sLecture

↳ Topic 2: Tomographic Reconstruction

↳ Intro

↳ CT

↳ PET

↳ Co-ordinate Rotation

↳ Radon Transform

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.

Fig 1: bottom left: ccw rotation of vector; top right: cw rotation of coordinate axes

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.

Fig 1: Geometric Interpretation

• 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]

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