Adaptive Control for Weakly Minimum Phase Linear Infinite-Dimensional Systems in Hilbert Space Using a Zero Filter

Abstract

Given a linear continuous-time infinite-dimensional plant on a Hilbert space and disturbances of known waveform but unknown amplitude and phase, we show that there exists a stabilizing direct model reference adaptive control law with persistent disturbance rejection and robustness properties. The plant is described by a closed, densely defined linear operator that generates a continuous semigroup of bounded operators on the Hilbert space of states. For this paper, the plant will be weakly minimum phase, i.e., there will be a finite number of unstable zeros with real part equal to zero. All other zeros will be exponentially stable.

The central result will show that all errors will converge to a prescribed neighborhood of zero in an infinite-dimensional Hilbert space even though the plant is not truly minimum phase. The result will not require the use of the standard Barbalat’s lemma which requires certain signals to be uniformly continuous. This result is used to determine conditions under which a linear infinite-dimensional system can be directly adaptively controlled to follow a reference model. In particular we examine conditions for a set of ideal trajectories to exist for the tracking problem. Our principal result will be that the direct adaptive controller can be compensated with a zero filter for the unstable zeros which will produce the desired robust adaptive control results even though the plant is only weakly minimum phase. Our results are applied to adaptive control of general linear infinite-dimensional systems described by self-adjoint operators with compact resolvent.

Appendices

Appendix 1: Proofs of Lemmae 1, 2, and Theorem 1

Proof of Lemma 1

Consider $\begin{array}{c}{P}_1^2=\left(B{(CB)}^{-1}C\right)\left(B{(CB)}^{-1}C\right)\\ {}=B{(CB)}^{-1}C\equiv {P}_1\end{array}$ _.

Hence P ₁ is a projection.

Clearly, $R\left({P}_1\right)\subseteq R(B)$ and $z=Bu\in R(B)$ which implies $\begin{array}{c}{P}_1z\kern0.5pt\,{=}\kern-0.5pt\left(\!B{(CB)}^{-1}C\!\right)\!Bu\\ {}=Bu=z\in R\left({P}_1\right)\end{array}$.

Therefore $R\left({P}_1\right)=R(B)$.

Also $N\left({P}_1\right)=N(C)$ because $N(C)\subseteq N\left({P}_1\right)$ and $z\in N\left({P}_1\right)$ implies that ${P}_1z\,{=}\,0$ which implies that $C{P}_1z=CB{(CB)}^{-1}Cz=0$ or $N\left({P}_1\right)\subseteq N(C)$.

So P ₂ is a projection onto R(B) along N(C) but ${P}_2^{*}\ne {P}_2$ so it is not an orthogonal projection in general. We have $X=R\left({P}_1\right)\oplus N\left({P}_1\right)$; hence $X=R(B)\oplus N(C).$

Since ${b}_i\in D(A)$, we have $R(B)\subset D(A)$.

Consequently $D(A)=\left(R(B)\cap D(A)\right)\oplus \left(N(C)\cap D(A)\right)=R(B)\oplus \left(N(C)\cap D(A)\right)$.

The projection P ₁ is bounded since its range is finite-dimensional, and the projection P ₂ is bounded because $\left\Vert {P}_2\right\Vert \le 1+\left\Vert {P}_1\right\Vert <\infty .$

This completes the proof of Lemma 1.

Proof of Lemma 2

Since X is separable, we can let $N(C)=\overline{sp}{\left\{{\theta}_k\right\}}_{k=1}^{\infty }$ be an orthonormal basis.

Define ${W}_2:X\to {l}_2$ by ${W}_2x\equiv \left[\begin{array}{c} \left({\theta}_1,{P}_2x\right) \\ {} \left({\theta}_2,{P}_2x\right) \\ {} \left({\theta}_3,{P}_2x\right) \\ {} \dots \\ {} \end{array}\right]$.

Note that ${\left\Vert {W}_2x\right\Vert}^2={\displaystyle \sum_{k=1}^{\infty }{\left|\left({\theta}_k,{P}_2x\right)\right|}^2}={\left\Vert {P}_2x\right\Vert}^2<\infty$ which implies ${W}_2x\in {l}_2$.

So W ₂ is a bounded linear operator, and an isometry of W ₂ N(C) into l ₂.

Consequently ${W}_2{W}_2^{*}=I$ on N(C).

Then we have ${W}_2^{*}{W}_2={P}_2$ and the retraction: ${z}_2={W}_2{P}_2x\in {l}_2$.

Also ${W}_2^{*}{z}_2={W}_2^{*}\left({W}_2{P}_2x\right)={P}_2x$.

Now, using $x={P}_1x+{P}_2x$ from Lemma 1, we have

$$ \begin{aligned}\overset{.}{y}&=C{P}_1\overset{.}{x}\\ &=C{P}_1A\left({P}_1x+{P}_2x\right)+C{P}_1Bu\\ &=C\left(B{(CB)}^{-1}C\right) AB{(CB)}^{-1}y+C\left(B{(CB)}^{-1}C\right)A\left({W}_2^{*}{z}_2\right)+C\left(B{(CB)}^{-1}C\right)Bu\\ &={\overline{A}}_{11}y+{\overline{A}}_{12}{z}_2+CBu\end{aligned} $$

and

$$ \begin{aligned}{\overset{.}{z}}_2&={W}_2{P}_2\overset{.}{x}\\ &=W{P}_2\left[A\left({P}_1x+{P}_2x\right)+Bu\right]\\ &={W}_2{P}_2A\left(B{(CB)}^{-1}y+{W}_2^8{z}_2\right)+{W}_2{P}_2Bu\\ &={W}_2\left(I-B{(CB)}^{-1}B\right) AB{(CB)}^{-1}y+{W}_2\left(I-B{(CB)}^{-1}B\right)A{W}_2^{*}{z}_2\\ &={\overline{A}}_{21}y+{\overline{A}}_{22}{z}_2.\vspace*{-3pt}\end{aligned} $$

This yields the normal form (11).

Choose $W\equiv \left[\begin{array}{c} C \\ {} {W}_2{P}_2 \end{array}\right]$ which is a bounded linear operator. Then W has a bounded inverse explicitly stated as ${W}^{-1}\equiv \left[\begin{array}{cc} B{(CB)}^{-1} & {W}_2^{*} \end{array}\right]$.

This gives

$$ \begin{aligned}W{W}^{-1}&=\left[\begin{array}{cc} CB{(CB)}^{-1} & C{W}_2^{*} \\ {} {W}_2{P}_2B{(CB)}^{-1} & {W}_2{P}_2{W}_2^{*} \end{array}\right]\\ {}&=\left[\begin{array}{cc} {I}_m & 0 \\ {} 0 & {W}_2{W}_2^{*} \end{array}\right]=\left[\begin{array}{cc} {I}_m & 0 \\ {} 0 & I \end{array}\right]=I\end{aligned}\vspace*{-5pt} $$

because $R\left({W}_2^{*}\right)\subseteq N(C)$.

Furthermore, ${W}^{-1}W={P}_1+{W}_2^{*}{W}_2{P}_2={P}_1+{P}_2=I$ because ${W}_2{W}_2^{*}=I$ on N(C).

Also direct calculation yields

$$ \left\{\begin{array}{l}\overline{B}\equiv WB=\left[\begin{array}{c} CB \\ {} {W}_2{P}_2B \end{array}\right]=\left[\begin{array}{c} CB \\ {} 0 \end{array}\right]\\ {}\overline{C}\equiv C{W}^{-1}=\left[\begin{array}{cc} CB{(CB)}^{-1} & C{W}_2^{*} \end{array}\right]=\left[\begin{array}{cc} {I}_m & 0 \end{array}\right]\\ {}\ \overline{A}\equiv WA{W}^{-1}=\left[\begin{array}{cc} CAB{(CB)}^{-1} & CA{W}_2^{*} \\ {} {W}_2{P}_2 AB{(CB)}^{-1} & {W}_2{P}_2A{P}_2{W}_2^{*} \end{array}\right]\end{array}\right. $$

This completes the proof of Lemma 2.

Proof of Theorem 1

Define ${\overline{S}}_1\equiv {W}^{-1}{S}_1=\left[\begin{array}{c} {\overline{S}}_a \\ {} {\overline{S}}_b \end{array}\right]\ \mathrm{and}\ {\overline{H}}_1\equiv W{H}_1=\left[\begin{array}{c} {\overline{H}}_a \\ {} {\overline{H}}_b \end{array}\right]$.

From (10), we obtain

$$ \left\{\begin{array}{l}\overline{A}{\overline{S}}_1+\overline{B}{S}_2={\overline{S}}_1{L}_m+{\overline{H}}_1\\ {}\overline{C}{\overline{S}}_1={H}_2\end{array}\right. $$

where $\left(\overline{A},\overline{B},\overline{C}\right)$ is the normal form (11).

From this we obtain

$\left\{\begin{array}{l}{\overline{S}}_a={H}_2\\ {}{S}_2={(CB)}^{-1}\left[{H}_2{L}_m+{\overline{H}}_a-\left({\overline{A}}_{11}{H}_2+{\overline{A}}_{12}{\overline{S}}_b\right)\right]\\ {}{\overline{A}}_{22}{\overline{S}}_b-{\overline{S}}_b{L}_m={\overline{H}}_b-{\overline{A}}_{21}{H}_2\end{array}\right.$.

We can rewrite the last of these equations as

$\left(\lambda I-{\overline{A}}_{22}\right){\overline{S}}_b-{\overline{S}}_b\left(\lambda I-{L}_m\right)={\overline{A}}_{21}{H}_2-{\overline{H}}_b\equiv \overline{H}$ for all complex λ.

Now assume that L _m is simple and therefore provides a basis of eigenvectors ${\left\{{\emptyset}_k\right\}}_{k=1}^L\ \mathrm{f}\mathrm{o}\mathrm{r}\ {\Re}^L$. This is not essential but will make this part of the proof easier to understand. The proof can be done with generalized eigenvectors and the Jordan form. So we have

$$ \left({\lambda}_kI-{\overline{A}}_{22}\right){\overline{S}}_b{\phi}_k-{\overline{S}}_b\underset{=0}{\underbrace{\left({\lambda}_kI-{L}_m\right){\phi}_k}}={\overline{A}}_{21}{H}_2-{\overline{H}}_b\equiv \overline{H} $$

which implies that

$$ {\overline{S}}_b{\phi}_k={\left({\lambda}_kI-{\overline{A}}_{22}\right)}^{-1}\overline{H}{\phi}_k $$

because ${\lambda}_{\mathrm{k}}\in \sigma \left({L}_m\right)\subset \rho \left({\overline{A}}_{22}\right)$.

Thus we have ${\overline{S}}_bz={\displaystyle \sum_{k=1}^L{\alpha}_k{\left({\lambda}_kI-{\overline{A}}_{22}\right)}^{-1}\overline{H}{\phi}_k}\forall z={\displaystyle \sum_{k=1}^L{\alpha}_k{\phi}_k\in {\Re}^L}$.

Since ${\lambda}_{\mathrm{k}}\in \sigma \left({L}_m\right)\subset \rho \left({\overline{A}}_{22}\right)$, all ${\left({\lambda}_kI-{\overline{A}}_{22}\right)}^{-1}$ are bounded operators.

Also $\overline{H}\equiv {\overline{A}}_{21}{H}_2-{\overline{H}}_b$ is a bounded operator on ℜ ^L.

Therefore ${\overline{S}}_b$ is a bounded linear operator, and this leads to S ₁ also bounded linear.

If we look at the converse statement and let ${\lambda}_{*}\in \sigma \left({L}_m\right)\cap \sigma \left({\overline{A}}_{22}\right)=\emptyset$.

Then there exists ${\phi}_{*}\ne 0$ such that $\left({\lambda}_{*}I-{\overline{A}}_{22}\right){\overline{S}}_b{\phi}_{*}-{\overline{S}}_b\underset{=0}{\underbrace{\left({\lambda}_{*}I-{L}_m\right){\phi}_{*}}}=\left({\lambda}_{*}I-{\overline{A}}_{22}\right){\overline{S}}_b{\phi}_{*}=\overline{H}$.

In this case three things can happen when ${\lambda}_{*}\in \sigma \left({\overline{A}}_{22}\right)$:

• (1) $\left({\lambda}_{*}I-{\overline{A}}_{22}\right)$ can fail to be one to one so multiple solutions of ${\overline{S}}_b$ will exist

• (2) $R\left({\lambda}_{*}I-{\overline{A}}_{22}\right)$ can fail to be all of X so no solutions ${\overline{S}}_b$ may occur, or

• (3) ${\left({\lambda}_{*}I-{\overline{A}}_{22}\right)}^{-1}$ can fail to be a bounded operator so solutions ${\overline{S}}_b$ may be unbounded.

In all cases these three alternatives lead to a lack of unique bounded operator solutions for S ₁.

The proof of Theorem 1 is complete.

Appendix 2: Proof of Theorem 2

From (15) and Pazy Cor 2.5 p. 107 [1], we have a well-posed system in (16) where A _c is a closed operator, densely defined on $D\left({A}_C\right)\subseteq X$ and generates a C ₀ semigroup on X, and all trajectories starting in D(A _C ) will remain there. Hence we can differentiate signals in D(A _C ).

Consider the positive definite function,

$$ V\equiv \frac{1}{2}\left(Pe,e\right)+\frac{1}{2}\mathrm{t}\mathrm{r}\left[\varDelta G{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right] $$

(21)

where $\varDelta G(t)\equiv G(t)-{G}^{\ast }$ and P satisfies (13).

Taking the time derivative of (21) (this can be done $\forall e\in D\left({A}_C\right)$) and substituting (2a) into the result yields $\overset{.}{V}=\frac{1}{2}\left[\left(P{A}_c\;e,e\right)+\left(e,P{A}_ce\right)\right]+\left(PBw,e\right)+\mathrm{t}\mathrm{r}\left[\varDelta \overset{.}{G}{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]+\left(Pe,v\right);w\equiv \varDelta Gz$.

Invoking the equalities in Definition 2 of strict dissipativity, using x ^T y = tr[yx ^T], and substituting (16) into the last expression, we get (with $\left\langle {e}_y,w\right\rangle \equiv {e}_y^{*}w$),

$$ {\fontsize{9}{11}\selectfont\left\{\begin{array}{ll}\overset{.}{V}&=\mathrm{R}\mathrm{e}\left(P{A}_ce,e\right)+\left\langle {e}_y,w\right\rangle -a\cdot \mathrm{t}\mathrm{r}\left[G{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]-\underset{\left\langle {e}_y,w\right\rangle }{\underbrace{\mathrm{tr}\left({e}_y{z}^T\varDelta {G}^T\right)}}+\left(Pe,v\right)\\ {}&\le -{q}_{\min }{\left\Vert e\right\Vert}^2-a\cdot \mathrm{t}\mathrm{r}\left[\left(\varDelta G+{G}^{\ast}\right){\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]+\left(Pe,v\right)\\ {}&\le -\left({q}_{\min }{\left\Vert e\right\Vert}^2+a\cdot \mathrm{t}\mathrm{r}\left[\varDelta G{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]\right)+a\cdot \left|\mathrm{t}\mathrm{r}\left[{G}^{\ast }{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]\right|+\left|\left(Pe,v\right)\right|\\ {}&\le -\left(\frac{2{q}_{\min }}{p_{\max }}\cdot \frac{1}{2}\left(Pe,e\right)+2a\cdot \frac{1}{2}\mathrm{t}\mathrm{r}\left[\varDelta G{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]\right)+a\cdot \left|\mathrm{t}\mathrm{r}\left[{G}^{\ast }{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]\right|+\left|\left(Pe,v\right)\right|\\ {}&\le -2aV+a\cdot \left|\mathrm{t}\mathrm{r}\ \left[{G}^{\ast }{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]\right|+\left|\left(Pe,v\right)\right|\end{array}\right. } $$

Now, using the Cauchy-Schwarz inequality

$$ \left|\mathrm{t}\mathrm{r}\left[{G}^{\ast }{\gamma}^{-1}\varDelta {G}^{\mathrm{T}}\right]\right|\le {\left\Vert {G}^{*}\right\Vert}_2{\left\Vert \varDelta G\right\Vert}_2 $$

and

$$ \left|\left(Pe,v\right)\right|\le \left\Vert {P}^{\frac{1}{2}}\nu \right\Vert\ \left\Vert {P}^{\frac{1}{2}}e\right\Vert =\sqrt{\left(P\nu, v\right)}\cdot \sqrt{\left(Pe,e\right)} $$

We have

$$ \begin{array}{l}\overset{.}{V}+2aV\le a\cdot {\left\Vert {G}^{*}\right\Vert}_2{\left\Vert \varDelta G\right\Vert}_2+\sqrt{p_{\max }}\left\Vert \nu \right\Vert \sqrt{\left(Pe,e\right)}\\ {}\le a\cdot {\left\Vert {G}^{*}\right\Vert}_2{\left\Vert \varDelta G\right\Vert}_2+\left(\sqrt{p_{\max }}{M}_{\nu}\right)\sqrt{\left(Pe,e\right)}\\ {}\le \left(a{\left\Vert {G}^{*}\right\Vert}_2+\sqrt{p_{\max }}{M}_{\nu}\right)\sqrt{2}\underset{V^{\frac{1}{2}}}{\underbrace{{\left[\frac{1}{2}\left(Pe,e\right)+\frac{1}{2}{\left\Vert \varDelta G\right\Vert}_2^2\right]}^{\frac{1}{2}}}}\end{array} $$

Therefore,

$$ \frac{\overset{.}{V}+2aV}{V^{\frac{1}{2}}}\le \left(a{\left\Vert {G}^{*}\right\Vert}_2+\sqrt{p_{\max }}{M}_{\nu}\right)\sqrt{2} $$

Now, using the identity $\mathrm{t}\mathrm{r}\left[ABC\right]=\mathrm{t}\mathrm{r}\left[ CAB\right]$,

$$ \begin{array}{ll}{\left\Vert {G}^{*}\right\Vert}_2&\equiv {\left[\mathrm{t}\mathrm{r}\left({G}^{*}{\gamma}^{-1}{\left({G}^{*}\right)}^T\right)\right]}^{\frac{1}{2}}={\left[\mathrm{t}\mathrm{r}\left({\left({G}^{*}\right)}^T{G}^{*}{\gamma}^{-1}\right)\right]}^{\frac{1}{2}}\\ {}&\le {\left[{\left(\mathrm{t}\mathrm{r}\left({\left({G}^{*}\right)}^T{G}^{*}{\left({G}^{*}\right)}^T{G}^{*}\right)\right)}^{{}^{\frac{1}{2}}}{\left(\mathrm{t}\mathrm{r}\left({\gamma}^{-1}{\gamma}^{-1}\right)\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}\\ {}&={\left[\mathrm{t}\mathrm{r}\left({G}^{*}{\left({G}^{*}\right)}^T\right)\right]}^{\frac{1}{2}}{\left[\mathrm{t}\mathrm{r}\left({\gamma}^{-1}\right)\right]}^{\frac{1}{2}}\\ {}&\le \frac{M_{\nu }}{a{M}_G}\cdot {M}_G=\frac{M_{\nu }}{a}\end{array} $$

which implies that

$$ \frac{\overset{.}{V}+2aV}{V^{\frac{1}{2}}}\le \left(1+\sqrt{p_{\max }}\right){M}_{\nu}\sqrt{2} $$

(22)

From

$$ \begin{aligned}\frac{d}{dt}\left(2{e}^{at}{V}^{\frac{1}{2}}\right)&={e}^{at}\frac{\overset{.}{V}+2aV}{V^{\frac{1}{2}}}\\ {}&\le {e}^{at}\left(1+\sqrt{p_{\max }}\right){M}_{\nu}\sqrt{2}\end{aligned} $$

Integrating this expression we have ${e}^{at}V{(t)}^{1/2}-V{(0)}^{1/2}\le \frac{\left(1+\sqrt{p_{\max }}\right)\;{M}_{\nu }}{a}\left({e}^{at}-1\right)$.

Therefore,

$$ V{(t)}^{1/2}\le V{(0)}^{1/2}{e}^{- at}+\frac{\left(1+\sqrt{p_{\max }}\right)\;{M}_{\nu }}{a}\left(1-{e}^{- at}\right) $$

(23)

The function V(t) is a norm function of the state e(t) and matrix G(t). So, since V(t)^1/2 is bounded for all t, then e(t) and G(t) are bounded. We also obtain the following inequality:

$$ \sqrt{p_{\min }}\left\Vert e(t)\right\Vert \le V{(t)}^{1/2} $$

Substitution of this into (23) gives us an exponential bound on state e(τ):

$$ \left\Vert e(t)\right\Vert \le \frac{e^{- at}}{\sqrt{p_{\min }}}V{(0)}^{1/2}+\frac{\left(1+\sqrt{p_{\max }}\right)\;{M}_{\nu }}{a\sqrt{p_{\min }}}\left(1-{e}^{- at}\right) $$

(24)

Taking the limit superior of (24), we have

$$ \overline{\underset{\tau \to \infty }{ \lim }}\left\Vert e(t)\right\Vert \le \frac{\left(1+\sqrt{p_{\max }}\right)\;}{a\sqrt{p_{\min }}}{M}_{\nu}\equiv {R}_{*} $$

(25)

And the proof is complete.