ece302hw6solns_Fa11

ece302hw6solns_Fa11 - 4; H W (W t) r;...

Info iconThis preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4; H W (W t) r; extrzg‘ié‘fiycrs‘fif? (1&3{3‘i‘ii’iikflfi arm wig: Kgééflll’ Eit’zfirxinfif w 15%;; :11»; fine” {WEEK}? mt, £23m, (:ngtfifibfiigfi‘m i v w» L: mew‘zm : fifig? a 9Y2ié§3 _ ” §EPCXW $33 @fivmwvuxsnéfif r Problem 2 (text 5.96) We have X1, X2 and X3 are uniform in [—1,1] so that 1 —1 < :c < 1 -— r : ) Z 27 _ — , le — 132(93) 133(56) { 07 otherwise a. Since X1 and X2 are independent, we have that fX1,X2($1, $2) : fX1($1)fX2($2)- We have then that the CDF of Y : X1 + X2 is FM?» : Pfly S 1/] = P7“[X1 + X2 S y] : / fX1,X2($1;$2)d$1d$2 {($1,$2)2$1+x2§y} 00 (11—231 2 / / fX1,X2(f131;$2)de‘2d$1 0 y < —2 Jag/+1 fy—ml id$2d$1 : (y+1)2 _ y2 y E Z —1 —1 4 8 > 1— fyfll fyfim :dxgdml = 1 ~ (2‘32 1/ e (0, 2] 1 y > 2 We then have that the PDF of Y is d ;+Z, yE[—2,0] fy(y) : d Fy(y) : 5— Z, 116(072] y 0, else. b. We find the PDF fz(z) first. Since Z 2 X3 + Y and X3 and Y are independent, from the notes we will have that fZ may be expressed as a convolution: fZ(Z) : fX3*fY(Z) /_ fX3<y>fy<z—y)dx % g§1é+idy +1 2116(z2+6zu9), ze[—3,~1) : éfz2—1:+Zdy+;0 éwidy =§(3*z2)a Nit—1:1) ifznli—Zdy: = 116(22~62~~9), 26 [1,3) 0, else We then have that CDF M2) = z MAMA fZOOOdA ‘ :0 z< —3 116 fj3A2+6A+9dA = 116(233 +322+92+9), z E [~37—1) : 3;“ Us? — A26” 2 2:4(9z- 5’ +12): z E [—171) g“ 16 f1 )2 _5)\+9d)\ : 48(23 —922+27z+21), 2 6 [1,3) 1 z>3. 2’) {‘6 z I / a? Xi)?‘ Problem ‘fiQtext 5.113) Since X and Y are zero-mean, unit—variance independent Gaussian random variables, :> fX,y(f13>y)= mmwy) : 21” exp {—ng + ya} a. From problem 5.108, we had that W :— X2+Y2 and 9 = tan‘1 Y/X. Since W represents the square distance from the origin, we use the result from problem 5.108 to find fig; (11)): 2W1 fWW) Z fW,e(w,9)d€ : /0 4” exp d6 1 extw) 2 p 2 ' Thus the probability that (X, Y) falls inside a circle of radius 7" is equivalent to II Pr[\/X2 + Y2 g 7"] = Pr[X2 + Y2 3 r2] : Pr[W 3 r2]. We want to find 7’ such that this probability is equal to ;, i.e. 7"2 T2 1 w 2 l < 2 : 2 — I — —T /2 Z: - PflW _ 7*] /_OO fW(w)dw /0 2 exp ( 2) dw l e 2 Solving for r: 7“ : \/21n(2). b. Let M denote the event that (X , Y) does not fall inside a ring with inner radius 7’1 and outer radius 7‘2. The complementary event, M, then is the event that (X , Y) do fall in this ring. Let R denote the set of (X,Y) within the ring, so that PT[M] = Pr[(X,Y) g2 By the definition of conditional pdfs, we have fX,Y(93:y) Jami/1M) = ny($,yl(w,y) ¢ 1%) = PTKWRP (W ¢ R ’ ’ 0, (Jamel? We now need only find PT[(X, Y) gé Pr[(X,Y)§éR] = 1—137” (X,Y)ER] : 1—P7“[7"1§VX2+Y2<7"2] = 1~Pr[r§gX2+Y2<r§} :1—Pr[r§§W<r§] 7% = 1M/ pW(w)dw 7’? 2 T 1 2 1—/2 6—w/2dw 7.2 1 — 1+X —rg ~—eX —T% _ ep 2 p 2 Therefore, fx,y($,yiM)={ gjeXM—éwufl}, Where 2 2 Z: 27r [1+exp —eXp . x p . ‘ in} 55.! -.\ a. (,0 S1“ AU) xflmfl- VA? "WW “‘9 “Q my P M 0%: Fjwi Cb)? » w) x. A A. RAW?“ AN . 3/ 4 0 , A .7 ! ‘ _ ‘ Q 1 J’A mm M. m4 “LgfiA . a. :3 : M21 31-kan i Lv' 6 $33 1 W.“ Mn"; N 6K4)" (Ag-7‘ ( {3. 5 {V’V‘A‘ x) if» « MM 8 C, , I; E ,r (A A 33:“; E ‘% £ :2 NC“? K 3»: \ E L \J‘W} “7’1. .5. :2. x: N M L x g E : a) i E $ GM“): “>2” ( (\ /\ A , Y ang m :5; {15: ‘x‘ & \ /°‘“1 952% “\{L m “I; {MS "5%" VA ‘0 (200,19. 3 A S EU», K J )w (l | M MW} WWW I ‘ A M {a " “it We (, g: B We (, < M3“ M i ...
View Full Document

Page1 / 13

ece302hw6solns_Fa11 - 4; H W (W t) r;...

This preview shows document pages 1 - 13. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online