part34

part34 - b). U 1 0 0 X1 I 4.53a)z= V = 110 X2 gm=1 W 1 1 1...

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Unformatted text preview: b). U 1 0 0 X1 I 4.53a)z= V = 110 X2 gm=1 W 1 1 1 >X3 .4 Y1 = = [1/ X2 = i/ —.X1=i/ —(.’ X3 = LV—X-l— \2=iV—V 12017125117) = bag) = fx(u,v — u.w — v) —A-1£ _ l _E: _(u_u)2/2 —(w—v)2/2 Jig-(wuw) = (“27.1% 2 e e = 1 e_%[2u2+2y2+w2—2uU-2le ( 2M3 .-= ( ; )3e‘[“2+"2+%w2_w_uwl ' 7r 4.76 The solution involves matching the coefficients of the polynomials in the exponent of the Gaussian pdf. If the exponent is given as: —{a:132-.Lby2 +cxy+d$ +ey+f} then from Eqn. 4.79 we must have 1 ‘ 1 2 __ I 2 __ __ 1. coef‘fofx =>_—g(1_p2)a§ -a:>1 '71 _ 2(1—p2)a 2coeflof2=>_1—_b:>2l U2—_1— - y 2(1— WI? ‘ 2 ~ 2(1” pm 3. coeflofmyi—“i‘—=C 2(1—p2)0'102 0' (l 1&2=>—.2 — of b ._2 ‘ 1,-&3: pa=c=>p=— ” g '2 ab To find m1 and m; we solve the following two equations in m1 and m2: d 2m1 V ‘2me 9 = -——\—.-:- —-—(—- = —_am1 — cm? . 2 . 2(1— p2)01 2(1 — p2)0102 "m '7 m .. 2 a ‘1 e = +—p-—-—=—22bm2—cm1 —2(1 — {flag 2(1— P2l010’2 Not all choices of a, b, c, d, e and f lead to legitimate Gaussian pdf’s. An unfortunate error in the first printing resulted in an invalid choice. We have 1 a=— 6:1 0:2 7 then 2 2 1/2 =45 )2— But a valid p must satisfy |p| < l. 50 we see the above choice is invalid. If instead we had been given the exponent 2 2 S ., l —- +—' —.“—i..—S S {316 3y +3”; r 51+} we would have: p _ c _ l _ _) —'—' 77::‘3 -\/ab ah? - 02 1 1 , 1 l : = 0',— :- ‘ 2(1—pzia - 2(1—p2)b i and 2 J: g 4 => m — <3) = 2- H 4.92 €[(X3 —aX1 — ng)2] £[X3(X3 — axl — bX2)] —a — 0X1 -' -bE[X2(X30(1X~1 — , \___—v——I y—v—fi 0 0 since error and observations are orthogonal glxgl — “ElX1X3l—bngX3X2l 2 2 92-1 2 Mil-1’2) 2 = - t a————— 1-9?” 1-9? M = 2_p§-_p‘lp +Pi"PiP22 1-12? _ 2 93-?Pipz+pl+pi-pi‘2 — 0' ———-————-——0' 1—12? 02{1_p¥;(Pi_P2l21 ‘/ 4.95 a) Let X1 and X2 be the zero-mean, unit-variance Gaussian RVs. 1. Generate a Gaussian RV with mean m1 and variance 0%, that is. let Z] = crle + m1. ‘2. Generate a Gaussian RV with conditional mean and variance given in (4.89) . . , a, . Z2 :: ill—pZO‘Q‘XQ+pU—:l21—7n1)+m2 \/l — p202‘X2 + 02pX1+ m2 0'1 0 ‘X1 + ml (72]? 02W 4Y2 m2 “my ll CL ('0 H >7 l y :1 II A2 — 6A 4- 7 — 3 :l: eigenvalues , _. >« m l The orthonormal eigenvectors are: l. l 1 l ‘ _ .38268 .92388 P=[£1-£2l- .92388 —.38268 .30401 1.16342 _ 1/2— A ‘ PD ‘l1.94107 —.48190l 4 _ .80401 1.16342 ‘ ‘ 1.94107 —.48190 Check + .80401 1.16342 00401 1.04107 = 21 I, M 2 1.94107 —.48190 1.16342 —.48190 1 4 V ______________________—————————- 4.102 R2 = X2 + Y2 a) When signal 0 is present R2 X 2 Y 2 ~2=<—> +> 0'0 0'0 0'0 X Y —, a are independent, zero-mean, unit-variance RV s. 122/03 is a chi-square RV with 2 riggrees of freedom. The pdf of R2/ag is uoe—ufl e—u/Z ‘2’r(l) ‘2 The pdf of R3, or 03 . is 1 e—u/2ag e-T7/2ag U0 2 201] Similarly, the pdf of R2 when signal 1 is present —R2 [217% 2 200 8 fmiRZIU = MHZ) = fR2(R210)P(0)+fR2(R2|1)P(l) 6-32/203 e—R’fla? = 203 p+ 20% (l‘p) H b) p: MR2 > “0111(0) + MRQ < THPU) w (Hz/203 T e—R2/2a—12 I 2+ ————1——'dR2T 903 MR [0 20% < p) .a II ...
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part34 - b). U 1 0 0 X1 I 4.53a)z= V = 110 X2 gm=1 W 1 1 1...

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