This post will focus on gain margins for LQR. It turns out that the feedback controller which comes out as a solution to the LQR problem has some robustness built in. The discussion is based on these notes. $ \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}} $

LQR Review

Here we briefly recall the continuous-time infinite horizon LQR problem formulation and solution. Consider the LTI system $G$ described as $\dot{x} = A x + B u$ with $(A, B)$ stabilizable. Suppose we want to find a $u$ which minimizes $$ J(u) = \int_0^\infty (x^\T Q x + u^\T R u) \: dt \:, $$ where $Q \succeq 0$ and $R \succ 0$. The solution is to set $u = - K x$, where $K = R^{-1} B^\T P$ and $P$ solves the continuous-time ARE (CARE) $$ \begin{align} A^\T P + P A - PB R^{-1} B^\T P + Q = 0 \:. \label{eq:care} \end{align} $$ We will assume that a solution $P$ exists and is positive definite. Sufficient conditions to ensure this are $(A, B)$ stabilizable and $(A, Q)$ observable.

LQR Robustness

Suppose we use the feedback law $u = -Kx$ with a modified system $\dot{x} = A x + {B}_1 u$, where ${B}_1 = B N$. This is equivalent to a system $\widehat{G}$ which places a static gain block $N$ in front of the original system $G$. What conditions on $N$ ensure that the resulting closed loop system remains stable? Letting $P$ denote a positive definite solution to the CARE, let us use a simple Lyapunov argument.

Proposition: Consider the closed loop system $\dot{x} = (A - B_1 K) x$, where $K = R^{-1} B^\T P$ and $P$ is a positive definite solution to the CARE $\eqref{eq:care}$. Suppose that $$ \begin{align} - Q + P B (R^{-1} - R^{-1} N^\T - N R^{-1} ) B^\T P \prec 0 \:. \label{eq:stability_lmi} \end{align} $$ Then the closed loop system is stable.

Proof: Define $V(x) = x^\T P x$. Then, $$ \begin{align*} \dot{V}(x) &= \dot{x}^\T P x + x^\T P \dot{x} \\ &= x^\T (A^\T - K^\T {B}_1^\T) P x + x^\T P (A - {B}_1 K) x \\ &= x^\T (A^\T P + P A - K^\T B_1^\T P - P B_1 K) x \\ &= x^\T (A^\T P + P A - K^\T N^\T B^\T P - P B N K) x \\ &= x^\T (- Q + P B R^{-1} B^\T P - K^\T N^\T B^\T P - P B N K ) x \\ &= x^\T (- Q + P B (R^{-1} - R^{-1} N^\T - N R^{-1} ) B^\T P ) x \\ &< 0 \:. \end{align*} $$ Hence, $V$ is a valid Lyapunov function for the closed loop system, from which we conclude its stability. $\square$

Note that a sufficient condition for $\eqref{eq:stability_lmi}$ to hold is that $Q \succ 0$ and $$ N R^{-1} + R^{-1} N^\T - R^{-1} \succeq 0 \:. $$ For simplicity, let us suppose that both $N$ and $R$ are diagonal. Then this equation reduces to $$ n_i r_i^{-1} + r_i^{-1} n_i \geq r_i^{-1} \:, \:\: i = 1, ..., n \:. $$ or equivalently that $n_i \geq 1/2$ for all $i=1, ..., n$. Hence, in this case we see that we have an infinite gain margin, and can tolerate a 50% reduction in gain while remaining stable.

Concluding Remarks

It turns out there is a more general robustness theory around LQR. See for instance Safonov and Athans, where they show that a generalization of $\eqref{eq:stability_lmi}$ ensures stability. This more general result can be used to show that LQR also has a 60 degree phase margin.