Acoustics

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15.16. Frequency response

A simplified on-axis frequency response of the electrostatic loudspeaker using linear approximations is shown in Fig. 15.18, although the full response from the analogous circuit of Fig. 15.16 together with Eq. (15.92) is given by

Figure 15.18 Frequency response using linear approximations. Solid curve is for no delay and dashed curve is for constant directivity (virtual oscillating sphere), which is achieved using concentric ring electrodes fed from tappings along a delay line, as described in Sections 15.4–15.6 .

\tilde{p} (r) = j ε_{0} k a^{2} ς \cdot \frac{E_{P}}{d} \cdot \frac{{\tilde{e}}_{in}}{2 d} \cdot \frac{e^{- j k r}}{2 r},

$\tilde{p} (r) = j ε_{0} k a^{2} ς \cdot \frac{E_{P}}{d} \cdot \frac{{\tilde{e}}_{in}}{2 d} \cdot \frac{e^{- j k r}}{2 r},$

(15.93)

where ζ is the force factor, which is given by

ς = \frac{S_{D} {\tilde{p}}_{0}}{{\tilde{f}}_{D}} = \frac{2 S_{D}^{2} Z_{A R}}{2 S_{D}^{2} Z_{A R} + Z_{M D}} .

$ς = \frac{S_{D} {\tilde{p}}_{0}}{{\tilde{f}}_{D}} = \frac{2 S_{D}^{2} Z_{A R}}{2 S_{D}^{2} Z_{A R} + Z_{M D}} .$

(15.94)

The acoustic radiation impedance is given by

Z_{A R} = {(\frac{1}{j ω M_{A R}} + \frac{1}{R_{A R} + \frac{1}{j ω C_{A R}}})}^{- 1},

$Z_{A R} = {(\frac{1}{j ω M_{A R}} + \frac{1}{R_{A R} + \frac{1}{j ω C_{A R}}})}^{- 1},$

(15.95)

and the mechanical membrane impedance by

Z_{M D} = j ω M_{M T} + R_{M S} + \frac{1}{j ω} (\frac{1}{C_{M D}^{'}} - \frac{β^{2}}{C_{E}^{'}}) .

$Z_{M D} = j ω M_{M T} + R_{M S} + \frac{1}{j ω} (\frac{1}{C_{M D}^{'}} - \frac{β^{2}}{C_{E}^{'}}) .$

(15.96)

To ensure the displacement does not exceed the gap width d, it is useful to calculate the peak displacement, which from Eq. (15.75) is given by

η_{peak} = \frac{α_{0}}{2 J_{1} (α_{0})} \sqrt{2} {\tilde{η}}_{a v} = 2.316 \sqrt{2} {\tilde{η}}_{a v},

$η_{peak} = \frac{α_{0}}{2 J_{1} (α_{0})} \sqrt{2} {\tilde{η}}_{a v} = 2.316 \sqrt{2} {\tilde{η}}_{a v},$

(15.97)

where the average displacement is given by

{\tilde{η}}_{a v} = \frac{{\tilde{u}}_{D}}{j ω (Z_{M D} + 2 S_{D}^{2} Z_{A R})} .

${\tilde{η}}_{a v} = \frac{{\tilde{u}}_{D}}{j ω (Z_{M D} + 2 S_{D}^{2} Z_{A R})} .$

(15.98)

Now let us simplify the circuit of Fig. 15.16 over specific frequency ranges to gain a better understanding of how the frequency response is determined.

The mechanical impedance is plotted in Fig. 15.19, also using linear approximations, which can be divided into five regions as follows:

1. Membrane stiffness control, f < f ₀. In the region f < f ₀, the membrane mass M _MD and stator resistance R _MS have negligible effect and the radiation load is almost a pure mass. Hence, after omitting the electrical part and referring the negative capacitance and acoustical elements to the mechanical side, the circuit of Fig. 15.16 simplifies to that of Fig. 15.20. This is a simple second-order high-pass filter, but when combined with the increasing amount of cancellation between the rear and front radiation with decreasing frequency, it gives rise to the third-order slope shown in Fig. 15.18.
2. Stator resistance control, f ≈ f ₀. At the fundamental membrane resonance frequency f ₀, the reactance of the combined positive and negative compliances in Fig. 15.20 cancels that of the radiation mass, so that in Fig. 15.16 we are left with just the stator resistance R _MS, as shown in Fig. 15.21.
Figure 15.19 Plot of mechanical impedance Z _M versus frequency f for the analogous circuit shown in Fig. 15.16 using linear approximations.
Figure 15.20 Simplified analogous circuit for mechanical impedance when f < f ₀.
Ignoring the mass of the membrane, the resonance frequency is given by
$f_{0} = \frac{1}{2 π S_{D}} \sqrt{\frac{1}{2 M_{A R}} (\frac{1}{C_{M D}} - \frac{1}{C_{M E}})}$ $f_{0} = \frac{1}{2 π S_{D}} \sqrt{\frac{1}{2 M_{A R}} (\frac{1}{C_{M D}} - \frac{1}{C_{M E}})}$

(15.99)
with a Q factor of
$Q = \frac{2 π f_{0} S_{D}^{2} 2 M_{A R}}{R_{M S}} = \frac{S_{D}}{R_{M S}} \sqrt{2 M_{A R} (\frac{1}{C_{M D}} - \frac{1}{C_{M E}})}$ $Q = \frac{2 π f_{0} S_{D}^{2} 2 M_{A R}}{R_{M S}} = \frac{S_{D}}{R_{M S}} \sqrt{2 M_{A R} (\frac{1}{C_{M D}} - \frac{1}{C_{M E}})}$

(15.100)
where M _AR, C _ME, and C _MD are given by Eqs. (15.89), (15.71), and (15.80), respectively. If we set the tension according to Eq. (15.86) so that − C _ME = −1/(2πT), then
Figure 15.21 Simplified analogous circuit for mechanical impedance when f ≈ f ₀.
$f_{0} = \frac{1}{8 a} \sqrt{\frac{5 (α_{0}^{2} - 2) T}{π a ρ_{0}}} = \frac{1}{3.26 a} \sqrt{\frac{T}{a ρ_{0}}}$ $f_{0} = \frac{1}{8 a} \sqrt{\frac{5 (α_{0}^{2} - 2) T}{π a ρ_{0}}} = \frac{1}{3.26 a} \sqrt{\frac{T}{a ρ_{0}}}$

(15.101)
$Q = \frac{4 a}{R_{M S}} \sqrt{\frac{π (α_{0}^{2} - 2) a ρ_{0} T}{5}} = \frac{6.17 a \sqrt{a ρ_{0} T}}{R_{M S}}$ $Q = \frac{4 a}{R_{M S}} \sqrt{\frac{π (α_{0}^{2} - 2) a ρ_{0} T}{5}} = \frac{6.17 a \sqrt{a ρ_{0} T}}{R_{M S}}$

(15.102)
To prevent excessive membrane excursion at the resonance frequency yet maintain an optimum bass response, the ideal value of Q is in the range 0.7–1.0. Hence, we choose the dust screen to have a specific acoustic resistance of
$R_{S} = \frac{R_{M S}}{S_{D}} = \frac{1}{Q} \sqrt{\frac{2 M_{A R} (C_{M E} - C_{M D})}{C_{M E} C_{M D}}} = \frac{2}{Q} \sqrt{\frac{ρ_{0} T}{a}}$ $R_{S} = \frac{R_{M S}}{S_{D}} = \frac{1}{Q} \sqrt{\frac{2 M_{A R} (C_{M E} - C_{M D})}{C_{M E} C_{M D}}} = \frac{2}{Q} \sqrt{\frac{ρ_{0} T}{a}}$

(15.103)
Note that when we turn off the polarizing voltage, both f ₀ and Q increase by 24% because C _ME = 2.89C _MD.
3. Radiation mass control, f ₀ < f < f ₁ . In the region f ₀ < f < f ₁, the membrane impedance is small compared to the radiation load, which is still mainly mass, although the radiation resistance is increasing with frequency. Note that the radiation mass is much greater than the mechanical mass of the membrane. Hence, the circuit of Fig. 15.16 simplifies to that of Fig. 15.22.
Here the driving pressure ${\tilde{p}}_{0}$ ${\tilde{p}}_{0}$ is constant, but the wavelength is still larger than the half-circumference of the membrane, so that the directivity pattern is a reasonably constant figure-of-eight; and we still have an increasing amount of cancellation between the rear and front waves with decreasing frequency. Hence, we have a first-order slope in Fig. 15.18 and Eq. (15.20) for the on-axis pressure is valid in this region. Notice that the effect of C _AR in Fig. 15.16 is to make the net radiation resistance proportional to f ⁴ in this region. Combined with the predominantly massive radiation impedance (due to M _AR), this causes the velocity ${\tilde{u}}_{D}$ ${\tilde{u}}_{D}$ to be inversely proportional to f. Hence the radiated power is proportional to f ², i.e., 6 dB/octave. The radiated power could only be leveled by mounting the membrane in an infinite baffle or an extremely large enclosure, in which case C _AR would vanish from Fig. 15.16.
Figure 15.22 Simplified analogous circuit for mechanical impedance when f ₀ < f < f ₁.
4. Radiation resistance control, f ₁ < f < f ₂ . In the region f ₁ < f < f ₂, the wavelength is smaller than the half-circumference, but membrane impedance is still small compared to the radiation load; and the radiation load has now become resistive so that the radiated power is constant. However, the on-axis pressure in Fig. 15.18 continues to rise at a rate of 6 dB/octave because of the narrowing directivity pattern without a delay, according to Eq. (15.20). However, if a delay line is used to imitate an oscillating sphere, the on-axis pressure is constant as given by Eq. (15.19). The simplified circuit is shown in Fig. 15.23.
Only in this region, where the membrane velocity is constant and the radiation load resistive, is the total radiated power constant. The frequency f ₁ is given by
$f_{1} = \frac{c}{π a}$ $f_{1} = \frac{c}{π a}$

(15.104)
The radiation resistance effectively damps the diaphragm modes, whereas below f ₁, we rely on the viscous flow resistance of a dust screen.
Figure 15.23 Simplified analogous circuit for mechanical impedance when f ₁ < f < f ₂.
Figure 15.24 Simplified analogous circuit for mechanical impedance when f > f ₂.
5. Membrane and stator mass control, f > f ₂ . In the region f > f ₂, the combined mass reactance of the membrane and stator perforations is greater than the radiation resistance so that we have the circuit shown in Fig. 15.24. Here, the radiated power decreases with increasing frequency. Hence, without a delay, the on-axis pressure response flattens off. As the directivity pattern continues to shrink, the radiated power becomes increasingly on-axis concentrated. Otherwise, if a delay line is used to keep the pattern constant, the on-axis pressure falls at a rate of 6 dB/octave with increasing frequency.
The frequency f ₂ is given by
$f_{2} = S_{D}^{2} R_{A R} / (π M_{M T})$ $f_{2} = S_{D}^{2} R_{A R} / (π M_{M T})$

(15.105)
where M _MT is the total moving mass of the membrane and stator perforations
$M_{M T} = M_{M D} + 2 M_{M S}$ $M_{M T} = M_{M D} + 2 M_{M S}$

(15.106)
and R _AR is the acoustic radiation resistance given by Eq. (15.91). At high frequencies, the membrane moves almost uniformly as a piston, unlike at very low frequencies where the displacement is almost parabolic. Hence, the effective mass of the membrane is
$M_{M D} = S_{D} ρ_{D} h,$ $M_{M D} = S_{D} ρ_{D} h,$

(15.107)
where ρ _D and h are the density and thickness of the membrane respectively. From Eq. (4.26), the mass of the air in the stator perforations is
$M_{M S} = S_{D} ρ_{0} (t + 0.85 a_{h} (1 - 1.28 \sqrt{ϕ_{h}})) / ϕ_{h},$ $M_{M S} = S_{D} ρ_{0} (t + 0.85 a_{h} (1 - 1.28 \sqrt{ϕ_{h}})) / ϕ_{h},$

(15.108)
where t is the thickness of the stator, a _h is the radius of the holes, and ϕ _h is the porosity, that is, ratio of open area to total area. The end correction factor of 0.85 is for the outside of the stator only. On the inside, the distance d between the stator and membrane is usually so small that there will be little in the way of extra mass extending beyond the hole entrance. Notice that the end correction factor depends on the square root of the porosity. For a triangular array of holes of radius a _h and hole pitch b _h, the porosity is given by
$ϕ_{h} = \frac{2 π a_{h}^{2}}{\sqrt{3} b_{h}^{2}} .$ $ϕ_{h} = \frac{2 π a_{h}^{2}}{\sqrt{3} b_{h}^{2}} .$

(15.109)
Hence,
$f_{2} = \frac{ρ_{0} c}{π (ρ_{D} h + 2 ρ_{0} (t + 0.85 a_{h} (1 - 1.2 \sqrt{ϕ_{h}})) / ϕ_{h})} .$ $f_{2} = \frac{ρ_{0} c}{π (ρ_{D} h + 2 ρ_{0} (t + 0.85 a_{h} (1 - 1.2 \sqrt{ϕ_{h}})) / ϕ_{h})} .$

(15.110)
Notice how this cut-off frequency is independent of the membrane area. This is a fairly simple stator model and a more comprehensive one can be found in Ref. [9].