Control of Manufacturing Processes

Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004

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

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

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

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

Input-Output Transfer Function Y(s) R(s) = c τs +1+ c 1 1 + c 1 1 + c = c 1 + c τ s +1 1+ 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

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

Disturbance Transfer Function Y(s) D(s) = τs +1 1 + c τ s +1 1 + 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

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

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

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

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

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

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

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

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

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

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

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

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

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

Position Servo Block Diagram θ r + e u Ω θ m c 1/s τms + 1 - 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 2 + 1 τ m s + c m τ m 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 28

Position Servo Block Diagram θ r + e u Ω θ m c 1/s τms + 1 - 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 2 + 1 τ m s + c m τ m 4/15/04 Lecture 18 D.E. Hardt, all rights reserved 29

Position Servo Transfer Function θ θ r = c m τ m s 2 + 1 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

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

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

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

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

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 2 + 1 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

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