integration point $\left(x',y',z'\right)$.

The solution of (\[eq:A\]) for differential mode (DM) transmission has been reported in [@ianconescu], where the authors show that the DM radiated power of a finite length TL is $$P_{rad}^{DM}=\frac{\eta_{0}\left(ka\right)^{2}}{2\pi}\left(\left|I_{f}^{DM}\right|^{2}+\left|I_{b}^{DM}\right|^{2}\right)\left(1-\operatorname{sinc}\left(4kL\right)\right),\label{eq:praddm}$$ where $a$ is the distance between the conductors, $I_f$ and $I_b$ are the forward and backward traveling currents respectively. The paper demonstrates that a TL excited by a DM radiates only from a region around its terminations. To confirm this, (\[eq:praddm\]) shows that if $L$ is much greater than the current wavelength $\lambda$, $P_{rad}^{DM}$ tends to a constant value, which means that the central part of a long cable is not a source of significant radiation. Conversely, $P_{rad}^{DM}$ is negligible when $L\ll\lambda$, rapidly increases and can be already considered stable at $L\simeq\lambda$. Given the fact that the cable radiates symmetrically from the terminations, a model for the PUL radiation resistance $R^{DM}(l)$ can be found by considering how half of $P_{rad}^{DM}$ changes up to the mid point of the line, which results in $$\begin{gathered}
R^{DM}(l)=\frac{P_{rad}^{DM}}{2\left(\left|I_{f}^{DM}\right|^{2}+\left|I_{b}^{DM}\right|^{2}\right)}=\\=\frac{\eta_{0}\left(ka\right)^{2}}{4\pi l}\left(\operatorname{sinc}\left(4kl\right)-\cos\left(4kl\right)\right),\label{eq:rdm}\end{gathered}$$ which is function of the distance $l$ from the termination point.

For what concerns the common mode (CM) transmission, a reference plane has to be taken into account in the system to 