Joint Design of Transmission Sequences and Receiver Filters Based on the Generalized Cross Ambiguity Function
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摘要: 具有良好相关特性的正交波形集可以提高多输入多输出(MIMO)雷达系统的目标检测性能和抗干扰能力。为此，该文提出主瓣增益和动态范围约束下最小化广义互模糊函数积分旁瓣(ISL)的发射序列集和接收滤波器组联合设计方法。该方法采用最大块改进算法(MBI)将非凸优化问题分解为多个子问题，再利用连续凸近似方法(SCA)迭代求解子问题。为了进一步降低运算量，该文还提出了具有并行实现潜力的交替方向惩罚法(ADPM)求解SCA的子问题。最后，通过仿真实验从收敛速度、积分旁瓣电平等方面验证了该方法的有效性。Abstract:
Objective A set of orthogonal waveforms with favorable correlation properties enhances target detection and anti-jamming performance in Multiple-Input Multiple-Output (MIMO) radar systems. Jointly designing the transmit sequence set and receive filter bank introduces additional degrees of freedom, reducing auto- and cross-correlation. However, research on their joint design based on the Generalized Cross Ambiguity Function (GCAF) is limited and primarily focuses on reducing the peak sidelobe level. Since a low Integrated Sidelobe Level (ISL) is also critical for radar imaging and target detection, this study formulates the joint design problem with the objective of minimizing the ISL of the GCAF, subject to mainlobe gain and dynamic range constraints. Methods This paper proposes a Maximum Block Improvement–Successive Convex Approximation (MBI-SCA) method for the nonconvex optimization problem involving High-Order Polynomials (HOP). The MBI algorithm decomposes the nonconvex problem into multiple subproblems, which are then solved iteratively using the SCA method. To further reduce computational cost, an Alternating Direction Penalty Method (ADPM) is introduced. This algorithm, which supports parallel implementation, dynamically updates the penalty factor in each iteration, ensuring the penalty term gradually converges to zero. This guarantees algorithm convergence and accelerates the search for a better feasible solution. Results and Discussions The proposed MBI-SCA algorithm converges in approximately 12 iterations, while the MBI-ADPM algorithm achieves faster convergence in about 10 iterations ( Fig. 1 ). The running time of the MBI-ADPM algorithm increases as the sequence length varies from 16 to 512, whereas the MBI-SCA algorithm exhibits an overall increase, with a decrease at ${2^8}$, likely due to an decrease in the number of iterations when the SCA method solves the subproblem (Fig. 2 ). Both algorithms demonstrate strong performance, with GCAF values in the locally optimized region significantly lower than those in the unoptimized region, all below –200 dB (Fig. 3 ). However, MBI-ADPM achieves better local optimization, reducing GCAF values to –320 dB, whereas MBI-SCA reaches only –260 dB (Fig. 4 ). The parameter $K$ determines the optimization interval’s range, and as $K$ increases from 5 to 35, the ISL values of both methods also increase. For MBI-SCA, the optimal range of the parameter $K$ is $5 \le K \le 15$, where the integral sidelobe levels remain below –50 dB, meeting the low sidelobe requirement. In contrast, MBI-ADPM performs best values of $K$ are 5 and 10, achieving an objective function value close to –300 dB (Fig. 7 ).Conclusions This paper proposes a joint design method for transmit waveforms and receive filters that minimizes the GCAF ISL under mainlobe gain and dynamic range constraints, addressing the reduction of auto- and cross-correlation integral levels in MIMO radar waveform sets. To solve the quartic nonconvex optimization problem, the original problem is first decomposed into manageable subproblems using the MBI algorithm, which are then solved iteratively with the SCA algorithm. To further reduce computational complexity, the ADPM algorithm is introduced to solve the SCA subproblems. Simulation results demonstrate that the MBI-ADPM algorithm converges faster and achieves a lower ISL than MBI-SCA for shorter distance intervals of interest. -
1 MBI算法求解问题式(8)
输入： ${\boldsymbol x}_m^{(0)}$, ${\boldsymbol h}_{m'}^{(0)}$，计算${\omega ^{(0)}} = F({\boldsymbol x}_1^{(0)},{\boldsymbol x}_2^{(0)}, \cdots ,{\boldsymbol x}_M^{(0)},{\boldsymbol h}_1^{(0)},{\boldsymbol h}_2^{(0)}, \cdots ,{\boldsymbol h}_M^{(0)})$, $t = 0$ 输出：式(8)的解$ {\boldsymbol x}_m^{(t + 1)} $, ${\boldsymbol h}_{m'}^{(t + 1)}$ 步骤1 (块更新).对$m = 1,2, \cdots ,M$，求解式(9)，对$m' = 1,2, \cdots ,M$，求解式(10)。令 ${\boldsymbol{u}}_m^{(t + 1)} = \arg \mathop {\min }\limits_{{{\boldsymbol x}_m}} F({\boldsymbol x}_1^{(t)},{\boldsymbol x}_2^{(t)}, \cdots ,{{\boldsymbol x}_m},{\boldsymbol x}_{m + 1}^{(t)},{\boldsymbol x}_{m + 2}^{(t)}, \cdots ,{\boldsymbol x}_M^{(t)},{\boldsymbol h}_1^{(t)},{\boldsymbol h}_2^{(t)}, \cdots ,{\boldsymbol h}_M^{(t)})$ ${\boldsymbol{u}}_{m'}^{(t + 1)} = \arg \mathop {\min }\limits_{{{\boldsymbol h}_{m'}}} F({\boldsymbol x}_1^{(t)},{\boldsymbol x}_2^{(t)}, \cdots ,{\boldsymbol x}_M^{(t)},{\boldsymbol h}_1^{(t)},{\boldsymbol h}_2^{(t)}, \cdots ,{{\boldsymbol h}_{m'}},{\boldsymbol h}_{m' + 1}^{(t)},{\boldsymbol h}_{m' + 2}^{(t)}, \cdots ,{\boldsymbol h}_M^{(t)})$ $ y_m^{(t + 1)} = F({\boldsymbol x}_1^{(t)},{\boldsymbol x}_2^{(t)}, \cdots ,{\boldsymbol{u}}_m^{(t + 1)},{\boldsymbol x}_{m + 1}^{(t)},{\boldsymbol x}_{m + 2}^{(t)}, \cdots ,{\boldsymbol x}_M^{(t)},{\boldsymbol h}_1^{(t)},{\boldsymbol h}_2^{(t)}, \cdots ,{\boldsymbol h}_M^{(t)}) $ $ y_{m'}^{(t + 1)} = F({\boldsymbol x}_1^{(t)},{\boldsymbol x}_2^{(t)}, \cdots ,{\boldsymbol x}_M^{(t)},{\boldsymbol h}_1^{(t)},{\boldsymbol h}_2^{(t)}, \cdots ,{\boldsymbol{u}}_{m'}^{(t + 1)},{\boldsymbol h}_{m' + 1}^{(t)},{\boldsymbol h}_{m' + 2}^{(t)}, \cdots ,{\boldsymbol h}_M^{(t)}) $ 步骤2 (最大块更新).令 ${y^{(t + 1)}} = {\min _{1 \le m,m' \le M}}(y_m^{(t + 1)},y_{m'}^{(t + 1)})$ ${\boldsymbol x}_{\hat m}^{(t + 1)} = \arg \min {y^{(t + 1)}}$ 或 ${\boldsymbol h}_{\hat m}^{(t + 1)} = \arg \min {y^{(t + 1)}}$ 更新 ${\boldsymbol x}_m^{(t + 1)} = {\boldsymbol x}_m^{(t)},\forall m \ne \hat m$ $ {\boldsymbol h}_{m'}^{(t + 1)} = {\boldsymbol h}_{m'}^{(t)},\forall m' \ne \hat m' $ ${\boldsymbol x}_m^{(t + 1)} = {\boldsymbol x}_{\hat m}^{(t + 1)}$ 或 ${\boldsymbol h}_{m'}^{(t + 1)} = {\boldsymbol h}_{\hat m'}^{(t + 1)}$ ${\omega ^{(t + 1)}} = {y^{(t + 1)}}$ 步骤3 (停止准则) $|{\omega ^{(t + 1)}} - {\omega ^{(t)}}| \le {10^{ - 3}}{\omega ^{(t + 1)}}$ ，停止。否则 令$t = t + 1$，返回1。 
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2 SCA算法求解问题式(9)和式(10)
输入：${\boldsymbol x}_m^{(t)}$, ${\boldsymbol h}_{m'}^{(t)}$, ${\omega ^{(t)}}$ 输出：式(9)和式(10)的解$ {\boldsymbol{u}}_m^{(t + 1)} $, $ {\boldsymbol{u}}_{m'}^{(t + 1)} $ 步骤1 ${\boldsymbol x}_m^{(i)} = {\boldsymbol x}_m^{(t)}$和${\boldsymbol h}_{m'}^{(i)} = {\boldsymbol h}_{m'}^{(t)},$及${\omega ^{(i)}} = {\omega ^{(t)}}$, $i = 0$ 步骤2 CVX工具箱分别求解式(14)和式(15)，得到${\boldsymbol x}_m^{(i + 1)}$,
${\boldsymbol h}_{m'}^{(i + 1)}$计算${\omega ^{(i + 1)}} = F({\boldsymbol x}_1^{(i)},{\boldsymbol x}_2^{(i)}, \cdots ,{\boldsymbol x}_m^{(i + 1)} $,
${\boldsymbol x}_{m + 1}^{(i)},{\boldsymbol x}_{m + 2}^{(i)}, \cdots ,{\boldsymbol x}_M^{(i)},{\boldsymbol h}_1^{(t)},{\boldsymbol h}_2^{(t)}, \cdots ,{\boldsymbol h}_M^{(t)})$
或${\omega ^{(i + 1)}} = F({\boldsymbol x}_1^{(t)},{\boldsymbol x}_2^{(t)}, \cdots ,{\boldsymbol x}_M^{(t)},{\boldsymbol h}_1^{(i)},{\boldsymbol h}_2^{(i)}, \cdots ,{\boldsymbol h}_{m'}^{(i + 1)}, $
${\boldsymbol h}_{m' + 1}^{(i)},{\boldsymbol h}_{m' + 2}^{(i)}, \cdots ,{\boldsymbol h}_M^{(i)})$步骤3 若$|{\omega ^{(i + 1)}} - {\omega ^{(i)}}| \le {10^{ - 3}}{\omega ^{(i + 1)}}$，输出
${\boldsymbol{u}}_m^{(t + 1)} = {\boldsymbol x}_m^{(i + 1)}$, ${\boldsymbol{u}}_{m'}^{(t + 1)} = {\boldsymbol h}_{m'}^{(i + 1)}$，否则，${\boldsymbol x}_m^{(i)} = {\boldsymbol x}_m^{(i + 1)}$,
${\boldsymbol h}_{m'}^{(i)} = {\boldsymbol h}_{m'}^{(i + 1)}$, ${\omega ^{(i)}} = {\omega ^{(i + 1)}}$，返回步骤2。
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3 ADPM算法求解问题式(14)
输入：${\boldsymbol x}_m^{(i)} = {\boldsymbol x}_m^{(t)}$, ${\boldsymbol h}_{m'}^{(t)}$, ${\boldsymbol{s}}_1^{(0)}$, ${\boldsymbol{s}}_2^{(0)}$, ${\boldsymbol{s}}_3^{(0)}$, $\mu _{{{\boldsymbol{s}}_1}}^{(0)}$, $\mu _{{{\boldsymbol{s}}_2}}^{(0)}$, $\mu _{{{\boldsymbol{s}}_3}}^{(0)}$计
算$\rho _{{{\boldsymbol{s}}_d}}^{(0)}$，设置$i = 0$输出：问题(14)的解${\boldsymbol x}_m^ * $ 步骤1 $i = i + 1$ 步骤2 分别通过求解问题式(18)–式(24)，更新${\boldsymbol{s}}_1^{\left( {i + 1} \right)}$, ${\boldsymbol{s}}_2^{\left( {i + 1} \right)}$,
${\boldsymbol{s}}_3^{\left( {i + 1} \right)}$步骤3 通过求解问题(27)，更新${\boldsymbol x}_m^{(i + 1)}$ 步骤4 若$ {e}_{{\mathrm{pri}}}^{(i+1)}\le {\varepsilon }_{{\mathrm{pri}}}^{(i+1)} $, $ {e}_{{\mathrm{dual}}}^{(i+1)}\le {\varepsilon }_{{\mathrm{dual}}}^{(i+1)} $，输出
${\boldsymbol x}_m^ * = {\boldsymbol x}_m^{(i + 1)}$，否则，返回步骤1。
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