Finding image forward projection and its transpose matrix

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Finding image forward projection and its transpose matrix

California State University, Fullerton, Summer 2008 Compiled on January 30, 2024 at 6:49am

Problem

Write the matrix which implements the forward projection and its transpose.

A simple case would be to consider a 2-D object made up of only 4 pixels and one projection. After that think about an object with many pixels and many projections.

Answer

I will use the convention used by the radon transform in Matlab in setting up the coordinates system which is as shown below (diagram from Matlab documentation page).

pict — Figure 1: radon transform convention

In our case, we need to perform the following projection, which is at angle $θ = - 90^{0}$ as follows

The equation for the above mapping is $g = H f$ , hence we write $(\begin{matrix} g_{1} \\ g_{2} \end{matrix}) = (\begin{matrix} h_{11} & h_{12} & h_{13} & h_{14} \\ h_{21} & h_{22} & h_{23} & h_{24} \end{matrix}) (\begin{matrix} f_{1} \\ f_{2} \\ f_{3} \\ f_{4} \end{matrix})$ Hence $\begin{aligned} g_{1} & = h_{11} f_{1} + h_{12} f_{2} + h_{13} f_{3} + h_{14} f_{4} \\ g_{2} & = h_{21} f_{1} + h_{22} f_{2} + h_{23} f_{3} + h_{24} f_{4} \end{aligned}$

But $g_{1} = f_{1} + f_{2}$ from the line integral at the above projection and $g_{2} = f_{3} + f_{4}$ , hence the above 2 equations becomes $\begin{aligned} f_{1} + f_{2} & = h_{11} f_{1} + h_{12} f_{2} + h_{13} f_{3} + h_{14} f_{4} \\ f_{3} + f_{4} & = h_{21} f_{1} + h_{22} f_{2} + h_{23} f_{3} + h_{24} f_{4} \end{aligned}$

By comparing coeﬃcients on the LHS and RHS for each of the above equations, we see that for the ﬁrst equation we obtain $h_{11} = 1, h_{12} = 1, h_{13} = 0, h_{14} = 0$ For the second equation we obtain $h_{21} = 0, h_{22} = 0, h_{23} = 1, h_{24} = 1$ Hence the $H$ matrix is $H = (\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{matrix})$ Taking the transpose $H^{T} = (\begin{matrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix})$ Hence if we apply $H^{T}$ operator onto the image $g$ , we obtain back a $2 \times 2$ image, which is written as $(\begin{matrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} g_{1} \\ g_{2} \end{matrix}) = (\begin{matrix} k_{1} \\ k_{2} \\ k_{3} \\ k_{4} \end{matrix})$ Hence $k_{1} = g_{1}, k_{2} = g_{1}, k_{3} = g_{2}, k_{4} = g_{2}$ . In other words, the image is a 4 pixels $[\begin{matrix} g_{1} & g_{1} \\ g_{2} & g_{2} \end{matrix}]$

$H^{T}$ can now be viewed as back projecting the image $g$ into a plane by smearing each pixel $g_{i}$ value over the plane along the line of sight as illustrated below

1 Case of 45 degree

We repeat the above for $θ = 45^{0}$

The equation for the above mapping is $g = H f$ , hence we write $(\begin{matrix} g_{1} \\ g_{2} \\ g_{3} \end{matrix}) = (\begin{matrix} h_{11} & h_{12} & h_{13} & h_{14} \\ h_{21} & h_{22} & h_{23} & h_{24} \\ h_{31} & h_{32} & h_{33} & h_{34} \end{matrix}) (\begin{matrix} f_{1} \\ f_{2} \\ f_{3} \\ f_{4} \end{matrix})$ Therefore $\begin{aligned} g_{1} & = h_{11} f_{1} + h_{12} f_{2} + h_{13} f_{3} + h_{14} f_{4} \\ g_{2} & = h_{21} f_{1} + h_{22} f_{2} + h_{23} f_{3} + h_{24} f_{4} \\ g_{3} & = h_{31} f_{1} + h_{32} f_{2} + h_{33} f_{3} + h_{34} f_{4} \end{aligned}$

We see from projection diagram that $f_{1} = g_{1}, f_{3} + f_{2} = g_{2}$ and $f_{4} = g_{3}$ , hence the above 3 equations become $\begin{aligned} f_{1} & = h_{11} f_{1} + h_{12} f_{2} + h_{13} f_{3} + h_{14} f_{4} \\ f_{3} + f_{2} & = h_{21} f_{1} + h_{22} f_{2} + h_{23} f_{3} + h_{24} f_{4} \\ f_{4} & = h_{31} f_{1} + h_{32} f_{2} + h_{33} f_{3} + h_{34} f_{4} \end{aligned}$

By comparing coeﬃcients, we obtain from the ﬁrst equation $h_{11} = 1, h_{12} = 0, h_{13} = 0, h_{14} = 0$ and from the second equation $h_{21} = 0, h_{22} = 1, h_{23} = 1, h_{24} = 0$ and from the last equation $h_{31} = 0, h_{32} = 0, h_{33} = 0, h_{34} = 1$ . Hence the $H$ matrix is $H = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix})$ Using $H^{T}$ to project the image $g$ we obtain $(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} g_{1} \\ g_{2} \\ g_{3} \end{matrix}) = (\begin{matrix} k_{1} \\ k_{2} \\ k_{3} \\ k_{4} \end{matrix})$ Hence $k_{1} = g_{1}, k_{2} = g_{2}, k_{3} = g_{2}, k_{4} = g_{3}$ , hence the back projection plane is $K = [\begin{matrix} g_{1} & g_{2} \\ g_{2} & g_{3} \end{matrix}]$ This also can be interpreted as back projecting the image $g$ on a $45^{0}$ onto a plane by smearing each pixel value $g_{i}$ on each pixel along its line of sight as illustrated below