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Controllability of state space equation

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dabargo
#1
May6-10, 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_1-t)}W_c^{-1}(t_1)[\exp^{At_1}x_0-x_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
A = [-0.5 0;0 1];
B = [0.5; 1];
C = [1 0];
D = 0;
syms t;
Wc = int(expm(A*t)*B*transpose(B)*expm(transpose(A)*t),0,2);
u = -transpose(B)*expm(transpose(A)*(2-t))*inv(Wc)*(expm(A*2)*[10;-1]-[0;0]);
sys = ss(A,B,C,D);
t = 0:0.1:2;
u = subs(u,t);
lsim(sys,u,t)
But the result i get with lsim doesn't fulfill my expectations at all.
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?
Attached Thumbnails
Timeresponse1.jpg   Timeresponse2.jpg  
Attached Images
File Type: bmp lsimresults.bmp (230.7 KB, 4 views)
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Eidos
#2
Nov8-10, 01:00 AM
P: 108
Quote Quote by dabargo View Post
2. Relevant equations
[tex]

u(t) = -B^T\exp^{A^T(t_1-t)}W_c^{-1}(t_1)[\exp^{At_1}x_0-x_1]

[/tex]
[tex]

W_c(t_1) = \int_0^{t_1} \exp^{A\gamma}BB^T\exp^{A^T\gamma} d\gamma

[/tex]
Where did you get those equations, your system is not doing what you want it to do I agree.

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](sI-A+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(sI-A+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 co-efficients 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 0-63% 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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