ECE146B_Problem_set2_solns

ECE146B_Problem_set2_solns - Problem Set 2 Solutions April,...

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Unformatted text preview: Problem Set 2 Solutions April, 2011 5.10 Part a The ensemble average, H E [X E [2 sin (207115 + 9)] 2E [sin (207rt) cos 9 + cos (20st) sin 9] 2 sin (2O7rt) E cos 9 v 2 cos (207%) E sin 9 2 sin (207rt) X 0 » 2 cos (20st) X 0 0 H The auto—correlation function, H E [4 sin (207d. + 9) sin (207m + 9)] 2E [cos (207T (t — 5)) 7 cos (207rt + 207m + 29)] 2 cos (207? (t i 3)) e 2E [cos (207d + 207m) cos 29 i sin (207rt w 207Ts),sin 29] ' 2 cos (207T (t i 8)) i 2 cos (20m + 207rs) Ecos 29 + 2 sin (207d es 2071’s) Esi1129 2 cos (20% (t ~ 3)) E [X (t) X (8)] H H Its is easy to see that Esin9 : Ecos9 : Esin29 : Ecos29 : 0 Part b EX (t) is not a function of t and RX (1375) : E [X (t) X (5)] is just a function oft ~ 5. Therefore, X (t) is a WSS process. Part c Strict Sense Stationary means that ALL the joint statistics of X(t) over any time index set t E T should be invariant to time shifts. Let us examine the first order statistics of X (t): fX (X (t) 2 :13) : 211-6 (x i 2sin (2O7rt)) + £6 (a: i 2 cos (207Tt)) + 56 (a: + 2 sin (207rt)) + %6 (cc + 2 cos (2O7rt)), which clearly depends on t. So clearly X (t) is not stationary. Though the first and second order moments are stationary(as the process is WSS) it is not even first order stationary as the PDF of X (it) depends on t. Part (:1 The time average of X (t), X (t) : liman % f 2sin(207rt + 9) dt 1im7_>00 201W E cos (20m + 9)]:T linle00 201W (cos (—2O7r7 + 9) 7 cos (2071'7' + 9)) iirnTLoO fizg; sin (207FT) cos 9 H ‘As‘ limT_,OO wig??? :7707vvehave that X (t) : EX The time averaged auto—correlation function T X (t) X (t ~ 7') limTH00 % f 4sin (207W? + 9) sin (207T (t i T) + 9) dt ~T T = limT_)DO % f (cos (207T7') + cos (40m ~ 207m" + 29)) dt 7T . r T : limT_)DO % <2T cos(207r7') + [—~—-—Sln(407rt;02737w+26)] T) - ' 40 T—ZO 29 A ‘ ~40 T—ZO 29 : 2COS (207”) + hmTfiw SIM—Wmfim.) a hmqu W 2 cos (207W) Clearly the time averaged mean and auto—correlation functions are independent of the realization of 9. Part e Yes, X (t) is ergodic in mean and auto—correlation as X (t) : EX (t) and X (t) X (t i T) = EX (t) X (t i ’7'). 5.11 _ I Part a Yes clearly lS as f1<7’) 2 [[70505] (T) * I[_O.5y0.5] Part b f2 (T) is not a valid auto—correlation function as it is not even and it does not satisfy the Cauchy—Schwartz inequality: f2 (0) = maXTeR f2 Part c f3 (7) is even and has its global maximum at 7 = 0. So we need to look at its PSD to determine whether it is a valid auto—correlation function or not. 53 (f) : 0.5 (2s‘inc2 (f) # sinc2 (2f) exp (ij27rf) i sinc2 (2f) exp (j27rf)) : (sine2 (f) sinc2 (2f) cos (27Tf)) : 2sinc2 (f) sin2 71]”. Clearly 53 (f) > 0 for all f E R and hence f3 (’1') is a valid auto—correlation function. Part (1 f4 (7) <fi> 2sinc (2f). Clearly 84 (f) < 0 for some f. Hence f4 (7) cannot be a valid auto—correlation function. 5.16 V. Part a H1(f)=j27rf- 51m : 1H1 (m2 Sn (f) :' 492f21[_1,11(f)- Part V H2 (f) : 53W 1 exp( ,77Ffd) V MIPLH gU—OCSEIS)1C4‘D , ' 2 _ (expuflfd)#epr—jvvfd» 1 Zsingwfdyexpkjfifd). 520‘) : 1H2 (f)\ Sn (f) — d 91%) 4m? 5.17 (a) RXm = e-“Wa > 0 2a a2 +47r2f2 00 SX :I::RX age-127116]? : J'O’r e are*./2/’!frdz_+J-fweare—12/gfrdf : 2a SY (f) = 8X (f) I Hm V = JIM] (f) PY : J‘:SY = a2 +i:2f2 df = :arctan(27:lW) (b) Similar as Example 2.6.1 in chapter 2 £3ng (f)df = 09913, = 0.9919(0) = 0.99 B 8 2a 2 2E8 j_BSX(f)df : a2 + 4fl2f2 df = ” arctan( a )= 0.99 [ B B 2a 2 27:13 LBSX (f)df z [3 a2 + M72 df = fl alum a )= 0.99 B=IO.13a ...
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ECE146B_Problem_set2_solns - Problem Set 2 Solutions April,...

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