This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 closedloop system where the plant transfer function is 1 GP(3) : 82 _ 4 A compensator is needed so that the unitstep 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 phaselag compen
sator GAS)? Justify your answer in one or two sentences. oCoos\8or the. root foch usma ago) =l< .20: In a Regerless 7! ﬁlm preparing/lag Can‘eral gab: Va.le m
m cloSaQ’Qoq 5d5Lam Ls pa'f: $460 mue ( 41$
~A’s plNJﬁ Quf Car ’2) (onerol leafs ti“ ro’b goua’ﬁ’ék'w) J 11': w.“ no'h raylé 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 phaselead 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 hat49' pall— ggab‘m aria; ‘f = l’a‘cmfﬂ /l (ExU’tPDLnv" 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 unitstep input. 8.333 _ _ __
K 5 696p: ’ 50‘ go ( —‘ Z)
P s—ao Pl   .1. 3.
€55 = —L .. 1° 4. (8 points) Augment the phaselead 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?) _ % iaL
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 21025 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“ (“$00 Wynn [51Hsz “’1’” “7%., o (5 points) Estimate the peak overshoot of the closed—loop system to a unitstep input. Because PM (C)J ‘Ht dual—ﬂoor; wok—m ;$
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 "VIran} ‘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 2l + (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. 19191928, 1985. 10 (c) (6 points) Assuming that both states are available for measurement, design a fullstate feedback controller i(t)=K(Z). to place the closed—loop poles at —1 i jl. 7R0. (JOSOQ — 6d 8 '52 m zRo thev 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 + BakuH
(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 alumvex error sayE45923 é : (Fl 1H) 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 ﬂri '7' nz+C3’LI)7 +3‘L5‘LL :o ebva'y‘ﬁg Cot'c‘c‘ClQﬂbs 0. _ L :50 ==) L2“: "10
n':3—L.=lo=> 23;; 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 closedloop response is given by _ Ym(3) _ wi __
Gm(s) _ UC(8) _ 32+26wns+w3 ‘ ’gm; Determine a continuoustime 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) (53440)“ « 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" Zfwnp
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 [unaideda (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 $604172. Coe ICIonﬁS 30'. 40 + 5‘50 5a” ano'QO
2 .. 9 /
5: ’1 awn—q So
Si:ql1~b3‘340n$ 3" TL b ...
View
Full Document
 Fall '07
 SCHIANO

Click to edit the document details