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Interpretation of the inverse Radon transform

Analysis of the Fourier inversion formula

$ \frac{1}{2}\left (\frac{1}{2\pi}  \right )^{n-1}\int_{\mathbb{S}^{n-1}}\int_{-\infty}^{\infty}
\mathfrak{F}\left [ f \right ]\left ( r\boldsymbol{\omega}  \right)\left | r \right |^{n-1}
e^{ir\mathbf{x} \boldsymbol{\omega }}dr d\boldsymbol{\omega }=
\frac{1}{2}\left (\frac{1}{2\pi}  \right )^{n-1}\int_{\mathbb{S}^{n-1}}
\mathfrak{F}^{-1}_{r} \left [ \mathfrak{F}_{t}\mathfrak{R}f \right ] \left | r \right |^{n-1}
 d\boldsymbol{\omega }
Let us look at the terms of the inversion formula again.

The adjoint to the Radon-transform

For the outermost integral $ \left (\frac{1}{2\pi}  \right )^{n-1}\int_{\mathbb{S}^{n-1}} d\boldsymbol{\omega } we have already introduced the notation of $\mathfrak{R^{+}} . To an $\mathfrak{A} operator the definition of the adjoint $\mathfrak{A^{+}} operator reads:
$ \left \langle \mathfrak{A}f,g \right \rangle=\left \langle f,\boldsymbol{\mathfrak{A}^{+}}g \right \rangle
here <f,g> is the scalar product of f and g . The adjoint to the Radon transform is the backprojection operator:
$ \int \int \mathfrak{R}f\left ( t,\boldsymbol{\omega }\right )h\left ( t,\boldsymbol{\omega } \right )dtd\boldsymbol{\omega }=\int\int \int f\left ( t\boldsymbol{\omega }+\mathbf{y}\right )h\left (t, \boldsymbol{\omega } \right )dtd\boldsymbol{\omega }d\mathbf{y}=\int\int \int f\left ( \mathbf{x}\right )h\left (\boldsymbol{\omega }\mathbf{x}, \boldsymbol{\omega } \right )d\mathbf{x}d\boldsymbol{\omega }=\int \int f\left ( \mathbf{x}\right )\mathfrak{R}^{+}h\left (\mathbf{x} \right )d\mathbf{x}
Here we have changed variables from the original t and y to the rotated x variables in the integration.

The Hilbert transform hidden in the Radon inversion formula

In the next step let us look at the terms disregarding the backprojection operator:
$ \mathfrak{F}^{-1}_{r} \left [ \mathfrak{F}_{t}\mathfrak{R}f \right ] \left | r \right |^{n-1}=\mathfrak{F}^{-1}_{r}\left ( -ir \right)^{n-1}\left ( i sgn(r)\right)^{n-1} \left [ \mathfrak{F}_{t}\mathfrak{R}f \right ]
In general it is true to a funcion g that
$\mathfrak{F}^{-1}_{r} \left ( -ir  \right )^{n-1} \mathfrak{F}_{t} g= \partial _{t}^{n-1}g
We have shown about the Hilbert Transform that
$ \mathfrak{F}^{-1}_{r}  i sgn\left ( r \right )  \mathfrak{F}_{t} g=  \mathfrak{H}_{t} g
from this:
$ f=\frac{1}{2}\left ( 2\pi  \right )^{n-1}\mathfrak{R}^{+}\mathfrak{H}^{n-1}\partial_{t}^{n-1}\mathfrak{R}f
As the sgn function is present, the formula
$\mathfrak{H}\mathfrak{H}g\left ( x \right )=-g\left ( x \right ) behaves differently depending on whether n is even or odd.

Now we can also write the odd and even terms separately:
$ f=\frac{1}{2}\left ( 2\pi  \right )^{n-1}\left\{\begin{matrix}
 \left (-1  \right )^{\left (n-2  \right )/2}\mathfrak{R}^{+}\mathfrak{H}^{n-1}\partial_{t}^{n-1}\mathfrak{R}f & \text{, if n even}\\ 
 \left (-1  \right )^{\left (n-1  \right )/2}\mathfrak{R}^{+}\partial_{t}^{n-1}\mathfrak{R}f & \text{, if n odd}

The lack of the Hilbert transform in the even dimensions have a fundamental impact:

  • when the Hilbert transform is missing from the formula, for the reconstruction of a certain point of the distribution we need the Radon transform on hyperplanes going through only the small neighborhood of the point
  • when the Hilbert transform is present, for the reconstruction at a point we need the whole sinogram

Note that at n=2 we obtain Radon's inversion formula. There are further analytical solutions to the inverse Radon problem, we will be dealing with that in the next section.

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