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### 243 Control Systems RL_Tut

Course: ENGINEERIN EG 243, Spring 2010
School: Swansea UK
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Word Count: 1496

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of University Wales Swansea SCHOOL OF ENGINEERING EG 243: Control Systems Root Locus 1 Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K: a) 2 K ; s (s + 1) b) K (s + 2 ) ; s (s + 1) c) K (s + 2 ) s(s + 1)(s + 5) Use the root locus technique to determine the range of values of K for which the system with open-loop transfer function Go(s) is...

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of University Wales Swansea SCHOOL OF ENGINEERING EG 243: Control Systems Root Locus 1 Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K: a) 2 K ; s (s + 1) b) K (s + 2 ) ; s (s + 1) c) K (s + 2 ) s(s + 1)(s + 5) Use the root locus technique to determine the range of values of K for which the system with open-loop transfer function Go(s) is closed-loop stable. Go (s ) = K (s - 1)(s + 2)(s + 8) 3 Plot the root locus of Go (s ) = s (s + 9 ) 2 K (s + 1) as a function of K. 4 A position control system has plant transfer function G (s ) = 4 . s (s + 4 ) A cascade proportional plus derivative compensator with transfer function D(s) = KP + TDs is added to the system so that the closed-loop poles of the compensated system have a natural frequency of n = 10 rad s-1 and damping ratio = 0.5. Determine the proportional gain, KP and derivative time, TD required to achieve this response. Calculate the steady-state velocity error of the compensated system. Hint: use the root locus angle criterion to find the location of the compensator zero when the closed-loop poles are positioned as specified. 1 1 Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K: a) K ; s (s + 1) b) K (s + 2 ) ; s (s + 1) c) K (s + 2 ) s(s + 1)(s + 5) See L17-20. Root locus sketched using 5 rules. L19, S7, S10, S12, S14. a) See L17, S14-17, which is a similar problem. Go (s ) = K ; s(s + 1) Gc (s ) = K s2 + s + K = K (s + 0.5)2 + (K - 0.25) At K = 0.25 there are concurrent poles at = -0.5, which is the break-away point Rule 1,2: there are 2 branches; n=2; poles at s = 0 and 1. Symmetry Rule 3: the root locus is on the real axis for 1 < s < 0 Rule 4 there are no finite zeros; m=0 2 infinite zeros The branch start and end points are: 1 at [0, - 0.5 + j]; 2 at [-1, -0.5 - j] Rule 5 using L19, S14/15 the asymptotes are at 90 and 270 with real-axis intercept at 0 = finite poles - finite zeros = (- 1 + 0) - (0) = -0.5 n-m 2-0 1 numKGH=1; denKGH=conv([1,0],[1,1]) rlocus(numKGH,denKGH) denKGH = 1 0 2 b) Go (s ) = K (s + 2 ) ; s (s + 1) Gc (s ) = s 2 + (1 + K )s + 2 K K (s + 2 ) Rule 1,2: there are 2 branches; n=2; poles at s = 0 and 1. Symmetry Rule 3: the root locus is on the real axis for 1 < s < 0 Rule 4 there is 1 finite zero; m = 1 1 infinite zero The branch start and end points are: 1 at [0, ]; 2 at [-1, -2] Rule 5 at 0 = finite poles - finite zeros = (- 1 + 0) - (- 2) = +1 n-m 2 -1 using L19, S14/15 the asymptote is at 180 with real-axis intercept It is possible to calibrate the sketch by finding the break-away [K maximum on real-axis section] and break-in [K minimum on real-axis section] points. For points on the root locus, 1 + K G(s) H(s) = 0 K G(s) H(s) = -1 - x2 + x K (x + 2) ; u sin g x for typing ease = -1; K = K G (s )H (s ) = x( x + 1) x+2 dK - x 2 + x *1 - ( x + 2 )[- (2 x + 1)] - x 2 - 4 x - 2 = = dx ( x + 2)2 ( x + 2)2 dK = 0, giving max and min values, when x 2 + 4 x + 2 = 0; dx x = -2 2 = -3.414 or - 0.586 Checked by transition method x = -2 - 2 : x = -2 + 2 : ( ) [( )] (6 + 4 2 ) + (- 2 - 2 ) = 4 + 3 2 = 5.828 (- 2 - 2 ) + 2 2 (6 - 4 2 ) + (- 2 + 2 ) = - 4 - 3 2 = 0.172 K =- (- 2 + 2 ) + 2 2 K =- Previous solution was x = -2 3 3 b) numKGH=[1,2]; denKGH=conv([1,0],[1,1]); rlocus(numKGH,denKGH) c) numKGH=[1,2]; denKGH=conv([1,1,0],[1 5]) denKGH = 1 6 5 0 rlocus(numKGH,denKGH) 4 c) Go (s ) = Gc (s ) = s 3 + 6 s 2 + (5 + K )s + 2 K K (s + 2 ) ; s(s + 1)(s + 5) K (s + 2 ) Gc (s ) = K (s + 2 ) s (s + 1)(s + 5) + K (s + 2) Rule 1,2: there are 3 branches; n=3; poles at s = 0, 1and -5. Rule 3: the root locus is on the axis for 1 < s < 0 and 5 < s <-2 Rule 4 there is 1 finite zero; m = 1 2 infinite zeros The branch start and end points are: 1 at [0, -j] 2 at [-1, +j] 3 at [-5, -2] Rule 5 intercept at 0 = using L19, S14/15 the asymptotes are at 90 and 270 with real-axis finite poles - finite zeros = (- 5 - 1 + 0) - (- 2) = -2 ; the zero! n-m 3 -1 m The breakaway point can be found using the transition method. n 1 1 U sin g = ; x + zi j =1x + p j i =1 1 1 1 1 = + + x + 2 x x +1 x + 5 Use Matlab to find roots of the resulting cubic. First construct each term. a=conv([1 1 0],[1 2]) a= 1 3 2 0 b=conv([1 1 0],[1 5]) b= 1 6 5 0 c=conv([1 2 0],[1 5]) c= 1 7 10 0 d=conv([1 3 2],[1 5]) d= 1 8 17 10 sum=a-b+c+d sum = 2 12 24 10 roots(sum) ans = -2.7211 + 1.2490i; Sketch on previous page -2.7211 - 1.2490i; -0.5578 break-away 5 2 Use the root locus technique to determine the range of values of K for which the system with open-loop transfer function Go(s) closed-loop is stable. Go (s ) = K (s - 1)(s + 2)(s + 8) Rule 1,2: there are 3 branches; n=3; poles at s = +1, -2 and -8. Rule 3: the root locus is on the axis for s < -8 and 2 < s <+1 3 infinite zeros 3 at [-8, -] Rule 4 there is no finite zero; m = 0 The branch start and end points are: 1 at [1, +] 2 at [-2, -] Rule 5 using L19, S14/15 the asymptotes are at 180 and 60 with realaxis intercept at 0 = finite poles - finite zeros = (+ 1 - 2 - 8) = -3 n-m 3-0 The break-away point between poles at +1 and 2 is determined by transition. n 1 1 U sin g = ; x + zi j =1x + p j i =1 m 0= 1 1 1 + + x -1 x + 2 x + 8 0 = ( x + 2 )( x + 8) + ( x - 1)( x + 8) + ( x - 1)( x + 2 ) 0 = x 2 + 10 x + 16 + x 2 + 7 x - 8 + x 2 + x - 2 0 = 3 x 2 + 18 x + 6 x= ( ) ( ) ( ) - 6 36 - 8 = -3 7 = -3 2.65 2 Break-away is at 0.35. [-5.65 is outside the range]. The j crossings are determined from the Routh Array. See L12, P8. Nise, p337 Closed-loop characteristic equation [CLCE] is: CLCE = 1 + Go (s )H (s ) = 1 + i.e. K =0 (s - 1)(s + 2)(s + 8) s3 + 9s2 + 6s2 16 + K = 0 6 s 3: s 2: s 1: s 0: a3 a2 b1 c1 a1 a0 b2 c2 1 9 70 - K 6 -16+K 0 0 a1 a2 - a3 a0 6 * 9 - (- 16 + K ) = 70 - K ; b2 = 0 a2 9 K = 70 makes a row of zeros at s1. Then the s2 row is a factor of the original CLCE polynomial and the roots provide the j crossings. b1 = i.e. 9s2 +(-16 + 70) = 9s2 + 54 = 0;s2 = -6;s = j6 = j2.45 The root locus also crosses the imaginary axis s = 0 when K = 16 Stability range is 16 < K < 70 numKGH=1; a=conv([1 -1],[1 2]) denKGH=conv([a],[1 8]) >> rlocus(numKGH,denKGH) a = 1 1 -2 denKGH = 1 9 6 -16 7 3 Plot the root locus of Go (s ) = s 2 (s + 9 ) K (s + 1) as a function of K. Rule 1: Rule 3: Rule 4 Rule 5 intercept at 0 = there are 3 branches; n=3; 2 poles at s = 0, and -9. the root locus is on the axis for 9 < s <-1 there is one finite zero; m=1 2 infinite zeros using L19, S14/15 the asymptotes are at 90 with real-axis n-m 3 -1 2 The branch start and end points are: 1 at [+0, -4 - j] 2 at [-0, -4 + j] 3 at [-9, -1] The break-away point is determined by transition: n 1 1 U sin g = ; x + zi j =1 x + p j i =1 m finite poles - finite zeros = (- 0 - 0 - 9) - (- 1) = - 8 = -4 1 1 1 1 = + + x +1 x x x + 9 x x - = 2; x +1 x + 9 x 2 + 6 x + 9 = 0; x( x + 9 ) - x( x + 1) = 2( x + 1)( x + 9 ) ( x + 3)2 = 0; x = -3 twice There is also break-away at s = 0 from 2 poles at origin. Characteristic equation = 1 + Go(s) = 1 + i.e. s +9s + K(s + 1) = 0; 27 + 81 + K(-3 + 1) = 0; 3 2 s (s + 9 ) Put s = -3 to find K. K = 27 2 K (s + 1) =0 When K = 27 the CLCE s3 +9s2 + 27s + 27 = (s + 3)3 = 0 Thus there are 3 closed-loop poles at s = -3, one on each branch. 8 There are no crossings of the imaginary axis so the system is stable for all values of gain, K. 9 4 A position control system has plant transfer function G (s ) = 4 . s (s + 4 ) A cascade proportional plus derivative compensator with transfer function D(s) = KP + TDs is added to the system so that the closed-loop poles of the compensated system have a natural frequency of n = 10 rad s-1 and damping ratio = 0.5. Determine the proportional gain, KP and derivative time, TD required to achieve this response. Calculate the steady-state velocity error of the compensated system. Hint: use the root locus angle criterion to find the location of the compensator zero when the closed-loop poles are positioned as specified. Kp 4TD s + 4 K p + TD s TD OLTF Go (s ) = G (s )D(s ) = = s (s + 4 ) s (s + 4 ) There are 2 open-loop poles [p1 and p2] and 1 zero [z]; place the zero. ( ) n = 10 rad s-1 and = 0.5 d = n = 5 d = n(1- 2) = 100.75 = 53 = cos-1 = 0.5; = 60 n d d For closed-loop poles to be at 5 53 the root locus angle criterion must be satisfied. See L18, S1-3. A closed-loop pole O j53 z O 2 -5 X X 1 origin O Follow L18, Ex9.2. Angle criterion is: -p1 -p2 + z = (2r + 1)*180 Take r = 0 Ap1 = n = 10 p1 = 180- 1 = 180 = 120; -1 p2 = 180 - 2 = 180 -tan (53) = 180 - 83.4 = 96.6; Ap2 = 76 z = - 180+ 120 + 96.6 = + 36.6 Az = (6.432 + 75) = 10.78 10 Oz = d + d tanz = 5 + 53 tan 36.6 = 5 + 6.43 = 11.43 = From L18, E7-8: Kp TD (s + p j ) n K = 4TD = j =1 m i =1 (s + zi ) = pole lengths = Ap1 * Ap2 = 10 * 76 = 8.09 Az 10.78 zero lenghts Kp = Oz*TD = 11.43*2.025 = 21.35 TD = K = 2.025 s; Go (s ) = 8.09(s + 11.43) s (s + 4 ) K v = lim s 0 sGo (s ) = 1 1 = = 0.043 K v 23.12 1 4 0 8.09 *11.43 = 23.12 4 Steady-state velocity error = numKGH=[1 11.43]; denKGH=conv([1 0],[1 4]) rlocus(numKGH,denKGH) [4.3 %] denKGH = This solution has branches between [0, -11.43] and [-4, -] This solution differs from previous work. 11
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Swansea UK - ENGINEERIN - EG 243
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University of Arkansas Community College at Morrilton - MATH - 246
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University of Arkansas Community College at Morrilton - MATH - 246
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University of Toronto - MATH - 246
Math 246 Project 3 Sample KEY Use complete sentences when appropriate; use complete mathematical sentences when appropriate. Failure to do so may result in a lower score. Show all your work. Indicate clearly the methods you use, because you will be graded
University of Toronto - MATH - 246
KEYProject Two _ Chapter 6 Use complete sentences when appropriate; use complete mathematical sentences whenappropriate. Failure to do so may result in a lower score. Show all your work. Indicate clearly the methods you use, because you will be graded o
University of Toronto - MATH - 246
KEY P roject F iveP roblem Talk T ime On a Cell Phone: Suppose the talk t ime in digital mode on a MotorolaTimeport P8160 is approximately normally distributed with mean 324 minutes and standard deviation of 24 minutes. A.) What proportion of the t ime
University of Toronto - MATH - 246
KEYName _ ID#_ Project 6 Chapter Eight Sampling Submit copies of the graphical summary for each question. This is an assignment dealing with verifying the idea of sampling distribution of the sample mean, and demonstrating the Central Limit Theorem. INST
University of Toronto - MATH - 246
KEY Project 7Chapter 9Use complete sentences when appropriate; use complete mathematical sentences when appropriate. Failure to do so may result in a lower score. Show all your work. Indicate clearly the methods you use, because you will be graded on th
University of Toronto - MATH - 246
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KEYProject 9You will be graded on completeness. Power is the probability that we reject the null hypothesis when the alternative hypothesis is assumed to be true. We will construct a Power Curve. Null: Alternative: =105. A simple random sample of size 3
University of Toronto - MATH - 246
KEYProject 10Chapter Ten: Hypothesis Testing A. Recently a friend of mine claimed that the summer of 2000 in Houston, Texas, was hotter than usual. To test this claim I went to AccuWeather.com and randomly selected 12 days in the summer of 2000. I recor
University of Toronto - MATH - 246
Math 246.KEYProject Eleven _/20 Chapter 11:READ EVERYTHING CAREFULLY: Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation. Us
University of Toronto - MATH - 246
Project 12Math 246KEYProject Twelve _/20_Chapter 12READ EVERYTHING CAREFULLY: Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy of your results and expla
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Project 14 _/20 Chapter 14READ EVERYTHING CAREFULLY: Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation. Use complete sentence
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-2 (0xx);Prob.1 Prob.2 Prob.3 TotalName :Section :I.D.Answer all the questions. All questions carry equal marks.1 Write an effective assembly language program
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-1 (0xx)Prob.1 Prob.2 Prob.3 TotalName :Section :I.D.Answer all three questions. All questions carry equal marks.1(a). Assume BX=1FF8H. Write a 1-line assembl
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-2(081); 14th January, 2009Prob.1 Prob.2 Prob.3 TotalName :Section :1I.D.Answer all three questions, which carries equal marks.Exam TIME: 8:00 to 9:30 PM1(a
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-1Prob.1 Prob.2 Prob.3 TotalAnswer the THREE question. All questions carry equal marks. Name : Section : I.D.1(a) If the data segment is starts from PA=1234H, wh