a?試對概率分別為：0?20,0?19,0.18,0.17,0.15,0.10,0.01的信源用費(fèi)諾編碼，求其二進(jìn)制代碼組及其編碼效率。解：(費(fèi)諾碼編碼)先將消息符號(hào)按概率大小排列，再按步驟進(jìn)行子集分解，本題經(jīng)過4次分解完成編碼，整個(gè)過程列于下表消息符號(hào)就第一次口分解口第二次」第三次u第四次二a制代碼組'長度也0.20^%(0.57)^(0.20)Op002p(11卯(0.37)Ip010^3p0.18^(0.185W口11口3p0.17^(0.43)p(0.17)Op102p0.15^(0.26)(0.15)0im3p0.10^iTl1111q(0.10)0^1110P4口0.01^l.LL11J1=(o.onimp4p編碼效率為H(X)2.61心

77=\'==95.3%b2.74b.0.1,信心信源包滿個(gè)符號(hào)消息，它們雄現(xiàn)概率分別為褊戰(zhàn)擊違5漏道基本符號(hào)為二進(jìn)制碼元，試用哈夫曼編碼方法對該信源的60.1,信解符號(hào)概率解符號(hào)概率bi0.30.20.150.150.10.120.30.20.150.150.10.1/由編碼結(jié)果，求得平均碼長為=2.5碼元/符號(hào)信源熵為H(X)=—Zp3)lbp(x)=2.471比特/碼元時(shí)間iiI編碼效率為門=H(X)/b=2.471/2.5=0.988/由此可得其編碼效率為98.8%，接近于最佳編碼。0.5

0.250.20.5

0.250.25

0.50.1250.1250.1250.125求信道容量CJ7(0.5,0.25,0.125,0.125)=-[0.51og0.5+0.251og0.25+2x0.1251og0.125]=1.75bit/f?i方法一：由輸入X等概率分布求輸出Y的分布，然后計(jì)算(F)]=[0.5I=[0.375i0.375Q.1250.1250.125"Id.1250.125J0.1251-[0.375log0.375+0.125Log0.125]x21.8113bit/符號(hào)=[0.5I=[0.375i0.375Q.1250.1250.125"Id.1250.125J0.1251信道拒陣可劃分p=F0-5425]_[0.1250325J成兩個(gè)子匣陣:]-|_0.250.5」為=0.125D.125N、=45+0.25=0+75N?=0.125■+0.125=0.25_VT■0.5+0."-0.75山j(luò)=0,125+0,125=°一"C■Log2-H(0.5,0.25+0.12S,0.125)-￡N*logAf*1?|=1-1,75-(0,75log0.75+0.25log0.25)=0061Z3bhF符號(hào)綜合題一9一已知ArYe(0f1}?XT構(gòu)成的聯(lián)合摭率空間為二X1[00011011-_P(頁F)」=|_0.10.20.30.4_試計(jì)算信息愉H(A').聯(lián)合隔日3F)及條件慨H(A'|Y)解：先求讖瓣率分布：P-{X=0}=P應(yīng)(0)=”00)4-p(Ol)=0.1+0.2=0.3P，{X=1}=八(1)=p(l0)*P(11)=D.3+0.4=0.7P，0=0}=(0)=p(00)十?(10)=0.1+0.3=0.4P-{F=1}=夕F(1)=夕(01)十mil)=0.2十0.4=0.6XY一00011011X~01Y~010.10.20.30.40.30.70.40.6再求條件概率分布：p(0I0)=g(00)/>F(0)=01/0-4=1/47?(1|0)=(10)/^F(0)=0.3/0.4-=3/4P(0I1)=夕(01)/Pf(1)=0-2/0.6=1/3夕(1|1)=夕(11)/夕『(1)=0.4/0=2/3信息郭:sH(X)[px(0)Logp*(0)+pr(1)LogpT(1)]=-[0.3Log0.3+0.7Log0.7]=0.881bit/w號(hào)聯(lián)合嫡：H(XY)=-[0.1Log0.1+0.2L(}g0.2+0.3L(}g0.3+0.4Log0.4]jk1.84-6bit/符號(hào)叩。0110"ILo.i0.20.30.4」條件螭：H(X|F)叩。0110"ILo.i0.20.30.4」條件螭：H(X|F)=—[P1(00)log+p(01)1。菖點(diǎn)0|0)=1/4/?(1|0)=3/4

^(0I1)=1/3p(l\l)=2/3|0)+/>(!0)log/>(!|0)p(0|1)*p(l1)1。.(1|1)]綜合題二,寫出此(n，，求：碼的所有碼字。設(shè)線性分組碼的生成矩陣為(1)(2)(3,寫出此(n，，求：碼的所有碼字。設(shè)線性分組碼的生成矩陣為(1)(2)(3)(4)求其對應(yīng)的一致校驗(yàn)矩陣H。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)誤圖樣和對應(yīng)的伴隨式。若接收碼字為000110，用伴隨式法求譯碼結(jié)果。000000，001011，010110，011101，100101,101110,110011,111000101110nioiooi0110102000000，001011，010110，011101，100101,101110,110011,111000101110nioiooi0110102)此碼是系統(tǒng)碼，由G知，on，則101001錯(cuò)誤圖樣E伴隨式1000001010100001100010000110001