## 2 Current and Voltage Transformations of Delta-Connected Machine
### Voltage Transformations  
Figure 2 shows a three-phase two-level voltage-source inverter supplying a delta-connected IM. The interconnection of phases is equal to that defined in Figure 1. Table 1 then shows the phase voltages and the resulting phase voltage space vectors for the individual inverter switching combinations (\"1\" corresponds to the upper switch being on and \"0\" corresponds to the lower switch being on, the sequence corresponds to the inverter leg numbering from Figure 2). Figure 2: Inverter supplying a delta-connected machine. In a star connection, the active phase voltage vectors can be expressed in a general form as  
\[\mathbb{E}_{\kappa(\Upsilon)}=\frac{2}{3}U_{\text{DC}}\text{e}^{\text{i}(k-1) \frac{\pi}{3}}, \tag{12}\]  
where \(\kappa\in\{1,\;2,\;3,\;4,\;5,\;6\}\). Therefore, the relationship between the star and delta voltage vectors for the same switching combination is given by  
\[\mathbb{E}_{\kappa(\Delta)}=\sqrt{3}\text{e}^{\frac{\pi}{3}}\mathbb{E}_{k( \Upsilon)}. \tag{13}\]  
A comparison of the six active voltage vectors for a delta and star winding configuration and a constant DC-link voltage is shown in Figure 3.  
Voltage reconstruction from the switching signals can be obtained with the help of Figure 4. First, the motor phase voltages are expressed in terms of the inverter phase voltages (defined in Figure 2) as  
\[u_{\text{a}} =u_{\text{a}0}-u_{\text{b}0}, \tag{14}\] \[u_{\text{b}} =u_{\text{b}0}-u_{\text{c}0}, \tag{15}\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Switching** & \(u_{\text{a}}\) & \(u_{\text{b}}\) & \(u_{\text{c}}\) &
\begin{tabular}{c} **Corresponding Phase** \\ **Voltage Vector (K=2/3)** \\ \end{tabular} \\ \hline
000 & 0 & 0 & 0 & \(\mathbb{E}_{0(\Delta)}=0\) \\ \hline
100 & \(+U_{\text{DC}}\) & 0 & \(-U_{\text{DC}}\) & \(\mathbb{E}_{1(\Delta)}=\frac{2}{\sqrt{3}}\,U_{\text{DC}}\text{e}^{\frac{\pi}{ 3}}\) \\ \hline
110 & 0 & \(+U_{\text{DC}}\) & \(-U_{\text{DC}}\) & \(\mathbb{E}_{2(\Delta)}=\frac{2}{\sqrt{3}}\,U_{\text{DC}}\text{e}^{\frac{\pi}{ 3}}\) \\ \hline
010 & \(-U_{\text{DC}}\) & \(+U_{\text{DC}}\) & 0 & \(\mathbb{E}_{3(\Delta)}=\frac{2}{\sqrt{3}}\,U_{\text{DC}}\text{e}^{\frac{\pi_{ 3}}{3}}\) \\ \hline
011 & \(-U_{\text{DC}}\) & 0 & \(+U_{\text{DC}}\) & \(\mathbb{E}_{4(\Delta)}=\frac{2}{\sqrt{3}}\,U_{\text{DC}}\text{e}^{-\frac{1}{ 3}}\) \\ \hline
001 & 0 & \(-U_{\text{DC}}\) & \(+U_{\text{DC}}\) & \(\mathbb{E}_{5(\Delta)}=\frac{2}{\sqrt{3}}\,U_{\text{DC}}\text{e}^{-\frac{1}{ 3}}\) \\ \hline
101 & \(+U_{\text{DC}}\) & \(-U_{\text{DC}}\) & 0 & \(\mathbb{E}_{6(\Delta)}=\frac{2}{\sqrt{3}}\,U_{\text{DC}}\text{e}^{-\frac{1}{ 3}}\) \\ \hline
111 & 0 & 0 & 0 & \(\mathbb{E}_{7(\Delta)}=0\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Switching combinations and corresponding voltage vectors for delta-connected machine. Figure 3: Six active voltage vectors in the stator-fixed \(\alpha\beta\) system: (**a**) delta-connected winding; (**b**) star-connected winding. \[u_{\rm c}=u_{\rm c0}-u_{\rm a0}, \tag{16}\]  
or more conveniently in a matrix form  
\[\left[ \begin{array}{c}u_{\rm a}\\ u_{\rm b}\\ u_{\rm c}\end{array} \right]=\left[ \begin{array}{ccc}1&-1&0\\ 0&1&-1\\ -1&0&1\end{array} \right]\left[ \begin{array}{c}u_{\rm a0}\\ u_{\rm b0}\\ u_{\rm c0}\end{array} \right]. \tag{17}\]  
Neglecting the dead-time effects and finite switching times, the inverter phase voltages can take two values: \(+U_{\rm DC}/2\) if an upper switch is on, and \(-U_{\rm DC}/2\) if a lower switch is on. This can be summarized as  
\[u_{\rm x0}=\frac{U_{\rm DC}}{2}(2S_{x}-1)x={\rm a,b,c}. \tag{18}\]  
Finally, the motor phase voltages can be reconstructed using (14)-(16) and (18) as  
\[u_{\rm a}=(S_{\rm a}-S_{\rm b})U_{\rm DC}, \tag{19}\]  
\[u_{\rm b}=(S_{\rm b}-S_{\rm c})U_{\rm DC}, \tag{20}\]  
\[u_{\rm c}=(S_{\rm c}-S_{\rm a})U_{\rm DC}. \tag{21}\]  
Transferring (19)-(21) into matrix form then gives  
\[\left[ \begin{array}{c}u_{\rm a}\\ u_{\rm b}\\ u_{\rm c}\end{array} \right]=U_{\rm DC}\left[ \begin{array}{ccc}1&-1&0\\ 0&1&-1\\ -1&0&1\end{array}