examI_s01

# examI_s01 - EE 428 EXAM I 27 February 2001 i ID 1 1 Name...

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Unformatted text preview: EE 428 EXAM I 27 February 2001 i ID#: 1 1 Name: SOIU‘LI‘O t DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO ) E I score - - - - - This test consists of four problems. Answer each problem on the exam itself; if you use additional paper, repeat the identifying information above, and staple it to the rest of your exam when you hand it in. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentence should convey what you are doing. Problem 1: (25 points) A 5180 LTI continuous-time system with output 11(3) and input u(t) is described by the transfer function _ Y(s) _ a: + 1 H") ‘ ms) '" 53 +—'"'23 +1 1. (5 points) Find an ODE description of the system. 53m) + 2:. m) + m) = 5‘ um + was.) 3‘+zé+}=ﬁ.+w 2. ( 12 points) Represent the system by an all-integrator block diagram that uses the mallest number of integrators. ﬁg? = Lac—as “K5 53+2\$+‘ ( ) .. qn'L 6.:H-P'l‘f’ 15' = “350-? H- Problem 2: (25 points) Fluid ﬂows play an important role in many physical systems. As an example, hydraulic ﬂuid ﬂows are used to move the control surfaces on airplanes. In this problem you will analyze the water tank system shown in Figure 1. Water enters at the top of the tank with a mass ﬂow rate of win(t) [Kg/sec] and exits at the bottom of the tank with a mass ﬂow rate wout(t) [Kg/sec]. The ﬂuid level Mt) satisﬁes Mt) = [winm — woman . where AT [1112] is the cross-sectional area of the tank and p [Kg/m3] is the density of water. This continuity relation is simply a statement of the conservation of matter. If the input and output ﬂow rates are equal, then the ﬂuid level remains constant. 0n the other hand, if win > wont, then the ﬂuid level Mt) increases. The ﬂuid exits the tank through a discharge nossle with cross-sectional area AN. The output ﬂow rate is given by Toricelli’s equation wou.(t)=pAN 29A, i/hltlu 1—:s '1‘ where g = 9.8 m/sec2 (use the symbol 9 in your analysis, do not substitute the numeric value for g). a): _ ATl m V W) k ANl (“out Figure 1: Elementary system for studying ﬂuid ﬂow dynamics. 1. (6 points) Find a state-space representation using 3(t) = h(t) as the state variable, u(t) = win(t) as the system input, and y(t) = h(t) as the system output. 2. 3‘66: Air [uﬂﬁ— e19” ——2'-rl‘ m R] 2. (7 points) Suppose that the input ﬂow rate is a constant value u(t) = up = mfn. What is the static equilibrium height h. and output mass ﬂow mg“ in terms of the constant mass ﬂow input ufn ? 2. o -3” :iE‘V—E'i‘jﬁw‘" 3. (12 points) Suppose the input flow is perturbed by a. small amount 6min from wfn, that is, win(t) = wﬁl + Gavin“). Find a. linear state-space model 5W) = f5==(t)+96u(t) 614*) NEW that describes the dynamic behavior of the ﬂuid system for small perturbations from the static equi— librium state 3,, where f, g, and h are constant scalars. Problem 3: (25 points) 1. (10 points) You are asked to design a. second-order system, with no—finite zeros, that meets the following design speciﬁcations. 0 A DC gain of 10. o A rise-time of 0.45 sec and a. peak overshoot of 16.3 % for the zero-state unit-step response. Specify the system transfer function. in“ 045‘ x "57w" =5 “M = :1};— (f: - gum? irr’w-aitnvp «49:46.3 Km. tug“ HG) = .s’-+ szns +w3' 2. (15 points) A 3130 system is described by the transfer function H(s)= 200 (3/100 + 1)(2.«:z + 45 + 200)" Estimate the rise-time, settling-time, peak overshoot (96), and peak time of the zero-state unit-step ' response. Carefully justify any assumption that you make. Hts)= -—-—-'°—‘5—-——"“ (shoe + I) (32‘4- 2.5 + loo) Complex- Fob! 0v"? 2- .. = we 3:: 3'”; .21.. 3 aw WM Z‘PWO" -’ '1 T— chause the Pole- avé: 5;: -mo 35 far- to the, lei-i: a: "H‘e— COrnpi-C‘ﬁ Conmgtl'ﬁu chI‘r’ 'H‘IQ- 60mph» Pole; Marnch the “heme. wronge- charm‘Ler-Lgtics. Problem 4: (25 points) Consider the feedback control system shown in Figure 2. Figure 2: Feedback control system with plant Gp(s) and controller Gc(s). 1. (4 points) Is the open-loop plant Gp(s) BIBO stable ? Justify your answer. 6,3, m; poles at £1. henna?! the. Eoﬂeaaé s=+b the. 5 J‘tﬂn 6 c5) :5 no“: 8160 s‘quOle. 2. (8 points) For What values of K, if any, will the proportional (P) controller G',(s) = K make the closed-loop system in Figure 2 BIBO stable '2' F"- eggs _-_ K) Y0) _, 4354-92.. __. K RC5) J + KGpCS) 52—..- Kr: :42 K 'there, w." be. a... pQ..Q S: + 1—K «~9— So 1310. sax-hem T5 unshble‘ F‘or- K A l the P195 up,” he. on tile. dub axis OWL :o the. ant-hem is why-table. ﬂlfkrnw‘hluglgj 05:9, “U0. gO-Jt‘) arm} For K. < I ikere. is a, S“ n 65‘4" "‘ ‘ 7'7 52‘ l '4- "l l I}. 3th,; ,1 1,4, shat!- p (1:19.. For! k lJ a k (a a a ‘ i; but” the. a!qu 5| .3“ Qﬂt ant? (K.,) I: (as; we, m {U e K- 'l “N %. tar-us [ it,th "bug loci” J a ‘3 \$115. For k, = 1) {two rows Earn} :0 (Cat til“ Pd? du'b' 5:0. % The. ClO‘Sng-vﬂoof «Sanka/n :5 pasta-HQ. ‘Q/ 4&- uaJK-QS ; K. / _—_____—____________—___\ 3. (8 points) Suppose we use a proportional plus derivative (PD) controller G=(s) = K + K D s. For what range of K and KD will the closed-loop system in Figure 2 be BIBO stable ? Ya) atmcem =_ K-I-KoS ‘ 7‘33 7‘ li-QQQB) .s"‘+ koS *‘ K" Raﬁ?) ATV-our} 4. (5 points) Suppose that the closed-loop system must have a steady-state error leu| < 0.1 for a unit- step input r(t) = 1(t). Using a. PD controller, determine the range of proportional gain K for which this design speciﬁcation is met. Rssvmma_ the. .yJ-Lem 75 3160 sire-blag, ‘Fw a. ware—slur Inpa‘: “4%; I - K = ' L5) (5 =' ‘ T“ 955 1 1 + M: J P .ggrét 6’ ) Ho 5 “I . = _K ...
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