final_f07 - EE 428 Last Name(Print Final Exam 17 December...

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Unformatted text preview: EE 428 Last Name (Print): Final Exam 17 December 2007 Sng [Lions First Name (Print): ID number (Last 4 digits): Section: 3. 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 three 8.5” x 11” note sheets. 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. NO 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) Figure 1 shows a closed-loop system where the plant transfer function is 1 GP(3) : 82 _ 4 A compensator is needed so that the unit-step response has a 15% overshoot and a risetime of 180 msec. R(s) + Gc(s) Gp(S) Y(s) Figure 1: Feedback control system. 1. (5 points) Can the desired performance specifications be met using either a proportional or phase-lag compen- sator GAS)? Justify your answer in one or two sentences. oCoos\8-or the. root foch usma ago) =l< .20: In a Regerless 7! film preparing/lag Can‘eral gab: Va.le m m cloSaQ’Qoq 5d5-Lam Ls pa'f: $460 mue- ( 41$ ~A’s plN-Jfi- Quf Car ’2) (onerol leafs ti“ ro’b goua’fi’ék'w) J 11': w.“ no'h rayl-é- in a— 543.4(2. ckyflagoap .C/r'lcm. FL 32.- -m owl; to ban}; the wt,— flaws 1’50 the gay,— on? P0 or d ,a gal. Co A i: no I. 2. (10 points) Using the root locus method, design a phase-lead compensator 8+Zl s+p1 Gc(s) = K so that the desired transient response characteristics are met. Clearly read item 6 on page 1. Carefully describe how you chose numeric values for the controller parameters K, 21, and p1. ‘ T/Rfl- hat-49' pall— gga-b‘m aria; ‘f = l’a‘cmffl /l (ExU’tPDL-nv" P‘ m 1h“ and5 nHer ac m s a any‘e’ Cofloa (‘él'cn twé SQ S“... 5‘46 “it”. Ecflsflb 6‘,st = I80+ 3602, -:— 3. (2 points) Using your phaselead control design in part 27 determine the steady—state error for a unit-step input. 8.333 _ _ __ K 5 696p: ’ 50‘ go ( —‘ Z) P s—ao Pl - - .1. 3-. €55 = -—L--- .. 1° 4. (8 points) Augment the phase-lead controller in part 2 with a lag component 8+zls+22 s+pls+p2’ Gc(s) = K and choose the parameters 22 and p2 to reduce the steady—state error calculated in part 3 by an order of magnitude. Carefully describe how you chose numeric values for the parameters 22 and 132. From pfir't 3 two, can Pave/ca 6gb b5 W [D e“ “c the?) _ % ia-L -Q,m606p"K~7—\§IPL(‘1 (whose elm/PL :- lo ‘50 "screwba- Valve, abfdfnecfl- In part '3. Ohoasa {71 close, +0 5 so) .So that‘s the Jag'éwflaa [Ja- Ebro {74;} 0600.5 flit rodf3 210-25 ngQJ and hence 47maslcm': res 00% ~ ' Problem 2: (25 Points) 1. (15 points) Figure 3 shows the Bode magnitude and phase plot of the loop transfer function for the feedback control system shown in Figure 2. R(s) + e G(s) Y(s) II(s) Figure 2: Feedback control system. 0 (5 points) Determine the phase margin of the system, and specify the frequency pr at which the phase margin is measured. Indicate the phase margin and pr in Figure 3. PM“; “2.09 } 8.6“- (“$00 Wynn [51Hsz “’1’” “7%., o (5 points) Estimate the peak overshoot of the closed—loop system to a unit-step input. Because PM (C)J ‘Ht dual—floor; wok—m ;$ Una labia, mp '5m 0 (5 points) What is the system Type number? Justify your answer by citing appropriate features of the Bode magnitude and phase plots in Figure 3. 145 0’90: Bode Diagram 200 WW 1 50 1 00 g 50 w E o '5 En g ~50 —100 —150 ‘200 180 Phase (deg) rlo m U‘I _270 I}: :::::::'W‘:.fl':> 1‘ : :‘r’w' 1o 10 1o 10' 103 103 1o 10 Frequency (rad/sec) Figure 3: Bode magnitude and phase plot of the loop transfer function. 2. (10 points) Figure 5 shows the Nyquist diagram for the feedback control system shown in Figure 4 when the proportional control gain 00(3) 2 K is set to unity. Rs) + Gas) Gas) Y(s) Figure 4: Feedback control system. 0 (5 points) Determine the phase margin of the system and indicate the phase margin Figure 5. ~ 724 ohm "VI-ran} ‘5 a—l’),’)fa)uflflu.£€(?_ “(5" (See. Eu araS') o (5 points) Determine the gain margin of the system in dB. Problem 3: (25 Points) 1. (9 points) Consider the following 8180 system where u is the control, a: = (561,1122)T is the state, and y is the output: . - ’l O; I 1121 = —a:1+a:1:2+u x: o ,>x i2 = 1172 + [3U y = 331 + 7:132 ; C I .1 ) K For what values of the parameters a, [3, 'y, if any, is this system % (a) (2 points) BIBS stable? (b) (3 points) BIBO stable? (c) (2 points) completely controllable? (d) (2 points) completely observable? m M03143: l W” "‘ Imaginary Axis Nyquist Diagram 2 WW“ d2 ‘4 _L. J, .J #1.. _. L 1 ~25 ~2 —1.5 —-1 —-0,5 0 0.5 1.5 Real Axis Figure 5: Nyquist diagram for K = 1. 2.5 2. (16 points) Many important technologies are governed by the equations of quantum mechanics. It is therefore surprising that few attempts have been made to enhance the performance of these technologies by employing control concepts. For this reason, control engineers, physicists and chemists have begun to pursue joint research in the area of quantum control. This work has application to laser physics (semiconductor linewidth reduction and the generation of squeezed state light), molecular dissociation (laser chemistry), nuclear magnetic reso— nance (NMR) (both imaging and spectroscopy) and a host of other technologically important processes whose dynamics are governed by the equations of quantum mechanics. This problem will explore the application of state space techniques to semiconductor laser controllfl. The small signal model of a semiconductor laser is given by the state equations ,—~———1p ’ 7 “a. x: (psi -2-l + (03”, o p=—2p+n x= -,-I)‘ I = n = — —n+7l .. J P P a y — (u o) x where p is the deviation of the photon density (total number of photons / volume of active region) from an equilibrium value Po, n is the deviation in carrier density (total number of carriers / volume of active region) from an equilibrium value No and 71 is the deviation in injection current from a nominal value Ia. In this system 1) and n are state variables and i is the control input. (NOTE: time has been scaled to nanoseconds, i.e. 7'1 2 dn/dT where 7' has units of nanoseconds. Without time scaling, the elements of the F matrix would be exceedingly large due to the short time constants associated with laser dynamics.) The variation in the photon density, p, can be measured by observing the laser output with a photomultiplier tube. On the other hand, the carrier density, n, cannot be directly measured. (a) (2 points) Is the linearized model of the semiconductor laser controllable? P: (3 F6) = (1’ ~'.)"’> MC?) =- 0 “I 1’0 ’9 “e' Sachem Is Compifl'l‘d; cat—brain“?- (b) (2 points) Is the linearized model of the semiconductor laser observable? Q= .2.» = r 0w was—«w w => édss’bu" is complete? obSeVuacHQo. 1M. Ohtsu and S. Kotajima, “Linewidth Reduction of a Semiconductor Laser by Electrical Feedback,” IEEE J. Quantum Electron, vol. QE—21, pp. 1905—1912, 1985. 2Y. Yamamoto, O. Nilsson and S. Saito, “Theory of a Negative Frequency Feedback Semiconductor Laser,” IEEE J. Quantum Electron, vol. QE—21, pp. 1919-1928, 1985. 10 (c) (6 points) Assuming that both states are available for measurement, design a full-state feedback controller i(t)=K(Z). to place the closed—loop poles at —1 i jl. 7R0. (JOSOQ -— 6d 8 '52 m zR-o the-v wa‘ge ~Jp1c¢ re pmsen‘hz'L'Ie/I 3? = (F + 6m x we, 0% '50 Clnoog‘ k. 5-. film/6 F + 5L has at -| til, or Mme Sachs? F + 6k z <le ~ll+ (7)0“ m z hit) 7‘ +2. -'* 7‘14- 23+Z=0 M (m {1344319)} ’- l_k| = 79+ (s—tm + Baku-H (fios‘rfl Auraggnyél‘c EDA) 37'+— 27"" 2- 11 (d) (6 points) Suppose that only 1) is measurable. If possible7 design a full—order observer, with poles at —5:|:j5 (these poles already take into account the time scaling discussed earlier), to provide estimates of n and 1). Draw an all—integrator block diagram of your observer. The alum-vex error say-E45923 é :- (F-l- 1-H) e—J wkerea 9" Choose... L- 80 the}? 71 2' + to 2 +50 Jot (.7lI.“ CF+L»))= Cfi+f+dflhfrj3$ 2 L. (1 a) __ LHZ- () F+LH= ('1 'l‘)+(‘4> " (L2," '1 MOI—Chm) = I‘L‘u " l‘Ll fl-ri '7' nz+C3’LI)7 +3‘L5‘LL :o ebva'y‘fig Cot'c‘c‘ClQfl-bs 0. _ -L :50 ==) L2“:- "10 n':3—L.=lo=> 2-3;;- L :3 = [0 290b50ruov- gain ueu'Edo" s+ubn—Jpa.w. mpmian‘kw‘bm \‘< The, tha \fva- —— r (at: -2><:;>+(7>~(::>w M? g: A _. U 0)(9,> 12 Problem 4: (25 Points) 1. (12 points) The linearized model for a pendulum, where the input is the acceleration of the pivot point and the output is the pendulum angle, is Y(s) ba A G = = ——~—. = C» = P(S) 32 + an a @ ) 2 It is desired to damp the process so that the closed-loop response is given by _ Ym(3) _ wi __ Gm(s) _ UC(8) _ 32+26wns+w3 ‘- ’gm; Determine a continuous-time controller 3(3) U(S) = T(s) Uc(8) - 5(8) Y(8) so that the specification is fulfilled. a N0 acrofio Car)an 5.3% A9:— 409m. I _ C4 .- 1 9 c For MlfllMaQ ensign“, 3).!th 54:43 i? Un=gyc53 ’“cn/ ) o Fmfl. 3.? (Ab «ma Spam? I49 -' a = 1'25] mu 83: 0,9,. =2, gym; flycmagm flew.) 2+ Sgt IAOCS) :- S+ A; the, 1001100144” 6,051 'E‘Wp p3,“ (yum A") = - fwn fawn ‘_fz-’ A. 3909 chance U. are. 9L 5 ‘SDLQ For R «Amo- S: E AM no 8 S P—-—-—1 a + S‘S'A'Su) = (.57? 'LanaS 1"”3? I)“ + p) (53-440)“ «- a 3 L 1 ‘55 1- 9w: 33+ Voszi- (dawns): + (as. +5.»)? 2 s + (P + zf...,.\s M7. mew». t5ou‘blha (99%6103'653 SI: (‘0: p+qum 3‘: a.+b.s,= “I; 4" Z-fwnp 5°: a“; 4- base 77w" 1.3 2. (13 points) Consider a 8180 plant with transfer function _ Y(s) _ bls — U(s) _ s2+als+ao that is controlled by R(s)U(s) = T(s)Uc(s) — S(s)Y(s). Following the approach introduced during an earlier laboratory session, determine R(s), 8(3), and T(s) such that the controller has integral action and the transfer function from Uc(s) to Y(s) is bml 3 + bmo G s :—-——————. m<> 52+am13+amo The observer polynomial in the characteristic polynomial should be (8 + p)7, where the positive integer 7 is as small as possible. . we 4m replwy— me Wm“é zero: Lab 3+3'S (&?~’Pr£e the ECL— t“- fire is unbttb‘e\ drug; 8— 5" bl c K. . T'o [unaided-a (Kim/rout! “15—105 R nos‘L' Inciufl— a- 2% In “Qfltfiog R muS‘lT Incucab 8+;s‘, amfl So we can c case. RCO ‘-'- LVRI 3 S R! ckfi.(@-7=-J’e 09')" 7' (J “‘4’ 8° W =| Cab (us) Is monto)~ K68) = s ' Fungi aefiCpa) 9R +- as ‘ 9.6 1 A09" 6+ --3 £0 [PO-:0) chowe'lso — 12¢ (140‘; +— (Aq) +— egg/(6*) ~ 5; flagm + fiflm — a». 2 I l 5 *‘50 C30 C8) >dfl— “(5:1) L ‘ ~> (quqls W) + 13\CS\‘S""$~') z 5 fails + qMo ‘- 8 n ——v H‘ £+f’ + £4 3 3 $604172. Coe ICIonfi-S 30'. 40 + 5‘50 5a” ano'QO 2- .. 9 / 5: ’1 awn—q So Si:ql1~b|3‘340n$ 3" TL b ...
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