Adaptive iterative learning control based on IF–THEN rules and data-driven scheme for a class of nonlinear discrete-time systems

Abstract

An adaptive iterative learning controller (ILC) is designed for a class of nonlinear discrete-time systems based on data driving control (DDC) scheme and adaptive networks called fuzzy rules emulated network (FREN). The proposed control law is derived by using DDC scheme with a compact form dynamic linearization for iterative systems. The pseudo-partial derivative of linearization model is estimated by the proposed tuning algorithm and FREN established by human knowledge of controlled plants within the format of IF–THEN rules related on input–output data set. An on-line learning algorithm is proposed to compensate unknown nonlinear terms of controlled plant, and the controller allows to change desired trajectories for other iterations. The performance of control scheme is verified by theoretical analysis under reasonable assumptions which can be held for a general class of practical controlled plants. The experimental system is constructed by a commercial DC motor current control to confirm the effectiveness and applicability. The comparison results are addressed with a general ILC scheme based on DDC.

Appendices

Appendices

1.1 Proof of Lemma 2

Proof

Recall the relation (39); thus, it is clear that the sequence is a convergence sequence if and only if

$$\begin{aligned} -1\le 1-\eta \frac{e_p^2(i,k)}{\mu +e^2_p(i,k)}\varphi (i-1,k) \le 1. \end{aligned}$$

(42)

Let us rearrange (42); thus, it can be obtained

$$\begin{aligned} 0\le \eta \varphi (i-1,k) \le 2 \frac{\mu +e_p^2(i,k)}{e^2_p(i,k)}. \end{aligned}$$

(43)

A constant $\mu$ is small positive and $|\varphi (i-1,k)|\le l_b$; thus, the relation (43) can be rewritten as

$$\begin{aligned} 0\le \text{ sign }\left\{ \varphi (i-1,k)\right\} \eta \le \frac{2}{l_b} \le \frac{2}{|\varphi (i-1,k)|}. \end{aligned}$$

(44)

By using the result in (32), both relations (44) and (40) are clearly identical. $\square$

1.2 Proof of Lemma 3

Proof

Let us recall (39) as $|\tilde{\varTheta }(i,k)|^2$ and subtract with $|\tilde{\varTheta }(i-1,k)|^2$ on both sides of equation; thus, we obtain

$$\begin{aligned}&|\tilde{\varTheta }(i,k)|^2-|\tilde{\varTheta }(i-1,k)|^2\nonumber \\&\quad =\left[ 1-\eta \frac{e_p^2(i,k)}{\mu +e^2_p(i,k)}\varphi (i-1,k)\right] ^2|\tilde{\varTheta }(i-1,k)|^2 \nonumber \\&\qquad -|\tilde{\varTheta }(i-1,k)|^2, \nonumber \\&\quad =\left[ -2\eta \varphi (k)\frac{e_p^2(i,k)}{\mu +e^2_p(i,k)}\right. \nonumber \\&\qquad +\left. \,\left[ \eta \varphi (k)\frac{e_p^2(i,k)}{\mu +e^2_p(i,k)}\right] ^2\right] |\tilde{\varTheta }(i-1,k)|^2. \end{aligned}$$

(45)

To simplify $\varphi (k)$ denotes as $\varphi (i-1,k)$. By using one step back iteration, the relation (36) can be rewritten as

$$\begin{aligned} \tilde{\varTheta }(i-1,k)=\frac{1}{e_u(i-1,k)\varphi (k)}e(i-1,k+1). \end{aligned}$$

(46)

Substitute (46) into (45); thus, it can be obtained

$$\begin{aligned}&|\tilde{\varTheta }(i,k)|^2-|\tilde{\varTheta }(i-1,k)|^2\nonumber \\&\quad =\frac{\eta }{\varphi (k)} \left[ -2+\eta \varphi (k)\frac{e_p^2(i,k)}{\mu +e^2_p(i,k)}\right] \nonumber \\&\qquad \times \frac{e_p^2(i,k)}{\mu +e^2_p(i,k)} \frac{e^2(i-1,k+1)}{e^2_u(i-1,k)}. \end{aligned}$$

(47)

Regarding (20), it is clear that $e_p(i,k)=e_u(i-1,k)$; thus, the relation in (47) can be rewritten as

$$\begin{aligned}&|\tilde{\varTheta }(i,k)|^2-|\tilde{\varTheta }(i-1,k)|^2\nonumber \\&\quad =\frac{\eta }{\varphi (k)} \left[ \eta \varphi (k)\frac{e_p^2(i,k)}{\mu +e^2_p(i,k)}-2\right] \nonumber \\&\qquad \frac{e^2(i-1,k+1)}{\mu +e^2_p(i,k)}. \end{aligned}$$

(48)

Summing (48) for $i=1$ to any i, we obtain

$$\begin{aligned}&|\tilde{\varTheta }(i,k)|^2\nonumber \\&\quad =|\tilde{\varTheta }(0,k)|^2+\sum ^i_{j=1}\frac{\eta }{\varphi (k)}\left[ \eta \varphi (k)\frac{e_p^2(j,k)}{\mu +e^2_p(j,k)}-2\right] \nonumber \\&\qquad \times \frac{e^2(j-1,k+1)}{\mu +e^2_p(j,k)}. \end{aligned}$$

(49)

It is clear that $\Big [\eta \varphi (k)\frac{e_p^2(j,k)}{\mu +e^2_p(j,k)}-2\Big ]<0$ when $\eta$ given by (40) and $|\tilde{\varTheta }(0,k)|^2$ is bounded but $|\tilde{\varTheta }(i,k)|^2$ must be nonnegative; thus, it leads to

$$\begin{aligned} \lim _{i\rightarrow \infty }\sum ^i_{j=1}\frac{e^2(j-1,k+1)}{\mu +e^2_p(j,k)} < \infty . \end{aligned}$$

(50)

It implies that

$$\begin{aligned} \lim _{i\rightarrow \infty }\frac{e(i,k+1)}{\sqrt{\mu +e^2_p(i,k)}} =0. \end{aligned}$$

(51)

Regarding the definition of error $e_p$, the relation (20) can be reformulated as

$$\begin{aligned} |e_p(i,k)|\le & {} |r(i-1,k+1)|+|y(i-2,k+1)|, \nonumber \\\le & {} |r(i-1,k+1)|+|r(i-2,k+1)|\nonumber \\&+\,|e(i-2,k+1)|. \end{aligned}$$

(52)

Generally, the reference signal r has been bounded as $|r(k)|\le l_r$ for all k; thus, the error in (52) can be obtained as

$$\begin{aligned} |e_p(i,k)| \le 2l_r+ \max _{i=2}^{i_{\mathrm {max}}}|e(i-2,k+1)|. \end{aligned}$$

(53)

According to the result in (53) and the convergence in (49), it implies that the asymptotic convergence of e(i, k) for the long run of iteration number can be obtained as

$$\begin{aligned} \lim _{i\rightarrow \infty }e(i,k+1)=0, \end{aligned}$$

(54)

over the finite sampling time interval as $k=\{1,2, \ldots , k_{\mathrm {max}}\}$. $\square$