7 ms·

Thanks to the sampling theorem, if there is a frequency limit in a continuous signal, then you merely have to sample at twice that frequency. The eye clearly h

Thanks to the sampling theorem, if there is a frequency limit in a continuous signal, then you merely have to sample at twice that frequency. The eye clearly has frequency limitations, so we just need to find the sampling strategy which accurately codes all perceptible frequencies.

rleigh 10y ago
It's going to be more than twice--Nyquist-Shannon is for 1D; 2D images would have a lower resolution on the diagonals, so you'd want a minimum of twice the frequency on the diagonals. In an ideal world, we would also be using hexagonal grids rather than square, which gives a resolution increase for the same area and a given pixel size, and also reduces off-axis aliasing.
- microcolonel 10y ago
  Very astute, thank you for bringing this to my attention. So it'd be about 2 and 58/70 times the maximum frequency, assuming two-dimensional sampling. As for hexagonal grids, there are a number of complications introduced by that, especially when it comes to using general algorithms based on regular 2D euclidean space.
  - rleigh 10y ago
    With a sampling rate of 2 on the diagonal, you'd have a rate of 2.828 on the axes, with a square grid (2 * sqrt(2)). Unless I'm missing something. Agreed about the complications of the hexagonal grid. We make implicit assumptions everywhere. Though I'm not sure they are insurmountable for some purposes; with graphics APIs like OpenGL, texture coordinates and samplers could be using a hexagonal grid relatively transparently. That would allow use of hexagonal displays even if the texture data is square.
acchow 10y ago
The eye clearly has frequency limitations No...the eye doesn't work that way. It is not a machine. It is organic and analog. It doesn't have some frequency limit. In any case, you're abusing the Shannon-Nyquist theorem. It absolutely does not imply that if some device is sampling at x hz then you can sample at 2x hz and produce a sufficient reproduction for the 1x hz sample. This is a common misconception.

rleigh 10y ago

It's going to be more than twice--Nyquist-Shannon is for 1D; 2D images would have a lower resolution on the diagonals, so you'd want a minimum of twice the frequency on the diagonals. In an ideal world, we would also be using hexagonal grids rather than square, which gives a resolution increase for the same area and a given pixel size, and also reduces off-axis aliasing.

microcolonel 10y ago
Very astute, thank you for bringing this to my attention. So it'd be about 2 and 58/70 times the maximum frequency, assuming two-dimensional sampling. As for hexagonal grids, there are a number of complications introduced by that, especially when it comes to using general algorithms based on regular 2D euclidean space.
- rleigh 10y ago
  With a sampling rate of 2 on the diagonal, you'd have a rate of 2.828 on the axes, with a square grid (2 * sqrt(2)). Unless I'm missing something. Agreed about the complications of the hexagonal grid. We make implicit assumptions everywhere. Though I'm not sure they are insurmountable for some purposes; with graphics APIs like OpenGL, texture coordinates and samplers could be using a hexagonal grid relatively transparently. That would allow use of hexagonal displays even if the texture data is square.

acchow 10y ago

The eye clearly has frequency limitations No...the eye doesn't work that way. It is not a machine. It is organic and analog. It doesn't have some frequency limit. In any case, you're abusing the Shannon-Nyquist theorem. It absolutely does not imply that if some device is sampling at x hz then you can sample at 2x hz and produce a sufficient reproduction for the 1x hz sample. This is a common misconception.