# ch_12 - β1 Β(t = 10 30 Rv(t = 2 6(1 E_2(t = 0 E{9T E{Β(t...

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Unformatted text preview: CHAPTER 12 12β1 Β§(t) = 10 + 30:) Rv(t) = 2 6(1) E{\_2(t)} = 0 E{9T} - E{Β§(t)} - 10 Cx(r) - 25(1) From (12-5) T Li T LL 2 1 1' 1 1β 1 an; E! Cx(-r)(1 - 2T )d't - Tr- J 26(1') (1 - 2T)dr - T -'I' -'r 12β2 The process Β§(t) is normal (note correction) and such that- F(x,x;βr)βr F2(x) as 1' + w (i) We shall show that it is mean-ergodic. It suffices to show that [see (12-10)] C(t)-->0 as 1+Β» Proof. We can assume (scaling and centering) that n: 0 C(0) = 1. With this assumption, the RVs 1.:(1: +1) andAΒ§(t) are N(0,0;1,1;r) where r a r(r) = C(r) is the autocovariance of Β§(t). Hence, 2 2 f(x1,x2;1) = __..L___ exp, - 1 (x - 2rx x +x2) 2n/1-r2 2(1βr2) 1 1 2 2 1 1 2 "β2β2 = -β- exp - (xl - rxz) e 2n/IIEZ 2(1-r2) Clearly, f(x,y) - f(y,x), hence, (see figure) x F(x+dx, x+dx;r) - F(x,x,r) = 2 I f(Β£,x) 1 Ix 3 1 2β ~x2/2 β exp - .__..2. (E-xr) d5 e d V E 2(1βr ) x 11 l-r 173 12-3 12β4 12-5 Furthermore , F2(x+dx) - F2(x) - 2 F(x)f(x)dx Fro- the above and (1) it follows that Ix-rx ___* β(J-173%β Gβ) Hence, r(1:) β->0 as T -> 0 If x(t) is normal, then [see (12-27)] Rβu) can) - RΒ§A+1)RΒ§A-r)+ x 50:) - Β§(t+x)Β§(t) If, therefore,Rx(t) - 0 for [1| >a, then sz(βt) 8 0 for lβlβl > 1+a. If x(t) - a ej (βn+9 then the time-average T * 2 -1- X(t+T)X (t)dt = ejmT Ial 21β ~ ~ .. -T -βββ_..-β--β_β__β__β___β_____-__..--__β-_β-__-_--_ If 3(t) = Β§(t+A)z(t), then szh) - E{Β§(t+l+T)z(t+r)Β§(t+})z(t)} β Riyο¬) and the result follows from (12β5) . ββ__β__ββββ-__β---_-___-β_-β-___.._β_-_β_-_-______-_ 174 12β6 The process 1::(t) = *0: -9) is stationary with mean a and covariance ' 6(1) given by [see (10-176) and (10β177)] T T %Jn(t)dt 5(1) = %β J C(t+r,t)dt 0 O n If R(t+t,t) +n2(t) as 1+Β°Β° (note correction), then C(t +T,t) -β> 0 hence E(r)-β+O T+m 1+Β» This shows that [see (12-10)] . 30:) is ergodic, therefore, c c+Β§ 1 - 1 - a: J Β§(t)dt a; I Β§(t)dt-ββ> n -c βc+Β§ This yields the desired result because for a specific outcome, 60;) - 6 is a constant and c+6 c 1 . 1 2.1m 26 I Β§(t)dt lim 2c J Β§(t)dt (β1:- _c+e c940 _c 12-7 From (9-38) it follows that T T 2 2 4T 0T - I I C(t1,t2)dt1,dt2 = J J C(t+'r,t)d'rdt -T βT D where D is the parallelogram in the figure. Given a > O, we can find a constant to such that 175 IC(t+-r,t)l <e for I-rl >10 (uniform continuity). Furthermore, if C(t,t) < P then 2 2 C (t1,t2) _<_C(t1,t1)C(t2,t2) < P Thus, lcl <e in 1:1 and |c| <P in D2 The area of D1 is less than 6T2; the area of D2 is less than 4101'. Hence 2 To 0T < 5+? β-> e 'r-n- And since 6 is arbitrary, we conclude that 01, + 12β8 It follows from (6-234) with x(t) = x, x(t+A) = y '71 = '72 = 0 012 = 022 = R(0) 1171,02 = R0) (b) The proof is based on the identity EIZIM} = E {ElzlfllM} M =. {5(060} (1) Proof Suppose first that D consists of the union of open intervals. In this case, if xED, then for small 6 the interval (x,x+6) is a subset of D, hence {xsx<x+dx,M}=(xsx<x+dx) for xeD and {ο¬} otherwise. This yields f(le) dx = W = l P(M) F f(x)dx p = P(M) for xeD and 0 otherwise. Similarly, f(x,y|M) = f(x,y)/p for xED and 0 otherwise. From the above it follows that 00 EIEQIyMn = In [La yf(ylx)dYJ fx(le)dx 176 12-9 Β°Β° yf(X.y)f(x) Β°Β° = βββ-β d = f , M (1 =13 M In Lo {mp ydx ID Loytxyl )dy x {y} } If D has isolated points, we replace each xED by an open interval (x-e, x+e) forming an open set Dβ. Clearly, Dβ -+ D as a -βo 0 and (i) follows if at the isolated points xi of D, E{y|xi} is interpreted as a limit. Since E(Β§(t+A)lΒ§(t)} = R(A)Β§(t)/R(O), (i) yields R(A) ; R A E{Β§(t+A)lMl = E{E{Β§(t+A)IΒ§(t)llMl = E { RβEO; gmlM } = m (c) We select for D the interval (a,b) and we form the samples x(nT), x(nT+A) of a single realization of x(t) retaining only the pairs x(t;). x(ti+A) such that a< x(ti)'<b. Using (5-51), we obtain E{Β§(t+,\)| a < in) < b} = 59-) Ir.- .1β N i=1 where )T= E{x(t)|a<x(t)<b}. This approximation is satisfactory if N is large and R(r)=0 (a) From (7-61) with E{w(t)} = cxyo): R",(r) = cx,(A+r)c,,(,\-r) + Cn(r)ny(r) + anu) = C"(r) + cum) (b) It follows from (a) that if one) βΒ» 0 CW0) β+ 0 ny('a) -Β» 0 then C"(r) βΒ» 0 as M βv 00; hence [see (12-10)] the process x(t) and y(t) are covariance ergodic. -β-β--βββ--------__-___β_-__-_-ββββ-β-ββββ_--β---ββ--------ββββ-_ββ--ββ_-βββ-β 12-11 We use as estimate of n the time average in in (12-1): As we know (see Example 12-4) : 2 ββ' β- Evy} n or H We wish to find 1-: such that P(q-e<nT<n+e)=0.95 (a) From (5-88): 0.95 = PUZTβYIISE) S l ' 82β 5 = 53 5 M5 0T2 (7T T (b) If V(t) is normal, then nT is normal; hence, 0.95 = 20 [i] - 1 G [i] a 0.975 is 20975 0T 01' This yields a s 5,, 2 JETS = e. .-_-β----------~-----------------β-ββββ__--β-_β-β------β--β--β-----_β-~β------ 12-12 (a) It follows from the convolution theorem for Fourier series (b) βIM l βNWT [1" ' 1 l0 0 m- 1 With pn as above, wn I 1-i- pu 9n 5 -jnTw sin 5.5m 1 2 P(w)- { e -β wowβPaΒ») sin 0.5mT 11 n--5 12-13 T 1 β3β: s '2 (m) I β Jx(t)e dt g (m) IXT(m) β5 Β«β2? _T and T T - (i) Iβ(u,v) = -21βTJ JR(t1 β t2)e βu t1 +vt2) dtldtz -T -T as in (9-173) and (9-174) yield E{Β§ (w)} - P(w. -w) Var Β§T(m) . |F(m, -w)|2 + |r(w,u) 2 3_E2{Β§T(w)} Var Β§T(0) - 2|r(o,0)|2 - 2{E2{Β§T(0)} The remaining part of the problem is more difficult. We outline the proof (For details see Papoulis, Signal Analysis). _ From (1) and the convolu- tion theorem it follows that Nu v) . sin Tgu +v -u2 sin Ta β nT(u+v-a) a S(v - a)da -w If 8(a)) is nearly constant in an interval of length 1/T, then it can be taken outside the integral sign. Hence, ___Β£__l _βJββl = soΒ» I 3:22:13 β2 1ββ W βa: 2:)β -oo This yields sin 2 Tu P(w,-w) = S(w) < T(w.w) = 3(w) Tm β o wT-W Var Β§ (m) _<_2 E2{Β§ (w)}*βE2{Β§ (m)} wT-NI' 179 12-14 The function T Xc(w) - I ccc)Β§(t)e'3ββtdc -r is the nude: transform of the product 1 Itl <-'I' β(β)Β§r(t) ESTβ) - g o ltl >T Hence, the function 2T8 (m) - |x (Β«0|2 ~T ~c is the Fourier transform of g(t)Β§.r(t) * c(-t))~iT(-t) T-Itl/Z I c(t+ gmβ; +Β§ )c(t β -%)g(t - gm: ~T+|rI/2 _β.--ββ_-..β_β__-β_β-β_β__β-.__βββ-β-β-β--ββ-_-β--.._ββ-β-_ββ_β-_-βββ______β__β--_ 12-15 Since C(βr) = C(r), it follows from (12-28) that for large T, 00 ' Var 13m) 2 iii.β [C(A+T)C(AβT) + elm] d7 Since S(w) is real, it follows from Parsevalβs formula and the pairs C(AH) H 21*βS(w) C(A-r) H eβ5*βS(w) that the above integral equals Iβ [elm S(w)ej"ββS(w) + 520.0] dw _β--_--_-β_-_β-_-β-β_-ββ_-β_--_βββ_~ββ_--β__β__-__β_β_β_β-β_ββββ-β-βββ--β-ββ-- 180 12-16 With c = T - III/2 T r _ .L'Ll. 5(t) = 15(t + 3):}(t - 5) E{5T(r)} - R(T)(1 - T ) (7β37) yields E{Β§(tl)Β§(t2)} β E{g(t1)}E{z(t2)} - 2 _ __ _ _ - R (t1 t2) + R(t1 t2 + r)R(t:1 t2 1) c c 2 2 lo '1β Var Btu) =- J I [R (:1 β :2) + R0;1 - Β£2 + T)R(tl t2 1)]dtldt2 -c -c 2c = [[R2(u) + R(a + T)R(a - 1)](2'1β - ITI - Ia|)da β2c 12-17 Equating coefficients of z" in (12-98), we obtain = + 12-18 R[0] - 8 RH] = 4 From (13-67) 5-8 al K1 0 5 P1 (1 K1)PO 6 -l β6 1 Islamβ)! 181 12-19 From page (13-67) . β1- β; ._1_4_l1 Pa 13 a1 K1 13 P1 13 P K - R[2] - atall] K . .l. 1 2 1 2 144 2 55 2 1 1595 31 m a2 17:: P2 37 _ 1595J< 144 Smlw) From (12-119) 13- q 5 2 s 13-q 5 = o cg = 14-/Β§I = 6.86 2 5 13-q Inserting the modified data 6.1β{, 5, 2 into the Yule-Walker equations (12-82) , we obtain β 2 " 2 , _ β1 β2 a1 ' 4'07 32 1 E2(z) 1 4.072 + 2 32(2) - 1 - 4.072β1 + 2β2 z z e150.62 1,2 Solving (12-91β) we obtain RLhn] - 6.86 6[m] + 3.07 cos 0.62m SL(w).- 6.86 + g; )(3.07 [6(m - 0.62) + 6(m + 0.62)] 12.20 (a) Let 2 2 6901 represent one of the roots of the Levinsion Polyβ nomial Pn(z) that lie on the unit circle. In that case Pm (β¬901) : O and substituting this into the recursion equation (12-177) we get RH (ejai) 8 : ββA/ββ-βββββ~T-β __ l n: PM W) so that . 5n = 630β. Let RH (630) 2 3(9) ewe) and since [371-1(2) is free of zeros in g 17 we have R(6) > 07 O < 6 < 2777 and once again substituting these into (12β177) we obtain 1 _ 53113β (690) : 53(9) 8mm) _ 8j(t9+o<)ej(nβ1)9 Mg) 5M0) = 3(9) [awe _ new) WW] : 23' RM) e-7(β9+0β)/2 sin (2M0) β β . Due to the strict Hurwitz nature of Pn-1(z), as 0 varies from 0 to 27?, there is no net increment in the phase term 2M0)7 and the entire argument of the sine term above increases by mr. Consequently 13710330) equals zero atleast at n distinct points 61702, - ~67Z7 O < 01- < 271 However Pn(z) is a polynomial 0d degree n in z and can have atmost n zeros. Thus all the above zeros are simple and they all lie on the unit circle. (b) Suppose Pn(2) and inο¬z) has a common zero at z 2 20. Then |z0| > 1 and from (12β137), we get 20 8n Pn_1(20) βββ O which gives Sn : 07 since Pn_1(30) 7Γ© 07 (124(2) has all its zeros in |zl < 1). Hence 3n 7E 0 implies Putz) and Pn-1(z) do not have a common zero. 183 12.21 Substituting 8n 2 pβ, [pl < 1 in (12-177) we get V 1 β pgβ PM) = [Duβ1(2) β (2/3)" Rial/5*) Let m 2 2p and 121(2) = Pans/p) Γ© Auras) so that the above iteration reduces to MAM) = Awe) -Β» Ami/m = An_1(m) β (rillnnο¬x) From problem (12-20)7 the polynomial An(:c) has all its zeros on the unit circle (since 5n 2 1). i.e.7 \$16 = 636ββ = 2k p. Hence the zeros 2k : (1/p)ej9k or : 1/p. (The zeros of lie on a circle of radius 1 / ,0). 12.22 The Levinson Polynomials Pn(z) satisfy the recusion in (12-177 Deο¬ne s;I : Ans", {AI = 17 and replacing 8β by and Pn(z) by in (12β177) we get 1372(2) = Iiiβ1(2) β 28% Pβn~1(zl = 132.103) β (3)0" 8n PΓ©βi1(2)(1/Z*) Let y = 2/\ and deο¬ne A) = An(y) so that the above recursion Simpliο¬es to any) β yβ 8n A2w1(1/βy*) n~1<y> '- ysn 1424(9) AM) I l A A and on comparing with (12β177), we notice that : : Pn()\z). Thus Pn()\Z) represents the new set of Levinsion Polynomials. 184 12.23 (a) In this case 8(6) = IH<ej9>|2 : l1 β W = 2 β β = 2(1 _ 0056) so that To 2 27m 2: β1,7βk : O7 2 2. Substituing these values into (9β196) and taking the determinant of the tridiagonal matrix Tn we obtain the recursion : : 2An~1 M ATL-Z where A0 = 2,A1 z 3. Let 0(2) 2 22:0 A712" so that the above recursion gives 2β2, 1 1 00 n 13(2): <1_Z)2:1_Z ! (1_z)2=Z(n+2)z and hence we get An = n + 2, n 2 0. Using (12β192) and (9β196) we get AH) (_1)n~1(_1)n 1 n: β1β*1 n z : , k>1. S ( ) An_1 n+1 n+l' _ (b) The new set of reο¬ection coefficient : βsk switches around the Levinson Polynomials and 62(3), and hence it follows that they correspond to the positiveβreal function which gives r6 : 2,77; 2 1,19 2 1. 185 ...
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## This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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ch_12 - β1 Β(t = 10 30 Rv(t = 2 6(1 E_2(t = 0 E{9T E{Β(t...

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