| Line 10: | Line 10: | ||
::↳ [[ECE637_tomographic_reconstruction_PET_S13_mhossain|PET]] | ::↳ [[ECE637_tomographic_reconstruction_PET_S13_mhossain|PET]] | ||
::↳ [[ECE637_tomographic_reconstruction_coordinate_rotation_S13_mhossain|Co-ordinate Rotation]] | ::↳ [[ECE637_tomographic_reconstruction_coordinate_rotation_S13_mhossain|Co-ordinate Rotation]] | ||
| + | ::↳ [[ECE637_tomographic_reconstruction_radon_transform_S13_mhossain|Radon Transform]] | ||
---- | ---- | ||
| Line 38: | Line 39: | ||
---- | ---- | ||
| − | + | <math>A_{\theta}</math> is the orthogonal counterclockwise rotation matrix given by <br/> | |
| − | + | ||
| − | <math>A_{\theta}</math> is the counterclockwise rotation matrix given by <br/> | + | |
<math>A_{\theta}=\begin{bmatrix} | <math>A_{\theta}=\begin{bmatrix} | ||
\cos(\theta) & -\sin(\theta) \\ | \cos(\theta) & -\sin(\theta) \\ | ||
| Line 50: | Line 49: | ||
[[Image:CR_fig1_mh.jpeg|800px|thumb|left|Fig 1: bottom left: ccw rotation of vector; top right: cw rotation of coordinate axes]] | [[Image:CR_fig1_mh.jpeg|800px|thumb|left|Fig 1: bottom left: ccw rotation of vector; top right: cw rotation of coordinate axes]] | ||
| + | |||
Let <math>{P_0}</math> be the unit vector shown in the top left corner of figure 1. <br/> | Let <math>{P_0}</math> be the unit vector shown in the top left corner of figure 1. <br/> | ||
| Line 90: | Line 90: | ||
* Inverse Transformation | * Inverse Transformation | ||
| + | The rotation can be inverted by multiplying the rotated vector with matrix <math>A_{-\theta}</math>, where <br/> | ||
| + | <math>A_{-\theta}=\begin{bmatrix} | ||
| + | \cos(-\theta) & -\sin(-\theta) \\ | ||
| + | \sin(-\theta) & \cos(-\theta) | ||
| + | \end{bmatrix} = \begin{bmatrix} | ||
| + | \cos(\theta) & \sin(\theta) \\ | ||
| + | -\sin(\theta) & \cos(\theta) | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | and we have that <br/> | ||
<math>\begin{bmatrix} | <math>\begin{bmatrix} | ||
r \\ | r \\ | ||
| Line 98: | Line 108: | ||
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| + | |||
| + | Notice that rotating a vector counterclockwise by angle <math>-\theta</math> is the same as rotating the vector clockwise by angle <math>\theta</math>. In fact, <math>A_{-\theta}</math> is the orthogonal clockwise rotation matrix. Since <math>A_{\theta}</math> is an orthogonal matrix, it's inverse is the same as its transpose, i.e. | ||
| + | |||
| + | <math> | ||
| + | A_{\theta}^{-1} = A_{\theta}^T = \begin{bmatrix} | ||
| + | \cos(\theta) & \sin(\theta) \\ | ||
| + | -\sin(\theta) & \cos(\theta) | ||
| + | \end{bmatrix} = A_{-\theta} | ||
| + | </math> | ||
| + | |||
| + | The result is obvious since rotating the action of rotating a point counterclockwise by angle <math>\theta</math> can be inverted by rotating it clockwise by angle <math>\theta</math>. | ||
| + | |||
---- | ---- | ||
Revision as of 11:36, 23 May 2013
- ↳ Topic 2: Tomographic Reconstruction
- ↳ Intro
- ↳ CT
- ↳ PET
- ↳ Co-ordinate Rotation
- ↳ Radon Transform
The Bouman Lectures on Image Processing
A sLecture by Maliha Hossain
Subtopic 3: Co-ordinate Rotation
© 2013
Contents
Excerpt from Prof. Bouman's Lecture
Accompanying Lecture Notes
$ A_{\theta} $ is the orthogonal counterclockwise rotation matrix given by
$ A_{\theta}=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $
The matrix rotates vector $ v_0 $ in a 2-dimensional real space by angle $ \theta $ in a fixed coordinate system. Notice that this is equivalent to keeping the vector fixed and rotating the coordinate system clockwise by $ \theta $. This equivalence is illustrated in figure 1.
Let $ {P_0} $ be the unit vector shown in the top left corner of figure 1.
$ P_0 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $
Rotating $ P_0 $ $ 90 $° counterclockwise produces the unit vector $ P_1 $ shown in the bottom left
$ P_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $
The result of rotating the coordinate axes clockwise by $ 90 $° is shown in the top right. We have that
$ P_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $
So we see that vectors $ P_1 $ and $ P_2 $ are equivalent. In other words, rotating a vector counterclockwise by angle $ \theta $ is the same as rotating the coordinate axes clockwise by angle $ \theta $.
Let us define a new coordinate system $ (r,z) $ where
$ \begin{bmatrix} x \\ y \end{bmatrix} = A_{\theta}\begin{bmatrix} r \\ z \end{bmatrix} $
i.e. vector $ [r,z]' $ is rotated counterclockwise angle $ \theta $ to produce vector $ [x,y]' $
Figure 1 shows the geometric interpretation of the rotation.
- Inverse Transformation
The rotation can be inverted by multiplying the rotated vector with matrix $ A_{-\theta} $, where
$ A_{-\theta}=\begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} $
and we have that
$ \begin{bmatrix} r \\ z \end{bmatrix} = A_{-\theta}\begin{bmatrix} x \\ y \end{bmatrix} $
Notice that rotating a vector counterclockwise by angle $ -\theta $ is the same as rotating the vector clockwise by angle $ \theta $. In fact, $ A_{-\theta} $ is the orthogonal clockwise rotation matrix. Since $ A_{\theta} $ is an orthogonal matrix, it's inverse is the same as its transpose, i.e.
$ A_{\theta}^{-1} = A_{\theta}^T = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} = A_{-\theta} $
The result is obvious since rotating the action of rotating a point counterclockwise by angle $ \theta $ can be inverted by rotating it clockwise by angle $ \theta $.
References
- C. A. Bouman. ECE 637. Class Lecture. Digital Image Processing I. Faculty of Electrical Engineering, Purdue University. Spring 2013.
- E. W. Weisstein, "Rotation Matrix." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/RotationMatrix.html. May 8th, 2013 [May 21st, 2013]
Questions and comments
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