Neural-based adaptive fixed-time prescribed performance control for the flexible-joint robot with actuator failures

Abstract

In this paper, a fixed-time prescribed performance fault-tolerant control scheme is presented for the n-link flexible joint robot with actuator failures. Firstly, a modified prescribed performance control method is proposed to enhance the robustness of the system against input perturbations and to ensure that the tracking error converges in a predetermined time, and the constrained system is transferred into an unconstrained system. Secondly, an adaptive-based passive fault-tolerant controller is constructed to counteract the actuator failures in the system. Then, the uncertainty problem in the flexible-joint robot system is solved by incorporating the radial basis function neural networks and adaptive techniques into the fixed-time backstepping framework. After that, the “complexity explosion” issue is well handled by creating the fixed-time second-order filter, in which the filtering errors are eliminated by the devised compensation mechanism. The stability analysis proves that the closed-loop system is fixed-time stable, and the tracking error is limited to the predefined range. Finally, simulations have been performed on a two-link FJR and a three-link flexible joint robot, respectively. Via the three conditions: actuators operating normally, actuators losing 50% of effectiveness instantaneously, and actuators losing 50% of effectiveness gradually, the results show that the tracking error of each joint of the system is less than 0.2 rad, and the tracking time is limited to the specified time (0.8 s), which proves the effectiveness of the proposed control scheme.

Appendices

Appendix I: Proof of lemma 7

Proof

Calculating the left and right limits of the function $\overline{\phi }(t)$ for $T$ leads to

$$ \left\{ \begin{gathered} \mathop {\lim }\limits_{{t \to T^{ - } }} \overline{\phi }(t) = \overline{\phi }_{\infty } + \pi_{1} \tanh (\pi_{2} \dot{x}_{d}^{2} ) \hfill \\ \mathop {\lim }\limits_{{t \to T^{ + } }} \overline{\phi }(t) = \overline{\phi }_{\infty } + \pi_{1} \tanh (\pi_{2} \dot{x}_{d}^{2} ). \hfill \\ \end{gathered} \right. $$

(100)

According to assumption 1, we have $\mathop {\lim }\limits_{{t \to T^{ + } }} \dot{x}_{d} = \mathop {\lim }\limits_{{t \to T^{ - } }} \dot{x}_{d}$, which means $\mathop {\lim }\limits_{{t \to T^{ - } }} \overline{\phi }(t) = \mathop {\lim }\limits_{{t \to T^{ + } }} \overline{\phi }(t)$ holds. Obviously, $\overline{\phi }(t)$ is continuous at $t = T$. We know that $\overline{\phi }(t)$ is separately continuous in the time intervals $t \in \left[ {0,T} \right)$ and $t \in \left[ {T,\infty } \right)$ from (23). Furthermore, $\mathop {\lim }\limits_{t \to \infty } \overline{\phi }(t) \le \overline{\phi }_{\infty } + \pi_{1}$. As a result, $\overline{\phi }(t)$ is continuous and bounded for all $t \ge 0$.

Furthermore, based on the definition of the derivation, one obtains

$$ \mathop {\lim }\limits_{{t \to T^{ - } }} (\overline{\phi }(t) - \overline{\phi }(T))/(t - T) = \mathop {\lim }\limits_{{t \to T^{ - } }} \pi_{1} (\tanh (\pi_{2} \dot{x}_{d} (t)^{2} ) - \pi_{1} \tanh (\pi_{2} \dot{x}_{d} (T)^{2} ))/(t - T), $$

(101)

$$ \mathop {\lim }\limits_{{t \to T^{ + } }} (\overline{\phi }(t) - \overline{\phi }(T))/(t - T) = \mathop {\lim }\limits_{{t \to T^{ + } }} \pi_{1} (\tanh (\pi_{2} \dot{x}_{d} (t)^{2} ) - \pi_{1} \tanh (\pi_{2} \dot{x}_{d} (T)^{2} ))/(t - T). $$

(102)

The function $\dot{x}_{d}$ is derivable for $t \ge 0$, so that $\pi_{1} (\tanh (\pi_{2} \dot{x}_{d} (t)^{2} )$ is derivable for $t \ge 0$. Then, we have

$$ \mathop {\lim }\limits_{{t \to T^{ - } }} (\overline{\phi }(t) - \overline{\phi }(T))/(t - T) = \mathop {\lim }\limits_{{t \to T^{ + } }} (\overline{\phi }(t) - \overline{\phi }(T))/(t - T). $$

(103)

Hence, it can be concluded that $\overline{\phi }(t)$ is differentiable at time $T$. According to the definition of $\overline{\phi }(t)$, it is derivable in the time intervals $t \in \left[ {0,T} \right)$ and $t \in \left[ {T,\infty } \right)$. As a result, $\overline{\phi }(t)$ is differentiable for all $t \ge 0$.

Next, taking the time derivative of $\overline{\phi }(t)$ generates

$$ \dot{\overline{\phi }}(t) = \left\{ \begin{aligned} - &(\overline{\phi }_{0} + e_{1} (0) - \overline{\phi }_{\infty } )\left(\frac{T - t}{T}\right)^{c - 1} {\text{e}}^{ - \beta t} \left(c + \beta \left(\frac{T - t}{T}\right)\right) + 2\pi_{1} \pi_{2} (1 - \tanh^{2} (\pi_{2} \dot{x}_{d}^{2} ))\dot{x}_{d} \ddot{x}_{d} ,t \in \left[ {0,T} \right) \hfill \\ &2\pi_{1} \pi_{2} (1 - \tanh^{2} (\pi_{2} \dot{x}_{d}^{2} ))\dot{x}_{d} \ddot{x}_{d} ,t \in \left[ {T,\infty } \right). \hfill \\ \end{aligned} \right. $$

(104)

Because $\dot{x}_{d}$ and $\ddot{x}_{d}$ are bound for $t \ge 0$, $\pi_{1} \pi_{2} (1 - \tanh^{2} (\pi_{2} \dot{x}_{d}^{2} ))2\dot{x}_{d} \ddot{x}_{d}$ is bounded. From the second equation of (104), we know that $\dot{\overline{\phi }}(t)$ is bounded for $t \in \left[ {T,\infty } \right)$. Furthermore, the function $- (\overline{\phi }_{0} + e_{1} (0) - \overline{\phi }_{\infty } )((T - t)/T)^{3} \exp ( - \beta t)(4 + \beta ((T - t)/T))$ is bounded for $t \in \left[ {0,T} \right)$, which indicates that $\dot{\overline{\phi }}(t)$ is bounded for $t \in \left[ {0,T} \right)$. As a result, $\dot{\overline{\phi }}(t)$ is bounded for $t \ge 0$.

Up to now, the proof of Lemma 7 has been finished.

Appendix II

$$ {\mathbf{M}}({\mathbf{q}}) = \left[ {\begin{array}{*{20}c} {2d_{1} + d_{4} C_{2} + d_{5} C_{23} } & {2d_{2} + d_{4} C_{2} + d_{6} C_{3} } & {2d_{5} + d_{5} C_{23} + d_{6} C_{3} } \\ {d_{4} C_{2} + d_{5} C_{23} } & {2d_{2} + d_{6} C_{3} } & {2d_{3} + d_{6} C_{3} } \\ {d_{5} C_{23} } & {d_{6} C_{3} } & {2d_{3} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} } \right] $$

(105)

$$ {\mathbf{C}}({\mathbf{q}},\mathop {\mathbf{q}}\limits^{.} ) \, = {\mathbf{C}}_{m} ({\mathbf{q}},\mathop {\mathbf{q}}\limits^{.} )\mathop {\mathbf{q}}\limits^{.} = \left[ {\begin{array}{*{20}l} {C_{m11} } \hfill & {C_{m12} } \hfill & {C_{m13} } \hfill \\ {C_{m21} } \hfill & {C_{m22} } \hfill & {C_{m23} } \hfill \\ {C_{m31} } \hfill & {C_{m32} } \hfill & {C_{m33} } \hfill \\ \end{array} } \right]\mathop {\mathbf{q}}\limits^{.} $$

(106)

$$ {\mathbf{G}}({\mathbf{q}}) = \left[ {\begin{array}{*{20}c} {\frac{1}{2}l_{1} C_{1} } & {l_{1} C_{1} + \frac{1}{2}l_{2} C_{12} } & {l_{1} C_{1} + l_{2} C_{12} + \frac{1}{2}l_{3} C_{123} } \\ 0 & {\frac{1}{2}l_{2} C_{12} } & {l_{2} C_{12} + \frac{1}{2}2l_{3} C_{123} } \\ 0 & 0 & {\frac{1}{2}l_{3} C_{123} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {m_{1} g} \\ {m_{2} g} \\ {m_{3} g} \\ \end{array} } \right] $$

(107)

in which,

$$ \begin{aligned} C_{m11} &= - \dot{q}_{2} d_{4} {\mathbf{S}}_{{\mathbf{2}}} - \dot{q}_{2} d_{5} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{3} d_{5} {\mathbf{S}}_{{{\mathbf{23}}}} - \dot{q}_{3} d_{6} {\mathbf{S}}_{{\mathbf{3}}} \hfill \\ C_{m12} &= - \dot{q}_{2} d_{4} {\mathbf{S}}_{{\mathbf{2}}} - \dot{q}_{2} d_{5} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{3} d_{6} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{3} d_{5} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{1} d_{4} {\mathbf{S}}_{{\mathbf{2}}} - \dot{q}_{1} d_{5} {\mathbf{S}}_{{\mathbf{3}}} \hfill \\ C_{m13} &= - \dot{q}_{2} d_{5} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{3} d_{5} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{3} d_{6} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{1} d_{5} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{1} d_{6} {\mathbf{S}}_{{\mathbf{3}}} - \dot{q}_{2} d_{6} {\mathbf{S}}_{{\mathbf{3}}} \hfill \\ C_{m21} &= - \dot{q}_{3} d_{6} {\mathbf{S}}_{{\mathbf{3}}} + \dot{q}_{1} d_{4} {\mathbf{S}}_{{\mathbf{2}}} + \dot{q}_{1} d_{5} {\mathbf{S}}_{{\mathbf{3}}} \hfill \\ C_{m22} &= - \dot{q}_{3} d_{6} {\mathbf{S}}_{{\mathbf{3}}} \hfill \\ C_{m23} &= - d_{6} {\mathbf{S}}_{{\mathbf{3}}} \left( {\dot{q}_{1} + \dot{q}_{2} + \dot{q}_{3} } \right) \hfill \\ C_{m31} &= \dot{q}_{1} d_{5} {\mathbf{S}}_{{\mathbf{3}}} + \dot{q}_{1} d_{6} {\mathbf{S}}_{{\mathbf{3}}} + \dot{q}_{2} d_{6} {\mathbf{S}}_{{\mathbf{3}}} \hfill \\ C_{m32} &= d_{6} {\mathbf{S}}_{{\mathbf{3}}} \left( {\dot{q}_{1} + \dot{q}_{2} } \right) \hfill \\ C_{m33} &= 0 \hfill \\ \end{aligned} $$

and

$$ \begin{aligned} d_{1} &= \frac{1}{2}\left[ {\left( {\frac{1}{4}m_{1} + m_{2} + m_{3} } \right)l_{1}^{2} + I_{o1} } \right] \hfill \\ d_{2} &= \frac{1}{2}\left[ {\left( {\frac{1}{4}m_{2} + m_{3} } \right)l_{2}^{2} + I_{o2} } \right] \hfill \\ d_{3} &= \frac{1}{2}\left[ {\left( {\frac{1}{4}m_{3} } \right)l_{3}^{2} + I_{o3} } \right] \hfill \\ d_{4} &= \left( {\frac{1}{2}m_{2} + m_{3} } \right)l_{1} l_{2} \hfill \\ d_{5} &= \frac{1}{2}m_{3} l_{1} l_{3} \hfill \\ d_{6} &= \frac{1}{2}m_{3} l_{2} l_{3} . \hfill \\ \end{aligned} $$