Non-blind and Blind Deconvolution Under Poisson Noise Using Fractional-Order Total Variation

Abstract

In a wide range of applications such as astronomy, biology, and medical imaging, acquired data are usually corrupted by Poisson noise and blurring artifacts. Poisson noise often occurs when photon counting is involved in such imaging modalities as X-ray, positron emission tomography, and fluorescence microscopy. Meanwhile, blurring is also inevitable due to the physical mechanism of an imaging system, which can be modeled as a convolution of the image with a point spread function. In this paper, we consider both non-blind and blind image deblurring models that deal with Poisson noise. In the pursuit of high-order smoothness of a restored image, we propose a fractional-order total variation regularization to remove the blur and Poisson noise simultaneously. We develop two efficient algorithms based on the alternating direction method of multipliers, while an expectation-maximization algorithm is adopted only in the blind case. A variety of numerical experiments have demonstrated that the proposed algorithms can efficiently reconstruct piecewise smooth images degraded by Poisson noise and various types of blurring, including Gaussian and motion blurs. Specifically for blind image deblurring, we obtain significant improvements over the state of the art.

Appendix

Based on the blind deconvolution model (23), we define the objective function

$$\begin{aligned} Q(u,h):=\Vert \nabla ^\alpha u\Vert _1+ \beta \langle h\circ u-{\bar{f}}\log (h\circ u), \mathbf {1}_{\varOmega }\rangle . \end{aligned}$$

(45)

The proof of Theorem 1 relies on the following two lemmas. Lemma 1 shows the decrease of the objective function values for the sequence generated by the EM algorithm (10). By switching u and h, we can show that $Q(u,h^{k+1})\le Q(u,h^k)$ for any u. For a fixed h, Lemma 2 guarantees the FOTV-regularized objective function decreases with respect to u. Therefore, we have

$$\begin{aligned} Q(u^{k+1},h^{k+1})\le Q(u^k,h^{k+1})\le Q(u^k,h^k). \end{aligned}$$

It is obvious that the objective function $Q(u,h)\ge 0$. Then it follows from the Bolzano–Weierstrass Theorem that there exists a convergent subsequence.

For the notational convenience, we assume u, h are column vectors, define a matrix H such that $Hu:=h\circ u$, and denote $(Hu)_i$ as the ith element of Hu. Therefore, we can rewrite the EM update (10) as

$$\begin{aligned} u^{k+1}_j = \frac{u_j^k}{\sum _i H_{ij}}\sum _i \frac{H_{ij}f_j}{(Hu^k)_i}. \end{aligned}$$

(46)

Lemma 1

Given f and H,  define

$$\begin{aligned} F(u):=\sum _i \Big ((Hu)_i - f_i \log (Hu)_i\Big ) \end{aligned}$$

(47)

For the sequence generated by (46), we have $F(u^{k+1})\le F(u^k).$

Proof

The proof follows the majorize-minimization (MM) framework [32, 41]. In particular, we define a surrogate function of F, denoted by $G(u,u^k)$, which satisfies the following two conditions:

$$\begin{aligned} \begin{aligned}&F(u^k) = G(u^k,u^k)\\&F(u) \le G(u,u^k), \quad \forall u. \\ \end{aligned} \end{aligned}$$

(48)

If $u^{k+1} = {{\,\mathrm{argmin}\,}}_u G(u,u^k),$ then

$$\begin{aligned} F(u^{k+1}) \le G(u^{k+1},u^k) \le G(u^k,u^k) = F(u^k), \end{aligned}$$

where the first inequality and the last equality are by definition in (48), and the second inequality is because that $u^{k+1}$ minimizes $G(u,u^k).$

We consider

$$\begin{aligned} G(u,u^k) = \sum _{i} (Hu)_i - \sum _{i}f_i \sum _j\frac{H_{ij}u^k_j}{(Hu^k)_i} \log \left( \dfrac{H_{ij}u_j}{\frac{H_{ij}u^k_j}{(Hu^k)_i}}\right) .\nonumber \\ \end{aligned}$$

(49)

We will show G is a surrogate function for F and the EM update (46) minimizes $G(u,u^k)$.

Simple calculations lead to $G(u^k,u^k) = F(u^k)$. To show $F(u)\le G(u,u^k)$, we define $a_{ij} = \frac{ H_{ij} u_j^k}{(Hu^k)_i}.$ It is straightforward that $a_{ij}\ge 0$ and $\sum _j a_{ij}=1,\ \forall i.$ Then by Jensen’s inequality, we have $\forall i,$

$$\begin{aligned} \begin{aligned} \sum _j\frac{H_{ij}u^k_j}{(Hu^k)_i} \log \left( \dfrac{H_{ij}u_j}{\frac{H_{ij}u^k_j}{(Hu^k)_i}}\right) =&\sum _j a_{ij} \log \left( \dfrac{H_{ij}u_j}{a_{ij}}\right) \\ \le&\log (\sum _j H_{ij}u_j).\\ \end{aligned} \end{aligned}$$

(50)

Therefore, by definitions in (49) and (47), we have $F(u)\le G(u,u^k)$, and hence, G is a surrogate function of F.

To show that the EM update (46) minimizes $G(u,u^k)$, we take the gradient of $G(u,u^k)$ with respect to u and set equal to zero, thus leading to

$$\begin{aligned} \sum _{i}H_{ij}- \frac{\sum _i f_i a_{ij}}{u_j} = 0. \end{aligned}$$

(51)

Therefore, we get

$$\begin{aligned} u_j = \frac{\sum _i f_i a_{ij}}{\sum _i H_{ij}} = \frac{u_j^k}{\sum _i H_{ij}}\sum _i \frac{H_{ij}f_j}{(Hu^k)_i}, \end{aligned}$$

(52)

which is equivalent to (46). By the MM framework, we have $F(u^{k+1})\le F(u^k)$.$\square$

Lemma 2

Define an FOTV regularized objective function

$$\begin{aligned} E(u):=\Vert \nabla ^\alpha u\Vert _1+\beta \sum _i (Hu)_i - f_i \log ((Hu)_i), \end{aligned}$$

(53)

and consider the following iterative scheme

$$\begin{aligned} u^{k+1}= & {} {{\,\mathrm{argmin}\,}}_u \Vert \nabla ^\alpha u\Vert _1\nonumber \\&+ \beta \sum _j (\sum _i H_{ij}) (u_j - u_j^{k+\frac{1}{2}} \log u_j), \end{aligned}$$

(54)

with $u^{k+\frac{1}{2}}$ defined by (46). We can show that $E(u^{k+1})\le E(u^k).$

Proof

Denote $R(u) = \Vert \nabla ^\alpha u\Vert _1$. Since $u^{k+1}$ is the optimal solution for (54), we have

$$\begin{aligned} \begin{aligned}&R(u^{k+1}) + \beta \sum _j (\sum _i H_{ij}) (u^{k+1}_j - u_j^{k+\frac{1}{2}} \log u_j^{k+1})\\&\quad \le R(u^k) +\beta \sum _j (\sum _i H_{ij}) (u_j^k - u_j^{k+\frac{1}{2}} \log u_j^k), \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \begin{aligned}&R(u^{k+1})-R(u^k)\\&\quad \le \beta \sum _j (\sum _i H_{ij}) \left( u_j^k-u_j^{k+1}+u_j^{k+\frac{1}{2}}\log \frac{u_j^{k+1}}{u_j^k}\right) \\&\quad =\beta \Big ( \sum _i\big ((Hu^k)_i-(Hu^{k+1})_i\big )\\&\qquad + \sum _j (\sum _i H_{ij})u_j^{k+\frac{1}{2}}\log \frac{u_j^{k+1}}{u_j^k}\Big ). \end{aligned} \end{aligned}$$

(55)

Using the definition of (46) and the inequality (55), we can compute

$$\begin{aligned} \begin{aligned}&\frac{1}{\beta }(E(u^{k+1}) - E(u^k))\\&\quad \le \sum _i f_i \log \left( \dfrac{(Hu^k)_i}{(Hu^{k+1})_i}\right) +\sum _{ij} f_i a_{ij}\log \frac{u_j^{k+1}}{u_j^{k}} \\&\quad \le \sum _i f_i \log \left( \dfrac{(Hu^k)_i}{(Hu^{k+1})_i}\right) +\sum _i f_i \log \left( \sum _j a_{ij}\frac{u_j^{k+1}}{u_j^k}\right) , \end{aligned}\nonumber \\ \end{aligned}$$

(56)

where the last inequality is from Jensen’s inequality and $a_{ij} = \frac{ H_{ij} u_j^k}{(Hu^k)_i}$. Simple calculations show that

$$\begin{aligned} \sum _j a_{ij} \frac{u_j^{k+1}}{u_j^k} = \sum _j \frac{H_{ij}u_j^{k+1}}{(Hu^k)_i} = \frac{(Hu^{k+1})_i}{(Hu^k)_i}. \end{aligned}$$

(57)

Since $f_i\ge 0,$ we get $E(u^{k+1})-E(u^k)\le 0$ by plugging (57) into (56).$\square$