Vibroacoustic Simulation. Alexander Peiffer

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e Superscript minus j k sine theta x Baseline plus normal upper Phi Superscript left-parenthesis upper R right-parenthesis Baseline e Superscript minus j k sine theta Super Superscript left-parenthesis upper R right-parenthesis Superscript x Baseline right-parenthesis"/> (2.100)

       StartLayout 1st Row 1st Column bold-italic v Subscript z Baseline left-parenthesis x comma z equals 0 right-parenthesis 2nd Column equals minus StartFraction partial-differential normal upper Phi Over partial-differential z EndFraction 2nd Row 1st Column Blank 2nd Column equals j k left-parenthesis cosine theta normal upper Phi e Superscript minus j k sine theta x Baseline minus cosine theta Superscript left-parenthesis upper R right-parenthesis Baseline normal upper Phi Superscript left-parenthesis upper R right-parenthesis Baseline e Superscript minus j k sine theta Super Superscript left-parenthesis upper R right-parenthesis Superscript x Baseline right-parenthesis EndLayout (2.101)

      We certainly shall not be able to match the impedance z2=p/vz at every surface position unless the arguments of the exponential functions are equal, hence

theta equals theta Superscript left-parenthesis upper R right-parenthesis

      So, we get from the surface impedance condition

       bold-italic z 2 left-parenthesis x comma z equals 0 right-parenthesis equals StartFraction bold-italic p left-parenthesis x comma z equals 0 right-parenthesis Over bold-italic v Subscript z Baseline left-parenthesis x comma z equals 0 right-parenthesis EndFraction equals StartFraction z 0 left-parenthesis normal upper Phi plus normal upper Phi Superscript left-parenthesis upper R right-parenthesis Baseline right-parenthesis Over cosine theta left-parenthesis normal upper Phi minus normal upper Phi Superscript left-parenthesis upper R right-parenthesis Baseline right-parenthesis EndFraction (2.102)

      With z0=ρ0c0 and rearranging the above equation, the reflection factor is given by

      The ratio between irradiated power to reflector power is the squared reflection factor called the.

       StartLayout 1st Row 1st Column r Subscript s Baseline left-parenthesis theta right-parenthesis 2nd Column upper R squared left-parenthesis theta right-parenthesis equals StartFraction left-parenthesis bold-italic z 2 cosine theta minus z 0 right-parenthesis squared Over left-parenthesis bold-italic z 2 cosine theta plus z 0 right-parenthesis squared EndFraction EndLayout (2.104)

       StartLayout 1st Row 1st Column alpha Subscript normal s Baseline left-parenthesis theta right-parenthesis 2nd Column 1 minus r Subscript s Baseline left-parenthesis theta right-parenthesis equals left-parenthesis 1 minus StartAbsoluteValue upper R left-parenthesis theta right-parenthesis EndAbsoluteValue squared right-parenthesis EndLayout (2.105)

      Note that those coefficients are exclusively described by the impedance of fluid and surface and not density or speed of sound. Thus, the impedance is the relevant quantity here.

      2.6 Reflection and Transmission of Plane Waves

      Region 1 of the incoming wave has two wave components, the incoming and the reflected wave, and region 2 the transmitted wave. Thus, both velocity potentials read

       StartLayout 1st Row 1st Column normal upper Phi 1 left-parenthesis bold r 1 right-parenthesis 2nd Column equals normal upper Phi 1 e Superscript minus j bold k 1 bold r 1 Baseline plus normal upper Phi 1 Superscript left-parenthesis upper R right-parenthesis Baseline e Superscript minus j bold k 1 Super Superscript left-parenthesis upper R right-parenthesis Superscript bold r 1 Baseline 2nd Row 1st Column normal upper Phi 2 left-parenthesis bold r 2 right-parenthesis 2nd Column equals normal upper Phi 2 e Superscript minus j bold k 2 bold r 2 EndLayout (2.106)

      Using the given angles as sketched in Figure 2.8 the wavenumber space vector products are given by

       StartLayout 1st Row 1st Column bold k 1 bold r 1 Superscript left-parenthesis upper R right-parenthesis 2nd Column equals k 1 left-parenthesis sine theta 1 Superscript left-parenthesis upper R right-parenthesis Baseline x plus cosine theta 1 Superscript left-parenthesis upper R right-parenthesis Baseline z right-parenthesis EndLayout (2.108)

       StartLayout 1st Row 1st Column bold k 2 bold r 2 2nd Column k 2 left-parenthesis sine theta 2 x minus cosine theta 2 z right-parenthesis equals l a b e l less-than 2.107 greater-than slash l a b e l less-than EndLayout (2.109)

p 1 left-parenthesis x comma z equals 0 right-parenthesis equals p 2 left-parenthesis x comma z equals 0 right-parenthesis

      gives

       rho 1 normal upper Phi 1 e Superscript j k 1 sine theta 1 x Baseline plus rho 1 normal upper Phi 1 Superscript left-parenthesis upper R right-parenthesis Baseline e Superscript j k 1 sine theta 1 Super Superscript left-parenthesis upper R right-parenthesis Superscript x Baseline equals rho 2 normal upper Phi 2 e Superscript j k 2 sine theta 2 x (2.110)

      A solution for any x is only possible if the arguments of the exponential functions are equal.

       k 1 sine theta 1 equals k 1 sine theta 1 Superscript left-parenthesis upper R right-parenthesis (2.111)

      So

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