How to calculate A and B matricies in Control Theory for my system given the ODE describing the system?

I have an ODE describing my system like so:

$\ddot{x} = -\Gamma_0\dot{x} - (\Omega_0 + u)x$

Where u is my feedback control actuator and $\Gamma_0$ and $\Omega_0$ are constants.

I want to find the $\mathbf{A}$ and $\mathbf{B}$ matrix for this system such that I have

$\dot{\vec{x}} = \mathbf{A}\vec{x} + \mathbf{B}\vec{u}$

Where $\vec{x}$ is my state vector

$\vec{x} = \begin{bmatrix} x \\ \dot{x} \\ \end{bmatrix}$

and $\vec{u} = u$

I am unsure how to proceed to get $\mathbf{A}$ and $\mathbf{B}$.

I understand that if my dynamics were non-linear I would find the fixed points of the non-linear dynamics and then linearise about that point by calculating the Jacobian of the dynamics and using this as my A matrix however I am not sure how to proceed as the actuator introduces the non-linearity.

Written as 2 first order differential equations I get

$$\begin{bmatrix} \dot{x} \\ \ddot{x} \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -(\Omega_0 + u) & -\Gamma_0 \\ \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \\ \end{bmatrix}$$

and I can split these up into the following 2 terms.

$$\begin{bmatrix} \dot{x} \\ \ddot{x} \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -(\Omega_0) & -\Gamma_0 \\ \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \\ \end{bmatrix} + \begin{bmatrix} 0\\ -x \\ \end{bmatrix} u$$

But it is my understanding that to implement control using closed-loop feedback control using $u=-\mathbf{K}\vec{x}$ the $\mathbf{A}$ and $\mathbf{B}$ must be constant matrices.

• I've added the representation of the system as a matrix of first order differential equations and the issue with doing this to implement control. – SomeRandomPhysicist Oct 23 '17 at 13:18
• My aim is to calculate the controllability matrix to check I can control the system and then define my $\mathbf{K}$ matrix in order to change the dynamics to change the position of the eigenvalues of the dynamics matrix in order to drive it to a particular desired state $\vec{x}_i$ – SomeRandomPhysicist Oct 23 '17 at 13:21
• This system is uncontrollable, its controllability matrix is zero. After linearization we obtain $B=(0,0)^T$, $U=(B,AB)=0_{2\times 2}$ – AVK Oct 23 '17 at 13:39
• Could you work through the process of linearizing of this system? – SomeRandomPhysicist Oct 23 '17 at 13:46

Denote $\bar x=(x,\dot x)$; the system is $$\dot{\bar x} = f(\dot{\bar x}),\quad f(\dot{\bar x})= \begin{bmatrix} \dot x\\ -(\Omega_0 + u)x -\Gamma_0\dot x \\ \end{bmatrix}$$ $$A=\left.\frac{\partial f}{\partial x}\right|_{\bar x=0,u=0}= \left.\left(\begin{array}{cc} 0&1\\ -(\Omega_0 + u)&-\Gamma_0 \end{array}\right)\right|_{\bar x=0,u=0}= \left(\begin{array}{cc} 0&1\\ -\Omega_0&-\Gamma_0 \end{array}\right)$$ $$B=\left.\frac{\partial f}{\partial u}\right|_{\bar x=0,u=0}= \left.\left(\begin{array}{c} 0\\ -x \end{array}\right)\right|_{\bar x=0,u=0}= \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$ The controllability matrix is equal to $$U=(B,AB)=\left(\begin{array}{cc} 0&0\\ 0&0 \end{array}\right)$$
• If you define a new input to the system as $v = x\,u$, then that system would be controllable. So if you find a control law for $v$ you can find the actual input with $u = v / x$, which indeed has a singularity at $x=0$. However if you can ensure that your system never reaches such a state, then it wouldn't be a problem. Or if $\dot{x}\neq0$ whenever $x=0$ you could briefly set $u=0$ whenever that happens, assuming that staying at $x=0$ is not your goal. Note, this also assumes perfect knowledge of the state. – Kwin van der Veen Oct 23 '17 at 18:12
$$\Phi(\ddot{x},\dot{x},x,u)=\ddot{x} +\Gamma_0\dot{x} + (\Omega_0 + u)x=0.$$
In order to linearize the ODE we need to evaluate the Taylor series of $\Phi$. It is given by:
$$\Phi(\ddot{x},\dot{x},x,u)\approx\left.\Phi(\ddot{x},\dot{x},x,u)\right|_{\text{eq}}+\left.\dfrac{\partial \Phi}{\partial \ddot{x}}\right|_{\text{eq}}\left[\ddot{x}-\left.\ddot{x}\right|_{\text{eq}} \right]+\left.\dfrac{\partial \Phi}{\partial \dot{x}}\right|_{\text{eq}}\left[\dot{x}-\left.\dot{x}\right|_{\text{eq}} \right]$$ $$+\left.\dfrac{\partial \Phi}{\partial x}\right|_{\text{eq}}\left[x-\left.y\right|_{\text{eq}} \right]+\left.\dfrac{\partial \Phi}{\partial u}\right|_{\text{eq}}\left[u-\left.u\right|_{\text{eq}} \right]=0.$$