Question

1. investigate the controllability of the system

\\\dot{\mathbf{x}} = \mathbf{a}\mathbf{x} + \mathbf{b}\mathbf{u}\\
if \\(\mathbf{u}(t)\\) is a scalar and
(a) \\(\mathbf{a} = \

$$\begin{bmatrix} 0 & 1 \\\\ -2 & -3 \\end{bmatrix}$$

\\), \\(\mathbf{b} = \

$$\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix}$$

\\);
(b) \\(\mathbf{a} = \

$$\begin{bmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 2 \\end{bmatrix}$$

\\), \\(\mathbf{b} = \

$$\begin{bmatrix} 0 \\\\ 1 \\\\ 1 \\end{bmatrix}$$

\\);
(c) \\(\mathbf{a} = \

$$\begin{bmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ -1 & -2 & -3 & -4 \\end{bmatrix}$$

\\), \\(\mathbf{b} = \

$$\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}$$

\\).

Explanation:

🆕 New Concept Discovered: Controllability of Linear Systems
*Determining if a system can be steered to any state using its inputs.*

Step 1: Understand the Controllability Criterion

A linear time-invariant system \(\dot{\mathbf{x}} = \mathbf{A}\mathbf{x} + \mathbf{B}\mathbf{u}\) with state dimension \(n\) is completely state controllable if and only if the controllability matrix \(\mathcal{C}\) has full row rank \(n\):

\[\mathcal{C} = \begin{bmatrix} \mathbf{B} & \mathbf{A}\mathbf{B} & \mathbf{A}^2\mathbf{B} & \dots & \mathbf{A}^{n-1}\mathbf{B} \end{bmatrix}\]

If \(\det(\mathcal{C}) eq 0\) (for square \(\mathcal{C}\)), the system is controllable.

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Step 2: Investigate System (a)

Given:
\[\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\]

Here, \(n = 2\). The controllability matrix is:
\[\mathcal{C} = \begin{bmatrix} \mathbf{B} & \mathbf{A}\mathbf{B} \end{bmatrix}\]

Calculate \(\mathbf{A}\mathbf{B}\):
\[\mathbf{A}\mathbf{B} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} (0)(1) + (1)(-1) \\ (-2)(1) + (-3)(-1) \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\]

Construct \(\mathcal{C}\):
\[\mathcal{C} = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\]

Calculate the determinant of \(\mathcal{C}\):
\[\det(\mathcal{C}) = (1)(1) - (-1)(-1) = 1 - 1 = 0\]

Since \(\det(\mathcal{C}) = 0\), the rank of \(\mathcal{C}\) is \(1 < 2\).

**Conclusion for (a):** The system is **uncontrollable**.

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Step 3: Investigate System (b)

Given:
\[\mathbf{A} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\]

Here, \(n = 3\). The controllability matrix is:
\[\mathcal{C} = \begin{bmatrix} \mathbf{B} & \mathbf{A}\mathbf{B} & \mathbf{A}^2\mathbf{B} \end{bmatrix}\]

Calculate \(\mathbf{A}\mathbf{B}\):
\[\mathbf{A}\mathbf{B} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\]

Calculate \(\mathbf{A}^2\mathbf{B}\):
\[\mathbf{A}^2\mathbf{B} = \mathbf{A}(\mathbf{A}\mathbf{B}) = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 4 \end{bmatrix}\]

Construct \(\mathcal{C}\):
\[\mathcal{C} = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{bmatrix}\]

Calculate the determinant of \(\mathcal{C}\) by expanding along the first row:
\[\det(\mathcal{C}) = 0 - 1 \cdot \det\begin{bmatrix} 1 & 1 \\ 1 & 4 \end{bmatrix} + 2 \cdot \det\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\]
\[\det(\mathcal{C}) = -1(4 - 1) + 2(2 - 1) = -3 + 2 = -1\]

Since \(\det(\mathcal{C}) = -1 eq 0\), the matrix has full rank (\(\text{rank} = 3\)).

**Conclusion for (b):** The system is **controllable**.

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Step 4: Investigate System (c)

Given:
\[\mathbf{A} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -2 & -3 & -4 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\]

Here, the system is in **controllable canonical form**. For any system where \(\mathbf{A}\) is in companion form:
\[\mathbf{A} = \begin{bmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -a_0 & -a_1 & -a_2 & \dots & -a_{n-1} \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix}\]
the system is always completely state controllable. Let's verify this by constructing \(\mathcal{C}\):

Calculate the vector sequence:
\[\mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\]

\[\mathbf{A}\mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ -4 \end{bmatrix}\]

\[\mathbf{A}^2\mathbf{B} = \mathbf{A}(\mathbf{A}\mathbf{B}) = \begin{bmatrix} 0 \\ 1 \\ -4 \\ -3(1) - 4(-4) \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -4 \\ 13 \end{bmatrix}\]

\[\mathbf{A}^3\mathbf{B} = \mathbf{A}(\mathbf{A}^2\mathbf{B}) = \begin{bmatrix} 1 \\ -4 \\ 13 \\ -2(1) - 3(-4) - 4(13) \end{bmatrix} = \begin{bmatrix} 1 \\ -4 \\ 13 \\ -42 \end{bmatrix}\]

Construct \(\mathcal{C}\):
\[\mathcal{C} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -4 \\ 0 & 1 & -4 & 13 \\ 1 & -4 & 13 & -42 \end{bmatrix}\]

This is an upper-triangular-like matrix (specifically, a counter-triangular matrix) with \(1\)s along the anti-diagonal. Its determinant is:
\[\det(\mathcal{C}) = (-1)^{\frac{4(4-1)}{2}} (1 \cdot 1 \cdot 1 \cdot 1) = 1 eq 0\]

Since \(\det(\mathcal{C}) eq 0\), the matrix has full rank (\(\text{rank} = 4\)).

**Conclusion for (c):** The system is **controllable**.

Answer:

• (a) Uncontrollable (rank of \(\mathcal{C}\) is 1)
• (b) Controllable (rank of \(\mathcal{C}\) is 3)
• (c) Controllable (rank of \(\mathcal{C}\) is 4)