If you know the desired closed-loop pole locations, you can use the MATLAB commands place or acker. can measure) all four of the state variables. The next step in the design process is to find the vector of state-feedback control gains assuming that we have access (i.e. This should confirm your intuition that the system is unstable Sys_ss = ss(A,B,C,D, 'statename',states, 'inputname',inputs, 'outputname',outputs) Īs you can see, there is one right-half plane pole at 5.5651. P = I*(M+m)+M*m*l^2 %denominator for the A and B matrices After execution in the MATLAB command window, the output will list the open-loop poles (eigenvalues of ) as shown below. Enter theįollowing lines of code into an m-file. The first step in designing a full-state feedback controller is to determine the open-loop poles of the system. To view the system's open-loop response please refer to the Inverted Pendulum: System Analysis page. vertical) and theĬart should move to its new commanded position. Reference is given to the system, the pendulum should be displaced, but eventually return to zero (i.e. We want to design a controller so that when a step The output contains both the position of the cart and the angle of the pendulum. The 4 states represent the position and velocity of the cart and theĪngle and angular velocity of the pendulum. In this problem, represents the step command of the cart's position. Note that here we feedback all of the system's states, rather than using the system's outputs The schematic of this type of control system is shown below where is a matrix of control gains. This problem can be solved using full-state feedback. Well suited to the control of multiple outputs as we have here. Pendulum vertical while controlling the cart's position to move 0.2 meters to the right. In this example, we are attempting to keep the We did not attempt to control the cart's position. In the other examples we were attemping to keep the pendulum vertical in response to an impulsive disturbanceįorce applied to the cart. Steady-state error of less than 2% for andĪs you may have noticed if you went through some of the other inverted pendulum examples, the design criteria for this exampleĪre different.Pendulum angle never more than 20 degrees (0.35 radians) from the vertical.Settling time for and of less than 5 seconds.Its in the same link you provided wikipedia link. The first good reason that comes to mind for the use of C and D matrices, is to perform the Observability and Controlability Tests. However, it might be easier to see, that it is easier to construct the equations for $x_1$ and $x_2$ because their odes are similar (while constructring the ode for the spring deformation will be different).īottom Line: state representation makes much more sense in more complex systems. If you are only interested in the deformation of the string you might create a $C=$ and you are done. However, there are other reasons, which I believe you suspect as linear combination of solutions to obtain a transformed solutionĮxample: think of the following system with two mass springs you can select either the absolute $x_1$ and $x_2$ displacements of the mass.Īnother equivalent representation is $x_1$ and $x_2-x_1$ (essentially the deformation of the spring). \dot$ as a way to perform a "boil down to essentials". In the state space representation, the state equation for a linear time-invariant system is:
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