Question

Solve the question by using MATLAB.

EXERCISE 1: Consider a process given below to be controlled by a PID controller, 400 G,(s) = S(S + 48.5) a) Obtain the unit step response of Gp(s). b) Try PI controllers with (Kp=2, 10, 100), and Ki=Kp/10. Investigate the unit step response in each case, compare the results and comment. c) Let Kp=100, Ki=10, and add a derivative term with (Kd=0.1, 0.9, 2). Investigate the unit step response in each case, compare the results and comment. Based on your results in parts b) and c) above what do you conclude as a suitable PID controller for this process and give your justification.

Answer 1

Solution: a) Obtain the unit step response of Gp(s). Command Window 70 >> num=[400]; 60 den=[1 48.5 0); 50 plant-tf (num, den): step (plant) 30 40 Amplitude 20 10 b) Try PI controllers with (Kp =2, 10, 100), and Ki=Kp/10. Investigate the unit step response in each case, compare the results and comment. Command Window Step Response 1.5 » num=[400]; den=[1 48.5 0]: plant=tf(num, den); step (plant) kp=2; ki=kp/10; contr=tf ([kp ki], [1 0]); sys_cl=feedback (contr*plant, 1); t=0:0.01:2; step (sys_ci,t) Amplitude 0.5 0 0.5 1:5 2

Step Response Command Window >> num=[400]: den-[1 48.5 01? plant-tf(num, den): step (plant) kp=10: ki-kp/10: contr=tf(tkp kit, 11 0); sys_cl-feedback (contraplant, 1): t=0:0.01:2: step (sys_cl,t) 05 05 15 Step Response Command Window >> 14001: den=[1 48.5 0); plant-tf (num, den): step (plant) kp=100; kiekp/10; contret ( [lep ki], [1 0]): aya_cl-feedback (contraplant, 1); t-0:0.01:2: step (ays_c1,t) 0.5 Conclusion: By comparing all above responses we can conclude that the proportional controller (kp) will have the effect of expand the rise time and will expand, but never eliminate the steady state error. An integral controller (ki) will have the effect of eliminating the steady state error, but it may make the transient response worse. Effect of each controller kp & ki on the closed-loop system are summarized below: CL Response Rise Time Overshoot Settling Time S-S Error kp Increase Increase Small Change Decrease ki Decrease Increase Increases Eliminate c) Let Kp =100, Ki=10, and add a derivative term with (Kd=0.1, 0.9, 2). Investigate the unit step response in each case, compare the results and comment.

Step Response 2 1.5 System sys_cl Time (seconds): 0.119 Amplitude: 0.998 Command Window >> num=[400]; den=[1 48.5 0]; plant=ti (num, den); step (plant) kp=350; ki=100; kd=0.1; contr=tf([kd kp ki], [1 0]); sys_cl=feedback (contr*plant, 1); t=0:0.01:2; step (sys_cl,t) 0.5 0 o 0.5 15 2 Step Response System sys_cl Time (seconds): 0.0458 Amplitude: 1 Command Window >> num=[400); den=[1 48.5 0); plant-tf(num, den); step (plant) kp=350; ki=100; kd=0.9; contr=tf([kd kp ki], [1 0]); sys_ci=feedback (contr*plant, 1); t=0:0.01:2; step (sys_ci,t) 0.51 0.5 1.5

1.5 Step Response System sys_cl Time (seconds): 0.0316 Amplitude: 1 Command Window >> num=[400]; den=[1 48.5 0); plant-tf (num, den); step (plant) kp=350; ki=100; kd=2; contratf([kd kp ki], [10]); sys_cl=feedback (contr*plant, 1); t=0:0.01:2; step (Sys_c1,1) 0.5 0 0 0.5 1.5 2 Conclusion: By comparing all above responses we can conclude that the derivative control (kd) will have the effect of increasing the stability of the system, reducing the overshoot and improving the transient response... Effect of controller kd closed-loop system are summarized below: CL Response Rise Time Overshoot Settling Time S-S Error kd Small Change Decrease Decrease Small Change Based on your results in parts b) and c) above what do you conclude as a suitable PID controller for this system and give your justification.

Conclusion: By comparing all above responses we can conclude that the proportional controller (kp) will have the effect of expand the rise time and will expand, but never eliminate the steady state error. An integral controller (ki) will have the effect of eliminating the steady state error, but it may make the transient response worse. The derivative control (kd) will have the effect of increasing the stability of the system, reducing the overshoot and improving the transient response... Effect of each controller kp, ki & kd on the closed-loop system are summarized below: CL Response Rise Time Overshoot Settling Time S-S Error kp Increase Increase Small Change Decrease ki Decrease Increase Increases Eliminate kd Small Change Decrease Decrease Small Change