# Steady State Response of Linear Time Invariant Systems Modeledby Multibond Graphs
## 5 Case Study
### Three-Phase Electrical System
\begin{array}{c} \varepsilon_{4}^{d}(t)\\ \varepsilon_{4}^{d}(t)\\ \varepsilon_{4}^{d}(t)\end{array} \right]\text{; }\underline{f}_{4}(t)=\left[ \begin{array}{c}f_{4}^{d}(t)\\ f_{4}^{d}(t)\\ f_{4}^{d}(t)\end{array} \right]\]  
with the voltage inputs given by:  
\[\underline{e}_{1}(t)=\left[ \begin{array}{c}v_{1}^{d}(t)\\ v_{1}^{d}(t)\\ v_{1}^{d}(t)\end{array} \right]\text{; }\underline{e}_{2}(t)=\left[ \begin{array}{c}v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\end{array} \right]\]  
By substituting (A46) into (71), the steady state responses in terms of the previous physical variables are obtained by:\[\underline{f_{3}}(t)=\left[ \begin{array}{c}f_{4}^{d}(t)\\ f_{3}^{d}(t)\\ f_{3}^{0}(t)\end{array} \right]=\frac{1}{\Lambda}\left[ \begin{array}{cccc} \frac{R_{4}^{d}}{(wt_{1})^{2}}&\frac{-(1+k)}{wL_{1}}&0&\frac{-R_{4}^{d}k_{1}}{( wL_{1})^{2}}&\frac{(1+k)k_{1}}{wL_{1}}&0\\ \frac{(1+k)}{wL_{1}}&\frac{k_{1}^{d}}{(wL_{1})^{2}}&0&\frac{-(1+k)k_{1}}{wL_{1} }&\frac{-R_{4}^{d}k_{1}}{(wL_{1})^{2}}&0\\ 0&0&0&0&0\end{array} \right]\left[ \begin{array}{c}v_{1}^{d}(t)\\ v_{1}^{d}(t)\\ v_{1}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\end{array} \right] \tag{72}\]  
\[\underline{f_{4}}(t)=\left[ \begin{array}{c}f_{4}^{d}(t)\\ f_{4}^{d}(t)\\ f_{4}^{d}(t)\\ f_{4}^{0}(t)\end{array} \right]=\frac{1}{\Lambda}\left[ \begin{array}{cccc} \frac{-R_{4}^{d}k}{w^{2}L_{1}L_{2}}&\frac{(1+k)k}{wL_{2}}&0&\frac{-R_{4}^{d}k_ {1}}{w^{2}L_{1}L_{2}}&+\frac{\frac{(1+k)k_{1}}{wL_{1}}}{wL_{2}}&0\\ \frac{-(1+k)k}{wL_{2}}&\frac{-R_{4}^{d}k}{w^{2}L_{1}L_{2}}&0&\frac{-(1+k)kk_{ 1}}{wL_{1}}&-\frac{R_{4}^{d}k_{1}}{w^{2}L_{1}L_{2}}&0\\ 0&0&0&0&0\end{array} \right]\left[ \begin{array}{c}v_{1}^{d}(t)\\ v_{1}^{d}(t)\\ v_{1}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ v_{2}^{d}(t)\\ \end{array} \right] \tag{73}\]  
Furthermore, the resistances and inductances on the \((d,q,0)\) axis are defined by:  
\[\mathbf{R}^{dq0} = \mathbf{P}\mathbf{R}^{abc}\mathbf{p}^{-1} \tag{74}\] \[\mathbf{L}^{dq0} = \mathbf{P}\mathbf{L}^{abc}\mathbf{p}^{-1} \tag{75}\]  
Unbalanced conditions may be simulated by appropriate modification of the parameters of the multiport system. Unbalanced conditions such as unbalanced voltages, unsymmetrical inductances, and unsymmetrical resistors with appropriate changes can be analyzed. However, these possible unbalanced conditions are not considered in this paper and can be treated for future works. In order to demonstrate the effectiveness of the proposed methodology, the steady state response will be obtained from the multibond graph in an integral causality assignment that represents the dynamic system and the direct substitution in (72) and (73). Considering a balanced system, the numerical parameters of the elements are: \(\mathbf{R}^{abc}=diag\{10\,\Omega,10\,\Omega,10\,\Omega\},\mathbf{L}_{1}^{abc }=diag\{0.1\,\mathrm{H},0.1\,\mathrm{H},0.1\,\mathrm{H}\},\mathbf{L}_{2}^{ abc}=diag\{0.15\,\mathrm{H},0.15\,\mathrm{H},0.15\,\mathrm{H}\},\)\(\mathbf{L}_{3}^{abc}=diag\{0.5\,\mathrm{H},0.5\,\mathrm{H},0.5\,\mathrm{H}\},\)\(F=60\) Hz and the supply voltages  
\[v_{1}^{abc}(t)=\left[ \begin{array}{c}200\cos(377t)\\ 200\cos(377t-120)\\ 200\cos(377t+120)\end{array} \right]V\;\text{and}\;v_{2}^{abc}(t)=\left[ \begin{array}{c}100\cos(377t)\\ 100\cos(377t-120)\\ 100\cos(377t+120)\end{array} \right]V.\]  
When a system is balanced \(\mathbf{R}^{dq0}=\mathbf{R}^{abc}\), \(\mathbf{L}^{dq0}=\mathbf{L}^{abc}\) and the voltages are:  
\[v_{1}^{dq0}(t)=\left[ \begin{array}{c}-0.5438\\ 244.9483\\ 0\end{array} \right]V;\;v_{2}^{dq0}(t)=\left[ \begin{array}{c}-0.2719\\ 122.4741\\ 0\end{array} \right]V\]  
From (72) and (73), the steady state for the electrical currents is given by:  
\[\left[ \begin{array}{c}\left(f_{3}^{d}\right)_{ss}\\ \left(f_{3}^{d}\right)_{ss}\\ \left(f_{3}^{d}\right)_{ss}\end{array} \right]=\left[ \begin{array}{c}-1.8291\\ 0.2227\\ 0\end{array} \right]A;\;\left[ \begin{array}{c}\left(f_{4}^{d}\right)_{ss}\\ \left(f_{4}^{d}\right)_{ss}\\ \left(f_{4}^{0}\right)_{ss}\end{array}