<p>It sounds exactly like that!</p>

<ul>
<li><span class="math-container">$\Omega$</span> is the image domain, it can be either discrete or continuous. No idea how it is defined in that particular book.</li>
<li><span class="math-container">$x=(x_1, x_2)$</span> are the 2D image coordinates</li>
<li><span class="math-container">$u_0, u, n$</span> are functions that return intensity values at the given location <span class="math-container">$x$</span>.</li>
</ul>

CC BY-SA 4.0
2020-06-09T18:30:44.320Z
2020-06-09T19:07:03.113Z
2020-06-09T19:07:03.113Z
Isolin
<p>I am reading "Image Processing and Analysis" by Chan and Shen, c 2005 SIAM. They introduce some notation I'm not 100% sure how to interpret:</p>

<p><span class="math-container">$$ u_0(x)=u(x)+n(x),  x=(x_1,x_2) \in \Omega $$</span></p>

<p>They state <span class="math-container">$u_0$</span> represents a noisy image, <span class="math-container">$u$</span> is the clean image and <span class="math-container">$n$</span> is Gaussian white noise. I assume that <span class="math-container">$ x=(x_1,x_2) $</span> is meant to convey that x is a two-dimensional vector indexing both image height and width. I don't know how to interpret <span class="math-container">$\Omega$</span>. Is it meant to denote the entire image index range? </p>

CC BY-SA 4.0
2023-03-14T02:39:46.033Z
$\Omega$ is just the image space. So $\Omega = [0,X -1] \times [0,Y - 1]$ assuming your image has size $X\times Y$.
user8469759
<image-processing><mathematics>
Interpreting Image Processing Math Equation
Ron Jensen
# Interpreting Image Processing Math Equation

(Asked by: username_0 on 2020-06-09)

I am reading "Image Processing and Analysis" by Chan and Shen, c 2005 SIAM. They introduce some notation I'm not 100% sure how to interpret:

$$ u\_0(x)=u(x)+n(x), x=(x\_1,x\_2) \in \Omega $$

They state $u\_0$ represents a noisy image, $u$ is the clean image and $n$ is Gaussian white noise. I assume that $ x=(x\_1,x\_2) $ is meant to convey that x is a two-dimensional vector indexing both image height and width. I don't know how to interpret $\Omega$. Is it meant to denote the entire image index range?

**Tags:** image-processing, mathematics

> **Comments:**
> **username_1**: $\Omega$ is just the image space. So $\Omega = [0,X -1] \times [0,Y - 1]$ assuming your image has size $X\times Y$.  

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## 1 answers

### Answer by username_2 on 2020-06-09. Score: 3 | Accepted answer

It sounds exactly like that!

* $\Omega$ is the image domain, it can be either discrete or continuous. No idea how it is defined in that particular book.
* $x=(x\_1, x\_2)$ are the 2D image coordinates
* $u\_0, u, n$ are functions that return intensity values at the given location $x$.


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