一致校验矩阵计算
目录
- T1
- T2
- T3
T1
设二元(7,4)码的生成矩阵为:
G = [ 1 0 0 0 1 1 1 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 ] G=\begin{bmatrix}1&0&0&0&1&1&1\\0&1&0&0&1&0&1\\0&0&1&0&0&1&1\\0&0&0&1&1&1&0\end{bmatrix} G= 1000010000100001110110111110
(1) 求该码的所有码字;
(2) 求该码的一致校验矩阵;
(3)作出该码的标准阵列。
解:
(1)所有码字如下表:
0000000 | 0100101 | 1000111 | 1100010 |
0001110 | 0101011 | 1001001 | 1101100 |
0010011 | 0110110 | 1010100 | 1110001 |
0011101 | 0111000 | 1011010 | 1111111 |
(2) H = [ 110110 101101 111001 ] H=\begin{bmatrix}110110\\101101\\111001\end{bmatrix} H= 110110101101111001
(3) G = [ 1 0 0 0 1 1 1 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 ] = G’ G=\begin{bmatrix}1&0&0&0&1&1&1\\0&1&0&0&1&0&1\\0&0&1&0&0&1&1\\0&0&0&1&1&1&0\end{bmatrix}=\text{G'} G= 1000010000100001110110111110 =G’
T2
设一个(8,4)系统码,其一致校验方程为:
{ c 1 = m 4 + m 3 + m 2 c 2 = m 3 + m 2 + m 1 c 3 = m 4 + m 2 + m 1 c 4 = m 4 + m 3 + m 1 \begin{cases}c_1=m_4+m_3+m_2\\c_2=m_3+m_2+m_1\\c_3=m_4+m_2+m_1\\c_4=m_4+m_3+m_1\end{cases} ⎩ ⎨ ⎧c1=m4+m3+m2c2=m3+m2+m1c3=m4+m2+m1c4=m4+m3+m1
式中, m 1 , m 2 , m 3 , m 4 m_1,m_2,m_3,m_4 m1,m2,m3,m4是信息位, c 1 , c 2 , c 3 , c 4 c_1,c_2,c_3,c_4 c1,c2,c3,c4是校验位。求该码的 G G G 和 H H H矩阵。
解:根据: C ‾ = M ‾ G \overline{C}=\overline{M}G C=MG 即 ( c 8 c 7 c 6 c 5 c 4 c 3 c 2 c 1 ) = ( m 4 m 3 m 2 m 1 ) G (c_8c_7c_6c_5c_4c_3c_2c_1)=(m_4m_3m_2m_1)G (c8c7c6c5c4c3c2c1)=(m4m3m2m1)G 得:
G = [ 1 0 0 0 1 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 1 0 ] G=\begin{bmatrix}1&0&0&0&1&1&0&1\\0&1&0&0&1&0&1&1\\0&0&1&0&0&1&1&1\\0&0&0&1&1&1&1&0\end{bmatrix} G= 10000100001000011101101101111110
G = [ I k P ] , H = [ P T I n − k ] G=[I_kP],H=[P^TI_{n-k}] G=[IkP],H=[PTIn−k] ,所以:
H = [ 1 1 0 1 1 0 0 0 1 0 1 1 0 1 0 0 0 1 1 1 0 0 1 0 1 1 1 0 0 0 0 1 ] H=\begin{bmatrix}1&1&0&1&1&0&0&0\\1&0&1&1&0&1&0&0\\0&1&1&1&0&0&1&0\\1&1&1&0&0&0&0&1\end{bmatrix} H= 11011011011111101000010000100001
T3
(15,5)循环码的生成多项式为: g ( x ) = x 10 + x 8 + x 5 + x 4 + x 2 + x + 1 g( x) = x^{10}+ x^{8}+ x^{5}+ x^{4}+ x^{2}+ x+ 1 g(x)=x10+x8+x5+x4+x2+x+1,试:
(1) 写出该码的校验多项式,
(2)写出该码的系统形式的生成矩阵和一致校验矩阵
解 :
(1)
h ( x ) = x 15 + 1 g ( x ) h( x) = \frac {x^{15}+ 1}{g( x) } h(x)=g(x)x15+1
h ( x ) = x 5 + x 3 + x + 1 h( x) = x^{5}+ x^{3}+ x +1 h(x)=x5+x3+x+1
(2)
G = [ 100001010011011 010001111010110 001000111101011 000101001101110 000010100110111 ] G=\begin{bmatrix}100001010011011\\010001111010110\\001000111101011\\000101001101110\\000010100110111\end{bmatrix} G= 100001010011011010001111010110001000111101011000101001101110000010100110111
H = [ 110101000000000 011010100000000 11100001000000 011100001000000 001110000100000 110010000010000 101100000001000 010110000000100 111110000000010 101010000000001 ] H=\begin{bmatrix}110101000000000\\011010100000000\\11100001000000\\011100001000000\\001110000100000\\110010000010000\\101100000001000\\010110000000100\\111110000000010\\101010000000001\end{bmatrix} H= 11010100000000001101010000000011100001000000011100001000000001110000100000110010000010000101100000001000010110000000100111110000000010101010000000001