习 题 二
1、求证:Rx(ti,tj)?Cx(ti,tj)?mximxj。 证明:Rx(ti,tj)?E(xi,xj)???xxp(x,x,t,t)dxdxijijijij
Cx(ti,tj)?E[(xi?mxi),(xj?mxj)]???(xi?mxi),(xj?mxj)p(xi,xj,ti,tj)dxidxj???(xixj?ximxj?xjmxi?mximxj)p(xi,xj,ti,tj)dxidxj ?Rx(ti,tj)?mximxj?mxjmxi?mximxj?Rx(ti,tj)?mximxj2、令x(n)和y(n)不是相关的随机信号,试证:若w(n)?x(n)?y(n),则mw?mx?my和?w??x??y。 证明:(1)
222m??E[?(n)]?E[x(n)?y(n)]?E[x(n)]?E[y(n)] ?mx?my2???E[(?(n)?m?)2]?E{[x(n)?y(n)?(mx?my)]2}(2)
?E[(x(n)?mx)?(y(n)?my)]2?E[(x(n)?mx)2]?E[(y(n)?my)2]?2E[(x(n)?mx)(y(n)?my)]22??x??y?2[mxmy?mxmy?mxmy?mxmy]22??x??y
即????x??y
3、试证明平稳随机信号自相关函数的极限性质,即证明: ①当??0时,Rx(0)?Dx,Cx(0)??x;
2②当???时,Rx(?)?mx,Cx(?)?0。
2222证明:(1)
Rx(?)?E[x(t)x(t??)]???x(t)x(t??)p(x1,x2,?)dx1dx22Cx(?)?Rx(?)?mx
Rx(0)??x2(t)p(x,?)dx
?E[x]?Dx2
2Cx(0)?Rx(0)?mx2?Dx?mx2??x
(2)
Rx(?)?limRx(?)?limE[xixj]??????2?E(xi)E(xj)?mx2Cx(?)?lim[Rx(?)?mx]???2?limRx(?)?mx
???
22?mx?mx?04、设随机信号x(t)?Acos?0t?Bsin?0t,?0为正常数,A、B为相互独立的随机变量,且E(A)?E(B)?0,D(A)?D(B)??.试讨论x(t)的平稳性。 解:(1)均值为
2mx?E[x(t)]?E[Acos?0t?Bsin?0t]?E[Acos?0t]?E[Bsin?0t] ?0(2)自相关函数为
Rx(t,t??)?E[x(t),x(t??)]?E[(Acos?0t?Bsin?0t)(Acos?0(t??)?Bsin?0(t??))]?E[A2cos?0tcos?0(t??)?ABcos?0tsin?0(t??)?ABsin?0tcos?0(t??)?B2sin?0tsin?0(t??)]?E[A2cos?0tcos?0(t??)]?E[ABcos?0tsin?0(t??)]?E[ABsin?0tcos?0(t??)]?E[B2sin?0tsin?0(t??)]QA、B相互独立
?EAB?EAEB?0
2故:Rx(t,t??)??cos?0?与起始时间无关 2(3)Dx?Rx(0)????
可见,该信号均值为一常数,自相关函数与起始时间无关,方差有限,故其为一个广义平稳的随机信号。
5、设随机信号x(t)?At?Bt,A、B是两个相互独立的随机变量,且
2E(A)?4,E(B)?7,D(A)?0.1,D(B)?2。求x(t)的均值、方差、相关函数和协方差函数。
解:(1)
mx(t)?E[x(t)]?E[At?Bt2]?E[At]?E[Bt2] ?4t?7t2(2)
Dx?E[x2(t)]?E[(At?Bt2)2]?E[A2t2?B2t4?2ABt3]?E[A2t2]?E[B2t4]?2E[ABt3]?0.1t2?56t3?2t4(3)
2?x2?Dx?mx?0.1t2?56t3?2t4?(4t?7t2)2 ??15.9t2?47t4Rx(t,t??)?E[x(t),x(t??)]?E[(At?Bt2)(A(t??)?B(t??)2)]?E[A2t(t??)?B2t2(t??)2?ABt(t??)2?ABt2(t??)]?0.1t(t??)?2t2(t??)2?28t(t??)2?28t2(t??)Cx(t,t??)?Rx(t,t??)?mx(t)mx(t??)
mx(t??)?E[x(t??)]?E[A(t??)?B(t??)2] ?4(t??)?7(t??)2Cx(t,t??)?0.1t(t??)?2t2(t??)2?28t(t??)2?28t2(t??)?(4t?7t2)[4(t??)?7(t??)2]??15.9t(t??)?47t2(t??)26、若两个随机信号x(t),y(t)分别为x(t)?A(t)cost,y(t)?B(t)sint,其中A(t),B(t)是各自平稳、零均值相互独立的随机信号,且具有相同的自相关函数。试证明
z(t)?x(t)?y(t)是广义平稳的。
证明:
E[z(t)] = E[A(t) cos t + B(t) sin t] = E[A(t)] cos t + E[B(t)] sin t = 0
Rz(t , t +?)= E[z(t)z(t+?)]= E{[A(t) cos t + B(t) sin t][A(t+?) cos(t+?) + B(t+?) sin(t+?)]}= E[cos t cos(t + ?)A(t)A(t+?) + sin t sin(t + ?)B(t)B(t + ?)]
= cos t cos(t + ?)RA(?) + sint sin(t+?)RB(?)= cos?RA(?)D(z) = Rz(0) = RA(0) = D(A) < ?
均值为零、自相关函数与时间t无关、方差有限,故其是广义平稳的
7、设随机信号x(t)?Acos(?0t??),式中A、?为统计独立的随机变量,?在[0,2?]上均匀分布。试讨论x(t)的遍历性。
解:(1)首先讨论x(t)的平稳性
????p(?)??1?,0???2??p(x)?p(?)?x?2?? ? ?0,其它???dx??x??d?m?x(t)?E[x(t)]????x(t)p(x)dx???Acos(?0t??)p(?)???x???x??d???2?10Acos(?0t??)2?d??A2?2?sin(?0t??)0?0m?x(t)?E[x(t)]????x(t)p(x)dx?????Acos(?0t??)p(?)d???2?0Acos(?0t??)12?d? ?E[A]?0?0Rx(t,t??)?E[x(t),x(t??)]???Acos(?0t??)Acos(?0(t??)??)p(?)???x???x??d??E?A2??2???1102?2[cos(?0(2t??)?2?)?cos?0?]d??E??A2??4?2?cos?0??D?A?2cos?0?Dx?R?A?x(0)?D2??
故x(t)是平稳随机信号 (2)遍历性
t无关 与
mxT?lim1Tx(t)dt??TT??2T 1T?limAcos(?0t??)dt?0?mx(t)T??2T??TRxT(?)?lim1Tx(t)x(t??)dtT??2T??T1T?lim[Acos(?0t??)Acos(?0(t??)??)]dt T??2T??T1TA2?lim[cos(2?0t?2???0?)?cos?0?]dtT??2T??T2A2?cos?0? 2?Rx(?)故x(t)不具有广义遍历性
8、随机序列x(n)?cos(?0n??),?在[0,2?]上均匀分布,x(n)是否是广义平稳的?
解:由已知得
?1?,0???2? p(?)??2???0,其它①
mx?n??E?x(n)???cos(?0n??)p(?)d?????
???2?01cos(?0n??)d?2?
12??02 ○
?2?0[cos?0ncos??sin?0nsin?]d?Rx(m,n)?E[x(m),x(n)]??cos(?0m??)cos(?0n??)p(?)??????xd??x????2?011[cos(?0m??0n??)?cos?0(m?n)]d?2?2
12??cos?0(m?n)d??04?1?cos?0(m?n)21?cos?0?2
现代数字信号处理课后习题解答
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