# Two-Wire Power Line Microwave Communication Using Low-Loss Modes
## 3 The Spectral Solution
### Evaluation of the Current  
Given the singularities of \(\widetilde{D}_{\pm}(\gamma)\), (30) can now be evaluated by deforming the original contour around the lower branch cut entirely on the proper Riemann sheet, as illustrated in Figures 2 and 3. Given factor \(e^{-j\gamma z}\) in the integrand of (30), the integrations along the lower infinite semicircle are zero. Hence, the original integrations reduce to integrations along both sides of the branch cut, as shown. However, because the pole at \(\gamma_{p}\) is enclosed by the original and new contours, its residue must be added to each integration.  
Equation (30) is integrated around the branch cut using the identities [24]  
\[H^{(2)}_{0}(-z)=-H^{(1)}_{0}(z) \tag{37}\]  
Figure 3: Contour deformation for evaluating (4) for the differential mode along with a definition of the relevant branch cut integration and the quasi-TEM pole.  
\[H_{0}^{(1)}(\zeta a)=J_{0}(\zeta a)+jY_{0}(\zeta a) \tag{38}\]  
\[H_{0}^{(2)}(\zeta a)=J_{0}(\zeta a)-jY_{0}(\zeta a) \tag{39}\]  
where \(H_{0}^{(1)}(q)\), \(J_{0}(q)\)and \(Y_{0}(q)\) are, respectively, the Hankel function of the first kind and Bessel functions of the first and second kind or order zero and argument \(q\). After multiplying the numerator and denominator by \(\zeta^{2}\) to avoid infinity in the denominator at \(\gamma=k_{0}\), the horizontal portion of the branch cut integral between \(0\) and \(k_{0}\) is (Note that the integrations on the two sides of the branch cut do not cancel because the argument of \(\zeta\) is different on the each side due to the fact that it is defined as a square root. More specifically, the different arguments of \(\zeta\) are indicated in Figures 2 and 3).  
\[\hat{I}_{h}^{c/d}(z)=\frac{-4\omega\varepsilon_{0}\hat{V}}{\pi}\int\limits_{0 }^{k_{0}}\left(\frac{\zeta^{2}(J_{0}(\zeta a)\pm J_{0}(\zeta a))e^{-j\gamma z }d\gamma}{M_{\pm}^{\text{h}}(\gamma)}\right) \tag{40}\]  
where  
\[\begin{array}{c}M_{\pm}^{\text{h}}(\gamma)=\zeta^{4}\Big{(}\big{(}J_{0}( \zeta a)\pm J_{0}(\zeta d)\big{)}^{2}+\big{(}Y_{0}(\zeta a)\pm Y_{0}(\zeta d) \big{)}^{2}\Big{)}\\ +j(2\delta_{iw}\zeta^{2})\big{(}Y_{0}(\zeta a)\pm Y_{0}(\zeta d)\big{)}-\delta _{iw}^{2}.\end{array} \tag{41}\]  
The vertical portion of the branch cut integral from \(\gamma=-j\infty\) to \(\gamma=0\) can be written (after using the substitutions \(\gamma=-ju\), \(d\gamma=-jdu\)) and defining \(\zeta_{i}=\big{(}k_{0}^{2}+u^{2}\big{)}^{1/2}\) as  
\[\hat{I}_{v}^{c/d}(z)\frac{-j4\omega\varepsilon_{0}\hat{V}}{\pi}\int\limits_{0 }^{\infty}\left(\frac{\zeta_{i}^{2}(J_{0}(\zeta_{i}a)\pm J_{0}(\zeta_{i}d))e^ {-uz}du}{M_{\pm}^{\text{v}}(\gamma)}\right) \tag{42}\]  
where  
\[\begin{array}{c}M_{\pm}^{\text{v}}(\gamma)=\zeta_{i}^{4}\Big{(}\big{(}J_{0} (\zeta_{i}a)\pm J_{0}(\zeta_{i}d)\big{)}^{2}+\big{(}Y_{0}(\zeta_{i}a)\pm Y_{0} (\zeta_{i}d)\big{)}^{2}\Big{)}\\ +j(2\delta_{iw}\zeta_{i}^{2})\big{(}Y_{0}(\zeta_{i}a)\pm Y_{0}(\zeta_{i}d) \big{)}-\delta_{iw}^{2}\end{array} \tag{43}\]  
The subscripts \"\(h\)\" and \"\(v\)\" in (40) and (42) indicate the contribution from the horizontal and vertical portions of the branch cut integration, respectively. If \(\delta_{iw}\) is finite, (40) and (42) are relatively straightforward to integrate numerically.  
As noted earlier in Figures 2 and 3, there is a pole on the proper Riemann sheet in the common mode case (i.e., the mSG pole) and in the differential mode case (i.e., the Quasi-TEM pole) located between the original contour and integration contour used in this work. Hence, the appropriate residue must be added to the branch cut integral in each case. This residue can be found directly from (31) by expanding \(\widetilde{D}_{\pm}(\gamma)\) as a Taylor series around \(\gamma_{p}\). The result is  
\[\hat{I}_{res}(z)=\frac{-2\pi j\,N_{V}}{\widetilde{D}_{\pm}^{\prime}(\gamma_{p })}e^{-j\gamma_{p}z} \tag{44}\]  
where \(D_{\pm}^{\prime}\big{(}\gamma_{p}\big{)}\) can be found in (35).  
The total induced modal currents then are  
\[\hat{I}^{c}(z)=\hat{I}_{h}^{c}(z)+\hat{I}_{v}^{c}(z)+\hat{I}_{res}^{m