(1)
![](//img.examk.com/exam/n/v/Eermi7ioSy399iDF7AEgGiaE5cbaD7Qq.jpg)
比特/信源符号;
(2)由各态历经定理,有,
![](//img.examk.com/exam/j/e/hjjF6Cgf4im2arpMhJotHMr2ue3zTL1q.jpg)
即
P(白)=p(白)p(白/白)+p(黑)p(白/黑)=0.9p(白)+0.2p(黑)
P(黑)=p(白)p(黑/白)+p(黑)p(黑/黑)=0.1p(白)+0.8p(黑)
解方程组得:p(白)=2p(黑),又由于p(白)+p(黑)=1,
所以p(白)=2/3,p(黑)=1/3
![](//img.examk.com/exam/e/a/mypSPTDEcfioh7LvkgxNmHQhPoPNABKp.jpg)
比特/符号;
(3)H
0(X)=log
22=1,无关联信源剩余度为1-0.8813/1=11.87%,
一阶马尔可夫信源剩余度为1-0.5533/1=44.67%
这说明马尔可夫信源比无相关信源的冗余度大,编码时可以获得更高的压缩比。