MR image reconstruction via guided filter

Abstract

Magnetic resonance imaging (MRI) reconstruction from the smallest possible set of Fourier samples has been a difficult problem in medical imaging field. In our paper, we present a new approach based on a guided filter for efficient MRI recovery algorithm. The guided filter is an edge-preserving smoothing operator and has better behaviors near edges than the bilateral filter. Our reconstruction method is consist of two steps. First, we propose two cost functions which could be computed efficiently and thus obtain two different images. Second, the guided filter is used with these two obtained images for efficient edge-preserving filtering, and one image is used as the guidance image, the other one is used as a filtered image in the guided filter. In our reconstruction algorithm, we can obtain more details by introducing guided filter. We compare our reconstruction algorithm with some competitive MRI reconstruction techniques in terms of PSNR and visual quality. Simulation results are given to show the performance of our new method.

Appendix

1.1 The derivation from Eqs. 7 and 8 to Eqs. 9 and 10

We first must be precise about our notation. In the following, we have ∥∇u − ∇u _E $\|_{2}^{2}$ = ∥∇ _x u − ∇ _x u _E $\|_{2}^{2}$ + ∥∇ _y u − ∇ _y u _E $\|_{2}^{2}$, where ∇ _x represents the horizontal difference operator, and ∇ _y represents the vertical differential operator.

To find the optimal value of u _I , we must solve the optimization problem

$$ u_{I}\!=\arg\min\limits_{u}\{\lambda\|\nabla_{x} u-\nabla_{x} u_{E}\|_{2}^{2}+\|\nabla_{y} u-\nabla_{y} u_{E}\|_{2}^{2}+\|P \mathcal{F}u -f\|_{2}^{2}\} $$

(12)

Because this problem is differentiable, the optimality conditions for u _I are easily derived. By differentiating with respect to u and setting the result equal to zero, we get the update rule

$$\begin{array}{@{}rcl@{}} &&[\lambda ({\nabla_{x}^{T}}\nabla_{x}+ {\nabla_{y}^{T}}\nabla_{y})+ \mathcal{F}^{T} P^{T} P\mathcal{F}]u\\ &&\quad=\lambda ({\nabla_{x}^{T}}\nabla_{x}+{\nabla_{y}^{T}}\nabla_{y})u_{E} + \mathcal{F}^{T} P^{T} f \end{array} $$

(13)

We now take advantage of the identities ∇^T∇ = −△ and $\phantom {\dot {i}\!}\mathcal {F}^{T}=\mathcal {F}^{-1}$ to get

$$ (\mathcal{F}^{T} P^{T} P\mathcal{F}-\lambda \triangle)u= \mathcal{F}^{T} P^{T} f -\lambda \triangle u_{E} $$

(14)

Therefore, the system that must be inverted to solve for u _I is circulant. Because of the circulant structure of this system, we can solve for the optimal value of u _I using only two Fourier transform. Through the Eq. 14, we can get the Eq. 9

$$ \mathcal{F} u_{I}=\frac{P^{T} f-\lambda \mathcal{F} \triangle \mathcal{F}u_{E}}{|P|^{2}-\lambda \mathcal{F} \triangle \mathcal{F}^{-1}} $$

(15)

Similarly, the problem (8) is differentiable. By differentiating with respect to u and setting the result equal to zero, we get the update rule

$$ (\beta+ \mathcal{F}^{T} P^{T} P \mathcal{F})u=\beta u_{E}+ \mathcal{F}^{T} P^{T} f $$

(16)

which is

$$ \mathcal{F}^{T}(\beta+ P^{T} P)\mathcal{F}u=\mathcal{F}^{T}(\beta \mathcal{F}u_{E}+ P^{T} f) $$

(17)

Thus, we can get the Eq. 10

$$ \mathcal{F}u_{p}=\frac{P^{T} f+\beta \mathcal{F}u_{E}}{|P|^{2}+\beta} $$

(18)