**2. Materials and Methods**

### *2.1. Mathematical Model of Small Signal Problems*

Synchronizing and Damping Momentums: When a short circuit occurs, the momentum of a synchronous machine is divided into two components. In order for a power system to exist, both components must be present. The lack of synchronizing momentums in a power system leads to instability that is dependent on the rotor angle and is not of the fluctuation type. The lack of damping momentums causes fluctuation instability. If one generator runs temporarily faster than the other does, the angular position of the rotor will increase in connection to that of the slow machine. The resulting angular difference transfers a portion of the load from the slow machine to the fast machine based on the theoretically known power angle relationship. This tends to reduce the speed difference and therefore the angular aperture. Further angular aperture may lead to a decrease in power transfer, leading to greater instability [19].

$$
\Delta T\_{\rm c} = T\_{\rm s} \Delta \delta + \text{TD} \Delta \omega \tag{1}
$$

TsΔδ: Synchronous momentum;

Ts: Coefficient of synchronous momentum;

TDΔω: Damping momentum (has the same phase as Δω);

TD: Coefficient of damping momentum.

This involves the protection of predetermined bus voltages by a power system to reach a stable state after a fault or short circuit occurs [20]. Therefore, the main reason for the instability in voltage is that the power system fails to provide reactive power. In other words, because the reactive power is directly proportional to the voltage, the electric power system has not been able to provide the reactive power required in its network well [1,21]. If for some reason (such as the input or output of a large production unit) the voltage drops in a part of the network and other generators or systems that compensate for the reaction power return to the current system, the voltage returns to its normal state [22]. Otherwise, the voltage drop will reach an unacceptable value and cause a power failure in another part of the power system network, which is called a voltage collapse [3,23,24].

#### *2.2. Definition of the Single Machine Infinite Bus (SMBI)*

The infinite bus is a source of voltage with constant voltage and frequency. Due to the infinite bus, the generator dynamics will not change EB voltage and frequency. In terms of small signal stability, both space- and block-type display methods are used to represent the small signal. In topics related to stability, the classical model is used to model the generator [25]. In this model shown as Figure 1, the generator is modeled as a voltage source connected to the reactance; however, all resistances in the generator or synchronous machine are ignored.

**Figure 1.** Classic model diagram in a Single Machine Infinite Bus (SMIB) system.

Calculation of complex and active forces in the network using flow equations:

$$
\lambda\_{\rm T} = \lambda\_{\rm d}^{\prime \prime} + \lambda\_{\rm E} \tag{2}
$$

$$\widetilde{\mathbf{I}\_{\rm t}} = \frac{\mathbf{E'}\_{\rm \gamma 0} - \mathbf{E\_{B\angle -\delta}}}{\mathbf{J\chi\_{\rm T}}} = \frac{\mathbf{E'} - \mathbf{E\_{B}}\widehat{\left(\mathbf{C}\text{os}\delta - \mathbf{J\hat{S}\text{in}}\delta\right)}}{\mathbf{J\chi\_{\rm T}}}\tag{3}$$

'p' is the power of air distance. p does not indicate the voltage of the terminal generator. However, since the stator resistance is ignored while the generator model is conventionally shown, P is also shown as the power of the terminal generator so that other analyzes can be performed.

$$\mathbf{S} = \mathbf{P} + \mathbf{J}\mathbf{Q} = \overleftrightarrow{\mathbf{E}}\ \overleftrightarrow{\mathbf{I}}\tag{4}$$

$$\mathbf{S} = \frac{\mathbf{E}'\mathbf{E\_B}}{\chi\_{\Gamma}}\mathbf{S}\mathbf{\hat{n}}\delta + \mathbf{J}\frac{\mathbf{E}'(\mathbf{E}' - \mathbf{E\_B}\mathbf{C}\mathbf{os}\boldsymbol{\delta})}{\chi\_{\Gamma}}\tag{5}$$

Another important point is that the air distance momentum is equal to the air distance power. The air distance momentum is as in Equation (6).

$$\text{Te} = \text{P} = \frac{\text{E}^{\prime}\text{E}\_{\text{B}}}{\chi\_{\text{T}}} \text{Sim}\delta \tag{6}$$

One of the most important equations related to rotor angle stability is the air distance momentum equation, and the other is the fluctuation equation (motion equation) [26].

#### 2.2.1. Linearization Model of Equations in Small Signal Stability (Sinδ, Cosδ)

In order to achieve small signal stability, the air distance momentum must be converted to a linear equation around the working point δ = δ0. To achieve this goal, first the sine in the air distance equation is linearized, so that the entire air distance equation can then be linearized [27]. Sinδ, Cosδ linearization around δ = δ0 equilibrium: a small deviation indicated by Δδ can be linearized as in the following equations:

$$
\Delta \delta = \delta - \delta\_{\text{\textquotedblleft}} 0 \Rightarrow \delta = \Delta \delta + \delta\_{\text{\textquotedblleft}} 0 \tag{7}
$$

δ0 = rotor angle at equilibrium

> Based on Sinδ, Cosδ equations and trigonometric relations:

$$\text{Cost}\,\delta = \text{Cost}(\delta\_0 + \Delta\delta) = \text{Cost}\,\delta\_0 \cdot \text{Cost}\,\Delta\delta - \text{Sim}\,\delta\_0 \cdot \text{Sim}\,\Delta\delta \tag{8}$$

$$\text{Sim}\delta = \text{Sim}(\delta\_0 + \Delta\delta) = \text{Sim}\delta\_0 \cdot \text{Cos}\Delta\delta + \text{Cos}\delta\_0 \cdot \text{Sim}\Delta\delta \tag{9}$$

The most important result obtained from the above equations is Δ δ = 1.

$$\begin{cases} \text{Cos}(\delta\_0 + \Delta \delta) = \text{Cos}\delta\_0 + \Delta \delta \cdot \text{Sim}\delta\_0\\ \text{Cos}(\delta\_0 + \Delta \delta) - \text{Cos}\delta\_0 = \Delta \delta \cdot \text{Sim}\delta\_0\\ \text{Sim}(\delta\_0 + \Delta \delta) - \text{Sim}\delta\_0 = \Delta \delta \cdot \text{Cos}\delta\_0 \end{cases} \tag{10}$$

Considering the above equations, when Te becomes linear at the working point δ = δ0, it becomes as follows in Equation (11).

$$\mathbf{T\_e} = \frac{\partial \mathbf{T\_e}}{\partial \delta} \Delta \delta = \frac{\mathbf{E} \mathbf{E'\_B}}{\chi\_{\Gamma}} \mathbf{Cost\_0}(\Delta \delta) \tag{11}$$

$$\begin{cases} \frac{d\Delta\omega\_{\rm r}}{dt} = \frac{1}{2\text{Fl}} (\mathbf{T\_{m}} - \mathbf{T\_{e}} - \mathbf{K\_{D}}\Delta\omega\_{\rm r})\\ \frac{d\delta}{dt} = \omega\_{0}\Delta\omega\_{\rm r} \end{cases} \tag{12}$$

$$\Rightarrow \begin{cases} \frac{\text{d}\Delta\omega\_{\text{r}}}{\text{d}\text{t}} = \frac{1}{2\text{H}} (\text{T}\_{\text{m}} - \text{K}\_{\text{s}}\Delta\text{\delta} - \text{K}\_{\text{D}}\Delta\omega\_{\text{r}})\\ \frac{\text{d}\delta}{\text{dt}} = \omega\_{0}\Delta\omega\_{\text{r}} \end{cases} \tag{13}$$

T m: Mechanical momentum;

Ks: Synchronous momentum coe fficient. *Energies* **2019**, *12*, 3412

$$\mathbf{K}\_{\mathsf{F}} = \left[\frac{\mathbf{E}^{\prime}\mathbf{E}\_{\mathsf{B}}}{\chi\_{\mathsf{T}}}\right] \mathbf{Cost}\_{0} \mathbf{T} \mathbf{e} = \mathbf{P} = \frac{\mathbf{E}^{\prime}\mathbf{E}\_{\mathsf{B}}}{\chi\_{\mathsf{T}}} \mathbf{Sim\delta} \tag{14}$$

Small signal stability can be analyzed by obtaining the master data. The representation of the linearized equation in the space matrix is as in Equation (15).

Ks: Synchronous momentum coefficient;

KD = Stabilizer momentum coefficient (stabilizer);

$$\begin{aligned} \stackrel{\bullet}{\mathbf{X}} &= \mathbf{A}\mathbf{X} + \mathbf{B}\mathbf{U} \\ \stackrel{\bullet}{\left[ \begin{array}{c} \boldsymbol{\Delta}\boldsymbol{\omega} \\ \boldsymbol{\Delta}\boldsymbol{\delta} \end{array} \right] &= \underbrace{\begin{bmatrix} -\frac{\mathbf{k}\mathbf{D}}{2\mathbf{H}} & -\frac{\mathbf{k}\mathbf{s}}{2\mathbf{H}} \\ \hline \omega\mathbf{u} & \mathbf{0} \end{bmatrix}}\_{\boldsymbol{\Delta}} \begin{bmatrix} \boldsymbol{\Delta}\boldsymbol{\omega}\mathbf{r} \\ \hline \boldsymbol{\Delta}\boldsymbol{\delta} \end{bmatrix} + \underbrace{\begin{bmatrix} \frac{1}{2\mathbf{H}} \\ \hline \boldsymbol{0} \end{bmatrix}}\_{\mathbf{B}} \boldsymbol{\Delta}\mathbf{T}\mathbf{m} \end{aligned} \tag{15}$$

H = Coefficient of inertia.

Equations belonging to the system are determined by considering the input and output of the system; then the system's main data is found using these equations and the system's small signal stability can be examined [28].

$$
\Delta\delta = \frac{\omega\_0}{\mathcal{S}} \left[ \frac{1}{2\text{HS}} (-\text{K}\_\delta \Delta \delta - \text{K}\_\mathcal{D} \Delta \omega\_\mathbf{r} + \Delta \text{T}\_\mathbf{m}) \right] \tag{16}
$$

$$
\Delta\delta = \frac{\omega\_0}{S} \left[ \frac{1}{2\text{FS}} \left( -\mathbf{K}\_\text{s} \Delta\delta - \mathbf{K}\_\text{D} \mathbf{S} \frac{\Delta\delta}{\omega\nu\_0} + \Delta \mathbf{T}\_\text{m} \right) \right] \tag{17}
$$

$$\rm S^2(\Delta \delta) + \frac{kD}{2\mathcal{H}} \rm S(\Delta \delta) + \frac{k\kappa}{2\mathcal{H}} \omega\_0(\Delta \delta) = \frac{\omega\_0}{2\mathcal{H}} \Delta T\_{\rm m} \tag{18}$$

$$\frac{\Delta\delta}{\Delta\mathbf{T\_m}} = \frac{\frac{\text{W}\_0}{\text{2H}}}{\text{S}^2 + \frac{\text{kD}}{\text{2H}}\text{S} + \frac{\text{ks-}\omega\_0}{\text{2H}}} \tag{19}$$

At this stage, in order to find the main data, the Equation (20) must be equal to zero [29].

$$\mathbf{S}^2 + \frac{\mathbf{k}\mathbf{D}}{2\mathbf{H}}\mathbf{S} + \frac{\mathbf{k}\mathbf{s} \cdot \boldsymbol{\omega}\mathbf{u}\_0}{2\mathbf{H}} = \mathbf{0} \tag{20}$$

The general representation of second-order equations will be as in Equation (21): ζ = Stability ratio

$$\mathbf{S}^2 + 2\xi\omega\_\mathbf{n}\mathbf{S} + \omega\_\mathbf{n}^2 = 0\tag{21}$$

The equations belonging to ωn ve ζ can also be found using the equations in Equations (20) and (21):

$$
\omega\_n = \sqrt{\mathbf{k}\_s \frac{\omega\_0}{2H}} \Big(\frac{\text{rad}}{\text{s}}\Big) \tag{22}
$$

$$\pounds = \frac{1}{2} \frac{\text{kD}}{2\text{H}\omega\_{\text{n}}} = \frac{1}{2} \frac{\text{kD}}{\sqrt{\text{k}\_{\text{s}} 2\text{H}\omega\_{0}}} \tag{23}$$

a. As Ks increases; the natural frequency increases, the stability rate decreases.


### 2.2.2. Control of Dynamic Systems

In a dynamic system, xi variables of the system affect each other. Each variable of the system is a function in terms of time. The aim of the analysis of dynamic systems is to examine the future ⎡⎢⎢⎢⎢⎣

behavior of the system. These behaviors include the determination of critical points and limit circles, examination of system stability, chelation, and chaos control.

If there is no change in rotor angle and speed, i.e., Δω (0) =Δ (0) = 0, and no change in mechanical momentum, the system will remain stable as per ΔTm (t) =0. For example, ΔTmt = 0, Δω0 = 0, Δδ0 = 5o ≈ 0.0875 Rad; in this case, KD = 10, KS = 0.757, H = 3.5, ω0 = 377.0δ

$$
\begin{array}{c}
\begin{array}{c}
\Delta\dot{\psi} \\
\Delta\dot{\delta}
\end{array}
\end{array}\Big|\_{}=\begin{bmatrix}
\frac{-10}{7} & \frac{-0.757}{7} \\
377.0 & 0 \\
&= \begin{bmatrix}
\frac{-10}{7} & \frac{-0.757}{7} \\
377.0 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta\dot{\omega} \\
\Delta\dot{\omega} \\
\Delta\dot{\delta}
\end{bmatrix}
+
\begin{bmatrix}
\frac{1}{7} \\
0 \\
\Delta\dot{\omega}
\end{bmatrix}
\end{array}\Big|\_{}+
$$

$$
\det\Big(\begin{bmatrix}
\frac{-10}{7} - \lambda & \frac{-0.757}{7} \\
377.0 & 0 - \lambda
\end{bmatrix}\Big) = 0
$$

$$
\begin{cases}
\begin{array}{c}
\lambda\_1 = -0.7143 + 6.3450 \text{i}; \\
\lambda\_2 = -0.7143 - 6.3450 \text{i};
\end{cases}
\end{array}
$$

The real portions of both quantities are negative, so the system is stable shown as Figure 2.

**Figure 2.** Diagram of the simulation of the rotor angle and speed changes up to 10 s.

If the damper coefficient becomes 0 (KD = 0) in the above example, the response and master data of the system will be as follows in Figure 3.

$$\begin{cases} \lambda\_1 = -0.7143 + 6.3450 \text{i}; \\ \lambda\_2 = -0.7143 - 6.3450 \text{i}; \end{cases}$$

**Figure 3.** Response diagram of the system fluctuating between stability and instability.

The real part of both master data is equal to zero. Therefore, the system does not go towards zero (stability) or infinity (instability). Instead, it continues its own fluctuation state. In another case shown as Figure 4, if the damping coefficient always changes as KD = −10, the master data and the response diagram of the system change as follows:

$$\begin{aligned} \lambda\_1 &= 0.7143 + 6.3450 \text{i} \\ \lambda\_2 &= 0.7143 - 6.3450 \text{i} \end{aligned}$$

**Figure 4.** The system response in which there is a linear relationship between instability and time.

The response of the system, as can be seen, is positive for both quantities. Therefore, the system is unstable, as the diagram shows, and the fluctuation ranges of the system increase as time goes on. When the excitation system is taken into account, the matrix of the system is per Equation (24).

$$
\begin{array}{c}
\Delta\dot{\boldsymbol{\omega}} \\
\Delta\dot{\boldsymbol{\delta}} \\
\Delta\dot{\boldsymbol{\psi}}\_{\rm fd} \\
\Delta\dot{\boldsymbol{\psi}}\_{1}
\end{array}
\left[
\begin{array}{c}
\mathbf{a}\_{11} & \mathbf{a}\_{12} & \mathbf{a}\_{13} & 0 \\
\mathbf{a}\_{21} & 0 & 0 & 0 \\
0 & \mathbf{a}\_{32} & \mathbf{a}\_{33} & \mathbf{a}\_{34} \\
0 & \mathbf{a}\_{42} & \mathbf{a}\_{43} & \mathbf{a}\_{44}
\end{array}
\right]
\left[
\begin{array}{c}
\Delta\boldsymbol{\omega} \\
\Delta\boldsymbol{\delta} \\
\Delta\boldsymbol{\psi}\_{\rm fd} \\
\Delta\boldsymbol{\upsilon}\_{1}
\end{array}
\right] + \left[
\begin{array}{c}
\mathbf{b}\_{1} \\
0 \\
0 \\
0
\end{array}
\right] \Delta\mathbf{T}\_{\rm m} \tag{24}
$$

The dynamics of a synchronous generator with PSS will be as in Equation (25) below.

$$
\begin{bmatrix}
\Delta\dot{\boldsymbol{\alpha}} \\
\Delta\ddot{\boldsymbol{\delta}} \\
\Delta\dot{\boldsymbol{\psi}}\_{\delta\dot{1}} \\
\Delta\dot{\boldsymbol{\psi}}\_{1} \\
\Delta\dot{\boldsymbol{\psi}}\_{2} \\
\Delta\dot{\boldsymbol{\psi}}\_{\ast}
\end{bmatrix} = \begin{bmatrix}
\mathbf{a}\_{11} & \mathbf{a}\_{12} & \mathbf{a}\_{13} & \mathbf{0} & \mathbf{0} & \mathbf{0} \\
\mathbf{a}\_{21} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & \mathbf{a}\_{32} & \mathbf{a}\_{33} & \mathbf{a}\_{34} & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & \mathbf{a}\_{42} & \mathbf{a}\_{43} & \mathbf{a}\_{44} & \mathbf{0} & \mathbf{0} \\
\mathbf{a}\_{51} & \mathbf{a}\_{52} & \mathbf{a}\_{53} & \mathbf{0} & \mathbf{a}\_{56} & \mathbf{0} \\
\mathbf{a}\_{61} & \mathbf{a}\_{62} & \mathbf{a}\_{63} & \mathbf{0} & \mathbf{a}\_{65} & \mathbf{a}\_{66} \\
\end{bmatrix} \begin{bmatrix}
\Delta\mathbf{u} \\
\Delta\boldsymbol{\delta} \\
\Delta\mathbf{v}\_{\delta\delta} \\
\Delta\mathbf{v}\_{1} \\
\Delta\mathbf{v}\_{2} \\
\Delta\mathbf{v}\_{3}
\end{bmatrix} \tag{25}
$$
