For today, we will tackle the problem of computing the $\mathcal{H}_{\infty}$ norm of a linear system. $\newcommand{\abs}[1]{| #1 |} \newcommand{\ind}{\mathbf{1}} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\ip}[2]{\langle #1, #2 \rangle} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\T}{\mathsf{T}} \newcommand{\Proj}{\mathcal{P}} \newcommand{\Tr}{\mathrm{Tr}} \newcommand{\A}{\mathcal{A}} \newcommand{\Hinf}{\mathcal{H}_{\infty}}$

$\Hinf$ norm for LTI systems

In general, the $\Hinf$ space and norm can be defined for matrix valued functions which are analytic on subsets of $\C$. See Hardy space for more details. We focus on the continuous LTI case, which can be thought of as defining a linear operator on $L^2 \longrightarrow L^2$.

Let $G = \left[ \begin{array}{c|c} A & B \\ \hline C & 0 \end{array} \right]$ denote a state space realization of an LTI. We abuse notation and let $G$ refer to both the operator and the transfer function $G(s)$. Assume that $A$ is stable (all its eigenvalues are in the open left-half plane). Recall that the transfer function is given as $$G(s) = C(sI - A)^{-1}B \:.$$ What is $\norm{G}_{L^2 \rightarrow L^2}$? By a standard calculation involving Parseval's theorem, $$\norm{G}_{L^2 \rightarrow L^2} = \sup_{\omega \in \R} \norm{ G(j\omega) } \:.$$ The notation $\norm{G(j \omega)}$ refers to the maximum singular value of the matrix $G(j \omega)$. The RHS quantity is important enough we denote it $\norm{G}_{\infty}$. Our goal for this post is to prove a very nice characterization of the norm $\norm{G}_{\infty}$.

Theorem: Consider the transfer function $G(s) = C (sI - A)^{-1} B$ and suppose that $A$ is stable. Let $\gamma > 0$. Then, $$\begin{align} \norm{G}_{\infty} < \gamma \Longleftrightarrow \begin{bmatrix} A & \gamma^{-2} BB^\T \\ -C^\T C & - A^\T \end{bmatrix} \text{ has no imaginary axis eigenvalues} \label{eq:main_equiv} \:. \end{align}$$

Proof: This is a nice proof which is based off of Lemma 4.7 in Zhou, Doyle and Glover and Andy Packard's notes. It uses some key ideas from linear systems theory. We first note we can assume without loss of generality that $\gamma = 1$ by the change of variables $B \gets \gamma^{-1} B$.

Recall that for an arbitrary complex matrix $M$, $\norm{M} < 1$ iff $I - M^* M \succ 0$. Fix any $\omega \in \R$ and define $\Phi(s) = I - G(s)^* G(s)$. Using this fact, we have that $$\norm{G(j\omega)} < 1 \Longleftrightarrow \Phi(j\omega) \succ 0 \:.$$ Hence, $$\begin{align} \sup_{\omega \in \R} \norm{G(j\omega)} < 1 &\Longleftrightarrow \Phi(j \omega) \succ 0 \text{ for all } \omega \in \R \cup \{\pm \infty\} \nonumber \\ &\stackrel{(a)}{\Longleftrightarrow} \Phi(j \omega) \text{ is non-singular for all } \omega \in \R \nonumber \\ &\Longleftrightarrow \Phi^{-1}(s) \text{ has no imaginary axis poles } \label{eq:poles_condition} \:. \end{align}$$ The equivalence (a) requires some justification. First, we dropped the condition at $\omega = \pm \infty$ since $\Phi(j (\pm \infty)) = I \succ 0$ holds. The $\Longrightarrow$ direction is clear. For the $\Longleftarrow$ direction, suppose that $\Phi(j \omega)$ is non-singular for all $\omega$, but there exists an $\omega'$ such that $\Phi(j\omega') \not\succ 0$. There are two cases: (a) either $\lambda_{\min}(G(j \omega')) = 0$ or (b) $\lambda_{\min}(G(j \omega')) < 0$. For case (a), that means $G(j \omega')$ is singular, violating the hypothesis. On the other hand, for case (b), since $\lambda_{\min}(\Phi(j(\pm \infty))) = 1$, by continuity of the map $\omega \mapsto \lambda_{\min}(\Phi(j \omega))$, there must exist a $\omega''$ such that $\lambda_{\min}(G(j \omega'')) = 0$, which also violates the hypothesis. Hence, we have reduced our problem to checking the poles of the inverse map $\Phi^{-1}(s)$.

To continue, we now define the LTI system $H$ with state space realization $$H = \left[ \begin{array}{cc|c} A & 0 & B \\ C^\T C & - A^\T & 0 \\ \hline 0 & B^\T & I \end{array} \right] \:.$$ We can compute the transfer function $H(s)$ by using the formula for block matrix inverse, $$\begin{bmatrix} sI - A & 0 \\ -C^\T C & sI + A^\T \end{bmatrix}^{-1} = \begin{bmatrix} (sI - A)^{-1} & 0 \\ (s I + A^\T)^{-1} C^\T C (sI - A)^{-1} & (sI + A^\T)^{-1} \end{bmatrix} \:.$$ Hence, $$\begin{align} H(s) &= \begin{bmatrix} 0 & B^\T \end{bmatrix}\begin{bmatrix} (sI - A)^{-1} & 0 \\ (s I + A^\T)^{-1} C^\T C (sI - A)^{-1} & (sI + A^\T)^{-1} \end{bmatrix}\begin{bmatrix} B \\ 0 \end{bmatrix} + I \nonumber \\ &= B^\T (sI + A^\T)^{-1} C^\T C (sI - A)^{-1} B + I \label{eq:h_transfer_function} \:. \end{align}$$ Next, we expand out $\Phi(j\omega)$, $$\begin{align} \Phi(j \omega) &= I - G(j\omega)^* G(j\omega) \nonumber \\ &= I - B^{\T} (j\omega I - A)^{-*} C^{\T} C(j\omega I - A)^{-1} B \nonumber \\ &= I - B^{\T} (-j\omega I - A^{\T})^{-1} C^{\T} C(j\omega I - A)^{-1} B \nonumber \\ &= I + B^{\T} (j\omega I + A^{\T})^{-1} C^{\T} C(j\omega I - A)^{-1} B \label{eq:phi_equiv} \:. \end{align}$$ Equation $\eqref{eq:h_transfer_function}$ combined with $\eqref{eq:phi_equiv}$ shows that for any $\omega \in \R$, $$H(j \omega) = \Phi(j \omega) \:.$$ We now use the following fact: if $G = \left[ \begin{array}{c|c} A & B \\ \hline C & I \end{array} \right]$ is a realization for an LTI system, then $$G^{-1} = \left[ \begin{array}{c|c} A - BC & - B \\ \hline C & I \end{array} \right]$$ is a realization for the inverse system $G^{-1}$. We use this formula and write the realization for $H^{-1}$ as $$H^{-1} = \left[ \begin{array}{cc|c} A & - BB^\T & -B \\ C^\T C & -A^\T & 0 \\ \hline 0 & B^\T & I \end{array} \right] \:.$$

We next check that $H^{-1}(s)$ does not have any pole-zero cancellations along the imaginary axis. This is equivalent to checking for an uncontrollable or unobservable mode for some $j \omega$. By the PBH test, suppose there is an $\omega \in \R$ and $(w_1, w_2) \neq 0$ such that $$j \omega \begin{bmatrix} w_1^* & w_2^* \end{bmatrix} = \begin{bmatrix} w_1^* & w_2^* \end{bmatrix} \begin{bmatrix} A & -BB^\T \\ C^\T C & -A^\T \end{bmatrix} \:, \:\: 0 = \begin{bmatrix} w_1^* & w_2^* \end{bmatrix} \begin{bmatrix} - B \\ 0 \end{bmatrix} \:.$$ Since $A$ is assumed to be stable, both $(j \omega I + A^\T)$ and $(j \omega I - A)$ are non-singular, and hence these equations imply $w_1 = w_2 = 0$, a contradiction. A nearly identical argument asserts there are no unobservable $j\omega$ modes. This means that $$H^{-1}(s) \text{ has an imaginary axis pole} \Longleftrightarrow \begin{bmatrix} A & -BB^\T \\ C^\T C & -A^\T \end{bmatrix} \text{ has an imaginary axis eigenvalue} \:.$$ But by a similarity transform, $$\begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix} \begin{bmatrix} A & -BB^\T \\ C^\T C & -A^\T \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix} = \begin{bmatrix} A & BB^\T \\ -C^\T C & -A^\T \end{bmatrix} \:.$$ Therefore, the eigenvalues of $\begin{bmatrix} A & -BB^\T \\ C^\T C & -A^\T \end{bmatrix}$ co-incide with the eigenvalues of $\begin{bmatrix} A & BB^\T \\ -C^\T C & -A^\T \end{bmatrix}$. Hence, we have proven $$H^{-1}(s) \text{ has an imaginary axis pole} \Longleftrightarrow \begin{bmatrix} A & BB^\T \\ -C^\T C & -A^\T \end{bmatrix} \text{ has an imaginary axis eigenvalue} \:.$$ Combining $\eqref{eq:poles_condition}$ with the fact that $H^{-1}(j\omega) = \Phi^{-1}(j\omega)$ yields the claimed result $\eqref{eq:main_equiv}$. This concludes the proof.