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基于正交时频空系统的低复杂度最大比合并接收机算法

王震铎 季天治 孙溶辰

王震铎, 季天治, 孙溶辰. 基于正交时频空系统的低复杂度最大比合并接收机算法[J]. 电子与信息学报, 2025, 47(5): 1392-1401. doi: 10.11999/JEIT241056
引用本文: 王震铎, 季天治, 孙溶辰. 基于正交时频空系统的低复杂度最大比合并接收机算法[J]. 电子与信息学报, 2025, 47(5): 1392-1401. doi: 10.11999/JEIT241056
WANG Zhenduo, JI Tianzhi, SUN Rongchen. Low-complexity MRC Receiver Algorithm Based on OTFS System[J]. Journal of Electronics & Information Technology, 2025, 47(5): 1392-1401. doi: 10.11999/JEIT241056
Citation: WANG Zhenduo, JI Tianzhi, SUN Rongchen. Low-complexity MRC Receiver Algorithm Based on OTFS System[J]. Journal of Electronics & Information Technology, 2025, 47(5): 1392-1401. doi: 10.11999/JEIT241056

基于正交时频空系统的低复杂度最大比合并接收机算法

doi: 10.11999/JEIT241056
基金项目: 国家自然科学基金(62001138)
详细信息
    作者简介:

    王震铎:男,副教授,研究方向为物理层波形理论与关键技术

    季天治:男,硕士生,研究方向为OTFS接收算法

    孙溶辰:男,副教授,研究方向为信道测量与建模

    通讯作者:

    孙溶辰 rongchensun@hrbeu.edu.cn

  • 中图分类号: TN914.3

Low-complexity MRC Receiver Algorithm Based on OTFS System

Funds: The National Natural Science Foundation of China (62001138)
  • 摘要: 正交时频空(OTFS)因其在高速移动场景下优异的误码率性能而受到广泛研究。针对OTFS接收机运算复杂度较高问题,该文提出一种基于最大比合并(MRC)的低复杂度接收机算法。首先,其核心思想在时延多普勒域利用最大比合并算法进行迭代,对接收的多径分量进行提取和相干组合,以提高组合信号的信噪比。其次,通过引入交织器和解交织器,信道矩阵转化为稀疏的上三角海森伯矩阵,有利于后续进行矩阵分解。再次,针对符号决策过程中矩阵求逆计算量大的问题,提出一种低复杂度的LDLH分解算法。最后,在此基础上进一步改进,提出了一种复杂度进一步降低的下三角矩阵求逆算法,以降低下三角矩阵求逆的复杂度。仿真结果表明,一方面该算法误码率与最大比合并算法相同,另一方面性能显著优于高阶调制的线性最小均方误差估计(LMMSE)线性均衡器与高斯−赛德尔(GS)迭代均衡算法。
  • 图  1  信道矩阵${{\boldsymbol{H}}_{{\text{DD}}}}$

    图  2  引入交织器与解交织器后的OTFS系统框架

    图  3  低复杂度MRC接收机在M=16,N=4下的复杂度分析

    图  4  MRC迭代接收机在时延-多普勒信道下的误码率性能(m=16,N=4)

    图  5  MRC迭代接收机在时延-多普勒信道下的误码率性能(m=32, N=16)

    图  6  多普勒频移不同时MRC迭代均衡器的误码性能表现

    图  7  不同路径数对低复杂度MRC算法误码性能的影响

    图  8  低复杂度MRC接收机在ETU信道下的误码率性能

    1  低复杂度${\text{LD}}{{\text{L}}^{\text{H}}}$分解算法

     输入:信道矩阵$ {{{\boldsymbol{D}}}_m} $,零矩阵$ {{\boldsymbol{G}}} $
     输出:下三角矩阵L,下三角矩阵$ {{{\boldsymbol{L}}}^{ - 1}} $,对角矩阵D
     (1) $ {{\boldsymbol{D}}}(1,1) = {{{\boldsymbol{D}}}_m}(1,1) $
     (2) for $ {\text{ }}{i}{ = 1:}{N} $
     (3)  for $ j = 1:i - 1 $
     (4)   for $ k = 1:j - 1 $
     (5)    $ {\text{tem}}{{\text{p}}_{(i)}} = \displaystyle\sum\limits_{j = 1}^{i - 1} {\displaystyle\sum\limits_{k = 1}^{j - 1} {{{\boldsymbol{G}}}(i,j){{{\boldsymbol{L}}}^ * }(j,k)} } $
     (6)   end
     (7)   $ {{\boldsymbol{G}}}(i,j) = {\boldsymbol{A}}(i,j) - {\text{tem}}{{\text{p}}_{(i)}} $
     (8)   $ {{\boldsymbol{L}}}(i,j) = {{\boldsymbol{G}}}(i,j)/{{\boldsymbol{D}}}(j,j) $
     (9)  end
     (10) for $ k = 1:i - 1 $
     (11)   $ {\text{tem}}{{\text{p}}_{(i)}} = \displaystyle\sum\limits_{k = 1}^{i - 1} {{{\boldsymbol{G}}}(i,k){{{\boldsymbol{L}}}^*}(i,k)} $
     (12) end
     (13) $ {{\boldsymbol{D}}}(i,i) = {{\boldsymbol{A}}}(i,i) - {\text{tem}}{{\text{p}}_{(i)}} $
     (14) end
     (15) for $ i = 1:MN $
     (16) $ {{{\boldsymbol{L}}}^{ - 1}}(i,i) = {{\boldsymbol{L}}}(i,i) $
     (17) for $ u = 1:i - 1 $
     (18)   $ {{{\boldsymbol{L}}}^{ - 1}}(i,u) = - {{\boldsymbol{L}}}(i,i) \cdot \displaystyle\sum\limits_{k = u}^{i - 1} {{L}(i,k){{{\boldsymbol{L}}}^{ - 1}}(k,u)} $
     (19) end
     (20) end
    下载: 导出CSV

    表  1  MRC迭代均衡算法计算复杂度

    步骤复杂度
    1$ O({M^3}{N^3}) $
    2$ O({M^3}{N^3} - N{M^3}) $
    3$ O((2{M^3}{N^3} + 3{M^2}{N^2} + MN)/12) $
    4$ O((2{M^3}{N^3} - 3{M^2}{N^2} + MN)/12) $
    下载: 导出CSV

    表  2  系统仿真参数

    参数
    子载波个数(M) 16, 32
    符号个数(N) 4,16
    载波频率(GHz) 4
    子载波间隔(kHz) 15
    用户移动速度(km/h) 350,700
    调制方案 4 QAM, 16 QAM, 64 QAM
    最大时延抽头数 4
    下载: 导出CSV
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出版历程
  • 收稿日期:  2024-12-02
  • 修回日期:  2025-05-05
  • 网络出版日期:  2025-05-10
  • 刊出日期:  2025-05-01

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