LMIs in Control/Click here to continue/Observer synthesis/Full-order state Observer

Full-Order State Observer
The problem of constructing a simple full-order state observer directly follows from the result of Hurwitz detectability LMI's, Which essentially is the dual of Hurwitz stabilizability. If a feasible solution to the first LMI for Hurwitz detectability exist then using the results we can back out a full state observer $$ L $$ such that $$ A + LC $$ is Hurwitz stable.

The System


\begin{align} \dot x(t)&=Ax(t)+Bu(t),\\ y(t)&=Cx(t)+Du(t)\\ x(0)&=x_0\\ \end{align}$$

where $$x(t)\in \R^n$$, $$y(t)\in \R^m$$, $$u(t)\in \R^q$$, at any $$t\in \R$$.

The Data

 * The matrices $$ A,B,C,D $$ are system matrices of appropriate dimensions and are known.

The Optimization Problem
The full-order state observer problem essential is finding a positive definite $$ P $$ such that the following LMI conclusions hold.

The LMI:
1) The full-order state observer problem has a solution if and only if there exist a symmetric positive definite Matrix $$ P $$ and a matrix $$ W $$ that satisfy Then the observer can be obtained as $$ L = P^{-1}W $$ 2) The full-state state observer can be found if and only if there is a symmetric positive definite Matrix $$ P $$ that satisfies the below Matrix inequality In this case the observer can be reconstructed as $$ L = -\frac{1}{2}P^{-1}C^{T} $$. It can be seen that the second relation can be directly obtained by substituting $$ W= -\frac{1}{2}C^{T} $$ in the first condition.
 * $$ A^TP + PA + W^TC + C^TW < 0. $$
 * $$ A^TP + PA - C^TC < 0 $$

Conclusion:
Hence, both the above LMI's result in a full-order observer $$L$$ such that $$ A + LC $$ is Hurwitz stable.