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Description: 利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),-use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G * H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),
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Size: 4757 |
Author: 心海 |
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Description: 利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),-use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G * H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),
Platform: |
Size: 1179 |
Author: 心海 |
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Description: gen_H:利用随机pi矩阵排列而成的LDPC稀疏校验矩阵产生；
girthlenth：计算产生的矩阵H的围长-gen_H : pi using random matrix arranged at the LDPC have sparse matrix calibration; girthlenth : calculation of the matrix H Girth
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Size: 2048 |
Author: 喻小星 |
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Description: 利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),-use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G* H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),
Platform: |
Size: 4096 |
Author: 心海 |
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Description: 利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),-use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G* H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),
Platform: |
Size: 1024 |
Author: 心海 |
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Description: 利用函数MadHG生成规则LDPC码的校验矩阵H,其行重为6,列重为3(Rules of LDPC codes generated by function MadHG check matrix H, the adults is 6, column 3)
Platform: |
Size: 3072 |
Author: OHQD@70629
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