信息论与编码习题答案.pdf

2.1 试问四进制、八进制脉冲所含信息量是二进制脉冲的多少倍？解：四进制脉冲可以表示 4 个不同的消息，例如：0,1,2,3 八进制脉冲可以表示 8 个不同的消息，例如：0,1,2,3,4,5,6,7 二进制脉冲可以表示 2 个不同的消息，例如：0,1 假设每个消息的发出都是等概率的，则：四进制脉冲的平均信息量symbolbitnXH/24loglog)(1=八进制脉冲的平均信息量symbolbitnXH/38loglog)(2=二进制脉冲的平均信息量symbolbitnXH/12loglog)(0=所以：四进制、八进制脉冲所含信息量分别是二进制脉冲信息量的 2 倍和 3 倍。2.2 居住某地区的女孩子有 25%是大学生，在女大学生中有 75%是身高 160 厘米以上的，而女孩子中身高 160 厘米以上的占总数的一半。假如我们得知“身高 160 厘米以上的某女孩是大学生”的消息，问获得多少信息量？解：设随机变量 X 代表女孩子学历 X x1（是大学生）x2（不是大学生）P(X)0.25 0.75 设随机变量 Y 代表女孩子身高 Y y1（身高160cm）y2（身高 log6 不满足信源熵的极值性。=17.016.017.018.019.02.0)(654321xxxxxxXPX解：2 585.26log)(/657.2 )17.0log17.016.0log16.017.0log17.018.0log18.019.0log19.02.0log2.0()(log)()(26=+=XHsymbolbitxpxpXHiii 不满足极值性的原因是。107.1)(6=iixp 2.7 证明：H(X3/X1X2)H(X3/X1)，并说明当X1,X2,X3是马氏链时等式成立。证明：0log1)/()(log)()/()(log1)/()/()()/()/(log)()/(log)()/(log)()/(log)()/(log)()/()/(2123132121233211231321123221313321123213133211231332112321332113133112321332113213=+=+=exxpxxpexxxpxxpxxpexxxpxxpxxxpxxxpxxpxxxpxxpxxxpxxxpxxxpxxpxxpxxxpxxxpXXHXXXHiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii 氏链是马等式等等的等等是时等式等等当_,)/()/()/()()/()/()()()/()/()()/()/(01)/()/()/()/(321132131232113121212131321213132131313213XXXxxxpxxpxxpxxxpxxpxxpxpxxpxxxpxxpxxpxxxpxxpxxxpxxpXXHXXXHiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii=2.8 证明：H(X1X2。Xn)H(X1)+H(X2)+H(Xn)。证明：.)/()(0);()/()(0);()./(.)/()/()().(21332131221212121312121XXXHXHXXXIXXHXHXXIXXXXHXXXHXXHXHXXXHnnn+=3 )(.)()()().()./()(0).;(32121121121nnnNNnNXHXHXHXHXXXHXXXXHXHXXXXI+2.9 设有一个信源，它产生 0，1 序列的信息。它在任意时间而且不论以前发生过什么符号，均按 P(0)=0.4，P(1)=0.6 的概率发出符号。(1)试问这个信源是否是平稳的？(2)试计算H(X2),H(X3/X1X2)及H；(3)试计算H(X4)并写出X4信源中可能有的所有符号。解：(1)这个信源是平稳无记忆信源。因为有这些词语：“它在任意时间而且不论以前发生过什么符号”(2)symbolbitXHXXXXHHsymbolbitxpxpXHXXXHsymbolbitXHXHNNNNiii/971.0)()./(lim/971.0)6.0log6.04.0log4.0()(log)()()/(/942.1)6.0log6.04.0log4.0(2)(2)(12132132=+=+=(3)1111111011011100101110101001100001110110010101000011001000010000的所有符号：/884.3)6.0log6.04.0log4.0(4)(4)(44XsymbolbitXHXH=+=2.10 一阶马尔可夫信源的状态图如下图所示。信源X的符号集为0,1,2。(1)求平稳后信源的概率分布；(2)求信源的熵H。解：(1)4 =+=+=+=+=+=+=+=3/1)(3/1)(3/1)(1)()()()()()()()()()()()()()()()/()()/()()()/()()/()()()/()()/()()(321321321133322211131333332322222121111epepepepepepepepepeppeppepeppeppepeppeppepeepepeepepepeepepeepepepeepepeepepep=+=+=+=+=+=+=+=+=+=3/123/113/10)(3/13/)()()()/()()/()()(3/13/)()()()/()()/()()(3/13/)()()()/()()/()()(131313333323232222212121111XPXppeppeppexpepexpepxpppeppeppexpepexpepxpppeppeppexpepexpepxp(2)()symbolbitppppppppppppppppeepeepeepeepeepeepeepeepeepeepeepeepeepeepeepeepeepeepeepeepepHijijiji/loglog log31log31log31log31log31log31 )/(log)/(31)/(log)/(31)/(log)/(31 )/(log)/(31)/(log)/(31)/(log)/(31 )/(log)/(31)/(log)/(31)/(log)/(31 )/(log)/()(33333232313123232222212113131212111133+=+=+=2.11 黑白气象传真图的消息只有黑色和白色两种，即信源X=黑，白。设黑色出现的概率为P(黑)=0.3，白色出现的概率为P(白)=0.7。(1)假设图上黑白消息出现前后没有关联，求熵H(X)；(2)假设消息前后有关联，其依赖关系为P(白/白)=0.9，P(黑/白)=0.1，P(白/黑)=0.2，P(黑/黑)=0.8，求此一阶马尔可夫信源的熵H2(X);(3)分别求上述两种信源的剩余度，比较H(X)和H2(X)的大小，并说明其物理含义。解：(1)symbolbitxpxpXHiii/881.0)7.0log7.03.0log3.0()(log)()(=+=(2)5 symbolbiteepeepepHepepepepepepepepepepepepeepepeepepepeepepeepepepijijiji/553.0 9.0log9.0321.0log1.0322.0log2.0318.0log8.031 )/(log)/()(3/2)(3/1)(1)()()(2)()(2.0)(9.0)()(1.0)(8.0)()/()()/()()()/()()/()()(21211212221112122222121111=+=+=+=+=+=+=(3)%7.442log553.02log%9.112log881.02log001001=HHHHHH H(X)H2(X)表示的物理含义是：无记忆信源的不确定度大与有记忆信源的不确定度，有记忆信源的结构化信息较多，能够进行较大程度的压缩。2.12 同时掷出两个正常的骰子，也就是各面呈现的概率都为 1/6，求：(1)“3 和 5 同时出现”这事件的自信息；(2)“两个 1 同时出现”这事件的自信息；(3)两个点数的各种组合（无序）对的熵和平均信息量；(4)两个点数之和（即 2,3,12 构成的子集）的熵；(5)两个点数中至少有一个是 1 的自信息量。解：(1)bitxpxIxpiii 170.4181log)(log)(18161616161)(=+=(2)bitxpxIxpiii 170.5361log)(log)(3616161)(=(3)两个点数的排列如下：11 12 13 14 15 16 6 21 22 23 2425 26 31 32 33 3435 36 41 42 43 4445 46 51 52 53 5455 56 61 62 63 6465 66 共有 21 种组合：其中 11，22，33，44，55，66 的概率是3616161=其他 15 个组合的概率是18161612=symbolbitxpxpXHiii/337.4181log18115361log3616)(log)()(=+=(4)参考上面的两个点数的排列，可以得出两个点数求和的概率分布如下：symbolbitxpxpXHXPXiii/274.3 61log61365log365291log912121log1212181log1812361log3612 )(log)()(36112181111211091936586173656915121418133612)(=+=(5)bitxpxIxpiii 710.13611log)(log)(3611116161)(=2.13 某一无记忆信源的符号集为0,1，已知P(0)=1/4，P(1)=3/4。(1)求符号的平均熵；(2)有 100 个符号构成的序列，求某一特定序列（例如有m个“0”和（100-m）个“1”）的自信息量的表达式；(3)计算(2)中序列的熵。解：(1)symbolbitxpxpXHiii/811.043log4341log41)(log)()(=+=(2)bitmxpxIxpmiimmmi 585.15.4143log)(log)(434341)(100100100100100+=7 (3)symbolbitXHXH/1.81811.0100)(100)(100=2.14 对某城市进行交通忙闲的调查，并把天气分成晴雨两种状态，气温分成冷暖两个状态，调查结果得联合出现的相对频度如下：若把这些频度看作概率测度，求：(1)忙闲的无条件熵；(2)天气状态和气温状态已知时忙闲的条件熵；(3)从天气状态和气温状态获得的关于忙闲的信息。解：(1)根据忙闲的频率，得到忙闲的概率分布如下：symbolbitxpxpXHxxXPXiii/964.010340log1034010363log10363)(log)()(1034010363闲忙)(221=+=(2)设忙闲为随机变量 X，天气状态为随机变量 Y，气温状态为随机变量 Z symbolbitYZHXYZHYZXHsymbolbitzypzypYZHsymbolbitzyxpzyxpXYZHjkkjkjijkkjikji/859.0977.1836.2)()()/(/977.1 10328log1032810332log1033210323log1032310320log10320 )(log)()(/836.2 10312log103121035log103510315log103151038log1038 10316log1031610327log103271038log103810312log10312 )(log)()(=+=+=(3)symbolbitYZXHXHYZXI/159.0859.0964.0)/()();(=8 2.15 有两个二元随机变量X和Y，它们的联合概率为 Y X x1=0 x2=1 y1=0 1/8 3/8 y2=1 3/8 1/8 并定义另一随机变量Z=XY（一般乘积），试计算：(1)H(X),H(Y),H(Z),H(XZ),H(YZ)和H(XYZ)；(2)H(X/Y),H(Y/X),H(X/Z),H(Z/X),H(Y/Z),H(Z/Y),H(X/YZ),H(Y/XZ)和H(Z/XY)；(3)I(X;Y),I(X;Z),I(Y;Z),I(X;Y/Z),I(Y;Z/X)和I(X;Z/Y)。解：(1)symbolbitypypYHyxpyxpypyxpyxpypsymbolbitxpxpXHyxpyxpxpyxpyxpxpjjjiii/1)(log)()(218183)()()(218381)()()(/1)(log)()(218183)()()(218381)()()(22212121112212221111=+=+=+=+=+=+=+=+=Z=XY 的概率分布如下：symbolbitzpZHzzZPZkk/544.081log8187log87)()(818710)(221=+=symbolbitzxpzxpXZHzpzxpzxpzxpzpzxpzpzxpzxpzxpzpxpzxpzxpzxpzxpxpikkiki/406.181log8183log8321log21)(log)()(81)()()()()(835.087)()()()()()(5.0)()(0)()()()(2222221211112121111112121111=+=+=+=+=9 symbolbitzypzypYZHzpzypzypzypzpzypzpzypzypzypzpypzypzypzypzypypjkkjkj/406.181log8183log8321log21)(log)()(81)()()()()(835.087)()()()()()(5.0)()(0)()()()(2222221211112121111112121111=+=+=+=+=symbolbitzyxpzyxpXYZHyxpzyxpyxpzyxpzyxpzyxpyxpzyxpyxpzyxpzyxpzyxpzxpzyxpzxpzyxpzyxpyxpzyxpyxpzyxpzyxpzyxpzyxpzyxpijkkjikji/811.181log8183log8383log8381log81 )(log)()(81)()()()()(0)(83)()()()()(838121)()()()()()(8/1)()()()()(0)(0)(0)(22222222222122122121121221211211111121111111211111111211111212221211=+=+=+=+=+=(2)symbolbitXYHXYZHXYZHsymbolbitXZHXYZHXZYHsymbolbitYZHXYZHYZXHsymbolbitYHYZHYZHsymbolbitZHYZHZYHsymbolbitXHXZHXZHsymbolbitZHXZHZXHsymbolbitXHXYHXYHsymbolbitYHXYHYXHsymbolbityxpyxpXYHijjiji/0811.1811.1)()()/(/405.0406.1811.1)()()/(/405.0406.1811.1)()()/(/406.01406.1)()()/(/862.0544.0406.1)()()/(/406.01406.1)()()/(/862.0544.0406.1)()()/(/811.01811.1)()()/(/811.01811.1)()()/(/811.181log8183log8383log8381log81)(log)()(2=+=(3)10 symbolbitYZXHYXHYZXIsymbolbitXZYHXYHXZYIsymbolbitYZXHZXHZYXIsymbolbitZYHYHZYIsymbolbitZXHXHZXIsymbolbitYXHXHYXI/406.0405.0811.0)/()/()/;(/457.0405.0862.0)/()/()/;(/457.0405.0862.0)/()/()/;(/138.0862.01)/()();(/138.0862.01)/()();(/189.0811.01)/()();(=2.16 有两个随机变量X和Y，其和为Z=X+Y（一般加法），若X和Y相互独立，求证：H(X)H(Z),H(Y)H(Z)。证明：)()()/()()()(log)()()/(log)/()()/(log)()/()(0)()()()/(2YHZHXZHZHYHypypxpxzpxzpxpxzpzxpXZHYxzYxzypxzpxzpYXZijjjiikikikiikikkiikikjikik=+=QQ 同理可得。)()(XHZH 2.17 给定声音样值X的概率密度为拉普拉斯分布+=+=+=+=+Q 2.18 连续随机变量X和Y的联合概率密度为：+=其他01),(2222ryxryxp，求H(X),H(Y),H(XYZ)和I(X;Y)。（提示：=20222log2sinlogxdx）解：12 +=+=+=202020220220202222022022222222222222222222222222222sinlog22cos1422cos1log4sinlogsin4logsin4sinlogsin4sinlogsin4)cos(sinlogsin4coslog4log2log)(/log21log log211log2log log)(2log log)(2log)(2log)()(log)()()(21)()(22222222ddrdrddrdrrrrdrrrrxdxxrxrrdxxrrxrdxxrxpsymbolbitererrdxxrxprdxxrxpdxrxpdxrxrxpdxxpxpXHrxrrxrdyrdyxypxprrrrrrrrrrrrrrrcxrxrxrxr令其中：13 eeededededededdderdrddrrdddrdr220222022022022022202202020202202022020202020log212sinlog21log212coslog1log122cos1log2coslog2sinlogcoscossin21sinlog2sinsinlog2sin12sinsinlog1sinlog2cos2log211logsinlog2cos21logsinlog2cos2)2log2(22sinlog1logsinlog2cos2sinlog22coslog2log2=+=+=+=+=其中：bit/symbolererXYHYHXHYXIbit/symbolrdxdyxyprdxdyrxypdxdyxypxypXYHbit/symbolerXHYHxpypryrryrdxrdxxypypccccRRRcCCyryryryr loglog logloglog2 )()()();(log )(log 1log)()(log)()(log21log)()()()()(21)()(22222222222222222222222=+=14 2.19 每帧电视图像可以认为是由 3?105个像素组成的，所有像素均是独立变化，且每像素又取 128 个不同的亮度电平，并设亮度电平是等概出现，问每帧图像含有多少信息量？若有一个广播员，在约 10000 个汉字中选出 1000 个汉字来口述此电视图像，试问广播员描述此图像所广播的信息量是多少（假设汉字字汇是等概率分布，并彼此无依赖）？若要恰当的描述此图像，广播员在口述中至少需要多少汉字？解：1)symbolbitXNHXHsymbolbitnXHN/101.27103)()(/7128loglog)(65=2)symbolbitXNHXHsymbolbitnXHN/13288288.131000)()(/288.1310000loglog)(=3)158037288.13101.2)()(6=XHXHNN 2.20 设是平稳离散有记忆信源，试证明：NXXXX.21=)./(.)/()/()().(12121312121+=NNNXXXXHXXXHXXHXHXXXH。证明：)./(.)/()/()()./(log).(.)/(log)()(log)()./(log).(.)/(log).(.)(log).(.)./()./()(log).(.).(log).(.).(1212131212111122112122111111122112122111221121112121122121+=NNiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiNXXXXHXXXHXXHXHxxxpxxxpxxpxxpxpxpxxxpxxxpxxpxxxpxpxxxpxxxpxxpxpxxxpxxxpxxxpXXXHNNNNNNNNNNNNNNNNNNN 2.21 设是N维高斯分布的连续信源，且XNXXXX.21=1,X2,XN的方差分别是，它们之间的相关系数22221,.,N),.,2,1,(0)(jiNjiXXji=。试证明：N维高斯分布的连续信源熵 15 =NiiNcceXXXHXH2212log21).()(证明：相关系数()()jiNjixxji=,.,2,1,0，说明是相互独等的。NXXX.21=+=+=+=NiiNNcccciicNcccNcceeeeXHXHXHXHeXHXHXHXHXXXHXH122222121221212log21 2log21.2log212log21 )(.)()()(2log21)()(.)()().()(Q 2.22 设有一连续随机变量，其概率密度函数=其他00)(2axbxxp(1)试求信源X的熵Hc(X)；(2)试求Y=X+A(A 0)的熵Hc(Y)；(3)试求Y=2X的熵Hc(Y)。解：1)symbolbiteabXHbaaFbxxFeababxdxxbbdxxxfdxxfbdxbxxfdxxfxfXHcXXRRRRRc/log32log)(13)(,3)(log92log log2log log)()(log log)()(log)()(33333222=Q 2)=+=+RRRRRcAyAYAydAyAybbdyAyyfdyyfbdyAybyfdyyfyfYHAybyFyfAybdxbxAyXPyAXPyYPyFAayAaAyax)()log()(2log )log()()(log )(log)()(log)()()()()()(3 )()()()(00222232Q 16 symbolbiteabYHbaAaFAybyFsymbolbiteababcYY/log32log)(13)(,)(3)(/log92log 33333=+=Q 3)symbolbiteabYHbaaFybyFbaeababeababydyybbdyyyfdyyfbdyybyfdyyfyfYHybyFyfybdxbxyXPyXPyYPyFayayaxcYYRRRRRcyY/1log32log)(13)2(,24)(329log92log 8log928log log48log log)()(8log 8log)()(log)()(8)()(24 )2()2()()(202003333333322223202+=+=QQ 17 17 3.1 设信源=4.06.0)(21xxXPX通过一干扰信道，接收符号为 Y=y1,y2，信道转移矩阵为43416165，求：(1)信源X中事件x1和事件x2分别包含的自信息量；(2)收到消息yj(j=1,2)后，获得的关于xi(i=1,2)的信息量；(3)信源X和信宿Y的信息熵；(4)信道疑义度H(X/Y)和噪声熵H(Y/X)；(5)接收到信息Y后获得的平均互信息量。解：1)bitxpxIbitxpxI 322.14.0log)(log)(737.06.0log)(log)(22222121=2)bitypxypyxIbitypxypyxIbitypxypyxIbitypxypyxIxypxpxypxpypxypxpxypxpyp 907.04.04/3log)()/(log);(263.16.04/1log)()/(log);(263.14.06/1log)()/(log);(474.06.06/5log)()/(log);(4.0434.0616.0)/()()/()()(6.0414.0656.0)/()()/()()(222222221212122212221211121122212122121111=+=+=+=+=3)symbolbitypypYHsymbolbitxpxpXHjjjiii/971.010log)4.0log4.06.0log6.0()(log)()(/971.010log)4.0log4.06.0log6.0()(log)()(22=+=+=4)symbolbitYHXYHXHYXHYXHYHXYHXHsymbolbitxypxypxpXYHijijiji/715.0971.0715.0971.0 )()/()()/()/()()/()(/715.0 10log)43log434.041log414.061log616.065log656.0()/(log)/()()/(2=+=+=+=+=+=Q 18 5)symbolbitYXHXHYXI/256.0715.0971.0)/()();(=3.2 设二元对称信道的传递矩阵为32313132(1)若P(0)=3/4,P(1)=1/4，求H(X),H(X/Y),H(Y/X)和I(X;Y)；(2)求该信道的信道容量及其达到信道容量时的输入概率分布；解：1)symbolbitYXHXHYXIsymbolbitXYHYHXHYXHXYHYHYXHXHYXIsymbolbitypYHxypxpxypxpyxpyxpypxypxpxypxpyxpyxpypsymbolbitxypxypxpXYHsymbolbitxpXHjjijijijiii/062.0749.0811.0)/()();(/749.0918.0980.0811.0)/()()()/()/()()/()();(/980.0)4167.0log4167.05833.0log5833.0()()(4167.032413143)/()()/()()()()(5833.031413243)/()()/()()()()(/918.0 10log)32lg324131lg314131lg314332lg3243()/(log)/()()/(/811.0)41log4143log43()()(222221212221221211112111222=+=+=+=+=+=+=+=+=+=+=+=2)21)(/082.010log)32lg3231lg31(2loglog);(max222=+=imixpsymbolbitHmYXIC 3.3 设有一批电阻，按阻值分 70%是 2K，30%是 5 K；按瓦分 64%是 0.125W，其余是 0.25W。现已知 2 K阻值的电阻中 80%是 0.125W，问通过测量阻值可以得到的关于瓦数的平均信息量是多少？解：对本题建立数学模型如下：);(求：2.0)/(,8.0)/(36.064.04/18/1)(瓦数 3.07.052)(阻值12112121YXIxypxypyyYPYxxXPX=以下是求解过程：19 ()()()symbolbitXYHYHXHYXIsymbolbityxpyxpXYHsymbolbitypYHsymbolbitxpXHyxpypyxpyxpyxpypyxpypyxpyxpyxpypxypxpyxpxypxpyxpijjijijjii/186.0638.1943.0881.0)()()();(/638.1 22.0log22.008.0log08.014.0log14.056.0log56.0 )(log)()(/943.036.0log36.064.0log64.0)()(/881.03.0log3.07.0log7.0)()(22.014.036.0)()()()()()(08.056.064.0)()()()()()(14.02.07.0)/()()(56.08.07.0)/()()(22222222212222221211112121111212111111=+=+=+=+=+=+=+=QQ 3.4 若X,Y,Z是三个随机变量，试证明(1)I(X;YZ)=I(X;Y)+I(X;Z/Y)=I(X;Z)+I(X;Y/Z)；证明：)/;();()/()/(log)()()/(log)()/()()/()/(log)()()/(log)();()/;();()/()/(log)()()/(log)()/()()/()/(log)()()/(log)();(ZYXIZXIzxpzyxpzyxpxpzxpzyxpzxpxpzxpzyxpzyxpxpzyxpzyxpYZXIYZXIYXIyxpzyxpzyxpxpyxpzyxpyxpxpyxpzyxpzyxpxpzyxpzyxpYZXIijkkikjikjiijkikikjiijkkiikikjikjiijkikjikjiijkjikjikjiijkijikjiijkjiijikjikjiijkikjikji+=+=+=+=(2)I(X;Y/Z)=I(Y;X/Z)=H(X/Z)H(X/YZ)；证明：=ijkkjkikjkjikjiijkkikjikjizypzxpzypzyxpzyxpzxpzyxpzyxpZYXI)()/()()/(log)()/()/(log)()/;(20 )/()/()/()/(log)()/()/(log)()/(log)()/(log)()/()/(log)()/;()/;()/()/(log)()/()()(log)()/()()(log)()/()()/()(log)(YZXHZXHYZXHzxpzxpYZXHzxpzyxpzyxpzyxpzxpzyxpzxpzyxpzyxpZYXIZXYIzypzxypzyxpzypzxpzyxpzyxpzypzxpzyxpzyxpzypzpzxpzyxpzyxpikkikiikkijkjiijkkjikjiijkkikjiijkkikjikjiijkkjkijkjiijkkjkikjikjiijkkjkikjikjiijkkjkkikjikji=+=(3)I(X;Y/Z)0，当且仅当(X,Y,Z)是马氏链时等式成立。证明：0)/;(0 log1)/(log1)/()(log)()/()/()(log1)/()/()()/()/(log)()/;()/()/(log)()/;(2222=ZYXIezxpezxpzypezyxpzyxpzxpzyxpezyxpzxpzyxpzyxpzxpzyxpZYXIzxpzyxpzyxpZYXIikiikijkkjijkkjiijkkjikikjiijkkjikikjiijkkjikikjiijkkikjikjiQ 当01)/()/(=kjikizyxpzxp时等式成立 21 )/()/()/()(/)()/()/()()/()/()()()/()/()()/()/(kjikikjkkjikikjkjikikjkkjkjikikjkjikizyxpzxpzypzpzyxpzxpzypzyxpzxpzypzpzypzyxpzxpzypzyxpzxp=所以等式成立的条件是 X,Y,Z 是马氏链 3.5 若三个随机变量，有如下关系：Z=X+Y，其中X和Y相互独立，试证明：(1)I(X;Z)=H(Z)-H(Y)；(2)I(XY;Z)=H(Z)；(3)I(X;YZ)=H(X)；(4)I(Y;Z/X)=H(Y)；(5)I(X;Y/Z)=H(X/Z)=H(Y/Z)。解：1)()()/()();()()(log)()()/(log)/()()/(log)()/()(0)()()()/(222YHZHXZHZHZXIYHypypxpxzpxzpxpxzpzxpXZHYxzYxzypxzpxzpYXZijjjiikikikiikikkiikikjikik=+=Q 2)(0)()/()();(0 )/(log)/()()/(log)()/()(0)(1)/(22ZHZHXYZHZHZXYIyxzpyxzpyxpyxzpzyxpXYZHzyxzyxyxzpYXZijkjikjikjiijkjikkjikjikjijik=+=+=+=Q 3)22 )(0)()/()();(0 )/(log)/()()/(log)()/(0 1)/(22XHXHYZXHXHYZXIzyxpzyxpzypzyxpzyxpYZXHyzxyzxzyxpYXZjkikjikjikjijkkjikjijkijkikji=+=Q 4)(0)()/()/()/;(0 )/(log)/()()/(log)()/(0 1)/(22YHYHXZYHXYHXZYIzxypzxypzxpzxypzyxpXZYHxzyxzyzxypYXZikjkijkijkiijkkijkjiikjikjkij=+=Q 5)/(0)/()/()/()/;(0 )/(log)/()()/(log)()/(0 1)/(22ZXHZXHYZXHZXHZYXIzyxpzyxpzypzyxpzyxpYZXHyzxyzxzyxpYXZjkikjikjikjijkkjikjijkijkikji=+=Q)/(0)/()/()/()/;(0 )/(log)/()()/(log)()/(0 1)/(22ZYHZYHXZYHZYHZYXIzxypzxypzxpzxypzyxpXZYHxzyxzyzxypYXZikjkijkijkiijkkijkjiikjikjkij=+=Q 23 3.6 有一个二元对称信道，其信道矩阵为98.002.002.098.0。设该信源以 1500 二元符号/秒的速度传输输入符号。现有一消息序列共有 14000 个二元符号，并设P(0)=P(1)=1/2，问从消息传输的角度来考虑，10 秒钟内能否将这消息序列无失真的传递完？解：信道容量计算如下：symbolbitHYHXYHYHYXICmi/859.0 )02.0log02.098.0log98.0(2log )()/()(max);(max222max=+=也就是说每输入一个信道符号，接收到的信息量是 0.859 比特。已知信源输入 1500 二元符号/秒，那么每秒钟接收到的信息量是：sbitsymbolbitssymbolI/1288/859.0/15001=现在需要传送的符号序列有 140000 个二元符号，并设 P(0)=P(1)=1/2，可以计算出这个符号序列的信息量是 bitI 14000 )5.0log5.05.0log5.0(1400022=+=要求 10 秒钟传完，也就是说每秒钟传输的信息量是 1400bit/s，超过了信道每秒钟传输的能力（1288 bit/s）。所以 10 秒内不能将消息序列无失真的传递完。3.7 求下列各离散信道的容量（其条件概率P(Y/X)如下：）(1)Z 信道 (2)可抹信道 (3)非对称信道 (4)准对称信道 ss101 2121212111ssssssss 43412121 3161616131316131 解：1)Z 信道 这个信道是个一般信道，利用一般信道的计算方法：a.由公式=jjijjijijxypxypxyp)/()/(log)/(2，求 j=+=+=+=ssssssssssssss122221212212)1(log)1(loglog10)1