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面向多目标多用户的雷达通信一体化恒模波形设计

朱锦锟 金添 唐宇航 宋勇平 刘伟

温郑铨, 陈玉燕. 含特性曲线族器件的规范化分段线性化动态网络的算法[J]. 电子与信息学报, 1994, 16(3): 275-283.
引用本文: 朱锦锟, 金添, 唐宇航, 宋勇平, 刘伟. 面向多目标多用户的雷达通信一体化恒模波形设计[J]. 电子与信息学报, 2023, 45(11): 3821-3829. doi: 10.11999/JEIT230827
Wen Zhengquan, Chen Yuyan. AN ALGORITHM FOR CANONICAL PIECEWISE-LINEAR DYNAMIC NETWORKS CONTAINING ELEMENTS WITH CHARACTERISTIC FAMILY[J]. Journal of Electronics & Information Technology, 1994, 16(3): 275-283.
Citation: ZHU Jinkun, JIN Tian, TANG Yuhang, SONG Yongping, LIU Wei. Multi-User and Multi-Target Constant Waveform Design of Dual-Function Radar and Communication System[J]. Journal of Electronics & Information Technology, 2023, 45(11): 3821-3829. doi: 10.11999/JEIT230827

面向多目标多用户的雷达通信一体化恒模波形设计

doi: 10.11999/JEIT230827
详细信息
    作者简介:

    朱锦锟:男,博士生,研究方向为雷达通信一体化

    金添:男,教授,研究方向为超宽带成像与目标识别

    唐宇航:男,硕士生,研究方向为雷达通信一体化

    宋勇平:男,讲师,研究方向为穿墙成像、雷达成像、雷达目标检测

    刘伟:男,副教授,研究方向为通信信号处理、5G/6G通信技术

    通讯作者:

    金添 tianjin@nudt.edu.cn

  • 中图分类号: TN959.1

Multi-User and Multi-Target Constant Waveform Design of Dual-Function Radar and Communication System

  • 摘要: 双功能雷达通信一体化系统可以使硬件和频谱资源得到有效利用,是解决当前无线频谱资源紧张问题的一种有效途径。该文针对同时多目标探测与多用户通信场景,以雷达接收回波的信干噪比(SINR)为指标保障雷达的探测性能,采用通信多用户干扰(MUI)作为通信指标保证通信传输性能。与此同时,为保证发射端功率放大器工作在饱和区域,增加了波形恒模的约束条件。该文通过对波形以及滤波器的联合优化,提出了一种在满足通信MUI功率一定的情况下,最大化雷达多目标探测回波最小信干噪比(SINR)的波形设计优化模型。针对此优化问题,采用了交替迭代优化的方法来求解此问题。仿真结果表明,所设计的波形在多目标探测以及多用户通信场景下,通过调整通信MUI功率门限,可以在保证通信MUI功率性能前提下,实现对多目标回波最小SINR的优化。
  • 图  1  双功能雷达通信系统模型示意图

    图  2  不同目标数量时算法实现的目标函数值随着迭代次数变化图

    图  3  发射波束图

    图  4  不同η值下的SINR变化图

    算法1 求解问题(21)算法流程
     输入:初始化变量s(0), e(0), β(0), δsδβ
     输出:问题式(38)的解s
     第1步:j=0
     第2步:s(j1) = s(0),e(j1) = e(0)
     第3步:求解问题式(38)得到(s(j),e(j),β(j))
     第4步:若,则输出 {\boldsymbol{s}} = {{\boldsymbol{s}}^{(j)}} ,否则继续下一步
     第5步:{{\boldsymbol{s}}^{(0)}}{\text{ = }}{{\boldsymbol{s}}^{(j)}},\;{e^{(0)}}{\text{ = }}{e^{(j)}},\;j{\text{ = }}j + 1返回第2步
    下载: 导出CSV
    算法2 交替迭代优化算法流程
     输入:初始化变量{s^{(0)}}, {e^{(0)}}, {\beta ^{(0)}}, {\delta _s}, {\delta _\beta } \kappa
     输出:问题式(16)的解{s^{(*)}},\left\{ {{\boldsymbol{w}}_m^*} \right\}_{m = 1}^M
     第1步:c = 0,构 {{\chi }_m}\left( {{{\boldsymbol{s}}^{(0)}}} \right) , {{\boldsymbol{\varLambda}} _m}\left( {{{\boldsymbol{s}}^{(0)}}} \right)
     第2步:求解问题式(19)得到\left\{ {{\boldsymbol{w}}_m^{(0)}} \right\}_{m = 1}^M
     第3步:c = c + 1
     第4步:构造 {{{\boldsymbol{\varOmega}} }_m}\left( {{\boldsymbol{w}}_m^{(c - 1)}} \right) , {{\boldsymbol{\varGamma}}_m}\left( {{\boldsymbol{w}}_m^{(c - 1)}} \right) ,求解问题式(21)
     得到{{\boldsymbol{s}}^{(c)}}
     第5步:构造 {{\chi }_m}\left( {{{\boldsymbol{s}}^{(c)}}} \right) , {{\boldsymbol{\varLambda}} _m}\left( {{{\boldsymbol{s}}^{(c)}}} \right) ,求解问题式(19),得到
     \left\{ {{\boldsymbol{w}}_m^{(c)}} \right\}_{m = 1}^M
     第6步:计算{\min _{\forall m}}{\text{SIN}}{{\text{R}}_m}\left( {{\boldsymbol{s}},{{\boldsymbol{w}}_m}} \right)
      若 |\min _{_{\forall m}}^j{{{\mathrm{SINR}}} _m}\left( {{{\boldsymbol{s}}^{(c)}},{\boldsymbol{w}}_m^{(c)}} \right) - \min _{_{\forall m}}^{(c - 1)}{{{\mathrm{SINR}}} _m}
      \left( {{{\boldsymbol{s}}^{({c} - 1)}},{\boldsymbol{w}}_m^{(c - 1)}} \right)|{\text{ }} < {\text{ }}\kappa
      输出{s^{(*)}}={s^{(c)}}, \left\{ {{\boldsymbol{w}}_m^*} \right\}_{m = 1}^M = \left\{ {{\boldsymbol{w}}_m^{(c)}} \right\}_{m = 1}^M,否则返回第3步。
    下载: 导出CSV
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出版历程
  • 收稿日期:  2023-08-20
  • 修回日期:  2023-11-02
  • 网络出版日期:  2023-11-14
  • 刊出日期:  2023-11-28

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