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라돈 변환과 곱 적분, 컨볼루션

정리¹

$\mathcal{R}$을 라돈 변환이라고 하자.

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

$\mathcal{R}_{\boldsymbol{\theta}}f(s) = \mathcal{R} f (s, \boldsymbol{\theta})$라고 하자. 다음의 공식들이 성립한다.

라돈 변환과 곱 적분

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

따름정리

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

라돈 변환과 컨볼루션

$$\mathcal{R}_{\boldsymbol{\theta}} (f \ast g) = \mathcal{R}_{\boldsymbol{\theta}}f \ast \mathcal{R}_{\boldsymbol{\theta}}g$$

증명

라돈 변환과 곱 적분

$s \boldsymbol{\theta} + \mathbf{u} \equiv \mathbf{x}$라고 치환하면,

$$\begin{align*} \int\limits_{-\infty}^{\infty} \mathcal{R}_{\boldsymbol{\theta}}f(s)g(s) ds =&\ \int\limits_{-\infty}^{\infty} \int\limits_{\boldsymbol{\theta}^{\perp}} f(s \boldsymbol{\theta} + \mathbf{u})g(s) d \mathbf{u} ds \\ =&\ \int \limits_{\mathbb{R}^{n}} f(\mathbf{x}) g(\mathbf{x} \cdot \boldsymbol{\theta}) d \mathbf{x} \end{align*}$$

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따름정리

$$\begin{align*} \int\limits_{-\infty}^{\infty} \mathcal{R}_{\boldsymbol{\theta}}f(t-s)g(s) ds =&\ \int\limits_{-\infty}^{\infty} \mathcal{R}f(t-s, \boldsymbol{\theta})g(s) ds \\ =&\ \int\limits_{-\infty}^{\infty} \int\limits_{\boldsymbol{\theta}^{\perp}} f((t-s) \boldsymbol{\theta} + \mathbf{u})g(s) d \mathbf{u} ds \\ \end{align*}$$

$(t-s)\boldsymbol{\theta} + \mathbf{u} \equiv \mathbf{x}$라 치환하면, $s = -\mathbf{x} \cdot \boldsymbol{\theta} + t$이고,

$$\int\limits_{-\infty}^{\infty} \int\limits_{\boldsymbol{\theta}^{\perp}} f((t-s) \boldsymbol{\theta} + \mathbf{u})g(s) d \mathbf{u} ds = \int \limits_{\mathbb{R}^{n}} f(\mathbf{x}) g(-\mathbf{x} \cdot \boldsymbol{\theta} + t) d \mathbf{x}$$

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라돈 변환과 컨볼루션

라돈 변환의 평행 불변성 $\mathcal{R}T_{\mathbf{a}}f (s, \boldsymbol{\theta}) = T_{\mathbf{a} \cdot \boldsymbol{\theta}}\mathcal{R}f(s,\boldsymbol{\theta})$에 의해,

$$\begin{align*} \mathcal{R}_{\boldsymbol{\theta}} (f \ast g) (s) =&\ \mathcal{R} (f \ast g) (s, \boldsymbol{\theta}) \\ =&\ \int \limits_{\boldsymbol{\theta}^{\perp}} f \ast g (s \boldsymbol{\theta} + \mathbf{u} ) d \mathbf{u} \\ =&\ \int \limits_{\boldsymbol{\theta}^{\perp}} \int \limits_{\mathbb{R}^{n}} f (s \boldsymbol{\theta} + \mathbf{u} - \mathbf{y} ) g(\mathbf{y}) d \mathbf{y} d \mathbf{u} \\ =&\ \int \limits_{\boldsymbol{\theta}^{\perp}} \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) g(s \boldsymbol{\theta} + \mathbf{u} - \mathbf{x}) d \mathbf{x} d \mathbf{u} \\ =&\ \int \limits_{\mathbb{R}^{n}} f (\mathbf{x})\int \limits_{\boldsymbol{\theta}^{\perp}} g(s \boldsymbol{\theta} + \mathbf{u} - \mathbf{x}) d \mathbf{u} d \mathbf{x} \\ =&\ \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) \int \limits_{\boldsymbol{\theta}^{\perp}} T_{\mathbf{x}} g(s \boldsymbol{\theta} + \mathbf{u}) d \mathbf{u} d \mathbf{x} \\ =&\ \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) \mathcal{R} T_{\mathbf{x}} g(s, \boldsymbol{\theta}) d \mathbf{x} \\ =&\ \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) T_{\mathbf{x} \cdot \boldsymbol{\theta}} \mathcal{R} g(s, \boldsymbol{\theta}) d \mathbf{x} \\ =&\ \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) \mathcal{R} g(s - \mathbf{x} \cdot \boldsymbol{\theta}, \boldsymbol{\theta}) d \mathbf{x} \\ =&\ \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) \mathcal{R}_{\boldsymbol{\theta}} g(s - \mathbf{x} \cdot \boldsymbol{\theta}) d \mathbf{x} \\ \end{align*}$$

위의 따름정리에 의해,

$$\begin{align*} \int \limits_{\mathbb{R}^{n}} f (\mathbf{x}) \mathcal{R}_{\boldsymbol{\theta}} g(s - \mathbf{x} \cdot \boldsymbol{\theta}) d \mathbf{x} =&\ \int\limits_{-\infty}^{\infty} \mathcal{R}_{\boldsymbol{\theta}} f(s-t) \mathcal{R}_{\boldsymbol{\theta}} g(t) dt \\ =&\ (\mathcal{R}_{\boldsymbol{\theta}}f \ast \mathcal{R}_{\boldsymbol{\theta}}g )(s) \end{align*}$$

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