Control of Manufacturing...

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

Transcript of Control of Manufacturing...

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Control of Manufacturing Control of Manufacturing ProcessesProcesses

Subject 2.830Subject 2.830Spring 2004Spring 2004Lecture #18Lecture #18

“Basic Control Loop Analysis"“Basic Control Loop Analysis"April 15, 2004April 15, 2004

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 2

Revisit Temperature Revisit Temperature Control ProblemControl Problem

τ dydt

+ y = Ku

u (heater current)

y (temperature)τ = time constantK = gain

yss = Ku∆K: heater resistance

ambient temperatureoven capacitance

∆yss = (∆K)u

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 3

The Transfer FunctionThe Transfer Functionfrom Laplace transform:

ddt

( f ( t)) ⇒ s ⋅ f (t)

τÝ y (t) + y(t) = Ku(t) ⇒ τsY(s) + Y(s) = KU (s)

⇒ Y (s)(τs + 1) = KU(s)and with the definition of the Transfer function:

OutputInput

=Y(s)U(s)

=K

(τs +1)

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

Inputs, Disturbances and NoiseInputs, Disturbances and Noise

K(τs +1)

Y(s)U(s)Kc-

R(s)

D(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)

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 5

InputInput--Output Transfer FunctionOutput Transfer FunctionK

(τs + 1)U(s) Y(s)

Kc-

R(s)

D(s)

N(s)

Y(s) =KcK

(τs +1)E(s); Y(s) 1+

KcK

(τs + 1)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

KcK

(τs + 1)R(s);E = R −Y

Y (s)R(s)

=

KcK

(τs + 1)

1 +K

cK

(τs + 1)

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=KcK

τs +1 + KcKτs +1τs +1

⎛ ⎝

⎞ ⎠

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 6

InputInput--Output Transfer FunctionOutput Transfer Function

=

KcK1 + KcKτ

1+ KcK⎛ ⎝ ⎜

⎞ ⎠ ⎟ s +1

11 + KcK

11 + KcK

⎜ ⎜ ⎜

⎟ ⎟ ⎟

=Kcl

τcl s +1Y (s)R(s)

=KcK

τs +1+ KcK

As Kc>>1

Kcl ->1

τcl-> 0

Kcl= closed loop gain

τcl = closed loop time constant

Y(s)R(s) Kcl

τcls +1

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InputInput--Output Transfer FunctionOutput Transfer FunctionK

(τs + 1)U(s) Y(s)

Kc-

R(s)

D(s)

N(s)

As Kc>>1 ;

Kcl ->1 ; Y(s) ->R(s)

τcl-> 0 ; ts->0

Y (s)R(s)

=Kcl

τ cls +1

System gets much faster and has less error!!!!

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 8

ClosedClosed--Loop Step ResponseLoop Step Response

1.0

Kc

ts ts ts

YR

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 9

Disturbance Transfer FunctionDisturbance Transfer FunctionK

(τs + 1)U(s) Y(s)

Kc-

R(s)

D(s)

N(s)

⇒Y (s)D(s)

=1

1 +K

cK

(τs + 1)

Y(s) = −KcK

(τs +1)Y (s) + D(s)

=τs +1

τs +1 + KcK

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Disturbance Transfer FunctionDisturbance Transfer Function

Y (s)D(s)

=

τs +11 + KcK

τ1 + KcK

⎛ ⎝ ⎜

⎞ ⎠ ⎟ s +1

Same τcl (dynamics) as before

What is Steady State ?

As Kc >>K

Y/D|ss 0YD ss

=1

1+ KcK

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Noise Transfer FunctionNoise Transfer Function

K(τs + 1)

U(s) Y(s)Kc

-

R(s)

D(s)

N(s)

Notice that N(s) and R(s) look like the “same” inputsif the systems tends to follow R(s) well,it will also follow N(s) well

As Kc>>1

Y/N -> 1.0 !!Y (s)N(s)

= −Kcl

τ cls +1

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SummarySummaryD(s)

K(τs +1)

U(s) Y(s)Kc-

R(s)

N(s)

As the controller (or loop) gain Kc 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

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Some InterpretationsSome Interpretations

τcl =τ

1 + Kc KThe closed-loop time constant decreases as Kc increases

At Kc = 0, τcl = τ ->The original open loop value

As Kc => ∞, τcl => 0 ”Infinitely Fast”

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Some InterpretationsSome Interpretationsτcl =

τ1 + Kc K

Note also that the characteristic equation for this first order system is given by:

τcl s+1=0

• And the root of this equation s1 = -1/τcl• Thus as Kc => ∞, s1 => −∞ τs +1 = 0

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Graphical InterpretationsGraphical InterpretationsAt Kc = 0, s1 = -1/τ the “open loop root”

As Kc => ∞, s1 => −∞ τcl =τ

1 + Kc KAnd for 0<Kc<∞ the locus of s1 is between -1/τ and -∞

-1/τ-∞

s-plane

Root Locus for Proportional control of the Plant TransferFunction G(s) = K/(τs+1)

Kc=0Kc>0

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Root Root -- Performance Performance RelationshipsRelationships

s-planeIm

Kc=0

-1/τ-∞ Re

Slow; large errorFast, small error

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SummarySummary

•• Feedback has Several ApplicationsFeedback has Several Applications•• Most Frequent is Most Frequent is ∆α∆α Reduction (disturbance Reduction (disturbance

rejectionrejection•• Need Analysis of ClosedNeed Analysis of Closed--Loop to Assess How Loop to Assess How

and Whyand Why•• Our Main Tool will be Our Main Tool will be Evans Root LocusEvans Root Locus•• Will Need to Describe Stochastic Will Need to Describe Stochastic

Performance EventuallyPerformance Eventually

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The Velocity and Position ServosThe Velocity and Position Servos•• Examples from the labExamples from the lab

CNC Mill

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Velocity Servo ProblemVelocity Servo Problem•• Spindle SpeedSpindle Speed•• Rolling Mill SpeedRolling Mill Speed•• Wafer Spinning Speed (Coating)Wafer Spinning Speed (Coating)•• Injection SpeedInjection Speed•• ......

ControllerReferenceSpeedΩr

Amplifier DC Motor

Load Inertia

Tachometer

u I

PowerDisturbance Torque

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Velocity Servo Block DiagramVelocity Servo Block Diagram

Gp(s)

1/Kt

I (s)

Td

++

U (s)E (s)Ωr (s)+ Kc KI

-Controller Amplifier Motor

Ω (s)

Tachometer

T (s)Ωr (s) Ω (s

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DC Motor ModelDC Motor ModelTTmm = = KKtt IIwhere where TTmmis the motor torqueis the motor torque

Σ T = JΩ = Tm - bΩ motor torque - bearing damping

J Ý Ω + bΩ = KtI Bearings with Damping bTm Ω

MotorInertiaJ

TdJb

Ý Ω + Ω =Kt

bI

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DC Motor ModelDC Motor Model

Jb

Ý Ω + Ω =Kt

bI⎛

⎝ ⎞ ⎠ =

Jb

(s +1)Ω(s) =Kt

bI(s)L

Ω(s)I(s)

=Kt / b

J / b( )s +1= Gp (s)

G p(s) =K

τs +1Same as heater model!

Same Results Apply!

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Closed Loop ResultsClosed Loop Results•• Motor T.F. same as Heater T.F.Motor T.F. same as Heater T.F.•• Loop (without Disturbance) Is the SameLoop (without Disturbance) Is the Same•• ClosedClosed--Loop Input Loop Input -- Output Performance is Output Performance is

the Samethe Same

K(τs + 1)

U(s) Ω(s)Kc

-

Ωr(s)

Gain/Amp Motor As Kc>>1 ;

Kcl ->1 ; Ω->Ωr

τcl-> 0 ; ts->0Ω(s)Ωr(s)

=Kcl

τcls +1

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SteadySteady--State ErrorState ErrorΩ(s)Ωr(s)

=Kcl

τcls +1

τclÝ Ω + Ω = KclΩr (t) Ωr( t) = Constant R

At Steady State all time derivatives = 0

=KcK

1+ KcK<1Ωss

R= KclΩss = KclR

Thus Error never goes to zero!

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The Position Servo Problem,The Position Servo Problem,

•• NC ControlNC Control•• RobotsRobots•• Injection Molding ScrewInjection Molding Screw•• Forming Press DisplacementForming Press Displacement•• ……. .

Controller Actuator “Load”dreference

position position

DC motorHydraulic cylinder

MassSpringDamper

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DC Motor Based Position ServoDC Motor Based Position Servo

Controller

Reference (θr)

Amplifier DC Motor

LoadMeasuremeTransducer θ)u I

Power

Controller Amplifier Motor/Loade u

-

+ Iθr θ

Now Measure θ not Ω

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Motor Transfer FunctionMotor Transfer Function

(s) = GGp(s)Ω (s)I (s)

Ω p(s) I (s)

Gp (s) =Kt

Js + b=

Kt / b(J / b)s +1

=Km

τ ms +1

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Position Servo Block DiagramPosition Servo Block Diagram

Controller Motor/Load PositionTransducer

Kceθr u Ω θ

1/sKmτms + 1

+

-

θ =1s

Km

τms +1u

Encoder

u = Kc(θr − θ)

θθr

=

KcKm

τm

s2 +1

τ m

s +KcKm

τ m

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Position Servo Block DiagramPosition Servo Block Diagram

Controller Motor/Load PositionTransducer

Kceθr u Ω θ

1/sKmτms + 1

+

-

Encoder

θ =1s

Km

τms +1u u = Kc(θr − θ)

θθr

=

KcKm

τm

s2 +1

τ m

s +KcKm

τ m

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Position Servo Transfer FunctionPosition Servo Transfer Function

θθr

=

KcKm

τm

s2 +1

τ m

s +KcKm

τ m

Using the canonical variable definitions for a 2nd order system

2ζωn =1

τ mθθr

=ωn

2

s2 + 2ζωns + ωn2

ωn2 =

KcKm

τ m

ζ =1

τm

12ωn

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General 2General 2ndnd Order System Order System Time ResponseTime Response

Ω(t) = ΩSS (1− Be −ζω nt sin(ωd t + φ))

B =1

1−ζ 2

ωd = ωn 1−ζ 2

φ = tan−1 1−ζ 2

ζ

⎝ ⎜

⎠ ⎟

A sinusoid of frequency ωd

with a magnitude envelope of

e−ζω nt

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Second Order Step ResponseSecond Order Step Response

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

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Overshoot and DampingOvershoot and Damping

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

ζ=1

ζ=0.5 ζ=0.707

ζ=0.2

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Overshoot and DampingOvershoot and Damping

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

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Step Response as a Function of Step Response as a Function of Controller GainController Gain

2

Kc = 1

Kc = 5

Kc = 10

θθr

1.5

1

ts0.5

0

ω n t

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Key Features of ResponseKey Features of Response

•• Settling Time Is InvariantSettling Time Is Invariant

•• Overshoot Increases with GainOvershoot Increases with Gain

•• Error is always Zero!Error is always Zero!

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Settling TimeSettling Time•• Basic form of Oscillatory Response:Basic form of Oscillatory Response:

y(t) = Ae−ζω nt sin(ωd t +φ)

exponential envelope sinusoid of frequency ωd

Time to fully decay?Time constant of envelope = 1/ζωn 4/ζωn

ζωn =

12τ m

= constant2ζωn =1

τ mAnd from above

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SteadySteady--State ErrorState Error

θθr

=ωn

2

s2 + 2ζωns + ωn2

Ý Ý θ + 2ζωnÝ θ +ωn

2θ = ωn2θrL-1

all derivatives 0

Independent of

Controller Gain Kcθ = θrωn

2θ = ωn2θr

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Zero Error for Velocity ServoZero Error for Velocity Servo

•• Add Integrator in Controller Instead of Add Integrator in Controller Instead of MeasurementMeasurement

Gp(s)∑U (s)E (s)Ωr (s)

+

-

ControllerMotor

Ω (s)

Tachometer

Kc/s

Gc(s) =Kc

s

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ClosedClosed--Loop Transfer FunctionLoop Transfer Function

T(s) ≡Gp

Kcs

1 + GpKcs

tand subs ituting for Gp(s):

T(s) =

KcKmτm

s 2 + 1τm

s + KcKmτm

Same form as Position Transfer Function

Thus same properties

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Step Response of Integral Step Response of Integral Controller as a Function of GainController as a Function of Gain

1.5

2

Kc = 1

Kc = 5

Kc = 10

ΩΩr

1

ts0.5

0

ω n t

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ConclusionsConclusions•• Velocity Servo has First Order ClosedVelocity Servo has First Order Closed--Loop Loop

DynamicsDynamics•• Better Response and Error with GainBetter Response and Error with Gain•• Never Zero ?errorNever Zero ?error

•• Position Servo has 2nd Order Closed Loop DynamicsPosition Servo has 2nd Order Closed Loop Dynamics•• Zero errorZero error•• Fixed Settling time Fixed Settling time •• Oscillatory Response as Gain IncreasesOscillatory Response as Gain Increases

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4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 43

ConclusionsConclusions

•• Zero Error Can be Achieved with IntegratorZero Error Can be Achieved with Integrator•• BUT AT A PRICE!BUT AT A PRICE!

•• We Need More OptionsWe Need More Options•• Root Locus for Higher Order SystemsRoot Locus for Higher Order Systems