## DC Motor Speed: PID Controller DesignKey MATLAB commands used in this tutorial are: ## ContentsFrom the main problem, the dynamic equations in the Laplace domain and the open-loop transfer function of the DC Motor are the following. (1) (2) (3) The structure of the control system has the form shown in the figure below. For the original problem setup and the derivation of the above equations, please refer to the DC Motor Speed: System Modeling page. For a 1-rad/sec step reference, the design criteria are the following. - Settling time less than 2 seconds
- Overshoot less than 5%
- Steady-state error less than 1%
Now let's design a controller using the methods introduced in the Introduction: PID Controller Design page. Create a new m-file and type in the following commands. ```
J = 0.01;
b = 0.1;
K = 0.01;
R = 1;
L = 0.5;
s = tf('s');
P_motor = K/((J*s+b)*(L*s+R)+K^2);
``` Recall that the transfer function for a PID controller is: (4) ## Proportional controlLet's first try employing a proportional controller with a gain of 100, that is, Kp = 100; C = pid(Kp); sys_cl = feedback(C*P_motor,1); Now let's examine the closed-loop step response. Add the following commands to the end of your m-file and run it in the command window. You should generate the plot shown below. You can view some of the system's characteristics by right-clicking on the figure and choosing t = 0:0.01:5; step(sys_cl,t) grid title('Step Response with Proportional Control') From the plot above we see that both the steady-state error and the overshoot are too large. Recall from the Introduction: PID Controller Design page that increasing the proportional gain This fact can be verified by experimenting with different values of ## PID controlRecall from the Introduction: PID Controller Design page adding an integral term will eliminate the steady-state error to a step reference and a derivative term will often reduce the overshoot. Let's try a PID controller with small ```
Kp = 75;
Ki = 1;
Kd = 1;
C = pid(Kp,Ki,Kd);
sys_cl = feedback(C*P_motor,1);
step(sys_cl,[0:1:200])
title('PID Control with Small Ki and Small Kd')
``` Inspection of the above indicates that the steady-state error does indeed go to zero for a step input. However, the time it takes to reach steady-state is far larger than the required settling time of 2 seconds. ## Tuning the gainsIn this case, the long tail on the step response graph is due to the fact that the integral gain is small and, therefore, it takes a long time for the integral action to build up and eliminate the steady-state error. This process can be sped up by increasing the value of Kp = 100; Ki = 200; Kd = 1; C = pid(Kp,Ki,Kd); sys_cl = feedback(C*P_motor,1); step(sys_cl, 0:0.01:4) grid title('PID Control with Large Ki and Small Kd') As expected, the steady-state error is now eliminated much more quickly than before. However, the large Kp = 100; Ki = 200; Kd = 10; C = pid(Kp,Ki,Kd); sys_cl = feedback(C*P_motor,1); step(sys_cl, 0:0.01:4) grid title('PID Control with Large Ki and Large Kd') As we had hoped, the increased
all of our design requirements will be satisfied. |

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