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 SteadyState 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 ClosedLoop 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 ClosedLoop 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 InProcess 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 11/2 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 SPlane  Response Relationships Same ζ Im Different ζω n Re 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 27
28 SPlane  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 closedloop 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 NonConstant Disturbances Random Inputs 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 44
45 Summary of InProcess Feedback Control: Feedback of Machine State InProcess Lumped Position, Force, Speed Feedback of Material States InProcess Point locations or Distributed forces, temperatures, pressures, etc. Feedback of Output InProcess Direct (or estimated) Geometry Feedback 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 45
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