信号检测与估值

这是信号检测与估值课后习题答案，是第一章后面的答案，希望对大家有所帮助because, conditioned on g(m) is a constant. We can then writeElg(xy=E(Eg(x)y c =eg(aEy=Eg(x)(19and the desired result follows from Eg. 18Problem 1.7By definition, the joint pdf of x and y+yexpand we have the change of random variable defined by the two equationsu=x cos 0- y sin g6+y cos 6It follows that the Jacobian is unity for any 6. The inverse functions are日+ w sin e0 wcos 6Substituting thesc last two cquations into the joint pdf of x and y rcsults in the joint pdf of u and w as2Problem 1.8Since the mean is zero, the joint characteristic function is重p(u1,u2,u3,4)=expwl, .,waIT is a column vector of variables and R is a 4x 4 autocorrelation matrix whose ithrow and th column IS llij. We know we can find the first moment of the first component by differentiatingp(w by w'l()=重(词)227)where m=1:1, 22, 473, Wi4. To find the second moment such as F(x1x2, we start, hy taking the secondpartial derivativeOw, dwNote thatt hatOw aw重[-m(d)]u)-/12For w-0, this is 412. To find the third moment, we take another partial derivative, resulting in)3()(-y)重()[m1123+m213+m3112一m工示2m3Thc fourth momcnt isdwl dw2dw3aΦ(了)Φ(d)112134+13124+14/12]plus terms which disappear when w=0. This gives the desired result