# Hybrid Harmonic Suppression Method at DC Link of Series-Connected 18-Pulse Rectifier
## 3 Injection Transformer Turned Ratio Calculation  
The following analysis takes combination 5 as an example. In combination 5, \(i_{\text{a2}}>0\), \(i_{\text{b2}}=0\), \(i_{\text{c2}}<0\), \(i_{\text{a3}}>0\), \(i_{\text{b3}}=0\), \(i_{\text{c3}}<0\). Therefore, voltage \(u_{\text{a2c2}}\) and voltage \(u_{\text{a3c3}}\) are calculated as  
\[\left\{ \begin{array}{l}u_{\text{a2c2}}=\dfrac{(N_{8}+4N_{7})}{3N_{8}}u_{ \text{o}}\\ \\ u_{\text{a3c3}}=\dfrac{(N_{8}-2N_{7})}{3N_{8}}u_{\text{o}}\end{array} \right. \tag{15}\]  
From Figures 2 and 3, it is obtained that  
\[u_{\text{a1c1}}=N_{2}u_{\text{n2b2}}+N_{1}u_{\text{a2n2}}+N_{1}u_{\text{n2c2}} +N_{2}u_{\text{a2n2}} \tag{16}\]  
where \(N_{1}=N_{11}/N_{21}=N_{31}/N_{21}\), \(N_{2}=N_{12}/N_{21}=N_{32}/N_{21}\). From above analysis, \(u_{\text{b2c2}}\) can be expressed as  
\[u_{\text{b2c2}}=\dfrac{(N_{8}-N_{7}-N_{1}N_{8}-4N_{1}N_{7})}{3N_{2}N_{8}}u_{ \text{o}} \tag{17}\]  
Then, \(u_{\text{a2b2}}\) is expressed as  
\[u_{\text{a2b2}}=\dfrac{(N_{2}N_{8}-2N_{2}N_{7}-N_{8}-2N_{7}+N_{1}N_{8}-2N_{1}N_ {7})}{3N_{2}N_{8}}u_{\text{o}} \tag{18}\]  
According to Equation (18) and the structure of the isolation transformer, voltage \(u_{\text{AN}}\) can be obtained as follows:  
\[u_{\text{AN}}=\dfrac{5(N_{2}N_{8}-2N_{2}N_{7}-N_{8}-2N_{7}+N_{1}N_{8}-2N_{1}N_ {7})}{3\sqrt{3}N_{2}N_{8}}u_{\text{o}} \tag{19}\]  
Similarly, the expressions of voltage \(u_{\text{AN}}\) in one supply cycle can also be obtained, as shown in Table 1.  
Figure 7: Operating waveforms of the proposed rectifier. \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Voltage** & **Level** & **Voltage** & **Level** \\ \hline \(u_{1}\) & \(\frac{5(N_{h}+4N_{\gamma})}{3\sqrt{3N_{h}}}\)\(u_{0}\) & \(u_{7}\) & \(\frac{5(N_{h}+2N_{\gamma}-N_{h}N_{h}+2N_{\gamma}N_{\gamma})}{3\sqrt{3N_{h}}N_{h}}\)\(u_{0}\) \\ \(u_{2}\) & \(\frac{5(N_{h}+2N_{\gamma})}{3\sqrt{3N_{h}}}\)\(u_{0}\) & \(u_{8}\) & \(\frac{5(N_{h}-N_{h}-2N_{\gamma}N_{\gamma})}{3\sqrt{3N_{h}}N_{h}}\)\(u_{0}\) \\ \(u_{3}\) & \(\frac{5(N_{h}-2N_{\gamma})}{3\sqrt{3N_{h}}}\)\(u_{0}\) & \(u_{9}\) & \(\frac{5(N_{h}-N_{h}N_{h}-4N_{h}N_{\gamma})}{3\sqrt{3N_{h}}N_{h}}\)\(u_{0}\) \\ \(u_{4}\) & \(\frac{5(N_{h}+2N_{\gamma}N_{\gamma}-N_{h}N_{\gamma})}{3\sqrt{3N_{h}}(N_{h}^{ \dagger}-N_{\gamma})}\)\(u_{0}\) & \(u_{10}\) & \(\frac{5(N_{h}N_{h}+2N_{\gamma}N_{\gamma}-N_{h}N_{\gamma}+2N_{\gamma}N_{\gamma} )}{3\sqrt{3N_{h}}N_{h}}\)\(u_{0}\) \\ \(u_{5}\) & \(\frac{5(N_{h}N_{h}+2N_{\gamma}N_{\gamma}-N_{h}N_{\gamma}-2N_{\gamma}N_{\gamma} )}{3\sqrt{3N_{h}}(N_{h}^{\dagger}-N_{\gamma}^{2})}\)\(u_{0}\) & \(u_{11}\) & \(\frac{5(N_{h}N_{h}-2N_{\gamma}N_{\gamma}-N_{\gamma}N_{\gamma}-2N_{\gamma}N_{ \gamma}+N_{\gamma}N_{\gamma}-2N_{\gamma}N_{\gamma})}{3\sqrt{3N_{h}}N_{h}}\)\(u_{0}\) \\ \(u_{6}\) & \(\frac{5(N_{h}^{2}N_{h}-N_{\gamma}N_{\gamma}N_{\gamma}-2N_{\gamma}N_{\gamma} )}{3\sqrt{3N_{h}}N_{h}(N_{h}^{\dagger}-N_{\gamma}^{2})}\)\(u_{0}\) & \(u_{12}\) & \(\frac{10N_{\gamma}}{3\sqrt{3N_{h}}(N_{h}-N_{\gamma})}\)\(u_{0}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Expressions of Voltage \(u_{\text{AN}}\) with the harmonic suppression method. Figure 8: Voltage \(u_{\text{AN}}\) in one supply cycle. Figure 9: The relationship between the THD of input voltage and the turn ratio of injection transformers. From Figure 9, when the turns ratio (\(n\)) is 0.0118, the THD of input voltage reaches the minimum value 0.0588. From Table 1 and Figure 9, the expressions of input voltages harmonics with harmonic suppression circuit can be obtained, written as follows:  
\[u_{\text{Har - 48}}=\sum_{n=6k\pm 1}^{\infty}\frac{1}{0.2806N_{8}\pi}\left\{