等长码基本问题 例3:设信源 S S,)P(S)P()P(S3)P(s)l ,∑P)=1h2 code/sig H P(S,1S1)=P(s, 1s2)=P(S41S3)=P(S3154)=1 其余PS|)=0 则二次扩展信源为: L2=2 code/2 sigs 15 F=1 code/sig P(S S, ) P(S, 2)P(25, P(S3S4)P(SS3) ∑P()=∑P(s)PS)|s)=1例3：设信源 4 1 2 3 4 1 2 3 4 1 , , , , ( ) 1 ( ) ( ) ( ) ( ) ( ) i i i S s s s s P s P s P s P s P s P s =     = =          2 1 1 2 4 3 3 4 ( | ) ( | ) ( | ) ( | ) 1 ( | ) 0 j i P s s P s s P s s P s s P s s = = = = = 且 其余 则二次扩展信源为： 2 1 2 2 1 3 4 4 3 1 2 2 1 3 4 4 3 , , , , ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( | ) 1 i j i j i j i ij ij S s s s s s s s s P s s P s s P s s P s s P s s P s s P s P s s       =          = = l=2 code/sig L2=2 code/2_sigs l=1 code/sig 信源有记忆时，采用N长源编码，l减小 一、等长码基本问题