Designing and application of modified SSA based ADRC controller for overhead crane systems

Abstract

In this paper, a modified salp swarm algorithm (SSA) based active disturbance rejection control (ADRC) controller is designed to improve anti-interference and anti-sway control performance for overhead crane systems. Aiming at the ADRC parameter setting problem, the modified SSA is introduced with several methods including the parameter adjustment scheme, adaptive evaluation mobile strategy, opposition-based learning (OBL) strategy and disengagement strategy. By implementing numeric experiment tests, the effectiveness of the modified SSA is confirmed by benchmark functions. Finally, by implementing the simulation experiments under different conditions, the results show that the modified SSA algorithm based ADRC controller has significantly improved the suppression of load swing and position control as well as the anti-interference.

Appendix A. Analysis of the stability of the proposed ADRC

1.1 Appendix A.1. Preliminary

In order to analyse the stability of the proposed ADRC, we first simplify overhead crane system.

By appropriately selecting TD and ESO, as well as nonlinear functions and parameters in nonlinear combinations, ADRC can control a broad class of uncertain objects:

$$\begin{aligned} x^{4}=f\left( x, \dot{x}, \cdots , \dot{x}^{4}\right) +b u \quad y=x \end{aligned}$$

(21)

The state-space design of the overhead crane is simplified as follows:

$$\begin{aligned} \left\{ \begin{array}{l} \dot{x}_{1}=x_{2} \\ \dot{x}_{2}=x_{3} \\ \dot{x}_{3}=x_{4} \\ \dot{x}_{4}=f\left( x_{1}, x_{2}, x_{3}, x_{4}\right) +b u \\ y=x_{1} \end{array}\right. \end{aligned}$$

(22)

Constructing the ESO, where $g_{i}\left( z_{1}-y\right)$ Is a nonlinear function:

$$\begin{aligned} \left\{ \begin{array}{l}\dot{z}_{1}=z_{2}-g_{1}\left( z_{1}-y\right) \\ \dot{z}_{2}=z_{3}-g_{2}\left( z_{1}-y\right) +bu \\ \dot{z}_{3}=z_{4}-g_{3}\left( z_{1}-y\right) +bu \\ \dot{z}_{4}=z_{5}-g_{4}\left( z_{1}-y\right) +b u \\ \dot{z}_{5}=-g_{5}\left( z_{1}-y\right) \end{array}\right. \end{aligned}$$

(23)

Specifically, the nonlinear function is:

$$\begin{aligned} f a l(x, \alpha , \delta )=\left\{ \begin{array}{l} |x|^{\alpha } {\text {sign}}(x),|x| \geqslant \delta \\ \frac{x}{\delta ^{1-\alpha }},|x|<\delta \end{array}\right. \end{aligned}$$

(24)

and

$$\begin{aligned} g_{k}\left( z_{1}-y\right) =\beta _{k} \text{ fal } \left( z_{1}-y, \alpha _{k}, \delta \right) \end{aligned}$$

(25)

The nonlinear state error feedback is:

$$\begin{aligned} u=\sum _{i=1}^{4} k_{i}{\text {fal}}\left( v_{1}-z_{1}, \alpha _{i}, \delta \right) -\frac{z_{5}}{b} \end{aligned}$$

(26)

1.2 Appendix A.2. Analysis of the stability of the proposed ADRC

The ADRC is simplified as follows:

• 1. Let the input be zero ($y_{0}=0$), then the output of TD is zero.

• 2. Change the nonlinear state error feedback to linear error feedback:

  $$\begin{aligned} u=\sum _{i=1}^{4} k_{i} z_{i}-\frac{z_{5}}{b} \end{aligned}$$

  (27)

• 3. Choose the same non-linear function in ESO $\varphi (x)$, where $x \varphi (x)>0$, $x \ne 0$ ;and let $\varphi \in F\left( \mu _{1}, \mu _{2}\right)$ where $\varphi (0)=0$ and $\forall x \ne 0$, $\varvec{\mu }_{1} x^{2}<x \varphi (x)<\mu _{2} x^{2}$ then $x \varphi (x)>0$, namely $\varphi (x) \in F(0,+\infty ]$, then:

  $$\begin{aligned} \left\{ \begin{array}{l}\dot{z}_{1}=z_{2}-\beta _{1} \varphi \left( z_{1}-y\right) \\ \dot{z}_{2}=z_{3}-\beta _{2} \varphi \left( z_{1}-y\right) +b u \\ \dot{z}_{3}=z_{4}-\beta _{3} \varphi \left( z_{1}-y\right) +b u \\ \dot{z}_{4}=z_{5}-\beta _{4} \varphi \left( z_{1}-y\right) +b u \\ \dot{z}_{5}=-\beta _{5} \varphi \left( z_{1}-y\right) \end{array}\right. \end{aligned}$$

  (28)

• 4. Let the overhead crane be a linear time-invariant object:

  $$\begin{aligned} \begin{aligned} x^{(4)}=-a_{4} x-a_{3} \dot{x} -a_{2} \dot{x}^{(2)} -a_{3} \dot{x}^{(3)}+bu \\ y=x \end{aligned} \end{aligned}$$

  (29)

Substitute Eqs. (27) and (29) into Eq. 22, let $X=\left[ \varvec{x}_{1}, \varvec{x}_{2},\varvec{x}_{3} , \varvec{x}_{4}\right] ^{\mathrm {T}}$ , $Z=\left[ \varvec{z}_{1}, \varvec{z}_{2},\varvec{z}_{3} , \varvec{z}_{4}\right] ^{\mathrm {T}}$, we get:

$$\begin{aligned} \left\{ \begin{array}{l}\dot{X}=A_{11} X+A_{12} Z+a_{13} z_{5} \\ y=x_{1}\end{array}\right. \end{aligned}$$

(30)

where $a_{13}=[0, 0, 0,-1]^{\mathrm {T}} \in R^{4}$, $A_{11}=\left[ \begin{array}{cccccc}0 &{} 1 &{} 0 &{} &{} 0 \\ 0 &{} 0 &{} 1 &{} &{}0 \\ \ 0 &{}0 &{} 0 &{} &{} 1 \\ -a_{4} &{} -a_{3} &{}-a_{2} &{} &{} a_{1}\end{array}\right]$, $A_{12}=\left[ \begin{array}{cccccc}0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{}0 \\ \ 0 &{}0 &{} 0 &{} 0 \\ -bk_{1} &{} -bk_{2} &{}-bk_{3} &{} -bk_{4}\end{array}\right]$ Substituting Eq. (27) into Eq. (28), we can get:

$$\begin{aligned} \left\{ \begin{array}{l}\dot{Z}=A_{22} Z+b_{2} u \\ \dot{z}_{5}=\beta _{5} u \\ u=-\varphi \left( z_{1}-y\right) \end{array}\right. \end{aligned}$$

(31)

where $b_2=\left[ \varvec{\beta }_{1}, \varvec{\beta }_{2},\varvec{\beta }_{3} , \varvec{\beta }_{4}\right] ^{\mathrm {T}}$, $A_{22}=\left[ \begin{array}{cccccc}0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0&{} 1 &{}0 \\ \ 0 &{}0 &{} 0 &{} 1 \\ -bk_{1} &{} -bk_{2} &{}-bk_{3} &{} -bk_{4}\end{array}\right]$

By combining Eq. (30) with Eq. (31), we can get:

$$\begin{aligned} \left\{ \begin{array}{l}\dot{X}=A_{11} X+A_{12} Z+a_{13} z_{5} \\ \dot{Z}=A_{22} Z+b_{2} u \\ \dot{z}_{5}=\beta _{5} u \\ u=-\varphi (y) \\ y=c_{1}^{\mathrm {T}} X+c_{2}{ }^{\mathrm {T}} Z\end{array}\right. \end{aligned}$$

(32)

where $c_{1}=[-1, 0, 0,0]^{\mathrm {T}} \in R^{4}$, $c_{2}=[1, 0, 0,0]^{\mathrm {T}} \in R^{4}$ let $Y=A_{11} X+a_{13} z_{5}, z_{5}^{\prime }=z_{5} / \beta _{5}$ we get

$$\begin{aligned} \left\{ \begin{array}{l}\dot{Y}=A_{11} Y+A_{11} A_{12} Z+a_{13} \beta _{5} u \\ \dot{Z}=A_{22} Z+b_{2} u \\ \dot{z}_{5}=u \\ u=-\varphi (y) \\ y=c_{1}^{\mathrm {T}} A_{11}^{-1} Y+c_{2}^{\mathrm {T}} Z-c_{1}^{\mathrm {T}} A_{11}^{-1} a_{13} \beta _{5} z_{5}^{\prime }\end{array}\right. \end{aligned}$$

(33)

Equation (33) conforms to a standard form of the first critical case of absolute stability, namely:

$$\begin{aligned} \left\{ \begin{array}{l}\dot{x}=A x+b u \\ \xi =u \\ u=-\varphi (y) \\ y=c^{\mathrm {T}} x=\rho \xi \end{array}\right. \end{aligned}$$

(34)

where $A=\left[ \begin{array}{cc}A_{11} &{} A_{11} A_{12} \\ 0 &{} A_{22}\end{array}\right] \quad b=\left[ \begin{array}{l}a_{13} \beta _{5} \\ \quad b_{2}\end{array}\right]$, $c^{\mathrm {T}}=\left[ \begin{array}{ll}c_{1}^{\mathrm {T}} A_{11}^{-1}&c_{2}^{\mathrm {T}}\end{array}\right]$, $\rho =-c_{1}^{\mathrm {T}} A_{11}^{-1} a_{13} \beta _{5}=\frac{\beta _{5}}{a_{4}}$

The necessary condition for the absolute stability of the solution of $F(0,+\infty ]$ zero for system expression (34) is: $\text{ Re } \lambda (A)<0, \rho >0$.

If the linear time-invariant object is asymptotically stable, that is, all eigenvalues of $A_{11}$ have A negative real part, and the selected parameter $k_{I}$ guarantees that all eigenvalues of $A_{22}$ have A negative real part, then the eigenvalues of A have A negative real part, so the following conclusion can be drawn:

The necessary conditions for the absolute stability of ADRC system Eq. (33) are that the linear steady-state object is asymptotically stable and the controller parameter is greater than zero and that $\text {Re} \lambda \left( A_{22}\right) <0$. A Lyapunov function is constructed, which has the form of quadratic plus nonlinear integral:

$$\begin{aligned} U(x, y)=x^{\mathrm {T}} P x+\alpha \left( y-c^{\mathrm {T}} x\right) ^{2}+\beta \int _{0}^{y} \varphi (y) \mathrm {d} y \end{aligned}$$

(35)

If P is a positive definite matrix, $\alpha , \beta \geqslant 0$, then U(x, y)is a positive definite function.

Equation (35) by taking the total derivative of Eq. (34) with respect to t, we can get:

$$\begin{aligned} \begin{aligned}-\dot{U}(x, y)=&x^{\mathrm {T}}\left( -P A-A^{\mathrm {T}} P\right) x+\\&2\left( P b-\alpha \rho c-\frac{1}{2} \beta A^{\mathrm {T}} c\right) ^{\mathrm {T}} x \varphi (y)+\\&\left( \beta \rho +\beta c^{\mathrm {T}} b\right) \varphi ^{2}(y)+2 \alpha \rho y \varphi (y) \end{aligned} \end{aligned}$$

(36)

let $Q=-P A-A^{\mathrm {T}} P$, $d=P b-\left( \alpha \rho c+\frac{1}{2} \beta A^{\mathrm {T}} c\right)$,$\gamma =\beta \left( \rho +c^{\mathrm {T}} b\right)$. Because $y \varphi (y)> 0, \rho = \left( \beta _ {5} / a_ {4} \right) > 0$, as long as $\alpha \geqslant 0$and the first three entries of Eq. (36) are positive definite quadratic of X and N +1 variables of $\varphi$, then $\dot{U}$must be negative definite, thus ensuring the global asymptotic stability of the zero solution of Eq. (34).