信息论与编码原理-第4、5章课后习题-20140604-23点-自己整理

《信息论与编码原理》四、五章习题信息符号对应码字000001011100101111011111s4.1对信源概率空间为进行二元编码，编码方案如表4.19所示（1）计算平均码长。（2）编码后信息传输率。（3）编码后信源信息率。（4）编码效率。1234567(s)0.20.190.180.170.150.100.01SsssssssPLR'R2s3s4s5s6s7s表4.19编码方案（1）平均码长1()3*(0.20.190.180.170.15)4*(0.100.01)3.11qiiiLPsL()HSRL()(0.2,0.19,0.18,0.17,0.15,0.10,0.01)2.609HSH码元/信源符号（2）编码后信息传输率bit/符号0.839Rbit/码元（3）编码后信源信息率'logRLr2log1rr'log3.11RLrbit/信源符号（4）编码效率max()0.839logHSRLRr进行二元编码，5种不同的编码方案如表4.20所示。表4.205种不同的编码方案信源符号00000010010110100000100111101100010110111111010010101000111111110110011010101111111111010110011s2s3s4s5s6s1C2C3C4C5C（1）这些码中哪些是唯一可译码？（2）这些码中哪些是即时码(异前缀码)？（3）计算即时码的平均码长和编码效率。4.3某信源概率空间为123456()0.30.250.20.150.060.04SssssssPs（1）这些码中是唯一可译码。123CCC（2）这些码中是即时码。13CC161()3*(0.30.250.20.150.060.04)3ciiiLPsL码元/信源符号（3）()(0.3,0.25,0.2,0.15,0.06,0.04)2.325HSHbit/符号11max()0.775logccHSLRRr361()1*0.32*0.253*0.24*0.155*0.066*0.042.54ciiiLPsL码元/信源符号33max()0.915logccHSLRRr进行次扩展，采用二元霍夫曼编码。当时的平均码长和编码效率为多少？1,2,3,NN4.7设离散无记忆信源的概率空间为，对信源12()0.70.3SssPs（1）时，将编成0，编成1，则1N1s2s11L又因为信源熵()(0.7,0.3)0.881HSHbit/符号所以编码效率11()0.881HSL（2）如果对长度的信源序列进行霍夫曼编码，编码结果如表4.8所示2N信源序列霍夫曼码0.4910.21010.210000.09001表4.8时的编码结果2N111ss12ss21ss22ss()iP此时，信源序列的平均码长210.4920.213(0.210.09)1.81L二元码符号/信源符号序列则单个符号的平均码长20.9052LL二元码符号/信源符号所以对长度为2的信源序列进行变长编码，编码后的编码效率2()0.8810.9730.905HSL用同样的方法进一步将信源序列的长度增加，对的序列进行最佳编码，可得平均码长和编码效率为3N30.9690.909L二元码符号/信源符号（4）时，由香农第一定理可知，必然存在唯一可译码，使Nlim()NrNLHSN而霍夫曼编码为最佳码，即平均码长最短的码，故lim1NN4.9已知离散无记忆信源的概率空间为用香农编码和霍夫曼编码法编成二进制变长码，计算平均码长和编码效率。12345()0.250.20.20.20.15SsssssPs解：香农编码信息符号符号概率累积分布码字长度码字0.25022000.20.252.3230100.20.452.3230110.20.652.3231010.150.852.743110isP()isiFlogP()is1s2s3s4s5s20.253(0.20.20.20.15)2.75L码源/信源符号(S)(S)logHHLrL()(0.25,0.2,0.2,0.2,0.15)2.305HSHbit/符号0.838霍夫曼编码信源符号码字码长01210211200030013is1s2s3s4s5s20.250.220.220.230.1532.35L码元/符号(S)2.3050.982.35logHLr（1）若信道输入符号，求、、和。（2）求该信道的信道容量及达到信道容量的最佳输入概率分布。（3）如果信道输入符号时，计算信道剩余度。03/4,(1)1/4PP（）()HX(|)HXY(|)HYX(;)IXY03/4,(1)1/4PP（）0.80.20.20.85.1设二进制对称信道的传递矩阵为（1）31()(,)0.811bit/44HxH符号的联合分布概率为：XY0101XY35320120153311()(,,,)1.533bit/520205HXYH符号137()(,)0.934bit/2020HYH符号(|)()()0.599bit/HXYHXYHY符号(|)()()0.722bit/HYXHXYHX符号(;)()()(,)()(|)()(|)0.212bit/IXYHXHYHXYHYHYXHXHXY符号（2）由对称信道的信道容量公式得：log()log2(0.8,0.2)0.278bit/CSHpH的行矢量符号且当信道输入是等概率分布时才能达到这个最大值,即:1(0)(1)2PP（3）信道剩余度为：(;)(;)10.237CIXYIXYCC信道剩余度5.3设某对称离散信道的信道矩阵为（1）求其信道容量。（2）写出该信道的二次扩展信道的信道矩阵，并计算信道容量。00.50.50000.50.50.5000.50.50.500P（1）log()log4(0.5,0.5)1bit/CSHpH的行矢量符号20001020310111213202122233031323300000000.250.25000.250.25000000100203101112132021222330313233P000000.250.25000.250.25000000000.25000.250.25000.25000000000.250.25000.250.250000000000000000.250.25000.250.25000000000000.250.25000.250.25000000000.25000.250.25000.25000000000.250.25000.250.250000.250.2500000000000.250.250000.250.2500000000000.250.250.25000.25000000000.25000.250.250.2500000000000.250.250000.250.25000.250.25000000000000.250.25000.250.25000000000.25000.250.25000.25000000000.250.25000.250.250000000000（2）二次扩展信道的信道矩阵为：则信道容量为：log()log16(0.25,0.25,0.25,0.25)2bit/CSHpH的行矢量符号5.4设某信道的转移矩阵为求其信道容量。11pqqppqpqP将划分为两个对称的矩阵：11pqqpPpqpq1211pqpPppqqPq11222,1,1,,2,2rNqMqNqMqn'''1231loglog(,,)log2((1)log(1)log(2))(1,,)log2[(1)log(1)(log2log)](1,,)(1)log2(1,)(1,,)nkkkCrNMHPPPqqqqHpqpqqqqqHpqpqqHqqHpqpq图5.14题5.10中的串联信道5.10有二个信道的信道矩阵分别为和，它们的串联信道如图5.14所示，求证111333110221002103312033(;)(;)IXZIXY证明：[(|)][(|)][(|)]10011111121333333011113300122222033PzxPyxPzy[(|)][(|)]PzxPyx(;)(;)IXZIXY