Sliding Mode Control with Tanh Function for Quadrotor UAV Altitude and Attitude Stabilization

Abstract

This paper addresses the problem of robust altitude (z), and attitude (ϕ, θ, ψ) control of ‘ ×’ mode configuration quadrotor UAV using a sliding mode control (SMC) with tanh function. The dynamic quadrotor model is derived by considering the nonlinearity factor. The dynamic model is simulated in MATLAB Simulink without and with the presence of external disturbance to test the robustness of the control method. A comparison is made with three others sliding mode control laws, such as reaching law, exponential reaching law and saturation function. The sliding condition is verified and guaranteed by the Lyapunov stability function. The result confirms that the sliding mode control using the tanh function shows tremendous performance and robustness against disturbance without being affected by the chattering phenomenon. This study deals with the nonlinear quadrotor control for attitude and altitude control using robust sliding mode control, where the tanh function eliminates the chattering phenomenon.

Appendix

Lemma 1.1: [28, 29] For every given scalar $x$ and positive scalar $\epsilon$, the following inequality holds:

$$ \begin{array}{*{20}c} {x.tanh\left( {\frac{x}{\epsilon}} \right) = \left| {x.tanh\left( {\frac{x}{\epsilon}} \right)} \right| = \left| x \right|\left| {\tanh \left( {\frac{x}{\epsilon}} \right)} \right| \ge 0} \\ \end{array} $$

Lemma 1.1 can be proved as follows: according to the definition of tanh function, we have

$$ \begin{array}{*{20}c} {x.\tanh \left( {\frac{x}{\epsilon}} \right) = s\frac{{e^{{\frac{x}{\epsilon}}} - e^{{ - \frac{x}{\epsilon}}} }}{{e^{{\frac{x}{\epsilon}}} + e^{{ - \frac{x}{\epsilon}}} }}{ } = \frac{1}{{e^{{2\frac{x}{\epsilon}}} + 1}}x\left( {e^{{2\frac{x}{\epsilon}}} - 1} \right)} \\ \end{array} $$

Since

$e^{{2\frac{x}{\epsilon}}} - 1 \ge 0{ }\,\,\,if{ }x \ge 0$

$e^{{2\frac{x}{\epsilon}}} - 1 < 0{ }\,\,\,if{ }x < 0$

Then

$$ x\left( {e^{{2\frac{x}{\epsilon}}} - 1} \right) \ge 0 $$

Therefore

$$ x.\tanh \left( {\frac{x}{\epsilon}} \right) = \frac{1}{{e^{{2\frac{x}{\epsilon}}} + 1}}x\left( {e^{{2\frac{x}{\epsilon}}} - 1} \right) \ge 0 $$

And

$$ x.\tanh \left( {\frac{x}{\epsilon}} \right) = \left| {x.tanh\left( {\frac{x}{\epsilon}} \right)} \right| = \left| x \right|\left| {\tanh \left( {\frac{x}{\epsilon}} \right)} \right| \ge 0 $$

Lemma 1.2, [28, 30] Let $f$, $V:\left[ {0,\infty } \right] \in R$, then $\dot{V} \le - \alpha V + f$, $\forall t \ge t_{0} \ge 0$ implies that

$$ V\left( t \right) \le e^{{ - \alpha \left( {t - t_{0} } \right)}} V\left( {t_{0} } \right) + \mathop \smallint \limits_{{t_{0} }}^{t} e^{{ - \alpha \left( {t - \tau } \right)}} f\left( \tau \right)d\tau { } $$

For any finite constant $\alpha .$

According to [28, 30], we have the proof as follows:

Let $\omega \left( t \right) \triangleq \dot{V} + \alpha V - f$, we have $\omega \left( t \right) \le 0$, and

$$ \dot{V} = - \alpha V + f + \omega $$

Implies that

$$ V\left( t \right) = e^{{ - \alpha \left( {t - t_{0} } \right)}} V\left( {t_{0} } \right) + \mathop \smallint \limits_{{t_{0} }}^{t} e^{{ - \alpha \left( {t - \tau } \right)}} f\left( \tau \right)d\tau + \mathop \smallint \limits_{{t_{0} }}^{t} e^{{ - \alpha \left( {t - \tau } \right)}} \omega \left( \tau \right)d\tau $$

Because $\omega \left( t \right) < 0$ and $\forall t \ge t_{o} \ge 0,$ we have

$$ V\left( t \right) = e^{{ - \alpha \left( {t - t_{0} } \right)}} V\left( {t_{0} } \right) + \mathop \smallint \limits_{{t_{0} }}^{t} e^{{ - \alpha \left( {t - \tau } \right)}} f\left( \tau \right)d\tau $$

Moreover, if we choose $f = 0$, then we have $\dot{V} \le - \alpha V$, implies that

$$ V\left( t \right) = e^{{ - \alpha \left( {t - t_{0} } \right)}} V\left( {t_{0} } \right) $$

If $\alpha$ is a positive constant value, $V\left( t \right)$ will tend to zero exponentially with $\alpha$ value.