**4. Problem setting: stabilization through dynamic output feedback**

This chapter considers the stabilization problem via dynamic output feedback laws and the synthesis of the stabilizing feedback laws. This section states the problem setting.

The approach presented here is based on the SDLMI approach, which derives the sufficient conditions of the existence of stabilizing feedback laws as the SDLMI conditions. We can obtain stabilizing feedback control laws and Lyapunov functions by solving the SDLMI conditions using numerical solvers.

Consider a nonlinear system given as

where *z* ∈ ℝ. Apparently, this polynomial is expressed as the sum of squares polynomial

The SOS decomposition can be solved by some numerical solvers, such as YALMIP [18] and SOSTOOLS [19]. When some coefficients of polynomials are decision variables in an SOS decomposition, by using the numerical solvers, we can find the feasible solutions such that the SOS decomposition holds. Therefore, we can adapt the SOS decomposition to the design of

With the relation to the stability theory presented in Section 2, the sufficient condition of the

**Theorem 3.** [2] Consider system (1). If there exist a positive definite function and an

Theorem 3 shows a direct application of the SOSs to the analysis of the stability. This implies that the SOS decomposition can be applied to the synthesis of the stabilizing feedback laws. This chapter develops a method to design dynamic output feedback laws based on the SDLMI

are state-dependent symmetric matrices. The constraint should be satisfied for any *z*

∈ ℝ*<sup>n</sup>*. This differs from standard LMIs and is the derivation of the word, state-dependent.

A relation of the SDLMIs and the SOS decompositions is shown as follows.

are the decision variables, the matrix functions *Fi*

(2)

:

Regarding Theorem 2, the polynomial is also expressed as

324 Nonlinear Systems - Design, Analysis, Estimation and Control

and the matrix in the right-hand side of (2) is positive definite.

feedback laws in control problems.

SOS polynomial , such that

stability is given as the SOS conditions.

then the equilibrium *x* = 0 is asymptotically stable.

where *ai* ∈ ℝ are the fixed coefficients, *ci*

approach [8]. The SDLMI is defined as the optimization problem:

$$\begin{aligned} \dot{\mathbf{x}} &= f(\mathbf{x}, \mathbf{u}), \qquad &\mathbf{x}(\mathbf{f}\_0) = \mathbf{x}\_0 \\ \mathbf{y} &= h(\mathbf{x}), \end{aligned} \tag{3}$$

where *x* ∈ ℝ*<sup>n</sup>* is the state, *u* ∈ ℝ*nu* is the input, *y* ∈ ℝ*ny* is the output, *f*: ℝ*<sup>n</sup>* × ℝ*nu* → ℝ*<sup>n</sup>*, *h*: ℝ*<sup>n</sup>* → ℝ*ny* , and *x*0 is the initial state. For the nonlinear systems given by (3), we assume that system (3) is expressed as

$$\begin{aligned} \dot{\mathbf{x}} &= A(\mathbf{x})Z(\mathbf{x}) + B(\mathbf{x})u, \quad \mathbf{x}(t\_0) = \mathbf{x}\_0 \\ \mathbf{y} &= C(\mathbf{x})Z(\mathbf{x}), \end{aligned} \tag{4}$$

where , *A*: ℝ*<sup>n</sup>* → ℝ*n×N*, *B*: ℝ*<sup>n</sup>* → ℝ*n×nu* , *C*: ℝ*<sup>n</sup>* → ℝ*ny×N*. Further, we assume that *Z*(*x*) = 0, if and only if *x* = 0. We consider the output stabilization of system (4) using a dynamic feedback law in the form of

$$\begin{aligned} \dot{\hat{\boldsymbol{x}}} &= \boldsymbol{A}\_c \left( \hat{\boldsymbol{x}}, \boldsymbol{y} \right) \hat{\boldsymbol{x}} + \boldsymbol{B}\_c \left( \hat{\boldsymbol{x}}, \boldsymbol{y} \right) \boldsymbol{y}, \quad \hat{\boldsymbol{x}} \left( t\_0 \right) = \hat{\boldsymbol{x}}\_0 \\ \boldsymbol{\mu} &= \boldsymbol{C}\_c \left( \hat{\boldsymbol{x}}, \boldsymbol{y} \right) \hat{\boldsymbol{x}} + \boldsymbol{D}\_c \left( \hat{\boldsymbol{x}}, \boldsymbol{y} \right) \boldsymbol{y}, \end{aligned} \tag{5}$$

where is the state of the dynamic feedback law,

$$A\_{\mathcal{C}} \colon \mathbb{R}^{n\_{\widehat{X}}} \times \mathbb{R}^{n\_{\widehat{Y}}} \to \mathbb{R}^{n\_{\widehat{X}} \times n\_{\widehat{X}}},\\B\_{\mathcal{C}} \colon \mathbb{R}^{n\_{\widehat{X}}} \times \mathbb{R}^{n\_{\widehat{Y}}} \to \mathbb{R}^{n\_{\widehat{X}} \times n\_{\widehat{Y}}},\\\mathcal{C}\_{\mathcal{C}} \colon \mathbb{R}^{n\_{\widehat{X}}} \times \mathbb{R}^{n\_{\widehat{Y}}} \to \mathbb{R}^{n\_{\widehat{X}} \times n\_{\widehat{X}}},\\\text{and}\\B\_{\mathcal{C}} \colon \mathbb{R}^{n\_{\widehat{X}}} \to \mathbb{R}^{n\_{\widehat{X}}} \times \mathbb{R}^{n\_{\widehat{X}}}$$

:ℝ × ℝ <sup>ℝ</sup> × , and 0 is the initial state.

We have the closed-loop system of (4) with the dynamic output feedback law (5), given by

$$\begin{aligned} \dot{\mathbf{x}} &= \left\{ A\begin{pmatrix} \mathbf{x} \end{pmatrix} + B\begin{pmatrix} \mathbf{x} \end{pmatrix} D\_c\begin{pmatrix} \hat{\mathbf{x}}, y \end{pmatrix} \mathbf{C}\begin{pmatrix} \mathbf{x} \end{pmatrix} \right\} Z\begin{pmatrix} \mathbf{x} \end{pmatrix} + B\begin{pmatrix} \mathbf{x} \end{pmatrix} \mathbf{C}\_c\begin{pmatrix} \hat{\mathbf{x}}, y \end{pmatrix} \hat{\mathbf{x}},\\ \dot{\hat{\mathbf{x}}} &= A\_c\begin{pmatrix} \hat{\mathbf{x}}, y \end{pmatrix} \hat{\mathbf{x}} + B\_c\begin{pmatrix} \hat{\mathbf{x}}, y \end{pmatrix} \mathbf{C}\begin{pmatrix} \mathbf{x} \end{pmatrix} Z\begin{pmatrix} \mathbf{x} \end{pmatrix} \end{aligned} \tag{6}$$

We consider the stabilization of the closed-loop system (6). To this end, we give a method to design the matrix functions , , , , , , , in the next section.

**Remark 1.** We obtain a system in the form of (4) as an expression of a nonlinear affine system

$$\begin{aligned} \dot{x} &= f(x) + g(x)u\_r \\ y &= h(x), \end{aligned}$$

by choosing *Z(x)* properly. Note that the choice of *Z(x)* is not unique in general. The systems in the form of (4) can be seen as a generalization of linear systems, given as

$$\begin{array}{l} \dot{\mathfrak{x}} = Ax + Bv\\ \mathfrak{y} = Cx, \end{array}$$

where the matrices *A*, *B*, and *C* are with the appropriate dimensions.

## **5. Design of dynamic output feedback laws through SOSs**

This section provides a design method of dynamic feedback laws (5) for the output stabilization of system (4). We show stability conditions of the closed-loop system of (6) as SDLMI conditions. We can obtain the stabilizing laws by solving the SDLMI conditions via SOS decomposition using numerical solvers.

The main idea of the proposed method is as follows. Instead of the dynamic feedback law (5), assume that there exists a static state feedback law

$$u = k(x),\tag{7}$$

where , such that the feedback law asymptotically stabilizes the origin of system (4). Then, according to the converse Lyapunov theorem, we have a Lyapunov function *U*1 (*x*). Then, we consider the design of the dynamic output feedback law (5) so that a function , given by

$$U(\mathbf{x}, \mathfrak{k}) = U\_1(\mathbf{x}) + (k(\mathbf{x}) - \mathfrak{k})^T \Sigma (k(\mathbf{x}) - \mathfrak{k}) \tag{8}$$

becomes the Lyapunov function of the closed-loop system (6) with some positive definite matrix *Σ*. When we design the output feedback laws, so that the function , of (8) is a Lyapunov function of the closed-loop system, the value of of the designed output feedback laws in (8) will estimate the value of *k*(*x*). A design procedure discussed here can be seen in reference [16], and is called as the direct design. As shown in the following, when we obtain the static feedback law (7) in polynomial forms, we can obtain the SDLMI conditions where the stability of the closed-loop system (6) is guaranteed by function (8).

In the following, if the matrix *B*(*x*) of (4) has rows all the elements of which are zero, we denote the corresponding row indices as . We also employ the notation

As discussed above, we design a stabilizing state feedback law as the first step. The state feedback law also can be designed by using SDLMIs. We introduce the following result shown in reference [8].

**Theorem 5.** ([**8**]) Suppose that there exist a symmetric polynomial matrix *P*: ℝ<sup>n</sup> → ℝ*<sup>N</sup>*×*<sup>N</sup>*, a polynomial matrix a parameter and an SOS polynomial such that

$$\begin{split} \boldsymbol{\upsilon}^{\mathsf{T}}(\mathcal{P}(\boldsymbol{\hat{x}}) - \epsilon\_{1}\boldsymbol{I})\boldsymbol{\upsilon}, \\ \boldsymbol{\upsilon} - \nu \left\{ \boldsymbol{P}(\boldsymbol{\hat{x}})\boldsymbol{A}(\boldsymbol{x})^{\mathsf{T}}\boldsymbol{M}(\boldsymbol{x})^{\mathsf{T}} + \boldsymbol{M}(\boldsymbol{x})\boldsymbol{A}(\boldsymbol{x})\boldsymbol{P}(\boldsymbol{\hat{x}}) + \boldsymbol{K}(\boldsymbol{x})^{\mathsf{T}}\boldsymbol{B}(\boldsymbol{x})^{\mathsf{T}}\boldsymbol{M}(\boldsymbol{x})^{\mathsf{T}} \\ \quad + \boldsymbol{M}(\boldsymbol{x})\boldsymbol{B}(\boldsymbol{x})\boldsymbol{K}(\boldsymbol{x}) - \sum\_{j \neq j} \frac{\partial \boldsymbol{P}}{\partial \boldsymbol{x}\_{j}}(\boldsymbol{\xi}) \left\{ \boldsymbol{A}\_{j}(\boldsymbol{x})\boldsymbol{Z}(\boldsymbol{x}) \right\} + \epsilon\_{2}(\boldsymbol{x})\boldsymbol{I} \right\} \boldsymbol{v} \end{split} \tag{9}$$

are SOS polynomials, where is the *j*-th row of *A*(x), and

:ℝ × ℝ <sup>ℝ</sup>

:ℝ × ℝ <sup>ℝ</sup>

× , :ℝ × ℝ <sup>ℝ</sup>

ˆˆ ˆ

*c c*

= +

, and 0 is the initial state.

( ) ( ) () ()

design the matrix functions , , , , , , , in the next section.

in the form of (4) can be seen as a generalization of linear systems, given as

where the matrices *A*, *B*, and *C* are with the appropriate dimensions.

**5. Design of dynamic output feedback laws through SOSs**

= + +

ˆ ,, .

*x A xy x B xyC xZx*

×

326 Nonlinear Systems - Design, Analysis, Estimation and Control

&

sition using numerical solvers.

by

assume that there exists a static state feedback law

×

We have the closed-loop system of (4) with the dynamic output feedback law (5), given by

ˆ ˆ ˆ

*c c*

We consider the stabilization of the closed-loop system (6). To this end, we give a method to

**Remark 1.** We obtain a system in the form of (4) as an expression of a nonlinear affine system

by choosing *Z(x)* properly. Note that the choice of *Z(x)* is not unique in general. The systems

This section provides a design method of dynamic feedback laws (5) for the output stabilization of system (4). We show stability conditions of the closed-loop system of (6) as SDLMI conditions. We can obtain the stabilizing laws by solving the SDLMI conditions via SOS decompo-

The main idea of the proposed method is as follows. Instead of the dynamic feedback law (5),

where , such that the feedback law asymptotically stabilizes the origin of system (4). Then, according to the converse Lyapunov theorem, we have a Lyapunov function *U*1 (*x*). Then, we consider the design of the dynamic output feedback law (5) so that a function , given

, , ,

& (6)

{ () () ( ) ()} () () ( )

*x Ax BxD xyC x Zx BxC xy x*

, :ℝ × ℝ <sup>ℝ</sup>

× , and

(7)

$$\mathcal{M}(\mathbf{x}) = \frac{\partial \mathcal{Z}}{\partial \mathbf{x}}(\mathbf{x}). \tag{10}$$

Then, the origin of (4) is asymptotically stabilized by a state feedback given by

$$u = k(\boldsymbol{x}) = K(\boldsymbol{x})P(\boldsymbol{\hat{x}})^{-1}Z(\boldsymbol{x}).\tag{11}$$

For the design of the output feedback laws, we show the following theorem as the main result, which gives a design condition of the feedback law (5) in terms of state-dependent matrix inequalities.

**Theorem 6.** Suppose that there exist a symmetric matrix a polynomial matrix a parameter and an SOS polynomial such that

$$\begin{aligned} v^{\top}(\mathbb{P}\_1 - \epsilon\_1 l)v, \\ -v^{\top}\{\mathbb{P}\_1 A(\mathbf{x})^{\top} M(\mathbf{x})^{\top} + \mathcal{M}(\mathbf{x})A(\mathbf{x})\mathbb{P}\_1 + K(\mathbf{x})^{\top}B(\mathbf{x})^{\top}M(\mathbf{x})^{\top} + \mathcal{M}(\mathbf{x})B(\mathbf{x})K(\mathbf{x}) + \epsilon\_2(\mathbf{x})l\}v \end{aligned} \tag{12}$$

are SOS polynomials, where and *M*(*x*) is given as 10. Further, suppose that there exist a symmetric matrix 2:ℝ ( + )×(+) , and an SOS polynomial such that

$$-\boldsymbol{\nu}^{\boldsymbol{r}}\left(\begin{pmatrix}\Lambda\_{11}(\boldsymbol{\mathfrak{x}},\boldsymbol{\hat{x}}) & \Lambda\_{12}(\boldsymbol{\mathfrak{x}},\boldsymbol{\hat{x}})\\\Lambda\_{12}^{\boldsymbol{r}}(\boldsymbol{\mathfrak{x}},\boldsymbol{\hat{x}}) & \Lambda\_{22}(\boldsymbol{\mathfrak{x}},\boldsymbol{\hat{x}})\end{pmatrix} + \boldsymbol{e}\_{3}\{\boldsymbol{\mathfrak{x}},\boldsymbol{\hat{x}}\}I\right)\boldsymbol{w}\tag{13}$$

is an SOS polynomial where ℝ +, and

where the matrices , , , , , , , , are given in (5). Then, the dynamic output feedback law (5) globally asymptotically stabilizes the origin of the system (4). **Proof.** According to Theorem 6, the function

$$U\_1(x) = Z(x)^T P\_1^{-1} Z(x)$$

is the Lyapunov function of the closed-loop system of (4) with the state feedback law

$$u = k(x) = K(x)P\_1^{-1}Z(x).$$

Then, to consider a dynamic output feedback law in the form of (5), we consider a function given by

$$V\left(\mathbf{x}, \hat{\mathbf{x}}\right) = U\_1\left(\mathbf{x}\right) + U\_2\left(\mathbf{x}, \hat{\mathbf{x}}\right),\tag{14}$$

where the function is given by

$$U\_2\left(\mathbf{x}, \hat{\mathbf{x}}\right) = \left(k\left(\mathbf{x}\right) - \hat{\mathbf{x}}\right)^{\top} P\_2\left(k\left(\mathbf{x}\right) - \hat{\mathbf{x}}\right).$$

Then, the time derivative of function (14) along the trajectory of the closed-loop system (6) is given as

$$\mathcal{V}(\mathbf{x}, \hat{\mathbf{x}}) = U\_1(\mathbf{x}) + U\_2(\mathbf{x}, \hat{\mathbf{x}}),$$

where

(12)

(13)

(14)

are SOS polynomials, where and *M*(*x*) is given as 10. Further, suppose that there exist a

where the matrices , , , , , , , , are given in (5). Then, the dynamic output feedback law (5) globally asymptotically stabilizes the origin of the system (4).

is the Lyapunov function of the closed-loop system of (4) with the state feedback law

Then, to consider a dynamic output feedback law in the form of (5), we consider a function

+, and

, and an SOS polynomial such that

( + )×(+)

symmetric matrix 2:ℝ

is an SOS polynomial where ℝ

328 Nonlinear Systems - Design, Analysis, Estimation and Control

**Proof.** According to Theorem 6, the function

where the function is given by

given by

$$\begin{split} U\_{1}(\mathbf{x}) &= Z(\mathbf{x})^{\top} P\_{1}^{-1} Z(\mathbf{x}) + Z(\mathbf{x})^{\top} P\_{1}^{-1} Z(\mathbf{x}) \\ &= Z(\mathbf{x})^{\top} \operatorname{He} \left( P\_{1}^{-1} M\left( \mathbf{x} \right) \left\{ A(\mathbf{x}) + B\left( \mathbf{x} \right) D\_{c} \left( \hat{\mathbf{x}}, \mathbf{y} \right) C(\mathbf{x}) \right\} \right) Z(\mathbf{x}) \\ &+ \hat{\mathbf{x}}^{\top} C\_{c}(\hat{\mathbf{x}}, \mathbf{y})^{\top} B(\mathbf{x})^{\top} M(\mathbf{x})^{\top} P\_{1}^{-1} Z(\mathbf{x}) \\ &+ Z(\mathbf{x})^{\top} P\_{1}^{-1} M\left( \mathbf{x} \right) B\left( \mathbf{x} \right) C\_{c}(\hat{\mathbf{x}}, \mathbf{y}) \hat{\mathbf{x}}, \end{split} \tag{15}$$

and

(16)

Therefore, the time derivative of the function , along the solution of system (6) is given as

$$\begin{split} \dot{V}'(\mathbf{x}, \hat{\mathbf{x}}) &= \dot{U}\_{\uparrow}(\mathbf{x}) + \dot{U}\_{\downarrow}(\mathbf{x}, \hat{\mathbf{x}}) \\ &= Z(\mathbf{x})^{\top} \Lambda\_{11} \left( \mathbf{x}, \hat{\mathbf{x}} \right) Z \left( \mathbf{x} \right) + Z(\mathbf{x})^{\top} \Lambda\_{12} \left( \mathbf{x}, \hat{\mathbf{x}} \right) \hat{\mathbf{x}} + \hat{\mathbf{x}}^{\top} \Lambda\_{12} \left( \mathbf{x}, \hat{\mathbf{x}} \right)^{\top} Z \left( \mathbf{x} \right) \\ &+ \hat{\mathbf{x}}^{\top} \Lambda\_{22} \left( \mathbf{x}, \hat{\mathbf{x}} \right) \hat{\mathbf{x}} \\ &= \left[ Z(\mathbf{x})^{\top} \qquad \hat{\mathbf{x}}^{\top} \right] \begin{bmatrix} \Lambda\_{11} \left( \mathbf{x}, \hat{\mathbf{x}} \right) & \Lambda\_{12} \left( \mathbf{x}, \hat{\mathbf{x}} \right) \\ \Lambda\_{12} \left( \mathbf{x}, \hat{\mathbf{x}} \right)^{\top} & \Lambda\_{22} \left( \mathbf{x}, \hat{\mathbf{x}} \right) \end{bmatrix} \begin{bmatrix} Z \left( \mathbf{x} \right) \\ \mathbf{x} \end{bmatrix}. \end{split} \tag{17}$$

Then, condition (13) of the theorem and Theorem 4 imply that

$$
\begin{bmatrix}
\Lambda\_{11}\left(\mathbf{x}, \hat{\mathbf{x}}\right) & \Lambda\_{12}\left(\mathbf{x}, \hat{\mathbf{x}}\right) \\
\Lambda\_{12}\left(\mathbf{x}, \hat{\mathbf{x}}\right)^{\mathsf{T}} & \Lambda\_{22}\left(\mathbf{x}, \hat{\mathbf{x}}\right)
\end{bmatrix} < 0, \quad \forall \left(\mathbf{x}, \hat{\mathbf{x}}\right) \in \mathbb{R}^{n\_{\boldsymbol{\epsilon}}} \times \mathbb{R}^{n\_{\boldsymbol{\epsilon}}}.\tag{18}
$$

From (17) and (18), we can conclude that ˙ , is negative definite. Therefore, according to Theorem 1, we can conclude that the origin of the closed-loop system is globally asymptotically stable. This completes the proof.

When we design the dynamic output feedback law (5) according to Theorem 6, we first solve the SOS decomposition of condition (12) to find the matrix *P*1. Then, if we can obtain the feasible solutions of the matrix *P*1 and the function *K*(*x*) satisfying condition (12), we try to find the matrix functions , , , , , , , , the matrix *P*2, and the SOS polynomial 3 > 0 satisfying condition (13). At this time, because the decision variables do not enter in (13) linearly, we set *P*2 = *I* in general. Then, we can consider the SOS decomposition for (13). If we can find the feasible solution of condition (13), we will obtain the stabilizing feedback laws in the form of (5).

**Remark 2.** The condition of (12) in Theorem 6 corresponds to the condition of (9) in Theorem 5. Note that the matrix *P*1 in Theorem 6 is a constant matrix, although the matrix *P*(*x*) in Theorem 5 is the function of *x*. This is due to the fact that the inverse of the matrix *P*1 appears in (16). If the matrix *P*1 is the polynomial matrix in Theorem 6, we cannot employ the SOS decomposition. Therefore, we limit ourselves to the case of the constant matrices in Theorem 6.

## **6. Numerical examples of dynamic output feedback stabilization**

#### **6.1. Numerical example 1**

This section shows some numerical examples of the dynamic output feedback stabilization by the proposed method shown in Section 5.

We show the first example of the stabilization. Consider a system given by

$$\begin{aligned} \dot{\boldsymbol{x}}\_1 &= 0.5 \,\boldsymbol{x}\_1 - 0.1 \boldsymbol{x}\_1^3 + \boldsymbol{u}, \\ \dot{\boldsymbol{x}}\_2 &= \boldsymbol{x}\_1^2 - \boldsymbol{x}\_2, \\ \boldsymbol{y} &= \boldsymbol{x}\_1, \end{aligned} \tag{19}$$

where *x* = (*x*1, *x*2) *<sup>T</sup>* is the state, *y* ∊ ℝ is the output, and *u* ∊ ℝ is the input. In order to design a dynamic output feedback law for the stabilization of system (19) based on the result presented in the previous section, we choose *Z*(*x*)=(*x*1, *x*2) *<sup>T</sup>*. Then, we have the expression of system (19) in the form of (4), where

$$A(\boldsymbol{\chi}) = \begin{pmatrix} 0.5 - 0.1\boldsymbol{x}\_1^2 & 0\\ \boldsymbol{x}\_1 & -1 \end{pmatrix}, \quad B(\boldsymbol{\chi}) = \begin{pmatrix} 1\\ 0 \end{pmatrix}, \quad C(\boldsymbol{\chi}) = \begin{pmatrix} 1 & 0 \end{pmatrix}.$$

We consider the output feedback stabilization of system (19) using the dynamic feedback law (5). We consider a low-dimensional dynamic feedback, and we assume that = 1. According to Theorem 6, by choosing *P*2 = *I*, we obtained the matrix *P*1 and the function *K*(*x*) by solving the SOS decomposition of (12) using YALMIP. We consider the function *K*(*x*) with zero degree. The obtained matrix *P*1 and the function *K*(*x*) are given as

(18)

(19)

From (17) and (18), we can conclude that ˙ , is negative definite. Therefore, according to Theorem 1, we can conclude that the origin of the closed-loop system is globally asymptotically

When we design the dynamic output feedback law (5) according to Theorem 6, we first solve the SOS decomposition of condition (12) to find the matrix *P*1. Then, if we can obtain the feasible solutions of the matrix *P*1 and the function *K*(*x*) satisfying condition (12), we try to find the matrix functions , , , , , , , , the matrix *P*2, and the SOS polynomial 3 > 0 satisfying condition (13). At this time, because the decision variables do not enter in (13) linearly, we set *P*2 = *I* in general. Then, we can consider the SOS decomposition for (13). If we can find the feasible solution of condition (13), we will obtain the stabilizing feedback laws in

**Remark 2.** The condition of (12) in Theorem 6 corresponds to the condition of (9) in Theorem 5. Note that the matrix *P*1 in Theorem 6 is a constant matrix, although the matrix *P*(*x*) in Theorem 5 is the function of *x*. This is due to the fact that the inverse of the matrix *P*1 appears in (16). If the matrix *P*1 is the polynomial matrix in Theorem 6, we cannot employ the SOS decomposition.

This section shows some numerical examples of the dynamic output feedback stabilization by

dynamic output feedback law for the stabilization of system (19) based on the result presented

*<sup>T</sup>* is the state, *y* ∊ ℝ is the output, and *u* ∊ ℝ is the input. In order to design a

*<sup>T</sup>*. Then, we have the expression of system (19)

Therefore, we limit ourselves to the case of the constant matrices in Theorem 6.

**6. Numerical examples of dynamic output feedback stabilization**

We show the first example of the stabilization. Consider a system given by

stable. This completes the proof.

330 Nonlinear Systems - Design, Analysis, Estimation and Control

the form of (5).

**6.1. Numerical example 1**

where *x* = (*x*1, *x*2)

in the form of (4), where

the proposed method shown in Section 5.

in the previous section, we choose *Z*(*x*)=(*x*1, *x*2)

$$\begin{aligned} P\_1 &= \begin{bmatrix} 1.2306 \times 10^{-2} & -9.8824 \times 10^{-11} \\ -9.8824 \times 10^{-11} & 5.2061 \times 10^{-2} \end{bmatrix}, \\ K\left(\boldsymbol{x}\right) &= \begin{bmatrix} -6.9660 \times 10^{-3} & -4.9775 \times 10^{-6} \end{bmatrix}. \end{aligned}$$

**Figure 1.** Time responses of *x*, , and u of (19) with dynamic output feedback law (5) with degree zero one.

Then, by using *P*1 and *K*(*x*), we found the feasible solution , , , , , , , , which are two degree, to the SOS decomposition of condition (13). Therefore, we obtain the dynamic output feedback laws that stabilizes system (19), given by

(20)

**Figure 1** shows the time responses of the state variables *x*(*t*), and *u*(*t*) of the closed-loop system (19) with the designed dynamic output feedback (20). The initial values are chosen as *x*(0) = (3,-1),and 0 = 0.5. In **Figure 1**, the states *x*(*t*) and converge to the origin.

Then, we also obtain a dynamic output feedback control law in the case where the elements of *K*(*x*) are degree zero, and the elements of , , , , , , and , are degree three with respect to and *y*. Again, by solving the SOS decomposition following Theorem 6, we obtain the value of the matrix *P*1 and the function *K*(*x*) as same as above.

We also obtain the values of , , , , , , and , as

$$\begin{aligned} A\_c(\hat{\mathbf{x}}, \mathbf{y}) &= -0.04345107574 - 1.9588 \times 10^{-7} \,\mathrm{y} - 0.044197 \,\mathrm{y}^2 - 8.8208 \times 10^{-8} \,\hat{\mathbf{x}} \\ &- 4.5391 \times 10^{-7} \,\mathrm{y} \,\hat{\mathbf{x}} - 0.043558 \,\hat{\mathbf{x}}^2 - 2.1533 \times 10^{-5} \,\mathrm{y}^3 + 3.9213 \times 10^{-9} \,\mathrm{y}^2 \mathbf{\hat{x}} \\ &- 1.5638 \times 10^{-5} \,\mathrm{y} \,\hat{\mathbf{x}}^2 - 2.1580 \times 10^{-9} \,\hat{\mathbf{x}}^3 \end{aligned}$$

$$\begin{split} B\_{\rm c} \{ \hat{\mathbf{x}}, \mathbf{y} \} &= -9.486666512 \times 10^{-5} + 2.8266 \times 10^{-7} \,\mathrm{y} + 1.2691 \times 10^{-4} \,\mathrm{y}^2 - 4.6792 \times 10^{-7} \,\mathrm{\hat{x}} \\ &- 8.6168 \times 10^{-7} \,\mathrm{y} \,\mathrm{\hat{x}} + 2.8204 \times 10^{-5} \,\mathrm{\hat{x}}^2 - 0.0031033 \,\mathrm{y}^3 + 5.6498 \times 10^{-7} \,\mathrm{y}^2 \,\mathrm{\hat{x}} \\ &- 0.0022537 \,\mathrm{y} \,\mathrm{\hat{x}}^2 - 3.1099 \times 10^{-7} \,\mathrm{\hat{x}}^3, \end{split}$$


**Figure 2.** Time responses of *x*, , and u of (19) with dynamic output feedback law (5) with degree zero one.

The obtained feedback control law also stabilizes system (19). **Figure 2** shows the time responses of the state , and the input *u*(*t*) of the closed-loop systems with the initial values *x*(0) = (3,-1), and 0 = 0.5. The state converges to the origin, and the value of *u*(*t*) also converges to zero.

## **6.2. Numerical example 2**

**Figure 1** shows the time responses of the state variables *x*(*t*), and *u*(*t*) of the closed-loop system (19) with the designed dynamic output feedback (20). The initial values are chosen as

Then, we also obtain a dynamic output feedback control law in the case where the elements of *K*(*x*) are degree zero, and the elements of , , , , , , and , are degree three with respect to and *y*. Again, by solving the SOS decomposition following Theorem 6,

*x*(0) = (3,-1),and 0 = 0.5. In **Figure 1**, the states *x*(*t*) and converge to the origin.

**Figure 2.** Time responses of *x*, , and u of (19) with dynamic output feedback law (5) with degree zero one.

we obtain the value of the matrix *P*1 and the function *K*(*x*) as same as above.

We also obtain the values of , , , , , , and , as

332 Nonlinear Systems - Design, Analysis, Estimation and Control

We consider the following example, which models a circuit with negative-resistance oscillator, taken from reference [17] and modified. Consider a system given by

$$\begin{aligned} \dot{\varkappa}\_1 &= \varkappa\_2, \\ \dot{\varkappa}\_2 &= -\varkappa\_1 + \varkappa\_2 - \frac{1}{3}\varkappa\_2^3 + u, \\ y &= \varkappa\_2, \end{aligned} \tag{21}$$

where *x* = (*x*1, *x*2) *<sup>T</sup>* is the state, *u* ∊ ℝ is the input, and y is the output. To design the dynamic output feedback laws, we express system (21) of form (4) as

$$
\begin{bmatrix}
\dot{\mathbf{x}}\_1 \\
\dot{\mathbf{x}}\_2
\end{bmatrix} = \begin{bmatrix}
0 & 1 \\
1 & 1
\end{bmatrix} \begin{bmatrix}
\mathbf{x}\_2
\end{bmatrix} + \begin{bmatrix}
0 \\
1
\end{bmatrix} \boldsymbol{\mu}\_{\prime\prime}
$$

$$
\mathbf{y} = \begin{bmatrix}
0 & 1
\end{bmatrix} \begin{bmatrix}
\mathbf{x}\_2
\end{bmatrix}.
$$

Following the design procedure in the previous section, we design the dynamic feedback control law with = 1.First, we obtain the constant matrix *P*1 and the polynomial matrix *K*(*x*) with degree zero. The matrix *P*1 and *K*(*x*) with zero degree are obtained as

$$\begin{array}{rcl} P\_1 = \begin{bmatrix} 0.0155674 & 0.0012554 \\ 0.0012554 & 0.0155125 \end{bmatrix} \\ K(\mathbf{x}) = \begin{bmatrix} -0.0013102 & -0.0167686 \end{bmatrix} \end{array}$$

Then, we solve the SOS decomposition (13) to find the matrices , , , , , , and , with degree one. By choosing *P*2 = *I*, the feasible solutions are obtained as

$$\begin{split} A\_{\boldsymbol{\varepsilon}} \left( \hat{\boldsymbol{\boldsymbol{x}}}, \boldsymbol{y} \right) &= -0.1675653629 - 9.5745 \times 10^{-9} \, \boldsymbol{y} + 8.1304 \times 10^{-11} \, \hat{\boldsymbol{\boldsymbol{x}}}, \\ B\_{\boldsymbol{\varepsilon}} \left( \hat{\boldsymbol{\boldsymbol{x}}}, \boldsymbol{y} \right) &= -0.1582768369 + 5.2303 \times 10^{-8} \, \boldsymbol{y} - 3.9643 \times 10^{-10} \, \hat{\boldsymbol{\boldsymbol{x}}}, \\ C\_{\boldsymbol{\varepsilon}} \left( \hat{\boldsymbol{\boldsymbol{x}}}, \boldsymbol{y} \right) &= 6.655333476 \times 10^{-5} + 3.0370 \times 10^{-11} \, \boldsymbol{y} + 3.9639 \times 10^{-12} \, \hat{\boldsymbol{\boldsymbol{x}}}, \\ D\_{\boldsymbol{\varepsilon}} \left( \hat{\boldsymbol{\boldsymbol{x}}}, \boldsymbol{y} \right) &= -1.081399695 - 4.5067 \times 10^{-11} \, \boldsymbol{y} + 4.1856 \times 10^{-12} \, \hat{\boldsymbol{\boldsymbol{x}}}. \end{split}$$

**Figure 3** shows the time responses of the states *x*, and the input u of the closed-loop systems. The figure shows that the states *x* and converge to the origin. Also, the figure shows that the input values converge to zero as the states converge to the origin.

**Figure 3.** Time responses of *x*, , and u of (21) with dynamic output feedback law (5) with degree zero one.
