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Parallel Compensator for Continuous and Relay Control Systems with Difficult Plants

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Data as in Example 1. insignificant increase. of settling time. For. For ... then both the systems are equivalent, i.e. for the same reference values r(t) we ... – PowerPoint PPT presentation

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Title: Parallel Compensator for Continuous and Relay Control Systems with Difficult Plants


1
Parallel Compensator for Continuous and Relay
Control Systems with Difficult Plants
  • Ryszard Gessing
  • Silesian University of Technology Gliwice, Poland

2
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

3
Introduction
  • Smith predictor (compensator), 1958
  • Deng, Iwai and Mizumoto 1999 for minimum phase
    plants
  • Gessing ACC 2004 for nonminimum phase plants

4
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

5
Plant
is stable
Parallel Compensator
Replacement Plant
6
const
It should be
In steady state
7
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

8
Regulator (for a large k)
Closed loop System
9
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

10
Replacement plant
is chosen so that
n degree of M(s), T is chosen so that
then the CL system is stable for very large k
11
CL system description
Characteristic equation
or
which has (n-1)-multiple root
12
Example 1
Plant
For the replacement
plant is
and parallel compensator
Assume k1000 which gives
13
For r1(t-1) we obtain
Settling time
for t 1, u 1000 for t 1.15, u -140
Acceptable responses for
14
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

15
(No Transcript)
16
Example 2
Data as in Example 1
For
insignificant increase of settling time
For
settling time 5
17
Comparison with regulator PID
Settling time 12.2 about 4-times longer.
18
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

19
During fast switchings we have
20
Equivalent systems
Relay system
Continuous system
switched amplitude
then both the systems are equivalent, i.e. for
the same reference values r(t) we obtain the
same outputs y(t) in both the systems.
21
Example 3
The plant and parallel comensator the same as in
previous examples.
Relay system h0.01, H10 solid line
Continuous system k1000, dotted line
For h0.005 both the outputs are
nondistinguishable.
22
Relay system h0.005, H20
Continuous system
The outputs y for both the systems are
nondinguishable.
23
Comparison
Sliding mode control
Parallel compensator
Ass minimum phase plant
without this assuption
Characteristic equation
24
Outline of Presentation
  • Introduction
  • Parallel Compensator
  • Approximate Description of the CL System
  • Choice of the Replacement Plant
  • Accounting of the Control Saturation
  • Relay Implementation
  • Final Conlusions

25
  • Easy for desinging
  • Some freedom in choosing the replacement plant
    and shaping dynamics
  • Shortening of transients
  • It has some robustness with respect to change of
    the plant parameters.
  • Implementation of sliding mode control without
    necessity of applying of higher order derivatives
  • Equivalence of continuous and relay systems
  • Plant must be stable
  • Regulator has a high order
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