Inversions of the windowed ray transform

A Numerical implementation

Here we discuss the results of 2-dimensional numerical implementation to compare the reconstructions obtained from formulas (2.1) and (2.4) and Theorem 2.3.

From formula (2.1), we obtain the regular X-ray transform Xf. There are several inversion formulas for the regular X-ray transform Xf. One of them can be obtained by converting Xf to the regular Radon transform R⁢f⁢(s,θ), defined by

R⁢f⁢(s,θ)=∫θ⊥f⁢(s⁢θ+y)⁢𝑑y,

setting R⁢f⁢(u⋅θ,θ)=X⁢f⁢(u,θ) and applying the inversion formula for the Radon transform.

In the experiments presented here we use the phantom shown in Figure 1 (a) and set h⁢(t)=e-11-t2⁢χ|t|≤1⁢(t), where χA⁢(t) is the characteristic function whose value is 1 when t belongs to A⊂ℝ and is zero when t∉A. Then h^⁢(0) is clearly not zero. This phantom, supported within the rectangle [-1,1]×[-1,1], is the sum of three characteristic functions of disks, whose values are 1.5, 1, and 0.5, with radii 0.05, 0.15 and 0.3, respectively. The windowed ray transform Ph⁢f is computed by numerical integration. All images presented in Figure 1 have 256×256 pixels.

For v, the range is from -16 to 16 and the size of one square is 1/32×1/32. When reconstructing from equations (2.1) and (2.4), we reconstruct for |v|=16⁢2,32⁢2 and |v|=1/16,1/32, respectively. To implement Theorem 2.3, we set a=0 and v2=1. Then we have

|σ|⁢∫ℝPh⁢f^⁢(σ,u2,v1,1)⁢𝑑v1=2⁢π⁢e-1⁢f^⁢(σ,u2).

When integrating with respect to v1, we integrate from -16 to 16 with the distance 1/32 as the range of v1. The image in Figure 1 (b) was reconstructed from 256×256 projections for u1 and u2. The images in Figure 1 (c) and (d) were reconstructed from equation (2.1). For these images, 256×256 projections for s and θ were used and R⁢f⁢(s,θ) is obtained by X⁢f⁢(s⁢(sin⁡θ)-1,0,θ). To use equation (2.1), we need the value of Ph⁢f at one point v with the largest value |v|. We average the four values of Ph⁢f at the four points with largest absolute value, e.g., v=(16,16), (-16,16), (16,-16), and (-16,-16). Taking the average value is better than taking one value, because dependence on only one value may increase the chance of having higher error. (In fact, we have the numerical results to show that taking the average value provides the better reconstruction. However, the difference of results between the average value and one value is not big and this difference is not a main issue; hence we do not insert those results.) In Figure 1 (c) and (d), we set |v|=16⁢2 and |v|=32⁢2, respectively. Lastly, the images in Figure 1 (e) and (f) were reconstructed according to equation (2.4). For these images, 256×256 projections for u1 and u2 were used. Also, we set |v|=1/16 and |v|=1/32 in Figure 1 (e) and (f), respectively. We take the average values in these reconstructions, too.

Figure 1

Reconstruction in two dimensions: (a) the phantom, (b) reconstruction using Theorem 2.3, (c) and (d) reconstructions from equation (2.1), and (e) and (f) reconstructions from equation (2.4).

(a)

(b)

(c)

(d)

(e)

(f)

In Figure 2, the magnified images of the red rectangular regions in Figure 1 are presented. Our reconstruction in Figure 2 (b) shows clearer boundaries than the others in Figure 2 (c)–(f).

Figure 2

The magnified images of the red rectangular regions in Figure 1.

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(b)

(c)

(d)

(e)

(f)