Multivariable Binary MRAC with Guaranteed Transient Performance Using Non-homogeneous Robust Exact Differentiators

Abstract

In this paper, we propose an extension of the binary model reference adaptive control (BMRAC) for uncertain multivariable linear plants with non-uniform arbitrary relative degree. The BMRAC is a robust adaptive strategy which has good transient properties and robustness of sliding mode control with the important advantage of having a continuous control signal free of chattering. The relative degree obstacle is circumvented using a multivariable version of a hybrid estimation scheme, named global robust exact differentiator (GRED). The hybrid estimator switches between robust exact differentiators (RED) based on higher-order sliding modes and lead filters in a way that the exact derivatives are globally obtained in finite time. To improve the robustness and the transient performance of the GRED, we propose a modification of the switching scheme replacing the conventional RED with a non-homogeneous one. Global exact output tracking is obtained with robustness and guaranteed transient performance without requiring stringent symmetry assumptions on the plant high-frequency-gain matrix.

Appendix A Proofs of Theorems

1.1 A.1 Proof of Lemma 1

Applying the change of variables $\upsilon _i = \zeta _i - f^{(i)}(t)$, $i=0,1,\ldots ,n$ to the system (30), it follows that

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{\upsilon }}_0 = -\lambda _0|\upsilon _0|^{n/(n+1)}{\text {sgn}}\left( \upsilon _0\right) - \mu _0\upsilon _0 + \upsilon _1 \\ \vdots \\ {\dot{\upsilon }}_i = -\lambda _i|\upsilon _i - \dot{\upsilon }_{i-1}|^{(n-i)/(n-i+1)}{\text {sgn}}\left( \upsilon _i - \dot{\upsilon }_{i-1}\right) - \\ + \mu _i(\upsilon _i - \dot{\upsilon }_{i-1}) + \upsilon _{i+1} \\ \vdots \\ {\dot{\upsilon }}_n = -\lambda _n{\text {sgn}}\left( \upsilon _{n} - \dot{\upsilon }_{n-1}\right) - \mu _n(\upsilon _{n} - \dot{\upsilon }_{n-1}) - f^{(n+1)} \end{array} \right. \end{aligned}$$

Considering that $x = |x|{\text {sgn}}\left( x\right)$ and ${\text {sgn}}\left( \upsilon _i - \dot{\upsilon }_{i-1}\right) = {\text {sgn}}\left( \upsilon _{i-1} - \dot{\upsilon }_{i-2}\right)$ for $i=2,\ldots ,n$, and thus ${\text {sgn}}\left( \upsilon _i - \dot{\upsilon }_{i-1}\right) = {\text {sgn}}\left( \upsilon _0\right)$, for $i=1,2,\ldots ,n$, the system can be rewritten as

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{\upsilon }}_0 = -\left[ \lambda _0|\upsilon _0|^{n/(n+1)} + \mu _0|\upsilon _0|\right] {\text {sgn}}\left( \upsilon _0\right) + \upsilon _1 \\ \vdots \\ {\dot{\upsilon }}_i = -\left[ \lambda _i|\upsilon _i - \dot{\upsilon }_{i-1}|^{\frac{(n-i)}{(n-i+1)}} + \mu _i|\upsilon _i - \dot{\upsilon }_{i-1}|\right] {\text {sgn}}\left( \upsilon _0\right) + \\ + \upsilon _{i+1}\\ \vdots \\ {\dot{\upsilon }}_n = -\left[ \lambda _n + \mu _n|\upsilon _{n} - \dot{\upsilon }_{n-1}|\right] {\text {sgn}}\left( \upsilon _0\right) - f^{(n+1)} \end{array} \right. \end{aligned}$$

It follows by mathematical induction that the non-recursive form of the system is given by

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{\upsilon }}_0 = -\left[ \phi _0(|\upsilon _0|) + \psi _0 |\upsilon _0|\right] {\text {sgn}}\left( \upsilon _0\right) + \upsilon _1 \\ \vdots \\ {\dot{\upsilon }}_i = -\left[ \phi _i(|\upsilon _0|) + \psi _i |\upsilon _0|\right] {\text {sgn}}\left( \upsilon _0\right) + \upsilon _{i+1}\\ \vdots \\ {\dot{\upsilon }}_n = -\left[ \phi _n(|\upsilon _0|) + \psi _n |\upsilon _0|\right] {\text {sgn}}\left( \upsilon _0\right) - f^{(n+1)} \end{array} \right. \end{aligned}$$

(43)

where

$$\begin{aligned} \left\{ \begin{array}{l} \phi _0 = \lambda _0|\upsilon _0|^{n/(n+1)}, \\ \psi _0 = \mu _0 \\ \phi _i = \lambda _i(\phi _{i-1} + \psi _{i-1}|\upsilon _0|)^{(n-i)/(n-i+1)} + \mu _i \phi _{i-1}, \\ \psi _i = \mu _i \psi _{i-1}, \quad i=1,2,\ldots ,n \end{array} \right. \end{aligned}$$

Note that $\psi _i$ is constant, and $\phi _i(|\upsilon _0|)$ obeys the following conditions

$$\begin{aligned} \left\{ \begin{array}{ll} \phi _i(|\upsilon _0|) \le \kappa _1^{[i]} &{} \text {se } |\upsilon _0| \le 1 \\ \displaystyle \frac{\phi _i(|\upsilon _0|)}{|\upsilon _0|} \le \kappa _2^{[i]} &{} \text {se } |\upsilon _0| > 1 \end{array} \right. \end{aligned}$$

for $i=0,1,\ldots ,n$ and some positive constants $\kappa _1^{[i]}$ and $\kappa _2^{[i]}$. The first inequality follows from the fact that $\phi _i$ is bounded, for $|\upsilon _0| \le 1$, if $\phi _{i-1}$ is also bounded. Since $\phi _0$ is bounded under these conditions, then the first inequality follows by mathematical induction. The second inequality can be demonstrated considering that

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle {\frac{\phi _0}{|\upsilon _0|} = \lambda _0\frac{1}{|\upsilon _0|^{\frac{1}{n+1}}}} \\ \displaystyle {\frac{\phi _i}{|\upsilon _0|} = \frac{\lambda _i}{|\upsilon _0|^{\frac{1}{n-i+1}}}\left( \frac{\phi _{i-1}}{|\upsilon _0|} + \psi _{i-1}\right) ^{\frac{n-i}{n-i+1}} + \mu _i \frac{\phi _{i-1}}{|\upsilon _0|}}, \\ \qquad i=1,2,\ldots ,n \end{array} \right. \end{aligned}$$

and the fact that, for $|\upsilon _0| > 1$, $\frac{\phi _i}{|\upsilon _0|}$ is bounded if $\frac{\phi _{i-1}}{|\upsilon _0|}$ is also bounded. Since $\frac{\phi _0}{|\upsilon _0|}$ is bounded under these conditions, then the second inequality follows by mathematical induction.

The equations ${\dot{\upsilon }}_i = -\left[ \phi _i(|\upsilon _0|) + \psi _i |\upsilon _0|\right] {\text {sgn}}\left( \upsilon _0\right) + \upsilon _{i+1}$, $i=0,1,\ldots ,n-1$ can be rewritten as

$$\begin{aligned} \dot{\upsilon }_i = -a_i(\upsilon _0)\upsilon _0 - b_i(\upsilon _0) + \upsilon _{i+1} \end{aligned}$$

where

$$\begin{aligned} a_i(\upsilon _0)&= \left\{ \begin{array}{ll} \psi _i, &{} |\upsilon _0| \le 1 \\ \psi _i + \displaystyle \frac{\phi _i(|\upsilon _0|)}{|\upsilon _0|}, &{} |\upsilon _0|> 1 \end{array} \right. , \\ b_i(\upsilon _0)&= \left\{ \begin{array}{ll} \phi _i(|\upsilon _0|){\text {sgn}}\left( \upsilon _0\right) , &{} |\upsilon _0| \le 1 \\ 0, &{} |\upsilon _0| > 1 \\ \end{array} \right. \end{aligned}$$

The equation ${\dot{\upsilon }}_n = -\left[ \phi _n(|\upsilon _0|) + \psi _n |\upsilon _0|\right] {\text {sgn}}\left( \upsilon _0\right) - f^{(n+1)}(t)$ can be rewritten as

$$\begin{aligned} \dot{\upsilon }_n = -a_n(\upsilon _0)\upsilon _0 - b_n(\upsilon _0) \end{aligned}$$

where

$$\begin{aligned} a_n(\upsilon _0)&= \left\{ \begin{array}{ll} \psi _n, &{} |\upsilon _0| \le 1 \\ \psi _n + \displaystyle \frac{\phi _n(|\upsilon _0|)}{|\upsilon _0|}, &{} |\upsilon _0|> 1 \end{array} \right. , \\ b_n(\upsilon _0)&= \left\{ \begin{array}{ll} \phi _n(|\upsilon _0|){\text {sgn}}\left( \upsilon _0\right) + f^{(n+1)}(t), &{} |\upsilon _0| \le 1 \\ f^{(n+1)}(t), &{} |\upsilon _0| > 1 \\ \end{array} \right. \end{aligned}$$

Note that, once $\left| f^{(n+1)}(t) \right| \le K_{n+1} \,\, \forall t$, then $\left| a_i(\upsilon _0) \right| \le K_{a_i}$ and $\left| b_i(\upsilon _0) \right| \le K_{b_i}$, for $i=0,1,\ldots ,n$ and some positive constants $K_{a_i}$ and $K_{b_i}$. Defining the complete state vector as $\varUpsilon = \left[ \begin{array}{cccc} \upsilon _0&\upsilon _1&\ldots&\upsilon _n \end{array}\right] ^T$ , the system (43) can be rewritten as

$$\begin{aligned} \dot{\varUpsilon } = A(\varUpsilon )\varUpsilon + b(\varUpsilon ) \end{aligned}$$

where

$$\begin{aligned} A(\varUpsilon ) = \left[ \begin{array}{ccccc} -a_0(\upsilon _0) &{} 1 &{} 0 &{} \ldots &{} 0 \\ -a_1(\upsilon _0) &{} 0 &{} 1 &{} \ldots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ -a_{n-1}(\upsilon _0) &{} 0 &{} 0 &{} \ldots &{} 1 \\ -a_n(\upsilon _0) &{} 0 &{} 0 &{} \ldots &{} 0 \\ \end{array} \right] \ \ b(\varUpsilon ) = \left[ \begin{array}{c} -b_0(\upsilon _0)\\ -b_1(\upsilon _0)\\ \vdots \\ -b_{n-1}(\upsilon _0)\\ -b_n(\upsilon _0)\\ \end{array} \right] \end{aligned}$$

and it follows that $\Vert b(\varUpsilon ) \Vert \le c_1$ and $\Vert A(\varUpsilon ) \Vert \le c_2$ for some positive constants $c_1$ and $c_2$.

Consider the function

$$\begin{aligned} V(\varUpsilon (t)) = \varUpsilon ^T(t) \varUpsilon (t) \end{aligned}$$

It can be verified that

$$\begin{aligned} \dot{V}(\varUpsilon ) \le 2c_2 V(\varUpsilon ) + 2c_1 \sqrt{V(\varUpsilon )} \end{aligned}$$

Consider the comparison equation $\dot{V_c}(\varUpsilon ) = 2c_2 V_c(\varUpsilon ) + 2c_1 \sqrt{V_c(\varUpsilon )}$. If $V_c(0) = V(0)$, then $V(t) \le V_c(t), \forall t \ge 0$. Introducing the new variable $\chi ^2 = V_c$, it follows that

$$\begin{aligned} \chi {\dot{\chi }} = c_2 \chi ^2 + c_1 \chi \end{aligned}$$

Considering $\chi \ne 0$, then $\frac{\mathrm{d}\chi }{\mathrm{d}t} = c_2 \chi + c_1$. In this case, it can be shown that

$$\begin{aligned} \frac{1}{c_2}\ln {\left( \frac{2c_2 \sqrt{V_c(t)} + 2c_1}{2c_2 \sqrt{V_c(0)} + 2c_1}\right) } = t \end{aligned}$$

and

$$\begin{aligned} V(t) \le \left[ \left( \sqrt{V(0)} + \frac{c_1}{c_2} \right) e^{c_2t} - \frac{c_1}{c_2} \right] ^2 \end{aligned}$$

Then the function V(t) cannot escape in finite time for any finite constant $K_{n+1}$, and thus $\upsilon \in {\mathcal {L}}_{\infty e}$. Furthermore, since the signals $f^{(i)}(t)$, $i=0,1,\ldots ,n,$ are bounded, then one can conclude that the state $\zeta$ cannot escape in finite time. $\square$

1.2 A.2 Proof of Theorem 2

The estimate given by the MIMO lead filter and the MIMO RED could be related to $\xi _y$ in (19) as follows:

$$\begin{aligned} {\hat{\xi }}_{l} = \xi _y + \varepsilon _{l}, \qquad {\hat{\xi }}_{r} = \xi _y + \varepsilon _{r}\,, \end{aligned}$$

(44)

where $\varepsilon _{l}$ and $\varepsilon _{r}$ are estimation errors. From (44), equation (38) can be rewritten as

$$\begin{aligned} {\hat{\xi }}_{g}&= \xi _y + \varepsilon _g, \quad \varepsilon _g&= \alpha ({\tilde{\nu }}_{rl}) \varepsilon _{l} + \left[ 1 - \alpha ({\tilde{\nu }}_{rl}) \right] \varepsilon _{r}\,. \end{aligned}$$

(45)

From (39), the estimation error $\varepsilon _g(t)$ can be rewritten as:

$$\begin{aligned} \varepsilon _g = \varepsilon _{l} + \beta _{\alpha }({\tilde{\nu }}_{rl}(t))\,, \end{aligned}$$

(46)

where by design $\beta _{\alpha }({\tilde{\nu }}_{rl}(t))$ is uniformly bounded by

$$\begin{aligned} \left| \! \left| \beta _{\alpha }({\tilde{\nu }}_{rl}(t)) \right| \! \right| < \varepsilon _M, \ \text {with} \ \varepsilon _M = \tau K_R\,. \end{aligned}$$

Substituting (45),(46) into (40)–(42), it can be seen that GRED adaptive law is equivalent to lead adaptive law (26)–(28) with an output disturbance $\left| \! \left| \beta _{\alpha }({\tilde{\nu }}_{rl}(t)) \right| \! \right| \le \varepsilon _M$.

Therefore, Theorem 1 holds if all signals of the GRED-BMRAC system belong to $L_{\infty e}$. In order to demonstrate that the condition is true, we only have to show that all signals in the MIMO RED system are $L_{\infty e}$. This property can be proved by contradiction. Suppose that the maximal interval of finiteness of the signals in the MIMO RED is $[0,T_M)$. During this interval, all conditions of Theorem 1 hold, and thus, all signals of the remaining subsystems of the GRED-BMRAC are bounded by a constant, and in particular, $\left| y_j^{(i)}(t)\right| , i=0,\dots ,\rho _j, \ j = 1,\dots ,M$, from Corollary 2. This leads to a contradiction with Lemma 1, whereby the signals in the MIMO RED could not diverge unboundedly as $t \rightarrow T_M$. As a consequence of the continuation theorem for differential equations (in Filippov’s theory), $T_M$ must be $\infty$, which means that all signals are defined $\forall t \ge 0$. Thus, Theorem 1 is valid for the GRED-MRAC system and the closed-loop error system with state z is GEpS with respect to a residual set.

Now, we will analyze the ultimate convergence of the GRED-BMRAC. According to Corollary 1, for sufficiently small $\tau$ and sufficiently large $\gamma$ the error state z is steered to an invariant compact set $D_R := \{ z : \left| z(t)\right| < R \}$ in some finite time $T_1 \ge 0$. Consider the following Lyapunov candidate

$$\begin{aligned} V = x_\varepsilon ^T P_2 x_\varepsilon \end{aligned}$$

(47)

whose time derivative is

$$\begin{aligned} \dot{V}= & {} - \frac{1}{\tau } x_{\varepsilon }^T Q_2 x_{\varepsilon } + 2x_{\varepsilon }^T P_2 B_{\varepsilon } \dot{\xi _y} \end{aligned}$$

following the previous steps

$$\begin{aligned} \dot{V}&{=}&- \frac{1}{\tau } x_{\varepsilon }^T Q_2 x_{\varepsilon } {+} 2x_{\varepsilon }^T Q_3 x_e + 2x_\varepsilon ^T Q_4 X_m {+} 2x_\varepsilon ^T Q_5 r \end{aligned}$$

(48)

$$\begin{aligned}&\quad \dot{V} \le - \frac{k_2}{\tau } \left| \! \left| x_{\varepsilon } \right| \! \right| ^2 + k_3 \left| \! \left| x_\varepsilon \right| \! \right| \left| \! \left| x_e \right| \! \right| + k_4 \left| \! \left| x_\varepsilon \right| \! \right| \end{aligned}$$

(49)

Within $D_R$, $x_e$ can be upper bounded by $\left| \! \left| x_e \right| \! \right| \le R$ such that

$$\begin{aligned} \dot{V} \le - \frac{k_2}{\tau } \left| \! \left| x_{\varepsilon } \right| \! \right| ^2 + \tau k_5 \end{aligned}$$

(50)

Thus, it is possible to show that

$$\begin{aligned} \left| \! \left| x_\varepsilon (t) \right| \! \right| \le c_\varepsilon e^{-a(t - t_0)}\left| \! \left| x_\varepsilon (t_0) \right| \! \right| + \tau k_6 \end{aligned}$$

(51)

Since $\left| \! \left| \varepsilon _l \right| \! \right| \le \left| \! \left| x_\varepsilon \right| \! \right|$, it is straightforward to show that for some finite $T_2 \ge T_1$, $\left| \! \left| \varepsilon _l \right| \! \right| \le {{\bar{\varepsilon }}}_l$, where ${{\bar{\varepsilon }}}_l = \tau K_{l}$.

Since the MIMO RED is time-invariant, its initial conditions can be considered to be at $t=T_1$. From Lemma 1, the initial conditions are finite. If the parameters $\lambda _i^j$ are adjusted properly, then from Levant (2009) the estimation error $\varepsilon _{r}(t)$ converges to zero in some finite time $T_3 > T_1$.

Since $K_R$ is chosen such that $\varepsilon _M > {\bar{\varepsilon }}_{l} + \varDelta$ and from (39), it follows that after some finite time ${{\bar{T}}} = \max \{ T_2, T_3\}$ the estimation of $\sigma$ becomes exact and being made exclusively by the MIMO RED ($\alpha ({\tilde{\nu }}_{rl}) = 0$), which implies that $\varepsilon _g(t) = 0, \forall t \ge {\bar{T}}$.

In this case, the overall error system can be described by

$$\begin{aligned} \dot{x}_e = A_c {x_e} + B_c K_p [u-u^*] \, , \quad {\bar{e}}_L = {\bar{L}} {\bar{H}} {x_e}\,, \end{aligned}$$

(52)

since this system has uniform relative degree one we can apply the result obtained in Yanque et al. (2012). Thus, it is possible to conclude that $e(t) \rightarrow 0$. $\square$