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This lecture is devoted to another controller for PMDC motor: Linear control.

This lecture is devoted to another controller for PMDC motor: Linear control.

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This lecture is devoted to another controller for PMDC motor: Linear control.

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  1. This lecture is devoted to another controller for PMDC motor: Linear control. ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013

  2. Linear control theory has been predominantly concerned with Linear Time Invariant (LTI) systems of the form • with x being a vector of states and A being the system matrix. LTI systems have quaite simple properties such as • A linear system has a unique equilibrium point if A is nonsingular; • The equilibrium point is stable if all eigenvalues of A have negative real parts, regardless of initial conditions; • In the presence of an external input u(t), i.e., with • the system has a number of interesting properties. For example a sinusoidal input leads to a sinusoidal output of the same frequency. ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Slotine, Li, 1993.

  3. Never forget THERE IS NO LINEAR SYSTEMS IN NATURE ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013

  4. One of the characteristic properties of nonlinear systems is “Multiple Equilibrium Points” Nonlinear systems frequently have more than one equilibrium point (an equilibrium point is a point where the system can stay forever without moving). This can be seen by the following simple examples. Consider the first order linear system ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Solution of this differential equation is Following figure shows the time variation of this solution for various initial conditions. The system clearly has a unique equilibrium point at x=0. Slotine, Li, 1993.

  5. Now consider the following nonlinear systems: Solution of this differential equation is ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Following figure shows the time variation of this solution. The system has two equilibrium points, x=0 and x=1, and its qualitative behavior strongly depends on its initial condition.

  6. Linearization of Nonlinear Systems General form of an n-dimensional nonlinear system is ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 and of an n-dimensional linear system is The linearized form of a nonlinear system can be found as

  7. Example: Linearize System ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Eigenvalues = 1,1 Origin is unstable

  8. Let’s linearize PMDC Motor model now. Dynamic model of the PMDC motor is as follows: mechanical subsystem ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 electrical subsystem

  9. mechanical subsystem electrical subsystem By selecting the state variables as ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 we get the state-space representation as follows:

  10. By setting the control input signal, u, to zero, let’s find the equilibrium points ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 From the first equation, we get Then from the last equation, we find Finally, from the second equation, we get

  11. Then the equilibrium points are ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Let’s linearize the system around

  12. Remember that the system dynamics is ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Then

  13. ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 The Jacobian is

  14. Since we want to linearize the system around the equilibrium point ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 then

  15. Then the linearized form of the system is ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 Note that this dynamical model is a general LTI system of the form Let’s design a simple linear state feedback controller in the form of so that we get In this way, by properly selecting the entries of K vector, we will be able to locate the eigenvalues of newly-created system matrix, (A-BK), to the left-half plane to get stability.

  16. But this stability result will be valid only around the small neighborhood of the linearization point, ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 and we will not have a global stability result. The question is how small this neighborhood is? The neighborhood region is called “Region of Attraction” and it varies from system to system.

  17. There are many algorithms in the literature proposed to find the entries of K vector. The most conventional algorithm can be implemented in MATLAB as follows . ECE 893 Industrial Applications of Nonlinear Control Dr. UgurHasirciClemson University, Electrical and Computer Engineering Department Spring 2013 % Desired closed-loop poles p1 = -10 + 10i; p2 = -10 - 10i; p3 = -50; % Entries of K K = place(A,B,[p1 p2 p3]);

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