多層中繼網(wǎng)絡(luò)上的分布式LT碼
doi: 10.11999/JEIT180804
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云南大學(xué)軟件學(xué)院 ??昆明 ??650500
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云南大學(xué)信息學(xué)院 ??昆明 ??650500
Distributed LT Codes on Multiple Layers Networks
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School of Software, Yunnan University, Kunming 650500, China
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School of Information, Yunnan University, Kunming 650500, China
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摘要: 目前對分布式LT碼(DLT)的研究僅限于信源數(shù)量較少、且只有1層中繼的情況,該文提出一種能夠部署在多層中繼網(wǎng)絡(luò)上的分布式LT碼,即多層分布式LT碼(MLDLT)。該碼將信源進行分組,將中繼進行分層,通過分層后的中繼群,可以將多達幾十個乃至上百個信源連接到同一個接收端,從而實現(xiàn)眾多信源通過多層中繼對同一個接收終端的分布式LT碼通信。通過對MLDLT碼進行與或樹分析,得出其中繼度分布的線性優(yōu)化方程。分別在無損和有損鏈路上計算該碼的漸進性能并進行數(shù)值仿真,結(jié)果證明MLDLT碼在無損和有損鏈路上的錯誤平臺都比較低。MLDLT碼非常適合于信源數(shù)量較多的多層中繼網(wǎng)絡(luò)。Abstract: Present researches on Distributed Luby’s Transmission (DLT) codes are restricted on several-sources and one-layer-relay networks, thus the Multiple Layers Distributed LT (MLDLT) code for multiple-layers-relays networks is proposed. In MLDLT, sources are grouped and realys are layered in order that scores of sources can be connected to the only destination through the layered relays. By this scheme, the distributed communication between scores of sources and the destination can be performed. Through the and-or tree analysis, the linear procedures for the optimization of the relays' degree distributions are derived. On both lossless and lossy links, asymptotic performances of MLDLT are analized and the numberical simulations are experimented. The results demonstrate that MLDLT can achieve satisfying erasure floors on both lossless and lossy links. MLDLT is a feasible solution for the scores-sources and multiple-layers-realys networks.
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表 1 中繼選擇和編碼算法
步驟1 從${t_1}$條鏈路上接收0級編碼包; 步驟2 統(tǒng)計實際接收到的0級編碼包數(shù)量$t_1^*$; 步驟3 中繼$R_{}^{\rm{I}}$以概率$\varGamma _h^{\rm{I}}$選擇度數(shù)$h$; 步驟4 if $t_1^* > h$ then 從$t_1^*$個0級編碼包中任選$h$個進行混合; else then 將$t_1^*$個0級編碼包進行混合。 end if 下載: 導(dǎo)出CSV
表 2
${\text{Γ}^{\text{I}}}\text{(}\text{x}\text{)}$ 和${\text{Γ}^{{\text{II}}}}\text{(}\text{x}\text{)}$ 的線性規(guī)劃結(jié)果T=50 ${\varGamma ^{\rm{I}}}(x)$ 0.7101x+0.2899x5 ${\varGamma ^{{\rm{II}}}}(x)$ 0.8850x+0.0117x2 + 0.1033x10 T=100 ${\varGamma ^{\rm{I}}}(x)$ 0.7101x+0.2899x5 ${\varGamma ^{{\rm{II}}}}(x)$ 0.8625x+0.0534x2+0.0841x20 下載: 導(dǎo)出CSV
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