A maximum likelihood filter using non-local information for despeckling of ultrasound images

Abstract

This work presents a new ultrasound image despeckling method based on the maximum likelihood principle that effectively exploits non-local information for estimating noise-free pixels. First, a new maximum likelihood filter is proposed which uses neighborhood information to despeckle images. For this purpose, the general speckle model is used in the log-likelihood function and despeckled pixels are obtained by maximizing this function. Second, the proposed filter is developed to use non-local information such that the distribution of each noisy pixel is weighted according to the statistical distance between the patch of the noisy pixel and that of the target pixel. Because it is optimally designed for ultrasound images, the Pearson distance is used to measure the statistical distance between the patches. A series of experiments are conducted on three different ultrasound images and one synthetic image. Subjective evaluations show that the proposed method is able to preserve edges and structural details of the image and objective evaluations using equivalent number of looks, natural image quality evaluator, peak signal-to-noise ratio, mean preservation, and structural similarity confirm that the proposed method can achieve superior performance.

Appendix

Firstly, the derivation of equation (14) is described. Substituting Eq. (12) into (13) and setting the derivative of Eq. (13) equal to zero yield:

$$\begin{aligned}&-\frac{\left( {2t_2 +1} \right) ^{2}}{\hat{{\sigma }}_u }+\frac{\hat{{\sigma }}_u^{\left( {1/\gamma } \right) -3} }{\gamma \sigma _n^{1/\gamma } }\sum _{p=1}^{\left( {2t_2 +1} \right) ^{2}} {\left( {u^{(p)}(x)-(\hat{{\sigma }}_u /\sigma _n )^{\frac{1}{\gamma }}} \right) } \nonumber \\&\quad +\frac{1}{\hat{{\sigma }}_u^3 }\sum _{p=1}^{\left( {2t_2 +1} \right) ^{2}} {\left( {u^{(p)}(x)-(\hat{{\sigma }}_u /\sigma _n )^{\frac{1}{\gamma }}} \right) ^{2}} =0 \end{aligned}$$

(A.1)

As already mentioned, the parameter $\gamma$ should be set to 0.5. After some manipulations, we obtain:

$$\begin{aligned}&\frac{\left( {2t_2 +1} \right) ^{2}}{\sigma _n^4 }\hat{{\sigma }}_u^4 +\left( {2t_2 +1} \right) ^{2}\hat{{\sigma }}_u^2 \nonumber \\&\quad -\sum _{p=1}^{\left( {2t_2 +1} \right) ^{2}} {\left( {u^{(p)}} \right) ^{2}(x)} =0 \end{aligned}$$

(A.2)

which is a quadratic function of $\sigma _u^2$. Equation (A.2) has two solutions:

$$\begin{aligned} \hat{{\sigma }}_u^2 =\frac{\sigma _n^4 }{2}.\left( {\pm \sqrt{1+4\frac{\sum _{p=1}^{\left( {2t_2 +1} \right) ^{2}} {\left( {u^{(p)}} \right) ^{2}(x)} }{\left( {2t_2 +1} \right) ^{2}\sigma _n^4 }}-1} \right) \end{aligned}$$

(A.3)

Note that $\hat{{\sigma }}_u^2$ is always a negative (positive) value for the minus (plus) sign. For the plus sign, Eq. (A.3) is equal to Eq. (14).

Secondly, the derivation of equation (17) is explained. Considering Eq. (16), we should maximize the log-likelihood function:

$$\begin{aligned} \hat{{\sigma }}_u= & {} \arg \mathop {\max }_{\sigma _{u(r)} } \left( {\ln \left( {L\left( {\sigma _u |u(x)} \right) } \right) } \right) \nonumber \\= & {} \arg \mathop {\max }_{\sigma _{u(r)} } \left( {\sum _{\forall x\in \Omega } {\alpha (x)\ln \left( {p_{u(r)} (u(x))} \right) } } \right) \end{aligned}$$

(A.4)

Substituting $p_{u(r)} (u(x))$ yields (the parameter $\gamma$ is set to 0.5):

$$\begin{aligned}&\sum _{\forall x\in \Omega } {\alpha (x)\ln \left( {\frac{1}{\sqrt{2\pi }\hat{{\sigma }}_{u(r)} }} \right) } \nonumber \\&\quad -\frac{1}{2\hat{{\sigma }}_{u(r)}^2 }\sum _{\forall x\in \Omega } {\alpha (x)\left( {u(x)-(\hat{{\sigma }}_{u(r)} /\sigma _n )^{2}} \right) ^{2}} \end{aligned}$$

(A.5)

Setting the derivative of Eq. (A.5) equal to zero yields:

$$\begin{aligned}&\frac{\sum _{\forall x\in \Omega } {\alpha (x)} }{\sigma _n^4 }\hat{{\sigma }}_u^4 +\sum _{\forall x\in \Omega } {\alpha (x)} \hat{{\sigma }}_u^2 \nonumber \\&\quad -\sum _{\forall x\in \Omega } {\alpha (x)u^{2}(x)} =0 \end{aligned}$$

(A.6)

Equation (A.6) has two solutions:

$$\begin{aligned} \hat{{\sigma }}_{u(r)}^2 =\frac{\sigma _n^4 }{2}.\left( {\pm \sqrt{1+4\frac{1}{\sigma _n^4 }\frac{\left( {\sum _{\forall x\in \Omega } {\alpha (x)u^{2}(x)} } \right) }{\left( {\sum _{\forall x\in \Omega } {\alpha (x)} } \right) }}-1} \right) \nonumber \\ \end{aligned}$$

(A.7)

To obtain a positive value for $\hat{{\sigma }}_{u(r)}^2$, the plus sign is selected [see Eq. (17)].