Approximate continuous fixed-time terminal sliding mode control with prescribed performance for uncertain robotic manipulators

Abstract

This paper studies an approximate continuous fixed-time terminal sliding mode control (CFTSMC) with prescribed performance for uncertain robotic manipulators. A transformation concerning tracking error using a fixed-time prescribed performance function is proposed to guarantee the transient and steady-state performance of trajectory tracking control for uncertain robotic manipulators within fixed time. Utilizing the transformed error, a smooth fixed-time sliding mode surface is designed. Then, based on the proposed sliding mode surface, an approximate CFTSMC scheme is presented to achieve inherent chattering-free control for uncertain robotic manipulators. According to the Lyapunov stability theory, it is proved that the position tracking error can be bounded in the prescribed performance boundaries and globally converges to a defined small region within fixed time and then approaches exponentially to the origin. Several numerical simulation results demonstrate the effectiveness and superiority of the proposed control strategy for uncertain robotic manipulators.

Appendices

Appendix A: The proof of Lemma 2

Proof

For inequality (6), with a positive constant $\alpha$, we can obtain that

$$\begin{aligned} \lim _{x\rightarrow 0^+}\frac{d|x |^{\alpha +1}}{dx}&=\lim _{x\rightarrow 0^+}\frac{dx^{\alpha +1}}{dx}\nonumber \\&=\left( \alpha +1 \right) x^{\alpha } =\left( \alpha +1 \right) |x |^{\alpha }\text {sgn} \left( x \right) \end{aligned}$$

(A1)

$$\begin{aligned} \lim _{x\rightarrow 0^-}\frac{d|x |^{\alpha +1}}{dx}&=\lim _{x\rightarrow 0^-}\frac{d\left( -x \right) ^{\alpha +1}}{dx} =-\left( \alpha +1 \right) \left( -x \right) ^{\alpha }\nonumber \\&=-\left( \alpha +1 \right) |x |^{\alpha } =\left( \alpha +1 \right) |x |^{\alpha }\text {sgn} \left( x \right) . \end{aligned}$$

(A2)

Then, for inequality (7), we have

$$\begin{aligned} \lim _{x\rightarrow 0^+}\frac{d|x |^{\alpha +1}\text {sgn} \left( x \right) }{dx}&=\lim _{x\rightarrow 0^+} \frac{dx^{\alpha +1}}{dx}\nonumber \\&=\left( \alpha +1 \right) x^{\alpha } =\left( \alpha +1 \right) |x |^{\alpha } \end{aligned}$$

(A3)

$$\begin{aligned} \lim _{x\rightarrow 0^-}\frac{d|x |^{\alpha +1}\text {sgn} \left( x \right) }{dx}&=\lim _{x\rightarrow 0^-} \frac{d\left[ -\left( -x \right) ^{\alpha +1} \right] }{dx}\nonumber \\&=\left( \alpha +1 \right) \left( -x \right) ^{\alpha } =\left( \alpha +1 \right) |x |^{\alpha }. \end{aligned}$$

(A4)

Hence, inequalities (6) and (7) hold. $\square$

Appendix B: The proof of Proposition 3

Proof

The time derivative of $\rho _i\left( t \right)$ is

$$\begin{aligned} \dot{\rho }_i\left( t \right) ={\left\{ \begin{array}{ll} -\frac{\sigma _i}{T_i}\left( \rho _{0i}-\rho _{\infty i} \right) \left( 1-\frac{t}{T_i} \right) ^{\sigma _i-1},\ 0\leqslant t<T_i\\ 0, t\geqslant T_i\\ \end{array}\right. }. \end{aligned}$$

(B5)

When $0\leqslant t<T_i$, it can be obtained that $\dot{\rho }_i<0$ with $\rho _{0i}>\rho _{\infty i}$. In light of (20) and (B5), it has

$$\begin{aligned} \underset{t\rightarrow T_{i}^{-}}{\lim }\rho _i\left( t \right)&=\left( \rho _{0i}-\rho _{\infty i} \right) \left( 1-\frac{T_i}{T_i} \right) ^{\sigma _i}\nonumber \\&\quad +\rho _{\infty i}=\rho _{\infty i} =\underset{t\rightarrow T_{i}^{+}}{\lim } \rho _i\left( t \right) \end{aligned}$$

(B6)

$$\begin{aligned} \underset{t\rightarrow T_{i}^{-}}{\lim }\dot{\rho }_i\left( t \right)&=-\frac{\sigma _i}{T_i}\left( \rho _{0i}-\rho _{\infty i} \right) \left( 1-\frac{T_i}{T_i} \right) ^{\sigma _i-1}=0\nonumber \\&=\underset{t\rightarrow T_{i}^{+}}{\lim }\dot{\rho }_i\left( t \right) . \end{aligned}$$

(B7)

Hence, $\rho _i\left( t \right)$ is monotone decreasing bounded smooth positive function. Due to the fact $\rho _i\left( 0 \right) =\rho _{0i}$ and $\rho _i\left( \infty \right) =\rho _{\infty i}$, we can get $0<\rho _{\infty i}\leqslant \rho _i\left( t \right) \leqslant \rho _{0i}$. With (B6), we have $\underset{t\rightarrow T_i}{\lim }\rho _i\left( t \right) =\rho _{\infty i}$, The proof of Proposition 3. $\square$

Appendix C: The proof of Proposition 4

Proof

The derivative of $\psi \left( x \right)$ with respect to x is

$$\begin{aligned} \dot{\psi }\left( x \right) =\frac{-\underline{b}_{i}\bar{b}_i \exp \left( x \right) \left( \bar{b}_i-\underline{b}_{i} \right) }{\left( \underline{b}_{i}\exp \left( x \right) -\bar{b}_i \right) ^2}. \end{aligned}$$

(C8)

It is easy to have $\psi \left( 0 \right) =0$. According to $\underline{b}_i<0<\bar{b}_i$, we have $\dot{\psi }\left( x \right) >0$, hence $\psi \left( x \right)$ is a monotonically increasing function. It has $\underset{\lim x\rightarrow \infty }{\psi \left( x \right) }=\frac{\underline{b}_{i}\bar{b}_i\left( \exp \left( \infty \right) -1 \right) }{\underline{b}_{i}\exp \left( \infty \right) -\bar{b}_i}=\bar{b}_i$ and $\underset{\lim x\rightarrow -\infty }{\psi \left( x\right) }=\frac{\underline{b}_{i} \bar{b}_i\left( \exp \left( -\infty \right) -1 \right) }{\underline{b}_{i}\exp \left( -\infty \right) -\bar{b}_i}=\underline{b}_i$, so we can have $\psi \left( x \right) \in \left( \underline{b}_{i}, \bar{b}_i \right)$. According to (12), it can obtain $e_i\in \left( \underline{b}_{i} \rho _i(t), \bar{b}_i\rho _i(t) \right)$, so Proposition 4 has been proved. $\square$

Appendix D: The proof of Proposition 5

Proof

From Proposition 3, we have $\rho _i>0$ and $\dot{\rho }_i\leqslant 0$. Considering $\alpha _i$ defined in (26), it can obtain $\alpha _i\geqslant 0$. From Proposition 4, we can have $\frac{e_i\left( t \right) }{\rho _i\left( t \right) }-\underline{b}_{i}>0$ and $\bar{b}_i-\frac{e_i\left( t \right) }{\rho _i\left( t \right) }>0$. According to the triangle inequality $a+b\geqslant 2\sqrt{ab}$ with $a\geqslant 0$ and $b\geqslant 0$, it has

$$\begin{aligned} \beta _i&=\left( \frac{1}{e_i/\rho _i-\underline{b}_{i}} +\frac{1}{\bar{b}_i-e_i/\rho _i} \right) /\rho _i\nonumber \\&\geqslant \frac{2}{\rho _i} \sqrt{\frac{1}{\left( e_i/\rho _i-\underline{b}_{i} \right) \left( \bar{b}_i-e_i/\rho _i \right) }}. \end{aligned}$$

(D9)

If and only if $\frac{1}{e_i/\rho _i-\underline{b}_{i}} =\frac{1}{\bar{b}_i-e_i/\rho _i}$, the equality in (D9) holds, which means that $\beta _i$ can get the minimum value when $\frac{e_i}{\rho _i}=\frac{\underline{b}_{i}+\bar{b}_i}{2}$. As a consequence, $\beta _i$ satisfies

$$\begin{aligned} \beta _i\geqslant \frac{4}{\rho _i\left( \bar{b}_i -\underline{b}_{i} \right) }. \end{aligned}$$

(D10)

From (13), it has $\bar{b}_i-\underline{b}_i<2$, then we can have $\beta _i>2\rho _{0i}^{-1}$. Proposition 5 has been proved. $\square$