## 2 Design Equations  
### Even and Odd Mode Analysis without DGS Elements  
The conventional WPD is comprised of two transmission lines having an electrical length of quarter wavelength with line impedance of \(\sqrt{2}Z_{0}\) and an isolation resistor having a resistance of \(2Z_{0}\) between these transmission lines. Figure 2a shows the proposed circuit, which substitutes these two lines by transmission lines with line impedances \(Z_{1}\) and \(Z_{2}\) and electrical lengths \(\theta_{1}\) and \(\theta_{2}\), respectively. The admittances of the transmission lines with the impedances \(Z_{1}\) and \(Z_{2}\) are \(Y_{1}\) and \(Y_{2}\), respectively. where  
\[Y_{1}=\frac{1}{Z_{1}},Y_{2}=\frac{1}{Z_{2}},\quad\text{and}\quad Y_{0}=\frac{ 1}{Z_{0}}\]  
A lumped resistor 2R is introduced between the transmission lines to achieve maximum isolation. A schematic diagram of the proposed divider, which realizes an equal power division at an arbitrary frequency, is illustrated in Figure 2b. The reference impedance of the circuit is \(Z_{0}\). The modeling starts with the even mode analysis by referring to Figure 2b, where the loading impedances \(Z_{0}\) are considered purely resistive to simplify the circuit evaluation. The circuit given in Figure 2b is symmetric across the midplane, i.e., along the dotted line.Therefore, there is no current flow through the 2R resistor. Thus, the circuit in Figure 2b can be bisected with open circuits at these points to achieve the circuit shown in Figure 3a. Then, the input admittance at Port 2 is obtained as:  
\[Y_{in}^{e}=Y_{1}\Bigg{(}\frac{\frac{Y_{0}}{2}+jY_{1}\tan\theta_{1}}{Y_{1}+j \frac{Y_{0}}{2}\tan\theta_{1}}\Bigg{)}=Y_{1}\frac{\frac{\left(Y_{1}Y_{0} \right)}{2}+\frac{\left(Y_{1}Y_{0}\right)}{2}\tan^{2}\theta_{1}}{Y_{1}^{2}+ \left(\frac{Y_{0}}{2}\tan\theta_{1}\right)^{2}} \tag{1}\]  
Figure 2: Proposed WPD circuit with \(\lambda/4\) transmission lines of characteristic impedances of \(Z_{1}\) and \(Z_{2}\), respectively. By applying conjugate matching in Figure 2(a), the input admittance at Port 2 is obtained:  
\[Y_{in}^{e}=Y_{0}-jY_{2}\tan\theta_{2} \tag{2}\]  
By comparing Equations (1) and (2) with respect to their real and complex parts, Equations (3) and (4) are obtained:  
\[\frac{\left(\frac{1}{2}\right)\sec^{2}\theta_{1}}{1+\left(\frac{Y_{0}}{2Y_{1}} \tan\theta_{1}\right)^{2}}=1 \tag{3}\]  
\[Y_{1}\frac{\tan\theta_{1}\bigg{(}Y_{1}^{2}-\left(\frac{Y_{0}}{2}\right)^{2} \bigg{)}}{Y_{1}^{2}+\left(\frac{Y_{0}}{2}\tan\theta_{1}\right)^{2}}=-Y_{2}\tan \theta_{2} \tag{4}\]  
In the case of odd-mode excitation, because of the symmetry, there is null voltage along the mid-section of the circuit in Figure 2. Considering Port 2, the resistance of the isolation register is realized as half, i.e., R, as shown in Figure 2(b). The input admittance at Port 2 can be calculated as:  
\[Y_{in}^{o}=Y_{2}\Bigg{(}\frac{\frac{1}{R}+jY_{2}\tan\theta_{2}}{Y_{2}+j\frac{ 1}{R}\tan\theta_{2}}\Bigg{)}=Y_{2}\frac{\left(\frac{Y_{2}}{R}+\frac{Y_{2}}{R} \tan^{2}\theta_{2}\right)+j\tan\theta_{2}\bigg{(}Y_{2}^{2}-\left(\frac{1}{R} \right)^{2}\bigg{)}}{Y_{2}^{2}+\left(\frac{1}{R}\tan\theta_{2}\right)^{2}} \tag{5}\]  
By applying the conjugate matching in Figure 2(b), the input admittance at Port 2 is obtained:  
\[Y_{in}^{o}=Y_{0}+jY_{1}\cot\theta_{1} \tag{6}\]  
Figure 3: Even and odd mode Analysis of the proposed circuit.