把n个二进制对称信道串接起来，每个二进制对称信道的错误传输概率为p(0

道串接如下图所示：

XBSC 0 X1 I 解：

BSC X2 II BSC Xn … N 用数学归纳法证明:当n?2时由:p??1?pp??2p?2p21?2p?2p2??1?p[P2]????????22?p1?pp1?p1?2p?2p2p?2p??????1?p2?2p?2p2?[1?(1?2p)2]2假设n?k时公式成立,则?1k[1?(1?2p)]?2[Pk?1]??1k?[1?(1?2p)]?2?1k?1[1?(1?2p)]?2 ??1k?1?[1?(1?2p)]?21?Pk?1?[1?(1?2p)k?1]21故Pn?[1?(1?2p)n]21?[1?(1?2p)k]??1?pp?2???p1?p?1k?[1?(1?2p)]??2?1?[1?(1?2p)k?1]?2?1k?1[1?(1?2p)]?2?11?1?2p?1?limPn?lim[1?(1?2p)n]?n??n??22设输入信源空间X0:p(X0?0)?a,p(X0?1)?1?a(其中0?a?1)则输出信源X?:p(X??0)?p(X0?0)?p(X??0X0?0)?p(X0?0)?p(X??0X0?1)?12?p(x?x0)?p(x?)(x0、x?取0或1) p(X??1)??limI(X0;Xn)???p(X0iX?j)logn??i?1j?1222212p(X?jX0i)p(X?j)???p(X0iX?j)logi?1j?122p(X?jX0i)p(X?j) ???p(X0iX?j)log1?0i?1j?1

试求下列各信道矩阵代表的信道的信道容量：

(1)

b1 b2 b3 b4 a1?0010?a2?1000?

?[P1]??a3?0001???a4?0100?(2)

b b b 123a1?100??a2100[Pa???010?? 2]?3a?010?4??a5?a?001?001?6??(3)

b1 b2 b3 b4 b5 b6 b7 b8 b9 a1?0.10.20.30.400000[P3]?a2?a?00000.30.70003??0000000.40.20.1解：

(1)信道为一一对应确定关系的无噪信道?C?logr?log4?2 bit/symble(2)信道为归并性无噪信道?C?logs?log3?1.585 bit/symble

(3)信道为扩张性无噪信道:?C?logr?log3?1.585 bit/symble

设二进制对称信道的信道矩阵为：

0 1[P]?0?3/41/41?? ?1/43/4??(1) 若p(0)=2/3，p(1)=1/3，求H(X)，H(X/Y)，H(Y/X)和I(X;Y)； (2) 求该信道的信道容量及其达到的输入概率分布。

b10 0?0?0.3???

解：

2211(1)H(X)???p(xi)logp(xi)??(?log??log)?0.9183 bit/symble3333i?1py(0)??p(xi)p(y?0xi)?i?122223117????34341221135????3434127755?log??log)?0.9799 bit/symble1212121222py(1)??p(xi)p(y?1xi)?i?12

H(Y)???p(yj)logp(yj)??(j?122H(YX)????p(xiyj)logp(yjxi)????p(xi)p(yjxi)logp(yjxi)i?1j?1i?1j?1233111211133 ??(?log??log??log??log)?0.8113 bit/symble344344344344?I(X;Y)?H(Y)?H(YX)?0.9799?0.8113?0.1686 bit/symbleH(XY)?H(X)?I(X;Y)?0.9183?0.1686?0.7497 bit/symble(2)本信道为强对称信道?C?logr?H(?)??log(r?1)?log2?H(0.25)?0.25log1?0.1887bit/symble1信源输入为等概分布,即p(X?0)?p(X?1)?时达到信道容量C.2

设某信道的信道矩阵为

b1 b2 b3 b4 b5 a1?0.60.10.10.10.1?a2?0.10.60.10.10.1???

[P]?a3?0.10.10.60.10.1???a4?0.10.10.10.60.1?a5??0.10.10.10.10.2??试求：

(1) 该信道的信道容量C； (2) I(a3;Y)； (3) I(a2;Y)。 解: