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Image registration is recovering an affine transformation (rotation + translation) between two images.
Image registration is recovering an affine transformation (rotation + translation) between two images.
==Problem Statement==
We are given two images <math>I_1</math> and <math>I_2</math>.<br>
Let <math>(x,y)</math> be uv coordinates within the image.<br>
We want to find a rotation and translation from <math>(x,y)</math> to <math>(x',y')</math> such that <math>I_1(x,y) = I_2(x', y')</math>.<br>
This is represented as:<br>
<math> x' = a_1 x + a_2 y + a_3</math><br>
<math> y' = a_4 x + a_5 y + a_6</math><br>
This can also be written as:<br>
<math>
\begin{pmatrix}
x' \\ y' \\ 1
\end{pmatrix}
=
\begin{pmatrix}
a_1 & a_2 & a_3\\
a_4 & a_5 & a_6\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ 1
\end{pmatrix}
</math>
==Log-Polar Transformation==
==References==

Revision as of 12:33, 15 May 2020

Image registration is recovering an affine transformation (rotation + translation) between two images.

Problem Statement

We are given two images \(\displaystyle I_1\) and \(\displaystyle I_2\).
Let \(\displaystyle (x,y)\) be uv coordinates within the image.
We want to find a rotation and translation from \(\displaystyle (x,y)\) to \(\displaystyle (x',y')\) such that \(\displaystyle I_1(x,y) = I_2(x', y')\).
This is represented as:
\(\displaystyle x' = a_1 x + a_2 y + a_3\)
\(\displaystyle y' = a_4 x + a_5 y + a_6\)
This can also be written as:
\(\displaystyle \begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix} = \begin{pmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} \)


Log-Polar Transformation

References