#  SCIENCE CHINA Information Sciences, Volume 63 , Issue 9 : 192203(2020) https://doi.org/10.1007/s11432-019-2712-7

## A parameter formula connecting PID and ADRC More info
• ReceivedJul 30, 2019
• AcceptedNov 1, 2019
• PublishedJul 24, 2020
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### Abstract ### Acknowledgment

This work was supported by National Key RD Program of China (Grant No. 2018YFA0703800), National Center for Mathematics and Interdisciplinary Sciences, Chinese Academy of Sciences, and National Natural Science Foundation of China (Grant No. 11688101).

### Supplement

Appendix

Proof of Theorem 1

Substituting (5) into (4), by Laplace transform, we obtain \begin{equation} \hat F(s)={\omega_o k_{ad}}({X_1(s)}-R(s))+ \omega_o s({X_1(s)}-R(s))+ \frac{\omega_o k_{ap}({X_1(s)}-R(s))}{s} , \tag{16}\end{equation} where $X_1(s),R(s)$ are the Laplace transforms of the state $x_1(t)$ and the transient process $r(t)$, respectively. Taking the inverse Laplace transform for (16), we have \begin{equation} \hat f={\omega_o k_{ad}}({x_1}-r)+ \omega_o ({x_2}-\dot r)+\omega_o k_{ap} \int_{0}^{t} ({x_1(\tau)}-r(\tau)) {\rm d}\tau . \tag{17}\end{equation} Hence, the control law (5) can be rewritten as \begin{equation} u=-{k_{p}}({x_1}-r)-{k_{d}}({x_2}-\dot r)-k_{i}\int_{0}^{t} ({x_1(\tau)}-r(\tau)) {\rm d}\tau+\ddot r , \tag{18}\end{equation} where $~k_p~=~k_{ap}+\omega_o~k_{ad},~k_d=k_{ad}+\omega_o,~k_i~=~\omega_o~k_{ap}.$

Because $~\lim\nolimits_{t~\to~\infty~}{w(t)}~\text{~exists}$, which can be denoted by a constant $c$, we have $$e_i(t)= \int_{0}^{t} e(\tau) {\rm d}\tau + \frac{h(y^{**},0)+c}{k_i}, e_d(t)= \dot e(t), g(e,e_d)= -h(y^{**}-e,-e_d)+h(y^{**},0).$$ Based on the definition of $\mathcal{F}$, it can be seen that $g\in~\mathcal{F}$ and $g(0,0)=0$. Then the closed-loop system (1) and (18) turns into \begin{equation} \begin{cases} \dot {e_i}= e, \\ \dot e=e_d, \\ \dot e_d= -k_i e_i -k_p e -k_d e_d +g(e,e_d)+\Delta(t) , \end{cases} \tag{19}\end{equation} where $\Delta(t)=g(e+y^{**}-r,e_d-\dot~r)~-g(e,e_d)+c-w(t).$ By the mean value theorem, it can be obtained that, for any $t\in~\mathbb{R}^+,$ there is \begin{equation} g(e+y^{**}-r,e_d-\dot r) -g(e,e_d)=\frac{\partial g}{\partial e} \Big | _ {(\bar e,\bar e_d)}(y^{**}-r)-\frac{\partial g}{\partial e_d} \Big | _ {(\bar e,\bar e_d)} \dot r , \tag{20}\end{equation} where $\bar~e~=~e+\theta~(y^{**}-r),~\bar~e_d~=~e_d-\theta~\dot~r,~\theta~\in~(0,1)$. Because $f\in~\mathcal{F}$, it can be deduced that $|\Delta|~\leq~L_1~|y^{**}-r|+L_2~|\dot~r|+c+L_3.$ Moreover, $(0,0,0)$ is an equilibrium of (19), when $t$ approaches infinity.

Following the analysis in [4,5], we let $$b(e)= \begin{cases} \dfrac{g(e,0)}{e},& e\neq 0,\\ \dfrac{\partial g}{\partial e}(0,0),& e=0, \end{cases} {\rm and} a(e,e_d)= \begin{cases} \dfrac{g(e,e_d)-g(e,0)}{e_d},& e_d\neq 0,\\ \dfrac{\partial g}{\partial e_d}(e,0),& e_d=0, \end{cases}$$ and then $g(e,e_d)$ can be expressed as $$g(e,e_d)= b(e) e + a(e,e_d) e_d.$$ By the mean value theorem again and the definition of $\mathcal{F}$, we have $|~b(e)|~\leq~L_1,~|~a(e,e_d)|~\leq~L_2$ for all $e,e_d$.

Hence, the closed-loop system (19) can be rewritten as \begin{equation} \begin{cases} \dot {e_i}= e, \\ \dot e=e_d, \\ \dot e_d= -k_i e_i -\phi (e) e - \psi (e,e_d) e_d +\Delta(t), \end{cases} \tag{21}\end{equation} where $\phi~(e)=k_p~-~b(e),~\psi~(e,e_d)=k_d~-~a(e,e_d).$ By the fact that $\omega_o~>~\frac{L_1-~k_{ap}}{k_{ad}},~\omega_o~>~L_2-k_{ad},$ there exist $\phi~(e)\geq~k_p-L_1>0$ and $\psi~(e,e_d)~\geq~k_d-L_2>0$.

To construct a Lyapunov function, we consider the following matrix $P$: \begin{equation} P=\frac{1}{2}\begin{bmatrix} \mu k_i & k_i & \delta \\ k_i & k_p-L_1 + \mu k_d & \mu \\ \delta & \mu & 1 \end{bmatrix}, \mu = \frac{2((k_p-L_1)(k_d-L_2) + k_i)}{4(k_p-L_1) + L_2^2}, \tag{22}\end{equation} where $\delta$ satisfies $0<\delta~<~\frac{2\mu~k_i}{(k_p-L_1)~+~\mu~k_d}$. We will first show that the matrix $P$ is positive definite.

Based on the definition of $\omega_o^*$ and the assumption $\omega_o>\omega_o^*,$ it can be obtained that \begin{equation}\begin{split} (k_p-L_1)(k_d-L_2)-k_i>L_2\sqrt{k_i(k_d-L_2)}; \end{split} \tag{23}\end{equation} thus, \begin{equation}\begin{split} &(k_p-L_1)(k_d-L_2)-k_i>0, \\ &[(k_p-L_1)(k_d-L_2)-k_i]^2>L_2^2{k_i(k_d-L_2)}. \end{split} \tag{24}\end{equation} Because \begin{equation}\begin{split} \mu-k_d+L_2= \frac{-2(k_p-L_1)(k_d-L_2) + 2k_i-L_2^2 (k_d-L_2)}{4(k_p-L_1) + L_2^2}<0, \end{split} \tag{25}\end{equation} and \begin{equation}\begin{split} 4(-\mu+k_d-L_2)(-k_i+\mu(k_p-L_1))-\mu^2 L_2^2= \frac{4[[(k_p-L_1)(k_d-L_2)-k_i]^2-L_2^2{k_i(k_d-L_2)}]}{4(k_p-L_1) + L_2^2}>0, \end{split} \tag{26}\end{equation} we have \begin{equation}\begin{split} -k_i+\mu(k_p-L_1)>0. \end{split} \tag{27}\end{equation} Then, based on (25)–(27), the following three inequalities can be verified: \begin{equation}\begin{split} \mu k_i>0, \left|\begin{array}{cccc} \mu k_i & k_i \\ k_i & k_p-L_1 + \mu k_d \end{array}\right| =k_i[\mu( k_p-L_1 + \mu k_d )-k_i]>0, \end{split} \tag{28}\end{equation} and \begin{equation}\begin{split} \left|\begin{array}{cccc} \mu k_i & k_i & \delta \\ k_i & k_p-L_1 + \mu k_d & \mu \\ \delta & \mu & 1 \end{array}\right| >k_i(\mu(k_p-L_1)+\mu^2 k_d -k_i -\mu^3)>0. \end{split} \tag{29}\end{equation} Thus, the matrix $P$ is positive definite.

We are now in a position to consider the following Lyapunov function [4,5]: \begin{equation} V(e_i,e,e_d)=[e_i,e,e_d]P[e_i,e,e_d]^{\rm T} + \int_{0}^{e} (L_1-b(s))s {\rm d} s. \tag{30}\end{equation} Because $0\leq~\int_{0}^{e}~(L_1-b(s))s~{\rm~d}~s~\leq~L_1~e^2$, from (30), we have \begin{equation} [e_i,e,e_d]P[e_i,e,e_d]^{\rm T}\leq V(e_i,e,e_d)\leq [e_i,e,e_d]P_0[e_i,e,e_d]^{\rm T}, \tag{31}\end{equation} where $$P_0=\frac{1}{2}\begin{bmatrix} \mu k_i & k_i & \delta\\ k_i & k_p+L_1 + \mu k_d & \mu\\ \delta & \mu & 1 \end{bmatrix}.$$ It can be deduced that \begin{equation} \lambda_{\min}(P) \left\Vert [e_i,e,e_d] \right\Vert^2 \leq {V} \leq\lambda_{\max}(P_0) \left\Vert {[e_i,e,e_d]}\right\Vert^2, \tag{32}\end{equation} where $\lambda_{\min}(\cdot)$ and $\lambda_{\max}(\cdot)$ are the minimum and maximum eigenvalues of the corresponding matrix, respectively.

The time derivative of $V(e_i,e,e_d)$ along the trajectories of (21) is \begin{equation} \dot V(e_i,e,e_d)=-[e_i,e,e_d]W(e,e_d) [e_i,e,e_d]^{\rm T}+ [\delta , \mu, 1][e_i,e,e_d]^{\rm T}\Delta , \tag{33}\end{equation} where \begin{align*}W(e,e_d)=\begin{bmatrix} \delta k_i & \dfrac{\delta \phi (e)}{2} & \dfrac{\delta \psi (e,e_d)}{2} \\ \dfrac{\delta \phi (e)}{2} & -k_i + \mu \phi (e) & -\dfrac{\mu a(e,e_d)+\delta}{2} \\ \dfrac{\delta \psi (e,e_d)}{2} & -\dfrac{\mu a(e,e_d)+\delta}{2} & -\mu+\psi(e,e_d) \end{bmatrix}. \end{align*} Let \begin{align*}Q(e,e_d)=\begin{bmatrix} -k_i + \mu \phi (e) & -\dfrac{\mu a(e,e_d)}{2} \\ -\dfrac{\mu a(e,e_d)}{2} & -\mu+\psi(e,e_d) \end{bmatrix}. \end{align*} From (25)–(27), it is easy to verify that $Q(e,e_d)$ is positive definite. Because $\phi~(e),~\psi(e,e_d)$ are bounded, based on the same analysis as in , there exists a constant $\delta^*$, such that the matrix $W(e,e_d)$ is positive definite when $\delta<\delta^*$. Thus $\delta$ can be defined by $\delta<~\min\{\frac{2\mu~k_i}{k_p-L_1~+~\mu~k_d},\delta^*\}.$

From (33), we have \begin{equation} \dot V\leq -\lambda_{\min}(W)\|[e_i,e,e_d]\| ^2+ c_0 \|[e_i,e,e_d]\| |\Delta |\leq -c_1 V +c_2 \sqrt V| \Delta| , \tag{34}\end{equation} where $c_0~=~\max\{\delta~,~\mu,~1\},~c_1~=~\frac{\lambda_{\min}(W)}{\lambda_{\max}(P_0)},~c_2~=~\frac{c_0}{\sqrt{\lambda_{\min}(P)}}.$ Because $|\Delta|$ is bounded, there exists a constant $M_0$, such that $|\Delta|~\leq~M_0$. Then, it can be obtained that \begin{equation} \sqrt V \leq {\rm e }^{\frac{-c_1 t}{2}}\sqrt {V(e_i(0),e(0),e_d(0))} + \frac{c_2 M_0}{c_1} \left(1-{\rm e }^{\frac{-c_1 t}{2}}\right)\leq M_1, \tag{35}\end{equation} where $M_1$ is a constant. Thus, $e_i,e$ and $e_d$ are bounded. Because $~\lim\nolimits_{t~\to~\infty~}{r(t)=y^{**}},~\lim\nolimits_{t~\to~\infty~}{\dot~r(t)=0}$ and $~\lim\nolimits_{t~\to~\infty~}{w(t)=c},$ we know that $~\lim\nolimits_{t~\to~\infty~}{\Delta~=0}$. Thus, for any $\varepsilon>0$, there exists $T>0$, such that for any $t>T$, there is $\dot~V\leq~-c_1~V~+~\varepsilon$.

In conclusion, $~\lim\nolimits_{t~\to~\infty~}{x_1(t)=y^{**}}$ and $\lim\nolimits_{t~\to~\infty~}{x_2(t)=0}$. The proof of Theorem ` is completed.

Proof of Theorem 2

Because the dynamic equations of $e$ and $e_d$ can be written as follows: \begin{equation} \begin{cases} \dot e=e_d, \\ \dot e_d= -k_{ap} e -k_{ad} e_d +e_f, \end{cases} \tag{36}\end{equation} we get $|\ddot~e(t)~+~k_{ad}~\dot~e(t)~+~k_{ap}e(t)|=~|e_f(t)|$. Based on the proof of Theorem 1, when $\omega_o>\omega_o^*,$ $e_f,e$ and $e_d$ are bounded. Let \begin{align*}P_2=\begin{bmatrix} \dfrac{1+k_{ap}}{2k_{ad}} + \dfrac{k_{ad}}{2k_{ap}}& \dfrac{1}{2k_{ap}} \\ \dfrac{1}{2k_{ap}} & \dfrac{1+k_{ap}}{2k_{ad}k_{ap}}\end{bmatrix}. \end{align*} Consider the following Lyapunov function: \begin{equation} V_2(e,e_d)=\begin{bmatrix}e&e_d\end{bmatrix} P_2 \begin{bmatrix} e \\ e_d \end{bmatrix}. \tag{37}\end{equation} The time derivative of $V_2~(e,e_d)$ along the trajectories of (36) is \begin{equation} \dot V_2=-e^2 -e_d^2 + \begin{bmatrix}e&e_d\end{bmatrix} \begin{bmatrix} \dfrac{1}{k_{ap}} \\ \dfrac{1+k_{ap}}{k_{ad}k_{ap}}\end{bmatrix}e_f \leq - \frac{V_2}{\lambda_{\max}(P_2)}+\frac {\max\left(\frac{1}{k_{ap}},\frac{1+k_{ap}}{k_{ad}k_{ap}}\right)\sqrt V_2|e_f|} {\sqrt{\lambda_{\min}(P_2)}}. \tag{38}\end{equation} From (38), it can be seen that $$\sqrt {V_2} \leq \frac {\max\left(\frac{1}{k_{ap}},\frac{1+k_{ap}}{k_{ad}k_{ap}}\right)\lambda_{\max}(P_2)\mathop{{\rm sup}}|e_f|} {\sqrt{\lambda_{\min}(P_2)}} ,$$ i.e., $$\| [e,e_d] \| \leq \frac {\max\left(\frac{1}{k_{ap}},\frac{1+k_{ap}}{k_{ad}k_{ap}}\right)\lambda_{\max}(P_2)\mathop{{\rm sup}}|e_f|} {{\lambda_{\min}(P_2)}}.$$ If $\sqrt~{V_2}~>~\frac~{\max\left(\frac{1}{k_{ap}},\frac{1+k_{ap}}{k_{ad}k_{ap}}\right)\lambda_{\max}(P_2)\mathop{{\rm~sup}}|e_f|}~{\sqrt{\lambda_{\min}(P_2)}}~$, we know that $\dot~V_2<0.$

From (4), the estimation error $e_f$ satisfies the following equation: \begin{equation} \dot e_f = -\omega_o e_f - \dot f, \tag{39}\end{equation} where $$\dot f= \frac{\partial f}{\partial x_2} k_{ap} e +\left(\frac{\partial f}{\partial x_2} k_{ad}-\frac{\partial f}{\partial x_1}\right)e_d-\frac{\partial f}{\partial x_2}e_f+\frac{\partial f}{\partial x_1 }\dot r+\frac{\partial f}{\partial x_2} \ddot r +\frac{\partial f}{\partial t}.$$ Because $f\in~\mathcal{F}$, we have \begin{equation}|\dot f| \leq \gamma_1 |e| + \gamma_2 |e_d| +L_2 |e_f|+ \gamma_3, \tag{40}\end{equation} where $\gamma_1=L_2~k_{ap},\gamma_2=|L_2~k_{ad}-L_1|,\gamma_3=L_1\dot~r+L_2~\ddot~r+L_3.$

Consider the following Lyapunov function: \begin{equation} V_1(e_f)=\frac{1}{2}e_f^2, \tag{41}\end{equation} the time derivative of $V_1~(e_f)$ along the trajectories of (39) is \begin{equation} \dot V_1=-\omega_o e_f^2- e_f \dot f \leq -\omega_o e_f^2 +|e_f| |\dot f|\leq -(\omega_o-L_2) e_f^2 +|e_f|(\gamma_1 |e| + \gamma_2 |e_d| + \gamma_3). \tag{42}\end{equation}

Let $$\gamma_4=\frac {\max\left(\frac{1}{k_{ap}},\frac{1+k_{ap}}{k_{ad}k_{ap}}\right)\lambda_{\max}(P_2)} {{\lambda_{\min}(P_2)}}, \omega_{o1}^*=L_2+\gamma_4(\gamma_1+\gamma_2)+1.$$ Next, it will be proved that when $\omega_o~>~\omega_{o1}^*,~\mathop{\rm~sup}|e_f|~\leq~{\rm~max}~\{~e_f(0),\gamma_3\}.$ From (42), it can be seen that \begin{equation} \dot V_1< -(\gamma_4(\gamma_1+\gamma_2)+1) e_f^2 +|e_f|(\gamma_4(\gamma_1+\gamma_2)\mathop{\rm sup}|e_f| + \gamma_3). \tag{43}\end{equation} When $|e_f|~>~{\rm~max}~\{~e_f(0),\gamma_3\},$ there is $\dot~V_1~<0;$ thus, $|e_f|~\leq~{\rm~max}~\{~e_f(0),\gamma_3\}.$ Moreover, there exists a constant $\gamma_5$, which does not depend on $\omega_o$, such that $|~\dot~f|~\leq~\gamma_5.$

If $\omega_o^*~\leq~\omega_o~\leq~\omega_{o1}^*,$ then by Theorem 1 and (34), we know that there exists a constant $\gamma_6$, such that $\gamma_6=\mathop{\rm sup} \nolimits_{\omega_o} | \dot f|$.

Let $M_{~\dot~f}={\rm~max}\{~\gamma_5,\gamma_6\}.$ Based on the above analysis, it can be deduced that \begin{equation} \sqrt V_1 \leq {\rm e }^{-\omega_o t}\sqrt {V_1(e_f(0))} + \frac{\sqrt 2 M_{ \dot f}}{2\omega_o} (1-{\rm e }^{-\omega_o t}); \tag{44}\end{equation} thus \begin{equation} |e_f| \leq {\rm e }^{-\omega_o t}|e_f(0)| + \frac{ M_{ \dot f}}{\omega_o} (1-{\rm e }^{-\omega_o t}) \leq \eta_1 {\rm e }^{-\omega_o t} + \frac{\eta_2}{\omega_o} , \tag{45}\end{equation} where $\eta_1=~|e_f(0)|~,\eta_2~=~M_{~\dot~f}$, which are irrelevant to $\omega_o$. Thus, Eq. (13) is obtained and the proof of Theorem 2 is completed.

Proof of Theorem 3

Based on the proofs of Theorems 1 and 2, when $\omega_o~>~\omega_o^*$, both $f$ and $\dot~f$ are bounded. Because the PID controller defined by (3) and (14) is equivalent to the ADRC (4) and (5), the total disturbance $f$ of the closed-loop system defined by (1), (3) and (14) is the same as that of the closed-loop system (1), (4) and (5). Based on (36), we have \begin{equation} \ddot e +k_{ad} \dot e +k_{ap} e = e_f. \tag{46}\end{equation} Taking the Laplace transform for (46), we get \begin{equation} E(s)= \frac{1}{s^2+k_{ad}s+k_{ap}} E_f (s), \tag{47}\end{equation} where $E(s)$ is the Laplace transform of $e(t)$ and $~E_f~(s)$ is the Laplace transform of $e_f(t)$. Based on (39), we know that \begin{equation} \dot e_{f_{\rm I}}=-k_i e + \dot e_f +\omega_o e_f. \tag{48}\end{equation} Taking the Laplace transform for (48), it can be obtained from (47) that \begin{equation} {E_{f_{\rm I}} (s)} = \frac{s^2 + k_d s + k_p }{s^2+k_{ad} s + k_{ap} } {E_f (s)} , \tag{49}\end{equation} where $~E_{f_{\rm~I}}~(s)$ is the Laplace transform of $e_{f_{\rm~I}}(t)$. Taking the Laplace transform for (39), it follows that \begin{equation} E_f (s) =G_{e_f}(s) F(s), G_{e_f}(s)=\frac{s}{s+\omega_o }, \tag{50}\end{equation} where $~F(s)$ is the Laplace transform of $f$. Thus, \begin{equation} E_{f_{\rm I}} (s) = G_{e_{f_{\rm I}}}(s)F(s), G_{e_{f_{\rm I}}}(s)=\frac{s^3 + k_d s^2 + k_p s}{(s+\omega_o)(s^2+k_{ad} s + k_{ap})}. \tag{51}\end{equation}

From (50) and (51), we obtain \begin{equation} { \frac{|G_{e_f} (i\omega)|^2}{|G_{e_{f_{\rm I}}} (i\omega)|^2} }=\frac{(k_{ap}- \omega^2)^2 + k_{ad}^2 \omega^2 }{(k_{p}- \omega^2)^2 + k_{d}^2\omega^2 }<1. \tag{52}\end{equation} Because $\lim\nolimits_{t~\to~\infty~}~{\frac~{e_f(t)}{e_{f_{\rm~I}}(t)}}=\lim\nolimits_{s~\to~0~}{\frac{s~E_f(s)}{sE_{f_{\rm~I}}(s)}}$, it is easy to see that $$\lim\limits_{t \to \infty }{\frac{e_f(t)}{e_{f_{\rm I}}(t)}} = \frac{k_{ap} }{k_{ap}+\omega_o k_{ad} }.$$ Therefore, the property (1) of Theorem 3 is obtained.

Based on (1) and (5), it can be obtained that \begin{equation} \dddot {\hat f_{\rm I}}= -k_d {\ddot {\hat {f_{\rm I}}}} -k_p {\dot {\hat {{f}_{\rm I}}}} -k_i \hat f_{\rm I} + k_i f. \tag{53}\end{equation} Taking the Laplace transform for (53), we have \begin{equation} \hat F_{\rm I} (s) =\frac{\omega_o k_{ap}}{(s+\omega_o)(s^2+k_{ad} s + k_{ap})} F(s), \tag{54}\end{equation} where $~\hat~F_{\rm~I}~(s)$ is the Laplace transform of $\hat~f_{\rm~I}(t)$ and $~F(s)$ is the Laplace transform of $~f.$

Based on (4), the dynamical equation of $\hat~f$ can be written as follows: \begin{equation} \dot {\hat {f}} = -\omega_o (\hat f - f). \tag{55}\end{equation} Taking the Laplace transform for (55), we have \begin{equation} \hat F(s) =\frac{\omega_o}{s+\omega_o}F(s), \tag{56}\end{equation} where $~\hat~F(s)$ is the Laplace transform of $\hat~f(t)$. Thus, $$\hat F_{\rm I}(s)=\frac{ k_{ap}}{s^2+k_{ad} s + k_{ap}}\hat F(s),$$ which is the property (2) of Theorem 3. Hence, the proof of Theorem 3 is completed.

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• Figure 1

The response curves of the state (a) $x_1$ and (b) $x_2$.

• Figure 2

The estimations (a) and the estimation errors (b) of the disturbance $f$.

• Figure 5

The curves of $\ddot~e(t)~+~k_{ad}\dot~e(t)~+~k_{ap}e(t)$ under (a) C1, (b) C2, (c) C3, and (d) C4.