📂Tomography

Radon Transformation

Definition

Let’s assume that a function $f :D \to \mathbb{R}$ is defined on some 2D domain $D\subset \mathbb{R}^{2}$. The Radon transform $\mathcal{R}f$ of $f$ is defined as follows, for $s \in \mathbb{R}$, $\boldsymbol{\theta} = (\cos \theta, \sin \theta) \in S^{1}$,

$$\begin{align*} \mathcal{R} f(s, \boldsymbol{\theta}):=&\ \int \limits_{t=-\infty}^{\infty} f ( s \boldsymbol{\theta} + t \boldsymbol{\theta}^{\perp} )dt \\ =&\ \int \limits_{t=-\infty} ^{\infty} f \left( s\cos\theta-t\sin\theta, s\sin\theta + t\cos\theta \right)dt \end{align*}$$

Explanation

Radon transform is a type of integral transform named after the Austrian mathematician Johann Radon (1887-1956).

The radioactive element radon is not named after the mathematician Radon, but rather its name comes from the word ‘radioactive’ with the inert gas suffix ‘-on’ added.

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

While $f$ is a function of Cartesian coordinates $(x, y)$, the Radon transform $\mathcal{R}f$ is a function of polar coordinates $(s, \theta)$.

The Radon transform is one of the core principles of CT and is based on the physical law known as Beer-Lambert’s law. This law states that the intensity of X-rays decreases differently depending on the type of medium they pass through. A reduction in the intensity of X-rays means that the medium has absorbed the X-rays. The extent to which a medium absorbs light is referred to as the attenuated coefficient, absorption coefficient, or absorbance. The fact that different media have different attenuation coefficients is utilized in non-destructive inspection using X-rays in CT. The reason why bones appear white in X-ray images is because bones absorb more X-rays than other materials.

Another Expression of the Definition

When $l_{s, \theta}$ is considered as a line determined by polar coordinates $(s,\theta)$,

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

Thinking about the geometric meaning,

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

When defined as $\boldsymbol{\theta}^{\perp} := \left\{ \mathbf{u} : \mathbf{u} \cdot \boldsymbol{\theta} = 0 \right\}$,

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

Regarding the Dirac delta function $\delta$,

$$\mathcal{R} f (s, \boldsymbol{\theta}) = \int\limits_{\mathbb{R}^{2}} f( \mathbf{x} ) \delta ( \mathbf{x} \cdot \boldsymbol{\theta} - s) d \mathbf{x}$$

Generalization

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.

Derivation¹

Let’s designate $x$ as the position, $I(x)$ as the intensity of X-rays, and $A(x)$ as the attenuation coefficient of the medium.

Beer-Lambert Law

The rate of change of the intensity of X-rays is as follows.

$$\begin{equation} \frac{ dI }{ dx } = -A(x)I(x) \end{equation}$$

Let $x_{0}$ and $x_{1}$ represent the starting and ending positions of the X-rays, respectively, and $I_{0}$ and $I_{1}$ represent the intensity of the X-rays at each point. When we separate the variables in $(1)$ and integrate both sides, we obtain the following.

$$\begin{align*} && \int_{x_{0}}^{x_{1}} \frac{1}{I(x)}dI &= - \int_{x_{0}}^{x_{1}}A(x)dx \\ \implies && \ln \left( I_{1} \right) - \ln \left( I_{0} \right)&= -\int_{x_{0}}^{x_{1}}A(x)dx \\ \implies && \ln \left( \frac{I_{1}}{I_{0}}\right) &= -\int_{x_{0}}^{x_{1}}A(x)dx \\ \implies && \ln \left( \frac{I_{0}}{I_{1}}\right) &= \int_{x_{0}}^{x_{1}}A(x)dx \end{align*}$$

Looking at this equation, $I_{0}$ is the intensity when the X-rays are shot, which is a known value. $I_{1}$ is the intensity after the X-rays have passed through the object, and this value is measured by the detector located at $x_{0}$. Therefore, the left side is a known value.

The range of integration on the right side is the path traveled by the X-rays we shot, so it is known. Hence, given the path $L$ of the X-rays and the intensities $I_{0}$, $I_{1}$ at both ends, we can obtain the value obtained by integrating $A(x)$ over the path $L$. This is referred to as the Radon transform of $A(x)$.

$$\mathcal{R}f (L) := \int_{L} f(x) dx = \ln \left( \frac{I_{0}}{I_{1}}\right)$$

Properties

The basic properties of the Radon transform are as follows.

• Linearity

  $$\mathcal{R} \left( \alpha f + \beta g \right) = \alpha \mathcal{R}f + \beta \mathcal{R}g$$

• Shift Invariance

  $$\mathcal{R}T_{\mathbf{a}}f (s, \boldsymbol{\theta}) = T_{\mathbf{a} \cdot \boldsymbol{\theta}}\mathcal{R}f(s,\boldsymbol{\theta})$$

• Rotation Invariance

  $$RAf = ARf$$

• Dilation Invariance

  $$RD_{r}f = \dfrac{1}D_{r}Rf$$