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7.3.1 $ \mathcal L_2$ gain

We consider a linear time-invariant system

$\displaystyle \dot x=Ax+Bu,\qquad y=Cx$ (7.17)

and the cost functional

$\displaystyle J(u)=\int_{t_0}^\infty \Big(\gamma u^T(t)u(t)-\frac 1\gamma y^T(t)y(t)\Big)dt$ (7.18)

with $ \gamma>0$ . In this cost, the squared norms of the input and the output inside the integral are multiplied by scalar weights of the opposite signs (and we assume the two weights to have been normalized so that their product equals $ -1$ ). Consequently, it is clear that the optimal cost will now be nonpositive (just set $ u\equiv 0$ ). Nevertheless, as we will see, the form of the optimal solution is very similar to the one we saw in Section 6.2 and can be established using similar calculations. Suppose that there exists a matrix $ P$ with the following three properties:

  1. $ P=P^T\ge 0$ .
  2. $ P$ is a solution of the ARE

    $\displaystyle PA+A^TP+\frac1\gamma C^TC+\frac1\gamma PBB^TP=0.$ (7.19)

  3. The matrix $ A+\frac1\gamma BB^TP$ is Hurwitz.7.2

Then, we claim that the optimal cost is

$\displaystyle V(x_0)=-x_0^TPx_0$ (7.20)

and the optimal control is the linear state feedback

$\displaystyle u^*(t)=\frac1\gamma B^TPx^*(t).$ (7.21)

(Notice the ``wrong" signs in the formulas (7.19)-(7.21) compared to Section 6.2; this sign difference could be reconciled by working with $ -P$ instead of $ P$ here.)

To prove this claim, let us define the function $ \widehat V (x):=x^TP x$ . Its derivative along solutions of the system (7.17) is $ \frac d{dt}\widehat V (x(t))=x^T(t)(PA+A^TP)x(t)+2x^T(t)PBu(t)
$ , which is easily checked to be equivalent to

\begin{displaymath}\begin{split}\frac d{dt}\widehat V (x(t))&=x^T(t)\Big(PA+A^TP...
...t)\Big)+\gamma u^T(t)u(t)-\frac 1\gamma y^T(t)y(t). \end{split}\end{displaymath} (7.22)

We now introduce the auxiliary finite-horizon cost

$\displaystyle \bar J^{t_1}(u):=\int_{t_0}^{t_1}\Big(\gamma u^T(t)u(t)-\frac 1\gamma y^T(t)y(t)\Big)dt-x^T(t_1)Px(t_1).$ (7.23)

Using the formula (7.22) and noting that the first term on its right-hand side vanishes by (7.19), we can rewrite this cost as

$\displaystyle \bar J^{t_1}(u)$ $\displaystyle =\int_{t_0}^{t_1}\gamma\Big\vert u(t)-\frac1\gamma B^TPx(t)\Big\vert^2dt+\left.x^T(t)Px(t)\right\vert _{t_0}^{t_1}-x^T(t_1)Px(t_1)$    
  $\displaystyle = \int_{t_0}^{t_1}\gamma\Big\vert u(t)-\frac1\gamma B^TPx(t)\Big\vert^2dt-x^T_0Px_0$    

which makes it clear that (7.20) and (7.21) are the optimal cost and optimal control for the cost functional (7.23). We want to show that they are also optimal for the original cost functional (7.18). To this end, we first note that since $ P \ge 0$ , the following bound holds for all $ u$ :

$\displaystyle J( u)$ $\displaystyle =\lim_{t_1\to\infty}\int_{t_0}^{t_1} \Big(\gamma u^T(t)u(t)-\frac 1\gamma y^T(t)y(t)\Big)dt$    
  $\displaystyle \ge\lim_{t_1\to\infty}\left(\int_{t_0}^{t_1} \Big(\gamma u^T(t)u(t)-\frac 1\gamma y^T(t)y(t)\Big)dt-x^T(t_1)Px(t_1)\right)$    
  $\displaystyle = \lim_{t_1\to\infty}J^{t_1}(u)\ge \lim_{t_1\to\infty}(-x_0^TPx_0)=-x_0^TPx_0.$    

On the other hand, we can pass to the limit as $ t_1\to\infty$ in the already established relation

$\displaystyle \bar J^{t_1}(u^*)=\int_{t_0}^{t_1}\Big(\gamma (u^*)^T(t)u^*(t)-\frac 1\gamma (y^*)^T(t)y^*(t)\Big)dt-(x^*)^T(t_1)Px^*(t_1)=-x_0^TPx_0

where $ y^*:=Cx^*$ . In view of the fact that $ A+\frac1\gamma BB^TP$ is a Hurwitz matrix, the closed-loop system $ \dot x^*=Ax^*+Bu^*=(A+\frac1\gamma BB^TP)x^*$ is exponentially stable. We thus obtain $ J(u^*)=-x_0^TPx_0$ , and the desired result is proved.

While the formulas appearing here and in Section 6.2 are similar, the meanings of the two problems are very different. The cost (7.18) no longer reflects the objective of keeping both $ y$ and $ u$ small. Instead, this cost is small when $ y$ is large relative to $ u$ . We can regard $ u$ here not as a control that regulates the output but as a disturbance that tries to make the output large, with the optimal input being in some sense the worst-case disturbance. Let us try to formulate this idea in more precise terms. The fact that (7.20) is the optimal cost for the functional (7.18) implies that the inequality

$\displaystyle -x_0^TPx_0\le\int_{t_0}^\infty \Big(\gamma \vert u(t)\vert^2-\frac 1\gamma \vert y(t)\vert^2\Big)dt$ (7.24)

holds for all $ u$ , with the equality achieved by the optimal control $ u^*$ . From now on we focus on the case when $ x_0=0$ . Specializing (7.24) to this case, after simple manipulations we reach

$\displaystyle \sup_{u\in\mathcal L_2\setminus\{0\}}\frac{\sqrt{\int_{t_0}^\infty\vert y(t)\vert^2dt}}{\sqrt{\int_{t_0}^\infty \vert u(t)\vert^2dt}} \le\gamma.$ (7.25)

The fraction on the left-hand side of (7.25) is the ratio of the $ \mathcal L_2$ norms7.3 of the input and the output, and the supremum is being taken over all nonzero inputs with finite $ \mathcal L_2$ norms. If we view the system (7.17), with the zero initial condition, as an input/output operator from $ \mathcal L_2$ to $ \mathcal L_2$ , then (7.25) says that the induced norm of this operator does not exceed $ \gamma$ . This induced norm is called the $ \mathcal L_2$ gain of the system.

We see that if, for a given value of $ \gamma$ , we can find a matrix $ P$ with the three properties listed at the beginning of this subsection, then the system's $ \mathcal L_2$ gain is less than or equal to $ \gamma$ . (We do not know whether $ \gamma$ is actually achieved by some control; note that the optimal control (7.21) is excluded in (7.25) because it is identically 0 when $ x_0=0$ .) A converse result also holds: if the $ \mathcal L_2$ gain is less than $ \gamma$ , then a matrix $ P$ with the indicated properties exists. If we sidestep the original optimal control problem and only seek sufficient conditions for the $ \mathcal L_2$ gain to be less than or equal to $ \gamma$ , then it is not hard to see from our earlier derivation that the conditions on the matrix $ P$ can be relaxed. Namely, it is enough to look for a symmetric positive semidefinite solution of the algebraic Riccati inequality

$\displaystyle PA+A^TP+\frac1\gamma C^TC+\frac1\gamma PBB^TP\le 0.$ (7.26)

The formula (7.22) then yields

$\displaystyle \frac d{dt}\widehat V (x(t))\le\gamma \vert u(t)\vert^2-\frac 1\gamma \vert y(t)\vert^2.

Integrating both sides from $ t_0$ to an arbitrary time $ T$ , rearranging terms, and using the definition of $ \widehat V$ and the fact that $ P \ge 0$ , we have

$\displaystyle \frac1\gamma\int_{t_0}^T \vert y(t)\vert^2dt\le\gamma\int_{t_0}^T...
...rt^2dt+\widehat V(0)-\widehat V(x(t))\le\gamma\int_{t_0}^T
\vert u(t)\vert^2dt

and in the limit as $ T\to\infty$ we again arrive at (7.25).

In the frequency domain, the system (7.17) is characterized by the transfer matrix $ G(s)=C(Is-A)^{-1}B$ . Using Parseval's theorem, it can be shown that the $ \mathcal L_2$ gain equals the largest singular value of $ G(j\omega)$ supremized over all frequencies $ \omega\in\mathbb{R}$ ; for systems with scalar inputs and outputs, this is just $ \sup_{\omega\in\mathbb{R}} \vert g(j\omega)\vert$ where $ g$ is the transfer function. In view of this fact, the $ \mathcal L_2$ gain is also called the $ \mathcal H_\infty $ norm.

next up previous contents index
Next: 7.3.2 control problem Up: 7.3 Riccati equations and Previous: 7.3 Riccati equations and   Contents   Index
Daniel 2010-12-20