Example 13.1. Consider two first order systems with the same control acting on both:

$$\begin{array}{r}{\dot{x}}_1+x_1=u,\\ {\dot{x}}_2+\alpha x_2=u,\end{array}$$ where $\alpha >0$. This gives

$$A=\left[\begin{array}{cc}-1& 0\\ 0& -\alpha \end{array}\right],B=\left[\begin{array}{c}1\\ 1\end{array}\right].$$

If $\alpha =1$, then it is easy to see from the definition that the system is not controllable. Choose $x^0=\left[\begin{array}{c}0\\ 0\end{array}\right]$. Since the equations are identical (and we have chosen identical initial conditions), we then have $x_1(t)=x_2(t)$ for all $t$, no matter what $u$ is. With $x^1=\left[\begin{array}{c}1\\ 0\end{array}\right]$ we then have that $x^1$ is not reachable from $x^0$: assume that it is reachable in time $T$, then $1=x_1(T)=x_2(T)=0$, which is a contradition.

The Kalman controllability matrix is

$$\left[\begin{array}{cc}B& AB\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 1& -\alpha \end{array}\right],$$

which has determinant

$$-\alpha +1,$$

which is zero if and only if $\alpha =1$. We see that the system is controllable if and only if $\alpha \ne 1$. If $\alpha =1$, then the Kalman controllability matrix equals

$$\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]$$

and we see that its image equals the multiples of

$$\left[\begin{array}{c}1\\ 1\end{array}\right],$$

so that the reachable subspace consists of vectors in ${\mathbb{R}}^2$ which have their first and second component the same.

The Hautus controllability matrix is

$$\left[\begin{array}{cc}sI-A& B\end{array}\right]=\left[\begin{array}{ccc}s+1& 0& 1\\ 0& s+\alpha & 1\end{array}\right].$$

Selecting the first and last column we obtain a matrix with determinant $s+1$ which therefore is invertible as long as $s\ne -1$. Therefore the Hautus controllability matrix is surjective for $s\ne -1$. For $s=-1$ the Hautus controllability matrix equals

$$\left[\begin{array}{ccc}0& 0& 1\\ 0& -1+\alpha & 1\end{array}\right].$$

Since the first column is zero, we only have to consider the last two columns. The matrix formed of the last two columns has determinant $\alpha -1$ and is therefore invertible if and only if $\alpha \ne 1$. We then obtain (using that the neglected column is zero) that the Hautus controllability matrix is invertible for all $s\in \mathbb{C}$ if and only if $\alpha \ne 1$. Therefore we again see the system is controllable if and only if $\alpha \ne 1$.

The control Lyapunov equation is (using that we know that $Q$ is symmetric)

$$\left[\begin{array}{cc}-1& 0\\ 0& -\alpha \end{array}\right]\left[\begin{array}{cc}{Q}_1& Q_0\\ Q_0& Q_2\end{array}\right]+\left[\begin{array}{cc}{Q}_1& Q_0\\ Q_0& Q_2\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -\alpha \end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]\left[\begin{array}{cc}1& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right],$$

which is

$$\left[\begin{array}{cc}-2Q_1+1& Q_0(-1-\alpha )+1\\ Q_0(-1-\alpha )+1& -2\alpha Q_2+1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Solving this (we really just have three separate equations in the three unknowns $Q_1$, $Q_2$ and $Q_0$) gives

$$Q=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{1+\alpha }\\ \frac{1}{1+\alpha }& \frac{1}{2\alpha }\end{array}\right].$$

The determinant of $Q$ equals

$$\frac{1}{4\alpha }-\frac{1}{(1+\alpha {)}^2},$$

which is zero if and only if $(1+\alpha {)}^2=4\alpha$, i.e. when $(1-\alpha {)}^2=0$, i.e. when $\alpha =1$. So we again see that the system is controllable if and only if $\alpha \ne 1$. When $\alpha =1$ we have

$$Q=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right],$$

which has image the scalar multiples of

$$\left[\begin{array}{c}1\\ 1\end{array}\right],$$

so that we again obtain the reachable subspace as obtained using a different method above.

Example 13.2. Consider the second order scalar differential equation

$$\ddot{q}+2\zeta \dot{q}+q=u,$$

where $\zeta \ge 0$, with state $x=\left[\begin{array}{c}q\\ \dot{q}\end{array}\right]$. We then have

$$A=\left[\begin{array}{cc}0& 1\\ -1& -2\zeta \end{array}\right],B=\left[\begin{array}{c}0\\ 1\end{array}\right].$$

The Kalman controllability matrix is

$$\left[\begin{array}{cc}B& AB\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& -2\zeta \end{array}\right].$$

The determinant equals $-1$ which is nonzero so that the Kalman controllability matrix is invertible and therefore surjective and therefore the system is controllable.

Example 13.3. We again consider the second order scalar differential equation from Example 13.2 (now with $\zeta >0$) and aim to find the infinite-time controllability Gramian. The control Lyapunov equation (using that $Q$ is symmetric) is

$$\left[\begin{array}{cc}0& 1\\ -1& -2\zeta \end{array}\right]\left[\begin{array}{cc}{Q}_1& Q_0\\ Q_0& Q_2\end{array}\right]+\left[\begin{array}{cc}{Q}_1& Q_0\\ Q_0& Q_2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& -2\zeta \end{array}\right]+\left[\begin{array}{c}0\\ 1\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right],$$

which is

$$\left[\begin{array}{cc}2Q_0& Q_2-Q_1-2\zeta Q_0\\ Q_2-Q_1-2\zeta Q_0& -2Q_0-4\zeta Q_2+1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

From the top-left corner we obtain $Q_0=0$ and subsequently from the bottom-right corner we obtain $Q_2=\frac{1}{4\zeta }$ and from the off-diagonal entry we obtain $Q_1=\frac{1}{4\zeta }$. Hence

$$Q=\frac{1}{4\zeta }\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].$$

Since this is invertible, we see (once again) that the system is controllable.

Example 13.4. Consider the undamped second order scalar differential equation

$$\ddot{q}+q=u,$$

with state $x=\left[\begin{array}{c}q\\ \dot{q}\end{array}\right]$, so that

$$A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right],B=\left[\begin{array}{c}0\\ 1\end{array}\right].$$

The aim is to, for a given $\zeta >0$, find a $F$ such that the characteristic polynomial of $A$ equals $s^2+2\zeta s+1$. From our knowledge of second order scalar differential equations, we know that $u=-2\zeta \dot{q}$ will achieve this. We want to check that Ackermann’s formula also leads to this.

Ackermann’s formula tells us that

$$F=-\left[\begin{array}{cc}0& 1\end{array}\right]{\left[\begin{array}{cc}B& AB\end{array}\right]}^{-1}(A^2+2\zeta A+I),$$

will achieve the objective. From Example 13.2 we have

$$\left[\begin{array}{cc}B& AB\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],$$

so that

$${\left[\begin{array}{cc}B& AB\end{array}\right]}^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$$

We further have

$$A^2=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right],$$

so that Ackermann’s formula is

$$\begin{array}{rl}F& =-\left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right](\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]+2\zeta \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])\\ & =-\left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]2\zeta \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\\ & =-\left[\begin{array}{cc}0& 1\end{array}\right]2\zeta \left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]\\ & =-2\zeta \left[\begin{array}{cc}0& 1\end{array}\right],\end{array}$$ which indeed is equivalent to the formula $u=-2\zeta \dot{q}$ that we obtained above.