Control of Manufacturing Processes

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1 Control of Manufacturing Processes Subject Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004

2 The Position Servo Problem, reference position NC Control Robots Injection Molding Screw Forming Press Displacement. Controller Actuator Load d DC motor Hydraulic cylinder Mass Spring Damper position 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 2

3 DC Motor Based Position Servo Reference θ r () Controller u Power Amplifier I DC Motor Load Measureme Transducer θ) Now Measure θ not Ω θ r + - e u I Controller Amplifier Motor/Load θ 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 3

4 Motor Transfer Function I (s) Gp(s) Ω (s) Ω (s) = Gp(s) I (s) G p (s) = K t Js + b = K t / b (J / b)s +1 = K m s +1 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 4

5 Position Servo Block Diagram θ r + e u Ω θ K K m c 1/s τms Controller Position Motor/Load Transducer Encoder θ = 1 K m s s +1 u θ θ r = u = K c (θ r θ) K c K m s s + K ck m 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 5

6 Position Servo Block Diagram θ r + e u Ω θ K K m c 1/s τms Controller Position Motor/Load Transducer Encoder θ = 1 s K m s +1 u u = K c (θ r θ) θ θ r = K c K m s s + K ck m 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 6

7 Position Servo Transfer Function θ θ r = K c K m s s + K ck m Using the canonical variable definitions for a 2nd order system θ θ r = ω n 2 s 2 + 2ζω n s + ω n 2 ω n 2 = K ck m 2ζω n = 1 ζ = 1 1 2ω n 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 7

8 General 2 nd Order System Time Response Ω(t) =Ω SS (1 Be ζω n t sin(ω d t + φ)) A sinusoid of frequency ω d with a magnitude envelope of e ζω n t 1 B = 1 ζ 2 ω d = ω n 1 ζ 2 1 ζ 2 φ = tan 1 ζ 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 8

9 Second Order Step Response /22/04 Lecture 19 D.E. Hardt, all rights reserved 9

10 Overshoot and Damping ζ=0.5 ζ=0.2 ζ= ζ= /22/04 Lecture 19 D.E. Hardt, all rights reserved 10

11 Overshoot and Damping /22/04 Lecture 19 D.E. Hardt, all rights reserved 11

12 Step Response as a Function of Controller Gain 2 θ θ r 1.5 Kc = 1 Kc = 5 Kc = t s 0 ω n t 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 12

13 Key Features of Response Settling Time Is Invariant Overshoot Increases with Gain Error is always Zero! 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 13

14 Settling Time Basic form of Oscillatory Response: y(t) = Ae ζω n t sin(ω d t +φ) exponential envelope sinusoid of frequency ω d Time constant of envelope = 1/ζω n Time to fully decay? 4/ζω n And from above 2ζω n = 1 ζω n = 1 2 = constant 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 14

15 Steady-State State Error θ = θ r ω n 2 s 2 + 2ζω n s + ω n 2 L -1 θ + 2ζω n &θ + ω n2 θ = ω n2 θ r all derivatives 0 ω n 2 θ = ω n 2 θ r θ = θ r Independent of Controller Gain K c 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 15

16 Zero Error for Velocity Servo Add Integrator in Controller Instead of Measurement G c (s) = K c s Ω r (s) + E (s) - K c /s Controller U (s) G p (s) Motor Ω (s) Tachometer 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 16

17 Closed-Loop Transfer Function T(s) G p K c s 1+G p K c s and subs tituting for G p (s): T(s) = KcKm s s + KcKm Same form as Position Transfer Function Thus same properties 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 17

18 Step Response of Integral Controller as a Function of Gain Ω Ω r Kc = 1 Kc = 5 Kc = t s 0 ω n t 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 18

19 Velocity Servo has First Order Closed-Loop Dynamics Better Response and Error with Gain Never Zero?error Conclusions Position Servo has 2nd Order Closed Loop Dynamics Zero error Fixed Settling time Oscillatory Response as Gain Increases 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 19

20 Conclusions Zero Error Can be Achieved with Integrator BUT AT A PRICE! We Need More Options Root Locus for Higher Order Systems 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 20

21 Outline Root Locus Interpretation of Servo Crossing the Line: 3rd Order Systems The effect of Different Controllers : PI Zeros and Integrators Summary of In-Process Control Application to Material Control: Roll Bending Application to Output Control: Welding 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 21

22 Root Locus Interpretation θ r + e u Ω θ K K c m 1/s τms Controller Motor/Load Position Transducer θ θ r = K c K m s s + K ck m Characteristic Equation: s s + K c K m = 0 or s 2 + 2ζω n s +ω n 2 = 0 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 22

23 Root Locus Interpretation s 2 + 2ζω n s +ω n 2 = 0 Solution s 1,s 2 = ζω n ±ω n ζ 2 1 ζ>1 roots are real ζ<1 roots are complex ζ=1 roots repeat for K m = =1 s 1,s 2 = 1 2 ± 1 4K c 4 K c <0.25 roots are real K c >0.25 roots are complex K c =0.25 roots repeat 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 23

24 Root Locus Interpretation s 1,s 2 = 1 2 ± 1 4K c 4 Im K c s 1,s 2 ζ ω n 0 0, , , , ? Re -1-1/2 1.5 ± j ± j2.95 1/ /22/04 Lecture 19 D.E. Hardt, all rights reserved 24

25 Root Locus Interpretation Recall Ω(t) = Ω SS (1 Be ζω n t sin(ω d t + φ)) s 1,s 2 = ζω n ± ω n ζ 2 1 Im for ζ < 1 ω d = ζω n ± jω n 1 ζ 2 ζω n = ζω n ± jω d Re -1 Real part Imaginary Part 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 25

26 Root Locus Interpretation For any root on the root locus: Im ω d = (ζω n ) 2 + ( 1 ζ 2 ω ) 2 n = ω n = cos(β) = ζω n ω n β = ζ Re ζω n The root location tells us: Settling time and natural frequency 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 26

27 S-Plane - Response Relationships Same ζ Im Different ζω n Re 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 27

28 S-Plane - Response Relationships Greater angle β Im Lower ζ More Overshoot and Oscillation Re 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 28

29 And for the Servo Root Locus K=1 Im K=0.5 K=0 Re K=0.25 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 29

30 Conclusions: Root Locus Locus of All Possible Roots of CE as K c changes Locations(s) Indicate Characteristic Response Change in System will change Root Locus What are System - Locus Relationships? Go to Higher Order systems 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 30

31 The Limit to Feedback: Instability θ r + e u K K Ω θ c m 1/s τms Controller Motor/Load Position Transducer K c K m s( s+1) a s+a feedback filter K=1; τ=1/a 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 31

32 System Closed Loop Roots K m K c s( s+1) a s+a θ θ r = K c K m (s + a) s 3 + ( 1/ + a)s 2 + (a / )s + K ck m a Roots of this C.E? Look at Root Locus

33 Open - Loop Roots? G(s) H(s) T(s) = G(s) 1+ G(s)H(s) Thus the C.E. = 1+GH(s) = 0 For Our System GH(s) = K c K m a s( s +1)(s + a) s( s +1)(s + a) = 0 s 1, s 2, s 3 =[1/, 0, -a] 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 33

34 Root locus? Start at Open Loop Poles Lie on Real Axis to left of odd root starting from + + Are Attracted by Zeros End at Zeros or Infinity Repelled by Poles 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 34

35 Root Locus Without Filter Same System as Before Flows to zeros at -1/ 0 Balanced Repulsion from Poles

36 Root Locus with Filter -a -1/ 0 starts at open loop roots real part lies to the left of odd number of roots 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 36

37 Effect of the Filter Pole? Poles repel the root locus -a -1/ 0

38 Effect of the Filter Pole roots repel the root locus and closer is stronger -a -1/ 0

39 Instability Crossover to positive real roots -a -1/ 0 Conclusion: At some value of K c the closed-loop system will go unstable 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 39

40 The Remedy?: Basic rules: roots repel the root locus zeros attract the root locus -α -1/ 0

41 Zeros Applied Here -a -α -1/ 0 if the zero is closer it wins!

42 Applied Here Basic rules: roots repel the root locus zeros attract the root locus -α -a -1/ 0 if the pole is closer it wins!

43 Making Zeros:The PI controller E(s) K p +K I /s U(s) U(s) E(s) = K p + K I s = K ps + K I s = K p(s +α) s α = K I K p zero integrator 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 43

44 Other Considerations Effect of Parameter Variations Consider 2nd Order System Effect of Non-Constant Disturbances Random Inputs 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 44

45 Summary of In-Process Feedback Control: Feedback of Machine State In-Process Lumped Position, Force, Speed Feedback of Material States In-Process Point locations or Distributed forces, temperatures, pressures, etc. Feedback of Output In-Process Direct (or estimated) Geometry Feedback 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 45

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