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examI_s07

# examI_s07 - EE 429 EXAM I 21 February 2007 Last Name(Print...

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Unformatted text preview: EE 429 EXAM I 21 February 2007 Last Name (Print): 5 OL'EA on} First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score] 1 35 l 2 35 L_ | 3 30 Total 100 _11_\ISTRUCTIONS 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: (35 Points) 1. (15 points) A difference equation may be used to model the temperature of a cup of coffee cooling to room temperature. Let T(0) represent the initial temperature and T(k) the temperature after k minutes of time. Using Newton’s law of cooling, the change in temperature over time can be represented by the difference equation T(k + 1) # T(k) = a {T(k) — TB}, where a is a constant and TB represents the temperature of the room. Suppose that the temperature of a lecture hall is TH : 72°F, the initial temperature of my cup of coffee at the beginning of lecture is T(0) : 180°F, and that a = —0.05. Using Z-transform techniques7 determine the temperature T(50) of the coffee at the end of a ﬁfty minute lecture. T<K+O — (or. +I)T(t=3 = "°°TR 10‘) z-rm- arm * WWW?“ = a-. (%~l-oc)TC%) = 27% " x‘Tp‘ 2—: THE) —_; i— T(o) -— OLT‘K 2' a—I-oc (a-l-th-H Expamﬂ the. seamﬂ- term 60 al.- Jvm or far-Eli [fab’é'ons l .. - c :. (e/o ) — \ __ l a l C; . +- _C2_ (El ©0195 2:, (Z‘l—olb(%'l) _. £4 24”” l : i if.“ ‘. ' 2 CL ' (2/60) (;'|-m)(2'0 2:14-01— T( )- __i‘—— 1-0») + ‘TR A - TR 1‘ i- - a—lvoc 2-1 z"‘“ J... 0:. 2. (20 points) On January lst Jenna purchases two newborn bunnies7 a male, Patrick7 and a female7 Patricia. Every month a pair rabbits can produce a new pair which becomes productive from the second month on. Let p(k) be the number of pairs of rabbits in the km month7 then IOU»? + 2) = 190»: +1) +p(k) for k 2 0. The initial conditions are 19(0) : 19(1) 2 1 because Patrick and Patricia cannot produce a pair of rabbits for the ﬁrst two months. Using Z—transform techniques, determine how many pairs of bunnies will be living in Jenna’s apartment after one year7 that is7 determine p(11). PCK+ Z3 — PU’x +0 -p(|o) =0 21PM)» aszs—zem - a mg + 2W0) ' W29 = O . =p = emigre—2 — x} = 229m- (o)1+ 22H) , “"3 ml Poles Sat/{25C} iZ'E—l :0) £2— — _L 4~ .5" H21): in?! l 2: JZ—t ﬁnal ~L—\) 7 Equngz pCE) cub 0L. sum o-l; Par‘lmovl ﬁfacbroﬂs Z 6* l 01 + Mu z—v— .— Z __ z’:+Ez-: Problem 2: (35 points) 1. (25 points) Once again consider the difference equation Mk + 2) = 1106 + 1) +1106), where p(k) represents the number of the pairs of rabbits in the kth month, and 11(0) 2 12(1) : 1 are the initial conditions. i (k) specify the matrices A and C in the state—space representation (a) (5 points) Using the state vector \$(k—l—1) : Aw(k) 1909) = 050(k) of the difference equation. (b) (10 points) Compute the state transition matrix ¢(k7) using the Z—transform technique (c) (10 points) Use the state transition matrix to determine p(11). ' o A (0') Y‘CK) : anﬂ— X;(")=— PCK'“). It 4:0“ was "tan x‘ (Kw) :— XLU") X2(K+I) = XIC'L) 1” XLUL) , W W yCK+7—) f0?) F“ +0 XMW) : x04) 5 (had) PCILH l ' —[ z | 2* 7(21—9 " = Z = 7.‘ , ) (I am) 2 a ' vino-l ‘FroLan/n erpanIu/‘I From Page, 3 we, whiuuo? know “thq- pa " r‘ ’L-ﬂ 2L EFL]? f? + ’5‘ S— %'4—+E— . 7- 2 I = IO {vb-’5”; '° " 7. a __ _ pow CamwQW C. C1. ‘3 ___._L..-—~ + -.—-——/‘ 2 ZZ—E’l] _. %’L_; éajz+g #/ 7v 2* E — “’ch ? I”? #2 r’ (A) — '0 - -4. 5* *Z'E" ENE—’32 '0 Z 2“: Uslﬂg Q.bJW 'uﬂS QA‘Q W12 Can eyfq/kﬂ/ remain/y 05¢. I Ubecb3 K—‘a 22"Z‘ 22' _ & zz_£,l ~ EL-E_1 gzd‘f'l 5 F" % - I0 2-J—+_i_ lo 2—4-7131 2. 7, 3' L @rn aJ :93 av cMeuL) no‘tz/ tka£ gen» :03? (o (C) {90‘} = cxm : C¢(I~’—)7((o)») mo): ’10:“) 3% F — l W9 “ C‘ 03¢“) . - 5+5“ L+E>H LEO??? L20 )0 2- L )o W : sq» ¢ ﬁe ﬂaw/mp) T/PQ’ QF/(QLSIOH [:5 [joﬂ—£:(c£ ﬁg Oyec‘una‘ﬁ I)! prob/em / [par-é 5‘ ,3: r” h - l__ H M: "7—r(é+%> + if (#525 (DUI) ‘: )V‘rf 2. (10 points) Consider a 8180 LTI discrete—time systenl represented by the transfer function Y(z) 4z — 1 G z : : ——————-—. ( ) U(Z) 823+1222+6z1+ 1 Find a state—Space representation I05 +1) : A\$U€> + Bu(k) y(k:) : Cm(k)+Du(k) of the system by specifying the matrices A7 B7 C, and D. p Ex 1- 4A.. :1 "‘ l ) g F e“ 6 Z) 6C?) 325 V6) ” 2833+lZ-Zzz4'6-E 1w Yz'l} - From P( l 5-) [/(} : ————-~—-———-———-—--'— / 3 833+IZEL+625 —H we. a b‘Eq 1h 8z3f’ca) : ﬁzz; (’(aj) ~ 62 We) -PI&> + uh?) 3/7, 3/1 J—~(A'(L) fﬂs-rSD : w? F(K+z3 '%F(K+l) —{?—lo{)c) + \$9 955:5“ varuiaiej as (Leg? in "NJ, thug" £146 Cﬁ’olce 75 m‘é vmﬁueﬁ: x)(i‘) 5 X2 (F) =x,()c+.) = f (KM) X3 (F) = 2.041?) : F0441) 06M tie Ow’E‘Qr Ckolce/ % 542646 varlq gléJ a“ mp) - 3990:) + 1: MW p—J Lp—J flkﬂ) PC “11) Problem 3: (30 points) Consider the discrete—time feedback control system in Figure 1 where the transfer function of the controller Gc(z) and the plant Gp(z) are 42 Gr) 27 : r—-— ‘( ) z” + z + 0.25 G42) : KP. controller plant Y(Z) Figure 1: Diseeret—time closed—loop system with controller Gc(z) and plant Gp(z). l. (5 points) Find the closed—loop transfer function representation of the system, and express your answer in the form: Y(z)\ bmz’"+---+bo Biz) zn+anilzn'1+-~+ao' m) - 6c 69 kt, _._"i_____ Mt) l+6 6p 2' +3 *0“ _____L/_i_‘_l‘:L_..a—— o _______________________._____.—. = ’- L1 11. 1: 1+ k? '12- E+z-+ozs+ P 2-14.3-9-0-13‘ YCE) _ 72‘ kp zz‘+ 210+ lephroz: R621) 2. (10 points) For a unit—stop input Mk) : 11,0(k)7 determine the steady-state error, em : lim kax and specify the condition under which your result will hold. RCE);@_ E'CE) ER) _(___ __ \ __ 22+ £+o.z.s“ 1-7 " l+6c6 _ 7% 7‘ 13’“ N p |+ k? gnaw-K z + (qKP +01- +0 «C(naQ, VQJVQ, 13460er 9-55 3 Quin (2—0503) 22)! E‘CiL) v 1 QM (it'd Egg) («a 2‘5' _ \ 27'4- Z + 0. 25‘ a? - Q'm ( A jﬂw , 22’7‘ Z + (‘1'[<P+l)£.1-O.Z§“ €35 = 2, ZS" ZZY-I- Kp s/ Ah ‘ I 7‘ 3J9 a” d"?— QarE re>JH3 093 ponfj “UL the“ [a J 3. (15 points) Once again consider the discrete—time feedback control system in Figure l where 4z G z :,————. pl) z2+z+025 Suppose that the controller is changed to z z—l‘ For a unit-step input : uo(/€)7 determine the steady—state error, Gee) : KP + K1 ess : lirn k—ix and specify the condition under which your result will hold. (K +-K::\L"‘l< G (a) : k? LE '0 4 K: L: ‘ P P) C 24 2' 2'75" than) 0—57 r" part 2-) 1 29:1. — ‘6 6 = —-———’—”l”"‘.,:/ ' '+ o L " < / 2—1 2- (%-—I‘)(2:'7-+ -z. +013 w) + [(Kpmﬁz—kpj “1% H (Z:~n(z:"+ a w- €ss ’ 2—7! . ) (arm (54% +0 25 ( "i — r z— : gig)! C 43 (z—QCEH 27w“) *[WPH‘LH ‘91 V 1 .52., ‘iKz €55 3 0 U2— on 900-4 7/92, Quyl: rest/If u/Ju WI, (20994 % ogﬂEca are m’ﬂﬂ/ﬂ t 011%. 74 Clﬂ/‘z’% tCL 10 11 ...
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