姚天任现代数字信号处理习题解答第二章答案

12．1已知x是一平稳随机信号，取1、0、-1三个值的概率相等。用x对载波)(nc进行调制后在噪声信道中传输。接受信号为Mnnvnxcny,,1,0),()()(式中)(nv是方差为2v的零均值白色高斯噪声，与x相互独立。上式用矢量表示为vcxy（1）求条件概率函数)/()/(xyfyxf和。（2）由y求x的四种估计：最大后验概率估计xMAPˆ，最大似然估计xMLˆ，最小均方误差估计xMSˆ，最小线性均方误差估计xLMSˆ。并用图形对它们进行比较。解：（1）先求)/(xyf，显然在这种情况下，y是一个1M的正态随机矢量，,][/cxvcxEmxyImmMvTTTxyxyxyvvEcxvcxcxvcxEyyE12///][]))([(]))([()]()(1exp[)2()](1)(21exp[][)2(1)/(222/)1(21221)1(221cxycxycxycxyvxyfTvMvMvTMMI求)/(yxf。)/(yxP=)()()/()(),(yfxPxyfyfyxf已知)1(31)(31)1(31)(xxxxP简记)/()/(ayfaxyf根据全概率公式，得：2)]1/()0/()1/([31)1()1/()0()0/()1()1/()()(xyYPxyYPxyYPxPxyYPxPxyYPxPxyYPyYPyF)]1/()0/()1/([31)()(yfyfyfydydFyf记)1/()0/()1/(ˆyfyfyfA，则AyfyxPAyfyxPAyfAyfyxP)1/()/1(,)0/()/0()1/(31)1/(31)/1(同理：由)/(yxP的分布律，我们可以容易得到)/(yxfAxyfxyfxyfyxf/)]1()1/()()0/()1()1/([)/(（2）求最大似然估计xMLˆ已知：0ˆ)/(ln(xxxyfMLxyccycccxycccxycxycxcxycxyxcxycxyTTMLTTvTTvTvTvMvxˆ0)(1])()([21)]()(21[)]}()(21exp[)2ln{(ˆ2222212解得：求最小均方误差估计xMSˆ3)2(2)2(2]2exp[]2exp[]exp[]2exp[]2exp[2,2,]exp[]exp[]exp[]exp[]exp[]21exp[)]2(21exp[)]2(21exp[)]2(21exp[)]2(21exp[)]2(21exp[1)]2(21exp[1)]1/()1/([1)]1()1/()()0/()1()1/([)/(22222222222222yachyashyayaayayayaycccaccycyccycycycycyyycccyyycccyyycccyyycccyyAycccyyAycccyyAyfyfAAxyfxyfxyfxdxyxxfexavTvTTTvTvTvTvTvTTvTTTvTTTvTTTvTTTvTTTvTTTvML则原式则令代入将求线性均方误差最小估计xLMSˆ已知)]([)])[var(,cov()(1ˆyEyyyxxExLMS①0)(xE，②TxTTTTTcvxcxxEyExEyxEyEyxExEyx2)]([)()()(]))())(([(),cov(③IMvTxTTTTccvcxvcxEyyEyEyyEyEy122)])([()(]))())(([()var(将IIMˆ14212221121][])1[()][var(vTxxvTxxvvTxxvIcIcIccIIcIcIy利用矩阵反演公式④yyEy)(yccccccyccccyccccccyccccccycccccyEyyyxxExvTTTvTxvvxTvTxvTxvTxxTTvTxvvxvTxvTxvTvxvTxLMSx22222222224222222222222242221)()(][][]1[)]([)])[var(,cov()(ˆ题2。2解：以知).()()(,)(4sin)(2nvnsnxnvnnsv的白噪声是方程差为，设)]3(),2(),1(),([)()],3(),2(),1(),0([)(nxnxnxnxnhhhhnxhIssvvssxxR2)]()([)]()([)]()([)]()([vTTTTnnEnnEnnEnnETTnsnsnsnsnsnsnsnsEnnsE)]3()(),2()(),1()(),()([)]()([sP]4sin4)3(sin,2sin21,4sin4)1(sin),4([sin21)(221nnnnnnnvRPh取]0,21,21,21[21)(32vnnh，则题2．3TTTRRRnsnsnsnsnsnsEnvnsnvnsnvnsnsEnnsE)]2(),1(),0([)]2()(),1()(),()([])]2()2(),1()1(),()()[([)]()([xP5vvvRRRRRRRRRnvnsnvnsnvnsnvnsnvnsnvnsEnnETTx22200000)0()1()2()1()0()1()2()1()0(])]2()2(),1()1(),()([)]]2()2(),1()1(),()([[)]()([xRToptmRhvmR]0914.0,2057.0,5358.0[45.0)(128.0P，所以、已知２．４答案：mmsssssszzzmRmRzzS115.05.0115.05.01zzzSzSsssx15.05.045.1zzzSzSzSwssxx设)1)(1()(12fzfzzSixx5.045.1)1(222iiff25.14.02if14.01)(zzBzzB4.01)(1zzzzzzzBzSsx4.0118.05.08.04.015.05.01)()(111115.08.0)()(zzBzSsx4.064.04.0)4.01(25.15.08.0)()()(1)(1112zzzzzBzSzBzHsxil２．５解：由信号模型可得系统传输函数：)(zH=16.011zSss)(z=)6.01)(6.01(82.011zzzHzHzSww)()()()(zzSSSssvsszsxS61)6.01)(6.01(82.0)()()()(1))((zzzzzzSSSSwssvsvsxx对zSxx进行谱分解:由)6.01)(6.01()1)(1(1)6.01)(6.01(82.0)(1121zzfzfzzzzSxx得18.26.0118.2)1(222fff解得可行解21032f)6.011031)(6.011031(2)(11zzzzzSxx116.013.01)(zzzB对)()(1zBzSsx进行因果和逆因果分解：zzzzzBzSsx3.016.01)6.01)(6.01(82.0)()(11zz3.013.06.0111因果部分116.011zzBzSsx..110nmin1111111221|)()(3.05.03.05.0)(3.05.06.0116.013.0121)()()(1)(cusxcssncsxcdzzzzzjnnzzZzZzzzzzzBzzBzSHSHSH=..11116.016.0182.03.015.06.016.0182.021cudzzzzzzzj=..13.06.015.082.021cudzzzzj7=5.0|6.015.082.03.0zz若用nx作ns的估计，则估计误差为dBneEnnunxnsne35.01lg1012２．６解:已知卡尔曼滤波标准形式为：)1|1()()1|1()|(^^^nnSacnxGnnSannS由模型可知：a=0.6c=1G与f的关系为：f=a(1-cG)=0.35.0G将数值代入得：)1|1(6.0)(5.0)1|1(6.0)|(^^^nnSnxnnSnnS物理解释：(1)式中第一部分)1|1(6.0^nnS是对)|(^nnS的预测(2)式中第二部分)1|1(6.0)(^nnSnx是在取得第n时刻的观测值，计算观测值和预测值的误差。(3)系数0.5是对预测误差的修正，以期滤波误差能在最小均方差意义下最小。２．７解：由题意得：a=0.95c=10975.0Q1R二次方程为Q=PPcRPRa22（P取正解）解得：P=0.31225235.031225.31225.2ioPcRcPG)1|1()()1|1()|(^^^nnSanxGnnSannS)1|1(95.0)(235.0)1|1(95.0^^nnSnxnnS：２．１０答案1101)(nA1000)(nQ801)(nC)(mR10001011)1(1001)()()1()()(nSnQnAnSnAnpT1101)(0101)()()()()()()()(npnpnRnCnpnCnCnpnGTT)()(01)(nPnGInS100010)0(S由上述递推公式和初始条件，可得n012Pn111010202.77.66.61.8Gn48.095.06.073.0n10001024.048.048.095.019.381.181.191.2n345pn19.400.500.500.969.219.219.259.3