examII_f07 - EE 350 EXAM II 18 October 2007 Last Name(Print...

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Unformatted text preview: EE 350 EXAM II 18 October 2007 Last Name (Print): AS 0 [Vi-.1 on First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO ‘Problem Weight Score 1 I 25 2 i 25 l 3 . 25_L 4 25 j Total L 100 j Test FormA INSTRUCTIONS 1. You have 2 hours to complete this exam. 2. This is a closed book exam. You may use one 8.5” x 11” note sheet. 3. Calculators are not allowed. 4. Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example7 Problem 1.2 Continued. NO credit will be given to solutions that do not meet this requirement. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. 6. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. Problem 1: (25 Points) 1. (10 points) A LTI system with input f(t) and output y(t) has the ODE representation (Pg 513_ .dt—2 + dt — f(t). Represent the system as a signal by determining the impulse response function h(t). If your expression contains a term with a Dirac delta functional7 simplify that term. Find. the. armada; unI'E-afiep mpomJ the” Alex =Jy/JH2. Q0): 7% + ’A = ’AC'AHB :0 =9 7"”; 7‘"' —t 35cm .1. c. + Cze. t 20 \Saklo, ‘CUP at. ‘- or. 1) no . f... t: éf+q%=¢ '9 6%? -—-L-.- gm; 515'94'JPH‘) .. Q,‘+cLe. (3‘0er C.‘ “N9" C2, 60 Said 5' 5K0): o: C'"+ CL 3"“) obs-NJ Cisvcz=-( ,“ l jco\:0* 0‘2“" (t3 : (.1 +0. 2. (9 points) A LTI system, different from the one in part 1, has the impulse response representation where [11(1) : 6(t — 3) mm : e—tua + 2) o (4 points) Is the system causal? o (5 points) Is the system BIBO stable? Justify your answers. - (+7 33 03 on W 3 at 3 - $35 3 ~ _ —e = e. <00 [unlit = J9} 1+.- = e. Xe 1+; " e. I -0, a ' 0° {:em :3 < (L < co) the, 5d 5 Because. I HH‘Hl‘US A in L -00 L {um-£2. con) " [3 H30 5 'L’db‘e. 3. (6 points) Figure 1 shows the block diagram representation of a system with input f(t) and output y(t). The block diagram contains three LTI subsystems represented by the impulse response functions hula), mm, and mm. Derive an expression for an overall impulse response function h(t), in terms of h1(t), mm, and h3(t), so that W) I f”) * Mt)- Figure 1: Block diagram representation of a system. w H.) M &: 4%l‘2‘ + (-F’kk."‘F):"l‘3 g= -F-rlhz.'ll.3+h‘*h33 W th) Problem 2: (25 points) 1. (12 points) The signal f(t) = etu(3 — t) is applied to a system whose impulse response function is h(t) = e3t [u(t — 1) — u(t — 2)]. Determine the zero-state response Mt) : fit) * Mt) : k (t) «4366) = 3.25.3);C-t "tMQ'C using the graphical convolution approach. Do got sketch y(t). In order to receive credit, clearly specify the regions of integration and, for each region, provide a sketch of f and h. $553ng l<-3+4= <2. or 74'5‘5- 7. 7.. 4'. 7.1 ‘C a e no zeta t t-‘c e 0Q? H warm? JCtl 3 S 9-3 Q ~3-H: -6 7.4.: t 't at Z 3 _€_: all —e + J I ‘3+'& 2 = £1. Q 2. 2. ~3+L R€|n3 -3+‘b>2 OY‘ £75“ amtangsa fig ~3-—-° Mam—c) '0 ’ » when t )5 t 7 3 . an fl?) -0 who“ / :\ a” = J W’ 1 -oo \ 2 ~3+-b 1: t1 ‘L t 4 ‘1 Sum «Him a —-e m / '5: [a 1 ¢e1-§:‘e‘6 "l<‘t45- 6mg» ‘-‘- 2- 12 7 5' O 2. (13 points) The signal fit) : u(t _ 1) - W — 3) is applied to a system Whose impulse response function is Mt) : 4 ['u(t) — u(t — 1)]. Specify the interval [t1 7 #2] over which the zero—state response y(t) : f(t) ac h(t) achieves its maximum value, along with the value of the response y 011 that interval. 409 h H.) We ”L l 4: h l .3 I The mammwn wwluo. a (4:) : he’ll“ ‘cL‘CU (M52; '0“: —'C) «ample-Le? overluf’J -l+t +- J°°+LQM£ -C)£C oCcura when tAereLf Maxi/ma”; the. drew: LCt‘r) MaMMUm auerlwr) 9“,er 1%,. Ht) 2 < 1: <3 ‘C 1/73 -H-t- *9 ct) '3 Th0. (airmspomgify “Wt/'2' ‘% y 47 4: t I 1 ’ —t = ‘l 5 Vft W‘- l. JM‘Z “0‘ hi: Hui: .— = ‘114: -C—I+t)] =7 m alimgm vdvlJQ. Problem 3: (25 points) 1. (10 points) Consider the network in Figure 2 that has input voltage f(t) and output voltage y(t). Figure 2: Passive RC bridge circuit. (a) (6 points) Find an expression for the frequeHCy response function H (3w) and place your answer in the standard form bm(]w)m + bm_1(]w)m‘1 + ‘ ' ' + b1(]a}) + b0 HOW) : (9a))" + (In—data)“1 + ' ‘ ‘ + “1“”) + 00 U.Smai 0 olive. SJV‘ISIOD I _. R "’ link. AF; R E W F l\ (b) (4 points) Find and simplify expressions for the magnitude, |H(]w)|, and phase angle, 1H (3a)), of the frequency response function. lac “l“ -&-—————l’ “WI/m = W :3”:- é‘” ‘ Irv/Act me u 4 ('W “ T“”"( 7/“) I [P9 2. (8 points) Consider a system with the frequency response function Y(Jw) mow) H(]w) : F(]w) : 0.1(Jw) + 1’ where ()1 is a positive, real—valued parameter. In response to the input @500» i: + e - ‘(O°) f(t) : cos(10t + 25°) + SUM, the sinusoidal-steady state response is ‘ 0 005 ((000 4: .. *1 go) y(t) = Acos(10t + (15) + 10 SinM 45°). Determine the numeric value of the parameters ()1, A, 9 and 45, and state any numeric approximations that were made. Specify the angles in units of degrees. we) __. ”LOU”! ($05004: + 15%, 3 Ham) + luca_.°.o)\cos(mt + 9—734 4H9coooi) 7‘ n = 5‘ == 10 SJ“. ‘Cor 5. «Min I HCdJOooUl 1' IO, ‘ Ibi+10aol ‘ loco L, 2 Iooobi : l0 )ucamal - : ""00 lo.la_looo+|l ’looy-Hl 6N8!) bl) 'Fln-Q. 9" ‘0 ‘ L'EE— / ' r2? r7: Hzll-Kouo)’: W = /}+// Fmg. d; %_ first «Qe'lrewmmral tag-Ham} & i‘o’ : low), 421-15 = our __ 0 ¢-.-_ 25°+ AH’é/“fl 1'") ¢‘7° LH’CdKQ = I000) boa—lemme e L? Lat “LIA; LH—Ld moo :VX. l0 =1‘7OO ziwdmoa 1 #zwj] 5 54;!er / OJ [000 70° ‘1 so = e ‘_ 700 + “#106003 fl) 3. (7 points) For a certain system it is known that the frequency response function has the form 520w)? + 510w) + 50 HOW) : a2(jw)2 + a1(jw) + do, but the six parameters b0, b1, b2, (10, a1, and (12 are unknown. In order to determine the value of the unknown parameters, the magnitude |H(jw)| and phase angle AHQw) of the frequency response function is experimentally measured for N different frequencies (U1 through am that span the frequency range over which the system operates. For each frequency w, the experimenter applies the input W) = E- co.s<wit + 02-) and observes the sinusoidal steady—state response y(t) = Y,- cos(wit + $2). This data is entered into MATLAB as the vectors w = [wthy‘ LUNl F = [FlaF2a' ,FNl Y = [Y1,Y2,'“,YNl theta_F = [017 92, ' 101V] theta_Y = [(1)17 $2, ‘ ’ ' 7 ¢Nl 7 where the angles are in units of degrees. Write a MATLAB m—file that: o (4 points) Estimates the values of the unknown parameters b0, b1, b2, (10, a1, and a2. 0 (3 points) Using the estimated parameters, calculates the magnitude and phase response of the system from 0.1 Hz to 100 kHz using one thousand frequencies equally spaced on a logarithmic scale. Store the estimated magnitude, in dB, in the vector est_mag, and store the phase angle, in degrees, in the vector est-phase. ; H: (Yo/F) . :3 ex? (A ‘7' (thatu.-7~ the'ku..-F’> §‘F-~-/Ls{0>3 n:- 2.3 {73:23 [13ng : mu‘prefis (H, w, mJ n); 'l‘ 3 2:355ch0. ("I3 S“J loco); Cat-“‘27) gamma]: Leora (go, asp; *4); es-L—ma/ = 2.0 4:: ch/IO Cert-M4043; 70 Shaw lo Coe gnclcnfis P 70 show a, Q 5.,e-ACC lent 10 Problem 4: (25 points) Consider approximating the signal f(t) as N) : 01¢1(t)+ 02952”) + 6U), where e(t) is the approximation error. The signals f (t), C(t), Q51“): and @(t) are real~valued and defined on the interval [t1,t2]. Table 1 specifies several relevant inner products. For example, (451(t), ¢)1(t)) : 2, while (fit), gbg(t)) : 8. Table 1: Table of inner products. 1. (5 points) Are the signals (1)] (t) and ¢2(t) mutually orthogonal, mutually orthonormal, or neither? Justify your answer in a short sentence. < 4):, Sb; ) :o the. signals gbl (.5) “"90 90L”) Be— C “0.80. Ora. Met/q“? 0/ th 06 of] “Q. Bfittxvfl <¢|g¢n> #l aan— <¢2fl ¢L7#/J o «I. the, 5 la nmlS (1), H3) «mQ @196) are, rig-E MV‘LUqu (,th oarm . 2. (8 points) Determine the numeric value of the coefficients (:1 and (:2 so that the energy of the approximation error 6(t) is minimized over the interval [1‘ 1, t2]. Because. the Signal) 515,96) dnaQ (fiblt) are, mu‘L‘de; OVt'l‘Odono/Q, the, cae-Lfitcmob C, any, C2, film/E,- eL-e) ave, the, energy, 0‘F the. array- mnm'mfiée. 11 3. (9 points) Suppose that <¢1(t),¢2(t)) : 2 instead of zero, as specified in Table 1. Using a standard calculus technique, determine the numeric value of the coefficients c] an Cg so that the energy ES : /t 2 [m —{C1¢i(t)+02®2(t))} F dt of the approximation error 60) is minimized. 44) (1;? ___....———-v- . Begausq, <¢U ¢L> #:O‘) W can 00+ 'F'anvL thQ. CL US)? CC : <¢LJ¢L7 Cheese. 0‘ omJL CL 30 {ska}: dEe/flcq: Allie/fllLsog J. J(-7— .1%: -_ L. fif—l—D - {Ca‘bo H.) + szLIt)3]l£/@)££ :o 4: 2' “b7- - . {7 :0 j: 4160 qbdil-aJ-e - Q: L4 (Wt) dyi—thQ-l: C7. 4:.ch It) (19%»? (419(1)) ‘ C, <¢IJ¢')‘ C2,. (dz) ¢-\) :0 (1..) a U” 7. ppfllél Ma e169 w‘lzwn m flog, twc unlit—flown) 7/ Z ZC —- l0 (2.) <¥/¢’§-c.<d> ¢,)—cz_<d>qu7: =2) 201* 7, ) /J 3 7” ‘1 +2_<1,-I‘1cz = 9 L3 4. (3 points) Comparing the solution methods in parts 2 and 3, what is the advantage, if any, of using mutually orthogonal basis signals 9571“)? Explain your reasoning in one or two short sentences. when the Slanuls (Phi-b) are Iris/bog“? orthoaoflurq we, can Jolve, ‘Fur the, CL “Agapenooant? MSW? C; "' <‘FJ (b; >/4¢LJ do). J3 whon the 5‘57?qu ¢nltl are, no'& rug/bulb, orthyon )‘LLIVIGDUJVIYJ we must b Olfi 6" .5 Cit, 7 n 650 aft/on) in CC 5 [mu '7‘.) Wkan the (fin RV?- mv't’afla arthoflonuflj [-5 LS chD-ef to Solve W the CL 12 ...
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