第八章 限失真信源编码

2002 Copyright EE Lab508

第八章 限失真信源编码

8.1设信源X的概率分布P(X):{p(?1), p(?2), …,p(?r) },失真度为d (?i, ?j)≥0,其中 (i=1,2,…,r;j=1,2,…,s).试证明:

Dmin??p(ai){mind(ai,bj)}

i?1jr并写出取得Dmin的试验信道的传输概率选取的原则,其中

mind(ai,bj)?min{p(b1/ai),p(b2/ai),?,p(bS/ai))}

jj(证明详见:p468-p470)

8.2设信源X的概率分布P(X):{p(?1), p(?2), …,p(?r) },失真度为d(?i, ?j)≥0,其中 (i=1,2,…,r;j=1,2,…,s).试证明:

Dmax?min{?p(ai)d(ai,bj)}

ji?1r并写出取得Dmax的试验信道传递概率的选取原则. (证明详见:p477-p478)

8.5设二元信源X的信源空间为:

0 1 ?X [X?P]:?

1-??P(X) ? 令ω≤1/2,设信道输出符号集Y:{0,1},并选定汉明失真度.试求:

(1) Dmin,R(Dmin); (2) Dmax,R(Dmax);

(3) 信源X在汉明失真度下的信息率失真函数R(D),并画出R(D)的曲线; (4) 计算R(1/8). 解：

(1)最小允许失真度：Dmin??p(ai)?minjd(ai,bj)??p(0)?0?p(1)?0＝0i?12则满足保真度D?Dmin?0的信道矩阵 0 10?10? [P]??1?01??p(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?(2)Dmax?Dmin?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{p(1);p(0)}?p(1)??j?此时I(X;Y)?0?R(Dmax)?R(?)?0?H.F.

2002 Copyright EE Lab508

(3)离散信源在汉明失真度下,R(D)?H(X)?H(D)?Dlog(r?1)?对此信源R(D)?H(X)?H(D)?H(?)?H(D)?H(?)?H(D) 0?D??即R(D)?? D???0

由上，可得R(D)曲线如下:

R(D) H(ω) D

0

(4)R(1/8)=H(ω)-H(1/8)= H(ω)-0.5436 bit/symble 8.6一个四进展等概信源

Dmax=ω

0 1 2 3 ?U ?[U?P]:?1111

P(U) ?4444?接收符号集V:{0,1,2,3},其失真矩阵为:

?0?1[D]???1??1101111011?1?? 1??0?(1) Dmin,R(Dmin); (2) Dmax,R(Dmax);

(3) 试求R(D), 并画出R(D)的曲线(去4到5个点). 解:

(1)设输出符号集Y:{b1,b2}最小允许失真度：Dmin??p(ui)?minjd(ui,bj)??p(0)?0?p(1)?0?p(2)?0?p(3)?0＝0i?14则满足保真度D?Dmin?0的信道矩阵?1000??0100??[P]???0010????0001?p(bj/ui)?0或p(bj/ui)?1(i?1,2,3,4),故此时H(U/Y)?01111?R(Dmin)?R(0)?min?I(U;Y)??min?H(U)?H(U/Y)??H(U)?H(,,,)?2bit/symble4444?H.F.

2002 Copyright EE Lab508

?4??3333?3(2)Dmax?Dmin?min??p(ui)d(ui,bj)??min?,,,??jj?4444?4?i?1?此时U、Y相互独立,故I(X;Y)?0?R(Dmax)?R(?)?0(3)离散信源在汉明失真度下,R(D)?H(X)?H(D)?Dlog(r?1)?对此信源R(D)?H(U)?H(D)?Dlog3?2?H(D)?Dlog33?2?H(D)?Dlog3 0?D???4即R(D)??3?0 D??4?可计算得:D?0,R(0)?2bit/symble;11 D?,R()?1.258bit/symble;8811 D?,R()?0.792bit/symble;4411 D?,R()?0.208bit/symble;2233 D?,R()?0bit/symble44?可得R(D)曲线如下:

2 1.258 0.792 0.208 0

8.7某二进制信源:

R(D) (bit/bymble) D

1/8 1/4 1/2 3/4

0 1 ?U ?[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);

?H.F.

2002 Copyright EE Lab508

(1)设输出符号集Y;{b1,b2}最小允许失真度：Dmin??p(ui)?minjd(ui,bj)??p(0)?0?p(1)?0＝0i?12则满足保真度D?Dmin?0的信道矩阵?10? [P]????01?p(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?(2)Dmax?Dmin?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?a ?min{a?p(1);a?p(0)}?j2a此时U、Y相互独立,I(U;Y)?0?R(Dmax)?R()?02(3)平均失真度D???p(ui)p(bj/ui)d(ui,bj)?a?p(ai)p(bj/ai)i?1j?1i?j22??pei??p(bj/ui)?D?a?p(ui)pei?aPe,当失真度满足保真度准则时,D?D?aPei?ji?jDD由费诺不等式:H(U/Y)?H(Pe)?Pelog(r?1)?H()?log(r?1)aaDDI(U;Y)?H(U)?H(U/Y)?H(U)?H()?log(r?1)aaDD?在D定义域中选取适当值可得R(D)?min{I(U;Y)}?H(U)?H()?log(r?1)ddDD?对此信源R(D)?H(U)?H()?1?H()aaDa?1?H() 0?D???d2即R(D)??a?0 D??2?

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:

a1 a2 ?X [X?P]:?

P(X) ? 1-??其中ω<1/2,其失真矩阵为:

?0d?[D]???

d0???H.F.

2002 Copyright EE Lab508

(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:

(1)最小允许失真度：Dmin??p(ai)?minjd(ai,bj)??p(0)?0?p(1)?0＝0i?12则满足保真度D?Dmin?0的信道矩阵?10? [P]????01?p(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?(2)Dmax?Dmin?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(?)?0(3)平均失真度D???p(ai)p(bj/ai)d(ai,bj)?d?p(ai)p(bj/ai)i?1j?1i?j22?pei??p(bj/ai)?D?d?p(ai)pei?dPe,当失真度满足保真度准则时,D?D?dPei?ji?jDD由费诺不等式:H(X/Y)?H(Pe)?Pelog(r?1)?H()?log(r?1)ddDDI(X;Y)?H(X)?H(X/Y)?H(X)?H()?log(r?1)ddDD?在D定义域中选取适当值可得R(D)?min{I(X;Y)}?H(X)?H()?log(r?1)ddD?对此信源R(D)?H(?)?H()dD?H(?)?H() 0?D?d??即R(D)??d? D?d??0

?H.F.

本文来源：https://www.bwwdw.com/article/ohrh.html