On tuning of ADRC with competing design indices: a quantitative study

Abstract

Controller optimization has mostly been done by minimizing a certain single cost function. In practice, however, engineers must contend with multiple and conflicting considerations, denoted as design indices (DIs) in this paper. Failure to account for such complexity and nuances is detrimental to the applications of any advanced control methods. This paper addresses this challenge heads on, in the context of active disturbance rejection controller (ADRC) and with four competing DIs: stability margins, tracking, disturbance rejection, and noise suppression. To this end, the lower bound for the bandwidth of the extended state observer is first established for guaranteed closed-loop stability. Then, one by one, the mathematical formula is meticulously derived, connecting each DI to the set of controller parameters. To our best knowledge, this has not been done in the context of ADRC. Such formulas allow engineers to see quantitatively how the change of each tuning parameter would impact all of the DIs, thus making the guesswork obsolete. An example is given to show how such analytical methods can help engineers quickly determine controller parameters in a practical scenario.

Appendix

Proof of Theorem 1 Considering unknown gain $b\in [\underline{b}, \overline{b}]$, based on the analyses about the PID parameters selection manifold $\varOmega _{pid}$ in [35] and the ESO bandwidth in [16], define

$$\begin{aligned} \varOmega _{b{\hat{b}}}=\{\omega \in R \mid {\tilde{h}}_0\omega ^4 +{\tilde{h}}_1\omega ^3 +{\tilde{h}}_2\omega ^2 +{\tilde{h}}_3\omega +{\tilde{h}}_4=0\},\nonumber \\ \end{aligned}$$

(A.1)

where ${\tilde{h}}_0{=}\frac{b^4k_{d}^2}{{\hat{b}}^4}$, ${\tilde{h}}_1{=}2\frac{b^3k_{d}}{{\hat{b}}^3}\left[ k_{d}\left( \frac{bk_{d}}{{\hat{b}}}{-}M_2\right) {+}k_{p}\left( \frac{b}{{\hat{b}}}{-}1\right) -M_1\right] ,$ ${\tilde{h}}_2=\left[ \frac{bk_{d}}{{\hat{b}}}\left( \frac{bk_{d}}{{\hat{b}}}-M_2\right) +\frac{bk_{p}}{{\hat{b}}}\left( \frac{b}{{\hat{b}}}-1\right) -\frac{bM_1}{{\hat{b}}}\right] ^2+ 2\frac{b^2k_{d}}{{\hat{b}}^2}\left( \frac{bk_{p}}{{\hat{b}}}-M_1\right) \left( \frac{bk_{d}}{{\hat{b}}}-M_2\right) -\frac{b^2M_2^2k_{p}}{{\hat{b}}^2},$ ${\tilde{h}}_3{=}2\left[ \frac{bk_{d}}{{\hat{b}}}\left( \frac{bk_{d}}{{\hat{b}}} -M_2\right) {+}\frac{bk_{p}}{{\hat{b}}}\left( \frac{b}{{\hat{b}}}{-}1\right) {-}\frac{bM_1}{{\hat{b}}}\right] \left( \frac{bk_{p}}{{\hat{b}}}-M_1\right) \left( \frac{bk_{d}}{{\hat{b}}} -M_2\right) -\frac{bM_2^2k_{p}}{{\hat{b}}}\left( \frac{bk_{d}}{{\hat{b}}}-M_2\right) ,$ ${\tilde{h}}_4=\left( \frac{bk_{p}}{{\hat{b}}}-M_1\right) ^2\left( \frac{bk_{d}}{{\hat{b}}}-M_2\right) ^2$. Then, define

$$\begin{aligned} \begin{aligned} \omega _{ob{\hat{b}}}^*&=\max \{0, \frac{\frac{{\hat{b}}M_1}{b}-k_{p}}{k_{d}}, \frac{{\hat{b}}M_2}{b}-k_{d}, {\bar{\omega }}_{ob{\hat{b}}}\},\\ {\bar{\omega }}_{ob{\hat{b}}}&= \left\{ \begin{array}{ll} 0, &{}~ \varOmega _{b{\hat{b}}}=\emptyset \text { or } \max _{\omega \in \varOmega _{b{\hat{b}}}}\omega \le 0,\\ \max _{\omega \in \varOmega _{b{\hat{b}}}}\omega , &{}~ \max _{\omega \in \varOmega _{b{\hat{b}}}}\omega >0. \end{array} \right. \end{aligned}\nonumber \\ \end{aligned}$$

(A.2)

According to Theorem 1 in [16], consider the ADRC controlled systems (1), (5) and (6), for any given $M_1$, $M_2$, $k_{p}$, $k_{d},$ b, ${\hat{b}}$, when the ESO parameter $\omega _{o}>\omega _{ob{\hat{b}}}^*$, the states $x_1$ and $x_2$ will satisfy $\lim \nolimits _{t \rightarrow \infty }x_1(t)=c$, $\lim \nolimits _{t \rightarrow \infty }x_2(t)=0,$ for any $(x_1(0), x_2(0)) \in R^2$ and constant c. And then, for any $b\in [\underline{b},\overline{b}]$, $\omega _{o\underline{b}{\hat{b}}}^*\ge \omega _{ob{\hat{b}}}^*$ will be proved.

Define $k(b,{\hat{b}})=(b{\hat{k}}_p-M_1)(b{\hat{k}}_d-M_2)-b{\hat{k}}_i- M_2\sqrt{b{\hat{k}}_i(b{\hat{k}}_d+M_2)},$ where ${\hat{k}}_p=\frac{1}{{\hat{b}}}\left( k_{p}+\omega k_{d}\right)$, ${\hat{k}}_d=\frac{1}{{\hat{b}}}(\omega + k_{d})$, ${\hat{k}}_i=\frac{1}{{\hat{b}}}\omega k_{p}.$ Thus, $\varOmega _{b{\hat{b}}}=\{\omega \in R \mid k(b,{\hat{b}})=0\}.$

If $\omega _{o}>\omega _{o\underline{b}{\hat{b}}}^*$, it can be obtained that $k(\underline{b},{\hat{b}})>0$, $\underline{b}{\hat{k}}_p$ $>M_1$ and $\underline{b}{\hat{k}}_d>M_2$. Therefore, there is $2b{\hat{k}}_p{\hat{k}}_d{-}$ $(M_1{\hat{k}}_d{+} M_2{\hat{k}}_p){-} {\hat{k}}_i {\ge }\left( b{\hat{k}}_p{\hat{k}}_d{-}\frac{M_1M_2}{\underline{b}}\right) {+} \frac{1}{\underline{b}} \left[ (\underline{b}{\hat{k}}_p{-}M_1)(\underline{b}{\hat{k}}_d{-}M_2){-} \underline{b}{\hat{k}}_i\right] {\ge } \sqrt{{\hat{k}}_i}M_2\frac{ \sqrt{\underline{b}( \underline{b}{\hat{k}}_d{+} M_2)}}{\underline{b}}$.

The following equation can be obtained by taking the partial derivative of $k(b,{\hat{b}})$. $\frac{\partial k(b,{\hat{b}})}{\partial b} =2b{\hat{k}}_p{\hat{k}}_d- {\hat{k}}_i- M_2\frac{2b{\hat{k}}_i{\hat{k}}_d+ {\hat{k}}_iM_2}{2\sqrt{b{\hat{k}}_i(b{\hat{k}}_d+M_2)}} {-}(M_1{\hat{k}}_d{+} M_2{\hat{k}}_p) {\ge } \sqrt{{\hat{k}}_i}M_2\left( \sqrt{{\hat{k}}_d{+} M_2/\underline{b}}\right.$$\left. - \sqrt{{\hat{k}}_d+ M_2/\underline{b}} \right) \ge 0$.

In summary, $k(b,{\hat{b}})$ is an monotone nondecreasing function with b, thus, ${\bar{\omega }}_{ob{\hat{b}}}\le {\bar{\omega }}_{o\underline{b}{\hat{b}}}$. At the meantime, it can be easily proved that $\frac{\frac{{\hat{b}}M_1}{b}-k_{p}}{k_{d}}\le \frac{\frac{{\hat{b}}M_1}{\underline{b}}-k_{p}}{k_{d}}$, $\frac{{\hat{b}}M_2}{b}-k_{d}\le \frac{{\hat{b}}M_2}{\underline{b}}-k_{d}.$ Hence $\omega _{o\underline{b}{\hat{b}}}^*\ge \omega _{ob{\hat{b}}}^*$, $\forall b\in [\underline{b},\overline{b}]$. $\square$

Proof of Theorem 2 The tracking error e(t) and its derivative $e_d$ satisfy:

$$\begin{aligned} \left\{ \begin{aligned}&\dot{e}=e_d,\\&\dot{e}_d=-k_{p}e-k_{d}e_d+e_f, \end{aligned} \right. \end{aligned}$$

thus, $|\ddot{e}(t) + k_{d}\dot{e}(t) + k_{p}e(t)|= |e_f (t)|$. From Theorem 1, when $\omega _{o}>\omega _{o\underline{b}{\hat{b}}}^*$, $e_f$, e and $e_d$ are bounded.

Consider the Lyapunov function:

$$\begin{aligned} V_1=[e,e_d]P \left[ \begin{array}{lcr} e \\ e_d \end{array} \right] , \text { where } P= \left[ \begin{array}{lcr} \frac{1+k_{p}}{2k_{d}}+\frac{k_{d}}{2k_{p}} &{} \frac{1}{2k_{p}} \\ \frac{1}{2k_{p}} &{} \frac{1+k_{p}}{2k_{d}k_{p}} \end{array} \right] . \end{aligned}$$

Thus, $\dot{V}_1(e,e_d)\le -\frac{V_1}{\lambda _{\max }(P)}+\frac{\max \left( \frac{1}{k_{p}},\frac{1+k_{p}}{k_{d}k_{p}}\right) \sqrt{V_1}|e_f|}{\sqrt{\lambda _{\textrm{min}}(P)}}.$ When $\sqrt{V_1}\ge \frac{\max \left( \frac{1}{k_{p}},\frac{1+k_{p}}{k_{d}k_{p}}\right) \lambda _{\max }(P)\textrm{sup}|e_f|}{\sqrt{\lambda _{\textrm{min}}(P)}}$, there is $\dot{V}_1 < 0$. Hence, $\sqrt{V_1}\le \frac{\max \left( \frac{1}{k_{p}},\frac{1+k_{p}}{k_{d}k_{p}}\right) \lambda _{\max }(P)sup|e_f|}{\sqrt{\lambda _{\textrm{min}}(P)}},$ i.e., $\Vert [e,e_d]\Vert \le \frac{\max \left( \frac{1}{k_{p}},\frac{1+k_{p}}{k_{d}k_{p}}\right) \lambda _{\max }(P)\textrm{sup}|e_f|}{\lambda _{\textrm{min}}(P)}.$

Moreover, the estimation error $e_f$ of the ESO (5) satisfies the equation $\dot{e}_f=-\omega _oe_f-\dot{{\tilde{f}}}.$ Since $\dot{f}=\frac{\partial f}{\partial x_2}k_{p}e+\left( \frac{\partial f}{\partial x_2}k_{d}- \frac{\partial f}{\partial x_1}\right) e_d- \frac{\partial f}{\partial x_2}e_f+ \frac{\partial f}{\partial x_1} \dot{v}+\frac{\partial f}{\partial x_2}\ddot{v}$, $e_f$ satisfies $\dot{e}_f=-\left[ \frac{\partial f}{\partial x_2}k_{p}-\left( \frac{b}{{\hat{b}}}-1\right) k_{d}k_{p}\right] e -\left[ \left( \frac{b}{{\hat{b}}}-1\right) k_{d}+\frac{b}{{\hat{b}}}\omega _o-\frac{\partial f}{\partial x_2}\right]$$e_f -\frac{\partial f}{\partial x_1} \dot{v}-\frac{\partial f}{\partial x_2}\ddot{v}-(\frac{b}{{\hat{b}}}-1){\dddot{v}}-b\frac{\partial d(t)}{\partial t}-[\frac{\partial f}{\partial x_2}k_{d}- \frac{\partial f}{\partial x_1}+(\frac{b}{{\hat{b}}}-1)(k_{p}-k_{d}^2)]e_d.$

Denote ${\hat{h}}=\max \{|\frac{\underline{b}}{{\hat{b}}}-1|,|\frac{\overline{b}}{{\hat{b}}}-1|\}$, $\gamma _1= {\hat{h}}k_{p}k_{d}+M_2k_{p}, \gamma _2= M_2k_{d}+M_1+{\hat{h}}|k_{p}-k_{d}^2|, \gamma _3= \sup _{t>0}\{|M_1\dot{v}|+ |M_2\ddot{v}|+ |{\hat{h}}\dddot{v}|+|\overline{b}M_3|\}$. According to (3), $\gamma _3<\infty$.

Consider the Lyapunov function $V_2(e_f)=\frac{1}{2}e_f^2,$ for all $f\in \mathcal {F}$ and $d(t) \in \mathcal {D}$, there is

$$\begin{aligned} \dot{V}_2(e_f)\le & {} -(-{\hat{h}}k_{d}+\frac{\underline{b}}{{\hat{b}}}\omega _o- M_2)e_f^2\nonumber \\{} & {} +|e_f|(\gamma _1|e|+ \gamma _2|e_d|+ \gamma _3). \end{aligned}$$

(B.1)

Denote $\gamma _4= \frac{\max \left( \frac{1}{k_{p}},\frac{1+k_{p}}{k_{d}k_{p}}\right) \lambda _{\max }(P)}{\lambda _\textrm{min} (P)}$, $\omega _{o1}^*=\max \{\omega _{o\underline{b}{\hat{b}}}^*,$ $[{\hat{h}}k_{ad}+ M_2+\gamma _4(\gamma _1+\gamma _2)+1]\frac{{\hat{b}}}{\underline{b}}\}$, $\gamma _5= \max \{e_f(0), \gamma _3\}$. Next, it will be proved that $\textrm{sup}|e_f|\le \gamma _5$ when $\omega _o> \omega _{o1}^*$.

When $\omega _o> \omega _{o1}^*$, from (B.1), $\dot{V}_2(e_f)\le -\left[ \gamma _4(\gamma _1+\gamma _2)\right.$$\left. {+}1\right] |e_f|^2 {+} |e_f|\left[ \gamma _4(\gamma _1{+} \gamma _2) sup|e_f|{+} \gamma _3\right]$, when $\textrm{sup}|e_f|>\gamma _5$, there is $\exists |{\hat{e}}_f|\in (\gamma _5, \textrm{sup}|e_f|),$ $\dot{V}_2({\hat{e}}_f)<0$, thus, $\forall t>0$, $|e_f(t)|$ $\le$ $|{\hat{e}}_f|<\textrm{sup}|e_f|$. This contradicts the definition of $\textrm{sup}|e_f|$, thus, $\textrm{sup}|e_f|\le \gamma _5$.

On the other hand, based on Theorem 1, when $\omega _{o\underline{b}{\hat{b}}}^*\le \omega _o\le \omega _{o1}^*$, $\exists \gamma _6 \in R$, such that $|e_f(t)|\le \gamma _6, |e(t)|\le \gamma _6$ and $|e_d(t)|\le \gamma _6$.

Denote $\gamma _7= [\gamma _4(\gamma _1+\gamma _2)+{\hat{h}}k_{d}+M_2 ]\max \{\gamma _5, \gamma _6\}+\gamma _3$. It can be deduced that from the above analyses:

$$\begin{aligned} \begin{aligned} \dot{V}_2(e_f)&\le -\frac{\underline{b}}{{\hat{b}}}\omega _oe_f^2+({\hat{h}}k_{d}+M_2)e_f^2\\&\quad +|e_f|(\gamma _1|e|+ \gamma _2|e_d|+ \gamma _3)\\&\le -\frac{2\underline{b}}{{\hat{b}}}\omega _oV_2+ \sqrt{2}\sqrt{V_2}\gamma _7,t\ge 0. \end{aligned} \end{aligned}$$

(B.2)

Thus, $\sqrt{V_2}\le e^{-\frac{\underline{b}}{{\hat{b}}}\omega _ot}\sqrt{V_1(e_f(0))}+ \frac{\sqrt{2} {\hat{b}}\gamma _7}{2\underline{b}\omega _o} (1-e^{-\frac{\underline{b}}{{\hat{b}}}\omega _ot})$. Therefore, $|e_f|\le (|e_f(0)|-\frac{\eta _0}{\omega _o})e^{-\frac{\underline{b}}{{\hat{b}}}\omega _ot}+ \frac{\eta _0}{\omega _o}$, where $\eta _0= \frac{ {\hat{b}}\gamma _7}{\underline{b}}$, which is irrelevant to $\omega _o$.

Furthermore, from Theorem 1, $\forall \varepsilon >0$, $\exists T>0$, such that $\dot{{\tilde{f}}}=(\frac{b}{{\hat{b}}}-1)\omega _oe_f+O(\varepsilon )$ as long as $t>T$. Thus, $\forall t>T$, $\dot{V}_2(e_f) \le -\frac{2\underline{b}}{{\hat{b}}}\omega _oV_2+ \sqrt{2}\sqrt{V_2}O(\varepsilon ),t\ge 0$. Therefore, $|e_f|\le (|e_f(0)|-\frac{O(\varepsilon )}{\omega _o})e^{-\frac{\underline{b}}{{\hat{b}}}\omega _ot}+ \frac{O(\varepsilon )}{\omega _o}.$

Define $\eta (t)= \left\{ \begin{array}{ll} \eta _0, &{}t\le T,\\ O(\varepsilon ), &{} t>T. \end{array} \right.$ Theorem 2 is obtained. $\square$

Proof of Theorem 3 When parameters in ADRC $k_p$, $k_d$, $\omega _o$, ${\hat{b}}$ satisfy:

$$\begin{aligned} \frac{{\hat{b}}M_1}{\underline{b}(k_p+ \omega _ok_d)}=\textrm{o}(1), \frac{{\hat{b}}M_2}{\underline{b}(\omega _o+ k_d)}=\textrm{o}(1), \omega _o>{\hat{b}}, \end{aligned}$$

(C.1)

there is $G_v(s)\thickapprox \frac{(s+\omega _o)(s^2+k_ds+k_p)}{\frac{{\hat{b}}}{b}s^3+ (\omega _o+k_d)s^2+ (\omega _ok_d+ k_p)s +\omega _ok_p}$,

(1) If $b={\hat{b}}$, $G_v(s)=\frac{1}{1+\textrm{o}(1)} \thickapprox 1$.

(2) If $b\ne {\hat{b}}$, replacing s with $j\omega$, it can be obtained that $G_v(j\omega )=\frac{\omega _ok_p- (\omega _o+ k_d)\omega ^2+ j\omega (-\omega ^2+\omega _ok_d+ k_p)}{\delta (j\omega )}$, $\delta (j\omega )=\omega _ok_p- [\omega _\textrm{o}(1+\textrm{o}(1)) + k_d]\omega ^2 + j\omega [\omega _ok_d(1+\textrm{o}(1))+ k_p- \frac{{\hat{b}}}{b}\omega ^2]$.

(i) If $\overline{b}\ge b> {\hat{b}}$, when $\omega \le (\frac{\omega _ok_d+k_p}{11-10*\frac{{\hat{b}}}{\overline{b}}})^{\frac{1}{2}}\triangleq A_1(\omega _o,{\hat{b}})$, ${\hat{b}}\in [\underline{b}, \overline{b}]$, there is $\delta (j\omega )=\omega _ok_p- [\omega _{\textrm{o}}(1+\textrm{o}(1)) + k_d]\omega ^2 + j\omega [\omega _ok_d(1+\textrm{o}(1))+ k_p-\omega ^2](1+\textrm{o}(1))$, thus, $G_v(j\omega )=\frac{1}{1+\textrm{o}(1)+\textrm{o}(1)j} \thickapprox 1$.

(ii) If $\underline{b}\le b< {\hat{b}}$, when $\omega \le (\frac{\omega _ok_d+k_p}{10*\frac{{\hat{b}}}{\underline{b}}-9})^{\frac{1}{2}}\triangleq A_2(\omega _o,{\hat{b}})$, ${\hat{b}}\in [\underline{b}, \overline{b}]$, there is $G_v(j\omega )=\frac{1}{1+\textrm{o}(1)+\textrm{o}(1)j} \thickapprox 1$.

Denote $\omega _{b}(\omega _o,{\hat{b}})= \min \{A_1(\omega _o,{\hat{b}}),A_2(\omega _o,{\hat{b}})\}, {\hat{b}}\in [\underline{b},\overline{b}].$ In summary, for any $b \in [\underline{b}, \overline{b}]$, if $\omega \le \omega _{b}(\omega _o,{\hat{b}})$, there is $G_v(j\omega )=\frac{1}{1+\textrm{o}(1)+\textrm{o}(1)j} \thickapprox 1$.

Next, the effect of ${\hat{b}}$ on $\min \{A_1(\omega _o,{\hat{b}}),A_2(\omega _o,{\hat{b}})\}$ is discussed. Given $\underline{b}$, $\overline{b}$, $\omega _o$, $k_d$, $k_p$, according to $A_1(\omega _o,{\hat{b}})$ and $A_2(\omega _o,{\hat{b}})$, when ${\hat{b}}$ decreases, $A_1(\omega _o,{\hat{b}})$ decreases and $A_2(\omega _o,{\hat{b}})$ increases; when ${\hat{b}}$ increases, $A_1(\omega _o,{\hat{b}})$ increases and $A_2(\omega _o,{\hat{b}})$ decreases. If $A_1(\omega _o,{\hat{b}})= A_2(\omega _o,{\hat{b}})$, there is ${\hat{b}}=b^*=\frac{2\underline{b}\overline{b}}{ \underline{b}+ \overline{b}}$, by a simple calculation, there is $\underline{b} \le b^*\le \frac{\underline{b}+\overline{b}}{2}$, take “=” if and only if $\underline{b}=\overline{b}$, at the same time, $A_1(\omega _o,b^*)= A_2(\omega _o,b^*)= \left[ \frac{(\omega _ok_d+k_p)(\underline{b}+\overline{b})}{11\overline{b} -9\underline{b}}\right] ^{\frac{1}{2}}\triangleq A_3(\omega _o)$. It is easy to get $A_3(\omega _o)= \omega _b(\omega _o,b^*) \ge \omega _b(\omega _o,{\hat{b}})$ for any ${\hat{b}}\in [\underline{b},\overline{b}]$.

According to the above analyses, when ${\hat{b}}=b^*$, $G_v(j\omega )=\frac{1}{1+\textrm{o}(1)+\textrm{o}(1)j} \thickapprox 1$ for $0< \omega \le A_3(\omega _o)$. Moreover, $\omega _{b}(\omega _o,{\hat{b}})$ can be rewritten as follows:$\omega _{b}(\omega _o,{\hat{b}}) = \left\{ \begin{aligned} A_1(\omega _o,{\hat{b}})&, {\hat{b}}\in [\underline{b},b^*],\\ A_2(\omega _o,{\hat{b}})&, {\hat{b}}\in [b^*,\overline{b}]. \end{aligned} \right.$

Therefore, it indicates that the reference input could be well tracked by the controlled output in the low frequency band. $\square$

Proof of Theorem 4 First of all, solve for the crossover frequency $\omega _c$: $|L(j\omega _c)|=1.$ Replacing s with $j\omega$, it can be obtained that $L(j\omega )=\frac{b}{{\hat{b}}}\frac{\omega _ok_p-(\omega _o+k_d)\omega ^2 +j(k_p+\omega _ok_d)\omega }{a_2\omega ^2-j(a_1+\omega ^2)\omega }$. When parameters in ADRC ($k_p$, $k_d$, $\omega _o$) satisfy:

$$\begin{aligned}{} & {} \frac{M_1}{k_p+ \omega _ok_d}=\textrm{o}(1), \frac{M_2}{\omega _o+ k_d}=\textrm{o}(1), \nonumber \\{} & {} \frac{k_p}{k_d^2}=\textrm{o}(1), \frac{k_d}{\omega _o}\in \left( \frac{1}{15},1\right) , \end{aligned}$$

(D.1)

there is $|L(jk_d)|>\frac{b}{{\hat{b}}}, |L\left( j\frac{bk_d}{{\hat{b}}}\right) |>\frac{{\hat{b}}}{b}.$ It implies that $\omega _{c}> \min \left\{ k_d,\frac{bk_d}{{\hat{b}}}\right\}$. Therefore, $\omega _{c}$ satisfies ${\tilde{f}}(\omega _{c})= 0$, where

$$\begin{aligned} {\tilde{f}}(\omega _{c})= & {} \omega _{c}^4- \left( \frac{b}{{\hat{b}}}\right) ^2(\omega _o+k_d)^2(1+2\textrm{o}(1))\omega _{c}^2\nonumber \\{} & {} \quad -\left( \frac{b}{{\hat{b}}}\right) ^2\omega _o^2k_d^2(1+4\textrm{o}(1)). \end{aligned}$$

(D.2)

Denote $v_1=\left[ \frac{{\hat{b}}\omega _{c}}{b(\omega _o+k_d)}\right] ^2$, (D.2) can be rewritten as ${\tilde{f}}_1(v_1)=\frac{{\tilde{f}}(\omega _{c})}{(\frac{b}{{\hat{b}}})^2 (\omega _o+k_d)^4} = \left( \frac{b}{{\hat{b}}}\right) ^2v_1^2 -\left( \frac{b}{{\hat{b}}}\right) ^2(1+2\textrm{o}(1))v_1-\textrm{o}(1)=0$. It can be obtained that $\omega _c=\frac{b}{{\hat{b}}}(\omega _o+k_d)(1+\textrm{o}(1))$.

Denote $\gamma =\frac{\omega _c}{\omega _o}=\frac{b}{{\hat{b}}} (1+\frac{k_d}{\omega _o})(1+\textrm{o}(1))$, when parameters in ADRC ($k_p$, $k_d$, $\omega _o$) satisfy (D.1), the phase margin can be obtained $\textrm{PM}=\frac{\pi }{2}+\arctan \frac{(k_p+\omega _ok_d)\omega _c}{\omega _ok_p- (\omega _o+k_d)\omega _c^2}+ \arctan \frac{-a_2\omega _c}{\omega _c^2+ a_1} =\frac{\pi }{2}-\arctan \frac{\frac{k_d}{\omega _o}(1+ \textrm{o}(1)\frac{k_d}{\omega _o})}{(1+\frac{k_d}{\omega _o}- \textrm{o}(1)\frac{ k_d^2}{\omega _o^2\gamma ^2})\gamma }-arctan\frac{\textrm{sign}(a_2)\sqrt{\frac{k_d}{\omega _o}\textrm{o}(1)}}{\gamma (1- \textrm{o}(1)\frac{k_d}{2\omega _o\gamma ^2})}$. Thus, $\textrm{PM}=\frac{\pi }{2}-\frac{ {\hat{b}}}{b}\Big (\frac{\frac{k_d}{\omega _o}}{(1+\frac{k_d}{\omega _o})^2} +\frac{\textrm{sign}(a_2)\frac{1}{3}\sqrt{ \frac{k_d}{\omega _o}}}{1+\frac{k_d}{\omega _o}}+ \textrm{o}(1)\Big )$. $\square$

Proof of Corollary 1 The permitted time delay $\tau = \frac{\textrm{PM}}{\omega _c}$. According to the analyses in Proof of Theorem 4, $\tau (\omega _o,{\hat{b}})\approx \frac{\frac{\pi }{2}-\frac{\frac{k_d}{\omega _o}}{\frac{ b}{{\hat{b}}}(1+\frac{k_d}{\omega _o})^2}-\frac{\textrm{sign}(a_2)\frac{1}{3}\sqrt{ \frac{k_d}{\omega _o}}}{\frac{b}{{\hat{b}}}(1+\frac{k_d}{\omega _o})}}{\frac{b}{{\hat{b}}}(\omega _o+k_d)}$.

First, it can be concluded that $\tau$ is a quadratic function of ${\hat{b}}$, denote $\tau {=}\frac{{\bar{A}}{\hat{b}}^2{+}{\bar{B}}{\hat{b}}{+}{\bar{C}}}{{\bar{D}}},$ where ${\bar{A}}{=}{-}\frac{\frac{k_d}{\omega _o}+\textrm{sign}(a_2)\frac{1}{3}(1+\frac{k_d}{\omega _o}) \sqrt{ \frac{k_d}{\omega _o}}}{b(1+\frac{k_d}{\omega _o})^2}$, ${\bar{B}}=\frac{\pi }{2}$, ${\bar{C}}=0$, ${\bar{D}}=b(\omega _o+k_d)$. The sign of ${\bar{A}}$ is determined as follows: (1) If $\textrm{sign}(a_2)>0$, ${\bar{A}}<0$; (2) If $\textrm{sign}(a_2)<0$ and $\frac{k_d}{\omega _o}\in (0, \frac{7-\sqrt{45}}{2}(\approx 0.146))$, ${\bar{A}}>0$; (3) If $\textrm{sign}(a_2)<0$ and $\frac{k_d}{\omega _o}\in (\frac{7-\sqrt{45}}{2}(\approx 0.146), 1)$, ${\bar{A}}<0$.

To sum up, when $\frac{k_d}{\omega _o}\in (\frac{1}{6}, 1)$, for any $|a_2|<M_2$, there is ${\bar{A}}<0$.

When ${\bar{A}}>0$, $\tau$ increases with increase of ${\hat{b}}$. When ${\bar{A}}<0$, the quadratic function is a parabola going downwards with two roots, one of which is 0 and another root is positive. The axis of symmetry is $E(b)\triangleq -\frac{B}{2A}=\frac{\frac{\pi }{4}b(1+\frac{k_d}{\omega _o})^2}{\frac{k_d}{\omega _o}+\textrm{sign}(a_2)\frac{1}{3}(1+\frac{k_d}{\omega _o}) \sqrt{\frac{k_d}{\omega _o}}}$, thus, when ${\hat{b}} \in (\frac{\underline{b}}{2}, E(\underline{b}))$, $\tau$ increases with increase of ${\hat{b}}$. When ${\hat{b}}> E(\overline{b})$ and $\omega _o<6k_d$, $\tau$ decreases with increase of ${\hat{b}}$. Furthermore, $E(b)\ge \frac{\frac{\pi }{4}\underline{b}(1+\frac{k_d}{\omega _o})^2}{\frac{k_d}{\omega _o}+\frac{1}{3}(1+\frac{k_d}{\omega _o}) \sqrt{ \frac{k_d}{\omega _o}}}\ge \underline{b}$. It can be obtained that when ${\hat{b}}<\underline{b}$, the permitted time delay $\tau$ decreases with decrease of ${\hat{b}}$.

On the other hand, the permitted time delay $\tau$ can be rewritten as follows:

$$\begin{aligned} \tau (\varpi ){\approx }\frac{\frac{\pi }{2}\frac{b}{{\hat{b}}}\varpi (1{+}\varpi )^2 {-}\textrm{sign}(a_2)\frac{1}{3}\varpi ^{\frac{3}{2}}(1{+}\varpi ){-}\varpi ^2 }{(\frac{b}{{\hat{b}}})^2k_d(1{+}\varpi )^3}, \end{aligned}$$

where $\varpi =\frac{k_d}{\omega _o}$, $\varpi \in (0,1)$. Since the derivative of $\tau$ can be written as follows: $\frac{\partial \tau }{\partial \varpi }=\frac{\chi (\varpi )}{(\frac{b}{{\hat{b}}})^2k_d(1+\varpi )^4}$, $\chi (\varpi )=\textrm{sign}(a_2)\frac{1}{6}\varpi ^{\frac{1}{2}}(\varpi -3)(\varpi +1) +\frac{\pi }{2}\frac{b}{{\hat{b}}}(1+\varpi )^2 +\varpi ^2- 2\varpi$.

If $\textrm{sign}(a_2)<0$, it can be obtained that $\chi (\varpi )\ge -\frac{1}{6}\varpi ^{\frac{1}{2}}(\varpi -3)(\varpi +1) +\frac{\pi }{6}(1+\varpi )^2 +\varpi ^2- 2\varpi >0, \varpi \in (0,1)$. Thus, $\frac{\partial \tau }{\partial \varpi }>0$ for all $b\in [\underline{b},\overline{b}]$.

If $\textrm{sign}(a_2)>0$, when $\varpi \in (0,\frac{1}{4})\cup (\frac{4}{5},1)$ and $\frac{b}{{\hat{b}}}\ge \frac{1}{3}$, it can be obtained that $\chi (\varpi )\ge \frac{1}{6}\varpi ^{\frac{1}{2}}(\varpi -3)(\varpi +1) +\frac{\pi }{6}(1+\varpi )^2 +\varpi ^2- 2\varpi >0$. Therefore, $\frac{\partial \tau }{\partial \varpi }>0$ for all $b\in [\underline{b},\overline{b}]$, $\varpi =\frac{k_d}{\omega _o}\in (0,\frac{1}{4})\cup (\frac{4}{5},1)$ and $\frac{b}{{\hat{b}}}\ge \frac{1}{3}$.

In conclusion, when $\omega _o> 4k_d$ and ${\hat{b}}\le 3\underline{b}$, the permitted time delay $\tau$ decreases with increase of $\omega _o$. $\square$

Proof of Theorem 5 When parameters in ADRC ($k_p$, $k_d$, $\omega _o$, ${\hat{b}}$) satisfy (C.1), there is $G_d(s)\thickapprox$ $\frac{{\hat{b}}s}{\frac{{\hat{b}}}{b}s^3+ (\omega _o+ k_d)s^2+ (\omega _ok_d+ k_p)s +\omega _ok_p}.$

Replacing s with $j\omega$, it can be obtained that $G_d(j\omega )\approx$$\frac{{\hat{b}}j\omega }{\omega _ok_p- (\omega _o+ k_d)\omega ^2+ j\omega (\omega _ok_d+ k_p- \frac{{\hat{b}}}{b}\omega ^2)},$ thus, $|G_d(j\omega )|^2\approx$$\frac{{\hat{b}}^2}{f_1(\omega )+ f_2(\omega )},$ where $f_1(\omega )=\frac{\omega _o^2k_p^2}{\omega ^2}- 2\omega _ok_p(\omega _o+ k_d) + (\omega _o+ k_d)^2\omega ^2$, $f_2(\omega )=(\omega _ok_d+ k_p)^2- 2(\omega _ok_d+ k_p)\frac{{\hat{b}}}{b}\omega ^2+ (\frac{{\hat{b}}}{b})^2\omega ^4.$

From the properties of inequality of arithmetic and geometric means and quadratic functions, it can be obtained that $f_1(\omega )=f_2(\omega )$ and $\frac{df_1(\omega )}{d\omega }*\frac{df_2(\omega )}{d\omega }<0,$ there is a unique solution to the above equation, which is denoted as ${\hat{\omega }}(b,{\hat{b}})$. Hence, $f_1(\omega )+ f_2(\omega )\ge f_1({\hat{\omega }})= \frac{\omega _o^2k_p^2}{{\hat{\omega }}^2}- 2\omega _ok_p(\omega _o+ k_d) + (\omega _o+ k_d)^2{\hat{\omega }}^2.$

When parameters in ADRC ($k_p$, $k_d$, $\omega _o$) satisfy conditions (3)–(4) in Theorem 4 and $\frac{\underline{b}}{{\hat{b}}}\ge \frac{1}{3}$, it can be obtained that ${\hat{\omega }}^2(b,{\hat{b}})\in \left\{ \frac{\omega _ok_p}{\omega _o+k_d}, \frac{b}{{\hat{b}}}(\omega _ok_d+k_p)\right\}$.

Furthermore,

$f_1\left( \sqrt{\frac{(\alpha +1)\omega _ok_p}{\omega _o{+}k_d}}\right) {=} \frac{\alpha ^2\omega _ok_p(\omega _o{+}k_d)}{\alpha {+}1}{\le }$ $f_2\left( \sqrt{\frac{(\alpha {+}1)\omega _ok_p}{\omega _o{+}k_d}}\right) ,$

where $\alpha {=} \left\{ \begin{aligned}&[\frac{\omega _o}{k_d}],{} & {} \omega _o\in (k_d,4k_d),\\&4,{} & {} \omega _o\in [4k_d,6k_d),\\&5,{} & {} \omega _o\in [6k_d,9k_d),\\&6,{} & {} \omega _o\in [9k_d,13k_d),\\&7,{} & {} \omega _o\in [13k_d,15k_d)\!. \end{aligned} \right.$

Therefore, ${\hat{\omega }}^2(b,{\hat{b}})\in \left\{ \frac{(\alpha +1)\omega _ok_p}{\omega _o+k_d}, \frac{b}{{\hat{b}}}(\omega _ok_d+k_p)\right\} .$ Hence, $f_1({\hat{\omega }})\ge f_1\left( \sqrt{\frac{(\alpha +1)\omega _ok_p}{\omega _o+k_d}}\right) = \frac{\alpha ^2\omega _ok_p(\omega _o+k_d)}{\alpha +1}.$

Define $A_4(\omega _o,{\hat{b}})=\sqrt{\frac{(\alpha +1){\hat{b}}^2}{\alpha ^2\omega _ok_p(\omega _o+k_d)}},$ thus, when parameters in ADRC $k_p$, $k_d$, $\omega _o$ and ${\hat{b}}$ satisfy conditions in Theorem 6, $|G_d(j\omega )|\le A_4(\omega _o,{\hat{b}})<1$ for any $\omega >0$. In particular, $\lim \nolimits _{\omega \rightarrow 0}|G_d(j\omega )|=0, \lim \nolimits _{\omega \rightarrow \infty }|G_d(j\omega )|=0.$

Moreover, when parameters in ADRC ($k_p$, $k_d$, $\omega _o$, ${\hat{b}}$) satisfy (C.1), there is $G_{du}(s){\thickapprox }{-}\frac{(\omega _o{+}k_d)s^2{+}(k_p{+}\omega _ok_d)s{+}\omega _ok_p}{\frac{{\hat{b}}}{b}s^3{+} (\omega _o+ k_d)s^2{+} (\omega _ok_d{+} k_p)s {+}\omega _ok_p},$ i.e. $G_{du}(j\omega ){\approx }{-}\frac{\omega _ok_p{-} (\omega _o{+} k_d)\omega ^2{+} j\omega (\omega _ok_d{+} k_p)}{\omega _ok_p{-} (\omega _o{+} k_d)\omega ^2{+} j\omega (\omega _ok_d{+} k_p{-} \frac{{\hat{b}}}{b}\omega ^2)},$ thus, $|G_{du}(j\omega )|^2\approx \frac{g_1^2+ g_2^2}{g_1^2+ g_3^2},$ where $g_1(\omega )=\omega _ok_p- (\omega _o+ k_d)\omega ^2$, $g_2(\omega )=(\omega _ok_d+ k_p)\omega$, $g_3(\omega )=[(\omega _ok_d+ k_p)-\frac{{\hat{b}}}{b}\omega ^2]\omega .$ Hence, $|G_{du}(j\omega )|^2\approx 1+\frac{g_4}{g_5},$ where $g_4(\omega )=-(\frac{{\hat{b}}}{b})^2\omega ^6+2\frac{{\hat{b}}}{b}(\omega _ok_d+k_p)\omega ^4,$ $g_5(\omega )=(\frac{{\hat{b}}}{b})^2\omega ^6+[(\omega _o+k_d)^2- 2\frac{{\hat{b}}}{b}(\omega _ok_d+k_p)]\omega ^4 +[(\omega _ok_d+k_p)^2- 2\omega _ok_p(\omega _o+k_d)]\omega ^2+ \omega _o^2k_p^2.$

Set $\varOmega _1=\{\omega >0\mid g_6(\omega )=0\}$, where $g_6(\omega )=11(\frac{{\hat{b}}}{b})^2\omega ^6 +[(\omega _o+k_d)^2 -22\frac{{\hat{b}}}{b}(\omega _ok_d+k_p)]\omega ^4+ [(\omega _ok_d+k_p)^2- 2\omega _ok_p(\omega _o+k_d)]\omega ^2+ \omega _o^2k_p^2$. Define

$$\begin{aligned} \begin{aligned}&\omega _{du1}(\omega _o, b,{\hat{b}})= \left\{ \begin{aligned}&\sqrt{2\frac{b}{{\hat{b}}}(\omega _ok_d+k_p)},{} & {} \varOmega _1=\emptyset ,\\&\min _{\omega \in \varOmega _1}\omega ,{} & {} \varOmega _1\ne \emptyset , \end{aligned} \right. \\&\omega _{du}(\omega _o,{\hat{b}})=\min _{b\in [\underline{b},\overline{b}]} \omega _{du1}(\omega _o,b,{\hat{b}})\le \sqrt{2\frac{\underline{b}}{{\hat{b}}}(\omega _ok_d+k_p)}. \end{aligned} \end{aligned}$$

there is $|G_{du}(j\omega )|=1+\textrm{o}(1), \forall \omega <\omega _{du}(\omega _o,{\hat{b}}).$ In particular, $\lim \nolimits _{\omega \rightarrow 0}G_{du}(j\omega )=-1, \lim \nolimits _{\omega \rightarrow \infty }|G_{du}(j\omega )|=0.$ $\square$

Proof of Theorem 6 The noise suppression performance of the closed-loop systems (1), (5) and (6) with (10) is considered below.

When parameters in ADRC ($k_p$, $k_d$, $\omega _o$, ${\hat{b}}$) satisfy (C.1), there is $G_{n1}(s)\thickapprox -\frac{k_p(s+\omega _o)}{\frac{{\hat{b}}}{b}s^3+ (\omega _o+ k_d)s^2+ (\omega _ok_d+ k_p)s +\omega _ok_p},$ $G_{n2}(s)\thickapprox -\frac{(\omega _o+k_d)s+ \omega _ok_d}{\frac{{\hat{b}}}{b}s^3+ (\omega _o+ k_d)s^2+ (\omega _ok_d+ k_p)s +\omega _ok_p}.$ Replacing s with $j\omega$, it can be obtained that $|G_{n1}(j\omega )|^2\approx \frac{k_p^2(\omega ^2+\omega _o^2)}{(f_1(\omega )+ f_2(\omega ))\omega ^2},$ $|G_{n2}(j\omega )|^2\approx \frac{(\omega _o+k_d)^2\omega ^2+ \omega _o^2k_d^2}{(f_1(\omega )+ f_2(\omega ))\omega ^2},$ where $f_1(\omega )=\frac{\omega _o^2k_p^2}{\omega ^2}- 2\omega _ok_p(\omega _o+ k_d) + (\omega _o+ k_d)^2\omega ^2$, $f_2(\omega )=(\omega _ok_d+ k_p)^2- 2(\omega _ok_d+ k_p)\frac{{\hat{b}}}{b}\omega ^2+ (\frac{{\hat{b}}}{b})^2\omega ^4.$

Denote ${\tilde{A}}=(\frac{{\hat{b}}}{b})^2$, ${\tilde{B}}=(\omega _o+ k_d)^2 -2\frac{{\hat{b}}}{b}(\omega _ok_d+ k_p)$, ${\tilde{C}}=\omega _o^2k_d^2- 2\omega _o^2k_p$. Set $\varOmega _2=\{\omega \in R\mid f_3(\omega )=0\}$, where $f_3(\omega )=(\frac{{\hat{b}}}{b})^2\omega ^6 +[(\omega _o+k_d)^2 -2\frac{{\hat{b}}}{b}(\omega _ok_d+k_p)]\omega ^4+ [(\omega _ok_d+k_p)^2- (\omega _o+k_d)^2- 2\omega _ok_p(\omega _o+k_d)]\omega ^2+ \omega _o^2k_p^2-\omega _o^2k_d^2$.

Define

$A_5(\omega _o,{\hat{b}}){=}\sqrt{\max _{b\in [\underline{b},\overline{b}]}\{A_{51}(\omega _o,b,{\hat{b}}), A_{52}(\omega _o,b,{\hat{b}})\}}$, where

$$\begin{aligned} \begin{aligned}&A_{51}(\omega _o,b,{\hat{b}}){=} \left\{ \begin{aligned}&0,{} & {} {\tilde{B}}^2{-} 4{\tilde{A}}{\tilde{C}}<0,\\&\frac{{-}{\tilde{B}} {+}({\tilde{B}}^2{-} 4{\tilde{A}}{\tilde{C}})^{\frac{1}{2}}}{2{\tilde{A}}},{} & {} {\tilde{B}}^2{-} 4{\tilde{A}}{\tilde{C}}{\ge }0. \end{aligned} \right. \\&A_{52}(\omega _o,b,{\hat{b}}){=} \left\{ \begin{aligned}&0,{} & {} \varOmega _2{=}\emptyset \text { or } \max _{\omega \in \varOmega _{2}}\omega \le 0,\\&\omega ^2,{} & {} \max _{\omega \in \varOmega _{2}}\omega {>}0. \end{aligned} \right. \end{aligned} \end{aligned}$$

From the properties of quadratic and cubic functions, there is $\forall \omega > A_{5}(\omega _o,{\hat{b}})$, $|G_{n1}(j\omega )|<1$ and $|G_{n2}(j\omega )|<1$. Particularly, $\lim \nolimits _{\omega \rightarrow \infty }|G_{n1}(j\omega )|=0, \lim \nolimits _{\omega \rightarrow \infty }|G_{n2}(j\omega )|=0.$

Moreover, when parameters in ADRC ($k_p$, $k_d$, $\omega _o$, ${\hat{b}}$) satisfy (C.1), there is $G_{n1u}(s)\thickapprox$ ${-}\frac{k_ps^3{+}k_p\omega _os^2{-}k_p(a_1{+}\omega _oa_2)s-a_1k_p\omega _o}{{\hat{b}}s^3{+} b(\omega _o{+} k_d)s^2{+} b(\omega _ok_d{+} k_p)s {+}b\omega _ok_p},$$G_{n2u}(s){\thickapprox }{-}$ $\frac{{\bar{\xi }}(s)}{{\hat{b}}s^3{+} b(\omega _o{+} k_d)s^2{+} b(\omega _ok_d{+} k_p)s {+}b\omega _ok_p},$ where ${\bar{\xi }}(s){=}(\omega _o{+}k_d)s^3$ ${+}[\omega _ok_d-a_2(\omega _o+k_d)]s^2 -[a_1(\omega _o+k_d){+}a_2\omega _ok_d]s-a_1\omega _ok_d.$ Replacing s with $j\omega$, it can be obtained that $\lim \nolimits _{\omega \rightarrow 0}|G_{n1u}(j\omega )|=|\frac{a_1}{b}|$, $\lim \nolimits _{\omega \rightarrow \infty }|G_{n1u}(j\omega )|=\frac{k_p}{{\hat{b}}}$, $\lim \nolimits _{\omega \rightarrow 0}|G_{n2u}(j\omega )|=|\frac{a_1k_d}{bk_p}|$, $\lim \nolimits _{\omega \rightarrow \infty }|G_{n2u}(j\omega )|=\frac{\omega _o+k_d}{{\hat{b}}}.$ $\square$