Robust image-based control of the quadrotor unmanned aerial vehicle

Abstract

This paper proposes an image-based visual servo controller for the quadrotor vertical takeoff and landing unmanned aerial vehicle (UAV). The controller utilizes an estimate of flow of image features as the linear velocity cue and assumes angular velocity and attitude information available for feedback. The image features are selected from perspective image moments and projected on a suitably defined image plane, providing decoupled kinematics for the translational motion. A nonlinear observer is designed to estimate the flow of image features using outputs of visual information. The controller for the translational dynamics is bounded which helps to keep the target points in the field of view of the camera. A smooth asymptotic controller, using the robust integral of the sign of the error method, is designed for the rotational dynamics in order to compensate for the unmodeled dynamics and external disturbances. Furthermore, the proposed approach is robust with respect to unknown image depth through an adaptive scheme and also the yaw information of the UAV is not required. The complete Lyapunov-based stability analysis is presented to show that all states of the system are bounded and the error signals converge to zero. Simulation examples are provided in both nominal and perturbed conditions which show the effectiveness of the proposed theoretical results.

Appendix

1.1 Appendix 1: The time derivative of $\dot{\bar{\varvec{\psi }}}$

From (2), the time derivative of $\dot{\psi }$ can be written as

$$\begin{aligned} \dot{\psi }=\frac{s_{\phi }}{c_{\theta }}\varOmega _2+\frac{c_{\phi }}{c_{\theta }}\varOmega _3. \end{aligned}$$

(46)

Using (4), the time derivative of (46) can be written as

$$\begin{aligned} \ddot{\psi }=&\frac{\dot{\phi }c_{\phi }c_{\theta }+\dot{\theta }s_{\theta }s_{\phi }}{c^2_{\theta }}\varOmega _2+\frac{s_{\phi }\left( J_{zz}-J_{xx}\right) }{c_{\theta }J_{yy}}\varOmega _1\varOmega _3\\&+\frac{s_{\phi }}{c_{\theta }J_{yy}}U_3+\frac{\dot{\theta }s_{\theta }c_{\phi }-\dot{\phi }s_{\phi }c_{\theta }}{c^2_{\theta }}\varOmega _3\\&+\frac{c_{\phi }\left( J_{xx}-J_{yy}\right) }{c_{\theta }J_{zz}}\varOmega _1\varOmega _2+\frac{c_{\phi }}{c_{\theta }J_{zz}}U_4. \end{aligned}$$

Therefore, the time derivative of $\dot{\bar{\varvec{\psi }}}$ can be obtained as

$$\begin{aligned} \ddot{\bar{\varvec{\psi }}}=\left[ 0\;0\;\ddot{\psi }\right] ^{\top }=\varvec{\varGamma }_1+\varvec{\varGamma }_2\varvec{\tau } \end{aligned}$$

where the vector $\varvec{\varGamma }_1$ and the matrix $\varvec{\varGamma }_2$ are defined as in the following:

$$\begin{aligned} \varvec{\varGamma }_1= & {} \left[ \begin{array}{ccc} 0 \\ 0 \\ \varGamma _{13} \end{array}\right] \\ \varvec{\varGamma }_2= & {} \left[ \begin{array}{lll} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0\\ 0 &{} \frac{s_{\phi }}{c_{\theta }} &{} \frac{c_{\phi }}{c_{\theta }} \end{array}\right] \end{aligned}$$

in which $\dot{\phi }$ and $\dot{\theta }$ are defined in (2) and

$$\begin{aligned} \varGamma _{13}=&\frac{\dot{\phi }c_{\phi }c_{\theta }+\dot{\theta }s_{\theta }s_{\phi }}{c^2_{\theta }}\varOmega _2+\frac{s_{\phi }\left( J_{zz}-J_{xx}\right) }{c_{\theta }J_{yy}}\varOmega _1\varOmega _3\\&+\frac{\dot{\theta }s_{\theta }c_{\phi }-\dot{\phi }s_{\phi }c_{\theta }}{c^2_{\theta }}\varOmega _3+\frac{c_{\phi }\left( J_{xx}-J_{yy}\right) }{c_{\theta }J_{zz}}\varOmega _1\varOmega _2. \end{aligned}$$

1.2 Appendix 2: Proof of Lemma 3

The following Lyapunov function is considered:

$$\begin{aligned} L_1=\frac{1}{2}{\mathbf {x}}_2^{\top }{\mathbf {x}}_2+l_1b\left( t\right) \sqrt{1+{\mathbf {x}}_1^{\top }{\mathbf {x}}_1}. \end{aligned}$$

Using (39) and Property 2, the time derivative of $L_1$ will be

$$\begin{aligned} \dot{L}_1=&-l_2b\left( t\right) {\mathbf {x}}_2^{\top }{\mathbf {h}}\left( {\mathbf {x}}_2\right) +{\mathbf {x}}_2^{\top }\varvec{\epsilon }+l_1\dot{b}\left( t\right) \sqrt{1+{\mathbf {x}}_1^{\top }{\mathbf {x}}_1}. \end{aligned}$$

(47)

First it is shown that $L_1$ and hence ${\mathbf {x}}_1$ and ${\mathbf {x}}_2$ do not have a finite escape time. Since $\varvec{\epsilon }$ and $\dot{b}\left( t\right)$ are bounded and converge to zero, one will have

$$\begin{aligned} \dot{L}_1\le a_1\left\| {\mathbf {x}}_2\right\| +a_2\left\| {\mathbf {x}}_1\right\| +a_3 \end{aligned}$$

where $a_1$, $a_2$ and $a_3$ are positive constants. Since $\left\| {\mathbf {x}}_2\right\| ^2\le 2L_1$ and $\left\| {\mathbf {x}}_1\right\| \le \frac{1}{l_1b\left( t\right) } L_1$, then one has $\dot{L}_1\le a_1\sqrt{2L_1}+\frac{a_2}{l_1b\left( t\right) } L_1+a_3$. Under the condition that $L_1\ge 1$, one has $\dot{L}_1\le \left( a_1\sqrt{2}+\frac{a_2}{l_1b\left( t\right) }+a_3\right) L_1$ which can be rewritten as $\frac{{\text {d}}L_1}{L_1}\le \left( a_1\sqrt{2}+\frac{a_2}{l_1b\left( t\right) }+a_3\right) {\text {d}}t$. Since $b\left( t\right)$ is positive and continuous, integrating this relation in a finite time implies that $L_1$ and therefore ${\mathbf {x}}_1$ and ${\mathbf {x}}_2$ do not have a finite escape time. Now, if $L_1<1$, one will have $\dot{L}_1\le \left( a_1\sqrt{2}+\frac{a_2}{l_1b\left( t\right) }+a_3\right) \sqrt{L}_1$ and by applying the same analysis, it can be concluded that $L_1$ and hence ${\mathbf {x}}_1$ and ${\mathbf {x}}_2$ do not have a finite escape time. Next it is shown that ${\mathbf {x}}_1$ and ${\mathbf {x}}_2$ are bounded and converge to zero. According to (47), under the condition that

$$\begin{aligned} l_2b\left( t\right) \frac{\left\| {\mathbf {x}}_2\right\| ^2}{\sqrt{1+{\mathbf {x}}_2^{\top }{\mathbf {x}}_2}}>\left\| {\mathbf {x}}_2\right\| \left\| \varvec{\epsilon }\right\| +l_1\left| \dot{b}\left( t\right) \right| \sqrt{1+{\mathbf {x}}_1^{\top }{\mathbf {x}}_1} \end{aligned}$$

(48)

one has $\dot{L}_1< 0$. Since $\varvec{\epsilon }$ and $\dot{b}\left( t\right)$ are bounded and converge to zero, and ${\mathbf {x}}_1$ and ${\mathbf {x}}_2$ do not have a finite escape time, then there is a finite time $t_1$ such that for $t\ge t_1$ the condition (48) is satisfied, and hence ${\mathbf {x}}_1$ and ${\mathbf {x}}_2$ are bounded. It should be noted that these signals are also bounded in the interval $\left[ 0, t_1\right)$ since they do not escape in a finite time. Therefore, it can be concluded that $\dot{L}_1$ is negative for all $t\ge t_1$ which means that out of the following set:

$$\begin{aligned} \varvec{\varPsi }=&\left\{ {\mathbf {x}}_2|\frac{\left\| {\mathbf {x}}_2\right\| ^2}{\sqrt{1+{\mathbf {x}}_2^{\top }{\mathbf {x}}_2}}\le \frac{1}{l_2b\left( t\right) }\left\| {\mathbf {x}}_2\right\| \left\| \varvec{\epsilon }\right\| \right. \\&\left. \qquad \;+\frac{l_1}{l_2b\left( t\right) }\left| \dot{b}\left( t\right) \right| \sqrt{1+{\mathbf {x}}_1^{\top }{\mathbf {x}}_1}\right\} \end{aligned}$$

${\mathbf {x}}_2$ is bounded, and since this region converges to zero, then ${\mathbf {x}}_2$ will be driven to zero. Consequently, applying Lemma 2 to (39) shows that $\dot{{\mathbf {x}}}_2$ converges to zero, which indicates that ${\mathbf {x}}_1$ also converges to zero.