# Medical Imaging

\(\newcommand{\R}{\mathbb{R}}\)

\(\newcommand{\RR}{\mathbb{R}}\)

\(\newcommand{\C}{\mathbb{C}}\)

\(\newcommand{\N}{\mathbb{N}}\)

For the application of medical imaging, you have the following setup. Assume you want to map out the internal density of some 2D region, i.e. find the density function \(\mu(x) \in \R\), for \(x\in \R\).

Characterize a line through the region by its angle and offset, respective \((\rho, phi)\), for \(\rho \in \R\), \(\phi \in [0,\pi]\).

We can also have a 2D line \(\delta\) for \((\rho,\phi)\), written \(\delta(\rho - x_1\cos\phi-x_2\sin\phi)\) which turns out, when integrated on \(dx_1\wedge dx_2\) against a function \(f\), to give the line integral of \(f\) along that line.

You know that material density affects the speed of the rays you can send through the region, so we get the line integral for a given line, we have:

Set \(\mathcal{R}\mu=M\). \(\mathcal{R}\) is then called the Radon transform, and the goal of imaging is to find its inverse.

As it turns out, it's not hard to show that, for \(\mathcal{F}\) the 2D Fourier transform, \(\mathcal{F}\_{\rho}\) the 1D Fourier transform in \(\rho\), and \((\xi_1,\xi_2) = (r\cos\theta, r\sin\theta)\):

So we have a way of recovering \(\mu\) from the mangling it undergoes by \(\mathcal{R}\).