examI_f02 - EE 350 ‘ EXAM I 3 October 2002 Last Name Sol...

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Unformatted text preview: EE 350 ‘ EXAM I 3 October 2002 Last Name: Sol U'iZlO n S First Name: ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Instruct ions 1. You have two hours to complete this exam. 2. This is a closed—book exam. You are allowed one 8.5” by 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 example, “Problem 2.1) Continued.” No credit will be given to a solution that does 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. (9 points) The zero-state response y(t) of a. given system to the input f(t) : 'u,(t — 2) y(t) = e—t Mt +1)— u<t — 1)]. o (3 points) Is the signal y(t) causal or noncausal ? Justify your answer in a short sentence. 30L) Sacwsa. #61:) #0 ‘F-er fl ta») <7“) is mews-40' o (3 points) Is the system causal or noncausal ? Justify your answer in a short sentence. fie) 8:2.ch 'th maps dCt) OCCMJ .bz'gmz fie 1'an ‘ 'i: ‘FC'CD 7) “ff"e‘a) “the. 2. fisher" is floflCawSaQ o (3 points) Is the system instantaneous or dynamic ? Justify your answer in a short sentence. $¢cow5c 'He 85km 75 0‘” C‘KVSwOJ tAO— avifv't oi fie. Vex/vb $1042 Qwrm knewioflyg #fi‘éme [AW—L7) affix $70 the. 3&5?!” >3 Ogl‘aml.‘ . 2. (8 points) 0 (4 points) Is the input m) = w — 2) specified in part 1.1 a power signal, an energy signal, or neither ? If the signal is either an energy signal or a power signal, find Ef or Pf. °° z i “in > m s mm E‘F 25.004: (‘5) £- = j: ‘- aneryy $.7nwL I + I’ Pg=flmlf 2¥7/£)1£=)€<LA.#:SEJ-b :JZO; JT-Hl T943 ~35 2 2 Taco T500 o (4 points) Is the zero-state response Mfi=€4M6+D—uU—DL specified in part 1.1 a power signal, an energy signal, or neither ? If the signal is either an energy signal or a power signal, find E, or Pu- cr» l 2 _ E; :: §.iZH\Jé = 3‘.(’e“lygl_t —_—. X'e 2+0€£ 3. (8 points) Once again consider the LTI system in part 1.1) that produces the zero-state response y(t) = 6’t [u(t + 1) — u(t — 1)] for the input N) = u(t — 2). In terms of y(t), find an expression for the zero-state response y1(t) of the same system for the input f1(t) shown in Figure 1. f1“) Ht) Figure 1: The input f1 (t) results in a zero-state response y1(t). We. Mei in Emma: filt) .46 av (um # way out 'bme. shrcflxfl Info-£5 ~Flt), - 2%. +2.) 14 #(tw 4“ 6*) gas 2 ‘ “E” * "if—at + " * 4‘ — '2 I -z .2 L Mefl Problem 2: (25 points) 1. (18 points) The location of the roots of the characteristic equation for a certain LTI system are shown in Figure 2. Suppose that the forcing function for the ODE representation is P(D)f(t) = 3 + 2e-2t t 2 0. Im( A.) Figure 2: Location of the characteristic roots. 0 (7 points) State the form of the natural solution yn (t) for t Z 0. I + . Ck4radzens'bic. Tait-S 2 7i. '3 o) 7‘2 :‘2') as.) ‘1 : if 3501:) : Cl + (Zena: +- e't (C5rost +- 95:01:) #20 o (7 points) State the form of the particular solution yp(t) for t Z 0. Bacavse. fig = 0 any. AL = ’g ~z-L- grtt) .: 5,4: + cute 7529 o (4 points) Is the system unstable, marginally stable, or asymptotically stable ? Justify your answer in a short sentence. f/lc. say-Len 75 may-fimy quélg because 7\, : o) wlmle, 32.1 73) an! 7H are. in fie LAW, 2. (7 points) A certain LTI system is represented by the ODE (D5 + 2D2 + 4D)y(t) : (2D — 1) f(t). Write a MATLAB m-file that o calculates the roots of the characteristic equation, and 0 displays the zero-state unit—step response over the interval 0 S t g 4 using 500 points. 70 59:th Poqu Q Q : El) 0)&) Z) ‘1’ 0:]; P = EzJ ~12); 7a Fmfl rad-(:5 % tic, cXmC’a-dzenj‘éf: 25w» (57/: radio ( 6?) ‘70 flat the Zeravrlsm'ée, vm'A-S-éeP r05fanJ€ ”Z; 7 mam/£2 fima Vale-W t :: 11(69ch (.0) ‘1) 50°); ,S'lsef (P) QJ't) Problem 3: (25 points) 1. (9 points) Find an ODE representation of the circuit shown in Figure 3 with input f(t) and output y(t). If you introduce a node voltage (mesh current) to determine the ODE, clearly indicate this variable in the circuit and label its polarity (direction). State your answer in the standard form dny dn—ly — dmf dm_1 +" +aay-‘bmfi'7‘n‘4'bm—1 W+a"‘1dtn-1 dtm_1 +---+bof- Figure 3: RL circuit with input voltage f(t) and output voltage y(t). KVLz'F:iR.-\—a_$ C=——'£—;—-T:—:—. 0) L :, R -R ——-' 73:4 73—; mg: Mt 2. (7 points) The circuit in Figure 4 represents a single data line on a computer memory bus. The data source is represented as an independent voltage source f(t) with source resistance R1 = 20 Q. The bus is terminated at the receiving end by a termination resistance R; = 20 (Z and a stray capacitance C,. The relationship between the input voltage f(t) and the output voltage y(t) of the bus is specified by the ODE . R1 + R: _ 1 11+ R1320; y _ Rica f. (U) Because of the short spacing between adjacent data pulses, it is necessary that the rise-time at the receiving end be less than 5 ln(9) ns for a step change in the input voltage f (t) What is the maximum value of stray capacitance C, that can be tolerated ? Figure 4: Representation of a computer data bus as at RC circuit. .. «,1? #L :1 fl = .. R1+flL QC?) ~ 7‘ + RszCS :3 ’ R. “1C5 Choose fr- flit-R2. C6 5-" A? 11.17. “IO 4 5.0/51 )i 10‘ I10 " SX\0 (7) H00 3. (9 points) Consider a first-order system whose output y(t) relates to its input f(t) via the ordinary differential equation dy __ d—t + 4y— 2f(t) Suppose that the system is driven by the input fit) 2 4+2e‘4t for t 2 0, and that the initial condition is 1.40) = — o (3 points) Calculate the zero-input response for t 2 0. ~ ‘14: (9(7): 7s+¢1 :0 =9 7\.-‘-"4 =? 3‘5”” z 0‘3 a" ‘I 4.- 0 ~ yacéb) '—' ah ('b) +%(‘b) :— C\€ “£20 ylo}: ‘2. 3C1 o (5 points) Calculate the zero-state response for t Z 0. a“ 1+.) = 9e” 9* t 2 0 as “5°“? ’- ‘l'b JvLs'b’izu‘LQ if m4}; fie 00C) gechse’ 7/; ~be . ~‘H: ‘H’ o ' ewe” » at 43/" +%/ ”f" 10 Problem 4: (25 points) 1. (9 points) Derive an ODE that relates the input current f(t) to the output voltage y(t) for the circuit shown in Figure 5. 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). 11 2. (9 points) A circuit, different from the one considered in part 1, is described by the ODE fi+y=2flfl Given the input f(t) = u(t) and the initial conditions y(0) = y(0) = 0, find y(t) for t Z 0. 0(7)) == ?\7‘ + 7‘ =0 :9 7‘13“!) 7.110 -t gut) - 0.62 + c; t 20 BCCAUSQ £05) 1 (any): ‘RY 1720 dflj (1L :0) vaS‘bt/{Q C7”, ln'EO ODL ‘L‘a oiJ'qun 73* y" 6-; :3 2 =9 0 ‘i" 6' = 2‘ f P 2’? (t) :: Zt 7320 :— C4€~t 4’ (Z. + 2t tzo C: U59. 9(0) 2(7QD 3—6 “1:0 4“"‘Q C‘ 9' 2’ .. C =. —C2 (0)20 :- CI+CZ ’9 l 9 ’ 7—) C‘zz 7) C2 "'2' 12 3. (7 points) A 12-V battery is sitting in a hut on a deserted island somewhere in the Pacific. The positive terminal of the battery is connected to one end of a 1/(21r) uF capacitor in series with a 1/(21r) [AH inductor. An earthquake in the Bonin Islands of Japan triggers a tsunami that crashes into the hut, spilling salt water onto a rag that connects the other end of the series inductor/ capacitor combination to the negative terminal of the battery so that a series RLC circuit is formed. The resulting oscillation is detected by a nearby ship monitoring a radio beacon signal at 600 kHz. What is the resistance of the damp rag ? Post: 'bswmmi. Circuit: C: if“: L= 5,74% VS (I. L K V; =‘-l2.\/ cg WQ. . file. says-hem I”: unferiamflel (304!) (£6.62. (We '50”. ft [Orofluczs am oscillu‘éofl Slam aflfl to o m (it (In ven( 7.: éooKHal 7:96 94 fl? C3 2‘ F~J~ 3 all. :- 2.77'- @oox IO ”5%: ' 6Nen Lang. C the. cum)» alteéermmey (all) 25 m ”mfg/224. It Inp'b V5 Copstfler “the. .S‘lmflL—plo any "’ J 13 139% cLarmLQns’bc, efivav‘E'tc/I 1') @m: 752- + LCM J— : aZ+ZFwnA ”(7b, thd£. 0"“ : "VJ—'5' 75 thaw/i" drug. ‘Cvo 2. 0.6 = [—207— z) 0 3C 1 "" Ami So 70:0 8 Fumg, L “UT xyo‘ {2: 2f - :— ZC°-8> W 14 ...
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