Controllability of state space equationby dabargo Tags: control theory, controllability, matlab, state space 

#1
May610, 11:46 AM

P: 2

1. The problem statement, all variables and given/known data
I have calculated the following mechanical system [tex] \left( \begin{array}{c}\dot{\mathbf{x_1}}(t) & \dot{\mathbf{x_2}}(t) \end{array} \right) = \left( \begin{array}{cc}0.5 & 0 & 0 & 1\end{array} \right) \cdot \left( \begin{array}{c}x_{1}(t) & x_{2}(t)\end{array} \right) + \left( \begin{array}{c}0.5 & 1\end{array} \right)\cdot u(t) [/tex] [tex] \left( \begin{array}{c}y_{1}(t) & y_{2}(t)\end{array} \right) = \left( \begin{array}{cc}1 & 0\end{array} \right) \cdot \left( \begin{array}{c}x_{1}(t) & x_{2}(t)\end{array} \right) [/tex] The question is to find the expression for u(t) that brings the system to its restposition in 2 seconds. Afterwards i have to simulate this time response in matlab with the function lsim. The initial conditions of the system are: [tex] x_1(0) = 10 [/tex] [tex] x_2(0) = 1 [/tex] 2. Relevant equations [tex] u(t) = B^T\exp^{A^T(t_1t)}W_c^{1}(t_1)[\exp^{At_1}x_0x_1] [/tex] [tex] W_c(t_1) = \int_0^{t_1} \exp^{A\gamma}BB^T\exp^{A^T\gamma} d\gamma [/tex] 3. The attempt at a solution I calculated u(t) with matlab with the following code
The result i get is given in the lsimresults.bmp. While i expect it to curve a bit down/up the original time responses (1st and 2nd thumbnails) where u(t) = 0, so that the amplitude of the system becomes 0 at 2 seconds. So, if you there is anybody who can give me some help where it goes wrong. Or if i interpretate something wrong maybe? 



#2
Nov810, 01:00 AM

P: 108

I've tried using state feedback to solve your problem, here is what I got: [tex]u=Kx[/tex] where [tex]x=[x_1 \, x_2]^{T}[/tex] and [tex]K=[K_1 \, K_2] [/tex] You then choose the eigenvalues of the closed loop system [tex](sIA+BK)[/tex] by varying K such that they have the properties you want. In this case you want a settling time of 2s, so I would make both of my eigenvalues equal to 2, which gives us the desired characteristic polynomial of [tex](s+2)(s+2)=s^2+4s+4[/tex] Now to go about solving for [tex]K[/tex], the controllers characteristic polynomial is [tex]det(sIA+BK)=\alpha_2s^2+\alpha_1s+\alpha_0[/tex] All you do now is compare the desired characteristic polynomial with the actual characteristic polynomial and solve for [tex]K_1, K_2[/tex] such that the coefficients of the two characteristic polynomials are equal i.e. [tex]\alpha_2 = 1[/tex] [tex]\alpha_1 = 2[/tex] [tex]\alpha_0 = 4[/tex] For this system there is a pole that no matter what feedback you give the system, will not move higher the pole is located at approx. 0.74 If the settling time we are talking about is 063% of final value then you are ok. The other pole you can shift around at whim using this technique. The gains I used are [tex]K=[1 \, 2][/tex] to give poles at 0.73 and 5.76 I don't have access to matlab now but try its "place" command maybe you can have better luck that way 


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