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Introduction to Acoustics

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k y<br />

k<br />

k<br />

Radiation circle<br />

k x<br />

Acoustic Holography 26.2 Acoustic Holography: Measurement, Prediction and Analysis 1081<br />

Exponentially decaying<br />

Prediction<br />

plane (z)<br />

Fig. 26.6 Propagating and exponentially waves in acoustic holography<br />

P(x, y, z h; f )<br />

P(x, y, z; f )<br />

Space domain Wave number domain<br />

P(k x, ky, z h; f )<br />

Fourier<br />

transform<br />

Inverse<br />

fourier<br />

transform<br />

Propagation<br />

P(k x, ky, z; f )<br />

« «<br />

�e ik z (z–z h )<br />

Fig. 26.7 The data-processing procedure for acoustic<br />

holography<br />

linear inhomogeneous wave equation, or the inhomogeneous<br />

Helmholtz equation in the frequency domain.<br />

That is,<br />

∇ 2 G(x |xh; f ) + k 2 G(x |xh; f )<br />

=−δ(x− xh) . (26.3)<br />

Therefore, we can select a Green’s function in such a way<br />

that we can eliminate one of the terms on the right-hand<br />

side of (26.1); (26.2) is one such case.<br />

To see what essentially happens in the prediction<br />

process, let us consider (26.2) when the measurement<br />

and prediction plane are both planar. Planar acoustic<br />

holography assumes that the sound field is free from<br />

reflection (Fig. 26.5); then we can write (26.2)as<br />

Propagating<br />

Hologram<br />

plane (z h)<br />

�<br />

P(x, y, z; f ) =<br />

Source<br />

plane (z s)<br />

Sh<br />

P(xh, yh, zh; f )<br />

× KPP(x − xh,y − yh,z − zh; f )dSh ,<br />

(26.4)<br />

KPP(x, y, z; f ) = 1 z<br />

(1 − ikr)exp(ikr) , (26.5)<br />

2π r3 �<br />

where r = x2 + y2 + z2 ,<br />

2π f<br />

k = ,<br />

c<br />

x = (x, y, z) ,<br />

xh = (xh, yh, zh) .<br />

KPP can be readily obtained by using two free-field<br />

Green’s functions that are located at zh and −zh, so<br />

that it satisfies the Dirichlet boundary condition.<br />

This is a convolution integral, and therefore we can<br />

write this in the wave-number domain as<br />

where<br />

ˆP(kx, ky, z; f ) =<br />

ˆP(kx, ky, zh; f )exp[ikz(z − zh)] , (26.6)<br />

ˆP(kx, ky, z; f )<br />

=<br />

kz =<br />

�∞<br />

�∞<br />

−∞ −∞<br />

P(x, y, z; f )e −i(kx x+ky y) dx dy , (26.7)<br />

�<br />

k 2 − k 2 x − k2 y .<br />

Part H 26.2

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