(s',s)=p(i,=s,y=s)is the metric of the trellis branch connecting =s'to S=s with u=i. Equation (4.11)follows from the Markov property of the encoder,i.e.,if S=s is known, events after time k are independent of the historyy up to S. n(s'.s) (s,s) Define r(s.s)=p(S.=s,yS=s) (4.13) as the state transition probability fromS=s'toS=s.If there exit parallel transitions between S=s'and S=s in the trellis diagram,then (s'.s)is equal to the sum of the metrics of the corresponding parallel branches. Substituting (4.12)into (4.10),we have 4w1=h三aee0gOn (4.14) (s)(s'.s)B.(s)/p(y) 政 下面我们来求a,(s),(s)和y(s',s)。By definition,.the term a,(s)can be computed via the recursion a.)=∑pS=sS=sX) “名=8 =g=)p8=S=的 ->a-(s)-n(ss) (4.15) with the initial value(s)=P(S.=s). Similarly,B(s)can be obtained via the following backward recursion 4204-20   1 ( ', ) , , | ' i k k k kk  s s pu iS s S s     y is the metric of the trellis branch connecting Sk-1 = s’ to Sk = s with uk = i. Equation (4.11) follows from the Markov property of the encoder; i.e., if Sk = s is known, events after time k are independent of the history 1 k y up to Sk. Define   1 ( ', ) , | ' k k kk  ss pS s S s    y ( ', ) ( ', ) k k i k b B ss  s s     (4.13) as the state transition probability from Sk-1 = s’ to Sk = s. If there exit parallel transitions between Sk-1 = s’ and Sk = s in the trellis diagram, then ( ', ) k  s s is equal to the sum of the metrics of the corresponding parallel branches. Substituting (4.12) into (4.10), we have 1 0 1 1 1 ( ', ) 0 1 1 ( ', ) ( ') ( ', ) ( ) / ( ) ( ) ln ( ') ( ', ) ( ) / ( ) k k N kk k ss B k N kk k ss B s ss s p L u s ss s p                  y y (4.14) 下面我们来求  kk k ( ), ( ) ( ', ) s s ss 和 。By definition, the term ( ) k  s can be computed via the recursion 1 1 ' ( ) ( , ', ) k k kk s  s pS sS s      y  1 1 11 11 ' ( ', ) ( , | ', ) k k k k kk s pS s pS s S s           y yy  1 11 1 ' ( ', ) ( , | ') k k k kk s p S s pS s S s          y y  1 ' ( ') ( ', ) k k s   s ss       (4.15) with the initial value 0 0  () ( ) s PS s   . Similarly, k ( ) can be obtained via the following backward recursion s Sk-1 Sk uk / ck ( ', ) i k  s s ( ', ) k  s s