Lecture Control system design: The performance of feedback control systems - Nguyễn Công Phương

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This chapter presents the following content: Test input signals, performance of second – order systems, effects of a third pole & a zero on the second – order system response, the s – plane root location & the transient response, the steady – state error of feedback control systems, performance indices, the simplification of linear systems, system performance using control design software.

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Nội dung Text: Lecture Control system design: The performance of feedback control systems - Nguyễn Công Phương

Nguyễn Công Phương CONTROL SYSTEM DESIGN The Performance of Feedback Control Systems
Contents I. Introduction II. Mathematical Models of Systems III. State Variable Models IV. Feedback Control System Characteristics V. The Performance of Feedback Control Systems VI. The Stability of Linear Feedback Systems VII. The Root Locus Method VIII.Frequency Response Methods IX. Stability in the Frequency Domain X. The Design of Feedback Control Systems XI. The Design of State Variable Feedback Systems XII. Robust Control Systems XIII.Digital Control Systems sites.google.com/site/ncpdhbkhn 2
The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 3
Introduction Performance Performance measure, M1 measure, M2 pmin Parameter, p sites.google.com/site/ncpdhbkhn 4
The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 5
Test Input Signals (1) • If the system is stable, the response to a specific input signal will provide several measures of the performance. • But because the actual input signal of a system is usually unknown, a standard test input signal is normally chosen. • Using a standard input allows the designer to compare several competing designs. • Many control systems experience input signals that are very similar to the standard test signals. • 4 types: – Unit impulse, – Step, – Ramp, – Parabolic. sites.google.com/site/ncpdhbkhn 6
Test Input Signals (2) r (t ) r (t )   A  2 2 0 t 0 t 1    ,   t  ,  0  A, t  0 A r (t )    2 2 ; R( s )  1 r (t )   ; R( s )  0, otherwise 0, t  0 s Unit impulse Step Ramp Parabolic r (t ) r (t ) A 0 t 0 t  At , t  0 A  At 2 , t  0 2A r (t )   ; R( s )  2 r (t )   ; R( s )  3 0, t0 s 0, t0 s sites.google.com/site/ncpdhbkhn 7
The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 8
Performance of Second – Order Systems (1) R( s) n2 Y ( s) G( s)  G( s) s( s  2n ) Y ( s)  R( s ) ( ) 1  G( s) n2  2 R( s ) s  2n s  n 2 1 n2 R( s )   Y ( s )  s s( s 2  2n s  n2 ) 1  y (t )  1  e nt sin(n 1   2 t  cos1  ) 1 2 1 nt 1 e sin(n  t   )  sites.google.com/site/ncpdhbkhn 9
Performance of Second – Order Systems (2) R( s) n2 Y ( s) G( s)  s( s  2n ) ( ) 1 1 nt r (t )  1  y (t )  1  e nt sin(n 1   2 t  cos1  )  1  e sin(n  t   ) 1 2  1.8 y (t )  = 0.1 1.6  = 0.2  = 0.4 1.4  = 0.7  = 1.1 1.2  = 2.0 1 0.8 0.6 0.4 0.2 t 0 0 0.5 1 1.5 2 2.5 3 sites.google.com/site/ncpdhbkhn 10
Performance of Second – Order Systems (3) R( s) n2 Y ( s) G( s)  G( s) s( s  2n ) Y ( s)  R( s ) ( ) 1  G( s) n2  2 R( s ) s  2n s  n 2 n2 R( s )  1  Y ( s )  2 s  2n s  n2 n  y (t )  e nt sin(n 1   2 t ) 1 2 1 nt 1 e sin(n  t )  sites.google.com/site/ncpdhbkhn 11
Performance of Second – Order Systems (4) R( s) n2 Y ( s) G( s)  s( s  2n ) ( ) n 1 nt r (t )   (t )  y (t )  e nt sin(n 1   2 t )  1  e sin(n  t ) 1 2  4 y (t )  = 0.10  = 0.25 3  = 0.50  = 1.1 2 1 0 -1 -2 t -3 0 0.5 1 1.5 2 2.5 3 sites.google.com/site/ncpdhbkhn 12
Performance of Second – Order Systems (5) 1 1 nt y (t ) r (t )  1  y (t )  1  e nt sin(n 1   2 t  cos 1  )  1  e sin(n  t   ) 1 2  1.6 M pt M pt  final value Percent overshoot  100% 1.4 final value Overshoot 1.2  100e  / 1 2 1.0   1 0.9 1.0   0.8 0.6  Peak time Tp  0.4 n 1   2 4 Rise time Tr Settling time Ts  0.2 n Tr1 0.1 t 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 sites.google.com/site/ncpdhbkhn 13
Performance of Second – Order Systems (5) 120 Percent overshoot Peak time 100 80 60 40 20 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Damping ratio,  sites.google.com/site/ncpdhbkhn 14
Performance of Second – Order 1.6 Systems (6)  = 10rad/s n 1.4  = 1rad/s n 1.2 1 Amplitude 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time (s) sites.google.com/site/ncpdhbkhn 15
Performance of Second – Order Ex. Systems (7) Find K & p so that the transient response to a step should be as R( s) K Y ( s) fast as attainable while retaining an overshoot of less than 5%, s( s  p ) and the settling time should be less than 4 seconds. ( ) K G( s) s( s  p ) K n2 T ( s)    2  2 1  G( s) 1  K s  ps  K s  2n s  n2 s( s  p )  / 1 2 s1,2  n  jn 1   2 Percent overshoot  100e 2 ) / 1(1/ 2 )2  100e  (1/ 4 Ts   4  n  1  4.32% n n  1 s1,2  1  j1   n 1    1 2  p  2n  2(1/ 2) 2  2   1/ 2   K  n  ( 2)  2 2 2  n  2 sites.google.com/site/ncpdhbkhn 16
Performance of Second – Order Ex. Systems (8) Find K & p so that the transient response to a step should be as R( s) K Y ( s) fast as attainable while retaining an overshoot of less than 5%, s( s  p ) and the settling time should be less than 4 seconds. ( ) K  2; p  2 1.4 1.2 1 0.8 y(t) 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time (s) sites.google.com/site/ncpdhbkhn 17
The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 18
Effects of a Third Pole & a Zero on the Second – Order System Response (1) 1 T ( s)  ( s 2  2 s  1)( s  1) 1.4 1.2 1 0.8 y(t) 0.6 0.4  = 2.25  = 1.5  = 0.9 0.2  = 0.4  = 0.05  = 0.001 0 0 2 4 6 8 10 12 14 Time (s) sites.google.com/site/ncpdhbkhn 19
Effects of a Third Pole & a Zero on the Second – Order System Response (2) 1 T ( s)  2 ( s  2 s  1)( s  1) 1 Percent Settling   overshoot time 2.25 0.444 0 9.63 1.50 0.666 3.9 6.30 0.90 1.111 12.3 8.81 0.40 2.50 18.6 8.67 0.05 20.0 20.5 8.37   0.45 sites.google.com/site/ncpdhbkhn 20