PDF Print

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.


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 )


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

Shifting theorem

$ h\left ( x,y \right )=g\left ( x-x_{0},y-y_{0} \right )
$  \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:
$\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
\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.


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 )

$\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:

$\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 )


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

$\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 )


$  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:
\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}:
\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.

Site Language: English

Log in as…