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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
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
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)
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)
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
⎛ ⎝
⎞ ⎠
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 7
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!!!!
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
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 10
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 11
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
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 13
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”
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 14
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 15
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 16
Root Root -- Performance Performance RelationshipsRelationships
s-planeIm
Kc=0
-1/τ-∞ Re
Slow; large errorFast, small error
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 17
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 18
The Velocity and Position ServosThe Velocity and Position Servos•• Examples from the labExamples from the lab
CNC Mill
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 19
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 20
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 21
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 22
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!
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 23
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 24
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!
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 25
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 26
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 27
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 28
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 29
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 30
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 31
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 32
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 33
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 34
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 35
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 36
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!
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 37
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 38
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 39
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 40
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 41
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
4/15/04 Lecture 18 © D.E. Hardt, all rights reserved 42
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
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