# On Modeling Antennas Using MoM-Based Algorithms: Wire-Grid versus Surface Triangulation
## 2 Materials and Methods
### Wire-Grid
#### 2.2.2 Wire-Grid: Algorithm
\begin{subarray}{c}along\\ the\ axis\end{subarray} }\sigma(l)\frac{e^{-jkR}}{4\pi R}dl,\] (20) \[\sigma(l)=-\frac{1}{j\omega}\frac{dI(l)}{dl}\] (21) and \(l\) is the length varying along the axis of the wire. 5. Divide the wire into \(N\) segments (Figure 1b) and for (18)-(21) use the pulse functions and the Dirac delta functions as test ones. In this case, the integrals in them are approximated by the sum of \(N\) integrals over segments. The current and charge on each segment are assumed constant, and the derivatives are approximated by finite differences on the same segments. Then, these equations will take the form of \[-E_{l}^{inc}(m)\approx-j\omega A(m)-\frac{\Phi(m^{+})-\Phi(m^{-})}{\Delta l_{m }},\] (22)  
\[A(m)\approx\mu\sum\limits_{n=1}^{N}\,I(n)\int\limits_{\Delta l_{n}}\frac{e^{- jkR}}{4\pi R}dl, \tag{23}\]  
\[\Phi(m^{+})\approx-\frac{1}{j\omega\epsilon}\sum\limits_{n=1}^{N}\left(\frac {I(n+1)-I(n)}{\Delta l_{n^{+}}}\right)\int\limits_{\Delta l_{n^{+}}}\frac{e^{ -jkR}}{4\pi R}dl, \tag{24}\]  
\[\Phi(m^{-})\approx-\frac{1}{j\omega\epsilon}\sum\limits_{n=1}^{N}\left(\frac {I(n)-I(n-1)}{\Delta l_{n^{-}}}\right)\int\limits_{\Delta l_{n^{-}}}\frac{e^{ -jkR}}{4\pi R}dl \tag{25}\]  
where \(m=1,\,\ldots,\,N,\,n^{-}\) and \(n^{+}\) are the start and end points of segment \(n\) respectively, \(\Delta l_{n}\) is its length (the increment between \(n^{-}\) and \(n^{+}\)), and \(\Delta l_{n}{}^{-}\) and \(\Delta l_{n}{}^{+}\) are the increments \"shifted\" by \(\pm 1/2\) segment \(n\) along \(l\), i.e., additional segments to ensure zero current at the wire ends. 6. Apply (22)-(25) to two separate segments (\(n\) and \(m\) from Figure 2) and obtain their impedance. In other words, for these segments, one should get the general notation of the integrals from (23)-(25) as \[\psi(n,m)=\frac{1}{\Delta I_{n}}\int\limits_{\Delta I_{n}}\frac{e^{-j\Re R_{mn }}}{4\pi R_{mn}}dl_{n}.\] (26) This function can be solved using either the Maclaurin series, as [PERSON] did in [9], or using any other numerical integration methods such as the Newton-Cotes formulas. In this way, according to (23), the vector potential at point \(m\), created by the current \(I(n)\) flowing in segment \(n\), is defined as \[A(m)=\mu\Delta I_{n}I(n)\psi(n,m).\] (27)
7. Determine the scalar potentials. To do this, assume that segment \(n\) consists of a current filament \(I(n)\) and two charge filaments are connected with the first one as \[q(n^{+})=\frac{1}{j\omega}I(n),\] (28) \[q(n^{-})=-\frac{1}{j\omega}I(n)\] (29) where \(q=\sigma\Delta I\). Then, the scalar potentials according to (24), (25), and Figure 2 are defined as \[\Phi(m^{+})=\frac{1}{j\omega\epsilon}\big{(}I(n)\psi(n^{+},m^{+})-I(n)\psi(n^ {-},m^{+})\big{)},\] (30) \[-E_{l}^{inc}(m)\approx-j\omega A(m)-\frac{\Phi(m^{+})-\Phi(m^{-})}{\Delta l_{m }},\]0 (31)
8. Substitute (27), (30), and (31) into (22). Then \[-E_{l}^{inc}(m)\approx-j\omega A(m)-\frac{\Phi(m^{+})-\Phi(m^{-})}{\Delta l_{m }},\]1 (32)
9. Calculate the impedance of the two segments using the obtained expression \[-E_{l}^{inc}(m)\approx-j\omega A(m)-\frac{\Phi(m^{+})-\Phi(m^{-})}{\Delta l_{m }},\]2 (33) or \[-E_{l}^{inc}(m)\approx-j\omega A(m)-\frac{\Phi(m^{+})-\Phi(m^{-})}{\Delta l_{m }},\]3 (34)
10. Use (34) to calculate all matrix elements and form an SLAE of form (15), in which the voltage matrix in the right part is defined by the applied field as \[\mathbf{V}= \begin{bmatrix}E_{l}^{inc}(1)\Delta I_{1}\\ E_{l}^{inc}(2)\Delta I_{2}\\ \ldots\\ \ldots\\ E_{l}^{inc}(N)\Delta I_{N}\end{bmatrix} .\] (35)
11. Solve SLAE (15) or calculate the total conductance matrix which is the inverse of the impedance matrix \(\mathbf{Z}\). 12. Determine the current distribution across the wire by multiplying the matrices of total conductivity and applied voltage. 13. Make adjustments regarding the wire excitation to solve the antenna problem. In this way, an antenna radiator is obtained by exciting a wire at one or more points along its length with a voltage source.