## 3 Transient Analysis of PES System Using Fourier Integral Transform
### Case of Passive Resistive-Inductive Load Without Back-Electromotive Force (emf)  
The connection of the electrical circuit is shown in Figure 7a, and the principal courses of the input and output (load) voltages in Figure 7b. For AC symmetric waveform is valid  
\[f(t)=a_{1}\mathrm{cos}\,\omega t+\dots+a_{v}\mathrm{cos}\,\
u\omega t+b_{1} \mathrm{sin}\,\omega t+\dots+b_{v}\mathrm{sin}\,\
u\omega t. \tag{18}\]  
Fourier coefficients for the v-harmonics type of cosine (see Appendix A)  
\[ \begin{split} a_{v}=U_{m}\frac{2}{\gamma}\int\limits_{0}^{T} \mathrm{cos}\,\omega t.\mathrm{cos}\,\
u\omega t\ \mathrm{d}t\\ =U_{m}\frac{1}{\pi}\Big{[}\frac{\mathrm{sin}(1-\
u)\beta-\mathrm{ sin}(1-\
u)\alpha}{(1-\
u)}+\frac{\mathrm{sin}\,(1+\
u)\beta-\mathrm{sin}\,(1+ \
u)\alpha}{(1+\
u)}\Big{]}.\end{split}  \tag{18a}\]  
Figure 7: Schematics (**a**) and courses of network and load voltages (**b**). As \(=2k+1\), then \(1-v=2k\) and \(1+v=2k+2\)  
\[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]  
And since  
\[\underset{k=0}{\text{lim}}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}= \beta-\alpha.\]  
Then  
\[a_{1}=U_{m}\frac{1}{\pi}\bigg{[}(\beta-\alpha)-\frac{\text{sin}\,2\beta-\text{ sin}\,2\alpha}{2}\bigg{]}. \tag{18c}\]  
Similarly for sine coefficients (see Appendix B)  
\[b_{v}=U_{m}\frac{2}{\pi}\int\limits_{0}^{T}\cos\omega t.\text{sin}\,\
u \omega t\text{ d}t \tag{19a}\] \[=U_{m}\frac{1}{\pi}(-1)^{\
u}\bigg{[}\frac{\cos(n+1)\beta-\cos(n-1) \alpha}{\
u-1}-\frac{\cos(n+1)\beta-\cos(n+1)\alpha}{\
u+1}\bigg{]}.\]  
As \(=2k+1\), then \(1-\
u=2k\) and \(1+\
u=2k+2\)  
\[b_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{cos}2k\beta-\text{cos}2k\alpha}{2 k}+\frac{\cos 2(k+1)\beta-\cos 2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{19b}\]  
And since  
\[\underset{k=0}{\text{lim}}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}=0,\]  
\[b_{1}=-U_{m}\frac{1}{\pi}\frac{\cos 2\beta-\text{cos}\,2\alpha}{2}. \tag{19c}\]  
Thus, the input voltage expressed by the Fourier series is defined using Equation (1). Then, for the load current using Equations (18b) and (19)  
\[i(t)=U_{m}\sum_{\
u=1}^{\infty}c_{\
u}\frac{\cos\left(\
u\omega t-\theta_{v}- \varphi_{\
u}\right)-\cos\left(\theta_{v}-\varphi_{\
u}\right)e^{-t/\tau_{\
u} }}{\sqrt{R^{2}+\left(\
u\omega L\right)^{2}}}\] (20a) where \[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]0 and \[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]1. Note: with a purely inductive load ( \[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]2 ), the transient component \[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]3 will be zero  
\[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]4  
Note: if \(\alpha=0\) and \(R=0\), the load current will be a purely sinusoidal one, i.e., for the fundamental harmonic  
\[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]5  
If the load current \(i(t)\) will be zero, then \(\omega t=\beta\) and so  
\[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]6  
from where it is possible to obtain the angle \(\beta\), e.g., in an iterative way. If \(\beta=\pi-\alpha\), it means that the load is purely inductive.