This post shows how to convert a frequency norm bound IQC into a hard time-domain IQC. I was made aware of this trick by Andy Packard. $ \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{\Z}{\mathbb{Z}} \newcommand{\T}{\mathsf{T}} \newcommand{\Proj}{\mathcal{P}} \newcommand{\Tr}{\mathrm{Tr}} \newcommand{\A}{\mathcal{A}} $

Let $G : \ell_{2e} \longrightarrow \ell_{2e}$ denote a causal LTI operator, and let $G(z)$ denote its $z$-transform. Suppose that $G$ is analytic for all $\abs{z} \geq 1$, and let $\norm{G}_\infty$ denote its $H_\infty$-norm. Also, let $P_T$ denote the truncation operator $$ (P_T u)(t) = \begin{cases} u_t & t \leq T \\ 0 &t > T \end{cases} \:. $$

Lemma: Put $y=Gu$ for $u \in \ell_{2e}$. Then for all $T = 0, 1, 2, ...$, we have that $$ \begin{align} \norm{ P_T y }_2 \leq \norm{G}_\infty \norm{P_T u}_2 \:. \label{eq:hard_iqc} \end{align} $$

Proof: Recall that causality of $G$ means that $P_T G P_T = P_T G$. Hence, $$ \begin{align*} \norm{P_T y}_2^2 = \norm{P_T G u}_2^2 = \norm{P_T G P_T u}_2^2 \leq \norm{P_T G}_\infty^2 \norm{P_T u}_2^2 \leq \norm{G}_\infty^2 \norm{P_T u}_2^2 \:. \end{align*} $$ Taking the square root of both sides yields the result. $\square$

Frequency norm bound IQC

Recall that a norm bound on $G$ gives rise to a special kind of sector IQC. Suppose that $\norm{G}_\infty \leq \gamma$. Specifically, putting $y = Gu$, we have for all $z \in \mathbb{T}$, $$ \begin{bmatrix} \widehat{u}(z) \\ \widehat{y}(z) \end{bmatrix}^* \begin{bmatrix} \gamma & 0 \\ 0 & -\gamma^{-1} \end{bmatrix}\begin{bmatrix} \widehat{u}(z) \\ \widehat{y}(z) \end{bmatrix} \geq 0 \:. $$ Integrating both sides yields the IQC $$ \int_{\mathbb{T}} \begin{bmatrix} \widehat{u}(z) \\ \widehat{y}(z) \end{bmatrix}^* \begin{bmatrix} \gamma & 0 \\ 0 & -\gamma^{-1} \end{bmatrix}\begin{bmatrix} \widehat{u}(z) \\ \widehat{y}(z) \end{bmatrix} \; dz \geq 0 \:. $$ By applying Parseval's identity to the IQC, we conclude that $\norm{y}_2 \leq \gamma \norm{u}_2$, which is the soft version of $\eqref{eq:hard_iqc}$. It is quite interesting that by using the causality of $G$, this soft IQC can be cast as a hard IQC as well.