examI_f00 - BB 350 EXAM I 2 October 2000 Last Name First...

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Unformatted text preview: BB 350 EXAM I 2 October 2000 Last Name: First Name: ID#: 5Q); [L‘xggs Section: Test Form B This test consists of four problems. Answer each problem on the exam itself; if you use additional paper, repeat the identifying information above, and staple it to the rest of your exam when you hand it in. 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. (9 points) In response to the input f1“) = "(0: a LTI system produces the zero—state response y1(t) = 2te"u(t). A new input, ffit), shown in Figure 1, is applied to the same system. Find an expression for the resulting zero-state response y2(t). £20) I -1 0 1 Figure 1: The input f2(t) results in a. zero-state response 3;; (t). Express £1 ('9 055 (A... June % 5 CUJEQ dag 423mg .rhfliteg. In abs '9" HA 1?an = swam —- Hut-t) fiat) 2' thaw: +0 — "i'FIC-b-t) BQ-Cawse. the. {QIS'bem is LTI MAL but) 3:. the. £210 ~5‘l3ufv3- Fawn-fink £0 "E. (’6‘) It 4:0”ij thw'é 2. (16 points) The zero-state responses of a certain system for three separate inputs are shown in Figure 2 on the next page. Using this information, answer each of the following questions in one or two complete sentences. 0 (3 points) Is the system causal or noncausal ? o (3 points) Is the system dynamic or instantaneous '? o (4 points) Is the system time-invariant or time varying ? o (6 points) Is the system zero-state linear or nonlinear ? ’ Tfie .sdrtem Is noncwgafl. . The respoose. @300 b%‘ms bQ-Cuen. the. 5th get) 1-; “Mlle . ' 733 35122;!” 75 op namtc. beta-2.03% the— WM 0042an Gr? (ltd/rod Value: % the, gnpu't. (see Qnrwer above.) - Tfle .s m is have—way}? . ObSeNQ, 5J6: '52} Ct) ='- 'F; (t+Z)J however) at“? ("a 7-9;, {t-I-ZD, 9’ 7J9, “gale”. 75 Dog/0156a)“ £39m the aflfiév} Quintin—y.) drug Fence, Saferflqsz‘fimr} Jag; not boil. Observe, thOJE fl (t) = 3 1C, (t3 + 3 4:364; but: a; ct) :7.“ 35,616+ E/BHQ. 012345 012345 gm 012345 012345 Figure 2: The zero-state response of the system to the input f1(t) is y1(t). Similarly, the input f2(t) produces the response y2(t) while f3(t) yields y3(t). Problem 2: (25 points) 1. (6 points) An input-output description of a certain system can be expressed in the form of an ordinary differential equation as 11221 d3; __ df Determine if the system is asymptotically stable or unstable. Justify your answer. QC?) = 67sL+57s+1=o 2. (9 points) The output y(t) of a second-order system is related to its input f(t) by (D2 + a1) + 64)y(t) = (D — 4)f(t), where a. is an adjustable real parameter. For what ranges of the real parameter a is the system underdamped, critically damped and overdamped ? Comfar.nd~ C905 : 31+ W!) 4— 6‘4 2. to 62055 :bL—t— afwnb 1—Wn fax/254k: WX-T’ 6% =3) W0 =9 5L 21PM”) —' a” '7') F“:- m -: 7% e For “Life, fiber; ")3 be. crrétoag, germ/flag 70:6 db 50 WQ Meg at. -.= [6. ' For am aueFQufiI/IPQ rasflgnge, we, mar! (“Va f0>lj N inlvqiefi'gi at > [6 3. ( 10 points) During an experiment the zero-state response y(t) of a LTI first-order system to a unit-step input f(t) = u(t) was recorded at various time instances and the results are Ihown in Table 1. Using these data1 find 0 (3 points} the system rise time t,, o (4 points) the time constant 1' of the system, and e (3 points) the root of the characteristic equation. Table 1: Recorded step response of a first-order LTI system. From Tabie. I.) as; 2&9 (105:3 ‘—‘ ‘40 State 5° 707° 55' 75 36 win/Q. ’07:: 55 Is ‘I, I": {Eta/s? {In}? 7/ ; USmdi the, rnwéronslnp I 7 ”Fins'iT—ovpaer _ Sigemg) Problem 3: (25 points) Consider a second-order system whose output y(t) relates to its input f(t) via the ordinary differential equation d’y dy _ 4 — 4 t = . a“, + d, + y() 3 fit) Suppose that the system is driven by the input f(t) = 8u(t), and that the initial conditions are y(0) = 2, 32(0) = 9- 1. (6 points) Find the form of homogeneous solution yh (t) (do not solve for any undetermined coefficients). 2. (5 points) Determine the particular solution yp(t). 3. (3 points) Evaluate the unknown constants in the homogeneous solution, and determine the total response of the system. 4. (5 points) By directly solving the ODE, determine the zero—input response yn-(t). 5. (6 points) By directly solving the ODE, determine the zero-state response y,,(t). l. 2. &cme, 4&3) is Coas'Eu/FE ‘L-‘F 't 20 ghee .fl __ ~z-t: a .. 5108 ‘ 2Cle + Clea E - 24: C20. 2t . «vi yCfi : "' 7% ‘" 52. 1‘1 = C2. -.—_ #4:): —‘i€2’t+ +25% +6 t .20 Se}: ‘FGf-T) :- 0 "Fur t- ZC’JJ “the my ér‘Q Cong/(£0005 own. d_(cb=z. 0090 9(0):? 60 191L411 BQ—Cocuse, ’dfle cLurch-Lercslstc. ang—é‘w, ,0,” mi- k C/anggfi athft) =' C.e + Cz'lse, i320. ”Covet/QT J :15 40(7) :0 J gr) (1'33 = 0. This can be. Seen 3 62.7%; a (4:) :1: ’4 then p 2 fl¢/*‘ ‘féé 1—~I%pr :- 0 =9 H-‘éO. 20 the. 3,1;13mn 3.! J 95 JP”) 2» 0 2f: 21!: 4" C1178. 10 5H.) :: 35H») + 319“" 3‘ C,€~Z‘b+ cz-LQZt + 6 "b'Ff-E.) A‘Ie’ “en Hip-.4 Uélfla tl‘v. [nl'b'uga aunjnlztans 01(0): 0':- Cl+é ==> Q="é ”'6 9’0, 50 “at —z-£: 012560 2/03: “66 “lz-te. +6 1230 24$ a, aka/'9 [Io-la, flag- - .- t: g—aLC-fi) +/£.,-Hs) —.—_ 282t+ {343a —ee:“— mg” +6 tzo Problem 4: (25 points) 1. (8 points) Derive an ODE that relates the input voltage f(t) to the output voltage y(t) for the circuit shown in Figure 3. If you introduce a node voltage (mesh current) to determine the ODE, clearly indicate this variable in the circuit schematic and label its polarity (direction). Figure 3: Circuit with input f (t) and output y(t). KCL a£ :1er I9- finfiej (,2 4.. W9- (-H , (. —-—— + 3,... yo 9. ) For 1h?- Walcr" Ll {(-blflntf '1‘" t J 2. - 2.0)4‘ czgwfimwflmcgc (a) «Subsitu'krfi (2) "7-123 CI) film t D .1. ‘LQ 4“ L7, (vfirryfirfllw Kg: 4— 012— :0 C33 b1 fieren'EKW-Lm b-H‘ 5‘ (0&5 “Jar I? i 2. (7 points) Once again consider the circuit from part 1, redrawn in Figure 4. Suppose that fit) ~_— 6 u(—~t) + 3 an) and R1 : R; = 10 k9. Determine the initial conditions y(0+) = m5 V d'y _ V Et:o+ — 5 Figure 4: Circuit with input f(t) and output y(t). , Rt 17:0" the. mJuCEuc-s are. ~55de C‘I'tw'él' “’“QSO a,co-)-_-. *CO‘) = .622... = 0.6,” My. R‘ IOKJL ‘JHRL {JUI- ‘ 6H: t =01} MOW”): 0.5m” amo- ‘9 lath“): R. C109“): 6v? ' Eacavge, (3(0"): CL (0*) "' C, (0') = .6»: fl") “’9' at” (gnu-4.2 shook vkwr): e\(.= 319L030. ' USIflé v, :: L (01L- _ %% 0mg We): Vfl'lfify [“5 3. (10 points) A circuit, difi'erent from the one considered in parts 1 and 2, is described by the ODE 3'] + 8g + 32 y(t) : 128 f(t). Given the input f(t) = u(t) and the initial conditions y(0) = 9(0) 2 0, find y(t) for t 2 0. CumyarmU. Q55) =3 7&2'1'8? +32. to 'I. an» =— n1+szm +Wo reueqig W76— 3:- 32— % Wm = 432.. = .1 (—2—: _..L_. ZFW 7" 3’ =7 19: £21: 5 anfl. so the, sdJ-Exm is uanQhQumflQcQ arm/Q #11 ('5) =- e- {Wt-{E (AC3WOQA: + éfran-é) “i320 WKWQ— Faun: ‘I amfl Woo—:Wndl“1°2'= "i. 8% Se. 43H) 1) COnJ‘an’E "Rr 'b 2.0 0mg 73, ¢ a gag. Cflu 7‘), %0J a? (t) L: #. Subs-L‘Evé,y y? {0% m 005. (7‘9‘05 .. A— W + 3% 1— 3L/ 512.? 4 [Si-:1 9mg. .50 at, Ct) = ‘1 t 2 o. The, 1’59"er Sea/Jam is g (t) = /h (1:) 47,93) 3 e“ Ht (9705 ‘115 + 8!"? V16) + 7. Us“: '5“?- tm‘EMQ ca L (on) fiycazo:n+y ==) 61:”? jfib>~o :-‘/lq+'4t3 =9 65192-7 9mg, 5 13 l4 ...
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