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随着通信技术的发展,传统的半双工通信已不能满足用户的需求。全双工通信作为5G通信中的一项关键技术,允许信号接收和发送同时同频进行,可以有效地利用频谱资源,降低系统时延[1-2]。大规模MIMO同样作为5G关键技术之一,通过在基站端布置成百上千根天线,可以有效提升数据速率和链路可靠性[3-4]。在新一代无线通信系统中,多种技术相结合已经成为一种趋势[5-6]。
由于接收天线可以收到自身节点发送的信号,全双工通信面临严重的自干扰问题。现有的自干扰消除技术主要分为两大类:主动干扰消除和被动干扰消除。主动干扰消除通过在通信链路部分模块加入有源信号消除自干扰的影响。文献[7]提出了一种射频端快速自适应干扰抵消算法。文献[8]提出了一种变窗长全双工自干扰信道估计算法。被动天线消除技术主要是通过天线自身特性和信号传播特点来设计天线模式,达到干扰消除的目的。近几年,天线分离技术作为一种被动自干扰消除技术,受到越来越多的关注[9]。文献[5]提出了一种全双工大规模天线系统中天线分配算法。文献[10]分析了全双工大规模MIMO系统迫零线性处理下的最优天线比。文献[11]分析了每个用户使用半双工模式,并且装备多个天线的天线分配。
在之前的文献中,天线分配问题的研究大都基于理想信道状态信息,然而当天线数目变大时,很难获取理想的信道状态信息。本文研究了在存在信道估计误差的情况下,最优的天线分配问题。首先推导出系统和速率表达式,同已有的文献相比,该表达式具有更紧下界。基于和速率下界,将原始的天线分配问题转化为凸优化问题,以系统和速率最大化为优化目标,通过拉格朗日标准算法,获取最优的天线分配。
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本文考虑单小区大规模MIMO系统,包括一个宏基站BS和K个用户。所有用户和BS工作在全双工模式。假定基站配置M根天线,其中发送天线数目为Nt,接收天线数目为Nr。用户配备双天线,一根为发送天线,另一根为接收天线。
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基站端的接收信号可以表示为:
$$ {\mathit{\boldsymbol{y}}^{\rm{u}}} = \sqrt {{P^u}} {\mathit{\boldsymbol{F}}^H}{\mathit{\boldsymbol{G}}_u}{\mathit{\boldsymbol{x}}^u} + \sqrt {{P^d}} {\mathit{\boldsymbol{F}}^H}{\mathit{\boldsymbol{W}}_B}(\mathit{\boldsymbol{P}}{\mathit{\boldsymbol{x}}^d} + \mathit{\boldsymbol{t}}) + {\mathit{\boldsymbol{F}}^H}{\mathit{\boldsymbol{n}}^u} $$ (1) 式中,$ {\mathit{\boldsymbol{G}}_u} = [\mathit{\boldsymbol{g}}_1^u, \mathit{\boldsymbol{g}}_2^u, \cdots , \mathit{\boldsymbol{g}}_K^u] \in {{\mathbb{C}}^{{N_r} \times K}} $代表上行信道矩阵;F为基站端信号检测矩阵;$ {\mathit{\boldsymbol{W}}_B} \in {{\mathbb{C}}^{{N_r} \times {N_t}}} $代表基站端的自干扰信道矩阵。
用户端接收到的信号可以表示为:
$$ {\mathit{\boldsymbol{y}}^d} = \sqrt {{P^d}} \mathit{\boldsymbol{G}}_d^H(P{x^d} + t) + \;\sqrt {{P^u}} {\mathit{\boldsymbol{W}}_u}{\mathit{\boldsymbol{x}}^u} + {n^d} $$ (2) 式中,$ {\mathit{\boldsymbol{G}}_d} = [\mathit{\boldsymbol{g}}_1^d, \mathit{\boldsymbol{g}}_2^d, \cdots , \mathit{\boldsymbol{g}}_K^d] \in {{\mathbb{C}}^{{N_t} \times K}} $代表下行信道矩阵;$ {\mathit{\boldsymbol{W}}_u} \in {{\mathbb{C}}^{K \times K}} $代表用户端的自干扰信道矩阵;上/下行信道矩阵$ {\mathit{\boldsymbol{G}}_i}, i \in \{ d, u\} $可以表示为$ {\mathit{\boldsymbol{G}}_i} = {\mathit{\boldsymbol{H}}_i}\mathit{\boldsymbol{D}}_i^{1/2} $,其中$ {\mathit{\boldsymbol{H}}_i} \in {{\mathbb{C}}^{{N_i} \times K}} $代表小尺度衰落,$ {\mathit{\boldsymbol{D}}_i} $代表大尺度衰落对角矩阵,其第k个对角元素代表大尺度衰落因子$ {\beta _{i, k}} $;不失一般性,$ {\mathit{\boldsymbol{W}}_B} $为独立同分布变量,$ {\mathit{\boldsymbol{W}}_B} \sim {\rm{CN}}(0, {\beta _B}) $,其元素$ {\beta _B} $反映了自干扰等级;$ {\mathit{\boldsymbol{W}}_u} $中的元素可以表示为$ w_{ij}^u $,代表用户i和用户j之间的信道,满足$ w_{ij}^u\sim{\rm{CN}}(0, {\beta _{ij}}) $。
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在信道估计阶段,所有用户和基站同时同频发送导频信号,则基站端的发送天线和接收天线接收到的导频信号可以分别表示为:
$$ \begin{array}{l} {\mathit{\boldsymbol{Y}}^R} = \sqrt {\tau P_p^r} {\mathit{\boldsymbol{G}}_{RR}}{\mathit{\boldsymbol{X}}_R} + \sqrt {\tau P_p^t} {\mathit{\boldsymbol{G}}_u}{\mathit{\boldsymbol{X}}_T} + {\mathit{\boldsymbol{N}}_R}\\ {\mathit{\boldsymbol{Y}}^T} = \sqrt {\tau P_p^t} {\mathit{\boldsymbol{G}}_{TT}}{\mathit{\boldsymbol{X}}_T} + \sqrt {\tau P_p^r} {\mathit{\boldsymbol{G}}_d}{\mathit{\boldsymbol{X}}_R} + {\mathit{\boldsymbol{N}}_T} \end{array} $$ (3) 式中,$ \tau $为导频信号长度;$ P_p^r $和$ P_p^t $为下行和上行导频信号功率。根据MMSE信道估计,信道矩阵可以表示为$ {\mathit{\boldsymbol{G}}_i} = {\mathit{\boldsymbol{\hat G}}_i} + {\mathit{\boldsymbol{\tilde G}}_i} $,其中,$ {\hat G_i} $为信道估计误差。根据MMSE信道估计特性,可以认为$ {\mathit{\boldsymbol{\tilde G}}_i} $和$ {\hat G_i} $为独立同分布变量,其每一列相互正交,第$ k $个对角线元素可以分别表示为$ \hat \beta _k^i = \tau P_p^i{(\beta _k^i)^2}/(\tau P_p^i\beta _k^i + 1) $和$ \tilde \beta _k^i = \beta _k^i/(\tau P_p^i\beta _k^i + 1) $。
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基站端采用MMSE信号检测准则,上行波束赋形矩阵P和下行预编码矩阵F可以表示为:
$$ \begin{array}{l} \mathit{\boldsymbol{F}} = {{\mathit{\boldsymbol{\hat G}}}_u}{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})^{ - 1}}\\ \mathit{\boldsymbol{P}} = {{\mathit{\boldsymbol{\hat G}}}_d}{({{\mathit{\boldsymbol{\hat G}}}_d}^H{{\mathit{\boldsymbol{\hat G}}}_d})^{ - 1}}u \end{array} $$ (4) 式中,$ u = {\rm{ Tr}}({({\mathit{\boldsymbol{\hat G}}_d}^H{\mathit{\boldsymbol{\hat G}}_d})^{ - 1}}){)^{ - 1/2}} $。
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将式(4)代入式(1),基站接收到的第k个用户的信号可以表示为:
$$ \begin{array}{c} y_k^u = \sqrt {{P^u}} x_k^u + e_k^T{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})^{ - 1}}{({{\mathit{\boldsymbol{\hat G}}}_u})^H}{n^u} + \\ \sqrt {{P^u}} e_k^T{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})^{ - 1}}{({{\mathit{\boldsymbol{\hat G}}}_u})^H}\sum\limits_{i = 1}^K {\tilde g_i^ux_i^u} + \\ \sqrt {{P^d}} e_k^T{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})^{ - 1}}{({{\mathit{\boldsymbol{\hat G}}}_u})^H}\mathit{\boldsymbol{W}}{{\hat G}_d}{({{\mathit{\boldsymbol{\hat G}}}_d}^H{{\mathit{\boldsymbol{\hat G}}}_d})^{ - 1}}u{x^d} \end{array} $$ (5) 式中,等式右边第1项为所需要的信号,功率为$ {P^u} $;第2项为系统噪声$ {I_n} $;第3项为信道估计误差噪声$ {I_e} $;第4项为自干扰$ {I_s} $。根据文献[12],有如下引理。
引理 1 给定矩阵$ \mathit{\boldsymbol{A}} \in {{\mathbb{C}}^{M \times M}} $,$ x \in {{\mathbb{C}}^M} $,其中x中元素为独立同分布随机变量,满足$ {\rm{CN}}(0, {\sigma _1}) $,则$ \frac{1}{M}{x^H}\mathit{\boldsymbol{A}}x - \frac{1}{M}{\rm{Tr}}({\sigma _1}\mathit{\boldsymbol{A}})\mathop \to \limits_{}^{a.s.} 0. $
根据引理1和Lindeberg-Ledvy中心极限定理,$ {I_e} $可以表示为如下形式:
$$ \begin{array}{c} {I_e} = {P^u}\sum\limits_{i = 1}^K \begin{array}{l} {\left| {e_k^T{{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})}^{ - 1}}\hat G_u^H\tilde g_i^u} \right|^2}\\ \end{array} \\ \mathop \to \limits^{a.s.} \frac{{{P^u}}}{{N_r^2}}\sum\limits_{i = 1}^K {{{\left| {e_k^T\hat D_u^{ - 1}{{\mathit{\boldsymbol{\hat G}}}_u}\mathit{\boldsymbol{\tilde g}}_i^u} \right|}^2}} = \\ \frac{{{P^u}}}{{N_r^{}}}\sum\limits_{i = 1}^K {{{\left| {{{(\hat \beta _k^u)}^{ - \frac{1}{2}}}\frac{{{{(\mathit{\boldsymbol{\hat g}}_k^u)}^H}\mathit{\boldsymbol{\tilde g}}_i^u}}{{\sqrt {{N_r}} }}} \right|}^2}} \\ \mathop \to \limits^d {(\hat \beta _k^u)^{ - 1}}\frac{{{P^u}}}{{N_r^{}}}\sum\limits_{i = 1}^K {{{(q_i^e)}^2}} \end{array} $$ (6) 式中,$ q_i^e $为随机变量,满足$ {\rm{CN}} \sim (0, \hat \beta _k^u\tilde \beta _i^u) $分布。
采用相同方法,$ {I_s} $和$ {I_n} $可以分别表示为:
$$ \begin{array}{c} {I_s} = {P^d}{\left| {e_k^T{{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})}^{ - 1}}\hat G_u^HW{{\mathit{\boldsymbol{\hat G}}}_d}{{(\hat G_d^H{{\mathit{\boldsymbol{\hat G}}}_d})}^{ - 1}}u} \right|^2} = \\ \frac{1}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_d^{ - 1})N_r^2N_t^{}}}{\left| {{{(\mathit{\boldsymbol{\hat \beta }}_k^u)}^{ - 1}}{{(\mathit{\boldsymbol{\hat g}}_k^u)}^H}W{{\mathit{\boldsymbol{\hat G}}}_d}\hat D_d^{ - 1}} \right|^2} = \\ \frac{1}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_d^{ - 1})N_r^2}}{(\mathit{\boldsymbol{\hat \beta }}_k^u)^{ - 2}}\sum\limits_{i = 1}^K {{P^d}{{(\mathit{\boldsymbol{\hat \beta }}_i^d)}^{ - 2}}} \frac{{{{\left\| {\mathit{\boldsymbol{\hat g}}_i^d} \right\|}^2}}}{{{N_t}}} \times \\ {\rm{ }}{(\mathit{\boldsymbol{\hat g}}_k^u)^H}W\mathit{\boldsymbol{\bar g}}_i^d{(\mathit{\boldsymbol{\bar g}}_i^d)^H}{W^H}\mathit{\boldsymbol{\hat g}}_k^u \end{array} $$ (7) $$ {I_n} = {\left| {e_k^T{{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})}^{ - 1}}\mathit{\boldsymbol{\hat G}}_u^H} \right|^2}\mathop \to \limits^{a.s.} {({N_r}\mathit{\boldsymbol{\hat \beta }}_k^u)^{ - 1}} $$ (8) 式中,$ \mathit{\boldsymbol{\bar g}}_i^d = \mathit{\boldsymbol{\hat g}}_i^d/\left\| {\mathit{\boldsymbol{\hat g}}_i^d} \right\| $;$ {\rm{Tr}}(\mathit{\boldsymbol{\bar g}}_i^d{(\mathit{\boldsymbol{\bar g}}_i^d)^H}) = 1 $,$ {\rm{rank}}(\mathit{\boldsymbol{\bar g}}_i^d{(\mathit{\boldsymbol{\bar g}}_i^d)^H}) = 1 $。基于奇异值分解,$ \mathit{\boldsymbol{\bar g}}_i^d{(\mathit{\boldsymbol{\bar g}}_i^d)^H} $可以表示为$ \mathit{\boldsymbol{U}}{\rm{diag}}({e_1}){\mathit{\boldsymbol{U}}^H} $。定义$ \mathit{\boldsymbol{\tilde W}}\mathop = \limits^\Delta \mathit{\boldsymbol{WU}} $,则$ {I_s} $可以表示为如下形式:
$$ \begin{array}{c} {I_s} = \frac{1}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_d^{ - 1})N_r^2}}{(\mathit{\boldsymbol{\hat \beta }}_k^u)^{ - 2}}{P^d}\sum\limits_{i = 1}^K {{{(\mathit{\boldsymbol{\hat \beta }}_i^d)}^{ - 2}}} \frac{{{{\left\| {\mathit{\boldsymbol{\hat g}}_i^d} \right\|}^2}}}{{{N_t}}} \times \\ {\rm{ }}{(\mathit{\boldsymbol{\hat g}}_k^u)^H}\mathit{\boldsymbol{\tilde W}}{\rm{diag}}({e_1}){{\mathit{\boldsymbol{\tilde W}}}^H}\mathit{\boldsymbol{\hat g}}_k^u = \\ \frac{1}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_d^{ - 1})N_r^2}}{(\mathit{\boldsymbol{\hat \beta }}_k^u)^{ - 2}}{P^d}\sum\limits_{i = 1}^K {{{(\mathit{\boldsymbol{\hat \beta }}_i^d)}^{ - 2}}} \frac{{{{\left\| {\mathit{\boldsymbol{\hat g}}_i^d} \right\|}^2}}}{{{N_t}}}\frac{{{{\left\| {{{\mathit{\boldsymbol{\tilde w}}}_1}} \right\|}^2}}}{{N_r^{}}} \times \\ \frac{{{{(\mathit{\boldsymbol{\hat g}}_k^u)}^H}}}{{\sqrt {N_r^{}\mathit{\boldsymbol{\hat \beta }}_k^u} }}{{\bar w}_1}\bar w_1^H\frac{{\mathit{\boldsymbol{\hat g}}_k^u}}{{\sqrt {N_r^{}\mathit{\boldsymbol{\hat \beta }}_k^u} }}N_r^{}\mathit{\boldsymbol{\hat \beta }}_k^u\\ \mathop {®}\limits^{a.s.} {\rm{ }}\frac{{{P^d}\mathit{\boldsymbol{\hat \beta }}_W^{}{{(\mathit{\boldsymbol{\hat \beta }}_k^u)}^{ - 1}}}}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_\mathit{\boldsymbol{d}}^{ - 1})\mathit{\boldsymbol{N}}_r^2}}\sum\limits_{i = 1}^K {{{(\mathit{\boldsymbol{\hat \beta }}_i^d)}^{ - 1}} = {P^d}\mathit{\boldsymbol{\hat \beta }}_W^{}{{(\mathit{\boldsymbol{\hat \beta }}_k^u)}^{ - 1}}} /\mathit{\boldsymbol{N}}_r^{} \end{array} $$ (9) 式中,$ {\mathit{\boldsymbol{\bar w}}_l} = {\mathit{\boldsymbol{\tilde w}}_l}/\left\| {{{\mathit{\boldsymbol{\tilde w}}}_l}} \right\| $。
根据式(6)~式(9),可以获得用户在基站端上行链路信干噪比为:
$$ \bar \gamma _k^u = \frac{{{N_r}{P^u}\hat \beta _k^u}}{{{P^u}\sum\limits_{i = 1}^K {{{(q_i^e)}^2}} + {P^d}\beta _W^{} + 1}} $$ (10) 根据香农定理和杰森不等式,上行链路遍历容量可以表示为:
$$ R_M^u = E{\log _2}(1 + \bar \gamma _k^u) \ge \bar R_M^u = {\log _2}(1 + {\mathbb{E}}\bar \gamma _k^u) $$ (11) 式中,$ \bar R_M^u $是速率的下界;$ {\mathbb{E}}\bar \gamma _k^u $可表示为:
$$ {\mathbb{E}}\bar \gamma _k^u = \frac{{{N_r}{P^u}\hat \beta _k^u}}{{{P^u}\hat \beta _k^u\sum\limits_{i = 1}^K {\tilde \beta _i^u} + {P^d}\beta _W^{} + 1}} $$ (12) -
将式(4)代入式(1),第k个用户接收到的信号可以表示为:
$$ \begin{array}{c} y_k^d = \sqrt {{P^d}} \mu x_k^d + \sqrt {{P^u}} \sum\limits_{i = 1}^K {w_{ik}^u} x_i^u + \\ \sqrt {{P^d}} {(\mathit{\boldsymbol{\tilde g}}_k^d)^H}{{\mathit{\boldsymbol{\hat G}}}_d}{(\mathit{\boldsymbol{\hat G}}_d^H{{\hat G}_d})^{ - 1}}\mu {\mathit{\boldsymbol{x}}^d} + {n^d} \end{array} $$ (13) 式中,等式右边第1项为所需要的信号;第2项为信道估计误差干扰;第3项为上行信号产生的自干扰;第4项为噪声。
采用和上行链路同样的处理方法,下行链路用户端信干噪比为:
$$ \bar \gamma _k^d = \frac{{{N_t}{P^d}}}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_d^{ - 1})\left( {{P^d}\tilde \beta _k^d + {P^u}\sum\limits_{i = 1}^K {{{(w_{ik}^u)}^2} + 1} } \right)}} $$ (14) 下行链路遍历容量为:
$$ R_M^d = E{\log _2}(1 + \bar \gamma _k^d)\bar R_M^d = {\log _2}(1 + E\bar \gamma _k^d) $$ (15) 式中,$ \bar R_M^d $为下界;$ E\bar \gamma _k^d $可以表示为:
$$ E\bar \gamma _k^d = \frac{{{N_t}{P^d}}}{{{\rm{Tr}}(\mathit{\boldsymbol{\hat D}}_d^{ - 1})\left( {{P^d}\tilde \beta _k^d + {P^u}\sum\limits_{i = 1}^K {\beta _{ik}^{} + 1} } \right)}} $$ (16) -
文献[10, 13]采用Wishart矩阵[12]进行信号干噪比分析。基于Wishart矩阵特性,上行链路的和速率的下界可以表示为:
$$ \bar R_M^u = {\log _2}\left( {1 + \frac{{({N_r} - K){P^u}\hat \beta _k^u}}{{{P^u}\sum\limits_{i = 1}^K {\tilde \beta _i^u} + {P^d}\beta _W^{} + 1}}} \right) $$ (17) 证明:根据文献[11]引理2.10,式(5)中$ {I_n} $、$ {I_e} $、$ {I_s} $的期望可以分别表示为:
$$ \begin{array}{c} {\mathbb{E}}{\left| {{I_n}} \right|^2} = {\mathbb{E}}\left| {\mathit{\boldsymbol{f}}_k^H{\mathit{\boldsymbol{f}}_k}} \right| = {\mathbb{E}}\left| {{{\left[ {{{({{\mathit{\boldsymbol{\hat G}}}_u}^H{{\mathit{\boldsymbol{\hat G}}}_u})}^{ - 1}}} \right]}_{kk}}} \right| = \\ \frac{1}{{K{{\hat \beta }_k}}}{\mathbb{E}}\left| {{\rm{Tr}}({{({{\mathit{\boldsymbol{\hat H}}}^H}\mathit{\boldsymbol{\hat H}})}^{ - 1}})} \right| = \frac{1}{{({N_u} - K){{\hat \beta }_k}}} \end{array} $$ (18) $$ \begin{array}{c} {\mathbb{E}}{I_e} = {P^u}\sum\limits_{i = 1}^K {{\mathbb{E}}\left| {\mathit{\boldsymbol{f}}_k^H\mathit{\boldsymbol{\tilde g}}_i^u{{(\mathit{\boldsymbol{\tilde g}}_i^u)}^H}\mathit{\boldsymbol{f}}_k^{}} \right|} = \\ {P^u}\sum\limits_{i = 1}^K {\tilde \beta _i^u{\mathbb{E}}\left| {\mathit{\boldsymbol{f}}_k^H\mathit{\boldsymbol{f}}_k^{}} \right|} = \frac{{{P^u}\sum\nolimits_{i = 1}^K {\tilde \beta _i^u} }}{{({N_u} - K){{\hat \beta }_k}}} \end{array} $$ (19) $$ \begin{array}{c} {\mathbb{E}}{I_s} = \frac{{{P^d}{u^2}}}{{{\rm{Tr}}({{(\mathit{\boldsymbol{\hat G}}_d^H{{\mathit{\boldsymbol{\hat G}}}_d})}^{ - 1}})}}E\left| {\mathit{\boldsymbol{f}}_k^H\mathit{\boldsymbol{W}}{\mathit{\boldsymbol{W}}^H}{\mathit{\boldsymbol{f}}_k}} \right| = \\ \frac{{{P^d}}}{{{N_t}}}E\left| {\mathit{\boldsymbol{f}}_k^H\mathit{\boldsymbol{W}}{\mathit{\boldsymbol{W}}^H}{\mathit{\boldsymbol{f}}_k}} \right| = \\ {P^d}{\beta _W}E\left| {\mathit{\boldsymbol{f}}_k^H{\mathit{\boldsymbol{f}}_k}} \right| = \frac{{{P^d}{\beta _W}}}{{({N_u} - K){{\hat \beta }_k}}} \end{array} $$ (20) 综上,可以获得用户在基站端信干噪比,从而得到式(17)。证毕。
当用户数目K变大时,式(17)获取的上行和速率变小,对比式(11)和式(17)可以看出,本文所提算法同传统算法相比具有更紧下界。
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根据获取的上行链路和下行链路速率的下界,系统和速率可以表示为:
$$ R({N_r}, {N_t}) = \sum\limits_{k = 1}^K {{{\log }_2}(1 + {N_r}\tilde \gamma _k^u) + {{\log }_2}(1 + {N_t}\tilde \gamma _k^d)} $$ (21) 式中,$ \tilde \gamma _k^u\mathop = \limits^\Delta \gamma _k^u/{N_r} $,$ \tilde \gamma _k^d\mathop = \limits^\Delta \gamma _k^d/{N_t} $。以和速率最大化为优化目标,则全双工大规模MIMO网络中的天线分配问题可以转化为最优化问题P1:
$$ (N_r^*, N_t^*) = \arg \mathop {\max }\limits_{{N_r}, {N_t}} R({N_r}, {N_t}) $$ (22) 约束条件为:
$$ 1){\rm{ }}{N_r} + {N_t} = M;\;\;\;2){\rm{ }}{N_r} \ge K, {N_t} \ge K $$ (23) 定义a为接收天线比例,令$ {N_r} = aM $,$ {N_t} = (1 - a)M $,则最优化问题P1可以表示为:
$$ \left\{ \begin{array}{l} \mathop {\max }\limits_a \sum\limits_{k = 1}^K {{{\log }_2}(1 + aM\tilde \gamma _k^u) + {{\log }_2}(1 + (1 - a)M\tilde \gamma _k^d)} \\ {\rm{s}}{\rm{.t}}{\rm{. }}\frac{K}{M} \le a \le 1 - \frac{K}{M} \end{array} \right. $$ (24) 求解$ R(a) $关于a的二阶偏导可得:$ \frac{{{\partial ^2}R(a)}}{{{\partial ^2}(a)}} \ge 0 $,所以$ R(a) $是关于a的凹函数。P1存在最优解$ {a^*} $,则收发天线最优比可以表示为:
$$ a'\mathop = \limits^\Delta \frac{{{N_r}}}{{{N_t}}} = \frac{{{a^*}}}{{1 - {a^*}}} $$ (25) 不失一般性,当只有一个用户的时候,对R(a)进行求导,并等于0,可以得到:
$$ {a^*} = \frac{1}{{2M}}\left( {\frac{1}{{\tilde \gamma _1^d}} - \frac{1}{{\tilde \gamma _1^u}}} \right) + \frac{1}{2} $$ (26) 该结果可以很容易扩展到多用户场景。
根据式(26),可得如下结论:
1) 当$ \tilde \gamma _k^u < \tilde \gamma _k^d $时,$ {a^*} < 0.5 $,即系统为高信干噪比的链路分配更多的资源,以提升系统和速率;
2) 当$ \tilde \gamma _k^u > \tilde \gamma _k^d $时,$ {a^*} > 0.5 $;
3) 当天线数目趋于无限大时,$ {a^*}\mathop \to \limits^{a.s.} 0.5 $。该结论同文献[10-11]结论一致;
4) 当每个用户采用相同的大尺度衰落系数,$ {a^*} $正比于上行导频功率,反比于下行导频功率。
Optimal Antenna Allocation in Full-Duplex Massive MIMO Systems
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摘要: 在全双工大规模MIMO网络中,为了降低自干扰,提升频谱效率,提出了一种最优天线分配算法。该算法在基站采用迫零线性信号处理和存在信道估计误差的前提下,首先推导出系统和速率的紧下界。通过分析该下界,以系统和速率最大化为优化目标,得到最优的收发天线比。理论分析和仿真结果表明:1)收发天线比与上行导频信号发射功率成正比,与下行导频信号发射功率成反比;2)当天线数目无穷大时,最优天线比趋近于1。Abstract: In order to suppress the self-interference and improve spectrum efficiency in full duplex massive multiple input multiple output (MIMO) network, an optimal antenna allocation algorithm is proposed. In the full duplex system, the base station exploits zero-forcing linear processing and therefor has imperfect channel estimation. A tight lower bound on the sum-rate for the system with large number of antennas is derived. By resorting to the tight lower bound, a sum-rate maximization problem in terms of antenna ratio is formulated. The analysis and simulation results show that, 1) the ratio of receive antennas and transmit antennas is proportional to the training power of uplink, while it is inverse proportional to the training power of downlink; 2) the optimal antenna ratio converges to 1, when the total number of antenna goes to infinity.
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Key words:
- antenna allocation /
- channel estimation error /
- full duplex /
- massive MIMO /
- optimization problem
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