]> 6.4.1 Example 1

#### 6.4.1 Example 1

For a time–delay system governed by the equations

 $\left\{\begin{array}{ccccc}\hfill {ż}_{1}\left(t\right)& \hfill =\hfill & -{z}_{1}\left(t\right)+{z}_{2}\left(t-1\right),\hfill & \hfill t\ge 0& \hfill \\ \hfill {ż}_{2}\left(t\right)& \hfill =\hfill & -{z}_{2}\left(t\right)+u\left(t\right),\hfill & \hfill t\ge 0& \hfill \\ \hfill {z}_{1}\left(0\right)& \hfill =\hfill & {z}_{1}^{0}\hfill & \hfill & \hfill \\ \hfill {z}_{2}\left(0\right)& \hfill =\hfill & {z}_{2}^{0}\hfill & \hfill & \hfill \\ \hfill {z}_{2}\left(\theta \right)& \hfill =\hfill & {\phi }_{2}\left(\theta \right),\hfill & \hfill -1\le \theta \le 0,& {\phi }_{2}\in {\text{L}}^{2}\left(-1,0\right)\hfill \end{array}\right\}$ (6.31)

we wish to construct a stabilizing controller, minimizing the performance index

 $J={\int }_{0}^{\infty }\left[{z}_{1}^{2}\left(t\right)+{u}^{2}\left(t\right)\right]dt$ (6.32)

The investigated system represents a class of degenerated time–delay systems, which are reducible to a ﬁnite–dimensional system without retardation. This reduction can be achieved by introducing the new state variables

 ${\xi }_{1}\left(t\right)={z}_{1}\left(t+1\right),\phantom{\rule{2em}{0ex}}{\xi }_{2}\left(t\right)={z}_{2}\left(t\right)$ (6.33)

As a result the equations (6.31) take the form

 $\left\{\begin{array}{cccc}\hfill {\stackrel{̇}{\xi }}_{1}\left(t\right)& \hfill =\hfill & -{\xi }_{1}\left(t\right)+{\xi }_{2}\left(t\right),\hfill & \phantom{\rule{1em}{0ex}}t\ge 0\hfill \\ \hfill {\stackrel{̇}{\xi }}_{2}\left(t\right)& \hfill =\hfill & -{\xi }_{2}\left(t\right)+u\left(t\right),\hfill & \phantom{\rule{1em}{0ex}}t\ge 0\hfill \\ \hfill {\xi }_{1}\left(0\right)& \hfill =\hfill & {\xi }_{1}^{0}={z}_{1}\left(1\right)\hfill & \hfill \\ \hfill {\xi }_{2}\left(0\right)& \hfill =\hfill & {\xi }_{2}^{0}={z}_{2}^{0}\hfill & \hfill \end{array}\right\}$ (6.34)

and the performance index

 $J={\int }_{0}^{1}{z}_{1}^{2}\left(t\right)dt+{\int }_{0}^{\infty }\left[{\xi }_{1}^{2}\left(t\right)+{u}^{2}\left(t\right)\right]dt$ (6.35)

provided that the restriction of ${z}_{1}$ to an interval $\left[0,1\right]$ is the solution of the nonhomogeneous differential equation

 $\left\{\begin{array}{ccc}\hfill {ż}_{1}\left(t\right)& \hfill =\hfill & -{z}_{1}\left(t\right)+{\phi }_{2}\left(t-1\right),\phantom{\rule{1em}{0ex}}0\le t\le 1\hfill \\ \hfill {z}_{1}\left(0\right)& \hfill =\hfill & {z}_{1}^{0}\hfill \end{array}\right\}$ (6.36)

The solution of (6.36) is

${z}_{1}\left(t\right)={e}^{-t}{z}_{1}^{0}+{\int }_{0}^{t}{e}^{-\left(t-\tau \right)}{\phi }_{2}\left(\tau -1\right)d\tau ,\phantom{\rule{2em}{0ex}}t\in \left[0,1\right]$

and thus

 ${\xi }_{1}^{0}={z}_{1}\left(1\right)={e}^{-1}{z}_{1}^{0}+{\int }_{-1}^{0}{e}^{\theta }{\phi }_{2}\left(\theta \right)d\theta$ (6.37)

The ﬁrst integral in (6.35) does not depend on control $u$ and thus the synthesis of a stabilizing controller reduces to the analysis of the system (6.1) in a Hilbert space $\text{H}={\mathbb{ℝ}}^{2}$ with

$\mathcal{𝒜}x=Ax,\phantom{\rule{2em}{0ex}}A=\left[\begin{array}{cc}\hfill -1& \hfill 1\\ \hfill 0& \hfill -1\end{array}\right],\phantom{\rule{2em}{0ex}}b=\left[\begin{array}{c}\hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}c=\left[\begin{array}{c}\hfill \phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \end{array}\right].$

The pair $\left(A,b\right)$ is controllable, hence stabilizable. $\left(A,c\right)$ is observable, hence detectable. The construction of stabilizing controller problem is thus well–posed.

An approximate solution can be found with the aid of an iterative process (6.9), which in ${\mathbb{ℝ}}^{n}$ agrees with the Newton–Kleinman method for solving the matrix Riccati equation (6.15) – see . Applying the Matlab/Control Toolbox one obtains

$\mathcal{ℋ}=\left[\begin{array}{cc}\hfill 0.476489\dots \hfill & \hfill 0.216845\dots \hfill \\ \hfill 0.216845\dots \hfill & \hfill 0.197368\dots \hfill \end{array}\right],\phantom{\rule{1em}{0ex}}g=\mathcal{ℋ}b=\left[\begin{array}{c}\hfill 0.216845\dots \hfill \\ \hfill 0.197368\dots \hfill \end{array}\right]$

and a characteristic polynomial of the closed–loop optimal system

$det\left(sI-A-b{g}^{\ast }\right)={s}^{2}+\left(2.197368\dots \right)s+\left(1.414213\dots \right)$

having the roots $\left(-1.098684\dots \right)±j\left(0.455089\dots \right)$.

The exact solution will be found by applying the Callier–Winkin lemma. Since $Re\sigma \left(A\right)<0$ we can take $f=0$ in (6.16). Now $A=Ã$ ($\stackrel{̃}{\mathcal{𝒜}}x=Ãx$ with $Ã\in L\left({\mathbb{ℝ}}^{2}\right)$) and we have

$\pi \left(\omega \right)=1+{\left|{c}^{\ast }{\left(j\omega I-A\right)}^{-1}b\right|}^{2}=\frac{1}{{\left(1+{\omega }^{2}\right)}^{2}}+1=\frac{{\omega }^{4}+2{\omega }^{2}+2}{{\left(1+{\omega }^{2}\right)}^{2}}\ge 1\phantom{\rule{2em}{0ex}}\forall \omega \in \mathbb{ℝ}.$

Consequently, (6.24) is a polynomial spectral factorization problem. The factor satisfying (6.25) has the form

 $\phi \left(s\right)=\frac{{s}^{2}+\sqrt{2+2\sqrt{2}}s+\sqrt{2}}{{\left(1+s\right)}^{2}}$ (6.38)

The optimal stabilizing controller can be determined from (6.23)

${g}_{\infty }=\stackrel{̃}{g}=\left[\begin{array}{c}\hfill {g}_{1}\hfill \\ \hfill {g}_{2}\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}{g}_{1}=1+\sqrt{2}-\sqrt{2+2\sqrt{2}},\phantom{\rule{2em}{0ex}}{g}_{2}=-2+\sqrt{2+2\sqrt{2}}$

which agrees with numerical calculations. The optimal controller produces the steering

 $u\left(t\right)=-{g}_{1}{\xi }_{1}\left(t\right)-{g}_{2}{\xi }_{2}\left(t\right)=-{g}_{1}{z}_{1}\left(t+1\right)-{g}_{2}{z}_{2}\left(t\right)$ (6.39)

The diagram explaining a synthesis of the optimal control law (6.39) is depicted in Figure 6.2. The initial conditions resolver recovers ${\xi }_{1}^{0}$ from ${z}_{1}^{0}$, ${\phi }_{2}$ using (6.37). Figure 6.2: Synthesis of the optimal control law

Interchanging $t$ by $t+\theta +1$, $-1\le \theta \le 0$ in the ﬁrst equation of (6.31) which can be done since $t+\theta +1\ge 0$ and this equation holds for $t\ge 0$, yields

${ż}_{1}\left(t+\theta +1\right)=-{z}_{1}\left(t+\theta +1\right)+{z}_{2}\left(t+\theta \right).$

Multiplying both sides by ${e}^{\theta }$ and integrating the right–hand side with respect to $\theta$ we get

${z}_{1}\left(t+1\right)=\frac{1}{e}{z}_{1}\left(t\right)+{\int }_{-1}^{0}{e}^{\theta }{z}_{2}\left(t+\theta \right)d\theta .$

An equivalent representation of control can be derived by taking the above result into account in (6.39)

 $u\left(t\right)=-\frac{1}{e}{g}_{1}{z}_{1}\left(t\right)-{g}_{2}{z}_{2}\left(t\right)-{g}_{1}{\int }_{-1}^{0}{e}^{\theta }{z}_{2}\left(t+\theta \right)d\theta$ (6.40)

The representation (6.40) removes any doubts concerning the realizability of the control $u$ expressed in a form (6.39) where the current values of $u$ are determined by the future values of ${z}_{1}$.

In a Hilbert space ${\mathbb{𝕄}}^{2}={\mathbb{ℝ}}^{2}\oplus {\text{L}}^{2}\left(-1,0;{\mathbb{ℝ}}^{2}\right)$, (6.31) can be written as an abstract equation (6.1) with

$x\left(t\right)=\left[\begin{array}{c}{z}_{1}\left(t\right)\hfill \\ {z}_{2}\left(t\right)\hfill \\ {\psi }_{1}\left(t\right)\hfill \\ {\psi }_{2}\left(t\right)\hfill \end{array}\right]\in {\mathbb{𝕄}}^{2},\phantom{\rule{2em}{0ex}}b=\left[\begin{array}{c}0\hfill \\ 1\hfill \\ 0\hfill \\ 0\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}c=\left[\begin{array}{c}1\hfill \\ 0\hfill \\ 0\hfill \\ 0\hfill \end{array}\right],$

 $\begin{array}{c}\mathcal{𝒜}\left[\begin{array}{c}\hfill {z}_{1}\hfill \\ \hfill {z}_{2}\hfill \\ \hfill {\psi }_{1}\hfill \\ \hfill {\psi }_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -{z}_{1}+{\psi }_{2}\left(-1\right)\hfill \\ \hfill -{z}_{2}\hfill \\ \hfill {\psi }_{1}^{\prime }\hfill \\ \hfill {\psi }_{2}^{\prime }\hfill \end{array}\right],\hfill \\ D\left(\mathcal{𝒜}\right)=\left\{\left[\begin{array}{c}{z}_{1}\hfill \\ {z}_{2}\hfill \\ {\psi }_{1}\hfill \\ {\psi }_{2}\hfill \end{array}\right]\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}{\psi }_{1},{\psi }_{2}\in {\text{W}}^{1,2}\left(-1,0\right),\left[\begin{array}{c}\hfill {\psi }_{1}\left(0\right)\hfill \\ \hfill {\psi }_{2}\left(0\right)\hfill \end{array}\right]=\left[\begin{array}{c}\hfill {z}_{1}\hfill \\ \hfill {z}_{2}\hfill \end{array}\right]\right\}\hfill \end{array}$ (6.41)

The performance index is then described in a form appearing in (6.2). It is well–known, see [30, p. 139], that a necessary and sufficient condition for the semigroup generated on ${\mathbb{𝕄}}^{2}={\mathbb{ℝ}}^{n}\oplus {\text{L}}^{2}\left(-r,0;{\mathbb{ℝ}}^{n}\right)$ by

 $\begin{array}{c}\mathcal{𝒜}\left[\begin{array}{c}z\hfill \\ \psi \hfill \end{array}\right]=\left[\begin{array}{c}\hfill Az+B\psi \left(-r\right)\hfill \\ \hfill {\psi }^{\prime }\hfill \end{array}\right]\hfill \\ D\left(\mathcal{𝒜}\right)=\left\{x=\left[\begin{array}{c}\hfill z\hfill \\ \hfill \psi \hfill \end{array}\right]\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}\psi \in {\text{W}}^{1,2}\left(-r,0;{\mathbb{ℝ}}^{n}\right),\phantom{\rule{0ex}{0ex}}\psi \left(0\right)=z\right\}\hfill \end{array}$ (6.42)

to be exponentially stable EXS is that all zeros of the characteristic quasipolynomial $s↦det\left[sI-A-{e}^{-sr}B\right]$ which is an entire function should have negative real parts. In our example,

$n=2,\phantom{\rule{1em}{0ex}}r=1,\phantom{\rule{1em}{0ex}}A=\left[\begin{array}{cc}\hfill -1& \hfill 0\\ \hfill 0& \hfill -1\end{array}\right],\phantom{\rule{1em}{0ex}}B=\left[\begin{array}{cc}\hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\end{array}\right]\phantom{\rule{0ex}{0ex}}$

and thus $det\left[sI-A-{e}^{-sr}B\right]={\left(s+1\right)}^{2}$. Hence the semigroup generated by $\mathcal{𝒜}$, deﬁned in (6.41) is EXS. The pair $\left(\mathcal{𝒜},b\right)$ is clearly stabilizable and $\left(\mathcal{𝒜},c\right)$ detectable. One may apply also the general results in . The problem of optimal stabilizing controller synthesis has a solution. To ﬁnd it, we put $f=0\in \text{H}$ in (6.16) getting $\mathcal{𝒜}=\stackrel{̃}{\mathcal{𝒜}}$. From the equation $\left(sI-\mathcal{𝒜}\right)x=b$, $s\ne -1$ one obtains

${\left(sI-\mathcal{𝒜}\right)}^{-1}b=\frac{1}{{\left(s+1\right)}^{2}}\left[\begin{array}{c}\hfill {e}^{-s}\hfill \\ \hfill s+1\hfill \\ \hfill {e}^{s\left(\theta -1\right)}\hfill \\ \hfill \left(s+1\right){e}^{s\theta }\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}{c}^{\ast }{\left(sI-\mathcal{𝒜}\right)}^{-1}b=\frac{{e}^{-s}}{{\left(s+1\right)}^{2}}.$

Now, the solution of the factorization problem (6.24) is again expressed in the form (6.38). The identity (6.23) uniquely represents an optimal controller, provided that $\left(\mathcal{𝒜},b\right)$ is approximately controllable. We have

 $rk\left[sI-A-{e}^{-sr}B,b\right]=n\phantom{\rule{2em}{0ex}}\forall s\in \mathbb{ℂ}$ (6.43)
 $rk\left[B,b\right]=n$ (6.44)

and approximate controllability easily follows from this criterion due to Triggiani and Manitius – see [75, p. 133].

In virtue of the Riesz representation theorem we may seek the optimal controller in a vector form

 ${g}_{\infty }=\stackrel{̃}{g}=\left[\begin{array}{c}{\alpha }_{1}\hfill \\ {\alpha }_{2}\hfill \\ {\beta }_{1}\left(\cdot \right)\hfill \\ {\beta }_{2}\left(\cdot \right)\hfill \end{array}\right]\in \text{H}$ (6.45)

Hence, the identity (6.23) takes the form

$\begin{array}{cc}\hfill {\stackrel{̃}{g}}^{\ast }{\left(sI-\mathcal{𝒜}\right)}^{-1}b=& \frac{1}{{\left(s+1\right)}^{2}}〈\left[\begin{array}{c}\hfill {e}^{-s}\hfill \\ \hfill s+1\hfill \\ \hfill {e}^{s\left(\theta -1\right)}\hfill \\ \hfill \left(s+1\right){e}^{s\theta }\hfill \end{array}\right],\left[\begin{array}{c}\hfill {\alpha }_{1}\hfill \\ \hfill {\alpha }_{2}\hfill \\ \hfill {\beta }_{1}\hfill \\ \hfill {\beta }_{2}\hfill \end{array}\right]〉=\phi \left(s\right)-1=\hfill \\ \hfill =& \frac{\left(\sqrt{2+2\sqrt{2}}-2\right)s+\left(\sqrt{2}-1\right)}{{\left(1+s\right)}^{2}},\phantom{\rule{2em}{0ex}}s\ne -1.\hfill \end{array}$

In fact this is an identity for entire functions,

 $\begin{array}{c}\hfill {\alpha }_{1}{e}^{-s}+{\alpha }_{2}\left(s+1\right)+{\int }_{-1}^{0}{\beta }_{1}\left(\theta \right){e}^{s\left(\theta -1\right)}d\theta +\left(s+1\right){\int }_{-1}^{0}{\beta }_{2}\left(\theta \right){e}^{s\theta }d\theta =\hfill \\ \hfill =\left(\sqrt{2+2\sqrt{2}}-2\right)s+\left(\sqrt{2}-1\right)\phantom{\rule{2em}{0ex}}\forall s\in \mathbb{ℂ}\hfill \end{array}$ (6.46)

The ﬁrst integral in (6.46) is an entire function of $s$ having the growth exponent distinct from the exponents of other terms (accordingly to the Paley–Wiener Theorem its support is located outside the interval $\left[-1,0\right]$). Thus, we should assume ${\beta }_{1}=0\in {\text{L}}^{2}\left(-1,0\right)$. Another justiﬁcation is also possible. Namely, by (6.21) and the Paley–Wiener Theorem we have

$\text{R}\left({\mathcal{𝒜}}^{\ast },c\right)=\text{NO}{\left({c}^{\ast },\mathcal{𝒜}\right)}^{\perp }={\left\{x\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}〈{\left(sI-\mathcal{𝒜}\right)}^{-1}x,c〉=0\phantom{\rule{2em}{0ex}}\forall s\in {\rho }_{0}\left(\mathcal{𝒜}\right)\right\}}^{\perp }=$

$=\left\{x=\left[\begin{array}{c}\hfill z\hfill \\ \hfill \psi \hfill \end{array}\right]\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}\left(s+1\right){z}_{1}+{e}^{-s}{z}_{2}+\right\$

${+\left(s+1\right){\int }_{-1}^{0}{e}^{-s\left(\theta +1\right)}{\psi }_{2}\left(\theta \right)d\theta =0\phantom{\rule{2em}{0ex}}\forall s\in \mathbb{ℂ}}}^{\perp }=\left\{x\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}x=\left[\begin{array}{c}{z}_{1}\hfill \\ {z}_{2}\hfill \\ 0\hfill \\ {\psi }_{2}\hfill \end{array}\right]\right\}.$

The restriction of the system to its invariant subspace $R\left({\mathcal{𝒜}}^{\ast },c\right)$ arises by zeroing the ﬁrst function component. As ${g}_{\infty }\in R\left({\mathcal{𝒜}}^{\ast },c\right)$, this component will not participate in the optimal control law.

Now, assume for a while that ${\beta }_{2}\in {\text{W}}^{1,2}\left(-1,0\right)$. Integrating–by–parts and comparing coefficients of the terms with the same growth exponents in (6.46), we obtain

 ${\alpha }_{2}=\sqrt{2+2\sqrt{2}}-2$ (6.47)

and the two–point boundary value problem

 $\left\{\begin{array}{ccc}\hfill {\beta }_{2}^{\prime }\left(\theta \right)& \hfill =\hfill & {\beta }_{2}\left(\theta \right)\hfill \\ \hfill {\beta }_{2}\left(-1\right)& \hfill =\hfill & {\alpha }_{1}\hfill \\ \hfill {\beta }_{2}\left(0\right)& \hfill =\hfill & \sqrt{2}+1-\sqrt{2+2\sqrt{2}}\hfill \end{array}\right\}$ (6.48)

Problem (6.48) has a unique solution

 ${\alpha }_{1}=\frac{\sqrt{2}+1-\sqrt{2+2\sqrt{2}}}{e},\phantom{\rule{1em}{0ex}}{\beta }_{2}\left(\theta \right)=\left(\sqrt{2}+1-\sqrt{2+2\sqrt{2}}\right){e}^{\theta },\phantom{\rule{0ex}{0ex}}\theta \in \left[-1,0\right]$ (6.49)

Taking (6.47), (6.49) and (6.45) into account in the formula for the optimal control $u\left(t\right)=-{g}_{\infty }^{\ast }x\left(t\right)$ we get again (6.40). The uniqueness of the optimal controller justiﬁes the fact that ${\beta }_{2}$ was assumed to be absolutely continuous.