## 2 Modeling of Five-Phase Induction Motor
### Mathematical Model of Five-Phase Induction Motor in Normal Operation
Mathematical Expression of the Modification of a Four-Phase Operation for a Five-Phase Induction Motor  
This subsection provides a detailed mathematical derivation of the phase current adjustment for a sinusoidally distributed MMF for one-phase failure. If one phase is interrupted, either on the motor side, the converter side, or between the motor and the power converter, the resulting MMF changes. For this case is considered the failure of phase \(a\). The resulting force will be expressed as follows:  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]3  
As it can be seen from (20), the current through phase a is equal to zero. By expressing this equation and subsequent separation into the real and imaginary part we obtain (21) and (22):  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]4  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]5  
The condition for the MMF to be evenly distributed in the one-phase failure is as follows:  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]6  
In the following steps, a detailed procedure for deriving the phase \(i_{b}\) current calculation is given. Equations (24) and (25) represent a suitable substitution from (23) for the expression of current \(i_{b}\):  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]7  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]8Relationships (26)-(32) represent the multiplication and expression of currents from (24) and (25):  
\[U_{qs}=R_{s}i_{qs}-\omega_{a}\psi_{ds}+\rho\psi_{qs}; U_{qr}=R_{r}i_{qr}+(\omega_{a}-\omega)\psi_{dr}+\rho\psi_{qr}, \tag{9}\]9  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]0  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]1  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]2  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]3  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]4  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]5  
Thus, (33) is obtained using the formulas for \(cos\) (\(x\)) \(-\)\(cos\) (\(y\)) and \(sin\) (\(x\)) \(+\)\(sin\) (\(y\)), which is substituted into (32):  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]6  
By multiplying the (33) and using the formula \(cos\) (\(x\) \(+\)\(y\)), (34) is obtained:  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]7  
The resulting modifications give the final expression of the current \(i_{b}\) for the adjustment of the operation of the five-phase induction motor in a single-phase failure:  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]8  
In a similar manner, given in (21)-(35), the equations of the other three currents, \(i_{c}\), \(i_{d}\), and \(i_{e}\), are obtained, which are given in (36)-(38):  
\[U_{xs}=R_{s}i_{xs}+\rho\psi_{xs}; U_{xr}=R_{r}i_{xr}+\rho\psi_{xr}, \tag{10}\]9  
\[U_{ys}=R_{s}i_{ys}+\rho\psi_{ys}; U_{yr}=R_{r}i_{yr}+\rho\psi_{yr}, \tag{11}\]0  
\[U_{ys}=R_{s}i_{ys}+\rho\psi_{ys}; U_{yr}=R_{r}i_{yr}+\rho\psi_{yr}, \tag{11}\]1  
Graphical Representation of the Modification of the Four-Phase Operation for a Five-Phase Induction Motor  
Figure 4 depicts a graphical representation of the foregoing mathematical modification of a five-phase induction motor running in phase failure \(a\). This lowers torque ripple while increasing efficiency [34; 35]. The graphic shows that phase \(a\) is open, hence the current \(i_{a}\) is equal to zero.