基于快速码根检验的RS码综合识别算法

doi:10.11959/j.issn.1000-436x.2022219

Abstract

Abstract:

In order to solve the problem of high computation and high missed alarm probability of RS (Reed-Solomon) codes for recognition, comprehensive recognition algorithm of RS codes based on fast code root trial was proposed.Firstly, the check relationship was solved in binary equivalently and fast code root trial was used to check parameters in sequence.Secondly, according to distribution characteristics of the combined code roots, m-level primitive polynomial field and error correction ability was associatively determined.Finally, the short codes and long codes were given different confidence weights and the determined parameters were comprehensively analyzed.The optimal parameter was selected and the generate polynomial was calculated.The proposed algorithm did not need prior information such as signal-to-noise ratio (SNR), and had good adaptability.The simulation results show that the proposed algorithm can effectively reduce the missed alarm probability under the condition of low complexity.Compared with the conventional hard decision algorithm, the performance of the proposed algorithm is improved, and the parameter recognition of RS codes can be completed quickly.

Key words: RS code, fast code root trial, confidence weight, comprehensive recognition, missed alarm probability, blind recognition

CLC Number:

TN911.7

Xiaolin ZHANG, Xiuqiao LI, Rongchen SUN. Comprehensive recognition algorithm of RS code based on fast code root trial[J]. Journal on Communications, 2022, 43(11): 117-126.

Figures/Tables 13

n	k	本原多项式 p(x)	生成多项式g(x)
7	5	x ³+x+1	$x^{2} + α^{4} x + α^{3}$
31	27	x⁵+x²+1	$x^{4} + α^{24} x^{3} + α^{19} x^{2} + α^{29} x + α^{10}$
127	119	x⁷+x³+x²+x+1	$\begin{array}{l} x^{8} + α^{102} x^{7} + α^{28} x^{6} + α^{88} x^{5} + α^{19} x^{4} + \\ α^{97} x^{3} + α^{46} x^{2} + α^{2} x + α^{36} \end{array}$

m_x	n_x	本原多项式 p_x(x)	连续码根的非零偶数值分布向量 $d$	Th₁	Th₂	检验元素	阈值差值Δth_i	k_x
4	15	$p_{x} (x) = x^{4} + x + 1$	(4,1,0,0,0,0,0)	10	1	α,α ²,α³,α⁴	Δth ₁=d₄-Th₂=0	11
5	31	$p_{x} (x) = x^{5} + x^{4} + x^{3} + x^{2} + 1$	(5,0,0,0,0,0,0,0,0,0,0,0,0,0,0)	4	1	α,α ²	Δth ₂=d₂-Th₁=1	29
5	31	$p_{x} (x) = x^{5} + x^{2} + 1$	(1,4,0,0,0,0,0,0,0,0,0,0,0,0,0)	4	1	α,α ²,α³,α⁴	Δth ₃=d₄-Th₂=3	27

m	列大小	$w_{\bar{τ}, \bar{δ}}^{m}$
3	21	12
4	60	31.9
5	155	80
6	378	190.5
7	889	448
8	2 040	1 016.3

References 18

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