z-plane analysis
TRANSCRIPT
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2.14 Fall 20042.14 Fall 2004
Digital Control - Z-plane analysisDigital Control - Z-plane analysis
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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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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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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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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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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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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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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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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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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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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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MappingMapping
s
0
s = 0z = e0 = 1 z
+1
z = eTσ ∠z =ωT
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Mapping Mapping jjωω Axis Axis
s
0
s = + jωz = e jωT = unit length vector at angle ωT
z
+1
ωT
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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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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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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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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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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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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