Control of Manufacturing Processes

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Control of Manufacturing Processes Subject 2.

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 Input-Output 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 Input-Output 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 Input-Output 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

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 closed-loop 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 s-plane - -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

17 Summary Feedback has Several Applications Most Frequent is α Reduction (disturbance rejection Need Analysis of Closed-Loop 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

The Velocity and Position Servos Examples from the lab CNC

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

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 Closed-Loop 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 Steady-State 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

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

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 Steady-State 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 Closed-Loop 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

42 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 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 42

Control of Manufacturing Processes

Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection