2.2f LTI State Space

LTI State Space Representation

The state-space of an LTI system is a mathematical representation in terms of a set of first-order difference equations. It describes the system dynamics using state variables, an input, and an output, in a compact matrix form. For a discrete-time LTI system, the state evolution equation is given by

$$q(n+1)=Aq(n)+Bx(n),$$

\end{alignat}$whereastheoutputequationisgivenas$y(n) = Cq(n) + Dx(n),$$

where $q(n)={\left[\begin{array}{cccc}{q}_1(n)& q_2(n)& \dots & q_N(n)\end{array}\right]}^T\in {\mathbb{R}}^N$ is the $N$-dimensional state vector, $x(n),y(n)\in \mathbb{R}$ are the 1-dimensional input and output of the system, and $A\in \mathbb{N}\times \mathbb{N}$, $B\in {\mathbb{R}}^N$, $C\in {\mathbb{R}}^{1\times N}$, and $D\in \mathbb{R}$ are the coefficients of the state evolution equation.

Similarly, the state evolution equation of a continuous-time LTI system is given as
$\frac{d}{dt}q(t)=Aq(t)+Bx(t),$
whereas the output equation is given as
$y(t)=Cq(t)+Dx(t).$

Note!

Examples

(Discrete-Time) Find the State Space of the LCCDE

Suppose we have the LCCDE representation of a discrete-time LTI system, such that
$y(n)=x(n)-2x(n-1)+\frac{5}{2}y(n-1)-y(n-2).$
This LCCDE can be described using the DAG,
{INSERT DAG}

Each delay represents a single state space. As such, we can represent the two elements of the state space vector as

q_{1}(n + 1) = -2x(n)+\frac{5}{2}y(n)+q_{2}(n), \

\text{and} \
q_{2}(n + 1) = -y(n).
\end{alignat}$$
Here, we can also substitute $y(n)=x(n)+q_1(n)$, such that the elements of the state vector equate

q_{1}(n + 1) = -2x(n)+\frac{5}{2}x(n)+\frac{5}{2}q_{1}(n) + q_{2}(n), \

\text{and} \
q_{2}(n + 1) = -x(n)-q_{1}(n).
\end{alignat}$$

With the knowledge of the elements of the state vector $q(n+1)$, we can now find the matrices $A=\left[\begin{array}{cc}\frac{5}{2}& 1\\ -1& 0\end{array}\right]$, and $B=\left[\begin{array}{c}\frac{1}{2}\\ -1\end{array}\right]$ to solve for the state evolution equation.

We proceed with a similar approach to solve the output equation, and find that $C=\left[\begin{array}{cc}1& 0\end{array}\right]$ and $D=1$ (since $y(n)=q_1(n)+x(n)$ as substituted earlier).

\usepackage{amsmath}
 
\begin{document}
\begin{tikzpicture}[auto, node distance=2cm and 2cm]
 
    % Define block style
    \tikzstyle{block} = [rectangle, draw, minimum width=1.5cm, minimum height=1cm]
 
    % Nodes
    \node [coordinate] (input) {};
    \node [circle, draw, inner sep=0.2cm] (sum1) [right of=input] {$+$};
    \node [block, right of=sum1] (delay1) {$z^{-1}$};
    \node [block, right of=delay1] (delay2) {$z^{-1}$};
    \node [circle, draw, inner sep=0.2cm] (sum2) [above of=delay1, node distance=2cm] {$+$};
    \node [circle, draw, inner sep=0.2cm] (sum3) [right of=delay2, node distance=2.5cm] {$+$};
    \node [coordinate, right of=sum3, node distance=1.5cm] (output) {};
 
    % Gain blocks
    \node [block, above of=delay1, node distance=1.5cm] (gain1) {$-2$};
    \node [block, below of=delay1, node distance=1.5cm] (gain2) {$\frac{5}{2}$};
    \node [block, below of=delay2, node distance=1.5cm] (gain3) {$-1$};
 
    % Connections
    \draw [->] (input) -- node {$x(n)$} (sum1);
    \draw [->] (sum1) -- (delay1);
    \draw [->] (delay1) -- node [above] {$x(n-1)$} (delay2);
    \draw [->] (delay2) -- node [above] {$y(n-2)$} (sum3);
    \draw [->] (delay1) -- (gain1);
    \draw [->] (gain1) -| (sum1);
    \draw [->] (delay1) -- (gain2);
    \draw [->] (gain2) -| (sum2);
    \draw [->] (delay2) -- (gain3);
    \draw [->] (gain3) -| (sum3);
    \draw [->] (sum3) -- node {$y(n)$} (output);
    \draw [->] (sum2) -| (sum1);
 
\end{tikzpicture}
\end{document}

(Continuous-Time) Find the State Space of the LCCDE

Consider a standard series RLC circuit, which has the LCCDE (each dot represents an order of the function’s integral),
${\ddot{v}}_o(t)=-\frac{R}{L}{\dot{v}}_o(t)-\frac{1}{LC}{v}_o(t)+\frac{1}{LC}{v}_s(t)⟺v_o(t)=-\frac{R}{L}\int {\dot{v}}_o(t)-\frac{1}{LC}\int \int v_o(t)+\frac{1}{LC}\int \int v_s(t)$
We can model this system into its state space form, such that
${\dot{q}}_1(t)=-\frac{R}{L}{v}_o(t)+q_2(t),$
and
${\dot{q}}_2(t)=-\frac{1}{LC}{v}_o(t)+\frac{1}{LC}{v}_s(t).$
Substituting $v_o(t)=q_1(t)$, we get
${\dot{q}}_1(t)=-\frac{R}{L}{q}_1(t)+q_2(t),$
and
${\dot{q}}_2(t)=-\frac{1}{LC}{q}_1(t)+\frac{1}{LC}{v}_s(t).$
By definition, $A=\left[\begin{array}{cc}-\frac{R}{L}& 1\\ \frac{1}{LC}& 0\end{array}\right]$, $B=\left[\begin{array}{c}0\\ \frac{1}{LC}\end{array}\right]$. Then, $C=\left[\begin{array}{cc}1& 0\end{array}\right]$, and $D=0$.