examI_s01 - EXAMINATION I EE 350 Continuous Time Linear...

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Unformatted text preview: EXAMINATION I EE 350 Continuous Time Linear Systems February 5, 2001 Last Name (Print): @3— First Name (Print): 11). number (Last 4 digits) : Official Section number : Do not turn this page until you are told to do so Problem Weight _ u _ u - I. 5 _ 100 Test form B 25 25 25 2 Instructions 1. You have 2 hours to complete this exam. 2. This is a closed book exam. You are allowed one sheet of 8.5” x 11” of paper for notes as mentioned in the syllabus. 3. Solve each part of the problem in the space following the question. Place your final answer in the box when provided. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example, “Question 4.2) Continued”. N0 credit will be given to solution that does not meet this requirement. 4. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. 5. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences shall convey what you are doing. To receive credit, you must show work. Problem 1: (25 Points) 1.1 l ' i ' ' ‘ ( 6 pomta) Cons der 3. continuous—awe mgnal {(t), not: flute I; Hz) = O m) =(3—t)u(t—2). 3 gr cm? 4: < z o (3 points) Is 3‘ (t) causal or noncausal? Why? 4%) is causal. beeswax. kHz-0’ 1; <0. ( Lg. 43M) = a ) o (6 points) Determine the expression for the odd component, fo(t), and the even component, fe (t), of f (t)- flt’c) z: 1cth+ H4?) 2 == (3't)u(£'1)+ (“hut—17¢) W 2 «Com -.— We) *‘R‘H 2 = (34) M({1'2)- (3+{:)ul~t-2) 2 o (7 points) Sketch f (3 «— t) in the figure below. ¥[3"t} .: t ‘33-» ll (3 ~ (3~’c))u( (3—h-2) t M143) fl 1.2 (9 Points) For the half-wave rectifier circuit in Figure 1 determine the power Pu and the root mean square value yum. of the output y(t) due to the input f (t) = 4sin(t). Assume that the diode is ideal. You may need the trigonometric identities: c032(;1:) = $0 + 608(231» and sin2(;t) = %(l - cos(2sc)) Figure l: Half-wave rectifier circuit for Problem 1.2 wt) = { We) , Rt) 20 0 ¥Lt)<0 9 ~41 (ti t o , T 9-1" To a 11" p1? = #0 OS Wthl alt = E14} 5 iesim’t 0H: T “If g g 1-cos(2t)dt = 4 _. 3| it = 4 as .g 3, H W a; ll :1 It to < Problem 2: (25 Points) 2.1 (10 points) Consider a first-order system whose input f (t) relates to its output y(t) by 2% + y(t) = 10 f(t). If f(t) = eztu(t) and the initial condition y(0) : 1, o (4 points) find the natural response yn(t),t Z 0, o (4 points) find the forced response y¢(t), t Z 0, and o (2 points) determine if the system is asymptotically stable. 1) an): 23+1=o =3, fizz; , —t M maximal mpome is W 10" mt Igm 3mm: c1: /2 2) Simex. ht) = git , i324), fiJ-{MadeW All I'rnffu. 2t ¥o4xm «NEH: PX. mot 200499 + «Mm = 1010(t) HT:— 4szt+P£2t=1012t => )3‘2- 11W = 13n(t)+1j¢(t)_ (berth-t2th We): 1:- c+2 =9 c=~l W) = “’0 31¢th We) 3,. 113a Agate/mid M foHc flab/(i firmer, may <0. 2.2 (15 points) The switch in the circuit in Figure 2 has been opened for along time and is closed at t = 0. 120 3 mH Figure 2: RL Circuit for Problem 2.2 o (7 points) Derive an ordinary differential equation in terms of i(t) for t 2 0. PM i: >0 , H30. chew} bxcorrmo A vii) = LEE L in") R “M 12 (1:): wt) 5 god T R Eli? KCL at A: 21ch gm = 0 id) + = 9' Lou ea} 29;, is {m = QR. E} + 2000 NH :2 19000 Git o (5 points) Calcuiate the initial conditions i(0+) and v(0+) At ho‘ , Wu. u'nwfl' Meow, 129. tic") 125?- 51?. _ é: ito') 24V 657. 0 $24“ 54V ice”) = 54+24 _ n+6 At t=0 , o (3 points) From the ODE in Part 1, compute the time constant 1' and the rise time tr. Sim“, am) .1: h+2ooo =-_ o Problem 3: (25 Points) 3.1 (8 points) An RLC circuit in Figure 3, is described by the second—order ODE f(t) d, Figure 3: RL Circuit for Problem 3.1 J If L = 50 mH, Determine the value of the capacitor C and the resistor R such that the overall system is underdamped with damping ratio C = 0.4 and natural frequency w" = 20 rad/sec. 2. . “n a Llc ——-—-> c _‘_._ * glues F. I} L'W‘ 50x16 -2o‘"‘ 3 ll ll ._ _R_ ' a -- . - . zzwm‘ L e R 2391,01. — 2 o_4—-zo-50xlo 70 ll 0-8 :2- 3.2 (17 Points) For the circuit in Figure 3, if R = 2.5 Q, L = 0.5 H, C =: 0.5 F, y(0+) = 6 V, iL(O“') = "1.5 A and f(t) = 3u(t), 0 (3 points) determine the initial condition %§|t:o+ it! = (ii-(0+) = “11:5 : "3 V/S dt t=o+ C 0.5 ' I (6 points) compute the zero-input response yzg(t),t 2 0 Subsh'i'ufx. ma val/us a" R, 1.J a, 4&1. ODE for; Rm imam, 0123 OH:1 + 503.4 +43%): Hit) 0“: To carmFM’fx. aim-(t) , 31.15 19d") =0 . =9 ypm=o aw) = 7"“+5A+4 = (2+1)(2\+4) -= 0 => A,=-I, fig-4 O -t -4.t mod 30, gm = when rm g C”, + c2; . 4; ~41: "j [’c) = -C,2. -—4c2»e USL IFm'rLI'aL code'Hom -. «9(0) = 6 = c1+c “410) ="3 1 x. o (8 points) compute the zero-state response yz,(t), t 2 D . Simu hf): 3, ’0 >0, Wm: p, o o d’f atg _ . tL + 5‘ t + 4F - 4 3 =9 [/3 = 3! _t -4t gt“: 3+C'L +64 To condor)qu 0323M 31b , 11(0) == 0 , afflr- 0. 10 Problem 4: (25 Points) 4.1 (7 points) Consider a linear time-invariant system whose zero-state response y(t) due to the input f (t) is given in Figure 4. fit) y“) 3 2 L'II System -1 0 l t -l 0 1 t gm 3 -2 0 2 t Figure 4: A LTI system in Problem 4.1 o (3 points) Is the system instantaneous or dynamic? Explain in one sentence. m Wi'm M wow/ma smu «3m , t z I, IS nmjsmo bul' Wald 5-? mo II’YIPMI' fix. pd) =0, 75 >1 0 (4 points) Find an expression for the zero-state response yl (t) due to the input f1(t) in terms of y(t). Obsunm. Hial' £6) = _3_(1C(t+1)+ $rt-1)) 2 met sima +131. me Ala LTI, «041(k) z .3. («3min ytt-UJ 11 4.2 (6 points) Consider a certain system Whose zero-state response y(t) due to the input f(t) can be expressed as W) = f2(t- 2)- o (2 points) Is the system zero-state linear? Explain. 717a Wit/m «54 momb'rmow. Mam homey/midi? Mop/i}? {0.1923, 1%) ~——-a> get): 'FZCJc—Z) 01 Ht) -—-—> oz‘J-“(Jc—z) 4 agti) o (2 points) Is the system time invariant or time varying? Explain. H is hbou—I'oqku'omi' SfrnCX. Him-T) —~—-——> —Fz({+T~.2) = W’s—FF) - (2 points) Is the system causal or noncausal? Explain. 1+ is caudal m'rna Val) (may ottme cm 1%. Pauli WM of 1%. I'm/mi 73%» “Phi-2) 12 4.3 (12 points) 0 (8 points) We desire to plot y = coe2 (t) over the interval 0 < t < (it using 300 points with the Matlab m-file shown below. Carefully examine each line if corrections are needed, reunite the corrected line. Original m—file Your corrections t=iinspace(0,6pi,3); t = MSPa-CL (0 , 6 * 9 300) y=cos(t)*cos(t); fl? = GOSH?) - '1“ 00.9 (i?) J' l , ; . DOWN) ( t, q? ) , a (4 points) Write a complete Matlab command for finding the characteristic roots of the ODE 613:; dy _ dzf 2E + dt + 3w) .. 4m + 5 dfi. >> noots([2,0,1,3]) >> A00t3([2 o 13]) 13...
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