An Optimization Method for Transmitting Waveform of Polarimetric Radar Against Interrupted Sampling Repeater Jamming
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摘要: 作为一种新型的有源干扰样式，间歇采样转发干扰(ISRJ)引起了人们越来越多的关注。极化是表征电磁波矢量性的重要参数，其引入可以显著提高雷达在抗干扰方面的性能。为此，该文针对性地研究了全极化雷达的抗ISRJ方法，通过波形设计和优化以获取比传统单极化雷达更好的抗干扰性能。另外，针对宽带雷达条件下现有抗ISRJ问题表征中未考虑目标特性对信号调制作用这一短板，该文在信干比的数学表达式中加入了目标特性调制这一因素。在此基础上，提出了一种具有多普勒容忍的抗ISRJ全极化雷达波形设计方法。采用实测目标数据开展的实验表明：与单极化雷达相比，极化信息的引入显著提高了雷达对ISRJ的抑制性能；宽带条件下，考虑扩展目标对信号的调制作用在信干比的计算上具有必要性。Abstract: Interrupted Sampling Repeater Jamming (ISRJ) has attracted increasing attention as a new type of active jamming. Polarization is an important parameter for characterizing an electromagnetic wave vector and can substantially improve radar anti-jamming performance. Therefore, in this study, the anti-ISRJ method of polarimetric radar is studied. Here, better anti-jamming performance than that of traditional radar is obtained through waveform design and optimization. Additionally, because the modulation effect of target characteristics on a signal is neglected in the existing characterization of the anti-ISRJ problem under wideband radar conditions, this paper includes target characteristics factors to the mathematical expression of SJR. Finally, on this basis, a waveform design method of anti-ISRJ polarimetric radar with Doppler tolerance is proposed. Experiments using the measured target data show that compared with traditional radar, the introduction of polarization information improves substantially radar ISRJ suppression performance; moreover, under wideband conditions, the modulation effect of the extended target on the signal must be considered when calculating SJR.
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算法1 宽带全极化雷达抗ISRJ恒模互补波形设计流程 输入：目标脉冲响应矩阵$\overline {{\boldsymbol{H}}(\theta )}$、干扰特性矩阵${\boldsymbol{C}}$、初始${\boldsymbol{s}}$,${\boldsymbol{w}}$和
${\boldsymbol{\varepsilon}} $输出：最优发射波形序列${{\boldsymbol{s}}^\nabla }$、接收滤波器序列${{\boldsymbol{w}}^\nabla }$和Jones矢量${{\boldsymbol{\varepsilon}} ^\nabla }$ 1 重复 2 将${{\boldsymbol{w}}^{(i)} }$代入式(31)计算${{\boldsymbol{w}}_1}$，将${{\boldsymbol{w}}_1}$代入式(31)计算${{\boldsymbol{w}}_2}$ 3 ${{\boldsymbol{r}}_1} = {{\boldsymbol{w}}_1} - {{\boldsymbol{w}}^{(i)} }$, ${{\boldsymbol{v}}_1} = {{\boldsymbol{w}}_2} - {{\boldsymbol{w}}_1} - {{\boldsymbol{r}}_1}$$ , $ ${\alpha _1} = - {{\boldsymbol{r}}_1}/{{\boldsymbol{v}}_1}$ 4 将${ {\boldsymbol{w} }^{(i)} } - 2{\alpha _1}{ {\boldsymbol{r} }_1} + {\alpha _1}^2{{\boldsymbol{v}}_1}$代入式(31)计算${{\boldsymbol{w}}^{(i + 1)}}$ 5 若$ \varGamma \left( {{{\boldsymbol{s}}^{{\text{(}}i{\text{)}}}},{{\boldsymbol{w}}^{(i + 1)}},{{\boldsymbol{\varepsilon }}^{{\text{(}}i{\text{)}}}}} \right) > \varGamma \left( {{{\boldsymbol{s}}^{{\text{(}}i{\text{)}}}},{{\boldsymbol{w}}^{{\text{(}}i{\text{)}}}},{{\boldsymbol{\varepsilon}} ^{{\text{(}}i{\text{)}}}}} \right) $，循环 6 ${\alpha _1} \leftarrow \left( {{\alpha _1} - 1} \right)/2$ 7 将${{\boldsymbol{w}} ^{{\text{(}}i{\text{)}}}} - 2{\alpha _1}{{\boldsymbol{r}}_1} + {\alpha _1}^2{{\boldsymbol{v}}_1}$代入式(31)计算${{\boldsymbol{w}}^{{\text{(}}i + 1{\text{)}}}}$ 8 结束 9 将${{\boldsymbol{s}}^{{\text{(}}i{\text{)}}}}$代入式(38)计算${{\boldsymbol{s}}_1}$，将${{\boldsymbol{s}}_1}$代入式(38)计算${{\boldsymbol{s}}_2}$ 10 ${{\boldsymbol{r}}_2} = {{\boldsymbol{s}}_1} - {{\boldsymbol{s}}^{{\text{(}}i{\text{)}}}}$, ${{\boldsymbol{v}}_2} = {{\boldsymbol{s}}_2} - {{\boldsymbol{s}}_1} - {{\boldsymbol{r}}_1},{\text{ }}{\alpha _2} = - {{\boldsymbol{r}}_2}/{{\boldsymbol{v}}_2}$ 11 将${{\boldsymbol{s}}^{ {\text{(} }i{\text{)} } } } - 2{\alpha _2}{{\boldsymbol{r}}_2} + {\alpha _2}^2{{\boldsymbol{v}}_2}$代入式(38)计算${s^{{\text{(}}i + 1{\text{)}}}}$ 12 若$\varGamma \left( {{{\boldsymbol{s}}^{{\text{(}}i + 1{\text{)}}}},{{\boldsymbol{w}}^{{\text{(}}i + 1{\text{)}}}},{{\boldsymbol{\varepsilon}} ^{{\text{(}}i{\text{)}}}}} \right) > \varGamma \left( {{{\boldsymbol{s}}^{{\text{(}}i{\text{)}}}},{{\boldsymbol{w}}^{{\text{(}}i + 1{\text{)}}}},{{\boldsymbol{\varepsilon}} ^{{\text{(}}i{\text{)}}}}} \right)$，循环 13 ${\alpha _2} \leftarrow \left( {{\alpha _2} - 1} \right)/2$ 14 将${ {\boldsymbol{s} }^{ {\text{(} }i{\text{)} } } } - 2{\alpha _2}{{\boldsymbol{r}}_2} + {\alpha _2}^2{ {\boldsymbol{v} }_2}$代入式(38)计算${{\boldsymbol{s}}^{{\text{(}}i + 1{\text{)}}}}$ 15 结束 16 使用网格法迭代求解${{\boldsymbol{\varepsilon}} ^{{\text{(}}i + 1{\text{)}}}}$ 17 $i \leftarrow i + 1$ 18 达到收敛条件，结束 
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