A Lyapunov based Saturated Super-Twisting Algorithm

Abstract

Two different structures of saturated super-twisting algorithms are presented. Both structures switch from a relay controller to super-twisting algorithm through a switching law that is based on Lyapunov-level curves allowing the algorithms to generate bounded control signals. The relay controller provides a saturated control signal enforcing the system trajectories to reach a predefined neighborhood of the origin in which the super-twisting algorithm dynamics does not saturate, ensuring finite-time convergence to the origin. In order to increase the maximal admissible bound of the perturbations, the second algorithm also includes a perturbation estimator setting super-twisting’s integrator to the theoretically exact perturbation estimation. Experimental results are presented to validate the proposed algorithms.

Appendix

1.1 Proof of Theorem 1

This proof is based on Castillo et al. [2] in which it was shown that the RC forces the trajectories to the neighborhood of the origin in finite time. It is shown in Moreno et al. [11] that if the parameters $\alpha _1$ and $\alpha _2$ of the STA are designed as in (8), the function

$$\begin{aligned} V_{s}(x,z) = V_{s}(\xi _1,\xi _2) = \xi ^T P \xi , \ P = \begin{bmatrix} p_{11} &{} -p_{12} \\ -p_{12} &{} p_{22} \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 4\alpha _2 +\alpha _1^2 &{} -\alpha _1 \\ -\alpha _1 &{} 2 \end{bmatrix}>0, \end{aligned}$$

(26)

and vector $\xi ^T = \left[ \xi _1 \;\;\; \xi _2 \right] = \left[ \lceil x \rfloor ^{\frac{1}{2}} \;\;\; z \right]$ is a Lyapunov function for system (2) in closed loop with the STA (1). It ensures the finite-time convergence of the state to the origin and the exact compensation of the perturbation.

Next, the admissible range for threshold $\delta >0$ based on Positive Invariant Sets (PINS) for the closed loop with RC and STA is derived. If the control signal $u=\pm \rho$, one get the sets

$$\begin{aligned} \mathbb {U_+} = \left\{ (x,z)\in \mathbb {R}^2 \ \big \vert \ z = \rho + \alpha _1 \lceil x \rfloor ^{\frac{1}{2}} \right\} , \ \mathbb {U_-} = \left\{ (x,z)\in \mathbb {R}^2 \ \big \vert \ z = -\rho + \alpha _1 \lceil x \rfloor ^{\frac{1}{2}} \right\} . \end{aligned}$$

(27)

See the black dashed lines in the (x, z) plane shown in Fig. 10.

Fig. 10

(left) Nominal phase plane with the maximal PINS between the saturation sets. (right) phase plane with perturbation $|\phi (t)| \le \phi _{max}$. Switch of control law in a neighborhood of the origin $|x| \le \delta$ (green). System trajectory (blue)

The PINS contained between the two saturation sets (that only touches the saturation sets in only one point) is defined as $\Omega _s = \lbrace \xi \in \mathbb {R}^2 | V_s \le c_s \rbrace$. In order to find $c_s$, one evaluates Lyapunov function (26) at the (upper) saturation set $\mathbb {U_+}$ such that $V_{s}\left( \xi _1, \rho + \alpha _1 \xi _1\right) - c_s = 0$. Saturation set $\mathbb {U_-}$ is not considered due to symmetry. This relation can be written as quadratic equation in terms of $\xi _1$ as $a_N \xi _1^2 + b_N \xi _1 + c_N = 0$ where $a_N = p_{22} \alpha _1^2 - 2 p_{12} \alpha _1 + p_{11}$, $b_N = -2 \rho \left( p_{12} - \alpha _1 p_{22}\right)$, and $c_N = p_{22}^{2} \rho - c_s$. To achieve a unique solution that only touches the saturation set in one point, the discriminant $b_N^2 - 4 a_N c_N$ of this quadratic equation has to vanish. Solving for $c_s$ yields

$$\begin{aligned} c_s = \rho ^2 \mu \qquad \text {with}\qquad \mu = \frac{p_{11} p_{22} - p_{12}^2 }{p_{22} \alpha _1^2 -2p_{12} \alpha _1 + p_{11}} . \end{aligned}$$

(28)

As shown in Fig. 10(left), $V_s(\delta _{max},z=0)= c_s$ defines the maximum value of $\delta$ such that PINS does not exceed the saturation set in the nominal case, i.e., $\phi _{max}=0$, such that $V_s(x,0) = p_{11}|x| = c_s$. The maximum admissible value for |x| to guarantee that the trajectory stays in the PINS is given by

$$\begin{aligned} |x| = \frac{c_s}{p_{11}} , \qquad \text {resulting in} \qquad 0 \le \delta \le \frac{c_s}{p_{11}}. \end{aligned}$$

(29)

Subsequently, the case with perturbation is considered, where $\xi _2 = z + \phi (t)$ holds. To find the maximum bound for the perturbation (9) that can be rejected, a level curve from Lyapunov function (26) in presence of the perturbation is evaluated in one of the saturation sets, i.e., $V_{s}\left( \xi _1,\rho +\alpha _1 \xi _1-\phi _{max}\right) = \ell _{\rho }$, that can be represented as a quadratic equation $a_{\rho } \xi _1^2 + b_{\rho } \xi _1 + c_{\rho } = 0$. In contrast to the nominal case, the size and the center of the level curves now depend on the maximum perturbation as shown in Fig. 10(right). Setting the discriminant to zero, one can find the level curve that touches the boundary $|u| = \rho$ at only one point, i.e., $b_{\rho }^2 -4a_{\rho }c_{\rho } = 0$, and one gets $\ell _{\rho } = \left( \phi _{max} - \rho \right) ^2 \mu$.

In order to find the relation between $\phi _{max}$ and $\delta$, the level curve is evaluated at $x=\delta$ and $z=0$ resulting in

$$\begin{aligned} V_s\left( \sqrt{\delta }, -\phi _{max}\right) = \ell _\rho . \end{aligned}$$

(30)

• 1. Solving (30) for $\delta$ depending on $\phi _{max}$ yields a quadratic equation in $\sqrt{\delta }$ as $a_\delta \delta + b_\delta \sqrt{\delta } + c_\delta = 0$ where $a_\delta = p_{11}$, $b_\delta = 2 p_{12} \phi _{max}$, and $c_\delta = p_{22} \phi _{max}^2 - \mu \left( \phi _{max} - \rho \right) ^2$. Then, the parameter $\delta$ is chosen less or equal than the minimum quadratic absolute values of the roots $0 \le \delta \le \delta _{max}$,

  $$\begin{aligned} \begin{array}{rl} \delta _{max} =&\min \left[ \left( \frac{- p_{12} \phi _{max} + \sqrt{\Delta _\delta (\phi _{max})} }{ p_{11}} \right) ^2, \left( \frac{- p_{12} \phi _{max} - \sqrt{\Delta _\delta (\phi _{max})} }{ p_{11}} \right) ^2 \right] = \left( \frac{- p_{12} \phi _{max} + \sqrt{\Delta _\delta (\phi _{max})} }{ p_{11}} \right) ^2 \end{array} \end{aligned}$$

  (31)

  where $\Delta _\delta \left( \phi _{max}\right) = \left( p_{12}^2+ \mu p_{11} - p_{11} p_{22}\right) \phi _{max}^2- 2 \mu p_{11} \phi _{max} \rho + \mu p_{11} \rho ^2 = p_{12}^2 \phi _{max}^2 - p_{11} \left( p_{22} \phi _{max}^2 -\mu \left( \phi _{max}-\rho \right) ^2 \right) .$

• 2. Solving (30) for the maximum perturbation bound $\phi _{max}$ depending on the choice of parameter $\delta$, a quadratic equation in $\phi _{max}$ as $a_\phi \phi _{max}^2 + b_\phi \phi _{max} + c_\phi = 0$ where $a_\phi = p_{22}-\mu$, $b_\phi = 2 \mu \rho + 2 p_{12} \sqrt{\delta }$, and $c_\phi = p_{11} \delta - \mu \rho ^2$ is considered. Then, the maximum bound $\phi _{max}$ that the algorithm can suppress is given by the minimum absolute value of the roots

  $$\begin{aligned} \begin{array}{rl} \phi _{max} =&\min \left[ \left| \frac{\mu \rho + p_{12} \sqrt{\delta } - \sqrt{\Delta _{\rho }(\delta )}}{\mu - p_{22}} \right| , \left| \frac{\mu \rho + p_{12} \sqrt{\delta } + \sqrt{\Delta _{\rho }(\delta )}}{\mu - p_{22}} \right| \right] = \frac{\mu \rho + p_{12} \sqrt{\delta } - \sqrt{\Delta _{\rho }(\delta )}}{\mu - p_{22}}, \end{array} \end{aligned}$$

  where $\Delta _{\rho }\left( \delta \right) = \left( p_{12}^2 + \mu p_{11} - p_{11} p_{22} \right) \delta + 2 \mu p_{12} \rho \sqrt{\delta } + \mu p_{22} \rho ^2 = \left( \mu \rho +\right. \left. p_{12} \sqrt{\delta } \right) ^2 - \left( \mu \rho ^2 - p_{11} \delta \right) \,\left( \mu - p_{22}\right) .$ If $\delta$ is set to zero, $\phi _{max}$ reduces to

  $$\begin{aligned} \phi _{max} = \kappa \rho \qquad \text {with}\qquad \kappa = \frac{\mu - \sqrt{\mu }}{\mu -1} . \end{aligned}$$

  (32)

  Substituting (26) and (28) into (32) results in

  $$\begin{aligned} \kappa = \frac{\mu - \sqrt{\mu }}{ (\mu -1)} \cdot \frac{\mu + \sqrt{\mu }}{\mu + \sqrt{\mu }} = \frac{1}{1 + \frac{\sqrt{\mu }}{\mu } \cdot \frac{\sqrt{\mu }}{\sqrt{\mu }}} = \frac{1}{1 + \frac{1}{\sqrt{\mu }}} = \frac{1}{1 + \sqrt{1 + \frac{\alpha _1^2}{\alpha _1^2 + 8 \alpha _2}}} \end{aligned}$$

  (33)

  which is clearly less than $\frac{1}{2}$.

Taking the values of the Lyapunov function (26) depending on $\alpha _1$, $\alpha _2$, together with (28), (32), and (31) leads to (9)–(10), respectively. This completes the proof. $\blacksquare$

1.2 Proof of Theorem 2

The proof is performed in two steps. First, it is shown how the estimator has to be tuned to achieve convergence before the RC switches to STA. Then, conditions for the STA parameters are derived. The error dynamics of estimator (12) are

$$\begin{aligned} \begin{array}{rl} \dot{e}_1 = &{} -\beta _1 \lceil e_1 \rfloor ^{1/2} + e_2 \\ \dot{e}_2 =&{} -\beta _2 \text {sign}\left( e_1\right) + \dot{\phi } \end{array} \end{aligned}$$

(34)

with $e_2 = \hat{x}_2 + \phi (t)$. Therefore, there exist gains $\beta _1$ and $\beta _2$ depending on (3) and a time $T_e>0$ where $e_1 = e_2 = 0$ as shown in [11, 13]. This implies that $\hat{x}_2 = - \phi (t)$ for all future time $t>T_e$. For the SSTA, we design the estimator gains to make the time $T_e$ smaller than the minimum time of convergence of the state under relay control.

The estimation of the minimum reaching time $T_{cmin}$ of the RC is made considering the case when the perturbation helps the system trajectories to converge to a vicinity of the origin $|x(t)|\le \delta$, starting from an initial condition $x_0 = x(0)$ outside of the vicinity $|x_0|> \delta$. Then, using Lyapunov function $V_{c}(x) = c_1|x|, c_1 > 0$ from Castillo et al. [2] yields the time derivative

$$\begin{aligned} \dot{V}_c(x) = - c_1 \rho +c_1 \text {sign}\left( x\right) \phi (t), \ \text {resp.} \ \min _{|\phi (t)|\le \phi _{max}} \dot{V}_c(x) = - c_1 \left( \rho + \phi _{max}\right) . \end{aligned}$$

(35)

If one selects $c_1 = 1/\left( \rho + \phi _{max}\right)$, the Lyapunov function derivative becomes $\dot{V}_c \ge -1$. In order to estimate to time of convergence from the initial condition to the neighborhood $\delta$, distance $|x_0|-\delta$ is considered in the Lyapunov function, i.e.,

$$\begin{aligned} V_c(x) \ge V_c(x_0) - t = \frac{|x_0|-\delta }{\rho + \phi _{max}} -t \end{aligned}$$

(36)

for $t_0 = 0$. This shows that $V_c$ reduces to $c_1 \delta$ no earlier than $T_{cmin}$ given in (19).

For the second part of the proof, consider the Lyapunov function from Polyakov et al. [13]

(37)

where $\bar{e}= [e_1 \;\; e_2]^T$. The terms $k(\bar{e})$ and $k_0(\bar{e})$ depend on the state $e_1$ and $e_2$, and $\bar{k}$ is a design parameter depending on L and the gains $\beta _1$ and $\beta _2$. Expressions $s(\bar{e})$ and $m(\bar{e})$ are also nonlinear functions of the state.

Note that with the knowledge of the initial condition $x_0$ it is possible to set $\hat{x}_1(0)= x_0$, and therefore $e_1(0) = x_0 - \hat{x}_1(0) = 0$ and as a result to use the second case of (37). We choose a parametrization of the estimator gains

$$\begin{aligned} \beta _1 =\ 2\sqrt{(18L+\epsilon )}, \qquad \beta _2 =\ 14L+\epsilon , \end{aligned}$$

(38)

with $\epsilon >0$, such that the conditions of Theorem 1 in Polyakov et al. [13] hold, i.e.,

$$\begin{aligned} \beta _2 = 14L+\epsilon >5L, \end{aligned}$$

(39)

and

$$\begin{aligned} \begin{array}{rcccl} 64L &{}<&{} \beta _1^2 &{}<&{} 8(\beta _2-L)\\ 64L &{}<&{} (2\sqrt{18L+\epsilon })^2 &{}<&{} 8((14L+\epsilon )-L)\\ 64L &{}<&{} 72L+4\epsilon &{}<&{} 104L+8\epsilon . \end{array} \end{aligned}$$

(40)

In order to design $\bar{k}$ in (37), parameter $g = 8 \gamma /\beta _1^2$ with $\gamma = \beta _2 - L \text {sign}\left( e_1e_2\right)$ as shown in Polyakov et al. [13] is considered. It may take two possible values $g^-=8(\beta _2-L)/\beta _1^2$ and $g^+=8(\beta _2+L)/\beta _1^2$ depending on the values of $\gamma \in \left\{ \beta _2+L,\beta _2-L \right\}$. Taking into account (38), function $g = \frac{2(14L + \epsilon \pm L)}{18L + \epsilon }$ also varies depending on the selection of the parameter $\epsilon$. Note that g is monotone with respect to $\epsilon$ since its derivative with respect to $\epsilon$ is positive, i.e.,

$$\begin{aligned} \frac{d g}{d \epsilon } = \frac{2}{18L + \epsilon } - \frac{ 2(14L \pm L + \epsilon )}{(18L + \epsilon )^2} = \frac{ 2(4L \pm L) }{(18L + \epsilon )^2} >0 . \end{aligned}$$

(41)

Therefore, the limits of $g^-$ and $g^+$ when $\epsilon \rightarrow 0$ and $\epsilon \rightarrow \infty$ are taken

$$\begin{aligned} \begin{array}{cc} \lim \limits _{\epsilon \rightarrow 0} g^- = \frac{13}{9}, &{} \qquad \lim \limits _{\epsilon \rightarrow \infty } g^- = 2, \\ \lim \limits _{\epsilon \rightarrow 0} g^+ = \frac{15}{9}, &{} \qquad \lim \limits _{\epsilon \rightarrow \infty } g^+ = 2. \\ \end{array} \end{aligned}$$

(42)

The whole range of variation of g depending on $\gamma$ and $\epsilon$ is $g \in \left[ g_m,\; g_M\right] =\left[ \frac{13}{9},\; 2\right]$.

Parameter $\bar{k}$ should belong to a intersection set of the intervals $I\left( g_m\right) \cap I\left( g_M\right) \ne 0$, where the interval $I\left( g\right)$ is given by

$$\begin{aligned} I(g) = \left( \frac{2}{g} + \frac{\exp {\left( 1/\sqrt{g-1}\right) \left( -\pi /2-\arctan \left( 1/\sqrt{g-1}\right) \right) }}{\sqrt{g}}, \frac{\exp {\left( 1/\sqrt{g-1}\right) \left( \pi /2-\arctan \left( 1/\sqrt{g-1}\right) \right) }}{\sqrt{g}} \right) . \end{aligned}$$

(43)

Evaluating the endpoints of g yields $I(g^-) = [1.4027, \;\; 2.01]$ and $I(g^+) = [1.0670, \;\; 1.5509]$. Parameter $\bar{k}$ can be selected as $\bar{k}= 1.4768$.

Using Theorem 1 in Polyakov et al. [13], we ensure that the time derivative of (37) along the trajectories of the system satisfies

$$\begin{aligned} \dot{V}_e \le -k\sqrt{V_e} \le -k_{min}\sqrt{V_e} \end{aligned}$$

(44)

and if the bound for $|e_2(0)|= |\phi (0)| = \phi _{max}$, the reaching time estimate can be referred to as

$$\begin{aligned} T_e \le \frac{2}{k_{min}} \sqrt{V_e\left( 0,\phi _{max}\right) }, \end{aligned}$$

(45)

with

$$\begin{aligned} k_{min} = \frac{\beta _1}{\sqrt{8}}\ \min _{\begin{array}{c} g \in \{ g^-,g^+ \}\\ \epsilon \in \{ 0,\infty \} \end{array}} f(g,\epsilon ) \end{aligned}$$

(46)

and

$$\begin{aligned} f(g,\epsilon ) = \left| g \bar{k} -\sqrt{g}\exp {\left( \frac{arctan\left( \frac{-1}{\sqrt{g-1}} \right) +\left( \frac{\pi (\beta _1^2g-8\beta _2)}{16L} \right) }{\sqrt{g-1}} \right) } \right| . \end{aligned}$$

(47)

Evaluating f with the two limits $g_m$ and $g_M$ and the parametrization of g, $f(g,\epsilon ) \in \left[ f_m,\; f_M\right] = [0.1550,\; 2.8066 ]$, and $k_{min} = \frac{\beta _1}{\sqrt{8}}f_m$.

From (37), (45), and setting the reaching time of the estimator $T_e$ less than the minimum reaching time of the state $T_{cmin}$, one gets

$$\begin{aligned} T_e \le \frac{8 \bar{k}\phi _{max}}{\beta _1^2 f_m} < T_{cmin}. \end{aligned}$$

(48)

Substituting (19) in (48) and solving for $\beta _1$ result in

$$\begin{aligned} \beta _1 \ge \sqrt{\frac{8\bar{k}}{f_m} \frac{\left( \phi _{max}^2 + \rho \phi _{max} \right) }{|x_0|-\delta }}. \end{aligned}$$

(49)

From $\beta _1$ parametrization (38), we solve for $\epsilon$ such that

$$\begin{aligned} \epsilon = \frac{1}{4}\beta _1^2 -18L. \end{aligned}$$

(50)

Substituting (50) in $\beta _2$ parametrization (38) yields

$$\begin{aligned} \beta _2 = \frac{1}{4}\beta _1^2 - 4L . \end{aligned}$$

(51)

Using the equality case of (49) results in

$$\begin{aligned} \beta _2 \ge \frac{2\bar{k}}{f_m}\frac{\left( \phi _{max}^2 + \rho \phi _{max} \right) }{|x_0|-\delta } -4L . \end{aligned}$$

(52)

Note that the value of the gains $\beta _1$ and $\beta _2$ tends to zero and to $-4L$, respectively, as $x_0$ tends to infinity because the restriction $\epsilon >0$ disappeared in these expressions. Therefore, conditions (49) and (52) are expressed as the maximum of two values to get and (17).

As shown before and in Castillo et al. [2], the PINS for STA (red curves in Fig. 11) move up or down in the (x, z)-plane depending on $\phi _{max}$. In addition, they change its size depending on $\delta$.

With the exact estimate of the perturbation, the integral control is set to $z(t_1) = - \phi (t_1)$ when the trajectory enters the neighborhood $|x(t_1)|\le \delta$ at time $t_1$, then equation (30) becomes

$$\begin{aligned} V_s\left( \sqrt{\delta }, 0\right) - \ell _\rho = p_{11} \delta - \mu \left( \phi _{max} - \rho \right) ^2 = 0. \end{aligned}$$

(53)

The set of possible choices of the parameter $\delta$ depending on $\phi _{max}$ is then given by

$$\begin{aligned} 0 \le \delta \le \delta _{max}; \qquad \delta _{max} = \frac{\mu \left( \phi _{max} - \rho \right) ^2}{p_{11}}. \end{aligned}$$

(54)

The set of maximum perturbation bound $\phi _{max}$ depending on the choice of parameter $\delta$ is

$$\begin{aligned} \phi _{max} = \rho - \sqrt{\frac{p_{11} \delta }{\mu }}. \end{aligned}$$

(55)

Note that the maximal rejectable perturbation depends on $\delta$, i.e., the smaller $\delta$, the higher maximum perturbation up to $\phi _{max} < \rho$, when $\delta = 0$.

Fig. 11

With the exact estimation of the perturbation the trajectories can start into a PINS of a size depending on $\delta$

The big red region in Fig. 11 equals a PINS for the choice $\delta =\delta _1$. As a result, perturbations with maximum $|\phi _{max1}|$ can be rejected. A choice of $\delta =\delta _2<\delta _1$ gives a smaller red region as shown in Fig. 11. As a consequence, perturbations with higher magnitude $|\phi _{max2}|>|\phi _{max1}|$ can be eliminated.

Using the values of the Lyapunov function (26) depending on $\alpha _1$, $\alpha _2$, together with (28), (55), and (54), we get (15) and (16), respectively.

This completes the proof. $\blacksquare$