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蜂窝通信系统中,单小区边缘用户由于受到相邻小区多个边缘用户的较强干扰而导致通信质量差。干扰对齐可以有效消除干扰,保证边缘用户的通信质量[1]。文献[2]针对部分链接蜂窝系统中边缘用户的干扰问题,提出了盲干扰对齐方法,提高了信道容量和自由度。文献[3]提出了蜂窝网络上行链路中消除边缘用户干扰的机会干扰对齐和干扰消除方法,可以有效地消除相邻小区边缘用户的干扰,得到无干扰的信号。文献[4]提出了蜂窝网络中的有限反馈干扰对齐方法,通过反馈能够捕捉到吞吐量的下降,然后通过选择合适的干扰对齐方法消除边缘用户的干扰,提高系统容量。
以上研究大多数是在假设每个小区只有少数几个边缘用户的情况下进行的。然而,实际蜂窝系统中边缘用户数较多,基站所能同时服务的用户数严格受天线数的限制。因此,为了保证用户的正常通信,需要选择能够实现更高速率的部分用户组合来进行通信[5-6]。文献[7]研究了蜂窝系统中边缘用户较多时,通过联合优化干扰对齐、数据流数量和用户选择等问题,以达到最大化系统吞吐量的目的,但是联合优化多个变量复杂度很高。文献[8]针对认知蜂窝网络边缘用户较多的情况,提出用排序搜索算法选择对主用户产生干扰较大的信号,并对这些信号进行对齐,以提高系统的传输速率,但是这种方式计算量很高。文献[9]通过联合优化干扰对齐和边缘用户选择以提高信道容量,但是没有考虑边缘用户间干扰的消除问题。文献[10]针对小区内边缘用户间的干扰问题,在上行链路中提出了基于迫零准则的多用户编码和译码算法,但是在消除干扰信号的同时也削弱了有用信号。
综上分析,本文针对蜂窝系统中单小区边缘用户的干扰问题,首先根据设定的准则用布谷鸟搜索算法选择合适的边缘用户,然后通过干扰对齐方法消除小区间干扰,最后通过编码和译码得到无干扰的用户信号,提出基于布谷鸟搜索算法的用户选择和干扰对齐算法研究。
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基于用户选择的系统模型[11-12]如图 1所示,系统有3个小区,每个小区有一个基站,每小区边缘用户数为${K_l}\left( {l = 1, 2, 3} \right) $,l表示第几个小区。假设每个小区内选择边缘用户的集合是${A_l} \subseteq {K_l}\left( {l = 1, 2, 3} \right) $。
系统速率最大化问题可以写为:
$$ \begin{gathered} \mathop {\max }\limits_{\left\{ {{A_l}} \right\}} \sum\limits_{l = 1}^3 {\sum\limits_{k = 1, k \in {A_l}}^{\left| {{A_l}} \right|} {\log } } \left| {\left( {\underbrace {\left( {\sum\limits_{l \ne k, j = 1}^{{A_l}} {\mathit{\boldsymbol{H}}_k^{l, j}{S^{lj}}\mathit{\boldsymbol{H}}_k^{{{\left( {l, j} \right)}^{\rm{H}}}}} } \right)}_{小区间干扰} + } \right.} \right. \hfill \\ \left. {{{\left. {\underbrace {\sum\limits_{m = 1, m \ne k}^{{A_l}} {\mathit{\boldsymbol{H}}_k^{k, m}{S^{km}}\mathit{\boldsymbol{H}}_k^{{{\left( {k, m} \right)}^{\rm{H}}}}} }_{用户间干扰}} \right)}^{-1}}H_k^{k, {\rm{k}}}{S^{kk}}H_k^{{{\left( {k, {\rm{k}}} \right)}^{\rm{H}}}}} \right| \hfill \\ \end{gathered} $$ (1) $$ {\rm{s}}{\rm{.t}}{\rm{.}}\left\{ \begin{gathered} {\rm{Tr}}\left( {{S^l}} \right) = {\rm{Tr}}\left( {{S^{l1}} + \cdots + {S^{l\left| {{A_l}} \right|}}} \right) \leqslant {P_l}\;\;\;\;\;\;l = 1, 2, 3 \hfill \\ {S^{lk}} \geqslant 0\;\;\;\;\;l = 1, 2, 3\;\;\;k = 1, 2, \cdots, \left| {{A_l}} \right| \hfill \\ \end{gathered} \right. $$ (2) 式中,Hkl, m为第l小区第m用户到第k小区基站的信道矩阵;Slj表示第l小区第j用户发送的信号,每个用户发送d个独立的数据流到基站;Pl表示功率受限。
图 1中,第i(i=1, 2, 3)个基站的接收信号可以表示为:
$$ \begin{gathered} {y^i} = \sum\limits_{t = 1}^3 {\sum\limits_{j = 1}^{{K_l}} {\mathit{\boldsymbol{H}}_i^{tj}{\mathit{\boldsymbol{w}}^{tj}}{s^{tj}}} } + {n^i} = \hfill \\ \sum\limits_{j = 1}^{{K_i}} {\mathit{\boldsymbol{H}}_i^{ij}{\mathit{\boldsymbol{w}}^{ij}}{s^{ij}}} + \sum\limits_{t = 1, t \ne i}^3 {\sum\limits_{j = 1}^{{K_l}} {\mathit{\boldsymbol{H}}_i^{tj}{\mathit{\boldsymbol{w}}^{tj}}{s^{tj}}} } + {\mathit{\boldsymbol{n}}^i} \hfill \\ \end{gathered} $$ (3) 式中,第一项为期望接收信号;第二项为小区间干扰信号;第三项ni是信道噪声;wij为第i小区第j用户的发送预编码矩阵:
$$ {\mathit{\boldsymbol{w}}^{ij}} = {\mathit{\boldsymbol{W}}^{ij}}{\mathit{\boldsymbol{B}}^{ij}} $$ (4) 式中,Wij是用于实现小区间干扰信号对齐的矩阵;Bij是第i小区第j用户的预编码矩阵。
基站利用级联的接收滤波器解调来自i小区第j用户的信号为:
$$ {y^{ij}} = {\left( {{\mathit{\boldsymbol{p}}^{ij}}} \right)^{\rm{H}}}{\left( {{\mathit{\boldsymbol{V}}^i}} \right)^{\rm{H}}}\sum\limits_{t = 1}^l {\sum\limits_{j = 1}^{{K_l}} {{\mathit{\boldsymbol{H}}^{ij}}{\mathit{\boldsymbol{w}}^{ij}}{s^{ij}}} } + {\mathit{\boldsymbol{\tilde n}}^{ij}} $$ (5) 式中,Vi为小区间干扰抑制矩阵;pij是第i小区第j用户的译码矩阵;${\mathit{\boldsymbol{\tilde n}}^{ij}} = {\left( {{\mathit{\boldsymbol{p}}^{ij}}} \right)^{\rm{H}}}{\left( {{\mathit{\boldsymbol{V}}^i}} \right)^{\rm{H}}}{\mathit{\boldsymbol{n}}^{ij}} $为i小区j用户的等效高斯白噪声。
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让每个小区所有边缘用户的发送预编码矩阵级联,使得相邻小区的干扰信号对齐到同一方向,其公式为:
$$ {\rm{span}}\left( {{\mathit{\boldsymbol{G}}^{12}}{\mathit{\boldsymbol{W}}^2}} \right) = {\rm{span}}\left( {{\mathit{\boldsymbol{G}}^{13}}{\mathit{\boldsymbol{W}}^3}} \right) $$ (6) $$ {\rm{span}}\left( {{\mathit{\boldsymbol{G}}^{21}}{\mathit{\boldsymbol{W}}^1}} \right) = {\rm{span}}\left( {{\mathit{\boldsymbol{G}}^{23}}{\mathit{\boldsymbol{W}}^3}} \right) $$ (7) $$ {\rm{span}}\left( {{\mathit{\boldsymbol{G}}^{31}}{\mathit{\boldsymbol{W}}^1}} \right) = {\rm{span}}\left( {{\mathit{\boldsymbol{G}}^{32}}{\mathit{\boldsymbol{W}}^2}} \right) $$ (8) $$ \begin{gathered} {\mathit{\boldsymbol{G}}^{ij}} = \left[{\mathit{\boldsymbol{H}}_i^{j1}, \mathit{\boldsymbol{H}}_i^{j2}, \cdots, \mathit{\boldsymbol{H}}_i^{jK}} \right] \hfill \\ {\mathit{\boldsymbol{W}}^i} = {\left[{{\mathit{\boldsymbol{W}}^{i1}}, {\mathit{\boldsymbol{W}}^{i2}}, \cdots, {\mathit{\boldsymbol{W}}^{iK}}} \right]^{\rm{T}}} \hfill \\ \end{gathered} $$ (9) 将上式展开得:
$$ \begin{gathered} {\rm{span}}\left( {\mathit{\boldsymbol{H}}_1^{21}\mathit{\boldsymbol{W}}_k^{21} + \mathit{\boldsymbol{H}}_1^{22}\mathit{\boldsymbol{W}}_k^{22} + \cdots + \mathit{\boldsymbol{H}}_1^{2K}\mathit{\boldsymbol{W}}_k^{2K}} \right) = \hfill \\ {\rm{span}}\left( {\mathit{\boldsymbol{H}}_1^{31}\mathit{\boldsymbol{W}}_k^{31} + \mathit{\boldsymbol{H}}_1^{32}\mathit{\boldsymbol{W}}_k^{32} + \cdots + \mathit{\boldsymbol{H}}_1^{3K}\mathit{\boldsymbol{W}}_k^{3K}} \right) \hfill \\ \end{gathered} $$ (10) 因此,存在非零数a1, a2满足下式:
$$ \begin{gathered} {a_1}\left( {\mathit{\boldsymbol{H}}_1^{21}\mathit{\boldsymbol{W}}_k^{21} + \cdots + \mathit{\boldsymbol{H}}_1^{2K}\mathit{\boldsymbol{W}}_k^{2K}} \right) = \hfill \\ {a_2}\left( {\mathit{\boldsymbol{H}}_1^{31}\mathit{\boldsymbol{W}}_k^{31} + \cdots + \mathit{\boldsymbol{H}}_1^{3K}\mathit{\boldsymbol{W}}_k^{3K}} \right) \hfill \\ \end{gathered} $$ (11) 从而,
$$ \begin{gathered} \dim \left\{ {{\rm{span}}\left( {\mathit{\boldsymbol{H}}_1^{21}{\mathit{\boldsymbol{W}}^{21}} \cdots \mathit{\boldsymbol{H}}_1^{2K}{\mathit{\boldsymbol{W}}^{2K}}\mathit{\boldsymbol{H}}_3^{31}{\mathit{\boldsymbol{W}}^{31}} \cdots \mathit{\boldsymbol{H}}_1^{3K}{\mathit{\boldsymbol{W}}^{3K}}} \right)} \right\} = \hfill \\ \dim \left\{ {{\rm{span}}\left( {\mathit{\boldsymbol{H}}_1^{21}\mathit{\boldsymbol{W}}_1^{21} \cdots \mathit{\boldsymbol{H}}_1^{2K}\mathit{\boldsymbol{W}}_1^{2K}\mathit{\boldsymbol{H}}_3^{31}\mathit{\boldsymbol{W}}_1^{31} \cdots \mathit{\boldsymbol{H}}_1^{3K}\mathit{\boldsymbol{W}}_1^{3K}} \right)} \right\} + \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \cdots + \hfill \\ \dim \left\{ {{\rm{span}}\left( {\mathit{\boldsymbol{H}}_1^{21}\mathit{\boldsymbol{W}}_d^{21} \cdots \mathit{\boldsymbol{H}}_1^{2K}\mathit{\boldsymbol{W}}_d^{2K}\mathit{\boldsymbol{H}}_3^{31}\mathit{\boldsymbol{W}}_d^{31} \cdots \mathit{\boldsymbol{H}}_1^{3K}\mathit{\boldsymbol{W}}_d^{3K}} \right)} \right\} = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {2K-1} \right)d \hfill \\ \end{gathered} $$ (12) 式中,dim()表示信号所占的子空间数。由式(12)可得:假设每小区选择K个用户通信,则相邻小区的2K个用户的干扰信号可以对齐到2(K-1)d个子空间中。每个基站用来解调本小区内K个用户的Kd个数据流的空间维度M应满足:
$$ M > \left( {2K-1} \right)d + Kd $$ (13) 当M和d给定时,则每小区选择的用户数K应该满足:
$$ K < \left[{\frac{{M + d}}{{3d}}} \right] $$ (14) 由式(14)可知,单小区基站最多可同时为K个用户服务。
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假设每小区边缘用户数为Kl,由式(14)每小区选择的用户数为K,Al=1, 2, ..., K为一个选择的用户集合,各小区进行用户选择的目标是选择一组用户,使得本小区传输速率最大。
定义路径损耗模型[10]为:${\rho _k} = P/d_k^\alpha $,其中P代表发送功率,dk代表用户K到基站的距离,α代表路径损耗因子,则每个用户的通信质量Sli为:
$$ \begin{gathered} S_l^1 = \mathop {\arg \;\max }\limits_{k \in \left\{ {1, 2, \cdots, {K_l}} \right\}} {\rho _k}{\rm{Tr}}\left( {{{\left( {\mathit{\boldsymbol{H}}_i^{ik}} \right)}^{\rm{H}}}\mathit{\boldsymbol{H}}_i^{ik}} \right) \hfill \\ S_l^2 = \mathop {\arg \;\max }\limits_{k \in \left\{ {1, 2, \cdots, {K_l}-1} \right\}} {\rho _k}{\rm{Tr}}\left( {{{\left( {\mathit{\boldsymbol{H}}_i^{ik}} \right)}^{\rm{H}}}\mathit{\boldsymbol{H}}_i^{ik}} \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \hfill \\ S_l^{{K_l}} = \mathop {\arg \;\max }\limits_{k \in \left\{ {1, 2, \cdots, K-\left( {{K_l}-1} \right)} \right\}} {\rho _k}{\rm{Tr}}\left( {{{\left( {\mathit{\boldsymbol{H}}_i^{ik}} \right)}^{\rm{H}}}\mathit{\boldsymbol{H}}_i^{ik}} \right) \hfill \\ \end{gathered} $$ (15) 选择K个用户构成的集合满足下式:
$$ {S_L} = \mathop {\arg \;\max }\limits_{k \in \left\{ {1, 2, \cdots, {K_l}} \right\}} \left( {S_l^1 + S_l^2 + \cdots + S_l^K} \right) $$ (16) 布谷鸟搜索算法[13]CS(Cuckoo search)是一种新兴启发算法,通过模拟布谷鸟的寄生育雏来有效地求解最优化问题的算法。CS算法采纳相关的莱维飞行搜索机制,由于其参数较少,算法简单,易于实现,广泛应用于各个领域。并且与粒子群优化算法、遗传算法相比,布谷鸟搜索算法能够获得更好的解决方案[14-15],成为智能算法中的新亮点。
本文用布谷鸟搜索算法进行用户的选择,算法步骤如下。
1) 初始化。将Sli最大的K个用户作为第一个鸟巢,即随机产生Kl×1的向量,选择其中的K个组成K×1的向量作为一个个体。然后依次循环移位得到的鸟巢组成一个解的集合,每个鸟巢中都用K个二进制数代表用户,1代表该用户被选中,0代表没被选中。
2) 选择。根据式(15)计算每个鸟巢的适应度值,并记录当前的最优解。
3) 种群更新。保留上一代最优鸟巢位置,在此基础上进行鸟巢位置的更新。
4) 判断。现有鸟巢与上一代鸟巢进行对比,适应度值较大,则将其作为当前最优位置。
5) 迭代终止。当前解迭代设定的有限次数后无更新,记录当前最优解;否则,放弃,随机产生新的鸟巢,重复步骤2)。
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当各小区进行通信的用户选定后,需要对相邻小区的干扰信号进行对齐处理。设M=16,d=2时,由式(14)可得各小区选择的边缘用户数为2,通过两个用户协作,可得:
$$ \dim \left\{ {{\rm{span}}\left( {\mathit{\boldsymbol{H}}_1^{21}{\mathit{\boldsymbol{W}}^{21}}\mathit{\boldsymbol{H}}_1^{22}{\mathit{\boldsymbol{W}}^{22}}\mathit{\boldsymbol{H}}_1^{31}{\mathit{\boldsymbol{W}}^{31}}\mathit{\boldsymbol{H}}_1^{32}{\mathit{\boldsymbol{W}}^{32}}} \right)} \right\} = 3 $$ (17) $$ \dim \left\{ {{\rm{span}}\left( {\mathit{\boldsymbol{H}}_2^{11}{\mathit{\boldsymbol{W}}^{11}}\mathit{\boldsymbol{H}}_2^{12}{\mathit{\boldsymbol{W}}^{12}}\mathit{\boldsymbol{H}}_2^{31}{\mathit{\boldsymbol{W}}^{31}}\mathit{\boldsymbol{H}}_2^{32}{\mathit{\boldsymbol{W}}^{32}}} \right)} \right\} = 3 $$ (18) $$ \dim \left\{ {{\rm{span}}\left( {\mathit{\boldsymbol{H}}_3^{11}{\mathit{\boldsymbol{W}}^{11}}\mathit{\boldsymbol{H}}_3^{12}{\mathit{\boldsymbol{W}}^{12}}\mathit{\boldsymbol{H}}_3^{21}{\mathit{\boldsymbol{W}}^{21}}\mathit{\boldsymbol{H}}_3^{22}{\mathit{\boldsymbol{W}}^{22}}} \right)} \right\} = 3 $$ (19) 此时,由式(6)~式(8),可将信道等效为3对用户组成的干扰信道[12]:
$$ \mathit{\boldsymbol{E}} = {\left( {{\mathit{\boldsymbol{G}}^{31}}} \right)^{-1}}{\mathit{\boldsymbol{G}}^{32}}{\left( {{\mathit{\boldsymbol{G}}^{12}}} \right)^{-1}}{\mathit{\boldsymbol{G}}^{13}}{\left( {{\mathit{\boldsymbol{G}}^{23}}} \right)^{-1}}{\mathit{\boldsymbol{G}}^{21}} $$ (20) $$ \mathit{\boldsymbol{F}} = {\left( {{\mathit{\boldsymbol{G}}^{32}}} \right)^{-1}}{\mathit{\boldsymbol{G}}^{31}} $$ (21) $$ \mathit{\boldsymbol{C}} = {\left( {{\mathit{\boldsymbol{G}}^{23}}} \right)^{-1}}{\mathit{\boldsymbol{G}}^{21}} $$ (22) $$ {\rm{span}}\left( {{\mathit{\boldsymbol{W}}^1}} \right) = {\rm{span}}\left( {\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{W}}^2}} \right) $$ (23) $$ {\mathit{\boldsymbol{W}}^2} = \mathit{\boldsymbol{C}}{\mathit{\boldsymbol{W}}^1} $$ (24) $$ {\mathit{\boldsymbol{W}}^3} = \mathit{\boldsymbol{C}}{\mathit{\boldsymbol{W}}^1} $$ (25) 式中,{e1, e2, ..., e16}是E的特征向量;W1是其任意2个特征向量。
根据文献[16],小区间干扰抑制矩阵Vi为:
$$ {\mathit{\boldsymbol{V}}^1} \subseteq {\rm{null}}\left( {\mathit{\boldsymbol{H}}_1^{21}{\mathit{\boldsymbol{W}}^{21}}\mathit{\boldsymbol{H}}_1^{22}{\mathit{\boldsymbol{W}}^{22}}\mathit{\boldsymbol{H}}_1^{31}{\mathit{\boldsymbol{W}}^{31}}\mathit{\boldsymbol{H}}_1^{32}{\mathit{\boldsymbol{W}}^{32}}} \right) $$ (26) $$ {\mathit{\boldsymbol{V}}^2} \subseteq {\rm{null}}\left( {\mathit{\boldsymbol{H}}_2^{11}{\mathit{\boldsymbol{W}}^{11}}\mathit{\boldsymbol{H}}_2^{12}{\mathit{\boldsymbol{W}}^{12}}\mathit{\boldsymbol{H}}_2^{31}{\mathit{\boldsymbol{W}}^{31}}\mathit{\boldsymbol{H}}_2^{32}{\mathit{\boldsymbol{W}}^{32}}} \right) $$ (27) $$ {\mathit{\boldsymbol{V}}^3} \subseteq {\rm{null}}\left( {\mathit{\boldsymbol{H}}_3^{11}{\mathit{\boldsymbol{W}}^{11}}\mathit{\boldsymbol{H}}_3^{12}{\mathit{\boldsymbol{W}}^{12}}\mathit{\boldsymbol{H}}_3^{21}{\mathit{\boldsymbol{W}}^{21}}\mathit{\boldsymbol{H}}_3^{22}{\mathit{\boldsymbol{W}}^{22}}} \right) $$ (28) 因此,边缘用户的等效信道为:
$$ \mathit{\boldsymbol{\tilde H}}_j^{kj} = {\left( {{\mathit{\boldsymbol{V}}^j}} \right)^H}\mathit{\boldsymbol{H}}_j^{kj}{\mathit{\boldsymbol{W}}^{jk}}\;\;\;j \in \left\{ {1, 2, 3} \right\}, k \in \left\{ {1, 2} \right\} $$ (29) -
假设上行用户同步发送信号,各小区中的第k个用户经过预编码矩阵Bjk处理后,分别把信号发送出去。基站可以通过译码矩阵R恢复出所有用户的信号。基站端译码后所有用户的估计信号为:
$$ \mathit{\boldsymbol{\hat s}} = \mathit{\boldsymbol{R}}\left( {\sum\limits_{j = 1}^K {\mathit{\boldsymbol{\tilde H}}_j^{jk}{\mathit{\boldsymbol{B}}^{jk}}{s_j} + n} } \right) $$ (30) 式中,$ \mathit{\boldsymbol{\hat s}} = \left[{s_1^{\rm{H}}, s_2^{\rm{H}}, \cdots, s_k^{\rm{H}}} \right]$。
此时,对用户k的信道$\mathit{\boldsymbol{\tilde H}}_j^{jk} $可以进行奇异值分解[14]:
$$ {\left( {\mathit{\boldsymbol{\tilde H}}_j^{jk}} \right)^{\rm{H}}} = \left[{\mathit{\boldsymbol{U}}_k^1\;\mathit{\boldsymbol{U}}_k^0} \right]\left[\begin{gathered} {\mathit{\Lambda }_k} \hfill \\ 0 \hfill \\ \end{gathered} \right]\mathit{\boldsymbol{V}}_k^{\rm{H}} $$ (31) 式中,Uk1为左奇异矩阵;Vk为右奇异矩阵;${\mathit{\Lambda} _k} = {\rm{diag}}\left( {{\lambda _1}, {\lambda _2}, \cdots, {\lambda _N}} \right) $为对角矩阵。因此,式(30)可以写为:
$$ \mathit{\boldsymbol{\hat s = R}}\left( {\sum\limits_{k = 1}^K {\mathit{\boldsymbol{U}}_k^1{\mathit{\boldsymbol{ \boldsymbol{\varLambda} }}_k}{\mathit{\boldsymbol{B}}^{jk}}{s_k} + n} } \right) $$ (32) 式中,第k个用户的预编码矩阵设计为Bjk=Vk。利用VkHVk=I,式(32)可以写为:
$$ \mathit{\boldsymbol{\hat s = R}}\left( {\sum\limits_{k = 1}^K {\mathit{\boldsymbol{U}}_k^1{\mathit{\boldsymbol{ \boldsymbol{\varLambda} }}_k}{s_k} + n} } \right) = \mathit{\boldsymbol{RU \boldsymbol{\varLambda} }}s + \mathit{\boldsymbol{R}}n $$ (33) 引入矩阵:
$$ \mathit{\boldsymbol{U = }}\left[{U_1^1, U_2^1, \cdots, U_K^1} \right], \;\;\mathit{\boldsymbol{V = }}\left[{\mathit{\Lambda} _1^1, \mathit{\Lambda} _2^1, \cdots, \mathit{\Lambda} _K^1} \right] $$ (34) 为了在基站端联合译码出所有用户的数据,设计译码矩阵$\mathit{\boldsymbol{R}} = {\mathit{\boldsymbol{U}}^ + } = {\left[{\mathit{\boldsymbol{U}}{\mathit{\boldsymbol{U}}^{\rm{H}}}} \right]^{ -1}}{\mathit{\boldsymbol{U}}^{\rm{H}}} $。这样,所有用户的信号可分离出来,没有多用户干扰,则用户译码后的信号和第k个用户译码后的信号可写为:
$$ \mathit{\boldsymbol{\hat s = \boldsymbol{\varLambda} }}s + \mathit{\boldsymbol{R}}n $$ (35) $$ {\mathit{\boldsymbol{\hat s}}^{jk}}\mathit{\boldsymbol{ = }}{\mathit{\boldsymbol{ \boldsymbol{\varLambda} }}^{jk}}{s^{jk}} + {n^{jk}} $$ (36) 这种译码方法可能会引起噪声信号的放大,因此,本文采用MMSE准则设计的译码矩阵,估计译码后的第j小区第k用户的信号为:
$$ {\hat s^{jk}}\mathit{\boldsymbol{ = }}\sum\limits_{i = 1}^2 {{{\left( {{\mathit{\boldsymbol{p}}^{jk}}} \right)}^{\rm{H}}}\mathit{\boldsymbol{\tilde H}}_j^{jk}{\mathit{\boldsymbol{B}}^{ji}}{s^{ji}}} + {\left( {{\mathit{\boldsymbol{p}}^{jk}}} \right)^{\rm{H}}}{n_k} $$ (37) 式中,pjk是译码向量。在基站端,由MMSE准则对信号进行联合译码,从而最小化总的误差:
$$ \begin{gathered} \mathop {\min \;{\rm{mize}}}\limits_{{p^{j1}}, {p^{j2}}} \sum\limits_{k = 1}^2 {\rm{E}} \left\{ {\left\| {{{\hat s}^{jk}}-{s^{jk}}} \right\|} \right\} \hfill \\ {\rm{Tr}}\left( {{{\left( {{\mathit{\boldsymbol{p}}^{jk}}} \right)}^{\rm{H}}}{\mathit{\boldsymbol{p}}^{jk}}} \right) \leqslant {P_{jk}}\;\;\;\;k = 1, 2 \hfill \\ \end{gathered} $$ (38) 进一步,译码矩阵可通过式(39)获得:
$$ \begin{gathered} \mathop {\min }\limits_{{p^{j1}},{p^{j2}}} \sum\limits_{k = 1}^2 {\left[ {{{\left\| {{{\left( {{\boldsymbol{p}^{jk}}} \right)}^{\text{H}}}\boldsymbol{\tilde H}_j^{jk}{\boldsymbol{B}^{jk}} - 1} \right\|}^2}} \right.} + \hfill \\ \left. {\sum\limits_{_{i \ne j}^{i = 1}}^2 {{{\left\| {{{\left( {{\boldsymbol{p}^{jk}}} \right)}^{\text{H}}}\boldsymbol{\tilde H}_j^{ji}{\boldsymbol{B}^{ji}}} \right\|}^2} + {{\left\| {{\boldsymbol{p}^{jk}}} \right\|}^2}} } \right] \hfill \\ \end{gathered} $$ (39) 由文献[13],Bjk为已知,所以有:
$$ \mathit{\boldsymbol{\hat H}}_j^{jk} = \mathit{\boldsymbol{\tilde H}}_j^{jk}{\mathit{\boldsymbol{B}}^{jk}} $$ (40) $$ {\mathit{\boldsymbol{p}}^{jk}} = {\left( {\mathit{\boldsymbol{\tilde H}}_j^{jk}{\mathit{\boldsymbol{B}}^{jk}}} \right)^{\rm{H}}}{\left( {\mathit{\boldsymbol{\hat H}}_j^{jk}{{\left( {\mathit{\boldsymbol{\hat H}}_j^{jk}} \right)}^{\rm{H}}} + \mathit{\boldsymbol{I}}} \right)^{-1}} $$ (41) -
$$ T\left( n \right) = 3O\left( {n + f\left( n \right)} \right) + O\left( n \right) = O\left( {n + f\left( N \right)} \right) $$ (42) 布谷鸟搜索算法的时间复杂度取决于计算目标函数的时间复杂度f(n),当f(n)比n高阶运算时,复杂度为O(f(n)),当同阶或低阶时,复杂度为O(n)。本文所提算法中,适应度函数f(n)是一阶函数,与n同阶,则本文所提布谷鸟搜索算法的时间复杂度为O(n)。快速排序法是精准算法,平均复杂度为O(nlog(n))。由算法分析中常见复杂度的比较:
$$ O\left( 1 \right) < O\left( {\log n} \right) < O\left( n \right) < O\left( {n\log n} \right) < O\left( {{n^2}} \right) $$ (43) 可以得知,布谷鸟搜索算法的时间复杂度低于快速排序法的时间复杂度。
User Selection Based on Cuckoo Search Algorithm and Interference Alignment
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摘要: 为了实现蜂窝系统中单小区边缘用户正常通信,减少相邻小区间多个边缘用户对本小区边缘用户造成的干扰,提出了一种基于布谷鸟搜索算法的用户选择和干扰对齐算法。该算法首先用布谷鸟搜索算法对小区边缘用户进行选择,接着采用干扰对齐方法消除相邻小区间的干扰,最后通过预编码和基于最小均方差(MMSE)译码方法消除小区内用户间的干扰。该布谷鸟搜索算法与快速排序搜索算法相比具有更低的时间复杂度。数值分析表明与基于迫零算法的译码方法相比,该译码方法能够提高系统容量2 b·s-1·Hz-2,改善误码率4 dB。Abstract: In cellular system, to reduce the interference of the cell-edge users and ensure communication among the cell-edge users of a single cell, a user selection scheme based on cuckoo search algorithm and interference alignment algorithm is proposed.First, the cell-edge users are selected by the cuckoo search algorithm. Then interference alignment scheme is adopted to eliminate interference.Finally, an encoding and decoding based on MMSE criterion are applied to eliminate interference among users.Compared with the quick sort search algorithm, the proposed scheme based on cuckoo search algorithm has less time complexity.And numerical results show that, compared to zero-forcing decoding, the proposed algorithm will increase the system capacity by 2 b·s-1·Hz-2, and the bit error rate improvement is about 4 dB.
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