examI_f05

examI_f05 - EE 428 Last Name (Print): EXAM I 11 October...

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Unformatted text preview: EE 428 Last Name (Print): EXAM I 11 October 2005 015 First Name (Print): ID number (Last 4 digits): Section: “9"?” DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO INSTRUCTIONS . You have 2 hours to complete this exam. This is a closed book exam. You may use one 8.5” X 11” note sheet. Calculators are allowed. . 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. N 0 credit will be given to solutions that do not meet this requirement. . DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. 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. (10 points) A 8180 LTI system with input u(t) and output y(t) is represented by the transfer function Y(s) 34 +1 G 2 2 ———~——~—. AS) U(S) 34+482+28+9 Using; the smallest number of integrators, construct an all-integrator block diagram representation of the system, and then provide a state -space representation dc : Fat—l—Gu y = Hsc-l—Ju. Clearly indicate the assignment of state—variables in the all-integrator block diagram. l=£L=( I )(5‘1+1) U U P 5"+ ‘Is"+2.5+q a .... _. _ .. '2 __ + W SfiP+HSZP+ZsP+ ‘7qu :5) P —— "if P F )2 O I O 0 xi 0 _ O 0 l o :2. 1. 3 UL ~ "' 3 l a x L’__¢J xi 1 =6 '2’- “. x ' :: ('LlX3"2x ‘7)(11') + 1 Q—C H X .4 + _X 2. (15 points) Consider the closed—loop system in Figure 1 with reference input 12(5), disturbance input W(s), and controlled output Y(s). The PI controller gains K and K1 are free parameters for the control engineer to select. W(s) Y(s) Figure 1: Closed—loop System with PI control gains K and K1. (a) (5 points) Determine the transfer function from the disturbance input to the controlled output and express your answer in the standard form Y(s) bmsm + bm_1sm_1 + - ~ ~ + bis + be W(5) 8n+an_1sn_l+u-+als+ao lo _—-I——'—__‘ 7. 5 +5 + 20 was) You 1 : _,____.___,._———- w )0 K5 + KI H" furs-r20 s .L w 53+ S’- + (ac-Holds + lOKr (b) (10 points) What is the system type and error constants (K p, K D, and K a) with respect to disturbance rejection ? O Fn-st -FmQ. E“) 1C" «(5) =0. Egg): 9%3—7/5): '21)). W53 wcn wCs) .— (05 = - SECS) flow. sWCs} 65S 5-30 53 1-57-‘+ (zo+|ok.)s + IoKI' u Note, ‘41»- W (53 7' 3L5 ) ollsba r1: since, mpvfi} Fa» ‘ a. VflI‘L—J‘bp Inpvfi MIC-b) .: flJ‘b)J W(S): “J5— .- ess : Q'.M S J. — [OS ’ 5-30 5 3 2 ‘ O) 5 + 5 +(Zo+lo|c)5+/o}4:: ding. so ICP :1 co_ For 6L unfit—ramp Inpu’t W (19-) 3 t 1- [ix‘ ; 5 z 2 653 = %/?§(fi*£‘”—‘)= “’1‘ —- lo 5—}0 53+ 52' {— (20+IOk-33 +10 K1: Kr F ‘ or as parqIO‘ch, "77.1: (at/(t): i; 1653) W65):— 21?; ~Io_S €55 :. Zéms *1?“ { =¢o(-—1) 56o 5 53+ sz'+ (25+lok35 + “3": Ami SO Problem 2: (25 Points) 1. (10 points) Consider the closed—loop system in Figure 2 that has a PI controller with gains K and K1. Using the Routh array, determine the range of the control parameters for which the Closed—loop system is BIBO stable, and indicate this range by sketching the appropriate region of the control parameter plane in Figure 3. Pl Controller Plant ‘ Y(s) = :11; -2 3 Figure 3: PI controller gain plane. KS‘F Kr l I. _ .s (‘5 +1)“ +13 _ Aw“ R ~ ‘1_ “Hg: 1 53+- 357'4- (2+K35 +KI 5 (5-H) (S + 1) 53 | 2.1- K 52‘ 3 K1- * — E2 5 Z+K ,3 .So KL ‘9. (15 points) The closed—loop system in Figure 4 contains an adjustable gain parameter K. The desired system response to a unit—step input is specified to have a peak time 2%,, 5 1 sec and an overshoot 111;, S 5 percent. Determine whether both specifications can be met simultaneously by selecting the right range of K. and if possible, indicate the range of K. Sketch the region in the s-plane where both specifications are met. Figure 4: Feedback control system with proportional gain K. K 2. —Y— :. fiL— = K has the. "(arm w" R l+ K 51+zs+K 51+- 250L005 +w07“ 5(5+z) 2. l Efivq‘hig Coefith/HISS a.i€l‘94 w” = r? qnoQ- P: ZWO : r? ‘ lfiyfiflosfl _ 6 For mP 50.053 we neeg- flm «0.670 4t? regret" mJlCU'e-S mam." q thFQa ciosoao—Q-Jof [LI/e} > >Re'CS) Um. flesw-ag, gums/orb f2)/lO/‘y- Ch¢r¢9€y‘y£\d_ like flan/fig Lr¢id2fly J: ‘F‘ K 5. £0 “Vine/Q, ml; . W ‘9' L 0 5- 4 Ki), (“QQJWQvQ vaJve). 59 we ,- 4: 00)} Can/wt A ! Sec” Slow/{:anecusi gut—ng n’) é 0.05‘ own/Q 7‘3}; _ Problem 3: (25 Points) 1. (12 points) A mechanical system with output y(t) and input u(t) is represented by y+y+y2 =u(t). (a) (2 points) Using an : y and $2 2 y as state-variables, find the state—space representation of the system. = ‘F‘Cx\"1-,w) .. X +9“- : ‘FZCXUXZ.)W) - x| :_ JCX.)X7_)\~‘) 0=x§ '70 L 6 =9 xe=~ o=-Q<$)”1*“° 0 (b) (4 points) For the constant input u(t) = uo, identify all static equilibrium states are. - W O T 0 There 0"9- bVVo 26V\l.brium Sink-La, (of rap/Jim” ly) - 11:0 ya : {—01.0 or (c) (6 points) For the constant input ua : 2, find the linearized state-space representation (5i : Féz + Géu (5y : H617 at the operating point that yields a positive displacement ye, where m(t) : me+6$7 u(t) = ue+5u, and y(t) = ye+6y. F = a F 34 : = p 75x1 AX], “a: '2X? ---l -15 ,‘ at. :0 x9 = 5— a‘lé w 0 6‘ : r. r. 3 72(4— 9 l .. 0 (as/M w « U > O 2. (13 points) A certain system has the state—space representation ¢:<_2 _:)nfi+((1))u :_ Fx+6u.. y z (1 1);; : #x (3:0) (a) (7 points) Find the transfer function representation of the system and express your answer in the standard form __ _ bmsm +bm—IS’m—l + ‘ ‘ ‘ +1918 + [)0 [7(3) s"+an_1s"_1+ ~~+a1s+ao ‘ 'i (no = (l,1)____L-__ (“7 ')('> :. (I) |)____L..’—- (“(+7) U Gfg) = Hal—F3" 6, + 3- .s’w‘is-m “1 3 0 ‘Z‘W‘H'H “1 WWW-J C 5:: - P) 1 (b) (6 points) Compute the state—transition matrix (Mt). S + ‘1 ’L’. —1 —, "' (s n)‘ (“’5‘ ¢ H.) = j i s I v P) = I ( ;1.. ,5...— Z. Problem 4: (25 Points) The closed—loop system in Figure 5 has adjustable parameters A and K D.The design specifications for the system are 1. a steady—state error of less than 10% to a unit-ramp reference input7 2. a maximum overshoot to a unit—step input less than 5%. and 3. a 1% settling time of less than 3 sec. R(s) Figure 5: Feedback control system with adjustable parameters A and K D. 1. (5 points) Find the closed—loop transfer function and express your answer in the standard form bmsm + bmrlsm‘l +-~+bls+bo 3(8) s" +an713n—1 I +---+ais+ao fl _________fl_____________-a-‘ 1+ [/+ kg 9 52-:- (2+-:€H<Qs + 1‘9 (A Q + N \./ ll 7>I< 2. (5 points) Sketch the region in the complex plane where the closed—loop poles may lie. loo m. < 57. => 302 M T. 0.670 ’ m we Closegafloo/ {Io/0) I’m/S“é L0“ ' hu'Es—APoQ My”; big-6° flchLQ‘Q, «n £0 dukk'ofl; flex\{a£ fldflJho/ré t ‘ J VeTUAB‘L GMI“"£€r” 'C 3. (5 points) What does specification (1) imply about the possible values of the parameter A. ‘- AK Ecs) = Kay-m) = «,9 [hr/R] : ‘10)} .s + (1+ 95 } 57— + (2+AKDSS +9 For R67: 3'; has): 1; 1m} ) - L+Aa esszsflgngsgcg = E 5’ (3+ 2+»ko) i): 0 SO 5L+CZ+AKD\S +fi ’9 65$ = #fifi. :04 => fiz 20440349) :9 AO-IOKQ 22,0 4. (5 points) What does specification (3) imply about the closed—loop poles. t5 = 4.6 C‘ 3 fl) PW” > [HS-'3‘ ,4; "Lhe ~ _ W film. "9Q far'b % the JOJerQOvp [0/0) I5 70 n, pogéfi mJ‘g—E £0 07L; #2 (115$? : 4.53 in the, S’plqne: :mCS\ \\> \ 5. (5 points) Calculate the error due to a unit-ramp input in terms of the parameters A and K D. From {1 av t 3 «.1: co 2) 2+ Alf—D H 10 ...
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examI_f05 - EE 428 Last Name (Print): EXAM I 11 October...

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