通信系统(西蒙赫金)答案.pdf )

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通信系统(西蒙赫金)答案.pdf ) 英文版
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<0 x(t)(x)={1 2 0<x<A A<X and the corresponding probability density function is x(t)(x)=56(x)+.(x-A) which are illustrated below Xe) 1.0 于(x) x比) 0 A (b) By ensemble-aver aging, we have E[X(t)]=了xx(t) (x)dx =了x[6(x)+ 2 o(x-A)] dx The autocorrelation function of x(t)is x(T)=EtX(t+t)X(t)] Define the square function Sq (t)as the square-wave shown below: sa (t) 0 0 0 T 2 2 0 Then, we may write R (τ)=E[Asq(t-t τ T t AS Sqr (t-t +t)Sqr (t-t)fr(ta)dt T/2 A了 To/2T(t t+τ)Sqr〔t d t d 0 (1-2),Irl Since the wave is per ic with period To, Ry(T)must al so be periodic with period (c) On a time-aver aging basis, we note by ins pection of Fig, P/,s that the mean is <x(t)>= A Next, the autocorrelation function 2 <x(t+τ)x(t)〉= x(t+τ)x(t)dt 0-T。/2 0 has its maximum val ue of a/2 at t =0, and decreases linearly to zero at t T/2. Therefore, x(t+)x(t)>=分-(1 Again, the autocorrelation must be periodic with period T (d)We note that the ensemble-averaging and time-aver aging procedures yield the same set of results for the mean and autocorrelation functions. Therefore, x(t)is ergodic in both the mean and the autocorrelation function. since ergodicity implies wide-sense stationarity, it follows that x(t)must be wide-sense stationary Problem 1.7 (a) For It>T, the random variables x(t)and X(t+t )occur in different pul se intervals and are there fore inde pendent. Thus, E[X(t〕x〔t+τ)]=E[X(t)]E[x(t+τ)], τ!>T。 Since both amplitudes are equally likely, we have E [X(t)] =E[x(t+)] A/2. Therefore forlτl>T, R〔τ)=" For It I< T, the random variables occur in the same pul se interval ift.<T -itl.If they do occur in the same pul se interval, Ex(+xGt+=2k2+202 We thus have a condition al ex pectation: E[x(t)X(t+τ)]=A/2, tA<T-hτl A/4, otherwi se vver aging over t,, we get T Rx(r)=∫ T-|τ 2 dt,+∫ 0 T-lt 4T dt d (b) The po wer spectral density is the Fourier tr ansform of the autocorrelation function The Fourier transform of (2)8 T<T =0 other wi se v G(f)=T sinc (fT). Therefore Sx(f)=46(f+“4sinc() We next note that A 6(fdf a A Tsinc (fT)df.A' s(f)df:R、(0)= x It follows therefore that half the po wer is in the dc component Problem 1.8 Since Y(t)=gn(t)+x(t)+73/2 and g(t)and x(t)are uncorrel ated, then p (τ) Where gP T)is the aut oc ov ariance of the periodic com po nent and (τ) is the autocov ariance of the rand om component. C\(r)is the plot in figure P/ 8 shifted down by 3/2, remov ing the dc com ponent gp r)and C(r)are plotted below C(7) ep T 0.5 C( Both g〔t)andx(t) have zero ean。 verage (a) The, power of the period ic component g_(t)is there fore, g(t) dt= (0) 0-T/2p 2 (b)The po wer of the random component x(t)is E[x2(t)=C,(0)=1 Problem 1.9 (a)xr (t)=ErX(t+t )Y(t) Re placingτwith- XY(-T)=ELX(t-t )Y(t) 【实例截图】
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