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ch_13 - -1 (t =a(t L(t-T gang[a(c bg-T)1lg(t.gum R(T/2...

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Unformatted text preview: CHAPTER 13 13-1 §(t -%) =a§(t)+L§_(t-T) gang-)-[a§(c)+bg(:-T)1lg(t).gum R(T/2) - a R(0) +b R(T) a!“ R‘('I'/2) 3 «$1,2 R(0) +R(T) 1+e-1 R(T/2) - a R(T)-+b R(0) r-z{[g(c-§)- §(c-§)1§(c-§>} 2 =R0-RT2- = -_;!:_<_T_/.2_>_=_I_ ( ) a ( / ) bR(T/2) R(0) R(0)-+R(T) l-+e-1 13-2 T J§(t)dt- [8 §(0)+b§(T)].L §(0). §(T) 0 T T J R(t)dt = aR(O) +bR(T) J R(T-t)dt = aR(T) +bR(0) o 0 The above two integrals are equal. Hence, T of R(t)dt a " b ‘ R(0-)_+R(T) 186 13—3 é'm - 81(t)+b§(t-T) §'(t) - [a §(t)+b3§(t-r)_l 350:). 5(t-r) Rs'xm) = a Rxxm) + b Rn“) Rs'x(o) - Rs,s(0) = 0 Rs'xfi) - a Rxxh) + 1: 113(0) R3“) =- Rssh) + RW(T) For small I V .. ’ ~ 2 u Rs,x('r) - 35.50:) - Rash) - rR'ssm) Rum -Rxx(0) + 'r Rum/2 Hence, a s ~b+O(1-2) 1R" (0) - a 12 R" (CD/2 +0(T3) ss xx 13-4 It suffices to show that, for any :11, ot-n E{[§(t) - X MEI-I)- x(nT)]§(mT)} = 0 nus—m The left side equals on R(t-mT) - 2 51“§:t_;‘:r“') R(nT -mT) n:— m From the sampling theorem (10-140) it follows that this is zero because the Fourier transform e‘jst (w) of R(t-'mT) is zero for he] >0. 187 13-5 13-6 Since £{§(u+x)lg(c)} = as(t) a . law/Rm) it follows from the assumption that §(t +1) - a §(t)_L§(t -1') Hence no”) - £795} R(1') (1) The only continuous function satisfying the above is an exponential. This is easily shown if we assume that RO) is differentiable for A >0. Differentiating“) with respect to A and setting 18 0+, we obtain mm + «11(1) - o a. = -R'(0+)/R(O) 1 >0 This yields R(r) - Ie-M for r >0. Given: E{Zn}=o x szl+ooo+y ax Furthermore the RVs yn are independent. Hence, yn is independent of Kai-1, o o o ’ 51. This §131ds E{xnlxn_1. s o o .51} . E{§n—1 +3°n|§n—lg o o o ’51} .—---—----—-—--—---——___..-—-—-—-—-——-__—__———_-_..--—-_-__._-—_——————-—_— —— _ 188 13-7 13-8 (3) If E{§nlgh_1,...,§1} - §n_1. then ' 5n ‘ §n-1’-l-§n—1"'°'§1 From this it follows that the RVs y -x -x are orthogonal and -n -n ~n—1 .fn'zn+§n-l'zn+zn-l+.”+zl 0') Conversely, if (i) is true and the RVs yn are orthogonal, then 25n - En-l g Xn-l-§n-1""’§l 2 2 (b) E{§n} = E{[(§n-§h,)+fn_1] } x2 -n—1 2 2 = E{(§n-§n_1) }+ E£§n_1}3E{ } for any n. The Fourier transform Ss(w) of the function sinzar 2 1 Rs(1) - A is a triangle as shown 189 And since Sv(m) = N,(13-16) yields H (w) a Ss(m) = Aan(1- w /Za) 1 Ss(w) + Sv(m) Aan(1-— w 2a)-+N We show next that 32(w) = ij1(w) Proof gm = §(t) §'(t)_1_§(g) all he. Hence Rum = o. This yields [see (9-131)] R€.x(1) = R;x(r) = 0, hence §'(t)'-§'(t)_l_§(i) ——I-.—~—I——» t) '(t) ' §(t) §(t) Qt) ( s - In other words, the estimate of §'(t) equals the derivative of the estimate §(t) of §(t). This follows from Prob. 13—9 with T(w) = jw. 13—9 We wish to show that the estimator of y(t) = I p(t—a)g(a)da p(t)+-* T0») equals 2(t) 3 I p(t-—a)§(o)da where s(t) is the estimator of s(t). Proof. Clearly mgm —§(t)]§(E)} = 0 all t,£ Hence mm) -§<c)1§<a)} on - J p(t—a)E{[§(a) -§(a)]§(z)}da =- 0 ___.._--_—___..---—_______._-~_-—--———_—__—_.._—._——_--_—--_-.._-_--—_—--—-—-_-_--_. 13—10 [See (13—46) and beyond] 1 1 (a) 8(3) - - 34+]. (32+5s+1)(32-53+1) (1:) Ms) - 1 2 2 Mr) - % e‘“ sinBtU(t) a - B - i (s+a) +3 ’2 (c) h1(t) - % 3‘“ e‘“ sinB(t+A)U(t) 1 «:1 (s +u)sinBA+8cosBA H (s) - - e i B (s +a)2+ 82 b0 - Eax(cosek+ gsinfl) Hi(s) 11(8) 3 W 3 bo+b1 S b _ sin“ «:1 1 e e 191 13-11 (a) The given equation is the Wiener-Hopf equation (13-40) for the prediction problem with A- znZ. We can, therefore, use the method described after (13—46): 2 3 22 49 — 253 8(5) 3 + - —-—-———-— 1-s2 9-.«32 (l-sz)(9-sz) L(s) - 7 + 53 m) - (e't +4e-3t)U(t) (1+8)(3+S) -2.n2 -t -3£n2 —-3t h1(t) - (e e +4e e )U(t) 1/2 4/8 2+5 2+s “1(3) " 1+5 + 3.+s " (l+$)(3+s) 3(3) ’ 7+Ss (b) 5 N(3) N(s) -2 D s) _ : 11(5) = _D(s) Ws-J— L(S)L( S) Y(S) Since Y(s) is analytic for I53 3 < 0, all roots of D(s) must be cancelled by the zeros of L(s), hence, D(s) = 7+SS. Similarly the poles s a---l and s *- 3 of L(s) must be cancelled by the zeros of the term N(s) -ZSD(s). With N(s) =As+B, this yields N(—1)-2_1D(-1) -—A+B—2"1(7-5) = o A . 1 N(-3) -2‘3n(-3) =-3A+n-2"3(7-15) = o B - 2 2+5 “(5) ' 7+Ss (c) The Laplace transform of the function R(t) in (a) equals 49 - 25:;2 9 - 1032 + 34 Hence (convolution theroem), the inverSe transform y(t) of Y(s) equals U y(t) = J h(a)R(t-o)do-R(t+ 1:12) 0 From the analyticity of Y(s) for 35 s < 0 it follows that y(t) .0 for t >0. Therefore, (b) gives a direct method for solving the Wiener—Hopf equation (13-40) . 13—12 (a) The given equation is identical with equation (13-22) for the prediction problem with r81. We can, therefore, use the method in (13-31) - (13-33): 3 8 70-25»: _ l 3(2) 5-2w +1o-3w“ <‘s-‘2w)(10-3w) “4+2 3 I /3—0-+ /5— 27.75 -1 MI) _ 01 -b2 —1 -1 (2-2 )(3-z ) b./3_o_/5‘=3.25 -1 -2 . £2 . - £[o z 00412 ‘001672 JLIO] 6 1.3 11(2) 1 —lL(z) 1 -0'4224. (b) _ N z) N(z) - zD(z) -l _ H(z) D(z) I—_-—-D(Z) L(Z)L(z ) Y(z) Since Y(z) is analytic for [2‘ <1, all roots of D(z) must be cancelled by the zeros of L(z), hence, D(z) = 1-0.42 2-1. Similarly, the poles z=1/2 and z =1/3 of L(z) must be cancelled by the zeros of the term N(z) -zD(z). With. N(z) = A+Bz-l, this yields 1 1 1 ~ '3 ~ 11(3) - 3 11(5)- A+ZB - 0.08 o A—3.42 ml) -lD(-1-)=A+3B+o.09=o B=-O.17 3 3 3 0.42 -’_‘.l7z_1 “(2) = 1-0.422‘1 The 2 transform of the sequence Rm in (3) equals 270-25w w-z+z- 6w - 35w + 50 1 Hence, the inverse transform yu of Y(z) equals n y'Z R_ 'R nk-O nk n+1 From the analyticity of Y(z) for Izl<1 it follows that yn=0 for n30. Therefore, (b) gives a direct method for solving (13-22) . 13-13, A predictor is a stable function H(z) vanishing at 00. Since H(z) —o 0 l as 2 -+ 00, we 13-14 conclude that EN(z) = l-H(z) —+ l and EN(z)H,('z) -» l as 2 —+ 00. From this and (13-25) it follows that the difference 1 - EN(z)H,(z) is a predictor and the MS error equals ' °° IE (eimzswm-P ml-” N because lAa(e5"’)| = 1. As we know, if §[n] = a §[n-1] + ... + amskx-m] + EIn] where §[n] is white noise, then the one-step predictor of s[n] equals A §1[n] =8 iIn-l] +0" +am§ [n-m] We wish to show that the sum §2[n] - a1§1[n-1]+ az§[n-2]+---+am§[n-m] is its two-step predictor. It suffices to show that h gln] - §2[n]_Ls[n-k] k_>_2 Proof sInl - §2[n] = al(§1[n-1] -§1[n-1]) + gln] This completes the proof because s[n-1]-;[n-1] 3(n-k], k > 2 and §[n] s[n-l<] k>1. -1 -1 .. _ .. - 194 13-15 The Nth order MS estimation error PN equals [see (13-66)] This tends to the MS estimation error in (13—34) . Hence, 0 A 21m 2n PN . 5;- I la S(w)'dw - 21m in NA” N—y-oo O O N-vw N To complete the proofs, we use (14-129) N A 2.1m % {tn—“Zil- 9.1m 2n ANA” N+°° nsl n N-Nv N and the result follows because ! N lnb 2.91/sz fix (2“ An+1 ' 2“ An) ' NN+1 " N n-1 and the last term tends to zero as N+°. 195 13-16 Po - RIO] I 15 R[1] I 10 R[2] I 5 RD] = 0 We use Levinson's algorithm [see (13-67)] 1 2 2 25 PoKl - RU] K1 81 I '5 P1 (1 —k1)P0 3 1 5 1 196 13-17 Po I R[O] I 5 K '- 0.4 K 3 0.6 K I' 0.8 1 2 3 1 RH] - 90x1 - 2 a1 - 0.4 P1 - 4.2 1 R[2] - R[1]al + P1K2 - 3.32 2 2 31 - 0.16 9.2 - 0.6 P2 - 2.688 m3] - M2132 + Rlllaz + p K -3 8816 1 2 2 3 ' 3 2 2 3 2 2 3 a3 - o a p = o 953 3 . 3 C N(s+c) AN and (13—104) yields ~21 _ Hx(s) - cs” hx(t) - (c_-21)e “um 197 13-19 (a) ;M[n+m] lab-k] k --m+1,...,0,...,N A eulnl - gln] - algln-ll - ---- aN§[n-N] (b) EMIB-fllig[fik] k --m+1,¢oo,o.ou.,N znln] - gln] - l18[n+1]--oo -aN§[n+N] (c) Emu—M] _L §[n+k] k--N-m+1,...,-N,...,0 Eula] - gln] - algtn-ll - «ugh-N] 13-20 2 m +0.04 w +0.04 x s . (a) From (13-16) 2 no») - -—-2--—- 5w +2.2 (b) From (13-104) 0.1.6 “,5” ' 1 ' “5-25” 'm (C) Using (13-48) Ls“) - “/32 mm - 55"“ um i1( _ ,5 -o.2(c+2) a 2 Jo" 1 t) e U“) 1(5) " a+0.2 Ms) - e'°"' §(c+2) - e’°"‘s(t) 198 (d) [see (13—99) and beyond] 1 8+0 2 0.66-5 . __ I‘ .._._'__ S ( ) = /20 Ssx(s) 0.04-52 x(s) /5_(sl0.66) six 8 s+0.2 Rsi (T) - J56 [f(t) +- 0'86] hi (1) = 0.85/35 e-o'2(t+2)U(t) x x s+0.2 ____0-86 20 e'°" 1_7_2__’°_"1 “1x(s) ' s+0.2 “x(s) ’ s+0.66 13-21 As in Example 13-2 with N 0 58(2) ’ (1-0.8 :Li)(1-0.82) Ls“) I fl '————--“ii:3:: $122232: =-——’§:i;?;:fll (a) H(z) - 9 h[n] - 2 x2‘|“| 40(1-0.52 2'1)(1-o.52) (b) From(13-ll4) with 2x[o] - v’E 3/8 1-o.5 z‘ 1 hxm - 3x0.5 um Hx(2) - 8 (c) From (13-33) with £[0] - V1.8 20 -1 H(z)'1-E-E;}-0.82 199 6 You] El [.9 43 (“v-139.] 3 [m + 1] (d) The power spectrmn of the estimate gob] of §[n] obtained in (b) equals -1 9/8 ).._...__._______._— s“ -S()H<)n(z 30(2) “2 "z x (1-0.8: 1)(1-0.82) Hence ' L; (z) - Therefore, the pure predictor of soln] equals [see (13-33)] A .— 111(z)=-1-i‘-[3]--o.821 L(z) And (13-117) yields -1 A 111(2) . gonna“) . .2:§_E_.:i. " " 1-0.5 z 200 13-22 | RJm1-5xm8mI«—> L8 (1-0.8 z'l)(1-o.8 z) Hence, as in (13-135) with Vn - 1.8, N - 5. And (13-143Dy1'e1ds :1 Fa - 0.64 1='n_1 + vIn cn_l F0 - VONO =- 9 5 Gn a 0.64 Fn-l +6.5Gnu1 G0 a v0+k0 = 6.8 Solving, we obtain Fn - 12(1.6)“ - 3(0.4)“ on = 5.4(1.6)“ + o.4<o.4)“ F n 12 Pa -G— -—-> 3.7- 1.875 :1 11+ “3 This agrees with Prob. 13—210 because the MS error of the Wiener filter equals 9 - 118(0) -2 Rslk]hx[k] - s -{ 5 ><o.3"‘><3xo.5m . 1.875 k-o k-O 8 201 13—23 115(1) - 5 e'o'2|T| RC1) - 359 6(1) 2 10 s (m) - A(t) - 0.2 V(t) - 2 N(t) - -— s m2+0.22 3 From (13-159) F'(t) - 0.2 F(t) +ZG(t) G'(t) - 0.3 F(t) + O.ZG(t) Case 1. If §(0) - 0, then P(0) - F(0) - 0, 6(0) . 1 Solving , we obtain pm .. :92 : W G“) 0.8: -0.8t 0.625 e +0.375e Case 2. If §(t) is stationary, then F(0) - P(0) - 118(0) - 5 60:) 2.5 ems: _ 0.9 e-0.8t 13 -24 The sequences aN[n] and Elfin] are the responses of the filters A N V A EN(z) = 1 - Z asz‘k 12,,(2) = z'NENU/z) k=l respectively, with input R[m] (see Fig. 13-113). Hence, N aN[m]= R[m] - Z R[m-k]akN k=1 N 2mm] = aNm-m] = R[m-N] - )3. R[m-N+k]akN k=1 From this and the Yule—Walker equation (13-65) it follows that A V qN[m] = qN[N-m] = 0 for l 5 m s N -l V ante] = qNIN1= PN This completes the proof. ...
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