Event-triggered adaptive output-feedback neural-networks control for saturated strict-feedback nonlinear systems in the presence of external disturbance

Abstract

An event-triggered (ET) neural-networks (NNs) adaptive output-feedback control approach is proposed for a class of input-saturated strict-feedback nonlinear systems with external disturbance. Compared with overall existing event-triggered control strategies, which are free from control input nonlinearities and suffer from the explosion of complexity, the proposed event-triggered-based NNs controller is able to handle system input saturation along with completely avoiding the complexity explosion problem. First, by introducing alternative state variables, and by implementing a low-pass filter, the difficulty arising from the cascading of the input-saturated strict-feedback system has been avoided. Thus, the system is converted to the normal canonical system, for which the controller synthesis is much simpler without resort to traditional back-stepping approach. Then, an observer is adopted to estimate the unknown states of the newly derived canonical system based on strictly positive real theory. In the design procedure, the unknown nonlinear functions are approximated by NNs to design a baseline controller, for which an additional robust term is embedded to deal with the input saturation nonlinearity, unknown disturbances and approximation errors using only two adaptive parameters. The proposed ET adaptive NNs control scheme is shown to guarantee the convergence of the output of the system to a small neighborhood of the origin along with the boundedness of all signals in the closed loop. Finally, simulation examples are presented to show the effectiveness of the proposed controller.

Appendix

Proof of Lemma 2

Let $f\left( \cdot \right)$ be a new functional of the independent variables $\overline{x}_{n - 1 } ,x_{n}$ and $s$ such as

$f\left( {\overline{x}_{n - 1 } ,x_{n} ,s} \right) = \varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right) - s$.

The equation $f\left( {\overline{x}_{n - 1 } ,x_{n} ,s} \right) = 0$ has the trivial explicit solution $s = \varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right)$ for which $f\left( {\overline{x}_{n - 1 } ,x_{n} ,\varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right)} \right) = 0$. However, we can show that this solution can, alternatively and implicitly, be given by $x_{n} = h\left( {\overline{x}_{n - 1 } ,s} \right)$ for which $f\left( {\overline{x}_{n - 1 } ,h\left( {\overline{x}_{n - 1 } ,s} \right),s} \right) = 0$.

First, for every fixed values $\overline{x}_{{\left( {n - 1} \right)_{0} }}$ and $s_{0}$, $f\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,x_{n} ,s_{0} } \right) = 0$ becomes a function of $x_{n}$ only. Since $\frac{{\partial f\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,x_{n} ,s_{0} } \right)}}{{\partial x_{n} }} = \frac{{\partial \varphi \left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,x_{n} } \right)}}{{\partial x_{n} }} > 0$, which means that $f\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,x_{n} ,s_{0} } \right)$ is strictly increasing, so using the implicit function theorem [45], for every point $\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,s_{0} } \right)$ the equation $f\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,x_{{n_{0} }} ,s_{0} } \right) = 0$ determines implicitly one and only one value $x_{{n_{0} }} = h\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,s_{0} } \right)$ for which $f\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,h\left( {\overline{x}_{{\left( {n - 1} \right)_{0} }} ,s_{0} } \right),s_{0} } \right) = 0$.

To generalize, the conditions of the implicit function theorem are fulfilled since $\frac{\partial s}{{\partial x_{n} }} = 0$, and we have that $\partial \overline{x}_{n - 1 } /\partial x_{n} = 0$ and $\partial \varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right)/\partial x_{n} > 0$, then

$\frac{{\partial f\left( {\overline{x}_{n - 1 } ,x_{n} ,s} \right)}}{{\partial x_{n} }} = \frac{{\partial \varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right)}}{{\partial x_{n} }} + \frac{{\partial \varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right)}}{{\partial \overline{x}_{n - 1 } }}\frac{{\partial \overline{x}_{n - 1 } }}{{\partial x_{n} }} - \frac{\partial s}{{\partial x_{n} }} = \frac{{\partial \varphi \left( {\overline{x}_{n - 1 } ,x_{n} } \right)}}{{\partial x_{n} }} > 0$, which means that $f\left( {\overline{x}_{n - 1 } ,x_{n} ,s} \right)$ is strictly increasing for every $x_{n}$; then, using the implicit function theorem, for every point $\left( {\overline{x}_{n - 1 } ,s} \right)$ the equation $f\left( {\overline{x}_{n - 1 } ,x_{n} ,s} \right) = 0$ determines a unique implicit function $x_{n} = h\left( {\overline{x}_{n - 1 } ,s} \right)$ for which $f\left( {\overline{x}_{n - 1 } ,h\left( {\overline{x}_{n - 1 } ,s} \right),s} \right) = 0$.