# ch_09 - CHAPTER 9 9'1(a EM“ “an-W“ 1/f2‘ c 0.25 0.5...

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Unformatted text preview: CHAPTER 9 9'1 (a) EM“? “an-W“ 1/f2‘ c - 0.25 0.5 30:, heads) = sinut = 1 t - 0.5 350:, tails) 8 2t = 1 O t - 1 2 F(><,0.25') F591) ‘1. 0.5 a r/t Mr; x 0 7- 9—2 as“) a egt man: a at1 at2 n(t) = Ia fa(a)da R(t1,t2) = I e e fa(a)da From (5-16) with x = g(a) = eta g’(a) = t eta = tx f(x,c) = xllt fa (-i- 2n x)U(x) ' - ——-—--.-—.--—--——-—---——-——--——.---—-.~—.—————————.——-——— 129 As we know, E(x(t)} = At and var x(t) = )th [see (9-18)]. But E{x(9) = 6} by assumption, hence, A = 2/3 (a) E(§(8)} = 24 var 520) = 242 (b) The RV x(2) is Poisson distributed with parameter 2) = 6. Hence, 3 k P(§(2) s 3} = e'” Z (—3—)— k=0 (c) The RVs z x(2) and w= x(4) - x(2) are independent and Poisson distributed with parameter 2A. Hence, - (2A)k _ 2) k 2) m P{E=k)=ezA k! p{£=k,y=m}=e4k(_m)_(_nj)_ P , 5- 3 mm s s I x(2) s 3) = 35—25—21 m2 5 3} = z p{z=k} ~ ~ P(zs3} ~ ~ ~ k=0 3 5-k P{zs3,w55-z}=z Z P{z=k,w=m} k=0 m=0 ~ ~ 15(1'.) = U(t-g) x(t) = 5(t-g) = §'(t) For t1 or t2<0, R(t1,t2) a 0; for t1 and t2>T, R(t1,t2) = 1. Otherwise, 1 aRx 1 E)sz 1 Rx(t1't2)=fmi“(‘1"2) ‘3???”“1' ‘2) " atlatz = 'T' “‘1 ' ‘2) From this and (9-105) it follows that TRy(t1-t )= 6(tl-t2) for 0 <t1,t <T and 0 otherwise. 2 a-bt=0 iff t=t =§/t_>. in (6-63), we obtain 1 ' _ l J. No < E1 <T} = E+Earc tanT (24" are tan 0) 130 9 — 6 The equations 3"(t) _—_ y(t)U(t) 3(0) - y'(0) = 0 specify a system with input v(t)U(t:) and impulse response h(t) = tU(t). Hence [see (9—100)] t E{gz(t)} = th)U(t) * 12100:) = J (t - T)1q(t)dr O ._-—---—._—--_—_--_---—-_-_-—-_-—-———.._-—-—-____-—__-_-_-_----..-_—_—_-___-—--__ 9-7 (a) From (5-88) with 5 - :50: + r) - 15(0) and (8-101): 2 P{|§(C+T) “ 15(t)‘ in} E W 2 a - 2[R(0) - R<r)1/a2 \. V 5 Nb \\\ § The above probability 7, , M _i_‘ ' equals the mass in the \ . 0. \ regions (shaded) ‘ Xith-O‘ xz-x1>a and xz-x1<-a : \ 0' \\ ’ Hence , §V O \\\\\‘ x‘ § ‘ \ X >X +a.\ \\\ \\'\\i \ \ P{|§(t+r) - §(t)l 3a} .. xzq a a .J I f(x1,x2;t)dx1dx2+I J ﬁxrxzzﬁdxldxz -6“ .4” -“ x2+a 131 9—8 (a) The RV x(t) is normal with zero mean and variance E{x2(t)} = R(0)=4, hence it is N(0,2) and P{§(t)s3} = F(3) = G(l.5) = 0.933 (b) E{[§(t+l) - §(t-l)]} = 2[R(0)-R(2)] = 8(1-e-4) 9—9 If x(t) = cej(“’“‘) and 17c = 0 then "x(t) = ’75“an = 0 Rxx(t+7.t) = aczejw" hence, x(t) is WSS. We shall prove the converse: If the process x(t) = cw(t) is WSS, then ﬂc=0 and w(t) = cum”) within a constant factor. Proof nx(t) = ncw(t) is independent of t; hence, nc=0. The function Rn(tl,t2) = ac2w(t1)w‘(t2) depends only on r=tl-tz; hence, w(t+r)w‘(t)=g(r). With r=0 this yields |W(t)|2 = 3(0) = constant w(t) = aemt) w(t+r)w‘(t) = azei[¢(t+f)~¢(t)] Hence the difference ¢(t+r)-¢(t) depends only on r. ¢<t+r)-¢(t) = rm 0) From this it follows that, if ¢(t) is continuous then, ¢(t) is a linear function of t. To simplify the proof, we shall assume that ¢(t) is differentiable. Differentiating with respect to t, we obtain ¢’(t+r) = ¢’(t) for every 7. With t=O this yields ¢'(T) = ¢’(0) = constant ¢(r) = anb 9—10 We shall show that if x(t) is a normal process with zero mean and z(t) = x2(t), then C"(r) = 2an(r). From (7-61): If the RVs xk are normal and E(xk)=0, then 132 Elxixzxsxd = Elflfz} Eifsfd + Elflfsi Elizfd + E‘flid Elfszs} ~~~~ With x1=xz=x(t+r) and x3 = x4 = x(t), we conclude that the autocorrelation of £(t) equals E{x2(t+r)x2(t)) = E2{x2(t+r)} + 2E2 (x(t+r)x(t)) = Rxxz(0) + 211,030) And since Rxx(r)=Cxx(r), and E{z(t)} = Rum), the above yields Cu”) = Rn“) ‘ Ez{£(t)} = 2Cxx2(7) y”(t) + 4y'(t) + l3y(t) = x(t) all t The process y(t) is the response of a system with input x(t) = 26 + y(t) and _ 1 _ 1 m. Since nx = 26, this yields ny = an(0) = 2. The centered process 31(t) = y(t)-n, is the response due to y(t). Hence [see (9-100)] E{'2(t)} - °°h2(t)dt — '—° Z ‘ q L, ' 104 With b=4 and c=l3 it follows that (see Example 9-276) ~1—0e'2'" [ cos3r - —§— sin 3M] + 4 Ram = 104 If u is normal, then y(t) is normal with mean 2 and variance Ryy(0) - 4 = lO/104; hence, R . Ergo» = 0 Ryy(t1.tz)= \$3 = wot-t2) R ' , = o Ru(t1at2) = = 6(tl't2) because Q(t1)6(t1-t2) = 6(t1-t2). 9-14 From (9-181) and the identity 48b: (a+b)2 it follows that 2 1 2 lkxym I s Rxxwmyym) s; [Rxx(o) +Ryy(o)1 Clearly (stationarity assumption) * mm) - g mlz} - Eilyo) - z<o)|2} =- 0 Furthermore , E{§(t+T)[§*(t) - y*<t)1} - Rum - nym and [see (9-177 IE{x(t+r)[§*(t) -2*(c)1}i25u|§(t+1)|2}E{|§*(c)-Z*(c)12} = 0 Hence, Rxx(1') - nyh) - 0; similarly, Ryyh) - nyh) 2 * * E{I§(t+r) - x(t)l }= E{[§(t+r) - gong (Hr) - :5 (t)]} - R(0) - Rm — 2*(1) + 11(0) = 211(0) — 2 E am From 0(1) = N2) = 0 it follows that E{cos?}=E{sin\$} = E{c052§}=E{sin 2\$}= 0 Hence, E{2_c(t)}- cosmt E{cos 9}- sinmt E{sing}= O and as in Example 9-14 2 cos [m(t+r) +g]cos(mt+\$) - cos wr+cos(2mt+wr+22) 2Rx('r) - coswt If g is uniform in (-n,1r), then sinnm a ¢(l) - 45(2) = 0 134 9-17 (at) §(tl)§(t2) =- [15(t1)-§(0)][§(t2)-§(t1)+§(t1)-x(0)] new) - mycl) -:_c(o)12}= 2113(5)} - R<c1.t1) (b) If tl+e <3:2 t1<t2<t1+e then E{[§(t1+E) - 5(t1)][§(t2+e) - 5(t2)]} = q(t1+e -t2) , then Ry(t1,t2) - 0; if Hence, €2Ry(1.') - q(£- ITI) for Irl altz-t1|§e W G E{§(t)y(t)} = J E{§(t)§(t-r)}h(r)dr -@ 9-18 m G = J Rxx(t,t—T)h(r)dt = J q(t)6(r)h(t)dr = h(0)q(t) a —Q W 9—197 As in Prob. 5-14, g(x) = 6+3 Fx(x). In this case, E{§2(t)} = 4, hence, 5(c) is N(O,2) and Fx(x) = G(x/2) __________________________________._____________ 9-20 150:) is SSS, hence, P{x(t) 5y} = Fx(y) does not depend on t. The RVs g and 350:) are independent, hence, [see (6—238)] Fy(y) = P{§(t—§) :yl §=e}= P{:_<(t-e) :yl §=e} = P{§(t -e) <y} = Fx(y) is independent of t. Similarly for higher order distributions. 135 9-21 E{x(t)} = n= constant, hence, [see(9-102)] E{x'(t)} = 0 Furthermore, R (-1) = R (‘1'). hence, R' (0) = O and (10-97) yields xx xx xx E{x(t)x'(t)} = Rxx.(0) = 0 9—22 (3) mg} = Rx(2) = 4e‘4 r152} = H32} = Rx<o) = 4 E{(5+Y)2} = Rx(0) + Rx(0) + 2Rx(2) = 8(1+e"‘) (b) g is N(0,2) P{§ <1} = 172(1) = C(1/2) rm = e“, fzw(z,w) : N(o,o;2,2;e“‘) 9-23 The RV §'(t) is normal with zero mean and variance ' = = - " EH1: ml . Rx.x. (o) R (0) Hence, P{x'(t) :a} = Fx,(a): R"(0)l] 9-24 The function are sin x is odd, hence. it can be expanded into a sine series in the interval (-1.1): Q o(x)5arc sinx=2 b sin nwrx lx!.<.l n=1 n 1 1 b = I u(x)sin n11 xdx = - i- J a(x)d cos nnx n mt -1 -l 1 1 1 a _ W + —. J cos nnx dc (x) mt nn -1 “1 1r/2 — cos n" + —l- ! cos(nﬂ'sinx)dx n n1r ~n/2 and the result follows because [see (9—81)] ‘ "/2 2 Rx(r) 1 r Ry(T) = :T- are sin R (O) Jo(z) 1* ~17 J cos(z sin x)dx x —n/2 As we know [see (5-100) and (6-193)] E{ejw~c(t)} = exp{- M20) (92} jiw1§(t+1)+w 150)] l 2 2 E{e }= exp {- -2- [R(0)w1+2R(T)mlw2+R(0)Lu2]} Hence, with jw = a 2 EH ea§(t)} = exp{§:2- Rx(0)} I E{1ea§(t+‘)1ea’~‘(t’} = Izexp{a [Rxm +Rxm1} (a) Ry(1) = a2E{§[c(t+T)]§(ct)] = a2R(c1) (b) If 56(t) = Elﬁn) then Rz (T) =ERX(ET) [as in (a)]. e If 5 >0 is sufficiently small and ¢(t) is continuous at the origin, then 6 6 J Rz (T)¢('r)d't = ¢ (0) J eRx(et) d1 -5 e —cS 56 w =- ¢(o) J “ware—3.9m) J -e'6 R(T)dT = q M0) .13 Hence, Rz (T)-+q 6(1') as e + 0°. ' e 137 9-27 t Y(t) ‘ J §(T)h(t-T)d1 ' t—T Hence, y(t1) and y(f}) depend linearly on the values of x(t) in the intervals (t1-T, t1) and (t2-T, t2) respectively. If It1-t2I >T then these intervals do not overlap and since E{§(Tl)§(12)} = 0 for 11 # T , it follows that 2 E{y(t1)z(t2)} - 0. 9-28 (a) t t J h(c.a)§<a)h(t.e)§(s) dude} 0 tt t = J J h(t,a)h(t,a)q(a)6(a-8)dud8 - J h2(t,u)q(o)du o o o (b) If y'(t) + c(t)y(t) = §(t), then y(t) is the output of a linear time-varying system as in (a) with impulse response h(t,a) such that 3§§§L91 + c(t)h(t,a) = 6(t-a) h(a‘,a)=() or equivalently 35§%L114-c(c)h(t,a) = o t>() h(a+,o) = 1 This yields t _ c(1)dt h(t,o) - e a Hence, if t 1(t) = I h2(t,u)q(a)da then I'(t) + 2c(t)I(t) s q(t) 0 because the impulse response of this equation equals 1: -2 I c(1)d1 2 e a = h (t,a) 138 (a) If g'(t) + 210:) = 50:). then 2(t) = §(t)*h(t) where h(t) = e-ZtU(t) and with q(t) = 5, (10-90)yie1ds E{y2(t)} = 5*e'4‘um = 5 I e'4Tdr =% o (b) As in (a) with q(t) - 5U(t). Hence, for t >0 4!: t “fun - 511(c)*e"‘u<t) a 5 J e-tnd'r -% (1-e‘ ) O T. {- 0 T t From (9-90) with q(t) - N[U(t) - U(t-T)] ‘ —2 ( ) AN 2 ANIe “*1 dr=T(1-e‘“t) 05t<T 0 0. E{Y2(t)) s '1‘ AN I —2<l(t-T)d . €31 ( 2aT_l)e-2at t >1. 0 a 139 Since x(t) is WSS, the moments of 8 equal the moments of S g - I §(t)dt -5 Hence, (see Fig. 9—5) 5 5 E{§2} - J J to Rxﬁrgugdgzﬂlo- I 1 l )Rx('r)d1' -5 -5 “I0 2 10 —21 E{s} = 80 as = 2 J (lO-T)10e dT y(t) = x(t)*h(t) h(t) a e' 4 (a) E{y2(t)} = 5*e' t U(t) - 5/4 -2t , 2 : _ * = nytt1,t2) 5 6(t1 t2) e U(t2) 5e -2(t -t ) 2 1 U(t2-t1) -2(t -t1) 2 -2t . l " it ‘ = _ k A \ 1,t2) 5e U(t2 :1) e yy U(t1) . ie-zltl-t2 4 The first equation follows from (9-10CD with q(t) - 5; the second frmm (9-94) with Rxx(t1,t2) I 56(t1—t2), and the third from (9-96). (b) With Rxx(t1,t2) - 56(t1-t2)u(t1)u(t2), (9-94) and. (9-96) yield the following: For t1 or :2 <0, ny(t1,t2) = Ryy(t1,t2) = 0. For C <t1 <tq ' -2t I - * ny(t1,t2) 56(1:1 t2) e t1 -2(t1-1) -2(tl-r) Ryy(t1,t2) - J S e e -2t 2 - 5 e 2 140 2 -ar -51 2 2 '— I e e d1 . e 5 Hm I e-a(1'+s/2a) deJE 8 32/40: This yields e-OLTZ 1 e-mZ/lou (1 11(1) at E{§(t+t)§(t)} = J J x1):2 f(x1,x2;'r)dx1dx2 5(a)) - I R(1)e_jmd‘r =- I e-jmJ I x11:2 f(xl,x'2;1')dx1 d):sz —n —m _@ The process y(t) = §(t+a) - x(t-—a) is the output of a system with input x(t) and system function H(u) = ejaw-e-jaw a 2j sinaw Hence [see (9-150)] Sy(m) = 4 sinzaw Sx(w) = (Z-ejzaw - e_j28w)Sx(w) Ryh) = 2 Rx(r) - Rx(’t+2&) - Rx('r-2a) 141 Since S(m) :0, we conclude with (9—136) that D 11(0) - am a 51; JS(w)(l-cos (or) dm -‘D Q I _>_ S(m) (1 -cos 2am) dw =%— [R(O) — R(2r)] -cc and the result: follows for n=1. Repeating the above, we obtain the general result. From (6—197) E{§2(t+r)§2(t)} = E{§2(t+1)}E{§2(t)} + 2E2{§2(t+1)§2(t)} Hence , Kym - Rim) + 2 Rim a 12(1+e‘2""Tl +e'zal‘lcos 281) 4a 2a 2a _2_‘2' + + 4a +m 4u2+(w-Zp)2 402+(w+2P)2 Sy(w) - 21r6(w) 4- Furthermore, =1-:{2 }=R(0) C()=2R2() 1"y f (t) x y T x T m jam 2 m * ]w(Tl-Tk) S(w) a e 1 dw = J S(m) a,a e du) _l E i _m LR 1 k 142 1 1 (3) 8(5) = =———-—-—-——-— 1+54 (32+V25+1)(32—v’25+1) A special case of example 9-27b with b = ff, c = 1. Hence, 1 e-‘Tl/E I III R = ._....+ ' (1') 375‘ (cos 5 Sin vIi.) -2|T| (b) From the pair e H 4/(4+w2) and the convolution theorem it follows that e-2|r| * e-2|1:| “ 162 2 (6+u ) Hence, for 1 >0 °° 0 16 Rm . I e--2|x| e-ZIt-xIdx III e21: e-2(1—x)dx T a ‘22 + J e.2x e-2(T-x) dx + I e“2x euT—x) dx 8%9. (1+ 2") 0 I And since R(-r) - 11(1), the above yields e-Zl‘rl 1+§|1| 1 2 2 3 (4+w ) a: * * jw'f Hl(-s)s=jw=H(jm) H(1/z)z‘eij=H(e ) Hence H(s)ﬁ*(—s ) =- IH(jm)I2 H(z)H*(1/z*) = |H(ej“’T:-|2 s=jm z=ij 143 9—41 From (6-197) Rym - E{§2(t+T)§2(t)} - E{§2(t+1)}z{§2(c)}t+2 E2{§(t+r)§(t)} - R:(0) +2 Rim From the above and the frequency convolution theorem it follows that syn») - 2nR:(o)s<m) + % 5x0») * sxm) 51“!) I ﬂy: 0 u; «I l-Zw; 0 2w; u l ) a S’UJ, 5“”) 2(uz-w.)/Ir U ‘0’: -u‘ 0 a, w: -2wz -2u. 0 2w, 2143 (M 9—42 y(t) = 2x(t) + 3m) nx = 5 Cum = 4e-2M The process y(t) is the output of the system H(s) = 2+35 with input x(t). Hence, ny-5H(O)=10 S c(w) = S,mc(w)|2+3'w|2 = ——216 (4+9w2) -- 144 ——2512 = S w 21r 26 w W J 4+0) — 4+w ’y( )— ﬂy ( ) 144 9—43 (a) y'(t) + 3y(t) = x(t), Rum = 56(1). The process y(t) is the output of the system H(s) = 8—1—3 h(t) = e'3‘U(t) Hence, [see (9-100) and (9-150)] calm E{X2(t)) = 5 Jje‘atdt = 5 5 S¥y(w) = W Ryy(T) = *6— 9'3“ (b) As in Example 9-18: 5 6 E{Zz(t)} = 5 I; e‘sada (l—e'“) t > 0 ny(t1,t2) = 5e‘2't2't1|U(t1)U(t2)U(t2-t1) 5 -§~13(1: t2.) 5 Rxﬁﬁﬁ) 9—44 We shall show that: If x(t) is a complex process with autocorrelation R(r) and |R(rl)|=R(0) for some 11, then R(r)=ej“’o'w(r) where w(r) is’ a periodic function with period 71. Furthermore, the process y(t) = e'jwo‘x(t) is MS periodic. Proof Clearly, R01) = R(0)ej¢. With wo = 43/11, Ryym = E{§(t+r)e-i~o(*+’)§‘(t)eiwo') = R(‘r)e'5“” Hence, Ryy(rl)=e'j“’o’lR(rl) = R(0) = Ry,(0). From this and (10-168) it follows that the function w(r) = RWU) is periodic. ————-_-_——————-—__-..-—--_--———————_---——---—--_—-——-_-_---__—_--_-_—---—--____ 9—45 (a) The cross spectrum Six(w):.-jsgnw Sxx(w) is an odd function. Hence, E{§(t)%'(t)} = %% J sgnwsxx(w)dw = 0 (b) The process i(t) is the output of the system (-1 sgnw) (-1 33m»): - 1 with input §(t). Hence, 3(t) - - §(t). 9—46 In general ngm} = 2% I Sx(w)lH(w) lzdw 5 lH(wm)l2 2—11; I Sx(w)dw = E{§2(t)}|n(mm)|2 -G where IH(wm)| is the maximum of IH(w)I. In our case, ‘H(m)|2 = _____l;_____ is maximum fortn- /§ (5—w)2-+4w2 and IH(wm)l2 = 1/16. Hence E{12(t)}'510/16 with equality if Rx(10) - 10 cos f3" r (Fig. b). “4(W)l 9%(fwj 0 vif u: ..q37 0 V; (a La) (4') 146 jw r 9 47 If Rx(r) :: e o , then Sx(w) = 2w6(w-w°), hence, the integral of 5x0») equals zero in any interval not including the point w= 000. From (9-182) it follows that the same is true for the integral of Sxy(w). This shows that Sxy(w) is a line atln= we for any 2(t). 9—48 (3) As in (9-147) and (9-149) f .. Ryx(r) - Rxx(r) *hlr) - J eja(T Y)hhr) dY - ejaTH(a) .ﬂ - - 2 Ryy(T) = Rxx(r) *P(T) ‘ J eja(1 Y)?(Y)dY = eJaT|H(a)l (b) As in (9-94) and (9—95) -jet ” ja(t -v) j(atg-8t ) R (t1,t2) = e 2 I e 1 h(y)dy = e 1 I 2 H(a) .61 - 8(t ‘Y) 3(at -Bt ) j 2 1 2 H(G)H*(8) —jut1 Ryy(tl,t2) = e H(a) I e * because h(t) is real and H(-B):: H (8). 9—49 If S (w)S (m) E 0 then S (w) = 0 or S (w) = 0 in any interval xx yy xx 1 yy (a,b). From this and (10468) it follows that the integral of Sxy(w) in any interval equals zero, hence, Sxy(w) E 0. 147 9-50 This is the discrete-time version of theorem (9-162). From (9—163) E2{(x[n+m+1] - x[n+m])x[n]} SE{|}£[n+m+1] — §[n+m]}2}E{l§[n]l2} (R[m+11 - R[m])2§2(R{0] - ammo] = 0 Hence, R[m-+l] 8 R[m] for any m. 9-51 We shall show that 2 2 Eé-gd} - R[O] 5M2] inc] (1) The covariance matrix of the RVs §[n], §[n-+1], and §[n~+2] is non- negative [see (7-29)]: R[0] R[1] R[2] R[1] R[0] R['l] _>_ o R[2] R[l] M0] This yields 2 2 3 2 R[0}R [2] - 2 R [1]R[2] - R [0]-+2R[0]R [ll :0 The above is a quadratic in R[2] with roots M0] and -R[0] +2 RZIII/R [01 Since it is nonpositive,R[2] must be between the roots as in (i) 9—52 If §[n] s Aejmi’T then 0 _ f 'Rxnn] a A23{ej(m+“)9Te j“‘~‘-’T} - A2 J em"T f(w)dw ‘0' But [see (9-19A) ] U R[m] = a}; I sx(m)ejm“’Tdm 0 hence, A2f(w) = Sx(w)/20 WM 148 (a) If y(0) = y'(0) = 0, then y(t) is the output of a system with input 3(t)U(t) and impulse response h(t) such that h"(t) + 7h'(t) + 10h(t) = 6(t) h(0') =11'(o') = o t __ e—St)U(t) h(t) =% (tan2 and with q(t) = 5U(t) 9 (9-100) yields t 2 -2: -5 H: (m -%J (e -e T 0 (b) If y[-l] 8 y[-2] =0 , then y[n] is the Output of a system with input §[n]U[n] and delta response h[n] such that 2 )dr 8h[n] - 6h[n—l]+h[n—2] 8 Mn] h[-l] = h[-2] = 0 _ 1 1 hln] - '2n+2 — 22114-3 Uln] and with qIn] = SUIn]. (10-1 6) yields 2 “ 1 1 2 Eiy [n]} = 5 { ———— - r - RIO 2k+2 22k+3 yln] - {[n]*h[n] (a) E{y2[n]} = s*2"‘fnn1 s o ' -m2 -(m2-m1) nylmlmzl - 56[m1—m2]*2 U[m2] = 5 2 U[m2-m1] -(m2-m1) -m Ryy[m1,m2] - sxz U[m2-m1]*2 1U[m1] The first equation follows from (9-190) with q[n] = 5; the second and third from (9-191) with 3n[m1,m2] - 5 “ml—1112]. (b) ' With Rxx[m1,m2] - 5 6[ml-m2]U[m1]U[mz]_, Prob. 9-258 yields the followmg: For 1111 or 1:12 <0, nylmlmzl - Ryy[m1,m2] = o. For 0 < 1111 < m2 -m -m 2 s 5'x2 2 ny[m1,m2] = 5 5[m1-m2]*2 m1 -(m2-k) -(m1-k) 5x2 2 149 (a) Rxlm1.m2] ' q[m1]6[ml—m2] 2 g g a] I 1 1 E{s } = a E{x n x[k } ‘ n-o k-O “ ‘ ‘ ' N N N 2 . go kzo anak qtnlstn-kl = 20 an qtnl n3 3 n: (b) Rx(t1,t2) = q<c1>s(t1-g2) 2 TT ' E{s } = J J a(t)a(r) E{x(t)x(t)}drdt 00 '1‘ TT = I I a(t)a(r)q(t)6(t-1)drdt 8 I 32(t)q(t)dt 90 0 150 ...
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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_09 - CHAPTER 9 9'1(a EM“ “an-W“ 1/f2‘ c 0.25 0.5...

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