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For spherical symmetry, $r\tilde{p}$ satisfies the one-dimensional wave equation, so an outgoing solution has the form $A(t-r/c)/r$. To relate $A$ to the source, use a velocity potential $\Phi$ with $u_r={\partial }_r\Phi$ and $\tilde{p}=-\rho {\partial }_t\Phi$. The outgoing monopole potentialhas $4\pi r^2\rho u_r\to q(t)$ as $r\to 0$, fixing its mass-flux normalization. It follows thatFor an element with displacement $f_0{e}^{i\omega t}$, its volume outflow is $i\omega f_0{h}^2{e}^{i\omega t}$ and mass outflow is $\rho$ times this. Thus $q^{\prime }=-\rho {\omega }^2{f}_0{h}^2{e}^{i\omega t}$. A rigid wall imposes zero normal velocity away from the source. The method of images uses a source of the same sign on the other side; a source on the plane and its image coincide, doubling the full-space monopole field. Equivalently its mass flux spreads over a hemisphere rather than a sphere. HenceThe small patch relative to wavelength justifies the point source, and the small displacement justifies linear acoustics.

Superposition over the aperture gives the Rayleigh integral for baffled acoustic radiationFor $r\gg L$, $R=r-(y{y}^{\prime }+z{z}^{\prime })/r+O(L^2/r)$ and $R^{-1}=r^{-1}[1+O(L/r)]$. The additional condition $r\gg \omega L^2/c$ makes the omitted phase small, so with the printed Fourier transform conventionThe minus signs in $m,n$ arise because the outgoing phase contributes $+i\omega (y{y}^{\prime }+z{z}^{\prime })/(rc)$.

For the rectangle the two integrals factor, givingwhere $\mathrm{sinc}u=\sin u/u$ and its zero-argument limit is one. If $k=\omega /c$, $kb\ll 1$ implies $|bn|\le kb\ll 1$, so the $z$-direction factor is nearly one throughout the radiation hemisphere. With $ka$ of order one the $y$-direction factor varies appreciably: the long dimension gives stronger directivity in that angular direction, while the short dimension radiates broadly. The forward amplitude is proportional to the area $4ab$. Nulls have $kay/r$ equal to nonzero integer multiples of $\pi$ when those values are physically accessible; order-one $ka$ does not automatically imply that the first null lies in the hemisphere.