![]() To determine the precise characteristics of the step response you can use the It looks like when = 0.15, we can meet our design requirements. Title( 'Step Response with K_p = 21, K_i = 500 and Different values of K_d') Let's see what happened to the step disturbance response, change the following commands in your m-file and re-run at the command Title( 'Step Response with K_p = 21, K_i = 500 and Different Values of K_d') Go back to the m-file and make the following changes. ![]() ![]() We will investigateĭerivative gains ranging from 0.05 to 0.25. ![]() We will attempt to reduce the settling time and overshoot by adding a derivative term toĪdding a derivative term to the controller means that we now have all three terms of the PID controller. We will choose = 500 because the error due to the disturbance decays to zero quickly, even though the response to the reference has a longer Specifically, the larger the value of employed, the faster the error decays to zero. However, the response due to the disturbance changes significantly as the integral gain is changed. Oscillation increasing slightly as is made larger. For the response to the step reference, all of the reponses look similar with the amount of The integral control has reduced the steady-state error to zero, even when a step disturbance is present that was the goalįor adding the integral term. Title( 'Response to a Step Disturbance with K_p = 21 and Different Values of K_i') You should generate a plot like the one shown in the figure below. Change the following commands in your m-file and re-run in theĬommand window. Now let's see what happened to the step disturbance response. Title( 'Response to a Step Reference with K_p = 21 and Different Values of K_i') Change your m-file to the following and run in the command window. We will set = 21 and test integral gains ranging from 100 to 500. Let's first try a PI controller to get rid of the steady-state error due to the disturbance. The DC Motor Position: System Modeling page that adding an integral term will eliminate the steady-state error and a derivative term can reduce the overshoot and Larger values of has the adverse effect of increasing the overshoot and settle time as can be seen from the step reference plot. The larger the value of the smaller the steady-state error is due to the disturbance, but it never reaches zero. Therefore, to have zero steady-state error in the presence of a disturbance, we need the disturbance This follows from the property of superposition that holds for linear systems. Simultaneously is equal to the sum of the two graphs shown above. Specifically, the response due to the reference and disturbance applied Steady-state error when the disturbance is added. This is due to the fact that the plant has an integrator, that is, the system is type 1. The above plots show that the system has no steady-state error in response to the step reference by itself, no matter theĬhoice of proportional gain. Title( 'Response to a Step Disturbance with Different Values of K_p') Add the following to the end of your m-file and run it in the command window. Refer back to the block diagram at the top of this page to see the structure of Now only the plant transfer function P( s) is in the forward path and the controller C( s) is considered to be in the feedback path. The feedback command can still be employed for generating the closed-loop transfer function where there is still negative feedback, however, In this case, we will assume a reference of zero and lookĪt how the system responds to the disturbance by itself. Let's also consider the system's response to a step disturbance. Title( 'Response to a Step Reference with Different Values of K_p') You should generate the plot shown in the figure below. Add the following code to the end of your m-file and again run it in theĬommand window. Now let's see what the step responses look like. Add the following code to the end of your m-file and run it in the MATLAB command window: The closed-loop transfer functions can be generated using the feedback command. An array of LTI models, each with a different proportional gain, can be built using a for loop. Let's first try using a proportional controller with gain ranging from 1 to 21. Recall that the transfer function for a PID controller has the following form. First create a new m-file and type in the following commands (refer to main problem for the details of getting these commands). Now let's design a PID controller and add it into the system. No steady-state error, even in the presence of a step disturbance input.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |