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Consider the LTI system \begin{equation}\label{e1} \dot{\mathbf{x}}(t) =A \mathbf{x}(t)+B \mathbf{u}(t) \end{equation} Assume that the system is controllable. it is well known that, if we want to steer the system from $\mathbf{x}(0)=0 $ to a certain target state $\mathbf{x}(t_f)=\mathbf{x}_f$, the control $\mathbf{u}(t)$ that does that and minimizes the energy functional $$E=\int_{0}^{t_{f}} \|\mathbf{u}(t) \|_2^2 \: \mathrm{d} t $$ is given by \begin{equation}\mathbf{u}^{*}(t)=B^{T} \mathrm{e}^{A^{T}\left(t_{f}-t\right)} W ^{-1}\mathbf{x}_{\mathrm{f}}\end{equation} where $$ W=\displaystyle\int_{0}^{t_{\mathrm{f}}} \mathrm{e}^{A\tau} B B^{T} \mathrm{e}^{A^{T}\tau} \: \mathrm{d} \tau $$ is the controllability Gramian matrix.

Now, I would like to write the optimal control $\mathbf{u}^{*}(t)$ in a feedback form, i.e. something like: $$ \mathbf{u}^{*}(t) = -K(t)\mathbf{x}(t)$$

Can anyone show me how to do it? What is the gain $K(t)$ in this case? Can I write it in terms of the Gramian matrix?

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In order to do this you can use a slightly more general expression. Namely, the expression for the input that drives the state from $x(t_i) = x_i$ to $x(t_f) = x_f$, which is given by

\begin{align} W(t) &= \int_0^t e^{A\,\tau} B\,B^\top e^{A^\top \tau} d\tau, \tag{1} \\ u(t) &= B^\top e^{A^\top (t_f - t)}\,W(t_f-t_i)^{-1}\left(x_f - e^{A\,(t_f-t_i)} x_i\right). \tag{2} \end{align}

Note that one gets your equation when one uses $t_i = 0$ and $x_i = 0$ in $(2)$.

In order to get a feedback policy one can replace $t_i$ and $x_i$ in $(2)$ with $t$ and $x(t)$ respectively, yielding

$$ u(t) = B^\top e^{A^\top (t_f - t)}\,W(t_f-t)^{-1}\left(x_f - e^{A\,(t_f-t)} x(t)\right). \tag{3} $$

The expression from $(3)$ can be seen as evaluating $(2)$ using $t_i$ equal to the current value of $t$ and only evaluating $u(t)$ at this current time.

It can be noted that $W(0) = 0$, thus in the limit of $t$ to $t_f$ the expression for $W(t_f-t)^{-1}$ in $(3)$ blows up.

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  • $\begingroup$ According to your formulation, it seems that $u(t_f)$ blows up and this does not seem correct. Am I missing something? $\endgroup$ May 19, 2020 at 19:49
  • $\begingroup$ @LeonardoMassai it only blows up if $x(t_f)\neq x_f$. Namely, the closer $t$ gets to $t_f$ then when $x(t)$ doesn't get closer to $x_f$ (possibly due to some external disturbance) a larger input would be required in order to achieve $x(t_f)=x_f$. Note that the initial equation (with $x_i$ instead of $x(t)$) is openloop, so that same disturbance would very likely result in $x(t_f) \neq x_f$. $\endgroup$ May 19, 2020 at 20:27
  • $\begingroup$ Thanks. And these two control laws are equivalent energy-wise? I know that for the open-loop formulation I get $E(u^*)=x_f^T W^{-1}x_f$, does the same hold for the closed-loop law? $\endgroup$ May 19, 2020 at 20:34
  • $\begingroup$ @LeonardoMassai It can be shown that $(2)$ also minimizes $\int_{t_i}^{t_f} \|u(t)\|^2\,dt$ while assuring that $x(t_f)=x_f$. It can be noted that $(3)$ can be seen as $(2)$ but repeatedly re-calculating the optimal control from $t$ to $t_f$. However, if both $(2)$ and $(3)$ are optimal means that they both should yield the same control input and $x(t)$ (assuming that $\dot{x}(t) = A\,x(t) + B\,u(t)$). $\endgroup$ May 19, 2020 at 23:30

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