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HWCE5_Sol - HW/CE 5 Solution(Where applicable use the...

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Unformatted text preview: HW/CE# 5 Solution (Where applicable, use the “more exact” tableon page 2 of the ref sheets to answer the following.) 100 1. Su oseGs =— pp () .qs+10) is the open-loop gain in a unity feedback system. a) What is the phase margin PM and crossover frequency we for the system? 0 ,mp/0 ‘§=05’—9/w=fl3 %=J%0 = $7na4/é b) Sketch the Nyquist plot for the system for a) > 0 in the vicinity of the -1 point. Ail c) Suppose we consider the effects of delay in the feedback path such that H (s) = e"""' where t,, = 0.041 sec Note that this will not change wc. What is the new phase margin for the system? . ..s[ao%/ CaL¥a1= _fiEL, a, ) {(51410) (1) Make a rough sketch for the Nyquist plot for the system with delay added. This need not be detailed but should show generally what will happen in the vicinity of the -1 point A3 and zero point on the plot. Excel generated plot for 7.5 5 w 5 432 rad/sec ( 1/07} 1?; ‘3’ > '3' Nyquist PlotG 2 _ 0.2 .0]. ._ e) Estimate Mp, tr, and t5 for the system with delay. Note that although we did not change, a)" will change in that the phase margin ( and C) changed. 0 My: 33.8 -—-7 f? 0.3 ——> HP = 34.237. Mn wh if: {Lg-Vi: alga Sec. “n K4? *fi .2155: =ité£m = 417,4“, 1) Use MATLAB, Excel, or a program of your own choosing to simulate the response of ‘. the system (with 0.041 sec delay) to a step input r(t) = u(t). What does your simulation f yield for Mp, t,, and t5 ? ‘ 671/; We ,D/9f ”{14' Excel Simulation plot: Step Response y(t) Time in Seconds Mp tr ts From 1e) above 37.23% 0.152 sec 1.79 sec Excel Simulation 39.5% 0.136 sec 1.71 sec MATLAB Simulink approach: Block Diagram: Transport Delay Simulation Results: a} Rmpe «I #191331- QIIMHEIW’S g .4 *2” :x", ”A . 63* g. :2- ,. ‘..I p: 2. Suppose we have a plant G(s): 21.4 Gm = s(s +1.85) in a unity feedback system. a) Determine for this system: settling time t5, overshoot for a step input Mp, crossover frequency we, and phase margin PM. 2 , . __ = . .' ' = . n. . My - N’v‘ .. La)? — (£626, 1-9;)” Ms" 39 0/“ 02. ts= 4.43 ._. 5;”:5‘fl‘o g) ”P —— 572.43% Z‘Jn a ,wb = (9,74% = 44%,...14 ) P7: 22.4 .—_-=:_._—-. ____’ K(Ts+1) b) Choose a controller: DC (5) = with a selected for a lead compensator. (aTs + 1) Find K, T, and a to realize a new design phase margin of 65.20 and a new crossover frequency of we = 9.2 rad/sec. 5w L/é[zd&)/r / = K {Z/fiP)” > k= 44a”— (ZN/m“ m (a ”cum : MW — (—18» = we», bar 0 IA 0 . / a » _. mg? )0 M (.5 . Wm /H =- (CZ—H.370 = 53.83" See 7/(152’ 94:} C. °¢= /— {“43” .3 (ix/M69 M A) = (J; 3 “7,2 = l...— Ww “TC/at .'. '7'— ’ __.L.—. — 0, 3329 22 .ma fl/JMM M : /¢_ w.» 64.03% __ 54434 “ /Dc (3' i2)/ = /.3/-7- . _ , 3 4 g 44:27,» #2:» 2— A50?) x o./od 7‘ _- a . '5’ 32 7 c) What would you estimate for overshoot Mp and settling time tS for the compensated system? 0 P” t 6 5" Z’ ) ”(a = g 2 MP : 4, 4 70 ’ I - J 1 f i = ’9 Rf, t ‘4)" 0‘ l r 7”,“!!! {1&5 /€ ' t! 5,??? ) g j "I: I a 5:)" n 50 g = +4“ 2 0/74?8 m 95m. ; ”f- d) Check your design done in part b) above by: 1) Doing a MATLAB generated Bode plot for Dc(s)G(s). Use MATLAB’s “margin” function to check the values of we and phase margin for the compensated system. How did these values compare to your design target values? Bode Diagram Prob 2 PM = 65.2 deg [at 9.2 radrsec) Magnitude (dB) -135 P hase (deg) 4 8|] - 10" 100 101 10‘ 10 Frequency [radisec] ) (a Design target values matched. 2) Finding closed-loop T (s) for the compensated system then doing a MATLAB generated step response to determine overshoot Mp and settling time? How did these values compare to the values you found in c) above? Amplitude 1.4 1.2 0.8 0.4 0.2 Original Uncompensated System: Design Goals: From 2c T(s) for Compensated System Mp 8.04 ‘36 0.5 Step Response Prob 2 settling time 0.954 D .B 1 Time (sec) Mp 52.7% 4.6% 8.04% 12 1.4 5.0 sec 0.5 sec 0.954 3. Suppose we have a plant G(s): 21.4 ' Gm = s(s+l.85) in a unity feedback system. (Same plant as for problem #2 above.) a) What is the steady state error for a unit step input to the closed loop system for this system? We / fig/em, r, 3;. arm #4.? 2; a(Ts +1) (aTs +1) input r(t) = t u(t) to the closed loop system is 5 0.05. In essence, find a and T for the lag compensator. b) Add a lag compensator: Dc (3) = so that the steady state error for a ramp /: ’_ / I ' P -:. ’1’; -: a .w.. - -_— W J 3 #136- / +. 404+.) (1.4 p _, V («72 +1) 5 (he/:5?) [in-v f5 7‘ an.” (df+l>(+/.8f7 : Aar‘ e ,0: 3-90 590 §(.(;3+,)(5f/lgf)+0((7§¢I)KZI.V flat x 2 A72? ~r sd 0(= A72? {9W .1 <4 NC “I: 35‘ __ 3&9 W.) 7’: <,{§‘DJ— r 7 2:0 20 c) Check your design done in part b) above by: 1) Doing a MATLAB generated Bode plot for Dc(s)G(s). Use MATLAB’s “margin” function to check the values of we and phase margin for the compensated system. These should be the same or close to the values you determined in problem 2a) above. Bode Diagram Gm = Int 9113 (at Int radlsec) , Pm = 21 .4 deg (at 4.45 radlsec) 100 50 6‘ B l.) E O 4: 3| 2" -50 4 00 ~90 a a: B a, -135 w a 1: 1 -130 ‘ 10' Frequency (radfsec) PM we Problem 2a 22.6° 4.44 rad/s Above Plot 21 .40 4.45 rad/s 2) Finding closed-loop T(s) for the compensated system then doing a MATLAB generated step response to determine t5 and Mp. These values should be the same or close to the values you determined in problem 2a) above. Step Response Prob 3 Mp = 55.3% settling time = 5.09 sec Amplitude U 1 2 3 4 5 6 Time (sec) Mp ts Problem 2a 52.7% 5.004 sec Plot above 55.3% 5.09 sec 3) Doing a MATLAB generated ramp response for the closed loop system to be sure that you met the steady state error specification. Ramp Response Prob 3 Ramp Error = 0.05 at 9.21 sec System: T Time (sec): 9.21 Amplitude: 9.15 Amplitude Time (sec) Steady state ramp error = r(t) — y(t) = t — y(t) as t 9 °° = 9.21 — 9.16 = 0.05 at t = 9.21 sec ...
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