Recovering initial conditions from observation of y(t)

Question

Consider the following state-space representation of a plant:

$$ x\dot(t) = Ax(t) + Bu(t) $$ $$ y(t) = Cx(t) $$

If the pair (A,C) is observable, is it true that under any causal control law that yields continuous $u(t)$ as a function of current and past values of $y(t)$ the initial condition of the plant $x(0)$ can be recovered from observation of $y(t)$ for $t \geq 0 $? Here you are allowed to use time-varying and even nonlinear control laws.

I guess this is really and output feedback problem. I'm assuming you can define $u(t)$ to be some value that incorporates a Moore-Penrose pseudo inverse and the observability gramian. Not sure where to start though...

Answer 1

If $(C,A)$ is observable then the observability gramian $W_O(t):=\int_0^t{e^{A^T\tau}C^TCe^{A\tau}d\tau}$ is positive definite for all $t>0$. The output response is given by $$y(t)=Cx(t)=Ce^{At}x(0)+\int_0^t{e^{A(t-s)}Bu(s)ds}$$ If we now define the known signal $$z(t):=y(t)-\int_0^t{e^{A(t-s)}Bu(s)ds}$$ then $$Ce^{At}x(0)=z(t)$$ Multiplying the above identity from the left with $e^{A^Tt}C^T$ and integrating over $[0,t]$ we obtain $$W_O(t)x(0)=\int_0^t{e^{A^T\tau}C^Tz(\tau)d\tau}$$

If $(C,A)$ is observable then the above equation has the unique solution $$x(0)=W_O^{-1}(t)\int_0^t{e^{A^T\tau}C^Tz(\tau)d\tau}$$

Answer 2

Yes. If $(C,A)$ is observable, the initial state can be recovered from knowledge of input, output, and the triple $(C,A,B)$. The input $u(t)$ has to be known but is otherwise arbitrary.