examII_s02 - EE 350 Exam 2 25 February 2002 Last Name 5...

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Unformatted text preview: EE - 350 Exam 2 25 February 2002 Last Name 5 0’ o'lll'ons First Name Student # Section DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO — — _ — — Test Form B Instructions You have 2 hours to complete the exam. Calculators are not allowed. This exam is closed book. You may use one 8.5" x 11" note sheet. 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 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. If you introduce a voltage or current in the analysis of a circuit, you must clearly label the new variable in the circuit diagram and indicate the voltage polarity or current direction. If you fail to clearly define the voltages and currents used in your analysis, you will receive ZERO credit. 7. 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. PP’Nt‘ Problem 1 (25 points) 1.1 (15 points) Use the graphical convolution method to determine an expression for zero-state response y(t) = f (t)* h(t) with h(t) = 4e" (u(t—2)—u(t—3)) f(t) = 3e'2‘u(t) . In order to receive credit, clearly specify the regions of integration. Also, for each region, provide a sketch off and h. Place your expression for y(t) in the box on page 3. 3 1 ,_.___.————*————v) "i: t 2. 3 5mg: 1cm Mica: mafia.) : j MQHJVCMQC 4w—o when i: <2. —F(-l:~r3hCC)=-'Ow 'F'Of‘ Valves cg: “C anti 3° 5MB : §_:h(t3-¢[t't\cac =0. pflgm'l 2<t<<3 t ‘. Ck # (£3 :1 5 50C) "Cd m It) _ K r m fi-b 1?) 2. fie Urbeamnfl ,5 flea/IE <1. 44: z. 1: 3 T’ 9W“ Z. 1.1) Cont. Region 3 Kt) 1.2 (10 points) Consider a LTI system with'irnpulse response h(t) and input f(t) shown below, and consider the zero-state response y(t) = f (t) * h(t). 0 (5 points) Assuming a at 0 , what is the duration (width) of the response y(t)? o (5 points) For what value of a is J: y(2')a'2' = 0 ? Justify your answer in a sentence or 0 burw‘bon OLGA -_-. vaw’hon -FH:3 + Dora. (on hC-E) :3 5 + é [marathon an» : u 5 Area. #61:) = [firew -F(-t) 1' 1: fine“. “(153—3 ‘To ob‘EoCm Dwayne,» = f ytronC= J Chane. 00 50 that thQ- otr‘odb 5% {-(t) 75 zero. Fram tho. V rat/7k % "C( t) fired, -F(b\ T— flrew L°l?¢\ Problem 2 (25 points) 2.1 (8 points) The input 6u(t) is applied to the system f(t) = (D2 +30 + 2) y(t). The system’s zero-state response is y (t) = (3 — 6e" + 3e“2‘ )u (t). Determine the impulse response of the system. As ' It ~$onows thm'lf.‘ swat) ' é‘aéh“) 14049- So 00 .L 59;. ’3 — ee’t-k 39:24:)ufis) MUD-"- :6J£E( .3 t “NH-9 .. .1— .. = .1. (6e: *eez*‘)v~(+b + o <3 6% +33, 6 :0 930°" 3° 2.2 (9 points) The block diagram of a system is shown below. Determine the system’s impulse response h(t), such that y(t) = f (t) * h(t). f(t) Figure: Block diagram for problem 2.2. we» _— has) av]: kdaa‘kuahcca - h3(-e)a=~f—C-ex3 3,03 = (but) *[lnlt)* MO —— 113ml] 91: {(123 2.3 (8 points) A LTI system has the impulse response h (t) = 6 (t) + e'3‘u (t) . Determine if the system is bounded input —— bounded output stable. mutt = :| 8(a) + éka-E)‘0Q‘E z \ 8(9 + aflt‘km 10¢": = Sio(g(t)+€—t3&'e 1' 3C6 it + 634; am? ; 1 + v Jgégt‘: Problem 3 (25 points) 3.1 (10 points) For the circuit shown below, assume that L = 250 mH, C = 8 11F, and R = 2 K9. Y Determine the frequency response function, = ? . Express your answer in the standard form b 'w "' +b '(o "'"+...+b 'a) +b How) (f 3 l’ ,2] ‘F’ ’ °- (1a)) +an_l (1w) +...+al (16c))+a0 (Hint: As an intermediate step, it may be useful to sketch the circuit in the phasor domain.) L C fa) R lye) Phéor bomm Cum/2*:- mgflmm N “i; &wt_ 4,. J, Y ~ 3.2 (8 points) A system’s frequency response, H(jw) =% , is shown in the Bode plot on the following page. Let f (t) = 12 cos (100t — 25°) + 25/5 cos (lOOOt + 75°), determine the sinusoidal steady state response of the system, y(t). 1‘; 1H(dw3\w_‘ .—. 041B 97> l (ZoQogwtzocQBB O 8: =|oo 3' O W I ,1— ‘300‘5 :3 77a; €379:wa “3 23.H{d®lw= looo '750 __‘_’___,_______________———-—-——-—-‘ C: H‘Hd‘alu/‘élooo =- H 12. 1&9-1003xCos (1001‘. ‘25" + 1.991003) (t3 :: 3 [H9 203)\ C05 (1000+- + 75° + M969) + 253 305-3 '3 l‘2. CosClOo‘L‘ZS") 'r chosOooot 1‘30) 7" Tho. Pom’tS IQ, 13/ CJ ahg/ D a.an shown ,5 the. Pvt/r2, an 074% 10. Magnitude [dB] Phase [Degrees] O The frequency response of a system. 1 1o2 10 Frequency [rad/sec] 10 Figure. A Bode plot of H(jco) =% for problem 3.2. 10 10 3.3 (7 points) Consider a system modeled as (15D+25)f(t) = (D2 +30D+5)y(t). Write a Matlab script that generates a graph of the frequency response H( j (o) = y?— . Use the frequency range 0.01 S 0) 3100-36? with 500 points spaced logarithmically. ¥ : flogspacfl (—2, 2.) .5003} hallo. < 135,251) £530,521)“; 11 Problem 4 (25 points) 4.1 (10 points) Find numeric values for the real-valued parameters a > 0 and ,3 > 0 so that {x1 (t),x2 form an orthonormal signal set on the time-interval 0 S t S 1. X1“) X2“) 12 4.2 (5 points) In problem set 5, you showed that (ejnabt ejmab’): 0,m¢n ’ 72,, m = n 21: . . . where 0) = Suppose that a s1gnal f(t) may be represented over the time-interval J;- S t < TT“ 0 as f(t) = Se‘j’" +5ejm . Given that 1}) = 0.1 , determine the energy of the signal f(t) over the time-interval ———Tz9- .<. t < Usmfl Parsevwhi “themer 4) g e‘d'Tl-L CV15. 2. .- ..f_7T-£ : <41) 7- Z. C: <¢nJ¢n> { (b; ’ a C2 5. HI'LQVAUL‘EIVQ. solJSlon: .. ~ TYL- Wt q+rrb+ saint) geiL +3~eii > < we, a = <5 : 254671767”) + 2S<e0mtj e~7nt> O 0.1 477.5 /7T£ e 1T1: Q at a 7L ) + aS’Ze , > +2§<ef 0 ’ 0.) 2.3 + 2.3“ :5“. H 13 4.3 (10 points) Consider the following fianctions defined over the time-interval 0 S t S 2 : f (t) = t2 x(t) = 2 2(t) = 3 — 6u(t — 1). Find the real—valued scalars a and b that minimize the energy of the signal f(t)—ax(t)—bz(t). Choose. _:_ <‘Frx3 b = <F, a) a» <><,><§ <32?) Erw‘yql tut <X,%>=OL I 7- 7. Cheek ‘. <XJ i>: j —- ‘Za‘Lt—lglé z ’l'LSJf 0 .. 12-412, 20/ .- 2' z. b _ 5 (Bi-.1“ 613M903th 35 L52“: --68 t ii: __‘:_____.____._———————--*= fi.7___~——:——~ 51. <3- 6w€t*l))z~i‘é S 7Ji~ + [(4)3345 o o 2- 2 o a 3 l0 3 I a ’3.” b + 3 ‘ - MM .. M .I' 7 t" 18 ‘8 14 ...
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