k-Wave

1. scottalm
   

   Member
   Joined: Jul '16
   Posts: 2

   I am using the SSE3 Linux binaries for this work, but I have seen the same error in the Windows AVX2 binaries.

   I'm not sure if this is a real bug or just me missing something somewhere, but it seems that when I try to use a plane wave source with the C++ version of k-wave I get an error reading in the HD5 file.

   As an example, running the following works:

   % create the computational grid
   Nx = 108;
   Ny = 108;
   Nz = 108;
   
   dx = .05;
   dy = .05;
   dz = .05;
   kgrid = makeGrid(Nx, dx, Ny, dy, Nz, dz);
   
   model = ones(Nx, Ny, Nz);
   % density
   rho_v=[1e3];
   % speed of sound
   c_v=[1500];
   
   medium.sound_speed = c_v(model); % [m/s]
   medium.density     = rho_v(model);  % [kg/m^3] 
   
   medium.alpha_coeff = 0;     % [dB/(MHz^2 cm)]
   medium.alpha_power = 0.999;
   
   sensor.mask = ones(Nx, Ny, Nz);
   
   % record the maximum pressure
   sensor.record = {'p_max'};
   
   source.u_mask =  ones(Nx, Ny, Nz);
   
   source.uz = zeros(Nx, Ny, Nz);
   source.uz(:, :, 3) = 1;
   source.uz = smooth(kgrid, source.uz, true);
   source.uz = reshape(source.uz, [], 1);
   
   input_args = {'PlotScale', 'auto', ...
       'PMLInside', false, 'PlotPML', false};
   
   sensor_data = kspaceFirstOrder3D(kgrid, medium, source, sensor, input_args{:})

   But when I replace kspaceFirstOrder3D with kspaceFirstOrder3DC, it errors out giving me

   K-Wave panic: Data loading was not successful!
   Error: Dataset "uz_source_input"  has wrong dimension sizes!

   The size is appropriate for Matlab, but doesn't seem to work for C++. Am I doing something incorrectly?

   Posted 8 years ago #

2. Jiri Jaros
   

   Developer
   Joined: Feb '12
   Posts: 115

   Hi scottalm,

   I think the matlab script is not correct - you mixed the source spatial definition (position and size) with the driving signal definition.

   source.u_mask - this should be the spatial definition of the transducer
   source.uz - this is the driving signal

   I think you tried to define the shape of the source in the uz component of the driving signal.
   Jiri

   Posted 8 years ago #

3. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,643

   Hi scottalm,

   If you add a second time point to your source, e.g., something like

   source.uz(:, 2) = 0;

   Does it work? I would need to check, but it's possible the C++ code explicitly checks for a 2D input signal.

   Brad.

   Posted 8 years ago #

4. scottalm
   

   Member
   Joined: Jul '16
   Posts: 2

   The problem ended up being how I was defining the source (along the time instead of the spatial dimension):

   source.uz = reshape(source.uz, [], 1);

   Should have been:

   source.uz = reshape(source.uz, 1, []);

   The Matlab code will execute this, but the C++ code won't (since it is improperly constructed). It would be nice if the Matlab code would do a check for dimension sizes, but at least there is a forum post if someone comes across this problem.

   Posted 8 years ago #

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