control of manufacturing...
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Control of Manufacturing Control of Manufacturing ProcessesProcesses
Subject 2.830 Spring 2004Spring 2004Lecture #22Lecture #22
““Discrete System ClosedDiscrete System Closed--Loop Dynamics"Loop Dynamics"May 3, 2004May 3, 2004
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 2
Difference Equations and DynamicsDifference Equations and Dynamics
• Consider the Discrete Transfer Function:
G(z) =Y (z)U (z)
=K
(z − p)Y(z)(z − p) = KU (z)
The corresponding Difference Equation is:
yi+1 − pyi = Kui ⇒ yi+1 = pyi + Kui
next output = p*current output + K* current input
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 3
Step DynamicsStep Dynamicsyi+1 = pyi + pui for p=0.5 and K=0.5 and u=unit step:
⇒ yi+1 = 0.5(yi + ui)I ui y i
1 1 0.0002 1 0.5003 1 0.7504 1 0.8755 1 0.9386 1 0.9697 1 0.9848 1 0.9929 1 0.996
10 1 0.99811 1 0.99912 1 1.00013 1 1.000
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 4
Step ResponseStep Response
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
First Order
Transfer Function!G(z) =
Y (z)U (z)
=K
(z − p)=
0.5z −0.5
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 5
Effect of Root Effect of Root pp
Time (samples)
TextEnd
Step Response
0 2 4 6 8 100
0.02
0.04
0.06
0.08
0.1
Time (samples)
Ampl
itude
TextEnd
Step Response
0 2 4 6 8 100
0.02
0.04
0.06
0.08
0.1
p=0.1 p=0.25
Ampl
itude
Time (samples)
TextEnd
Step Response
Time (samples)
TextEnd
Step Response
0 50 100 1500
50
100
150
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
Ampl
itude
Ampl
itude
p=0.5p=1.0
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 6
Effect of Root Effect of Root pp
Time (samples)
Am
plitu
de
TextEnd
Step Response
0 2 4 6 8 100
0.05
0.1
0.15
0.2
0.25
Time (samples)
Ampl
itude
TextEnd
Step Response
0 2 4 6 8 100
0.02
0.04
0.06
0.08
0.1
p=-0.1 p=-0.25
Time (samples)
TextEnd
Step Response
Time (samples)
TextEnd
Step Response
0 10 20 30 400
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10 120
0.1
0.2
0.3
0.4
0.5
Ampl
itude
p=-1.0
Am
plitu
de
p=-0.5
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 7
What’s Next?What’s Next?
•• ZZ-- domain modeling of control systemdomain modeling of control system•• Root locus in zRoot locus in z--domain (it’s the same!)domain (it’s the same!)•• Cycle to Cycle Stability LimitsCycle to Cycle Stability Limits
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 8
Analysis of Dynamics of Analysis of Dynamics of Discrete Feedback SystemsDiscrete Feedback Systems
•• Given the Discrete Transfer Function Given the Discrete Transfer Function For the Process For the Process GGpp(z(z))
•• What are the Dynamics of the resulting What are the Dynamics of the resulting Closed Loop System?:Closed Loop System?:
Gc(z) Gp(z)
H(z)
R(z) Y(z)
-
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 9
ClosedClosed--Loop DynamicsLoop Dynamics
Gc(z) Gp(z)
H(z)
R(z) Y(z)
-
Y (z )R(z)
= T (z) =GcGp
1+ GcGp H
and the characteristic equation is:
1 + Gc(z)Gp (z)H(z) = 0
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 10
ZZ--Domain Root LocusDomain Root Locus
1+GcGp H(z) = 0
In general the CE:
will be a polynomial in z:
anzn + an−1z
n −1 + an −2 zn− 2 + ...a1z + a0 = 0
and the roots of this polynomial will define the dynamics of our system
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 11
SS--Z Plane MappingZ Plane Mapping
z = esTRecall the Definition:
s = σ + jωAnd that
then z in polar coordinates is given by:
e(σ + jω )T = eσTejωT
z = eTσ ∠z = ωT
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MappingMapping
s
0
s = 0
z = e0 = 1 z
+1
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Mapping Mapping jjωω AxisAxis
s
0
s = + jω
z = e jωT = unit length vector at angle ωTz
+1
ωT
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 14
Mapping Mapping jjωω AxisAxis
jω z
0 +1
Note: when ωT = π, we have reached -1 on z-plane
π /T
− π /T
And when ωT = -π, we have come around the other way
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Settling Time MappingSettling Time Mappings = −p + jω
z = e− pT + jωT = e− pTe jωT
S-plane lines of constant settling:
0
ωT
s
ts=4/p
-p
z
+1
circle of radius e-pT
r=e-pT
5/4/04
ts = −4T
1ln(r)
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Settling Time MappingSettling Time Mapping
s z
0 +1
ωT
circle of radius e-pT
ts
CLOSER TO Z=0 = FASTER SETTLING
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Damping Ratio MappingDamping Ratio Mapping
s z
0
β
∠s = β = constant0 ≤ s ≤ ∞
s
+1
5/4/04
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Stability?Stability?
jw axis maps to the “unit circle”zs
0
π /T
−π /T
+1
roots inside the unit circle are stable, outside are unstable
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Mapping: Negative Real AxisMapping: Negative Real Axis
s z
0+1
+jω
∠z = ωT = 0
0 ≤ s ≤ −∞⇒ e0 ≤ z ≤ e−∞ ⇒1 ≤ z ≤ 0
-jω
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 20
Mapping: What’s leftMapping: What’s left
s z
0
π /T
−π /T
−∞
oscillatory region
real-root
+1
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ZgridZgrid from Matlabfrom MatlabLines of constant damping ratio (ζ=0.7 shown)
Lines of constant natural frequency-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
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-1 -0.5 0 0.5 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
ZZ--Plane Plane -- ReponseReponseRelationshipRelationship
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 23
Simple First Order Control System Simple First Order Control System
KC Gp(z)-R(z) Y(z)
G p(z) =K
(z − p) open loop root = +p
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 24
Root LocusRoot LocusROOTS OF 1+KCG-P(z) = 0 ?
Same rules as before:
Starts at open loop pole = +p
Real axis part to left of
odd number of poles
Ends at zero at infinityp
z
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 25
Root LocusRoot Locus-- InterpretationInterpretation
p
z
+1
slow, no oscillation
faster, no oscillation
fastest w/o overshoot
begins oscillating
very oscillatory
unstable
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Now for Now for
-1 -0.5 0 0.5 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Imag
Axi
s
k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
5/4/04
G(z) =K
(z − p)=
0.5z − 0.5 K=0.5
p=0.5
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Response?Response?k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
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Response?Response?k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
5/4/04
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Response?Response?k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
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Response?Response?k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
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Response?Response?k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
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Response?Response?k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
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5/4/04 Lecture 22 © D.E. Hardt, all rights reserved 33
Steady State Unit Step Error?Steady State Unit Step Error?
unit step
For s - domain y(∞) = lims→0
s G(s) 1s
For z - domain y(∞) = limz→1
(z −1) G(z) zz −1
T (z) =
KcK
z − p
1 +K
cK
z − p
=KcK
z − p + KcKOur closed-loop TF =
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Steady State Error?Steady State Error?
Kc p y(∞)
0.22 0.39 0.18
0.73 0.13 0.44
1 0 0.5
1.5 -0.27 0.6
2.8 -0.92 0.74
3 -1 0.75
Steady State Error is Minimized as K increases
limz→1
(z −1) T (z) z
z −1= (z −1)
KcKz − p + KcK
z
(z −1)
=KcK
1− p + KcK
K=0.5p=0.5