Finite-time estimator with enhanced robustness and transient performance applied to adaptive problems

Abstract

In this paper, a dynamic regressor extension and mixed estimator is proposed with finite-time convergence and freedom to choose its time-varying adaptation gain and its derivation order. This freedom is exploited to enhance the transient and robustness performance of the estimation by analytically establishing the effects of both variables. The proposed estimator is used to design adaptive controllers and observers for nonlinear systems, which exhibit exponential order of convergence at an arbitrary rate of decay with robust and improved transient properties. These results are illustrated in a tracking control of nonlinear systems with parametric uncertainty.

Appendix: Proof of Proposition 1

(Sufficiency) Notice that by sending $\mu \rightarrow 0^+$ —recall that $\mu$ is a designer-chosen parameter— condition (11) can be restated as $\big ( I^{\alpha }_{0^+} \lambda \Delta ^2 \big ) (t=\delta ) >0$. Since $\lambda >0$, m is continuous and the filters map continuous functions to continuous functions, this is satisfied whenever $\Delta$ is not the constant function with constant equals to zero. (Necessity) If $\Delta \equiv 0$, then $Y \equiv 0$ according to (5) and no information of $\theta$ can be extracted from the measurement of Y.

1.1 Proof of Proposition 2

Let $I=(a,b)$ be the interval where the components of m are linearly independent and $t_{0}$ a point in the interior of I. Since the filters $H_{i}$ are arbitrarily chosen operators, we can choose the shift operators $H_{i}(x)(t)=x(t - \delta _{i})$ for some $\delta _{i} >0$ such that $t_{0} - \delta _{i} \in I$. It is easy to see that these shift operators are linear and map continuous functions to continuous functions, and that the matrix M defined in the second paragraph of Sect. 3 takes the following form

$$\begin{aligned} M(t)= \begin{pmatrix} m_{1}(t_{1}) &{}\quad \cdots &{}\quad m_q(t_{1}) \\ \vdots &{}\quad \ddots &{}\quad \vdots \\ m_{1}(t_q) &{}\quad \cdots &{}\quad m_q(t_q) \end{pmatrix}, \end{aligned}$$

where $t_{i}=t - \delta _{i}$ for $i=1,\ldots , q$. According to the choice of $\delta _{i}$, we have $t_{1}, \ldots , t_q \in I$ when M is computed at $t = t_{0}$. Since the components of m are linearly independent in I, there exists a choice of $\delta _{i}$ for $i=1,\ldots ,q$ such that $M(t_{0})$ is invertible. To see this, note that the span of the range of the function $g: t \in (a,t_{0}) \rightarrow (m_{1}(t),\ldots ,m_q(t))$ is ${\mathbb {R}}^q$ because its orthogonal set is $\{0\}$ as the functions $m_{i}$ are linearly independents in $(a,t_{0})$. Hence, there exist $t_{1}, \ldots , t_q \in (a,t_{0})$ such that $\{ g(t_{i}) \}_{i=1,\ldots ,q}$ are linearly independents vectors, which entails the existence of $\delta _{i}$ for $i=1,\ldots ,q$. Therefore, $\Delta (t_{0}) \ne 0$ and the claim follows from the continuity assumption.

1.2 Proof of Proposition 3

Since ${\tilde{\theta }}^{FT}$ satisfies $D^{\alpha }_{0^+} {\tilde{\theta }}^{FT} = - \lambda \Delta ^2 {\tilde{\theta }}^{FT}$ when $t < t_f$, it follows that ${\tilde{\theta }}^{FT}$ does not change its sign (see, for example, [4], Chapter 7]). Hence, $\lambda _{1} \Delta ^2 {\tilde{\theta }}^{FT} \le \lambda _{2} \Delta ^2 {\tilde{\theta }}^{FT}$ when ${\tilde{\theta }}^{FT} (0) \ge 0$. A comparison argument (see, for example, [4], Chapter 6]) and reversing the inequalities when ${\tilde{\theta }} (0) \le 0$ yields $|{\tilde{\theta }}^{FT}_{1}| \ge |{\tilde{\theta }}^{FT}_{2}|$. For $t>t_f$, this inequality is true because of the finite-time convergence.

If $\Delta$ is continuous and not the zero function, it satisfies (11) for some $\mu$ whenever $\lambda >0$ according to Proposition 2. Then, the increase in $\lambda$ reduces the time $\delta$ when $\mu$ is fixed as can be seen from expression (11) and the fact that $\Delta$ is continuous. Since $t_f < \delta$, the second claim follows. This means that in the above paragraph we take $t_f = {t_f}_{1} > {t_f}_{2}$ to yield the first claim.

1.3 Proof of Propositions 4

Since $\lambda (0) \Delta (0) \ne 0$, the equation $D_{0^+}^{\alpha }w (t) = - \lambda (t) \Delta (t) w(t)$ implies $|\frac{d}{dt}w(t)| = {\mathcal {O}} (t^{\alpha -1})$ when $t \rightarrow 0^+$ and $\alpha < 1$ (see [9]). Moreover, the sign of $\frac{d}{dt}w(t)$ is non-positive for a small enough interval $[0,\epsilon )$ as can be seen from the fact that $D_{0^+}^{\alpha }w (0) = - \lambda (0) \Delta (0) <0$ since $w(0) =1$, and from the definition of the fractional derivative (2).

Using again that $w(0) =1$, the differentiability of the solution for $t>0$ (see [9]) and the mean value theorem (MVT), we can find a small enough number $\varepsilon$ such that $w_{\alpha _{1}}(t) < w_{\alpha _{2}} (t)$ for any $t \in [0,\varepsilon )$, where $w_{\alpha }$ denotes the solution of the equation of w for derivation order $\alpha$ with $w(0) =1$. This is because the MVT ensures that $w_{\gamma }(t)-w_{\gamma }(0^+)= w_{\gamma }(t)-1 = \dot{w}_{\gamma }(\xi ) t$ for $\xi \in (0,t)$ and any $\gamma \in (0,1]$, where $w(0^+)$ is the right-hand limit at $t=0$. This and the fact that $\dot{w}_{\gamma }(\xi )$ grows as $\gamma$ decreases for $\xi$ near of $0^+$, according to the above paragraph, yield the inequality. Since ${\tilde{\theta }}(t)=w(t) {\tilde{\theta }}(0)$, the claim follows.