(完整word版)信息论与编码-曹雪虹-课后习题答案

1/21/2 p1/302/31/32/301W1W1WW213233125 WWW1W6325=0.5，p(1|01)=0.5，0.80.20

00.50 00.20.8设各状态00，01，10，11的稳态分布概率为W51140.2W0.5WW

W1计算得到27i1i0.5W0.8WW

6666186636111213141516212223242526313233343536414243444546515253545556616263646566其中11，22，33，44，55，66的概率是11 36

X23456789101112 P(X)11115151111361812936636912183621log121log121log121log125log51log136366

高>160cm）<160cm） Xx10x21x32x43P3/8

1/8 021032011223210}，求该序列的自信息量1log81.415bit (1)I(●)=Log(4)2 1Log(4)3Log40.811(2)P(黑/黑)=H(Y/黑)=(3)P(黑/白)=H(Y/白)=(4)P(黑)=P(白/黑)=P(白/白)=P(白)= 12log38218log381.24bit/符号p(yj)38

385.25bit/符号2.12两个实验X和Y，X={x1x2 r11r12

r137/241/24

21r22r231/241/41/2431r32

r331/247/24 =2.3bit/符号x17/241/24 0x21/241/41/24 0 1/247/24bit/符8/248/248/24

8/248/248/24 (2)H(X/Y),H(Y/X),H(X/Z),H(Z/X),H(Y/Z),H(Z/Y),H(X/YZ),H(Y/XZ)H(Z/XY)；(3)I(X;Y),I(X;Z),I(Y;Z),I(X;Y/Z),I(Y;Z/X)和Zz170z211 j

P(i/j)= =＝0.3，白色出现的概率p(白)＝0.7。为：P(白|白)＝0.9143，P(黑|白)＝0.0857，P(白|黑)100.7log100.8813bit/符号P(黑|白)=P(黑)10.08570.7log

10.20.3log1＝0.512bit/符号 NH(X)2.1101580372.20给定语音信号样值X的概率密度为 1e(x)dx 1dy2r2x2(rxr)rp(x)log2rxdxrp(x)log2dxrp(x)logr2x2dxlogrrp(x)logr2x2dx p(x)logr2x2dxr2rxlogr2x2dx4rr2x2logr2x2dxr20r22r22 02

022sincoscoslog1loged1logecos2d 1dx2r2y2(ryr)r2y21/4，P(1)=3/4。p(xi)44

1 2 3ji1 2 0 3 0(1)求(X,X24

44

1 2 3i1

j

2

0

1

2

33

1/61/12

0

14/245/245/24212

212=1.209bit/符号1 2 31 2 0 3 0H(X3|X2)7log27log41log45log35log35log35log3=1.26bit/符号

272

272转移概率距阵为2

1W2W2W1 W4213233 4133

W3314244

33 (1)求平稳概率P(j/i)=P(j/i)= W1=,W2=,W3=2.32一阶马尔可夫信源的状态图如图2－13所示，1pp/2p/2p/2p/21p (1p)WpW

pWW

W(1p)W pWW 21

i1i 22 p2p0,当

p=2/3 22(1p) 0<p<2/3时H

2/3<p<1时

所以当p=2/3时H得p(0)=p(1)=p(2)= 012

0101110

01

10 1 1 0 2

0

1

0 10 01 0 2 P(y1=0)=p(y1=1)=1/2p(y2=1)=p(y2=1)=1/2

01

01000=0.5bit/符号

24

P(y1y2x)00 0 1/40 0 01 0 0 1/402 0 1/40 P(y1y2|x)0001 10110 1 0 0 01 0 0 1 02 0 1/20 y1y200 p 1/41/41/41/4=1.5bit/符号

=1.5-1=0.5bit/符号 3333 3，转移概率矩阵为1/2 1/20 1/21/41/4 解：1/2 1/20 1/21/41/4i j0 41alog4 3log21log12alog1a 24

取e为底2

41a41a1ln212a21ln1aa(11)41a41a41a1a2(1a2)41a41a1ln1aa22=01a441ac13log21log113log125 12545 1620 410 已知P(0)=P(1)=1/2，0.990.01 0.010.99 1e e1－e=(i=1,2,3) j

=log[20+2×2(1-

)log(1-)+log

]=log[1+21-H()]=log[1+2(1 ] 12(1

121

H()(1)12(1)(1

23CP(b2)(j=1,2,3) 12(1 (1)12(1)(1

=1/2 C1=logr-H(p1'p2'p3')-NklogMk其中r=2,N1=M1=1-2N2=2M2=4所以C1=log2-H(p,p-log4=log2+(p)log(p)+(p-og2=log2-2p)+(p-=(1-2中r=2,N1=M1=1-2N2=M2=2，所以C=logr-H(p-,p-=log2+(p-)log(p-)+(p-g2=log2-(1-2-=(1-2)+(p-3-6设有扰离散信道的传输情况分别如图3－17所 121200解：012120001110021C1bit/符号 P(x1/y2)=1，P(x2/y2)=1，P(x3/y2)=3 Pe=P(x1/y2)+P(x3/y2)=130.8 1Log(2)1Log(5)1Log52Log51Log(5)1Log51Log(10)3Log51.301215210 330

INH2.2510649106bitCtI9101.5105bit/sPPX0 a0某二元信源P(X)1/21/2其失真矩阵为0a aaaaX0 111011101110，求 an1aa 111011101110Dmax=min{3,3,3,3}R(Dmax)=0Dmin=0R(Dmin)=R(0)=H(X)=log(4)=2 p(y1)+p(y2)+p(y3)+p(y4)=1在[0,1]区间可以消概率 1/2000 0 0 0 1 1/4001 10000001u31/16010 1101101001100u41/16011 11101100010101u51/16100 111101001110110u61/161010111111111101111110111H(X)=log(4)=2 bit/ms=200bit/s=0.198bit/ms=198bit/s 111111 248163264128128H(U)=1Log(2)1Log(4)1Log(8)1Log(16)1Log(32)1Log(64)1Log(128)1Log(128)1.984 信源符号累加-Logp(xi)码长码字符号概率概率 0 1 1 0 2 2 3 3 0.8754 4 0.9385 5 0.9696 6 1/1280.9847 7 1/1280.9927 7 信符号第第第第第第第二元码源概率一二三四五六七符pix11/2x21/4x31/8

0001

0x41/16 1 0 1x51/32 1 0 x61/64 0 x71/128 1 0 1x81/128 111111110 信源符号累加-Logp(xi)码长码字符号概率概率 0 2 3 3 3 4 5 1 2 3 4 0 20 1 2 0 2 0 31 1 0 41 1 4 3222181612

38322218

403832

6040

02223445.16已知二元信源{0，1}，其p0=1/4,p1=3/4,试用P(s=11111100)=