信息論與編碼第五章答案

1、5.1設(shè)信源_(_ai a?a a? I0.2 0.19 0.18 0.17 0.15 0.1 0.01(1) 求信源熵H(X);(2) 編二進制香農(nóng)碼；(3) 計算平均碼長和編碼效率解：(1)7H (X) - > p(a)log2 p(aji亠=-0.2 log? 0.2_0.19 log 0.19-0.18 log20.18-0.17 log2 0.17-0.15 log20.15_0.1 log20.1-0.01 log2 0.01=2.609bit/symbolaiP(ai)Pa(aj)ki碼字a10.203000a20.190.23001a30.180.393011a40.17

2、0.573100a50.150.743101a60.10.8941110a70.010.9971111110_7£ 八 kjP(N)=0.2 3 0.19 3 0.18 3 0.17 3 0.15 3 i呂0.1 4 0.01 7-3.141r=H(X)/R =2.609, 3.14仁83.1%R5.2 對習(xí)題5.1的信源編二進制費諾碼，計算編碼效率.解：aiP(ai)編碼碼字kia10.200002a20.19100103a30.1810113a40.1710102a50.15101103a60.11011104a70.01111114=2M)=2x0J+3x0W+3x<,J

3、8+2x0-17+3x0J5+4x®-1+4x0®1= 2.745.3Hg HX 2609 * = 95%K X74對信源彳峰巧巧斗形畔工/_AAD_102 019 018 017 015 01 0.01dJ編二進制和三進制哈夫曼碼，計算各自的平均碼長和編碼效率解：二進制哈夫曼碼：XiP(Xi)編碼碼字kiS61S50.610S40.391S30.350S20.261X10.20102X20.191112X30.1800003X40.1710013X50.1500103Si0.111X60.1001104X70.01101114=2；Jt_p()=2xOJ+2xO 19+3

4、x0.18+3xO_17+3xOJ5 + 4xOJ+4x0.01= 272j/(JQ2.609R i _ 2 72三進制哈夫曼碼:Xip(Xi)編碼碼字kiS31S20.540S10.261X10.2221X20.190002X30.181012X40.172022X50.150102X60.11112X70.012122rKX = lx 022+2x(019+018+017 +015+DJ+O_Or)i= 1_85.4設(shè)信源EE2609l_8xlog23= 914%X求信源熵H(X);(2) 編二進制香農(nóng)碼和二進制費諾碼；(3) 計算二進制香農(nóng)碼和二進制費諾碼的平均碼長和編碼效率;(4) 編

5、三進制費諾碼；(5) 計算三進制費諾碼的平均碼長和編碼效率；解：(1)=-刀 pG0Le 工 p)iA.64xlog,1281 123128=xlog3 2+xlog3 4+xlpg38+一xlDg216 + 一xog332-l-一xlog364 + 248163264= 1_984 bU I symbol二進制香農(nóng)碼:XiP(Xi)Pa(Xi)ki碼字X10.5010X20.250.5210X30.1250.753110X40.06250.87541110X50.031250.9375511110X60.0156250.968756111110X70.00781250.98437571111

6、110X80.00781250.992187571111111二進制費諾碼:XiP(Xi)編碼碼字kiX10.5001X20.2510102X30.125101103X40.06251011104X50.0312510111105X60.015625101111106X70.00781251011111107X80.0078125111111117香農(nóng)編碼效率:=y*X)=-xl + -x2+-x3+Ax4+-!-x5+J-x6+x7 + x7 V24816326412812S= 1_M4£W = S)= 14=100%R K 1_984費諾編碼效率：疋辺俎心滬卜“扣十扣十護4十=

7、1_M4132128X7 +£W = S)= 14=100% 盤 K 1J984基護蝕寺“存十”存2噲x4 +ix4Xip(Xi)編碼碼字kiX10.5001X20.25111X30.12520202X40.06251212X50.03125202203X60.01562512213X70.00781252022204X80.0078125122214= 12328RKbg 工麗 132«xkjg23ci5.5設(shè)無記憶二進制信源1_他32嘰先把信源序列編成數(shù)字0, 1, 2,8,再替換成二進制變長碼字，如下表所示(1) 驗證碼字的可分離性；(2) 求對應(yīng)于一個數(shù)字的信源序列

8、的平均長度心；(3) 求對應(yīng)于一個碼字的信源序列的平均長度心；5(4) 計算更！，并計算編碼效率；(5)若用4位信源符號合起來編成二進制哈夫曼碼，求它的平均碼長K，并計算編碼效率.序列數(shù)字二元碼字101000011100100131010000131011000014110000000151101000000161110000000017111100000000805.6有二元平穩(wěn)馬氏鏈，已知p(0/0) = 0.8，p(1/1) = 0.7，求它的符號熵.用三個符號合成一個來編寫二進制哈夫曼碼，求新符號的平均碼字長度和編碼效率5.7對題5.6的信源進行游程編碼若“0”游程長度的截至值為 16，“ 1”游程長度的截至值 為8,求編碼效率.5.8選擇幀長N = 64(1) 對