# A New Modified Non-Approximative Method for Dynamic Systems Direct Calculation
## 3 Problem Solution and Simulation Example
### Case of Constant Excitation Function  
Furthermore, assume that \(u_{1}(t)=U\), i.e., constant (equal to \(100\ (V)\)), and \(u_{2}(t)=0\). Regarding time domain investigation, the Laplace-Carson operator method has been used [3; 6]. Applying the Laplace-Carson transform, we obtain the operator form of variables the \(X_{1}(p),\ X_{2}(p)\) of Equation (10a,b)  
\[pX_{1}(p)=a_{11}X_{1}(p)+a_{12}X_{2}(p)+b_{11}U \tag{11a}\]  
\[pX_{2}(p)=a_{11}X_{1}(p)+a_{12}X_{2}(p)+0 \tag{11b}\]  
with the initial conditions \(X_{1}(0=0),\ X_{2}(0)\). After the separation of \(X_{1}(p)\) and \(X_{2}(p)\)  
\[X_{1}(p)=U\frac{A(p)}{B(p)};\ X_{2}(p)=U\frac{C(p)}{D(p)} \tag{12}\]and using the inverse transform and lemma of residua [4; 6],  
\[x_{1}(t)=U\bigg{[}\frac{A(0)}{B(0)}+\sum_{k=1}^{N}\frac{A(p_{k})}{p_{k}B\prime(p_ {k})}\mathrm{e}^{p_{k}t}\bigg{]} \tag{13a}\] \[x_{2}(t)=U\bigg{[}\frac{A(0)}{B(0)}+\sum_{k=1}^{N}\frac{C(p_{k})}{p_{k}D\prime(p_ {k})}\mathrm{e}^{p_{k}t}\bigg{]} \tag{13b}\]  
we obtain, under zero initial conditions,  
\[x_{1}(t)=[\left(1-\cos\Omega t.\mathrm{e}^{-\frac{t}{\Gamma_{C}}}\right)- \frac{1}{\Omega}.\bigg{(}\frac{1}{\tau_{C}}-\frac{1}{\tau_{L}}\bigg{)}\,\sin \Omega t.\mathrm{e}^{-\frac{t}{\Gamma_{C}}}]\,\,U/R \tag{14a}\] \[x_{2}(t)=\bigg{[}1-\bigg{(}\cos\Omega t+\frac{1}{\tau_{C}\Omega}.\sin\Omega t \bigg{)}\mathrm{e}^{-\frac{t}{\Gamma_{C}}}\bigg{]}U \tag{14b}\]  
where \(\tau_{C}(\equiv 2 RC)=20\times 10^{-3}[\mathrm{sec}]\), \(\tau_{L}(\equiv L/R)=0.1\times 10^{-3}[\mathrm{sec}]\), \(\Omega\Big{(}\equiv\sqrt{\frac{1}{LC}-\frac{1}{\tau_{C}^{2}}}\Big{)}=3154.36\)\([1/\sec]\). For the time waveforms of \(x_{1}(t)\) and \(x_{2}(t)\), see Figure 1.  
These waveforms are the response of the system to a unit step of \(U\). During the calculation, continuous time is substituted by the discrete time instants \(k.h\), where \(h\) is the integration step and \(k\) is the step order. The numerical values of the variables at the discrete time instant \(n.t\) are listed in Table 1 when the time constant \(t=RC\) equals 10 msec. \[x_{10}(t)=\bigg{[}\cos\Omega t+\frac{1}{\Omega\tau_{C}}\sin\Omega t\bigg{]} \mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{1}(0)+\frac{1}{\Omega C_{1}}\,\sin \Omega t.\mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{2}(0), \tag{15a}\] \[x_{20}(t)=\bigg{[}\cos\Omega t-\frac{1}{\Omega\tau_{C}}\,\sin\Omega t\bigg{]} \mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{2}(0)+\frac{1}{\Omega C_{2}}\,\sin \Omega t.\mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{1}(0). \tag{15b}\]  
\[pX_{2}(p)=a_{11}X_{1}(p)+a_{12}X_{2}(p)+0 \tag{11b}\]0 Time waveforms of variables \(x_{1}(t)\) and \(x_{2}(t)\) for one half-period (**a**) and steady state (**b**). \begin{table} \begin{tabular}{c c c c c c} \hline \(n\) & **0** & **1** & **2** & **3** & **\(\dots\)** & **\(\Psi\)** \\ \hline \(x_{1}(n.t)\) & 0 & 42.6942 & 52.5253 & 48.5001 & \(\dots\)** & \(\dots\)** & \(10\) \\ \(x_{2}(n.t)\) & 0 & 40.3978 & 65.9700 & 81.4927 & \(\dots\)** & \(\dots\)** & \(100\) \\ \hline \end{tabular}  Note: with respect to non-zero initial conditions, we obtained the response to internal state change. \[x_{10}(t)=\bigg{[}\cos\Omega t+\frac{1}{\Omega\tau_{C}}\sin\Omega t\bigg{]} \mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{1}(0)+\frac{1}{\Omega C_{1}}\,\sin \Omega t.\mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{2}(0),\] (15a) \[x_{20}(t)=\bigg{[}\cos\Omega t-\frac{1}{\Omega\tau_{C}}\,\sin\Omega t\bigg{]} \mathrm{e}^{-\frac{t}{\Gamma_{C}}}\,\,X_{2}(0)+\frac{1}{\Omega C_{2}}\,\sinThe total response is:  
\[x_{1}(t)=[\left(1-\cos\Omega t.\mathrm{e}^{-\frac{t}{\mathrm{C}}}\right)-\frac{1}{ \Omega}.\left(\frac{1}{\tau_{\mathrm{C}}}-\frac{1}{\tau_{\mathrm{L}}}\right) \sin\Omega t.\mathrm{e}^{-\frac{t}{\mathrm{C}}}]\;U/C_{3}+x_{10}(t), \tag{16a}\] \[x_{2}(t)=\left[1-\left(\cos\Omega t+\frac{1}{\tau\Omega}\sin\Omega t\right) \mathrm{e}^{-\frac{t}{\mathrm{C}}}\right]U+x_{20}(t), \tag{16b}\]  
where \(C_{1}=0.1\times 10^{-3}\), \(C_{2}=1\times 10^{-3}\), \(C_{3}=10\), and \(U=100\).