Low-complexity prescribed performance tracking control for uncertain high-order nonlinear systems considering input quantization

Abstract

Low-complexity prescribed performance control (PPC) is a control methodology whose strongest feature is its structural simplicity, namely, uncertainty can be handled without unknown parameter estimators or approximation structures, e.g., fuzzy logic or neural networks, considered in the design of the control scheme. This work shows that low-complexity PPC can tackle uncertain high-order odd-rational power nonlinear dynamics, i.e., an integrator chain where the power is the ratio of odd integers. State of the art has shown the feasibility of PPC for strict-feedback nonlinear dynamics (an integrator chain with a power equal to one) or high-order odd-integer-power nonlinear dynamics. In this paper, we show that the same structural simplicity of PPC can be retained for the more general class of high-order odd-rational-power nonlinear systems. Note that all the methods proposed in the literature for handling high-order nonlinearities cannot be used with rational powers (unless restricting the system’s nonlinearity growth). Therefore, a new lemma is proposed to handle the odd-rational power scenario without restricting the nonlinearities’ growth. Interestingly, the developed scheme generalizes the most advanced PPC and effectively handles the input quantization, and unknown power problems studied recently.

Appendix

Proof of Lemma 5

The goal is to find an upper and lower bound in the form

$$\begin{aligned} \underline{\vartheta }x_1^p+\underline{\mu } x_2^p\le (x_1+x_2)^p \le \bar{\vartheta } x_1^p+\bar{\mu } x_2^p \end{aligned}$$

(48)

for some appropriately bounded functions $\underline{\mu }(\cdot ,\cdot )$, $\underline{\vartheta }(\cdot ,\cdot )$, $\bar{\mu }(\cdot ,\cdot )$, and $\bar{\vartheta }(\cdot ,\cdot )$. By employing the binomial theorem [39, Sect. 3.1, p. 10], an upper bound can be derived for $\forall x_1$, $x_2\in {\mathbb {R}}$:

$$\begin{aligned}&(x_1+x_2)^p\le x_1^p+x_2^p+\sum _{k=1}^{p-1} \left( \begin{array}{c} p \\ k \\ \end{array}\right) |x_1|^k|x_2^{p-k}|\nonumber \\&\quad \le x_1^p+x_2^p+\sum _{k=1}^{p-1} \bigg (\frac{k}{p}\varepsilon ^{\frac{p}{k}}|x_1|^p\nonumber \\&\qquad +\frac{p-k}{p} \left( \begin{array}{c} p \\ k \\ \end{array}\right) \varepsilon ^{\frac{-p}{p-k}}|x_2|^p\bigg )\nonumber \\&\quad \le x_1^p+x_2^p+\sum _{k=1}^{p-1}\varepsilon _k|x_1|^p +\sum _{k=1}^{p-1}\omega _k|x_2|^p\nonumber \\&\quad \le \big [1+\bar{\epsilon }\cdot \textrm{sign}(x_1)\big ]x_1^p +\big [1+\omega \cdot \textrm{sign}(x_2)\big ]x_2^p, \end{aligned}$$

(49)

where the second inequality relies on Lemma 4, $\varepsilon _k=\frac{k}{p}\varepsilon ^{\frac{p}{k}}$, $\omega _k=\frac{p-k}{p}\left( \begin{array}{c} p \\ k \\ \end{array}\right) \varepsilon ^{\frac{-p}{p-k}}$, $\omega =\sum _{k=1}^{p-1}\omega _k$, and $\bar{\epsilon }=\sum _{k=1}^{p-1}\varepsilon _k$ satisfies $0<\bar{\epsilon }<1$ by appropriately selecting the small positive constant $\varepsilon$.

A lower bound will be sought along the following three situations.

Situation 1: When $x_1<0$ and $x_1+x_2\ge 0$, we immediately have $(x_1+x_2)^p\ge 0\ge x_1^p$ as p is a positive odd integer.

Situation 2: When $x_1<0$ and $x_1+x_2<0$, it follows that

$$\begin{aligned}&x_1^p+x_2^p\nonumber \\&\quad = 2\left[ \sum _{m=1}^{\frac{p-1}{2}}\left( \begin{array}{c} p \\ 2m-1 \\ \end{array}\right) \underbrace{\overbrace{\left( \frac{x_1+x_2}{2}\right) ^{2m-1}}^{<0} \overbrace{\left( \frac{x_1-x_2}{2}\right) ^{p-2m+1}}^{>0}}_{<0}\right. \nonumber \\&\qquad \left. +\left( \frac{x_1+x_2}{2}\right) ^p\right] \le 2^{1-p}(x_1+x_2), \end{aligned}$$

(50)

which indicates that $(x_1+x_2)^p\ge 2^{p-1}x_1^p+2^{p-1}x_2^p$.

Situation 3: When $x_1\ge 0$ and $x_2\in {\mathbb {R}}$, then following similar derivations to (49), it holds that $(x_1-x_2)^p=\big [x_1+(-x_2)\big ]^p\le \big [1+\bar{\epsilon }\cdot \textrm{sign}(x_1)\big ]x_1^p -\big [1-\omega \cdot \textrm{sign}(x_2)\big ]x_2^p.$ Besides, note that $(x_1+x_2)^p+(x_1-x_2)^p=2\Big [x_1^p+\sum _{k=1}^{\frac{p-1}{2}} \left( \begin{array}{c} p \\ 2k-1 \\ \end{array}\right) \overbrace{x_1^{2k-1}}^{\ge 0} \overbrace{x_2^{p-2k+1}}^{\ge 0}\Big ]\ge 2x_1^2$. Thus, we have $(x_1+x_2)^p\ge [1-\bar{\epsilon }\cdot \textrm{sign}(x_1)]x_1^p+[1+\omega \cdot \textrm{sign}(x_2)]x_2^p$.

Having derived all the necessary upper and lower bounds in the form of (48), we conclude that the equality $\big [x_1^p\vartheta (x_1,x_2)+x_2^p\mu (x_1,x_2)\big ]^{\frac{1}{q}}= (x_1+x_2)^{\frac{p}{q}}$ holds for any $x_1$, $x_2$, for some function $\vartheta (\cdot ,\cdot )\subset \big [1-\bar{\epsilon }, \max \{1+\bar{\epsilon }, 2^{p-1}\}\big ]$ with $\bar{\epsilon }\in (0,1)$ and $|\mu (\cdot ,\cdot )|\le \bar{\upsilon }$ with $\bar{\upsilon }>0$ being independent of $x_1$ and $x_2$. This completes the proof. $\square$