This post is inspired by a recent paper from Lidström and Rantzer, Optimal Distributed $H_\infty$ State Feedback for Systems with Symmetric and Hurwitz State Matrix. Their main result uses the KYP Lemma followed by a few tricks to show that a very simple state feedback controller solves the H-infinity problem whenever $A$ is symmetric and Hurwitz. Here, we relax the symmetric assumption and see what the resulting LMI is. $ \newcommand{\abs}[1]{| #1 |} \newcommand{\ind}{\mathbf{1}} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\ip}[2]{\langle #1, #2 \rangle} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\T}{\mathsf{T}} \newcommand{\Proj}{\mathcal{P}} \newcommand{\Tr}{\mathrm{Tr}} \newcommand{\A}{\mathcal{A}} \newcommand{\Hinf}{\mathcal{H}_{\infty}} $
We consider the following continuous-time LTI plant $P$ with state space realization $$ P = \left[ \begin{array}{c|cc} A & B_1 & B_2 \\ \hline C_1 & D_{11} & D_{12} \\ I & 0 & 0 \end{array} \right] \:. $$ The inputs to $P$ are $(w, u)$, where $w$ is the disturbance and $u$ is the input from the controller. The outputs of $P$ are $(z, y)$, where $z$ is the performance output and $y$ is the input to the controller. Observe in this case, $y = x$, so we are assuming perfect state observation. We assume that $(A, B_2)$ is stabilizable.
Given a static feedback matrix $F$, we close the lower loop with the state feedback law $u = F x$. The resulting closed loop system $T_{w \mapsto z}$ has state space realization $$ T_{w \mapsto z} = \left[ \begin{array}{c|c} A + B_2 F & B_1 \\ \hline C_1 + D_{12} F & D_{11} \end{array} \right] \:. $$
Goal: Given $P$, find a feedback matrix $F$ such that (a) $T_{w \mapsto z}$ is stable and (b) the norm $\norm{T_{w \mapsto z}}_{\infty}$ is minimized.
We note that this goal is remarkably simpler than the general H-infinity synthesis problem, where we are not necessarily given state feedback, and we instead search over stabilizing LTI controllers. As we will see, a sufficient condition for this problem falls almost immediately out of the KYP lemma.
KYP Lemma
We now state the KYP lemma. See Rantzer's paper for the proof.
Lemma: Let $A$, $B$, and $M$ be real-valued matrices. Suppose that $M$ is symemtric and $A$ has no $j\omega$-axis eigenvalues. The following are equivalent:
- For all $\omega \in \R \cup \{\infty\}$, $$ \begin{bmatrix} (j\omega I - A)^{-1} B \\ I \end{bmatrix}^* M \begin{bmatrix} (j\omega I - A)^{-1} B \\ I \end{bmatrix} \prec 0 \:. $$
- There exists a real-valued symmetric $P$ such that $$ \begin{bmatrix} A^\T P + P A & P B \\ B^\T P & 0 \end{bmatrix} + M \prec 0 \:. $$
Let us use the KYP lemma to quickly derive the Bounded Real Lemma. Fix a $\gamma > 0$. Let $G$ be an LTI system with transfer function $G(s) = C(sI - A)^{-1} B$ (not to be confused with the state space matrices of $P$). Now observe that $$ \begin{align*} \norm{G(j\omega)} < \gamma &\Longleftrightarrow G(j\omega)^* G(j\omega) \prec \gamma^2 I \\ &\Longleftrightarrow \begin{bmatrix} (j\omega I - A)^{-1} B \\ I \end{bmatrix}^* \begin{bmatrix} C^\T C & 0 \\ 0 & -\gamma^2 I \end{bmatrix} \begin{bmatrix} (j\omega I - A)^{-1} B \\ I \end{bmatrix} \prec 0 \:. \end{align*} $$ Now applying the KYP lemma, we have that as long as $A$ has no $j\omega$-axis eigenvalues, $\norm{G}_{\infty} < \gamma$ iff there exists a real-valued symmetric $P$ satisfying $$ \begin{align} \begin{bmatrix} A^\T P + P A + C^\T C & P B \\ B^\T P & -\gamma^2 I \end{bmatrix} \prec 0 \:. \label{eq:bounded_real} \end{align} $$ Note that if $A$ is Hurwitz, then any feasible $P$ for $\eqref{eq:bounded_real}$ must be positive definite.
This equivalence is known as the Bounded Real Lemma, and gives a semidefinite characterization of the H-infinity norm. We now apply a Schur complement argument to $\eqref{eq:bounded_real}$ and conclude that $\norm{G}_\infty < \gamma$ iff there exists a real-valued symmetric $P$ satisfying $$ \begin{align} \begin{bmatrix} A^\T P + P A & P B & C^\T \\ B^\T P & -\gamma^2 I & 0 \\ C & 0 & -I \end{bmatrix} \prec 0 \:. \label{eq:bounded_real_schur} \end{align} $$
Controller synthesis
The equivalence $\eqref{eq:bounded_real_schur}$ will be our starting point. Suppose that $A + B_2 F$ is Hurwitz. Plugging the state space matrices for $T_{w \mapsto z}$ into $\eqref{eq:bounded_real_schur}$, we get that $$ \norm{T_{z \mapsto w}}_\infty < \gamma \Longleftrightarrow \begin{bmatrix} (A + B_2F)^\T P + P (A + B_2 F) & P B_1 & C_1^\T + F^\T D_{12}^\T \\ B_1^\T P & -\gamma^2 I & 0 \\ C_1 + D_{12} F & 0 & -I \end{bmatrix} \prec 0 \text{ for some } P \succ 0 \:. $$ The expression on the right hand side is not quite an LMI in $(P, F)$ since it depends on terms such as $P B_2 F$. To work around this, we use the idea in Theorem 1 from Lidström and Rantzer and conjugate both sides of the relation by the matrix $\begin{bmatrix} P^{-1} & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & I \end{bmatrix}$ to obtain the equivalence $$ \begin{bmatrix} P^{-1} (A + B_2 F)^\T + (A + B_2 F) P^{-1} & B_1 & P^{-1}(C_1^\T + F^\T D_{12}^\T) \\ B_1^\T & - \gamma^2 I & 0 \\ (C_1 + D_{12} F) P^{-1} & 0 & - I \end{bmatrix} \prec 0 \text{ for some } P \succ 0 \:. $$ Using the change of variables $Q \gets P^{-1}$ and $R \gets F P^{-1}$, we get the following LMI in $(Q, R)$, $$ \begin{align} \begin{bmatrix} A Q + Q A^\T + B_2 R + R^\T B_2^\T & B_1 & Q C_1^\T + R^\T D_{12}^\T \\ B_1^\T & -\gamma^2 I & 0 \\ C_1 Q + D_{12} R & 0 & -I \end{bmatrix} \prec 0 \text{ for some } Q \succ 0 \:, \:\: R \:. \label{eq:lmi_hinfsyn} \end{align} $$
Algorithm
We now have the following algorithm for a fixed $\gamma$:
- Solve the LMI $\eqref{eq:lmi_hinfsyn}$ for a positive definite $Q$ and $R$. If the LMI is not feasible, return None.
- Set $P = Q^{-1}$ and $F = R P$.
To find the smallest $\gamma$, wrap the procedure in an outer loop of binary search.