Эротические рассказы

Vibroacoustic Simulation. Alexander PeifferЧитать онлайн книгу.

Vibroacoustic Simulation - Alexander Peiffer


Скачать книгу
(2.137)

      The first term on the right-hand side is the volume integral over all sources fq(r) in the volume. So given a known source distribution we can calculate the according sound field. The two terms in the surface integral take care of the boundary condition. The pressure gradient in the first can be converted into the normal velocity using (2.35). The second surface integral allows establishing the correct surface impedance. Equation (2.137) is called the constant frequency version of the .

      2.7.2 Rayleigh integral

       upper G left-parenthesis bold r comma bold r 0 right-parenthesis equals StartFraction 1 Over 4 pi l EndFraction e Superscript minus j k l Baseline plus StartFraction 1 Over 4 pi l Superscript prime Baseline EndFraction e Superscript minus j k l Super Superscript prime Superscript Baseline with l equals StartAbsoluteValue bold r minus bold r 0 EndAbsoluteValue l Superscript prime Baseline equals StartAbsoluteValue bold r minus bold r 0 prime EndAbsoluteValue (2.138)

      Figure 2.10 Half space in front of a rigid wall. Source: Alexander Peiffer.

      We enter this version of the Green’s function in Equation (2.137) and we get

       StartLayout 1st Row 1st Column bold-italic p left-parenthesis bold r right-parenthesis 2nd Column equals integral Underscript upper V Endscripts upper G left-parenthesis bold r 0 comma bold r right-parenthesis f Subscript q Baseline left-parenthesis bold r 0 right-parenthesis d bold r 0 2nd Row 1st Column Blank 2nd Column integral Subscript negative normal infinity Superscript normal infinity Baseline integral Subscript negative normal infinity Superscript normal infinity Baseline upper G left-parenthesis bold r 0 comma bold r right-parenthesis StartFraction partial-differential bold-italic p left-parenthesis bold r 0 right-parenthesis Over partial-differential z EndFraction minus bold-italic p left-parenthesis bold r 0 right-parenthesis StartFraction partial-differential upper G left-parenthesis bold r 0 comma bold r right-parenthesis Over partial-differential z EndFraction d x 0 d y 0 period EndLayout (2.139)

      We assume a source-free half space so fq(r)=0, and due to the mirror source symmetry ∂G(r0,r)∂z=0 is also true. By clever selection of the Green’s function we fulfilled the boundary condition automatically. For the surface integral the contributions from the half sphere with infinite radius are supposed to be zero. From Equation (2.35) the first expression can be converted into an expression for the surface velocity vz. Performing the limit process z0→0 we get

       upper G left-parenthesis bold r 0 comma bold r right-parenthesis equals StartFraction 2 Over 4 pi l EndFraction e Superscript minus j k l Baseline l equals StartRoot z squared plus left-parenthesis x minus x 0 right-parenthesis squared plus left-parenthesis y minus y 0 right-parenthesis squared EndRoot (2.140)

      and with this Green’s function we can derive the Rayleigh integral that allows the calculation of infinite half space sound fields excited by a rigid vibrating plane with arbitrary velocity distribution vz(x0,y0).

      2.7.3 Piston in a Wall

      A cylindrical loudspeaker in a wall can be modelled by a piston of radius R vibrating with velocity vz located in a rigid wall. For convenience the surface integral will be expressed in cylindrical coordinates r0 and φ0. The receiver coordinates are given as spherical coordinates r and ϑ(Figure 2.11). Without loss of generality the azimuthal angle φ is set to zero.

       StartLayout 1st Row 1st Column bold-italic p left-parenthesis bold r comma theta right-parenthesis 2nd Column equals integral Subscript 0 Superscript 2 pi Baseline integral Subscript 0 Superscript normal infinity Baseline StartFraction j omega rho 0 Over 2 pi l EndFraction e Superscript minus j k l Baseline bold-italic v Subscript z Baseline left-parenthesis r 0 right-parenthesis r 0 d r 0 d phi 0 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 2 pi Baseline integral Subscript 0 Superscript upper R Baseline StartFraction j omega rho 0 Over 2 pi l EndFraction e Superscript minus j k l Baseline bold-italic v Subscript z Baseline r 0 d r 0 d phi 0 EndLayout (2.142)

      In the far field approximation we assume l≈r and get

       l equals r plus r 0 sine theta cosine phi 0 period (2.143)

      So, the approximate result is

       bold-italic p left-parenthesis r comma theta right-parenthesis equals StartFraction j omega rho 0 Over 2 pi r bold-italic v Subscript z Baseline EndFraction integral Subscript 0 Superscript 2 pi Baseline integral Subscript 0 Superscript upper R Baseline e Superscript j k r 0 sine theta cosine phi 0 Baseline d phi 0 r 0 d r 0 (2.144)

      The integral is the Bessel function of first order

       bold-italic p left-parenthesis r comma theta right-parenthesis equals StartFraction j omega rho 0 Over 2 pi r bold-italic v Subscript z Baseline EndFraction left-parenthesis StartFraction 2 upper J 1 left-parenthesis k upper R sine theta right-parenthesis Over k upper R sine theta EndFraction right-parenthesis period (2.145)


Скачать книгу
Яндекс.Метрика