http://www.k-wave.org/forum/topic/input-for-2d-and-3d-simulations-and-units Support for the k-Wave MATLAB toolbox en-US Tue, 12 May 2026 23:39:23 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/input-for-2d-and-3d-simulations-and-units#post-7601 Sat, 13 Jun 2020 12:53:10 +0000 Bradley Treeby 7601@http://www.k-wave.org/forum/ <p>Hi Frank,</p> <p>Yes, the results will be different in 2D and 3D. This is not an idiosyncrasy of k-Wave, but is built into the wave equation in 2D (the gradients are zero in the third-dimension, meaning everything is the same or repeated). There are some details in the k-Wave manual about how the source terms appear in the wave equation which might help.</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/input-for-2d-and-3d-simulations-and-units#post-7580 Mon, 08 Jun 2020 13:22:51 +0000 s148275 7580@http://www.k-wave.org/forum/ <p>Hi Brad,</p> <p>Thanks for the quick reply! I understand that the intensity is dimensionally incorrect in this way, so indeed P0_2D is in W/m.</p> <p>However, I would like to validate a 2D simulation of a point source. In this case I apply a pressure of 1 MPa with a point source transducer. With the equation: p = sqrt(P0*rho*c0/(2pi*x)) I can find the pressure at position x. However, how can I relate this P0 to the input pressure of the script?</p> <p>And what if I would go to 3D? Could it be that the results are not the same in 2D and 3D for a point source, because the intensity scales with 1/x^2 in 3D and because you use in 2D this infinite third dimension?</p> <p>Thanks in advance!</p> <p>Frank </p> http://www.k-wave.org/forum/topic/input-for-2d-and-3d-simulations-and-units#post-7572 Sat, 06 Jun 2020 20:55:21 +0000 Bradley Treeby 7572@http://www.k-wave.org/forum/ <p>Hi s148275</p> <p>In 2D, k-Wave solves the 2D wave equation. This inherently assumes that everything in the third dimension is infinitely extended / repeated. For example, a point source in 2D corresponds to an infinite line source in 3D, a circle in 2D corresponds to an infinite cylinder in 3D, and so on. </p> <p>With this in mind, the equation <code>I = P0/(2pi*x)</code> is not dimensionally correct (this would give intensity in units of W/m). While it's correct that the intensity decays with 1/x in 2D, P0 in this case should be the power per unit length (again think of the 2D domain infinitely repeated in the out-of-plane dimension). Nothing is needed to account for this as it's "built-in" if you like to the wave equation in different dimensions.</p> <p>The 1/N scaling used for the mass source terms is used to divide the source amongst the components of the density, which is artificially split to apply a perfectly matched layer. </p> <p>Hope that helps,</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/input-for-2d-and-3d-simulations-and-units#post-7539 Tue, 02 Jun 2020 10:36:57 +0000 s148275 7539@http://www.k-wave.org/forum/ <p>I am doing focused ultrasound simulations in 2D. Could you explain to me what you do in the third dimension is this case? Do you set the pressure to zero in the third dimension?</p> <p>From the manual it is clear that the input source.p0 has to be put in the units N/m^2 (independent of the amount of dimensions). By conversion the source term Ms is found using Equation 2.19. In this equation the parameter N is used to describe the amount of dimensions. What is the reason for the use of this parameter N?</p> <p>When looking at a point source in 2D the intensity decreases with distance according to: I = P0/(2pi*x), with P0 the input power in Watt and x the distance from the point source in meters. Using that I = p*v = p^2/(rho*c0) it is found that: p = sqrt(P0*rho*c0/(2pi*x)). When looking at the units of this pressure, it is not in N/m^2. When doing the same derivation of the pressure in 3D you actually do get the correct units, since then I_3D = P0/(4pi*x^2). So I am wondering if this mismatch in units here is also something which is accounted for in the k-wave simulation? And is so, how is this done precisely?</p> <p>Thanks in advance! </p>