=Aep4E0i+Yc】 (4.18) 。 where is a constant that is independent of code bits.As aforementioned,let Le=4E,INo (4.19 which is called the reliability value of the channel.Then (4.18)can be written as p(y IS:=s,S=s',m=i)=A exp(Lyici+Lyfcf (4.20) Therefore, )P(1)4+ye if P(S=s1=1 (4.21) 10 otherwise P(u =i).A exp(Lyici+Lyc(s'.s) where (s',s)=P(S,=s1S-=s',u =i). 至此，如果将式(4.14)分子分母中的p(y)约掉，L(4)的求解己基本完成。然而， 由于式(4.20)是从连续随机变量的概率密度计算得到，(s,)的值可能大于1，这会使 得式(4.15八、(4.16)产生上溢出，导致整个算法不稳定。另外，由于下列原因，也可能导 致下溢出： Some state metric becomes vanishingly small as k increases,resulting in errors due to precision limitations.This is evident by considering the summation Ea,)=pi的 The probability of receiving a particular sequence,p(y),becomes very small after a few time instantsk. 因此，有必要对a:(s),B(s)进行归一化处理。 度 a(s)=a(s)/p(y) B(s)=B.(s)/p(y) (4.22) 因为py)=∑p(S4=s,y),所以 a(s)=a()/∑a() (4.23) 将式(4.15)代入上式，并且分子分母同除以py),得到 ∑a-(s)y(d,s)1py- 立(9)= 2Σa,p 名4-22 0 4( ) exp ss pp s kk k k k E yc y c A N         (4.18) where Ak is a constant that is independent of code bits. As aforementioned, let 0 4 / L EN c s  (4.19) which is called the reliability value of the channel. Then (4.18) can be written as 1 ( | , ', ) kk k k p y S sS s u i    exp  s s pp   A Lyc Lyc k c kk c k k (4.20) Therefore, 1 ( ) exp{ }, if ( | ', ) 1 ( ', ) 0, otherwise ss pp i k k c kk c k k k k k k Pu i A Lyc Ly c PS s S s u i  s s            (4.21) ( ) exp{ } ( ', ) ss pp i Pu i A Lyc Lyc s s k k c kk c k k k     where 1 ( ', ) ( | ', ) i k kk k  s s PS s S s u i    . 至此，如果将式(4.14)分子分母中的 p N ( ) y1 约掉， L uk ( ) 的求解已基本完成。然而， 由于式(4.20)是从连续随机变量的概率密度计算得到， k ( ', ) s s 的值可能大于 1，这会使 得式(4.15)、(4.16)产生上溢出，导致整个算法不稳定。另外，由于下列原因，也可能导 致下溢出： Some state metric becomes vanishingly small as k increases, resulting in errors due to precision limitations. This is evident by considering the summation 1 () ( ) k k s  s p    y  The probability of receiving a particular sequence, 1 ( ) k p y , becomes very small after a few time instants k. 因此，有必要对 k k ( ), ( ) s s  进行归一化处理。 令 ~   ( ) ( )/ ( ) k k k s sp  y1 ~   ( ) ( )/ ( | ) kk k N k s sp  y y 1 1 (4.22) 因为 p pS s k k k s () ( ,) y y 1 1    ，所以 ~   ( ) ( )/ ( ) kk k s ss s   (4.23) 将式(4.15)代入上式，并且分子分母同除以 1 1 ( ) k p  y ，得到 ~ ( ) ( ') ( ', ) / ( ) ( ') ( ', ) / ( ) ' '      k k k k s k k k s s s s ss p s ss p        1 1 1 1 1 1 y y