New Class of Attitude Controllers Guaranteed to Converge within Specified Finite-Time

Abstract

This paper introduces a new class of finite-time feedback controllers for rigid-body attitude dynamics subject to full actuation. The control structure is Lyapunov-based and is designed to regulate the configuration from an arbitrary initial state to any prescribed final state within user-specified finite transfer-time. A salient feature here is that the synthesis of the control structure is explicit, i.e., given the transfer-time time, the feedback-gains are explicitly stated to satisfy the convergence specifications. A major contrast between this work and others in the literature is that instead of resorting to feedback-linearization (to get to the so-called normal form), our approach efficiently marries the process of designing time-varying feedback gains with the logarithmic Lyapunov function for attitude kinematics based on the Modified Rodrigues Parameters representation. Saliently, this finite-time solution extends nicely for accommodating trajectory tracking objectives and possesses robustness with respect to bounded external disturbance torques. Numerical simulations are performed to test and validate the performance and robustness features of the new control designs.

Appendices

Appendix A

In this section, we prove that if:

$$ \mu^{\lambda}(t) \ln \left( \begin{array}{llll} 1 + \boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) \end{array}\right) \leq V_{0}(0)\exp \left[\begin{array}{llll} -kt_{f}\cdot\bar{\mu}(t) \end{array}\right], $$

(55)

then $\mu ^{\alpha _{1}} \boldsymbol {\sigma } \in L_{\infty }, \forall \alpha _{1} \in \mathbb {R}$.

Starting from Eq. 55, we use the definition $\bar {\mu }(t) \triangleq \mu (t) - 1$ to get:

$$ \begin{array}{@{}rcl@{}} \mu^{\lambda}(t) \ln \left( \begin{array}{llll} 1 + \boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) \end{array}\right) &\leq& V_{0}(0)\exp \left[\begin{array}{llll} -kt_{f}\cdot(\mu(t) - 1) \end{array}\right] \\ &= &V_{0}(0)e^{kt_{f}} \exp \left[\begin{array}{llll} -kt_{f} \cdot \mu(t) \end{array}\right] \\ &= &\alpha_{2} \exp \left[\begin{array}{llll} -kt_{f} \cdot \mu(t) \end{array}\right], \end{array} $$

(56)

where $\alpha _{2} \triangleq V_{0}(0)e^{kt_{f}} > 0$.

Defining $\beta (t) \triangleq \exp \left [\begin {array}{llll} -kt_{f}\cdot \mu (t) \end {array}\right ]$, it follows that:

$$ \begin{array}{@{}rcl@{}} \mu^{\lambda}(t)\ln\left( \begin{array}{llll} 1 + \boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) \end{array}\right) &\leq& \alpha_{2} \beta(t) \\ \ln \left( \begin{array}{llll} 1 + \boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) \end{array}\right) &\leq& \alpha_{2} \mu^{-\lambda}(t)\beta(t) \\ 1 + \boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) &\leq& \exp \left[\begin{array}{llll} \alpha_{2} \mu^{-\lambda}(t)\beta(t) \end{array}\right] \\ \boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) &\leq &\exp \left[\begin{array}{llll} \alpha_{2} \mu^{-\lambda}(t)\beta(t) \end{array}\right] - 1 \\ \mu^{2\alpha_{1}}(t)\boldsymbol{\sigma}^{T}(t) \boldsymbol{\sigma}(t) &\leq& \mu^{2\alpha_{1}}(t) \left[\begin{array}{llll} \exp \left[\begin{array}{llll} \alpha_{2} \mu^{-\lambda}(t)\beta(t) \end{array}\right] - 1 \end{array}\right] \\ ||\mu^{\alpha_{1}}(t)\boldsymbol{\sigma}(t)||^{2} &\leq& \frac{\exp \left[\begin{array}{llll} \alpha_{2} \mu^{-\lambda}(t)\beta(t) \end{array}\right] - 1} {\mu^{-2\alpha_{1}}(t)}. \end{array} $$

(57)

In order to show that the signal $f(t) \triangleq ||\mu ^{\alpha _{1}}(t)\boldsymbol {\sigma }(t)||^{2}$ is bounded, we need to evaluate the limit as t → t_f. Taking the limit on both sides:

$$ \lim\limits_{t \to t_{f}}f(t) \leq \lim\limits_{t \to t_{f}}\frac{\exp \left[\begin{array}{llll} \alpha_{2} \mu^{-\lambda}(t)\beta(t) \end{array}\right] - 1} {\mu^{-2\alpha_{1}}(t)}. $$

(58)

The above limit can be rewritten as:

$$ \lim\limits_{t \to t_{f}}f(t) \leq \lim\limits_{\mu \to \infty}\frac{\exp \left[\begin{array}{llll} \alpha_{2} \mu^{-\lambda} \exp \left[\begin{array}{llll} -k\cdot t_{f}\cdot\mu \end{array}\right] \end{array}\right] - 1} {\mu^{-2\alpha_{1}}}. $$

(59)

Assuming that λ < 2α₁, Lemmas 1 and 2 are used to prove that the right-hand side of Eq. 59 is equal to zero, implying that $||\mu ^{\alpha _{1}}(t)\boldsymbol {\sigma }(t)||^{2} \in L_{\infty }, \forall \alpha _{1} > \lambda /2$. In addition, since $||\mu ^{\eta _{1}}(t)\boldsymbol {\sigma }(t)||^{2} \leq ||\mu ^{\eta _{2}}(t)\boldsymbol {\sigma }(t)||^{2}$, for η₁ ≤ η₂, then $||\mu ^{\alpha _{1}}(t)\boldsymbol {\sigma }(t)||^{2} \in L_{\infty }, \forall \alpha _{1} \in \mathbb {R}$.

Lemma 1

For any finite real constantsα₁≠ 0,α₂ > 0,γ₁ > 0,γ₂ > 0, then:

$$ \lim\limits_{x \to 0+} \alpha_{1} x^{-\gamma_{1}}exp\left[\begin{array}{llll} -\alpha_{2} x^{-\gamma_{2}} \end{array}\right] = 0. $$

(60)

Proof

Making the substitution $y = x^{-\gamma _{2}}$, then:

$$ \lim\limits_{x \to 0+} \alpha_{1} x^{-\gamma_{1}}exp\left[\begin{array}{llll} -\alpha_{2} x^{-\gamma_{2}} \end{array}\right] = \lim\limits_{y \to \infty} \alpha_{1} y^{\gamma_{3} + \gamma_{4}}e^{-\alpha_{2} y}, $$

(61)

where $\gamma _{3} \in \mathbb {N} \triangleq \lfloor \gamma _{1}/\gamma _{2} \rfloor$ and $\gamma _{4} \in [0, 1) \triangleq \gamma _{1}/\gamma _{2} - \gamma _{3}$.

One can notice that the limit in Eq. 61 is a product of zero with ∞, which can be solved for by using L’Hospital’s rule. Defining $\gamma _{5} \triangleq \gamma _{3} + \gamma _{4}$, we apply L’Hospital’s rule γ₃ times, leading to:

$$ \begin{array}{@{}rcl@{}} \lim_{y \to \infty} \alpha_{1} y^{\gamma_{5}}e^{-\alpha_{2} y} &= &\lim\limits_{y \to \infty} -\alpha_{1} \alpha_{2} \gamma_{5} y^{\gamma_{5}-1}e^{-\alpha_{2} y} \\ &= &\lim\limits_{y \to \infty} \alpha_{1} {\alpha_{2}^{2}} \gamma_{5}(\gamma_{5}-1) y^{\gamma_{5}-2}e^{-\alpha_{2} y} \\ &= &\cdots \\ &= &\lim\limits_{y \to \infty} (-1)^{\gamma_{3}}\alpha_{1} \alpha_{2}^{\gamma_{3}} \gamma_{5}(\gamma_{5}-1)\cdots(\gamma_{5}-\gamma_{3}) y^{\gamma_{4}}e^{-\alpha_{2} y}. \end{array} $$

(62)

If γ₄ = 0, then the proof is complete. However, if γ₄ ∈ (0, 1), then we need to use L’Hospital’s rule one more time:

$$ \begin{array}{@{}rcl@{}} \lim\limits_{y \to \infty} \alpha_{1} y^{\gamma_{5}}e^{-\alpha_{2} y} &= &\lim\limits_{y \to \infty} (-1)^{\gamma_{3}+1}\alpha_{1} \alpha_{2}^{\gamma_{3}+1} \gamma_{5}(\gamma_{5}-1)\cdots(\gamma_{5}-\gamma_{3})\gamma_{4} y^{\gamma_{4}-1}e^{-\alpha_{2} y} = 0. \end{array} $$

(63)

Lemma 2

For any finite real constantsα₁ > 0,α₂ > 0, and0 < γ₁ ≤ γ₂ < γ₃,then$\lim _{x \to \infty }f(x) = 0$,where:

$$ f(x) = \frac{\exp \left[\begin{array}{llll} \alpha_{1} x^{-\gamma_{2}}\exp\left[\begin{array}{llll} -\alpha_{2} x^{\gamma_{1}} \end{array}\right] \end{array}\right] - 1} {x^{-\gamma3}}. $$

(64)

Proof

Defining $y \triangleq x^{-\gamma _{3}}$, we have that $y^{-\gamma _{4}} = x^{\gamma _{1}}$, and $y^{\gamma _{5}} = x^{-\gamma _{2}}$, where $\gamma _{4} \triangleq \frac {\gamma _{1}}{\gamma _{3}}$ and $\gamma _{5} \triangleq \frac {\gamma _{2}}{\gamma _{3}}$. Since 0 < γ₁ ≤ γ₂ < γ₃, then 0 < γ₄ ≤ γ₅ < 1. The limit can be rewritten as:

$$ \lim\limits_{x \to \infty}f(x) = \lim\limits_{y \to 0^{+}}f(y) = \frac{\exp \left[\begin{array}{llll} \alpha_{1} y^{\gamma_{5}}\exp\left[\begin{array}{llll} -\alpha_{2} y^{-\gamma_{4}} \end{array}\right] \end{array}\right] - 1} {y}. $$

(65)

For notation simplicity, we define $\beta (y) \triangleq \exp \left [\begin {array}{llll} -\alpha _{2} y^{-\gamma _{4}} \end {array}\right ]$, leading to:

$$ \lim\limits_{y \to 0^{+}}f(y) = \lim\limits_{y \to 0^{+}}\frac{\exp \left[\begin{array}{llll} \alpha_{1} y^{\gamma_{5}} \beta(y) \end{array}\right] - 1} {y}. $$

(66)

It is straightforward to see that $\lim _{y \to 0^{+}}\beta (y) = 0$ and that $\lim _{y \to 0^{+}}e^{\alpha _{1} y^{\gamma _{5}} \beta (y)} = 1$. Since this limit is a ratio of zero with zero, we can use L’Hospital’s rule to show that the right-hand side of Eq. 66 converges to zero as y → 0⁺. We define the numerator signal as:

$$ n(y) \triangleq e^{\alpha_{1} y^{\gamma_{5}}\beta(y)} - 1. $$

(67)

Clearly, it is sufficient to prove that if $\lim _{y \to 0^{+}}\frac {dn(y)}{dy} = 0$, then

$$ \lim\limits_{y \to 0^{+}}\frac{e^{\alpha_{1} y^{\gamma_{5}}\beta(y)} - 1} {y} = 0, $$

(68)

implying that $\lim _{y \to 0^{+}}f(y) = 0$. Using the notation $f^{\prime } \triangleq \frac {\partial f}{\partial y}$ the derivative of n(y) is given by:

$$ n^{\prime}(y) = \alpha_{1} \left( \begin{array}{llll} \frac{\gamma_{5}}{y^{\gamma_{6}}}\beta(y) + y^{\gamma_{5}}\beta^{\prime}(y) \end{array}\right)e^{\alpha_{1} y^{\gamma_{5}}\beta(y)}, $$

(69)

where γ₆ > 0 is defined as $\gamma _{6} \triangleq 1 - \gamma _{5}$. Given that

$$ \beta^{\prime}(y) = \alpha_{2} \gamma_{4} y^{-\gamma_{4} - 1}\exp \left[\begin{array}{llll} -\alpha_{2} y^{-\gamma_{4}} \end{array}\right] = \frac{\alpha_{2} \gamma_{4}}{y^{\gamma_{7}}}\beta(y), $$

(70)

for $\gamma _{7} \triangleq 1 + \gamma _{4} > 1$, we can substitute Eq. 70 into Eq. 69:

$$ n^{\prime}(y) = \alpha_{1} \left( \begin{array}{llll} \frac{\gamma_{5}}{y^{\gamma_{6}}}\beta(y) + \frac{\alpha_{2} \gamma_{4}}{y^{\gamma_{7} - \gamma_{5}}}\beta(y) \end{array}\right)e^{\alpha_{1} y^{\gamma_{5}}\beta(y)}. $$

(71)

One should note that since γ₇ > 1 and 0 < γ₅ < 1, then γ₇ − γ₅ > 0. Using Lemma 1 and the definition $\beta (y) \triangleq \exp \left [\begin {array}{llll} -\alpha _{2} y^{-\gamma _{4}} \end {array}\right ]$, then:

$$ \left\{\begin{array}{llll} \lim_{\xi \to 0^{+}} \frac{\gamma_{5}}{y^{\gamma_{6}}}\beta(y) = 0, \\ \lim_{\xi \to 0^{+}} \frac{\alpha_{2} \gamma_{4}}{y^{\gamma_{7} - \gamma_{5}}}\beta(y) = 0 \end{array}\right.. $$

(72)

Remembering that $\lim _{y \to 0^{+}}\beta (y) = 0$, and that $\lim _{y \to 0^{+}}e^{\alpha _{1} y^{\gamma _{5}}\beta (y)} = 1$, then $\lim _{y \to 0^{+}} n^{\prime }(y) = 0$, which implies that $\lim _{x \to \infty }f(x) = 0$.

Appendix B

In this section, we show that the inequality

$$ \mu^{\lambda}(t)\ln(1 + \boldsymbol{\sigma}^{T}(t)\boldsymbol{\sigma}(t)) \leq \bar{V}, $$

(73)

for a constant $\bar {V} > 0$, implies that μ^λ/2σ ∈ L_∞ and that $\lim _{t \to t_{f}}\mu ^{\rho }(t)\boldsymbol {\sigma }(t) = 0, \forall \rho < \lambda /2$.

Starting from Eq. 73, we have that:

$$ \begin{array}{@{}rcl@{}} \boldsymbol{\sigma}^{T}(t)\boldsymbol{\sigma}(t) &\leq& \exp \left[\begin{array}{llll} \bar{V}\mu^{-\lambda}(t) \end{array}\right] - 1 \end{array} $$

(74)

$$ \begin{array}{@{}rcl@{}} \mu^{\lambda}(t)\boldsymbol{\sigma}^{T}(t)\boldsymbol{\sigma}(t) &\leq& \mu^{\lambda}(t) \left[\begin{array}{llll} \exp \left[\begin{array}{llll} \bar{V}\mu^{-\lambda}(t) \end{array}\right] - 1 \end{array}\right] \end{array} $$

(75)

$$ \begin{array}{@{}rcl@{}} ||\mu^{\lambda/2}(t)\boldsymbol{\sigma}(t)||^{2} &\leq& \frac{\exp \left[\begin{array}{llll} \bar{V}\mu^{-\lambda}(t) \end{array}\right] - 1}{\mu^{-\lambda}(t)}. \end{array} $$

(76)

In order to show that the signal $f(t) \triangleq ||\mu ^{\lambda /2}(t)\boldsymbol {\sigma }(t)||^{2}$ is bounded, we need to evaluate the limit as t → t_f. Taking the limit on both sides:

$$ \lim\limits_{t \to t_{f}}f(t) \leq \lim\limits_{t \to t_{f}} \frac{\exp \left[\begin{array}{llll} \bar{V}\mu^{-\lambda}(t) \end{array}\right] - 1}{\mu^{-\lambda}(t)}. $$

(77)

Assuming λ > 0, the above limit can be rewritten as:

$$ \lim\limits_{t \to t_{f}}f(t) \leq \lim\limits_{\mu \to \infty} \frac{\exp \left[\begin{array}{llll} \bar{V}\mu^{-\lambda} \end{array}\right] - 1}{\mu^{-\lambda}}. $$

(78)

Defining $\xi (t) \triangleq \mu ^{-\lambda }(t)$, we have that:

$$ \lim\limits_{t \to t_{f}}f(t) \leq \lim\limits_{\xi \to 0^{+}} \frac{\exp \left[\begin{array}{llll} \bar{V}\xi \end{array}\right] - 1}{\xi}. $$

(79)

Since the above limit is a ratio of zero with zero, we can use L’Hospital’s rule:

$$ \begin{array}{@{}rcl@{}} \lim\limits_{t \to t_{f}}f(t) &\leq& \lim\limits_{\xi \to 0^{+}} \frac{\frac{d}{d\xi} \left[\begin{array}{llll} \exp \left[\begin{array}{llll} \bar{V}\xi \end{array}\right] - 1 \end{array}\right]}{\frac{d}{d\xi}\xi} \end{array} $$

(80)

$$ \begin{array}{@{}rcl@{}} &=& \lim\limits_{\xi \to 0^{+}} \bar{V} \exp \left[\begin{array}{llll} \bar{V}\xi \end{array}\right] \end{array} $$

(81)

$$ \begin{array}{@{}rcl@{}} &=& \bar{V} \end{array} $$

(82)

Therefore, $f(t) \triangleq ||\mu ^{\lambda /2}(t)\boldsymbol {\sigma }(t)||^{2}$ is bounded, i.e., μ^λ/2σ ∈ L_∞. Also, since $||\mu ^{\lambda /2}(t)\boldsymbol {\sigma }(t)||^{2} \leq \bar {V}$, then for any 𝜖 > 0 and constant ρ such that 2𝜖 + ρ = λ/2 we have that $||\mu ^{\rho }\boldsymbol {\sigma }(t)||^{2} \leq \mu ^{-\epsilon }(t)\bar {V}$, implying that $\lim _{t \to t_{f}} ||\mu ^{\rho }\boldsymbol {\sigma }(t)||^{2} = 0$.