State Space Realization

Let $T_{uy}$ be real, rational, and proper. Then a state space realization (or just realization) of that $T_{uy}$ is an LTI state space model of the form

$$\begin{array}{r}\dot{x}=Ax+Bu\\ y=Cx+Du\end{array}$$

such that $C(sI-A{)}^{-1}B+D=T_{uy}(s)$.

Note: State space realizations are NOT unique!

Mathematical tool: Inverse of a diagonal matrix

Recall that the inverse of a diagonal matrix is given by:

$${\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]}^{-1}=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& \frac{1}{b}\end{array}\right]$$

Example

Suppose we have a transfer function

$$\begin{array}{rl}{T}_{uy}(s)& =\frac{9s^2+34s+29}{(s+1)(s+2)(s+3)}=\frac{2}{s+1}+\frac{3}{s+2}+\frac{4}{s+3}\\ & =\left[\begin{array}{ccc}\frac{1}{s+1}& \frac{1}{s+2}& \frac{1}{2+3}\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 1& 1\end{array}\right]\left[\begin{array}{ccc}\frac{1}{s+1}& 0& 0\\ 0& \frac{1}{s+2}& 0\\ 0& 0& \frac{1}{s+3}\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 1& 1\end{array}\right]{\left[\begin{array}{ccc}s+1& 0& 0\\ 0& s+2& 0\\ 0& 0& s+3\end{array}\right]}^{-1}\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]\\ & =\underset{C}{\left[\begin{array}{ccc}1& 1& 1\end{array}\right]}{\left(\underset{sI}{\left[\begin{array}{ccc}s& 0& 0\\ 0& s& 0\\ 0& 0& s\end{array}\right]}-\underset{A}{\left[\begin{array}{ccc}-1& 0& 0\\ 0& -2& 0\\ 0& 0& -3\end{array}\right]}\right)}^{-1}\underset{B}{\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]}+\underset{D}{0}\\ & =C(sI-A{)}^{-1}B+D\end{array}$$

Then, we can write

$$\begin{array}{rl}\dot{x}& =\left[\begin{array}{ccc}-1& 0& 0\\ 0& -2& 0\\ 0& 0& -3\end{array}\right]x+\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]u\\ y& =\left[\begin{array}{ccc}1& 1& 1\end{array}\right]x\end{array}$$

General Approach

This approach works for any transfer function $T_{uy}(s)$ with only simple poles.

Assume $u(t)=0\forall t\ge 0$, such that $\dot{x}=Ax$. We’re just looking at the dynamics of the system without any control input.

Assume $A$ is diagonalizable (sufficient condition: all eigenvalues of $A$ are distinct).

Let ${\lambda }_1,\dots ,{\lambda }_n$ be the eigenvalues of $A$.

Let $v_1,\dots ,v_n$ be their associated eigenvectors:

$$\Lambda =\left[\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right],V=\left[\begin{array}{cccc}|& |& & |\\ v_1& v_2& \dots & v_n\\ |& |& & |\end{array}\right]$$

Then:

$$\begin{array}{rl}A& =V\Lambda V^{-1}\text{[def. of diagonalizable]}\\ \Lambda & =V^{-1}AV\end{array}$$

Suppose:

$$\begin{array}{rl}\dot{x}=\left[\begin{array}{c}{\dot{x}}_1\\ ⋮\\ x_n\end{array}\right]& =\left[\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right]\left[\begin{array}{c}{x}_1\\ ⋮\\ x_n\end{array}\right]\\ \dot{x}& =\Lambda x\end{array}$$

Then:

$$\begin{array}{rl}{\dot{x}}_1& ={\lambda }_1{x}_1⟹x_1(t)=e^{{\lambda }_{1}t}{x}_1(0)\\ {\dot{x}}_2& ={\lambda }_2{x}_2⟹x_2(t)=e^{{\lambda }_{2}t}{x}_2(0)\\ & ⋮\\ {\dot{x}}_n& ={\lambda }_n{x}_n⟹x_n(t)=e^{{\lambda }_{n}t}{x}_n(0)\end{array}$$

Note that the dynamics become decoupled and are trivial to solve.

Let $z=V^{-1}x$ such that $x=Vz$. Then:

$$\dot{z}=V^{-1}x=V^{-1}Ax=V^{-1}V\Lambda V^{-1}x=\Lambda V^{-1}x=\Lambda z$$

Then:

$$\dot{z}=\Lambda z$$

such that

$$\begin{array}{rl}{\dot{z}}_1& ={\lambda }_1{z}_1⟹z_1(t)=e^{{\lambda }_{1}t}{z}_1(0)\\ {\dot{z}}_2& ={\lambda }_2{z}_2⟹z_2(t)=e^{{\lambda }_{2}t}{z}_2(0)\\ & ⋮\\ {\dot{z}}_n& ={\lambda }_n{z}_n⟹z_n(t)=e^{{\lambda }_{n}t}{z}_n(0)\end{array}$$

Then:

$$\begin{array}{rl}x(t)=Vz(t)& =\left[\begin{array}{ccc}|& & |\\ v_1& \dots & v_n\\ |& & |\end{array}\right]\left[\begin{array}{c}{z}_1(t)\\ ⋮\\ z_n(t)\end{array}\right]\\ & =\sum _{i=1}^n{z}_i(t)v_i\\ & =\sum _{i=1}^n{e}^{{\lambda }_i(t)}{z}_i(0)v_i\end{array}$$

Eigenvalue Stability

Eigenvalue stability

An eigenvalue $\lambda$ of $A$ is stable if $\lambda \in {\mathbb{C}}^{-}$ and unstable if $\lambda \notin {\mathbb{C}}^{-}$.

Example:

$$\dot{x}=\frac{1}{3}\left[\begin{array}{cc}4& 5\\ 10& -1\end{array}\right]x$$

Then:

$$\begin{array}{rl}{\lambda }_1& =-2,v_1=\left[\begin{array}{c}-1\\ 2\end{array}\right]\\ {\lambda }_2& =3,v_2=\left[\begin{array}{c}1\\ 1\end{array}\right]\end{array}⟹x(t)=e^{{\lambda }_1(t)}{z}_1(0)v_1+e^{{\lambda }_{2}t}{z}_2(0)v_2$$

and $x(0)=z_1(0)v_1+z_2(0)v_2$.

Let’s look at some possible initial conditions:

• Case 1:

$$\begin{array}{rl}& z_2(0)=0,z_1(0)\ne 0⟹x(0)=z_1(0)v_1\\ ⟹& x(t)=e^{{\lambda }_{1}t}{z}_1(0)v_1\\ ⟹& x(t)\text{ never escapes }\text{span}(v_1)\\ ⟹& \underset{t\to \infty }{\lim }x(t)=0\text{ because }{\lambda }_1\text{ is stable}\end{array}$$

• Case 2:

$$\begin{array}{rl}& z_1(0)=0,z_2(0)\ne 0⟹x(0)=z_2(0)v_2\\ ⟹& x(t)=e^{{\lambda }_{2}t}{z}_2(0)v_2\\ ⟹& x(t)\text{ never escapes }\text{span}(v_2)\\ ⟹& \underset{t\to \infty }{\lim }x(t)=\infty \text{ because }{\lambda }_2\text{ is stable}\end{array}$$

• Case 3:

$$\begin{array}{rl}& z_1(0)\ne 0,z_2(0)\ne 0⟹x(0)=z_1(0)v_1+z_2(0)v_2\\ ⟹& x(t)=e^{{\lambda }_{1}t}{z}_1(0)v_1+e^{{\lambda }_{2}t}{z}_2(0)v_2\\ ⟹& x(t)\text{ goes to 0 along }\text{span}({\lambda }_1)\text{ and }\infty \text{ along }\text{span}({\lambda }_2)\end{array}$$

Essentially, we’ve decomposed some state space $x$ into its eigenvector components, which allows each of them to be decoupled/independent. $z$ is just $x$ expressed in eigenvector coordinates, where each $z_i$ is the strength of motion along eigenvector $v_i$. The stability of the system depends on if the eigenvectors ${\lambda }_i\in \mathbb{C}$ .