http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points Support for the k-Wave MATLAB toolbox en-US Wed, 13 May 2026 01:49:53 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1493 Mon, 18 Mar 2013 18:51:58 +0000 Bradley Treeby 1493@http://www.k-wave.org/forum/ <p>If you are working with experimental data, you should directly set dt and Nt based on your acquisition settings (rather than letting k-Wave estimate a suitable value). In other words, after you create <code>kgrid</code> using <code>makeGrid</code>, use something like:</p> <p><code>kgrid.t_array = 0:dt:(Nt - 1)*dt;</code> </p> http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1472 Mon, 18 Mar 2013 12:06:37 +0000 ha5438 1472@http://www.k-wave.org/forum/ <p>Hi, thanks to your help. </p> <p>Now I have some similar reconstructed image.</p> <p>But, there still have a problem I can't understand. </p> <p>In the Kgrid, how the function calculate the dt? </p> <p>Because I set Nx = 500, x = 150e-3, dx = x/Nx = 3e-4 and the medium assumed 1500. </p> <p>I thought the dt is dx/medium speed = 3e-4/1500 = 2e-7. But in the matlab, the calculated dt is 6e-9. </p> <p>I can 't sure it will affect the data that I loaded into the program, because the sinogram data (360*5000)from my experiment is acquired at 250Mhz sampling rate. </p> <p>The dt sould be 1/250M = 4e-9. I guess this parameter must be the same between program and exp, right? </p> http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1377 Tue, 12 Mar 2013 13:40:20 +0000 Bradley Treeby 1377@http://www.k-wave.org/forum/ <p>The example is in the toolbox under Photoacoustic Image Reconstruction -&gt; Automatic Sound Speed Selection. The details of how it works can be found <a href="http://www.medphys.ucl.ac.uk/research/mle/pdf_files/JBO-Treeby-2011.pdf">here</a>. </p> http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1376 Tue, 12 Mar 2013 13:26:55 +0000 ha5438 1376@http://www.k-wave.org/forum/ <p>I assume the speed of sound is 1500 m/s, it's the velocity of soft tissue. Actually, I </p> <p>have programmed a tomography by using labview, and it's work for simple source like (a) </p> <p>pencil lead(s). But I don't know why it's not available by k-wave. Also I can't find the</p> <p>keywords "autofocus" in the help?! Which example has this function ? </p> http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1375 Tue, 12 Mar 2013 11:45:39 +0000 Bradley Treeby 1375@http://www.k-wave.org/forum/ <p>It sounds like the sound speed you are using in the reconstruction doesn't match the one in the experiment (or alternatively, the radius you have set is incorrect). Do you know the sound speed accurately? You could try using the autofocus routine if not (there is an example of how to do this in the toolbox). </p> http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1374 Tue, 12 Mar 2013 10:12:27 +0000 ha5438 1374@http://www.k-wave.org/forum/ <p>Sorry, I have solved this problem when I check the code. I think it will be available as </p> <p>the x is greater 2 times than the sensor radius. Now, the program can run successfully. </p> <p>But...here comes another problem. Because I already get the sinogram(360*12000) from my </p> <p>photoacoustic experiment previously. So I replaced this data to "sensor_data" , set "p0 </p> <p>=0" and delete some syntax. Also, I have modified the Cartesian circular sensor code </p> <p>just like my experiment setup. As follow:</p> <p>sensor_radius = 37.5e-3; % [m]<br /> sensor_angle = 2*pi; % [rad]<br /> sensor_pos = [0,0]; % [m]<br /> num_sensor_points = 360;</p> <p>In the end the result of reconstructed image is not well, the image shows like a </p> <p>concentric circle(◎ &lt;- like this, and it's red in the middle ). Could you give some </p> <p>advice? </p> http://www.k-wave.org/forum/topic/the-relationship-between-kgrid-and-cartesian-points#post-1373 Tue, 12 Mar 2013 07:29:47 +0000 ha5438 1373@http://www.k-wave.org/forum/ <p>Hi </p> <p>Thanks your hard work in k-wave tool. I have a problem about the kgrid. I think the </p> <p>kgrid size must be greater than the Cartesian circular sensor, isn't it? In my case, I </p> <p>assigning x and y = 50e-3, Nx and Ny = 100. Another hand, defining sensor_radius = 40e-3 </p> <p>,sensor_angle = 2*pi. But it doesn't work... The error message shows "Cartesian points </p> <p>must lie within the k-space grid defined by kgrid." How can I debug this problem? </p>