**5.1 Problem statement**

Consider the following non-linear system class of *n*-th order:

$$\mathbf{x}^{(n)} = f(\mathbf{x}) + b(\mathbf{x})u,\tag{33}$$

where *u* is the control input, *x*ð Þ *<sup>n</sup>* , is the *n*-th order derivative of the interested scalar output variable *<sup>x</sup>* with respect to time *<sup>t</sup>*∈½ Þ 0, <sup>∞</sup> . Here, also, *<sup>x</sup>* <sup>¼</sup> *<sup>x</sup>*, *<sup>x</sup>*\_, … , *<sup>x</sup>*ð Þ *<sup>n</sup>*�<sup>1</sup> � �*<sup>T</sup>* represents the system state vector, and both *<sup>f</sup>*(*x*) and *<sup>b</sup>*(*x*), such that *<sup>f</sup>*, *<sup>b</sup>* : *<sup>R</sup><sup>n</sup>* ! *<sup>R</sup>*, denote nonlinear functions.

The following assumptions will be made in terms of the dynamic system presented in (33).

**Assumption 1.** *f is an unknown function such that it is bounded by a known function <sup>x</sup>, i.e.,* ^*f*ð Þ� *<sup>x</sup> <sup>f</sup>*ð Þ *<sup>x</sup>* � � � � �≤*F*ð Þ *<sup>x</sup> , where* ^*f is an estimate of f .*

� **Assumption 2.** *Input gain b*ð Þ *x is an unknown function such that it is positive and bounded, i.e.,* 0<*bmin* ≤*b*ð Þ *x* ≤ *bmax.*

In the proposed state space control problem, the *x* state vector must be able to follow a desired trajectory *<sup>x</sup><sup>d</sup>* <sup>¼</sup> *xd*, *<sup>x</sup>*€*d*, … , *<sup>x</sup>*ð Þ *<sup>n</sup>*�<sup>1</sup> *<sup>n</sup>* h i, even under the presence of parametric uncertainties and unmodulated dynamics.

The following assumptions should also be made during the development of the control law.

**Assumption 3.** *The state vector x has availability.*

**Assumption 4.** *The desired trajectory x<sup>d</sup> is differentiated once in time. Moreover, each element of the vector x<sup>d</sup> as well as x*ð Þ *<sup>n</sup> <sup>d</sup> , is available and has known bounds.*

Now, let *x*~ ¼ *x* � *x<sup>d</sup>* be defined as the tracking error for the variable *x*, and *<sup>x</sup>*<sup>~</sup> <sup>¼</sup> *<sup>x</sup>* � *<sup>x</sup><sup>d</sup>* <sup>¼</sup> *<sup>x</sup>*~, *<sup>x</sup>*~\_, … , *<sup>x</sup>*~ð Þ *<sup>n</sup>*�<sup>1</sup> h i as the tracking error vector.

Let us define a sliding surface *S* in the state space by the equation *s*ð Þ¼ *x*~ 0 in which *s* is the function mapping from *n*-dimensional real space *R<sup>n</sup>* to onedimensional real space *<sup>R</sup>*, i.e., *<sup>s</sup>*: *<sup>R</sup><sup>n</sup>* ! *<sup>R</sup>*, and satisfying the following equation:

$$s(\tilde{\mathfrak{x}}) = \left(\frac{d}{dt} + \lambda\right)^{n-1} \tilde{\mathfrak{x}},$$

which can be plainly rewritten as

$$\mathbf{s}(\tilde{\mathfrak{x}}) = \mathbf{c}^T \tilde{\mathfrak{x}},\tag{34}$$

where *<sup>c</sup>* <sup>¼</sup> *cn*�<sup>1</sup>*λ<sup>n</sup>*�<sup>1</sup> <sup>þ</sup> … <sup>þ</sup> *<sup>c</sup>*1*λ*,*c*<sup>0</sup> � � with *ci* representing binomial coefficients as follows:

$$c\_i = \binom{n-1}{i} = \frac{(n-1)!}{(n-i-1)!i!}, i = 0, 1, \dots, n-1\tag{35}$$

which makes *cn*�<sup>1</sup>*λ<sup>n</sup>*�<sup>1</sup> <sup>þ</sup> … <sup>þ</sup> *<sup>c</sup>*1*λ*,*c*<sup>0</sup> a Hurwitz polynomial.

It can be easily verified from (35) that *c*<sup>0</sup> ¼ 1, for ∀*n*≥1. Therefore, the time derivative of *s* will be expressed in the following form:

$$\dot{\boldsymbol{\alpha}} = \boldsymbol{\sigma}^{T}\dot{\boldsymbol{\tilde{x}}} = \underbrace{\begin{bmatrix} c\_{n-1}\boldsymbol{\lambda}^{n-1} \\ \vdots \\ c\_{1}\boldsymbol{\lambda} \\ c\_{0} \end{bmatrix}}\_{\boldsymbol{\epsilon}^{T}} \underbrace{\begin{bmatrix} \ddot{\boldsymbol{x}}, \ddot{\ddot{\boldsymbol{x}}}, \dots, \ddot{\boldsymbol{x}}^{(n)} \end{bmatrix}}\_{\dot{\boldsymbol{\tilde{x}}}} = \underbrace{\begin{bmatrix} \mathbf{0} \\ c\_{n-1}\boldsymbol{\lambda}^{n-1} \\ \vdots \\ c\_{2}\boldsymbol{\lambda}^{2} \\ c\_{1}\boldsymbol{\lambda} \end{bmatrix}}\_{\boldsymbol{\tilde{\Delta}} \in \boldsymbol{\tilde{\epsilon}^{T}}} \underbrace{\begin{bmatrix} \ddot{\boldsymbol{x}}, \dot{\ddot{\boldsymbol{x}}}, \dots, \ddot{\boldsymbol{x}}^{(n-1)} \end{bmatrix}}\_{\dot{\boldsymbol{\tilde{\Delta}}}} + \ddot{\boldsymbol{x}}^{(n)} \underbrace{\begin{bmatrix} \ddot{\boldsymbol{\tilde{\varepsilon}}}, \ddot{\boldsymbol{\tilde{\varepsilon}}}, \dots, \ddot{\boldsymbol{\tilde{\varepsilon}}}^{(n-1)} \end{bmatrix}}\_{\boldsymbol{\tilde{\Delta}} \in \boldsymbol{\tilde{\varepsilon}^{T}}}$$

i.e.,

$$\dot{\boldsymbol{s}} = \mathbf{c}^{T}\dot{\tilde{\mathbf{x}}} = \tilde{\mathbf{x}}^{(n)} + \bar{\mathbf{c}}^{T}\tilde{\mathbf{x}} \tag{36}$$

where, here, as used for the first time above, there is a definition in the form of �*<sup>c</sup>* <sup>¼</sup> 0,*cn*�1*λn*�<sup>1</sup> , ⋯,*c*1*λ* � �. At this point, let us evaluate Eqs. (34) and (36) for *n* ¼ 3, i.e.,

$$s(\ddot{\mathfrak{x}}) = \left(\frac{d}{dt} + \lambda\right)^2 \ddot{\mathfrak{x}} = \ddot{\ddot{\mathfrak{x}}} + 2\lambda \dot{\breve{\mathfrak{x}}} + \lambda^2 \ddot{\mathfrak{x}},$$

from which *<sup>c</sup>* appears to be as *<sup>c</sup>* <sup>¼</sup> *<sup>c</sup>*2*λ*<sup>2</sup> ,*c*1*λ*,*c*<sup>0</sup> � �. Then, *<sup>s</sup>*ð Þ¼ *<sup>x</sup>*<sup>~</sup> *<sup>c</sup><sup>T</sup>x*<sup>~</sup> or to create a polynomial, *cx*~*<sup>T</sup>* <sup>¼</sup> *<sup>c</sup>*2*λ*<sup>2</sup> ,*c*1*λ*,*c*<sup>0</sup> � � *x*~ *x*~\_ *x*€~ 2 6 4 3 7 <sup>5</sup> <sup>¼</sup> *<sup>c</sup>*2*λ*<sup>2</sup> *<sup>x</sup>*<sup>~</sup> <sup>þ</sup> *<sup>c</sup>*1*λx*~\_ <sup>þ</sup> *<sup>c</sup>*0*x*€~ with *<sup>c</sup>*<sup>0</sup> <sup>¼</sup> 1 as always,

and *<sup>c</sup>*2*λ*<sup>2</sup> <sup>þ</sup> *<sup>c</sup>*1*<sup>λ</sup>* <sup>þ</sup> *<sup>c</sup>*<sup>0</sup> is a Hurwitz polynomial. That is, it is defined as the polynomial with its coefficients (i.e., *ci*) that are positive real numbers, and its zeros are located in the left half-plane –i.e., the real part of every zero is negative– of the complex plane.

Now, Let the problem of controlling the uncertain nonlinear system expressed by (33) be handled for review through the classical sliding mode approach that defines a control rule consisted of an equivalent control *<sup>u</sup>*^ <sup>¼</sup> ^ *b* �1 �^*<sup>f</sup>* <sup>þ</sup> *<sup>x</sup>*ð Þ *<sup>n</sup> <sup>d</sup>* � �*c<sup>T</sup>x*<sup>~</sup> � � and a discontinuous term �*Ksgn s*ð Þ as follows:

$$
\mu = \hat{b}^{-1} \left( -\hat{f} + \varkappa\_d^{(n)} - \bar{\mathfrak{a}}^T \tilde{\mathfrak{x}} \right) - \text{Ksgn}(s). \tag{37}
$$

where ^ *<sup>b</sup>* <sup>¼</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi *bmaxbmin* <sup>p</sup> represents the estimated value of *<sup>b</sup>*, and *<sup>K</sup>* represents a positive gain. Furthermore, the sign or signum function represented by *sgn s*ð Þ above

$$\text{sgn}\,(s) = \begin{cases} -1, \,\,\sharp f\,s < 0\\ 0, \,\,\sharp f\,s = 0.\\ 1, \,\,\sharp f\,s > 0 \end{cases}$$

Based on Assumptions 1 and 2 given above and taking into account the fact that *β*�<sup>1</sup> ≤ ^ *<sup>b</sup>=b*<sup>≤</sup> *<sup>β</sup>*, where *<sup>β</sup>* <sup>¼</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi *bmax=bmin* p , the gain *K* must be determined in such a way as to ensure the following inequality:

$$K \ge \beta \hat{b}^{-1}(\eta + F) + (\beta - \mathbf{1}) \left| \hat{b}^{-1} \left( -\hat{f} + \boldsymbol{x}\_d^{(\eta)} - \bar{\mathbf{c}}^T \tilde{\mathbf{x}} \right) \right|, \tag{38}$$

where *η* is a strictly positive constant of the reaching time. Now, in this step, let us reaffirm the validity of the lower and upper bounds of *b* using the ^ *b* and *β*

*An In-Depth Analysis of Sliding Mode Control and Its Application to Robotics DOI: http://dx.doi.org/10.5772/intechopen.93027*

definitions given: first of all, Let the definitions of ^ *b* and *β* be placed in the expression *β*�<sup>1</sup> ≤ ^ *b=b*≤*β* given above. In this case,

$$\frac{1}{\sqrt{b\_{\max}/b\_{\min}}} \le \frac{\sqrt{b\_{\max}b\_{\min}}}{b} \le \sqrt{b\_{\max}/b\_{\min}}.$$

If each side is multiplied by 1*=* ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi *bmaxbmin* <sup>p</sup> , the following inequality is obtained:

$$\frac{1}{b\_{\max}} \le \frac{1}{b} \le \frac{1}{b\_{\min}}.$$

In this inequality, if inversion is applied to all terms, inequalities will be completely displaced, that is to say, it will become *bmax* ≥*b*≥*bmin*. This is a necessary initial acceptance. Therefore, when we turn *b*'s upper and lower bounds into inequality, we once again confirm the correctness of the definitions for ^ *b* and *β*. Since the control rule will be designed to be robust against the inequality *β*�<sup>1</sup> ≤ ^ *b=b*≤ *β*, that is, a bounded multiplicative uncertainty, taking advantage of the similarity to the terminology used in linear control, we can call *β* the gain margin of the design.

In order to ensure *x* � *xd* system tracking, we define a sliding surface *s* ¼ 0 according to *<sup>s</sup>*ð Þ¼ *<sup>x</sup>*; *<sup>t</sup> <sup>d</sup> dt* <sup>þ</sup> *<sup>λ</sup>* � �*<sup>n</sup>*�<sup>1</sup> *x*~, that is,

$$
\mathfrak{s} = \left(\frac{d}{dt} + \lambda\right)\breve{\mathfrak{x}} = \dot{\breve{\mathfrak{x}}} + \lambda\breve{\mathfrak{x}}.
$$

When we derive the expression *s*, we obtain the following:

$$
\dot{\mathfrak{x}} = \ddot{\tilde{\mathbf{x}}} + \lambda \dot{\tilde{\mathbf{x}}} = \ddot{\mathbf{x}} - \ddot{\mathbf{x}}\_d + \lambda \dot{\tilde{\mathbf{x}}} = f + u - \ddot{\mathbf{x}}\_d + \lambda \dot{\tilde{\mathbf{x}}}.
$$

For *<sup>s</sup>*\_ <sup>¼</sup> *<sup>f</sup>* <sup>þ</sup> *<sup>u</sup>* � *<sup>x</sup>*€*<sup>d</sup>* <sup>þ</sup> *<sup>λ</sup>x*~\_ <sup>¼</sup> 0 to be realized, other terms outside of *<sup>u</sup>* must be determined equal to the opposite sign of *u*^, which is the best approximation of a continuous control rule *u* that can implement *s*\_ ¼ 0, that is,

$$
\hat{u} = -\hat{f} + \ddot{\mathbf{x}}\_d - \lambda \dot{\vec{\mathbf{x}}}.
$$

In fact, to see this result, the first thing to do is to draw *u* from the equation *<sup>s</sup>*\_ <sup>¼</sup> *<sup>f</sup>* <sup>þ</sup> *<sup>u</sup>* � *<sup>x</sup>*€*<sup>d</sup>* <sup>þ</sup> *<sup>λ</sup>x*~\_ <sup>¼</sup> 0. Hence, *<sup>u</sup>* ¼ �*<sup>f</sup>* <sup>þ</sup> *<sup>x</sup>*€*<sup>d</sup>* � *<sup>λ</sup>x*~\_ is obtained. Then, from here, in order to obtain the approximate value of *u*, searching for the approximation of the function on the right side of the equation, and representing this approximated function by ^*f* symbolically are sufficient to lead us to the correct result, as seen above.

The control rule *<sup>u</sup>* <sup>¼</sup> ^ *b* �1 ½ � *u*^ � *ksgn s*ð Þ with predefined *s* and *u*^, and *k* defined by the inequality *k*≥*β*ð Þþ *F* þ *η* ð Þ *β* � 1 j j *u*^ —as will be explained little below—meets the sliding condition. Indeed, when we substitute this control rule in the expression *s*\_ ¼ *<sup>f</sup>* <sup>þ</sup> *bu* � *<sup>x</sup>*€*<sup>d</sup>* <sup>þ</sup> *<sup>λ</sup>x*~\_ by choosing the use of *<sup>x</sup>*€ <sup>¼</sup> *<sup>f</sup>* <sup>þ</sup> *bu*, which is more specific to this type of structure, instead of *x*€ ¼ *f* þ *u* used only just two above, we obtain the following,

$$\dot{s} = f + b\dot{b}^{-1}[\dot{u} - k \text{sgn}(s)] - \ddot{\varkappa}\_d + \dot{\varkappa}\dot{\bar{\varkappa}}.$$

Once the previously determined *u*^ is replaced above, the following equation is reached:

$$\dot{\varkappa} = f + b\dot{b}^{-1} \left[ -\hat{f} + \ddot{\varkappa}\_d - \lambda \dot{\hat{\varkappa}} - k \text{sgn}(s) \right] - \ddot{\varkappa}\_d + \lambda \dot{\hat{\varkappa}}.$$

$$\dot{s} = \left(\boldsymbol{f} - b\boldsymbol{b}\boldsymbol{\hat{b}}^{-1}\boldsymbol{\hat{f}}\right) + \left(\mathbf{1} - b\boldsymbol{\hat{b}}\boldsymbol{\hat{b}}^{-1}\right)\left(-\ddot{\boldsymbol{x}}\_d + \lambda\dot{\hat{\mathbf{x}}}\right) - b\boldsymbol{\hat{b}}\boldsymbol{\hat{b}}^{-1}\boldsymbol{k}\text{sgn}(\boldsymbol{s})\dots$$

$$k \ge \left| \hat{b}b^{-1}f - \hat{f} + \left( \hat{b}b^{-1} - \mathbf{1} \right) \left( -\ddot{\boldsymbol{x}}\_d + \lambda \dot{\hat{\mathbf{x}}} \right) \right| + \eta \hat{b}b^{-1} \tag{39}$$

$$\dot{s} = \mathbf{0} = \left(\boldsymbol{f} - b\boldsymbol{b}^{-1}\hat{\boldsymbol{f}}\right) + \left(\mathbf{1} - b\boldsymbol{b}^{-1}\right)\left(-\ddot{\boldsymbol{x}}\_d + \lambda\dot{\dot{\boldsymbol{x}}}\right) - b\boldsymbol{b}^{-1}\mathbf{k}\text{sgn}(s) \Longrightarrow$$

$$b\dot{\boldsymbol{b}}^{-1}\text{ksp}(s) = \left(\boldsymbol{f} - b\dot{\boldsymbol{b}}^{-1}\hat{\boldsymbol{f}}\right) + \left(\mathbf{1} - b\dot{\boldsymbol{b}}^{-1}\right)\left(-\ddot{\boldsymbol{x}}\_d + \lambda\dot{\dot{\boldsymbol{x}}}\right) \Longrightarrow$$

$$k\text{sgn}(s) = \hat{\boldsymbol{b}}\,\boldsymbol{b}^{-1}\boldsymbol{f} - \hat{\boldsymbol{f}} + \left(\hat{\boldsymbol{b}}\,\boldsymbol{b}^{-1} - \mathbf{1}\right)\left(-\ddot{\boldsymbol{x}}\_d + \lambda\dot{\dot{\boldsymbol{x}}}\right) \Longrightarrow$$

$$k\text{sgn}(s) = \hat{\boldsymbol{b}}\,\boldsymbol{b}^{-1}\left[\hat{\boldsymbol{f}} + \left(\boldsymbol{f} - \hat{\boldsymbol{f}}\right)\right] - \hat{\boldsymbol{f}} + \left(\hat{\boldsymbol{b}}\,\boldsymbol{b}^{-1} - \mathbf{1}\right)\left(-\ddot{\boldsymbol{x}}\_d + \lambda\dot{\dot{\boldsymbol{x}}}\right).$$

$$k \ge \hat{b}b^{-1}F + \eta \hat{b}b^{-1} + \left| \hat{b}b^{-1} - \mathbf{1} \right| \left| \hat{f} - \ddot{\varkappa}\_d + \lambda \dot{\breve{\mathbf{x}}} \right|.$$

$$k \ge \beta (F + \eta) + |\beta - \mathbf{1}||\hat{u}|.\tag{40}$$

$$\frac{1}{2}\frac{d}{dt}s^2 \le -\eta|s|\_\ast$$

*An In-Depth Analysis of Sliding Mode Control and Its Application to Robotics DOI: http://dx.doi.org/10.5772/intechopen.93027*

which indeed guarantees the convergence of the tracking error vector to the sliding surface *S* and consequently its exponential stability in a finite-time. In response to the uncertainty of *f* on dynamics, we add a discontinuous term to *u*^ across the surface *s* ¼ 0 to meet the slip condition given above, that is:

$$u = \hat{u} - k \text{sgn}(s).$$

Here we can now guarantee that the sliding condition will be verified by choosing *k* ¼ *k x*ð Þ , *x*\_ sufficiently large. Indeed,

$$\frac{1}{2}\frac{d}{dt}\mathbf{s}^2 = \dot{\mathbf{s}}\mathbf{s} = \left[f + \hat{\boldsymbol{\mu}} - \text{kg}\mathbf{n}(\boldsymbol{s}) - \ddot{\boldsymbol{\kappa}}\_d + \lambda\dot{\dot{\mathbf{c}}}\right]\mathbf{s} = \left[f - \hat{f} - \text{kg}\mathbf{n}(\boldsymbol{s})\right]\mathbf{s}.$$

This last operation is important; because we have reached this point by using the equations *<sup>u</sup>* ¼ �*<sup>f</sup>* <sup>þ</sup> *<sup>x</sup>*€*<sup>d</sup>* � *<sup>λ</sup>x*~\_, *<sup>u</sup>*^ ¼ �^*<sup>f</sup>* <sup>þ</sup> *<sup>x</sup>*€*<sup>d</sup>* � *<sup>λ</sup>x*~\_, and *<sup>u</sup>* <sup>¼</sup> *<sup>u</sup>*^ � *ksgn s*ð Þ as follows:

$$\begin{split} \dot{s} &= \left[ f + \mathfrak{u} - \ddot{\mathfrak{x}}\_d + \dot{\lambda}\dot{\breve{\mathfrak{x}}} \right] = f + \hat{\mathfrak{u}} - k \text{sgn}(\mathfrak{s}) - \ddot{\mathfrak{x}}\_d + \dot{\lambda}\dot{\breve{\mathfrak{x}}} \\ &= f + \left( -\hat{f} + \ddot{\mathfrak{x}}\_d - \dot{\lambda}\dot{\breve{\mathfrak{x}}} \right) - k \text{sgn}(\mathfrak{s}) - \ddot{\mathfrak{x}}\_d + \dot{\lambda}\dot{\breve{\mathfrak{x}}} = f - \hat{f} - k \text{sgn}(\mathfrak{s}). \end{split}$$

If we continue where we were,

$$\frac{1}{2}\frac{d}{dt}\mathbf{s}^2 = \left[f-\hat{f}-k\text{sgn}(\mathbf{s})\right]\mathbf{s} = \left(f-\hat{f}\right)\mathbf{s} - k\text{sgn}(\mathbf{s})\mathbf{s}.$$

Therefore, since *sgn s*ð Þ*s* ¼ j j*s* , the following expression is reached,

$$\frac{1}{2}\frac{d}{dt}\mathbf{s}^2 = \dot{\mathbf{s}}\mathbf{s} = \left[f - \hat{f} - k\text{sgn}(\mathbf{s})\right]\mathbf{s} = \left(f - \hat{f}\right)\mathbf{s} - k|\mathbf{s}|\mathbf{s}$$

So that, when *k* ¼ *F* þ *η* is selected, the above statement follows,

$$\frac{1}{2}\frac{d}{dt}\mathbf{s}^2 = \dot{\mathbf{s}}\mathbf{s} = \left(f - \hat{f}\right)\mathbf{s} - F|\mathbf{s}| - \eta|\mathbf{s}|.$$

However, although the definition of ^*<sup>f</sup>* � *<sup>f</sup>* � � � � � �≤*<sup>F</sup>* is given at the beginning of the section, recalling that we prefer the form *<sup>f</sup>* � ^*<sup>f</sup>* � � � � � �<sup>≤</sup> *<sup>F</sup>* to be used in the case study below, Let us give the statement its final form:

$$\frac{1}{2}\frac{d}{dt}\mathbf{s}^2 = \dot{\mathbf{s}}\mathbf{s} = \left(f - \hat{f}\right)\mathbf{s} - \left|f - \hat{f}\right|\left|\mathbf{s}\right| - \eta|\mathbf{s}|.\tag{41}$$

In fact, note that here the expression *<sup>f</sup>* � ^*<sup>f</sup>* � � � � � �, which we substitute for *<sup>F</sup>* represents the smallest value that *F* can take. We generally know that *F* is greater than this value.

We will now carry out the following case studies for the Eq. (41): **Case 1**. *If f* � ^*<sup>f</sup>* � � *and s are both negative or positive, as such, f* � ^*<sup>f</sup>* � �*<sup>s</sup>* � *<sup>f</sup>* � ^*<sup>f</sup>* � � � � � �j j*<sup>s</sup>* <sup>¼</sup> <sup>0</sup>*. However, it is known to be F* <sup>≥</sup> *<sup>f</sup>* � ^*<sup>f</sup>* � � � � � �*, hence, f* � ^*<sup>f</sup>* � �*<sup>s</sup>* � *F s*j j≤0*, i.e., it will always be negative.*

**Case 2**. *However, if f* � ^*<sup>f</sup>* � � *and s opposite signs; f* � ^*<sup>f</sup>* � �*s will always be negative.* �*F s*j j *will also be negative. Hence, f* � ^*<sup>f</sup>* � �*<sup>s</sup>* � *F s*j j *will always be more negative as compared to Case 1.*

*As a result,*

$$\frac{1}{2}\frac{d}{dt}s^2 \le -\eta|s|\_\ast$$

### *is always true.*

However, the presence of a discontinuous term (i.e., �*Ksgn s*ð Þ) in the control rule leads to the well-known *chattering* effect. To prevent these unwanted high-frequency oscillations of the controlled variable, Slotine had proposed the idea of adopting a thin boundary layer *S*<sup>∅</sup> around the switching surface [1]:

$$\mathcal{S}\_{\mathcal{Q}} = \{ \tilde{\mathfrak{x}} \in \mathbf{R}^n \, | \, |s(\tilde{\mathfrak{x}})| \le \mathcal{Q} \}. \tag{42}$$

Here ∅ is an absolute positive constant, which represents the boundary layer thickness.

The boundary layer is accomplished by replacing the sign function with a continuous interpolation in *S*∅. It should be emphasized that this smooth approximation referring to the flatness or smoothness of the interpolating curve and its derivatives, which will be called *φ*ð Þ *s*, ∅ , here, will definitely act as a sign function outside the boundary layer.

Various options are available to smooth out the ideal switch. But the closest choices are the saturation function expressed by

$$sat\left(\frac{s}{\mathcal{Q}}\right) = \begin{cases} \text{sgn}\left(s\right), if\left|\frac{s}{\mathcal{Q}}\right| \ge 1\\ \begin{array}{l} s \\ \frac{s}{\mathcal{Q}}, if\left|\frac{s}{\mathcal{Q}}\right| < 1, \end{array} \end{cases} \tag{43}$$

and the hyperbolic tangent function expressed by *tanh <sup>s</sup>* ∅ � �. Thus, the smooth sliding mode control rule can be expressed as follows:

$$
\mu = \hat{b}^{-1} \left( -\hat{f} + \varkappa\_d^{(n)} - \bar{\mathbf{c}}^T \tilde{\mathbf{x}} \right) - K\rho(\mathbf{s}, \mathcal{Q}).\tag{44}
$$

#### **5.2 Convergence analysis**

The attractiveness and invariance properties of the boundary layer are introduced in the following theorem:

**Theorem 1.** Consider four previously made assumptions with the uncertain nonlinear system given in (33). Therefore, the smooth sliding mode controller defined by (38) and (44) provides the finite-time convergence of the tracking error vector to the boundary layer *S*<sup>∅</sup> defined by (42).

**Proof.** Let a positive-definite Lyapunov function candidate *V* be defined as,

$$V(t) = \frac{1}{2} s\_{\mathcal{Q}}^{-2}.\tag{45}$$

*An In-Depth Analysis of Sliding Mode Control and Its Application to Robotics DOI: http://dx.doi.org/10.5772/intechopen.93027*

Here, as a measure of the distance of the current error to the boundary layer, *s*<sup>∅</sup> can be computed as follows:

$$s\_{\mathcal{Q}} = s - \mathfrak{Q}sat\left(\frac{s}{\mathcal{Q}}\right). \tag{46}$$

Noting that *<sup>s</sup>*<sup>∅</sup> <sup>¼</sup> 0 in the boundary layer, it is shown that *V t* \_ ðÞ¼ 0 inside *<sup>S</sup>*∅. It is also possible to easily verify that *s*\_<sup>∅</sup> ¼ *s*\_ outside the boundary layer through (43) and (46), and in this case, *V*\_ can be written as follows:

$$\begin{split} \dot{\tilde{V}}(t) &= \mathbf{s}\_{\mathcal{Q}} \dot{\mathbf{s}}\_{\mathcal{Q}} = \mathbf{s}\_{\mathcal{Q}} \dot{\mathbf{s}} = \mathbf{s}\_{\mathcal{Q}} \Big( \mathbf{c}^{T} \dot{\tilde{\mathbf{x}}} \Big) = \mathbf{s}\_{\mathcal{Q}} \Big( \tilde{\mathbf{x}}^{(n)} + \bar{\mathbf{c}}^{T} \ddot{\tilde{\mathbf{x}}} \Big) = \left( \mathbf{x}^{(n)} - \mathbf{x}\_{d}^{(n)} + \bar{\mathbf{c}}^{T} \ddot{\tilde{\mathbf{x}}} \right) \mathbf{s}\_{\mathcal{Q}} \\ &= \left( f + bu - \mathbf{x}\_{d}^{(n)} + \bar{\mathbf{c}}^{T} \ddot{\tilde{\mathbf{x}}} \right) \mathbf{s}\_{\mathcal{Q}}. \end{split}$$

Next, considering that the control rule given by (44) is written as

$$u = \hat{b}^{-1} \left( -\hat{f} + \varkappa\_d^{(n)} - \bar{\mathbf{c}}^T \tilde{\mathbf{x}} \right) - \text{Ksgn}(s\_{\mathcal{Q}}), \tilde{\mathbf{x}}$$

outside the boundary layer and noting that *<sup>f</sup>* <sup>¼</sup> ^*<sup>f</sup>* � ^*<sup>f</sup>* � *<sup>f</sup>* � �, we get the following result:

$$\dot{V}(t) = -\left[\left(\hat{f} - f\right) - b\hat{b}^{-1}\left(-\hat{f} + \mathfrak{x}\_d^{(n)} - \bar{\mathfrak{c}}^T\tilde{\mathfrak{x}}\right) - \hat{f} + \mathfrak{x}\_d^{(n)} - \bar{\mathfrak{c}}^T\tilde{\mathfrak{x}} + b\text{Ksgn}(s\_\mathcal{O})\right]s\_\mathcal{O}\dots$$

Thus, by taking the Assumptions 1 and 2 into consideration, and defining *K* according to (38), *V*\_ can be written as follows,

$$
\dot{V}(t) \le -\eta |s\_{\mathcal{O}}| \,.
$$

Because the Lyapunov function candidate, which we initially defined with (45) as positive definite, essentially inspired by the inequality in the form of <sup>1</sup> 2 *d dt s* <sup>2</sup> <sup>≤</sup> � *η*j j*s* which we have always correctly demonstrated above, may well be represented by a similar structure to the form, <sup>1</sup> 2 *d dt s*<sup>∅</sup> <sup>2</sup> <sup>≤</sup> � *<sup>η</sup>*j j *<sup>s</sup>*<sup>∅</sup> *:* It will be seen from here that *V t* \_ ðÞ¼ *<sup>s</sup>*∅*s*\_<sup>∅</sup> <sup>≤</sup> � *<sup>η</sup>*j j *<sup>s</sup>*<sup>∅</sup> , as well. Hence, the inequality *V t* \_ ð Þ<sup>≤</sup> � *<sup>η</sup>*j j *<sup>s</sup>*<sup>∅</sup> will imply that *V t*ð Þ≤*V*ð Þ 0 and therefore *s*<sup>∅</sup> is bounded. Moreover, from the definitions of *s* and *s*<sup>∅</sup> expressed in (35) and (46), respectively, it can be verified that *x*~ is bounded. Therefore, Assumption 4 and (36) imply that *s*\_ is also bounded.

Finite-time convergence of the tracking error vector to the boundary layer can be shown remembering the expression,

$$\dot{V}(t) = \frac{1}{2}\frac{d}{dt}\mathbf{s}\_{\mathcal{Q}}\,^2 = \mathbf{s}\_{\mathcal{Q}}\dot{\mathbf{s}}\_{\mathcal{Q}} \le -\eta|\mathbf{s}\_{\mathcal{Q}}|.$$

Then, dividing both sides into j j *s*<sup>∅</sup> above and integrating them between 0 and *t* will refer to the following result:

$$\int\_0^t \frac{s\_{\mathcal{Q}}}{|s\_{\mathcal{Q}}|} \dot{s}\_{\mathcal{Q}} d\tau \le -\int\_0^t \eta d\tau.$$

**Remark 3.** Here, considering the ratio *s*∅*=*j j *s*<sup>∅</sup> as the ratio of two numbers of the same size and therefore assuming it disappeared, that is, since it has no effect in

size, substantially it is a correct approach to consider the integral as an equivalent to Ð*t* <sup>0</sup>*s*\_∅*dτ*. This produces the result j jj *<sup>s</sup>*∅ð Þ*<sup>t</sup> <sup>t</sup>* <sup>0</sup>*:* Consequently, knowing the fact that in the situation before taking this approach, the product *s*∅*s*\_<sup>∅</sup> which appears in the numerator of the integral to the left of inequality is essentially equal to the derivative of the positive-definite *V* Lyapunov candidate function and is therefore positive again, it is essential to show the terms on the left side of the inequality with absolute value. That is to say, it is important to see that *<sup>s</sup>*∅*s*\_<sup>∅</sup> j j *<sup>s</sup>*<sup>∅</sup> <sup>&</sup>gt;0*:* Then, the next step to ensure this will turn into the form j j *s*∅ð Þ*t* � j j *s*∅ð Þ 0 ≤ � *ηt:* In this way, considering *treach* as the time required to reach *s*<sup>∅</sup> and noting that *s*<sup>∅</sup> *t* j j ð Þ *reach* ¼ 0, we have the expression,

$$t\_{reach} \le \frac{|s\_{\mathcal{Q}}(\mathbf{0})|}{\eta}$$

guaranteeing the convergence of the tracking error vector to the boundary layer in a time interval less than j j *s*∅ð Þ 0 *=η:*

**Remark 4.** If both sides of j j *s*∅ð Þ*t* � j j *s*∅ð Þ 0 ≤ � *ηt* are multiplied by �1, j j *s*∅ð Þ 0 � j j *s*∅ð Þ*t* ≥*ηt* is obtained, that is, briefly, the inequality is displaced. If *t* is left alone in the next step, *t*≤ j j� *<sup>s</sup>*∅ð Þ <sup>0</sup> j j *<sup>s</sup>*∅ð Þ*<sup>t</sup> <sup>η</sup>* is obtained. Hence, it is guaranteed to be *treach* ≤j j *s*∅ð Þ 0 *=η*. That is, the right-hand side will act as the largest value achievable for *treach*. In other words, it will appear as a guaranteed upper value. Then, the value of j j� *<sup>s</sup>*∅ð Þ <sup>0</sup> j j *<sup>s</sup>*∅ð Þ*<sup>t</sup> <sup>η</sup>* is expected to be less than this guaranteed value of j j *s*∅ð Þ 0 *=η:*

Therefore, to keep the reaching time, *treach*, as short as possible, the value of the positive constant *η* can be chosen appropriately. We clearly see from **Figure 7** that the time evolution of j j *s*<sup>∅</sup> is bounded by the linear equation j j *s*∅ð Þ*t* ¼ j j *s*∅ð Þ 0 � *ηt*.

Lastly, the proof of the boundedness of the tracking error vector is based on Theorem 2.

**Theorem 2.** Let the boundary layer *S*<sup>∅</sup> be defined according to (42). Then, once inside *S*∅, the tracking error vector will exponentially converge to an *n*-dimensional box defined according to *x*~ð Þ*<sup>i</sup>* � � � �≤*ζiλ<sup>i</sup>*�*n*þ<sup>1</sup> *ϕ*, *i* ¼ 0, 1, ⋯, *n* � 1, with *ζ<sup>i</sup>* satisfying

*An In-Depth Analysis of Sliding Mode Control and Its Application to Robotics DOI: http://dx.doi.org/10.5772/intechopen.93027*

$$\zeta\_i = \begin{cases} \mathbf{1}, for i = 0\\ \mathbf{1} + \sum\_{j=0}^{i-1} \binom{i}{j} \zeta\_j, for i = \mathbf{1}, 2, \dots, n-1. \end{cases} \tag{47}$$

**Proof**. Considering the fact that j j *s*ð Þ *x*~ ≤ ∅ can be rewritten as �∅ ≤ *s*ð Þ *x*~ ≤ ∅ with the definition of *s* given in (34), the expression below

$$\begin{split} s(\tilde{\boldsymbol{\pi}}) &= \mathbf{c}^{T} \tilde{\boldsymbol{\pi}} = \left[ \boldsymbol{c}\_{n-1} \boldsymbol{\lambda}^{n-1} + \dots + \boldsymbol{c}\_{1} \boldsymbol{\lambda}, \boldsymbol{c}\_{0} \right] \begin{bmatrix} \ddot{\boldsymbol{\pi}} \\ \dot{\tilde{\boldsymbol{\pi}}} \\ \vdots \\ \ddot{\tilde{\boldsymbol{\pi}}}^{(n-1)} \\ \ddot{\tilde{\boldsymbol{\pi}}}^{(n-1)} \end{bmatrix} \\ &= \boldsymbol{c}\_{0} \ddot{\boldsymbol{\pi}}^{(n-1)} + \boldsymbol{c}\_{1} \boldsymbol{\lambda} \ddot{\boldsymbol{\pi}}^{(n-2)} + \dots + \boldsymbol{c}\_{n-1} \boldsymbol{\lambda}^{n-1} \ddot{\boldsymbol{\pi}}. \end{split}$$

Thus,

$$- \mathfrak{Q} \le \mathfrak{s}(\tilde{\mathfrak{x}}) \le \mathfrak{Q} = -\mathfrak{Q} \le c\_0 \tilde{\mathfrak{x}}^{(n-1)} + c\_1 \lambda \tilde{\mathfrak{x}}^{(n-2)} + \dots + c\_{n-1} \lambda^{n-1} \tilde{\mathfrak{x}} \le \mathfrak{Q},\tag{48}$$

or the following,

$$- \mathfrak{Q} \le \left(\frac{d}{dt} + \lambda\right)^{n-1} \tilde{\kappa} \le \mathfrak{Q} \tag{49}$$

can be written. If (49) is multiplied by *eλ<sup>t</sup>* , the following statement is reached:

$$-\mathcal{Q}e^{\lambda t} \le \left(\frac{d}{dt} + \lambda\right)^{n-1} \|e^{\lambda t} \le \mathcal{Q}e^{\lambda t} \dots$$

In fact, this expression is equal to

$$-\mathcal{Q}e^{\lambda t} \le \frac{d^{n-1}}{dt^{n-1}} \left(\ddot{\lambda}e^{\dot{\lambda}t}\right) \le \mathcal{Q}e^{\lambda t}.\tag{50}$$

That is to say,

$$\left(\frac{d}{dt} + \lambda\right)^{n-1} \ddot{\mathbf{x}} e^{\lambda t} = \frac{d^{n-1}}{dt^{n-1}} \left(\ddot{\mathbf{x}} e^{\lambda t}\right). \tag{51}$$

We can confirm this form of (51) for small *n* values. Namely, if binomial expansion is applied for *<sup>d</sup> dt* <sup>þ</sup> *<sup>λ</sup>* � �*<sup>n</sup>*�<sup>1</sup> ,

$$\left(\frac{d}{dt} + \lambda\right)^{n-1} = \sum\_{k=0}^{n-1} \binom{n-1}{k} \frac{d^k}{dt^k} \lambda^{n-1-k}$$

is written. Hence,

$$\mathfrak{s}(\tilde{\mathfrak{x}}) = \left(\frac{d}{dt} + \lambda\right)^{n-1} \tilde{\mathfrak{x}} = \sum\_{k=0}^{n-1} \binom{n-1}{k} \frac{d^k \tilde{\mathfrak{x}}}{dt^k} \lambda^{n-1-k}.\tag{52}$$

At this point, we can make a confirmation by taking *n* ¼ 3:

For *<sup>n</sup>* <sup>¼</sup> 1, it becomes *<sup>s</sup>*ð Þ¼ *<sup>x</sup>*<sup>~</sup> <sup>0</sup> 0 � � *<sup>d</sup>*0*x*<sup>~</sup> *dt*<sup>0</sup> *<sup>λ</sup>*<sup>0</sup> <sup>¼</sup> *<sup>x</sup>*~. The binomial coefficient of this single term is 1, and this number is at the top of the Pascal triangle. For *n* ¼ 2, it becomes *<sup>s</sup>*ð Þ¼ *<sup>x</sup>*<sup>~</sup> <sup>1</sup> 0 � � *<sup>d</sup>*0*x*<sup>~</sup> *dt*<sup>0</sup> *<sup>λ</sup>*1�<sup>0</sup> <sup>þ</sup> 1 1 � � *dx*<sup>~</sup> *dt <sup>λ</sup>*1�<sup>1</sup> <sup>¼</sup> *<sup>λ</sup>x*<sup>~</sup> <sup>þ</sup> *dx*<sup>~</sup> *dt*. Here, the coefficients of both terms are 1. It gives the numbers of one-down row from the top of the Pascal triangle. For *n* ¼ 3, it becomes

$$s(\ddot{\mathbf{x}}) = \binom{2}{0} \frac{d^0 \ddot{\mathbf{x}}}{dt^0} \lambda^{2-0} + \binom{2}{1} \frac{d \ddot{\mathbf{x}}}{dt} \lambda^{2-1} + \binom{2}{2} \frac{d^2 \ddot{\mathbf{x}}}{dt^2} \lambda^{2-2} = \lambda^2 \ddot{\mathbf{x}} + 2\lambda \frac{d \ddot{\mathbf{x}}}{dt} + \frac{d^2 \ddot{\mathbf{x}}}{dt^2}. \tag{53}$$

Here, the coefficients of the three terms from left to right are 1, 2, 1. This gives the elements of the two-down row from the top of the Pascal triangle. If the expression �∅ ≤ *<sup>s</sup>*ð Þ *<sup>x</sup>*<sup>~</sup> ≤ ∅ is multiplied by *<sup>e</sup>λ<sup>t</sup>* ,

$$-\mathcal{Q}e^{\hat{\mathcal{U}}} \le s(\hat{\mathfrak{x}})e^{\hat{\mathcal{U}}} \le \mathcal{Q}e^{\hat{\mathcal{U}}}$$

is obtained. If the result for *n* ¼ 3 in Eq. (52), or its equivalent (53) is substituted above,

$$-\mathcal{Q}e^{\dot{\lambda}t} \le \left(\dot{\lambda}^2 e^{\dot{\lambda}t} \tilde{\boldsymbol{\kappa}} + 2\lambda e^{\dot{\lambda}t} \frac{d\tilde{\boldsymbol{\kappa}}}{dt} + e^{\dot{\lambda}t} \frac{d^2 \tilde{\boldsymbol{\kappa}}}{dt^2}\right) \le \mathcal{Q}e^{\dot{\lambda}t},$$

or the following expression is obtained:

$$-\mathcal{Q}e^{\lambda t} \le \frac{d^2}{dt^2} \left( e^{\lambda t} \tilde{\boldsymbol{x}} \right) \le \mathcal{Q}e^{\lambda t} \dots$$

This verifies the multiplication of *<sup>s</sup>*ð Þ *<sup>x</sup>*<sup>~</sup> with *<sup>e</sup>λ<sup>t</sup>* for *<sup>n</sup>* <sup>¼</sup> 3. In other words, the equation *<sup>d</sup> dt* <sup>þ</sup> *<sup>λ</sup>* � �<sup>2</sup> *xe* <sup>~</sup> *<sup>λ</sup><sup>t</sup>* <sup>¼</sup> *<sup>d</sup>*<sup>2</sup> *dt*<sup>2</sup> *xe* <sup>~</sup> *<sup>λ</sup><sup>t</sup>* � � is satisfied. Once this statement is generalized for *n*, the validity of Eq. (51) is proven.

If the inequality (50) is integrated between 0 and *t*,

$$-\int\_0^t \mathcal{Q}e^{\dot{\lambda}\tau}d\tau \le \int\_0^t \frac{d^{n-1}}{dt^{n-1}} \left(\ddot{\varkappa}e^{\dot{\lambda}t}\right)d\tau \le \int\_0^t \mathcal{Q}e^{\dot{\lambda}t}d\tau,$$

and one step later,

$$\left. -\frac{\mathcal{Q}}{\lambda} e^{\lambda \tau} \right|\_{0}^{t} \leq \frac{d^{n-2}}{dt^{n-2}} \left( \left. \tilde{\kappa} e^{\lambda \tau} \right| \right)\_{0}^{t} \leq \frac{\mathcal{Q}}{\lambda} e^{\lambda \tau} \Big|\_{0}^{t},$$

and finally the following expression is reached:

$$\frac{-\boxtimes}{\lambda}e^{\lambda t} + \frac{\boxtimes}{\lambda} \le \frac{d^{n-2}}{dt^{n-2}} \left( \check{\mathbf{x}} e^{\lambda t} \right) - \frac{d^{n-2}}{dt^{n-2}} \left( \check{\mathbf{x}} e^{\lambda t} \right) \Big|\_{t=0} \le \frac{\boxtimes}{\lambda} e^{\lambda t} - \frac{\boxtimes}{\lambda} \cdot \mathbf{1}$$

When the term *dn*�<sup>2</sup> *dtn*�<sup>2</sup> *xe* <sup>~</sup> *<sup>λ</sup><sup>t</sup>* � �� � � *<sup>t</sup>*¼<sup>0</sup> is added to each side of this expression, it takes the form below:

$$\left| \frac{-\mathcal{Q}}{\lambda} e^{\lambda t} + \frac{\mathcal{Q}}{\lambda} + \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{x} e^{\lambda t} \right) \right|\_{t=0} \leq \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{x} e^{\lambda t} \right) \leq \frac{\mathcal{Q}}{\lambda} e^{\lambda t} - \frac{\mathcal{Q}}{\lambda} + \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{x} e^{\lambda t} \right) \Big|\_{t=0}.\tag{54}$$

$$\left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\mathbf{x}} e^{\boldsymbol{\omega}} \right) \right|\_{t=0} \geq - \left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\mathbf{x}} e^{\boldsymbol{\omega}} \right) \right|\_{t=0} \text{ and } \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\mathbf{x}} e^{\boldsymbol{\omega}} \right) \Big|\_{t=0} \leq \left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\mathbf{x}} e^{\boldsymbol{\omega}} \right) \right|\_{t=0}$$

$$\left| \frac{-\mathcal{Q}}{\lambda} e^{\lambda t} + \frac{\mathcal{Q}}{\lambda} - \left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\chi} e^{\lambda t} \right) \right|\_{t=0} \leq \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\chi} e^{\lambda t} \right) \leq \frac{\mathcal{Q}}{\lambda} e^{\lambda t} - \frac{\mathcal{Q}}{\lambda} + \left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\chi} e^{\lambda t} \right) \right|\_{t=0}$$

$$\left|\frac{-\mathcal{Q}}{\lambda}e^{\lambda t} - \left(\left|\frac{d^{n-2}}{dt^{n-2}}\left(\ddot{\mathbf{x}}e^{\lambda t}\right)\right|\_{t=0} + \frac{\mathcal{Q}}{\lambda}\right) \leq \frac{d^{n-2}}{dt^{n-2}}\left(\ddot{\mathbf{x}}e^{\lambda t}\right) \leq \frac{\mathcal{Q}}{\lambda}e^{\lambda t} + \left(\left|\frac{d^{n-2}}{dt^{n-2}}\left(\ddot{\mathbf{x}}e^{\lambda t}\right)\right|\_{t=0} + \frac{\mathcal{Q}}{\lambda}\right).$$

$$\begin{split} & \underbrace{-\int\_{0}^{t} \frac{\mathcal{Q}}{\lambda} e^{\lambda t} dt}\_{\text{Parr}(a)} - \underbrace{\int\_{0}^{t} \left( \left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\boldsymbol{x}} e^{\lambda t} \right) \right|\_{t=0} + \frac{\mathcal{Q}}{\lambda} \right) dt}\_{\text{Parr}(b)} \leq \underbrace{\int\_{0}^{t} \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\boldsymbol{x}} e^{\lambda t} \right) dt}\_{\text{Parr}(c)} \leq \underbrace{\int\_{0}^{t} \frac{\mathcal{Q}}{\lambda} e^{\lambda t} dt}\_{\text{Parr}(a)} \\ & + \underbrace{\int\_{0}^{t} \left( \left| \frac{d^{n-2}}{dt^{n-2}} \left( \ddot{\boldsymbol{x}} e^{\lambda t} \right) \right|\_{t=0} + \frac{\mathcal{Q}}{\lambda} \right) dt}\_{\text{Parr}(b)}. \end{split}$$

$$\int\_0^t \frac{\mathcal{D}}{\lambda} e^{\lambda t} dt = \frac{\mathcal{D}}{\lambda^2} e^{\mu t} \Big|\_0^t = \frac{\mathcal{D}}{\lambda^2} e^{\mu t} - \frac{\mathcal{D}}{\lambda^2}, n = 2 \text{(for the 3rd integral)}$$

$$\frac{\mathcal{D}}{\lambda^3} e^{\lambda t} - \frac{\mathcal{D}}{\lambda^3}, n = 3 \text{(for the 3rd integral)}$$

$$\vdots$$

$$\vdots$$

$$\frac{\mathcal{D}}{\lambda^{n-2}} e^{\mu t} - \frac{\mathcal{D}}{\lambda^{n-2}}, n = n - 2 \text{(for the } (n-2) \text{th integral)}$$

$$\frac{\mathcal{D}}{\lambda^{n-1}} e^{\mu t} - \frac{\mathcal{D}}{\lambda^{n-1}}, n = n - 1 \text{(for the } (n-1) \text{th integral: Generalized form)}$$

$$\int\_0^t \frac{d^{n-1}}{dt^{n-1}} \left(\ddot{x}e^{\acute{u}t}\right) dt = \frac{d^{n-2}}{dt^{n-2}} \left(\ddot{x}e^{\acute{u}t}\right)\Big|\_0^t = \frac{d^{n-2}}{dt^{n-2}} \left(\ddot{x}e^{\acute{u}t}\right) - \frac{d^{n-2}}{dt^{n-2}} \left(\ddot{x}e^{\acute{u}t}\right)\Big|\_{t=0}, n = \mathtt{1} (\mathtt{1st integral})$$

$$\int\_0^t \frac{d^{n-2}}{dt^{n-3}} \left(\ddot{x}e^{\acute{u}t}\right) dt = \frac{d^{n-3}}{dt^{n-3}} \left(\ddot{x}e^{\acute{u}t}\right)\Big|\_0^t = \frac{d^{n-3}}{dt^{n-3}} \left(\ddot{x}e^{\acute{u}t}\right) - \frac{d^{n-3}}{dt^{n-3}} \left(\ddot{x}e^{\acute{u}t}\right)\Big|\_{t=0}, n = \mathtt{2} (\mathtt{2nd}\text{ integral})$$

$$\int\_0^t \frac{d^2}{dt^2} \left(\ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t}\right) dt = \frac{d}{dt} \left(\ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t}\right)\Big|\_0^t = \frac{d}{dt} \left(\ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t}\right) - \frac{d}{dt} \left(\ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t}\right)\Big|\_{t=0}, n = n - 2((n-2)th\text{ integral})$$

$$\int\_0^t \frac{d}{dt} \left(\ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t}\right) dt = \ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t}\Big|\_0^t = \ddot{\mathbf{x}} e^{\dot{\mathbf{x}}t} - \ddot{\mathbf{x}}(0), n - \mathbf{1}((n-1)th\text{ integral})$$

$$\begin{split} & \frac{-\mathscr{Q}}{\lambda^{n-1}} e^{\lambda t} + \frac{\mathscr{Q}}{\lambda^{n-1}} - \frac{\left( \left| \frac{d}{d t^{n-2}} \left( \check{\mathbf{x}} e^{\lambda t} \right) \right|\_{t=0} + \frac{\mathscr{Q}}{\lambda} \right) \* t^{n-2}}{(n-2)!} - \dots \leq \check{\mathbf{x}} e^{\lambda t} - \check{\mathbf{x}}(0) \leq \frac{\mathscr{Q}}{\lambda^{n-1}} e^{\lambda t} - \frac{\mathscr{Q}}{\lambda^{n-1}} \\ & + \frac{\left( \left| \frac{d}{d t^{n-2}} \left( \check{\mathbf{x}} e^{\lambda t} \right) \right|\_{t=0} + \frac{\mathscr{Q}}{\lambda} \right) \* t^{n-2}}{(n-2)!} + \dots \end{split}$$

$$\begin{split} & \frac{-\mathscr{Q}}{\lambda^{n-1}} e^{\mathsf{i}t} - \frac{\left(\left|\frac{d}{dt^{n-1}}\left(\check{\mathsf{x}}e^{\mathsf{i}t}\right)\right|\_{t=0} + \left|\frac{\mathscr{Q}}{\lambda}\right| \* t^{n-2}}{(n-2)!} - \dots - \left(\left|\mathring{\mathsf{x}}(\mathsf{0})\right| + \frac{\mathscr{Q}}{\lambda^{n-1}}\right) \leq \check{\mathsf{x}} e^{\mathsf{i}t} \leq \frac{\mathscr{Q}}{\lambda^{n-1}} e^{\mathsf{i}t} \\ & + \frac{\left(\left|\frac{d}{dt^{n-2}}\left(\check{\mathsf{x}}e^{\mathsf{i}t}\right)\right|\_{t=0} + \frac{\mathscr{Q}}{\lambda}\right) \* t^{n-2}}{(n-2)!} + \dots + \left(\left|\mathring{\mathsf{x}}(\mathsf{0})\right| + \frac{\mathscr{Q}}{\lambda^{n-1}}\right) \end{split} \tag{55}$$

$$\begin{split} & \frac{-\mathcal{Q}}{\lambda^{n-1}} - \frac{1}{e^{\mathsf{d}t}} \frac{\left(\left|\frac{d}{d\mathsf{r}^{n-2}}\left(\check{\mathsf{x}}e^{\mathsf{d}t}\right)\right|\_{t=0} + \frac{\mathcal{Q}}{\lambda}\right) \* t^{n-2}}{(n-2)!} - \dots - \frac{1}{e^{\mathsf{d}t}} \left(\left|\check{\mathsf{x}}(\mathsf{O})\right| + \frac{\mathcal{Q}}{\lambda^{n-1}}\right) \leq \check{\mathsf{x}}(\mathsf{t}) \leq \frac{\mathcal{Q}}{\lambda^{n-1}}, \\ & \qquad + \frac{\mathbf{1}}{e^{\mathsf{d}t}} \frac{\left(\left|\frac{d}{d\mathsf{r}^{n-2}}\left(\check{\mathsf{x}}e^{\mathsf{d}t}\right)\right|\_{t=0} + \frac{\mathcal{Q}}{\lambda}\right) \* t^{n-2}}{(n-2)!} + \dots + \frac{\mathbf{1}}{e^{\mathsf{d}t}} \left(\left|\check{\mathsf{x}}(\mathsf{O})\right| + \frac{\mathcal{Q}}{\lambda^{n-1}}\right). \end{split}$$

$$\frac{-\mathcal{Q}}{\lambda^{n-1}} \le \tilde{\varkappa}(t) \le \frac{\mathcal{Q}}{\lambda^{n-1}}.\tag{56}$$

$$\begin{split} \frac{1-\mathfrak{Q}}{\lambda^{n-2}}e^{\mathsf{i}t} - \frac{\left(\left|\frac{d}{d^{n-2}}\left(\check{\mathbf{x}}e^{\mathsf{i}t}\right)\right|\_{t=0}+\frac{\mathfrak{Q}}{\lambda}\right)\*t^{n-3}}{(n-3)!} - \dots - \left(\left|\mathring{\mathbf{x}}(\mathbf{0})\right|+\frac{\mathfrak{Q}}{\lambda^{n-2}}\right) \leq \frac{d}{dt}\left(\check{\mathbf{x}}e^{\mathsf{i}t}\right) \leq \frac{\mathfrak{Q}}{\lambda^{n-2}}e^{\mathsf{i}t}, \\ +\frac{\left(\left|\frac{d}{d^{n-2}}\left(\check{\mathbf{x}}e^{\mathsf{i}t}\right)\right|\_{t=0}+\frac{\mathfrak{Q}}{\lambda}\right)\*t^{n-3}}{(n-3)!} + \dots + \left(\left|\mathring{\mathbf{x}}(\mathbf{0})\right|+\frac{\mathfrak{Q}}{\lambda^{n-2}}\right), \end{split} \tag{57}$$

and the derivative expression,

$$d\left(\ddot{\chi}e^{\dot{\lambda}t}\right)/dt = \dot{\ddot{\lambda}}e^{\dot{\lambda}t} + \ddot{\varkappa}\lambda e^{\dot{\lambda}t},$$

by having (56)'s bounds accepted to (57) and dividing it back into *<sup>e</sup><sup>λ</sup><sup>t</sup>* for *<sup>t</sup>* ! <sup>∞</sup>,

$$\begin{array}{l} \frac{-\mathcal{Q}}{\lambda^{n-2}} - (\bullet) \frac{t^{n-3}}{(n-3)!e^{\lambda t}} - \dots - (\bullet) \frac{1}{e^{\lambda t}} \leq \dot{\tilde{x}}(t) + \tilde{x}(t)\lambda \leq \frac{\mathcal{Q}}{\lambda^{n-2}} + (\bullet) \frac{t^{n-3}}{(n-3)!e^{\lambda t}} + \dots \\\ + (\bullet) \frac{1}{e^{\lambda t}}, \end{array}$$

and finally from here,

$$\frac{-\mathcal{Q}}{\lambda^{n-2}} \le \dot{\tilde{\varkappa}}(t) + \lambda \tilde{\varkappa}(t) \le \frac{\mathcal{Q}}{\lambda^{n-2}} \tag{58}$$

is obtained. However, in order to determine the bounds of (58) based on only *x t* ~\_ð Þ, the bounds corresponding to the term *<sup>λ</sup>x t*ð Þ in addition to *x t* ~\_ð Þ must be determined exactly. For this, (56) is used and if each side in this inequality is multiplied by *λ*,

$$\frac{-\mathcal{Q}}{\lambda^{n-2}} \le \lambda \tilde{\omega}(t) \le \frac{\mathcal{Q}}{\lambda^{n-2}}\tag{59}$$

expression is obtained. Now then, if the effect of *λx t* ~ð Þ in the inequality of (58) is substituted by the bound determination ascertained by (59) above, the inequality (59) turns into

$$-2\frac{\mathcal{Q}}{\lambda^{n-2}} \le \dot{\hat{\mathbf{x}}}(t) \le 2\frac{\mathcal{Q}}{\lambda^{n-2}}.\tag{60}$$

Similarly, taking into account the ð Þ *n* � 3 *th* integral of (50),

$$\begin{split} \frac{-\mathcal{Q}}{\lambda^{n-3}}\boldsymbol{\epsilon}^{\rm{L}} - \frac{\left(\left|\frac{d}{d^{n-3}}\left(\ddot{\mathbf{x}}\boldsymbol{\epsilon}^{\rm{L}}\right)\right|\_{t=0} + \frac{\mathcal{Q}}{\lambda}\right) \boldsymbol{\epsilon}^{n-4}}{(n-4)!} - \dots - \left(\left|\ddot{\mathbf{x}}(\mathbf{0})\right| + \frac{\mathcal{Q}}{\lambda^{n-3}}\right) \leq \frac{d^{2}}{dt^{2}}\left(\ddot{\mathbf{x}}\boldsymbol{\epsilon}^{\rm{L}}\right) \leq \frac{\mathcal{Q}}{\lambda^{n-3}}\boldsymbol{\epsilon}^{\rm{L}} \\ + \frac{\left(\left|\frac{d^{n-3}}{d^{n-3}}\left(\ddot{\mathbf{x}}\boldsymbol{\epsilon}^{\rm{L}}\right)\right|\_{t=0} + \frac{\mathcal{Q}}{\lambda}\right) \boldsymbol{\epsilon}^{n-4}}{(n-4)!} + \dots + \left(\left|\ddot{\boldsymbol{\bar{x}}}(\mathbf{0})\right| + \frac{\mathcal{Q}}{\lambda^{n-3}}\right), \end{split} \tag{61}$$

and the derivative expression,

$$d\left(\dot{\tilde{\mathbf{x}}}e^{\mathbf{\hat{x}}} + \tilde{\mathbf{x}}\lambda e^{\mathbf{\hat{x}}}\right)/dt = \ddot{\tilde{\mathbf{x}}}e^{\mathbf{\hat{x}}} + \lambda e^{\mathbf{\hat{x}}}\dot{\tilde{\mathbf{x}}} + \dot{\tilde{\mathbf{x}}}\lambda e^{\mathbf{\hat{x}}} + \tilde{\mathbf{x}}\lambda^2 e^{\mathbf{\hat{x}}} = \ddot{\tilde{\mathbf{x}}}e^{\mathbf{\hat{x}}} + 2\dot{\tilde{\mathbf{x}}}\lambda e^{\mathbf{\hat{x}}} + \ddot{\mathbf{x}}\lambda^2 e^{\mathbf{\hat{x}}},$$

by imposing the bounds of (56) and (60) on (61) and dividing this expression once again to *<sup>e</sup><sup>λ</sup><sup>t</sup>* for *<sup>t</sup>* ! <sup>∞</sup>, the following step is obtained first:

$$\frac{-\mathcal{Q}}{\lambda^{n-3}}e^{\lambda t} \le \ddot{\tilde{\mathcal{X}}} + 2\dot{\tilde{\mathcal{X}}}\lambda + \ddot{\varkappa}\lambda^2 \le \frac{\mathcal{Q}}{\lambda^{n-3}}e^{\lambda t}.\tag{62}$$

Now, the bounds for *x*~*λ*<sup>2</sup> and 2*x*~\_ *λ* are respectively determined with,

$$\frac{-\mathcal{D}}{\lambda^{n-3}} \le \lambda^2 \tilde{\varkappa}(t) \le \frac{\mathcal{D}}{\lambda^{n-3}}\tag{63}$$

*An In-Depth Analysis of Sliding Mode Control and Its Application to Robotics DOI: http://dx.doi.org/10.5772/intechopen.93027*

by multiplying each side of the inequality of (56) by the term *λ*<sup>2</sup> , and with,

$$-4\frac{\mathcal{Q}}{\lambda^{n-3}} \le 2\dot{\tilde{\mathbf{x}}}\lambda \le 4\frac{\mathcal{Q}}{\lambda^{n-3}}\tag{64}$$

by multiplying each side of inequality of (60) by the term 2*λ*. Once these bounds determined by the inequalities (63) and (64) are imposed on (62), the expression,

$$
\frac{-\mathcal{Q}}{\lambda^{n-3}} - \frac{\mathcal{Q}}{\lambda^{n-3}} - 4\frac{\mathcal{Q}}{\lambda^{n-3}} \le \ddot{\tilde{\kappa}} \le \frac{\mathcal{Q}}{\lambda^{n-3}} + \frac{\mathcal{Q}}{\lambda^{n-3}} + 4\frac{\mathcal{Q}}{\lambda^{n-3}},
$$

and hence in brief, the result,

$$\left| \mathfrak{G} \frac{\mathfrak{D}}{\lambda^{n-3}} \leq \ddot{\tilde{\mathbf{x}}} \leq \mathfrak{G} \frac{\mathfrak{D}}{\lambda^{n-3}} \right. \tag{65}$$

is concluded. As in obtaining (56), (60) and (65), the following general conclusion is reached if the similar procedure is applied sequentially until the bounds of *x*~ð Þ *<sup>n</sup>*�<sup>1</sup> are achieved:

$$-\left(\mathbf{1} + \sum\_{i=0}^{n-2} \binom{n-1}{i} \zeta\_i\right) \mathcal{Q} \le \bar{\mathbf{x}}^{(n-1)} \le \left(\mathbf{1} + \sum\_{i=0}^{n-2} \binom{n-2}{i} \zeta\_i\right) \mathcal{Q}.\tag{66}$$

Here, the coefficients *ζi*ð Þ *i* ¼ 0, 1, ⋯, *n* � 2 are related to the pre-acquired bounds of each *x*~ð Þ*<sup>i</sup>* and are summarized in Theorem 2.

In this way, by examining Eqs. (56), (60), (65), and (66) and, as much as other skipped boundaries, the integrals of (50), the tracking error will be kept within the bounds of *x*~ð Þ*<sup>i</sup>* � � � �≤*ζiλ<sup>i</sup>*�*n*þ<sup>1</sup> ∅, *i* ¼ 0, 1, ⋯, *n* � 1, where *ζ<sup>i</sup>* is defined by (47).

**Figure 8.** *Convergence region Φ in the case of a second-order system.*

**Remark 5.** It should be noted that an *n*-dimensionally separated partition defined according to the boundaries mentioned earlier is not entirely within the boundary layer. Considering the attractiveness and invariant properties of *S*<sup>∅</sup> proved in Theorem 1, the convergence region can be expressed as the intersection of an *n*-dimensional separated partition and boundary layer defined in Theorem 2. Thus, the tracking error vector will converge exponentially to a closed region

<sup>Φ</sup> <sup>¼</sup> *<sup>x</sup>*∈*R<sup>n</sup>*jj j *<sup>s</sup>*ð Þ *<sup>x</sup>*<sup>~</sup> ≤ ∅ and *<sup>x</sup>*~ð Þ*<sup>i</sup>* � � � �≤*ζiλi*�*n*þ<sup>1</sup> ∅, *i* ¼ 0, 1, ⋯, *n* � 1 n o, where *<sup>ζ</sup><sup>i</sup>* is defined by (47).

**Figure 8** describes the Φ convergence region defined according to Remark 5 for a second-order (*n* ¼ 2) system.
