Filtered Repetitive Control with Nonlinear Systems: An Adaptive-Control-Like Method

Abstract

An adaptive-control-like method is the leading repetitive control (RC, or repetitive controller, also RC) method for nonlinear systems. Before using this method, a tracking problem for nonlinear systems needs to be converted into a rejection problem for nonlinear error dynamics subject to periodic disturbance. This conversion reduces the difficulties of the original problem. Furthermore, the periodic disturbance is considered as an unknown parameter that needs to be learned. As discussed in Chap. 4, the stability is enhanced by introducing a low-pass filter in RC for linear systems, resulting in a filtered repetitive controller (FRC, or filtered repetitive control, which is also designated as FRC). Similarly, it is also necessary to introduce a filter into RC for nonlinear systems. However, the theory on FRC proposed in [1] is derived in the frequency domain and can be applied only with difficulty, if at all, to nonlinear systems.

6.9 Appendix

6.1.1 6.9.1 Proof of Lemma 6.1

Let (6.6) be the original system. To apply additive state decomposition (see Sect. 2.2.3), select the primary system as follows:

$$\begin{aligned} \mathbf {A}_{0,\varepsilon }\dot{\mathbf {x}}_{\text {p}}\left( t\right)&=-\mathbf {x}_{\text {p}}\left( t\right) +\mathbf {A}_{1,\varepsilon } \mathbf {x}_{\text {p}}\left( {t}-{T}\right) +\varvec{\delta }\left( t\right) \nonumber \\ \mathbf {x}_{\text {p}}\left( \theta \right)&=\mathbf {v}\left( T+\theta \right) ,\theta \in \left[ -{T,0}\right] , \end{aligned}$$

(6.35)

where $\mathbf {A}_{0,\varepsilon },\mathbf {A}_{1,\varepsilon }\in \mathbb {R} ^{n\times n}$, $\varvec{\delta }\left( t\right) \in \mathbb {R} ^{n}$. From the original system (6.6) and the primary system (6.35), the secondary system is determined by additive state decomposition as follows:

$$\begin{aligned} -\mathbf {A}_{0,\varepsilon }\dot{\mathbf {x}}_{\text {p}}\left( t\right)&=-\mathbf {x}_{\text {s}}\left( t\right) +\mathbf {x}_{\text {s}}\left( {t} -{T}\right) +\left( {\mathbf {I}}-\mathbf {A}_{1,\varepsilon }\right) \mathbf {x}_{\text {p}}\left( {t}-{T}\right) -\varvec{\delta }\left( t\right) \nonumber \\ \mathbf {x}_{\text {s}}\left( \theta \right)&=\mathbf {0},\theta \in \left[ -{T,0}\right] . \end{aligned}$$

(6.36)

By additive decomposition,

$$\begin{aligned} \mathbf {x}\left( t\right) =\mathbf {x}_{\text {p}}\left( t\right) +\mathbf {x}_{\text {s}}\left( t\right) . \end{aligned}$$

(6.37)

Let $\varvec{\delta }\left( t\right) \equiv \mathbf {x}_{\text {s}}\left( {t-T}\right)$. Then

$$ \mathbf {v}\left( t\right) =\mathbf {x}_{\text {p}}\left( t\right) +\varvec{\delta }\left( t+T\right) . $$

Furthermore, using (6.35), model (6.6) can be written as follows:

$$\begin{aligned} \mathbf {A}_{0,\varepsilon }\dot{\mathbf {x}}_{\text {p}}\left( t\right)&=-\mathbf {x}_{\text {p}}\left( t\right) +\mathbf {A}_{1,\varepsilon } \mathbf {x}_{\text {p}}\left( {t}-{T}\right) +\varvec{\delta }\left( t\right) \nonumber \\ \mathbf {v}\left( t\right)&=\mathbf {x}_{\text {p}}\left( t\right) +\varvec{\delta }\left( t+T\right) \nonumber \\ \mathbf {x}_{\text {p}}\left( \theta \right)&=\mathbf {v}\left( T+\theta \right) ,\theta \in \left[ -{T,0}\right] . \end{aligned}$$

(6.38)

In the following sections, the bound on $\varvec{\delta }\left( t\right)$ will be discussed, $t\in \left[ 0,\infty \right) .$ From (6.36) and (6.37),

$$\begin{aligned} \mathbf {A}_{0,\varepsilon }\dot{\mathbf {x}}_{\text {p}}\left( t\right) =-\mathbf {x}_{\text {p}}\left( t\right) -\left( {\mathbf {I}}-\mathbf {A} _{1,\varepsilon }\right) \mathbf {x}_{\text {p}}\left( {t}-{T}\right) +\mathbf {v}\left( t\right) \end{aligned}$$

(6.39)

If $\mathbf {A}_{0,\varepsilon }>0$ and $\left\| \mathbf {A}_{1,\varepsilon }\right\| <1$, then $\mathbf {A}_{0,\varepsilon }\dot{\mathbf {x}}_{\text {p} }\left( t\right) =-\mathbf {x}_{\text {p}}\left( t\right) -\left( {\mathbf {I}}_{n}-\mathbf {A}_{1,\varepsilon }\right) \mathbf {x}_{\text {p}}\left( {t}-{T}\right)$ is exponentially stable [16]. Consequently, there exists a bounded and T-periodic function $\mathbf {x}_{\text {p}}\left( t\right)$ that satisfies (6.39) [17]. Note that $\varvec{\delta }\left( t\right) \equiv \mathbf {v}\left( t-{T}\right) -\mathbf {x}_{\text {p}}\left( {t} -{T}\right)$ is also a bounded and T-periodic function. Consequently, (6.38) can be rewritten in the form of (6.7). If $\mathbf {A}_{0,\varepsilon }=\mathbf {0}$ and $\mathbf {A}_{1,\varepsilon } =\mathbf {I}_{m}$, then (6.6) can be rewritten as (6.7) with $\varvec{\delta }\left( t\right) \equiv \mathbf {0}.$

6.1.2 6.9.2 Detailed Proof of Theorem 6.1

(i) If $\mathbf {A}_{0,\varepsilon }=\mathbf {0}$, then $\varvec{\delta }\left( t\right) \equiv \mathbf {0}$ by Lemma 6.1. Thus, (6.16) becomes

$$\begin{aligned} \dot{V}\left( t,\mathbf {z}_{t}\right) \le -2k_{\text {e}}\mathbf {e} ^{\text {T}}\left( t\right) \mathbf {M}\left( t\right) \mathbf {e}\left( t\right) . \end{aligned}$$

(6.40)

Then,

$$ \dot{V}\left( t,\mathbf {z}_{t}\right) \le -2c_{3}\left\| \mathbf {e} \left( t\right) \right\| ^{2}. $$

From the above inequality,

$$\begin{aligned} V\left( t,\mathbf {z}_{t}\right) \le V\left( 0,\mathbf {z}_{0}\right) . \end{aligned}$$

(6.41)

Further,

$$ \int _{0}^{t}\left\| \mathbf {e}\left( s\right) \right\| ^{2} \text {d}s\le \frac{1}{2c_{3}}V\left( 0,\mathbf {z}_{0}\right) . $$

Therefore, $\mathbf {e}\in \mathscr {L}_{\infty }\left[ 0,\infty \right) \cap \mathscr {L}_{2}\left[ 0,\infty \right)$. Next, it is proved that $\mathbf {e}\left( t\right)$ is uniformly continuous. If this is true, then $\underset{t\rightarrow \infty }{\lim }\left\| \mathbf {e}\left( t\right) \right\| =0$ by Barbalat’s Lemma. Given that $\mathbf {e}\in \mathscr {L} _{\infty }\left[ 0,\infty \right) ,$ $\mathbf {e}\left( t\right) \in K\subset \mathbb {R} ^{m}\$for a compact set K. By Assumption 6.1, there exists $b_{f},$ $b_{b}>0$ such that $\left\| \mathbf {f}\left( t,\mathbf {e}\left( t\right) \right) \right\| \le b_{f},$ $\mathbf {b}\left( t,\mathbf {e} \left( t\right) \right)$ for all $\left( t,\mathbf {e}\left( t\right) \right) \in \left[ 0,\infty \right) \times K.$ Let $t_{1}\$and$\ t_{2}$ be any real numbers such that $0<t_{2}-t_{1}\le h.$ Then,

$$\begin{aligned} \left\| \mathbf {e}\left( t_{2}\right) -\mathbf {e}\left( t_{1}\right) \right\|&=\left\| \int _{t_{1}}^{t_{2}}\dot{\mathbf {e}}\left( s\right) \text {d}s\right\| \nonumber \\&=\left\| \int _{t_{1}}^{t_{2}}\left( \mathbf {f}\left( s,\mathbf {e} \left( s\right) \right) +\mathbf {b}\left( s,\mathbf {e}\left( s\right) \right) \tilde{\mathbf {v}}\left( s\right) \right) \text {d}s\right\| \nonumber \\&\le b_{f}\left( t_{2}-t_{1}\right) +b_{b}\int _{t_{1}}^{t_{2}}\left\| \tilde{\mathbf {v}}\left( s\right) \right\| \text {d}s. \end{aligned}$$

(6.42)

Note that, by (6.41), $V\left( t,\mathbf {z}_{t}\right) \$is bounded for$\$all$\ t>0$ with respect to the bound $b_{V}>0$. Hence,

$$ \int _{t-T}^{t}\left\| \tilde{\mathbf {v}}\left( s\right) \right\| ^{2}\text {d}s\le b_{V} $$

for$\$all$\ t>0.$ Thus,

$$\begin{aligned} \int _{t_{1}}^{t_{2}}\left\| \tilde{\mathbf {v}}\left( s\right) \right\| ^{2}\text {d}s\le Nb_{V}, \end{aligned}$$

(6.43)

where $N=\left\lfloor \left( t_{2}-t_{1}\right) / T\right\rfloor +1\$and $\left\lfloor \left( t_{2}-t_{1}\right) / T\right\rfloor$ represents the floor integer of $\left( t_{2}-t_{1}\right) / T$. Using the Cauchy–Schwarz inequality,

$$ \int _{t_{1}}^{t_{2}}\left\| \tilde{\mathbf {v}}\left( s\right) \right\| \text {d}s\le \left( \int _{t_{1}}^{t_{2}}1^{2}\text {d}s\right) ^{\frac{1}{2} }\left( \int _{t_{1}}^{t_{2}}\left\| \tilde{\mathbf {v}}\left( s\right) \right\| ^{2}\text {d}s\right) ^{\frac{1}{2}}. $$

Consequently, $\left\| \mathbf {e}\left( t_{2}\right) -\mathbf {e}\left( t_{1}\right) \right\|$ in (6.42) is further bounded by

$$ \left\| \mathbf {e}\left( t_{2}\right) -\mathbf {e}\left( t_{1}\right) \right\| \le b_{f}h+h_{1}\sqrt{h}, $$

where (6.43) is utilized and $h_{1}=\sqrt{Nb_{V}}b_{b}$. Therefore, for any $\varepsilon >0$, there exists

$$ h=\left( \frac{-h_{1}+\sqrt{h_{1}^{2}+b_{f}\varepsilon }}{2b_{f}}\right) ^{2}>0 $$

such that $\left\| \mathbf {e}\left( t_{2}\right) -\mathbf {e}\left( t_{1}\right) \right\| <\varepsilon$ for any $0<t_{2}-t_{1}<h$. This implies that $\mathbf {e}\left( t\right)$ is uniformly continuous.

(ii) By considering the form of $V\left( t,\mathbf {z}_{t}\right)$ in (6.14), the inequality

$$\begin{aligned} \gamma _{1}\left\| \mathbf {z}\left( t\right) \right\| ^{2}\le V\left( t,\mathbf {z}_{t}\right) \le \gamma _{2}\left\| \mathbf {z}\left( t\right) \right\| ^{2}+\int _{t-T}^{t}{\left\| \mathbf {z}\left( s\right) \right\| ^{2}}\text {{d}}{s} \end{aligned}$$

(6.44)

is satisfied with $\gamma _{1}=\min \left( 2k_{\text {e}}c_{1},\lambda _{\min }\left( \mathbf {A}_{0,\varepsilon }\right) \right)$ and $\gamma _{2} =\max \left( 2k_{\text {e}}c_{2},\lambda _{\max }\left( \mathbf {A}_{0,\varepsilon }\right) \right) .$ If $\mathbf {A}_{0,\varepsilon }>0$, then $\varvec{\delta }\in \mathscr {L}_{\infty }\left[ 0,\infty \right)$ by Lemma6.1. Then, inequality (6.16) becomes

$$ \dot{V}\left( t,\mathbf {z}_{t}\right) \le -2\kappa _{1}\left\| \mathbf {z}\left( t\right) \right\| ^{2}+2\kappa _{2}\left\| \mathbf {z}\left( t\right) \right\| , $$

where $\kappa _{1}=\min \left( \underline{\lambda }_{M}k_{\text {e}},\frac{1}{2}\lambda _{\min }\left( \mathbf {A}_{0,\varepsilon }\right) \right) \$and $\kappa _{2}=\left\| \varvec{\delta }\right\| _{\infty }.$ Given that $\kappa _{1}>0$, the following inequality holds:

$$ -\kappa _{1}\left\| \mathbf {z}\left( t\right) \right\| ^{2}+2\kappa _{2}\left\| \mathbf {z}\left( t\right) \right\| -\kappa _{1}^{-1} \kappa _{2}^{2}\le 0. $$

Then,

$$\begin{aligned} \dot{V}\left( t,\mathbf {z}_{t}\right) \le -\gamma _{3}\left\| \mathbf {z}\left( t\right) \right\| ^{2}+\sigma \end{aligned}$$

(6.45)

is satisfied with $\gamma _{3}=\kappa _{1}$ and $\sigma =\kappa _{1}^{-1} \kappa _{2}^{2}.$ Then, for any value of $\mathbf {z}_{0},$ the solutions to Eq. (6.13) are uniformly ultimately bounded according to Corollary2.1. This can conclude this proof by noting that $\left\| \mathbf {e}\left( t\right) \right\| \le \left\| \mathbf {z}\left( t\right) \right\|$ based on the definition of $\mathbf {z}\left( t\right)$.

6.1.3 6.9.3 Proof of Lemma 6.2

Before proving Lemmas 6.2, 6.3 is first introduced.

Lemma 6.3

Given vectors $\mathbf {a},\mathbf {b\in } \mathbb {R} ^{m}$, suppose that $\mathbf {a}=$sat$\left( \mathbf {a}\right) .$ Then

$$\begin{aligned} \left( \mathbf {a}-\mathrm{sat}\left( \mathbf {b}\right) \right) ^{\mathrm{T} }\left( \mathbf {b}-\mathrm{sat}\left( \mathbf {b}\right) \right) \le 0. \end{aligned}$$

(6.46)

Proof

Let $\mathbf {a}$ $\mathbf {=}$ $\mathbf {[}a_{1}$ $\cdots$ $a_{m}\mathbf {]}^{\text {T}}$ and $\mathbf {b}$ $\mathbf {=}$ $\mathbf {[}b_{1}$ $\cdots$ $b_{m}\mathbf {]}^{\text {T}}.$ Then,

$$\begin{aligned} \left( \mathbf {a}-\text {sat}\left( \mathbf {b}\right) \right) ^{\text {T} }\left( \mathbf {b}-\text {sat}\left( \mathbf {b}\right) \right) =\underset{i=1}{\overset{m}{ {\displaystyle \sum } }}\left( a_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) ^{\text {T}}\left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) . \end{aligned}$$

(6.47)

In the following section, the term $\left( a_{i}-\text {sat}_{\beta _{i} }\left( b_{i}\right) \right) ^{\text {T}}\left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right)$ is considered. It is claimed that

$$\begin{aligned} \left( a_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) ^{\text {T} }\left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) \le 0. \end{aligned}$$

(6.48)

With it, inequality (6.46) holds according to (6.47). Next, inequality (6.48) will be proved. For the case $\left| b_{i}\right| \le \beta _{i},$ equality (6.48) holds due to $b_{i}-$sat$_{\beta _{i}}\left( b_{i}\right) =0$. There is a strict inequality when $\left| b_{i}\right| >\beta _{i}.$ It follows for the case $b_{i}<-\beta _{i}$ that

$$ \left( a_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) \left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) =\left( a_{i}+\beta _{i}\right) \left( b_{i}+\beta _{i}\right) . $$

Given that $a_{i}+\beta _{i}>0$ according to $\mathbf {a}=$sat$\left( \mathbf {a}\right)$ and $b_{i}+\beta _{i}<0,$ $\left( a_{i}-\text {sat} _{\beta _{i}}\left( b_{i}\right) \right) \left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) <0.$ For the case $b_{i}>\beta _{i}$, it follows that

$$ \left( a_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) \left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) =\left( a_{i}-\beta _{i}\right) \left( b_{i}-\beta _{i}\right) . $$

Given that $a_{i}-\beta _{i}<0$ according to $\mathbf {a}=$sat$\left( \mathbf {a}\right)$ and $b_{i}-\beta _{i}>0,$ $\left( a_{i}-\text {sat} _{\beta _{i}}\left( b_{i}\right) \right) \left( b_{i}-\text {sat}_{\beta _{i}}\left( b_{i}\right) \right) <0.$ Therefore, (6.48) holds regardless of $b_{i}$.       $\square$

With Lemma 6.3, the proof of Lemma 6.2 is now possible. For simplicity, denote $\tilde{\mathbf {v}}\left( t-T\right)$ by $\tilde{\mathbf {v}}_{-T}\left( t\right) .$ If $\mathbf {A}_{0,\varepsilon }=\mathbf {0}$ and $\mathbf {A}_{1,\varepsilon }=\mathbf {I}_{m},$ then $\mathbf {x}_{\text {p} }=\mathbf {v}\$and$\ \varvec{\delta }=\mathbf {0}.$ By (6.17),

$$ \mathbf {0}=-\hat{\mathbf {v}}^{*}\left( t\right) +\hat{\mathbf {v}}\left( t-T\right) +k_{\text {e}}\mathbf {R}^{\text {T}}\left( t\right) \mathbf {e} \left( t\right) . $$

On the other hand,

$$ \mathbf {0}=\mathbf {x}_{\text {p}}\left( t\right) -\mathbf {x}_{\text {p} }\left( t-T\right) . $$

Then,

$$\begin{aligned} \tilde{\mathbf {v}}_{-T}&=\tilde{\mathbf {v}}^{*}+k_{\text {e}} \mathbf {R}^{\text {T}}\mathbf {e}\\&=\tilde{\mathbf {v}}+\left( \left( \hat{\mathbf {v}}-\hat{\mathbf {v}} ^{*}\right) +k_{\text {e}}\mathbf {R}^{\text {T}}\mathbf {e}\right) . \end{aligned}$$

Consequently,

$$\begin{aligned} \tilde{\mathbf {v}}_{-T}^{\text {T}}\tilde{\mathbf {v}}_{-T}&=\tilde{\mathbf {v}}^{\text {T}}\tilde{\mathbf {v}}+\left( \hat{\mathbf {v}}-\hat{\mathbf {v}}^{*}+k_{\text {e}}\mathbf {R}^{\text {T}}\mathbf {e}\right) ^{\text {T} }\left( \hat{\mathbf {v}}-\hat{\mathbf {v}}^{*}+k_{\text {e}}\mathbf {R} ^{\text {T}}\mathbf {e}\right) +2\tilde{v}^{\text {T}}\left( \hat{\mathbf {v} }-\hat{\mathbf {v}}^{*}+k_{\text {e}}\mathbf {R}^{\text {T}}\mathbf {e}\right) \\&=\tilde{\mathbf {v}}^{\text {T}}\tilde{\mathbf {v}}+\left( \hat{\mathbf {v} }-\hat{\mathbf {v}}^{*}+k_{\text {e}}\mathbf {R}^{\text {T}}\mathbf {e}\right) ^{\text {T}}\left( \hat{\mathbf {v}}-\hat{\mathbf {v}}^{*}+k_{\text {e} }\mathbf {R}^{\text {T}}\mathbf {e}\right) \\&\text { }-2\left( \mathbf {v}-\text {sat}\left( \hat{\mathbf {v}}^{*}\right) \right) ^{\text {T}}\left( \hat{\mathbf {v}}^{*}-\text {sat} \left( \hat{\mathbf {v}}^{*}\right) \right) +2\tilde{\mathbf {v} }^{\text {T}}k_{\text {e}}\mathbf {R}^{\text {T}}\mathbf {e}. \end{aligned}$$

Furthermore, given that $\left( \mathbf {v}-\text {sat}\left( \hat{\mathbf {v} }^{*}\right) \right) ^{\text {T}}\left( \hat{\mathbf {v}}^{*}-\text {sat}\left( \hat{\mathbf {v}}^{*}\right) \right) \le 0$ by Lemma 6.3, it follows that

$$ \tilde{\mathbf {v}}^{\text {T}}\tilde{\mathbf {v}}-\tilde{\mathbf {v}} _{-T}^{\text {T}}\tilde{\mathbf {v}}_{-T}\le -2k_{\text {e}}\tilde{\mathbf {v} }^{\text {T}}\mathbf {R}^{\text {T}}\mathbf {e}. $$