Module 9 : State Feedback Control Design

Lecture 3 : State Estimators or Observers

Computation of observer gain matrix L

The task is to place the poles of $ \vert A-LC\vert$. Necessary and sufficient condition for arbitrary pole placement is that the pair should be controllable.

Assumption: The pair $ (A,\; C)$ is observable. Thus, from the theorem of duality, the pair $ (A^T,\; C^T)$ is controllable.

You should note that the eigenvalues of $ A^T-C^TL^T$ are same as that of $ A-LC$. It is same as a hypothetical pole placement problem for the system $ \mathbf{\bar x}(k+1)=A^T\mathbf{\bar x}(k)+C^T\mathbf{\bar u}(k)$, using a control law $ \bar u(k)=-L^T\mathbf{\bar x}(k)$.

Example:



The observability matrix

$\displaystyle U_O= \begin{bmatrix}C \\ CA \end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} $

is non singular. Thus the pair$ (A,\; C)$ is observable. The observer dynamics are

$\displaystyle \hat{\mathbf{x}}(k+1)=A\hat{\mathbf{x}}(k)+B{u}(k)+LC(\mathbf{x}(k)-\hat{\mathbf{x}}(k)) $

L should be designed such that the observer poles are at 0.2 and 0.3.

We design LT such that $ A^T-C^TL^T$ has eigenvalues at 0.2 and 0.3.

$\displaystyle A^T= \begin{bmatrix}0&20 \\ 1&0 \end{bmatrix}, \;\;\; C^T = \begin{bmatrix}1 \\ 0 \end{bmatrix} $

Using Ackermann's formula, $ L^T = [-0.5 \;\;\; 20.06]$. Thus

$\displaystyle L = \begin{bmatrix}-0.5 \\ 20.06 \end{bmatrix}$