基于迭代并行干擾消除的低復(fù)雜度大規(guī)模MIMO信號(hào)檢測(cè)算法
doi: 10.11999/JEIT180111 cstr: 32379.14.JEIT180111
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重慶郵電大學(xué)移動(dòng)通信重點(diǎn)實(shí)驗(yàn)室 ??重慶 ??400065
Low Complexity Iterative Parallel Interference Cancellation Detection Algorithms for Massive MIMO Systems
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Chongqing Key Laboratory of Mobile Communications, Chongqing University of Posts and Telecommunications, Chongqing 400065, China
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摘要: 基于干擾消除思想該文提出一種適用于大規(guī)模MIMO系統(tǒng)上行鏈路的低復(fù)雜度迭代并行干擾消除算法,在算法實(shí)現(xiàn)中避免了線性檢測(cè)算法所需的高復(fù)雜度
$({\cal O}({K^3}))$ 矩陣求逆運(yùn)算,將復(fù)雜度保持在$({\cal O}({K^2}))$ 。在此基礎(chǔ)上,引入噪聲預(yù)測(cè)機(jī)制,提出一種基于噪聲預(yù)測(cè)的迭代并行干擾消除算法,進(jìn)一步提高了硬判決檢測(cè)性能??紤]天線間殘留干擾,將干擾消除思想運(yùn)用到軟判決中,最后提出一種基于迭代并行干擾消除的低復(fù)雜度軟輸出信號(hào)檢測(cè)算法。仿真結(jié)果表明:提出的信號(hào)檢測(cè)方法的復(fù)雜度優(yōu)于MMSE檢測(cè)算法,經(jīng)過幾次簡(jiǎn)單的迭代,算法即快速收斂并獲得接近甚至優(yōu)于MMSE檢測(cè)算法的誤碼率性能。-
關(guān)鍵詞:
- 大規(guī)模MIMO /
- 低復(fù)雜度 /
- 迭代并行干擾消除 /
- 噪聲預(yù)測(cè) /
- 軟輸出
Abstract: Based on interference cancellation method, a low complexity Iterative Parallel Interference Cancellation (IPIC) algorithm is proposed for the uplink of massive MIMO systems. The proposed algorithm avoids the high complexity matrix inversion required by the linear detection algorithm, and hence the complexity is maintained only at$({\cal O}({K^2}))$ . Meanwhile, the noise prediction mechanism is introduced and the noise-prediction aided iterative parallel interference cancellation algorithm is proposed to improve further the detection performance. Considering the residual inter-antenna interference, a low-complexity soft output signal detection algorithm is proposed as well. The simulation results show that the complexity of all the proposed signal detection methods are better than that of the MMSE detection algorithm. With only a small number of iterations, the proposed algorithm achieves its performance quite close to or even surpassing that of the MMSE algorithm. -
表 1 基于迭代并行干擾消除算法(IPIC)
算法1 基于迭代并行干擾消除算法(IPIC) 輸入: ${{H}},{{y}},{\sigma ^2},K,{T_{{\rm{iter}}}};$ 初始化: (1) ${{G}} = {{{H}}^{\rm{H}}}{{H}},{} = {{{H}}^{\rm{H}}}{{y}},{{\hat{ s}}^{(0)}} = {{{D}}^{ - 1}}{{{H}}^{\rm{H}}}$ ${{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 輸出 ${\hat{ s}} = {{\hat{ s}}^{({T_{{\rm{iter}}}})}}$ 下載: 導(dǎo)出CSV
表 2 基于噪聲預(yù)測(cè)的迭代并行干擾消除算法(NP-IPIC)
算法2 基于噪聲預(yù)測(cè)的迭代并行干擾消除算法(NP-IPIC) 輸入: ${{H}},{{y}},{\sigma ^2},K,{T_{{\rm{iter}}}};$ 初始化: (1) ${{G}} = {{{H}}^{\rm{H}}}{{H}},{} = {{{H}}^{\rm{H}}}{{y}}$, ${{D}} = {\rm{diag}}({{G}} + {\sigma ^2}{{{I}}_K})$ ${{\hat{ s}}^{(0)}} = Q({{{D}}^{ - 1}}{{{H}}^{\rm{H}}}{{y}}) = \{ \hat s_1^{(0)},\hat s_2^{(0)}, ·\!·\!· ,\hat s_K^{(0)}\}$ (2) 對(duì) ${{H}}$列范數(shù)進(jìn)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) 判斷 $i$是否等于1,如果為1,則計(jì)算 $\bar s_{o(1)}^{(t)} = Q\left(\hat s_{o(1)}^{(t)}\right)$, 噪聲
采樣 $\hat n_{o(1)}^{(t)} = \hat s_{o(1)}^{(t)} - \bar s_{o(1)}^{(t)} = \hat s_{o(1)}^{(t)} - \mathbb{Q}\left(\bar s_{o(1)}^{(t)}\right)$, 如果 $i > 1$,跳過
本步驟,執(zhí)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) 根據(jù) ${{\hat{ s}}^{({T_{{\rm{iter}}}})}}$中下標(biāo)進(jìn)行重新排序得到 ${{\hat{ s}}^{{\rm{final}}}}$ 輸出 ${\hat{ s}} = {{\hat{ s}}^{{\rm{final}}}}$ 下載: 導(dǎo)出CSV
表 3 基于迭代并行干擾消除的軟輸出算法(S-IPIC)
算法3 基于迭代并行干擾消除的軟輸出算法(S-IPIC) 輸入: ${{H}},{{y}},{\sigma ^2},K,{T_{{\rm{iter}}}};$ 初始化: (1) ${G}={H}^{\rm H}{H}, ={H}^{\rm H}{y}$, ${{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) 估計(jì)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 估計(jì)發(fā)送信號(hào)并計(jì)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) 計(jì)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) 計(jì)算SINR ${{\rm Y}_i} = {{\mu _i^2} / {{{(\nu _i^{(t)})}^2}}}$ (9) $i = i + 1$ end for (10) $t = t + 1$ end for 輸出
${L_{i,b}} = {{\rm Y} _i}\left( {\mathop {\min }\limits_{a \in {\cal{O}}_b^0} {{\left| {\frac{{\hat s_i^{(t)}}}{{{\mu _i}}} - a} \right|}^2} - \mathop {\min }\limits_{a' \in {\cal{O}}_b^1} {{\left| {\frac{{\hat s_i^{(t)}}}{{{\mu _i}}} - a'} \right|}^2}} \right)$下載: 導(dǎo)出CSV
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