信息論-習(xí)題答案

1、Xxx12P(X)0.60.41.設(shè)信源通過一干擾信道，接收符號(hào)為Y=y,y,信道1251661344轉(zhuǎn)移矩陣為，求：(1)信源X中事件x和事件x分別包含的自信息量；12(2)收到消息y(j=1,2)后，獲得的關(guān)于x(i=1,2)的信息量；ji(3)信源X和信宿Y的信息嫡；(4)信道疑義度H(X/Y加噪聲嫡H(Y/X);(5)接收到信息Y后獲得的平均互信息量。解：1)l(x)logp(x)Iog0.60.737bitl212l(xjlogp(x)Iog0.41.322bit22222)51p(y)p(x)p(y/x)p(x)p(y/x)0.60.40.6111121264130.4p(y)p(

2、x)p(y/x)p(x)p(y/x)0.60.4212122246p(y/x)5/611l(x;y)loglog0.474bit1122P(y)0.61p(y/x)1/621Kx；y)bglog1.263bit1222p(y)0.42p(y/x)l/412l(x;yjloglog1.263bit2122p(y)0.61p(y/x)3/422l(x;yjlog0.907bitlog2222P(y)0.423)H(X)p(x)logp(x)(0.6log0,60.4log0.4jlogl00,971bit/symbolii2iH(Y)p(y)logp(y)(0.6log0,60.4log0.4)l

3、ogl00.971bit/symboljj2j4)H(Y/X)p(x)p(y/x)logp(y/x)ijijiij55111133log0.6log0.4Iog0.4log)logio(0.62666644440.715bit/symbolH(X)H(Y/X)H(YjH(X/Y)H(X/Y)H(X)H(Y/X)H(Y)0.9710.7150.9710.715bit/symbol5)l(X;Y)H(XH(X/Y)0,9710.7150.256bit/symbol2133122.設(shè)二元對(duì)稱信道的傳遞矩陣為P(0)P(1)H(X),H(X/Y),H(Y/X)I(X;Y)(1薩=3/4,=1/4,求和

4、；(2)求該信道的信道容量及其達(dá)到信道容量時(shí)的輸入概率分布；解:1)3311H(X)p(x)(loglog)0.811bit/symboli224444iH(Y/X)p(x)p(y/x)logp(y/x)ijijiij231111112232lglglglg)lOglO(23433433433430.918bit/symbol3211p(y)p(xy)p(xy)p(x)p(y/x)p(x)p(y/x)0.583311121111212434331120.4167P(y)p(xy)p(xy)p(x)p(y/x)p(x)p(y/x)212221212224343H(Y)p(y)(0.5833log

5、O.58330.4167Io段4167)0,980bit/symbolj22jl(X;Y)H(X)H(X/Y)H(Y)H(Y/X)H(X/Y)H(X)H(Y)H(Y/X)0,8110.9800.9180.749bit/symboll(X;Y)H(X)H(X/Y)0,8110.7490.062bit/symbol2)1122Cmaxl(X;Y)loglO0.082bit/symbolmHIog2(lglg)log2mi2233331p(x)i2X3.黑白氣象傳真圖的消息只有黑色和白色兩種，即信源=黑,白。設(shè)黑色出P(黑)P(白)現(xiàn)的概率為=0.3,白色出現(xiàn)的概率為=0.7。H(X)(1)假設(shè)圖上

6、黑白消息出現(xiàn)前后沒有關(guān)聯(lián)，求嫡；P（白/白）P（黑/白）（2）假設(shè)消息前后有關(guān)聯(lián)，其依賴關(guān)系為=0.9,=0.1,P(白/黑)P(黑/黑)H(X)=0.2,=0.8,求此一階馬爾可夫信源的嫡；2解：(1)H(X)p(x)logp(x)(03log030.7log0.7)0,881bit/symboliii(2)pp(e)p(e/e)p(e)p(e/e)1111212p(黑/黑)=0.8pp(e)p(e/e)p(e)p(e/e)2222121P0.8pO.lp(ej黑e1112P(e)0.9p(e)0.2p(e)221p(白/黑)=0.21.0闡2P回=)21白/p(e)p(e)1白12(pP1

7、/31e2白P(e)2/32p(白/白)=0.9Hp(e)p(e/e)logp(e/e)ijljiijl2210.8log0.80.2log0.2O.llogO.l0.9log0.933330.553bit/symbolP(0)P(1)4.某一無記憶信源的符號(hào)集為0,1,已知=1/4,=3/4。(1)求符號(hào)的平均嫡；mm(2)有100個(gè)符號(hào)構(gòu)成的序列，求某一特定序列(例如有個(gè)“陋(100-)個(gè)"1，'的自信息量的表達(dá)式；(3)計(jì)算(2)中序列的嫡。解：1331H(X)p(x)logp(x)loglog0.811bit/symbolii4444i(2)mlOOmlOOml33p

8、(x)i100444100m341.51.585mbitl(x)logp(x)logii1004(3)100H(X)100H(X)1000.81181.1bit/symbol5.已知一個(gè)高斯信道，輸入信噪比(比率)為3。頻帶為3kHz,求最大可能傳輸?shù)南⒙?。若信噪比提高?5,理論上傳送同樣的信息率所需的頻帶為多少？解：PX3000logCWlogl136000bit/st2PNC6000tW1500Hzlogl15P2XloglpN試求以下各信道矩陣代表的信道的容量：6.工0ICOClcOOlO010100000100010010100(1)P=(2)P=000000.1C,2030.40

9、00000.30,7000C0000000/0.23.10,3p=解：1)這個(gè)信道是一一對(duì)應(yīng)的無干擾信道Clogn1%42bit/symboi222)這個(gè)信道是歸并的無干擾信道C喻1%31385附加mb口I113)這個(gè)信道是擴(kuò)展的無干擾信道匚logn1c1以1.585bit/5ymn比飛祟面423456*P(X)O,2O,19C,130.170,150,10,017.設(shè)信源H(X)1)求信源嫡；(2)編二進(jìn)制香農(nóng)碼；(3)計(jì)算平均碼長和編碼效率。解：(1)7H兇曲:旗P：x£ii1戰(zhàn)1030.2C.19I。豺193.18I口乳柒0.1710,1722220.15C豺15口.1bgO.

10、l0.01碘01)2222.6口先腦炯麻1(2)xp(x)p(x)k碼字iiaiix0.2030001x0.190.230012x0.180.3930113x0.170.5731004x0.150.7431015x0.10.89411106x0.010.99711111107(3)Kkp(x)30.230.193018301730.1540.17O.Oliii3.14H即(X)2.60983.1%3.14RKXXXXXXXX2345671P(X)020190180.170.150.10.01編二進(jìn)制費(fèi)諾碼，計(jì)算8.對(duì)信源編碼效率。解:xp(x)k編碼碼字iiix0.200021x0.19000

11、10321x0.18101133x0.1701024x0.150110351x0,11011100.0111111kp(x)20,230,1930.1820.1730.1540.14O.Oliii2.74H(X|H(X)2.609952%R2.74KXxxxxxxx2345671P(X)0.20.190.180.170.150.10.019.對(duì)信源編二進(jìn)制和三進(jìn)制哈夫曼碼,計(jì)算各自的平均碼長和編碼效率。解:二進(jìn)制哈夫曼碼:xp(x)k編碼碼字iiis6s0.6105s0.3914s00.353s0.262x0.201021x0.191122x0.1800033x0.17100134x0.150

12、1035s0.111x0.1011046x0.011011147Kkp(X)20.220,1930.1830,1730.1540.14O.OLii272H(X)H(X)2.609959%R2.72KQ010Uk=?,寫出此（n,k）10.設(shè)線性分組碼的生成矩陣為101100G碼的所有碼字。（2）求其對(duì)應(yīng)的一致校驗(yàn)矩陣010110(1)此（n,k）碼的n=?Ho（3）確定最小碼距，問此碼能糾幾位錯(cuò)?列出其能糾錯(cuò)的所有錯(cuò)誤圖樣和對(duì)應(yīng)的伴隨式。（4）若接收碼字為000110,用伴隨式法求譯碼結(jié)果。解：1）n=6,k=3,由C=mG可得所有碼字為：000000,001011,010110,011101

13、,100101,101110,110011,111000101hPi位錯(cuò)。Q1101Q此碼是系統(tǒng)碼，由G知，則IP1Q011110100ioimi3）由h可知，其任意2列線性無關(guān)，而有3列線性相關(guān)，故有，能糾一d311nhi錯(cuò)誤圖樣e伴隨式TjEH100000101010000110001000011000100100000010010000001討對(duì)X001T4）由知E=010000,貝UK1（1）求該碼的編碼效率？RrE010110SrH110311.已知（7,4）循環(huán)碼的生成多項(xiàng)式，求:（2）求其對(duì)應(yīng)的一致校驗(yàn)多項(xiàng)式h（x）（3）寫出該碼的生成矩陣，校驗(yàn)矩陣。2(4)若消息碼式為，求其碼字。m(x)1xxk解：1)n技4,57,14%硒1422)h|x)643lOUOOOxx532O1O