examII_f07

examII_f07 - EE 428 EXAM II 8 November 2007 Last Name...

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Unformatted text preview: EE 428 EXAM II 8 November 2007 Last Name (Print): 8 Olu l o as First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO INSTRUCTIONS 1. You have 2 hours to complete this exam. 2. This is a closed book exam. You may use one 8.5” x 11” note sheet. 3. Calculators are 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 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. 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. (13 points) For the BIBO stable'systems with the transfer functions shown below, 0 determine the DC gain, 0 sketch the pole-zero map, 0 write an expression for the step response without solving for the response, 0 state the nature of the step response (critically damped, overdamped and so on), and 0 estimate the rise-time, settling time, time—to—peak, and percent peak overshoot of the step response. (a) (5 points) (1(3) 2 TL 5 +lls+ 10 - _ 72 p01nts)— m go no ‘Fma'te- EBTOSJ IL \ 618a 5! “a Ca.) GL5): Swabs“) { Poms “k _| anQ— ~16) ads em :5 7/RQ umt- S‘Lef response, LS w‘kluo‘) ayxg JLSIE‘mt (5) 6(3) 3 J: L Caz = K n ’- s -r 239w”; + w} w!) - ’9‘! =9 “In ; ’zrnJlbeo; ‘P: ——G_. = I —. 20m ‘1 No 'Cmrbe Ea.me Dabs «fl; 5:. ~f’wn «9;; thfl- = - 33: . [1.6 as“ 95 flu. sds-Lem is 6180 $424.36 be 8am ; 663‘s“) 3 K = t Po‘e, —- Zero "‘04 P T—‘W- an. -s+£p 09390052. ’Qus {*0 'C‘V‘m -3'h -3’t- J. 3.0a 1 { c‘e (10501.6 9+ ale: SmUl-é‘b) + Bud—t) BQC‘Wse, 0 4 ‘F 41 «ml flare, am. no fimfz Zero; the. ¥°"OW'? “flflvcanoc'é‘wn‘ aye, Vet‘th 2. (12 points) A closed-loop system transfer function has two dominant complex conjugate poles. Sketch the region in the s-plane where the complex poles should be located to meet the following specifications: (a) (6 points) a peak overshoot S 10% with a time-to-peak 5 10s (b) (6 points) a peak overshoot g 10% with a rise-time S lOOms (0‘) Mr: 40'”) :5 f > V'IT'L-i- ,ij'mp) Problem 2: (25 Points) 1. (13 points) For each of the following closed-loop transfer functions, determine 0 if the closed-loop system is BIBO stable, 0 the number of poles, if any, in the right-half plane, 0 the location of any complex conjugate roots on the jw axis, and o if there is an adjustable parameter K appearing in the transfer function, determine the range of K for which the closed-loop system is BIBO stable. - T Z 200 (a) (5 pomts) (3) 3I + 33 + 232 + 43 — 8 K(s + 4) (b) (8 Pom“) T“) = (a) Rov'tb 9 Frau;- 1 8| n .2qu In ‘1 I z -8 a f— S W I‘rs‘l.‘ Column 53 l ‘f anlcwla Oflev 51 —z_ —-3 unShh—Qz {20‘9’ 5° -3 % who édS'l—em 35 m4: 5%" ,‘l‘h‘bl’lo’ tle .QelrE—M‘é- plum m‘” 22,03 (9-170) .3 m (ow % Fran GLUXA‘uJ-rd Qfiu‘la‘hpn: c fla— Wot mules amplex (02.,vwe: Pole.) 1) ¢w «10.5- ,..- > .zst—Kao 9‘) 4' l! g 52’ l K * 7' A. 2" 3 .- K. >0 52' 3 Lu“ «F— 70 ’7 5‘ 3‘ ii.” K‘ = 2 "'g‘ 0 3 0 3 LlK n CoN/lex he ‘ y r.” v¥ téfad /nocccw'b z; e “:6!” ‘Uhan k 1 J t a .5 L5 Pram ‘bFa “V X ‘l “,7 v V av} an tif— w a“ __ 4.. '2, r;— 2. (12 points) This problem introduces a set of results useful for determining sensitivity functions. (a) (4 points) Suppose that T(s) is the transfer function of a closed-loop system that contains a dynamic block G(s, a), where a is a parameter within the block G. Show that $=$$. Ta-BT—g £=E&:ST\/ (b) (4 points) Often the closed-loop transfer function T is expressed as the ratio of polynomials N(s, a) T(3:a) : D(S a); where a is a parameter that may be subject to variation due to the environment. Show that 5,3 =ng—553. 'r' fill—'02 oC- b".-95 31:3“? ' MT ‘ Mi co b7" "5' up shaves—wage. Boo/V 'aooD .. b _ g" -56; x/ (c) (4 points) A certain closed-loop system has the transfer function T03, k) : G1(S) -- kG2(S) : “(SJ G3(3) “ “14(3), 1) (S) \c) where k is a parameter that may be subject to variation. Using the result from part b, show that 5T : k (G3 + kG4)(G1 + ng)’ 0 D .. N E. _ ab s_<,_ s‘;= sK—sn — %.-:~ 919 $1 me) ~ K@1(6|*K6L3 m K - A...“ m K = Gamma» sgcmtw’ (6\+\’~6;3C65 * 6‘ U Problem 3: (25 Points) Consider the closed-loop system in Figure 1 where 10 3+1. R(S) + K G(s) Y (s) Figure 1: Feedback control system with proportional gain K. G(s) = 1. (10 points) Calculate the steady—state error for a unit—step input in terms of the controller parameter K .Under what range of controller gain is your answer valid? _ I‘ K L; 1: 10K KP ' gar: 6 ) ’ The, er-presam 4w 2.1.: T: Y kg _ 10L ‘1 ‘ |+K6 S-rH'lO‘C 2. (15 points) Suppose that we desire a finite error for a ramp input. Replace the proportional controller with a PI controller, and determine the steady—state error in terms of the controller gains KP and K1. Under what range of controller gains is your answer valid? ecu): n+1: :. $321112 .5 .S $g+¥z m I _— - ’— :. 10K: K:- = 3 G669 6‘53 ’ 1&0; z“ 5“ tlQ, T/fiz fexl'b ‘p-J/ £33 7) we],th CD Qny a) 010199,,pr ova-Eon Is 6160 5+4le Gee next/0%) ‘ Tia, obseQ— goof “brdUS'EI’ gnc‘ém [.9 m) _ _...__—-—-— _‘. RC5) II |\ GCC5) 60) | +— atoms» 10 (S kp + k1) ____________________,.._./ gz' 4- 50+ [9 hp) + 10K: EH- Buao sfi‘o‘L ‘37; lo \‘I [4— \Okp 70 => arfl, loKI7o Problem 4: (25 Points) 1. (18 points) Consider the closed-loop system in Figure 2 where 1 3(32 + 63 + 12) R(s) + K G(s) Y (s) Figure 2: Feedback control system with proportional gain K. G’(s) = Neatly sketch by hand root loci for the plant transfer functions listed below as the proportional gain is varied from zero towards infinity. Be sure to specify breakaway and breakin points, asymptotes, arrival and departure angles, and imaginary-axis crossings, if any. Hint: An inflection point of K (s) is both a breakin and a breakaway point. A :: 3 Aopn—Qoof Poge; 5,4 5—; a) g: -313]; Athrac. brunch» In th' rod: 900:: m :0 .9 'H'H'BQ. 3% A ln‘pma‘t? °C 3 18° + 3602‘ ; :‘1‘ 600' angle efl «yap-£942. n” m a 6 mg 4.x») m kite/’47 V‘ :— 20; -- £3 : T : ‘_ 2 n -m m ‘0le dc— (the. - an 290 rte-Q «Jusl 15% M rous QM: 75" «2°02, J: 5,0 In .NeeQ ho ckmlu 'Cw- bf‘e‘tl‘v‘" 1 lyeakdm} pol'l‘ES . peep. 1,3. barman /“’ onus c r0593! VOIOJ‘ M. W‘Q 4,0 Ling “A/Q‘ ale/mare 42‘ 2k (cm-flex (DIQJJ PL 8' pl c B \c‘. n b u, awn . é; — ‘ red, /\ a Q. (Q ‘c # {erfts sun'hs‘cy 45 2":333" 1%Z’CSS-I-6SL-r1u33 :0 2. 'z. 5 +12 +- ‘ 2" — 3 3 12,0 =5 ,5 1‘15 +‘1— (5+2) :_0 .9 5 L—Zj—Z. dz’ .1. _ fig. 40 35:1.- a(5)Sl '1 65 +12, :0 9 S: -P— is an m 45’ (“Vb 8-:‘L S’J—‘L 0’? there—Lara, q, brealan ang— er‘th/y‘ pom-L, . Local-71') am»; can; traumas as)? 1-H. gawk“ array, The chrzc-éenjfic ezua-fiu‘, ‘ + K 555‘ :0 flied; '- s3 + 653 +125 + K. -,o_ Rot/1h array 3 You; % zefbk 'Rr K. "—‘= Fram fin, dulluluy S l ‘2. .L m . ‘ S7" 6 K ézva'bm 65 + 72, :01 dos/“fix”. Chi-N’an m4, s: 3925 weak-72.— .nnvfi. #1 ooepar-bam. gé Camflléx- pole. ,7; z; —3 +//§: Mum: (“+ng = new . “up *5 —xcs+_o) [to ~ 360 ; ’cw. *70" ~ 030° ‘ éx—wfi/filfl (aw) $9, .: ~60° Imb’s M=7Z 2. (7 points) Write an m—file that o constructs the root locus for the system in part 1, 0 draws lines of constant C = 1/ W2) and can 2 2rad/s on the plot, and 0 allows the user to determine the gain for a desired pole position by selecting a point in the graphics window. 070 avatar pla‘d: i=4: ( 1) T316311) 03)) leukg' roo'LS on rodb ; (0,. user- 98 «70 «Call, ya n Waging, (cats) 11 ...
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examII_f07 - EE 428 EXAM II 8 November 2007 Last Name...

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