发布时间 : 星期三 文章信息论与编码习题参考答桉1更新完毕开始阅读
2002 Copyright EE Lab508
?U 0 1 ?[U?P]:?11
?P(U) 22?其失真矩阵为:
0 10?0a? [D]???1?a0?(1) 试求Dmin,R(Dmin); (2) 试求Dmax,R(Dmax); (3) 试求R(D);
(1)设输出符号集最小允许失真度：则满足保真度Y;{b1,b2}2Dmin??i?1p(ui)?minjd(ui,bj)??p(0)?0?p(1)?0＝0D?Dmin?0的信道矩阵0??1??1 [P]???0p(bj/ui)?0或p(bj/ui)?1(i?1,2),故此时H(U/Y)?0?R(Dmin)?R(0)?min?I(U;Y)??min?H(U)?H(U/Y)??H(U)?log2?1bit/symble(2)Dmax?D?min?2??min??p(ui)d(ui,bj)??min?p(0)d(0,0)?p(1)d(1,0);p(0)d(0,1)?p(1)d(1,1)?jj?i?1?a2 ?min{a?p(1);a?p(0)}?ja此时U、Y相互独立,I(U;Y)?0?R(Dmax)?R()?0222(3)平均失真度?pei?D???i?1j?1p(ui)p(bj/ui)d(ui,bj)?a?p(ai)p(bj/ai)i?j?i?jp(bj/ui)?D?a?p(ui)pei?aPe,当失真度满足保真度准i?j则时,D?D?aPe由费诺不等式:H(U/Y)?H(Pe)?Pelog(r?1)?H(Da)?DaDa)?Dalog(r?1)I(U;Y)?H(U)?H(U/Y)?H(U)?H(?在D定义域中选取适当值可?对此信源R(D)?H(U)?H(log(r?1)Dd)?Ddlog(r?1)得R(D)?min{I(U;Y)}?H(U)?H(Da)?1?H(Da)Da?1?H() 0?D???d2即R(D)???0 D?a?2??H.F.
2002 Copyright EE Lab508
8.8对于离散无记忆信源U,其失真矩阵[D]中,如每行至少有一个元素为零,并每列最多只有一个元素为零,试证明R(D)=H(U).
8.9试证明对于离散无记忆信源,有RN(D)=NR(D),其中N为任意正整数,D>Dmin. 8.10某二元信源X的信源空间为:
[X?P]:?X a1 a?2 ?P(X) ? 1-?
其中ω<1/2,其失真矩阵为:
[D]??0d???d0? ?(1) 试求Dmin,R(Dmin); (2) 试求Dmax,R(Dmax);
(3) 试求R(D);
(4) 写出取得R(D)的试验信道的各传输概率;
(5) 当d=1时,写出与试验信道相对应得反向试验信道的信道矩阵. 解:
?H.F.
2002 Copyright EE Lab508
2(1)最小允许失真度：则满足保真度Dmin??i?1p(ai)?minjd(ai,bj)??p(0)?0?p(1)?0＝0D?Dmin?0的信道矩阵0??1??1 [P]???0p(bj/ai)?0或p(bj/ai)?1(i?1,2),故此时H(X/Y)?0?R(Dmin)?R(0)?min?I(X;Y)??min?H(X)?H(X/Y)??H(X)?H(?)(2)Dmax?D?min?2??min??p(ai)d(ai,bj)??min?p(0)d(0,0)?p(1)d(1,0);p(0)d(0,1)?p(1)d(1,1)?jj?i?1? ?min{d?p(1);d?p(0)}?d?p(1)?d?j此时X、Y相互独立,I(X;Y)?0?R(Dmax)?R(?)?022(3)平均失真度?pei?D???i?1j?1p(ai)p(bj/ai)d(ai,bj)?d?p(ai)p(bj/ai)i?j?i?jp(bj/ai)?D?d?p(ai)pei?dPe,当失真度满足保真度准i?j则时,D?D?dPe由费诺不等式:H(X/Y)?H(Pe)?Pelog(r?1)?H(Dd)?DdDd)?Ddlog(r?1)I(X;Y)?H(X)?H(X/Y)?H(X)?H(?在D定义域中选取适当值可?对此信源R(D)?H(?)?H(log(r?1)Dd)?Ddlog(r?1)得R(D)?min{I(X;Y)}?H(X)?H(Dd)D?H(?)?H() 0?D?d??即R(D)??d?0 D?d??
?H.F.
2002 Copyright EE Lab508
(4)I(X;Y)取得R(D)时的试验信道的信道矩??d2?Dd?D?d?D2?2?d?2D?d[PX]??2Dd??D?22??d??d?2Dd?2Dd?检验:22阵为:Dd?D?d?D2??2?d?2D?d222?d??d?2Dd?Dd??D?22d??d?2Dd?2Dd???22p(y1)?p(y2)??i?12p(xi)p(y1/xi)???d?Dd?D?d?D?d2?2D?d?(1??)Dd??Dd2??d2?2Dd?2Dd??d??Dd?2D?i?1p(xi)p(y2/xi)??D?(1??)(1?D)?d?d??Dd?2D2p(x1/y1)?p(x1)p(y1/x1)p(y1)???d2?Dd?D?d?D?d?2D?dd??Dd?2DDd??Dd222?1?Ddp(x2/y1)?p(x2)p(y1/x2)p(y1)(1??)?2D??d?2Dd?2Dd??d??Ddd?2Dp(x1)p(y2/x1)p(y2)??Dd?D?d?D2p(x1/y2)??d?2D?dd?d??Dd?2Dd22?Dd2p(x2/y2)?p(x2)p(y2/x2)p(y2)22(1??)???d22?2Dd?Dd??D2d??d?2Dd?2Dd?d?d??Dd?2D?1?DdH(X/Y)???i?1?j?1p(bj)p(ai/bj)logp(ai/bj) ??[p(y1)p(x1/y1)logp(x1/y1)?p(y1)p(x2/y1)logp(x2/y1)? p(y2)p(x1/y2)logp(x1/y2)?p(y2)p(x2/y2)logp(x2/y2)] ??[p(y1)(1?Dd)log(1?DdDd)?p(y1))log(1?DdDdlog)?DdDd?p(y2)(1?Dd]Dd)log(1?Dd)?p(y2)DdlogDd] ??[p(y1)?p(y2)]?[(1? ?H(Dd)log?I(X;Y)?H(X)?H(X/Y)?H(?)?H(实际上是先根据参数法或者直接按照课本Dd)矩阵.p487求出\反向信道\矩阵,再由反推正向信道传输?H.F.