## 3 Method and Results of Modeling
The alignment chart given in [35] can be used for this purpose, or one can use the direct calculation of the series to which the integral is reduced; in this case, to achieve a high calculation accuracy, it is sufficient to consider the 14 members of the series. Accounting for the delay in EMF propagation becomes more complicated when calculating the effects for harmonics as it becomes necessary to apply [PERSON]'s solutions to the intermediate and far ranges, and to use the long-line equations as steady-state equations. The Fazonord package forms an equivalent lattice circuit of a multi-conductor system based on expressions for the self- and mutual impedances of individual conductors. Therefore, we need a high-performance algorithm for implementing [PERSON]'s formulas that will provide sufficient accuracy for practical purposes in the near, intermediate, and far ranges. In [PERSON]'s approach where the actual earth is replaced by a plane homogeneous solid, the method of mirror images is adjusted for its finite conductivity; in this case, the external self-impedance of a rectilinear conductor of small diameter and the mutual impedance of two parallel conductors are calculated as per the following formula [34]:  
\[\underline{Z}_{\mathrm{ext}}=0.001f+jf[0.01148-0.001256\ln(r\sqrt{0.02\sigma f })],\ \mathrm{Ohm/km}; \tag{4}\]0  
where \(\omega\) is the angular frequency, s\({}^{-1}\); \(\mu_{0}=4\pi\cdot 10^{-7}\) H/m; \(r_{ik}\)\({}^{\prime}\) is the distance between the conductor \(k\) and the mirror image of the wire \(i\), measured in m; \(r_{ik}\) is the distance between conductors \(i\) and \(k\), measured in m.  
If we do not consider ground bias currents, the summand \(\underline{E}_{ik}\) determining the addition due to the finite ground conductivity can be found as follows:  
\[ \begin{array}{c}\underline{E}_{ik}=0.5-\ln(\lambda r)-j\frac{\pi}{4}-\\ -\sum\limits_{n=1}^{\infty}{(-1)^{n}\frac{(r/2)^{2n}e^{i\pi n/2}}{n!(n+1)!} \times\big{[}\big{(}\ln(\lambda r)-m_{n}+j\frac{\pi}{4}\big{)}\cos(2n\theta)- \theta\sin(2n\theta)\big{]}-}\\ -2\sum\limits_{n=1}^{\infty}{(-1)^{n}\frac{r^{2n-1}e^{i(2n-1)\pi/4}\cos(2n-1) \theta}{1^{2}\cdot 3^{2}\cdot 5^{2}\cdot\ldots(2n-1)^{2}\cdot(2n+1);}}\\ m_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n+1}-\frac{1}{2(n+1)};\\ r=r_{ik}\prime\sqrt{\frac{\omega\mu_{0}}{\rho}}=& 2.8099\cdot 10^{-3}\cdot r_{ik}\prime\sqrt{\frac{f}{\rho}};r_{ik} \prime=\sqrt{\big{(}x_{i}-x_{k}\big{)}^{2}+\big{(}y_{i}+y_{k}\big{)}^{2}};\\ \theta=\text{arctg}\frac{\pi-x_{k}}{y_{i}+y_{k}},\end{array}  \tag{14}\]  
where \(f\) is the frequency, measured in Hz; \(\rho\) represents ground resistivity, measured in Ohm-m; and \(\lambda=0.890536209\); \((x_{i},y_{i})\), \((x_{k},y_{k})\) represents the coordinates of conductors, m.  
In deriving expressions (13) and (14), the \(y\)-axis direction is chosen to be vertically upward, the \(x\)-axis is perpendicular to the conductor axes, and the \(z\)-axis is opposite to the positive current direction. The origin of the coordinates is on the surface of the earth. In the near range, when \(r\leq 0.25\) an approximate expression of the following form is used for Formula (14):  
\[\underline{E}_{ik}=-0.077216-\ln\frac{r}{2}+\frac{\sqrt{2}}{3}r\cos\theta-j \Bigg{(}\frac{\pi}{4}-\frac{\sqrt{2}}{3}r\cos\theta\Bigg{)} \tag{15}\]  
for the far range that meets the condition \(r\geq 5\), the following formula is used:  
\[\underline{E}_{ik}=\frac{\sqrt{2}}{r}\Bigg{[}(1-j)\cos\theta+j\frac{\sqrt{2} \cos 2\theta}{r}-(1+j)\frac{\cos 3\theta}{r^{2}}+(1-j)\frac{3\cos 5\theta}{r^{4}} \Bigg{]} \tag{16}\]  
when \(i=k\) and \(r\leq 0.2\), Expression (13) corresponds to the external impedance of the conductorru  
\[ \begin{array}{c}\underline{Z}_{k}neu=\frac{\omega\mu_{0}}{8}+j\frac{\omega \mu_{0}}{2\pi}\ln\Big{(}\frac{1}{r_{ik}}\frac{2}{\gamma}\sqrt{\frac{\rho\rho} {\omega\mu_{0}}}\Big{)}\\ =\frac{\omega\mu_{0}}{8}+j\frac{\omega\mu_{0}}{2\pi}\ln\Big{(}\frac{1.8514}{r _{ii}}\sqrt{\frac{\rho}{\omega\mu_{0}}}\Big{)},\text{ Ohm}/m,\end{array}