## 3 Transient Analysis of PES System Using Fourier Integral Transform
### Case of Passive Resistive-Inductive Load Without Back-Electromotive Force (emf)
The graphic interpretation of Equation (20b) is shown in \[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]7 The parameters of the system are given in \[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]8  
The solution to the transient phenomenon of the system is significant only when the current is continuous, with the turn-on angle equal to the phase shift angle, i.e., \(36.76-90=-53.24\) electrical degrees. Under this condition, the current will be purely sinusoidal. During the transient state, when switching occurs at \(\alpha=0\), the current will increase from zero to its steady-state value according to the modified form of relation Equation (11a) for the fundamental harmonic  
\[a_{v}=U_{m}\frac{1}{\pi}\bigg{[}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}+ \frac{\text{sin}\,2(k+1)\beta-\text{sin}\,2(k+1)\alpha}{2(k+1)}\bigg{]}. \tag{18b}\]9  
The waveform for \(\alpha=0\) is shown in \[\underset{k=0}{\text{lim}}\frac{\text{sin}2k\beta-\text{sin}2k\alpha}{2k}= \beta-\alpha.\]0 \begin{table}
\begin{tabular}{c c c c c c c} \hline
**Um** & \(f_{1}\) **R** & **L** & **Z** & \(\tau_{1}\) & \(\varphi_{1}\) \\
**[V]** & **[Hz]** & **[W]** & **[mH]** & **[W]** & **[ms]** & **[deg]** \\ \hline
325 & 50 & 18.4 & 43.93 & 23 & 2.3875 & 36.76 \\ \hline \end{tabular}
\end{table} 
Table 1: Parameters of the system. Figure 8: Time waveform of the load current and load and network voltages **(a)**; the decomposition of the total current (**b**). Figure 9: Time waveforms of input voltage and load current. Since the equation consists of steady- and transient-state components, we can write  
\[ \begin{array}{c}F_{i_{RL}}(j\omega)=\mathbf{F}_{st-st}(j\omega)+F_{trans}(j\omega)\\ =\frac{U_{m}}{Z}\int\limits_{0}^{\infty}\cos(\omega_{1}t-\varphi_{1})e^{-j \omega t}\mathrm{d}t+\frac{U_{m}}{Z}\int\limits_{0}^{\infty}\cos(-\varphi_{1} )e^{-t/\tau_{1}}e^{-j\omega t}\mathrm{d}t.\end{array}  \tag{22}\]  
The Fourier transform of \(i_{RL}(t)\) function Equation (21) can be derived (a simplified approach [1,11]) with the time-shift rule,  
\[f(t\pm t_{1})\leftrightarrow e^{j\omega t_{1}}\mathcal{F}\{f(t)\}, \tag{23}\]  
where \(t_{1}=\frac{\varphi_{1}}{\omega_{1}}\). Then, for steady-state component  
\[ \begin{array}{c}\mathbf{F}_{st-st}(j\omega)=e^{j\varphi_{1}}\mathcal{F}_{st-st} \{u(t)\text{cos}\omega_{1}t\}=\frac{U_{m}}{Z}e^{j\varphi_{1}}\mathcal{F}_{st- st}\Big{\{}\frac{1}{2}\big{[}e^{j\omega_{1}t}+e^{-j\omega_{1}t}\big{]} \Big{\}}\\ =\frac{U_{m}}{Z}\frac{1}{Z}e^{j\varphi_{1}}\{\mathbf{F}_{u}[j(\omega-\omega_{1}) +\mathbf{F}_{u}[j(\omega+\omega_{1})]\}\\ =\frac{U_{m}}{Z}\frac{1}{Z}e^{i\omega\frac{\varphi_{1}}{\omega_{1}}}\big{[} \tau\delta(\omega-\omega_{1})+\frac{1}{j(\omega-\omega_{1})}+\tau\delta( \omega+\omega_{1})+\frac{1}{j(\omega+\omega_{1})}\big{]}\\ =\frac{U_{m}}{Z}\bigg{\{}\frac{\pi}{2}[\delta(\omega-\omega_{1})+\delta( \omega+\omega_{1})]e^{j\varphi_{1}}+\frac{i\omega}{(\omega_{1}^{2}-\omega^{2} )}e^{j\varphi_{1}}\bigg{\}}.\end{array}  \tag{24}\]  
And, for transient component  
\[\mathbf{F}_{trans}(j\omega)=\mathcal{F}_{trans}\bigg{\{}\frac{U_{m}}{Z}\text{cos}( -\varphi_{1})e^{-t/\tau_{1}}\bigg{\}}=\frac{U_{m}}{Z}\text{cos}\ \varphi_{1}\frac{1}{j\omega+\frac{1}{\tau_{1}}}. \tag{25}\]  
The graphic interpretation of the amplitude spectra of both components is shown in Figure 10 in the relative p.u. units (i.e., \(U_{m}/Z\equiv 1\)). Given that the time constant \(\tau_{1}\) is relatively small (2.3875 ms), which is less than one-fourth of the time period, the predominant component of the response is the steady-state component.