**5. Simultaneous Sensor and Actuator Faults Reconstruction**

In this section, a more general case is investigated where the sensor and actuator faults occur simultaneously, i.e., *fs*(*t*) = 0 and *fa*(*t*) = 0. Before proceeding to the main results, some precalculations need to be done. As discussed earlier, the nonsingular transformation matrices *H* = [*H*1, *H*2] *<sup>T</sup>* and <sup>T</sup> exist such that:

$$\begin{aligned} \mathcal{T}A\mathcal{T}^{-1} &= \begin{bmatrix} A\_1 & A\_2 \\ A\_3 & A\_4 \end{bmatrix} \mathcal{T}B = \begin{bmatrix} B\_1 \\ 0 \end{bmatrix}, \\ \mathcal{T}F\_d &= \begin{bmatrix} F\_{a\_1} \\ 0 \end{bmatrix}, \mathcal{T}D = \begin{bmatrix} D\_1 \\ 0 \end{bmatrix}, \mathcal{T}M = \begin{bmatrix} M\_1 \\ 0 \end{bmatrix}, \\ H\mathcal{T}T^{-1} &= \begin{bmatrix} \mathcal{C}\_1 & 0 \\ 0 & \mathcal{C}\_4 \end{bmatrix}, H\mathcal{F}\_s = \begin{bmatrix} 0 \\ F\_{\mathcal{S}\_2} \end{bmatrix}. \end{aligned} \tag{48}$$

Then, the system (1) in the new coordinates *z*¯ = T *z* = *z*¯1 *z*¯2 and *y*¯ = *Hy* = *y*¯1 *y*¯2 is decomposed as

$$\begin{cases} \dot{\mathbb{E}}\_{1}(t) = A\_{1}\mathbb{E}\_{1}(t) + A\_{2}\mathbb{E}\_{2}(t) + B\_{1}u(t) \\ \qquad + F\_{\mathbb{E}\_{1}}f\_{d}(t) + D\_{1}d(t) + M\_{1}\partial(t,y,u) \\ \quad \bar{g}\_{1}(t) = C\_{1}\mathbb{E}\_{1}(t) \end{cases} \tag{49}$$

$$\begin{cases} \ ^{\sharp}\dot{\boldsymbol{z}}\_{2}(t) = \boldsymbol{A}\_{3}\boldsymbol{\overline{z}}\_{1}(t) + \boldsymbol{A}\_{4}\boldsymbol{\overline{z}}\_{2}(t) \\\ \ddot{\boldsymbol{y}}\_{2}(t) = \boldsymbol{\overline{C}}\_{4}\boldsymbol{\overline{z}}\_{2}(t) + \boldsymbol{F}\_{\boldsymbol{\overline{s}}\_{2}}\boldsymbol{f}\_{\boldsymbol{s}}(t) \end{cases} \tag{50}$$

By defining *z*¯3 = *z*¯2 *fs* and *<sup>C</sup>*<sup>5</sup> = *<sup>C</sup>*<sup>4</sup> *Fs*<sup>2</sup> , Equation (51) can be rewritten as:

$$\begin{array}{l}N\dot{\boldsymbol{\varepsilon}}\_{3} = \bar{A}\_{3}\boldsymbol{\varepsilon}\_{1} + \bar{A}\_{4}\boldsymbol{\varepsilon}\_{3} + \bar{F}\_{\boldsymbol{\varepsilon}\_{2}}\boldsymbol{f}\_{\boldsymbol{s}}\\\dot{g}\_{2} = \mathbf{C}\_{5}\boldsymbol{\varepsilon}\_{3}\end{array} \tag{51}$$

where

$$\begin{cases} \dot{z}\_1(t) = A\_1 \bar{z}\_1(t) + A\_2 \bar{z}\_2(t) + B\_1 u(t) \\ \qquad + F\_{a\_1} f\_a(t) + D\_1 d(t) + M\_1 \partial(t, y, u) \\ \quad \bar{y}\_1(t) = C\_1 \bar{z}\_1(t) \end{cases} \tag{52}$$

$$\begin{cases} \dot{\tilde{z}}\_2(t) = A\_3 \bar{z}\_1(t) + A\_4 \bar{z}\_2(t) \\ \dot{g}\_2(t) = \mathbb{C}\_4 \mathbb{Z}\_2(t) + F\_{s2} f\_s(t) \end{cases} \tag{53}$$

Considering *A*¯ <sup>2</sup> = [*A*2, 0], from (50) one obtains:

$$\begin{aligned} \dot{z}\_1(t) &= A\_1 \ddot{z}\_1(t) + A\_2 \ddot{z}\_3(t) + B\_1 u(t) + F\_{a\_1} f\_a(t) \\ &+ D\_1 d(t) + M\_1 \ddot{\partial}(t, y, u) \end{aligned} \tag{54}$$

Combining (50)–(54), we get:

$$\begin{cases} \dot{\boldsymbol{z}}\_1(t) = A\_1 \boldsymbol{\overline{z}}\_1(t) + \bar{A}\_2 \boldsymbol{\overline{z}}\_3(t) + B\_1 \boldsymbol{u}(t) \\ \qquad + \boldsymbol{F}\_{a\_1} \boldsymbol{f}\_a(t) + \boldsymbol{K}\_1 \boldsymbol{\psi}(t, \boldsymbol{y}, \boldsymbol{u}) \\ \quad \boldsymbol{g}\_1(t) = \mathbf{C}\_1 \boldsymbol{\overline{z}}\_1(t) \end{cases} \tag{55}$$

$$\begin{cases} N\dot{\mathbb{E}}\_3 = \bar{A}\_3 \bar{z}\_1 + \bar{A}\_4 \bar{z}\_3 + \bar{F}\_{s\_2} f\_s\\ \bar{y}\_2 = C\_5 \mathbb{E}\_3 \end{cases} \tag{56}$$

where *<sup>K</sup>*<sup>1</sup> = [*D*1, *<sup>M</sup>*1], *<sup>K</sup>*¯2 = [*D*¯ 2, *<sup>M</sup>*¯ <sup>2</sup>], *<sup>ψ</sup>*(*t*, *<sup>y</sup>*, *<sup>u</sup>*) = *<sup>d</sup>*(*t*) *∂*(*t*, *y*, *u*) . Based on the above results, the following theorem characterizes the proposed method for simultaneous reconstruction of the sensor and actuator faults in the presence of disturbances and uncertainties.

**Theorem 4.** *Consider the faulty system (55) and (56), and assume the observer structure as:*

$$\begin{array}{c} (N + V\mathbb{C}\_5) \dot{\mathbf{x}} = (\bar{A}\_4 - L\_1 \mathbb{C}\_5) \mathbf{x} + L\_2 (y\_1 - \mathbb{C}\_1 \dot{\mathbf{z}}\_1) \\ + \bar{A}\_3 \dot{\mathbf{z}}\_1 + \bar{A}\_4 (N + V\mathbb{C}\_5)^{-1} V y\_2 \end{array} \tag{57}$$

$$\mathcal{B}\_3 = \mathfrak{x} + (N + V\mathbb{C}\_5)^{-1} V y\_2 \tag{58}$$

$$
\dot{\vec{z}}\_1 = A\_1 \dot{\vec{z}}\_1 + \vec{A}\_2 \dot{\vec{z}}\_3 + B\_1 u(t) + \vec{G}\_n v(t) - \vec{G}\_l e\_{y\_1}.\tag{59}
$$

*Then, the observer error is bounded if there exist P*<sup>1</sup> = *P<sup>T</sup>* <sup>1</sup> *, <sup>P</sup>*<sup>3</sup> = *<sup>P</sup><sup>T</sup>* <sup>3</sup> *, and K satisfying the following LMIs*

$$
\underbrace{\begin{bmatrix} Q\_{11} & P\_1 A\_2 & P\_1 K \\ \ast & Q\_{22} & Q\_{23} \\ \ast & \ast & -I \end{bmatrix}}\_{\widetilde{\alpha}} < 0 \tag{60}
$$

$$\begin{cases} \text{Q} \\ P\_1 > 0, \text{ } P\_3 > 0 \end{cases} \tag{61}$$

*where* ∗ *denotes the transpose of each symmetric element, and*

$$\begin{array}{l} Q\_{11} = A\_1^T P\_1 + P\_1 A\_1 + \lambda^2 \left\| \left| \mathcal{T}^{-1} \right| \right\|^2 I\_{m \prime} \\ Q\_{23} = P\_3 (N + V \mathcal{C}\_5)^{-1} \\ Q\_{22} = Q\_{23} \bar{A}\_4 - K \mathcal{C}\_5 - \bar{A}\_4^T Q\_{23}^T - \mathcal{C}\_5^T K^T \\ + \lambda^2 \left\| \left| \mathcal{T}^{-1} \right| \right\|^2 I\_{n + p - 2m} \end{array} \tag{62}$$

*Furthermore, the observer error bound is given as follows*

$$||\boldsymbol{\varepsilon}|| < \sigma = 2\lambda\_{\text{min}}^{-1}(Q)\left\|\boldsymbol{P\_3}\begin{bmatrix} \boldsymbol{0} \\ \left(\boldsymbol{V\_2F\_{s\_2}}\right)^{-1} \end{bmatrix}\right\| + \sigma\_0\tag{63}$$

*where σ*<sup>0</sup> *is a small positive scalar.*

**Proof.** For the observer (57)–(59), with *ey*<sup>1</sup> = *y*ˆ1 − *y*<sup>1</sup> = *C*1*e*<sup>1</sup> we have:

$$\sigma(t) = -\rho\_0(\sharp\_1 - \sharp\_1) \left( \left| \left( \sharp\_1 - \sharp\_1 \right) \right| \right)^{-1} \tag{64}$$

$$\begin{array}{l} L\_2 = \bar{A}\_3 \mathbb{C}\_1^{-1} \\ L\_1 = (N + V \mathbb{C}\_5) P\_3 K \end{array} \tag{65}$$

where *<sup>ρ</sup>*<sup>0</sup> is a positive scalar. Defining *<sup>e</sup>*<sup>1</sup> = *<sup>z</sup>*ˆ¯1 − *<sup>z</sup>*¯1, *<sup>e</sup>*<sup>3</sup> = *<sup>z</sup>*ˆ¯3 − *<sup>z</sup>*¯3 and using (57)–(59) one obtains:

$$\begin{array}{l} \dot{e}\_{1} = (A\_{1} - \tilde{G}\_{l}\mathbb{C}\_{5})e\_{1} + \bar{A}\_{2}e\_{3} - F\_{a\_{1}}f\_{a}(t) \\ \qquad + \tilde{G}\_{l}v(t) - K\_{1}\psi(t, y\_{\prime}, u) \\ \dot{e}\_{3} = -\underbrace{Q\_{23}\bar{A}\_{4}(N + V\mathbb{C}\_{5})^{-1}}\_{Q\_{23}^{\prime}}e\_{3} - Q\_{23}F\_{s\_{2}}f\_{s} \end{array} \tag{66}$$

Defining *s*(*t*) = *e*1(*t*) and *V* = <sup>1</sup> <sup>2</sup> *<sup>s</sup>TP*1*s*, we get:

$$\dot{V} \le \|P\_1 e\_1\| \left( \begin{array}{l} \|(A\_1 - \vec{G}\_1 \mathbb{C}\_5)e\_1 + \vec{A}\_2 e\_3\| - \rho\_0 \vec{G}\_n \\ -\|F\_{a\_1} f\_a(t)\| - \|K\_1 \psi(t, y, u)\| \end{array} \right) \tag{67}$$

where *<sup>G</sup>*¯*nv*(*t*) <sup>=</sup> <sup>−</sup>*ρ*0*G*¯*ne*1*e*1−<sup>1</sup> . Choosing *<sup>ρ</sup>*<sup>0</sup> <sup>≥</sup> \$ \$*G*¯ <sup>−</sup><sup>1</sup> *<sup>n</sup>* \$ \$ - *l* \$ \$(*A*<sup>1</sup> <sup>−</sup> *<sup>G</sup>*¯ *lC*5) \$ \$ <sup>+</sup> \$ \$*A*¯ <sup>2</sup> \$ \$ − *Fa*<sup>1</sup> − *K*1 *ε*, implies that after a finite time, we have *e*1(*t*) = *e*˙1(*t*) = 0. Then, one obtains

$$0 = \bar{A}\_2 e\_3 - F\_{a\_1} f\_a(t) + \bar{G}\_{\mathcal{V}} \upsilon(t) - K\_1 \psi(t, y, u). \tag{68}$$

Now, the following actuator reconstruction signal is defined:

$$f\_a = \mathcal{W}\mathcal{G}\_n v\_{eq}(t) \tag{69}$$

where *<sup>W</sup>* = *<sup>F</sup>*−<sup>1</sup> *<sup>a</sup>*<sup>1</sup> . To preserve sliding motion, *<sup>v</sup>*(*t*) must take in the average *veq*(*t*) = −*ρ*0*e*1(*e*1 + *ε*) −1 . Then, multiplying (68) by *W* results:

$$
\hat{f}\_a = -\mathcal{W}\vec{A}\_2\mathbf{e}\mathbf{e} + f\_a(t) + \mathcal{W}\mathcal{K}\_1\psi(t, \mathbf{y}, \mu). \tag{70}
$$

Then, one obtains

$$\begin{aligned} \hat{f}\_a - f\_a &= -\mathsf{W}\bar{A}\_2 \mathsf{e}\_3 + \underbrace{\mathsf{W}K\_1}\_{Z} \psi(t, y, u) \\ &\rightarrow \left\| \hat{f}\_a - f\_a \right\| < \nu' + \left\| Z \right\| \psi(t, y, u) \end{aligned} \tag{71}$$

where *υ* > \$ \$*WA*¯ <sup>2</sup>*e*<sup>3</sup> \$ \$ is a small positive constant. Assuming *Z*<sup>∞</sup> < *υ* with *υ* > 0, results:

$$\left\| \left| \hat{f}\_a - f\_a \right| \right\| < \upsilon. \tag{72}$$

For small *υ*, it implies ˆ *fa* ≈ *fa*. Then, from (66) one obtains

$$
\dot{e}\_3 + Q\_{23}'e\_3 = \sigma = -Q\_{23}F\_{s\_2}f\_s. \tag{73}
$$

Now, the following sensor reconstruction signal is defined:

$$f\_s = \mathcal{W}\_s \sigma$$

where *Ws* <sup>=</sup> (−*Q*23*F*¯ *s*2 ) −1 . Then, one obtains

$$\left\|\left\|\mathbb{2}\_{3}-\mathbb{2}\_{3}\right\|\right\|\leq\sigma.\tag{75}$$

This implies ˆ *fs* ≈ *fs* and completes the proof.

**Remark 1.** *Regarding the design parameters tuning, it is worthwhile to mention some points. The parameter ε in (34) should be initially selected as a very small scalar and then gradually increased such that it approximates the output error injection v and fulfills the design requirements. The parameter ξ in (41) is a small constant that satisfies the LMI conditions (43).*

**Remark 2.** *In practice, both sensor and actuator faults may occur simultaneously. Therefore, unlike most existing approaches dealing with sensor and actuator faults separately, our proposed approach takes care of simultaneous sensor and actuator faults, which can be a critical issue in some systems such as aircraft, wind turbines, etc., which need some more technical cares to have much better performance and efficiency.*

The step-by-step procedure of applying the proposed design algorithm is summarized as follows:

