# Intuition behind using basis functions to optimize discretization of a 2D image obtained via projections

If I am not mistaken, the technique used to avoid a pixelated appearance of CAT scan images described as basis functions on page 4032 of this paper is not unique to CAT scan, and probably applied to many other signal processing methods.

The question I have is whether a good schematic intuition of local basis functions $$b(x,y)$$ could be considered the idea behind this image:

where I superimposed a 2D Gaussian distribution of identical parameters and with truncated domain (in the article linked:"We have truncated the Gaussian at a radius of 1.5 times its FWHM") on each point of the grid (of course just a few representative points are illustrated with bivariate Gaussian contour plots).

Background info:

$$\mathbf P$$ whose entry in the $$i$$-th row and $$j$$-th column is given by

$$p_{ij}=S(b_j;r_i, \theta_i,w)$$

[where $$S$$ is the integral of the attenuation of the x-ray coursing through the patient along a strip $$S$$ at an angle $$\theta,$$ and at a distance to the origin of the coordinate system of $$r_i,$$ and $$b_j$$ is one of the basis functions, such as the function controlling the attenuation of the x-ray beam at any point $$f(x,y)$$ can be estimated linearly as $$\hat f(x,y)= \sum_{j=1}^N a_jb_j(x,y).$$]

...

All basis functions in a set are non-negative and identical except for location. We pick the basis grid so that it is aligned with the $$x$$ and $$y$$ axes. Let us now consider basis functions that are local and repetitive. By local, we mean that the support of each $$b_j$$ is small compared with that of $$f.$$ In general, the basis functions are centered- on each point of an equally spaced 2-D grid with spacing $$\Delta.$$

$$b_{j}(x,y)=b(x-x_j, y-y_j)\tag{ 10}$$

$$p_{i,j}=S(b_j;r_i,\theta_i,w)=S(b;d_{ij},\theta_i,w)\tag{11}$$

Writing $$(x_j,y_j)$$ for the coordinates of the $$j$$-th grid point, we have for the corresponding basis function where $$b(x,y)$$ is a basis function centered at the origin, and $$d_{ij} = r_i - x_j \cos \theta_i - y_j \sin \theta_i$$ is the signed distance between the projection strip $$h_i$$ and the $$j$$-th grid point $$(x_j,y_j)$$ as shown in Fig. 2.

Since the basis function has finite support, at each angle $$\theta$$ there will be some minimum distance $$R_\theta$$ for which the right-hand side of Eq. (11) will be zero when $$\vert d_{ij}\vert > R_\theta.$$ Because the basis function is local, $$R_\theta$$ is much smaller than the width of the reconstruction region. Therefore, $$p_{ij}$$ is sparse, as mentioned above.

I have found an answer to the question, which I would be happy to include as a formal "answer" if it weren't for the fact that it would possibly preempt a more informed contribution on what is already a 'tumbleweed' question. So here:

1. The basis functions - in this case the Gaussians - are placed on a Cartesian grid, representing pixels.

2. Each Gaussian is centered with FWHM of one pixel. The FWHM is equal to the grid spacing. Consequently, there is overlap at the tails of these Gaussian curves (which is not represented on the schematic above).