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Finding image forward projection and its transpose matrix

Nasser M. Abbasi

California State University, Fullerton, Summer 2008 page compiled on July 1, 2015 at 9:15pm

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).

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In our case, we need to perform the following projection, which is at angle 𝜃 = − 900   as follows

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The equation for the above mapping is g =  Hf  , hence we write

                              (   )
(   )    (                  ) | f1|
  g1      h11  h12  h13  h14  | f2|
      =                       ||   ||
  g2      h21  h22  h23  h24  ( f3)
                                f
                                  4

Hence

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But g1 = f1 + f2   from the line integral at the above projection and g2 = f3 + f4   , hence the above 2 equations becomes

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By comparing coefficients on the LHS and RHS for each of the above equations, we see that for the first equation we obtain

h11 = 1, h12 = 1,h13 = 0,h14 = 0

For the second equation we obtain

h21 = 0, h22 = 0,h23 = 1,h24 = 1

Hence the H  matrix is

|-----(-----------)-|
|       1  1  0  0  |
|H =                |
|       0  0  1  1  |
---------------------

Taking the transpose

|------(-1--0)--|
|      ||     ||  |
|HT  = | 1  0|  |
|      |( 0  1|)  |
|               |
---------0--1---|

Hence if we apply   T
H  operator onto the image g  , we obtain back a 2 × 2  image, which is written as

(     )          (   )
  1  0  (   )      k1
|| 1  0||   g1     || k2||
||     ||       =  ||   ||
( 0  1)   g2     ( k3)
  0  1             k
                    4

Hence k1 = g1,k2 = g1,k3 = g2,k4 =  g2   . In other words, the image is a 4 pixels [      ]
 g1  g1

 g2  g2

HT  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

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1 Case of        0
𝜃 = 45

We repeat the above for       0
𝜃 = 45

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The equation for the above mapping is g =  Hf  , hence we write

(   )    (                  )  ( f1)
| g1|    | h11  h12 h13  h14|  || f2||
|( g2|) =  |( h21  h22 h23  h24|)  ||   ||
                               ( f3)
  g3       h31  h32 h33  h34     f
                                  4

Therefore

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We see from projection diagram that f1 = g1,f3 + f2 = g2   and f4 = g3   , hence the above 3 equations become

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By comparing coefficients, we obtain from the first equation h11 = 1,h12 = 0,h13 = 0,h14 = 0  and from the second equation h21 =  0,h22 = 1,h23 = 1,h24 = 0  and from the last equation h31 = 0,h32 = 0,h33 = 0,h34 = 1  . Hence the H  matrix is

|-----(-----------)-|
|       1  0  0  0  |
|     |           | |
|H =  |( 0  1  1  0|) |
|                   |
--------0--0--0--1---

Using HT  to project the image g  we obtain

( 1  0  0) (   )    ( k1 )
|| 0  1  0|| | g1|    || k2 ||
||        || |( g2|)  = ||    ||
( 0  1  0)          ( k3 )
  0  0  1    g3       k
                        4

Hence k1 = g1,k2 = g2,k3 = g2,k4 =  g3   , hence the back projection plane is

      [      ]
       g1  g2
K  =
       g2  g3

This also can be interpreted as back projecting the image g  on a 450   onto a plane by smearing each pixel value gi  on each pixel along its line of sight as illustrated below

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