Distributed time-varying formation tracking of multi-quadrotor systems with partial aperiodic sampled data

Abstract

In this article, a distributed formation tracking controller is proposed for Multi-agent systems (MAS) consisting of quadrotors. It is considered that each quadrotor in the MAS only shares its translation position information with its neighbors. Moreover, position information is transmitted at nonuniform and asynchronous time instants. The control system is divided into an outer-loop for the position control and an inner-loop for the attitude control. A continuous-discrete time observer is used in the outer-loop to estimate both position and velocity of the quadrotor and its neighbors using discrete position information it receives. Then, these estimated states are used to design the position controller in order to enable quadrotors to generate the required geometric shape. A finite-time attitude controller is designed to track the desired attitude as dictated by the position controller. Finally, a closed-loop stability analysis of the overall system including nonlinear coupling is performed.

Appendices

Appendix A

Proof

Let $v=\min \left( \frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)$, $\xi =2\frac{c\delta }{b}$ and $\kappa =1-\xi$. Since $\delta \in \left( 0,\varrho \min \left( \frac{b}{c},\frac{1}{\sigma }\right) \right)$, one has

$$\begin{aligned}&0< 2\frac{c\delta }{b}<2\frac{c}{b}\varrho \min \left( \frac{b}{c},\frac{1}{\sigma }\right) \le 2\varrho \nonumber \\&\quad \Rightarrow \quad 1-2\varrho<\underbrace{1-\xi }_{=\kappa }<1 \end{aligned}$$

(46)

$$\begin{aligned}&0<v\kappa \delta<v\delta <v\varrho \min \left( \frac{b}{c},\frac{1}{\sigma }\right) \le \sqrt{2}\varrho \end{aligned}$$

(47)

Then $\varrho >0$ can be chosen, independently of a, b, c, k such that for all $\delta \in \left( 0,\varrho \min \left( \frac{b}{c},\frac{1}{\sigma }\right) \right)$ we have

$$\begin{aligned}&\kappa \in \left( \frac{1}{\sqrt{2}},1\right) \end{aligned}$$

(48)

$$\begin{aligned}&e^{v\kappa \delta }\le 1+2v\kappa \delta \end{aligned}$$

(49)

Consider the following Lyapunov function

$$\begin{aligned} W(v_t)&=v_1^2(t)+v_2^2(t) \nonumber \\&\quad +\,c\int _0^\delta \int _{t-s}^te^{v\kappa (\mu -t+s)}v_2^2(\mu ){\mathrm{d}}\mu {\mathrm{d}}s \end{aligned}$$

(50)

where $v_t(s)=\left[ v_1(t+s),v_2(t+s)\right] ^{\mathrm{T}}$, $s\in [-\delta ,0]$. One has

$$\begin{aligned}&\dot{W}(v_t)= \frac{\mathrm{d}}{\mathrm{d}t}\left( v_1^2(t)+v_2^2(t)\right) \\&\qquad \qquad +c\int _0^\delta \frac{\mathrm{d}}{\mathrm{d}t}\int _{t-s}^te^{v\kappa (\mu -t+s)}v_2^2(\mu ){\mathrm{d}}\mu {\mathrm{d}}s. \end{aligned}$$

Applying Leibniz integration leads to

$$\begin{aligned} \dot{W}(v_t)=&-av_1^2(t)-bv_2^2(t)+g(t)(v_1^2(t)+v_2^2(t))\\&\quad -v\kappa c\int _0^\delta \int _{t-s}^te^{v\kappa (\mu -t+s)}v_2^2(\mu ){\mathrm{d}}\mu {\mathrm{d}}s \\&\quad +c\int _0^\delta e^{v\kappa s}v_2^2(t)-v_2^2(t-s){\mathrm{d}}s \\&\le -av_1^2(t)-bv_2^2(t)+c\int _{t-\delta }^tv_2^2(s){\mathrm{d}}s\\&\quad +g(t)(v_1^2(t)+v_2^2(t))+k \\&\quad -v\kappa c\int _0^\delta \int _{t-s}^te^{v\kappa (\mu -t+s)}v_2^2(\mu ){\mathrm{d}}\mu {\mathrm{d}}s \\&\quad +c\int _0^\delta e^{v\kappa s}v_2^2(t){\mathrm{d}}s-c\int _0^\delta v_2^2(t-s){\mathrm{d}}s \\ \le&-av_1^2(t)-bv_2^2(t)+g(t)W(v_t)+k \\&\quad +c\left( \frac{e^{v\kappa \delta }-1}{v\kappa }\right) v_2^2(t) \\&\quad -v\kappa \left( W(v_t)-v_1^2(t)-v_2^2(t)\right) \end{aligned}$$

Since $\frac{e^{v\kappa \delta }-1}{v\kappa }\le 2\delta$ and given the definition of v, the following inequalities are achieved

$$\begin{aligned}&\dot{W}(v_t)+\left( v\kappa -g(t)\right) W(v_t) \\&\le \left( -a+v\kappa \right) v_1^2(t)+(-b+2c\delta +v\kappa )v_2^2(t)+k\\&\dot{W}(v_t)+\left( v\kappa -g(t)\right) W(v_t) \\&\le -a\left( 1-\frac{1}{\sqrt{2}}\right) v_1^2(t)-b\left( 1-(1-\kappa )-\frac{\kappa }{\sqrt{2}} \right) v_2^2(t)+k \\&\dot{W}(v_t)+\left( v\kappa -g(t)\right) W(v_t) \\&\le -a\left( \frac{\sqrt{2}-1}{\sqrt{2}}\right) v_1^2(t)-b\kappa \left( \frac{\sqrt{2}-1}{\sqrt{2}} \right) v_2^2(t)+k \\&\dot{W}(v_t) \le \left( -v\kappa +g(t)\right) W(v_t)+k \end{aligned}$$

since $v\kappa \ge \sigma$, so

$$\begin{aligned} \dot{W}(v_t)\le & {} \left( -\sigma +g(t)\right) W(v_t)+k \end{aligned}$$

To get over-estimation of W(t), let us consider the following differential equation

$$\begin{aligned}&\dot{W}(v_t)+\left( \sigma -g(t)\right) W(v_t)=k \end{aligned}$$

The solution of the above differential equation can be given as

$$\begin{aligned} W(v_t)= & {} e^{\int _0^t {-\sigma +g(\mu ){\mathrm{d}}\mu }}W(0) \nonumber \\&+e^{\int _0^t {\sigma +g(\mu ){\mathrm{d}}\mu }}\int _0^te^{\int _0^s {\sigma -g(\mu ){\mathrm{d}}\mu }}k{\mathrm{d}}s\nonumber \\= & {} e^{\int _0^t {-\sigma +g(\mu ){\mathrm{d}}\mu }}W(0)+\int _0^te^{\int _s^t {-{\bar{\sigma }} +g(\mu ){\mathrm{d}}\mu }}k{\mathrm{d}}s\nonumber \\ \end{aligned}$$

(51)

Furthermore, one has

$$\begin{aligned} \int _0^te^{\int _s^t {-\sigma +g(\mu ){\mathrm{d}}\mu }}k{\mathrm{d}}s= & {} \int _0^te^{-\sigma (t-s)}e^{\int _s^t g(\mu ){\mathrm{d}}\mu }k{\mathrm{d}}s \end{aligned}$$

and since $g(t)=0$ for $t>t^*$,

$$\begin{aligned}&\int _0^te^{\int _s^t {-\sigma +g(\mu ){\mathrm{d}}\mu }}k{\mathrm{d}}s \\&= \int _0^{t^*}e^{-\sigma (t-s)}e^{\int _s^{t^*} g(\mu ){\mathrm{d}}\mu }k{\mathrm{d}}s+\int _{t^*}^te^{-\sigma (t-s)}k{\mathrm{d}}s \\&\le {\bar{\gamma }}\int _0^{t^*}e^{-\sigma (t-s)}k{\mathrm{d}}s+\int _{t^*}^te^{-\sigma (t-s)}k{\mathrm{d}}s \end{aligned}$$

where ${\bar{\gamma }}=e^{\int _0^{t^*} g(\mu ){\mathrm{d}}\mu }$. Hence, we have

$$\begin{aligned} \int _0^te^{\int _s^t {-\sigma +g(\mu ){\mathrm{d}}\mu }}k{\mathrm{d}}s&\le {\bar{\gamma }} e^{-\sigma t}\left[ \frac{e^{\sigma t^*}}{\sigma }k-\frac{k}{\sigma }\right] \nonumber \\&\quad +e^{-\sigma t}\left[ \frac{e^{\sigma t}}{{\bar{\sigma }}}k-\frac{e^{\sigma t^*}}{\sigma }k\right] \nonumber \\&\le \frac{{\bar{\gamma }} k}{\sigma }\left[ e^{\sigma (t^*-t)}-e^{-\sigma t}\right] \nonumber \\&\quad + \frac{k }{\sigma }\left[ 1-e^{\sigma (t^*-t)}\right] \end{aligned}$$

(52)

Using (52), (51) becomes

$$\begin{aligned} W(v_t)\le & {} {\bar{\gamma }} e^{-\sigma t}W(0)+\frac{{\bar{\gamma }} k}{{\bar{\sigma }}}\left[ e^{\sigma (t^*-t)}-e^{-{\bar{\sigma }} t}\right] \nonumber \\&\quad +&\frac{k }{{\bar{\sigma }}}\left[ 1-e^{\sigma (t^*-t)}\right] \nonumber \\\le & {} \gamma e^{-\sigma t}W(0)+\frac{{\bar{\gamma }} k}{\sigma }\left[ 1-e^{-\sigma t}\right] \nonumber \\\le & {} \left( {\bar{\gamma }} W(0)-\frac{\gamma k}{\sigma }\right) e^{-\sigma t}+\frac{\gamma k}{\sigma } \end{aligned}$$

(53)

where $\gamma =\max \{1,{\bar{\gamma }}\}$. Choosing ${\bar{\alpha }}= {\bar{\gamma }} W(0)-\frac{\gamma k}{\sigma }$ finishes the proof. $\square$

Appendix B

One has

$$\begin{aligned}&2 K_1 (e_{\xi })^T [\Omega \otimes QB]F_\Delta \\&\quad =2 K_1 (e_{\xi })^T [\Omega \otimes Q][\textit{I}_N \otimes B]F_\Delta \\&\quad \le 2 K_1 \sqrt{(e_{\xi })^T [\Omega \otimes Q]e_{\xi }} \\&\quad \sqrt{[(\textit{I}_N \otimes B]F_\Delta ]^{\mathrm{T}}[\Omega \otimes Q](\textit{I}_N \otimes B]F_\Delta } \end{aligned}$$

Using the Cauchy-Schwarz inequality one obtains

$$\begin{aligned}&K_1 (e_{\xi })^T [(\Omega \otimes (QB)]F_\Delta \\&\quad \le 2 K_1\sqrt{V_c}\sqrt{\lambda _{\max }(\Omega \otimes Q)}\Vert \textit{I}_N\otimes B\Vert \Vert F_\Delta \Vert \end{aligned}$$

and using Rayleigh inequality one has

$$\begin{aligned}&K_1 (e_{\xi })^T [(\Omega \otimes (QB)]F_\Delta \nonumber \\&\quad \le 2 K_1\sqrt{V_c}\sqrt{\omega _{\max }\lambda _{\max }( Q)}\Vert F_\Delta \Vert \end{aligned}$$

(54)

Over estimation of term $\Vert F_\Delta \Vert$ gives

$$\begin{aligned} \Vert F_\Delta \Vert \le \sum _{i=1}^N \left\| \frac{T_i}{m_i} H_i\right\| \end{aligned}$$

with

$$\begin{aligned} \left\| \frac{T_i}{m_i}H_i\right\|= & {} \frac{1}{m_i}\vert T_i(e_{\xi _i})\vert \Vert H_i \Vert \end{aligned}$$

(55)

$$\begin{aligned}= & {} \frac{1}{m_i}\vert T_i(e_{\xi _i})\vert \sqrt{h_1^2+h_2^2+h_3^2} \end{aligned}$$

(56)

where

$$\begin{aligned} \vert T_i(e_{\xi _i}) \vert= & {} m_i\Vert \mu _i +g e_3\Vert \nonumber \\= & {} m_i\sqrt{\mu _{x_i}^2+\mu _{y_i}^2+(\mu _{z_i}+g)^2}. \end{aligned}$$

(57)

Following the proof of [26, Theorem 1], one has

$$\begin{aligned} \Vert \mu _{i}\Vert =\Vert \mu _i^f\Vert\le & {} c_4 \sum _{i=1}^N\Vert e_{\xi _i} \Vert +c_5\sum _{k=0}^Ns_{i,k}\Vert \bar{x}_{i,k}\Vert \end{aligned}$$

(58)

where $c_4,c_5>0$. So, we obtain

$$\begin{aligned} \Vert T_i \Vert\le & {} m_i \left( g+c_6 \left\{ \sum _{i=1}^N\Vert e_{\xi _i} \Vert +\sum _{k=0}^Ns_{i,k}\Vert \bar{x}_{i,k}\Vert \right\} \right) \end{aligned}$$

(59)

$$\begin{aligned}\le & {} l_1\left( r_1 +\sum _{i=1}^N\Vert e_{\xi _i} \Vert +\sum _{k=0}^Ns_{i,k}\Vert \bar{x}_{i,k}\Vert \right) \end{aligned}$$

(60)

where $c_6=\max (c_4,c_5)$, $l_1= m_ic_6$ and $r_1 = m_i g/l_1$. From these inequalities, one can deduce that

$$\begin{aligned} \vert T_i(e_{\xi _i})\vert \le \left\{ \begin{array}{lll} r_2{\bar{e}}_i &{} \text {for} &{} {\bar{e}}_i\ge r_1\\ r_1r_2&{} \text {for} &{} {\bar{e}}_i< r_1 \end{array} \right. \end{aligned}$$

(61)

where ${\bar{e}}_i = \sum _{i=1}^N\Vert e_{\xi _i} \Vert +\sum _{k=0}^N\Vert \bar{x}_{i,k}\Vert$ and $r_2=2l_1$.

Now replacing $(\phi _i,\theta _i,\psi _i)$ with $(\phi _{di}+e_{\phi _i},\theta _{di}+e_{\theta _i},\psi _{di}+e_{\psi _i})$ and using the following trigonometric equalities

$$\begin{aligned} \sin (a+b)= & {} \sin (a)+\sin \left( \frac{b}{2}\right) \cos \left( a+\frac{b}{2}\right) \end{aligned}$$

(62)

$$\begin{aligned} \cos (a+b)= & {} \cos (a)-\sin \left( \frac{b}{2}\right) \sin \left( a+\frac{b}{2}\right) \end{aligned}$$

(63)

then $h_3$ can be written as

$$\begin{aligned} h_3= & {} c\phi _i c\theta _i -c\phi _{di}c\theta _{di} \\= & {} c(\phi _{di}+e_{\phi _i}) c(\theta _{di}+e_{\theta _i}) -c\phi _{di}c\theta _{di} \\= & {} [c\phi _{di}-s(e_{\phi _i}/2)s(\phi _{di}+e_{\phi _i}/2)]\\&[c\theta _{di}-s(e_{\theta _i}/2)s(\theta _{di}+e_{\theta _i}/2)]-c\phi _{di}c\theta _{di} \\= & {} -c\phi _{di}s(e_{\theta _i}/2)s(\theta _{di}+e_{\theta _i}/2) \\- & {} c\theta _{di}s(e_{\phi _i}/2)s(\phi _{di}+e_{\phi _i}/2) \\&+[-s(e_{\phi _i}/2)s(\phi _{di}+e_{\phi _i}/2)]\\&\times [s(e_{\theta _i}/2)s(\theta _{di}+e_{\theta _i}/2)] \end{aligned}$$

Using the following trivial inequalities

$$\begin{aligned}&\vert \sin (a)\vert \le \vert a\vert , \qquad \vert \sin (a)\vert \le 1, \qquad \vert \cos (a)\vert \le 1\nonumber \\&\vert a\vert \vert b\vert \le \frac{1}{2}(\vert a\vert +\vert b\vert )\;\text {for}\;\vert a\vert \le 1, \vert b\vert \le 1\nonumber \\&\vert a\vert \vert b\vert \vert c\vert \le \frac{1}{2}(\vert a\vert +\vert b\vert +\vert c\vert )\;\text {for}\; \nonumber \\&\vert a\vert \le 1, \vert b\vert \le 1, \vert c\vert \le 1 \end{aligned}$$

(64)

one gets

$$\begin{aligned} \vert h_3 \vert\le & {} \vert s(e_{\phi _i}/2)\vert + \vert s(e_{\theta _i}/2) \vert +\vert s(e_{\phi _i}/2)\vert \vert s(e_{\theta _i}/2) \vert \nonumber \\\le & {} \vert s(e_{\phi _i}/2)\vert + \vert s(e_{\theta _i}/2) \vert +\frac{1}{2}(\vert s(e_{\phi _i}/2)\vert + \vert s(e_{\theta _i}/2)\vert )\nonumber \\\le & {} \frac{3}{2}(\vert s(e_{\phi _i}/2)\vert + \vert s(e_{\theta _i}/2) \vert )\nonumber \\\le & {} \frac{3}{4}(\vert e_{\phi _i} \vert + \vert e_{\theta _i} \vert ) \end{aligned}$$

(65)

Therefore, the following inequality can be obtained

$$\begin{aligned} h_3^2\le & {} \frac{9}{16} ( e_{\phi _i}^2+e_{\theta _i}^2+2\vert e_{\phi _i} \vert \vert e_{\theta _i} \vert )\nonumber \\\le & {} \frac{9}{8} ( e_{\phi _i}^2+e_{\theta _i}^2)\nonumber \\\le & {} \varsigma _3 ( e_{\phi _i}^2+e_{\theta _i}^2+e_{\psi _i}^2) \end{aligned}$$

(66)

where $\varsigma _3=\frac{9}{8}$. Similarly, one can show that

$$\begin{aligned} h_2^2\le & {} \varsigma _2 ( e_{\phi _i}^2+e_{\theta _i}^2+e_{\psi _i}^2) \end{aligned}$$

(67)

$$\begin{aligned} h_1^2\le & {} \varsigma _1 ( e_{\phi _i}^2+e_{\theta _i}^2+e_{\psi _i}^2) \end{aligned}$$

(68)

Therefore, one gets

$$\begin{aligned}&\Vert H(e_{\xi _i},e_{\eta _i}) \Vert = \sqrt{h_1^2+h_2^2+h_3^2}\nonumber \\&\quad \le \sqrt{\varsigma _1 ( e_{\phi _i}^2+e_{\theta _i}^2+e_{\psi _i}^2)+\varsigma _2 ( e_{\phi _i}^2+e_{\theta _i}^2+e_{\psi _i}^2)+ \varsigma _3 ( e_{\phi _i}^2+e_{\theta _i}^2+e_{\psi _i}^2)} \le c_7 \Vert \eta _i-\eta _{d_i}\Vert \end{aligned}$$

(69)

with $c_7=\sqrt{\varsigma _1+\varsigma _2+\varsigma _3}$

From (61) and (69), one can show that for ${\bar{e}}_i\ge r_1$

$$\begin{aligned} \left\| \frac{T_i}{m_i}H_i\right\|\le & {} \frac{1}{m_i} r_2 {\bar{e}}_{i} c_7 \Vert e_{\eta _i}\Vert \\\le & {} c_8\Vert e_{\eta _i}\Vert {\bar{e}}_i \end{aligned}$$

with $c_8=\frac{1}{m_i} r_2 c_7$. So one has

$$\begin{aligned} \left\| \frac{T_i}{m_i}H_i\right\| \le c_8\Vert e_{\eta _i}\Vert \left[ \sum _{i=1}^N\Vert e_{\xi _i} \Vert +\sum _{k=0}^Ns_{i,k}\Vert \bar{x}_{i,k}\Vert \right] \end{aligned}$$

(70)

Since

$$\begin{aligned} \sum _{i=1}^N\Vert e_{\xi _i} \Vert \le \frac{\sqrt{N}}{\sqrt{\lambda _{\min }(Q)}\sqrt{\omega _{\min }}}\sqrt{V_c(e_\xi ^c)} \end{aligned}$$

and

$$\begin{aligned} \Vert \bar{x}_{i,k}\Vert \le \frac{1}{\sqrt{\lambda _{\min }(P)}} \sqrt{V_o({{\bar{x}}}_{i,k})} \end{aligned}$$

inequality (70) becomes

$$\begin{aligned} \left\| \frac{T_i}{m_i}H_i\right\| \le c_8c_9\Vert e_{\eta _i}\Vert \left[ \sqrt{V_c(e_\xi ^c)}+\sum _{k=0}^Ns_{i,k}\sqrt{V_o({{\bar{x}}}_{i,k})}\right] \end{aligned}$$

with $c_9=\max \left( \frac{\sqrt{N}}{\sqrt{\lambda _{\min }(Q)}\sqrt{\omega _{\min }}}, \frac{1}{\sqrt{\lambda _{\min }(P)}}\right)$. On can write the above inequality as

$$\begin{aligned} \left\| \frac{T_i}{m_i}H_i\right\| \le \chi \Vert e_{\eta _i} \Vert \left[ \sqrt{V_c(e_\xi ^c)}+\sum _{k=0}^Ns_{i,k}\sqrt{V_o({{\bar{x}}}_{i,k})}\right] \nonumber \\ \end{aligned}$$

(71)

where $\chi =c_8c_9$ which leads to

$$\begin{aligned} \Vert F_\Delta \Vert \le \sum _{i=1}^N \chi (\Vert e_{\eta _i} \Vert )\left[ \sqrt{V_c(e_\xi ^c)}+\sum _{k=0}^Ns_{i,k}\sqrt{V_o({{\bar{x}}}_{i,k})}\right] \nonumber \\ \end{aligned}$$

(72)

Hence, inequality (54) is achieved.