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![Page 1: Control of Manufacturing Processesdspace.mit.edu/.../7F9BD337-D0AC-4C49-A58B-7ED4A46FB7C6/0/lec… · Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 “Basic](https://reader030.vdocuments.net/reader030/viewer/2022040612/5edd3f0aad6a402d6668456f/html5/thumbnails/1.jpg)
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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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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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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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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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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ClosedClosed--Loop Step ResponseLoop Step Response
1.0
Kc
ts ts ts
YR
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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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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