k-Wave

1. prabodh_iitk
   

   Member
   Joined: Nov '13
   Posts: 6

   Dear Sir,

   I was running example_IVP_3D_simulation.m code with :
   % create the computational grid
   Nx = 64; % number of grid points in the x direction
   Ny = 64; % number of grid points in the y direction
   Nz = 64; % number of grid points in the z direction
   dx = 0.1e-3; % grid point spacing in the x direction [m]
   dy = 0.1e-3; % grid point spacing in the y direction [m]
   dz = 0.1e-3; % grid point spacing in the z direction [m]

   *****************************
   Running k-Wave simulation...
   start time: 03-Feb-2016 12:20:30
   reference sound speed: 1800m/s
   dt: 16.6667ns, t_end: 7.3833us, time steps: 444
   input grid size: 64 by 64 by 64 grid points (6.4 by 6.4 by 6.4mm)
   maximum supported frequency: 7.5MHz

   Here maximum supported frequency is 7.5MHz while the sampling frequency is ~ 60 MHz.

   I wish to know what is the meaning for maximum supported frequency?

   Does this mean that we need to filter out all the frequencies > 7.5MHz from the simulated photoacoustic signal?

   Thank you.

   --
   Regards
   Prabodh

   Posted 8 years ago #

2. Anthony
   

   Member
   Joined: Apr '13
   Posts: 96

   Hi Prabodh,

   The maximum supported frequency is computed based on the spatial grid step because you need at least two grid points per wavelength (Shannon-Nyquist theorem): see the user guide for more details.

   Therefore, if you need to propagate higher frequencies, you can simply decrease your spatial grid step (which will of course require more RAM and computational time but you can use the C versions of k-Wave or the GPU version if Matlab is too slow).

   Best regards,
   Anthony

   PS : You probably have a medium with a 1500m/s sound speed which gives you a wavelength of 0.2e-3m at 7.5 MHz, which corresponds to 2 points per wavelength.

   Posted 8 years ago #

3. prabodh_iitk
   

   Member
   Joined: Nov '13
   Posts: 6

   Dear Anthony,

   Thank you for your response. My major concern lies in the fact that if I take the Fourier spectra of the photoacoustic signal, It will have the frequency amplitudes corresponding to the sampling frequency that is ~ 60MHz. While the maximum supported frequency is 7.5 MHz. So I wish to know what the significance of the amplitudes corresponding to the frequencies greater than 7.5MHz are.

   Thank you

   --
   Regards
   Prabodh

   Posted 8 years ago #

4. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,640

   Hi Prabodh,

   If you look at the Fourier spectrum of your recorded time signal, it won't contain any energy above the maximum frequency spatially supported on the grid (e.g., see the second figure in this example). The time sampling is typically finer than the spatial sampling to meet stability requirements, i.e., c*dt < dx or CFL < 1. You can read more about this in the k-Wave manual if you would like more details. You will also encounter this situation in experimental data, where the temporal sampling frequency is often much higher than the maximum frequency in your data.

   Brad.

   Posted 8 years ago #

5. NealXu
   

   Member
   Joined: Jan '18
   Posts: 1

   Hi..as per my knowledge if you need to propagate higher frequencies, you can simply decrease your spatial grid step which will of course require more RAM and computational time but you can use the C versions of k-Wave or the GPU version if Matlab is too slow.

   Posted 6 years ago #

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