Dynamic Motion Backstepping Control of Underwater Autonomous Vehicle Based on Averaged Sub-gradient Integral Sliding Mode Method

Abstract

The aim of this study is to develop robust guidance laws for the control motion of an underwater autonomous vehicle (UAV) in a three-dimensional (3D) space. The control design is based on the use of Averaged Sub-Gradient (ASG) version of a class of dynamic integral sliding mode (ISM) method being sequentially applied to the subsystems of the complete model realizing, the so-called backstepping (or cascade) approach. The mathematical form of the UAV model induces a backstepping formulation for solving the tracking trajectory problem sequentially for the position, translation velocity, angular velocity and actuators (thrusters) dynamics. The solution of the trajectory tracking problem at each stage implements the ASG-version of the ISM method. This problem is treated as the optimization of a suitable convex (not obligatory strongly convex) cost functional, depending on the tracking error and reaching its minimal value at the origin of the error tracking space. This study shows that the minimization of the proposed functional leads to the optimal tracking regime under the presence of uncertainties in the mathematical model description. A numerical example proves the effectiveness of the suggested robust dynamic controller. The comparison between the obtained trajectory tracking results and the outcomes produced by a set of standard proportional integral derivative (PID) controllers, is presented. The proposed controller exhibits a better tracking of the reference trajectory compared with the PID version, showing a smaller mean square estimation for the tracking error.

Appendix

Proof Proof of Theorem 1

To prove that the designed pseudo-controllers my solve the proposed optimization problems, let consider the following general equivalent stabilization problem.

• a) Consider the Lyapunov function

  $$ V(\mathbf{s})=\frac{1}{2}\left\Vert \mathbf{s}\right\Vert^{2},\text{ }\rho >0, $$

  (48)

  Taking then s := s₁, we get

  $$ \left. \begin{array}{c} \dot{V}(\mathbf{s}_{1})=\mathbf{s}_{1}^{\top }\dot{\mathbf{s}}_{1}= \mathbf{s}_{1}^{\top }\left[ \ddot{\mathbf{\varphi}}_{1}+\frac{\dot{\mathbf{ \varphi}}_{1}}{t+\kappa }-\frac{\mathbf{\varphi }_{1}+\mathbf{\alpha } _{1}}{\left( t+\kappa \right)^{2}}-\frac{1}{t+\kappa }\mathbf{\Gamma }_{1}+ \frac{1}{t+\kappa }\partial J_{\mathbf{1}}(\mathbf{\varphi }_{1})\right] \\ =\mathbf{s}_{1}^{\top }\left[ \ddot{\mathbf{{\varkappa}}}-\ddot{{\varkappa}}^{\ast }+\frac{\dot{\mathbf{{\varkappa}}}-\dot{{\varkappa}}^{\ast }}{t+\kappa }- \frac{\mathbf{{\varkappa} -{\varkappa} }^{\ast }+\mathbf{\alpha }_{1}}{\left( t+\kappa \right)^{2}}-\frac{1}{t+\kappa }\mathbf{\Gamma }_{1}+\frac{1}{ t+\kappa }\partial J_{\mathbf{1}}(\mathbf{\varphi }_{1})\right] \\ =\mathbf{s}_{1}^{\top }\left[ \frac{d}{dt}\left( {{\varTheta}} \mathbf{u} _{1}^{\ast }\right) +\dot{\mathbf{\zeta}}_{\mathbf{{\varkappa} }}+\mathbf{g} _{1}\right] . \end{array} \right\} $$

  (2)

  Select the intermediate pseudo-control $\mathbf {\upsilon =u}_{1}^{\ast }$ satisfying (17). Then from (2) we get

  $$ \left. \begin{array}{c} \dot{V}(\mathbf{s}_{1})=\mathbf{s}_{1}^{\top }\left[ -k_{1}\text{sign} \left( \mathbf{s}_{1}\right) +\dot{\mathbf{\zeta}}_{\mathbf{{\varkappa} }} \right] \leq \left( -k_{1}\sum\limits_{i=1}^{3}\left\vert s_{1,i}\right\vert +\left\Vert \mathbf{s}_{1}\right\Vert \dot{\zeta}_{ \mathbf{{\varkappa} }}^{+}\right) \\ \leq \left\Vert \mathbf{s}_{1}\right\Vert \left( -k_{1}+\dot{\zeta}_{\mathbf{ {\varkappa} }}^{+}\right) =-\mathring{\rho}\left\Vert \mathbf{s} _{1}\right\Vert =-\mathring{\rho}\sqrt{2V(\mathbf{s}_{1})}, \end{array} \right\} $$

  which leads to the following relations

  $$ \left. \begin{array}{c} \frac{dV(\mathbf{s}_{1})}{\sqrt{V(\mathbf{s}_{1})}}\leq -\mathring{\rho} \sqrt{2}dt\rightarrow 2\left( \sqrt{V(\mathbf{s}_{1})}-\sqrt{V(\mathbf{s} _{1}(0))}\right) \leq -\mathring{\rho}\sqrt{2}t, \\ 0\leq \sqrt{V(\mathbf{s}_{1})}\leq \sqrt{V(\mathbf{s}_{1}(0))}-\frac{ \mathring{\rho}}{\sqrt{2}}t, \end{array} \right\} $$

  implying that $V(\mathbf {s}_{1}\left (t\right ) )=0$ for all

  $$ t\geq t_{reach}:=\frac{1}{\mathring{\rho}}\sqrt{2V(\mathbf{s}_{1}(0))}= \frac{\left\Vert \mathbf{s}_{1}\left( 0\right) \right\Vert }{\mathring{\rho} }. $$

  (3)

  But by (20), $\mathbf {s}_{1}\left (0\right ) =\mathbf {0}$, and hence from the beginning of the process

  $$ \mathbf{s}_{1}\left( t\right) =\dot{\mathbf{s}}_{1}\left( t\right) =\mathbf{0 }. $$

  (4)

• b) Let now show that the robust controller (17), (19), (20), providing property (4), solves the optimization problem (16) as in (21). Indeed, following [17] and defining μ(t) := t + κ, we represent (4) as

  $$ \begin{array}{@{}rcl@{}} \mu (t)\mathbf{s}_{1}&=&\mu (t)\dot{\mathbf{\varphi}}_{1}(t)+\mathbf{\varphi } _{1}(t)+\mathbf{\alpha }_{1}+\mathbf{\gamma }(t)=0,\\&&\text{ }\dot{\mathbf{ \gamma}}(t)=\partial J_{\mathbf{1}}(\mathbf{\varphi }_{1}(t)),\text{ } \mathbf{\gamma }(0)=0, \end{array} $$

  or, equivalently,

  $$ \mu (t)\dot{\mathbf{\varphi}}_{1}(t)+\mathbf{\varphi }_{1}(t)+\mathbf{\alpha }_{1}=-\mathbf{\gamma }(t), $$

  which gives

  $$ \begin{array}{c} \frac{d}{dt}\left[ \frac{1}{2}\left\Vert \mathbf{\gamma }\right\Vert^{2} \right] =\dot{\mathbf{\gamma}}^{\top }\mathbf{\gamma }=-\partial^{\top }J_{\mathbf{1}}(\mathbf{\varphi }_{1})\left[ \mu \dot{\mathbf{ \varphi}}_{1}+\mathbf{\varphi }_{1}+\mathbf{\alpha }_{1}\right] \\ =-\partial^{\top }J_{\mathbf{1}}(\mathbf{\varphi }_{1})\mathbf{\varphi }_{1}-\partial^{\top }J_{\mathbf{1}}(\mathbf{\varphi }_{1})\left( \mu \dot{\mathbf{\varphi}}_{1}+\mathbf{\alpha }_{1}\right) . \end{array} $$

Applying the inequality

$$ \partial^{\top }J_{\mathbf{1}}(\mathbf{\varphi }_{1})\mathbf{\varphi } _{1}\geq J_{\mathbf{1}}(\mathbf{\varphi }_{1})-J_{\mathbf{1}}(\mathbf{ \varphi }_{1}^{\ast })=J_{\mathbf{1}}(\mathbf{\varphi }_{1}),\text{ }J_{\mathbf{1}}(\mathbf{\varphi }_{1}^{\ast })=0 $$

(49)

to the first term in the right-hand side and using the identity

$$ \partial^{\top }J_{\mathbf{1}}(\mathbf{\varphi }_{1})\dot{\mathbf{ \varphi}}_{1}=\frac{d}{dt}J_{\mathbf{1}}(\mathbf{\varphi }_{1}), $$

we have

$$ \frac{d}{dt}\left[ \frac{1}{2}\left\Vert \mathbf{\gamma }\right\Vert^{2} \right] \leq -J_{\mathbf{1}}(\mathbf{\varphi }_{1})-\mu \frac{d}{dt}J_{ \mathbf{1}}(\mathbf{\varphi }_{1})-\partial^{\top }J_{\mathbf{1}}(\mathbf{\varphi }_{1})\mathbf{\alpha }_{1}. $$

Then, integrating this inequality on interval $\left [ 0,t\right ]$, we get

$$ \begin{array}{c} \int\limits_{\tau =0}^{t}J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau \leq \frac{1}{2}\left( \underset{0}{\underbrace{\left\Vert \mathbf{\gamma }\left( 0\right) \right\Vert^{2}}}-\left\Vert \mathbf{\gamma }\right\Vert^{2}\right) - \\ \int\limits_{\tau =0}^{t}\mu \left( \tau \right) \frac{d}{dt}J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau -\left( \int\limits_{\tau =0}^{t}\partial J_{\mathbf{1}}(\mathbf{\varphi }_{1})d\tau \right)^{\top }\mathbf{\alpha }_{1} \end{array} $$

Since $\dot {\mu }_{\tau }=1$, in the last inequality leads (using of the integration by parts) we have

$$ \left. \begin{array}{c} \int\limits_{\tau =0}^{t}\mu \left( \tau \right) \frac{d}{dt}J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau = \left[ \mu \left( \tau \right) J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )\right]_{\tau =0}^{\tau =t}-\int\limits_{\tau =0}^{t}\dot{\mu} \left( \tau \right) J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau \\ =\mu J_{\mathbf{1}}(\mathbf{\varphi }_{1})-\kappa J_{\mathbf{1}}(\mathbf{ \varphi }_{1}\left( 0\right) )-\int\limits_{\tau =0}^{t}J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau \\ =\mu J_{\mathbf{1}}(\mathbf{\varphi }_{1})-\kappa J_{\mathbf{1}}(\mathbf{ \varphi }_{1}\left( 0\right) )-\mathbf{\gamma }, \end{array} \right\} $$

which leads to

$$ \begin{array}{@{}rcl@{}} ~\int\limits_{\tau =0}^{t}J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau &\leq& -\frac{1}{2}\left\Vert \mathbf{\gamma }\right\Vert^{2}-\mu J_{\mathbf{1}}(\mathbf{\varphi }_{1})+\kappa J_{\mathbf{1}}(\mathbf{ \varphi }_{1}\left( 0\right) )\\ &&+\int\limits_{\tau =0}^{t}J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( \tau \right) )d\tau -\mathbf{\gamma }^{\top }\mathbf{\alpha }_{1} \end{array} $$

or equivalently,

$$ \left. \begin{array}{c} \mu J_{\mathbf{1}}(\mathbf{\varphi }_{1})\leq \kappa J_{\mathbf{1}}(\mathbf{ \varphi }_{1}\left( 0\right) )-\frac{1}{2}\left\Vert \mathbf{\gamma } \right\Vert^{2}-\mathbf{\gamma }^{\top }\mathbf{\alpha }_{1}=\kappa J_{ \mathbf{1}}(\mathbf{\varphi }_{1}\left( 0\right) )-\frac{1}{2}\left( \left\Vert \mathbf{\gamma }\right\Vert^{2}+2\mathbf{\gamma }^{\top } \mathbf{\alpha }_{1}\right) \\ =\kappa J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( 0\right) )-\frac{1}{2} \left( \left\Vert \mathbf{\gamma }\right\Vert^{2}+2\mathbf{\gamma } ^{\top }\mathbf{\alpha }_{1}+\left\Vert \mathbf{\alpha }_{1}\right\Vert^{2}\right) +\frac{1}{2}\left\Vert \mathbf{\alpha }_{1}\right\Vert^{2} \\ =\kappa J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( 0\right) )-\frac{1}{2} \left\Vert \mathbf{\gamma +\alpha }_{1}\right\Vert^{2}+\frac{1}{2} \left\Vert \mathbf{\alpha }_{1}\right\Vert^{2}\leq \kappa J_{\mathbf{1}}(\mathbf{\varphi }_{1}\left( 0\right) )+\frac{1}{2}\left\Vert \mathbf{\alpha }_{1}\right\Vert^{2} \end{array} \right\} $$

that gives (22). Theorem is proven. □

Proof Proof of Theorem 2

$$ \left. \begin{array}{c} \dot{V}(\mathbf{s}_{2})=\mathbf{s}_{2}^{\top }\dot{\mathbf{s}}_{2} \\ =\mathbf{s}_{2}^{\top }\left[ \ddot{\mathbf{\upsilon}}-\ddot{\mathbf{ \upsilon}}^{\ast }+\frac{\dot{\mathbf{\upsilon}}-\dot{\mathbf{\upsilon}} ^{\ast }}{t+\kappa }-\frac{\mathbf{\upsilon }-\mathbf{\upsilon }^{\ast }+ \mathbf{\alpha }_{2}}{\left( t+\kappa \right)^{2}}-\frac{1}{t+\kappa } \mathbf{\Gamma }_{2}+\frac{1}{t+\kappa }\partial J_{\mathbf{2}}(\mathbf{ \varphi }_{2})\right] \\ =\mathbf{s}_{2}^{\top }\left[ \frac{d}{dt}\left( \mathbf{B}_{2}\mathbf{ u}_{2}\right) +\dot{\mathbf{\zeta}}_{\upsilon }+\mathbf{g}_{2}\right] = \mathbf{s}_{2}^{\top }\left[ -k_{2}\text{Sign}\left( \mathbf{s}_{2}\right) +\dot{\mathbf{\zeta}}_{\upsilon }\right] , \end{array} \right\} $$

and then the proof exactly follows the proof of Theorem 1. □

Proof Proof of Theorem 3

$$ \left. \begin{array}{c} \dot{V}(\mathbf{s}_{3})=\mathbf{s}_{3}^{\top }\dot{\mathbf{s}}_{3} \\ =\mathbf{s}_{3}^{\top }\left[ \ddot{\mathbf{\omega}}-\ddot{\mathbf{ \omega}}^{\ast }+\frac{\dot{\mathbf{\omega}}-\dot{\mathbf{\omega}}^{\ast }}{ t+\kappa }-\frac{\mathbf{\omega }-\mathbf{\omega }^{\ast }+\mathbf{\alpha }_{3}}{\left( t+\kappa \right)^{2}}-\frac{1}{t+\kappa }\mathbf{\Gamma }_{3}+ \frac{1}{t+\kappa }\partial J_{\mathbf{3}}(\mathbf{\varphi }_{3})\right] \\ =\mathbf{s}_{2}^{\top }\left[ \frac{d}{dt}\left( \mathbf{B}_{3}\mathbf{ u}_{3}\right) +\mathbf{g}_{3}+\dot{\mathbf{\zeta}}_{\omega }\right] =\mathbf{ s}_{2}^{\top }\left[ -k_{3}\text{Sign}\left( \mathbf{s}_{3}\right) + \dot{\mathbf{\zeta}}_{\omega }\right] , \end{array} \right\} $$

and then the proof follows the proof of Theorem 1. □

Proof Proof of Theorem 4

$$ \left. \begin{array}{c} \dot{V}(\mathbf{s}_{4})=\mathbf{s}_{4}^{\top }\dot{\mathbf{s}}_{4} \\ =\mathbf{s}_{4}^{\top }\left[ \ddot{\mathbf{\tau}}-\ddot{\mathbf{\tau}}^{\ast }+\frac{\dot{\mathbf{\tau}}-\dot{\mathbf{\tau}}^{\ast }}{t+\kappa }- \frac{\mathbf{\tau }-\mathbf{\tau }^{\ast }+\mathbf{\alpha }_{4}}{\left( t+\kappa \right)^{2}}-\frac{1}{t+\kappa }\mathbf{\Gamma }_{4}+\frac{1}{ t+\kappa }\partial J_{\mathbf{4}}(\mathbf{\varphi }_{4})\right] \\ =\mathbf{s}_{4}^{\top }\left[ \frac{d}{dt}\left( \mathbf{B}_{4}\mathbf{ u}_{4}\right) +\mathbf{g}_{4}+Z_{E}\mathbf{g}\right] \leq \\ -k_{4}\mathbf{s}_{4}^{\top }\text{Sign}\left( \mathbf{s}_{4}\right) +\left\Vert \mathbf{s}_{4}\right\Vert \left\Vert Z_{E}\right\Vert \dot{g}^{+}\leq -\left\Vert \mathbf{s}_{4}\right\Vert \left( k_{4}-\left\Vert Z_{E}\right\Vert \dot{g}^{+}\right) , \end{array} \right\} $$

and then the proof follows the proof of Theorem 1. □