Small Target Detection Using Two-Dimensional Least Mean Square (TDLMS) Filter Based on Neighborhood Analysis

Abstract

TDLMS is a general adaptive filter algorithm and when applied to infrared small target detection, traditional structure and implementation of TDLMS may cause some problems in this field. This paper presents a new TDLMS filter structure and implementation incorporating neighborhood analysis and data fusion, which is capable of acquiring and analyzing more information from the vicinity of the target, leading to a more prominent detection result. This enables TDLMS filter to perform better and become more suitable in the field of small target detection. Experiments showed the efficiency of the proposed algorithm.

Appendix

In the four-point situation mentioned in section 3.1, the criterion function we use in the proposed algorithm has the form

$$ \begin{array}{*{20}l} {{J = S_{W} + S_{B} = S_{T} } \hfill} \\ {{ = \frac{1} {N}{\sum\limits_{l = 1}^N {{\left( {x_{l} - m} \right)}^{2} } }} \hfill} \\ {{ = \frac{1} {4}{\left[ {{\left( {\overline{Y} _{1} - m} \right)}^{2} + {\left( {\overline{Y} _{2} - m} \right)}^{2} + {\left( {\overline{Y} _{3} - m} \right)}^{2} + {\left( {\overline{Y} _{4} - m} \right)}^{2} } \right]}} \hfill} \\ \end{array} $$

where $m = \frac{{\overline{Y} _{1} + \overline{Y} _{2} + \overline{Y} _{3} + \overline{Y} _{4} }} {4}$, and $\overline{Y} \in {\left[ {\overline{Y} _{{\min ,}} \overline{Y} _{{\max }} } \right]}$.

Using the Lagrange multiplier method, and the constraint function is then

$$ \varphi {\left( {m,\overline{Y} _{1} ,\overline{Y} _{2} ,\overline{Y} _{3} ,\overline{Y} _{4} } \right)} = 4m - \overline{Y} _{1} - \overline{Y} _{2} - \overline{Y} _{3} - \overline{Y} _{4} = 0. $$

Let $f{\left( {\overline{Y} _{1} ,\overline{Y} _{1} ,\overline{Y} _{1} ,\overline{Y} _{1} ,m,\lambda } \right)} = {\left( {\overline{Y} _{1} - m} \right)}^{2} + {\left( {\overline{Y} _{2} - m} \right)}^{2} + {\left( {\overline{Y} _{3} - m} \right)}^{2} {\kern 1pt} + {\left( {\overline{Y} _{4} - m} \right)}^{2} + {\kern 1pt} \lambda {\left( {4m - \overline{Y} _{1} - \overline{Y} _{2} - \overline{Y} _{3} - \overline{Y} _{4} } \right)}$ and we have the following equations:

$$ \left\{ {\begin{array}{*{20}c} {f_{{x1}} = 2{\left( {\overline{Y} _{1} - m} \right)} - \lambda = 0} \\ {f_{{x2}} = 2{\left( {\overline{Y} _{2} - m} \right)} - \lambda = 0} \\ {f_{{x3}} = 2{\left( {\overline{Y} _{3} - m} \right)} - \lambda = 0} \\ {f_{{x4}} = 2{\left( {\overline{Y} _{4} - m} \right)} - \lambda = 0} \\ {f_{m} = 4m - \overline{Y} _{1} - \overline{Y} _{2} - \overline{Y} _{3} - \overline{Y} _{4} + 2\lambda = 0} \\ {f_{\lambda } = 4m - \overline{Y} _{1} - \overline{Y} _{2} - \overline{Y} _{3} - \overline{Y} _{4} = 0} \\ \end{array} } \right. $$

Solve these equations we get

$$ \overline{Y} _{1} = \overline{Y} _{2} = \overline{Y} _{3} = \overline{Y} _{4} $$

which means as long as $\overline{{Y_{i} }}$ equals to each other, J reaches its minimum 0. Starting from this extremum point, as $\overline{Y} _{i}$ moves more and more separated from each other, since J doesn’t have any other extremum points, the function will increase monotonously. Therefore, the maximum value will be reached when $\overline{Y} _{i}$ stops at the boundary of $\overline{Y} _{{\min }}$ or $\overline{Y} _{{\max }}$. Let $\overline{Y} _{1} = \overline{Y} _{{\min }}$ and $\overline{Y} _{4} = \overline{Y} _{{\max }}$, and there will be three possible combinations of $\overline{Y} _{i}$, namely

$$ \begin{array}{*{20}c} {}{\overline{Y} _{1} = \overline{Y} _{2} = \overline{Y} _{3} = \overline{Y} _{{\min }} ,\overline{Y} _{4} = \overline{Y} _{{\max }} } \\ {{\text{or}}}{\overline{Y} _{1} = \overline{Y} _{2} = \overline{Y} _{{\min }} ,\overline{Y} _{3} = \overline{Y} _{4} = \overline{Y} _{{\max }} } \\ {{\text{or}}}{\overline{Y} _{1} = \overline{Y} _{{\min }} ,\overline{Y} _{2} = \overline{Y} _{3} = \overline{Y} _{4} = \overline{Y} _{{\max }} } \\ \end{array} $$

Via simple calculation it is easily seen that when $\overline{Y} _{1} = \overline{Y} _{2} = \overline{Y} _{{\min }}$ and $\overline{Y} _{3} = \overline{Y} _{4} = \overline{Y} _{{\max }}$, J reaches its maximum value $\frac{{{\left( {\overline{Y} _{{\max }} - \overline{Y} _{{\min }} } \right)}^{2} }} {4}$.