信息論第五章答案

1、5.1設(shè)信源Xxxxx(X)b5).20.190.180.17xx560.150.1x0.01：(1)求信源熵H(X)；編二進制香農(nóng)碼；計算平均碼長和編碼效率。解：(1)H(X)：p(x)logp(x)i2ii：1：(0.2：log0.2：0.19：log0.19：0.18：log0.18：0.17：log0.172222：0.15：log0.15：0.1：log0.1：0.01：log0.01)222：2.609bit/symbol(2)碼字(3)K:kp(x)：3：0.2：3：0.19：3：0.18：3：0.17：3：0.15：4：0.1：7：0.01iii：3.145.2對信源K：XX3

2、.1483.1%xxxxx234560.190.180.170.150.1。為:二進制費諾碼，計算編碼效率。解：編碼碼字K:kp(x)：2：0.2：3：0.19：3：0.18：2：0.17：3：0.15：4：0.1：4：0.01iii：2.74：95.2%:H(X):H蘭).2.609RK2.745.3對信源Xxxxx(X)b5).20.190.180.17xx560.150.1X0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：iii2.72.H(X)H竺).2.609RK2.72.95.9%三進制哈夫曼碼：iii.1.8logm2.2.6091.8og3291.4%5

3、.4設(shè)信源XP(Xxxxxxx13141516171881632641281284x12(1)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；編三進制費諾碼；計算三進制費諾碼的平均碼長和編碼效率；解：(1)H(X)p(x)logp(x)i2ii-log2-log4-log8log16log32丄log64丄log128丄log128224&282162322642128212821.984bit/symbol=127/64bit/symbol(2)二進制費諾碼：香農(nóng)編碼效率：二進制香農(nóng)碼：碼字Kw1A28噲右噲占1.984127/

4、64H(X)H竺)J984.log%RK1.984費諾編碼效率：K)4略唁1.984loo%RK1.984(4)編碼碼字(5)Kkp(x)!1丄丄S丄4B4ii248163264128128i1.328H(X)H(X)1.98494.3%RKm1.328og322Xo15.5設(shè)無記憶二進制信源jp(x)J-J).9o.L先把信源序列編成數(shù)字0,1,2,率率，8,再替換成二進制變長碼字，如下表所示。驗證碼字的可分離性；求對應(yīng)于一個數(shù)字的信源序列的平均長度疋；1求對應(yīng)于一個碼字的信源序列的平均長度K；2計算莓并計算編碼效率；K1(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

5、算編碼效率。序列數(shù)字二元碼字10100001110010013101000013101100001411000000015110100000016111000000001711110000000080其它碼字的前2)序列長度、序列概率及二元碼長如下表所示：序列序列長度L.序列概率p；數(shù)字二元碼長L.二元碼字110.10410000120.1X0.914100100130.1X0.92341010000140.1X0.933410110000150.1X0.9444110000000160.1X0.95541101000000170.1X0.966411100000000180.1X0.9774

6、11110000000080.98810KpL5693信源符號/數(shù)字符號1iii1(3)KpL2.7086bi/數(shù)字符號2iii1K(4)0.4756bit/信源符號，此值表示無記憶二元信源采用游程長度編碼后每個二元信源需1要的平均碼長。H(X)p(x)logp(x)0469bit/信源符號,H(巴986%i2iK/Ki21(5)4位信源符號的聯(lián)合概率、Huffman編碼及碼長如下表：(碼字可以不同，但碼長一樣)S4P(Si)碼字Wi碼長LiS4P(s)碼字Wi碼長Li00000.65610110010.00811111010700010.0729110310100.0081111101170

7、0100.0729100311000.0081111111070l000.0729l0l30lll0.0009lllllll009l0000.0729lll04l0ll0.0009lllllll0l900ll0.008llllll06ll0l0.0009llllllll090l0l0.008lllll0007lll00.000llllllllll0l00ll00.008lllll00l7llll0.000llllllllllll0Kp(s)LI.970羽it/4Sym,KK/4O.492旳it/Sym4ii4i空952%K5.6有二元平穩(wěn)馬氏鏈，已矢中(0/0)=0.8p(l/l)=0.7,求

8、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。解：平穩(wěn)時馬爾科夫狀態(tài)的概率：解得：p(S).3/5P(S)2/5)08F(S)(l0.7)P(S)0l)P(S)llHl1/0一階馬氏信源的熵：lp(S)p(S|S)logP(S|S)TOC o 1-5 h ziji2ji蘭log5log3log70786bit/sym252252252丿p(SSS)p(S)p(SIS)p(SIS)l23l2l32p(S0)3/5,p(Sl)2/5,p(S01S0)08,p(SllS0)02,p(S01Sl)03,p(SlISl)07SSS123P(SlS2S3)LiWiS

9、SSl23P(SlS2S3)LiWi00048/l25ll0ll2l/250400lllll49/2503000ll02l/25040l0000ll2/l2530ll0l09/25050l0l0l00l2/l25400l0l0l3/l2550l0llTOC o 1-5 h zo48l249l22l2l93Kp(s)L3(9)4()5(-)iil25l25250l2525025025012532726l6bit/3syml25H90-l4%K/3i96/25000049/25011124/25024/25021/25021/2509/2506/25000110001111001010115/25

10、094/2501154/250O5.7對題5.6的信源進行游程編碼。若“0”游程長度的截止值為16，“1”效率。游程長度的截止值為8，求編碼解：一階馬氏信源的熵同上題，MJ曲閃曲二元平穩(wěn)一階記憶序列“0”游程的長度概率：p(l0).鮭P/0ip150/0l0.1,2,3.15il0.15i.p(l0).1il0.1i二元平穩(wěn)一階記憶序列“1”游程的長度概率：p(l1).ili1.1p1i/10/171/1l1.1,2,3.7il1.7i.p(l1).1l1.1i1”游程長度的熵：Hliplilogpliplilogplip7logp7ii2ii2ii/i2i/ilili.iipmlog(pvp

11、)p7logp7i/i0/i2i/i0/ii/i2i/i口Pl1plogp(lil)pl1plogpp7logp71/10/120/1i1/10/121/11/121/1liiip7)lOgpi/i20/iii口ipplogp)pip7logp71/10/121/1p1/11/121/1/ilil1件丿怡SiTOC o 1-5 h z HYPERLINK l bookmark28r1.p(ip7)!1 HYPERLINK l bookmark307p7logpi/ii/ilogpp7logp7/i2i/iip2i/ii/i2i/i1/1i件丿怡Sip(ip7)li/iilogp HYPERLI

12、NK l bookmark34p2i/i1/1(i.Pl7/)H(p,p)2696bit/信源符號p0/ii/i1/1同理，“o”游程長度的熵：Hlo(ip52H(p,p)3483bit/信源符號iip0/0i/oo/o分別對“0”和“i”游程序列進行Huffman編碼，并分別計算出它們的編碼效率?！?”游程序列的長度、對應(yīng)得概率、Huffman編碼的二元碼長及碼字序列序列長度L.1序列概率p數(shù)字二元碼長L.I二元碼字0iPi/010211002PP0/0i/0130010003P2P0/01/02301100004P3P0/01/0331010000,05P4P0/01/0440001000

13、0,006P5P0/01/05401010000,0007P6P0/01/06410010000,00008P7P0/01/075000000000,0000,09P8P0/01/085010010000,0000,00i0P9P0/01/095100010000,0000,000iiP10P0/01/0A60000100000,0000,0000i2P11P0/01/0B61000000000,0000,0000,013P12P0/01/0C61000010000,0000,0000,0014P13P0/01/0D700001100000,0000,0000,00015P14P0/01/0E

14、700001110000,0000,0000,000016P150/0F501000LpL2p3pp(1pp2)4pp4(1pp2)5pp7(1pp2)5p152ii11000100010000i16pp10(1pp2)7pp13(1p)10001002p(1pp2)(3pp4pp45pp76pp10)7pp13(1p)5p151001010101010002ppp(1pp2)(34p35p66p9)7pp13(1p)5p151100000010003.511信源符號/數(shù)字符號Hl0/L3483/351199.19%0i21”游程序列的長度、對應(yīng)得概率、Huffman編碼的二元碼長及碼字:序列

15、序列長度K1序列概率p1數(shù)字二元碼長KiI二元碼字11P0/10201112P0/1P1/112101113PP20/11P30/11/1331101111,15PP40/11116PP50/111117PP60/11/16411111111,11118P71/1740000TOC o 1-5 h z HYPERLINK l bookmark48KpKp(1p)(23p2)4pp4(1pp2)4p72ii0/11/11/10/11/11/11/11/1i1273二元碼字/數(shù)字符號Hl1/K2,6969875%1i22

16、73.HloHliHl0/Hl1/i0i1Hl0Hl1iLK22348326963.5112.7399%可見滿足，這里的“0”游程編碼效率高，因為游程長度長，而“1”游程編碼效率受游程01的長度限制顯得比“0”游程編碼效率略低一些，因此整體的編碼效率介于兩者之間。5.8選擇幀長N=63對00100000,00000000,00000000,00000000,01000000,00000000,00000000,0000000編L-D碼;對10000100,00101100,00000001,00100001,01001000,00000111,00000100,0000001編L-D碼再譯碼；

17、對0000000000000000000000000000000000000000000000000000000000000000編L-D碼；對10100011010111000110001110100110000111101100101000110101011010010編L-D碼；對上述結(jié)果進行討論。解：(1)本幀內(nèi)信息位數(shù)Q=2；各信息位位置值片=3,n2=34；幀長N=63。TQCjC1C22528530njL334jJQ位和T位需要的二進制自然碼位數(shù)分別是：Og(Nl)IOg64622CqC21953g195*11N632所以,L-D編碼結(jié)果：000010,01000010010解碼

18、：已知N=63，故前6位為Q的自然碼表示，所以Q=2；后11位為T的自然碼表示，得T=530尋找某一值K,使CqTCq,即：C2530C2,得K33再令TTC2530C25305282再令，1K33再次尋找某一值L,使CqTCq,即：C12C1,得L=2L1LLL得:L=2所以解碼出信息位的位置值是n1=3,n2=34對10000100,00101100,00000001,00100001,01001000,00000111,00000100,00000010編L-D碼本幀內(nèi)信息位數(shù)Q=15；各信息位位置值n=1，n=6，n=11，n=13，n=14，n=24，n=27，n=32，n=34,T

19、OC o 1-5 h z123456789n10=37,n11=46,n12=47,n13=48,n14=54,n15=63；幀長N=63。TQCjC1C2C3C4C5C6C7C8njL16111314242732JC9C10C11C12C13C14C1534374647485463010120495128710094疋65780578887238567100254186851015059590,3891061765,140676848452403979900093052749999209564676929470Q位和T位需要的二進制自然碼位數(shù)分別是:Hbg(N1)Hbg6462212213173426989547CQC1