# A Robust Control Algorithm of a Descent Vehicle Angular Motion in the Earth's Atmosphere
## 6 Robust Output Regulator
\begin{split}\mathbf{F}_{RY}=\mathbf{K}_{RY}\underbrace{C_{RY}^ {T}\Big{(}C_{RY}C_{RY}^{T}\Big{)}^{-1}}_{\mathbf{\tilde{C}}_{RY1}^{T}}= \mathbf{I}_{RY}\underbrace{\begin{bmatrix}\kappa_{12}\bar{\mu}_{2}-\kappa_{11 }\bar{\mu}_{1}\\ \bar{f}_{RY1,1}\end{bmatrix}}_{\mathbf{\tilde{f}}_{RY1,1}}&\underbrace{\bar{ \mu}_{2}+\kappa_{12}}_{\mathbf{\tilde{f}}_{RY1,2}}&\underbrace{-\mu_{1}- \kappa_{11}}_{\mathbf{\tilde{f}}_{RY1,3}}\\ \underbrace{\kappa_{22}\bar{\mu}_{2}-\kappa_{21}\bar{\mu}_{1}}_{\mathbf{ \tilde{f}}_{RY2,1}}&\underbrace{\bar{\mu}_{1}+\kappa_{22}}_{\mathbf{\tilde{f} }_{RY2,2}}&\underbrace{-\mu_{2}-\kappa_{21}}_{\mathbf{\tilde{f}}_{RY3,3}} \end{bmatrix}.\end{split}  \tag{31}\]  
This matrix describes the set of solutions to the modal output control problem (14), (15), characterized by the parameters \(\mu_{1},\mu_{2}\) and the poles \(f_{1},\,f_{2},\,f_{3},\,f_{4}\). Symbolic calculations in MATLAB confirm the validity of the equation  
\[\text{eig}(A_{RY}-\mathbf{B}_{RY}\mathbf{F}_{RY}\mathbf{C}_{RY})=\Lambda_{RY}\]  
written on the basis of expressions (14), (15) and (31). In order to reduce the mutual influence of control channels, let us set the problem of zeroing the cross coefficients \(\tilde{f}_{RY1,3}\), \(\tilde{f}_{RY2,1}\) and \(\tilde{f}_{RY2,2}\) between the \"Roll\" and \"Yaw\" channels from the record (31), having at our disposal the parameters \(\mu_{1}\), \(\mu_{2}\) obeyingconditions (26) and (29), as well as any ratios of the poles \(\phi_{1}\), \(\phi_{2}\), \(\phi_{3}\), \(\phi_{4}\), that do not violate the location of these poles in the left complex half-plane. This problem is described by the following system of equations and inequalities:  
\[\left\{ \begin{array}{lcl}\tilde{f}_{RY1,3}=0&\Leftrightarrow&\kappa_{11}+\mu_ {1}=0,\\ \tilde{f}_{RY2,1}=0|_{\tilde{f}_{RY2,2}=0}&\Leftrightarrow&\begin{bmatrix} \tilde{\mu}_{1}=0,\\ \kappa_{21}+\tilde{\mu}_{2}=0,\\ \end{bmatrix}&\begin{array}{l}\mu_{1}\
eq 0,\\ \text{Re}\,\phi_{2}<0,\\ \mu_{2}\
eq\frac{a_{52}}{a_{4,2}}\mu_{1},\end{array}&\begin{array}{l}\mu_{ 1}\
eq 0,\\ \text{Re}\,\phi_{3}<0,\\ \text{Re}\,\phi_{4}<0.\end{array}\\ \end{array} \right. \tag{32}\]  
Since \(a_{52}>0\), the solution to problem (32) is one of the systems  
\[\left\{ \begin{aligned} \begin{bmatrix}\mu_{2}=\phi_{1},\\ \mu_{2}=\phi_{2},\\ \mu_{1}=\frac{a_{4,2}}{\phi_{4}-\mu_{2}},\\ \mu_{2}\phi_{3}=a_{5,2},\\ \end{aligned} \right. \tag{33}\]  
Here the first system corresponds to the first equation \(\tilde{\mu}_{1}=0\) in the set from the record (32), and the second system corresponds to the second equation \(\kappa_{21}+\tilde{\mu}_{2}=0\) in the same set. Thus, system (32) has five qualitatively different solutions. The first solution  
\[\phi_{1}\
ot\in\left\{\phi_{4},\quad\phi_{4}-\phi_{3}\right\},\quad\phi_{1} \phi_{3}=a_{5,2},\quad\mu_{1}=\frac{a_{4,2}}{\phi_{4}-\phi_{1}},\quad\mu_{2}= \phi_{1}\]  
corresponds to matrices (30) and (27) with spectra (20) and (18) in the form  
\[\tilde{\Phi}_{RY0}\big{|}_{\phi_{1}\
eq\phi_{4}}= \begin{bmatrix}\phi_{4}&- \frac{a_{4,2}}{\phi_{4}-\phi_{1}}\\ 0&\phi_{3}\end{bmatrix} ,\quad\tilde{\Phi}_{RY1}\big{|}_{\phi_{1}\
eq\phi_{4} }= \begin{bmatrix}\phi_{2}&\frac{a_{4,2}}{\phi_{4}-\phi_{1}}\\ 0&\phi_{1}\end{bmatrix}  \tag{34}\]  
and the output regulator matrix (31) equal to  
\[\mathbf{F}_{RY}\big{|}_{\phi_{1}\
eq\phi_{3}=a_{5,2}}=\mathbf{I}_{RY} \begin{bmatrix} \phi_{2}\phi_{4}&-\phi_{2}-\phi_{4}&0\\ 0&0&-\phi_{1}-\phi_{3}\end{bmatrix} .\]  
The second solution  
\[\phi_{1}\
ot\in\left\{\phi_{3},\quad\phi_{3}-\phi_{4}\right\},\quad\phi_{1} \phi_{4}=a_{5,2},\quad\mu_{1}=\frac{a_{4,2}}{\phi_{3}-\phi_{1}},\quad\mu_{2}= \phi_{1}\]  
corresponds to matrices (30) and (27) with spectra (20) and (18) in the form  
\[\tilde{\Phi}_{RY0}\big{|}_{\phi_{1}\
eq\phi_{3}}= \begin{bmatrix}\phi_{3}&- \frac{a_{4,2}}{\phi_{3}-\phi_{1}}\\ 0&\phi_{4}\end{bmatrix} ,\quad\tilde{\Phi}_{RY1}\big{|}_{\phi_{1}\
eq\phi_{3} }= \begin{bmatrix}\phi_{2}&\frac{a_{4,2}}{\phi_{3}-\phi_{1}}\\ 0&\phi_{1}\end{bmatrix}