Robust iterative learning control with rectifying action for nonlinear discrete time-delayed systems

Abstract

This paper addresses the two-dimensional (2-D) linear inequalities based robust Iterative Learning Control (ILC) for nonlinear discrete systems with time delays. The proposed two-gain ILC rule has a rectifying action to iterative initial error and external disturbances. It guarantees a reduced bound of the ILC tracking error against the iteration-varying initial error and disturbances. For the iteration-invariant initial error and disturbances, the ILC tracking error can be driven to convergence, and a complete tracking to reference trajectory beyond the initial time point is even achieved as the control gain is specifically selected. An example is used to validate the proposed ILC method.

Appendix

1.1 Proof of Lemma 1

As $t=t_0^*$, (5b) is written as

$$\begin{aligned} Y_{k+1} (t_0^*)\le A3_k (t_0^*)\cdot X_k (t_0^*)+A4_k (t_0^*)\cdot Y_k (t_0^*)+\beta _k (t_0^*). \end{aligned}$$

(31)

From the definitions on matrix relation “$\le$” and matrix norm in Sect. 2, the fact is easily known that if $A,\;B$ are nonnegative matrices with $A\le B$, then, $\left\Vert A \right\Vert\le \left\Vert B \right\Vert$. Therefore, we have

$$\begin{aligned} \left\Vert {Y_{k+1} \left( {t_0^*} \right)} \right\Vert&\le \left\Vert {A3_k \left( {t_0^*} \right)\cdot X_k \left( {t_0^*} \right)+A4_k \left( {t_0^*} \right)\cdot Y_k \left( {t_0^*} \right)+\beta _k \left( {t_0^*} \right)} \right\Vert \nonumber \\&\le \left\Vert {A3_k \left( {t_0^*} \right)\cdot X_k \left( {t_0^*} \right)+\beta _k \left( {t_0^*} \right)} \right\Vert+\left\Vert {A4_k \left( {t_0^*} \right)} \right\Vert\cdot \left\Vert {Y_k \left( {t_0^*} \right)} \right\Vert \end{aligned}$$

(32)

From the assumptions, $A3_k (t_0^*), \,X_k \left( {t_0^*} \right), \,\beta _k \left( {t_0^*} \right)$ and $Y_0 (t_0^*)$ are finite, and $\left\Vert {A4_k (t_0^*)} \right\Vert\le \rho <1$. Let $d_k (t_0^*)=\left\Vert {A3_k (t_0^*)\cdot X_k (t_0^*)+\beta _k (t_0^*)} \right\Vert$, which is finite. It follows from (32) that

$$\begin{aligned} \left\Vert {Y_{k+1} (t_0^*)} \right\Vert\le \rho \left\Vert {Y_k (t_0^*)} \right\Vert+d_k (t_0^*) \end{aligned}$$

(33)

According to the Lemma in Appendix of Wang (1998), the following inequality for $k\ge 1$ is derived from the difference inequality (33)

$$\begin{aligned} \left\Vert {Y_k (t_0^*)} \right\Vert&\le \sum _{j=0}^{k-1} {\rho ^{k-1-j}\cdot d_j (t_0^*)} +\rho ^{k}\cdot \left\Vert {Y_0 (t_0^*)} \right\Vert \nonumber \\&\le d(t_0^*)\cdot \sum _{i=0}^{k-1} {\rho ^{i}} +\rho ^{k}\cdot \left\Vert {Y_0 (t_0^*)} \right\Vert, (\text{ Let}\, i=k-1-j) \end{aligned}$$

(34)

where $d(t_0^*)=\mathop {\sup }\nolimits _k d_k (t_0^*)$. Consequently, $\mathop {\limsup }\nolimits _{k\rightarrow +\infty } \left\Vert {Y_k (t_0^*)} \right\Vert\le \frac{d(t_0^*)}{1-\rho }$. That is, $\mathop {\sup }\nolimits _k \left( {{\begin{array}{c} {\left\Vert {X_k (t_0^*)} \right\Vert} \\ {\left\Vert {Y_k (t_0^*)} \right\Vert} \\ \end{array} }} \right)$ is finite.

Assume that for $t=l \quad (t_0^*\le l<t_0^*+n^{*}), \,\mathop {\sup }\nolimits _k \left( {{\begin{array}{c} {\left\Vert {X_k (l)} \right\Vert} \\ {\left\Vert {Y_k (l)} \right\Vert} \\ \end{array} }} \right)$ is finite. As a direct result of (5a), $\mathop {\sup }\nolimits _k \left\Vert {X_k \left( {l+1} \right)} \right\Vert$ is finite because of the boundedness of $A1_k (l), \,A2_k (l)$ and $\alpha _k (l)$. On the other hand, as $t=l+1$, (5b) is written as

$$\begin{aligned} Y_{k+1} (l+1)\le A3_k (l+1)\cdot X_k (l+1)+A4_k (l+1)\cdot Y_k (l+1)+\beta _k (l+1) \end{aligned}$$

(35)

Because of the nonnegative property of matrices/vectors in (35), it follows that

$$\begin{aligned} \left\Vert {Y_{k+1} \left( {l+1} \right)} \right\Vert&\le \left\Vert {A3_k \left( {l+1} \right)\cdot X_k \left( {l+1} \right)+A4_k \left( {l+1} \right)\cdot Y_k \left( {l+1} \right)+\beta _k \left( {l+1} \right)} \right\Vert \nonumber \\&\le \left\Vert {A3_k \left( {l+1} \right)\cdot X_k \left( {l+1} \right)+\beta _k \left( {l+1} \right)} \right\Vert+\left\Vert {A4_k \left( {l+1} \right)} \right\Vert\cdot \left\Vert {Y_k \left( {l+1} \right)} \right\Vert\nonumber \\ \end{aligned}$$

(36)

In (36), $A3_k (l+1), \,X_k (l+1), \,\beta _k (l+1)$ and $Y_0 (l+1)$ are finite, and $\left\Vert {A4_k \left( {l+1} \right)} \right\Vert\le \rho <1$. Let $d_k (l+1)=\left\Vert {A3_k (l+1)\cdot X_k (l+1)+\beta _k (l+1)} \right\Vert$, which is finite.

It follows from (36) that

$$\begin{aligned} \left\Vert {Y_{k+1} (l+1)} \right\Vert\le \rho \left\Vert {Y_k (l+1)} \right\Vert+d_k (l+1) \end{aligned}$$

(37)

Similar to the process of (33) to (34), for $k\ge 1$, we have

$$\begin{aligned} \left\Vert {Y_k (l+1)} \right\Vert\le d(l+1)\sum _{i=0}^{k-1} {\rho ^{i}} +\rho ^{k}\cdot \left\Vert {Y_0 (l+1)} \right\Vert, \end{aligned}$$

where $d(l+1)=\mathop {\sup }\nolimits _k d_k (l+1)$. Therefore, $\mathop {\limsup }\nolimits _{k\rightarrow +\infty } \left\Vert {Y_k (l+1)} \right\Vert\le \frac{d\left( {l+1} \right)}{1-\rho }$. That is, $\mathop {\sup }\nolimits _k \left( {{\begin{array}{c} {\left\Vert {X_k (l+1)} \right\Vert} \\ {\left\Vert {Y_k (l+1)} \right\Vert} \\ \end{array} }} \right)$ is finite. Based on the mathematical induction, we have $\mathop {\sup }\nolimits _k \left( {{\begin{array}{c} {\left\Vert {X_k (t)} \right\Vert} \\ {\left\Vert {Y_k (t)} \right\Vert} \\ \end{array} }} \right)$ is finite for $t=t_0^*, t_0^*+1, \ldots , t_0^*+n^{*}$.

Furthermore, if $\mathop {\lim }\nolimits _{k\rightarrow +\infty } \alpha _k (t)=\mathop {\lim }\nolimits _{k\rightarrow +\infty } \beta _k (t)=0$ for $t=t_0^*, t_0^*+1, \ldots , t_0^*+n^{*}$, and $\mathop {\lim }\nolimits _{k\rightarrow +\infty } X_k (t_0^*)=0, \,\mathop {\sup }\nolimits _k \left\Vert {Y_k (t)} \right\Vert$ is still finite for $t=t_0^*, t_0^*+1, \ldots , t_0^*+n^{*}$. Similar to the above steps, we apply the mathematical induction to the 2-D linear inequalities (5a) and (5b). In the inductive steps, taking supremum limit to (33) and (37), and considering that $\mathop {\lim }\nolimits _{k\rightarrow +\infty } d_k (t_0^*)=0$ and $\mathop {\lim }\nolimits _{k\rightarrow +\infty } d_k (l+1)=0$, it is easy to prove that $\mathop {\lim }\nolimits _{k\rightarrow +\infty } \left( {{\begin{array}{c} {\left\Vert {X_k (t)} \right\Vert} \\ {\left\Vert {Y_k (t)} \right\Vert} \\ \end{array} }} \right)=0$ for $t=t_0^*, t_0^*+1, \ldots , t_0^*+n^{*}$.