{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE302%20mid2_sol_fa06

EE302%20mid2_sol_fa06 - Sothmvs EE302 Midterm#2 TR...

Info iconThis preview shows pages 1–6. 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 Document Right 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 Document Right 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 Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Sothmvs EE302 Midterm #2 TR 7:30—8:45 AM, Prof. Gelfand Instructions: There are 10 true-false problems (5 pts each) and 2 work-out problems (25 pts each). Do all problems You must show work to receive any credit on work-out problems. Calculators but not laptops are allowed Cheating will result in failure of the course. Do not cheat! Put your name on every page of the exam and turn in everything when time is called. Useful Formula: Binomial: pK (k) = (71:)pk (1 ~p)"_k, 19 = 0, . . .,n, Where p is average arrival rate. Y? = np,a§{ =np(1 —p). Also (71:) = [cl—(71%,0!= 1. Geometric: pT(t) =p(l—p)t"1,t= 1,2,...,.T= i012; : 1—152. Negative Binomial order k: pT (t) : ill)“ (1 —p)t'k ,t = k, 19+ 1,. . .,. T = g, 0% = k if? . k _ .._ Poisson: pK (k) = W, k = O, l, . . . ,where A is the average arrival rate. K = At, 0% = /\t. Exponential: fT (t) = Ae‘dt, t 2 0. T = %, 0%, = i_ k— — __ Erlangorderk:fT(t)=%,t20_T=§, 033,2}? Uniform: fx (:3) = Fl?» (1 < :c < -— 2 Gaussian: fX (m) = flap (_% (mo—f) > . X (1) function: (13(32): ”5 —1— (—%2 dx. —oo ‘57; exp Questions 1 - 10 are true-false problems (5 pts each). Label each statement true or false to the left of the problem number. (Note: if statement is not always true, then it is false). T 1. Let X and Y be jointly continuous random variables. Then Pr (X + Y Z 1) = ff; fffy fxy (3:, y) dmdy T 2. Let X and Y be jointly continuous random variables, and let U = aX + bY where a and b are positive numbers. Then fay (u, y) = if)“, (”—j’y, y) ‘2 3. Let X and Y be jointly continuous random variables, and let Z = X + Y. Then fle (zlz) = fy (z — m) . 4. Let X and Y be jointly continuous independent random variables, and let 9 (w) and h (y) be functions. Then E[g (X) h(Y)] = E [g (X)] E [h (Y)] 5. Let X and Y be random variables and a, b, c, d be numbers. Then Cov [aX + bY, cX + dY] = ac Var [X] + bd Var [Y] . random variables. 7. Let X and Y be jointly Gaussian random variables. Then fXIy (xly) is a Gaussian density in T F T 6. Let X and Y be random variables. If iprI = 1 then X and Y cannot be jointly continuous T x. 8. In a signal quantization system, a uniformly distributed random variable on [~a, a] is input to a uniform quantizer with 2 levels. Then the root mean square error is f T 9. In a digital communications system, the received signal is Y = s+N when a “1” is transmitted, and Y = N when a “0” is transmitted, where the signal 3 > O is known, the noise N is Gaussian with mean 0 and variance 02, and the transmitted bits are equally likely and independent of the noise. The decision on the received signal is made according to: Y > 5 say a “1” is transmitted, and Y S gsay a “0” is transmitted. Then the probability of error is 1 — (I) (fi). F 10. In a sequence of independent coin flips with an unfair coin, the probability of a head on any [6‘1 -at flip is or. Then the probability that the k — th head occurs on flip t is 526371)?— (k = 1, 2, . . ., and t: 1,2,...). No as '- . b a E a Q S (““69 a 3;. CW3) .L , 1e: u;ji 30;,qu \{ o ‘ \l - ' \A t‘ X l‘l 3C1 t \j \ 9- ) . 3-}‘Xi’cl‘1l: SYUH—m) . coutmx Ha‘i mun] : MVwTX] A, gummy) + («“505 Cuv‘lJ‘X] ‘ Uni-CW Ii \ymw HAW ‘(~aX~l\3 Wt mm. KY th 3mm, ‘ ' 2 r , 1.. b; ,_ : E— . E (trim-V) \LWCX—(rvw W X'— 9‘1) 0%} 3? {‘E .. ‘2’ __> E if? s 9 .. 9. PE t liPy-(N‘r {Bi—{PFCHPK \- zxiiC\~5§(-:—G\\ 3 s Q 8 3: - h A». - - Questions 11 and 12 are work-out problems (25 pts each). You must show work to receive credit. 11. Let X and Y be jointly continuous random variables with pdf bra/(any) = 0, OSxSyS 1, 0, else. where c is a constant. () - (b) Find Cov [X , Y]. Are X and Y uncorrelated? ) Find fx (9:) , fy (31) . Are X and Y independent? as?" ) SHCM33=F sxcxugycj) C9 <~ 5709‘ 5d” = 2% “nu m m m l I I ‘ indeyen denl: . “x: 920—0-«61 := \-_g_ = y/ -> Cc Gov) l4 éxgg _ 36 :. X3: y am cameloled ?Mo‘o\em H COthvecl. §2C73= 0 ‘50 12. A circuit component burns out on average once a day. When a component burns out, it is immediately detected and replaced with a new component. Note that the burnout and subsequent replacement can happen at any time of the day. Assume that the replacement times follow a Poisson process, starting at day zero. (a) Find the pdf, mean and variance for the amount of time up to the k — th component replacement. (b) Find the pmf, mean and variance for the number of component replacements in t days. (0) Find the conditional pdf, mean and variance for the time of the first replacement given that the component which started at day zero has lasted at least 1 day. 0/») The h‘me between been outs is ex\>onen\-{0cl with Pmameiea A e i -. ’n‘me M1 the R“ bum—out: is cm KH‘ omdex-fmlanfi dislmwtgcj Pondmm mane, with W” K K-l _ §TKCJ°K>= ’< JCVe‘h CA=\) @ Q01”. E [TKQ ] : £2 1. _l_<_ => VQM C11): .14.. A = ‘ A3 K2 3“ A; (N) 5) Nvmbem oi lawn—outs“ is poissoh disinilov’red- PNCM = wt m @ “I EiN]: Kt VMCN)‘: '0: @ ‘1 ~ Ae'l‘t at = ' S—fl'rrb). S (It ‘10 Er/EO k7, | “1ka c k0 MEET] : .2 dogs. Vomiomce is uhckongect X< 9mm cu‘ns .L G) ...
View Full Document

{[ snackBarMessage ]}