The properties of the Radon transform

The basic properties of the Radon transform

The properties of the Radon transform to be stated here are also valid for more dimensions, we restrict ourselves to 2D cases as in the medical practice it is the most relevant.

Symmetry

The parameter set of $$ t\in \left \{ 0,\infty \right \}$ and $$ \vartheta\in \left \{ 0,2\pi \right \}$ describes every element of the Radon transform, since

$$ \mathfrak{R}f\left ( t,\vartheta \right )=\mathfrak{R}f\left ( -t,\vartheta+\pi \right )$

Linearity

From the definition of the Radon transform follows that for $$\alpha _{i}$ constants and f_i functions
$$\mathfrak{R}\left ( \sum_{i}\alpha _{i}f_{i} \right )= \sum_{i}\alpha _{i}\mathfrak{R}f_{i} \right )$

Shifting theorem

Let
$$ h\left ( x,y \right )=g\left ( x-x_{0},y-y_{0} \right )$
Then
$$ \mathfrak{R}h\left ( t,\vartheta \right )=\int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( x-x_{0}, y-y_{0} \right ) \delta \left ( t-x\cos \vartheta -y\sin \vartheta\right )dxdy$
let us introduce new variables:
$$x':=x-x_{0},y':=y-y_{0}$
then:
$$\mathfrak{R}h\left ( t,\vartheta \right )=\int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( x', y' \right ) \delta \left ( t-x_{0}\cos \vartheta -y_{0}\sin \vartheta - x'\cos \vartheta -y'\sin \vartheta \right )dx'dy'$
Let us introduce a new notation for t:
$$t'=t-x_{0}\cos \vartheta -y_{0}\sin \vartheta$
Now:
$$ \int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( x', y' \right ) \delta \left ( t' - x'\cos \vartheta -y'\sin \vartheta \right )dx'dy'=\mathfrak{R}g\left ( t', \vartheta\right )$

A shift thus will not alter the variable $$ \vartheta$ of the Radon transform, it only affects the affine t paramter by the transform of $$ \vartheta$ .

Rotation

For simpler notations let us work in polar coordiantes: $$(r,\varphi)$ ,
then a 2D Radon transform reads:
$$\mathfrak{R}f\left ( t, \vartheta\right )=\int_{-\infty }^{\infty}\int_{-\infty }^{\infty}f \left ( r, \varphi \right ) \delta \left ( t - r\cos \varphi\cos \vartheta -r\sin \varphi\sin \vartheta \right )\left | r \right |drd\varphi= \int_{-\infty }^{\infty}\int_{-\infty }^{\infty}f \left ( r, \varphi \right ) \delta \left ( t - r\cos \left (\varphi-\vartheta \right )\right )\left | r \right |drd\varphi$

Now let us choose an angle of rotation of $$ \varphi_{0}$ , now the function is:
$$ h\left ( r,\varphi\right )=g\left ( r,\varphi- \varphi_{0}\right )$

Thus:
$$\mathfrak{R}h\left ( t, \vartheta\right )=\int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( r, \varphi-\varphi_{0} \right ) \delta \left ( t - r\cos \left (\varphi-\vartheta \right ) \right )\left | r \right |drd\varphi$
Let us introduce new notations:
$$\varphi':=\varphi-\varphi_{0}$

Now:
$$\mathfrak{R}h\left ( t, \vartheta\right )=\int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( r, \varphi' \right ) \delta \left ( t - r\cos \left (\varphi'+\varphi_{0}-\vartheta \right ) \right )\left | r \right |drd\varphi' =\mathfrak{R}g\left ( t, \varphi_{0}-\vartheta\right )$

Scaling

Let us choose 0<a, 0<b constants, and scale the variables as follows:

$$ h\left ( x,y \right )=g\left ( \frac{x}{a},\frac{y}{b} \right )$

the Radon transform of "h" is:

$$\mathfrak{R}h\left ( t,\vartheta \right )=\int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( \frac{x}{a}, \frac{y}{b} \right ) \delta \left ( t-x\cos \vartheta -y\sin \vartheta\right )dxdy$

Again w choose new notations with the $$x':=\frac{x}{a}, y:=\frac{y}{b}$ and $$\gamma$ variables:
$$\mathfrak{R}h\left ( t,\vartheta \right )=\frac{ab}{\left |\gamma \right | } \int_{-\infty }^{\infty}\int_{-\infty }^{\infty}g \left ( x', y' \right ) \delta \left ( \frac{t}{\gamma }-x'\frac{a\cos \vartheta}{\gamma } -y'\frac{b\sin \vartheta}{\gamma }\right )dx'dy'$
Let us asssume that $$\gamma$ can be shosen such that $$ \vartheta$ is also replaceable by a $$ \vartheta'$ variable as follows:

$$ \cos\vartheta'=\frac{a\cos \vartheta}{\gamma }$ (1)
and
$$ \sin\vartheta'=\frac{b\sin \vartheta}{\gamma }$ (2)

For that we have to normalize first the squared sums:

$$\cos ^{2}\vartheta'+\sin ^{2}\vartheta'=1$
from that:
$$ a^{2}\cos ^{2}\vartheta+b^{2}\sin ^{2}\vartheta=\gamma^{2}$

We can express $$\gamma$ from Eq. (1) and (2) and equate to:
$$\frac{a\cos \vartheta}{\cos \vartheta'}=\gamma =\frac{b\sin \vartheta}{\sin \vartheta'}$
Thus:
$$\tan^{-1}\left ( \frac{a}{b}\tan \vartheta \right )=\vartheta'$

Now:
$$\mathfrak{R}h(t,\vartheta )=\frac{ab}{\sqrt{a^{2}\cos ^{2}\vartheta +b^{2}\sin ^{2}\vartheta}}\int_{-\infty }^{\infty }g(x',y')\delta \left ( t'-x'\cos \vartheta '-y\sin \vartheta ' \right )dx'dy'=\mathfrak{R}g\left ( t',\vartheta ' \right )$

here then
$$t'=\frac{t}{\sqrt{a^{2}\cos ^{2}\vartheta +b^{2}\sin ^{2}\vartheta}}$ and

$$\vartheta'=\tan^{-1}\left ( \frac{a}{b}\tan \vartheta \right )$

Convolution

Let
$$ h\left ( x,y \right )=f*^{x}*^{y}g=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }f\left ( x_{1},y_{1} \right )g\left ( x-x_{1},y-y_{1} \right ) dx_{1}dy_{1}$
where the $$*^{x}*^{y}$ notation is the convolution in x and y dimensions respectively.
The Radon transform ofh is:
$$ \mathfrak{R}h= \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }f\left ( x_{1},y_{1} \right )g\left ( x-x_{1},y-y_{1} \right )\delta \left ( t-x\cos \vartheta -y\sin \vartheta \right ) dx_{1}dy_{1}dxdy$

The dxdy integral is the Radon transform of the f function shifted to the points $$ \left ( x_{1},y_{1} \right )$ , using the shift theorem:
$$\mathfrak{R}h= \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }f\left ( x_{1},y_{1} \right )\mathfrak{R}g\left ( t-x_{1}\cos \vartheta -y_{1}\sin \vartheta,\vartheta\right ) dx_{1}dy_{1}$
Let us insert a new integration of t₁ with a Dirac delta function, and carry out the integration according to $$x_{1},y_{1}$ :
$$\mathfrak{R}h= \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }f\left ( x_{1},y_{1} \right )\mathfrak{R}g\left ( t-t_{1},\vartheta\right )\delta \left ( t_{1}-x_{1}\cos \vartheta -y_{1}\sin \vartheta \right ) dx_{1}dy_{1}dt_{1}= \int_{-\infty }^{\infty }\mathfrak{R}f\left ( t_{1},\vartheta \right )\mathfrak{R}g\left ( t-t_{1},\vartheta\right )dt_{1}$
This is just a convoluton according to the t affine parameter. With a more compact notation:
$$ \mathfrak{R}h=\mathfrak{R} \left [ f*^{x}*^{y}g \right ] =\left [\mathfrak{R}f \right ]*^{t}\left [\mathfrak{R}g \right ]$

So the convolution in the two real spatial dimensions is a convolution only in the affine parameter in the sinogram space. This has an important consequence on the reconstruction.

In the next chapter we will discuss the numerical implementation of the Radon transform and multi dimension Radon transform.

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