Nonlinear augmented observer design and application to quadrotor aircraft

Abstract

This paper presents a nonlinear augmented observer, and it is applied to an underactuated quadrotor aircraft. Not only the proposed augmented observer can estimate the velocity from the position measurement, but also the uncertainties can be obtained. The robustness in time domain and in frequency domain is analyzed, respectively. The merits of the presented augmented observer include its synchronous estimation of velocity and uncertainties, finite-time stability, ease of parameters selection, and sufficient stochastic noise rejection. Moreover, a simple control law based on the augmented observer is designed to stabilize the flight dynamics. Simulation results illustrate the effectiveness of the proposed method.

Appendix

Before giving the proof of Theorem 3, we present a Lemma on finite-time stability of a high-order continuous nonlinear system.

Lemma 1

Let $k_{1},\ldots , k_{n}>0$ be such that $s^{n}+k_{n}s^{n-1}+\cdots +k_{2}s+k_{1}$ is Hurwitz, and consider the system

$$\begin{aligned} \dot{x}_{i}&= x_{i+1}-k_{n-i+1}\left| x_{1}\right| ^{\alpha _{n-i+1}}\mathrm{sign}(x_{1});\quad i=1,\ldots , n-1\nonumber \\ \dot{x}_{n}&= -k_{1}\left| x_{1}\right| ^{\alpha _{1}}\mathrm{sign}(x_{1}) \end{aligned}$$

(44)

there exists $\xi \in (0,1)$ such that for every $\alpha \in (1-\xi ,1)$, the origin is globally finite-time-stable equilibrium for Eq. (44) where $\alpha _{1},\ldots , \alpha _{n}$ satisfy

$$\begin{aligned} \alpha _{n-i+1}=\frac{i\alpha +n-i}{n},\quad i=1,\ldots , n \end{aligned}$$

(45)

Proof of Lemma 1

From Definition 2 and Eq. (44), let

$$\begin{aligned}&\rho ^{r_{i+1}}x_{i+1}-k_{n-i+1}\left| \rho ^{r_{1}}x_{1}\right| ^{\alpha _{n-i+1}}\mathrm{sign}(x_{1})\nonumber \\&\quad =\rho ^{r_{i}+m}\left[ x_{i+1}-k_{n-i+1} \left| x_{1}\right| ^{\alpha _{n-i+1}}\mathrm{sign}(x_{1})\right] ; \nonumber \\&i =1,\ldots , n-1;\nonumber \\&-\,k_{1}\left| \rho ^{r_{1}}x_{1}\right| ^{\alpha _{1}}\mathrm{sign}(x_{i}) \!=\!\rho ^{r_{n}+m}\left[ -k_{1}\left| x_{1}\right| ^{\alpha _{1}}\mathrm{sign}(x_{1})\right] \nonumber \\ \end{aligned}$$

(46)

Therefore, from (46), we have

$$\begin{aligned} r_{i+1}&= \alpha _{n-i+1}r_{1}=r_{i}+m,\quad i=1,\ldots , n-1,\nonumber \\ \alpha _{1}r_{1}&= r_{n}+m \end{aligned}$$

(47)

Selecting $r_{1}=1$, it follows that

$$\begin{aligned}&m =\alpha _{n}-1,\alpha _{n-i+1}=\frac{i\alpha _{1}+n-i}{n},\nonumber \\&r_{n+1-i} =\frac{(n-i)\alpha _{1}+i}{n},\quad i=1,\ldots , n \end{aligned}$$

(48)

Let $k_{1},\ldots , k_{n}>0$ be chosen Hurwitz, and for each $\alpha >0$, let $f_{\alpha }$ denote the closed-loop vector field obtained in (44). For each $\alpha >0$, the vector field $f_{\alpha }$ is continuous. It is also easy to verify that for each $\alpha >0$, the vector field $f_{\alpha }$ is homogeneous of degree $\alpha _{n}-1=\frac{\alpha _{1}-1}{n}$, where $\alpha _{1}=\alpha$ and $\alpha _{1},\ldots , \alpha _{n}$ satisfy (45). Moreover, the vector field $f_{1}$ is linear with the Hurwitz characteristic polynomial $s^{n}+k_{n}s^{n-1}+\cdots +k_{2}s+k_{1}$. Therefore, by Theorem 6.2 in [31], there exists a positive-definite, radially unbounded, Lyapunov function $V:R^{n}\rightarrow R$ such that $L_{f_{1}}V$ is continuous and negative definite. Let $A=V^{-1}([0,1])$ and $S=b\hbox {d}A=V^{-1}(\{1\})$. Then, $A$ and $S$ are compact since $V$ is proper and $0\in S$ since $V$ is positive definite. Define $\varphi :(0,1]\times S\, \rightarrow R$ by $\varphi (\alpha , z)=L_{f_{\alpha }}V(z)$. Then, $\varphi$ is continuous and satisfies $\varphi (1,z)<0$ for all $z\in S$, that is, $\varphi (\{1\}\times S)\subset (-\infty , 0)$. Since $S$ is compact, it follows from Lemma 5.8 in [39, p. 169] that there exists $\xi >0$ such that $\varphi ((1-\xi ,1]\times S)\subset (-\infty , 0)$. It follows that for $\alpha \in (1-\xi , 1], \, L_{f_{\alpha }}V$ takes negative values on $S$. Thus, $A$ is strictly positively invariant under $f_{\alpha }$ for every $\alpha \in (1-\xi , 1]$. By Theorem 6.1 in [31], the origin is a globally asymptotically stable equilibrium under $f_{\alpha }$ for every $\alpha \in (1-\xi , 1]$. The result now follows from Theorems 7.1 and 7.3 in [31] by noting that for every $\alpha \in (1-\xi ,1)$, the degree of homogeneity of $f_{\alpha }$ is negative.$\square$

Proof of Theorem 3

The system error for Eqs. (11) and (10) is obtained as follow:

$$\begin{aligned} \dot{e}_{1}&= e_{2}-\frac{k_{3}}{\varepsilon }\left| e_{1}\right| ^{\alpha _{3}}\mathrm{sign}(e_{1})\nonumber \\ \dot{e}_{2}&= e_{3}-\frac{k_{2}}{\varepsilon ^{2}}\left| e_{1}\right| ^{\alpha _{2}}\mathrm{sign}(e_{1})\nonumber \\ \dot{e}_{3}&= -\frac{k_{1}}{\varepsilon ^{3}}\left| e_{1}\right| ^{\alpha _{1}}\mathrm{sign}(e_{1})-\eta (t) \end{aligned}$$

(49)

Thus, Eq. (49) can be rewritten as:

$$\begin{aligned}&\frac{\hbox {d}e_{1}}{\hbox {d}t/\varepsilon } = \varepsilon e_{2}-k_{3}\left| e_{1}\right| ^{\alpha _{3}}\mathrm{sign}(e_{1})\nonumber \\&\frac{\hbox {d}\varepsilon e_{2}}{\hbox {d}t/\varepsilon } = \varepsilon ^{2}e_{i+1}-k_{2}\left| e_{1}\right| ^{\alpha _{2}}\mathrm{sign}(e_{1}) \nonumber \\&\frac{\hbox {d}\varepsilon ^{2}e_{3}}{\hbox {d}t/\varepsilon } = -k_{1}\left| e_{1}\right| ^{\alpha _{1}}\mathrm{sign}(e_{1})-\varepsilon ^{3}\eta (t) \end{aligned}$$

(50)

Let a coordinate transformation be described as follow:

$$\begin{aligned}&\tau =t/\varepsilon ; z_{i}\left( \tau \right) =\varepsilon ^{i-1}e_{i}\left( t\right) ,\quad i=1,2,3;\nonumber \\&z=[ \begin{array}{ccc} z_{1}&z_{2}&z_{3} \end{array} ]^\mathrm{T}; \bar{\eta }(\tau )=\varepsilon ^{3}\eta (t) \end{aligned}$$

(51)

Therefore, we obtain $z=\varXi (\varepsilon )e$, and Eq. (50) can be written as

$$\begin{aligned} \frac{\hbox {d}z_{1}}{\hbox {d}\tau }&= z_{2}-k_{3}\left| z_{1}\right| ^{\alpha _{3}}\mathrm{sign}(z_{1})\nonumber \\ \frac{\hbox {d}z_{2}}{\hbox {d}\tau }&= z_{3}-k_{2}\left| z_{1}\right| ^{\alpha _{2}}\mathrm{sign}(z_{1})\nonumber \\ \frac{\hbox {d}z_{3}}{\hbox {d}\tau }&= -k_{1}\left| z_{1}\right| ^{\alpha _{1}}\mathrm{sign}(z_{1})-\bar{\eta }(\tau ) \end{aligned}$$

(52)

Therefore, we obtain

$$\begin{aligned} \delta =\underset{\tau \in [0,\infty )}{\sup }\left| \bar{\eta } (\tau )\right| \le \varepsilon ^{3}L_{d} \end{aligned}$$

(53)

From Lemma 1, Theorem 2 and Eq. (53), for Eq. (52), there exist positive constants $\mu$ and $\varGamma \left( z\left( 0\right) \right)$, such that

$$\begin{aligned} \left\| z\left( \tau \right) \right\| \le \mu \delta ^{\gamma }=\mu (\varepsilon ^{3}L_{d})^{\gamma },\forall \tau \in [\varGamma \left( z\left( 0\right) \right) , \infty )\nonumber \\ \end{aligned}$$

(54)

Therefore, from coordinate transformation (51), we obtain

$$\begin{aligned}&\Vert [\begin{array}{ccc} e_{1}&\varepsilon e_{2}&\varepsilon ^{2}e_{3} \end{array} ^\mathrm{T}\Vert \nonumber \\&\quad \le \mu (\varepsilon ^{3}L_{d})^{\gamma },\forall t\in [\varepsilon \varGamma \left( \varXi (\varepsilon )e\left( {0}\right) \right) ,\infty ) \end{aligned}$$

(55)

Thus, the following inequality holds:

$$\begin{aligned} \left| e_{i}\right| \le \mu (\varepsilon ^{3}L_{d})^{\gamma }\varepsilon ^{-i+1}=L\varepsilon ^{3\gamma -i+1} \end{aligned}$$

(56)

$\forall t\in [\varepsilon \varGamma \left( \varXi (\varepsilon )e\left( {0} \right) \right) , \infty )$, where $i=1,2,3$, and $L=\mu L_{d}^{\gamma }$. To make $3\gamma -i+1>1,\, i=1,2,3$, from Theorem 2, we let

$$\begin{aligned} \beta \in \left( 0,1/2\right) \end{aligned}$$

(57)

In fact, from Theorem 1, $\beta$ can be chosen to be arbitrarily small. Hence, the requirement that $\beta$ lie in $\beta \in \left( 0,1/2\right)$ is not restrictive. Accordingly, we can obtain $\gamma =\left( {1-}\beta \right) /\beta >1$. Thus, $3\gamma -3+1>1$ holds. Therefore, $3\gamma -i+1>1$ for $i=1,2,3$. The choice of $\beta$ leads to $3\gamma -i+1>1$ in (56) which implies that for $\varepsilon \in (0,1)$, the ultimate bound (56) on the estimation error is of higher order than the perturbation.

Finally, from the Routh–Hurwitz stability criterion, polynomial $s^{3}+k_{3}s^{2}+k_{2}s+k_{1}$ is Hurwitz if $k_{1}>0,k_{3}>0,k_{2}>k_{1}/k_{3}$. This concludes the proof. $\square$

Proof of Theorem 4

The system error for Eqs. (11) and (10) is given by:

$$\begin{aligned} \dot{e}_{1}&= e_{2}-\frac{k_{3}}{\varepsilon }\left| e_{1}\right| ^{\alpha _{3}}\mathrm{sign}(e_{1}-\sigma (t))\nonumber \\ \dot{e}_{2}&= e_{3}-\frac{k_{2}}{\varepsilon ^{2}}\left| e_{1}\right| ^{\alpha _{2}}\mathrm{sign}(e_{1}-\sigma (t))\nonumber \\ \dot{e}_{3}&= -\frac{k_{1}}{\varepsilon ^{3}}\left| e_{1}\right| ^{\alpha _{1}}\mathrm{sign}(e_{1}-\sigma (t))-\eta (t) \end{aligned}$$

(58)

Equation (58) can be rewritten as:

$$\begin{aligned} \frac{\hbox {d}e_{i}}{\hbox {d}t/\varepsilon }&= \varepsilon e_{2}-k_{3}\left| e_{1}-\sigma (t)\right| ^{\alpha _{3}}\mathrm{sign}(e_{1}-\sigma (t))\nonumber \\ \frac{\hbox {d}\varepsilon e_{2}}{\hbox {d}t/\varepsilon }&= \varepsilon ^{2}e_{3}-k_{2}\left| e_{1}-\sigma (t)\right| ^{\alpha _{2}}\mathrm{sign}(e_{1}-\sigma (t))\nonumber \\ \frac{\hbox {d}\varepsilon ^{2}e_{3}}{\hbox {d}t/\varepsilon }&= -k_{1}\left| e_{1}-\sigma (t)\right| ^{\alpha _{1}}\mathrm{sign}(e_{1}-\sigma (t))\nonumber \\&-\varepsilon ^{n}\eta (t) \end{aligned}$$

(59)

Let

$$\begin{aligned} \tau&= t/\varepsilon , z_{i}\left( \tau \right) =\varepsilon ^{i-1}e_{i}\left( t\right) , \quad i=1,2,3,\nonumber \\ z&= [ \begin{array}{ccc} z_{1}&z_{2}&z_{3} \end{array} ]^\mathrm{T},\bar{\sigma }(\tau )=d(t),\bar{\eta }(\tau )=\varepsilon ^{n}\eta (t) \end{aligned}$$

(60)

therefore, we have $z=\varXi (\varepsilon )e$. The Eq. (59) can be written as

$$\begin{aligned} \frac{\hbox {d}z_{1}}{\hbox {d}\tau }&= z_{2}-k_{3}\left| z_{1}-\bar{\sigma }(\tau )\right| ^{\alpha _{3}}\mathrm{sign}(z_{1}-\bar{\sigma }(\tau ))\nonumber \\ \frac{\hbox {d}z_{2}}{\hbox {d}\tau }&= z_{3}-k_{2}\left| z_{1}-\bar{\sigma }(\tau )\right| ^{\alpha _{2}}\mathrm{sign}(z_{1}-\bar{\sigma }(\tau ))\nonumber \\ \frac{\hbox {d}z_{3}}{\hbox {d}\tau }&= -k_{1}\left| z_{1}-\bar{\sigma }(\tau )\right| ^{\alpha _{1}}\mathrm{sign}(z_{1}-\bar{\sigma }(\tau ))-\bar{\eta }(\tau )\nonumber \\ \end{aligned}$$

(61)

Furthermore, Eq. (61) can be rewritten as

$$\begin{aligned} \frac{\hbox {d}z_{1}}{\hbox {d}\tau }&= z_{2}-k_{3}\left| z_{1}\right| ^{\alpha _{3}}\mathrm{sign}(z_{1})\nonumber \\&-\,k_{3}\{\left| z_{1}-\bar{\sigma }(\tau )\right| ^{\alpha _{3}}\mathrm{sign}(z_{1}-\bar{\sigma }(\tau ))\nonumber \\&-\left| z_{1}\right| ^{\alpha _{3}}\mathrm{sign}(z_{1})\}\nonumber \\ \frac{\hbox {d}z_{2}}{\hbox {d}\tau }&= z_{3}-k_{2}\left| z_{1}\right| ^{\alpha _{2}}\mathrm{sign}(z_{1})\nonumber \\&-\,k_{2}\{\left| z_{1}-\bar{\sigma }(\tau )\right| ^{\alpha _{2}}\mathrm{sign}(z_{1}-\bar{\sigma }(\tau ))\nonumber \\&-\,\left| z_{1}\right| ^{\alpha _{2}}\mathrm{sign}(z_{1})\}\nonumber \\ \frac{\hbox {d}z_{3}}{\hbox {d}\tau }&= -k_{1}\left| z_{1}\right| ^{\alpha _{1}}\mathrm{sign}(z_{1})\nonumber \\&-\,k_{1}\{\left| z_{1}-\bar{\sigma }(\tau )\right| ^{\alpha _{1}}\mathrm{sign}(z_{1}-\bar{\sigma }(\tau ))\nonumber \\&-\,\left| z_{1}\right| ^{\alpha _{1}}\mathrm{sign}(z_{1})\} -\bar{\eta }(\tau ) \end{aligned}$$

(62)

Let

$$\begin{aligned}&g\left( \tau , z\left( \tau \right) \right) \nonumber \\&\quad =[ \begin{array}{ccc} g_{1}\left( \tau , z\left( \tau \right) \right)&g_{2}\left( \tau , z\left( \tau \right) \right)&g_{3}\left( \tau , z\left( \tau \right) \right) \end{array} ]^\mathrm{T}\nonumber \\ \end{aligned}$$

(63)

where

$$\begin{aligned} g_{1}\left( \tau , z\left( \tau \right) \right)&= -k_{3}\{\left| z_{1}- \bar{\sigma }(\tau )\right| ^{\alpha _{3}}\mathrm{sign}(z_{1}\nonumber \\&\qquad -\,\bar{\sigma }(\tau ))-\left| z_{1}\right| ^{\alpha _{3}}\mathrm{sign}(z_{1})\}\nonumber \\ g_{2}\left( \tau , z\left( \tau \right) \right)&= -k_{2}\{\left| z_{1}- \bar{\sigma }(\tau )\right| ^{\alpha _{2}}\mathrm{sign}(z_{1}\nonumber \\&\qquad -\,\bar{\sigma }(\tau )-\left| z_{1}\right| ^{\alpha _{2}}\mathrm{sign}(z_{1}))\}\nonumber \\ g_{3}\left( \tau , z\left( \tau \right) \right)&= -k_{1}\{\left| z_{1}- \bar{\sigma }(\tau )\right| ^{\alpha _{1}}\mathrm{sign}(z_{1}\nonumber \\&\qquad -\,\bar{\sigma }(\tau ))-\left| z_{1}\right| ^{\alpha _{1}}\mathrm{sign}(z_{1})\}-\bar{\eta }(\tau )\nonumber \\ \end{aligned}$$

(64)

Therefore, from Assumption 1 and Remark 1, we obtain

$$\begin{aligned} \delta&= \underset{(\tau , z)\in R^{4}}{\sup }\left\| g\left( \tau ,z\left( \tau \right) \right) \right\| \nonumber \\&\le \sum \limits _{{i=1}}^{{3} }2^{1-\alpha _{3-i+1}}k_{3-i+1}L_{\sigma }^{\alpha _{3-i+1}}+\varepsilon ^{3}L_{d}\nonumber \\&\le \varepsilon ^{3}L_{d}+L_{\sigma }^{\alpha _{p}}\sum \limits _{{i=1}}^{{3 }}2^{1-\alpha _{3-i+1}}k_{3-i+1} \end{aligned}$$

(65)

where $L_{\sigma }^{\alpha _{p}}=L_{\sigma }^{\alpha _{1}}$ when $0<L_{\sigma }<1$, and $L_{\sigma }^{\alpha _{p}}=L_{\sigma }^{\alpha _{3}}$ when $L_{\sigma }>1$. In fact, we can calculate, respectively, the minimum and maximal values of the following expression [defined in Eq. (45)]:

$$\begin{aligned} \alpha _{3-i+1}=\frac{i\alpha _{1}+3-i}{3},\quad i=1,2,3 \end{aligned}$$

(66)

Defining the following function

$$\begin{aligned} \varPsi (s)=\frac{s\alpha _{1}+3-s}{3},s\in (0,+\infty ) \end{aligned}$$

(67)

and taking derivative of $\varPsi (s)$ with respect to variable $s$, we obtain

$$\begin{aligned} \frac{\hbox {d}\varPsi (s)}{\hbox {d}s}=\frac{\alpha _{1}-1}{3}<0,s\in (0,+\infty ) \end{aligned}$$

(68)

Because $\alpha _{1}\in (0,1)$, function $\varPsi (s)$ is monotone decreasing with respect to $s$. Moreover, the sequence $\{1,2,3\}$ is monotone increasing in $(0,+\infty )$. Therefore,

$$\begin{aligned} \underset{i\in \left\{ 1,2,3\right\} }{\min }\left\{ \alpha _{3-i+1}\right\} =\alpha _{1} \end{aligned}$$

(69)

and

$$\begin{aligned} \underset{i\in \left\{ 1,2,3\right\} }{\max }\left\{ \alpha _{3-i+1}\right\} =\alpha _{3} \end{aligned}$$

(70)

Therefore, $L_{\sigma }^{\alpha _{p}}=\underset{i\in \left\{ 1,2,3\right\} }{ \max }\left\{ L_{\sigma }^{\alpha _{3-i+1}}\right\} =L_{\sigma }^{\alpha _{1}}$ when $\left| L_{\sigma }\right| <1$; and $L_{\sigma }^{\alpha _{p}}=\underset{i\in \left\{ 1,2,3\right\} }{\max }\left\{ L_{\sigma }^{\alpha _{3-i+1}}\right\} =L_{\sigma }^{\alpha _{3}}$ when $\left| L_{\sigma }\right| >1$.

From Lemma 1, Theorem 2 and Eq. (65), for Eq. (62), there exist positive constants $\mu$ and $\varGamma \left( z\left( 0\right) \right)$, such that

$$\begin{aligned}&\left\| z\left( \tau \right) \right\| \le \mu \delta ^{\gamma }\nonumber \\&\quad \le \mu \left( \varepsilon ^{3}L_{d}+L_{\sigma }^{\alpha _{p}}\sum \limits _{{i=1}}^{{3} }2^{1-\alpha _{3-i+1}}k_{3-i+1}\right) ^{\gamma } \end{aligned}$$

(71)

for $\forall \tau \in [\varGamma \left( z\left( 0\right) \right) ,\infty )$. From coordinate transformation (60), we obtain

$$\begin{aligned}&\Vert \left[ \begin{array}{ccc} e_{1} \varepsilon e_{2} \varepsilon ^{2}e_{3} \end{array} \right] \Vert \nonumber \\&\quad \le \mu \left( \varepsilon ^{3}L_{d}+L_{\sigma }^{\alpha _{p}}\sum \limits _{{i=1}}^{{3}}2^{1-\alpha _{3-i+1}}k_{3-i+1}\right) ^{\gamma } \end{aligned}$$

(72)

for $\forall t\in [\varepsilon \varGamma \left( {\varXi (\varepsilon )e} \left( {0}\right) \right) , \infty )$. Thus, the following inequality holds:

$$\begin{aligned} \left| e_{i}\right| \le L(\delta _{di})^{\gamma },i\!=\!1,2,3,\forall t\!\in \! [\varepsilon \varGamma \left( {\varXi (\varepsilon )e}\left( {0} \right) \right) , \infty )\nonumber \\ \end{aligned}$$

(73)

where $L=\mu L_{d}^{\gamma }$, $\delta _{di}=\varepsilon ^{3-\frac{i-1}{ \gamma }}+\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}}\sum _{{i=1}}^{{3} }2^{1-\alpha _{3-i+1}}$ $k_{3-i+1}\varepsilon ^{-\frac{i-1}{\gamma }}$, $i=1,2,3$. If $\varepsilon \in \left( 0,1\right)$ and $L_{\sigma }<\left( \frac{1-\varepsilon ^{3}}{\sum _{{i=1}}^{{3}}2^{1-\alpha _{3-i+1}}k_{3-i+1}}L_{d}\right) ^{\frac{1}{\alpha _{p}}}$, then

$$\begin{aligned} 0<\varepsilon ^{3}+\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}}\sum \limits _{{i=1} }^{{3}}2^{1-\alpha _{3-i+1}}k_{3-i+1}<1 \end{aligned}$$

(74)

Furthermore, from Theorem 1, $\beta$ can be chosen to be arbitrarily small. Hence, the requirement that $\beta$ lies on

$$\begin{aligned} \beta \in \left( 0,\min \left\{ \frac{1}{\frac{3\log \varepsilon }{\log (\varepsilon ^{3}+\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}}\sum \limits _{{i=1} }^{{3}}2^{1-\alpha _{3-i+1}}k_{3-i+1})}+1},\frac{1}{2}\right\} \right) \nonumber \\ \end{aligned}$$

(75)

is not restrictive. Accordingly, we can obtain

$$\begin{aligned} \gamma&= (1-\beta )/\beta \nonumber \\&> \max \left\{ \frac{3\log \varepsilon }{\log \left( \varepsilon ^{3}+\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}}\sum _{{i=1}}^{{3} }2^{1-\alpha _{3-i+1}}k_{3-i+1}\right) },1\right\} \nonumber \\ \end{aligned}$$

(76)

Therefore,

$$\begin{aligned} \gamma \log \left( \varepsilon ^{3}+\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}} \sum \limits _{{i=1}}^{{3}}2^{1-\alpha _{3-i+1}}k_{3-i+1}\right) <3\log \varepsilon \nonumber \\ \end{aligned}$$

(77)

i.e.,

$$\begin{aligned} \varepsilon ^{3}+\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}}\sum \limits _{{i=1}}^{ {3}}2^{1-\alpha _{3-i+1}}k_{3-i+1}<\varepsilon ^{\frac{3}{\gamma }} \end{aligned}$$

(78)

Therefore, from $\varepsilon \in \left( 0,1\right)$ and $\gamma >3$, we can obtain

$$\begin{aligned} \varepsilon ^{\frac{3}{\gamma }}<\varepsilon ^{\frac{i-1}{\gamma }},\quad i=1,2,3 \end{aligned}$$

(79)

Then,

$$\begin{aligned} \delta _{di}\!=\!\varepsilon ^{3-\frac{i-1}{\gamma }}\!+\!\frac{L_{\sigma }^{\alpha _{p}}}{L_{d}}\sum \limits _{{i=1}}^{{3}}2^{1-\alpha _{3-i+1}}k_{3-i+1}\varepsilon ^{-\frac{i-1}{\gamma }}\!<\!1\nonumber \\ \end{aligned}$$

(80)

where $i=1,2,3$. The choice of $\phi$ leads to $\gamma >1$ in (73) which implies that for $\delta _{di}\in (0,1)$, the ultimate bound (73) on the estimation error is of higher order than the perturbation. Consequently, the presented augmented observer leads to perform rejection of low-level persistent disturbances. This concludes the proof. $\square$

Proof of Theorem 5

In the light of Corollary 1, for $t\ge t_s$, the observation signals $\left\| \widehat{p}_{p}-p_{p}\right\| \le L\varepsilon ^{3\gamma },\, \left\| \widehat{\dot{p}}_{p}\right.$ $\left. -\dot{p}_{p}\right\| \le L\varepsilon ^{3\gamma -1},\, \left\| \widehat{\delta }_{p}-\delta _{p}\right\| \le L\varepsilon ^{3\gamma -2}$, where

$$\begin{aligned}&p_{p}=\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] , \quad \dot{p}_{p}=\left[ \begin{array}{c} \dot{x} \\ \dot{y} \\ \dot{z} \end{array} \right] , \quad \widehat{p}_{p}=\left[ \begin{array}{c} \widehat{x} \\ \widehat{y} \\ \widehat{z} \end{array} \right] , \nonumber \\&\widehat{\dot{p}}_{p}=\left[ \begin{array}{c} \widehat{\dot{x}} \\ \widehat{\dot{y}} \\ \widehat{\dot{z}} \end{array} \right] \end{aligned}$$

(81)

Considering controller (39), the closed-loop error system for position dynamics is

$$\begin{aligned} \ddot{e}_{p}&= -k_{p1}e_{p}-k_{p2}\dot{e}_{p}\nonumber \\&-\,k_{p1}(\widehat{p} _{p}-\!p_{p})\!-k_{p2}(\widehat{\dot{p}}_{p}\!-\!\dot{p}_{p}) -m^{-1}(\widehat{\delta }_{p}\!-\!\delta _{p})\nonumber \\ \end{aligned}$$

(82)

For $t\ge t_{s}$ and sufficiently small $\varepsilon$, selecting the Lyapunov function be $V_{p}=k_{p1}e_{p}^{T}e_{p}+\frac{1}{2}\dot{e}_{p}^{T} \dot{e}_{p}$, we can obtain that $e_{p}\rightarrow 0$ and $\dot{e} _{p}\rightarrow 0$ as $t\rightarrow \infty$. This concludes the proof.$\square$

Proof of Theorem 6

In the light of Corollary 1, for $t\ge t_s$, the observation signals $\left\| \widehat{p}_{a}-p_{a}\right\| \le L\varepsilon ^{3\gamma },\, \left\| \widehat{\dot{p}}_{a}\right.$ $\left. -\dot{p}_{a}\right\| \le L\varepsilon ^{3\gamma -1},\, \left\| \widehat{\delta }_{a}-\delta _{a}\right\| \le L\varepsilon ^{3\gamma -2}$, where

$$\begin{aligned}&a_{a}=\left[ \begin{array}{c} \psi \\ \theta \\ \phi \end{array} \right] , \quad \dot{a}_{a}=\left[ \begin{array}{c} \dot{\psi } \\ \dot{\theta } \\ \dot{\phi } \end{array} \right] , \quad \widehat{a}_{a}=\left[ \begin{array}{c} \widehat{\psi } \\ \widehat{\theta } \\ \widehat{\phi } \end{array} \right] , \nonumber \\&\quad \widehat{\dot{a}}_{a}=\left[ \begin{array}{c} \widehat{\dot{\psi }} \\ \widehat{\dot{\theta }} \\ \widehat{\dot{\phi }} \end{array} \right] \end{aligned}$$

(83)

Considering controller (42), the closed-loop error system for attitude dynamics is

$$\begin{aligned} \ddot{e}_{a}&= -k_{a1}e_{a}-k_{a2}\dot{e}_{a}-k_{a1}(\widehat{a} _{a}-a_{a})-k_{a2}(\widehat{\dot{a}}_{a}-\dot{a}_{a})\nonumber \\&-J^{-1}(\widehat{\delta }_{a}-\delta _{a}) \end{aligned}$$

For $t\ge t_{s}$ and sufficiently small $\varepsilon$, selecting the Lyapunov function be $V_{a}=k_{a1}e_{a}^{T}e_{a}+\frac{1}{2}\dot{e}_{a}^{T} \dot{e}_{a}$, we can obtain that $e_{a}\rightarrow 0$ and $\dot{e} _{a}\rightarrow 0$ as $t\rightarrow \infty$. This concludes the proof. $\square$