BMEG302 Biomedical Sensors and Actuators Week 12: Analysis and Synthesis

BMEG302: Biomedical Sensors and Week 12: Analysis and Synthesis of Control SystemsActuators for Biomedical Engineering Applications I. Lecturer/Coordinator Dr. Emre Sariyildiz School of Mechanical, Materials, Mechatronic...

BMEG302: Biomedical Sensors and Week 12: Analysis and Synthesis of Control SystemsActuators for Biomedical Engineering Applications I. Lecturer/Coordinator Dr. Emre Sariyildiz School of Mechanical, Materials, Mechatronic and Biomedical Engineering, Building 8, Room 116. Phone: 02 4221 3319 E-mail: [email protected] Consultation Times: Available in SO. 1 Prepared by Dr. E. BMEG302: Biomedical Sensors and Actuators Outlin e 9th Week: Design, Modelling, and Control of Biomedical Systems Developed by using Sensors and Actuators. 10th Week: Analysis and Synthesis of Control Systems for Biomedical Engineering Applications I. 11th Week: Public Holiday. 12th Week: Analysis and Synthesis of Control Systems for Biomedical Engineering Applications II. 2 Prepared by Dr. E.  Analysis and Synthesis of Control Systems Analy Synthesi sis s PID controller Stability Analysis: synthesis Ziegler– Bode Plot, Root Nichols Locus, Nyquist Performance Pole Diagrams Analysis: Settling placement/assignmen time, Overshoot t methods, e.g., Robustness Analysis: Frequency-domain polynomials Stability and performance design approaches, 3 underbychanging Prepared Dr. E. conditions e.g., Bode Plots  Feedback Controller Synthesis High-gain Feedback Control: Energy d s  Consumption! e s  u s   y  s  Bandwidth r s   C s  G s  Limitations!   n(e.g., sampling  s  time: 1ms) Saturation  ! Cost! y s  C  s G  s   C s G s   1 r  s  1  C  s G  s  Other Prepared by Dr. E. components! 4  Speed Control of DC Motor l  m  ref Ki m 1 m C s CKs p    s  G  sG  s Js  b  ˆ m MMss 1 5 Prepared by Dr. E.  Speed Control of DC Motor l ref  m  Ki m 1 m C s   K p  G s    s  Js  b  ˆ m M s  1 L s  K pp s  K ii L Ts s C  s  G  s  M  s 2 2 1  L s  Js  Js b p  s  Ki Kbs 6 Prepared by Dr. E.  l  m mref K m 1 C s   K p  i G s    Js  b   s K p s  Ki ˆ m T s   Js 2  b  K p  s  K i M s  1 Pole Assignment Example Design a PI controller for G(s) to locate the closed loop des des p poles of the 1  p and system 1 p 2  p at 2 Desired characteristic Ch s  p1des s  p2des s 2  2n s  n2  functionbis K Ki 2 p 2n des des n p1 , p2 ???? J J  ? K p 2 J n  b K i  J n2 n ? 7 Prepared by Dr. E.  Velocity Controller Synthesis: Example 1 3 Design a controllerG sfor  to s 7 achieve P 7 Controller a) Steady-state error of step input is less than ess 0.01. 7  3K !. b) Settling time is 5 ms (or t 5  5ms faster). 3K T s   s  7  3K 8 Prepared by Dr. E.  Velocity Controller Synthesis: Example 1 3 Design a controllerG sfor  s 7 to achieve 7 ess s 0.01. a) Steady-state error of step input is less than   0.01 7  3K 7  0.07 K 231 0.03 9 Prepared by Dr. E.  Velocity Controller Synthesis: Example 1 3 3K Design a controller for G s   s 7 to T s   s  7  3K achieve K 331 7  0.07 b) Settling time is 5 ms (or K 231 Steady-state0.03 error faster). 3K K 331 7  3 Step K Response 1 1  0.07 T s      110 3 K 331 1 7  3K 0.03 s 1 7  3K K 331 10 Prepared by Dr. E.  Velocity Controller Synthesis: Example 2 Design a controllerG for s  3 T s  3to K s  K  p i s 7 achieve s2  7  3K  s  3K p i a) No steady-state error is observed for PI step input. Controller 1!. b) Settling time is 1s (ort 5  1s    =0.25 5 faster). 11 Prepared by Dr. E. T s  3 K s  K  p i s 2  7  3K  s  3K p i Velocity Controller Synthesis: Example 2 Let’s assume that we tune Kp and Ki so that the characteristic Ch s   s  1K p sfunction  K i 3 is K p s  Ki  1 T s  3  Ch s  3K p s  3  K p KK is   s  1K p s  Ki 3  s  1  2 p  sK 3K i i 1 3 2 3K p  3s K s  7  3K pp s13K i i1 7  3K p  0.25 1 7 Kp  Ki  3 3 1 7  1.33  9.333 Prepared by Dr. E. 0.75 0.75 12  Velocity Controller Synthesis: Example 3 Design a controllerG sfor  3 T s  3to K s  K  p i s 7 achieve s2  7  3K  s  3K p i a) No steady-state error is observed forPI step input. Controller 4 ts  b) Settling time is 1s (or 1; 2%  !. n 4 n faster). c) No overshoot is 1  n 4 observed. 13 Prepared by Dr. E.  T s  3 K s  K  p i s2  7  3K  s  3K Ch s  s  2n s   2 2 n p i Velocity Controller Synthesis: Example 3  1 T s   K p s  Ki   Ch s  s  8s  16 2 3  n 4 s 2  8s  16 2n  7 1 s  16 n  2 16 Kp   T s   K i  2  3 3 s  4  3 3 s  16  T s   2 s  4  14 Prepared by Dr. E.  Zero(s) of Dynamic Systems Overshoot due to dominant s  16  s  16    1 s  2 zero(s) closed-loop T s   2  T s   s  8s  16 s  4 2 2 s  4 2 No overshoot s 161 Do we observe s  2  any since T  s   2 T  s   overshoot ? s  4 2 s  4  15 Prepared by Dr. E.  Position Controller Synthesis Determine the gain values K p of K v and so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1s. Find the rise J time 1 Nms / radtime. Assume that 1 kgm 2 andbsettling and  ref e 1  1  Kp  Js  b s    ref  Kv  Kp Js  b  K p K v  s  K p 2 16 Prepared by Dr. E. Modern Control Engineering, 4th Ed., Ogata,   Kp  2 Js  b  K p K v  s  K p ref  Position Controller Synthesis Determine the gain values K p of K v and so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1s. Find the rise J time 1 Nms / radtime. Assume that 1 kgm 2 andbsettling and Consider the overshoot goal:   1  2  M p e 0.2   ln 0.2    0.456 1  2 17 Prepared by Dr. E. Modern Control Engineering, 4th Ed., Ogata,   Kp  2 Js  b  K p K v  s  K p ref  Position Controller Synthesis Determine the gain values K p of K v and so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1s. Find the rise J time 1 Nms / radtime. Assume that 1 kgm 2 andbsettling and Consider the peak time goal:  tp  1  n 3.53 2 n 1   18 Prepared by Dr. E. Modern Control Engineering, 4th Ed., Ogata,  Position Controller Synthesis Determine the gain values K p of K v and so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1s. Find the rise J time 1 Nms / radtime. Assume that 1 kgm 2 andbsettling and  Kp n2  0.456  2  2 where Js  b  K p K v  s  K p 2 s  2n s  n ref  n 3.53 2 K pJ  b K p  J n2 12.5 Kv  0.178 Kp 19 Prepared by Dr. E. Modern Control Engineering, 4th Ed., Ogata,  Position Controller Synthesis Determine the gain values K p of K v and so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1s. Find the rise J time 1 Nms / radtime. Assume that 1 kgm 2 andbsettling and Rise Settling 2% criterion Time: Time: 1  1   2  4 tr  tan  1   0.65 s ts  2.4850 s n 1   2    n   20 Prepared by Dr. E. Modern Control Engineering, 4th Ed., Ogata,  Force Controller Synthesis Example  1 Ch s  s 2  2n s  n2 Design a force controller to achieve  n 40 s 2  80s  1600 4 a) Settling time is 100 ms (or ts  0.1; 2%  n 40 n faster). b) No overshoot is 1  n 40 observed. when J = 0.1 kg, b = 0.25N/s, Ke = 100N/m and De = 0.5 Ns/m. Prepared by Dr. E. 21   1 Ch s  s 2  2n s  n2 n 40 s 2  80s  1600 Force Controller Synthesis Example 1 qe ref F  e  1 1 q  F K Js 2  bJs De s  K e    e  s  Kes Db   We can’t independently assign two poles using P controller! De s  K e De s  K e L  s  K 2 T s  K 2 Js  b  De  s  K e Js  b  1  K  De  s  1  K  K e 22 Prepared by Dr. E.   1 Ch s  s 2  2n s  n2 n 40 s 2  80s  1600 Force Controller Synthesis Example 2 qqee ref FF ref ee Ki  11 1 qq  DeessKKee FF Kp  D   s Js 2  b Js D eb s  K e s     We can’t independently assign three poles using PI controller! L s   K p s  Ki De s  Ke  T s   De K p s 2  K e K p  De K i  s  K e K i s  Js 2  b  De  s  K e      Js 3  b  1  K p  De s 2  De K i  1  K p  K e s  K e K i 23 Prepared by Dr. E.   1 Ch s  s 2  2n s  n2 n 40 s 2  80s  1600 Force Controller Synthesis Example 3 qe ref F e  1 1 q  F KKpp  KKddss Js 2  bJs De s  K e   e  s  Kes Db    We can independently assign two poles using PD However, controller!PD controller is very noise-sensitive in force control!  K  K d s De s  K e  De K d s 2  K e K d  De K p  s  K e K p L s   p T s   Js  b  De  s  K e 2  J  De K d  s 2  b  1  K p  De  K d K e  s  1  K p  K e 24 Prepared by Dr. E.   1 Ch s  s 2  2n s  n2 n 40 s 2  80s  1600 Force Controller with Velocity Feedback: Example 4 qqe e refref ref FF  ee  11 1 11 qq  FF KKF FK F 2   Js Js bJs v bKDv e  s  Kse s bK DD D es eess KK K e ee       We can independently assign two poles using P Kv controller with velocity feedback! De s  K e De s  K e L s  K F 2 T s  K F 2 Js  b  K v  De  s  K e Js  b  1  K F  De  K v  s  1  K F  K e 25 Prepared by Dr. E.   1 Ch s  s 2  2n s  n2 n 40 s 2  80s  1600 Force Controller with Velocity Feedback: Example 4 When J = 0.1 kg, b = 0.25N/s, Ke = 100N/m and De = 0.5 Ns/m. De s  K e 5s  1000 T s  K F 2 K F 2 Js  b  1  K F  De  K v  s  1  K F  K e s  7.5  5K F  10 K v  s  1  K F 1000 1  K F 1000 1600 7.5  5K F  10 K v 80 1600 80  7.5  5 0.6 KF   1 0.6 Kv  6.95 1000 10 26 Prepared by Dr. E.  e s  1 Ki 1  C s   K p   K d s G s   2 r  s  1  C  s G  s  s Js  bs Feedback Controller Synthesis u s  d s  r s  ? r s  e s  u s   y s   e s  C s  G s  d s  ?   n s   u s   ? n s  y s  C  s G  s  y s  G s  y s  C  s G  s     r  s  1  C  s G  s  d s  1  C  s G  s  n s  1  C  s G  s  27 Prepared by Dr. E.  Conclusion Analysis and Synthesis of Feedback Control Systems Analysis: Stability, Performance, Robustness….. Synthesis: P, PI, PID, Ziegler–Nichols, Pole Placement Pole Assignment by using Desired Characteristic Function Motion Controller Synthesis Velocity, Position and Force Controller Synthesis Robustness Against Disturbances and Noise 28 Prepared by Dr. E.  End Next Week: Analysis and Synthesis of Control Systems for Biomedical Engineering Applications II. 29 Prepared by Dr. E.  Pole Placement Example 25 Design a s   controllerGfor to achieve s 2  7.07 s  25 25 a) Steady-state error is less than 0.01 for ess lim se s   s 0 25  25 K p step input. 4 b) Settling time is 25 ms. ts  ; 2% n ln M p  c) Overshoot is    2  ln 2 M p  10%. 30 Prepared by Dr. E. Modern Control Engineering, 4 th Ed., Ogata,  Pole Placement Example 25 Design a s   controllerGfor to achieve s 2  7.07 s  25 25 a) Steady-state error is less than 0.01 ss lim se s   efor s 0 25  25 K p step input. 25  0.01 25  25 K p 25  0.25 Kp  99 0.25 31 Prepared by Dr. E.  Pole Placement Example K p  99 25 Design a s   controllerGfor to achieve s 2  7.07 s  25 4 4 b) Settling time is ts  ; 2% n  160 n 0.025 25 ms. 2n  7.07 Kd  25 =12.5172 32 Prepared by Dr. E.  Pole Placement Example K p  99 25 Design a s   controllerGfor to achieve s 2  7.07 s  25 K d 12.5172 ln M p  ln 0.1 c) Overshoot is   2 2  0.5912   ln M p   2  ln 2 0.1 10%. 4 If n  t 160 from settling time constrain, then the s natural frequency is 160 160 n   270.6566  0.5912 33 Prepared by Dr. E.  Pole Placement Example 25 Design a s   controllerGfor to achieve s 2  7.07 s  25 a) Steady-state error is less K p  99 than 0.01. K d 12.5172 b) Settling time is 25 ms. c) Overshoot is  0.5912 n 270.6566 10%. 2 n  25 Kp  25 2929.2 34 Prepared by Dr. E.  PoleMore Placement than Example 20% overshoot 25 C s   K p  K d s Design a s   2 controllerGfor to achieve s  7.07 s  25 C  s G  s  T s   a) Steady-state error is less 1  C  s G  s  than 0.01. b) Settling time is 25 ms. T s  25 K p  Kd s  c) Overshoot is s 2  7.07  25 K d  s  25  25 K p  10%. Settling time goal is achieved 312.93s +73230 C s  12.5172s +2929.2 T s   2 s  320s  73255 35 Prepared by Dr. E.  Pole Placement Example 25 Design a s   controllerGfor to achieve s 2  7.07 s  25 a) Steady-state error is less 312.93s +73230 T s   2 than 0.01. s  320s  73255 b) Settling time is 25 ms. 73230 CFF s   c) Overshoot is 312.93s +73230 10%. 73230 312.93s +73230 73230 CFF s T s   2  2 312.93s +73230 s  320s  73255 s  320s  73255 36 Prepared by Dr. E.  Feedforward Compensation d s  r s  e s  u s   y s   CFF s  C s  G s    y s  C  s G  s  CFF s  r s  1  C  s G  s  37 Prepared by Dr. E.