# Lecture Fundamentals of control systems: Chapter 5 - TS. Huỳnh Thái Hoàng

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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.

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## Nội dung Text: Lecture Fundamentals of control systems: Chapter 5 - TS. Huỳnh Thái Hoàng

1. 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
2. Chapter 5 ANALYSIS OF CONTROL SYSTEM PERFORMANCE 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 2
3. 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
4. Performance criteria 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 4
5. 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
6. 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
7. 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
9. 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
10. 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
11. 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
12. 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) ess0 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
13. Transient response 6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 13
14. 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
15. 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
16. 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
17. 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
18. Second--order oscillating system Second R(s) K Y(s) T 2 s 2  2Ts  1  The transfer function of the second-order oscillating system: K Kn2 1 G( s)  2 2  2 (n  , 0    1) T s  2Ts  1 s  2n s  n 2 T  The system has two complex conjugate poles: p1, 2  n  jn 1   2 1 Kn2  Transient response: Y ( s )  R ( s )G ( s )  . 2 s s  2n 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
19. 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
20. 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