免調(diào)度非正交多址接入上行鏈路的非2冪次長(zhǎng)度二元擴(kuò)頻序列
doi: 10.11999/JEIT210293
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燕山大學(xué)信息科學(xué)與工程學(xué)院 秦皇島 066004
Binary Spreading Sequences of Lengths Non-Power-of-Two for Uplink Grant-Free Non-Orthogonal Multiple Access
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School of Information Science & Engineering, Yanshan University, Qinhuangdao 066004, China
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摘要: 為了解決5G大規(guī)模機(jī)器類通信場(chǎng)景下大規(guī)模接入和如何提高頻譜效率的問(wèn)題,該文針對(duì)免調(diào)度非正交多址接入(NOMA)系統(tǒng)上行鏈路,通過(guò)采用插入函數(shù)在2元Golay序列上插入元素的方法,提出具有低峰均功率比(PAPR)且長(zhǎng)度為非2冪次的2元擴(kuò)頻序列集。仿真結(jié)果表明,得到的序列集具有低相干性,這為基于壓縮感知的活躍用戶檢測(cè)提供了可靠的性能。同傳統(tǒng)的Zadoff-Chu序列相比,新型2元序列集具有更小的字符集,便于實(shí)現(xiàn)。此外,所構(gòu)造的序列PAPR最大為4,低于高斯隨機(jī)序列和Zadoff-Chu序列,因此可以有效解決時(shí)域信號(hào)峰均功率比過(guò)高的問(wèn)題。
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關(guān)鍵詞:
- 大規(guī)模機(jī)器類通信 /
- 非正交多址接入 /
- 壓縮感知 /
- 擴(kuò)頻序列集
Abstract: In order to solve the problem of mass access and how to improve spectrum efficiency in 5G massive Machine-Type Communication (mMTC) scenario, for the uplink grant-free Non-Orthogonal Multi-Access(NOMA) system, new sets of non-orthogonal binary spreading sequences with low Peak Average Power Ratio(PAPR) and lengths non-power-of-two is proposed by inserting elements into binary Golay sequences using insertion function. Simulation results confirm that the resulting sequence sets has low coherence, which provides reliable performance for active user detection based on Compressed Sensing (CS). Compared with the traditional Zadoff-chu sequences, the new binary sequence sets has a smaller alphabet set, which is easy to implement. Moreover, the resultanted sequences exhibit the PAPR of at most 4, which is lower than those for Gaussian and Zadoff-Chu sequences. Therefore, the problem of high peak-to-power ratio in time domain can be solved effectively. -
表 1 使
${{\boldsymbol{\varPhi}} '}$ 達(dá)到最優(yōu)的置換集$ {M'}{\text{ = }}{{\text{2}}^m} $ 用戶過(guò)載因子 相干值 置換集$\varGamma$ 32 $ 2 \le L \le 8 $ 0.25 $(5,4,3,2,1),(3,4,2,5,1),(4,2,5,3,1),(4,3,5,1,2),(4,5,1,3,2),(5,3,1,4,2),(5,4,2,1,3),(4,1,2,5,3)$ 64 $ 2 \le L \le 5 $ 0.125 $(3,4,5,2,6,1),(6,3,2,4,1,5),(4,1,6,5,2,3),(6,5,3,1,2,4),(5,3,2,1,6,4)$ 128 $ 2 \le L \le 8 $ 0.125 $\begin{array}{l} {\text{(4,5,1,3,6,7,2),(4,2,5,1,6,7,3),(6,7,1,2,3,5,4),(5,3,6,4,1,7,2),(6,4,7,3,1,5,2),(4,3,6,7,5,2,1),} } \\ {\text{(6,1,3,2,7,4,5),(6,7,5,1,4,3,2)} } \\ \end{array}$ 256 $ 2 \le L \le 5 $ 0.0625 ${\text{(4,5,6,1,3,7,8,2),(7,6,8,2,3,1,4,5),(7,1,8,6,4,3,5,2),(6,7,2,3,8,4,1,5),(8,3,1,5,2,7,4,6)} }$ 512 $ 2 \le L \le 8 $ 0.0625 $\begin{array}{l} {\text{(8,3,7,4,9,2,5,1,6),(8,4,3,7,2,6,1,9,5),(9,5,4,1,6,8,3,7,2),(6,5,8,7,9,3,4,2,1),} } \\ {\text{(4,1,7,6,8,9,2,5,3),(4,8,2,6,9,7,5,3,1),(5,3,7,8,2,1,6,9,4),(5,6,9,3,7,1,8,2,4)} } \\ \end{array}$ 1024 $ 2 \le L \le 5 $ 0.03125 $\begin{array}{l} {\text{(9,1,6,3,2,8,5,4,10,7),(5,1,9,8,2,10,6,3,7,4),(6,3,8,10,9,7,1,5,4,2),(7,6,8,1,3,2,10,9,4,5),} } \\ {\text{(9,5,3,2,4,8,6,10,7,1)} } \\ \end{array}$ 下載: 導(dǎo)出CSV
表 2 擴(kuò)頻矩陣相干值
$\mu ({\boldsymbol{\varPhi}} )$ 擴(kuò)頻矩陣 序列長(zhǎng)度 $\mu ({\boldsymbol{\varPhi} } )$ 擴(kuò)頻矩陣 序列長(zhǎng)度 $\mu ({\boldsymbol{\varPhi} } )$ 擴(kuò)頻矩陣 序列長(zhǎng)度 $\mu ({\boldsymbol{\varPhi } })$ 擴(kuò)頻矩陣 序列長(zhǎng)度 $\mu ({\boldsymbol{\varPhi } })$ 本文 33 0.2727 本文 34 0.2941 文獻(xiàn)[16] 32 0.2500 基于ZC序列 31 0.1796 65 0.1385 66 0.1515 64 0.1250 61 0.1280 129 0.1318 130 0.1385 128 0.1250 127 0.0887 257 0.0661 258 0.0698 256 0.0625 257 0.0624 513 0.0643 514 0.0661 512 0.0625 509 0.0443 1025 0.0322 1026 0.0331 1024 0.03125 1021 0.0313 下載: 導(dǎo)出CSV
表 3 擴(kuò)頻矩陣?yán)镄蛄械淖畲驪APR
擴(kuò)頻矩陣 序列長(zhǎng)度 PAPR 擴(kuò)頻矩陣 序列長(zhǎng)度 PAPR 擴(kuò)頻矩陣 序列長(zhǎng)度 PAPR 擴(kuò)頻矩陣 序列長(zhǎng)度 PAPR 本文 33 2.4545 本文 34 2.9412 文獻(xiàn)[16] 32 2.0000 基于ZC序列 31 4.4066 65 2.2825 66 2.6172 64 1.9928 61 4.0922 129 2.2403 130 2.4923 128 2.0000 127 4.3376 257 2.1699 258 2.3265 256 1.9978 257 4.7396 513 2.1228 514 2.2490 512 2.0000 509 4.8785 1025 2.0834 1026 2.1713 1024 1.9993 1021 5.2751 下載: 導(dǎo)出CSV
表 4 幾種確定性擴(kuò)頻矩陣的參數(shù)
擴(kuò)頻矩陣 擴(kuò)頻序列長(zhǎng)度$ M $ $\mu ({\boldsymbol{\varPhi} } )$ PAPR上界 字符集大小 文獻(xiàn)[16] $ {2^m} $ $\sqrt{{1}/{ {2}^{m\text{-1} } } },\;\;m{\text{為奇數(shù)} }$
$\sqrt{{1}/{ {2}^{m} } },\;m{\text{為偶數(shù)} }$2 2 基于ZC序列 $ {M_{{\text{zc}}}} $為任意素?cái)?shù) $\sqrt {{1}/{ { {M_{ {\text{zc} } } } } } }$ >4 $ {M_{{\text{zc}}}} $ 本文 $ {2^m}{\text{ + }}1 $ $\dfrac{\sqrt{ {2}^{m+1} }+1}{ {2}^{m}+1},\;m{\text{為奇數(shù)} }$
$\dfrac{\sqrt{ {2}^{m} }+1}{ {2}^{m}+1},\;m{\text{為偶數(shù)} }$4 2 本文 $ {2^m}{\text{ + 2}} $ $ \dfrac{\sqrt{{2}^{m+1}}+2}{{2}^{m}+2},m{\text{為奇數(shù)}} $,$ \dfrac{\sqrt{{2}^{m}}+2}{{2}^{m}+2},m{\text{為偶數(shù)}} $ 4 2 下載: 導(dǎo)出CSV
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