[Fourier transform and its application study notes] 30. Radon Transform

X-ray tomography (tomography)

There are two different imaging methods in business: CT, MRI, and the two methods have some similarities in the method of realization, which is about Ct.

A body profile is assumed to contain a variety of sticky substances, such as bones, muscles, blood vessels, and spinal cords, described by variable density function $\mu (x_1,x_2) $. If we know what $\mu$ is, it means that we can know the condition of the body profile.

About $\mu$, in the notes Professor recommended us to read a book, "Naked to the Bone:medical Imaging in the twentieth Centry", the author is Bettyann Kevels, the book is described in a paragraph so $\mu$ Of

Dimmer and Dimmer What happens the light passes through murky water? It gets dimmer and dimmer the farther it goes, of Couse–this is not a trick question. If the water is the same murkiness throughout, meaning, for example, uniform density of stuff floating around in it, then It's natuarl to assume so the intensity of light decreases by the same percent amount per length of path traveled. Through absorption, scattering, etc, whatever intensity comes in, a certain percentage of this intensity goes out; Over a given distance the murky water removes a percentage of light, and this percentage depends only on the distance Trav Eled and not where the starting and stopping points is. We ' re assuming here, light are traveling in a straight line through the water.

Constant percent change characterizes exponential growth, or decay, and the attenuation of the intensity of light passing t Hrough a homogeneous medium is modeled by

$I =i_0e^{-\mu x}$

Where $I _0$ is the initial intensity, $x $ are the distance traveled, and $\mu$ is a (positive) "Murkiness constant", $x $ ha s dimension of length and $\mu$ has dimension 1/length and units "Murkiness/length". $\mu$ is constant because we assume, the medium is homogeneous. We know the value of $I _0$, and one measurement of $x $ and $I $ would determine $\mu$. In fact, what we do are to put a detector at a known distance $x $ and measure the intensity when it arrives at the detector .

What happens when the light goes through the turbid water? The farther the light goes through the muddy water, the darker it becomes, of course, the obvious question. What if the water has the same degree of turbidity? The same turbidity means that something with a uniform density floats in the water, so it is natural to assume that the intensity of the light will have the same percentage of attenuation without passing through the same unit length. Light in the incident, regardless of the intensity of the light, will be due to absorption, scattering and other reasons, in the exit is only equivalent to a certain percentage of the intensity of the incident light, in a certain distance of the muddy water, a certain proportion of the light will be eliminated, and this proportion depends only on the distance of light transmission rather than entry out point. Here we think that light travels in a straight line.

The change of light intensity is exponential growth, or attenuation, so the light intensity attenuation model is as follows when the light travels in a homogeneous medium.

$I =i_0e^{-\mu x}$

Where $i_0$ is the initial light intensity, $x $ is the propagation distance, $\mu$ is (positive) turbidity coefficient, $x $ is measured in length (length), $\mu$ in the inverse of length (murkiness/length), $\mu$ is a constant, Because we assume that the medium is homogeneous. We know $i_0$, $x $ can be measured, $I $ determines how much $\mu$ is. In fact, all we have to do is place a measuring instrument with a light intensity at a known distance of $x$.

Now suppose the water isn't uniformly murky, but rather the light passes through a number of layers, each layer of Unifor M murkiness. If the i ' th layer has murkiness costant $\mu_i$ and is length $\delta x_i$, and if there are $n $ layers, then the intensit Y of light that reaches the detector can is modeled by

$\displaystyle{i=i_0 exp\left (–\sum_{i=1}^{n}\mu_i\delta x_i \right)}$

Clearly, if the murkiness is describled by a function $\mu (x) $, then the intensity arriving at the detector are modeled by

$\displaystyle{i=i_0 Exp\left (–\int_l \mu (x) dx \right)}$

Where $L $ is the line the light travels along. It ' s common to call the number

$p =\displaystyle{\int_l \mu (x) dx =–ln\left (\frac{i}{i_0} \right)}$

Now suppose that the water is not uniformly cloudy, then we can make the passage of light as a lot of cloudy small pieces. If the turbidity of the $i$ block is $\mu_i$, and its length is $\delta x_i$, there are altogether $n$ small pieces, then the light intensity model becomes

$\displaystyle{i=i_0 exp\left (–\sum_{i=1}^{n}\mu_i\delta x_i \right)}$

If the turbidity of the water in this light is expressed in $\MU (x) $, then the light intensity model becomes

$\displaystyle{i=i_0 Exp\left (–\int_l \mu (x) dx \right)}$

$L $ represents the light (the path through which light passes), usually the following number is the attenuation factor

$p =\displaystyle{\int_l \mu (x) dx =–ln\left (\frac{i}{i_0} \right)}$

Radon transform (the Radon Transform)

When we are imaging, we need to know the density of the entire body profile, that is $\mu (x_1,x_2) $, then the above equation becomes

$\displaystyle{\mathcal{r}_l\mu = \int_l \mu (x_1,x_2) dx_1dx_2}$

This expression represents the $\mu$ transformation, where $\mathcal{r}$ represents the radon transformation, the variable is a straight line $l$, and the transformation of the different lines will result in different results.

So what does it mean to point the part on the line?

1. Set a formula to represent a straight line

Cartesian coordinate system:

$kx _1+b=x_2$

Different $k$ and $b$ can get different lines, but in fact this representation is not suitable for solving the current problem

Angular coordinate system:

The distance from the line to the origin is $\rho$, and the angle of the normal vector is $\phi$.

How do you represent a line with these two values?

The unit method vector is $\underline{n} (Cos\phi,sin\phi) $, and any point on the line is $x_1,x_2$.

They have a rule: all points on a line are projected on the unit normal vector as $\rho$, which is

$\rho=\underline{n}\cdot\underline{x} = x_1cos\phi + x_2 sin\phi$

2. Set a formula that represents the integral on the line

We need to show that the integral in the line, that is, need to extract the value of the line position from $\MU (x_1,x_2) $, can be achieved by the impulse function, imagine that the line is full of pulse function, expressed as follows

$\delta (\rho-x_1 cos\phi-x_2 sin\phi) $

The $\MU (x_1,x_2) $ in this line position can be obtained by using the sampling characteristics of the $\delta$.

$\mu (x_1,x_2) \delta (\rho-x_1 cos\phi–x_2 sin\phi) $

Then the radon transformation eventually becomes

$\mathcal{r}_l \mu=\mathcal{r}\mu (\rho,\phi) = \displaystyle{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mu (x_ 1,x_2) \delta (\rho-x_1cos\phi-x_2sin\phi) dx_1dx_2} \qquad (-\infty<\rho<\infty,0\leqslant \phi < \PI) $

expressed in vector form as

$\displaystyle{\mathcal{r}\mu (\rho,\underline{n}) =\int_{\mathbb{r}^2}\mu (\underline{x}) \delta (\rho-\underline{ X}\cdot\underline{n}) D\underline{x}}$

Radon Inverse transform (inverting the Radon Transform)

In medical image imaging, the X-ray through the human body to achieve, the ray will decay after passing through the human body, and then measured by the instrument, that is, we already know the results of the radon transformation, we need to use this result to restore the ray through the human profile, the process of restoration is called radon inverse transformation.

The Radon transform has two variables: the distance between the line and the origin is $\rho$, and the normal vector angle of the line is $\phi$.

Now that the $\phi$ is constant, that is, through the beam angle of the human profile, there is a unique variable $\rho$

At this point the $\mathcal{r}\mu$ is Fourier transformed (variable is $\rho$)

$\begin{align*}
\mathcal{f}_{\rho}\mathcal{r}\mu (R,\phi)
&=\int_{-\infty}^{\infty} E^{-2\pi i R\rho}\mathcal{r}\mu (\rho,\phi) d\rho \ \
&=\int_{-\infty}^{\infty}e^{-2\pi i r\rho}\left (\int_{-\ Infty}^{\infty}\int_{-\infty}^{\infty}\mu (x_1,x_2) \delta (\rho-x_1cos\phi-x_2sin\phi) dx_1dx_2 \right) d\rho\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mu (x_1,x_2) \left (\int_{-\infty}^{\infty}\delta (\rho-x_ 1cos\phi-x_2sin\phi) E^{-2\pi i r\rho}d\rho \right) dx_1dx_2\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\ Infty}\mu (x_1,x_2) e^{-2\pi ir (x_1cos\phi+x_2sin\phi)}dx_1dx_2 \qquad (\delta\ shift\ property) \ \
&=\int_{-\ Infty}^{\infty}\int_{-\infty}^{\infty}\mu (x_1,x_2) e^{-2\pi I (x_1rcos\phi+x_2rsin\phi)}dx_1dx_2\\
&=\int_ {-\infty}^{\infty}\int_{-\infty}^{\infty}\mu (x_1,x_2) e^{-2\pi I (x_1\xi_1+x_2\xi_2)}dx_1dx_2 \qquad \left (letting\ \begin{cases}
\xi_1=rcos\phi \ \
\xi_2=rsin\phi
\end{cases} \right)
\end{align*}$

expressed in vector form as

$\displaystyle{\mathcal{f}_{\rho}\mathcal{r}\mu (\underline{\xi}) =\int_{\mathbb{r}^2}e^{-2\pi i\underline{x}\ Cdot\underline{\xi}}\mu (\underline{x}) d\underline{x}}$

This becomes a two-dimensional Fourier transform on $\MU (x_1,x_2) $, i.e.

$\mathcal{f}_{\rho}\mathcal{r}\mu (R,\phi) =\mathcal{f}\mu (\xi_1,\xi_2) \qquad \begin{cases} \xi_1=rcos\phi \ \xi_2= Rsin\phi \end{cases}$

Then the Fourier inverse transformation of this result can get the original human profile picture $\mu$

Summary of medical image imaging process

1. Centered on the human body, each side has X-ray source and receiving sensor, and can rotate around the center, variable angle of $0\leqslant \phi < \pi$

2. A cluster angle of $\phi$ parallel X-ray emitted from the source to the sensor receiver, wherein the cross section is the human profile, the density is unknown, expressed in $\mu (X_1,x_2) $.

The data received by the sensor can be used to calculate the attenuation of X-rays, denoted by $g_{\phi} (\rho) $, wherein $\phi$ represents a fixed angle, and $\rho$ represents the distance from the center point of a beam of x-rays in parallel to the cluster.

3. Angle $\phi$ need to transform from $0$ to $\pi$, and for each angle $\phi$ need to calculate $\mathcal{f}g_{\phi} (r) $, which is the Fourier transform of $g_{\phi} (\rho) $

4. Since

$\mathcal{f}g_{\phi} (r) =\mathcal{f}\mu (\xi_1,\xi_2) \qquad \begin{cases} \xi_1=rcos\phi \ \xi_2=rsin\phi \end{cases }$

$r $ itself is a variable, with $-\infty<r<\infty$, we can get arbitrary $\xi_1,\xi_2$ by changing the angle $\phi$ (from $0$ to $\pi$), thus covering $\mathcal{f}\mu (\xi_1 , \xi_2) $ for all ranges.

5. Finally, the two-dimensional Fourier inverse transformation of the above results can restore the human profile image $\mu$

$\mu (\underline{x}) = \displaystyle{\int_{\mathbb{r}^2}e^{2\pi I\underline{x}\cdot\underline{\xi}}\mathcal{f}\mu ( \UNDERLINE{\XI}) D\underline{\xi}}$

Of course, the actual data operation is certainly discrete, so the use of DFT. There are also some other considerations that are interesting to see the notes of this course

[Fourier transform and its application study notes] 30. Radon Transform