**6.1. Feedback control for the model with hydrodynamic damping**

Considering the model (17) with the inclusion of control: *uF* .

$$\begin{aligned} \dot{\mathbf{x}}\_1 &= \mathbf{x}\_2\\ \dot{\mathbf{x}}\_2 &= -r\mathbf{x}\_2 - b\mathbf{x}\_1 - c\mathbf{x}\_1^3 - \frac{d}{\left(a + \mathbf{x}\_1\right)^2} + \frac{e}{\left(a + \mathbf{x}\_1\right)^8} + g\cos\Omega\mathbf{x}\_3 - \frac{p}{\left(a + \mathbf{x}\_1\right)^3}\mathbf{x}\_2 + F\_u \end{aligned} \tag{18}$$

and defining a periodic orbit as a function of *x t* ( ) . The desired regime is given by:

$$\ddot{\tilde{y}} = -r\dot{\tilde{y}} - b\tilde{y} - c\tilde{y}^3 - \frac{d}{\left(a + \tilde{y}\right)^2} + \frac{e}{\left(a + \tilde{y}\right)^8} + g\cos\Omega\,\tau - \frac{p}{\left(a + \tilde{y}\right)^3}\dot{\tilde{y}} + \tilde{u} \tag{19}$$

Since *u* control the system in the desired trajectory, and *x t* ( ) is a solution of (19), without the term control *uF* , then 0 *u* , resulting:

$$\tilde{y} = a\_0 + a\_1 \cos(\Omega \tau) + b\_1 \text{sen}(\Omega \tau) + a\_2 \cos(\Omega \Omega \tau) + b\_2 \text{sen}(\Omega \Omega \tau) + \dots \tag{20}$$

The feedforward control *u* is given by:

$$\tilde{u} = \ddot{\tilde{y}} + r\dot{\tilde{y}} + b\tilde{y} + c\tilde{y}^3 + \frac{d}{\left(a + \tilde{y}\right)^2} - \frac{e}{\left(a + \tilde{y}\right)^8} - g\cos\Omega\,\tau + \frac{p}{\left(a + \tilde{y}\right)^3}\dot{\tilde{y}}\tag{21}$$

Replacing (19) in (18) and defining the deviation from the desired trajectory as:

$$e = \begin{bmatrix} \boldsymbol{\chi}\_1 - \tilde{\boldsymbol{\chi}}\_1 \\ \boldsymbol{\chi}\_2 - \tilde{\boldsymbol{\chi}}\_2 \end{bmatrix} \tag{22}$$

On an Overview of Nonlinear and Chaotic Behavior and

, *R* 1 , and using the matlab(R)

Their Controls of an Atomic Force Microscopy (AFM) Vibrating Problem 57

11 1 12 <sup>1</sup> <sup>2</sup> *u ke ke e e* 3.1127 3.9293 (27)

controlled uncontrolled

, 0.17602 *b* , 2.6364

sin

(28)

sin *<sup>b</sup>*

 

 

*b*

1

*z x*

1

*y* ( ) *x x* (30)

, 2.5 *z* , 1*a* 1 ,

(29)

0 1 0.05 0.1 *<sup>A</sup>* 

shown.

, <sup>0</sup> 1 *<sup>B</sup>* , and defining 10 0

**Figure 12.** Displacement of the tip to the system without control and with control

3 2 11 21 2

where: *U uu* , *u* is the feedback control, *u* is the feedforward control, given by :

Replacing (29) in (28) and defining the deviation from the desired trajectory as:

where *x* is the desired orbit, and rewriting the system in deviations, results:

*x ax ax b U z x*

 

<sup>0</sup> <sup>10</sup> <sup>20</sup> <sup>30</sup> <sup>40</sup> <sup>50</sup> <sup>60</sup> <sup>70</sup> <sup>80</sup> <sup>90</sup> <sup>100</sup> -1.5

T

3 2 11 21 2

*u x ax ax b*

**6.2. Feedback control for a model with cubic spring** 

1 2

*x x*

Considering the following parameters: 0.14668



0

x1

0.5

1

1.5

<sup>2</sup>*a* 14.5 , and the control *U* in (16) results:

0 10 *<sup>Q</sup>*

the control *u* is obtained. In Figure 12 the tip displacement without and with control are

The system (18) can be represented as follows:

$$\begin{aligned} \dot{e}\_1 &= e\_2\\ \dot{e}\_2 &= -re\_2 - be\_1 - c\left(e\_1 + \tilde{\mathbf{x}}\_1\right)^3 + c\tilde{\mathbf{x}}\_1^{-3} - \frac{d}{\left(a + e\_1 + \tilde{\mathbf{x}}\_1\right)^2} + \frac{d}{\left(a + \tilde{\mathbf{x}}\_1\right)^2} + \frac{e}{\left(a + e\_1 + \tilde{\mathbf{x}}\_1\right)^8} - \frac{e}{\left(a + \tilde{\mathbf{x}}\_1\right)^8} - \frac{e}{\left(a + \tilde{\mathbf{x}}\_1\right)^8} \\ &- \frac{p\left(e\_2 + \tilde{\mathbf{x}}\_2\right)}{\left(a + e\_1 + \tilde{\mathbf{x}}\_1\right)^3} + \frac{p\tilde{\mathbf{x}}\_2}{\left(a + \tilde{\mathbf{x}}\_1\right)^3} + u \end{aligned} \tag{23}$$

with *<sup>u</sup> uF u* , and the feedback control: *u Ke* . The system (23) can be represented in deviations as:

$$
\dot{e} = Ae + g(\mathbf{x}) - g(\tilde{\mathbf{x}}) + Bu \tag{24}
$$

Representing the system (24) in the form:

$$
\begin{bmatrix}
\dot{e}\_1\\ \dot{e}\_2
\end{bmatrix} = \begin{bmatrix}
0 & 1\\ -b & -r
\end{bmatrix} \begin{bmatrix}
e\_1\\ e\_2
\end{bmatrix} + g(\mathbf{x}) - g(\tilde{\mathbf{x}}) + \begin{bmatrix}
0\\ 1
\end{bmatrix} \mu \tag{25}
$$

where

$$g(\mathbf{x}) - g(\tilde{\mathbf{x}}) = \begin{bmatrix} 0 \\ -c \left[ \left( e\_1 + \tilde{\mathbf{x}}\_1 \right)^3 - \tilde{\mathbf{x}}\_1^{\; 3} \right] - \frac{d}{\left( a + e\_1 + \tilde{\mathbf{x}}\_1 \right)^2} + \frac{d}{\left( a + \tilde{\mathbf{x}}\_1 \right)^2} \\\\ \begin{bmatrix} 0 \\ \frac{e}{\left( a + e\_1 + \tilde{\mathbf{x}}\_1 \right)^8} - \frac{e}{\left( a + \tilde{\mathbf{x}}\_1 \right)^8} - \frac{p\left( e\_2 + \tilde{\mathbf{x}}\_2 \right)}{\left( a + e\_1 + \tilde{\mathbf{x}}\_1 \right)^3} + \frac{p\tilde{\mathbf{x}}\_2}{\left( a + \tilde{\mathbf{x}}\_1 \right)^3} \end{bmatrix}$$

Defining the desired trajectory as the periodic orbit, with amplitude less than (a) and frequency equal to ( ), then:

$$
\tilde{\mathfrak{X}} = 1.3 \sin(t) \tag{26}
$$

Considering the parameters values: 1; r 0.1 ; b 0.05 ; c 0.35 ; d 4/27 ; e 0.0001 ; g 0.2 ; p 0.005 e a 1.6 the matrices A and B, are given by:

$$A = \begin{bmatrix} 0 & 1 \\ -0.05 & -0.1 \end{bmatrix}, B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \text{ and defining } \ Q = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}, \text{ } R = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \text{ and using the } \text{mathbb}(R) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

the control *u* is obtained. In Figure 12 the tip displacement without and with control are shown.

$$
\mu = -k\_{11}e\_1 - k\_{12}e = -3.1127e\_1 - 3.9293e\_2 \tag{27}
$$

**Figure 12.** Displacement of the tip to the system without control and with control
