《信息論與編碼》答案-第四章

1、信息論與編碼第四章 信息率失真函數(shù) 習(xí)題答案4.10 11 依題意可知：失真矩陣： ，轉(zhuǎn)移概率 ( | )p b ad 11 0ji平均失真：22D ( ) ( | ) ( , )p a p b a d a bijiiji1 j11/2 ) 0 1/2 1 1/2 1 1/2 ) 0 4.20 1依題意可知：失真矩陣： ，d2 0D p(x )mind(x ,y ) 1/201/200iijjiD minD min p(x )d(x ,y ) 1/201/211/221/20 舍去)jiijji0，R(D ) R(0) H(X)2當(dāng)D1 0P0 11/2，R(D ) 0當(dāng)Db)pb ) a b

2、 ,a b ,D : p因1212220 1P0 14.3111134D minD min p(x )d(x ,y ) 1 1 1 04444maxjiijji1111D p(x )mind(x ,y ) 0 0 0 004444miniijji0，R(D ) R(0) H(X)4當(dāng)D1 0 0 00 1 0 0P0 0 1 00 0 0 13/4，R(D ) 0當(dāng)DD (b ) p(b ) p(b ) p(b ) 0a b,a b,a b,a b: p1234112131411 0 0 01 0 0 0P1 0 0 01 0 0 04.40 1 1/4依題意可知：失真矩陣： ，d1 0 1/

3、4D p(x )mind(x ,y ) 1/201/20 0iijjiD minD min p(x )d(x ,y ) /21/41/21/4)1/4(其它個均為1/2)jiijji0，R(D ) R(0) H(X)2當(dāng)D1 0 0P0 1 01/4，R(D ) 0當(dāng)D(b ) 0, p(b ) 0, p(b ) 3，: pDmax1230 0 1b ,a bP即a0 0 13234.50 110依題意可知：失真矩陣： ，轉(zhuǎn)移概率為： Pdq 1q1 0nmD ( ) ( | ) ( , ) 10 01 ) 1 ) )0p x p y x d x yppp qpqijiiji1 j1q p)p

4、(x )mind(x ,y ) p0 p)0 0Diijji因為R(D)是 的遞減函數(shù)，所以D R(D R(D ) H(p)H(D )p p p) p)0max(R(DD 0當(dāng)q D p(x )d(x ,y ) p0 p1 p1 p ) D，maxjiijji因為R(D)是D的遞減函數(shù)，所以R(D) R(D ) H(p) H(D ) 01min(R(DD 1 p當(dāng)q)4.60 1 01u依題意可知：失真矩陣： ，信源d 1/2 1/2 0 1pu) D ( )min ( , ) 1/201/200，p x d x yiijjiD D p(x )d(x ,y ) /201/221/2/211/2

5、maxjiijji, 另二個 )，0 D1DR(D)lnnH a a1，nR(D)1D4.70 1 1d 1 0 1 1 1 04.8 1:2:3: D ；max R)X)RD*)0D*maxD信息論與編碼課后習(xí)題解答 1p(x )i 5I(x )log p(x )log225.581 bitii413p(x )Ci135213I(x ) p(x )13.208 Cii1352 XP( =P( =。12Y 、 )12p(y /x ) 11p(x )p(y /x )0.25I(x /y ) p(x /y )1 11p(y )0.511111Xx 0 x 1 x 2 x 3 1234P(X)3/81/4 1/41/8 個個個6個311 14 25 6p 848 p87.811 I I/n87.811/451.951 bit p(x )7%YI(x )logp(x )log0.073.837 bitYYp(x )93%NI(x )logp(x )log0.930.105 bitNN2H(X) p x ( )log ( )(0.07log0.070.93log0.366p x/bit symboliii2H(X) p x ( ) ( )(0.005p x /iii 和5 1 1 1 1 1 1p(x ) 6 6 6 6 i1I(x )p(x