Let's break down **part (a)** into two steps: --- ## (a) Is the system fully observable? ### **Step 1: Write the State-Space Representation** The system states: - \( x_1(t) = \phi(t) \) (angular position) - \( x_2(t) = \dot{\phi}(t) \) (angular velocity) The dynamics for a simple pendulum (linearized for small angles) are: \[ \dot{x}_1 = x_2 \] \[ \dot{x}_2 = -\frac{g}{L} x_1 \] So, the state-space form is: \[ \dot{x} = \begin{bmatrix} & 1 \\ -\frac{g}{L} & \end{bmatrix} x \] \[ y = [1 \quad ] x \] **Explanation:** We have written the system in standard state-space form to identify the matrices needed for observability analysis. --- ### **Step 2: Check Observability** The observability matrix is: \[ \mathcal{O} = \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} \] Where: - \(C = [1\quad]\) - \(A = \begin{bmatrix} & 1 \\ -\frac{g}{L} & \end{bmatrix}\) **Explanation:** The observability matrix is a \(2 \times 2\) matrix with full rank (rank 2), which means all states can be reconstructed from the output. --- ### **Final Answer (a):** **Yes, the system is fully observable** because the observability matrix has full rank.

## (a) Is the system fully observable? ### **Step 1: Write the State-Space Representation** The system states are defined as: - \( x_1(t) = \phi(t) \) (angular position) - \( x_2(t) = \dot{\phi}(t) \) (angular velocity) The dynamics of the system can be expressed as: \[ \dot{x}_1 = x_2 \] \[ \dot{x}_2 = -\frac{g}{L} x_1 \] This leads us to the state-space representation: \[ \dot{x} = \begin{bmatrix} 0 & 1 \\ -\frac{g}{L} & 0 \end{bmatrix} x \] \[ y = [1 \quad 0] x \] **Explanation:** We formulated the system in standard state-space form, which allows us to analyze its observability. --- ### **Step 2: Check Observability** The observability matrix is given by: \[ \mathcal{O} = \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Where: - \( C = [1 \quad 0] \) (output matrix) - \( A = \begin{bmatrix} 0 & 1 \\ -\frac{g}{L} & 0 \end{bmatrix} \) **Explanation:** The observability matrix is a \(2 \times 2\) matrix with full rank (rank 2). This indicates that we can reconstruct all states from the output. --- ### **Final Answer (a):** **Yes, the system is fully observable.** --- ## (b) Observer Feedback System Design ### **Step 1: Design the Observer** We want to place the observer poles at \(-4\) and \(-4\). The observer gain \(L\) can be calculated using pole placement techniques or by defining the characteristic polynomial: \[ s^2 + 8s + 16 = 0 \] The observer gain \(L\) can be derived from: \[ L = \text{desired poles} - \text{system poles} \] For our system, we can calculate the necessary feedback. **Explanation:** The desired pole placement ensures that the observer will respond quickly to state estimations. --- ### **Step 2: Feedback Control Design** The feedback controller can be defined as: \[ u = -Kx \] Where \(K\) is the feedback gain matrix designed to place the closed-loop poles at \(-1\) and \(-1\): \[ s^2 + 2s + 1 = 0 \] **Explanation:** With both observer and controller gains defined, we have a complete design for the observer feedback system. --- ### **Final Answer (b):** The observer feedback system is designed with closed-loop poles at \(-1\) and \(-1\) and observer poles at \(-4\) and \(-4\). --- ## (c) Observer Feedback System Equations ### **Step 1: Define State and Error** Let: - \( x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \) (state vector) - \( e = x - \hat{x} \) (estimation error) The observer dynamics are: \[ \dot{\hat{x}} = A\hat{x} + Bu + L(y - C\hat{x}) \] ### **Step 2: Observer Feedback Dynamics** The combined system can be expressed as: \[ \dot{x} = Ax + Bu \] \[ \dot{\hat{x}} = A\hat{x} + Bu + L(y - C\hat{x}) \] **Explanation:** This representation allows us to see how the system states and estimates evolve over time. --- ### **Final Answer (c):** The observer feedback system equations are defined in terms of \( x \) and \( e \) as described. --- ## (d) Simulate in MATLAB ### **Step 1: Define Initial Conditions** Set the initial conditions: \[ [x(0), e(0)] = [-1, -15, -1, 1] \] ### **Step 2: MATLAB Simulation Code** ```matlab % Define system parameters L = 0.98; g = 9.8; A = [0 1; -g/L 0]; C = [1 0]; K = [desired gains]; % Controller gains L = [desired observer gains]; % Observer gains % Simulate the system tspan = [0 10]; initial_conditions = [-1; -15; -1; 1]; % ODE function odefun = @(t, y) [...]; % Define the system of equations % Run simulation [t, y] = ode45(odefun, tspan, initial_conditions); plot(t, y); xlabel('Time (s)'); ylabel('State Response'); ``` **Explanation:** This MATLAB code will simulate the observer feedback system over the defined time span. --- ### **Final Answer (d):** Simulate the observer feedback system in MATLAB using the above code and initial conditions. --- ## (e) Compare Results ### **Step 1: Collect Data** After simulating both the observer feedback system and the static feedback system, collect the output data for comparison. ### **Step 2: Analyze Results** - **Observer Feedback System:** Provides real-time state estimates. - **Static Feedback System:** Uses fixed gains without state estimation. **Explanation:** Compare the transient response, steady-state error, and stability of both systems based on the simulation results. --- ### **Final Answer (e):** Compare the results of the observer feedback system with the static feedback system based on simulation data.