Control of Manufacturing Processes


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1 Control of Manufacturing Processes Subject Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004
2 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss = u y ss = ( )u : u (heater current) heater resistance ambient temperature oven capacitance y (temperature) 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 2
3 The Transfer Function from Laplace transform: d dt ( f (t)) s f (t) τý y (t) + y(t) = u(t) τsy(s) + Y(s) = U(s) Y(s)(τs + 1) = U(s) and with the definition of the Transfer function: Output Input = Y(s) U(s) = (τs +1) 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 3
4 Inputs, Disturbances and Noise D(s) R(s)  c U(s) (τs +1) Y(s) N(s) R(s) Reference Input (Y(s) should follow) D(s) Output Disturbance (Y(s) should not follow) N(s) Measurement Noise (Y(s) should not follow) 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 4
5 InputOutput Transfer Function R(s)  c U(s) (τs + 1) D(s) Y(s) N(s) Y(s) = c (τs +1) E(s); Y(s) 1+ c ;E = R Y (τs + 1) = c (τs + 1) R(s) Y(s) R(s) = c (τs + 1) 1 + c (τs + 1) τs +1 τs +1 = c τs +1 + c 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 5
6 InputOutput Transfer Function Y(s) R(s) = c τs +1+ c c c = c 1 + c τ s c = cl τ cl s +1 cl = closed loop gain τ cl = closed loop time constant As c >>1 cl >1 τ cl > 0 R(s) cl τ cl s +1 Y(s) 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 6
7 InputOutput Transfer Function R(s)  c U(s) (τs + 1) D(s) Y(s) N(s) Y(s) R(s) = cl τ cl s +1 As c >>1 ; cl >1 ; Y(s) >R(s) τ cl > 0 ; t s >0 System gets much faster and has less error!!!! 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 7
8 ClosedLoop Step Response 1.0 Y R c t s t s t s 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 8
9 Disturbance Transfer Function R(s)  c U(s) (τs + 1) D(s) Y(s) N(s) Y(s) = c Y(s) + D(s) (τs +1) Y(s) D(s) = c (τs + 1) = τs +1 τs +1 + c 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 9
10 Disturbance Transfer Function Y(s) D(s) = τs c τ s c Same τ cl (dynamics) as before What is Steady State? Y D ss = 1 1+ c As c >> Y/D ss 0 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 10
11 Noise Transfer Function R(s)  c U(s) (τs + 1) D(s) Y(s) N(s) Notice that N(s) and R(s) look like the same inputs if the systems tends to follow R(s) well, it will also follow N(s) well Y(s) N(s) = cl τ cl s +1 As c >>1 Y/N > 1.0!! 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 11
12 Summary D(s) R(s)  c U(s) (τs +1) Y(s) N(s) As the controller (or loop) gain c increases: The output better follows the input (good) The output disturbance is rejected (good) The measurement noise is more perfectly followed(bad 2 out of 3 isn t bad
13 τ cl = τ 1 + c Some Interpretations The closedloop time constant decreases as c increases At c = 0, τ cl = τ >The original open loop value As c =>, τ cl => 0 Infinitely Fast 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 13
14 Some Interpretations τ cl = τ 1 + c Note also that the characteristic equation for this first order system is given by: τ cl s+1=0 And the root of this equation s 1 = 1/τ cl Thus as c =>, s 1 => τs +1 = 0 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 14
15 At c = 0, s 1 = 1/τ Graphical Interpretations the open loop root As c =>,s 1 => τ cl = τ 1 + c And for 0< c < the locus of s 1 is between 1/τ and  c >0 c =0 splane  1/τ Root Locus for Proportional control of the Plant Transfer Function G(s) = /(τs+1) 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 15
16 Root  Performance Relationships Im splane c =01/τ  Re Fast, small error Slow; large error 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 16
17 Summary Feedback has Several Applications Most Frequent is α Reduction (disturbance rejection Need Analysis of ClosedLoop to Assess How and Why Our Main Tool will be Evans Root Locus Will Need to Describe Stochastic Performance Eventually 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 17
18 The Velocity and Position Servos Examples from the lab CNC Mill 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 18
19 Velocity Servo Problem Spindle Speed Rolling Mill Speed Wafer Spinning Speed (Coating) Injection Speed... Reference Speed u Power Controller Amplifier DC Motor Ω r I Load Inertia Disturbance Torque Tachometer 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 19
20 Velocity Servo Block Diagram T d 1/ t Ω r (s) E (s) U (s) I (s) c I G p (s)  Controller Amplifier Motor Ω (s) Tachometer Ω r (s) Ω (s T (s) 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 20
21 T m = t I where T m is the motor torque DC Motor Model Σ T = JΩ = T m  bω motor torque  bearing damping JΩ Ý T + bω= t I m Ω Bearings with Damping b J b Ý Ω + Ω = t b I Motor Inertia J T d 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 21
22 DC Motor Model L J b Ý Ω + Ω = t b I = J b (s +1)Ω(s) = t b I(s) Ω(s) I(s) = t / b ( J / b)s +1 = G (s) p G p (s) = τs +1 Same as heater model! Same Results Apply! 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 22
23 Closed Loop Results Motor T.F. same as Heater T.F. Loop (without Disturbance) Is the Same ClosedLoop Input  Output Performance is the Same Ω r (s)  c U(s) Gain/Amp (τs + 1) Ω(s) Motor As c >>1 ; Ω(s) Ω r (s) = cl τ cl s +1 cl >1 ; Ω>Ω r τ cl > 0 ; t s >0 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 23
24 SteadyState State Error Ω(s) Ω r (s) = cl τ cl s +1 τ Ý cl Ω + Ω = cl Ω r (t) Ω r (t) = Constant R At Steady State all time derivatives = 0 Ω ss = cl R Ω ss R = cl Thus Error never goes to zero! = c 1+ c <1 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 24
25 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 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 25
26 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 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 26
27 Motor Transfer Function I (s) Gp(s) Ω (s) Ω (s) = Gp(s) I (s) G p (s) = t Js + b = t / b (J / b)s +1 = m τ m s +1 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 27
28 Position Servo Block Diagram θ r + e u Ω θ m c 1/s τms Controller Position Motor/Load Transducer Encoder θ = 1 m s τ m s +1 u θ θ r = u = c (θ r θ) c m τ m s τ m s + c m τ m 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 28
29 Position Servo Block Diagram θ r + e u Ω θ m c 1/s τms Controller Position Motor/Load Transducer Encoder θ = 1 s m τ m s +1 u u = c (θ r θ) θ θ r = c m τ m s τ m s + c m τ m 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 29
30 Position Servo Transfer Function θ θ r = c m τ m s τ m s + c m τ 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 = c m τ m 2ζω n = 1 τ m ζ = 1 τ m 1 2ω n 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 30
31 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 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 31
32 Second Order Step Response /15/04 Lecture 18 D.E. Hardt, all rights reserved 32
33 Overshoot and Damping ζ=0.5 ζ=0.2 ζ= ζ= /15/04 Lecture 18 D.E. Hardt, all rights reserved 33
34 Overshoot and Damping /15/04 Lecture 18 D.E. Hardt, all rights reserved 34
35 Step Response as a Function of Controller Gain 2 θ θ r 1.5 c = 1 c = 5 c = t s 0 ω n t 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 35
36 ey Features of Response Settling Time Is Invariant Overshoot Increases with Gain Error is always Zero! 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 36
37 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 τ m ζω n = 1 2τ m = constant 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 37
38 SteadyState State Error θ = θ r ω n 2 s 2 + 2ζω n s + ω n 2 L 1 Ý θ + Ý 2ζω θ +ω 2 θ n n = ω 2 θ n r all derivatives 0 ω n 2 θ = ω n 2 θ r θ = θ r Independent of Controller Gain c 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 38
39 Zero Error for Velocity Servo Add Integrator in Controller Instead of Measurement G c (s) = c s Ω r (s) + E (s)  c /s Controller U (s) G p (s) Motor Ω (s) Tachometer 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 39
40 ClosedLoop Transfer Function T(s) G p c s 1+G p c s and subs tituting for G p (s): T(s) = cm τ m s τ m s + cm τ m Same form as Position Transfer Function Thus same properties 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 40
41 Step Response of Integral Controller as a Function of Gain Ω Ω r c = 1 c = 5 c = t s 0 ω n t 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 41
42 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 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 42
43 Conclusions Zero Error Can be Achieved with Integrator BUT AT A PRICE! We Need More Options Root Locus for Higher Order Systems 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 43
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