Let's break down the problem and **design a closed-loop estimator** (observer) for the given system. --- ## **Given System** \[ \dot{x} = \begin{bmatrix} & 1 \\ 2 & 4 \end{bmatrix} x + \begin{bmatrix} \\ 2 \end{bmatrix} u \] \[ y = \begin{bmatrix} 1 & \end{bmatrix} x \] --- ## **Step 1: Observer Structure** The standard observer (Luenberger observer) is: \[ \dot{\hat{x}} = A\hat{x} + Bu + L(y - C\hat{x}) \] Where - \( A = \begin{bmatrix} & 1 \\ 2 & 4 \end{bmatrix} \) - \( B = \begin{bmatrix} \\ 2 \end{bmatrix} \) - \( C = \begin{bmatrix} 1 & \end{bmatrix} \) - \( L \) is the observer gain to be designed --- ## **Step 2: Error Dynamics** Define the estimation error \( e = x - \hat{x} \): \[ \dot{e} = (A - LC)e \] We design \( L \) so that the eigenvalues of \( (A - LC) \) are at desired locations. --- ## **Step 3: Choose Observer Poles** Usually, observer poles are chosen to be **faster** than controller poles. Controller poles: \( s = -3 \pm 2i \) Let’s choose observer poles to be at \( s = -5 \pm 4i \) (arbitrary, but faster). --- ## **Step 4: Observer Gain Calculation** Let \( L = \begin{bmatrix} l_1 \\ l_2 \end{bmatrix} \). \[ A - LC = \begin{bmatrix} & 1 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} l_1 \\ l_2 \end{bmatrix} \begin{bmatrix} 1 & \end{bmatrix} = \begin{bmatrix} - l_1 & 1 \\ 2 - l_2 & 4 \end{bmatrix} = \begin{bmatrix} -l_1 & 1 \\ 2 - l_2 & 4 \end{bmatrix} \] --- ### **Characteristic Equation** Let’s set the characteristic equation of \( A - LC \) to match the desired observer poles: \[ \det\left(sI - (A - LC)\right) = \] \[ \det\left( \begin{bmatrix} s + l_1 & -1 \\ -(2-l_2) & s - 4 \end{bmatrix} \right) = \] \[ (s + l_1)(s - 4) - 1(2 - l_2) = \] \[ (s + l_1)(s - 4) - (2 - l_2) = \] \[ s^2 + (l_1 - 4)s - 4l_1 - (2 - l_2) = \] \[ s^2 + (l_1 - 4)s - 4l_1 - 2 + l_2 = \] --- ### **Desired Characteristic Equation** For poles at \( s = -5 \pm 4i \): \[ (s + 5 - 4i)(s + 5 + 4i) = (s + 5)^2 + (4)^2 = s^2 + 10s + 25 + 16 = s^2 + 10s + 41 \] --- ### **Matching Coefficients** Match with above: \[ s^2 + (l_1 - 4)s + (-4l_1 - 2 + l_2) = s^2 + 10s + 41 \] So, - \( l_1 - 4 = 10 \implies l_1 = 14 \) - \( -4l_1 - 2 + l_2 = 41 \implies l_2 = 41 + 4l_1 + 2 \) Substitute \( l_1 = 14 \): \[ l_2 = 41 + 4 \times 14 + 2 = 41 + 56 + 2 = 99 \] --- ## **Step 5: Final Observer** \[ L = \begin{bmatrix} l_1 \\ l_2 \end{bmatrix} = \begin{bmatrix} 14 \\ 99 \end{bmatrix} \] The observer is: \[ \boxed{ \dot{\hat{x}} = A\hat{x} + Bu + L(y - C\hat{x}), \quad L = \begin{bmatrix} 14 \\ 99 \end{bmatrix} } \] --- ### **Summary** - **Observer poles chosen:** \( s = -5 \pm 4i \) - **Observer gain:** \( L = \begin{bmatrix} 14 \\ 99 \end{bmatrix} \) - **Estimator equation:** \[ \boxed{ \dot{\hat{x}} = \begin{bmatrix} & 1 \\ 2 & 4 \end{bmatrix} \hat{x} + \begin{bmatrix} \\ 2 \end{bmatrix} u + \begin{bmatrix} 14 \\ 99 \end{bmatrix} (y - \begin{bmatrix} 1 & \end{bmatrix} \hat{x}) } \]