**3. Controller formulation and stability proof**

*BLf* <sup>¼</sup> <sup>1</sup>

*BLf* <sup>¼</sup> *<sup>D</sup><sup>γ</sup>* <sup>1</sup>

*BLf* <sup>¼</sup> *<sup>D</sup><sup>γ</sup>*

The voltage and the current dynamics across the fractional capacitor is written

capacitor, *Cf* is the fractional capacitance, and *D*�*<sup>γ</sup>* represents the fractional integrator. Now, the current–voltage dynamics across the fractional capacitor are

> *Cf* <sup>¼</sup> *Ic Dγ Vc*

*BCf* ¼ 2*πf Cf BCf* <sup>¼</sup> <sup>2</sup>*π<sup>f</sup> Ic*

*BCf* <sup>¼</sup> <sup>2</sup>*πf D<sup>γ</sup> Ic*

*Dγ Vc*

> *Dγ Vc* � �

*BL*ð2*<sup>π</sup>* � <sup>2</sup>*<sup>α</sup>* <sup>þ</sup> sin 2ð Þ *<sup>α</sup>* Þ þ *<sup>π</sup>D<sup>γ</sup>*

The modified state space representation of the power system with the fractional

*BL*ð2*<sup>π</sup>* � <sup>2</sup>*<sup>α</sup>* <sup>þ</sup> sin 2ð Þ *<sup>α</sup>* Þ þ *<sup>π</sup>D<sup>γ</sup>*

� �

*π* � �

1

*Ic*. Here, *Vc* represents the voltage, *Ic* is the current across the

By multiplying the operator *D<sup>γ</sup>* on both hand sides of Eq. (13) yields

*Dγ*

*BTCR f* <sup>¼</sup> *<sup>D</sup><sup>γ</sup>*

The fractional capacitive susceptance is written as

0 @

*Dγ*

By combining Eqs. (10), (15) and (18), one obtains

*BSVCf* <sup>¼</sup> *<sup>D</sup><sup>γ</sup>*

order SVC-based system configurations written as

� �

*<sup>X</sup>* � *Dw*

<sup>2</sup> *<sup>D</sup><sup>γ</sup>*

Using Eq. (14), and by equating *D<sup>γ</sup>*

is written as

*Control Theory in Engineering*

as *Vc* <sup>¼</sup> <sup>1</sup> *C <sup>f</sup> D*�*<sup>γ</sup>*

> \_ *δ* ¼ *w <sup>w</sup>*\_ <sup>¼</sup> <sup>1</sup>

*<sup>V</sup>*\_ <sup>2</sup> <sup>¼</sup> <sup>1</sup> *τ*

*BL* <sup>¼</sup> <sup>1</sup>

*Dγ*

**272**

expressed as [31].

Simplifying Eq. (17) yields

*<sup>M</sup> Pm* � *<sup>V</sup>*1*V*2sin*<sup>δ</sup>*

*<sup>τ</sup> Vref* � *<sup>V</sup>*<sup>2</sup> � �

�*V*<sup>2</sup> 2 *<sup>X</sup>* <sup>þ</sup> *<sup>V</sup>*<sup>2</sup> 2*πf VL Dγ IL*

> 2*πf VL Dγ IL*

*BL*ð Þ 2*π* � 2*α* þ sin 2ð Þ *α*

" #

(13)

(14)

(16)

(18)

*BL*, Eq. (9) in fractional order sense

*<sup>π</sup>* (15)

A (17)

*BCf <sup>π</sup>* (19)

*BCf*

þ

*V*1*V*2cos*δ*

*<sup>X</sup>* � *kPd*

(20)

The control objective is to formulate a robust control system for the fractional order SVC-based power system that must behave as insensitive to the disturbances and uncertainties, thus ensure a stabilized voltage over the transmission lines. Thus, the control objective is the regulation of the bus voltage *V*<sup>2</sup> regardless of the variation in the generation side and load side. Let *<sup>e</sup>* <sup>¼</sup> *<sup>V</sup>*2*<sup>r</sup>* � *<sup>V</sup>*2, *<sup>e</sup>*\_ <sup>¼</sup> *<sup>V</sup>*\_ <sup>2</sup>*<sup>r</sup>* � *<sup>V</sup>*\_ <sup>2</sup> and the control signal is *BSVCf* <sup>¼</sup> *<sup>D</sup><sup>γ</sup> BSVC*, then the fractional order sliding manifold is defined as

$$\mathbf{S} = \mathbf{C}\_1 \mathbf{D}^{-a/2} \mathbf{e} + \mathbf{C}\_2 \dot{\mathbf{e}} \tag{21}$$

In the above expressions *e* represents the error signal, *V*2*<sup>r</sup>* is the reference signal and *V*\_ <sup>2</sup> represents the first derivative of the command signal. Differentiating Eq. (21) with respect to *D<sup>α</sup>* yields

$$D^a \mathbf{S} = \mathbf{C}\_1 \mathbf{D}^{a/2} \mathbf{e} + \mathbf{C}\_2 \mathbf{D}^a \dot{\mathbf{e}} \tag{22}$$

By combining Eqs. (20) and (22), one obtains

$$D^a \mathbf{S} = \mathbf{C}\_1 \mathbf{D}^{a/2} e + \mathbf{C}\_2 \mathbf{D}^a \left( \dot{V}\_{2\tau} + \frac{\mathbf{1} \mathbf{V}\_2^2}{\tau} - \frac{\mathbf{1}}{\tau} \mathbf{V}\_2^2 \mathbf{D}^\gamma B\_{\text{SVC}} - \frac{V\_1 V\_2 \cos \delta}{\tau \mathbf{X}} + \frac{\mathbf{1}}{\tau} k P\_d \right) \tag{23}$$

The control law is derived as

$$D'B\_{\rm SVC} = -\frac{\tau}{V\_2^2} \left( -\frac{V\_2^{\prime 2}}{\tau X} + \frac{V\_1 V\_2 \cos \delta}{\tau X} - \frac{k}{\tau} P\_d - \dot{V}\_{2r} - \frac{C\_1}{C\_2} D^{-a/2} e - K\_r D^{-a} \text{sgn}(\mathcal{S}) \right) \tag{24}$$

The following inequality holds [30].

$$\left| \sum\_{j=1}^{\infty} \frac{\Gamma(1+a)}{\Gamma(1-j+a)\Gamma(1+j)} D^j \mathbb{S} D^{a-j} \mathbb{S} \right| \le \mathfrak{r} |\mathbb{S}| \tag{25}$$

Here *τ* is a positive constant. To prove the stability of the closed loop system, the Lyapunov function is chosen as *<sup>V</sup>* <sup>¼</sup> <sup>1</sup> 2 *S*2 . Applying operator *D<sup>α</sup>* to the Lyapunov function yields

$$D^a V = SD^a \mathbb{S} + \left| \sum\_{j=1}^{\infty} \frac{\Gamma(\mathbf{1} + a)}{\Gamma(\mathbf{1} - j + a)\Gamma(\mathbf{1} + j)} D^j SD^{a-j} \mathbb{S} \right| \tag{26}$$

Using Eq. (25), we can simplify Eq. (26) as

$$D^a V \le \mathcal{S} D^a \mathcal{S} + \tau |\mathcal{S}| \tag{27}$$

Using Eqs. (23) and (27), *D<sup>α</sup>V* is calculated as

$$D^a V \le S \left( C\_1 D^{a/2} e + C\_2 D^a \left( \dot{V}\_{2r} + \frac{1}{\tau} \frac{V\_2^2}{X} - \frac{1}{\tau} V\_2^2 D^t B\_{\text{SVC}} - \frac{V\_1 V\_2 \cos \delta}{\tau X} + \frac{1}{\tau} k P\_d \right) \right) + \tau |\mathbf{S}| \tag{28}$$

By combining Eqs. (24) and (28), one obtains

$$D^a V \le -K\_s |\mathbf{S}| + \tau |\mathbf{S}|\tag{29}$$

Eq. (35) can be expressed as

*DOI: http://dx.doi.org/10.5772/intechopen.79615*

*e t*ð Þ� *trD*ð Þ <sup>2</sup>�<sup>1</sup>

*D*�<sup>2</sup> *D*2

*t*¼*tr*

� *C*2 *C*1

*<sup>t</sup> e t*ð Þ h i

*tr* ≤*ts* < ∞. Using this concept Eq. (39) can be expressed as

*trD*ð Þ<sup>1</sup> *<sup>t</sup> e t*ð Þ h i

k½ � *e t* \_ð Þ *<sup>t</sup>*¼*tr*

**4. MATLAB/Simulink response optimization tool box**

Combination of Eqs. (37) and (38) yields

8 < :

<sup>k</sup>*e t*ð Þ� *trD*ð Þ<sup>1</sup>

From Eq. (40), one obtains

points in finite time.

descent methods [39, 40].

**275**

Using Lemma 1, Eq. (36) is expanded as

*<sup>t</sup> e t*ð Þ h i

ð Þ¼� *e*

*Static Var Compensator with Fractional Order Dynamics for Enhanced Stability and Control*

ð Þ *<sup>t</sup>* � *tr* <sup>2</sup>�<sup>1</sup>

Application of Lemma 2 to the right hand side of Eq. (37) yields

*D*�1�*α=*<sup>2</sup>

*t*¼*tr*

ð Þ *t* � *tr*

*t*¼*tr*

If *S t*ð Þ¼ ¼ *ts* 0 and *e t*ðÞ¼ 0, then the necessary condition of convergence is

*tr* <sup>≤</sup> *ts* � <sup>2</sup>k*e t*ð Þk *<sup>r</sup>* k½ � *e t* \_ð Þ *<sup>t</sup>*¼*tr*

From Eq. (41), it is proved that the error will converge near to the equilibrium

MATLAB/Simulink offers built-in response optimization tool box, which is extensively used for the optimal parameters selection of the control system [38]. To start the GUI, go to analysis tab of the simulink and click response optimization tool box. Define the parameters to be optimized in the MATLAB workspace. Then, use the response optimization tool box for importing the workspace defined parameters as design variables. The minimum and maximum search space of the design variables are adjusted. In the second step, the input reference signal is defined and imported using the toolbox. Finally, the output signal is imported from the simulink model that will follow the reference imported signal such that the error between the two signals is minimum. The error is minimized in an iterative manner and in this case, the error signal is the cost function. While minimizing the cost function, the parameters of the control system are adjusted online. The toolbox is associated with different optimization methods such as pattern search, simplex search, and gradient

*C*2 *C*1

<sup>2</sup> � *e t*ð Þ¼� *<sup>r</sup>*

*e t*ð Þ<sup>≤</sup> � *<sup>C</sup>*<sup>2</sup>

*C*1

<sup>2</sup> k�k*e t*ð Þk *<sup>r</sup>* <sup>≤</sup> � *<sup>C</sup>*<sup>2</sup>

ð Þk *ts* � *tr* ≤ 2k*e t*ð Þk *<sup>r</sup>*

ð Þk *tr* � *ts* ≤ � 2k*e t*ð Þk *<sup>r</sup>*

*D*�1�*α=*<sup>2</sup>

*C*2 *C*1

*D*�1�*α=*<sup>2</sup>

*C*1

9 = ;

*e* (36)

*K*k*e t*ð Þk (38)

<sup>k</sup> (41)

*e t*ð Þ (37)

*K*k*e t*ð Þk (39)

(40)

The first term of Eq. (29) is negative, so if the discontinuous gain *KS* > ∣*τ*∣, then it is shown that the fractional derivative of the Lyapunov function is always less than zero, that is, *D<sup>α</sup>V* ≤0, which means that reaching condition of the sliding surface is satisfied and *S* ¼ 0. The following lemmas are defined that will be used in the convergence proof.

*Lemma 1. If integral of the fractional derivative aD<sup>α</sup> <sup>t</sup> of a function f(t) exists then according to* [30].

$${}\_{a}D\_{t}^{-a} \left( {}\_{a}D\_{t}^{a}f(t) \right) = f(t) - \sum\_{J=1}^{K} \left[ {}\_{d}D\_{t}^{a-J}f(t) \right]\_{a=t} \frac{(t-a)^{a-J}}{\Gamma(a-J+1)} \tag{30}$$

Here *K* � 1 ≤ *α* < *K* and Γ represents the standard gamma function.

*Lemma 2. The fractional integral operator aD*�*<sup>α</sup> <sup>t</sup> with α* > 0 *is bounded such that the expression in Eq*. (31) *is valid* [29].

$$\left\| \left\| \_{d} D\_{t}^{-\alpha} f \right\| \right\|\_{P} \leq K \left\| f \right\| \left\| \_{P}; \mathbf{1} \leq P \leq \infty; \mathbf{1} \leq K \leq \infty \tag{31}$$

From Eq. (29) it is proved that the sliding surface is zero, that is, *S* ¼ 0. With this condition Eq. (31) can be simplified as

$$\begin{aligned} \left[ D^{-a/2} e = -\frac{\mathbf{C\_2}}{\mathbf{C\_1}} \dot{e} \right] \\ \left[ D^{(1+a/2)} e = -\frac{\mathbf{C\_1}}{\mathbf{C\_2}} e \right] \end{aligned} \tag{32}$$

Multiplying Eq. (32) by *D*�1�*α=*<sup>2</sup> and by further simplification, one obtains

$$\begin{cases} e = -\frac{\mathcal{C}\_2}{\mathcal{C}\_1} D^{-1-a/2} e \\\\ D^{-a/2} \left( D^{a/2} e \right) = -\frac{\mathcal{C}\_2}{\mathcal{C}\_1} D^{-1-a/2} e \end{cases} \tag{33}$$

Using Lemma 1, Eq. (33) can be expressed as

$$\left[e - \left[ {\_\_t}D\_t^{(a/2-1)}e \right]\_{t=t\_r} \frac{\left(t - t\_r\right)^{a/2-1}}{\Gamma(a/2)} = -\frac{C\_2}{C\_1}D^{-1-a/2}e\right] \tag{34}$$

At time *t* ¼ *tr*, the term under the fractional integral of Eq. (34) is equal to zero, that is, *trD*ð Þ *<sup>α</sup>=*2�<sup>1</sup> *<sup>t</sup> e* h i *t*¼*tr* ð Þ *<sup>t</sup>*�*tr <sup>α</sup>=*2�<sup>1</sup> <sup>Γ</sup>ð Þ *<sup>α</sup>=*<sup>2</sup> <sup>¼</sup> 0, then the remaining expression of Eq. (34) can be written as

$$e = -\frac{C\_2}{C\_1} D^{-1-a/2} e\tag{35}$$

*Static Var Compensator with Fractional Order Dynamics for Enhanced Stability and Control DOI: http://dx.doi.org/10.5772/intechopen.79615*

Eq. (35) can be expressed as

Using Eqs. (23) and (27), *D<sup>α</sup>V* is calculated as

*<sup>e</sup>* <sup>þ</sup> *<sup>C</sup>*2*D<sup>α</sup> <sup>V</sup>*\_ <sup>2</sup>*<sup>r</sup>* <sup>þ</sup>

By combining Eqs. (24) and (28), one obtains

*Lemma 1. If integral of the fractional derivative aD<sup>α</sup>*

*Lemma 2. The fractional integral operator aD*�*<sup>α</sup>*

*aD*�*<sup>α</sup> <sup>t</sup> <sup>f</sup>* � � � �

this condition Eq. (31) can be simplified as

*<sup>t</sup> f t*ð Þ � � <sup>¼</sup> *f t*ðÞ�<sup>X</sup>

1 *τ V*<sup>2</sup> 2 *<sup>X</sup>* � <sup>1</sup> *τ V*<sup>2</sup> 2 *Dγ*

� � � �

The first term of Eq. (29) is negative, so if the discontinuous gain *KS* > ∣*τ*∣, then it is shown that the fractional derivative of the Lyapunov function is always less than zero, that is, *D<sup>α</sup>V* ≤0, which means that reaching condition of the sliding surface is satisfied and *S* ¼ 0. The following lemmas are defined that will be used in the

*K*

*aD<sup>α</sup>*�*<sup>J</sup> <sup>t</sup> f t*ð Þ � �

*a*¼*t*

*<sup>P</sup>* ≤*K f* k k*P*; 1≤ *P*≤ ∞; 1≤ *K* ≤ ∞ (31)

*J*¼1

From Eq. (29) it is proved that the sliding surface is zero, that is, *S* ¼ 0. With

*<sup>e</sup>* ¼ � *<sup>C</sup>*<sup>2</sup> *C*1 *e*\_

> *C*2 *e*

*C*1

<sup>Γ</sup>ð Þ *<sup>α</sup>=*<sup>2</sup> ¼ � *<sup>C</sup>*<sup>2</sup>

*D*�1�*α=*<sup>2</sup> *e* 9 >>=

>>;

*D*�1�*α=*<sup>2</sup>

*e* (35)

*C*1

<sup>Γ</sup>ð Þ *<sup>α</sup>=*<sup>2</sup> <sup>¼</sup> 0, then the remaining expression of Eq. (34) can

*<sup>D</sup>*ð Þ <sup>1</sup>þ*α=*<sup>2</sup> *<sup>e</sup>* ¼ � *<sup>C</sup>*<sup>1</sup>

Multiplying Eq. (32) by *D*�1�*α=*<sup>2</sup> and by further simplification, one obtains

*D*�1�*α=*<sup>2</sup> *e*

*<sup>e</sup>* � � ¼ � *<sup>C</sup>*<sup>2</sup>

ð Þ *<sup>t</sup>* � *tr <sup>α</sup>=*2�<sup>1</sup>

At time *t* ¼ *tr*, the term under the fractional integral of Eq. (34) is equal to zero,

*D*�1�*α=*<sup>2</sup>

Here *K* � 1 ≤ *α* < *K* and Γ represents the standard gamma function.

*D*�*α=*<sup>2</sup>

*<sup>e</sup>* ¼ � *<sup>C</sup>*<sup>2</sup> *C*1

8 >><

>>:

Using Lemma 1, Eq. (33) can be expressed as

*<sup>e</sup>* � *trD*ð Þ *<sup>α</sup>=*2�<sup>1</sup> *<sup>t</sup> e* h i

ð Þ *<sup>t</sup>*�*tr <sup>α</sup>=*2�<sup>1</sup>

*D*�*α=*<sup>2</sup> *D<sup>α</sup>=*<sup>2</sup>

*t*¼*tr*

*<sup>e</sup>* ¼ � *<sup>C</sup>*<sup>2</sup> *C*1 *BSVC* � *<sup>V</sup>*1*V*2cos*<sup>δ</sup>*

*<sup>D</sup><sup>α</sup><sup>V</sup>* <sup>≤</sup> � *Ks*∣*S*<sup>∣</sup> <sup>þ</sup> *<sup>τ</sup>*∣*S*<sup>∣</sup> (29)

*τX* þ

*<sup>t</sup> of a function f(t) exists then*

*<sup>t</sup> with α* > 0 *is bounded such that the*

<sup>Γ</sup>ð Þ *<sup>α</sup>* � *<sup>J</sup>* <sup>þ</sup> <sup>1</sup> (30)

ð Þ *<sup>t</sup>* � *<sup>a</sup> <sup>α</sup>*�*<sup>J</sup>*

1 *τ kPd*

þ *τ*∣*S*∣

(28)

(32)

(33)

*e* (34)

*D<sup>α</sup>V* ≤ *S C*1*D<sup>α</sup>=*<sup>2</sup>

*Control Theory in Engineering*

convergence proof.

*aD*�*<sup>α</sup> <sup>t</sup> aD<sup>α</sup>*

*expression in Eq*. (31) *is valid* [29].

*according to* [30].

that is, *trD*ð Þ *<sup>α</sup>=*2�<sup>1</sup>

be written as

**274**

*<sup>t</sup> e* h i

*t*¼*tr*

$$D^{-2}D^2(e) = -\frac{C\_2}{C\_1}D^{-1-a/2}e\tag{36}$$

Using Lemma 1, Eq. (36) is expanded as

$$\left[\boldsymbol{\varepsilon}(t) - \left[\boldsymbol{t\_{r}}\boldsymbol{D}\_{t}^{(2-1)}\boldsymbol{\varepsilon}(t)\right]\_{t=t\_{r}} \frac{\left(t - t\_{r}\right)^{2-1}}{2} - \boldsymbol{\varepsilon}(t\_{r}) = -\frac{\mathbf{C}\_{2}}{\mathbf{C}\_{1}}\boldsymbol{D}^{-1-a/2}\boldsymbol{\varepsilon}(t) \tag{37}$$

Application of Lemma 2 to the right hand side of Eq. (37) yields

$$-\frac{C\_2}{C\_1}D^{-1-a/2}e(t) \le -\frac{C\_2}{C\_1}K||e(t)||\tag{38}$$

Combination of Eqs. (37) and (38) yields

$$\left| \left| e(t) - \left[ {\_{t}}D\_{t}^{(1)}e(t) \right]\_{t=t\_{r}} \frac{(t-t\_{r})}{2} \right| - \left| \left| e(t\_{r}) \right| \right| \leq -\frac{C\_{2}}{C\_{1}}K \left| \left| e(t) \right| \right|\tag{39}$$

If *S t*ð Þ¼ ¼ *ts* 0 and *e t*ðÞ¼ 0, then the necessary condition of convergence is *tr* ≤*ts* < ∞. Using this concept Eq. (39) can be expressed as

$$\left\{ \left. \left[ {\_{t}D\_{t}^{(1)}e(t)} \right]\_{t=t\_{r}} (t\_{s} - t\_{r}) \right| \leq 2 \left| |e(t\_{r})| \right| \right\} \tag{40}$$
 
$$\left\{ \left|| \dot{e}(t) \right|\_{t=t\_{r}} (t\_{r} - t\_{s}) \right| \leq -2 \left| |e(t\_{r})| \right| $$

From Eq. (40), one obtains

$$\mathbf{t}\_r \le \mathbf{t}\_s - \frac{2||\dot{\mathbf{e}}(t\_r)||}{||\dot{\mathbf{e}}(t)\_{\mathbf{t}=t\_r}||}\tag{41}$$

From Eq. (41), it is proved that the error will converge near to the equilibrium points in finite time.

### **4. MATLAB/Simulink response optimization tool box**

MATLAB/Simulink offers built-in response optimization tool box, which is extensively used for the optimal parameters selection of the control system [38]. To start the GUI, go to analysis tab of the simulink and click response optimization tool box. Define the parameters to be optimized in the MATLAB workspace. Then, use the response optimization tool box for importing the workspace defined parameters as design variables. The minimum and maximum search space of the design variables are adjusted. In the second step, the input reference signal is defined and imported using the toolbox. Finally, the output signal is imported from the simulink model that will follow the reference imported signal such that the error between the two signals is minimum. The error is minimized in an iterative manner and in this case, the error signal is the cost function. While minimizing the cost function, the parameters of the control system are adjusted online. The toolbox is associated with different optimization methods such as pattern search, simplex search, and gradient descent methods [39, 40].
