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The radiation impedance matrix

From section 3.6 the radiation impedance matrix for a cylindrical pipe terminated in an infinite baffle is:
$\displaystyle Z_{nm} = \frac{\rho c}{S} \int\limits_0^{\frac{\pi}{2}} \sin{\phi} D_n(\sin{\phi}) D_m(\sin{\phi}) d \phi$      
$\displaystyle + \frac{i\rho c}{S} \int\limits_0^\infty \cosh{\xi} D_n(\cosh{\xi}) D_m(\cosh{\xi}) d \xi$     (4.1)

where
\begin{displaymath} D_n(\tau) = \frac{-\sqrt{2} \tau J_1(\tau k R)} {(\frac{\gamma_n}{kR})^2 - \tau^2}. \end{displaymath} (4.2)


Jonathan Kemp 2003-03-24