Cognitive Radio Network Downlink Power Allocation and Beamforming Method with Imperfect Channel State Information
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摘要: 针对非理想信道状态信息(CSI)条件下工作于underlay模式的认知无线网络(CRN)多用户下行功率分配和波束赋形研究中普遍存在的问题,包括忽略主网络(PN)对认知用户(SU)的干扰、传统的凸优化SDR方法对约束条件的近似要求以及实现算法复杂、实用性受限等,首先建立CRN模型,增添PN对SU的干扰项,而后在非理想CSI的最差条件下形成优化问题。再通过Lagrange对偶对问题的约束条件进行变换,并基于变换后的问题形式,利用上行和下行的对偶特性,引入虚拟功率,将优化问题转换为上行功率分配和波束赋形问题,进一步得到简便、快速和实用的迭代算法。数值仿真显示,算法收敛很快。并且发现非理想CSI引起的误差不仅对下行功率影响明显而且还改变优化问题的可行解区域;PN基站(PBS)的发送功率的变化对可行解区域有显著的影响。Abstract: Some problems of multi-user downlink power allocation and beamforming in a underlay Cognitive Radio Network (CRN) with imperfect Channel State Information (CSI) are addressed. They include ignoring the interferences of the Primary Network (PN) to the Secondary Users (SU), conventional SDR algorithm of convex optimization needing the constraint approximation, the high complexity of the algorithm, and implemented with difficulty, etc. Firstly the term of interference of the PN to the SU is added to the CRN model. The optimization problem is formulated with the worst-case imperfect CSI. Next the constraints of the problem are transformed by means of Lagrange duality. Then, based on the form of the problem, the simple, fast and practical iterative algorithm is obtained by utilizing the duality of uplink-downlink, introducing virtual power, and transforming the optimization problem into the problem of uplink power allocation and beamforming. Numerical simulation results show that it converges faster. It is also found that the errors of the imperfect CSI not only influence the downlink power but also change the feasibility region. The variation of transmitting power of the PN Base Station (PBS) could affect the feasibility region notably.
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1. 引言
认知无线网络(Cognitive Radio Network, CRN)技术能有效地利用无线频谱资源。它是在不影响授权用户,即主用户(Primary User, PU)的前提下,通过认知无线网络用户,即次级用户(Secondary User, SU)采用通信技术接入频谱,完成与PU的无线资源共享。CRN有3种工作模式,分别是interweave, overlay和underlay。在interweave模式,SU只能和PU分时接入频谱,而一旦出现冲突,SU必须让出频谱;这样就有可能导致SU反复地接入和退出,无法保证SU正常工作。在overlay模式,CRN通过获知PU的码本和消息知识,采用编码方法来消除对PU的干扰;而在多数的场合,CRN的这种需求无法得到满足,因而会影响PU的工作。在underlay模式,CRN通过信道条件来控制自身的发送参数,实现与PU共享频谱。
下行功率分配和波束赋形是工作于underlay模式的多用户CRN的主要频谱接入方法。其中的波束赋形需要依据正确的信道状态信息(Channel State Information, CSI)才能有效实现。而在实际中,无线信道的状态是时刻在变化的,完全能够与信道匹配的CSI几乎是不可能得到的。非理想CSI条件下的下行功率分配和波束赋形问题是当前CRN技术研究的重点之一。目前的研究存在以下的问题:一是处理信道不确定的方法中包含一些近似条件[1,2],得到的是次优解;二是在求解优化问题中会产生高秩解,而实际的低秩解却不一定能得到[3,4] 。并且文献[3~5]的优化问题均使用传统的半正定松弛(Semi-Definite Relax, SDR)算法[6]进行求解,计算复杂度高,不便于实用。三是CRN网络模型中普遍略去了主网络(Primary Network, PN)对SU的影响[7-11],这会导致CRN的服务质量(Quality of Service, QoS)约束条件放宽,由此得到的结果不能确保CRN有效工作;文献[12]的网络模型中虽然考虑了PN对SU的影响,但只是用一随机数值表示,不能体现信道的作用,依旧不能确保CRN有效工作。
针对以上的问题,本文对在非理想CSI条件下工作于underlay模式的CRN的下行功率分配和波束赋形问题作进一步研究。首先建立CRN模型[13],增添PN对SU的干扰;而后在非理想CSI的最差条件下形成CRN下行功率分配和波束赋形的优化问题。再通过Lagrange对偶将问题的约束条件进行形式变换,并且在变换后的问题求解中,根据问题形式,采用上行和下行的对偶特性[14],引入虚拟功率,将下行功率分配和波束赋形问题转换为上行功率分配和波束赋形问题,得到波束赋形向量的更新方法和上行功率的迭代方法,再由对应的参量变换得到下行功率,由此形成简便、快速和实用的迭代算法。数值仿真显示,算法收敛很快;与SDR算法性能等同;非理想CSI引起的误差对下行功率影响明显;并且还影响优化问题的可行解区域;主网络基站(Primary Base Station, PBS)的发送功率的变化对可行解区域有显著的影响。
2. CRN网络模型及优化问题
2.1 网络模型
网络模型如图1所示,CRN网络由1个CRN基站CBS(Cognitive Base Station)和多个SU组成;PN网络由1个主网络基站PBS和多个PU组成。实际网络中PBS的数目不止1个,但其它PBS对SU的干扰远没有与CBS共享频谱的PBS的干扰效应显著,因此对分析结果几乎无影响,可以略去。SU的数目为
K , PU的数目为L , CBS的发送天线数目为Nt , PBS为单天线发送,PU和SU均是单天线接收。CBS与SU之间的信道向量
hk=(hk,1hk,2··· hk,Nt)H ,发送的功率为p=(p1p2···pK)T ,波束赋形矩阵U=(u1u2···uK) ,其中uk=(uk,1uk,2··· uk,Nt)H ;k=1,2,···,K 。PBS与SU
k 之间的信道向量hpk (维数为1),发送的功率为pp 。SU
k 的接收信号yk(t) :yk(t)=√pkhHkuksk(t)+K∑i=1,i≠k√pihHkuisi(t)+√pphpksp(t)+zk(t),k=1,2,···,K (1) 式中,
sj(t) (j=1,2,···,K )是CBS发送给SUj 的信号,E(|sj(t)|2)=1 ,zk(t) 是加性高斯白噪声,zk(t)∼ N(0,σ2k) ,sp(t) 是PBS的发送信号,E(|sp(t)|2)=1 。CBS对PU
l (l=1,2,···,L )的干扰信号为:∑Ki=1√pihHK+luisi(t) 。在实际应用中,很难得到信道的瞬时特性,一般是使用期望值。可设:SU
k 与CBS之间的信道相关矩阵Rk=E(hkhHk) , SUk 与PBS之间的信道相关矩阵Rpk=E(hpkhHpk) 。可得SUk 的信干噪比(Signal to Interference plus Noise Ratio, SINR):SINRk=pkuHkRkukK∑i=1,i≠kpiuHiRkui+ppRpk+σ2k (2) 设PU
l 与CBS之间的信道相关矩阵RK+l= E(hK+lhHK+l) ,可得CBS对PUl (l=1,2,···,L) 的干扰为Il=K∑i=1piuHiRK+lui (3) 式中,
Rk (k=1,2,···,K+L )是Hermitian矩阵,并且Rk≻_0 ,即Rk 是正定/半正定的。2.2 优化问题
非理想信道条件下,无线信道的变化特性导致CSI存在估计误差,因此
Rk ,Rpk (k=1,2,···,K )和RK+l (l=1,2,···,L )包含误差项,分别变为:¯Rk=ˆRk+Δk ,¯Rpk=ˆRpk+Δpk ,¯RK+l=ˆRK+l+ ΔK+l 。ˆRk ,ˆRpk 和ˆRK+l 是信道的估计值;Δk ,Δpk 和ΔK+l 是随机的误差项,分别满足:‖Δk‖≤αk ,‖Δpk‖≤ξk ,‖ΔK+l‖≤βl ;‖⋅⋅‖ 是Frobenius范数;αk ,ξk ,βl 是门限值。根据Rk 的性质,为确保优化问题有可行解,需要增加如下的约束条件:¯Rk≻_0 ;¯Rpk≻_0 ;¯RK+l≻_0 。考虑非理想信道状态的最差条件,下行功率分配和波束赋形的优化问题如式(4)—式(6):P1:min (4) \begin{array}{l}{\rm{s}}.{\rm{t}}.\;\;\mathop {\min }\limits_\begin{array}{l}\left\| {{{Δ} _k}} \right\| \le {\alpha _k}\\{\widehat {{R}}_k} + {{Δ} _k} \ \underline \succ \ {{0}}\\\left\| {{{Δ} _{{\rm{p}}k}}} \right\| \le {\xi _k}\\{\widehat {{R}}_{{\rm{p}}k}} + {{Δ} _{{\rm{p}}k}}\ \underline \succ \ {{0}}\end{array} \left\{ {\;\frac{{{p_k}{{u}}_k^{\mathop{\rm H}\nolimits} \left({{\widehat {{R}}}_k} + {{Δ}_k}\right){{{u}}_k}}}{{\left[ {\displaystyle\sum\limits_{i = 1,i \ne k}^K {{p_i}{{u}}_i^{\mathop{\rm H}\nolimits} \left({{\widehat {{R}}}_k} + {{Δ}_k}\right){{{u}}_i} + {p_{\rm{p}}}\left({{\widehat {{R}}}_{{\rm{p}}k}} + {{Δ}_{{\rm{p}}k}}\right) + \sigma _k^2} } \right]}}} \right\} \ge {\gamma _k},\\ \quad \quad\quad\quad\quad\quad\quad\; k = 1,2, ·\!·\!· ,K, {p_k} \ge 0;\;\;\left\| {{{{u}}_k}} \right\| = 1\end{array} (5) \begin{aligned}&\mathop {\max }\limits_\begin{array}{l}\Large{\left\| {{{Δ} _{ { K+ l} }}} \right\| \le {\beta _{ l}}\atop{\widehat {{R}}_{ K + l}} + {{Δ} _{{ K+ l} }} \ \underline \succ \ {{0}}} \end{array}\;\sum\limits_{i = 1}^K {{p_i}{{u}}_i^{\mathop{\rm H}\nolimits} ({{\widehat {{R}}}_{K + l}} + {{Δ}_{K + l}}){{{u}}_i}} \le \frac{1}{{{\gamma _{_{K + l}}}}},\;\\& \quad\quad\quad\quad \quad \quad \,l = 1,2, ·\!·\!· ,L\end{aligned} (6) 式中,
{\gamma _k} 和{1}/({{{\gamma _{_{K + l}}}}}) 分别是{\rm{SIN}}{{\rm{R}}_k} 和{I_l} 的门限。2.3 优化问题变换
取
{{{A}}_k} = {\gamma _k} \displaystyle\sum\nolimits_{i = 1,i \ne k}^K\!\! {{p_i}{{{u}}_i}{{u}}_i^{\rm{H}} - } {p_k}{{{u}}_k}{{u}}_k^{\rm{H}} ,{b_k} = {\gamma _k}{p_{\rm{p}}} ,{{C}} = \displaystyle\sum\nolimits_{i = 1}^K \!\!{{p_i}{{{u}}_i}{{u}}_i^{\rm{H}} } ,问题P1的约束条件式(5)可变换为\mathop {\min }\limits_\begin{array}{l}\left\| {{{Δ} _k}} \right\| \le {\alpha _k}\\{\widehat {{R}}_k} + {{Δ} _k} \ \underline \succ \ {{0}}\\\left\| {{{Δ} _{{\rm{p}}k}}} \right\| \le {\xi _k}\\{\widehat {{R}}_{{\rm{p}}k}} + {{Δ} _{{\rm{p}}k}}\ \underline \succ \ {{0}}\end{array} \!\!\! \!\!\!\!\!- \left({\rm{Tr}}\left\{ {{{Δ} _k}{{{A}}_k}} \right\} + {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}{{{A}}_k}} \right\} + {b_k}{\widehat {{R}}_{{\rm{p}}k}} \right. \\\quad\quad \quad\quad \left. + {b_k}{{Δ} _{{\rm{p}}k}} + {\gamma _k}\sigma _k^2\right) \ge 0 (7) 式中,
{\rm{Tr}}\left\{ {{B}} \right\} 表示矩阵{{B}} 的迹。式(7)的不等式左边等价于问题C1:
\;\;\;\left. \begin{array}{l}{\rm{C}}1: \quad \mathop {\min }\limits_{{{Δ}_k},{{Δ}_{{\rm{p}}k}}} - \left({\rm{Tr}}\left\{ {{{Δ}_k}{{{A}}_k}} \right\} + {\rm{Tr}} \left\{ {{{\widehat {{R}}}_k}{{{A}}_k}} \right\} \right. \\ \left. \quad \quad\quad \quad\quad+ {b_k}{\widehat {{R}}_{{\rm{p}}k}} + {b_k}{{Δ}_{{\rm{p}}k}} + {\gamma _k}\sigma _k^2 \right)\\\quad\quad \quad{\rm{s}}{\rm{.t}}{\rm{. }}\; \; \left\| {{{Δ}_k}} \right\| \le {\alpha _k}{\rm{,}}\;{\widehat {{R}}_k} + {{Δ}_k} \; \underline \succ \; {{0}}, \\\quad \quad \quad \quad \quad\left\| {{{Δ}_{{\rm{p}}k}}} \right\| \; \; \le {\xi _k}{\rm{, }}{\widehat {{R}}_{{\rm{p}}k}} + {{Δ}_{{\rm{p}}k}} \; \underline \succ \; {{0}}\end{array} \right\} (8) 由目标函数可知,对于给定的
{{{A}}_k} 和{b_k} ,它是凸函数;而约束条件也是凸函数,所以C1是凸问题[15]。因此,可得到Lagrange对偶为\begin{array}{l}\!\!\!\!\!\!g({\lambda _k},{\mu _k},{{{Y}}\!_k},{{{Z}}_k}) \\= \mathop {\inf }\limits_{{{Δ} _k},{{Δ} _{{\rm{p}}k}}} \Bigr( - {\rm{Tr}}\left\{ {{{Δ} _k}{{{A}}_k}} \right\} - {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}{{{A}}_k}} \right\} \\ - {b_k}{{\widehat {{R}}}_{{\rm{p}}k}} - {b_k}{{Δ} _{{\rm{p}}k}} - {\gamma _k}\sigma _k^2 \\+ {\lambda _k}\left({\left\| {{{Δ} _k}} \right\|^2} - \alpha _k^2\right) - {\rm{Tr}}\left\{ {\left({{\widehat {{R}}}_k} + {{Δ} _k}\right){{{Z}}_k}} \right\} \\\left. { + {\mu _k}\left({{\left\| {{{Δ} _{{\rm{p}}k}}} \right\|}^2} - \xi _k^2\right) - {\rm{Tr}}\left\{ {\left({{\widehat {{R}}}_{{\rm{p}}k}} + {{Δ} _{{\rm{p}}k}}\right){{{Y}}\!_k}} \right\}} \right)\end{array} (9) 由
\displaystyle\frac{{\partial g}}{{\partial {{Δ} _k}}} = {{0}} 和\displaystyle\frac{{\partial g}}{{\partial {{Δ} _{{\rm{p}}k}}}} = {{0}} ,分别可得:{Δ} _k^ * = \displaystyle\frac{{{{{A}}_k} + {{{Z}}_k}}}{{2{\lambda _k}}}, {Δ} _{{\rm{p}}k}^ * = \displaystyle\frac{{{b_k}{{I}} + {{{Y}}_k}}}{{2{\mu _k}}} 。把
{Δ} _k^ * 和{Δ} _{{\rm{p}}k}^ * 代入式(9)可得C2:\left. \begin{array}{l}{\rm{C}}2: \;\; \mathop {\max }\limits_{{\lambda _k},{{{Z}}_k},{\mu _k},{{{Y}}\!_k}} - \frac{{{{\left\| {{{{A}}_k} + {{{Z}}_k}} \right\|}^2}}}{{4{\lambda _k}}} - {\lambda _k} \; \alpha _k^2 \\\quad\quad\quad\quad- {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}({{{Z}}_k} + {{{A}}_k})} \right\} - {\gamma _k} \; \sigma _k^2\\\quad\quad\quad\quad - \frac{{{{\left\| {{b_k}{{I}} + {{{Y}}\!_k}} \right\|}^2}}}{{4{\mu _k}}} - {\mu _k} \; \xi _k^2 \\ \quad\quad\quad \quad- {\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}({{{Y}}\!_k} + {b_k}{{I}})} \right\}\\\quad\quad{\rm{s}}{\rm{.t}}{\rm{. }} \; {\lambda _k} \ge 0, {{{Z}}_k}\; \underline \succ \; {{0}}, {\mu _k} \ge 0, {{{Y}}_k}\; \underline \succ \;{{0}}\end{array} \right\} (10) C2分别对
{\lambda _k} 和{\mu _k} 取极值可得:\lambda _k^ * \! =\! \displaystyle\frac{{\left\| {{{{A}}_k} \!+\! {{{Z}}_k}} \right\|}}{{2{\alpha _k}}} ,\mu _k^ * = \displaystyle\frac{{\left\| {{b_k}{{I}} + {{{Y}}\!_k}} \right\|}}{{2{\xi _k}}} ,并代入,有\begin{align}\mathop {\max }\limits_{{{{Z}}_k},{{{Y}}\!_k}}& - {\alpha _k}\left\| {{{{A}}_k} + {{{Z}}_k}} \right\| - {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}({{{Z}}_k} + {{{A}}_k})} \right\} \\&- {\gamma _k}\; \sigma _k^2 - {\xi _k}\left\| {{b_k}{{I}} + {{{Y}}\!_k}} \right\| \\&- {\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}({{{Y}}_k} + {b_k}{{I}})} \right\} \end{align} (11) 由于问题C1是凸问题,因此存在误差项
{{Δ} _k} \!=\! {\widetilde \alpha _k}{{I}} ,0<{\widetilde \alpha _k} < {\alpha _k}/\sqrt {{N_t}} 和{{Δ} _{{\rm{p}}k}} = {\widetilde \xi _k}{{I}} ,0< {\widetilde \xi _k} < {\xi _k} ,是严格可行解。问题C1和C2满足Slater条件[15],因此式(11)和C1是强对偶,C1可由式(11)替换得到约束条件如式(12):\begin{align}& \mathop {\max }\limits_{{{{Z}}_k},{{{Y}}_k}} \left( - {\alpha _k}\left\| {{{{A}}_k} + {{{Z}}_k}} \right\| - {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}({{{Z}}_k} + {{{A}}_k})} \right\} \right. \\& \quad \quad - {\gamma _k}\sigma _k^2 - {\xi _k}\left\| {{b_k}{{I}} + {{{Y}}\!_k}} \right\| \\& \quad \quad \left.- {\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}({{{Y}}_k} + {b_k}{{I}})} \right\} \right) \ge 0\end{align} (12) 当有最优解时,
{{Z}}_k^* = {{0}} [3],{{Y}}\!_k^* = {{0}} 。{{Y}}_k^* = {{0}} 的证明:证明 可由式(12)中得到以下的问题:
\mathop {\max }\limits_{{{{Y}}_k}} \left( {\left. { - {\xi _k}\left\| {{b_k}{{I}} + {{{Y}}_k}} \right\| \!-\! {\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}({{{Y}}_k} \!+\! {b_k}{{I}})} \right\}} \right)} \right. (13) 因为
{\xi _k} >0,{b_k} >0,{\widehat {{R}}_{{\rm{p}}k}}\; \underline \succ \; {{0}} ,{{{Y}}_k}\; \underline \succ \; {{0}} 可得:{\xi _k}\left\| {{b_k}{{I}} + {{{Y}}_k}} \right\| > 0 。设:
{\widehat {{R}}_{{\rm{p}}k}} 和{{{Y}}_k} + {b_k}{{I}} 均是M \times M 的矩阵,其中:{\widehat {{R}}_{{\rm{p}}k}} = {{{U}}^{\rm{H}} }{{AU}} ,{{A}} = {\rm{diag}} \left\{ {{\lambda _{A1}},{\lambda _{A2}}, ·\!·\!· , {\lambda _{AM}}} \right\} ;{{{U}}^{\rm{H}} }{{U}} = {{U}}{{{U}}^{\rm{H}} } = {{I}} 。{{{Y}}_k} + {b_k}{{I}} = {{{V}}^{\rm{H}} }{{BV}} ,{{B}} = {\rm{diag}} \left\{ {{\lambda _{B1}},{\lambda _{B2}}, ·\!·\!· , {\lambda _{BM}}} \right\} ;{{{V}}^{\rm{H}} }{{V}} = {{V}}{{{V}}^{\rm{H}} } = {{I}} 。由{\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}({{{Y}}_k} + {b_k}{{I}})} \right\}\! =\! {\rm{Tr}}\left\{ {{{AB}}} \right\} \!=\! \displaystyle\sum\limits _{i = 1}^M {{\lambda _{Ai}}{\lambda _{Bi}}} \!\ge\! 0 可知,要使式(13)得到最大值,必有:{{Y}}\!_k^ * = {{0}} 。否则,如果有{{Y}}_k^1 \succ {{0}} 使式(13)最大,则有\begin{align}&\left( { - {\xi _k}\left\| {{b_k}{{I}} + {{Y}}\!_k^{\,1}} \right\| - {\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}\left({{Y}}_k^1 + {b_k}{{I}}\right)} \right\}} \right) \\& \ \ - \left( { - {\xi _k}\left\| {{b_k}{{I}} \!+\! {{Y}}\!_k^{\,*} } \right\| \!-\! {\rm{Tr}}\left\{ {{{\widehat {{R}}}_{{\rm{p}}k}}\left({{Y}}\!_k^{\,*} + {b_k}{{I}}\right)} \right\}} \right) < 0\end{align} (14)
这与假设冲突。所以
{{Y}}\!_k^{\,*} = {{0}} 成立。当{\widehat {{R}}_{{\rm{p}}k}} 和{{{Y}}\!_k} 均是1维,即M = 1 时,结论显然成立。证毕因此,问题P1的约束条件式(7)变换为
\begin{align}&- {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}{{{A}}_k}} \right\} - {\rm{Tr}}\left\{ {{b_k}{{\widehat {{R}}}_{{\rm{p}}k}}} \right\} \\ &\quad \quad\ge {\alpha _k}\left\| {{{{A}}_k}} \right\| + {\xi _k}{b_k} + {\gamma _k}\;\sigma _k^2\end{align} (15) 问题P1的约束条件式(6)通过与约束式(7)类似的变换可得
- {\gamma _{_{K + l}}}\; {\rm{Tr}}\left\{ {{{C}}{{\widehat {{R}}}_{{K + l}}}} \right\} \ge {\gamma _{_{K + l}}}\; {\beta _l}\left\| {{C}} \right\| - 1 (16) 所以问题P1可变换为P2:
{\rm{P}}2:\;\;\mathop {\min }\limits_{{{U}},{p_1}, ··· ,{p_{_K}}} \;\sum\limits_{k = 1}^K {{p_{_k}}} (17) \left. \begin{array}{l}\!\!\!{\rm{s}}{\rm{.t}}{\rm{.}} \ - {\rm{Tr}}\left\{ {{{\widehat {{R}}}_k}{{{A}}_k}} \right\} \ge {\alpha _k}\left\| {{{{A}}_k}} \right\| + {\xi _k}{b_k} \\\quad\quad\ + {\rm Tr}\left\{ {{b_k}{{\widehat {{R}}}_{{\rm{p}}k}}} \right\} + {\gamma _k}\; \sigma _k^2;\;k \!=\! 1,2, ·\!·\!· ,K\\\quad \ - {\gamma _{_{K + l}}} \; {\rm{Tr}}\left\{ {{{C}}{{\widehat {{R}}}_{K + l}}} \right\} \!\ge {\gamma _{_{K + l}}}\; {\beta _l}\left\| {{C}} \right\| - 1;\;\;\\ \quad\quad\ l = 1,2, ·\!·\!· ,L\\\quad \ {p_k} \ge 0;\;\;\left\| {{{{u}}_k}} \right\| = 1\end{array}\!\!\!\! \right\}\;\;\;\;\; (18) 从P2与P1对比中可以看到,由于误差项的引入,约束条件均发生了改变。
{{Δ} _k} 和{{Δ} _{{\rm{p}}k}} 的效应由{\alpha _k}\left\| {{{{A}}_k}} \right\| + {\xi _k}{b_k} 来表征;而{{Δ} _{K + l}} 的效应由{\gamma _{K + l}}\; {\beta _l}\left\| {{C}} \right\| 来表征。3. 基于上行和下行对偶的迭代算法
3.1 算法思路
P2可由SDR方法来求解[3],但存在两个问题:一个是SDR方法因为凸优化条件的需要,略去了秩1约束,是一种近似的方法;另一个是SDR算法实现比较复杂,实用性受限。
文献[14]基于上行和下行的对偶特性,引入虚拟上行功率和PU虚拟下行功率,将理想信道条件下不考虑主网络影响的优化问题(P2中的
{\alpha _k} ,{\xi _k} ,{b_k} 和{\beta _l} 均为0)转换为式(19),式(20)的形式:\mathop {\min }\limits_{{{U}},{q_{_1}}, \cdot \cdot \cdot ,{q_{_{K + L}}}} \; \sum\limits_{k = 1}^K {\left( {{\gamma _k}\sigma _k^2} \right){q_{_k}}} - \sum\limits_{l = 1}^L {{p_{_{K} + l}}{q_{_{K + l}}}} \quad\quad\quad (19) \begin{aligned}{\rm{SINR}}_k^V =& \frac{{{q_{_k}}{{u}}_k^{\rm{H}}{{{R}}_k}{{{u}}_k}}}{{{{u}}_k^{\rm{H}}\left( {\displaystyle\sum\limits_{i = 1,i \ne k}^{K + L} {{q_i}{\gamma _i}} {{{R}}_i} + {{I}}} \right){{{u}}_k}}} \ge 1,\\ & \,\, k = 1,2, ·\!·\!·,K ,\ {q_{_k}} \!\ge\! 0,\;\left\| {{{{u}}_k}} \right\| = 1,\\ & \;{p_{_{K + l}}} \ge 0,\;{p_{_{K + l}}} \!\le\! 1,\;{q_{_{K + l}}}\! \ge 0, \\ &\; l = 1,2, ·\!·\!· ,L\end{aligned} (20) 其中,
{q_1}, {q_2}, ·\!·\!· ,{q_{_{K + L}}} 是K 个SU和L 个PU的虚拟上行功率。{p_{_{K + l}}} (l = 1,2, ·\!·\!· ,L )是L 个PU的虚拟下行功率。{p_{_{K + l}}} \le 1 的引入是将约束条件- {\gamma _{_{K + l}}}\;{\rm{Tr}} \left\{ {{{C}}{{{R}}_{K + l}}} \right\}\ge - 1 ,变换为:- {\gamma _{_{K + l}}} {\rm{Tr}}\left\{ {{{C}}{{{R}}_{_{K + l}}}} \right\} \ge - {p_{_{K + l}}} 成为PU 的虚拟SINR,使它能和SU的SINR约束条件- {\rm{Tr}}\left\{ {{{{R}}_k}{{{A}}_k}} \right\} \ge {\gamma _k}\; \sigma _k^2 相一致。基于上行和下行对偶的迭代算法的思路是先求解上行虚拟功率和波束赋形向量,而后由变换关系,得到下行功率和波束赋形向量。可设:
\begin{aligned}{\eta _k}(t) =& {\alpha _k}\left\| {{{{A}}_k}(t)} \right\| + {\xi _k}{b_k} + {\gamma _k}\sigma _k^2 + {\rm{Tr}}\left\{ {{b_k}{{\widehat {{R}}}_{{\rm{p}}k}}} \right\} , \\ & \quad k = 1,2, ·\!·\!· ,K\end{aligned} (21) \begin{array}{l}{\eta _{K + l}}(t) \\= \left\{ \begin{array}{l}1 - {\gamma _{_{{{ K}} + l}}}\;{\beta _l}\left\| {{{C}}(t)} \right\|,\\\quad 1 - {\gamma _{_{{{ K}} + l}}}\;{\beta _l}\left\| {{{C}}(t)} \right\| \ge {{0}}\\1,\;{{其它}}\end{array} \right.,\;\; l = 1,2, ·\!·\!· ,L\end{array} (22) 式中,
t 是迭代次数。即可得到P2的基于上行和下行对偶问题P3:
\begin{aligned}& {\rm{P}}3: \quad\quad \mathop {\min }\limits_{{{U}},{q_1}, \cdot \cdot \cdot ,{q_{_{{ K}} + L}}} \;\sum\limits_{k = 1}^K {{\eta _k}{q_k}} - \sum\limits_{l = 1}^L {\frac{{{p_{{_{ K}} + l}}{q_{_{{ K}} + l}}}}{{{\eta _{_{{ K}} + l}}}}}\end{aligned} (23) \begin{aligned}{\rm{SINR}}_k^V =& \frac{{{q_k}{{u}}_k^{\mathop{\rm H}\nolimits} {{\widehat {{R}}}_k}{{{u}}_k}}}{{{{u}}_k^{\mathop{\rm H}\nolimits} \left( {\displaystyle\sum\limits_{i = 1,i \ne k}^K {{q_i}{\gamma _i}} {{\widehat {{R}}}_i} + \displaystyle\sum\limits_{l = 1}^L {\displaystyle\frac{{{q_{_{K + l}}}\; {\gamma _{{_K + l}}}}}{{{\eta _{_{K + l}}}}}} {{\widehat {{R}}}_{{_K + l}}} + {{I}}} \right) {{{u}}_k}}} \ge 1,k = 1,2, ·\!·\!· ,K,\\&{q_k} \ge 0,\;\left\| {{{{u}}_k}} \right\| = 1, {p_{{_K + l}}} \ge 0,\;{p_{_{K + l}}} \le 1,\;{q_{_{K + l}}} \ge 0;l = 1,2, ·\!·\!· ,L\end{aligned} (24) 3.2 变换参量
定义:
{{q}} = {[{q_1} \; {q_2} ·\!·\!· {q_{_{K + L}}}]^{\rm{T}} } 为K个SU和L 个PU的虚拟上行功率。{{p}} = {\left[{{p}}_1^{\rm{T}} \; {{p}}_2^{\rm{T}} \right]^{\rm{T}} } ,其中,{{{p}}_1} = {[{p_1} \; {p_2} } ·\!·\!· {p_{_K}}]^{\rm{T}} 是K 个SU的下行功率,{{{p}}_2} = {[{p_{_{K + 1}}} \; } {p_{{_K + 2}}} ·\!·\!· {p_{_{K + L}}}]^{\rm{T}} 是L 个PU的虚拟下行功率。\!\!\!\!\!{[{{D}}(t)]_{k,k}} = {{u}}_k^{\rm{H}} (t){\widehat {{R}}_k}{{{u}}_k}(t)\quad \quad \quad \quad \;\; (25) \!\!\!\!\!{[{{{G}}_2}(t)]_{l,j}} = \frac{{{\gamma _{_{K + l}}}}}{{{\eta _{_{K + l}}}(t)}}{{u}}_j^{\rm{H}} (t){\widehat {{R}}_{K + l}}\;{{{u}}_j}(t) (26) \!\!\!\!\!{[{{{G}}_1}(t)]_{i,j}} = \left\{ \begin{aligned}& 0, \quad \quad \quad \quad \quad \quad \; i = j\\& {\gamma _i}{{u}}_j^{\mathop{\rm H}\nolimits} (t){\widehat {{R}}_i}{{{u}}_j}(t), \; i \ne j \end{aligned} \right. (27) \begin{aligned}{η} (t) = &{[{\eta _1}(t) ·\!·\!· {\eta _{_K}}(t)]^{\rm{T}} } ; \; l = 1,2, ·\!·\!· ,L, \\& k,i,j = 1,2, ·\!·\!· ,K\end{aligned} (28) 3.3 迭代算法
迭代算法的步骤如下:
步骤 1 初始化:
{{{A}}_k}(0) = 0 ,{\eta _k}(0) = {\xi _k}{b_k} + {\gamma _k}\sigma _k^2 +{\rm{Tr}}\left\{ {{b_k}{{\widehat {{R}}}_{{\rm{p}}k}}} \right\} ,{q_k}(0) = 1 (k = 1,2, ·\!·\!· ,K );{{C}}(0) = 0 ,{\eta _{_{K + l}}}(0) = 1{;} 步骤 2 迭代循环:
t = 1,2, ·\!·\!· 直至收敛:(1)波束赋形因子更新:
\begin{aligned}{\phi _k}(t) =& \mathop {\max }\limits_{\left\| {{{{u}}_k}(t)} \right\| = 1} \frac{{{q_k}(t - 1){{u}}_k^{\rm{H}}(t){{\widehat {{R}}}_k}{{{u}}_k}(t)}}{{{{u}}_k^{\rm{H}}(t)\left( {{{Q}} + {{I}}} \right){{{u}}_k}(t)}},\\& \quad k = 1,2, ·\!·\!· ,K\end{aligned} (29) 其中,
\begin{align}{{Q}} =& \sum\limits_{i = 1,i \ne k}^K {{q_i}(t - 1){\gamma _i}} {\widehat {{R}}_i} \\& + \sum\limits_{l = 1}^L {\frac{{{q_{_{K + l}}}(t - 1){\gamma _{_{K + l}}}}}{{{\eta _{_{K + l}}}(t - 1)}}} {\widehat {{R}}_{_{K + l}}}\end{align} (30) 式中,
{\phi _k}(t) 是矩阵对\left({q_k}(t - 1){\widehat {{R}}_k}, {{Q}} + {{I}} \right) 的最大广义特征值,而{{{u}}_k}(t) 是对应的单位特征向量。它们可通过乘幂法求解[16]。(2)SU虚拟上行功率更新:
{q_k}(t) = \frac{{{q_k}(t - 1)}}{{{\phi _k}(t)}} (31) (3)转换为下行功率:
{{{p}}_1}(t) = {({{D}}(t) - {{{G}}_1}(t))^{ - 1}}{η} (t) (32) {{{p}}_2}(t) = {{{G}}_2}(t){{{p}}_1}(t)\quad\quad\quad\quad\;\; (33) (4)PU虚拟上行功率更新:
{q_{_{K + l}}}(t) \!=\!\! \left\{\!\!\!\! \begin{array}{l}{p_{_{K + l}}}(t){q_{{_K + l}}}(t - 1),\;\;\; \\ \quad\quad\quad\quad\quad \min \left\{ {{p_1}, ·\!·\!· ,{p_{_K}}} \right\} \ge 0\\{q_{_{K + l}}}(t - 1),\; {{其它}}\end{array} \right. (34) (5)参数更新:依据式(21)和式(22)更新
{\eta _k}(t) 和{\eta _{_{K + l}}}(t) ;步骤 3 收敛性检测:
\left| {{q_k}(t) - {q_k}(t - 1)} \right| \le {\delta _1} , \; \left| {{p_{_{K + l}}}\;(t) - 1} \right| \le {\delta _2} (35) 式中,
k = 1,2, ·\!·\!· ,K + L ,l = 1,2, ·\!·\!· ,L ;{\delta _1} 和{\delta _2} 是设定的收敛门限;步骤 4 满足收敛性条件,到步骤5,否则到步骤2;
步骤 5 当经
N 次迭代后,满足收敛条件时,可得最优值为:(1)下行功率分配:
{p_1}(N),{p_2}(N), ·\!·\!· ,{p_{_K}}(N) 。(2)波束赋形向量:
{{{u}}_1}(N),{{{u}}_2}(N), ·\!·\!·,{{{u}}_{_K}}(N) 。3.4 迭代算法和SDR算法复杂度比较
3.4.1 SDR算法
文献[3]的SDR优化算法可用凸优化软件CVX[17]来计算求解。SDR算法步骤如下:
步骤 1 初始化:
{{{A}}_k}(0) \!=\! 0 ,{\eta _k}(0) \!=\! {\xi _k}{b_k} \!+\! {\gamma _k}\sigma _k^2 \! + {\rm{Tr}}\left\{ {{b_k}{{\widehat {{R}}}_{{\rm{p}}k}}} \right\} , (k = 1,2, ·\!·\!· ,K );{{C}}(0) = 0 ,{\eta _{_{K + l}}}(0) = 1 (l = 1,2, ·\!·\!· ,L) ;步骤 2 迭代循环:
t = 1,2, ·\!·\!· 直至收敛:(1)波束赋形计算:用SDR求解,计算得到
{{{W}}\!_k} [3](k = 1,2, ·\!·\!· ,K )。(2)参数更新;依据式(21)和式(22)更新
{\eta _k}(t) 和{\eta _{_{K + l}}}(t) ;步骤 3 收敛性检测:
\left| {\sum\limits_{k = 1}^K {{\rm{Tr}}\left\{ {{{{W}}\!_k}(t)} \right\}} - \sum\limits_{k = 1}^K {{\rm{Tr}}\left\{ {{{{W}}\!_k}(t - 1)} \right\}} } \right| \le \delta (36) \delta 是一设定的收敛门限;步骤 4 满足收敛性条件,到步骤5,否则到步骤2;
步骤5 当经
N 次迭代后,满足收敛条件时,可得最优解为:{{{W}}_1}(N),{{{W}}_2}(N), ·\!·\!· ,{{{W}}_k}(N) 。(1)下行功率分配:
{p_1}(N),{p_2}(N), ·\!·\!· ,{p_{_K}}(N) 分别是{{{W}}_1}(N),{{{W}}_2}(N), ·\!·\!· ,{{{W}}_K}(N) 的最大特征值。(2)波束赋形向量:
{{{u}}_1}(N),{{{u}}_2}(N), ·\!·\!· ,{{{u}}_{_ K}}(N) 分别是对应于{p_1}(N),\;{p_2}(N), ·\!·\!· ,\;{p_{_K}}(N) 的{{{W}}_1}(N), {{{W}}_2}(N), ·\!·\!·,{{{W}}_K}(N) 的单位特征向量。3.4.2 算法复杂度比较
迭代算法与SDR算法的复杂度差别集中体现在迭代算法的步骤2的(1)~(4);SDR算法的步骤2的(1)。以乘和加作为基本运算单位[15],不考虑参加运算的数的类型,因为数的类型不影响复杂度的级别。迭代算法步骤2的(1)~(4)的复杂度如下:
(1)最大广义特征值和对应的特征向量求解:根据文献[16]的乘幂法算法,其中运算量最大的是
{N_t} ×{N_t} 矩阵求逆,因此,运算复杂度是O\left(N_t^3\right) 。当k = 1,2, ·\!·\!· ,K 时,运算复杂度是O\left(KN_t^3\right) 。(2)需
K 次运算。(3)
{{D}}(t) 需2 KN_t^2 +2K{N_t} - K 次运算;{{{G}}_2}(t) 需2LKN_t^2 +2 LK{N_t} 次运算;{{{G}}_1}(t) 需2 {K^2}N_t^2 + 2 {K^2}{N_t}-2 KN_t^2 -2 K{N_t} 次运算;{{{p}}_1}(t) 需{K^3} +2{K^2} 次运算;{{{p}}_2}(t) 需2LK 次运算。(4)需
L + K 次运算:步骤2-步骤4的复杂度是2 {K^2}N_t^2 +2LKN_t^2 +2{K^2}{N_t}+2LK{N_t} + {K^3} +2 {K^2}+ 2LK +L+ K ,通常情况下,K 和L 的数量级相同,可以略去{K^2}{N_t} ,LK{N_t} ,{K^3} ,{K^2} ,LK ,L 和K 等低阶项,并且L \approx K ,可得算法步骤2-步骤4的复杂度是O\left({K^2}N_t^2\right) 。由步骤1的复杂度O\left(KN_t^3\right) ,可得算法的复杂度是O\left(\max \left(KN_t^3,{K^2}N_t^2\right)\right) 。而由文献[14]知,SDR算法的复杂度是O\left({K^3}N_t^3\right) 。因此,迭代算法的复杂度要比SDR算法低很多。4. 数值仿真及结果分析
SU用户
K =5,PU用户L =5; CRN的BS的发送天线{N_t} =9; PU用户的干扰门限{1}/{{{\gamma _{_{K + l}}}}} =–6 dB,(l = 1,2, ·\!·\!· ,L ); SU的QoS约束{\gamma _k} (k = 1,2, ·\!·\!· ,K )在区间[0.1, 1.0]内取值,为便于分析,取{\gamma _k} = \gamma 。噪声\sigma _k^2 =–6 dB, (k = 1,2, ·\!·\!· ,K );信道是Rayleign衰落信道,对于由信道得到的结果,均进行100次Monte-Carlo运算后取均值。PBS的发送功率{p_{\rm{p}}} =30 dBm。取{\delta _1} = {\delta _2} = \delta =10–6。{{Δ} _k} ,{{Δ} _{{\rm{p}}k}} 和{{Δ} _{K + l}} 的值分别由在以0为球心,{\alpha _k} ,{\xi _k} ,{\beta _l} 为半径的球的范围内随机产生;{\alpha _k} = \alpha =0.2,{\xi _k} = \xi =0.2,{\beta _l} = \beta =0.1。0<{\alpha _k} <1, 0<{\xi _k} <1; (k = 1,2, ·\!·\!· ,K ), 0<{\beta _l} <1; (l = 1,2, ·\!·\!· ,L )。由Rayleign衰落信道得到{\overline {{R}} _k} ,{\overline {{R}} _{{\rm{p}}k}} ; (k = 1,2, ·\!·\!· ,K )和{\overline {{R}} _{K + l}} ; (l = 1,2, ·\!·\!· ,L )。如果{\widehat {{R}}_k} ,{\widehat {{R}}_{{\rm{p}}k}} 和{\widehat {{R}}_{K + l}} 中有负特征值,把负特征值用0代换。对于SDR算法,使用凸优化软件CVX来实现。以下图2和图5中的下行功率(W)表示\displaystyle\sum\nolimits_{k = 1}^K \!{{p_k}} ,单位是W。4.1 迭代算法的收敛特性
迭代算法的收敛门限和迭代次数如表1所示。
表 1 不同收敛门限下的迭代次数序号 1 2 3 4 5 收敛门限 \delta 10–3 10–4 10–5 10–6 10–7 迭代次数 N 11 17 25 47 67 由表1可知,算法收敛很快。如当收敛门限为10–6时,经过47次迭代即满足收敛条件。相对于SDR算法[3],迭代算法不仅降低了计算量,而且收敛速度快。
4.2 误差项对下行功率的影响
如图2所示,当各误差项的门限值增大时,下行功率也相应地增加。这是由于
\alpha 和\xi 的增加,使干扰信号增加,导致SU的QoS约束条件发生变化。如果要保证SU的QoS则必须增加下行功率。\beta 的变化,影响PU的干扰约束门限,在满足SU的QoS时,对下行功率的影响不明显。4.3 误差项对可行解区域的影响
可行解区域的度量指标是可解度百分比(feasibility percentage),是指在一定数量的随机信道实现条件下,优化算法有可行解的信道实现所占百分比。它是算法对信道变化适应性的一种度量。取
{\gamma _k} = \gamma (k = 1,2, ·\!·\!· ,K )。如图3所示,对应于确定的目标\gamma ,当各误差项的门限值增大时,可行解区域是缩小的。这是因为当误差项增大时,信道特性的变化也增大,在\gamma 增加到一定值时,优化问题的约束条件不能总是得到满足,可解度百分比下降,可行解区域减小。4.4 PBS发送功率对可行解区域的影响
如图4所示,PBS的发送功率
{p_{\rm{p}}} 的变化影响可行解区域。当{p_{\rm{p}}} 增大时,可行解区域减小。这是由于PBS的发送功率对于SU是干扰,当目标\gamma 增大时,SU的QoS条件对于干扰的容限也减小。当\gamma 达到某个值时,如图4中\gamma =0.5 ({p_{\rm{p}}} =30 dBm)及\gamma =0.7 ({p_{\rm{p}}} =27 dBm),可解度百分比就开始下降。但是可以看到当{p_{\rm{p}}} 由30 dBm减小到27 dBm时,可行解区域的变化还是较明显的,由此可知PBS发送功率对CRN的影响是不能忽略的。4.5 迭代算法与SDR算法的性能比较
(1)能量效率:取
{\alpha _k} = \alpha =0.1,{\xi _k} = \xi =0.1,{\beta _l} = \beta =0.05,迭代算法(ITE)和SDR算法的能量效率[18]如图5所示。当SINR\gamma 增加时,迭代算法和SDR算法的下行功率均增加,数值很接近;只是当\gamma 较大时,即\gamma =0.8时,才出现较细小的差异。这是由于\gamma 增加,约束条件导致优化问题的可解性变化,由此产生了下行功率的差异。因此,迭代算法和SDR算法有几乎相同的能量效率。(2)可行解区域:取
{\alpha _k} = \alpha =0.1,{\xi _k} = \xi =0.1,{\beta _l} = \beta =0.05,迭代算法(ITE)和SDR算法的可解度百分比如图6所示。当\gamma >0.7时,迭代算法和SDR算法的可解度百分比均是100%;而当\gamma >0.7,两种算法的可解度百分比都有下降,但有一些小的差异。这也是\gamma 增加致使优化问题的可解性发生变化而引起的。由(1)和(2)可知,迭代算法具有与SDR算法等同的性能。
5. 结论
通过对非理想信道条件下工作于underlay模式的CRN下行功率分配和波束赋形问题的研究,得到了包含PBS对SU干扰的非理想CSI最差条件下的最优化问题形式;给出了便于工程实现的鲁棒性迭代算法;为CRN的实际应用提供了一个有效的工具。由数值仿真发现,误差项的存在,会使下行功率增加,同时还影响可行解区域;PBS的功率变化对可行解区域的影响比较显著,不应忽视。尤其是在CRN工程组网中,更应当考虑这一因素。
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表 1 不同收敛门限下的迭代次数
序号 1 2 3 4 5 收敛门限 \delta 10–3 10–4 10–5 10–6 10–7 迭代次数 N 11 17 25 47 67 -
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