Acoustics

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13.20. Time reversal

Imagine that we place a planar array of microphones between a stage and an audience and then make a recording of a performance by musicians on the stage. We assume that the extent of the array is large enough to be considered infinite and that the microphones are small enough and far enough apart not to disturb the sound field produced. We also assume that the ratio of the smallest wavelength to the microphone pitch is large enough not to introduce spatial aliasing. If we play back the recording through an infinite array of either omnidirectional (monopole) or bidirectional (dipole) loudspeakers in place of the microphones (making the same assumptions as with the microphones), we will faithfully reproduce the concert when listening from the audience side of the array. However, if we listen from the stage side, we will not hear what was heard by the musicians, but that which was heard by the audience. Reproducing a sound field with sources in it is not so easy. Near-field acoustical holography, as described in the previous section, only provides a way to calculate the sound field on the stage side. In other words, we only have a virtual field, not a real one.

In time reversal, we play the recording backwards. Although this obviously makes no sense for music, it does have an interesting effect in the case of signals such as a continuous tone or an impulse. During the recording, an impulse arrives at the middle microphone first and then the ones on either side of that and works its way progressively toward the outermost microphones. During normal playback, the impulse leaves the loudspeakers in the same order as it arrived at the corresponding microphones, but if the recording is played backwards the sound emanates in reverse order starting from the outermost loudspeakers and finishing from the middle one. The effect of this is to focus the sound toward the source from which it was originally produced during the recording. If the sound was produced by a point source [see Eq. (4.71)], will the original source be faithfully reproduced?

To answer that question, let us now consider a simpler example. Suppose now that we have a spherical wave converging toward a point. If there is no source or sink at the focal point, the spherical wave will pass through the focal point and reemerge as a diverging wave. From Eq. (2.134) we have

\tilde{p} (r) = {\tilde{A}}_{+} \frac{e^{- j k r}}{r} + {\tilde{A}}_{-} \frac{e^{j k r}}{r},

$\tilde{p} (r) = {\tilde{A}}_{+} \frac{e^{- j k r}}{r} + {\tilde{A}}_{-} \frac{e^{j k r}}{r},$

(13.362)

where

{\tilde{A}}_{-}

${\tilde{A}}_{-}$

is the amplitude of the sound pressure in the incoming wave at unit distance from the center of the sphere and

{\tilde{A}}_{+}

${\tilde{A}}_{+}$

is the same for the outgoing wave. To meet the boundary condition of pressure continuity, or zero pressure gradient, at the center, we set

{\tilde{A}}_{+} = - {\tilde{A}}_{-}

${\tilde{A}}_{+} = - {\tilde{A}}_{-}$

so that

\tilde{p} (r) = 2 j {\tilde{A}}_{-} \frac{\sin k r}{r} .

$\tilde{p} (r) = 2 j {\tilde{A}}_{-} \frac{\sin k r}{r} .$

(13.362)

The incoming wave is reflected back out again as if there were a rigid termination point at the center. To absorb it we have to place a point sink

- {\tilde{A}}_{+} e^{- j k r} / r

$- {\tilde{A}}_{+} e^{- j k r} / r$

at the center. It has been pointed out that, when time reversing the waves produced by dropping a pebble into a pond, a pebble must rise out of the water at the end of the sequence [44]. In the case of a plane wave, it is relatively straight forward to absorb it using a ρ ₀ c termination as was shown in Section 2.4. In the case of a converging spherical wave, the characteristic impedance is only approximately ρ ₀ c at a distance of several wavelengths from the center, as demonstrated by Eq. (2.144). At closer distances, the impedance is mainly massive. Therefore, a sphere whose surface impedance is ρ ₀ c can only be used as an acoustic sink to absorb an incoming spherical wave if it has a diameter of several wavelengths.

If we now return to the problem of the planar loudspeaker array, the same principle applies. In the absence of any acoustic sink, the waves converge on the point from which the sound originally emanated and reemerge on the opposite side. As they pass through the focal point, there is a transition from the positive phase angle of the converging wave to the negative phase angle of the diverging one [45]. Hence the imaginary part of the pressure field is zero in the plane of the focal point where the converging and diverging waves meet. In this way, the singularity of the original point source is removed and we are left with an approximation of it.

Problem 13.1. In Section 13.17, we derived the far-field directivity of a rectangular piston in an infinite baffle using the Bridge product theorem. Show that Eq. (13.318) may also be derived using the far-field Green's function of Eqs. (13.45) and (13.46) in the Rayleigh integral of Eq. (13.6), taking into account the double-strength source as illustrated in Fig. 13.4 for a circular piston. The rectangular piston lies in the xy plane with its center located at the origin of the rectangular coordinate system and oscillates with a velocity

{\tilde{u}}_{0}

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

in the z-direction.

Hint: Use the integral

\int_{- l_{x} / 2}^{l_{x} / 2} e^{j k x x_{0} / R} d x_{0} = {[\frac{e^{j k x x_{0} / R}}{j k x / R}]}_{- l_{x} / 2}^{l_{x} / 2} = \frac{\sin (k l_{x} x / 2 R)}{k x / 2 R}

$\int_{- l_{x} / 2}^{l_{x} / 2} e^{j k x x_{0} / R} d x_{0} = {[\frac{e^{j k x x_{0} / R}}{j k x / R}]}_{- l_{x} / 2}^{l_{x} / 2} = \frac{\sin (k l_{x} x / 2 R)}{k x / 2 R}$

together with sin θ ₁ = x/R and sin θ ₂ = y/R. Show that the far-field pressure is of the form

\tilde{p} (x, y, z) = j k l_{x} l_{y} ρ_{0} c {\tilde{u}}_{0} \frac{e^{- j k R}}{2 π R} D (θ_{x}, θ_{y}) .

$\tilde{p} (x, y, z) = j k l_{x} l_{y} ρ_{0} c {\tilde{u}}_{0} \frac{e^{- j k R}}{2 π R} D (θ_{x}, θ_{y}) .$

Problem 13.2. In Section 13.17, we derived the radiation impedance of a rectangular piston in an infinite baffle from the far-field directivity using Bouwkamp's impedance theorem. Show that Eqs. (13.323) and (13.324) may also be derived using the Fourier Green's function of Eqs. (13.33) and (13.34) in the Rayleigh integral of Eq. (13.6), taking into account the double-strength source as illustrated in Fig. 13.4 for a circular piston. The rectangular piston lies in the xy plane with its center located at the origin of the rectangular coordinate system and oscillates with a velocity

{\tilde{u}}_{0}

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

in the z-direction.

Hint: Because the sinusoidal function is odd, the imaginary parts of the complex exponents cancel out over the positive and negative values of k _x and k _y in the infinite integrals. After evaluating the surface integrals to derive the expression for the specific impedance Z _s, use polar coordinates, where k _x = kt cos ϕ, k _y = kt sin ϕ, and dk _x dk _y = k ² tdtdϕ, to reduce the double infinite integral to a single infinite one over t and a single finite one over ϕ. Then separate the infinite integral into real and imaginary parts to produce the resistance R _s and reactance X _s, respectively.

Problem 13.3. Use the Bridge product theorem to derive the far-field directivity pattern of four pistons of radius a in an infinite baffle regularly spaced in a straight line at intervals of d, where the directivity of an array of N point sources is given by Eq. (4.85) and that of a circular piston in an infinite baffle is given by Eq. (13.102). Use the identity (sin 4x)/(sin x) = 4 cos x cos 2x and do not forget to include the ϕ dependency in Eq. (4.85) as was done in Section 13.8.

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13.20. Time reversal

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Acoustics