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第9章 差错控制编码

Dec 31, 2015

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tanisha-gilliam

第9章 差错控制编码. 9.1 引言 9.2 纠错编码的基本原理 9.3 常用的简单编码 9.4 线性分组码 9.5 循环码 9.6 卷积码. 9.1 引言. 差错控制编码的基本方法 在发送端被传输的信息序列上附加一些监督码元,这些多余码元与信息码元之间以某种确定的规则相互关联(约束),接收端按照既定的规则检验信息码元与监督码元之间的关系. 常用差错控制方法. 检错重发 前向纠错. 检错码. 发. 收. 应答信号. 纠错码. 发. 收. 混合纠错. 纠检错. 发. 收. 应答信号. - PowerPoint PPT Presentation
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  • 9 9.1 9.2 9.3 9.4 9.5 9.6

  • 9.1 ,(),.

  • : ,

  • 12331TI23ACKNAK,,.

  • 1 2 3 4 5 6 2 3 4 5 6 7 8 91 2 3 4 5 6 2 3 4 5 6 7 8 9NAK2 (,,)

  • 9.2

  • 3

  • 4nkn k n k = r 1 d0

  • d0 e d0 e + 1 t d0 2 t + 1 t e d0 e + t +1Bd0BA12BB345d0

  • AB1te01PP
  • n=7 P=10-3 P(1) 710-3 P(2) 2.110-5 P(3) 3.510-8

    ,,1 ~ 2,.

  • 9.3 ,,1().

    ,

  • ,,,

  • 1(0).10,.

    ,,,1.

  • 9.4 .,,,. (2). ()

  • ,S,. S=0,S=1 ., rr,2r-1,,2r-1.(n,k),2r-1n .

  • (n,k)k=4,,r3, n=k+r=7

  • n=2r 1 k= 2r 1 r r12

  • =1=

  • HH Pr k Ir r r [ P Ir ]H 2=

  • QGAG

  • A

  • AB=[bn-1 bn-2 b0] B A = E E= [en-1 en-2 e0]

    B=A+E

  • 3

    3

  • 10S0

    B1B2B32 S2=H 1 E=[1 0 0 0 0 0] 1

    S3=H 3 E=[0 0 1 0 0 0] 3

  • ,(0)

  • 9.5 9.5.1 ,,.,,,.n

  • F(x) = N(x) Q(x) + R(x) F(x) R(x) ( N(x) )

  • , T(x)n,xi T(x)xn +1,. T(x)

  • nk2k

    gxk-10k0 0k00k-1gx0n-k

  • g(x),xg(x),x2g(x), xk-1g(x),,G

    mk-1 mk-2 m0

  • T(x)g(x),,k-1g(x).g(X) T(x) = h(x) g(x) g(x), xkg(x)xn+1, =1

  • g(x) xn+1n k, g(x) .

  • 9.5.2 n,k)g(x),xn+1n-kg(x). T(x)g(x),m(x). n-k0

  • (7,3) m(x)=x2+x, g(x)= x4 +x2+ x+1

    r(x)

  • a+b+cd+mfeS73m a b c d e f0 0 0 0 0 0 01 1 1 1 0 1 11 1 0 0 1 1 10 1 0 1 0 1 00 0 1 0 1 0 00 0 0 1 0 1 10 0 0 0 1 0 00 0 0 0 0 1 1f=mf=e

  • R(x)g(x),, R(x)g(x), .

  • 9.6 ,.kn,kn,,.n k,N-1,Nn.NnN,(n,k,N)

  • 6 5 4 3 2 1+bicibicik=1, n=2, N=6

  • b6 b5 b4 b3 b2 b1++S6 S5 S4 S3 S2 S1+ (2,1,6)bicici13?

  • b1,(1)

  • (1)E(b1),12(b1~b6,c1~c6)2,E(b1)=1, (2)331. 3,,1,b1,E(b1).(2)

  • Hb10

  • 1 1 0 0 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 0 0 0 0 0 1 1

    H1

    nn-k

  • H170

  • In-k n-kPi n-kk0 n-k

    h=[ PN 0 PN-1 0 PN-2 0 P1 In-k ]

    hH1

  • G g g=[ IK Q1 0 Q2 0 Q3 0 QN ]Ik k Qi =PiT k n-k 0 k

  • 313 a m1m200 b m1m201 c m1m210 d m1m211 M3 M2 M1++ mjmjy1jY2j

  • 313a

  • 313

  • 313aabbccda111110101000011100001010abcd111000110101100001011010

  • a11010111

  • nkNkn2k2k(N-1)2k2k