*7.2. LRC Technique*

We have, .

$$
\dot{V} = \mathfrak{e}\_2 \dot{\mathfrak{e}}\_2.\tag{100}
$$

where,

$$\dot{c}\_2 = \left(\omega \mathbf{x}\_4 + \frac{1}{c} \mathbf{x}\_1 - \frac{1}{c} \frac{P\_0}{\mathbf{x}\_3} - \frac{1}{c} \mathbf{x}\_5 - \frac{1}{c} \mathbf{u}\_0 - \dot{\mathbf{x}}\_{3d}\right) + \left(\frac{1}{c} v + d\right),\tag{101}$$

And here,

$$w = -c \ast d \max \ast \text{sat}(\frac{d \max \ast w}{\mu}),\tag{102}$$

$$d = \Delta\omega n\_4 + \Delta\omega \mathbf{x}\_4 + \omega n\_4 + \frac{1}{c} \Delta \mathbf{x}\_1 - \frac{1}{c} (n\mathbf{\tilde{s}} + \Delta \mathbf{x} \mathbf{\tilde{s}}) - \frac{1}{c} dp \ , \tag{103}$$

And,

$$d\max = \frac{1}{c}\delta\_{\text{x1}} + \delta\_{\text{\omega}}\delta\_{\text{n4}} + \delta\_{\text{\omega}}\delta\_{\text{\textmathbf{x4}}} + \omega\delta\_{\text{n4}} - \frac{1}{c}\delta\_{\text{\textmathbf{5}}} - \frac{1}{c}\delta\_{\text{P}}/\,\delta\_{\text{\textmathbf{x3}}} \tag{104}$$

If we choose

$$u\_0 = -c \left[ \frac{1}{c} \frac{P\_0}{x\_3} - \omega \pounds\_4 + \frac{1}{c} x\_5 + \dot{x}\_{3d} - k\_1 e\_1 - k\_2 e\_2 \right],\tag{105}$$

Putting it all together,

$$\begin{split} \dot{V} &= e\_2 \Big( \Big( \omega \mathbf{x}\_4 + \frac{1}{\varepsilon} \mathbf{x}\_1 - \frac{1}{\varepsilon} \frac{p\_0}{x\_3} - \frac{1}{\varepsilon} \mathbf{x}\_5 - \frac{1}{\varepsilon} u\_0 - \dot{\mathbf{x}}\_{3d} \Big) &+ \Big( \frac{1}{\varepsilon} \Big( -\mathbf{c} \ast d \max \ast \mathrm{sat} \Big( \frac{d \max \ast \mathrm{aw}}{\mu} \Big) \Big) \\ &+ \Big( \Delta \omega n\_4 + \Delta \omega \mathbf{x}\_4 + \omega n\_4 + \frac{1}{\varepsilon} \Delta \mathbf{x}\_1 - \frac{1}{\varepsilon} (n\_5 + \Delta \mathbf{x}\_5) - \frac{1}{\varepsilon} d \Big) \Big) , \end{split} \tag{106}$$

$$\begin{aligned} \dot{V} &= c\_2 \Big( \Big( \omega \mathbf{x}\_4 + \frac{1}{c} \mathbf{x}\_1 - \frac{1}{c} \frac{P\_0}{\mathbf{x}\_3} - \frac{1}{c} \mathbf{x}\_5 - \frac{1}{c} \Big( -c \Big[ \frac{1}{c} \frac{P\_0}{\mathbf{x}\_3} - \omega \mathbf{x}\_4 + \frac{1}{c} \mathbf{x}\_5 + \dot{\mathbf{x}}\_{3d} - k\_1 \varepsilon\_1 - k\_2 \varepsilon\_2 \Big] \Big) \Big) - \dot{\mathbf{x}}\_{3d} \Big) \\ &+ \Big( \frac{1}{c} \Big( -c \ast \left( \frac{1}{c} \delta\_{x1} + \delta\_{\omega} \delta\_{\eta 4} + \delta\_{\omega} \delta\_{x4} + \omega \delta\_{\eta 4} - \frac{1}{c} \delta\_5 - \frac{1}{c} \delta\_{\eta} / \delta\_{x3} \right) \\ &\ast \text{sat} \left( \frac{\left( \frac{1}{c} \delta\_{x1} + \delta\_{\omega} \delta\_{\eta 4} + \delta\_{\omega} \delta\_{x4} + \omega \delta\_{\eta 4} - \frac{1}{c} \delta\_5 - \frac{1}{c} \delta\_{P} / \delta\_{x3} \right) \ast \mathbf{w}}{\mu} \right) \\ &+ (\Delta \omega n\_4 + \Delta \omega \ge\_4 + \omega n\_4 + \frac{1}{c} \Delta \mathbf{x}\_1 - \frac{1}{c} (n\_5 + \Delta \mathbf{x}\_5) - \frac{1}{c} d\_P)) \Big) \end{aligned} \tag{107}$$

Now let,

*ω* = 60 Hz, *x*<sup>3</sup> = 600 V, *x*<sup>4</sup> = 10 V, Δ*x*<sup>1</sup> = 200 A, Δ*x*<sup>2</sup> = 200 A, Δ*x*<sup>5</sup> = 10 A, *n*<sup>3</sup> = 50 V, *n*<sup>4</sup> = 50 V, *n*<sup>5</sup> = 30 A, *n*<sup>6</sup> = 30 A, Δ*ω* = 10 Hz, *dP* = 50 A, *dQ* = 20 A, *δx*<sup>1</sup> = 4000 A, *δx*<sup>3</sup> = 200 A, *δx*<sup>4</sup> = 100 V, *δω* = 70 Hz, *δ<sup>P</sup>* = 30 kW, *δ<sup>Q</sup>* = 20 Var, *δn*<sup>3</sup> = *δn*<sup>4</sup> = *δn*<sup>5</sup> = *δn*<sup>6</sup> = 100 A, *ρx*<sup>3</sup> = 200 V, and *μ* = 100, *δx*<sup>5</sup> = 50 A, *δx*<sup>6</sup> = 3 A, *δ*<sup>5</sup> = 150 A and *δ*<sup>6</sup> = 13 A, *c* = 10 μF.

Putting the values, we get:

$$\dot{V} = c\_2((-(100)(600 - 480) - (20)(0)) + \left(\frac{1}{10\mu}(-10\mu \ast \left(\frac{1}{10\mu}(4000) + (70)(100)\right) + \dot{V})(100)\right)$$

$$+ (70)(100) + (60)(100) - \frac{1}{10\mu}(150) - \frac{1}{10\mu} \left(\frac{30\text{k}}{200}\right)$$

$$\ast \text{sat} \left(\frac{\left(\frac{1}{10\mu}(4000) + (70)(100) + (70)(100) + (60)(100) - \frac{1}{10\mu}(150) - \frac{1}{10\mu}(\frac{300}{200})\right) + (60)}{100}\right)$$

$$+ ((10)(50) + (10)(10) + (60)(50) + \frac{1}{10\mu}(100) - \frac{1}{10\mu}(30 + 10) - \frac{1}{10\mu}(50))\right),$$

$$\dot{V} = c\_2[988900 - [370.02 \times 10^6] \ast \text{sat}(222.012 \times 10^6)] \tag{109}$$

As we are getting . *V* ≤ 0 from this equation, the system is globally stable.

#### **8. Conclusions**

A microgrid system has several advantages over the conventional utility grid system, such as unlimited renewable fuel resources, environment-friendly power generation, easy implementation, cost effectiveness, and so on. However, the maintenance of the microgrid electrification has been confronted by the challenge of continually increasing instability issues due to the growth of modern electronic devices. For improving the stability scenario of the microgrid system despite the presence of dense CPL loads, a storage-based load side compensation technique has been adopted in this paper. Besides that, Sliding Mode Controller (SMC) and Lyapunov Redesign Controller (LRC), two of the most prominent nonlinear control techniques, have been implemented individually to retain microgrid system stability. After that, SMC and LRC controller robustness analysis have been presented with the variation of CPL power. Next, the comparative analysis between the SMC controller and the LRC controller robustness has been illustrated which ascertains that Lyapunov Redesign Controller has a superior performance than the former one to retain microgrid stability in dense CPL-loaded conditions. Reasons for inferior SMC performance and ways to overcome them have been discussed afterwards, followed by numerical analysis of both of the control techniques to verify their performance in real microgrids. All the necessary results have been simulated in Matlab/Simulink platform with appreciable aftermath.

**Acknowledgments:** No funding has been received for this research project.

**Author Contributions:** All the authors contributed equally for the research article to be decimated in its current version.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


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