# Enhanced Pathological Element-Based Symbolic Nodal Analysis
## 4 Application Examples  
To demonstrate the usage of the presented symbolic NA method and the proposed active models, three circuit examples are given below. Considering the FDCCII-based current-mode biquad in Figure 2 of reference [25], its NGMFM equivalence is given in Figure 4. The multiple-output FDCCII model is derived from the model in Figure 3d by replacing the CMs with the grounded two-output CM in Figure 1f and the current replication cell with seven outputs (the extended version of current replication cell in Figure 2e). Based on the steps 1-2 in Section 3, the nodal admittance equations in matrix form can be obtained as:  
\[\left[ \begin{array}{c}I_{2}+I_{C2}-I_{C1}\\ I_{3}+I_{C2}\\ -I_{C1}\\ -I_{C2}\end{array} \right]=\left[ \begin{array}{cccc}sC_{2}&0&0&0\\ 0&sC_{1}&0&0\\ 0&0&G_{2}&0\\ 0&0&0&G_{1}\end{array} \right]\left[ \begin{array}{c}V_{1}\\ V_{2}\\ V_{3}\\ V_{4}\end{array} \right] \tag{2}\]  
By applying the 3 th to 4 th steps of the proposed analytic procedure, as given by (3), one equation has been removed since one grounded two-output CM incurs the deletion of one row. Applying the 5 th to 7 th steps, two unknown voltage variables have been removed since each differential cell incurs the deletion of two columns of the nodal admittance matrix, as given by (4). Figure 4: FDCCII-based current-mode biquad in Figure 2 of [23]. \[\left[ \begin{array}{c}I_{2}+I_{\text{C2}}\\ I_{3}+I_{\text{C2}}\\ -I_{\text{C2}}\end{array} \right]=\left[ \begin{array}{ccc}sC_{2}&0&-G_{2}&0\\ 0&sC_{1}&0&0\\ 0&0&0&G_{1}\end{array} \right]\left[ \begin{array}{c}V_{1}\\ V_{2}\\ V_{3}\\ V_{4}\end{array} \right] \tag{3}\]  
\[\left[ \begin{array}{c}I_{2}+I_{\text{C2}}\\ I_{3}+I_{\text{C2}}\\ -I_{\text{C2}}\end{array} \right]=\left[ \begin{array}{cc}sC_{2}&-G_{2}\\ 0&sC_{1}\\ G_{1}&0\end{array} \right]\left[ \begin{array}{c}V_{1,4}\\ V_{2,3}\end{array} \right] \tag{4}\]  
After applying the 8 th step, the nodal admittance matrix equations are obtained as (5). The current replication cell in Figure 4 incurs the deletion of one equation. By solving (5), we can obtain the outputted currents as (6)-(8). \[\left[ \begin{array}{c}I_{2}\\ I_{3}\end{array} \right]=\left[ \begin{array}{cc}sC_{2}+G_{1}&-G_{2}\\ G_{1}&sC_{1}\end{array} \right]\left[ \begin{array}{c}V_{1,4}\\ V_{2,3}\end{array} \right] \tag{5}\]  
\[I_{out2}=I_{\text{C2}}=-G_{1}V_{4}=\frac{-I_{2}sC_{1}G_{1}-I_{3}G_{1}G_{2}}{s^ {2}C_{1}C_{2}+sC_{1}G_{1}+G_{1}G_{2}} \tag{6}\]  
\[I_{out1}=I_{1}+I_{\text{C2}}=I_{1}-G_{1}V_{4}=\frac{\left[I_{1}(s^{2}C_{1}C_{2 }+sC_{1}G_{1}+G_{1}G_{2})-I_{2}(sC_{1}G_{1})-I_{3}G_{1}G_{2}\right]}{s^{2}C_{1 }C_{2}+sC_{1}G_{1}+G_{1}G_{2}} \tag{7}\]  
\[I_{out3}=-I_{\text{C1}}=G_{2}V_{3}=\frac{-I_{2}(G_{1}G_{2})+I_{3}(sC_{2}G_{2}+ G_{1}G_{2})}{s^{2}C_{1}C_{2}+sC_{1}G_{1}+G_{1}G_{2}} \tag{8}\]  
The results are in according with the expressions in Equations (10)-(12) of [25], so the validity of the proposed method is confirmed. The second application example, which takes into account the DDCC+-based filter in Figure 2a of [26], is a voltage-mode filter. Figure 5 depicts its NGMFM equivalence using the DDCC model in Figure 3e with omission of duplicated current at \(Z-\) terminal. According to steps 1-2 in Section 3 of the proposed method, the nodal admittance equations in matrix form can be obtained as (9). Applying steps 3 and 4 respectively incur the deletion of one equation, so (9) becomes (10). Based on steps 5-8, the nodal admittance equations can be obtained as (11) because of the voltage property (\(V_{4}=V_{3}-V_{2}+V_{1}\)) of the DDCC+. By applying step 9, the square nodal admittance matrix can be obtained, as given by (12). Thus the transfer function is computed from (12) and given by (13). The analyzed results are consistent with the expressions in Equation (11) of [26]. Figure 5: DDCC+-based voltage-mode filter in Figure 2(a) of [24]. \[\left[ \begin{array}{c}I_{i}\\ 0\\ -I_{C1}\\ -I_{C1}\end{array} \right]=\left[ \begin{array}{cccc}0&0&0&0\\ 0&y_{1}+y_{4}&-y_{1}&0\\ 0&-y_{1}&y_{1}+y_{2}+y_{3}&-y_{2}\\ 0&0&-y_{2}&y_{2}\end{array}