k-Wave

1. Anthony
   

   Member
   Joined: Apr '13
   Posts: 96

   Hello everyone,

   I have a question about the maximum frequency which can be propagated in several dimensions. In 1D, Shannon-Nyquist says that you need at least two points per wavelength.

   But, in 2D, if you wave does not propagate in the direction of your basis vectors, don't you need up to 2*sqrt(2) points per wavelength ? (and 2*sqrt(3) in 3D...) ? Am I right?

   Best regards,
   Anthony

   EDIT: Sorry, I just made the calculations on a rectangular grid with Peterson-Middleton theorem and it seems enough to have two points per wavelength :-) It seems counter-intuitive to me though... Anyway, sorry, I shouldn't have posted ;-)

   Posted 8 years ago #

2. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,640

   Hi Anthony,

   Interesting question! If you consider the grid geometrically, it would seem that the points are spaced further apart in the diagonal direction, and therefore would support a lower maximum frequency.

   However, if you think about the wave propagation in k-space, the wave field is broken into a sum of plane waves propagating in different directions. These directions are given by the values of kx and ky. Because the discrete k-space grid is square, the grid actually supports slightly higher maximum frequencies in the diagonal directions, as the corners of the k-space have a higher wavenumber than the limits of the vertical and horizontal axes.

   Hope that helps,

   Brad,

   Posted 8 years ago #

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