http://www.k-wave.org/forum/topic/how-to-understand-the-sensor_data-received-by-a-binary-sensor-mask Support for the k-Wave MATLAB toolbox en-US Tue, 12 May 2026 23:07:18 +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-understand-the-sensor_data-received-by-a-binary-sensor-mask#post-8191 Tue, 08 Jun 2021 17:04:01 +0000 Bradley Treeby 8191@http://www.k-wave.org/forum/ <p>Hi Zeng,</p> <p>The size of <code>sensor_data</code> will be <code>(number_sensor_points, number_time_points)</code>. In your case, yes, there are 141 grid points that form part of the sensor mask, so there will be 141 time series in the returned sensor data.</p> <p>Yes, you could sum them up in the way you mention. Then depending on what you're trying to achieve in the reconstruction, either use a smaller grid, or use the same grid/mask but assign the summed value to all of the grid points that form part of that detecting element.</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/how-to-understand-the-sensor_data-received-by-a-binary-sensor-mask#post-8173 Tue, 08 Jun 2021 06:53:35 +0000 zengqw2021 8173@http://www.k-wave.org/forum/ <p>Dear all,<br /> I'm try to understand 2D Time Reversal Reconstruction For A Line Sensor Example. If I define a binary line sensor by:</p> <p>sensor.mask = zeros(Nx, Ny);<br /> sensor.mask(60:200, 60) = 1; %1 for receiver, 0 for no receiver?</p> <p>we find the received sensor_data is a array 141×Ny. Does that mean there are 141 receivers?<br /> If so, can I sum up several receiver signals and reconstruct the image again?<br /> For example:<br /> NewSensor_data(1,:)= sensor_data(1,:)+sensor_data(2,:)+sensor_data(3,:);<br /> NewSensor_data(2,:)= sensor_data(6,:)+sensor_data(7,:)+sensor_data(8,:);<br /> ...<br /> The new detection point is the sum of the received signals of the original three detection points, and the distance between the new detection point is about the distance of the original three detection points.<br /> Is the above idea correct?<br /> Thanks<br /> Zeng </p>