**Appendix A**

*Detailed Mathematical Formulation*

> The mathematical formulation for the model in Figure 1

$$\left(-\frac{\Delta\omega(s)}{R} \times \frac{k(s-z)}{(s-p)} - \Delta P\_d(s)\right)\frac{1}{Ms+D} = \Delta\omega(s) \tag{A1}$$

$$\frac{\Delta\omega(s)}{\Delta P\_d(s)} = \frac{\frac{-1}{M}s + \frac{p}{M}}{s^2 + \left(\frac{k + DR - pMR}{RM}\right)s + \left(\frac{-pDR - kZ}{RM}\right)}\tag{A2}$$

Assuming, D = 0;

$$\frac{\Delta\omega(s)}{\Delta P\_d(s)} = \frac{\frac{-1}{M}s + \frac{p}{M}}{s^2 + \left(\frac{k - pMR}{RM}\right)s + \left(\frac{-kZ}{RM}\right)}\tag{A3}$$

Denominator can be compared with *s*2 + 2*ζ*Ω*n<sup>s</sup>* + <sup>Ω</sup>2*n*

$$\frac{\Delta\omega(s)}{\Delta P\_d(s)} = \frac{\frac{-1}{M}s + \frac{p}{M}}{s^2 + 2\zeta\Omega\_n s + \Omega\_n^2} \tag{A4}$$

$$\text{where } \Omega\_n = \sqrt{\frac{-kz}{RM}}, \zeta = \frac{k - pMR}{2RM\Omega\_n} \tag{A5}$$

Modeling disturbance as step response of magnitude *Ad*

$$
\Delta\omega(s) = \frac{-\frac{1}{M}s + \frac{p}{M}}{s^2 + 2\zeta\Omega\_n s + \Omega\_n^{-2}} \times \frac{A\_d}{s} \tag{A6}
$$

$$
\Delta\omega(s) = \frac{pA\_d}{M\Omega\_n^{-2}} \left[ \frac{1}{s} - \frac{s + \frac{\Omega\_n}{p} (\Omega\_n + 2p\tilde{\zeta})}{s^2 + 2\tilde{\zeta}\Omega\_n s + \Omega\_n^{-2}} \right] \tag{A7}
$$

By adding and subtracting *ζ*<sup>2</sup>Ω*n*<sup>2</sup> to the denominator

$$
\Delta\omega(\mathbf{s}) = \frac{p A\_d}{M \Omega\_n^{\frac{2}{2}}} \left[ \frac{1}{\mathbf{s}} - \frac{\mathbf{s} + \frac{\Omega\_\mathbf{n}}{p} \left(\Omega\_\mathbf{n} + 2p\zeta\right)}{\mathbf{s}^2 + 2\zeta\Omega\_\mathbf{n}\mathbf{s} + \zeta^2\Omega\_\mathbf{n}^{\frac{2}{2}} + \Omega\_\mathbf{n}^{\frac{2}{2}} - \zeta^2\Omega\_\mathbf{n}^{\frac{2}{2}}} \right] \tag{A8}
$$

$$
\Delta\omega(\mathbf{s}) = \frac{p A\_d}{M \Omega\_n^{-2}} \left[ \frac{1}{\mathbf{s}} - \frac{1}{\Omega\_d} \frac{\left(\mathbf{s} + \frac{\Omega\_n}{p} (\Omega\_n + 2p\zeta)\right) \Omega\_d}{(\mathbf{s} + \zeta \Omega\_n)^2 + \Omega\_d^{-2}} \right] \tag{A9}
$$

$$
\Omega\_d = \Omega\_\pi \sqrt{1 - \zeta^2} = \frac{\sqrt{-\left(k - pMR\right)^2 - 4kz}}{2RM} \tag{A10}
$$

Taking inverse Laplace of (A9)

$$\begin{split} \Lambda \omega(t) &= \frac{p A\_d}{M \Omega\_n^{\prime}} \Big[ 1 - e^{-\zeta \Omega\_n t} \Big( \cosh(j \Omega\_d t) - \\ &\frac{j \left( \frac{\Omega\_n^{\prime} + 2p \zeta \Omega\_n}{p} - \zeta \Omega\_n \right) \sinh(j \Omega\_d t)}{\Omega\_d} \Big) \Big] \end{split} \tag{A11}$$

$$\cosh(j\Omega\_d t) = \frac{e^{j\Omega\_d t} + e^{-j\Omega\_d t}}{2} = \cos(\Omega\_d t),\tag{A12}$$

$$\sinh(j\Omega\_d t) = \frac{e^{j\Omega\_d t} - e^{j\Omega\_d t}}{2} = j\sin(\Omega\_d t) \tag{A13}$$

Equation (A11) gets modified to (A14)

$$\begin{split} \Delta\omega(t) &= \frac{p A\_d}{M \Omega\_\text{n}^2} \left[ 1 - e^{-\frac{r}{\xi} \Omega\_\text{n} t} \left( \cos(\Omega\_d t) + \frac{\sin(\Omega\_d t)}{\Omega\_d} \right. \right. \\ & \left. \left( \frac{\Omega\_\text{n}^2 + 2p \zeta \Omega\_\text{n}}{p} - \zeta \Omega\_\text{n} \right) \right) \Bigg] \end{split} \tag{A14}$$

$$\begin{split} \Delta\omega(t) &= \frac{pA\_d}{M\Omega\_n^{-2}} \Big[ 1 - e^{-\frac{r}{\zeta}\Omega\_n t} \\ &\left( \cos(\Omega\_d t) + \frac{\Omega\_n^{-2} + p\_\gamma^r \Omega\_n}{p\Omega\_d} \sin(\Omega\_d t) \right) \Big] \end{split} \tag{A15}$$

To calculate the peak of <sup>Δ</sup>*ω*(*t*) given by <sup>Δ</sup>*ωpeak*,

$$\left. \frac{d\Delta\omega(t)}{dt} \right|\_{t=t\_{peak}} = 0 \tag{A16}$$

$$\begin{aligned} &\zeta\_s^\gamma \Omega\_n e^{-\zeta \Omega\_n t\_{pank}} \Big[ \cos(\Omega\_d t\_{pank}) + \frac{\Omega\_n^{-2} + p\zeta \Omega\_n}{p\Omega\_d} \sin(\Omega\_d t\_{pank}) \Big] \\ &- e^{-\zeta \Omega\_n t\_{pank}} \Big[ -\Omega\_d \sin(\Omega\_d t\_{pank}) + \\ &\frac{\Omega\_n^{-2} + p\zeta \Omega\_n}{p} \cos(\Omega\_d t\_{pank}) \Big] = 0 \end{aligned} \tag{A17}$$

$$\tan(\Omega\_d t\_{\text{peak}}) = \frac{\Omega\_d}{p + \mathcal{J}\Omega\_\text{n}}\tag{A18}$$

$$t\_{peak} = \frac{\tan^{-1}\left(\frac{\Omega\_d}{p + \xi \Omega\_n}\right)}{\Omega\_d} \tag{A19}$$

From (A18), sin(<sup>Ω</sup>*dtpeak*) and cos(<sup>Ω</sup>*dtpeak*):

$$\sin(\Omega\_d t\_{pck}) = \frac{\Omega\_d}{\sqrt{p^2 + 2p\zeta\Omega\_n + \Omega\_n^2}}\tag{A20}$$

$$\cos(\Omega\_d t\_{\text{peak}}) = \frac{p + \zeta \Omega\_n}{\sqrt{p^2 + 2p\zeta\Omega\_n + \Omega\_n^{-2}}} \tag{A21}$$

$$\text{s such that } \sin^2(\Omega\_d t\_{\text{peak}}) + \cos^2(\Omega\_d t\_{\text{peak}}) = 1 \tag{A22}$$

Replacing values of (A20) and (A21) in (A15)

$$\begin{split} \Delta\omega\_{\text{peak}} &= \frac{p A\_d}{M \Omega\_\text{n}^2} \Big[ 1 - e^{-\zeta \Omega\_\text{n} t\_{\text{peak}}} \\ &\left( \frac{p + \zeta \Omega\_\text{n}}{\sqrt{p^2 + 2p\zeta \Omega\_\text{n} + \Omega\_\text{n}^2}} + \frac{\Omega\_\text{n}^2 + p\zeta \Omega\_\text{n}}{p \Omega\_d} \frac{\Omega\_d}{\sqrt{p^2 + 2p\zeta \Omega\_\text{n} + \Omega\_\text{n}^2}} \right) \Big] \end{split} \tag{A23}$$

$$
\Delta\omega\_{\text{peak}} = \frac{p A\_d}{M \Omega\_n^{\prime}} \left[ 1 - e^{-\zeta \Omega\_n t\_{\text{peak}}} \left( \frac{\sqrt{p^2 + 2p \zeta \Omega\_n + \Omega\_n^{\prime}}}{p} \right) \right] \tag{A24}
$$

*p*2 + 2*pζ*Ω*n* + <sup>Ω</sup>*n*<sup>2</sup> calculated from (A5)

$$p^2 + 2p\zeta\Omega\_{\rm ll} + \Omega\_{\rm n}^{\prime 2} = \frac{k(p-z)}{MR} \tag{A25}$$

Substituting the values from (A25) into (A24)

$$
\Delta\omega\_{\text{peak}} = -\frac{p A\_d R}{k z} \left[ 1 - \frac{e^{-\zeta \Omega\_n t\_{\text{peak}}}}{p} \sqrt{\frac{k(p-z)}{MR}} \right] \tag{A26}
$$
