http://www.k-wave.org/forum/topic/how-to-determine-grid-point-spacing Support for the k-Wave MATLAB toolbox en-US Tue, 12 May 2026 23:35:32 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/how-to-determine-grid-point-spacing#post-390 Fri, 16 Mar 2012 11:35:29 +0000 bencox 390@http://www.k-wave.org/forum/ <p>Hi Chao,</p> <p>If you replace your &lt; signs with &lt;= signs, then I think you are proving that when the maximum frequency supported by the grid, fmax_grid = c * kmax / 2pi = c / (2 *dx), equals the maximum frequency supported temporally, fmax_time = fs/2 = 1/(2 * dt), then necessarily dx = c * dt.</p> <p>Of course, that need not be the case. You can sample in time much more often than dx / c if you want to, but if you're using a time-varying source you need to make sure that the grid can support the frequencies that are in your input signal. It is probably a good idea to have a somewhat oversampled source, ie. the highest frequency in the time-varying source is rather less than fs/2.</p> <p>A final point is that for an initial value problem when the medium is homogeneous you can take as large a timestep as you like and it will be exact. (This is the whole point of the k-space method, as opposed to simply a pseudospectral time domain (although as you get close to the edges of the domain the PML will need smaller timesteps to be effective.)</p> <p>Ben </p> http://www.k-wave.org/forum/topic/how-to-determine-grid-point-spacing#post-388 Fri, 16 Mar 2012 02:49:00 +0000 huangchao 388@http://www.k-wave.org/forum/ <p>Hi Dr. Cox and Dr. Treeby,</p> <p>Given the grid point spacing 'dx' and sound speed 'c' (assume c is constant), the time step 'dt' should satisfy the following condition</p> <p>dt &lt; dx/c (1)</p> <p>So I am wondering what's the condition for the grid point spacing 'dx' given time step 'dt' and sound speed 'c'. In other words, how to determine the maximum dx given 'dt' and 'c'.</p> <p>My thoughts are dx&lt;λ_min/2, where λ_min is the minimum acoustic wavelength, i.e. λ_min=c/f_max, where f_max is the maximum frequency of the acoustic wave. If the acoustic wave is wide band, and we neglect absorption and transducer frequency response, then f_max=f_s/2, where f_s=1/dt is the sampling rate.</p> <p>Put it together, we have</p> <p>dx &lt; λ_min/2 = c/(2*f_max) = c/f_s = c*dt (2)</p> <p>But this conclusion contradicts (1). I don't know where the problem is.</p> <p>Thanks in advanced!</p> <p>Best,<br /> Chao </p>