Arterial contour detectability in head CT angiography

Abstract

Purpose

Arterial contour extraction is essential for visualization and analysis of vasculature in CT angiography (CTA). A means for evaluating the detectability of artery contours CTA images is required. We developed and tested a new method for this purpose based on phase information from two-dimensional Fourier transforms of CTA images. The relationship between arterial contour detectability and a patient’s ocular lens dose was evaluated in CTA images obtained with various tube voltages and currents.

Methods

A head phantom was designed for use as a target object containing a simulated vascular tree, filled with dilute contrast medium (10 mg iodine/ml). The head phantom was scanned using a 64-multidetector CT scanner with tube voltages of 80–140 kV and tube currents corresponding to volume CT dose index $(\hbox {CTDI}_\mathrm{vol})$ ranging from 24.4 to 72.8 mGy. Lens doses were measured using the planar silicon PIN-photodiode system. The quality of artery contours in the CTA source images was assessed using a computed detectability index.

Results

Lens dose increased proportionally with tube voltage and current. The use of 80 kV provided the highest contour detectability. However, for each tube voltage, the detectability of artery contours was almost constant across the $\hbox {CTDI}_\mathrm{vol}$ values. These results were mostly consistent with the subjective recognition of artery contours on CTA images.

Conclusions

A CTA protocol using 80 kV and 420 mA can reduce the radiation exposure to ocular lens by approximately 40 %, and improve the artery contour detectability compared with a routine protocol.

Appendices

Appendix 1: Reconstruction algorithm of phase-only image

We will make a brief explanation here of the two-dimensional discrete Fourier transform, which was used to obtain phase-only images from CTA source images. The two-dimensional discrete Fourier transform is used to perform the spatial frequency analysis and digital signal processing of digital images. At first, we consider the one-dimensional discrete Fourier transform to simplify the explanation of two-dimensional discrete Fourier transform. The Fourier transform of a signal $s(t)$ is defined as

$$\begin{aligned} s(\omega )=\int \limits _{-\infty }^\infty {s(t)\exp (-i\omega {}t)dt} \end{aligned}$$

(6.1)

where $\omega$ is an angular frequency and $i=\sqrt{-1}$. Discrete Fourier transform is an operator to evaluate the Fourier transform of the sampled signal $s(n)$ with a finite number of samples, $N$. It is defined as

$$\begin{aligned} S(k)=\frac{1}{N}\sum _{n=0}^{N-1} {s(n)} \exp \left[ {-\frac{2\pi {}i}{N}nk} \right] ; 0\le k\le N-\hbox {1 }. \end{aligned}$$

(6.2)

This mathematical concept of one-dimensional discrete Fourier transform can be extended up to two-dimensional Fourier transform. Now, let $M \times M$ pixels be the ROI size in the original CTA source image, $f(j, k)$ $(j = 0, 1, 2,{\ldots }, M-1; k = 0, 1, 2,{\ldots }, M-1)$. Taking into account Eq. (6.2), the discrete two-dimensional Fourier transform of the ROI image, $F(u,v)$, is defined in series form as,

$$\begin{aligned} F(u,v)=\frac{1}{M^{2}}\sum _{j=0}^{M-1} {\sum _{k=0}^{M-1} {f\left( {j,k} \right) \exp \left[ {-\frac{2\pi {}i}{M}\left( {uj+vk} \right) } \right] } } \end{aligned}$$

(6.3)

where $u = 0, 1, 2,{\ldots }, M-1, v = 0, 1, 2,{\ldots }, M-1,$ and the indices $(u,v)$ are called the spatial frequency. $F(u,v)$ is a complex number in the frequency domain,

$$\begin{aligned} F(u,v)=R(u,v)+iI(u,v) \end{aligned}$$

(6.4)

where $R(u,v)$ and $I(u,v)$ are the real and imaginary parts of $F(u,v)$, respectively, and can also be expressed in magnitude and phase angle form,

$$\begin{aligned} F(u,v)=A(u,v)\exp \left[ {i\theta (u,v)} \right] \end{aligned}$$

(6.5)

where

$$\begin{aligned} A(u,v)=\sqrt{R(u,v)^{2}+I(u,v)^{2}} \end{aligned}$$

(6.6)

and

$$\begin{aligned} \theta (u,v)=\arctan \left( {\frac{I(u,v)}{R(u,v)}} \right) \quad . \end{aligned}$$

(6.7)

Then, the phase-only image $g(j, k)$ can be obtained from the normalization of $F(u, v)$ with the division by $A(u, v)$. That is,

$$\begin{aligned} g(j,k)&= \sum _{u=0}^{M-1} {\sum _{v=0}^{M-1} {\frac{F\left( {u,v} \right) }{A(u,v)}\exp \left[ {\frac{2\pi {}i}{M}\left( {uj+vk} \right) } \right] } }\nonumber \\&= \sum _{u=0}^{M-1} {\sum _{v=0}^{M-1} {\exp \left[ {i\theta (u,v)} \right] \exp \left[ {\frac{2\pi {}i}{M}(uj+vk)} \right] } }.\nonumber \\ \end{aligned}$$

(6.8)

In this study, the phase-only images were obtained using a fast Fourier transform algorithm.

Appendix 2: Normal probability plot

When a random variable $x$ is statistically characterized by a normal distribution, its cumulative probability function $GF(x)$ is expressed as,

$$\begin{aligned} GF(x)=\frac{1}{\sigma \sqrt{2\pi }}\int \limits _{-\infty }^x {\exp \left[ {\frac{-(x-\mu )^{2}}{2\sigma ^{2}}} \right] } {}dx \end{aligned}$$

(7.1)

where $GF(x)$, $\mu$ and $\sigma$ are the cumulative probability of $x$, the mean, and standard deviation of $x$, respectively. Here, let $\Phi ^{-1}(GF(x))$ denote the inverse standard normal distribution function for the random variable $x$. By using $\Phi ^{-1}(GF(x))$, the Eq. (6.1) can be transformed to

$$\begin{aligned} \Phi ^{-1}(GF(x))=\frac{x-\mu }{\sigma }. \end{aligned}$$

(7.2)

The normal probability plot based on this equation is a graphical technique to test where a parameter follows a normal distribution. The plot is constructed as follows: the observed random variables are ranked from smallest to largest; then these ordered observations are plotted along the horizontal axis against their estimated cumulative frequency along the vertical axis; here, the vertical axis is scaled according to the inverse standard normal distribution. Therefore, the pixel values distributed in a normal distribution will be plotted along a straight line in the normal probability plot. In this study, the inverse standard normal distribution function value for a random variable $x$ was calculated by Microsoft Excel (Office 2000; Microsoft, USA).

Here, the cumulative probability function $GF(x)$ can be estimated using the mean rank method based on order statistics [19–21]. In this study, we adopted this estimation method because it will achieve high accuracy in calculating cumulative probability [19]. For the pixel value, the estimated cumulative probability function $GF(x_{(i)})$ was derived as follows. The pixel values were arranged in ascending order, and let $x_{(1)} \le x_{(2)} \le {\cdots } \le x_{(n)}$ be these $n$ arranged values. Then, ${GF}(x_{(i)})$ was computed as

$$\begin{aligned} GF(x_{(i)} )=\frac{i}{n+1},\hbox { for } i= 1,\ldots ,n, \end{aligned}$$

(7.3)

where $n$ is a sampling size.