信源编码习题答案

1、5.1设信源1!X“1X2X3X4X5X6X7解:RX)0.20.190.180.170.150.10.01求信源嫡H(X)；编二进制香农码；计算平均码长和编码效率。7H(X)=，p(x)log2p(X)i1(-(0.2log20.20.19log20.190.18log20.180.17log20.170.15log20.150.1log20.10.01log20.01)=2.609bit/symbol(2)Xip(Xi)pa(xi)ki码字Xi0.203000X20.190.23001X30.180.393011X40.170.573100X50.150.743101X60.10.8941

2、110X70.010.9971111110K八kip(xi)=30.230.1930.1830.1730.1540.170.01=3.14H(X)H(X)2.60988883.1%RK3.145.2 对信源IX=,Xix2x3x4x5x6X7?编二进制费诺码，计算编码效率:P(X),0.20.190.180.170.150.10.01,解:Xip(Xi)编码码字kiX10.200002X20.19100103X30.1810113X40.1710102X50.15101103X60.11011104X70.01111114K=kp(Xi)=20,230,1930.1820.1730.1540.

3、140.01=2.74H(X)H(X)2.609=9=95.2%RK2.745.3 对信源XIdx2x3x4x5x6x71编二进制和三进制哈夫曼码，计算|P(X)_0.20.190.180.170.150.10,01各自的平均码长和编码效率。解：二进制哈夫曼码:xip(xi)编码码字kiS61S50.610S40.391S30.260.350S21x10.20102x20.191112x30.1800003x40.1710013x50.1500103S10.111x60.1001104x70.01101114Kkip(xi)=20,220,1930.1830.1730.1540.140.01=

4、2.72H(X)H(X)2.6091=-=-=95.9%RK2.72三进制哈夫曼码:xiP(xi)编码码字kiS31S20.540S10.261x10.2221x20.190002x30.181012x40.172022x50.150102x60.11112X70.012122K=，kiP(Xi)=10,22(0.190.180.170.150.10,01)=1.8H(X)RH(X)Klog2m2.6091.8log23=91.4%X3X4x5x6X7111118163264128X81128XX1X25.4设信源|X|=(11-p(x)-4求信源嫡H(X)；(2)编二进制香农码和二进制费诺码

5、；(3)计算二进制香农码和二进制费诺码的平均码长和编码效率;(4)编三进制费诺码；(5)计算三进制费诺码的平均码长和编码效率；解：8H(X)=，P(为)10g2p(Xi)i11,c1,1,c1,“1,cc1,c1,1,=-log22log24log28log216log232log264log2128loc248163264128128、=1.984bit/symbol(2)二进制香农码:XiP(Xi)Pa(Xi)ki码字X10.5010X20.250.5210X30.1250.753110X40.06250.87541110X50.031250.9375511110X60.0156250.9

6、68756111110X70.00781250.98437571111110X80.00781250.992187571111111二进制费诺码:XiP(Xi)编码码字kiX10.5001X20.2510102X30.125101103X40.06251011104X50.0312510111105X60.015625101111106X70.00781251011111107X80.0078125111111117香农编码效率:-11111111K八kiP(Xi)1-2-345677i248163264128128=1.984H(X)H(X)1.984=-:)=100%RK1.984费诺编码

7、效率:1/1c1c1,11八11)K=kip(Xj)12345677i248163264128128=1.984H(X)H(X)1.984八i100%RK1.984(4)XiP(Xi)编码码字kiXi0.5001X20.25111X30.12520202X40.06251212X50.0312502203X60.015625212213X70.00781252022204X80.0078125122214-,.11111111K=kip(xi)11223344i248163264128128=1.328H(X)_H(X)RKlog2m1.984c,94.3%5.5设无记忆二进制信源葭HUI1.

8、328log23,8,再替换成二进制变长码字，如下表所示。先把信源序列编成数字0,1,2,(1)验证码字的可分离性；(2)求对应于一个数字的信源序列的平均长度Ki;(3)求对应于一个码字的信源序列的平均长度K2;(4)计算K2,并计算编码效率；Ki(5)若用4位信源符号合起来编成二进制哈夫曼码，求它的平均码长K,并计算编码效率。序列数字二元码字101000011100100131010000131011000014110000000151101000000161110000000017111100000000805.6 有二元平稳马氏链，已知p(0/0)=0.8,p(1/1)=0.7,求它的符号嫡。用三个符号合成一个来编写二进制哈夫曼码，求新符号的平均码字长度和编码效率。5.7 对题5.6的信源进行游程编码。若“0”游程长度的截至值为16,“1”游程长度的截至值为8,求编码效率。5.8 选择帧长N=64对0010000000000000000000000000000001000000000000000000000000000000遍L-D码；(2)对1000010000101100000000010010000101001000000001110000010000000010遍L-D码再译码；对000000000000000000