I have this system.
It's a hydrulic piston which pushes a mass $M$ and that mass pushes a bending beam. The beam has a stiffness $k_3$ and a damping $b_2$ due to the rotation velocity and position.
And here is the equations for this system. Yes! I do not need to focus on the oil bulk modulus beacuse I going only to simulate a step function of the system. Not an impulse signal.
$$ \frac{1}{3}m_{2}L^{2}\ddot{\theta } = Lk_{_{2}}[y-Lsin(\theta )] -k_{3}\theta -b_{2}\dot{\theta } +\frac{1}{2}Lm_{2}gsin(\theta )\ $$
$$ \ M \ddot{y} = k_{1}(z - y) + b_{1}(\dot{z} - \dot{y}) - M g \mu \dot{y} -k_{2}[y-Lsin(\theta )] \ $$
$$ \ m_{1} \ddot{z} = -k_{1}(z-y) -b_{1}(\dot{z} - \dot{y}) + p_{p}A_{p} - p_{m}A_{m}\dot{z}\ $$
$p_m$ is the pressure at the piston's minus side. $p_p$ is the pressure at the piston's plus side. $\mu$ is the friction factor for the mass $M$. $\frac{1}{3}m_{2}L^{2}\ddot{\theta }$ is the mass rotation inertia with the weight in the center. That's why $\frac{1}{3}$ are used. $Lsin(\theta )$ is the length the top of the beam moves in horizontal direction
My question is....is this correct? I can simulate this system. It works! But this system only gives my NOT full rank when I try to check if this system is controllable. Do I miss an equation or equations?
I give you the matrix system writen in first order ODEs.
States are: $$ z = x_1 , \dot{z} = x_2 $$ $$ y = x_3 , \dot{y} = x_4 $$ $$ \theta = x_5 , \dot{\theta} = x_6$$
I have choosen the values:
$$b_1 = 3000; b_2 = 2000; m_1 = 10; m_2 = 7; M = 2000; g = 9.82; mu = 0.3; L = 0.1; A_p = 0.004; A_m = 0.002; P_m = 1.5*10^6; k_1 = 17777.77; k_2 = 41824.95; k_3 = 9170.76; $$
If i create a system A with a signalmatrix B $$\ A = \begin{bmatrix} 0& 1& 0& 0& 0& 0& \\ -1777.77& 600& 1777.77& 300& 0& 0&\\ 0& 0& 1& 0& 0& 0& \\ 8.88& 1.5& -29.80& -4.446& 2.09& 0& \\ 0& 0& 0& 0& 0& 1& \\ 0& 0& 179249.8& 0& -410810.17& -85714.25& \end{bmatrix}\ $$
$$\ B = \begin{bmatrix} 0\\ 0.0004\\ 0\\ 0\\ 0\\ 0\\ \end{bmatrix}\ $$
This will give $$rank(ctrb(A, B)) = 2 $$
But change A to
$$\ A = \begin{bmatrix} 0& 1& 0& 0& 0& 0& \\ -1777.77& 60& 1777.77& 300& 0& 0&\\ 0& 0& 1& 0& 0& 0& \\ 8.88& 1.5& -29.80& -4.446& 2.09& 0& \\ 0& 0& 0& 0& 0& 1& \\ 0& 0& 179249.8& 0& -410810.17& -857& \end{bmatrix}\ $$
Will give me
$$rank(ctrb(A, B)) = 6 $$
as it should be!
Question: Can I still develop a LQR regulator for this system, even if it's NOT controllable, but it's stable.
Try to run this code and you will see that the system is stable. If you don't have MATLAB, you can use Octave with the contol package. It's free and based on Matlab's control package.
A = [0 1 0 0 0 0; -3555.6 -1000 3555.6 600 0 0;0 0 0 1 0 0; 35.556 6 -119.206 -8.946 8.365 0; 0 0 0 0 0 1; 0 0 125475 0 -287560.2 -90000]
B = [0; 80; 0; 0; 0; 0];
C = [1 0 0 0 0 0; 0 0 1 0 0 0; 0 0 0 0 1 0]
D = 0;
sys = ss(A, B, C, D);
u = linspace(20, 20, 100);
t = linspace(0, 4, 100);
x0 = [0; 0; 0; 0; 0; 0]
lsim(sys, u, t', x0')
rank(ctrb(A, B)) % This will gives 3.
eig(A) % This will give all negative eigenvalues = Stable system
