153_Homework_No._3_Solutions_2005

153_Homework_No._3_Solutions_2005 - ECE '53 Harmonia!“...

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Unformatted text preview: ECE '53 Harmonia!“ 6.6 g(t) . 9(t—T) o 1‘ ' o T 1 T+1_’t a) P[Y(t)=1]=PLq(t—T)=1]V ForQ<t<1 P[Y(t)=0]=P[t<T]=1—t=1—P[Y(t)=1] For1<t<2 P[Y(t)=1] = P[t<T+1]=P[T>t-1]=1—(t—1) = 2—t ~ 'P[Y(t)=0] = 1—P[Y(t)=1]=t—1 'P[X(t)r= 0] P[X(t) = 1] __J:>yC:L_+t ___zé:::L_* o 1 2 o 1 2 Mflflm=bPWM=H=HH0=H E[Y(t1)Y(t2)] = ~ [01 E[_q(t1 — T)g(t2 — T)|T = Mfg-(AMA ' = [01 g(t1 — A)g(t2 — A)dA 9(t1 — 1‘) 9(t2 — I“) A t1 — 1 t1 t2 — 1 t2 g(t1 — A)g(t2 — A) = 0 for t1 <'t2 — 1 => RY(t1,t2) = 0 for ‘tg - t1 >1 Ift2—1<t,then ' ' 1 t —1 < X<t 9(11 —‘ A)g(t2 " A) = { 0 eisewhere . 1 Case 1 1 LA 0b;1 t1 1 Ry(t1,t2)=t1—(t2—1)=1—(t2—t1) t1<1, 0(t2—1, tg—t1<1 Case 2 / tg-l 0 t1 1 _ RY(t1,t2) = t1 t1 < 1, t2 < 1, t2 —t1 <1 Case 3 ______[:LA 0 Q—ll n Ry(t1,t2) = 2 —t2 t1 > 1, t2 < 2 6.7 a) We will use conditional probability: P[X(t) S :6] = PlgU-T) S :6] = -/01P[g(t- T) s zIT = «\1fT(»\)d»\ [01.19% — A) 5 mm ‘ since mi) =1 t = / P[g(u) S 1:]du after letting u = t — A t—1 g(u) (and hence P[g(u) S 3]) is a periodic function of u with period, so we can change the limits of the above integral to any full period. Thus ’ P[X(tl) g 1:] = [01 P[g(u) s z]du Note that g(u) is deterministic, so 1u:g(u)$z 0 u:g(u)>z Pmu) s z] ={ So finally 1 P[X(t) _<_:c] = Amwa du=_/;_X1 du=z . 1 b) mx(t) = E[X(t)] = /o :1: d3 = The correlation is again found using conditioning on T: 1 . E[X(.t)X(t + Tn = /0 Ema — T)y(t + r — T)IT. = A]fr(A)dr\ 1 = /0 g(t — A)g(t + 'r — A)dA t . = /t _1 g(u)g(u + r)du~ g(u)g(u + 7') is a. periodic function in u so we can change the limits to (0,1): E[X(t)X(t + 7)] = fol g(u)g(u + mu (u)=(1—u) i(u+*r)=2-—(u+‘r) 1— n+7) 0 1—7' 1 ~ here we assume 0 < 7' < 1 since E[X (t)X (t + 7)] is periodic in 7'. E[X(t)X(t + 7)] = [1H0 — u)(1— u — 'r)du + [1:0 — u)(2 — u — mu 17737-273 3 2+6+2 6 _ 1.1+: ‘ 3 2 2 Thus 1 T T2 1 01‘0"“) — 5747"; __ 1 7+72-. ‘ 12 2 2 6.15 £[Z(t)] = spa + Y] = £[X]t + £[Y] = tmx + my 2 ma) 51(th + Y)(Xt2 + Y)] -- mz(t1)mz(t;) tlt;£[Xz] + (tl + t;)£[XY] + 8Y2 — t1t2m§ —(t1 + t2)mxmy - mix = tltzafig + (t1 + t2)0x0YPXY + 0}” 02(t13 t2) b) From Example 4.32, Z (t) is a. Gaussian random process. az—tmx-mxlz v exp {— 2(t ax+2wxdypr+Vy)} 0.0 = ——_——-——-—_——-_—— 27r(tza§{ + 2t0'X0’YPXY + 012’) 6.18 Z(t) = X(t) cos wt + Y(t)sin Qt V a) £[Z(t)] = mx(t) coswt + my(t) sinwt = 0 Cz(t1,t;) £[(X(t1) cos wtl + Y(t1) sin wtl) (X (t;) cos wt; + Y(t;) sin wt;)] 8 [X (t1)X (t;)] cos wtl cos wt; + £[X(t1-)Y(t;)] costh sin wt; \—__‘,-—_.1 o + £[Y(t1)X(t;)] sin wtl cos wt; + €[Y(t1)Y(t;)] sin wtl sin wt; 0 . C(t1,t;) cos wtl cos wt; + C(tl, t;) sin wt] sin wt; C(tl, t;) cos w(t1 — t;) ‘ ‘ 1 t b) fz(t)(z) = m e—z‘ /20(t,t) 6.27 a) 4’5” (w) = Qx(w)" = (e—QIWI)" = e—MIWI na/1r => fs..(~'€) = m b) Since 5,, has independent and stat. increments fSn,Sn+h(yia 3/2) = f5» (y1)fsn+h—n (3/2 " yl) ‘ na/7r ' ’ lea/1r (1112 + "202) ((112 - 111,)2 + kzaz) 6.33 a) P[N(t) = 0] = e-M b) PlN (t) s 1] = e-’“ + Ate-M 6.40 Y(t) is a random telegraph process with transition rate pa, since if T = time till next transition, then T = 1'1 + + TN where N is geometric RV Ic=1 up 1" up = . ~——-—= . =>T . atea 0—]‘01—229551 01,—”) eXP r P P[Y(t) = +1] = g = P[Y(t) = —1] if P[Y(0)] = a. and CY (t1, t2) = e‘zatlt"‘1l 6.41 Let X (t) be the random telegraph process, then. P[Y(t) = 1] = P[X(t) = 1] =§ P[Y(t) = 0] = P[X(t) = —1]= % mm = ewes Cy(t1,t2) = £[Y(t1)Y(t2)] _ = P[Y(t1 = 1,Y(t2) = 11—; = P[Y(t1) = 1]P[even # transitions in t2 — t1] — i = $0 + Waltz-‘1'); -21. 1 -2alt2-f1| H m 6.49 As in the previous problem, we need only find the mean and variance of Z (t) = X (t) + X (t ,— s) ’ 8mm = 8[X(t)]+8[X(t—s)]=0 VAR[z(t)1 = amt) + m 4 W] = £[X’(t)] + 2£[X(t)X (t — 3)] + £[X2(t —- 3)] = at+2a(t—s)+a(t—s) ' = 4at — 3as e .___z’___ fz(:)(z) = -——-—-—xp{ “(m-3°” } 21r(4at - 3as) b) Set a = —1 in Problem 6.48. = €COS27rt mx(t) = 8M] cos 21rt = 0 Cx(t1, t2) = VARM] cos 2th cos 21rt2 from Example 6.6 ‘ = % cos 21rt1 cos 21rt2 Autocovariance does not depend only on t1 — t2 => X (t,§ ) not stationary, not Wide sense stationary 6.58 a) From Problem 6.18: £[Z(t)] o 02(t1, t2) = CX(t2 — t1.) COS w(t2 —- t1) => Z(t) is wss b) Z (t) is a. Gaussian RV with mean zero and variance Cx(0). 6.63 a) E[X(t)] E[As(t)] = E[A]s(t) RX(t1,t2) ; = E[A2]3(t1)3(t2) CX(t1,t2) = RX(t1‘, t2) — = b) ' X,(t) = X(t + o) I T E[X,(t)] = % f0 mx(t)dt % ATM/1140.122: E[A]§1,— /0T.9(t)dt l T . 121w) T/o Rx(t+7,t)dt % f: E[A2]s(t + T)3(t)3t ll E[A2]% /0Ts(t + T)3(t)dt Thus the mean and autocorrelation of X,(t) are determined by time averages of 3(t). If 3(t) is as below 1 l 0 % T then 1 T 1 1 T 1 T T . T [0 s(t)dt= 5 and T/o 3(t)3(t—T)dt - 5—? [7| < 2 6.77 mz(t) = 0 => mx(t) = 0. Proceeding as in Ex. 6.41: ‘ t sz(t1,t2) = ./t; a e_a(t2—T)Rz(t1,T)dT = ./(')3 c—a(t3-r)a.2e-fi|t1—1-|d1_ We note that: e-BIn—TI = c‘”“"’)' for t1 2 “r cam") for 1%; S 1' We now suppose that t1 5 t2 so the above integral becomes: t2 t1 sz (t1, t2) age—“‘7 cafe-5‘1 cBTdT + cafe/#1 c—firdT] 0 t1 (6““3”! — 1) + The autocorrelation of X is then RX (t1 a t2) ‘1 A ("4‘1")sz (1', t2)dT 2 _at1e_at2 [/41 eaf e'fi‘r(e(a+fi)‘r _ d1- : 0' e ‘ o a + ,6 t 1' (a—fi)t2 (cu—B)? 0 a — 1 1 1 = 2 —at1 —at2 _ 2011 _ _ (a-fi)t1 _ a' e e {a+fl [2a(c 1) a_ (c 1)] a2_'32 _e-at2 -I3¢1 _ c-Bh ‘ah _] Letting t2 = t1 + 1', we see that as t1 —> oo transient efl'ects die out and the autocorrelation becomes RZ(t1 a t2) _’ [c-fiag-41) _ Ec‘a(t2-t1 ...
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This note was uploaded on 11/11/2009 for the course ECE NA taught by Professor Najmabadi during the Spring '08 term at UCSD.

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153_Homework_No._3_Solutions_2005 - ECE '53 Harmonia!“...

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