bet188金宝搏

函数OutBits = CTCEncoder (InpBits, N, P)

CircularStateTable = [
0 6 4 2 7 1 3 5;
0 1 2 3 4 5 6 7;
0 5 3 2 6 7 1 4;
0 4 1 5 6 2 7 3;
0 2 5 7 1 3 4 6;
0 7 6 1 2 3 4 5);

InitState = (0 0 0);
[垃圾FinalState] = CTCRSCEncoder (InpBits N InitState);
国际扶轮= mod (N, 7);
Ci = FinalState (1) * 4 + FinalState (2) * 2 + FinalState (3);
c = CircularStateTable (Ri, Ci + 1);
InitState地板(1)= (CS / 4);
c = mod (CS, 4);
InitState(2) =地板(CS / 2);
c = mod (CS, 2);
InitState (3) = c;
[p1 FS] = CTCRSCEncoder (InpBits N InitState);
如果总和(FS = = InitState) ~ = 3
disp (P1不匹配)
返回
结束
IB = CTCInterleaver (InpBits, N, P);
InitState = (0 0 0);
[垃圾FinalState] = CTCRSCEncoder (IB, N, InitState);
国际扶轮= mod (N, 7);
Ci = FinalState (1) * 4 + FinalState (2) * 2 + FinalState (3);
c = CircularStateTable (Ri, Ci + 1);
InitState地板(1)= (CS / 4);
c = mod (CS, 4);
InitState(2) =地板(CS / 2);
c = mod (CS, 2);
InitState (3) = c;
[p2 FS] = CTCRSCEncoder (IB, N, InitState);
如果总和(FS = = InitState) ~ = 3
disp (P2不匹配)
返回
结束

OutBits = 0 (6 * N, 1);
k = 0: n - 1
OutBits (6 * k + 1) = InpBits (2 * k + 1);
OutBits (6 * k + 2) = InpBits (2 * k + 2);
OutBits (6 * k + 3) = p1 (2 * k + 1);
OutBits (6 * k + 4) = p2 (2 * k + 1);
OutBits p1 (6 * k + 5) = (2 * k + 2);
OutBits (6 * k + 6) = p2 (2 * k + 2);
结束

函数[ParityBits FinState] = CTCRSCEncoder (InpBits N InitState)
% (ParityBits FinState] = CTCRSCEncoder (InpBits N InitState)
% InitState =初始状态的二进制编码器(0 - 7)
% FinState =最终状态的二进制编码器(0 - 7)

负载的CTCEncTable
TNxt = CTCEncTable (:, 6:8);
兜售= CTCEncTable (:, 9:10);
ParityBits = 0 (2 * N, 1);
c = InitState;

k = 0: n - 1
一个= InpBits (2 * k + 1);
B = InpBits (2 * k + 2);
m = c (1) * 16 + CS (2) * 8 + CS (3) * 4 + * 2 + B;
:NS = TNxt (m + 1);
Pb =兜售(m + 1,:);
ParityBits (2 * k + 1) = Pb (1);
ParityBits k + (2 * 2) = Pb (2);
c = NS;
结束
FinState = NS;

函数InterleavedBits = CTCInterleaver (InpBits, N, P)
% InterleavedBits = CTCInterleaver (InpBits, N, P)
% N -输入一些成对的长度
% P - (p0 p1 p2 p3)衬垫参数

InterleavedBits = 0 (2 * N, 1);
ξ= 0 (N, 2);
k = 0: n - 1
m = CTCGetInterleavedAddress (k, N, P);
一个= InpBits (2 * m + 1);
B = InpBits (2 * m + 2);
如果国防部(k, 2) = = 1
InterleavedBits (2 * k + 1) =一个;
InterleavedBits k + (2 * 2) = B;
其他的
InterleavedBits (2 * k + 1) = B;
InterleavedBits k + (2 * 2) = A;
结束

结束