A 2Tf sin( 2f t) Therefore, sin( 2if t) ElY(t)]= 2nf E[A】=0 sin-(2丌ft) Var[Y(t)]= Var [A] (2Tf sin(2f t) (1) (2If Y(t) is Gaussian-distributed, and so we may ex press its probability density function as Y(t) y A sin(arf t exp[- sin (2Tf t) o A (b) From Eq. (1)we note that the variance of y(t)depends on time t, and so y(t)is nonstationary. (c)For a random process to be ergodic it has to be stationary. Since y(t) is nonstationary it follows that it is not ergodic Problem 1.4 (a) The expec ted value of Z(t)is EIZ(t4 )] cos( art,)E[X]+ sin(2t, )Ely] Since E[X] =E[Y] =0, we deduce that E[z(t,)]=0 Similarly, we find that E[Z(t,)]=0 Next, we note that Cov[Z(t,)Z(t )] E[Z(t, )z(t)] EtIX cos(2t1)+Y sin(2t, )]CX cos(2Tto)+Y sin(2t)J] cas(2t,)cos(2 t)EIX [cos( 2nt sin(2nt)+sin(2Tt )cos(2t)JE[XY + sin( 2it)sin(2It E[Y] Noting that E[X]=σy+E[]}-=1 E[Y ]=o++E[Y] E[XY ] 0 we obtain cov[Z〔t1)Z(t,)]=cos(2mt,)cos(2πt。)+in(2rt1)in(2mt2) cos[2,。)] (1) e七 he process Since ever y weighted sum of the samples, Z(t) is Gaussian, it follows that z(t)is a Gaussian process. Furthermore, we note that z〔t) E[z(t,)]=1 This result is obtained by putting t,=t2 in Eq.(1). Simil arly, o(t。)=E[z(t)]=1 Therefore, the correlation coefficient of Z(t,)and Z(t)1s Cov[z(t,) z(to ) d(t )°z(t) =c。s[2r(t-t。〕] Hence, the joint probability density function of Z(t, and z(t) 1,2(t\(z1, 22)=c explO(z,, 2 z〔t, )3 where 2r1-cos[2π(t,-t,)] I sini2r(t→t) Q(21,z 2 2cos[2π〔t 2sin[2m(七1。)] 1-2)]z122+z (b)We note that the covariance of z(t,)and z(t,) depends only on the time difference t1-t2. The process 2(t)is therefore wide-sense stationary. Since it is Gaussian it is al so strictly stationary Problem 1. 5 (a) Let X〔t)=A+Y(t) where A is a constant and y(t)is a zero-mean random process. The autocorrelation function of x(t)is Ry(τ)=E[X(t+τ)x(t)] EI[A+Y(t+τ〕][A+Y(t)] E[A-+AY(t+τ)+AY(t)+Y〔t+τ)Y(t)] Which shows that R, c X ( T contains a constant component equal to A (b) Let X(t)= A cos(2Tf t+0)+z(t where Ac cos(2f t+e)represents the sinusoid al component of x(t) and 6 is a random phase variable. The autocorrelation function of X(t)is Ry()=E[x(t+τ)x(t)] =E{ A cos(2mfot+2rfaτ+θ)+z(t+τ)[A,oos(2nmft+6)+Z(t)]} =E[Acos(2πft+2rπfτ+θ)cos(2rft+6)] +E[(t+T )A cos(2f t +0)] +ElA cos(2f t+ 2Tf t +6)Z(t)] +E[Z(t+τ)z(t)] A22)c(2nf、τ)+Rn(r) which shows that R(t)contains a sinusoidal com ponent of the same fr equency as x(t) Problem 1.6 (a) We note that the distribution function of X(t)is 5 X