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

李玉博; 王亞會(huì); 于麗欣; 劉凱

doi:10.11999/JEIT210293

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

燕山大學(xué)信息科學(xué)與工程學(xué)院秦皇島 066004

基金項(xiàng)目: 河北省自然科學(xué)基金(F2020203043, F2021203040)，河北省高等學(xué)?？茖W(xué)技術(shù)研究項(xiàng)目(ZD2020179, ZD2021105)

詳細(xì)信息

作者簡(jiǎn)介:
李玉博：男，1985年生，副教授，碩士生導(dǎo)師，研究方向?yàn)閴嚎s感知技術(shù)，序列設(shè)計(jì)與編碼理論

王亞會(huì)：女，1999年生，碩士生，研究方向?yàn)槊庹{(diào)度NOMA擴(kuò)頻序列設(shè)計(jì)

于麗欣：女，1998年生，碩士生，研究方向?yàn)閿U(kuò)頻序列設(shè)計(jì)

劉凱：女，1977年生，副教授，碩士生導(dǎo)師，研究方向?yàn)閿U(kuò)頻序列設(shè)計(jì)

通訊作者:
李玉博　liyubo6316@ysu.edu.cn

中圖分類號(hào): TN914.42
計(jì)量
- 文章訪問(wèn)數(shù): 1344
- HTML全文瀏覽量: 532
- PDF下載量: 107
- 被引次數(shù): 0
出版歷程
- 收稿日期: 2021-04-09
- 修回日期: 2021-08-24
- 網(wǎng)絡(luò)出版日期: 2021-09-08
- 刊出日期: 2022-04-18

Binary Spreading Sequences of Lengths Non-Power-of-Two for Uplink Grant-Free Non-Orthogonal Multiple Access

School of Information Science & Engineering, Yanshan University, Qinhuangdao 066004, China

Funds: The Natural Science Foundation of Hebei Province(F2020203043, F2021203040), The Science and Technology Research Project of Colleges and Universities in Hebei Province (ZD2020179, ZD2021105)

摘要

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。
- 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.
- Massive machine-type communications /
- Non-Orthogonal Multiple Access (NOMA) /
- Compressed Sensing (CS) /
- Sets of spreading sequences

HTML全文

圖 1 上行免調(diào)度NOMA的系統(tǒng)模型^[21]

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圖 2 $ M{\text{ = }}129 $時(shí)(${M_{{\rm{zc}}}} = 127$)每個(gè)設(shè)備的SNR上的基于CS的CE和MUD的性能

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圖 3 $ M{\text{ = }}130 $時(shí)(${M_{{\rm{zc}}}} = 131$)每個(gè)設(shè)備的SNR上的基于CS的CE和MUD的性能

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圖 4 $ M{\text{ = }}129 $時(shí)(${M_{{\rm{zc}}}} = 127$)用戶過(guò)載因子L上的基于CS的CE和MUD的性能

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圖 5 $ M{\text{ = }}130 $時(shí)(${M_{{\rm{zc}}}} = 131$)用戶過(guò)載因子L上的基于CS的CE和MUD的性能

下載: 全尺寸圖片幻燈片

圖 6 $ M{\text{ = }}129 $時(shí)(${M_{{\rm{zc}}}} = 127$)活躍概率P_a上的基于CS的CE和MUD的性能

下載: 全尺寸圖片幻燈片

圖 7 $ M{\text{ = 130}} $時(shí)(${M_{{\rm{zc}}}} = 131$)活躍概率P_a上的基于CS的CE和MUD的性能

下載: 全尺寸圖片幻燈片

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

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