# On Invertibility of an Interconnected System Composed of Two Dynamic Subsystems
## 4 On Computation of the Dynamics Inverse of the Interconnected System
\right] \tag{11}\]  
\(\Delta(\mathrm{x}_{H})\) is a nonsingular matrix with full rank:  
\[\mathrm{rank}\,\,\Delta(\mathrm{x}_{H})=\mathrm{m} \tag{12}\]  
In order to derive a function of states and output in (2) to represent \(u_{L}(t)\) as, the first step is to differentiate \(\,y_{i}\,\mathrm{i}=1,\ldots,\mathrm{m}\) to obtain the derivatives:  
Suppose \(r_{i}\,=1\), one can get:  
\[\mathrm{y}_{i}^{(1)}=\frac{\partial\mathrm{h}_{\mathrm{H}i}( \mathrm{x}_{H})}{\partial\mathrm{x}_{H}}\,\dot{x}_{H}(\mathrm{t})\] \[\quad=\frac{\partial\mathrm{h}_{\mathrm{H}i}(\mathrm{x}_{H})}{ \partial\mathrm{x}_{H}}\,\big{(}f_{\mathrm{H}}(\mathrm{x}_{H})+\mathrm{g}_{H}( \mathrm{x}_{H})\,\,\mathrm{u}_{\mathrm{L}}\big{)}\] \[\quad=\mathrm{L}_{\mathrm{f}_{\mathrm{H}}}^{\mathrm{l}}\mathrm{h} _{\mathrm{H}i}(\mathrm{x}_{H})+\sum_{j=1}^{m}\mathrm{L}_{\mathrm{g}_{\mathrm{H} j}}^{\mathrm{L}_{\mathrm{f}_{\mathrm{H}}}^{\mathrm{l}}}\mathrm{h}_{\mathrm{H}i}( \mathrm{x}_{H})\mathrm{u}_{\mathrm{l}j}\]If \(r_{i}\
eq 1\), then \(\mathrm{L}_{\mathrm{R}_{H}}^{1}\mathrm{L}_{I_{H}}^{0}\mathrm{h}_{\mathrm{H} \mathrm{H}}(\mathrm{x}_{H})=0\); \(1\leq\mathrm{j}\leq\mathrm{m}\)  
In this way,  
\[\mathrm{y}_{\mathrm{i}}^{(1)}=\mathrm{L}_{I_{H}}^{1}\mathrm{h}_{\mathrm{H} \mathrm{H}}(\mathrm{x}_{H}) \tag{13}\]  
Generally speaking, it needs to continue this differential procedure, for \(k<r_{i}\), one will have  
\[\mathrm{y}_{\mathrm{i}}^{(\mathrm{j})}=\mathrm{L}_{I_{H}}^{j}\mathrm{h}_{ \mathrm{H}\mathrm{H}}(\mathrm{x}_{H}) \tag{14}\]  
Until the relative degree \(\mathrm{r_{i}}\), it reaches  
\[\mathrm{y}_{\mathrm{i}}^{(\mathrm{r_{i}})}=\mathrm{L}_{I_{H}}^{\mathrm{r_{i}}} \mathrm{h}_{\mathrm{H}\mathrm{H}}(\mathrm{x}_{H})+\sum_{\mathrm{j=1}}^{ \mathrm{m}}\mathrm{L}_{\mathrm{R}_{H\mathrm{j}}}(\mathrm{L}_{I_{H}}^{\mathrm{ r_{i}}-1}\mathrm{h}_{\mathrm{H}\mathrm{H}}(\mathrm{x}_{H}))\ \mathrm{u}_{\mathrm{j}}\ \mathrm{i}=1,\ldots,\mathrm{m} \tag{15}\]  
If there are \(m\) relative order \(r_{1},\ldots,r_{m}\) related to the output \(y\), and the total relative degree satisfied (16):  
\[\mathrm{r}=\sum_{\mathrm{i=1}}^{\mathrm{m}}\mathrm{r_{i}}=\mathrm{n} \tag{16}\]  
then calculating expressions for their derivatives can be referred to as a one-step algorithm to obtain an inverse, and we get  
\[\left[ \begin{array}{c}\mathrm{y}_{\mathrm{1}}^{(\mathrm{r_{1}})}\\ \vdots\\ \mathrm{y}_{\mathrm{m}}^{(\mathrm{r_{m}})}\end{array} \right]=\left[ \begin{array} []{c}\mathrm{L}_{I_{H}}^{\mathrm{r_{1}}}\mathrm{h}_{\mathrm{H}\mathrm{I}}( \mathrm{x}_{H})\\ \vdots\\ \mathrm{L}_{I_{H}}^{\mathrm{r_{m}}}\mathrm{h}_{\mathrm{H}\mathrm{H}}(\mathrm{x}_ {H})\end{array} \right]+\left[ \begin{array}{ccc}\mathrm{L}_{\mathrm{R}_{H \mathrm{H}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{1}}-1}\mathrm{h}_{\mathrm{H} \mathrm{H}\mathrm{I}}(\mathrm{x}_{H})&\ldots&\mathrm{L}_{\mathrm{R}_{H \mathrm{m}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{1}}-1}\mathrm{h}_{\mathrm{H} \mathrm{H}\mathrm{I}}(\mathrm{x}_{H})\\ \ldots&\ldots&\ldots\\ \mathrm{L}_{\mathrm{R}_{H\mathrm{H}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{m}}-1} \mathrm{h}_{\mathrm{H}\mathrm{H}\mathrm{H}}(\mathrm{x}_{H})&\ldots&\mathrm{L} _{\mathrm{R}_{H\mathrm{m}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{m}}-1}\mathrm{h}_{ \mathrm{H}\mathrm{H}\mathrm{H}}(\mathrm{x}_{H})\end{array} \right]\mathrm{u}_{ \mathrm{L}} \tag{17}\]  
the Equation (18) can be solved for \(\mathrm{u}_{\mathrm{L}}\) to obtain  
\[\mathrm{u}_{\mathrm{L}}=\left[ \begin{array}{ccc}\mathrm{L}_{\mathrm{R}_{H \mathrm{L}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{1}}-1}\mathrm{h}_{\mathrm{H} \mathrm{I}}(\mathrm{x}_{H})&\ldots&\mathrm{L}_{\mathrm{R}_{H\mathrm{m}}} \mathrm{L}_{I_{H}}^{\mathrm{r_{1}}-1}\mathrm{h}_{\mathrm{H}\mathrm{I}}( \mathrm{x}_{H})\\ \ldots&\ldots&\ldots\\ \mathrm{L}_{\mathrm{R}_{H\mathrm{L}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{m}}-1} \mathrm{h}_{\mathrm{H}\mathrm{H}\mathrm{m}}(\mathrm{x}_{H})&\ldots&\mathrm{L }_{\mathrm{R}_{H\mathrm{m}}}\mathrm{L}_{I_{H}}^{\mathrm{r_{m}}-1}\mathrm{h}_{ \mathrm{H}\mathrm{H}\mathrm{m}}(\mathrm{x}_{H})\end{array} \right]^{-1}\cdot \left(\left[