http://www.k-wave.org/forum/topic/derivation-of-the-transducer-surface-pressure-from-the-total-radiated-power Support for the k-Wave MATLAB toolbox en-US Thu, 14 May 2026 00:04:28 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/derivation-of-the-transducer-surface-pressure-from-the-total-radiated-power#post-9176 Tue, 28 Jan 2025 10:28:40 +0000 Tomaubier 9176@http://www.k-wave.org/forum/ <p>Although I found the source of this problem a long time ago, I realize that I've left my post unanswered. For future reference or if anyone is faced with this situation, you should know that this has to do with improper "on-grid" definition of the source surface. This topic has already been covered extensively (<a href="https://doi.org/10.1121/1.5116132" rel="nofollow">https://doi.org/10.1121/1.5116132</a>) but the main takeaway is that builtin functions implementing "off-grid" strategies should be used for the definition of sources. In the specific example I gave, using <code>kWaveArray</code> together with <code>addArcElement</code> fixes the issue.<br /> Best,<br /> Tom </p> http://www.k-wave.org/forum/topic/derivation-of-the-transducer-surface-pressure-from-the-total-radiated-power#post-9105 Tue, 25 Jun 2024 09:08:16 +0000 Tomaubier 9105@http://www.k-wave.org/forum/ <p>Hello,<br /> I'm currently performing simulations of a spherical mono element transducer radiating in a multilayered axisymmetric domain. My goal is to evaluate temperature elevations produced by this transducer in the presence of those layers. The input of these simulations is the total acoustic power emitted by the transducer measured in water using a radiation force balance.<br /> To be able to run the simulation I have to go from this acoustic power <code>tx_emitted_ac_power</code> to pressure magnitudes <code>tx_p_mag</code> that can be applied to the transducer surface.<br /> Considering the plane wave assumption, a pressure magnitude can be obtained such that<br /> <code>tx_p_mag = sqrt(2) * sqrt((tx_emitted_ac_power * rho_tx_coupling_medium * c_tx_coupling_medium) / tx_surface_area)</code>.</p> <p>Regarding the transducer surface, my first instinct was to evaluate it as a section of a sphere such that<br /> <code>tx_surface_area = 2pi * r (r- sqrt(r^2 - a^2))</code> where <code>r</code> corresponds to the radius of curvature and <code>a</code> to half of the transducer aperture.</p> <p>After several validation with Rayleigh-based simulation models in homogeneous domains, I realized that the pressures obtained with kWave were consistently under-estimated.</p> <p>I later realized that evaluating <code>tx_surface_area</code> as the planar projection of the transducer aperture tx_surface_area = pi * (tx_aperture / 2)^2 led to an agreement of kWave with the reference Rayleigh simulations.</p> <p>Does anyone know what is going on here?</p> <p>Best,</p> <p>Tom </p>