## Appendix A
### Elliptic Filter  
The initial data for calculating the elliptic filter [38] are cut-off frequency \(\omega_{0}\), specified transition band \(\omega_{1}\), permissible distortion in the passband \(\mathrm{R}_{\mathrm{P}}\), and the required suppression in the stopband \(\mathrm{R}_{\mathrm{S}}\). The normalized elliptic low-pass filter was calculated based on the following corridor of its AFR: \(\omega_{\mathrm{P}}\) = 1 rad, \(\omega_{\mathrm{s}}\) = 1.37 rad, \(\mathrm{G}_{\mathrm{P}}\) = 0.9772 p.u., \(\mathrm{G}_{\mathrm{s}}\) = 0.01 p.u. The parameters of the filter corridor are as follows:  
\[ \begin{array}{rl}\varepsilon_{\mathrm{P}}=\sqrt{\frac{1}{\mathrm{G}_{ \mathrm{P}}^{2}}-1}=\sqrt{\frac{1}{0.9772^{2}}-1}=0.2173;\ \varepsilon_{\mathrm{s}}=\sqrt{\frac{1}{\mathrm{G}_{\mathrm{s}}^{2}}-1}= \sqrt{\frac{1}{0.01^{2}}-1}=99.95;\\ \mathrm{k}=\frac{\omega_{0}}{\omega_{1}}=\frac{1}{1.37}=0.7299;\ \mathrm{k}_{1}= \frac{\varepsilon_{\mathrm{P}}}{\varepsilon_{\mathrm{s}}}=\frac{0.2173}{99.95}= 0.0022\end{array}  \tag{20}\]  
We have checked the filter order compliance with the given AFR corridor of the elliptic low-pass filter. To achieve this, we first calculate the full and complementary elliptic integrals using the Landen transform [42]:\[\mathrm{K(k)=K(0.7299)=1.8829;\ K(k_{1})=K(0.0022)=1.5708;}\] \[\mathrm{K^{\prime}(k)=K\Big{(}\sqrt{1-k^{2}}\Big{)}=K\Big{(}\sqrt{1-0. 7299^{2}}\Big{)}=1.8272;} \tag{101}\] \[\mathrm{K^{\prime}(k_{1})=K\Big{(}\sqrt{1-k_{1}^{2}}\Big{)}=K\Big{(} \sqrt{1-0.0022^{2}}\Big{)}=7.518}\]  
Then, the filter order is as follows:  
\[\mathrm{n=\frac{K^{\prime}(k_{1})\cdot K(k)}{K(k_{1})\cdot K^{\prime}(k)}= \frac{7.518\cdot 1.8829}{1.5708\cdot 1.8272}=4.9319} \tag{102}\]  
Since it was assumed that all the filters under study in this paper should have \(\mathrm{n=5}\) order, the parameters of the elliptic low-pass filter AFR corridor were selected correctly. The elliptic filter module was recalculated. It was previously assumed that \(\mathrm{N=2.L+r}\), where \(\mathrm{L=2}\), \(\mathrm{r=1}\). In this case:  
\[\mathrm{u_{i}=\frac{2i-1}{n},\ \ \ i=1;2,\ \ \ u_{1}=\frac{1}{5}=0.2,\ \ \ u_{2}= \frac{3}{5}=0.6} \tag{103}\]  
The k module is recalculated as follows:  
\[\mathrm{k_{1p}=\sqrt{1-k_{1}^{2}}=\sqrt{1-0.0022^{2}}=0.99999516;} \tag{104}\]  
\[\mathrm{k_{p}=k_{1p}^{n}\cdot\prod_{i=1}^{L}sn^{4}\big{(}u_{i}\cdot K(k_{1p}),k_{1p}\big{)}=0.6725;} \tag{105}\]  
\[\mathrm{k=\sqrt{1-k_{p}^{2}}=\sqrt{1-0.6725^{2}}=0.7401} \tag{106}\]  
Calculation of the elliptic functions \(\mathrm{sn}\) was carried out using Landen transformation [42]. _Calculation of filter zeros and poles_. Filter zeros were calculated as follows:  
\[\mathrm{z_{1}=j\frac{1}{k\cdot cd(u_{1}\cdot K(k),k)}=j\frac{1}{0.7401\cdot cd (0.2\cdot K(0.7401),0.7401)}=j1.3976} \tag{107}\]  
\[\mathrm{z_{2}=j\frac{1}{k\cdot cd(u_{2}\cdot K(k),k)}=j\frac{1}{0.7401\cdot cd (0.6\cdot K(0.7401),0.7401)}=j2.0411} \tag{108}\]  
_Calculation of filter poles_. For this purpose, it was pre-calculated:  
\[\mathrm{K^{\prime}(k)=K\Big{(}\sqrt{1-k^{2}}\Big{)}=K\Big{(}\sqrt{1-0. 7299^{2}}\Big{)}=1.8272;} \tag{101}\]0  
Filter poles were calculated as follows:  
\[\mathrm{K^{\prime}(k)=K\Big{(}\sqrt{1-k^{2}}\Big{)}=K\Big{(}\sqrt{1-0. 7299^{2}}\Big{)}=1.8272;} \tag{101}\]1  
where \(\mathrm{\sigma}_{\mathrm{i}}=\mathrm{Re}(\mathrm{p}_{\mathrm{i}})\) and \(\mathrm{\omega}_{\mathrm{i}}=\mathrm{Im}(\mathrm{p}_{\mathrm{i}})\). Since it was not necessary to calculate the odd-order filter \((\mathrm{G}_{\mathrm{p}})^{1-\mathrm{r}}\), after substituting Equations (10), (11), (12)-(13) and (10) into Equation (12), we obtained the transfer function of the normalized elliptic low frequencies filter:  
\[\mathrm{K^{\prime}(k)=K\Big{(}\sqrt{1-k^{2}}\Big{)}=K\Big{(}\sqrt{1-0. 7299^{2}}\Big{)}=1.8272;} \tag{101}\]2  
Filter scaling was performed by substituting Equation (11) into Equation (12). The expression for calculating the AFR of high-pass elliptic filter was obtained according to Equation (12).