# A Frequency-Reconfigurable Dual-Band RF Crossover Based on Coupled Lines and Open Stubs
## 2 Theoretical Analysis  
The circuit schematic of the proposed dual-band frequency-reconfigurable crossover is shown in Figure 1. Four quarter-wavelength CLs are utilized, along with two open 90 stubs. There are two sets of CLs regarding the even and odd impedances (\(Z_{\rm e1},Z_{\rm o1}\)) and (\(Z_{\rm e2},Z_{\rm o2}\)). Whereas the stubs have an impedance of \(Z_{1}\) and each stub is connected through a varactor diode, so two varactor diodes are used. The design is symmetric along both the horizontal and vertical axes to ensure finding closed-form equations and simplify the design process. Consequently, even-odd-mode analysis is carried out [38], and Figure 2 shows the equivalent even-even, odd-even, even-odd, and odd-odd circuits. The equations corresponding to these circuits are as follows:  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]  
\[Z_{\rm b}=-\rm j\left(\frac{2Z_{1}}{\tan\theta}+\frac{1}{0.5\omega C}\right) \tag{2}\]  
\[Z_{\rm c1}=Z_{\rm e1}\frac{Z_{\rm b}+\rm jZ_{\rm e1}\tan\theta}{Z_{\rm e1}+\rm j Z _{\rm b}\tan\theta} \tag{3}\]  
\[Z_{\rm d1}=-\rm j\frac{Z_{\rm e2}}{\tan\theta} \tag{4}\]  
Figure 1: Circuit schematic of the dual-band frequency-reconfigurable crossover. Figure 2: Even–odd-mode equivalent circuits of the dual-band frequency-reconfigurable crossover: (**a**) even–even, (**b**) odd–even, (**c**) even–odd, (**d**) odd–odd. The even-even admittance:  
\[\mathrm{Y_{ee}}=\frac{1}{\mathrm{Z_{c1}}}+\frac{1}{\mathrm{Z_{d1}}}=\mathrm{j} \Bigg{(}\frac{\tan\theta}{\mathrm{Z_{e2}}}-\frac{1}{\mathrm{Z_{el}}}\frac{0.5 \omega\mathrm{CZ_{el}}\tan(\theta)+\mathrm{Z_{1}}\omega\mathrm{C}\tan(\theta)+ \left(\tan(\theta)\right)^{2}}{\mathrm{Z_{1}}\omega\mathrm{C}+\tan(\theta)-0.5 \omega\mathrm{CZ_{e1}}\left(\tan(\theta)\right)^{2}}\Bigg{)} \tag{5}\]  
\[\mathrm{Z_{c2}}=\mathrm{Z_{o1}}\frac{\mathrm{Z_{b}}+\mathrm{jZ_{o1}}\tan\theta} {\mathrm{Z_{o1}}+\mathrm{jZ_{b}}\tan\theta} \tag{6}\]  
\[\mathrm{Z_{d2}}=\mathrm{jZ_{o2}}\tan\theta \tag{7}\]  
The odd-even admittance:  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]0  
The even-odd admittance:  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]1  
The odd-odd admittance:  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]2  
To realize a crossover design, the following criteria should be met:  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]3  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]4  
\[Z_{\rm a}=-\rm j\frac{2Z_{1}}{\tan\theta} \tag{1}\]5  
where \(\mathrm{S_{11}}\) is the return loss, \(\mathrm{S_{31}}\) is the insertion loss, and \(\mathrm{S_{21}}\) and \(\mathrm{S_{41}}\) are isolation S-parameters. Since the crossover is a passive microwave device, it means that its scattering matrix is reciprocal:  
\[\mathrm{S}= \begin{bmatrix}|\mathrm{S_{11}}|&|\mathrm{S_{12}}|&|\mathrm{S_{13 }}|&|\mathrm{S_{14}}|\\ |\mathrm{S_{21}}|&|\mathrm{S_{22}}|&|\mathrm{S_{23}}|&|\mathrm{S_{24}}|\\ |\mathrm{S_{31}}|&|\mathrm{S_{32}}|&|\mathrm{S_{33}}|&|\mathrm{S_{34}}|\\ |\mathrm{S_{41}}|&\mathrm{S_{42}}|&|\mathrm{S_{43}}|&|\mathrm{S_{44}}|\\ \end{bmatrix} = \begin{bmatrix}0&0&1&0\\ 0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&1&0&0\\ \end{bmatrix}