Lecture Fundamentals of control systems: Chapter 5 - TS. Huỳnh Thái Hoàng
lượt xem 5
download
Lecture "Fundamentals of control systems - Chapter 5: Analysis of control system performance" presentation of content: Performance criteria, steady state error, transient response, the optimal performance index,... Invite you to reference.
Bình luận(0) Đăng nhập để gửi bình luận!
Nội dung Text: Lecture Fundamentals of control systems: Chapter 5 - TS. Huỳnh Thái Hoàng
- Lecture Notes Fundamentals of Control Systems Instructor: Assoc. Prof. Dr. Huynh Thai Hoang Department of Automatic Control Faculty of Electrical & Electronics Engineering Ho Chi Minh City University of Technology Email: hthoang@hcmut.edu.vn huynhthaihoang@yahoo.com Homepage: www4.hcmut.edu.vn/~hthoang/ 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1
- Chapter 5 ANALYSIS OF CONTROL SYSTEM PERFORMANCE 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 2
- Content Performance criteria Steady state error Transient response The optimal performance index Relationship between frequency domain performances and time domain performances. 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 3
- Performance criteria 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 4
- Performance criteria: Steady state error yfb(t) Y(s) ess R(s) E(s) r(t) +_ G(s) Yfb(s) e(t) ess H(s) t 0 Error: is the difference between the set-point set point (input) and the feedback signal. e(t ) r (t ) y fb (t ) E ( s ) R ( s ) Y fb ( s ) Steady-state error: is the error when time approaching infinity. ess lim li e(t ) ess lim li sE E ( s) t s 0 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 5
- Performance criteria – Percent of Overshoot (POT) Overshoot: refers to an output exceeding its steady-state steady state value. value y(t) y(t) overshoot ymax ymax yss yss yss yss No overshoot t t 0 0 Percentage of Overshoot (POT) is an index to quantify the overshoot of a system, POT is calculated as: y max y ss POT 100% y ss 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 6
- Performance criteria – Settling time and rise time Settling time (ts): is the time required for the response of a system to reach and stay within a range about the steady- state value of size specified by absolute percentage of the steady-state value (usually 2% or 5%) Rise time (tr): is the time required for the response of a system to rise from 10% to 90% of its steady-state steady state value. value y(t) y(t) (1+)yss yss yss (1)yss 0.9yss 0.1yss t t 0 0 ts tr 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 7
- Steady--state error Steady 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 8
- Steady--state error Steady R( ) R(s) E( ) E(s) Y( ) Y(s) +_ G(s) Yfb(s) ( ) H(s) R( s) Error expression: E (s) 1 G(s) H (s) sR( s ) Steady-state error: ess lim sE ( s ) lim s 0 s 0 1 G ( s ) H ( s ) Remark: Steady-state error not only depends on the structure and p parameters of the system y but also depends p on the input p signal. 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 9
- Steady--state error to step input Steady Step input: R( s ) 1 / s 1 Steady-state error: ess 1 Kp with K p lim G ( s ) H ( s ) (position constant) s 0 yfb(t) yfb(t) 1 1 t t 0 0 G(s)H(s) does not have G(s)H(s) has at least 1 any deal integral factor ideal integral factor 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 10
- Steady--state error to ramp input Steady Ramp input: R( s ) 1 / s 2 1 ess with K v lim sG ( s ) H ( s ) (velocity constant) Kv s 0 s yfb(t) yfb(t) yfb(t) r(t) r(t) r(t) ess 0 ess = 0 e(t) t t t 0 0 0 G(s)H(s) does not have G(s)H(s) has 1 ideal G(s)H(s) has at least deal integral factor integral factor 2 ideal integral factors 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 11
- Steady--state error to parabolic input Steady t R( s) 1 / s 3 P b li iinput: Parabolic 1 ess with K a lim s G ( s ) H ( s ) 2 (acceleration Ka s 0 constant ) yfb(t) yfb(t) yfb(t) r(t) r(t) r(t) ess0 ess= 0 e(t) t t t 0 0 0 G(s)H(s) has less than G(s)H(s) has 2 ideal G(s)H(s) has more than 2 ideal integral factors integral factors 2 ideal integral factors 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 12
- Transient response 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 13
- First--order system First R(s) K Y(s) Ts 1 K Transfer function: G (s) T 1 Ts 1 First order system has 1 real pole: p1 T 1 K Transient response: Y ( s ) R ( s )G ( s ) . s Ts 1 y (t ) K (1 e t /T ) 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 14
- First--order system (cont’) First Im s y(t) (1+).K K (1).K Re s 0 0.63K 1/T t 0 T ts Pole – zero plot Transient response of a first order system of the first order y (t ) K (1 e t /T ) 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 15
- First--order system – Remarks First First order system has only one real pole at (1/T), its transient response doesn’t have overshoot. Time constant Ti t t T: T is i the th time ti required i d for f the th step t response off the system to reach 63% its steady-state value. The further the pole (1/T) of the system is from the imaginary axis, the smaller the time constant and the faster the time response of the system. Settling time of the first order system is: 1 ts T ln l where = 0.02 0 02 (2% criterion) or = 0.05 0 05 (5% criterion) 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 16
- First--order system First The relationship between the pole and the time response The further the pole of the system is from the imaginary axis, the smaller the time constant and the faster the time response of the system. Im s y(t) K Re s 0 t 0 Pole – zero plot Transient response of a first order system of the first order 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 17
- Second--order oscillating system Second R(s) K Y(s) T 2 s 2 2Ts 1 The transfer function of the second-order oscillating system: K Kn2 1 G( s) 2 2 2 (n , 0 1) T s 2Ts 1 s 2n s n 2 T The system has two complex conjugate poles: p1, 2 n jn 1 2 1 Kn2 Transient response: Y ( s ) R ( s )G ( s ) . 2 s s 2n s n2 y (t ) K 1 e nt 1 2 sin (n 1 2 )t (cos ) 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 18
- Second--order oscillating system (cont’) Second y(t) Im s cos = j n 1 2 (1+).K n K ((1)).K R s Re n 0 j n 1 2 t 0 ts Pole – zero plot of a second Transient response of a second order oscillating system order oscillating system 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 19
- Second--order oscillating system – Remark Second A second order oscillation system has two conjugated complex poles, its transient response is a oscillation signal. If = 0, 0 transient t i t response is a stable oscillation signal =0 at the frequency n n = 0.2 is called natural oscillation = 0.4 frequency. If 0
CÓ THỂ BẠN MUỐN DOWNLOAD
-
Lecture Fundamentals of control systems: Chapter 1 - TS. Huỳnh Thái Hoàng
93 p | 71 | 11
-
Lecture Fundamentals of control systems: Chapter 2 - TS. Huỳnh Thái Hoàng
120 p | 98 | 8
-
Lecture Fundamentals of control systems: Chapter 3 - TS. Huỳnh Thái Hoàng
54 p | 68 | 7
-
Lecture Fundamentals of control systems: Chapter 4 - TS. Huỳnh Thái Hoàng
72 p | 69 | 7
-
Lecture Fundamentals of control systems: Chapter 8 - TS. Huỳnh Thái Hoàng
60 p | 60 | 7
-
Lecture Fundamentals of control systems: Chapter 9 - TS. Huỳnh Thái Hoàng
65 p | 61 | 7
-
Lecture Fundamentals of control systems: Chapter 6 - TS. Huỳnh Thái Hoàng
103 p | 81 | 6
-
Lecture Fundamentals of control systems: Chapter 7 - TS. Huỳnh Thái Hoàng
48 p | 61 | 6
Chịu trách nhiệm nội dung:
Nguyễn Công Hà - Giám đốc Công ty TNHH TÀI LIỆU TRỰC TUYẾN VI NA
LIÊN HỆ
Địa chỉ: P402, 54A Nơ Trang Long, Phường 14, Q.Bình Thạnh, TP.HCM
Hotline: 093 303 0098
Email: support@tailieu.vn