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基于迭代并行干扰消除的低复杂度大规模MIMO信号检测算法

申滨 赵书锋 金纯

申滨, 赵书锋, 金纯. 基于迭代并行干扰消除的低复杂度大规模MIMO信号检测算法[J]. 电子与信息学报, 2018, 40(12): 2970-2978. doi: 10.11999/JEIT180111
引用本文: 申滨, 赵书锋, 金纯. 基于迭代并行干扰消除的低复杂度大规模MIMO信号检测算法[J]. 电子与信息学报, 2018, 40(12): 2970-2978. doi: 10.11999/JEIT180111
Bin SHEN, Shufeng ZHAO, Chun JIN. Low Complexity Iterative Parallel Interference Cancellation Detection Algorithms for Massive MIMO Systems[J]. Journal of Electronics & Information Technology, 2018, 40(12): 2970-2978. doi: 10.11999/JEIT180111
Citation: Bin SHEN, Shufeng ZHAO, Chun JIN. Low Complexity Iterative Parallel Interference Cancellation Detection Algorithms for Massive MIMO Systems[J]. Journal of Electronics & Information Technology, 2018, 40(12): 2970-2978. doi: 10.11999/JEIT180111

基于迭代并行干扰消除的低复杂度大规模MIMO信号检测算法

doi: 10.11999/JEIT180111
基金项目: 重庆市科委重点产业共性关键技术创新专项(cstc2015zdcy-ztzx40008)
详细信息
    作者简介:

    申滨:男,1978年生,教授,研究方向为认知无线电、大规模MIMO等

    赵书锋:男,1991年生,硕士生,研究方向为大规模MIMO信号检测

    金纯:男,1966年生,教授,研究方向为无线通信信号处理、物联网等

    通讯作者:

    申滨  shenbin@cqupt.edu.cn

  • 中图分类号: TN929.5

Low Complexity Iterative Parallel Interference Cancellation Detection Algorithms for Massive MIMO Systems

Funds: The Innovation Project of the Common Key Technology of Chongqing Science and Technology Industry (cstc2015zdcy-ztzx40008)
  • 摘要: 基于干扰消除思想该文提出一种适用于大规模MIMO系统上行链路的低复杂度迭代并行干扰消除算法,在算法实现中避免了线性检测算法所需的高复杂度 $({\cal O}({K^3}))$ 矩阵求逆运算,将复杂度保持在 $({\cal O}({K^2}))$ 。在此基础上,引入噪声预测机制,提出一种基于噪声预测的迭代并行干扰消除算法,进一步提高了硬判决检测性能。考虑天线间残留干扰,将干扰消除思想运用到软判决中,最后提出一种基于迭代并行干扰消除的低复杂度软输出信号检测算法。仿真结果表明:提出的信号检测方法的复杂度优于MMSE检测算法,经过几次简单的迭代,算法即快速收敛并获得接近甚至优于MMSE检测算法的误码率性能。
  • 图  1  算法复杂度对比

    图  3  $128 \times 32$ MIMO系统下BER性能(硬判决)

    图  2  $128 \times 16$ MIMO系统下BER性能(硬判决)

    图  4  信道估计误差下算法BER性能(硬判决)

    图  6  $128 \times 16$ MIMO系统中各算法软输出BER性能

    图  5  $128 \times 32$ MIMO系统中各算法软输出BER性能

    表  1  基于迭代并行干扰消除算法(IPIC)

     算法1 基于迭代并行干扰消除算法(IPIC)
     输入: ${{H}},{{y}},{\sigma ^2},K,{T_{{\rm{iter}}}};$
     初始化:
     (1) ${{G}} = {{{H}}^{\rm{H}}}{{H}},{{b}} = {{{H}}^{\rm{H}}}{{y}},{{\hat{ s}}^{(0)}} = {{{D}}^{ - 1}}{{{H}}^{\rm{H}}}$ ${{y}} = \{ \hat s_1^{(0)},\hat s_2^{(0)}, ·\!·\!· ,\hat s_K^{(0)}\} $
      For $t = 1:{T_{{\rm{iter}}}};$
       For $i = 1:K$;
     (2) 更新 $\hat s_i^{(t)} = \hat s_i^{(t - 1)} + \frac{{{b_i} - \displaystyle\sum\nolimits_{j = 1}^{i - 1} {{G_{ij}}} \hat s_j^{(t)} - \displaystyle\sum\nolimits_{j = i}^K {{G_{ij}}} \hat s_j^{(t - 1)}}}{{{G_{ii}}}}$
     (3) 更新 ${{\hat{ s}}^{(t)}} = {\left[ {\hat s_1^{(t)},\hat s_2^{(t)}, ·\!·\!· ,\hat s_{i - 1}^{(t)}}, {Q(\hat s_i^{(t)})}, {\hat s_{i + 1}^{(t - 1)},\hat s_{i + 2}^{(t - 1)}, ·\!·\!· ,\hat s_K^{(t - 1)}}\right]^{\rm{T}}}$
     (4)   $i = i + 1$
    end for
     (5)   $t = t + 1$
       end for
     输出 ${\hat{ s}} = {{\hat{ s}}^{({T_{{\rm{iter}}}})}}$
    下载: 导出CSV

    表  2  基于噪声预测的迭代并行干扰消除算法(NP-IPIC)

     算法2 基于噪声预测的迭代并行干扰消除算法(NP-IPIC)
     输入: ${{H}},{{y}},{\sigma ^2},K,{T_{{\rm{iter}}}};$
     初始化:
     (1) ${{G}} = {{{H}}^{\rm{H}}}{{H}},{{b}} = {{{H}}^{\rm{H}}}{{y}}$, ${{D}} = {\rm{diag}}({{G}} + {\sigma ^2}{{{I}}_K})$
        ${{\hat{ s}}^{(0)}} = Q({{{D}}^{ - 1}}{{{H}}^{\rm{H}}}{{y}}) = \{ \hat s_1^{(0)},\hat s_2^{(0)}, ·\!·\!· ,\hat s_K^{(0)}\}$
     (2) 对 ${{H}}$列范数进行降序排序,
    $o = \arg {\rm{sort}}({\tau _1},{\tau _2}, ·\!·\!· ,{\tau _K}),\ {\tau _k} = \left\| {{{{h}}_k}} \right\|_2^2,\ \forall k = 1,2, ·\!·\!· ,K$
       For $t = 1:{T_{{\rm{iter}}}}$;
        For $i = 1:K$;
     (3) 更新
    $\hat s_{o(i)}^{(t)} = \hat s_{o(i)}^{(t - 1)} + \frac{{{b_{o(i)}} - \displaystyle\sum\limits_{j = 1}^{i - 1} {{G_{o(i)o(j)}}} \hat s_{o(j)}^{(t)} - \displaystyle\sum\limits_{j = i}^K {{G_{o(i)o(j)}}} \hat s_{o(j)}^{(t - 1)}}}{{{G_{o(i)o(i)}}}}$
     (4) 判断 $i$是否等于1,如果为1,则计算 $\bar s_{o(1)}^{(t)} = Q\left(\hat s_{o(1)}^{(t)}\right)$, 噪声
    采样 $\hat n_{o(1)}^{(t)} = \hat s_{o(1)}^{(t)} - \bar s_{o(1)}^{(t)} = \hat s_{o(1)}^{(t)} - \mathbb{Q}\left(\bar s_{o(1)}^{(t)}\right)$, 如果 $i > 1$,跳过
    本步骤,执行下一步;
     (5) 更新 ${\hat{ n}} = \frac{{{{a}}_{o(i - 1)}^{\rm{H}}}}{{{{\left\| {{{{a}}_{o(i - 1)}}} \right\|}^2}}}\hat n_{o(i - 1)}^{(t)}$
     (6) $\hat n_{o(i)}^{(t)} = {{{a}}_{o(i)}}{\hat{ n}}$, $\bar s_{o(i)}^{(t)} = Q\left(\hat s_{o(i)}^{(t)} - \hat n_{o(i)}^{(t)}\right)$
     (7) 更新
    ${{\hat{ s}}^{(t)}} = {[ {\hat s_{o(1)}^{(t)},\hat s_{o(2)}^{(t)}, ·\!·\!· ,\hat s_{o(i - 1)}^{(t)}}, {\bar s_{o(i)}^{(t)}}, {\hat s_{o(i + 1)}^{(t - 1)},\hat s_{o(i + 2)}^{(t - 1)}, ·\!·\!· ,\hat s_{o(K)}^{(t - 1)}}]^{\rm{T}}}$
     (8)    $i = i + 1$
        end for
     (9)   $t = t + 1$
       end for
     (10) 根据 ${{\hat{ s}}^{({T_{{\rm{iter}}}})}}$中下标进行重新排序得到 ${{\hat{ s}}^{{\rm{final}}}}$
     输出 ${\hat{ s}} = {{\hat{ s}}^{{\rm{final}}}}$
    下载: 导出CSV

    表  3  基于迭代并行干扰消除的软输出算法(S-IPIC)

     算法3 基于迭代并行干扰消除的软输出算法(S-IPIC)
     输入: ${{H}},{{y}},{\sigma ^2},K,{T_{{\rm{iter}}}};$
     初始化:
     (1) ${G}={H}^{\rm H}{H}, {b}={H}^{\rm H}{y}$,
    ${{D}} = {\rm{diag}}({{G}} + {\sigma ^2}{{{I}}_K}) {{\hat{ s}}^{(0)}} = {{{D}}^{ - 1}}{{{H}}^{\rm{H}}}{{y}} = \{ \hat s_1^{(0)},\hat s_2^{(0)}, ·\!·\!· ,\hat s_K^{(0)}\} $
     (2) 估计方差
       For $i = 1:K;$
     (3) $V_i^{(0)} = \sum\limits_{{\alpha _n} \in {\cal{Q}}} \Bigr| {\alpha _n} - \hat s_i^{(0)}{\Bigr|^2}P({s_i} = {\alpha _n})$
       end for
       估计发送信号并计算NPI方差
       For $t = 1:{T_{{\rm{iter}}}};$
        For $i = 1:K;$
     (4) 更新
         $\hat s_i^{(t)} = {\rm{ }}\hat s_i^{(t - 1)} + \frac{{{b_i} - \displaystyle\sum\limits_{j = 1}^{i - 1} {{G_{ij}}} \hat s_j^{(t)} - \displaystyle\sum\limits_{j = i}^K {{G_{ij}}} \hat s_j^{(t - 1)}}}{{{G_{ii}}}}$
     (5) 更新 ${{\hat{ s}}^{(t)}} = {\left[ {\hat s_1^{(t)},\hat s_2^{(t)}, ·\!·\!· ,\hat s_{i - 1}^{(t)}}, {\hat s_i^{(t)}}, {\hat s_{i + 1}^{(t - 1)},\hat s_{i + 2}^{(t - 1)}, ·\!·\!· ,\hat s_K^{(t - 1)}}\right]^{\rm T}}$
     (6) 更新
         $V_i^{(t)} = \sum\limits_{{\alpha _n} \in {\cal{O}}} | {\alpha _n} - \hat s_i^{(t)}{|^2}P({s_i} = {\alpha _n})$
     (7) 计算等效信道增益和NPI方差
       ${\mu _i} = 1$,
       ${(\nu _i^{(t)})^2}{\rm{ }} = \frac{1}{{G_{ii}^2}}\left( {\sum\limits_{j = 1}^{i - 1} | {G_{ij}}{|^2}V_j^{(t)} + \sum\limits_{j = i + 1}^K | {G_{ij}}{|^2}V_j^{(t - 1)}} \right) + \frac{{{\sigma ^2}}}{{{G_{ii}}}}$
     (8) 计算SINR ${{\rm Y}_i} = {{\mu _i^2} / {{{(\nu _i^{(t)})}^2}}}$
     (9)    $i = i + 1$
        end for
     (10)   $t = t + 1$
       end for
     输出
         ${L_{i,b}} = {{\rm Y} _i}\left( {\mathop {\min }\limits_{a \in {\cal{O}}_b^0} {{\left| {\frac{{\hat s_i^{(t)}}}{{{\mu _i}}} - a} \right|}^2} - \mathop {\min }\limits_{a' \in {\cal{O}}_b^1} {{\left| {\frac{{\hat s_i^{(t)}}}{{{\mu _i}}} - a'} \right|}^2}} \right)$
    下载: 导出CSV
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出版历程
  • 收稿日期:  2018-01-25
  • 修回日期:  2018-05-29
  • 网络出版日期:  2018-08-14
  • 刊出日期:  2018-12-01

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