Image gradient sampling grid
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
Hi, everyone. I am having a difficulty with my project in MATLAB. I really do not understand this concept of sampling grid and from the circle to R. Thank you for any help in advance! Here is the question:
Id is a discrete grayscale image of size N^2, thus an element of R N^N. This discrete image Id can be viewed as sampling on regular grid of a function I from the square [-1,1]^2 to R, with the grid spacing being h=2/(N-1) in both directions. How to compute I from Id?
0 comentarios
Respuestas (1)
Image Analyst
el 7 de En. de 2016
You never said what the badly-named "I" is, so I don't know how to compute it from Id.
All the rest of the question is just a bunch of mathematical mumbo jumbo that really doesn't amount to anything except to allow the author to impress himself. Trust me, you can safely ignore it.
2 comentarios
Image Analyst
el 8 de En. de 2016
Not sure what that means. A digital image is an array with rows and columns. A real image is just a bunch of light (or whatever). Not sure if you want to interpolate a different number of pixels, or if you want some kind of spatial calibration system that keeps track of the real-world (x,y) given the (column, row) indexes. For example element (1,1) corresponds to x=-1 and y=-1, and element (480, 1) (last row of a 480x640 image) corresponds to x=-1, y=+1, and element (480,640) corresponds to x=1, y=1. If so, you'll have to keep track of the spatial calibration factor(s) separately whenever you want to convert from real world coordinates to indexes.
See my spatial calibration demo, attached.
Ver también
Categorías
Más información sobre Image Processing Toolbox en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!