# Steady State Response of Linear Time Invariant Systems Modeledby Multibond Graphs
## Appendix D Calculation of \(\mathbf{P}_{X}\)  
From (69):  
\[\mathbf{P}_{x}^{-1}\mathbf{X}_{GY}^{d}=\mathbf{I}-\mathbf{J}_{12}^{21} \mathbf{L}^{d}\mathbf{J}_{21}^{21}\mathbf{X}_{GY}^{d}\] (A37)  
and substituting (65), (54), and (67) into (A37):  
\[\left(\mathbf{X}_{GY}^{d}\right)^{-1}\mathbf{P}_{x}=\left[ \begin{array}{ccc} \mathbf{I}+\mathbf{R}_{1}\mathbf{X}_{1}^{-1}(wL_{1})&\mathbf{0}&\mathbf{X}_{3} (wL_{3})\\ \mathbf{0}&\mathbf{I}&\mathbf{X}_{3}(wL_{3})\\ -\mathbf{X}_{1}^{-1}(wL_{1})&-\mathbf{X}_{2}^{-1}(wL_{2})&\mathbf{I}\end{array}  \right]^{-1}\] (A38)  
In order to obtain (A38), the following partition is done [44]:  
\[\left[ \begin{array}{ccc}\mathbf{I}&\mathbf{0}&\mathbf{X}_{3}(wL_{3})\\ \mathbf{0}&\mathbf{I}&\mathbf{X}_{3}(wL_{3})\\ -\mathbf{X}_{1}^{-1}(wL_{1})&-\mathbf{X}_{2}^{-1}(wL_{2})&\mathbf{I}\end{array}  \right]^{-1}=\left[ \begin{array}{ccc}\mathbf{R}_{11}&\mathbf{R}_{12}\\ \mathbf{R}_{21}&\mathbf{R}_{22}\end{array} \right]^{-1}\] \[=\left[ \begin{array}{ccc}\widehat{\Delta_{R}}^{-1}&-\widehat{ \Delta_{R}}^{-1}\mathbf{R}_{12}\mathbf{R}_{22}^{-1}\\ -\mathbf{R}_{22}^{-1}\mathbf{R}_{21}\widehat{\Delta_{R}}^{-1}&\mathbf{R}_{22}^ {-1}+\mathbf{R}_{22}^{-1}\mathbf{R}_{21}\widehat{\Delta_{R}}^{-1}\mathbf{R}_ {12}\mathbf{R}_{22}^{-1}\end{array} \right]\] (A39)  
where:  
\[\widehat{\Delta_{R}}=\mathbf{R}_{11}-\mathbf{R}_{12}\mathbf{R}_{21}^{-1} \mathbf{R}_{21}\] (A40)  
then:  
\[\mathbf{R}_{11} = \left[ \begin{array}{ccc}\mathbf{I}&\mathbf{0}\\ \mathbf{0}&\mathbf{I}\end{array} \right]\text{; }\mathbf{R}_{12}=\left[ \begin{array}{c} \mathbf{X}_{3}(wL_{3})\\ \mathbf{X}_{3}(wL_{3})\end{array} \right]\] (A41) \[\mathbf{R}_{21} = \left[ \begin{array}{ccc}-\mathbf{X}_{1}^{-1}(wL_{1})&-\mathbf{X }_{2}^{-1}(wL_{2})\end{array} \right]\text{; }\mathbf{R}_{22}=\mathbf{I}\]  
By substituting (A41) into (A40):  
\[\widehat{\Delta_{R}}=\left[ \begin{array}{ccc}\mathbf{I}+\mathbf{R}_{1} \mathbf{X}_{1}^{-1}(wL_{1})+\mathbf{X}_{3}(wL_{3})\mathbf{X}_{1}^{-1}(wL_{1})& \mathbf{X}_{3}(wL_{3})\mathbf{X}_{2}^{-1}(wL_{2})\\ \mathbf{X}_{3}(wL_{3})\mathbf{X}_{1}^{-1}(wL_{1})&\mathbf{I}+\mathbf{X}_{3}( wL_{3})\mathbf{X}_{2}^{-1}(wL_{2})\end{array} \right]\] (A42)  
To get the inverse of (A42), we apply (A39) one more time, with the partition:  
\[\left[ \begin{array}{ccc}\mathbf{I}+\mathbf{R}_{1}\mathbf{X}_{1}^{-1}(wL_{1 })+\mathbf{X}_{3}(wL_{3})\mathbf{X}_{1}^{-1}(wL_{1})&\mathbf{X}_{3}(wL_{3}) \mathbf{X}_{2}^{-1}(wL_{2})\\ \mathbf{X}_{3}(wL_{3})\mathbf{X}_{1}^{-1}(wL_{1})&\mathbf{I}+\mathbf{X}_{3}( wL_{3})\mathbf{X}_{2}^{-1}(wL_{2})\end{array} \right]=\left[ \begin{array}{ccc} \mathbf{W}_{11}&\mathbf{W}_{12}\\ \mathbf{W}_{21}&\mathbf{W}_{22}\end{array} \right]\] (A43)  
with:  
\[\widehat{\Delta_{W}}=\mathbf{W}_{11}-\mathbf{W}_{12}\mathbf{W}_{22}^{-1} \mathbf{W}_{21}\] (A44)  
From (A43) and (A44) \(\mathbf{W}_{22}^{-1}=\left[\mathbf{I}+\mathbf{X}_{3}(wL_{3})\mathbf{X}_{2}^{- 1}(wL_{2})\right]^{-1}\) and from (57) and \(\mathbf{X}_{i}^{-1}(wL_{i})=\left[ \begin{array}{ccc}0&-\frac{1}{wL_{i}}&0\\ \frac{1}{wL_{i}}&0&0\\ 0&0&0\end{array} \right]\), we have \(\mathbf{W}_{22}^{-1}=diag\left\{\frac{L_{2}}{L_{2}+L_{3}},\frac{L_{2}}{L_{2}+L_{ 3}},1\right\}\). For simplicity, we use the notation \(D\left(\frac{L_{2}}{L_{2}+L_{3}},\frac{L_{2}}{L_{2}+L_{3}},1\right)=diag\left\{ \frac{L_{2}}{L_{2}+L_{3}},\frac{L_{2}}{L_{2}+L_{3}},1\right\}\). \(\mathbf{W}_{12}=D\left(\frac{L_{3}}{L_{2}},\frac{L_{3}}{L_{2}},0\right)\), \(\mathbf{W}_{21}=D\left(\frac{L_{3}}{L_{1}},\frac{L_{3}}{L_{1}},0\right)\). \[\widehat{\Delta_{R}}^{-1}=\left[ \begin{array}{ccc}\widehat{\Delta_{W}}^{-1}&- \widehat{\Delta_{W}}^{-1}\mathbf{W}_{12}\mathbf{W}_{22}^{-1}\\ -\mathbf{W}_{22}^{-1}\mathbf{W}_{21}\widehat{\Delta_{W}}^{-1}&\mathbf{W}_{22}^{-1 }+\mathbf{W}_{22}^{-1}\mathbf{W}_{21}\widehat{\Delta_{W}}^{-1}\mathbf{W}_{12} \mathbf{W}_{22}^{-1}\end{array} \right]\] (A45)  
The circuit can be solved with the well-known mesh formulation described by:  
\[\left[ \begin{array}{cc}R_{1}+J2\pi F(L_{1}+L_{3})&J2\pi FL_{3}\\ J2\pi FL_{3}&J2\pi F(L_{2}+L_{3})\end{array}