EE 3530 TEST 2 Spring 2003

EE 3530 TEST 2 Spring 2003 - l ‘ a Miterm Test(1 EE3530...

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Unformatted text preview: ). l ' ‘ a?) Miterm- Test (1:), EE3530 ' pts). What is the transfer function of a. PD) centroller? (5 pts). What is the differential-integral equation describing the relation between the input and tput of a PID controller? ' (c) (10 pts). Design a PI controller using Quarter Decay Ratio method for a. system with the following step response. - m a I ..__.‘__.. n .. n 3 1n 1 l . an an m :3. 4|: 4a Figure 1: Problem 1 — Design PI Controller Problem 2 (15 pts) Consider a system with transfer function (3 + 2000) Gm = (33 + s + 1)(_s* + 103 + 10000)‘ Compute the the overshoot and settling time of the step response of the system. (Hint: Truncate the system as a second order system by eliminating non-dominant poles and zeros.) 'Problem 3 'COnsider the unit feedback system shown in Figure 2. Suppose that the controller is a proportional controller with gain'K. Let G(s} = 3%. (a) (5 pts). Let r(t) = 1(t) and K > 0. Determine the steady state error. (b) (10 pts). Let r(t) = (2 + t) 1(t}. Determine controller gain K so that thesteady state error is less that 0.001? (cl (5 pts}. Let r(t) = t3 1(t). Show that it is not possible-to specify If so that the steady state error I is finite. ' (d) (15 pts). Determine controller gain If so that the output has an overshoot of no more-than 12% in response to a unit step input signal. ' Problem 4 Consider a system shown in Figure 3. Assume that the controller is a PID controller-width transfer function C(s) = K(1 + 373; + Tps). Let C(s) be stable (i.e., all poles have negative real parts). (a) (15 pts). Show that E(s) = l—ficfifl— fin where R,D are the Laplace transforms of r(t). d(t} respectively. (b) (15 pts). Let the reference input signal be r(t] = a. 1(t) and the disturbance signal be d(r) = w 1 (t) where a. and w are unlmown constants. Show that the steady state error will be 0 for any constants c and w. Figure 2: Problem 3 Figure 3: Problem 4 _ $4 1.000 (514-54!) (5’3 ‘r' sous-{moo _ (JON H 1* 53+15wn5fL0n5é Mm ffi FROM Loomgucg RT (fiLUd/IC- ’Dlfi-LHZW :1. off: / 5:165) = (14.19) " W8) V155) r gcwgém) ' / (Om'gzué TH-gsg 'To (1 e7' Lit-’5'? “=— racs) ~- ETD £6165) ECU fl + Kt’qufl: gas) 5(5) fl Bag) I 1‘ K6113) . of #rND' 1% Vii—E“) = r H) <%—-—«—~—«-$> é :- 2(3) NOW "Fl/MD STEM)? gkafenemffl; £59337 ?. Lat/$555 ‘ 5am) 51.1w. \ .. «._- azlw' For: 833 mean. .-.--«._ / 0 rm) '1 t3 t Lt) t 2537'“ 0’ “To MSW-{‘72.} ||‘~—LL51ST Film/5 Tim? go AMT P05 3”; LE "{1} Cip/f’ewvt- L.“ ‘i’ J/x. @Dé‘firnu‘ufi k; iér‘r 18:75:) HP—é 0% -——p Hp { clef FWD TE..&-:\Js+‘rf7;> WM; [ Eek. .7 62: b ave Uaflm.) . £2.53-.. _ ,7“ Z75“) : gotcmsffls) "gf312}+'0!‘-' gm «4:; w. gzfljfflogf: . sis-12) " Tlf e” w”? 5z+25b§1n$ "J’UOWL E \ @CGV- K 1+ 4117123") EMS-i- I“: 37%“: .....7'£‘3 /‘ [@ V5.1) -—b ECO ‘ d“) 4;» DCS) Fm” momma (W 7:41: BLOCK DzAaflAM GIVEN W 73633 ,a o ...»~W E, J: i V I 4 CE; 5555} g 5(3): {265) ~— grams} {’5} = x + C-pcs‘fic’cs‘) IL. 3:419 C: CD) I 1-} {11¢ *‘tififl Y’C—{y —:»- a, i('{:) 5L (lard WM’C-Wl I GOV-gfiHZLL-L.§ chi-é) “3- CoifiéJ gL-LD‘LU vii-1a.? €+€5€ « #0591. mm {,{fiU Rafi 0.93m” amt-'7 a/bLX I I. Mwsrfil-WW Elm) “Lilo/L 35(5) 2: C3 r-n W W SFDD_ _ ..g(oo3 : _' hm .5. P] +UW-S { 5-90 + ac s—vo “QM-(Wham m @y . a. _ _ I_+ (“—7 5—90 I + (’10, «ff—.51) w E( ' rct)} 1;} ‘3; I. ,7... I' I --. - w m 5 U“ ——L ¢ [AM __.. C163}: assrmea w/aLQr.12_c\o ; 5470 0+ (51/ij 5—90 ’5/Ci-f’C‘Z‘l {€121 grads. «FL-r F0123, 3 . ~17: Haj j at J... ‘ H Um Zac) + _'.___ mo CH *7 ‘ A“ .u-— 'r ...
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EE 3530 TEST 2 Spring 2003 - l ‘ a Miterm Test(1 EE3530...

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