2. Unit 4 The Performance of Second Order System 中華技術學院電子系 副教授 蔡樸生 副教授 林盈灝

3. Open Loop & Close Loop Open Loop: Close Loop:

4. The Performance of Second Order System

5. The Response of Second Order System

6. Homework 1 1. Steady State Error : 2. Overshoot : 0.163 3. The Peak Time : 1.8138 s 4. The Rise Time : 0.9 s 5. The Setting Time : 4 s [Hint] : max

7. The Response of

8. Homework2 : The effect of damping ratio

9. P Controller This type of control action is formally known as proportional control (Gain) This type of control action is formally known as proportional control (Gain) Homework3 : K=1, K=4, K=8, K=12, K=36 Homework3 : K=1, K=4, K=8, K=12, K=36 Please explain the effect of P controller to the Please explain the effect of P controller to the second order system second order system

10. Solution of HW3

11. PD Controller

13. The Performance of P Controller : The error signal is positive, the torque : The error signal is positive, the torque is positive and rising rapidly. The large overshoot and oscillations in the output because lack of damping. : The error signal is negative, the torque : The error signal is negative, the torque is negative and slow down causes the direction of the output to reverse and undershoot. : The torque is again positive, thus tending : The torque is again positive, thus tending to reduce the undershoot, the error amplitude is reduced with each oscillations.

14. The contributing factors to the high overshoot The positive correcting torque in the interval The positive correcting torque in the interval is too large ( 抑制 ) is too large ( 抑制 ) Decrease the amount of positive torque Decrease the amount of positive torque The retarding torque in the interval The retarding torque in the interval is inadequate ( 增強 ) is inadequate ( 增強 ) Increase the retarding torque Increase the retarding torque

15. The Effect of PD Controller : is negative; this will reduce the original torque due to alone. : is negative; this will reduce the original torque due to alone. : both and is negative; the negative retarding torque will be greater than that with only P controller. : both and is negative; the negative retarding torque will be greater than that with only P controller. : and have opposite signs. : and have opposite signs. Thus the negative torque that originally contributes to the undershoot is reduced also. Thus the negative torque that originally contributes to the undershoot is reduced also.

16. Homework 4

17. Solution of PD Controller clear; clear; x1=0;x2=0;dt=0.01;r=1;step=2000; x1=0;x2=0;dt=0.01;r=1;step=2000; kp=36;kd=6;pe=r-x1; kp=36;kd=6;pe=r-x1; for k=1:step for k=1:step t(k)=k*dt; t(k)=k*dt; e=r-x1; e=r-x1; de=(e-pe)/dt; de=(e-pe)/dt; u=kp*e+kd*de; u=kp*e+kd*de; x1=x2*dt+x1; x1=x2*dt+x1; x2=(u-4*x2)*dt+x2; x2=(u-4*x2)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e; pos(k)=x1;vel(k)=x2;pe=e; end end

18. PI Controller

19. HW5 : The Effect of PI Controller Adds a zero at to the forward-path T.F. Adds a zero at to the forward-path T.F. Adds a pole at to the forward-path T.F. Adds a pole at to the forward-path T.F. This means that the steady-state error of the original system is improved by one order. This means that the steady-state error of the original system is improved by one order. a=2,b=8,k=1

20. Program of PID Controller clear; clear; x1=0;x2=0;dt=0.01;r=1;step=2000; x1=0;x2=0;dt=0.01;r=1;step=2000; kp=1;kd=6;ki=0.1;pe=r-x1;ie=(r-x1)*dt; kp=1;kd=6;ki=0.1;pe=r-x1;ie=(r-x1)*dt; for k=1:step for k=1:step t(k)=k*dt; t(k)=k*dt; e=r-x1; e=r-x1; de=(e-pe)/dt; de=(e-pe)/dt; ie=ie+e*dt; ie=ie+e*dt; u=kp*e+kd*de+ki*ie; u=kp*e+kd*de+ki*ie; x1=x2*dt+x1; x1=x2*dt+x1; x2=(u-2*x2-8*x1)*dt+x2; x2=(u-2*x2-8*x1)*dt+x2; pos(k)=x1;vel(k)=x2;pe=e; pos(k)=x1;vel(k)=x2;pe=e; end end

21. PID Controller

22. Homework6 針對以下系統, 憑藉經驗值調諧 PID 三個參數值, 使得系統響應為最佳化

23. Homework7 : Ziegler-Nichols Tuning Step 1 : Let until the occur of critical stable Step 2 : Optimal Parameter Tuning

24. Homework8: Pendulum System

25. Feedback Controller Design State Feedback Controller