Construct a linear system $$ \begin{cases} \dot x = Ax + Bu \\ y=Cx \end{cases} $$ that satisfies all the following requirements:
- 4th order
- 1 input & 1 output
- Unstable
- Stabilizable & detectable
- Rank of controlability matrix = Rank of obeservability matrix = 2
- $C(sI-A)^{-1}B$ is first order, i.e., can be written as $\frac{\alpha}{s+\beta}$ for some scalar constants $\alpha$ and $\beta$.
Attempt
Requirement 1&2: $$ A_{4 \times 4}= \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{bmatrix}, \hspace{0.6cm} B_{4 \times 1}= \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \end{bmatrix}, \hspace{0.6cm} C_{1 \times 4}= \begin{bmatrix} c_{1} & c_{2} & c_{3} & c_{4} \\ \end{bmatrix}, \hspace{0.6cm} x_{4 \times 1}= \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix}, \hspace{0.6cm} u_{1 \times 1}, \hspace{0.6cm} y_{1 \times 1}, $$
Requirement 3:
Real$\{ \lambda_i \}_{i=1,2,3,4}$ is zero or positive. Let $a$ be non-negative real number, and $b,c,d$ be any real numbers. $$ \det(A-\lambda I)=(a-\lambda)(b-\lambda)(c-\lambda)(d-\lambda)=0 $$
Requirement 4:
Stabilizable if $$ Rank( \begin{bmatrix} sI-A & B \\ \end{bmatrix} )=4 $$
Detectable if $$ Rank( \begin{bmatrix} sI-A \\ C \\ \end{bmatrix} )=4 $$
Requirement 5:
The controllability matrix is given by $$ \mathcal{C}= \begin{bmatrix} B & AB & A^2B & A^3B \end{bmatrix} $$
The obeservability matrix is given by $$ \mathcal{O}= \begin{bmatrix} C \\ CA \\ CA^2 \\ CA^3 \end{bmatrix} $$
$Rank(\mathcal{C})=Rank(\mathcal{O})=2$
Requirement 6:
The transfer function is given by $$ \frac{Y(s)}{U(s)}=\frac{\alpha}{s+\beta} $$ for some scalar constants $\alpha$ and $\beta$.
Question
I understand all the requirements and implications but I am not sure how to starting actually constructing. Any help?
Some Hints:
Start attempting with a 4x4 diagonal matrix for $A$ for simplicity(Requirement 1&2). You know that you need a positive eigen value for instability (Requirement 3), I would start with $A = \text{Diag}(1,-1, -1, -1)$ beacause it is the simplest. As $A² = I$ and $A³ = A$, you get your "Rank of controlability matrix = Rank of obeservability matrix = 2" for free (Requirement 5).
Now you can go to the apparent paradox of having a 1st degree TF with a 4th order system. That means some part of your state is not controllable/observable, e.g. with $C = B^t = [1~0~0~0]$ (Requirement 6). Because I have choose the $C$ and $B$ so that I can control and observe the unstable part of the state, it will be Stabilizable/Detectable (Requirement 4).
Ps: Not sure you are aware, but on your Stabilizable/Detectable definition, that must be true for those values of $s$ equal the unstable eigenvalues (greater or equal than zero).