Transform ODE to State-Space

185 Views Asked by At

Im currently training to transform ODEs into State-Space Form.

For ODEs in the Form of:

$\dfrac{d^n y}{dt^n}+a_{n-1}\dfrac{d^{n-1}y}{dt^{n-1}}+...+a_1\dfrac{dy}{dt}+a_0y=b_0\cdot u(t)$

there is no problem. But Im stucked with ODEs which have other forms. For example:

$a_3\dfrac{d^3 y}{dt^3}+a_1\dfrac{dy}{dt}+a_0y=b_0\cdot u(t).$

My first attempt was to isolate the highest Derivative:

$\dfrac{d^3 y}{dt^3} = \dfrac{1}{a_3}\left(-a_0 y-a_1\dot y+b_0u(t)\right)$

Defining: $x_1=y, x_2=\dot y$ but from now on, I dont know any further..

Thanks for Help!

Edit:

Im trying it:

Defining: $x_1=y, x_2=\dot y,x_3=\ddot y$.

Assume:

$\dot x_1 = \dot y = x_2\\\dot x_2 = \ddot y = x_3\\\dot x_3 = \dddot y =\dfrac{1}{a_3}\left(-a_0 y-a_1 \dot y+b_0u(t)\right)=\dfrac{1}{a_3}\left(-a_0x_1-a_1x_2+b_0u(t)\right)$

So the State-Space Representation would be:

$\begin{bmatrix} \dot x_1\\ \dot x_2\\ \dot x_3 \end{bmatrix}=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -\dfrac{a_0}{a_3} & -\dfrac{a_1}{a_3} & 0\end{bmatrix}\cdot\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}+\begin{bmatrix} 0\\ 0\\ \dfrac{b_0}{a_3} \end{bmatrix}\cdot u(t)$

Is that right?