2.14 Fall 2004 - MITweb.mit.edu/2.14/www/PSets/nov17.pdf11/10/04 2.14 Fall 2004 2 Difference...
Transcript of 2.14 Fall 2004 - MITweb.mit.edu/2.14/www/PSets/nov17.pdf11/10/04 2.14 Fall 2004 2 Difference...
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2.14 Fall 20042.14 Fall 2004
Digital Control - Z-plane analysisDigital Control - Z-plane analysis
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11/10/04 2.14 Fall 2004 2
Difference Equations and DynamicsDifference Equations and Dynamics
• Consider the Discrete TransferFunction:
The corresponding Difference Equation is:
next output = p*current output + K* current input
�
G(z) =Y (z)U (z)
=K
(z − p)
�
Y(z)(z − p) = KU (z)
�
yi+1 − pyi = Kui ⇒ yi+1 = pyi + Kui
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11/10/04 2.14 Fall 2004 3
Step DynamicsStep Dynamicsyi+1 = pyi + pui
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 0.99610 1 0.99811 1 0.99912 1 1.00013 1 1.000
⇒ yi+1 = 0.5(yi + ui)for p=0.5 and K=0.5 and u=unit step:
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11/10/04 2.14 Fall 2004 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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11/10/04 2.14 Fall 2004 5
Effect of Root Effect of Root pp
Time (samples)
Ampl
itude
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
Time (samples)
Ampl
itude
TextEnd
Step Response
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
Time (samples)
Ampl
itude
TextEnd
Step Response
0 50 100 1500
50
100
150
p=0.1 p=0.25
p=0.5p=1.0
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11/10/04 2.14 Fall 2004 6
Effect of Root Effect of Root pp
Time (samples)
Ampl
itude
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 10 120
0.1
0.2
0.3
0.4
0.5
Time (samples)
Ampl
itude
TextEnd
Step Response
0 10 20 30 400
0.2
0.4
0.6
0.8
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
p=-0.5p=-1.0
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11/10/04 2.14 Fall 2004 7
WhatWhat’’s Next?s Next?
•• Z- domain modeling of control systemZ- domain modeling of control system•• Root locus in z-domain (itRoot locus in z-domain (it’’s the same!)s the same!)•• Cycle to Cycle Stability LimitsCycle to Cycle Stability Limits
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11/10/04 2.14 Fall 2004 8
Analysis of Dynamics ofAnalysis of Dynamics ofDiscrete Feedback SystemsDiscrete Feedback Systems
•• Given the Discrete Transfer FunctionGiven the Discrete Transfer FunctionFor the Process GFor the Process Gpp(z)(z)
•• What are the Dynamics of the resultingWhat are the Dynamics of the resultingClosed Loop System?:Closed Loop System?:
Gc(z) Gp(z)
H(z)
-R(z) Y(z)
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11/10/04 2.14 Fall 2004 9
Closed-Loop DynamicsClosed-Loop Dynamics
Gc(z) Gp(z)
H(z)
-R(z) Y(z)
Y (z)R(z)
= T (z) =GcGp
1+GcGpH
and the characteristic equation is:
1 +Gc(z)Gp (z)H(z) = 0
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11/10/04 2.14 Fall 2004 10
Z-Domain Root LocusZ-Domain Root Locus
1 +GcGpH(z) = 0
In general the CE:
will be a polynomial in z:
anzn + an−1z
n −1 + an −2zn− 2 + ...a1z + a0 = 0
and the roots of this polynomial willdefine the dynamics of our system
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11/10/04 2.14 Fall 2004 11
S-Z Plane MappingS-Z Plane Mapping
Recall the Definition: z = esT
And that s = σ + jω
then z in polar coordinates is given by:
e(σ + jω )T = eσTejωT
z = eTσ ∠z =ωT
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11/10/04 2.14 Fall 2004 12
MappingMapping
s
0
s = 0z = e0 = 1 z
+1
z = eTσ ∠z =ωT
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11/10/04 2.14 Fall 2004 13
Mapping Mapping jjωω Axis Axis
s
0
s = + jωz = e jωT = unit length vector at angle ωT
z
+1
ωT
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11/10/04 2.14 Fall 2004 14
Mapping Mapping jjωω Axis Axis
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 Mapping
z
+1
S-plane lines of constant settling:
0
ωT
s
ts=4/p
-p
s = −p + jωz = e− pT+ jωT = e− pTe jωT
circle of radius e-pT
r=e-pT
11/10/04
ts ≈ −4Tln(r)
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Settling Time MappingSettling Time Mapping
s z
0 +1
ωT
circle of radius e-pT
CLOSER TO Z=0 = FASTER SETTLING
ts
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Damping Ratio MappingDamping Ratio Mapping
s z
0 +1
β
11/10/04
�
∠s = β = constant0 ≤ s ≤ ∞
�
s
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Stability?Stability?
s z
0 +1
jw axis maps to the “unit circle”
roots inside the unit circle are stable, outside are unstable
π /T
−π /T
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Mapping: Negative Real AxisMapping: Negative Real Axis
s z
0+1
+jω
-jω
∠z =ωT = 00 ≤ s ≤ −∞⇒ e0 ≤ z ≤ e−∞ ⇒1 ≤ z ≤ 0
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11/10/04 2.14 Fall 2004 20
Mapping: WhatMapping: What’’s lefts left
s z
0 +1
π /T
−π /T
−∞
oscillatory region
real-root
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Imag
Axis
TextEnd
Pole-zero mapzgrid zgrid from Matlabfrom Matlab Lines of
constantdamping ratio(ζ=0.7 shown)
Lines of constantnatural frequency
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Imag
Axis
TextEnd
Pole-zero map
Z-Plane - ResponseZ-Plane - ResponseRelationshipRelationship
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11/10/04 2.14 Fall 2004 23
Simple First Order Control SystemSimple First Order Control System
KC Gp(z)-R(z) Y(z)
�
Gp(z) =K
(z − p) open loop root = +p
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11/10/04 2.14 Fall 2004 24
Root LocusRoot LocusROOTS OF 1+KCG-P(z) = 0 ?
Same rules as before:Starts at open loop pole = +pReal axis part to left of
odd number of polesEnds at zero at infinity
p
z
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11/10/04 2.14 Fall 2004 25
Root Locus- InterpretationRoot Locus- Interpretation
p
z
+1
slow, no oscillation
faster, nooscillation
fastest w/oovershoot
beginsoscillating
very oscillatory
unstable
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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
Real Axis
Imag
Axis
Now forNow for
k r
0.22 0.39
0.73 0.13
1 0
1.5 -0.27
2.8 -0.92
3 -1
11/10/04
�
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
11/10/04
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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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11/10/04 2.14 Fall 2004 33
Steady State Unit Step Error?Steady State Unit Step Error?
Our closed-loop TF =
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 + KcK
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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 isMinimizedas K increases
�
limz→1
(z −1) T (z) zz −1
= (z −1) KcKz − p + KcK
z
(z −1)
�
=KcK
1− p + KcK
K=0.5p=0.5