Tuning of active disturbance rejection control for differentially flat systems under an ultimate boundedness analysis: a unified integer-fractional approach

Abstract

An adaptation of the active disturbance rejection control (ADRC) approach is applied to a fractional-order system with a flat output. Albeit the rather scarce information about the system, the conditions to establish an ultimate bound for a specific configuration of the system are found and compiled in a guideline for the tuning of the observer implemented in the ADRC.

Appendices

Appendix A: concepts on fractional-order systems

A generalization of the integer-order dynamic systems is found in fractional-order systems (FOS), providing better mathematical tools for the modeling of physical and engineering systems [22]. For this matter, it is necessary to give some definitions taken from the fractional calculus [27].

Definition 1

The Caputo fractional derivative of order $\alpha$ applied to the function f(t) is defined as

$$\begin{aligned} _{t_0}D^{\alpha }_{t} = \dfrac{1}{\varGamma (m - \alpha )} \int _{t_0}^{t} \dfrac{g^{(m)}(\tau )}{(t - \tau )^{\alpha - m +1}}\mathrm{d}\tau , \end{aligned}$$

(38)

where $m - 1< \alpha < m$, the term m is the first integer that is not less than $\alpha$, and $\varGamma$ is the Gamma function:

$$\begin{aligned} \varGamma (z) = \int ^{\infty }_{0} t^{z-1} \mathrm {e}^{-t} \mathrm{d}t. \end{aligned}$$

The Caputo fractional derivative is chosen, because its properties facilitates its application to dynamic systems [27].

Definition 2

The two-parameter Mittag–Leffler function is defined as

$$\begin{aligned} E_{\alpha , \beta } (z) = \textstyle \sum \limits _{k = 0}^{\infty } \dfrac{z^k}{\varGamma (\alpha k + \beta )},~~\alpha> 0, ~~ \beta > 0. \end{aligned}$$

(39)

The Laplace transform of Caputo fractional derivative takes the form:

$$\begin{aligned} \int _0^{\infty }\mathrm {e}^{-st}{}_{0}D_t^{\alpha } f(t)\mathrm{d}t&= s^{\alpha }F(s) - \textstyle \sum \limits _{k=0}^{n-1} s^{\alpha - k -1} f^{(m)}(0), \nonumber \\&\quad n-1 \leqslant \alpha < n. \end{aligned}$$

(40)

For zero initial conditions, (40) simplifies itself to:

$$\begin{aligned} {\mathcal {L}}\{ _{0}D_t^\alpha f(t) \} = s^\alpha F(s). \end{aligned}$$

(41)

For the sake of brevity, the $\alpha$-order Caputo derivative for a f(t) function will be represented as $f^{(\alpha )}$ with $t_0 = 0$ and $0<\alpha \leqslant 1$.

In the case of commensurate order systems, the transfer function takes the following form:

$$\begin{aligned} G(s) = K_0 \dfrac{\textstyle\sum \limits _{k=0}^{M}b_k(s^{\alpha })^k}{\textstyle\sum \limits _{k=0}^{N}a_k(s^{\alpha })^k} = K_0 \dfrac{N(s^{\alpha })}{D(s^{\alpha })}. \end{aligned}$$

(42)

If $N > M$, the function G(s) is named as a proper rational function in the variable $s^{\alpha }$.

Theorem 2

A commensurate order system under the mapping $w = s^{\alpha }$, and ${|{\text {arg}(w_i)}|} = {|{\phi _i}|}$, where $w_i$ is the ith root of $D(s^{\alpha } = w)$, is stable if and only if ${|{\phi _i}|}>\small {\dfrac{\pi }{2}\alpha }$.

According to [28], the final value theorem (FVT) is applicable under stability conditions, even when there is a branch point at $s=0$ for a FOS f(t):

$$\begin{aligned} \lim _{t \rightarrow \infty } f(t)= \lim _{s \rightarrow 0} sF(s). \end{aligned}$$

(43)

Appendix B: Proof of the positiveness of $g_j(t)$

A fundamental aspect for the proof of Theorem 2 is the positiveness of $g_j(t)$, i.e., $g_j(t) \geqslant 0$ for $t \geqslant 0$. The expression in (17) shows with ease that if every product $C_i\,f_i$ is non-negative, $g_i(t)$ is also non-negative. Therefore, the cases for the function $f_i$ and the constant $C_i$ to be non-negative are shown.

1.1 Positiveness of $f_i$

Consider the inverse Laplace transform $f_{k}(t)={\mathcal {L}}^{-1}\bigg \{ \dfrac{1}{\prod \nolimits _{i=0}^{k}(s^{\alpha }+\lambda _i)}\bigg \}$ with $k \in {\mathbb {N}}$ and $0<\alpha \leqslant 1$.

For $k=1$,

$$\begin{aligned} f_{1}(t)={\mathcal {L}}^{-1}\bigg\{ \dfrac{1}{\left( s^{\alpha }+\lambda _1\right) }\bigg\} =t^{\alpha -1}E_{\alpha ,\alpha }(-t^{\alpha }\lambda _1), \end{aligned}$$

(44)

where

$$\begin{aligned} E_{\alpha ,\alpha }(x)=\textstyle \sum \limits _{k=0}^{\infty }\dfrac{x^{k}}{\varGamma (\alpha k+{{\alpha }})}. \end{aligned}$$

(45)

The term $t^{\alpha -1}>0$ when $t>0$, and the Mittag–Leffler function $E_{\alpha ,\alpha }(-t^{\alpha }\lambda )$ is positive for all $t>0$ and zero for $t<0$ [29].

For $k=2$,

$$\begin{aligned} f_{2}(t)={\mathcal {L}}^{-1}\bigg\{ \dfrac{1}{(s^{\alpha }+\lambda _{1})(s^{\alpha }+\lambda _{2})}\bigg\} =f_{21}(t)*f_{22}(t), \end{aligned}$$

(46)

where $*$ is the convolution operator. Since $f_{21}(t)$ and $f_{22}(t)$ are positive for all $t>0$, (46) is positive for all $t>0$. The same procedure is used for $k = n + m$ concluding that $f_{k}(t)>0$ for all $t>0$.

1.2 Positiveness of $C_i$

Consider the transfer function $G_j(s)$ defined in (15). Its numerator

$$\begin{aligned} N_j(s) = s^{\left( j-1\right) \alpha }+l_{n+m-1}s^{\left( j-2\right) \alpha }+\cdots +l_{n+m-j+1} \end{aligned}$$

(47)

can be written as

$$\begin{aligned} \begin{aligned} N_{j}(s)&=c_{j-1}(s^{\alpha }+\lambda _{1})(s^{\alpha }+\lambda _{2})\dots (s^{\alpha }+\lambda _{j-2})(s^{\alpha }+\lambda _{j-1})\\&\quad +c_{j-2}(s^{\alpha }+\lambda _{1})(s^{\alpha }+\lambda _{2})\dots (s^{\alpha }+\lambda _{j-2})\\&\quad +\cdots +c_{1}(s^{\alpha }+\lambda _{1})\\&\quad +c_{0}, \end{aligned} \end{aligned}$$

(48)

where the coefficients $c_{j-k}$ are given by

$$\begin{aligned} c_{j-k}={\left\{ \begin{array}{ll} \sum \limits _{A_{k-1}\subset M_{k}}\Big[ \textstyle \prod \limits _{i\epsilon A_{k-1}}\lambda _{i}\Big] , &{} 2\leqslant k\leqslant j,\\ 1, &{} k=1, \end{array}\right. } \end{aligned}$$

(49)

where $M_{k}=\left\{ j-k+2,j-k+3,j-k+4,\ldots ,n+m\right\}$ and $A_{k-1}$ is a $k-1$ element subset of $M_k$, so the sum is made over all $k-1$ element subsets of $M_k$. Given that the Assumption 3 holds, all the $c_{j-k}$ are positive real numbers.

The proof of this is built by strong induction, matching the resulting $c_{j-k}$ coefficients with the respective coefficients $l_k$ seen in (14).

The coefficient $c_{j-1}$ is the only coefficient that companions the term $s^{(j-1)\alpha }$ in $N_j(s)$, which is a monic polynomial according to (14); therefore, $c_{j-1}$ is equal to 1.

The coefficient $c_{j-2}$ must be calculated to ensure that the coefficient of $s^{j-2}$ in $N_j(s)$ is equal to $l_{m+n-1}$. Therefore,

$$\begin{aligned} l_{m+n-1}=\textstyle\sum \limits _{a=1}^{n+m}\lambda _{a}=\sum \limits _{a=1}^{j-1}\lambda _{a}+c_{j-2}. \end{aligned}$$

(50)

Solving for $c_{j-2}$ from (50), there is

$$\begin{aligned} c_{j-2}=\textstyle \sum \limits _{a=1}^{n+m}\lambda _{a}-\textstyle \sum \limits _{a=1}^{j-1}\lambda _{a} =\textstyle \sum \limits _{a=j}^{n+m}\lambda _{a}=\textstyle \sum \limits _{A_{1}\subset M_{2}}\Big[ \prod \limits _{i\epsilon A_{1}}\lambda _{i}\Big] , \end{aligned}$$

(51)

where $M_{2}=\left\{ j,j+1,j+2,\ldots ,n+m\right\}$ and $A_{1}$ is a single element of $M_{2}$, resolving this way $c_{j-2}$.

To proof the validity of (49) for the coefficients $c_{j-k}$ with $k>2$, suppose that is true for $2\leqslant k\leqslant k_{0}-1$, and let prove it true for $k_{0}=k$.

First, the coefficient of $s^{j-k}$ in (48) is calculated and labeled as C resulting in

$$\begin{aligned} C=\textstyle \sum \limits _{k=1}^{k_{0}-1}c_{j-k}\Big[ \textstyle \sum \limits _{B_{k_{0}-k}\subset S_{k}} \Big[ \prod \limits _{i\epsilon B_{k_{o}-k}}\lambda _{i}\Big] \Big] +c_{j-k_{0}}, \end{aligned}$$

(52)

where $S_{k}=\left\{ 1,2,3,\ldots ,j-k\right\}$ and $B_{k_{0}-k}$ is a $k_{0}-k$ element subset of $S_{k}$.

Now, C must be equal to $l_{n+m-k+1}$. Therefore,

$$\begin{aligned} l_{n+m-k_{0}+1}=\textstyle \sum \limits _{A_{k_{0}-1}\subset E}\Big[ \textstyle \prod \limits _{i\epsilon A_{k_{0}-1}}\lambda _{i}\Big] =C, \end{aligned}$$

(53)

where $E=\left\{ 1,2,3,\ldots ,n+m\right\}$.

Solving for $c_{j-k_{0}}$, there is

$$\begin{aligned} c_{j-k_{0}}&=\textstyle \sum \limits _{A_{k_{0}-1}\subset E}\Big[ \textstyle \prod \limits _{i\epsilon A_{k_{0}-1}}\lambda _{i}\Big] \nonumber \\&\quad -\textstyle \sum \limits _{k=1}^{k_{0}-1}c_{j-k}\Big[ \textstyle \sum \limits _{B_{k_{0}-k}\subset S_{k}} \Big[ \textstyle \prod \limits _{i\epsilon B_{k_{0}-k}}\lambda _{i}\Big] \Big] . \end{aligned}$$

(54)

The task now is to prove that (54) is equivalent to (49). Consider first the product $\prod \nolimits _{i\epsilon A_{k_{o}-1}}\lambda _{i}$, which is generated by choosing a $k_{0}-1$ element subset of indexes from E and multiplying every $\lambda _{i}$ in it. That means that each $k_{0}-1$ element subset of E defines one and only one term in the sum $\textstyle \sum _{A_{k_{0}-1}\subset E}[\prod \nolimits _{i\epsilon A_{k_{0}-1}}\lambda _{i}]$. Thus, every term involved in this sum is generated by the set of all $k_{0}-1$ element subset of E.

By induction hypothesis, the term $c_{j-k}\Big [\sum \nolimits _{B_{k_{0}-k}\subset S_{k}}\big [\prod \nolimits _{i\epsilon B_{k_{0}-k}}\lambda _{i}\big ]\Big ]$ can be written as:

$$\begin{aligned}&c_{j-k}\Big[ \textstyle \sum \limits _{A_{k-1}\subset S_{k}}\Big[ \textstyle \prod \limits _{i\epsilon A_{k-1}}\lambda _{i}\Big] \Big] \nonumber \\& =\Big( \textstyle \sum \limits _{A_{k-1}\subset M_{k}}\Big[ \textstyle \prod \limits _{i\epsilon A_{k-1}}\lambda _{i}\Big] \Big) \Big( \Big[ \textstyle \sum \limits _{B_{k_{0}-k}\subset S_{k}}\Big[ \textstyle \prod \limits _{i\epsilon B_{k_{0}-k}}\lambda _{i}\Big] \Big] \Big) . \end{aligned}$$

(55)

Each term involved in the resulting sum of (55) can be generated by choosing $k-1$ elements from the set $M_{k}$ and $k_{0}-k$ elements from the set $S_{k}$. Then, the resulting set of indexes defines the elements that must be multiplied. Thus, each set of the form $N_{k}=A_{k-1}\cup B_{k_{0}-k}$ defines exactly one term in (55).

According to the term $\sum \nolimits _{k=1}^{k_{0}-1}c_{j-k}\Big [\sum \nolimits _{B_{k_{0}-k}\subset S_{k}}\big [\prod \nolimits _{i\epsilon B_{k_{0}-k}}\lambda _{i}\big ]\Big ]$, it is concluded that each term involved in this sum is defined by exactly one of the $N_{k}$ for some value of k, such that $1\leqslant k\leqslant k_{0}-1$ and some sets $A_{k-1}$ and $B_{k_{0}-k}$.

Based on the previous analysis, to prove that (54) is equivalent to (49), it is necessary to prove that

$$\begin{aligned} P_{M_{k_{0}}}=P_{E}-D_{k_{0}-1}, \end{aligned}$$

(56)

where

$$\begin{aligned} &P_{M_{k_{0}}}=\{ A_{k_{0}-1}\,|\, A_{k_{0}-1}\subset M_{k_{0}}\}, \\ &P_{E}=\{ A_{k_{0}-1}\,|\, A_{k_{0}-1}\subset E\},\\ &D_{k_{0}-1}=\{ N_{k}\,|\, N_{k}=A_{k-1}\cup B_{k_{0}-k}\} \end{aligned}$$

with $A_{k-1}\subset M_{k}$, $B_{k_{0}-k}\subset S_{k}$, and $1\leqslant k\leqslant k_{0}-1$.

This means that the set of all the $k_{0}-1$ element subsets of $M_{k_{0}}$ can be obtained by taking the set of the $k_{0}-1$ element subsets of E and removing the subsets that can be formed by taking $k-1$ elements from $M_{k}$, and $k_{0}-k$ elements from $S_{k}$, where $1\leqslant k\leqslant k_{0}-1$.

To prove the validity of (56), it will be shown that $P_{M_{k_{0}}}\subset P_{E}-D_{k_{0}-1}$ and $P_{E}-D_{k_{0}-1}\subset P_{M_{k_{0}}}$.

Consider first a set $A_{k_{0}-1}$ which belongs to $P_{M_{k_{0}}}$, this set has $k_{0}-1$ elements, and they all belong to the set $M_{k_{0}}=\left\{ j-k_{0}+2,j-k_{0}+3,\ldots ,n+m\right\}$. It is clear that this set also belongs to $P_{E}$, because $P_{E}$ is the set of all $k_{0}-1$ element subsets of E and $M_{k_{0}}\subset E$. Suppose now that $A_{k_{0}-1}$ belongs to $D_{k_{0}-1}$, and then, $A_{k_{0}-1}$ has $k_{0}-1$ elements that belong to $M_{k_{0}}$, $k-1$ elements that belong to $M_{k}$, and $k_{0}-k$ elements from $S_{k}$ for some $1\leqslant k\leqslant k_{0}-1$. Hence, it has $k_{0}-k$ elements from the set $\left\{ j-k_{0}+2,j-k_{0}+3,\ldots ,j-k+1\right\}$, but this set has exactly $k_{0}-k$ elements, so all its elements must belong to $A_{k_{0}-1}$. However, it is impossible, because $A_{k_{0}-1}$ must have $k_{0}-k$ elements from $S_{k}=\left\{ 1,2,3,\ldots ,j-k\right\}$, and this set excludes the element $j-k+1$. Therefore, $A_{k_{0}-1}$ does not belong to $D_{k_{0}-1}$, and $A_{k_{0}-1}\in P_{E}-D_{k_{0}-1}$. This implies that all the elements that belong to $P_{M_{k_{0}}}$ belong to $P_{E}-D_{k_{0}-1}$, and then, $P_{M_{k_{0}}}\subset P_{E}-D_{k_{0}-1}$.

Consider now a set $A_{k_{0}-1}$ which belongs to $P_{E}-D_{k_{0}-1}$, and suppose that $A_{k_{0}-1}\notin P_{M_{k_{0}}}$. Then, $A_{k_{0}-1}$ includes at least one element from the set $\{1, 2, 3, \ldots , j-k_{0}+1 \}$, but $A_{k_{0}-1}$ does not belong to $D_{k_{0}-1}$. Now, suppose that $A_{k_{0}-1}$ includes $l_{1}$ elements from $\left\{ 1,2,3,\ldots ,j-k_{0}+1\right\}$, and it includes all the elements from the set $\Omega =\left\{ j-k_{0}+l_{1}+1, j-k_{0}+l_{1}+2, \ldots , j-1\right\}$. In this case, $A_{k_{0}-1}$ has $k_{0}-1$ elements from $S_{1}=\left\{ 1, 2, 3,\ldots , j-1\right\}$ and no element from $M_{1}$, so it belongs to $D_{k_{0}-1}$, which is a contradiction. Therefore, $A_{k_{0}-1}$ must exclude at least one element from $\Omega$. Suppose that $j-k_{0}+l_{1}+\alpha$ is the first element in $\Omega$, which is not included in $A_{k_{0}-1}$, if $\alpha =1$, $A_{k_{0}-1}$ has $l_{1}$ elements from $S_{k_{0}-l_{1}}$ and $k_{0}-1-l_{1}$ elements from $M_{k_{0}-l_{1}}$, so $A_{k_{0}-1}\in D_{k_{0}-1}$, and this is a contradiction. If $2\leqslant \alpha \leqslant k_{0}-l_{1}-1$, then $A_{k_{0}-1}$ has $l_{1}+\alpha -2$ elements from $S_{k_{0}-l_{1}-\alpha +2}$ and $k_{0}-l_{1}-\alpha +1$ elements from $M_{k_{0}-l_{1}-\alpha +2}$, so $A_{k_{0}-1}\in D_{k_{0}-1}$ which is another contradiction. Therefore, it is concluded that $A_{k_{0}-1}\in P_{M_{k_{0}}}$, and subsequently $P_{E}-D_{k_{0}-1}\subset P_{M_{k_{0}}}$ and $P_{M_{k_{0}}}=P_{E}-D_{k_{0}-1}$, which is the expression in (56) that proves the validity of (54), and finishing the proof of (49) by strong induction.