Joint design of transmit beamforming and receive filter for transmit subaperturing MIMO STAP radar

Abstract

This paper deals with the joint design of transmit and receive beamforming for airborne transmit subaperturing (TS) multiple-input multiple-output (MIMO) radar to improve the moving target detection performance in the presence of signal-dependent clutter background. The considered system is referred to as an airborne TS–MIMO radar, whose transmit array is divided into several subarrays to form the steerable waveforms with directional gain. The joint design of transmit and receive beamforming problem is formulated by maximizing the signal-to-interference-plus-noise ratio under the energy constraint. To resolve the resulting non-convex joint design problem, we devise an efficient iterative optimization algorithm within the majorization-minimization framework. To obtain the desired transmit beampattern property, we extend the iterative optimization algorithm to resolve the energy and similarity constrained transmit beamforming design. Simulation results demonstrate that the proposed two methods are monotonically increasing and convergent, and exhibit superior property to the compared methods.

Appendix 1

Proof of Proposition 1

Since the problem (42) is convex, the obtained solution ${\mathbf{Z}}^{ * }$ and $\varsigma^{ * }$ should be the Karush–Kuhn–Tucker (KKT) points of problem (42) (Boyd & Vandenberghe, 2004). The Lagrangian function of (42) can be written as

$$ \begin{gathered} L\left( {{\mathbf{T}},\varsigma } \right) = - {\text{Tr}}\left( {{\mathbf{X}}\left( {{\mathbf{w}}^{\left( n \right)} } \right){\mathbf{Z}}} \right) + \eta_{1} \left( {{\text{ - Tr}}\left( {{\mathbf{T}}_{r} {\mathbf{Z}}} \right) + \varsigma \delta } \right) - {\text{Tr}}\left( {{\mathbf{YZ}}} \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; + \eta_{2} \left( {{\text{Tr}}\left( {{\mathbf{R}}_{cn} \left( {\mathbf{w}} \right){\mathbf{Z}}} \right) - 1} \right) + \eta_{3} \left( {{\text{Tr}}\left( {\mathbf{Z}} \right) - \varsigma } \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\; = {\text{Tr}}\left\{ {\left( { - {\mathbf{X}}\left( {{\mathbf{w}}^{\left( n \right)} } \right) - \eta_{1} {\mathbf{T}}_{r} - {\mathbf{Y}} + \eta_{2} {\mathbf{R}}_{cn} \left( {\mathbf{w}} \right) + \eta_{3} {\mathbf{I}}} \right){\mathbf{Z}}} \right\} \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; + \eta_{1} \varsigma \delta - \eta_{2} - \eta_{3} \varsigma \hfill \\ \end{gathered} $$

(48)

where $\eta_{1}$ and ${\mathbf{Y}}$ are the Lagrange multipliers of the inequality constraints $- {\text{Tr}}\left( {{\mathbf{T}}_{r} {\mathbf{Z}}} \right) + \varsigma \delta \le 0$ and $- {\mathbf{Z}} \le {\mathbf{0}}$, respectively; $\eta_{2}$ and $\eta_{3}$ are the Lagrange multipliers of the equality constraints ${\text{Tr}}\left( {{\mathbf{R}}_{cn} \left( {\mathbf{w}} \right){\mathbf{Z}}} \right) - 1 = 0$ and ${\text{Tr}}\left( {\mathbf{Z}} \right) - \varsigma = 0$, respectively. Thus, the KKT condition for problem (42) is given by

$$ - {\text{Tr}}\left( {{\mathbf{T}}_{r} {\mathbf{Z}}^{ * } } \right) + \varsigma^{ * } \delta \le 0, $$

(49a)

$$ {\mathbf{Z}}^{ * } \ge 0, $$

(49b)

$$ \eta_{1} \ge 0,\;\;{\mathbf{Y}} \ge {\mathbf{0}}, $$

(49c)

$$ {\text{Tr}}\left( {{\mathbf{R}}_{cn} \left( {\mathbf{w}} \right){\mathbf{Z}}^{ * } } \right) - 1 = 0, $$

(49d)

$$ {\text{Tr}}\left( {{\mathbf{Z}}^{ * } } \right) - \varsigma^{ * } = 0, $$

(49e)

$$ \eta_{1} \left( { - {\text{Tr}}\left( {{\mathbf{T}}_{r} {\mathbf{Z}}^{ * } } \right) + \varsigma^{ * } \delta } \right) = 0, $$

(49f)

$$ {\mathbf{YZ}}^{ * } = {\mathbf{0}}, $$

(49g)

$$ - {\mathbf{X}}\left( {{\mathbf{w}}^{\left( n \right)} } \right) - \eta_{1} {\mathbf{T}}_{r} - {\mathbf{Y}} + \eta_{2} {\mathbf{R}}_{cn} \left( {\mathbf{w}} \right) + \eta_{3} {\mathbf{I}} = {\mathbf{0}}. $$

(49h)

From (49 h), we have

$$ {\mathbf{Y}} = \eta_{2} {\mathbf{R}}_{cn} \left( {\mathbf{w}} \right) + \eta_{3} {\mathbf{I}} - {\mathbf{X}}\left( {{\mathbf{w}}^{\left( n \right)} } \right) - \eta_{1} {\mathbf{T}}_{r} , $$

(50)

Since ${\mathbf{R}}_{cn} \left( {\mathbf{w}} \right)$ and ${\mathbf{I}}$ are positive definite. From (49c) and (50), it can be concluded from ${\mathbf{Y}} \ge {\mathbf{0}}$ that at least one of $\eta_{2} > 0$ and $\eta_{3} > 0$ satisfies the condition. Without loss of generality, we assume that $\eta_{2} = 0$ and $\eta_{3} > 0$. Then, the rank of ${\mathbf{Y}}$ can be derived as

$$ {\text{rank}}\left( {\mathbf{Y}} \right) = {\text{rank}}\left( {\eta_{3} {\mathbf{I}} - {\mathbf{X}}\left( {{\mathbf{w}}^{\left( n \right)} } \right) - \eta_{1} {\mathbf{T}}_{r} } \right) \ge N^{\prime} - 1. $$

(51)

In addition, from (49 g), we have ${\mathbf{Z}}^{ * } \ne {\mathbf{0}}$ and ${\mathbf{YZ}}^{ * } = {\mathbf{0}}$. Then the rank of ${\mathbf{Y}}$ satisfies ${\text{rank}}\left( {\mathbf{Y}} \right) \le N^{\prime} - 1$, otherwise,${\mathbf{Z}}^{ * } = {\mathbf{0}}$. Thus, the rank of ${\mathbf{Y}}$ is ${\text{rank}}\left( {\mathbf{Y}} \right) = N^{\prime} - 1$.

From ${\text{rank}}\left( {\mathbf{Y}} \right) = N^{\prime} - 1$ and ${\mathbf{YZ}}^{ * } = {\mathbf{0}}$, we can conclude that ${\text{rank}}\left( {{\mathbf{Z}}^{ * } } \right) \le 1$. Since ${\mathbf{Z}}^{ * } \ne {\mathbf{0}}$, we infer that the rank of ${\mathbf{Z}}$ is ${\text{rank}}\left( {{\mathbf{Z}}^{ * } } \right) \ge 1$. Therefore, the rank of ${\mathbf{Z}}$ is ${\text{rank}}\left( {{\mathbf{Z}}^{ * } } \right){ = }1$. The proof is completed.