# Step 1: Checking Observability First, let's write the state-space representation of the system. Given: - State vector: \( x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \varphi(t) \\ \dot{\varphi}(t) \end{bmatrix} \) - Output: \( y = x_1 = \varphi(t) \) The linearized equation for a balancing pointer (inverted pendulum approximation) can be written as: \[ \dot{x}_1 = x_2 \\ \dot{x}_2 = \frac{g}{L}x_1 + u \] So, the state-space form is: \[ \dot{x} = \begin{bmatrix} & 1 \\ \frac{g}{L} & \end{bmatrix} x + \begin{bmatrix} \\ 1 \end{bmatrix} u \] \[ y = \begin{bmatrix} 1 & \end{bmatrix} x \] **Observability Matrix**: \[ \mathcal{O} = \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} \] This is a \(2 \times 2\) matrix with full rank (rank = 2). **Explanation:** Since the observability matrix has full rank, the system is **fully observable**. --- # Step 2: Observer Feedback System Design Given: - Closed-loop poles: -1, -1 - Observer poles: -4, -4 **State Feedback (controller):** Let the controller be \( u = -Kx \). The closed-loop system will have eigenvalues at -1, -1. **Observer:** The observer estimates the states as \( \hat{x} \). Observer dynamics: \[ \dot{\hat{x}} = A\hat{x} + Bu + L(y - C\hat{x}) \] Place the observer poles at -4, -4 (use pole placement method to find \(L\)). **Explanation:** With the system being observable, we can design a full-order observer (Luenberger observer) to estimate the states. With the given pole locations, you can use MATLAB's `place` function to find suitable \(K\) and \(L\). --- ## **Final Answer** **a. The system is fully observable** (since the observability matrix is full rank). **b. An observer feedback system can be designed with the specified closed-loop and observer poles using pole placement techniques.**