4. investigate the observability of the system
\ \dot{\mathbf{x}} = \mathbf{a}\mathbf{x} + \mathbf{b}\mathbf{u} \
\ \mathbf{y} = \mathbf{c}\mathbf{x} \
where
(a) \mathbf{a} = \begin{bmatrix} -1 & 2 \\ 3 & -4 \end{bmatrix}, \quad \mathbf{c} = \begin{bmatrix} 1 & 1 \end{bmatrix};
(b) \mathbf{a} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{bmatrix}, \quad \mathbf{c} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}.

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Accepted Answer

# Explanation:

🆕 New Concept Discovered: Observability Matrix
Determining if internal states can be reconstructed from outputs.

## Step 1: Understand Observability

A linear time-invariant system is completely state observable if and only if the observability matrix $ \mathcal{O} $ has full rank (i.e., rank equal to the number of states $ n $).

The observability matrix is defined as:
$$ \mathcal{O} = \begin{bmatrix} C \ CA \ CA^2 \ \vdots \ CA^{n-1} \end{bmatrix} $$

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## Step 2: Investigate System (a)

For system (a), the system matrices are:
$$ A = \begin{bmatrix} -1 & 2 \ 3 & -4 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 1 \end{bmatrix} $$

The number of states is $ n = 2 $. Thus, the observability matrix is:
$$ \mathcal{O} = \begin{bmatrix} C \ CA \end{bmatrix} $$

First, calculate $ CA $:
$$ CA = \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} -1 & 2 \ 3 & -4 \end{bmatrix} = \begin{bmatrix} 1(-1) + 1(3) & 1(2) + 1(-4) \end{bmatrix} = \begin{bmatrix} 2 & -2 \end{bmatrix} $$

Construct the observability matrix $ \mathcal{O} $:
$$ \mathcal{O} = \begin{bmatrix} 1 & 1 \ 2 & -2 \end{bmatrix} $$

Calculate the determinant of $ \mathcal{O} $:
$$ \det(\mathcal{O}) = (1)(-2) - (1)(2) = -2 - 2 = -4 $$

Since $ \det(\mathcal{O})
eq 0 $, the matrix $ \mathcal{O} $ has full rank ($ \text{rank}(\mathcal{O}) = 2 $). Therefore, system (a) is **observable**.

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## Step 3: Investigate System (b)

For system (b), the system matrices are:
$$ A = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ -6 & -11 & -6 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} $$

The number of states is $ n = 3 $. Thus, the observability matrix is:
$$ \mathcal{O} = \begin{bmatrix} C \ CA \ CA^2 \end{bmatrix} $$

First, calculate $ CA $:
$$ CA = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ -6 & -11 & -6 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} $$

Next, calculate $ CA^2 $:
$$ CA^2 = (CA)A = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ -6 & -11 & -6 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} $$

Construct the observability matrix $ \mathcal{O} $:
$$ \mathcal{O} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} $$

The matrix $ \mathcal{O} $ is the $ 3 \times 3 $ identity matrix, which has a determinant of $ 1 $. Since $ \det(\mathcal{O})
eq 0 $, the matrix has full rank ($ \text{rank}(\mathcal{O}) = 3 $). Therefore, system (b) is **observable**.

# Answer:

* **System (a)** is **observable** because its observability matrix $ \mathcal{O} = \begin{bmatrix} 1 & 1 \ 2 & -2 \end{bmatrix} $ has full rank ($ \text{rank} = 2 $).
* **System (b)** is **observable** because its observability matrix $ \mathcal{O} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} $ has full rank ($ \text{rank} = 3 $).

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QUESTION IMAGE

Question

Explanation:

Step 1: Understand Observability

Step 2: Investigate System (a)

Step 3: Investigate System (b)

Answer: