Unique Decomposition and a New Model for the Ground Moving Target Indication Problem

Abstract

In wide-area surveillance radar systems, ground moving target indication is the main task. The underlying mathematical problem is to decompose a complex matrix into a low rank matrix and a structured sparse matrix. In this paper, we show that such decomposition has a unique solution under reasonable assumptions. We propose a phase-based model to fully describe the special sparse structure. An alternating direction method of multipliers is implemented to solve the resulting nonconvex complex matrix problem. Simulation results verify the superior efficiency and the improvement of the new model.

Appendix

Proof of Theorem 2.2

First, we consider $n = 3$. For the sake of deriving a contradiction, suppose there exists another nontrivial decomposition $({\widetilde{C}} ,{\widetilde{T}} )$ which satisfies $({\widetilde{C}} ,{\widetilde{T}} )\ne ( C , T )$. Assume they take the following form

$$\begin{aligned} C= & {} (v_1,v_1,v_1), \ T =(z_1,z_2,z_3)=(r^{i\theta }, r^{i(\theta +\alpha )}, r^{i(\theta +2\alpha )}) ,\\ {\widetilde{C}}= & {} (v_2,v_2,v_2),\ {\widetilde{T}} =(y_1,y_2,y_3)=(t^{i\phi }, t^{i(\phi +\beta )}, t^{i(\phi +2\beta )}), \end{aligned}$$

where $v_1,v_2\in \mathbb {C}$ and $r,t>0$. Assume $\theta ,\phi \in [0,2\pi [$ and $\alpha ,\beta \in ]0,2\pi [.$ By $C + T ={\widetilde{C}} +{\widetilde{T}}$, we have $v_1+z_j = v_2 + y_j,\ j= 1,2,3$. Let $v=v_1-v_2$. There is $z_j+v =y_j,\ j= 1,2,3$. The problem is simplified to find a complex number $v\in \mathbb {C}$ and a phase $\beta \in ]0,2\pi [$ such that the following holds

$$\begin{aligned} |z_1+v| = |z_2+v|=|z_3+v|, \ z_2+v = (z_1+v)e^{i\beta }, \ z_3+v = (z_2+v)e^{i\beta }. \end{aligned}$$

(16)

Fig. 9

Geometrical interpretation of $n=3$

The proof below is geometrically illustrated in Fig.  9. Let $l_1$ be the bisector of the angle formed by $z_1$ and $z_2$. Let $l_2$ be the bisector of the angle formed by $z_2$ and $z_3$. By Lemma 2.1, every v satisfying $|z_1+v| = |z_2+v|$ belongs to line $l_1$. Similarly, every v satisfying $|z_2+v| = |z_3+v|$ lies in line $l_2$. As a result, every v satisfying both conditions must lie in the intersection of $l_1$ and $l_2$.

If the phase difference $\alpha \in ]0,\pi [$ or $\alpha \in ]\pi , 2\pi [$ (see Fig. 9a), then we obtain $l_1\bigcap l_2=\{0\}$. The solution of the first condition in (16) could only be the origin. It follows that another nontrivial solution $({\widetilde{C}} ,{\widetilde{T}} )$ does not exist. If $\alpha = \pi$, then $l_1$ coincides with $l_2$, i.e., $l_1=l_2$. It implies that any v in $l_1$ satisfies the modulus constraints. Then T must take the form $T= (z_1,-z_1, z_1)$. Without loss of generality, suppose $\theta = 0$, i.e., $z_1 = r>0$. According to Lemma 2.1, $v=p \cdot i$ and $p\in \mathbb {R}$, $p\ne 0$ (see Fig. 9b). The phase constraints in (16) yield

$$\begin{aligned} -r+p\cdot i = (r+p\cdot i)e^{i\beta },\ r+p\cdot i = (-r+p\cdot i)e^{i\beta } . \end{aligned}$$

Multiplying the second equation by $e^{i\beta }$ and substituting it into the first equation, we obtain $-r+p\cdot i = (-r+p\cdot i)e^{i2\beta }$. Then $\beta =\pi$. Together with $-r+p\cdot i = -(a+p\cdot i)e^{i\beta }$, one has $p=0$. This contradicts with the assumption that $p\ne 0$. In other words, there is no such $(v, \beta )$ that satisfies (16) with $v\ne 0$ and $\beta \in ]0,2\pi [$. It follows that the structure decomposition is unique.

For $n\ge 4$, there are more constraints besides (16). The unique decomposition is then implied by the case of $n=3$. The proof is finished.$\square$