0
$\begingroup$

I am trying to get a large sound pressure level by arranging individual horns, but I am curious how to design the arrangement structure to obtain a synthetic sound field with sharp directivity in the axial direction given the synthetic sound pressure level and the sound pressure level of the individual horns.

$\endgroup$

1 Answer 1

2
$\begingroup$

In the simplest case, where there is no gain shading (which effectively constitutes amplitude weighting, possibly frequency dependent) the response of an array consisting of omnidirectional sources (or receivers, due to reciprocity principle) is given by

$$Y(\psi) = \frac{1}{N} \frac{\sin \left( N \frac{\psi}{2}\right)}{\sin \left( \frac{\psi}{2}\right)}$$

with $N$ the number of elements in the array. Replacing $\psi$ with $\psi = \frac{2 \pi}{\lambda} d \cos (\theta)$ where $k$ is the wavenumber given by $k = \frac{\omega}{c} = \frac{2 \pi}{\lambda}$, with $\lambda$ being the wavelength, $c$ the speed of sound and $d$ the distance between the elements, you can get the response of the array for various angles of interest $\theta$ for any frequency of interest corresponding to a wavelength $\lambda$.

More information on this topic and a thorough explanation on arrays can be found in Optimum Array Processing: Part IV of Detection, Estimation, and Modulation Theory by Harry L. Van Trees (this is where I also got the information provided here). Of course there other resources that could provide some information. This is just a suggestion which had proved to be quite useful to me in the past.

$\endgroup$
3
  • 1
    $\begingroup$ thank for your excellent reply.. $\endgroup$
    – midas
    Jan 24 at 10:57
  • 1
    $\begingroup$ @ZaellixA, this is great, +1, glad to have you on board here. -NN $\endgroup$ Jan 24 at 19:06
  • $\begingroup$ Thanks a lot NN, really appreciated getting that from someone like you. I am extremely glad to be on the same board with people like you. $\endgroup$
    – ZaellixA
    Jan 24 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.