**4.** ð Þ� *α***,** *β* **pythagorean fuzzy numbers controller**

Now we continue to study the feedback control and stability of pythagorean fuzzy descriptor systems according to the traditional research path of the control systems.

Suppose.

Rule *<sup>i</sup>*: if *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *Pi* <sup>1</sup>ð Þ *<sup>x</sup>*1ð Þ*<sup>t</sup>* and ... and *xn*ð Þ*<sup>t</sup>* is *<sup>P</sup><sup>i</sup> <sup>n</sup>*ð Þ *xn*ð Þ*t* , then.

$$u(\mathbf{x}(t)) = \sum\_{i=1}^{r} h\_i(\mathbf{x}(t)) \mathbf{G}\_i \mathbf{x}(t) \tag{3}$$

where *Gi*(*i* = 1,2,..., *r*) are the state feedback-gains matrices.

$$h\_i(\boldsymbol{\kappa}(t)) = \frac{h\_{i(a,\boldsymbol{\beta})}(\boldsymbol{\kappa}(t))}{\sum\_{i=1}^r h\_{i(a,\boldsymbol{\beta})}(\boldsymbol{\kappa}(t))}, i = 1, 2, 3, \dots, r;$$

where

$$h\_{i(a,\beta)}(\mathbf{x}(t)) = \begin{cases} h\_i^1(\mathbf{x}(t)) & \text{when } h\_i^1(\mathbf{x}(t)) \ge a \text{ or } h\_i^2(\mathbf{x}(t)) \le \beta\\ 0 & \text{else} \end{cases}, a + \beta \le 1, i = 1, 2, 3, \cdots, r;$$

$$h\_i^1(\mathbf{x}(t)) = \frac{\mu\_{P'}(\mathbf{x}(t))}{\sum\_{i=1}^r \mu\_{P'}(\mathbf{x}(t))}, h\_i^2(\mathbf{x}(t)) = \frac{\nu\_{p'}(\mathbf{x}(t))}{\sum\_{i=1}^r \nu\_{p'}(\mathbf{x}(t))},$$

where *h*<sup>1</sup> *<sup>i</sup>*ð Þ *x t*ð Þ and *<sup>h</sup>*<sup>2</sup> *<sup>i</sup>*ð Þ *x t*ð Þ are respectively positive and negative membership functions.

$$\sum\_{i=1}^{r} h\_{i1}(\mathbf{x}(t)) = \mathbf{1}, \sum\_{i=1}^{r} h\_{i2}(\mathbf{x}(t)) = \mathbf{1};$$

$$\mu\_{p^i} \left(\mathbf{x}\_j(t)\right) = \prod\_{j=1}^{r} \mu\_{p^i\_j} \left(\mathbf{x}\_j(t)\right), \nu\_{p^i} \left(\mathbf{x}\_j(t)\right) = \prod\_{j=1}^{r} \nu\_{p^i\_j} \left(\mathbf{x}\_j(t)\right);$$

*μP j <sup>i</sup> <sup>x</sup> <sup>j</sup>*ð Þ*<sup>t</sup>* � � and *<sup>ν</sup>P<sup>i</sup> j <sup>x</sup> <sup>j</sup>*ð Þ*<sup>t</sup>* � � is the membership function value of *<sup>x</sup> <sup>j</sup>*ð Þ*<sup>t</sup>* that belongs and does not belong to the intuitionistic fuzzy numbers set *P<sup>i</sup> j :*

If we take (3) into (1, 2), we can get.

$$E\dot{\mathbf{x}}(t) = \sum\_{i=1}^{r} \sum\_{j=1}^{r} h\_i(\mathbf{x}(t)) h\_j(\mathbf{x}(t)) \left( A\_i + B\_i G\_j \right) \mathbf{x}(t) \tag{4}$$

$$\mathbf{y}(t) = \sum\_{i=1}^{r} \sum\_{j=1}^{r} h\_i(\mathbf{x}(t)) h\_j(\mathbf{x}(t)) \left(\mathbf{C}\_i + D\_i \mathbf{G}\_j\right) \mathbf{x}(t) \tag{5}$$

The system stability is guaranteed by determining the feedback gains *G <sup>j</sup>*.

Basic LMI-based stability conditions guaranteeing the stability of the above control system in the form of (4, 5) are given in the following theorem.

Theorem 4.1 The system (3) is asymptotically stable, if there exist matrices *<sup>N</sup> <sup>j</sup>* <sup>∈</sup> *Rm*�*<sup>n</sup>* (*<sup>j</sup>* = 1,2,3,..., *<sup>r</sup>*) and *<sup>K</sup>* <sup>¼</sup> *<sup>K</sup><sup>T</sup>* <sup>∈</sup> *<sup>R</sup><sup>n</sup>*�*<sup>n</sup>* such that the following LMIs are satisfied:

$$K > 0\tag{6}$$

$$E^T K = K^T E \ge \mathbf{0} \tag{7}$$

$$\mathbf{Q}\_{\dot{\mathbf{y}}} = \mathbf{A}\_{i}\mathbf{K}^{-1} + \mathbf{K}^{-1}\mathbf{A}\_{i}^{T} + \mathbf{B}\_{i}\mathbf{N}\_{\dot{\mathbf{y}}} + \mathbf{N}\_{\dot{\mathbf{y}}}^{T}\mathbf{B}\_{i}^{T} < \mathbf{0} \forall i, \mathbf{j} \tag{8}$$

where the feedback gains are defined as *G <sup>j</sup>* ¼ *N jK* for all *j:* Proof: Considering the quadratic Lyapunov function.

$$V(\mathfrak{x}(t)) = \mathfrak{x}^T(t)E^T\mathbb{K}\mathfrak{x}(t),$$

where 0 <sup>&</sup>lt;*<sup>K</sup>* <sup>¼</sup> *<sup>K</sup><sup>T</sup>* <sup>∈</sup>*R<sup>n</sup>*�*<sup>n</sup>*. then

$$\begin{split} \dot{V}(\mathbf{x}(t)) &= \dot{\mathbf{x}}^T(t)\mathbf{E}^T\mathbf{K}\mathbf{x}(t) + \mathbf{x}^T(t)\mathbf{E}^T\mathbf{K}\dot{\mathbf{x}}(t) = (\mathbf{E}\dot{\mathbf{x}}(t))^T\mathbf{K}\mathbf{x}(t) + \mathbf{x}^T(t)\mathbf{K}^T(\mathbf{E}\dot{\mathbf{x}}(t)) \\ &= \sum\_{i=1}^r \sum\_{j=1}^r h\_i h\_j \mathbf{x}^T(t)\mathbf{K}\mathbf{K}^{-1}\Big{\{\mathbf{A}^T\boldsymbol{\_i}\mathbf{K} + \mathbf{K}^T\mathbf{N}\_j^T\mathbf{B}\_i^T\mathbf{K} + \mathbf{K}^T\mathbf{A}\_i + \mathbf{K}^T\mathbf{B}\_i\mathbf{N}\_j\}\} \mathbf{K}^{-1}\mathbf{K}\mathbf{x}(t), \end{split}$$

*(α, β)*�*Pythagorean Fuzzy Numbers Descriptor Systems DOI: http://dx.doi.org/10.5772/intechopen.95007*

let *Z* = *Kx*(*t*), then

$$\begin{split} \dot{V}(\mathbf{x}(t)) &= \sum\_{i=1}^{r} \sum\_{j=1}^{r} h\_i h\_j \mathbf{x}^T(t) \mathbf{K} \mathbf{K}^{-1} \Big\{ \mathbf{A}^T \boldsymbol{i} \mathbf{K} + \mathbf{K} \mathbf{N}\_j^T \mathbf{B}\_i^T \mathbf{K} + \mathbf{K} \mathbf{A}\_i + \mathbf{K} \mathbf{B}\_i \mathbf{N}\_j \mathbf{K} \Big\} \|\mathbf{K}^{-1} \mathbf{K} \mathbf{x}(t) \|\_{2}^2 \\ &= \sum\_{i=1}^{r} \sum\_{j=1}^{r} h\_i h\_j \mathbf{Z} \Big( \mathbf{K}^{-1} \mathbf{A}^T \boldsymbol{i} + \mathbf{N}\_j^T \mathbf{B}\_i^T + \mathbf{A}\_i \mathbf{K}^{-1} + \mathbf{B}\_i \mathbf{N}\_j \Big) \mathbf{Z}. \end{split}$$

As *Qij* <sup>¼</sup> *AiK*�<sup>1</sup> <sup>þ</sup> *<sup>K</sup>*�<sup>1</sup> *Ai <sup>T</sup>* <sup>þ</sup> *BiN <sup>j</sup>* <sup>þ</sup> *<sup>N</sup> <sup>j</sup> TBi <sup>T</sup>* <0, so the system (3) is asymptotically stable.

### **5. Simulation example**

**Example 5.1:** Considering an inverted pendulum, subject to parameter uncertainties [12–15] as the nonlinear plant to be controlled. The dynamic equation for the inverted pendulum is given by.

$$\ddot{\theta}(t) = \frac{\text{g}\sin\left(\theta(t)\right) - am\_pL\dot{\theta}(t)^2\sin\left(2\theta(t)\right)/2 - a\cos\left(\theta(t)\right)\mu(t)}{4L/3 - am\_pL\cos^2\left(\theta(t)\right)}$$

Where *<sup>θ</sup>*ð Þ*<sup>t</sup>* is the angular displacement of the pendulum, *<sup>g</sup>* = 9.8 m/s<sup>2</sup> is the acceleration due to gravity, *mp* ∈[*mp*min ,*mp*max ] = [2,3]*kg* is the mass of the pendulum, *Mc* ∈ ½ � *M*min, *M*max = [8, 12].

*Kg* is the mass of the cart, *a* ¼ 1*= mp* þ *Mc* � �, 2 *L* = 1 m is the length of the pendulum, and *u t*ð Þis the force (in newtons) applied to the cart. The inverted pendulum is considered working in the operating domain characterized by *<sup>x</sup>*<sup>1</sup> <sup>¼</sup> *<sup>θ</sup>*ð Þ*<sup>t</sup>* <sup>∈</sup>½ � �5*π=*12, 5*π=*<sup>12</sup> and *<sup>x</sup>*<sup>2</sup> <sup>¼</sup> \_ *θ*ð Þ*t* ∈[�5,5].

Rule 1: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>1</sup> 1, *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>1</sup> 2, then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
10.0078 & 0
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
0 \\
\end{pmatrix} \mu(t);
$$

Rule 2: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>2</sup> 1, *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>2</sup> 2, then

$$
\begin{pmatrix}
\dot{\boldsymbol{x}}\_1(t) \\
\dot{\boldsymbol{x}}\_2(t)
\end{pmatrix} = \begin{pmatrix}
\mathbf{0} & \mathbf{1} \\
\mathbf{10.0078} & \mathbf{0}
\end{pmatrix} \begin{pmatrix}
\boldsymbol{x}\_1(t) \\
\boldsymbol{x}\_2(t)
\end{pmatrix} + \begin{pmatrix}
\mathbf{0} \\
\end{pmatrix} \boldsymbol{\mu}(t);
$$

Rule 3: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>3</sup> 1, *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>3</sup> 2, then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
18.4800 & 0
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
0 \\
\end{pmatrix} \mu(t);
$$

Rule 4: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>4</sup> <sup>1</sup> , *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>M</sup>*<sup>4</sup> <sup>2</sup> , then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
\mathbf{0} & \mathbf{1} \\
\mathbf{18.4800} & \mathbf{0}
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
\mathbf{0} \\
\end{pmatrix} \mu(t);
$$

Next, according to the ideas based on the principles of interpolation and interval coverage, we firstly change the interval-valued T-S fuzzy model of inverted

pendulum into the special (*α*,*β*)- pythagorean fuzzy descriptor systems of inverted pendulum as follows.

Rule 1: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>1</sup> <sup>1</sup>ð Þ *<sup>x</sup>*1ð Þ*<sup>t</sup>* , *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>1</sup> <sup>2</sup>ð Þ *x*2ð Þ*t* , then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
10.0078 & 0
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
0 \\
\end{pmatrix} \mu(t);
$$

Rule 2: If *<sup>x</sup>*1ð Þ*<sup>k</sup>* is *<sup>P</sup>*<sup>2</sup> <sup>1</sup>ð Þ *<sup>x</sup>*1ð Þ*<sup>t</sup>* , *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>2</sup> <sup>2</sup>ð Þ *x*2ð Þ*t* , then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
10.0078 & 0
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
0 \\
\end{pmatrix} \mu(t);
$$

Rule 3: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>3</sup> <sup>1</sup>ð Þ *<sup>x</sup>*1ð Þ*<sup>k</sup>* , *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>3</sup> <sup>2</sup>ð Þ *x*2ð Þ*t* , then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
18.4800 & 0
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
0 \\
\end{pmatrix} \mu(t);
$$

Rule 4: If *<sup>x</sup>*1ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>4</sup> <sup>1</sup> ð Þ *<sup>x</sup>*1ð Þ*<sup>k</sup>* , *<sup>x</sup>*2ð Þ*<sup>t</sup>* is *<sup>P</sup>*<sup>4</sup> <sup>2</sup> ð Þ *x*2ð Þ*t* , then

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t) \\
\dot{\varkappa}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
18.4800 & 0
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t) \\
\varkappa\_2(t)
\end{pmatrix} + \begin{pmatrix}
0 \\
\end{pmatrix} \mu(t);
$$

According to the theorem 4.1, we can get.

*<sup>K</sup>* <sup>¼</sup> <sup>1</sup>*=*<sup>7</sup> �1*=*<sup>7</sup> �1*=*7 8*=*7 � �,*K*�<sup>1</sup> <sup>¼</sup> 8 1 1 1 � �, *<sup>N</sup>*<sup>1</sup> <sup>¼</sup> *<sup>N</sup>*<sup>3</sup> <sup>¼</sup> ð Þ 100 100 , *<sup>N</sup>*<sup>2</sup> <sup>¼</sup> *<sup>N</sup>*<sup>4</sup> <sup>¼</sup> ð Þ 1000 1000 ,

So the above (*α*,*β*)-pythagorean fuzzy descriptor systems of inverted pendulum is asymptotically stable, and the state feedback-gains matrices *G*<sup>1</sup> ¼ *G*<sup>3</sup> ¼ ð Þ 0 100 , *G*<sup>2</sup> ¼ *G*<sup>4</sup> ¼ ð Þ 0 1000 .

**The first case**, suppose *<sup>x</sup>*1ð Þ¼� <sup>0</sup> <sup>11</sup>*<sup>π</sup>* <sup>29</sup> ,*x*2ð Þ¼� 0 0*:*88,*α* ¼ 0*:*3, *β* ¼ 0*:*25, then take the variable *x*1ð Þ*t* as the main factor of the control, and according to **Table 2** we can control in three steps, i.e. *<sup>x</sup>*1ð Þ¼� <sup>0</sup> <sup>11</sup>*<sup>π</sup>* <sup>29</sup> ! *x*1ð Þ! *t*<sup>1</sup> *x*1ð Þ *t*<sup>2</sup> ≈ 0 and 0 <*t*<sup>1</sup> ≤*t*2.

When *<sup>x</sup>*1ð Þ¼� <sup>0</sup> <sup>11</sup>*<sup>π</sup>* <sup>29</sup> ,*x*2ð Þ¼� 0 0*:*88, and *α* ¼ 0*:*30, *β* ¼ 0*:*25, according to **Table 2** we can get *μP*<sup>1</sup> 1 ð Þ *x*1ð Þ 0 = *μP*<sup>2</sup> 1 ð Þ *x*1ð Þ 0 = 0.69, *νP*<sup>1</sup> 1 ð Þ *x*1ð Þ 0 = *νP*<sup>2</sup> 1 ð Þ *x*1ð Þ 0 = 0.72,*μP*<sup>3</sup> 1 ð Þ *x*1ð Þ 0 = *μP*<sup>4</sup> 1 ð Þ *x*1ð Þ 0 = 0, *νP*<sup>3</sup> 1 ð Þ *x*1ð Þ 0 = *νP*<sup>4</sup> 1 ð Þ *x*1ð Þ 0 = 1, *μP*<sup>1</sup> 2 ð Þ *x*2ð Þ 0 = *μP*<sup>2</sup> 2 ð Þ *x*2ð Þ 0 = 0.02,*νP*<sup>1</sup> 2 ð Þ *x*2ð Þ 0 = *νP*<sup>2</sup> 2 ð Þ *x*2ð Þ 0 = 1, *μP*<sup>3</sup> 2 ð Þ *x*2ð Þ 0 = *μP*<sup>4</sup> 2 ð Þ *x*2ð Þ 0 = 0.40, *νP*<sup>3</sup> 2 ð Þ *x*2ð Þ 0 = *νP*4 2 ð Þ *x*2ð Þ 0 = 0.92, noteworthy, *μP*<sup>1</sup> 1 ð Þ *x*1ð Þ 0 +*νP*<sup>1</sup> 1 ð Þ *x*1ð Þ 0 = *μP*<sup>2</sup> 1 ð Þ *x*1ð Þ 0 +*νP*<sup>2</sup> 1 ð Þ *x*1ð Þ 0 = 1.41 > 1, *μP*<sup>3</sup> 2 ð Þ *x*2ð Þ 0 + *νP*<sup>3</sup> 2 ð Þ *x*2ð Þ 0 = *μP*<sup>4</sup> 2 ð Þ *x*2ð Þ 0 +*νP*<sup>4</sup> 2 ð Þ *x*2ð Þ 0 = 1.32 > 1. Then according to Definition 2.1, taking it one step further, we can get *h*<sup>1</sup> <sup>1</sup> = 0.49, *<sup>h</sup>*<sup>2</sup> <sup>1</sup> = 0.22, *<sup>h</sup>*<sup>1</sup> 2 = 0.49, *h*<sup>2</sup> <sup>2</sup> = 0.22, *<sup>h</sup>*<sup>1</sup> <sup>3</sup> = 0.01, *<sup>h</sup>*<sup>2</sup> <sup>3</sup> = 0.28, *<sup>h</sup>*<sup>1</sup> <sup>4</sup> = 0.01, *<sup>h</sup>*<sup>2</sup> <sup>4</sup> = 0.28, so *h*<sup>1</sup> = 0.5, *h*<sup>2</sup> = 0.5, *h*<sup>3</sup> = 0, *h*<sup>4</sup> = 0, according to 4.2, so the overall fuzzy model of the (0.30,0.25)- pythagorean fuzzy descriptor systems is.

$$
\begin{pmatrix}
\dot{\mathbf{x}}\_1(t) \\
\dot{\mathbf{x}}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 0.25 \\
2.5019 & -13.929
\end{pmatrix} \begin{pmatrix}
\mathbf{x}\_1(t) \\
\mathbf{x}\_2(t)
\end{pmatrix},
$$

$$
\text{The solution of the systems is } \mathbf{x}(t) = \begin{pmatrix}
0.2172 - 0.045\boldsymbol{t} - 1.0972\boldsymbol{\varepsilon}^{-13.93}
\end{pmatrix}.
$$

*(α, β)*�*Pythagorean Fuzzy Numbers Descriptor Systems DOI: http://dx.doi.org/10.5772/intechopen.95007*

When *x*1ð Þ 4 ≈ � 0*:*19103,*x*2ð Þ 4 ≈0*:*0372, and *α* ¼ 0*:*30, *β* ¼ 0*:*25, according to **Table 2**, we can get *μP*<sup>1</sup> 1 ð Þ *x*1ð Þ 4 = *μP*<sup>2</sup> 1 ð Þ *x*1ð Þ 4 = 0.03, *νP*<sup>1</sup> 1 ð Þ *x*1ð Þ 4 = *νP*<sup>2</sup> 1 ð Þ *x*1ð Þ 4 = 1,*μP*<sup>3</sup> 1 ð Þ *x*1ð Þ 4 = *μP*<sup>4</sup> 1 ð Þ *x*1ð Þ 4 = 0.20, *νP*<sup>3</sup> 1 ð Þ *x*1ð Þ 4 = *νP*<sup>4</sup> 1 ð Þ *x*1ð Þ 4 = 0.98, *μP*<sup>1</sup> 2 ð Þ *x*2ð Þ 4 = *μP*2 2 ð Þ *x*2ð Þ 4 = 0.50, *νP*<sup>1</sup> 2 ð Þ *x*2ð Þ 4 = *νP*<sup>2</sup> 2 ð Þ *x*2ð Þ 4 = 0.87, *μP*<sup>3</sup> 2 ð Þ *x*2ð Þ 4 = *μP*<sup>4</sup> 2 ð Þ *x*2ð Þ 4 = 0, *νP*3 2 ð Þ *x*2ð Þ 4 = *νP*<sup>4</sup> 2 ð Þ *x*2ð Þ 4 = 1, noteworthy, *μP*<sup>1</sup> 1 ð Þ *x*1ð Þ 4 +*νP*<sup>1</sup> 1 ð Þ *x*1ð Þ 4 = *μP*<sup>2</sup> 1 ð Þ *x*1ð Þ 4 + *νP*2 1 ð Þ *x*1ð Þ 4 = 1.03 > 1, *μP*<sup>1</sup> 2 ð Þ *x*2ð Þ 4 +*νP*<sup>1</sup> 2 ð Þ *x*2ð Þ 4 = *μP*<sup>2</sup> 2 ð Þ *x*2ð Þ 4 +*νP*<sup>2</sup> 2 ð Þ *x*2ð Þ 4 = 1.37 > 1.Then according to Definition 2.1, taking it one step further, we can get *h*<sup>1</sup> <sup>1</sup> = 0.49, *<sup>h</sup>*<sup>2</sup> 1 = 0.23, *h*<sup>1</sup> <sup>2</sup> = 0.49, *<sup>h</sup>*<sup>2</sup> <sup>2</sup> = 0.23, *<sup>h</sup>*<sup>1</sup> <sup>3</sup> = 0.01, *<sup>h</sup>*<sup>2</sup> <sup>3</sup> = 0.27, *<sup>h</sup>*<sup>1</sup> <sup>4</sup> = 0.01, *<sup>h</sup>*<sup>2</sup> <sup>4</sup> = 0.27, so *h*<sup>1</sup> = 0.50, *h*<sup>2</sup> = 0.50, *h*<sup>3</sup> = 0, *h*<sup>4</sup> = 0, then according to 4.2, so the overall fuzzy model of the (0.3,0.3)- pythagorean fuzzy descriptor systems is.

$$
\begin{pmatrix}
\dot{\varkappa}\_1(t-4) \\
\dot{\varkappa}\_2(t-4)
\end{pmatrix} = \begin{pmatrix}
0 & 0.25 \\
2.5019 & -13.929
\end{pmatrix} \begin{pmatrix}
\varkappa\_1(t-4) \\
\varkappa\_2(t-4)
\end{pmatrix},
$$

The solution of the systems is *x t*ðÞ¼ �0*:*<sup>19103</sup> <sup>þ</sup> <sup>0</sup>*:*25*<sup>t</sup>* <sup>0</sup>*:*<sup>03754</sup> � <sup>0</sup>*:*045*<sup>t</sup>* � <sup>0</sup>*:*00034*e*�13*:*93*<sup>t</sup>* 

When *x*1ð Þ 4*:*764 ≈ � 0*:*0000344, *x*2ð Þ 4*:*764 ≈0*:*00323, so the overall fuzzy model of the (0.30, 0.25)-pythagorean fuzzy descriptor systems is *E*-asymptotic stability. But it takes a shorter time (**Figure 1**).

**The second case**(interval-valued T-S fuzzy model of inverted pendulum), suppose *<sup>x</sup>*1ð Þ¼� <sup>0</sup> <sup>11</sup>*<sup>π</sup>* <sup>29</sup> ,*x*2ð Þ¼� 0 0*:*88, then take the variable *x*1ð Þ*t* as the main factor of the control, and according to **Table 1** we can control in three steps, i.e. *x*1ð Þ¼ 0 � <sup>11</sup>*<sup>π</sup>* <sup>29</sup> ! *x*1ð Þ!1 *x*1ð Þ*t* ≈0.

When *<sup>x</sup>*1ð Þ¼� <sup>0</sup> <sup>11</sup>*<sup>π</sup>* <sup>29</sup> <sup>∈</sup> � <sup>11</sup>*<sup>π</sup>* <sup>29</sup> , 0 , *<sup>x</sup>*2ð Þ¼� <sup>0</sup> <sup>0</sup>*:*88<sup>∈</sup> ½ Þ �0*:*88, 0 , and *<sup>λ</sup>*<sup>1</sup> <sup>¼</sup> *<sup>λ</sup>*<sup>2</sup> <sup>¼</sup> <sup>1</sup> 2 , according to **Table 1**, we can get *h*<sup>1</sup> = 0.26, *h*<sup>2</sup> = 0.58, *h*<sup>3</sup> = 0.05, *h*<sup>4</sup> = 0.11, according to theorem 4.1, so the overall interval-valued fuzzy model of the interval-valued fuzzy descriptor systems is

$$
\begin{pmatrix}
\dot{\boldsymbol{x}}\_1(t) \\
\dot{\boldsymbol{x}}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
\end{pmatrix} \begin{pmatrix}
\boldsymbol{x}\_1(t) \\
\boldsymbol{x}\_2(t)
\end{pmatrix},
$$

The solution of the systems is *x t*ðÞ¼ �1*:*2229*e*�0*:*82*<sup>t</sup>* <sup>þ</sup> <sup>0</sup>*:*0319*e*�59*:*1*<sup>t</sup>* <sup>1</sup>*:*0048*e*�0*:*82*<sup>t</sup>* � <sup>1</sup>*:*8848*e*�59*:*1*<sup>t</sup>* ;

When *x*1ð Þ¼� 1 0*:*5386 ∈½ Þ �0*:*5386, 0 , *x*2ð Þ¼ 1 0*:*4425 ∈½ Þ 0, 0*:*4425 , and *λ*<sup>1</sup> ¼ *<sup>λ</sup>*<sup>2</sup> <sup>¼</sup> <sup>1</sup> 2 , according to **Table 1**, we can get *h*<sup>1</sup> = 0.32, *h*<sup>2</sup> = 0.25, *h*<sup>3</sup> = 0.24, *h*<sup>4</sup> = 0.19, according to 4.2, the overall interval-valued fuzzy model of the interval-valued fuzzy descriptor systems is

**Figure 1.** *x*1ð Þ*t and x*2ð Þ*t under the (0.30,0.25)- pythagorean fuzzy descriptor systems.*


**Table 1.**

*The membership functions of the IT-2 T-S fuzzy model of inverted pendulum.*


#### **Table 2.**

*The membership functions and non-membership functions of (*α*,*β*)-pythagorean fuzzy descriptor systems of inverted pendulum.*

$$
\begin{pmatrix}
\dot{\mathbf{x}}\_1(t) \\
\dot{\mathbf{x}}\_2(t)
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
\end{pmatrix} \begin{pmatrix}
\mathbf{x}\_1(t) \\
\mathbf{x}\_2(t)
\end{pmatrix}
$$

$$
\text{The solution of the systems is } \mathbf{x}(t) = \begin{pmatrix}
0.4174e^{-0.78(t)} + 0.0251e^{-57.37(t)}
\end{pmatrix};
$$

,

*(α, β)*�*Pythagorean Fuzzy Numbers Descriptor Systems DOI: http://dx.doi.org/10.5772/intechopen.95007*

#### **Figure 2.**

*x*1ð Þ*t and x*2ð Þ*t under the IT2 T-S fuzzy descriptor systems.*

When *x*1ð Þ 9*:*6 ≈ � 0*:*0006, *x*2ð Þ 9*:*6 ≈ � 0*:*0005, so the IT2 T-S fuzzy descriptor system of the inverted pendulum will to be stable too.

Thus the stable control time of the (0.30,0.25)-pythagorean fuzzy descriptor systems of inverted pendulum is **4.836** second shorter than the interval-valued T-S fuzzy descriptor systems of the inverted pendulum (**Figure 2**).

**Remark 5.1:** In this way, the (0.30,0.25)-pythagorean fuzzy descriptor systems can get the better effect than the control effect of interval-valued T-S fuzzy model of inverted pendulum. It is easy to see that the (0.30,0.25)-pythagorean fuzzy descriptor systems has the best control, and can reduce the number of rules and thus reduce the amount of calculations.

In this way, it can get the better effect than the control effect of interval-valued T-S fuzzy model of inverted pendulum. Because the feedback more or less needs a little time, when the system carries out feedback instructions, but the time has gone, so the feedback that have been given are also lagging and out of date. ð Þ� *α*, *β* pythagorean fuzzy descriptor systems can be closer to the actual, and easy to control the error range. The new control method is more convenient and feasible!

#### **6. Conclusions**

In this paper, the new ð Þ� *α*, *β* pythagorean fuzzy descriptor systemsare firstly introduced, and more consistent with the human way of thinking and more likely to be set up and more convenient for popularization. The newð Þ� *α*, *β* pythagorean fuzzy descriptor systems is very simply and quickly. We can do not know the control principle, but we can directly achieve good control effect. The new theory can be studied in parallel to the basic framework of the original theories and easy to promote the old theories and achieve good results. In addition, we can judge the degree of weight in the control process according to the value of the positive and negative membership functions of the rules. By setting the values of*α*and*β*, we decide whether the rules will participate in the final calculation, thereby reducing the number of the rules and the calculation process, and improving the control efficiency and effectiveness. Otherwise, T-S fuzzy descriptor systems are the special examples of ð Þ� *α*, *β* pythagorean fuzzy descriptor systems. ð Þ� *α*, *β* pythagorean fuzzy controller and the stability ofð Þ� *α*, *β* pythagorean fuzzy descriptor systems are deeply researched. At last, a numbers example is given to show the corollaries are corrected.

But the theoretical part of the new systems need to be in-depth studied, and specific applications are also to be further developed. For example, ð Þ� *α*, *β* pythagorean fuzzy descriptor systems can also be used as the model of autonomous learning in order to establish intelligent control, and can be used well in unmanned driving in the future. Soð Þ� *α*, *β* pythagorean fuzzy descriptor systems is just to meet the reality requirements.
