### VFD's Supply-Side Rectifier Operating Mode as a Periodic Transient Process  
To evaluate the effect of PDL internal inductance on the operation of the VFD's supply-side rectifier, it is necessary to calculate the transient that occurs when charging the DC circuit capacitor begins (Figure 5a). In transient process calculations, the sinusoid of the PDL voltage is shifted through \(\alpha\) so that the process calculation starts from zero. Such a transient process for calculating \(i(t)\) is described by [PERSON]'s integral (Figure 5b). The transient function:  
\[g(t)=\frac{1}{R}\Bigl{(}1-e^{\frac{\pi R}{R}t}\Bigr{)} \tag{8}\]  
The voltage of the circuit:  
\[u(t)=U_{m}\sin\Bigl{(}\omega t+\frac{\pi}{2}-\alpha\Bigr{)}-u_{c} \tag{9}\]  
Figure 5: Calculation of the transient process: (**a**) voltages; (**b**) equivalent circuit. The integral of [PERSON] for the current \(i(t)\) of this circuit is written as follows:  
\[i(t)=\int_{0}^{t}u^{{}^{\prime}}\ (\tau)\ g(t-\tau)d\tau \tag{10}\]  
where \(\tau\) is the intermediate variable. After rearranging the equation:  
\[g(t-\tau)=\frac{1}{R}\Big{(}1-e^{\frac{-R}{L}(t-\tau)}\Big{)} \tag{11}\]  
\[u^{\prime}(\tau)=U_{m}\omega\ \cos\Big{(}\omega t+\frac{\pi}{2}-\alpha\Big{)} \tag{12}\]  
After inserting into the integral:  
\[i(t)=\frac{U_{m}\omega}{R}\int_{0}^{t}\cos\Big{(}\omega t+\frac{\pi}{2}- \alpha\Big{)}\Big{(}1-e^{\frac{-R}{L}(t-\tau)}\Big{)}d\tau \tag{13}\]  
This integral is easy to divide into two parts and integrate them separately:  
\[i_{1}(t)=\frac{U_{m}\omega}{R}\int_{0}^{t}\cos\Big{(}\omega t+\frac{\pi}{2}- \alpha\Big{)}d\tau \tag{14}\]  
\[i_{2}(t)=-\frac{U_{m}\omega}{R}\int_{0}^{t}\cos\Big{(}\omega\tau+\frac{\pi}{ 2}-\alpha\Big{)}\Big{(}e^{\frac{-R}{L}(t-\tau)}\Big{)}d\tau \tag{15}\]  
The integration of the first integral yields the equation:  
\[i_{1}(t)=\frac{U_{m}\omega}{R}\Big{(}\sin\Big{(}\omega t+\frac{\pi}{2}-\alpha \Big{)}-\sin\Big{(}\frac{\pi}{2}-\alpha\Big{)}\Big{)} \tag{16}\]  
The first integral does not depend on the circuit inductance \(L\). The second integral is much more complex to integrate:  
\[u(t)=U_{m}\sin\Bigl{(}\omega t+\frac{\pi}{2}-\alpha\Bigr{)}-u_{c} \tag{9}\]0  
Figure 6 shows the position of the current pulse of one half-period VFD input and the shape of the curve calculated according to Equations (16)-(18). Calculations were performed by varying the value of the internal inductance of the PDL. The position of the curves shows that at the lowest PDL inductance, the impulse of the VFD input current curve is almost symmetrical with respect to the point \(\pi/2\), so \(tg\phi\) and the reactive power \(Q_{\text{Fund}}\) are close to zero. This is confirmed by the calculation data in \[u(t)=U_{m}\sin\Bigl{(}\omega t+\frac{\pi}{2}-\alpha\Bigr{)}-u_{c} \tag{9}\]1 Mathematically, only the component \(i_{1}(t)\) predominates in this case, and the other component \(i_{2}(t)\) is equal to zero. As the internal inductance of the PDL increases, \(tg\phi\) increases, thus increasing the \(i_{2}(t)\) component and the reactive power drawn out by the VFD. This is evident from the displacement of the VFD input current pulses shown in Figure 6 with respect to the PDL voltage. At the maximum values of the internal inductance of the PDL, \(tg\phi\) reaches value 0.32, which means that the reactive power provided or drawn out by the VFD is about one-third of the active power of the VFD. \[u(t)=U_{m}\sin\Bigl{(}\omega t+\frac{\pi}{2}-\alpha\Bigr{)}-u_{c} \tag{9}\]2  
\[u(t)=U_{m}\sin\Bigl{(}\omega t+\frac{\pi}{2}-\alpha\Bigr{)}-u_{c} \tag{9}\]3 VFD input current pulse \(i(t)\) position at different PDL internal \(R/\omega L\) ratios calculated on the basis of Equations (16)–(18). Another component that can affect the reactive power provided by VFD is the THD reduction inductive choke.