機(jī)器類通信中基于NOMA短編碼塊傳輸?shù)母呖煽康瓦t延無線資源分配優(yōu)化方案
doi: 10.11999/JEIT190128
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重慶郵電大學(xué)通信與信息工程學(xué)院計算機(jī)網(wǎng)絡(luò)與通信重慶市重點實驗室 重慶 400065
Optimal Scheme of Resource Allocation for Ultra-reliable and Low-latency in Machine Type Communications Based on Non-orthogonal Multiple Access with Short Block Transmission
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School of Communication and Information Engineering, Chongqing University of Posts and Telecommunications Chongqing key Laboratory of Computer Networks and Communications, Chongqing 400065, China
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摘要: 針對機(jī)器類通信(MTC)應(yīng)用場景的業(yè)務(wù)特征和服務(wù)質(zhì)量(QoS)要求,該文考慮基于非正交多址(NOMA)的MTC中短分組/短編碼塊傳輸,探討MTC中基于NOMA的高可靠低遲延無線資源優(yōu)化問題。首先,上行傳輸是基于NOMA的MTC通信的瓶頸,考慮無線蜂窩網(wǎng)絡(luò)中支持NOMA和高可靠低遲延性能要求,該文建立了上行無線資源優(yōu)化的系統(tǒng)模型;然后,分析上行傳輸遲延,導(dǎo)出基于距離的鏈路可靠性函數(shù);進(jìn)一步,以遲延、可靠性和帶寬為約束下條件,提出一種最大化中心用戶和速率的無線資源分配算法,并給出算法的收斂性證明和復(fù)雜度分析;最后,實驗仿真驗證了所提算法的性能優(yōu)勢。
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關(guān)鍵詞:
- 機(jī)器類通信 /
- 高可靠低遲延 /
- 短編碼塊傳輸 /
- 非正交多址 /
- 無線資源優(yōu)化
Abstract: For the service characteristics and Quality of Service (QoS) requirements of Machine Type Communications (MTC), short-packet/short-coded block transmission in MTC based on Non-Orthogonal Multiple Access (NOMA) is considered in this paper, and the resource optimization problem of the Ultra-Reliable and Low-Latency (URLL) in MTC based on NOMA is discussed. Currently, uplink transmission is a bottleneck of MTC based on NOMA. Firstly, considering the performance requirements supporting NOMA and high reliability and low latency in wireless cellular networks, a system model for uplink wireless resource optimization is established. Then, the uplink transmission delay is analyzed and the link reliability function based on distance is derived. Further, with the constraints of delay, reliability and bandwidth, a wireless resource allocation algorithm for maximizing the sum rates of central users is proposed, and also the convergence proof and complexity analysis of the algorithm are given. Finally, the simulation results show the performance advantages of the proposed optimal scheme. -
表 1 算法1的具體流程
算法1 用戶2的和速率最大化算法 步驟1 輸入${N_k}$, ${r_k}$,當(dāng)${N_k} \le {N_{\max }}$時,初始值${B_{lb}} = 0$, ${B_{ub}} = {W_{\rm{c}}}$, ${B_0} = \frac{{{B_{lb}} + {B_{ub}}}}{2}$; 步驟2 當(dāng)${B_{ub}} - {B_{lb}} > {\delta _b}$,通過二分法計算得到使${f_u}\left( {{r_k},{N_k},{B_i},\varepsilon _i^1,\varepsilon _i^2} \right)$最小化的$\varepsilon _i^1$, $\varepsilon _i^2$; 步驟3 若${f_u}\left( {{r_k},{N_k},{B_i},\varepsilon _i^1,\varepsilon _i^2} \right) > {\varepsilon ^{\max }}$,更新初始值${B_{lb}} = {B_i}$, ${B_i} = \frac{{{B_{1b}} + {B_{ub}}}}{2}$,或者${B_{ub}} = {B_i}$, ${B_i} = \frac{{{B_{1b}} + {B_{ub}}}}{2}$;否則轉(zhuǎn)步驟4; 步驟4 若${f_u}\left( {{r_k},{N_k},{B_i},\varepsilon _i^1,\varepsilon _i^2} \right) < {\varepsilon ^{\max }}$, ${B_k}({N_k}) = {B^{\rm{opt}}}$或者${B_k}({N_k}) = {\rm{NaN}}$; 步驟5 求出$N_k^*{\rm{ = }}\mathop {\arg }\limits_{{N_k}} \min {N_k}{B_k}({N_k})$, $B_k^*{\rm{ = }}{B_k}(N_k^*)$; 步驟6 把$N_k^*$, $B_k^*$代入到$R_k^2$中求出最優(yōu)的$R_k^{2 * }$。 下載: 導(dǎo)出CSV
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