Question

1. consider the two dynamic systems

\\ s_1: \

$$\begin{cases} \\dot{x}_1 = -x_1 + u \\\\ y = \\alpha x_1 \\end{cases}$$

\\
\\ s_2: \

$$\begin{cases} \\dot{x}_2 = x_3 \\\\ \\dot{x}_3 = -2x_2 - 3x_3 + v \\\\ z = x_2 \\end{cases}$$

\\
where \\( \alpha \in \mathbb{r} \\) is a parameter.
\\( s_1 \\) has state \\( x_1 \\), control \\( u \\) and output \\( y \\).
\\( s_2 \\) has state \\( (x_2, x_3) \\), control \\( v \\) and output \\( z \\).
(a) write down the matrix form of the system equations for \\( s_1 \\) and \\( s_2 \\) and determine whether each system is controllable, observable.
(b) these two systems are connected in series as seen in the figure below (\\( u = z \\)). the resulting system is called \\( s_3 \\).
write down the matrix form of the system equations of \\( s_3 \\) and determine whether \\( s_3 \\) is controllable, observable.
(c) the systems are now connected in a feedback configuration as shown in the figure below to produce \\( s_4 \\) (\\( u = y \\)).
write down the matrix form of the system equations for \\( s_4 \\) and determine whether \\( s_4 \\) is controllable, observable.

Explanation:

🆕 New Concept Discovered: State-Space Representation and Controllability/Observability
Representing systems in matrix form and analyzing structural properties.

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Step 1: Matrix Form and Analysis of $S_1$

The system equations for $S_1$ are:
\[ \dot{x}_1 = -x_1 + u \]
\[ y = \alpha x_1 \]

In state-space matrix form:
\[ \dot{x}_1 = A_1 x_1 + B_1 u \]
\[ y = C_1 x_1 + D_1 u \]

where:
\[ A_1 = [-1], \quad B_1 = [1], \quad C_1 = [\alpha], \quad D_1 = [0] \]

Controllability of $S_1$:

The controllability matrix $\mathcal{C}_1$ is:
\[ \mathcal{C}_1 = B_1 = [1] \]
Since $\text{rank}(\mathcal{C}_1) = 1$ (full rank), $S_1$ is always **controllable** for any $\alpha \in \mathbb{R}$.

Observability of $S_1$:

The observability matrix $\mathcal{O}_1$ is:
\[ \mathcal{O}_1 = C_1 = [\alpha] \]
For $S_1$ to be observable, we require $\text{rank}(\mathcal{O}_1) = 1$, which means:
* If $\alpha
eq 0$, $S_1$ is **observable**.
* If $\alpha = 0$, $S_1$ is **not observable**.

---

Step 2: Matrix Form and Analysis of $S_2$

The system equations for $S_2$ are:
\[ \dot{x}_2 = x_3 \]
\[ \dot{x}_3 = -2x_2 - 3x_3 + v \]
\[ z = x_2 \]

Let the state vector be $x_{S2} = \begin{bmatrix} x_2 \\ x_3 \end{bmatrix}$. In matrix form:
\[ \dot{x}_{S2} = A_2 x_{S2} + B_2 v \]
\[ z = C_2 x_{S2} \]

where:
\[ A_2 = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}, \quad B_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C_2 = \begin{bmatrix} 1 & 0 \end{bmatrix} \]

Controllability of $S_2$:

The controllability matrix $\mathcal{C}_2$ is:
\[ \mathcal{C}_2 = \begin{bmatrix} B_2 & A_2 B_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -3 \end{bmatrix} \]
Since $\det(\mathcal{C}_2) = -1
eq 0$, $\text{rank}(\mathcal{C}_2) = 2$. Thus, $S_2$ is **controllable**.

Observability of $S_2$:

The observability matrix $\mathcal{O}_2$ is:
\[ \mathcal{O}_2 = \begin{bmatrix} C_2 \\ C_2 A_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Since $\det(\mathcal{O}_2) = 1
eq 0$, $\text{rank}(\mathcal{O}_2) = 2$. Thus, $S_2$ is **observable**.

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Step 3: Series Connection $S_3$

In the series connection, the output of $S_1$ becomes the input to $S_2$ with $u = z$ (as stated in the text, though the diagram shows $S_1$ feeding into $S_2$, meaning the output of $S_1$ is the input to $S_2$, i.e., $v = y = \alpha x_1$).

Let the state vector of the combined system $S_3$ be $x_{S3} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$.
The equations are:
\[ \dot{x}_1 = -x_1 + u \]
\[ \dot{x}_2 = x_3 \]
\[ \dot{x}_3 = -2x_2 - 3x_3 + \alpha x_1 \]
\[ z = x_2 \]

In matrix form:
\[ \dot{x}_{S3} = A_3 x_{S3} + B_3 u \]
\[ z = C_3 x_{S3} \]

where:
\[ A_3 = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ \alpha & -2 & -3 \end{bmatrix}, \quad B_3 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad C_3 = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} \]

Controllability of $S_3$:

The controllability matrix $\mathcal{C}_3$ is:
\[ \mathcal{C}_3 = \begin{bmatrix} B_3 & A_3 B_3 & A_3^2 B_3 \end{bmatrix} \]
\[ A_3 B_3 = \begin{bmatrix} -1 \\ 0 \\ \alpha \end{bmatrix} \]
\[ A_3^2 B_3 = A_3 \begin{bmatrix} -1 \\ 0 \\ \alpha \end{bmatrix} = \begin{bmatrix} 1 \\ \alpha \\ -\alpha - 3\alpha \end{bmatrix} = \begin{bmatrix} 1 \\ \alpha \\ -4\alpha \end{bmatrix} \]
\[ \mathcal{C}_3 = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & \alpha \\ 0 & \alpha & -4\alpha \end{bmatrix} \]

The determinant of $\mathcal{C}_3$ is:
\[ \det(\mathcal{C}_3) = 1 \cdot (0 - \alpha^2) = -\alpha^2 \]
* If $\alpha
eq 0$, $\text{rank}(\mathcal{C}_3) = 3$, so $S_3$ is **controllable**.
* If $\alpha = 0$, $\text{rank}(\mathcal{C}_3) < 3$, so $S_3$ is **not controllable**.

Observability of $S_3$:

The observability matrix $\mathcal{O}_3$ is:
\[ \mathcal{O}_3 = \begin{bmatrix} C_3 \\ C_3 A_3 \\ C_3 A_3^2 \end{bmatrix} \]
\[ C_3 A_3 = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \]
\[ C_3 A_3^2 = \begin{bmatrix} \alpha & -2 & -3 \end{bmatrix} \]
\[ \mathcal{O}_3 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \alpha & -2 & -3 \end{bmatrix} \]

The determinant of $\mathcal{O}_3$ is:
\[ \det(\mathcal{O}_3) = -1 \cdot (0 - \alpha) = \alpha \]
* If $\alpha
eq 0$, $\text{rank}(\mathcal{O}_3) = 3$, so $S_3$ is **observable**.
* If $\alpha = 0$, $\text{rank}(\mathcal{O}_3) < 3$, so $S_3$ is **not observable**.

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Step 4: Feedback Connection $S_4$

From Figure 2, the feedback loop is configured as follows:
* The external input is $w$.
* The input to the block $-1$ before $S_2$ is $w + y$, where $y$ is the output of $S_1$.
* Thus, the input to $S_2$ is $v = -(w + y) = -w - \alpha x_1$.
* The output of $S_2$ is $z = x_2$.
* The input to the block $-1$ before $S_1$ is $-z = -x_2$.
* Thus, the input to $S_1$ is $u = -x_2$.

Let's write the state equations using these relations:
\[ \dot{x}_1 = -x_1 + u = -x_1 - x_2 \]
\[ \dot{x}_2 = x_3 \]
\[ \dot{x}_3 = -2x_2 - 3x_3 + v = -\alpha x_1 - 2x_2 - 3x_3 - w \]

Let the state vector be $x_{S4} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ and the output be $y = \alpha x_1$.
In matrix form:
\[ \dot{x}_{S4} = A_4 x_{S4} + B_4 w \]
\[ y = C_4 x_{S4} \]

where:
\[ A_4 = \begin{bmatrix} -1 & -1 & 0 \\ 0 & 0 & 1 \\ -\alpha & -2 & -3 \end{bmatrix}, \quad B_4 = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}, \quad C_4 = \begin{bmatrix} \alpha & 0 & 0 \end{bmatrix} \]

Controllability of $S_4$:

The controllability matrix $\mathcal{C}_4$ is:
\[ \mathcal{C}_4 = \begin{bmatrix} B_4 & A_4 B_4 & A_4^2 B_4 \end{bmatrix} \]
\[ A_4 B_4 = \begin{bmatrix} 0 \\ -1 \\ 3 \end{bmatrix} \]
\[ A_4^2 B_4 = A_4 \begin{bmatrix} 0 \\ -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ -7 \end{bmatrix} \]
\[ \mathcal{C}_4 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 3 \\ -1 & 3 & -7 \end{bmatrix} \]

The determinant of $\mathcal{C}_4$ is:
\[ \det(\mathcal{C}_4) = 1 \cdot (0 - 1) = -1 eq 0 \]
Since $\det(\mathcal{C}_4)
eq 0$ for all $\alpha$, $S_4$ is always **controllable** for any $\alpha \in \mathbb{R}$.

Observability of $S_4$:

The observability matrix $\mathcal{O}_4$ is:
\[ \mathcal{O}_4 = \begin{bmatrix} C_4 \\ C_4 A_4 \\ C_4 A_4^2 \end{bmatrix} \]
\[ C_4 A_4 = \begin{bmatrix} -\alpha & -\alpha & 0 \end{bmatrix} \]
\[ C_4 A_4^2 = \begin{bmatrix} -\alpha & -\alpha & 0 \end{bmatrix} \begin{bmatrix} -1 & -1 & 0 \\ 0 & 0 & 1 \\ -\alpha & -2 & -3 \end{bmatrix} = \begin{bmatrix} \alpha & \alpha & -\alpha \end{bmatrix} \]
\[ \mathcal{O}_4 = \begin{bmatrix} \alpha & 0 & 0 \\ -\alpha & -\alpha & 0 \\ \alpha & \alpha & -\alpha \end{bmatrix} \]

The determinant of $\mathcal{O}_4$ is:
\[ \det(\mathcal{O}_4) = \alpha \cdot (\alpha^2 - 0) = \alpha^3 \]
* If $\alpha
eq 0$, $\text{rank}(\mathcal{O}_4) = 3$, so $S_4$ is **observable**.
* If $\alpha = 0$, $\text{rank}(\mathcal{O}_4) < 3$, so $S_4$ is **not observable**.

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Answer:

(a) Systems $S_1$ and $S_2$

• $S_1$ Matrix Form:

$$ \dot{x}_1 = [-1]x_1 + [1]u, \quad y = [\alpha]x_1 $$

• Controllability: Always controllable.
• Observability: Observable if and only if $\alpha

eq 0$.

• $S_2$ Matrix Form:

$$ \dot{x}_{S2} =
LATEXBLOCK0
x_{S2} +
LATEXBLOCK1
v, \quad z =
LATEXBLOCK2
x_{S2} $$

• Controllability: Always controllable.
• Observability: Always observable.

(b) Series System $S_3$

• $S_3$ Matrix Form:

$$ \dot{x}_{S3} =
LATEXBLOCK3
x_{S3} +
LATEXBLOCK4
u, \quad z =
LATEXBLOCK5
x_{S3} $$

• Controllability: Controllable if and only if $\alpha

eq 0$.

• Observability: Observable if and only if $\alpha

eq 0$.

(c) Feedback System $S_4$

• $S_4$ Matrix Form:

$$ \dot{x}_{S4} =
LATEXBLOCK6
x_{S4} +
LATEXBLOCK7
w, \quad y =
LATEXBLOCK8
x_{S4} $$

• Controllability: Always controllable.
• Observability: Observable if and only if $\alpha

eq 0$.