7.2设输入符号集与輸出符号集为X = Y= 1? 2: 3,且输入倍源的分布为(1 = 0. L 2, 3)设失真矩阵为001=:I1 1 10 1 11 0 1t 10求久丫和ZU及R(D). 解；口皿远=lninZ?, = min Y = xl + xl + Jx1+-x0D血二工尹(xjTnin%*, jj) = xO + -xD+-1Ar b=111-4 (| 1-tn I 严)-入2/reF 朮8- 2设有两牛高斯B&即变量X和Y的联合概率密度为p(x, y)=芒 T T F“y&上口 ZifJr来賄机变量7 XIV IX) H*J TH 二 p(y) = y=-= rU)P(J)J.随机応XN俪T J 肌叫 n I I. X -J Y独“因此，Z服从正态分布.且有E=拭 X + y)= E(X) + 尸(F) = Wr + tnrDZ) = D(X 十打=D(X) + Z(n = erYJ 斗 j/Z - NO宮 4-冈 y*b J + (.Ty 砒化=loji $2屁Q j 亠昇，尊列?. 2设祚两牛禹斯随机藍堆X和Y的底汁槪率密匪为P(A) =丄 f2jtCl）求随机变產Z=X+Y前槪率密度 2 J il算 II（Z解 $ P( r,)-) -e2jljT1757f -二 p&、U“F丿心TIi/A= 5丄严Ztff(E) bg2,reo lug 2B. J址随机变H:天和疋服从Ml卜&緘(/j .r y ud)汕別就 Hpf). H(Y). H(V).fll KX:Y).f十宀川）求 HpCY）解；小H住#二一丁（讷炖p（导肚创Il (应一utJ - c) fb 谢 K沖fx ciai x.h其它FX; =+- H(Xt Y) = 0曲？亍（孑-c）8, g说某-. Ifjfi 率.刈5.&lbitf 5 :嗓尹19率谱为N二5* W粗傅扌缶:/：册限B =逍中传验求无差错怙端需竖的用小限愉入山率P是窑少？-C = 5.6x|(T/Sr N = 5ZQp爪用丿屁、B=4k/=小叫心RTF；9. 1_100011r010010100100110001110试求该码组的校验矩阵n 解：_100011r0100101G =001001100010_1101100_H =1011010111000:1 1 11 0 10 1 11 1 09.2已知某系统汉明码的校验矩阵为1H - 01 试求其生成矩阵=当输入序列为11010111010D-H =0111010:10100100010g = Qt =00:001100001011将输人序列分三組分别是MO1 0110 1010 计算相应的码宇:110 10 01110 101 0 1 0 0 11DW时，求编码器编出的码序列n1110：0 1 1 11 1 D 1 I_10001Dr01001110010110000101100010010011100L0110000:0111000:0r01001110010:10000101100 1所以編岀的码序列Mj： 11010010110001 loioor.9. 3设线性分组码的校验矩阵为_100100110101010010011100001101011101H =试求该矩阵的标准校验矩阵和生成矩阵 解：_1D0D:爲1H011000010:01:10000011L0_-2 片0100100100010010a:00011010:1100_0010;1001010:1Q00001001000:00:G = 1=0Q:00:1000D010001J0D001010:1D01D0110巧=C +巧0D10100La6 =+ f g0101D0000D001:10:1LD0100o-C2 宀 C5001D101aa1a01a0ai0|_Q1011a0019. 4已知（6. 3J线性分组码的全部码字为1101001100110110100111011010U1000111101I10000000问该码能纠正单个错误吗？构造该码组的生成矩阵和校验矩阵* 解；最小汉明重堇是3,最小没明距禺为3,可以.纠正单个惜误_11010=捉一1 =击一尸一1店=& u = l)r=Guhen r = 42r =2-=16 A; -r-18-4-1=13n = k-r-4 = 121 1 11 000010001 0 00 11100100E =0 1 10 110100100 1 01 101100019.7试把生成矩阵其中厅为一个I线性码的校馥矩阵，且対应的码组最小重量d应为奇数=证副利用校验矩 阵丹-所形成的码为（nl切线性码且码组的最小重量为兔点. =宅十圧 仕点4_-衣|帀；方孑=门計件仁丿尸此“.孙咏-14 E亠几e1奇凯 肝 以对于任意才站列求和不均0,即任惹码険列线性无关,而存在川施列加最后一列为零”即日込十1列 绮性相关所以d.=d.+o所叹最小垣量. +1 .