Finite Time Observer Design for Teleoperation System

Abstract

The finite-time output feedback synchronization control problem is considered for a bilateral teleoperation system in the presence of the modeling error and disturbance. A new observer is designed for the velocity estimation by using neural network approximation and fast terminal sliding mode method. It is shown that the resulting velocity error system is semi-globally finite-time stable. The observer based output feedback finite-time controller is developed by employing a novel nonsingular fast integral terminal sliding mode. The closed-loop system is proved to be semi-globally stable and the master-slave synchronization error converges to zero in finite time. Compared with the existing controllers, the designed controller of this paper only uses the position information and renders the master-slave synchronization error reaching zero in the given finite time. Simulation and experiment are performed and the results demonstrate the effectiveness of the proposed method.

Appendix

Proof: The specific prove procedure for the FTSM finite-time velocity observer will be divided into three steps.

Step1: Let us consider the following Lyapunov candidate

$$\begin{aligned} U_{1}=\tilde{x}_{i}^{T}P_{i}\tilde{x}_{i}+\tilde{W}_{i}^{T}P_{i2}\varOmega _{i}^{-1}\tilde{W}_{i} \end{aligned}$$

(A.1)

Differentiating $U_{1}$ with respect to time yields

$$\begin{aligned} \begin{aligned} \dot{U}_{1} \le \,&\tilde{x}_{i}^{T}(P_{i}A_{i}+A_{i}^{T}P_{i})\tilde{x}_{i}+2\tilde{x}_{i}^{T}P_{i}H_{i}\\&+2\tilde{x}_{i}^{T}P_{i}\Delta Q_{i}+2\tilde{x}_{i}^{T}P_{i}\nu _{i}-2\tilde{W}_{i}P_{i2}\varOmega _{i}^{-1}\overset{\cdot }{\hat{W}}_{i} \end{aligned} \end{aligned}$$

(A.2)

With the Young inequality, we have $2\tilde{x}_{i}^{T}P_{i}H_{i}\le \frac{1}{\varepsilon _{i}}\tilde{x}_{i}^{T}P_{i}P_{i}\tilde{x}_{i}+\varepsilon _{i}H_{i}^{T}H_{i}$, $\varepsilon _{i}$ is a positive constant. Since $C_{i}(y_{i},\bar{x}_{i2})\bar{x}_{i2}$ is Lipschitz, thus $\varepsilon _{i}H_{i}^{T}H_{i}\le \varepsilon _{i}\bar{C}_{i}^{2}\left\| \widetilde{\bar{x}}_{i2}\right\| ^{2}$, where $\bar{C}_{i}$ is a positive constant. Furthermore, with the transformation $\tilde{x}_{i2}=\widetilde{\bar{x}}_{i2}-T_{i}\tilde{x}_{i1}$, we have $\bar{C}_{i}^{2}\left\| \widetilde{\bar{x}}_{i2}\right\| ^{2}\le 2\bar{C}_{i}^{2}\left\| \tilde{x}_{i2}\right\| ^{2}+2\bar{C}_{i}^{2}\left\| T_{i}\right\| ^{2}\left\| \tilde{x}_{i1}\right\| ^{2}$.

According to Remark 1 yields $\left\| W_{i}^{*T}\varphi _{i}(y_{i},\hat{x}_{i2})-W_{i}^{*T}\varphi _{i}(y_{i},x_{i2})\right\| \le b_{i}\left\| \tilde{x}_{i2}\right\|$, $b_{i}$ is a positive constant. Additionally, one has that $2\tilde{x}_{i}^{T}P_{i}N_{i}\le \dfrac{1}{z_{i}}\tilde{x}_{i}^{T}P_{i}P_{i}\tilde{x}_{i}+z_{i}N_{i}^{T}N_{i}\le \dfrac{1}{z_{i}}\tilde{x}_{i}^{T}P_{i}P_{i}\tilde{x}_{i}+z_{i}b_{i}^{2}\left\| \tilde{x}_{i2}\right\| ^{2}$, $z_{i}$ is a positive constant.

Afterwards, substituting the NNs adaptive tuning law $\overset{\cdot }{\hat{W}}_{i}=-\varOmega _{i}\varphi _{i}(y_{i},\hat{x}_{i2})$

$\tilde{x}_{i2}^{T}$ into (A.2) yields

$$\begin{aligned} \begin{aligned} \dot{U}_{1} \le&\tilde{x}_{i}^{T}(P_{i}A_{i}+A_{i}^{T}P_{i}+\frac{1}{\varepsilon _{i}}P_{i}P_{i}+\frac{1}{z_{i}}P_{i}P_{i}\\&+\varPsi _{i})\tilde{x}_{i}+2\tilde{x}_{i2}^{T}P_{i2}\epsilon _{i}^{*}+2\tilde{x}_{i}^{T}P_{i}\nu _{i} \end{aligned} \end{aligned}$$

(A.3)

With $O_{i}=P_{i}A_{i}+A_{i}^{T}P_{i}+\dfrac{1}{\varepsilon _{i}}P_{i}P_{i}+\dfrac{1}{z_{i}}P_{i}P_{i}+\varPsi _{i}<0$, $\Lambda _{i}=-\lambda _{\max }(O_{i})$, $\varPsi _{i}=diag(2\varepsilon _{i}\bar{C}_{i}^{2}\left\| T_{i}\right\| ^{2}I_{n},0_{n\times n})+$ $diag(0_{n\times n},2\varepsilon _{i}\bar{C}_{i}^{2}I_{n})+$ $diag(0_{n\times n},z_{i}b_{i}^{2}I_{n})$, it has that

$$\begin{aligned} \begin{aligned} \dot{U}_{1} <&-\Lambda _{i}\left\| \tilde{x}_{i}\right\| ^{2}-2k_{i1}\underset{j=1}{\overset{n}{\sum }}p_{i1j}\left| \tilde{x}_{i1j}\right| -2P_{ik1}\tilde{x}_{i}\\&~~~-2k_{i1}^{r_{i1}}P_{ik2}\tilde{x}_{i}-2k_{i1}^{r_{i2}}P_{ik3}\tilde{x}_{i}+2\tilde{x}_{i2}^{T}P_{i2}\epsilon _{i}^{*}\nonumber \\&\le -\left\| \tilde{x}_{i}\right\| (\Lambda _{i}\left\| \tilde{x}_{i}\right\| -2k_{i1}^{r_{i1}}\sqrt{\underset{j=1}{\overset{n}{\sum }}k_{i2}^{2}p_{i2j}^{2}}-2k_{i1}^{r_{i2}}\sqrt{\underset{j=1}{\overset{n}{\sum }}k_{i3}^{2}p_{i2j}^{2}}\nonumber \\&\ \ -2\sqrt{\underset{j=1}{\overset{n}{\sum }}p_{i2j}^{2}\chi _{ij}^{2}}-2\sqrt{\underset{j=1}{\overset{n}{\sum }}p_{i2j}^{2}\epsilon _{ij}^{*2}})\nonumber \end{aligned} \end{aligned}$$

(A.4)

where

$$\begin{aligned} P_{ik1}&=[0_{1\times n},p_{i21}\chi _{i1}sign(\tilde{x}_{i11})\cdots p_{i2n}\chi _{in}sign(\tilde{x}_{i1n})],\\ P_{ik2}&=[0_{1\times n},p_{i21}k_{i2}sign(\tilde{x}_{i11})\cdots p_{i2n}k_{i2}sign(\tilde{x}_{i1n})],\\ P_{ik3}&=[0_{1\times n},p_{i21}k_{i3}sign(\tilde{x}_{i11})\cdots p_{i2n}k_{i3}sign(\tilde{x}_{i1n})]. \end{aligned}$$

Furthermore, we can obtain that $U_{1}<0$ with $\left\| \tilde{x}_{i}\right\| >R_{i}$, where

$$ R_{i}=\frac{2k_{i1}^{r_{i1}}\sqrt{\underset{j=1}{\overset{n}{\sum }}k_{i2}^{2}p_{i2j}^{2}}+2k_{i1}^{\ r_{i2}}\sqrt{\underset{j=1}{\overset{n}{\sum }}k_{i3}^{2}p_{i2j}^{2}}+2\sqrt{\underset{j=1}{\overset{n}{\sum }}p_{i2j}^{2}\chi _{ij}^{2}}+2\sqrt{\underset{j=1}{\overset{n}{\sum }}p_{i2j}^{2}\epsilon _{ij}^{*2}}}{\Lambda _{i}}$$

This implies that if $\left\| \tilde{x}_{i}\right\|$ is outside the ball with radius $R_{i}$, then $U_{1}<0$, the $\left\| \tilde{x}_{i}\right\|$ will reduce, resulting in $\left\| \tilde{x}_{i}\right\|$ being bounded by $\left\| \tilde{x}_{i}\right\| \le R_{i}$. Therefore, there exists a moment $t_{0}$ such that $\left\| \tilde{x}_{i}\right\| =\bar{R}_{i}$, where $\bar{R}_{i}>R_{i}$.

Step2: Consider Lyapunov function $U_{2}=\frac{1}{2}p_{i1j}\tilde{x}_{i1j}^{2}$, differentiating $U_{2}$, we have

$$\begin{aligned} \dot{U}_{2}=p_{i1j}\overset{\cdot }{\tilde{x}}_{i1j} \end{aligned}$$

(A.5)

Applying Eq. (11.14), we can obtain that

$$\begin{aligned} \dot{U}_{2}\le -p_{i1j}(l_{i1j}-a_{i1j})\tilde{x}_{i1j}^{2}-p_{i1j}\left| \tilde{x}_{i1j}\right| (k_{i1}-\left| \tilde{x}_{i2j}\right| ) \end{aligned}$$

(A.6)

Set $l_{i1j}-a_{i1j}=\omega _{i}>0$ and $k_{i1}-\bar{R}_{i}=\varrho _{i}>0$, we have $\dot{U}_{2}\le -p_{i1j}\omega _{i}\tilde{x}_{i1j}^{2}-p_{i1j}\varrho _{i}\left| \tilde{x}_{i1j}\right| =-2\omega _{i}U_{2}-\sqrt{2p_{i1j}}\varrho _{i}U_{2}^{1/2}$. Hence the $\tilde{x}_{i1}$ and $\overset{\cdot }{\tilde{x}}_{i1}$will converge to zero in finite time with the convergence time $t_{j}\le \frac{1}{\omega _{i}}\ln (1+\dfrac{2\omega _{i}U_{2}^{1/2}(\bar{R}_{i})}{\sqrt{2p_{i1j}}\varrho _{i}})+t_{0}$ $(j=1,2,\ldots ,n)$.

Step3: When the sliding mode $\tilde{x}_{i1}=\overset{\cdot }{\tilde{x}}_{i1}=0$, which causes $\nu _{i1}=-\tilde{x}_{i2}$, then $\nu _{i2}=-k_{i2}\left| \tilde{x}_{i2}\right| ^{r_{i1}}sign(\tilde{x}_{i2})-k_{i3}\left| \tilde{x}_{i2}\right| ^{r_{i2}}sign(\tilde{x}_{i2})$. Consider the Lyapunov function $U_{3}=\tilde{x}_{i2}^{T}P_{i2}\tilde{x}_{i2}$, with the results obtained in Step1, the boundedness of $\hat{W}_{i}^{T}\varphi _{i}(y_{i},\hat{x}_{i2})-W_{i}^{*T}\varphi _{i}(y_{i},\hat{x}_{i2})$ can be achieved. Differentiating $U_{3}$ with Lemma 2 yields

$$\begin{aligned} \begin{aligned} \dot{U}_{3} <&\tilde{x}_{i2}^{T}(-2P_{i2}T_{i}+\frac{1}{\varepsilon _{i}}P_{i2}P_{i2}+\frac{1}{z_{i}}P_{i2}P_{i2}+\varepsilon _{i}\bar{C}_{i}^{2}I_{n}+z_{i}b_{i}^{2}I_{n})\tilde{x}_{i2}\\&~~~+2\underset{j=1}{\overset{n}{\sum }}p_{i2j}\tilde{x}_{i2j}\digamma _{i}-2\underset{j=1}{\overset{n}{\sum }}\chi _{ij}p_{i2j}\left| \tilde{x}_{i2j}\right| \nonumber \\&~~~-2k_{i2}\lambda _{\min }(P_{i2})n^{1-r_{i1}/2}(\tilde{x}_{i2}^{T}\tilde{x}_{i2})^{r_{i1}+1/2}\nonumber \\&~~~-2k_{i3}\lambda _{\min }(P_{i2})(\tilde{x}_{i2}^{T}\tilde{x}_{i2})^{r_{i2}+1/2}\nonumber \\&\le \tilde{x}_{i2}^{T}(-2P_{i2}T_{i}+\frac{1}{\varepsilon _{i}}P_{i2}P_{i2}+\frac{1}{z_{i}}P_{i2}P_{i2}+\varepsilon _{i}C_{i}^{2}I_{n}+z_{i}b_{i}^{2}I_{n})\tilde{x}_{i2}\nonumber \\&~~~+2\underset{j=1}{\overset{n}{\sum }}p_{i2j}\tilde{x}_{i2j}\digamma _{ij}-2\underset{j=1}{\overset{n}{\sum }}\chi _{ij}p_{i2j}\left| \tilde{x}_{i2j}\right| \nonumber \\&~~~-2k_{i2}n^{1-r_{i1}/2}\frac{\lambda _{\min }(P_{i2})}{\lambda _{\max }(P_{i2})^{r_{i1}+1/2}}U_{3}^{r_{i1}+1/2}\nonumber \\&~~~-2k_{i3}\frac{\lambda _{\min }(P_{i2})}{\lambda _{\max }(P_{i2})^{r_{i2}+1/2}}U_{3}{}^{r_{i2}+1/2}\nonumber \end{aligned} \end{aligned}$$

(A.7)

When $-2P_{i2}T_{i}+\frac{1}{\varepsilon _{i}}P_{i}P_{i}+\frac{1}{z_{i}}P_{i}P_{i}+\varepsilon _{i}\bar{C}_{i}^{2}I_{n}+z_{i}b_{i}^{2}I_{n}<0$ and $\chi _{ij}\ge \digamma _{i}$, then we have $\dot{U}_{3}<-2k_{i2}n^{1-r_{i1}/2}\frac{\lambda _{\min }(P_{i2})}{\lambda _{\max }(P_{i2})^{r_{i1}+1/2}}U_{3}^{r_{i1}+1/2}-2k_{i3}\frac{\lambda _{\min }(P_{i2})}{\lambda _{\max }(P_{i2})^{r_{i2}+1/2}}U_{3}{}^{r_{i2}+1/2}$

$=-\psi _{i1}U_{3}^{r_{i1}+1/2}-\psi _{i2}U_{3}{}^{r_{i2}+1/2}$, where $\psi _{i1}=2k_{i2}n^{1-r_{i1}/2}\frac{\lambda _{\min }(P_{i2})}{\lambda _{\max }(P_{i2})^{^{r_{i1}+1/2}}}$ and $\psi _{i2}=2k_{i3}\frac{\lambda _{\min }(P_{i2})}{\lambda _{\max }(P_{i2})^{r_{i2}+1/2}}$. Set $\psi =\sqrt{U_{3}}$, with Eq. (A.7) we have $\dot{\psi }=\frac{1}{2\sqrt{U_{3}}}\dot{U}_{3}=-\psi _{i1}\psi ^{r_{i1}}-\psi _{i2}\psi ^{r_{i2}}$. According to Lemma 1, it has that the $\psi$ will converge to zero within the settling time $T=\frac{1}{\psi _{i1}}\frac{1}{r_{i1}-1}+\frac{1}{\psi _{i2}}\frac{1}{1-r_{i2}}$. Based on the definitions of $\psi$ and $U_{3}$, we also can obtain that $\tilde{x}_{i2}$ will converge to zero in finite time within the settling time T. Therefore the settling time of the velocity observer is $T_{2}=T+\max (t_{1},\ldots ,t_{n})$. This completes the proof.