📂Tomography

n-Dimensional Radon Transform

Definition¹

For $s \in \mathbb{R}^{1}$, $\boldsymbol{\theta} \in S^{n-1}$, the Radon transform $\mathcal{R} : L^{2}(\mathbb{R}^{n}) \to L^{2}(Z_{n})$ is defined as follows.

$$ \mathcal{R} f (s, \boldsymbol{\theta}) = \int\limits_{\mathbf{x} \cdot \boldsymbol{\theta} = s} f(\mathbf{x}) d \mathbf{x} $$

Here, $Z_{n} := \mathbb{R}^{1} \times S^{n-1}$ is a unit cylinder in $n+1$ dimensions.

Description

The geometric meaning of $\mathcal{R} f (s, \boldsymbol{\theta})$ is to integrate over all points that are $s$ away from the origin and perpendicular to $\boldsymbol{\theta}$.

$$ \mathbf{x} \cdot (-\boldsymbol{\theta}) = -s \iff \mathbf{x} \cdot \boldsymbol{\theta} = s\quad \forall \mathbf{x} \in \mathbb{R}^{n} $$

Since the above equation holds, the Radon transform is an even function.

$$ \mathcal{R}f(-s, -\boldsymbol{\theta}) = \mathcal{R}f(s, \boldsymbol{\theta}) $$

Alternative Expression

For a fixed $\boldsymbol{\theta}$,

$$ \mathcal{R} f (s, \boldsymbol{\theta}) = \mathcal{R}_{\boldsymbol{\theta}}f (s) $$

Let $\boldsymbol{\theta} ^{\perp} := \left\{ \mathbf{u} : \mathbf{u} \cdot \boldsymbol{\theta} = 0 \right\}$. Then,

$$ \mathcal{R} f (s, \boldsymbol{\theta}) = \int\limits_{\boldsymbol{\theta}^{\perp}} f( s \boldsymbol{\theta} + \mathbf{u}) d \mathbf{u} $$

Regarding the Dirac delta function $\delta$,

$$ \mathcal{R} f (s, \boldsymbol{\theta}) = \int\limits_{\mathbb{R}^{n}} f( \mathbf{x} ) \delta ( \mathbf{x} \cdot \boldsymbol{\theta} - s) d \mathbf{x} $$

Theorem

Let $f \in L^{1}(\mathbb{R})$. Then $\mathcal{R}_{\boldsymbol{\theta}}f(s) \in L^{1}(\mathbb{R})$, and the following holds.

$$ \int\limits_{-\infty}^{\infty} \mathcal{R}_{\boldsymbol{\theta}}f(s) ds = \int\limits_{\mathbb{R}^{n}}f(\mathbf{x})d \mathbf{x} $$

Proof

$\mathcal{R}_{\boldsymbol{\theta}}f(s)$ is integrated over a plane that is $t$ away from the origin and perpendicular to $\boldsymbol{\theta}$. Integrating this over all $s \in \mathbb{R}$ is the same as integrating $f$ over all points of $\mathbb{R}^{n}$. Therefore,

$$ \int\limits_{-\infty}^{\infty} \mathcal{R}_{\boldsymbol{\theta}}f(s) ds = \int\limits_{\mathbb{R}^{n}}f(\mathbf{x})d \mathbf{x} \lt \infty $$

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