📂Tomography

Back Projection: The Dual of the Radon Transform

Definition^{1 2}

The dual operator $\mathcal{R}^{\#} : L^{2}(Z_{n}) \to L^{2}(\mathbb{R}^{n})$ of the Radon transform $\mathcal{R} : L^{2}(\mathbb{R}^{n}) \to L^{2}(Z_{n})$ is referred to as back projection.

$$\left\langle \mathcal{R}f ,g \right\rangle_{L^{2}(Z_{n})} = \left\langle f , \mathcal{R}^{\#}g \right\rangle_{L^{2}(\mathbb{R}^{n})}$$

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

Theorem

Formula

Specifically, back projection is as follows.

$$\mathcal{R}^{\#} g (\mathbf{x}) = \int_{S^{n-1}} g (\mathbf{x} \cdot \boldsymbol{\theta}, \boldsymbol{\theta}) d\boldsymbol{\theta}$$

Especially in 2 dimensions,

$$\mathcal{R}^{\#} g (x,y) = \int_{0}^{2\pi} g (x\cos\theta + y\sin\theta, \theta) d\theta$$

Back Projection of the Radon Transform

The following equation holds.

$$\mathcal{R}^{\#} \mathcal{R} f = \left| S^{n-2} \right| \dfrac{1}{\left| \mathbf{x} \right|} \ast f$$

Where $\ast$ is convolution, $\left| S^{n-1} \right|$ is the surface area of a sphere in $n$ dimensions. Especially in 2 dimensions,

$$\mathcal{R}^{\#} \mathcal{R} f = \dfrac{2}{\left| \mathbf{x} \right|} \ast f$$

Explanation

Since back projection is the dual of the Radon transform, it can be thought of as a candidate for the inverse Radon transform. However, the Radon transform is not unitary, so the following does not hold.

$$\mathcal{R}^{-1} \ne \mathcal{R}^{\#}$$

Looking at the second theorem, one can see that $\mathcal{R}^{\#}\mathcal{R}f$ is similar to $f$, but not the same. In fact, it appears to be a blur of the original.

Therefore, in order to accurately obtain $f$, it must go through another operator that acts as a filter, and such an inverse Radon transform is called a filtered back projection.

Geometric Meaning and Visualization

For understanding, consider 2 dimensions. The back projection of the Radon transform is as follows.

$$\mathcal{R}^{\#} \mathcal{R} f(\mathbf{x}) =\ \int_{0}^{2\pi} \mathcal{R}f(\mathbf{x} \cdot \boldsymbol{\theta}, \theta) d \theta ,\quad \boldsymbol{\theta} = (\cos \theta, \sin \theta)$$

Where $\mathcal{R}f(\mathbf{x} \cdot \boldsymbol{\theta}, \theta)$ is, $f$ integrated over a line $l_{\mathbf{x}\cdot \boldsymbol{\theta}, \theta}$ that is at a distance $\mathbf{x} \cdot \boldsymbol{\theta}$ from the origin and perpendicular to $\boldsymbol{\theta}$. This line is a line passing through point $\mathbf{x}$ at an angle perpendicular to $\theta$.

However, since back projection is summing (integrating) value $\mathcal{R}f(\mathbf{x} \cdot \boldsymbol{\theta}, \theta)$ over all $\theta \in [0,2\pi)$, $\mathcal{R}^{\#} \mathcal{R} f(\mathbf{x})$ becomes the average (divided by $2\pi$) of the line integrals of $f$ passing through point $\mathbf{x}$.

The following pictures show the process of accumulating the value of $\mathcal{R}f(\mathbf{x} \cdot \boldsymbol{\theta}, \theta)$ from $\theta = 0$ when calculating $\mathcal{R}^{\#} \mathcal{R} f(\mathbf{x})$.

Proof

Formula

$$\begin{align*} \left\langle \mathcal{R}f ,g \right\rangle_{L^{2}(Z_{n})} =&\ \int_{\mathbb{R}}\int_{S^{n-1}} \mathcal{R}f(s, \boldsymbol{\theta}) g(s, \boldsymbol{\theta}) d\boldsymbol{\theta} ds \\ =&\ \int_{\mathbb{R}}\int_{S^{n-1}} \int_{\mathbb{R}}f(s\boldsymbol{\theta} + t\boldsymbol{\theta}^{\perp})dt g(s, \boldsymbol{\theta}) d\boldsymbol{\theta} ds \\ =&\ \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{S^{n-1}} f(s\boldsymbol{\theta} + t\boldsymbol{\theta}^{\perp}) g(s, \boldsymbol{\theta}) d\boldsymbol{\theta} ds dt \end{align*}$$

Substituting with $s \boldsymbol{\theta} + t \boldsymbol{\theta}^{\perp} = \mathbf{x}$, we get $s = \mathbf{x} \cdot \boldsymbol{\theta}$, and the following holds.

$$\begin{align*} \left\langle \mathcal{R}f ,g \right\rangle_{L^{2}(Z_{n})} =&\ \int_{\mathbb{R}^{n}}\int_{S^{n-1}} f(\mathbf{x}) g(\mathbf{x} \cdot \boldsymbol{\theta}, \boldsymbol{\theta}) d\boldsymbol{\theta} d \mathbf{x} \\ =&\ \int_{\mathbb{R}^{n}} f(\mathbf{x}) \left( \int_{S^{n-1}} g(\mathbf{x} \cdot \boldsymbol{\theta}, \boldsymbol{\theta}) d\boldsymbol{\theta} \right) d \mathbf{x} \\ =&\ \left\langle f, \left( \int_{S^{n-1}} g(\left\langle \cdot, \boldsymbol{\theta} \right\rangle, \boldsymbol{\theta}) d\boldsymbol{\theta} \right) \right\rangle_{L^{2}(\mathbb{R}^{n})} \end{align*}$$

Therefore,

$$\mathcal{R}^{\#} g (\mathbf{x}) = \int_{S^{n-1}} g (\mathbf{x} \cdot \boldsymbol{\theta}, \boldsymbol{\theta}) d\boldsymbol{\theta}$$

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Back Projection of the Radon Transform³

$$\begin{align*} \mathcal{R}^{\#} \mathcal{R} f(\mathbf{x}) =&\ \int\limits_{S^{n-1}} \mathcal{R}f(\mathbf{x} \cdot \boldsymbol{\theta}, \boldsymbol{\theta}) d \boldsymbol{\theta} \\ =&\ \int\limits_{S^{n-1}} \int\limits_{\mathbf{y} \cdot \boldsymbol{\theta} = 0} f\big((\mathbf{x} \cdot \boldsymbol{\theta})\boldsymbol{\theta} + \mathbf{y} \big) d \mathbf{y} d \boldsymbol{\theta} \\ \end{align*}$$

Here, $(\mathbf{x} \cdot \boldsymbol{\theta})\boldsymbol{\theta} = \mathbf{x} - (\mathbf{x} \cdot \boldsymbol{\theta}^{\perp})\boldsymbol{\theta}^{\perp}$, and since the second term is included in $\mathbf{y}$ which is $\mathbf{y} \cdot \boldsymbol{\theta} = 0$, the integral is as follows.

$$\mathcal{R}^{\#} \mathcal{R} f(\mathbf{x}) =\ \int\limits_{S^{n-1}} \int\limits_{\mathbf{y} \cdot \boldsymbol{\theta} = 0} f(\mathbf{x} + \mathbf{y} ) d \mathbf{y} d \boldsymbol{\theta}$$

Auxiliary Theorem

$$\int \limits_{S^{n-1}} \int \limits_{\mathbf{y} \cdot \boldsymbol{\theta} = 0} f(\mathbf{x} + \mathbf{y}) d \mathbf{y} d \boldsymbol{\theta} = \left| S^{n-2} \right| \int \limits_{\mathbb{R}^{n}} \dfrac{f(\mathbf{y})}{\left| \mathbf{x} - \mathbf{y} \right|}d \mathbf{y}$$

Then, by the auxiliary theorem,

$$\mathcal{R}^{\#} \mathcal{R} f(\mathbf{x}) = \left| S^{n-2} \right| \int \limits_{\mathbb{R}^{n}} \dfrac{f(\mathbf{y})}{\left| \mathbf{x} - \mathbf{y} \right|}d \mathbf{y} = \left| S^{n-2} \right| \dfrac{1}{\left| \mathbf{x} \right|} \ast f$$

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