http://www.k-wave.org/forum/topic/simulating-3d-disk-transducer-with-thickness Support for the k-Wave MATLAB toolbox en-US Wed, 13 May 2026 04:30:40 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/simulating-3d-disk-transducer-with-thickness#post-7945 Thu, 26 Nov 2020 11:57:06 +0000 Bradley Treeby 7945@http://www.k-wave.org/forum/ <p>Hi raffaele84,</p> <p>(1) Yes you will need to run a 3D simulation. There are some examples for disc sources in 3D as part of the new <a href="http://www.k-wave.org/forum/topic/alpha-version-of-kwavearray-off-grid-sources">kWaveArray class</a>.</p> <p>(2) One way would be to define a disc in 2D, and then copy this into slabs of a 3D matrix to give you a cylinder of the required height.</p> <p>(3) This will be approximately right if you convert power -&gt; intensity -&gt; pressure (e.g., using plane wave relationships). However, keep in mind that in practice, transducers rarely act as ideal radiators, so using a uniform additive pressure source will not give the exact response.</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/simulating-3d-disk-transducer-with-thickness#post-7926 Thu, 19 Nov 2020 15:28:27 +0000 raffaele84 7926@http://www.k-wave.org/forum/ <p>Hi,</p> <p>I'm new to k-wave. Some basic question: I want to simulate the field and heat generated by a disk transducer that has a non-zero thickness, to see the effect of different thicknesses.</p> <p>1) I was following the "Example: Heat Diffusion In A Homogeneous Medium" using the makeDisk mask, but I don't know how to get the field along the axis perpendicular to the disk, do I need 3D simulation? The 2D example seems to generate the filed AROUND the disk.</p> <p>2) How do I generate a disk defining also the thickness?</p> <p>3) I see that the input and output are given in pressure, I generally know the power generated by my transducer (or received), but not the pressure. Is it correct to convert my "power divided by the transducer surface" (so I get the intensity) into pressure? And vice-versa, at the receiver location? </p>