Radon Transform
Published:
The Radon transform is the transform of our n-dimensional volume to a complete set of (n-1)-dimensional line integrals. Whereas the inverse Radon transform is the transform from our complete (n-1)-dimensional line integrals back to the original image.
Notations
$f(x, y)$：一个二维函数，表示被检体的吸收率（或称为物质密度）
Definition
直线$L$穿过$f(x, y)$，所对应的强度即函数$f(x, y)$在直线$L$上的线积分 \(\mathcal{R}_{L}=\int_{L}f(x, y)ds\)
$\delta$函数
\[\delta(x-x_{0})= \begin{cases} \infty,\ x=x_{0}\\ 0,\ x\ne x_{0} \end{cases}\]且
\[\int_{-\infty}^{+\infty}\delta(x)dx=1\]直线表示
直线$L$远离原点的法线方向为$\vec{n}=(\cos{\theta}, \sin{\theta})$，原点到直线的距离为$p$。则直线表示为
\[x\cos{\theta}+y\cos{\theta}=p\]方向由$\theta$唯一定义，该方向上的每一点放射源由$p$唯一定义。则根据$(\theta, p)$可以定义Radon变换
\[\begin{aligned} \mathcal{R}_{L}&=\int_{L}f(x, y)ds\\ \mathcal{R}(\theta, p)&=\int_{(\theta, p)}f(x, y)ds\\ \mathcal{R}(\theta, p)&=\int\limits_{x\cos{\theta}+y\cos{\theta}-p}f(x, y)ds\\ \mathcal{R}(\theta, p)&=\iint f(x, y)\cdot \delta(p-x\cos{\theta}-y\sin{\theta})dxdy \end{aligned}\]Radon逆变换
根据不同的$(\theta, p)$求出$f(x, y)$。在积分形式下得出函数表达式，第一想到的是Fourier变换
Fourier变换：卷积性质
对
\[f(t)\star g(t)=\int f(\tau)g(t-\tau)d\tau\]Fourier变换得
\[\begin{aligned} \mathcal{F}[f(t)\star g(t)]&=\int\left[\int f(\tau)g(t-\tau)d\tau\right]e^{-j\omega t}dt\\ &=\int\left[\int f(\tau)g(t-\tau)d\tau\right]e^{-j\omega (\tau+t-\tau)}dt\\ &=\int f(\tau)e^{-j\omega\tau}\left[\int g(t-\tau)e^{-j\omega (t-\tau)}dt\right]d\tau\\ &=\int f(\tau)e^{-j\omega\tau}\hat{g}(\omega)d\tau\\ &=\hat{f}(\omega)\hat{g}(\omega) \end{aligned}\]类似地，对$\mathcal{R}_{(\theta,p)}$代入Fourier变换得
\[\begin{aligned} \mathcal{F}[\mathcal{R}(\theta, p)] & =\int\left[\iint f(x, y) \delta(p-x \cos \theta-y \sin \theta) \mathrm{d} x \mathrm{~d} y\right] e^{-j w p} \mathrm{~d} p \\ & =\int\left[\iint f(x, y) \delta(p-x \cos \theta-y \sin \theta) \mathrm{d} x \mathrm{~d} y\right] e^{-j w(p-x \cos \theta-y \sin \theta + x\cos \theta+y\sin \theta) } \mathrm{~d} p \\ & =\iint f(x, y) e^{-j w(x \cos \theta+y \sin \theta)}\left[\int \delta(p-x \cos \theta-y \sin \theta) e^{-j w(p-x \cos \theta-y \sin \theta)}\mathrm{~d} p \right]\mathrm{~d} x \mathrm{~d} y \\ & =\iint f(x, y) e^{-j w(x \cos \theta+y \sin \theta)} \hat{\delta}(w) \mathrm{d} x \mathrm{~d} y \\ & =\hat{\delta}(w) \iint f(x, y) e^{-j w(x \cos \theta+y \sin \theta)} \mathrm{d} x \mathrm{~d} y \\ & =\hat{f}(w \cos \theta, w \sin \theta) \hat{\delta}(w) \end{aligned}\]$\delta$函数的Fourier函数变换为常数1，故最终的等式为
\[\hat{f}(\omega \cos \theta, \omega\sin\theta)=\mathcal{F}[\mathcal{R}(\theta, p)]\]Funk transform
在球的中心圆上对函数进行积分。令$f$为$\mathbb{S}^2$上的连续函数，那么对于单位向量$\mathbf{x}$，有
\[\mathcal{F}f(\mathbf{x})=\int\limits_{\mathbf{u}\in C(x)}f(\mathbf{u})\mathrm{~d}s(\mathbf{u})\]其中积分是在球的中心圆$C(x)$的弧长$\mathrm{~d}s$上进行的。球的中心圆$C(x)$包含所有垂直于$\mathbf{x}$的单位向量
\[C(x)=\{\mathbf{u}\in \mathbb{S}^{2}|\mathbf{u}\cdot\mathbf{x}=0\}\]