# Two-Wire Power Line Microwave Communication Using Low-Loss Modes
## 2 Problem Definition and Exact Solution  
The problem is illustrated in Figure 1. Here, two infinitely long parallel lossy (conductivity \(\sigma_{w}\)) conductors, each of radius a and spaced a distance d apart are placed in the z direction. A voltage source of amplitude V volts at \(z=0\) and/or a \(y\) oriented elementary electric dipole of length \(dl\) at a distance h from the origin along the \(y\)-axis are used to excite currents on both conductors. The approach here is to solve for the currents excited by the voltage source and then (later) to use the fixed relation between dipole and voltage sources derived in [22] to calculate those caused by the dipole source. The method is identical to that for the single-wire case except that there is an additional source of field (i.e., the second conductor). Hence, there will be two coupled equations (one for each conductor boundary condition) rather than a single one, as in [22]. A pair of homogeneous Fredholm integral equations of the second kind for the current \(\hat{I}_{1}(z)\) and \(\hat{I}_{2}(z)\) induced on this conductor (valid for \(-\infty<z<\infty\)) can be written as  
\[ \begin{split}\int\limits_{-\infty}^{\infty}&\big{(}g _{ez}^{11}(a,z-z^{\prime})\hat{I}_{1}(z\prime)+g_{ez}^{12}(d-a,z-z^{\prime}) \hat{I}_{2}(z\prime)\big{)}dz^{\prime}+\hat{E}_{ze}(-d/2+a,z)\\ &\quad=-\hat{V}_{1}\delta(z)+z_{iw}\hat{I}_{1}(z^{\prime})\\ \int\limits_{-\infty}^{\infty}&\big{(}g_{ez}^{21}( d-a,z-z^{\prime})\hat{I}_{1}(z^{\prime})+g_{ez}^{22}(a,z-z^{\prime})\hat{I}_{2}(z ^{\prime})\big{)}dz^{\prime}+\hat{E}_{ze}(+d/2-a,z)\\ &\quad=-\hat{V}_{2}\delta(z)+z_{iw}\hat{I}_{2}(z^{\prime})\end{split}  \tag{1}\]  
\(g_{ez}^{mm}(\rho,z-z^{\prime})\) is the axial electric field at \((\rho,z)=(a,z)\) of a unit current of length \(dz^{\prime}\) at \((0,z^{\prime})\), \(\hat{E}_{ze}(a,z)\) is the axial electric field at \((y,z)=(a,z)\) of the vertical electric dipole, and \(\delta(z)\) is the Dirac delta function. \(z_{iw}\) is the intrinsic impedance per unit length of the conductor, which for conductor sizes and higher frequencies of interest used here can be approximated as follows:  
\[z_{iw}\cong\bigg{(}\frac{\omega\mu_{o}}{2\sigma_{w}}\bigg{)}^{1/2}\frac{(1+j) }{2\pi a},\ |k_{w}a|>>1 \tag{2}\]  
For (1), it has been assumed that the thin wire boundary condition \(\hat{E}_{z}(a,z)=z_{1w}\hat{I}(z)\) is applied to the inside surface of each conductor radius. The time variation is \(\exp(j\omega t)\), and the caret notation, ', indicates a phasor quantity. A formal solution to (1) for the current can be developed given that the integral equation is valid over the entire range of \(z\) from \(-\infty\) to \(\infty\). The solution can be found by taking the spatial Fourier transform of both sides of (1). This transform and its inverse (i.e., \(\widetilde{F}(\gamma)\) and \(\widetilde{F}^{-1}(\gamma)\)) used here are defined as follows:  
\[\widetilde{F}(f(z))=\widetilde{F}(\gamma)=\int\limits_{-\infty}^{\infty}f(z)e ^{+j\gamma z}dz \tag{3}\]  
\[\widetilde{F}^{-1}\Big{(}\widetilde{F}(\gamma)\Big{)}=f(z)=\frac{1}{2\pi}\int \limits_{-\infty}^{\infty}\widetilde{F}(\gamma)e^{-j\gamma z}d\gamma \tag{4}\]  
Figure 1: Geometry of the Problem. Here, the symbol \(\sim\) indicates a spatial Fourier transform with respect to the wire direction z that is dependent on the transform variable \(\gamma\). Taking the Fourier transform of (1) using (3) and the convolution identity  
\[\widetilde{F}\left[\int\limits_{-\infty}^{\infty}f\big{(}z^{\prime}\big{)}h \big{(}z-z^{\prime}\big{)}dz^{\prime}\right]=\widetilde{F}(f(z))\widetilde{H}(h (z)) \tag{5}\]  
Results in the algebraic equations (Note: the physical dimensions of all terms are \"volts\" in the Fourier transform domain.)