[\begin{matrix} \tilde{f} \\ \tilde{u} \end{matrix}] = [\begin{matrix} 1 & Z_{M} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{f}}_{L} \\ {\tilde{u}}_{L} \end{matrix}] = M_{2} \cdot [\begin{matrix} {\tilde{f}}_{L} \\ {\tilde{u}}_{L} \end{matrix}],$

(3.53)

$[\begin{matrix} {\tilde{f}}_{L} \\ {\tilde{u}}_{L} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 1 / Z_{L} & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{f}}_{0} \\ {\tilde{u}}_{0} \end{matrix}] = M_{3} \cdot [\begin{matrix} {\tilde{f}}_{0} \\ {\tilde{u}}_{0} \end{matrix}] .$

(3.54)

The overall transmission matrix is then obtained as follows

$\begin{matrix} M = M_{0} \cdot M_{1} \cdot M_{2} \cdot M_{3} \\ = [\begin{matrix} 1 & Z_{E} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 0 & B l \\ 1 / B l & 0 \end{matrix}] \cdot [\begin{matrix} 1 & Z_{M} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 1 & 0 \\ 1 / Z_{L} & 1 \end{matrix}] \\ = [\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}], \end{matrix}$

(3.55)

where

$A_{11} = \frac{B l}{Z_{L}} + \frac{Z_{E}}{B l} (1 + \frac{Z_{M}}{Z_{L}}),$

(3.56)

$A_{12} = B l + \frac{Z_{E} Z_{M}}{B l},$

(3.57)

$A_{21} = \frac{1}{B l} (1 + \frac{Z_{M}}{Z_{L}}),$

(3.58)

$A_{22} = \frac{Z_{M}}{B l} .$

(3.59)

We see from Fig. 3.45 that

${\tilde{u}}_{o} = 0$

so that

$[\begin{matrix} {\tilde{e}}_{0} \\ {\tilde{i}}_{0} \end{matrix}] = [\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}] \cdot [\begin{matrix} {\tilde{f}}_{0} \\ 0 \end{matrix}]$

(3.60)

From this matrix, we can gather everything we need to know about the transducer. For example, the parameters at the interfaces between the circuit elements (voltages, currents, forces and velocities, etc.) can be obtained through a combination of the overall transmission matrix and elemental matrices. Straightaway we obtain the force exerted on the load

${\tilde{f}}_{L} = {\tilde{f}}_{0} = \frac{{\tilde{e}}_{0}}{A_{11}} = \frac{Z_{L} B l {\tilde{e}}_{0}}{Z_{E} (Z_{M} + Z_{L}) + {(B l)}^{2}}$

(3.61)

and hence also the load velocity

${\tilde{u}}_{L} = \frac{{\tilde{e}}_{0}}{A_{11} Z_{L}} = \frac{B l {\tilde{e}}_{0}}{Z_{E} (Z_{M} + Z_{L}) + {(B l)}^{2}}$

(3.62)

The latter is important for evaluating the radiated sound pressure, as will be explained in Chapters 4, 12, and 13. The total electrical impedance Z _ET as viewed from the voltage generator is found to be

$Z_{E T} = \frac{{\tilde{e}}_{0}}{{\tilde{i}}_{0}} = \frac{A_{11} {\tilde{f}}_{0}}{A_{21} {\tilde{f}}_{0}} = Z_{E} + \frac{{(B l)}^{2}}{Z_{M} + Z_{L}}$

(3.63)

The second term on the right-hand side is usually called the motional impedance because if the mechanical side is blocked there is no movement (that is, Z _L → ∞) and then Z _ET = Z _E, which is the static impedance. This equation illustrates a striking fact, viz., that the electromagnetic transducer is an impedance inverter. By an inverter we mean that a mass reactance on the mechanical side becomes a capacitance reactance when referred to the electrical side of the transformer, and vice versa. Similarly, an inductance on the electrical side reflects through the transformer as a mechanical compliance.

Impedance matrix for an electromagnetic-mechanical transducer

Refer to Fig. 3.44. Another type of matrix in common usage is the impedance matrix based on z-parameters:

$[\begin{matrix} {\tilde{e}}_{i n} \\ {\tilde{e}}_{o u t} \end{matrix}] = Z \cdot [\begin{matrix} {\tilde{i}}_{i n} \\ - {\tilde{i}}_{o u t} \end{matrix}] = [\begin{matrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{matrix}] \cdot [\begin{matrix} {\tilde{i}}_{i n} \\ - {\tilde{i}}_{o u t} \end{matrix}],$

(3.64)

where the z-parameters are given by

$z_{11} = \frac{{\tilde{e}}_{i n}}{{\tilde{i}}_{i n}} | {\tilde{i}}_{o u t} = 0,$

(3.65)

$z_{12} = \frac{{\tilde{e}}_{i n}}{- {\tilde{i}}_{o u t}} | {\tilde{i}}_{i n} = 0,$

(3.66)

$z_{21} = \frac{{\tilde{e}}_{o u t}}{{\tilde{i}}_{i n}} | {\tilde{i}}_{o u t} = 0,$

(3.67)

$z_{22} = \frac{{\tilde{e}}_{o u t}}{- {\tilde{i}}_{o u t}} | {\tilde{i}}_{i n} = 0$

(3.68)

In other words,

z ₁₁ is ratio of input voltage to applied input current measured with the output terminals open-circuited.
z ₁₂ is ratio of input voltage to applied output current measured with the input terminals open-circuited.
z ₂₁ is ratio of output voltage to applied input current measured with the output terminals open-circuited.
z ₂₂ is ratio of output voltage to applied output current measured with the input terminals open-circuited.

Comparing Eq. (3.64) with Eq. (3.46), we can solve for the following transmission parameter to z-parameter transformation equations

$z_{11} = a_{11} / a_{21},$

(3.69)

$z_{12} = \det (A) / a_{21},$

(3.70)

$z_{21} = 1 / a_{21},$

(3.71)

$z_{22} = a_{22} / a_{21},$

(3.72)

where

$\det (A) = a_{11} a_{22} - a_{12} a_{21} .$

(3.73)

Many passive networks, especially ones in which no energy is created or lost, have a determinant whose magnitude is unity, in which case the z-parameter matrix is symmetrical about the diagonal. That is, z ₁₂ = z ₂₁. However, we shall see that in the case of an electromagnetic-mechanical transducer, it turns out to be skew symmetrical, that is, with z ₁₂ = − z ₂₁, because det( A ) = −1, which in turn is because of the fact that the current flow in a wire resulting from movement through a magnetic field is in the opposite direction to that producing the same movement (Fleming's generator rule vs. motor rule). The reverse transformation equations are of the same form

$a_{11} = z_{11} / z_{21},$

(3.74)

$a_{12} = \det (Z) / z_{21},$

(3.75)

$a_{21} = 1 / z_{21},$

(3.76)

$a_{22} = z_{22} / z_{21},$

(3.77)

where

$\det (Z) = z_{11} z_{22} - z_{12} z_{21} .$

(3.78)

Applying the transformations of Eqs. (3.69–3.72) to the transmission parameters of Eq. (3.56–3.59), while noting that in this instance

${\tilde{u}}_{0} = 0$

and

${\tilde{f}}_{0} = {\tilde{f}}_{L}$

, yields the following z-parameter impedance matrix:

$[\begin{matrix} {\tilde{e}}_{0} \\ {\tilde{f}}_{L} \end{matrix}] = [\begin{matrix} Z_{E} + \frac{{(B l)}^{2}}{Z_{M} + Z_{L}} & \frac{- B l Z_{L}}{Z_{M} + Z_{L}} \\ \frac{B l Z_{L}}{Z_{M} + Z_{L}} & \frac{Z_{M} Z_{L}}{Z_{M} + Z_{L}} \end{matrix}] \cdot [\begin{matrix} {\tilde{i}}_{0} \\ 0 \end{matrix}],$

(3.79)

Not surprisingly, z ₁₁ = Z _ET as given by Eq. (3.63). If we remove the load impedance by letting Z _L → ∞, we obtain the following simple z-parameter impedance matrix for just the transducer without any external load:

$[\begin{matrix} {\tilde{e}}_{0} \\ {\tilde{f}}_{L} \end{matrix}] = [\begin{matrix} Z_{E} & - B l \\ B l & Z_{M} \end{matrix}] \cdot [\begin{matrix} {\tilde{i}}_{0} \\ - {\tilde{u}}_{L} \end{matrix}]$

(3.80)

Transmission matrix for an electrostatic-mechanical transducer

For the electrostatic-mechanical transducer of the type shown in Fig. 3.46, Z _E is the electrical impedance with the mechanical motion free

$(\tilde{f} = 0)$

$Z_{E}^{'} \equiv Z_{E} + \frac{1}{j ω C_{E}^{'}}$

is the electrical impedance with the mechanical motion blocked (ũ = 0).

Z _L is the mechanical impedance of the acoustical load on the diaphragm.

$Z_{M} \equiv R_{M} + j ω M_{M} + \frac{1}{j ω C_{M}}$

is the mechanical impedance of the mechanical elements in the transducer measured with the electrical terminals short-circuited

$({\tilde{e}}_{1} = 0)$

$Z_{M}^{'}$

is the mechanical impedance of the mechanical elements measured with the electrical terminals open-circuited

$({\tilde{i}}_{1} = 0)$

. It is defined by the same expression as that for Z _M above except that C _M is replaced by

$C_{M}^{'} = (1 - \frac{d_{31}^{2}}{C_{E} C_{M}}) C_{M}$

which is the mechanical compliance in the transducer with

${\tilde{i}}_{1}$

= 0.

The individual transmission matrices for each element can be written from inspection:

(3.81)

$[\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = [\begin{matrix} 1 & 0 \\ j ω C_{M}^{'} & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}_{2} \\ {\tilde{i}}_{2} \end{matrix}] = M_{1} \cdot [\begin{matrix} {\tilde{e}}_{2} \\ {\tilde{i}}_{2} \end{matrix}],$

(3.82)

Figure 3.46 Analogous circuit for an electrostatic-mechanical transducer. The mechanical side is of the impedance type.

$[\begin{matrix} {\tilde{e}}_{2} \\ {\tilde{i}}_{2} \end{matrix}] = [\begin{matrix} C_{M} / d_{31} & 0 \\ 0 & d_{31} / C_{M} \end{matrix}] \cdot [\begin{matrix} \tilde{f} \\ \tilde{u} \end{matrix}] = M_{2} \cdot [\begin{matrix} \tilde{f} \\ \tilde{u} \end{matrix}],$

(3.83)

$[\begin{matrix} \tilde{f} \\ \tilde{u} \end{matrix}] = [\begin{matrix} 1 & Z_{M} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{f}}_{L} \\ {\tilde{u}}_{L} \end{matrix}] = M_{3} \cdot [\begin{matrix} {\tilde{f}}_{L} \\ {\tilde{u}}_{L} \end{matrix}],$

(3.84)

$[\begin{matrix} {\tilde{f}}_{L} \\ {\tilde{u}}_{L} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 1 / Z_{L} & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{f}}_{0} \\ {\tilde{u}}_{0} \end{matrix}] = M_{4} \cdot [\begin{matrix} {\tilde{f}}_{0} \\ {\tilde{u}}_{0} \end{matrix}] .$

(3.85)

Using the relationship

$C_{E}^{'} C_{M} = C_{E} C_{M}^{'}$

from Eqs. (3.36) and (3.37), the overall transmission matrix is then obtained as follows:

$\begin{matrix} M = M_{0} \cdot M_{1} \cdot M_{2} \cdot M_{3} \cdot M_{4} \\ = [\begin{matrix} 1 & Z_{E} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 1 & 0 \\ j ω {C^{'}}_{E} & 1 \end{matrix}] \cdot [\begin{matrix} C_{M} / d_{31} & 0 \\ 0 & d_{31} / C_{M} \end{matrix}] \cdot [\begin{matrix} 1 & Z_{M} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 1 & 0 \\ 1 / Z_{L} & 1 \end{matrix}] \\ = [\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}], \end{matrix}$

(3.86)

where

$A_{11} = \frac{j ω C_{E}^{'} C_{M}}{d_{31}} (1 + \frac{Z_{M}^{'}}{Z_{L}}) (Z_{E}^{'} + \frac{d_{31}^{2}}{ω^{2} C_{E}^{' 2} C_{M}^{2} (Z_{M}^{'} + Z_{L})}),$

(3.87)

$A_{12} = \frac{j ω C_{E}^{'} C_{M}}{d_{31}} Z_{M}^{'} (Z_{E}^{'} + \frac{d_{31}^{2}}{ω^{2} C_{E}^{' 2} C_{M}^{2} Z_{M}^{'}}),$

(3.88)

$A_{21} = \frac{j ω C_{E}^{'} C_{M}}{d_{31}} (1 + \frac{Z_{M}^{'}}{Z_{L}}),$

(3.89)

$A_{22} = \frac{j ω C_{ME}^{'} C_{M}}{d_{31}} Z_{M}^{'} .$

(3.90)

Impedance matrix for an electrostatic-mechanical transducer

Applying the transformations of Eqs. (3.69–3.72) to the transmission parameters of Eqs. (3.87–3.90), while noting that in this instance

${\tilde{u}}_{0} = 0$

and

${\tilde{f}}_{0} = {\tilde{f}}_{L}$

, yields the following z-parameter impedance matrix:

$[\begin{matrix} {\tilde{e}}_{0} \\ {\tilde{f}}_{L} \end{matrix}] = [\begin{matrix} Z_{E}^{'} + {(\frac{d_{31}}{ω C_{E}^{'} C_{M}})}^{2} \frac{1}{Z_{M}^{'} + Z_{L}} & \frac{d_{31}}{j ω C_{E}^{'} C_{M}} \frac{Z_{L}}{Z_{M}^{'} + Z_{L}} \\ \frac{d_{31}}{j ω C_{E}^{'} C_{M}} \frac{Z_{L}}{Z_{M}^{'} + Z_{L}} & \frac{Z_{M}^{'} Z_{L}}{Z_{M}^{'} + Z_{L}} \end{matrix}] \cdot [\begin{matrix} {\tilde{i}}_{0} \\ 0 \end{matrix}],$

(3.91)

If we remove the load impedance by letting Z _L → ∞, we obtain the following simple z-parameter impedance matrix for just the transducer without any external load:

$[\begin{matrix} {\tilde{e}}_{0} \\ {\tilde{f}}_{L} \end{matrix}] = [\begin{matrix} Z_{E}^{'} & \frac{d_{31}}{j ω C_{E}^{'} C_{M}} \\ \frac{d_{31}}{j ω C_{E}^{'} C_{M}} & Z_{M}^{'} \end{matrix}] \cdot [\begin{matrix} {\tilde{i}}_{0} \\ - {\tilde{u}}_{L} \end{matrix}] .$

(3.92)

This matrix is symmetrical about the main diagonal, as for any ordinary electrical passive network. By contrast matrix (3.80) is skew symmetrical because the off-diagonal elements have opposite signs. For transient problems, replace jω by the operator s = d/dt [9].

The impedance matrix for the electrostatic transducer is almost identical in form to that for the electromagnetic transducer, the difference being that the mutual terms have the same sign, as contrasted to opposite signs for the electromagnetic case. This means that while electrostatic transducers are reciprocal, electromagnetic transducers are antireciprocal. For the electrostatic transducer, the total impedance is given from Eq. (3.91) as

$Z_{E T} = z_{11} = Z_{E}^{'} + {(\frac{d_{31}}{ω C_{E}^{'} C_{M}})}^{2} \frac{1}{Z_{M}^{'} + Z_{L}} .$

(3.93)

The first and second terms on the right-hand side are called the static and motional impedances, respectively, as before.

Again we see that the transducer acts as a sort of impedance inverter. An added positive mechanical reactance (+X _M) comes through the transducer as a negative electrical reactance.

Some interesting facts can be illustrated by assuming that we have an electrostatic and an electromagnetic transducer, each stiffness controlled on the mechanical side so that

$Z_{M} + Z_{L} = \frac{1}{j ω C_{M 1}} .$

(3.94)

Substitution of Eqs. (3.94) into (3.63) yields

$Z_{E T} = Z_{E} + j ω (B^{2} l^{2} C_{M 1}) .$

(3.95)

The mechanical compliance C _M appears from the electrical side to be an inductance with a magnitude B ² l ² C _M1. We now substitute Eqs. (3.94) into (3.63) to obtain

$Z_{E T} = z_{11} = Z_{E}^{'} + j {(\frac{d_{31}}{C_{E}^{'} C_{M}})}^{2} \frac{C_{M 1}^{'}}{ω} .$

(3.96)

The mechanical compliance

$C_{M}^{'}$

of this transducer appears from the electrical side to be a negative capacitance (see Fig. 3.37(b)), that is to say,

$C_{M 1}^{'}$

appears to be an inductance with a magnitude that varies inversely with ω ². The effect of this is simply to reduce the value of

$C_{M}^{'}$

. Another way of looking at this is to note from Fig. 3.46 that with R _M = M _M = 0 and Z _L = 1/jω

$C_{M L}^{'}$

, the total compliance is less than

$C_{M}^{'}$

because of the added compliance

$C_{M L}^{'}$

Analogous circuits for the two-port network using z-parameters [12]

The transmission matrix for a two-port network may be conveniently separated into three matrices

$[\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = [\begin{matrix} 1 & z_{11} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 0 & - z_{12} \\ 1 / z_{21} & 0 \end{matrix}] \cdot [\begin{matrix} 1 & z_{22} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}_{2} \\ {\tilde{i}}_{2} \end{matrix}] .$

(3.97)

These matrices multiply together to form a single transmission matrix containing the elements defined in Eqs. (3.74–3.77)

$[\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}_{2} \\ {\tilde{i}}_{2} \end{matrix}] .$

(3.98)

The first matrix of Eq. (3.97) contains only the self-impedance z ₁₁ of the input port, which is shown as a series impedance in Fig. 3.47(a) and (b). Similarly, the third matrix contains only the self-impedance z ₂₂ of the output port, also shown as a series impedance in Fig. 3.47(a) and (b). The second matrix contains the mutual impedances z ₁₂ and z ₂₁ between the input and output, and vice versa

Figure 3.47 Analogous circuit of the two-port network using (a) current-controlled voltage sources (b) voltage-controlled current sources, and (c) all-passive elements for case where z₁₂ = z₂₁.

$[\begin{matrix} {\tilde{e}}^{'}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = [\begin{matrix} 0 & - z_{12} \\ 1 / z_{21} & 0 \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}^{'}_{2} \\ {\tilde{i}}_{2} \end{matrix}] .$

(3.99)

This may be considered as a gyrator in which the polarity of the forward transconductance is reversed. However, it is more intuitive to represent it as a pair of current-controlled voltage sources, as shown in Fig. 3.47(a). Then the relationships of Eq. (3.99) are shown explicitly. Alternatively, we may represent it as a pair of voltage-controlled current sources, as shown in Fig. 3.47(b). If the mutual impedances are equal, that is z ₂₁ = z ₁₂, then we may use the purely passive scheme shown in Fig. 3.47(c). Here, the matrices divide up as follows

$[\begin{matrix} {\tilde{e}}_{1} \\ {\tilde{i}}_{1} \end{matrix}] = [\begin{matrix} 1 & z_{11} - z_{12} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 1 & 0 \\ 1 / z_{12} & 1 \end{matrix}] \cdot [\begin{matrix} 1 & z_{22} - z_{12} \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} {\tilde{e}}_{2} \\ {\tilde{i}}_{2} \end{matrix}] .$

(3.100)

We will use this for modeling a lossy tube in Section. 4.23.

Example 3.9. A moving-coil earphone, which is driven at frequencies above its first resonance frequency, may be represented by the circuit of Fig. 3.43(a). Its mechanical and electrical characteristics are

R _E = 8 Ω
B = 1 T (10⁴ G)
l = $3 / 4$ m
Figure 3.48 Analogous circuits for Example 3.9.
M _MD = 60 mg
Y _MR = jω2.7 × 10⁻³ m/N·s

where Y _MR is the admittance that the diaphragm sees when the earphone is on the ear (because of the stiffness of the air trapped in the ear cavity), M _MD is the mass of the diaphragm, R _E and l are the resistance and the length of wire wound on the voice coil, and B is the flux density cut by the moving coil. Determine the sound pressure level produced at the ear at 1000 Hz when the earphone is operated from a very low impedance amplifier with an output voltage of e _rms = 1 V. Assume that the area of the diaphragm is 1 cm².

Solution. The circuit diagram for the earphone with the element sizes given in SI units is shown in Fig. 3.48(a). Eliminating the transformer gives the circuit of Fig. 3.48(b). Solving, we get

$\begin{matrix} \tilde{u} = {\tilde{f}}_{2} Y_{M R} = (10^{- 4} \tilde{p}) j 6280 (2.7 \times 10^{- 3}) \\ = (j 1.7 \times 10^{- 3}) \tilde{p}, \\ {\tilde{f}}_{D} = j ω M_{M D} \tilde{u} = (- 6.4 \times 10^{- 4}) \tilde{p}, \\ \tilde{f} = {\tilde{f}}_{D} + {\tilde{f}}_{2} = (- 5.4 \times 10^{- 4}) \tilde{p}, \\ 1 \frac{1}{3} \tilde{e} = \tilde{u} + 14.22 \tilde{f} = (j 1.7 \times 10^{- 3} - 7.67 \times 10^{- 3}) \tilde{p}, \end{matrix}$

$| p_{rms} | = \frac{1.33 \times 10^{3}}{\sqrt{{1.7}^{2} + {7.67}^{2}}} \approx 170 Pa,$

Figure 3.49 Combined electrostatic-electromagnetic transducers. (a) Block mechanical diagram of the device. (b) Analogous circuit with impedances on mechanical side. (c) Same as (b), except that the electrical elements are referred to the mechanical side. (d) Because the mechanical part of circuit (c) has zero impedance, (c) simplifies to this form.

$SPL = 20 \log \frac{170}{2 \times 10^{- 5}} = 138.6 dB r e 20 μ Pa .$

Example 3.10. Two transducers, one a piezoelectric crystal and the other a moving coil in a magnetic field, are connected to a mass M _M2 of 6.27 g as shown in Fig. 3.49(a). Determine the total stored mechanical energy in the masses M _M1 and M _M2 at 10 kHz for the following constants:

e _rms = 1 V
R _E = 10 Ω
B = 1 T
l = 20 m
C _E = 1.3 × 10⁻⁹ F
M _M1 = 6.4 × 10⁻³ kg
M _M2 = 6.27 × 10⁻³ kg
C _M = 2 × 10⁻⁸ m/N
d ₃₃ = 10⁻⁹ C/N
ω = 62,800 rad/s

Solution. The transducers are shown schematically in (b) of Fig. 3.49. A further simplification of this diagram is shown in (c). Let us determine the value of Z _M first.

$\begin{matrix} Z_{M 1} = j ω (M_{M 1} + M_{M 2}) - j \frac{1}{C_{M} ω} \\ = j (402 + 394) - j 796 = 0. \end{matrix}$

In other words, the impedance is zero at 10 kHz. Hence, circuit (c) simplifies to that shown in (d). Then the velocity

${\tilde{u}}_{0}$

of the two masses M _M1 and M _M2 is

$\begin{matrix} {\tilde{u}}_{o} = (\frac{B l}{R_{E}} + \frac{C_{E}}{d_{33}}) \frac{R_{E}}{{(B l)}^{2}} \tilde{e} \\ = (\frac{1}{B l} + \frac{R_{E} C_{E}}{{(B l)}^{2} d_{33}}) \tilde{e} \\ = (\frac{1}{20} + \frac{10 \times 1.3}{20^{2}}) \tilde{e} = (82.5 \times 10^{- 3}) \tilde{e} \end{matrix}$

so that u _0rms = 82.5 × 10⁻³ m/s. The total mechanical stored energy in the two masses M _M1 and M _M2 is

$\frac{1}{2} (M_{M 1} + M_{M 2}) u_{0 r m s}^{2} = 0.5 \times (6.4 + 6.27) \times {82.5}^{2} \times 10^{- 9} = 4.31 \times 10^{- 5} W \cdot s .$

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for Acoustics

Create new playlist

Sign In

Sign Up

3.10. Transducer impedances

Transmission matrix for an electrical two-port network

Transmission matrix for an electromagnetic-mechanical transducer

Impedance matrix for an electromagnetic-mechanical transducer

Transmission matrix for an electrostatic-mechanical transducer

Impedance matrix for an electrostatic-mechanical transducer

Analogous circuits for the two-port network using z-parameters [12]

Table of Contents for
Acoustics