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Hilbert Transform 📂Tomography

Hilbert Transform

Buildup

Radon Inversion

f(x,y)=12B{F−1[∣S∣F(Rf)(S, θ)]>}(x,y) f(x,y)=\dfrac{1}{2} \mathcal{B} \left\{ \mathcal{F}^{-1} \Big[ |S|\mathcal{F} (\mathcal{R}f) (S,\ \theta) \Big]> \right\} (x,y)

This formula is for obtaining ff from Radon transform Rf\mathcal{R}f of ff. First, recall the following property of the Fourier transform.

F[f′](ξ)=iξF(ξ) \mathcal{F} [f^{\prime} ] (\xi) = i\xi \mathcal{F}(\xi)

Here, if we substitute Rf\mathcal{R}f for ff, we get the following.

F(∂(Rf)(t, θ)∂t)(S, θ)=iSF(Rf)(S, θ) \begin{equation} \mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) = i S \mathcal{F}(\mathcal{R}f)(S,\ \theta) \label{eq1} \end{equation}

And let’s represent it as ∣S∣=S⋅sgn(S)|S|=S\cdot \mathrm{sgn}(S). sgn\mathrm{sgn} is the sign function.

sgn(S):={1S>00S=0−1S<0 \mathrm{sgn}(S):=\begin{cases} 1 & S>0 \\ 0 & S=0 \\ -1 & S<0 \end{cases}

Then (eq1)\eqref{eq1} is as follows.

F(∂(Rf)(t, θ)∂t)(S, θ)=i∣S∣sgn(S)F(Rf)(S, θ) \mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) = i \dfrac{|S|}{\mathrm{sgn}(S)} \mathcal{F}(\mathcal{R}f)(S,\ \theta)

If we multiply both sides by i⋅sgn(S)i \cdot \mathrm{sgn}(S), we obtain the following.

i⋅sgn(S)F(∂(Rf)(t, θ)∂t)(S, θ)=−∣S∣F(Rf)(S, θ) i \cdot \mathrm{sgn}(S)\mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) =- |S|\mathcal{F}(\mathcal{R}f)(S,\ \theta)

The right-hand side of the above equation appears in the Radon inversion. Therefore, substituting it yields the following.

f(x,y)=−12B{F−1[i⋅sgn(S)F(∂(Rf)(t, θ)∂t)(S, θ)]}(x,y) f(x,y)=-\dfrac{1}{2} \mathcal{B} \left\{ \mathcal{F}^{-1} \Big[ i \cdot \mathrm{sgn}(S)\mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) \Big]\right\} (x,y)

Definition

For g:R→Rg: \mathbb{R} \to \mathbb{R}, we define the Hilbert transform Hg\mathcal{H}g of gg as follows.

Hg(t):=F−1[i⋅sgn(S)⋅Fg(ξ)](t) \mathcal{H} g(t) :=\mathcal{F}^{-1} \big[i \cdot \mathrm{sgn} (S) \cdot \mathcal{F}g(\xi) \big] (t)

When representing the Radon inversion through the Hilbert transform, it is as follows.

f(x,y)=−12B[H(∂(Rf)(t, θ)∂t)(S, θ)](x,y) f(x,y)=-\dfrac{1}{2} \mathcal{B} \left[ \mathcal{H} \left( \dfrac{\partial (\mathcal{R}f) (t,\ \theta)}{\partial t} \right) (S,\ \theta) \right] (x,y)