final_s00

final_s00 - EE 350 FINAL EXAM 5 May 2000 b Name: 9 l n3...

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Unformatted text preview: EE 350 FINAL EXAM 5 May 2000 b Name: 9 l n3 ID#: Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Test Form A Important Guidelines You have 1 hour and 50 minutes to complete the exam. This test consists of six problems. The exam score is calculated by adding the five highest problem scores. This is a closed book exam. You are allowed one sheet of 8.5” x 11” paper for notes as stated in the syllabus. Answer each problem on the exam itself. The quality of your written solutions is as important as your answers. Your reasoning must be precise and clear. Your grade will reflect the clarity of your presentation. V Problem 1: (20 points) A linear time-invariant system with input f(t) and output y(t) is described by the transfer function u “— — 3 _,__S....—_—- _ F(s) '— 33 + K s + 4 = (5+MS'H) K=s Where K is a real number. 1. (5 points) Determine the zero and pole locations of the tranfer function assuming that K = 5. 29m: Sachs-é; 4+0) :0 0’?— IS = O I 22m) - roles Poles Salas£d_ HLS) =°’ 0?— (S+:)LS+~¢):0 cm 2. (5 points) For the value of K in part 1, is the system asymptotically stable ? 82.01052, b°’l.'J- p.90 Rave. nOaanan. (‘qu Part5) "the... .Sflfl-Qm ‘o Stable. 3. (5 points) Determine the DC gain of the system. DC 60cm = Melba =0, 714.. mac"? :5 are, 4. (5 points) Determine the value of K so that the response of the system to a unit step input is critically damped. For as- ch't‘tqug (app/25¢. fl“?— mufitf 5Q- fleaabNQ-g [Gag cm (Jeni?ch fiéwnlmhla In this ca»- .32" 1- K5 +7 :— 5?— ... 2fwn5 +y/n?_' flafl sa cu}: Li ml 21%“ = K, sol/:2} Q» K (7M2: 2mm 2‘1, Nat thfli: The. Sash-n TS cation" 5 J1 9Q 312cm cl: ha; K- T— SL+ :- fin a, dc, “209/ -‘f ioqflnttc% / Problem 2: (20 Points) b 1. (10 points) Given a LTI system with impulse response Mt) "—— edt u(t) and input f0?) = 6—3' “(0 find an expression for the zero—state response y(t) using the method of Laplace trans- forms. . _ t I . U5.8 the trans-Lorna p44»- ea' “(-5) é—a fl: dew} I f Ugly fin, com/wa out) 5 4643* Mt) l .. YLSE =1 PC9N—5) :- m "' 5+2. 5+3 A (“‘9- 3 are, fun—Jen}: 2. (10 points) Given a LTI system with impulse response h(t) = Eng: 11.“) and input f0) = 6'“ RU) find an expression for the zero-state reaponse y(t) using graphical convolution. In order to receive credit, clearly specify the regions of integration and, for each region, provide a sketch of f and h. No credit will be given if you quote a solution from the Convolution Table (Table 2.1) provided in the text. 016:3 = -FH:):a~h(t) :- jsngpG—ficgt ~ZCt-r) Problem 3: (20 points) 1. (10 points) The operational amplifier in Figure 1 is ideal. Determine the transfer L function and express it in the form _ Y(s) _ 1313+!)D H(s)— F(s) _ 3+0”. The coefficients an, 60 and bi must be specified in terms of the component values R1, R2, R3, L, and Figure 1: ActiVe circuit with input f(t) and output y(t). L/ The trdnfggr ‘pv‘flc'blon ‘Ffim #(t) 1b 2.66..) is oé‘éa‘moog ()5: n& 001+?“ IV “Stun Fun «9%) a woe/h ECS) = I PC) YB) = (SL—t- R4422. -———-—~R‘ > 13(5)J Subs‘l'l'bvfion a (£385 2. (10 points) The actiVe circuit shown in Figure 1 is driven by a. periodic signal with period To = 1 and f(t) = 3—” for 0 S t S To. As a first tep in calculating the steady-state response y(t), determine the coefficents D“ of the complex exponential Fourier series representation of - coo-l: Dn ='- 2? Heed" OQ-i: 3, we :. z'n‘ Problem 4: (20 points) 1. (10 points) Find the transfer function of the closed-loop system hown in Figure 2 and show that it can be represented as Y(3) bm3m+b -1Sm"1+"'+bl 8+bo H(3) = Fm: s"+an_1s“'1+-~+a1s+ao ' 5-1 4» 3 q“; __——-—-— 5 _, “15+?— 57- «s + 3(1511.) 2. (10 points) For a. certain closed-loop system the transfer function from the reference input r(t) to closed-loop error e(t) is E(s)_ 32+403+4 R(s) _ W' For a. unit step input r(t) = u(t), choose the proportional gain K so that the absolute value of the steady-state error is less than 0.1 e“ = e(t) < 0.1. 05.? “Ll-2. fitflfl «9,1,2, fjeofem 855 3 SECS) = aflsfliew '5 5° 530 £65) II Q €55 ‘— ——~" 40.; :9 1+K > 70 => 34—: [Va-[2 thj; 4;»— flo. Lanna- Value. theorem £2) £0. «fling/9) 55(5) cunno'E ,Qd-UQ. any [9on in tie ry‘léflJLIg/Me or- M a“ iw aux—LS. ' 10 Problem 5: (20 points) 1. (10 points) A LTI system has the frequency response function .. 3"“ + 1 HUw) = : = 10 100 F .271 + 1 1000 Construct the Bode magnitude and phase plots using the semilog grids provided in Figure 3. In order to receive credit: 0 In both your magnitude and phase plots, indicate each term separately using dashed lines. 0 Indicate the slope of the straight-line segments and corner frequencies of the final magnitude and phase plots. 0 Do not show the 3 dB corrections in the magnitude plot. I DO NOT COPY THE DISPLAY FROM YOUR CALCULATOR !! 11 ' '-=-s;p---~+-+-+-2-+eéé-;-;-»s-we«é-eva-ee: Was; L 504: . . . . . . . i . . I . . . . , . _ . _ . ‘ . _ . . . . . . . . . . I . . . . . . _ . , $ ._ Phase deg I 4:. 01 C _90 . . . . ‘ 10° 10‘ 102 103 1o4 10 Frequency (rad/sec) Figure 3: Blank graphs for sketching the magnitude and phase response of the frequency reponse function. 2. (10 points) A second-order linear time-invariant system has a. frequency response func- tion H ( 3w) with the magnitude and phase plots shown in Figure 4. The input to the sytem is 15" 9/ f(t)=100+cos(700t+45°). =3 AF: —. I Find the steady-state reponse y(t) of the system. U From tl-o, 15°94, maaficbflI-‘L filo-E 4-: A 5') II “b +3 + —> 1.3"" a E) (EC + “f E T‘ T DC- ‘Uavm SJGuSotcfloa Dc {awn $.0v5b1flt—Q tw- 4:2!”00 fl 0‘- 90mm g! 2' ‘F. 2: ’00 {0—- __-_-_ (o 1H“ 740) ; V2 2‘ F1. l 7°01” Cg / _. o -" 70° ‘ I eat 75 Joe a “(5‘0 : [0 e at £063 : [o + [o €9.5(700'I': ’75“) 13 Phase (deg); Magnitude (dB) Bode Diagrams 3 1 02 70 o 10 Frequency (rad/sec) Figure 4: Bode magnitude and phase plots for the LTI system considered in part 2. 14 0 Problem 6: (20 points) A double sideband supressed carrier (DSB—SC) modulation system produces a signal 3(t) Whose spectrum is 5(w) = M(w +wo) + M(w — we). This signal is demodulated by the system shown in Figure 5 where H (w) an ideal lowpaas filter. Ideal Lowpass Filter oos(coot) Figure 5: Demodulator for the DSB-SC' signal 3(t). 15 1. (10 points) Assume that «1., = 100 rad/sec and M(w) = 34”” 10. Sketch the spectrum on the graph provided in Figure 6. 2 6;, 25 CE) COS «uon :. 2L1? {Si-B335" Q’ECo; «No-L] 7, 4? 5909 3k 1;- E Svfw 1-way 4i Sifwawa'] 2,! = JESLVMQ + Aida/Mo) guLs‘L'Ev'émfl 5(w\ :1 M (we-ouch -I~ M (w- we) E5“) =-‘ JgMCV’rz'UcB-F f’”@ 4— Jimfw) +— fMCw—Ywu) E (U?) = fMLw-FLWQ 4— Mfw) + tMCW—Zwa) -1 a 00° : \oo) Mfuh: a 9"1/! Figure 6: Blank graph for sketching 16 2. (10 points) If possible, determine the cutoff frequency we of the ideal lowpass filter so that the output of the demodulator matches the original message m(t). If we cannot be found to satisfy this condition, expalin why in one or two sentences. 14: “L5 065 FOSSIL‘Q '50 EIIMIkK/g- {‘9‘ Com/moné! .1; M [w fwo) dnj/ J}; M (w, W‘s ova—t1) (Do—E J¢S£Wg,y M (w). 7}“) 75 bQCMLSQ M(oc) :6 “at 0.1 ba JWAGQ' 17 ...
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final_s00 - EE 350 FINAL EXAM 5 May 2000 b Name: 9 l n3...

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