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# ch_11 - 11—1 I1—2 11-3 W 5/9 10/3(z 1/z 5-2(z 1h)‘...

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Unformatted text preview: 11—1 I1—2 11-3 W 5/9 10/3 - (z-+1/z) 5-2(z+1h)‘ 3x“) " 10-3(z+1/z) 2 —+ 3 - 3 -1 S( ): 34+64 a:s2+lcs+8 82—43+8 x s "4"" “‘2 " "2"" ""' '2' s -Mk +9 5 +45+3 s ~45+3 s2+45+8 s +4s+3 L(s) = —-_-__—___--——__-_—_-_-..-_-_———-——__-_—--_____---_..--.._-_..-___________-.___ First proof °° 2 °° 2 §[n] . X £[n]1[n-k] Eh: [11]} = X 9. [k] k=0 k=0 Second proof w 5(2) = L(z)L(1/z) le] =[[m] *[[-m] = X £[k]9~[k-m] k=0 atol- 2 £2th k'O -——--_—_——----—__-—-——-—————.——_-——_--—--——--—-_-__—-_—_--————--——--—--—---. 166 11—4 (a) This is a special case of (11-22) and (11-23). (b) From (a) it follows that R"1+3R' 1+2R r=61 yx( ) yx( ) yx( ) q ( ) Since Ranch) - 0 for 'r < O, the above shows that R 1' =0fr1<0- R' 0- =0 W() o __ yx() Furthermore , 2 S (s)=_.9____ yx s +33+2 hence (initial value theorem) + 2 R(0)-2.imss(s)=0 (0+)=9.ims S(s)=q yx 9-,.» am Ryx 5%“, yx Similarly, R;y(r) + 3 R;y('r) + 2 Ryy('r) = nyh) = Ryx(—'r) = 0 for 'r > 0 ”(5) = g _gs/12+g/4 +—g\$/12+g/4 (52 +3s+2)(s2 —35+2) 52+3s+2 32—35+2 S+y (s) a gs/1z+9/4 s2+3s+2 R (0+)=R (0)=£im s2 s+(s)=—‘L W W s +o= yy 12 II C 1133(0) = him 3 [5 8+ y(s) _1§2_] 5-H» 11-5 sxm - ssm + s,<z) - 1 q = 1+ D 2) 1.5 s (z) - ——-——-——— ’ 2.5-(2‘1+z) " 1-2.Sz +2 167 11-6 11—7 The process 1 n z[n] = ; 1‘21 §(nT+kT) is the output of a system with input gun] and system function n 3(2) .3: 2 zk “ k-l Furthermore, s - z[0] and n2|H(eij)|2 = E ejkwrlz , k=l 2 _ eij - e:l (”INT I sinzan/Z 1 - eij sinsz/Z Hence [see (9-51)] 2 1 °° sinzan/Z E{§ } - Ry[0] - 2 I Sx(w) -—-§-——- dm 21m a sian/Z -clt1-t2 Since R(t1,c2) - e , (12-58) yields t1 -c(tl-t2) c(t1-t2) I e ¢(t2)dt2 + I e ¢(t2)dt2 - A¢(t1) -a £1 Differentiating twice and using (1) we obtain (omitting details) It 2 M) (t) + (2c - Ac )¢(t) - 0 Hence,’ ¢(t) - Bcos wt and ¢(t) = B'cosw't To determine m, we insert into (i). This yields 2c + msinaw-CCosaw e-ac (ect + e-ct) a Zn Acoswt 2 2 2 2 c +m c +0.) 168 (i) 11-8 This yields 2:; usﬂnau —ccxaw =0 1 = n n n n 2 c+m n The constants 8n are determined from (normalization) a 1 = J 82 coszw tdt B = n n -3 Similarly for PI" sin mt'lt. As in (9-60) 2 T/2 -jm(tl-t2) E{ |§r(m)l } - J Ru:1 — t2)e dtldtz -T/2 T = J (T-|TI)R(T) e~jmtdt -'1‘ Differentiating with respect to‘Iandusing the fact that if t ¢(t) = I f(x;t)dx then _t .t d¢(t) , g; T— s f(t,t) - f(—t,t) +J 3t (x,t)dx -t we obtain 2 T 3E{I (m)' } _ 2 f; = I Rme jmdt = 13%;; I szi } -T The above approaches 5(w) as T + w. _--_-_-_—_—_—__-_-_-———-——-—_-_-_-—-——__-_-_—-—_—-__-_——...____———_-_--——-—_--_. 169 a 11—9 E{x(w)}-15cos 3te-jwtdt = Ssina(m-3) + 5sina(m+3) ~ on -3 w+3 -a Var. X0») - 2qa-4a. —____--—-_—___-_——__--—--____—-__——-__—___———___-_---_..-___..__-_-_————__-——-—- 11—10 E{§(u)§(v)}. 2 2 0:6[n-k1e‘j(““'k")'r “3.09 ken-m . E 02 e-jn(u-v)T -_---___—..___.—_-____..-_——-___—-__—-__——____-—__--—_-_--___.._——-_--————---—__-— 11—11 Shifting the origin, we set 1 T/z _ I T/2 . 0n = — I x(t)e"““’0‘dt ﬁn(a) = — I R(r-a)e"““’o7dr ~ T #172 ~ T -T/2 (a) We shall show that if A m A x(t) = z cneinwot then E(lx(t)-x(t)|2} = o for m < T/2 (i) n=-oo . l T/2 . . Proof 13(ch (0)} = — J. E{x(t)x (a)}e""“’0‘dt = Na) .. .. T 4/2 .. .. The functions ﬁn(a) are the coefficients of the Fourier expansion of R(r—a): R(1—a) = Z ﬁn(a)ej““’°’ m < T/2 (ii) 11:-” Hence mgmga» = 2: mgnfqopmot = Z mww n=-oo n=-oo 170 From (ii) it follows with r = a = t that the last sum equals R(0). Similarly, E{;t‘(t)x(t)) = R(O) and (i) results. "‘ l T/2 " 'nw t l T/Z 'nw t (b) Hing,“ ) =—T— I 15(an (t))e-‘ 0dt= I may:J Odt 'T/z T _T/2 (c) If T is sufficiently large, then 172 . . Tﬂnm) = J. R(T‘0)e’1n"’o"dr 4: S(nwo)e-anoa ~T/2 E{cnc;}- T2 3010’ o) fmexm"‘”°‘da~ 50“” J/T 111‘“ -m 0 maen Thus, for large T, the coefficients cn of an arbitrary WSS process are nearly orthogonal. 11—12 E{§(t1)§*(t2)} = 4—; E { I00 Juno E {~X(u)Z( .(v)}ej(“t1“'t2)dudv = 4&2 E { I00 [00 Q(u)6(u—v)e5(“¢1-V‘2)dudv =' 51? [00 Q(u)e5“(*l'*2)du -oo -00 ~00 This depends only on r = tl-tzz Rn(r) = ijf; Q(u)e5“'du Sum) = 29,522 11-13 Equations (11-79) can be written in the following form: Hf (11):} (v)} = Q(u)6(u-v) = H? (u)§ (v)} H501)? (V)} = 0 for u 20,,v 2 0. We shall show that if the above is true and E{A(w)} = E{~B(w)} = 0, then the process on {(0 = 71— In [§(w)coswt - E(w)sinwt] duo is WSS. Proof Clearly, E{x(t)) = 0 and 171 E(§(t+r)§(t)} 0 0 l °° . . = 17]“) Q(u) [cosu(t+r)cosu t+ smu(t+r)smut] du l 00 = 3—10 Q(u)cosurdu From this and (9-136) it follows that x(t) is WSS with Sxx(w) = Q(w)/7r. 11—14 E{z(t)} = 0 E(§ T((a))} = ITT f(t)e'thdt The above integral is the transform of the product f(t)pT(t), hence (frequency convolution theorem), it equals F(w).sinTw/1rw. T . Var XT(w) = E { I I u(t)e"“"dt ~ _T ~ 2} The integral is the transform of the nonstationary white noise V(t)pT(t). The autocorrelation of this process equals q(t1)6(t1-tz) where q(t) = qu(t). Hence, [see (11-69)] 172 ...
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ch_11 - 11—1 I1—2 11-3 W 5/9 10/3(z 1/z 5-2(z 1h)‘...

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