# Two-Wire Power Line Microwave Communication Using Low-Loss Modes
## 2 Problem Definition and Exact Solution
\begin{bmatrix}\widetilde{G}_{ez}^{11}-z_{iw}+\widetilde{G}_{ez}^{1 2}&0\\ 0&\widetilde{G}_{ez}^{22}-z_{iw}-\widetilde{G}_{ez}^{12}\end{bmatrix}\begin{bmatrix} \hat{\kappa}_{gm1}\\ \hat{\tilde{\chi}}_{gm2}\end{bmatrix} =\frac{-1}{\sqrt{2}} \begin{bmatrix}\hat{V }_{1}+\hat{V}_{2}+\hat{\tilde{E}}_{ez}^{1}+\hat{\tilde{E}}_{ez}^{2}\\ \hat{V}_{1}-\hat{V}_{2}+\hat{\tilde{E}}_{ez}^{1}-\hat{\tilde{E}}_{ez}^{2} \end{bmatrix}  \tag{23}\]  
Differential and Common ModesSince the matrix is diagonalized, the solutions to this equation for the component amplitudes can be obtained via a simple inspection, as follows:  
\[\hat{\tilde{I}}_{gm1}=-\sqrt{2}\frac{\hat{V}_{1}+\hat{V}_{2}+\hat{\tilde{E}}_ {ez}^{1}+\hat{\tilde{E}}_{ez}^{2}}{\widetilde{G}_{ez}^{11}-z_{iw}+\widetilde{ G}_{ez}^{12}} \tag{24}\]  
and  
\[\hat{\tilde{\tilde{I}}}_{gm2}=-\sqrt{2}\frac{\hat{V}_{1}-\hat{V}_{2}+\hat{ \tilde{E}}_{ez}^{1}-\hat{\tilde{E}}_{ez}^{2}}{\widetilde{G}_{ez}^{11}-z_{iw}- \widetilde{G}_{ez}^{12}} \tag{25}\]  
From these results, it is trivial to find formal solutions for the actual conductor currents, if, for example, the external electric fields \(\hat{\tilde{E}}_{ez}^{1}\) and \(\hat{\tilde{E}}_{ez}^{2}\) are assumed to be zero, and \(\hat{V}_{1}=\hat{V},\ \hat{V}_{2}=0\), wherein both the common and differential mode are excited. Given this,  
\[\hat{\tilde{I}}_{1}=\hat{\tilde{I}}_{2}=-\frac{\hat{V}}{\hat{G}_{ez}^{11}-z_{iw }+\widetilde{G}_{ez}^{12}} \tag{26}\]  
for the common mode and  
\[\hat{\tilde{I}}_{1}=-\hat{\tilde{I}}_{2}=-\frac{\hat{V}}{\widetilde{G}_{ez}^{1 1}-z_{iw}-\widetilde{G}_{ez}^{12}} \tag{27}\]  
for the differential mode. As above, with excitation on only one conductor, both geometric modes will be excited, and each has one or more distinct propagation modes, and the rate at which each mode is attenuated as it propagates along the wires will be different. Now using (8), the denominators of (26) and (27) can be written, respectively, as  
\[ \begin{split}\widetilde{G}_{ez}^{11}-z_{iw}\pm\widetilde{G}_{ez }^{12}&=\frac{1}{4\alpha c_{0}}\Big{\{}(\gamma^{2}-k_{0}^{2}) \Big{[}H_{0}^{(2)}\Big{(}\big{(}k_{0}^{2}-\gamma^{2}\big{)}^{1/2}a\Big{)}\pm H _{0}^{(2)}\Big{(}\big{(}k_{0}^{2}-\gamma^{2}\big{)}^{1/2}d\Big{)}\Big{]}- \delta_{iw}\Big{\}}\\ &=\frac{-1}{4\alpha c_{0}}\Big{\{}(k_{0}^{2}-\gamma^{2})\Big{[}H_{ 0}^{(2)}\Big{(}\big{(}k_{0}^{2}-\gamma^{2}\big{)}^{1/2}a\Big{)}\pm H_{0}^{(2)} \Big{(}\big{(}k_{0}^{2}-\gamma^{2}\big{)}^{1/2}d\Big{)}\Big{]}+\delta_{iw} \Big{\}}\\ &=\frac{-1}{4\alpha c_{0}}\Big{\{}\zeta^{2}\Big{[}H_{0}^{(2)}( \zeta a)\pm H_{0}^{(2)}(\zeta d)\Big{]}+\delta_{iw}\Big{\}}\\ &\text{where the $+$ sign is for the common mode and the $-$ sign is for the differential mode and}\\ \zeta=\Big{(}k_{0}^{2}-\gamma^{2}\Big{)}^{1/2},\text{Im}(\zeta)\leq 0\ \delta_{iw}=4 \omega\varepsilon_{0}z_{iw}\end{split}  \tag{29}\]Finally, the formal solution to the integral equations for common mode can be written using the inverse Fourier transform as  
\[\hat{I}_{1}^{c/d}(z)=\pm\hat{I}_{2}^{c/d}(z)=\frac{1}{2\pi}\int\limits_{-\infty}^ {\infty}\left(\frac{4\omega\epsilon_{0}\hat{V}e^{-\hat{\gamma}z}}{\tilde{D}_{\pm }(\gamma)}\right)d\gamma \tag{30}\]  
where  
\[\tilde{D}_{\pm}(\gamma)=\zeta^{2}\Big{[}H_{0}^{(2)}(\zeta a)\pm H_{0}^{(2)}( \zeta d)\Big{]}+\delta_{iw} \tag{31}\]  
and the individual currents in each wire can be written as  
\[\hat{I}_{1}(z)=\hat{I}_{1}^{c}(z)+\hat{I}_{1}^{d}(z) \tag{32}\]  
and  
\[\hat{I}_{2}(z)=\hat{I}_{2}^{c}(z)+\hat{I}_{2}^{d}(z)=\hat{I}_{1}^{c}(z)-\hat{I }_{1}^{d}(z) \tag{33}\]