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Unformatted text preview: THE OHIO STATE UNIVERSITY
DEPARTMENT or ELECTRICAL AND COMPUTER ENGINEERING
804: RANDOM SIGNAL ANALYSIS
Elif UysalBiyikoglu Midterm
Wednesday, November 9, 2005 Name: C  F' Got/LL51 Department:
Year: e—mail: The duration of the exam is 45 minutes. Questions carry different weights, which are noted
next to the question number. The problems have been carefully designed to be clear and unambiguous, and reading the
problem statement slowly and carefully should prevent any confusion. Make sure you un
derstand the problem before jumping to a solution. It is acceptable to leave answers containing factorials, integrals, summations, etc. in the
given parameters. Please write your solutions in the Spaces provided with sufﬁcient detail to evaluate your
arguments and reasoning. Put down your name, department, email address, and year above. Some possibly useful facts are provided in the next page. 1 Problem 1 (40 pts total  8 pts for each part) Consider an Internet router which has been receiving packets at times Ti, 2' = l, 2,3= . . . , starting with the
ﬁrst packet at time T1 = 0. Each arriving packet either belongs to ﬂow A (with probability p), or to ﬂow
B (with probability 1 — 12,) independently from packet to packet. 1. Let Z be the number of ﬂow A packets among the ﬁrst 20 packets received. Note that Z can be
expressed as a sum of independent 1a.nd0m variables in the form Z= 2201.11, and using this, ﬁnd
the expectation and variance of Z. 2:0 __ __
Z: Z— 5 W I; ,1; Tile ,bijiwcLQEJw W 974 9% “HM” NIL51):,” ?(I£:o):4—f , rigmncaﬂ
5(2): 75(23): Zena): 2979
\fWZ: ZUMCEZ) :: 20 (47:) W;”WML£ 70 2. Let W be the number of ﬂow B packets among the ﬁrst 20 packets received. Let D = 2Z + 5W + 50.
Find the variance of D, varD. (You may leave your anSWer in terms of varZ.) “b: 22rs(go,z)+so : 130.32
gaff) :QMCLFZ ngolf(4—/a) 3. Let N be the total number of packets that have been received by the time the third packet from ﬂow
A is received. Find E(N) and varN. Maia “fin/t N=M+NL+N3 W N, rs wf W MhWWW‘mMﬂW The amount of time from the arrival of the 1"”1 packet to the arrival of the (2' + 1)” is X; = Ti+1 — Ti.
We will model the X53 as independent, exponential random variables, of rate A. 4. Find the transform @T3(s) of T3, the arrival time of packet 3.
733X4+y1 %M‘Mﬁ/W K‘Ai—sz
(I'd WWWC/i)
I
_ () 2.. 3..
fit—5(5): ¢¥«(S)¢Lm”(¢xr S )’ (NS ) 5. Determine E(T§?) (in determining the second moment, you could choose to use the transform calcu
lated in the previous part or not to use it.) all 5 .— (11. )\ 2 g __ E(Tl
dsz¢r3()_ cis‘ ()“5 ;;_ __ 3) H 2 Problem 2 (60 pts total) X and Y are random variables whose joint PDF is given by C]_ (27,11) E A1
fXY($:y) = 62 (may) E A2
0 otherwise where the region A1 is the region enclosed by the lines :9 = x + 2, y = 3, and :1: = 3, and the region A2 is
the region enclosed by the lines y = :5 — 2, y : 3, and :1: = 3, as depicted in the following ﬁgure. We are given that P({(:c, y) 6 A1}) = 2P({(:1:, y) E A2}). 1. (5 pts) Determine the values of c1 and C2. 7({CX8)6A1\() #2/3 : Q4 M“(A4):C1'2 7;)C12A/3 4
?( {uraOGAal = (12‘? 3 Us 7’) C1: /r€ 2. (5 pts) Are X and Y independent? Explain with a one—line argument. NO l;Or warn/(£21., 3“ '3 75 “(>3 MX>3=>Y<3I In the following, it is acceptable to leave 01 and C2 as parameters. 3. (15 pts) Find and sketch the conditional density of X conditioned on Y: f le(£i7,y). 2H2
11(5): 3; czalx; egg—x) HUG ‘3'?
(xlwh’ﬁl: (l‘rimc‘l) : cal 1§$>l<6<3x3<1<a+2
*1 a) 7 ‘3" 4. (10 pts) Determine the minimum mean square error (MSE) estimator of X given Y. (Note that we
have shown that this is E (X {Y).) ¥lWMWJP¢UiOQ pun/{wit L3),
50M): ‘H‘ , 3<‘r<s T
1313.5 4<Y<3
'2. 5. (10 pts) Let W: X Y Q: X + 2 be two random variables. Find their joint density wa(w,q) in
terms of f xy F‘RLWW /}1 w‘: @(z/QJ: 7(a
ﬂ: Rhfraz): x+2 ”viPk “Hm Wm Wm M :qaz/ ”a“: 5):? O 3
snag): a? $2: ._ do a.“ :4, 1. 2—61
0 0 L8
'31» '34“ A 5 (15 pts) We 013561118 Z = Y + B7 where B is a. binary random variable, that is, P(B = 1) = P;
P(B = —1)— 1 —p for some 0 <p < 1. B is independent ofX and Y. Find E(X]Z_ — 3)_ £(XIZ= 3) =E(><Jz= 3 is: 0'” (.57 412=3)+E(ij g 3 WMH
:E( I —2)?Gl)2 3) +E(X[Y=4)(4—PU§=HZ #233”
?(@:)\Z 3:) {218(c‘3i921)?0?:1) .. /£\1(2)Aa 4‘02“”) ﬁrm/ff wow)
so
lax/12:3}: 35 AW 9+1 {4— 13m)?
7— an (you?) 2
_ '7_ J 4 1
a 2 i.— + ._( __ ( .7: ) ...
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 Fall '05
 UYSALBIYIKOGLU

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