Acoustics

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Development of Low-Frequency Pressure Response

It can be shown [2] that for $k a < 1 / 2$ , a loudspeaker behaves as a spherical source. Here, a represents the radius of the loudspeaker. For a 15? diameter loudspeaker in air, this low frequency limit is about 150 Hz. For smaller loudspeakers, this limit increases. This limit dominates the limit which ignores $L E$ , and is consistent with the limit that models $Z R A D$ by $M A 1$ .

Within this limit, the loudspeaker emits a volume velocity $U 0$ , as determined in the previous section. For a simple spherical source with volume velocity $U 0$ , the far-field pressure is given by [1]:

$p(r) \simeq j\omega\rho_0 U_0 \frac{e^{-jkr}}{4\pi r}$

It is possible to simply let $r = 1$ for this analysis without loss of generality because distance is only a function of the surroundings, not the loudspeaker. Also, because the transfer function magnitude is of primary interest, the exponential term, which has a unity magnitude, is omitted. Hence, the pressure response of the system is given by [1]:

$\frac{p}{V_{IN}} = \frac{\rho_0s}{4\pi}\frac{U_0}{V_{IN}} = \frac{\rho_0Bl}{4\pi S_DR_EM_AS}H(s)$

Where $H (s) = s G (s)$ . In the following sections, design methods will focus on $| H (s) | 2$ rather than $H (s)$ , which is given by:

$|H(s)|^2 = \frac{\Omega^8}{\Omega^8 + \left(a^2_3 - 2a_2\right)\Omega^6 + \left(a^2_2 + 2 - 2a_1a_3\right)\Omega^4 + \left(a^2_1 - 2a_2\right)\Omega^2 + 1}$

$\Omega = \frac{\omega}{\omega_0}$

This also implicitly ignores the constants in front of $| H (s) |$ since they simply scale the response and do not affect the shape of the frequency response curve.