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Fourier Slice Theorem 📂Tomography

Fourier Slice Theorem

Theorem

For f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R}, the following equation holds:

F2f(ξcos⁡θ, ξsin⁡θ)=F(Rf)(ξ, θ) \begin{equation} \mathcal{F}_2 f(\xi \cos\theta,\ \xi \sin\theta)=\mathcal{F}(\mathcal{R}f)(\xi ,\ \theta) \label{thm1} \end{equation}

Here, F\mathcal{F} represents the 1-dimensional Fourier transform, F2\mathcal{F}_2 represents the 2-dimensional Fourier transform, and R\mathcal{R} is the Radon transform.

Ff(y)=∫f(x)e−ixydxF2f(y1,y2)=∫∫f(x1,x2)e−i(x1,x2)⋅(y1,y2)dx1dx2Rf(s,θ)=∫t=−∞∞f(scos⁡θ−tsin⁡θ, ssin⁡θ+tcos⁡θ)dt \begin{align*} \mathcal{F}f (y) &= \int f(x) e^{-i xy } dx \\ \mathcal{F}_{2} f (y_{1}, y_{2}) &= \int \int f(x_{1}, x_{2}) e^{-i (x_{1}, x_{2}) \cdot (y_{1}, y_{2})} dx_{1} dx_{2} \\ \mathcal{R}f(s,\theta) &= \int_{t =-\infty}^{\infty} f\big( s\cos\theta-t\sin\theta,\ s\sin\theta + t\cos\theta \big) dt \end{align*}

Explanation

It is also called the projection slice theorem or the central slice theorem.

Proof

When calculating the left-hand side of (thm1)\eqref{thm1}, it turns out like this:

F2f(ξcos⁡θ, ξsin⁡θ)=∫−∞∞∫−∞∞f(x,y)e−i(ξcos⁡θ⋅x+ξsin⁡θ⋅y)dxdy=∫−∞∞∫−∞∞f(x,y)e−iξ(xcos⁡θ+ysin⁡θ)dxdy \begin{equation} \begin{aligned} \mathcal{F}_2 f(\xi \cos\theta,\ \xi \sin\theta) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)e^{-i(\xi \cos\theta \cdot x +\xi \sin\theta \cdot y) }dxdy \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)e^{-i \xi (x\cos\theta +y\sin\theta ) }dxdy \end{aligned} \label{eq1} \end{equation}

And to express a point on the plane in polar coordinates as shown below, let’s replace it as shown in the following figure.

5CF9E0040

s=xcos⁡θ+ysin⁡θt=−xsin⁡θ+ycos⁡θ s=x\cos\theta + y\sin\theta \\ t=-x\sin\theta+y\cos\theta

Then

x=scos⁡θ−tsin⁡θy=ssin⁡θ+tcos⁡θ x=s\cos\theta-t\sin\theta \\ y=s\sin\theta + t \cos \theta

And since dxdy=dsdtdxdy=dsdt, when substituting for (eq1)\eqref{eq1}

F2f(ξcos⁡θ, ξsin⁡θ)=∫−∞∞∫−∞∞f(scos⁡θ−tsin⁡θ, ssin⁡θ+tcos⁡θ)e−iξsdtds=∫−∞∞(∫−∞∞f(scos⁡θ−tsin⁡θ, ssin⁡θ+tcos⁡θ)dt)e−iξsds=∫−∞∞Rf(s, θ)e−iξsds=FRf(ξ, θ) \begin{align*} \mathcal{F}_2 f( \xi \cos\theta,\ \xi \sin\theta) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(s\cos\theta-t\sin\theta,\ s\sin\theta + t \cos \theta)e^{-i\xi s}dtds \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} f(s\cos\theta-t\sin\theta,\ s\sin\theta + t \cos \theta)dt \right)e^{-i \xi s} ds \\ &= \int_{-\infty} ^{\infty} \mathcal{R}f(s,\ \theta) e^{-i\xi s} ds \\ &= \mathcal{F} \mathcal{R}f(\xi,\ \theta) \end{align*}

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