# A Robust Control Algorithm of a Descent Vehicle Angular Motion in the Earth's Atmosphere
## 6 Robust Output Regulator
 \tag{35}\]  
and the output regulator matrix (31) equal to  
\[\mathbf{F}_{RY}\big{|}_{\phi_{1}\
eq\phi_{3}}=I_{RY} \begin{bmatrix}\phi_{2}\phi_{3} &-\phi_{2}-\phi_{3}&0\\ 0&0&-\phi_{1}-\phi_{4}\end{bmatrix} .\]The third solution  
\[\phi_{2}\
otin\{\phi_{4},\ \ \ \phi_{4}-\phi_{3}\},\ \ \ \phi_{2}\phi_{3}=a_{5,2},\ \ \ \mu_{1}=\frac{a_{4,2}}{\phi_{4}-\phi_{2}},\ \ \ \mu_{2}=\phi_{2}\]  
corresponds to matrices (30) and (27) with spectra (20) and (18) in the form  
\[\left.\begin{array}{ll}\mathbf{\Phi}_{RY0}\right|_{\phi_{2}\
eq\phi_{4}}= \begin{bmatrix}\phi_{4}&-\frac{a_{42}}{\phi_{4}-\phi_{2}}\\ 0&\phi_{3}\end{bmatrix} ,\ \ \ \mathbf{\Phi}_{RY1}\right|_{\phi_{2}\
eq\phi_{4}}= \begin{bmatrix}\phi_{1}&\frac{a_{42}}{\phi_{4}-\phi_{2}}\\ 0&\phi_{2}\end{bmatrix}  \tag{36}\]  
and the output regulator matrix (31) equal to  
\[\left. \begin{array}{ll}\mathbf{F}_{RY}\right|_{\phi_{2}\
eq\phi_{3}=a_{5,2} }=\mathbf{J}_{RY}\begin{bmatrix}\phi_{1}\phi_{4}&-\phi_{1}-\phi_{4}&0\\ 0&0&-\phi_{2}-\phi_{3}\end{bmatrix}.\end{array} \]  
The forth solution  
\[\phi_{2}\
otin\{\phi_{3},\ \ \ \phi_{3}-\phi_{4}\},\ \ \ \phi_{2}\phi_{4}=a_{5,2},\ \ \ \mu_{1}=\frac{a_{4,2}}{\phi_{3}-\phi_{2}},\ \ \ \mu_{2}=\phi_{2}\]  
corresponds to matrices (30) and (27) with spectra (20) and (18) in the form  
\[\left.\begin{array}{ll}\mathbf{\Phi}_{RY0}\right|_{\phi_{2}\
eq\phi_{3}}= \begin{bmatrix}\phi_{3}&-\frac{a_{42}}{\phi_{3}-\phi_{2}}\\ 0&\phi_{4}\end{bmatrix} ,\ \ \ \left. \begin{array}{ll}\mathbf{\Phi}_{RY1} \right|_{\phi_{2}\
eq\phi_{3}}=\begin{bmatrix}\phi_{1}&\frac{a_{42}}{\phi_{3}- \phi_{2}}\\ 0&\phi_{2}\end{bmatrix}\end{array} \right. \tag{37}\]  
and the output regulator matrix (31) equal to  
\[\left. \begin{array}{ll}\mathbf{F}_{RY}\right|_{\phi_{2}\
eq\phi_{4}=a_{5,2} }=\mathbf{J}_{RY}\begin{bmatrix}\phi_{1}\phi_{3}&-\phi_{1}-\phi_{3}&0\\ 0&0&-\phi_{2}-\phi_{4}\end{bmatrix}.\end{array} \]  
The fifth solution  
\[s_{12}\
otin\{-\phi_{3},\ \ -\phi_{4},\ \ \ s_{34}\},\ \ \ m_{12}=a_{5,2},\ \ \ \mu_{1}=\frac{a_{4,2}}{s_{12}-s_{34}},\ \ \ \mu_{2}=\frac{a_{5,2}-m_{34}}{s_{12}-s_{34}}-s_{12},\]  
where \(s_{12}=-\phi_{1}-\phi_{2}\), \(s_{34}=-\phi_{3}-\phi_{4}\), \(m_{12}=\phi_{1}\phi_{2}\), \(m_{34}=\phi_{3}\phi_{4}\), corresponds to matrices (30) and (27) with spectra (20) and (18) in the form  
\[\left.\begin{array}{ll}\mathbf{\Phi}_{RY0}\right|_{s_{12}\
eq s_{34}}= \begin{bmatrix}\frac{m_{12}-m_{34}}{s_{12}-s_{34}}-s_{34}&-\frac{a_{42}}{s_{12 }-s_{34}}\\ \frac{1}{a_{42}}\left(\frac{(m_{12}-m_{34})}{s_{12}-s_{34}}+s_{12}m_{34}-s_{34 }m_{12}\right)&-\frac{m_{12}-m_{34}}{s_{12}-s_{34}}\end{bmatrix} ,\\ \left. \begin{array}{ll}-\frac{m_{12}-m_{34}}{s_{12}-s_{34}}&\frac{a_{42}}{s_{12 }-s_{34}}\\ -\frac{1}{a_{42}}\left(\frac{(m_{12}-m_{34})}{s_{12}-s_{34}}+s_{12}m_{34}-s_{3 4}m_{12}\right)&\frac{m_{12}-m_{34}}{s_{12}-s_{34}}-s_{12}\end{array} \right], \tag{38}\]  
and the output regulator matrix (31) equal to  
\[\left. \begin{array}{ll}\mathbf{F}_{RY}\right|_{\phi_{1}\
eq\phi_{2}=a_{5,2} }=\mathbf{J}_{RY}\begin{bmatrix}\phi_{3}\phi_{4}&-\phi_{3}-\phi_{4}&0\\ 0&0&-\phi_{1}-\phi_{2}\end{bmatrix}.\end{array} \]  
Having compared the results of the five presented solutions, and also taking into account the fact that the spectra (18) and (20) can be swapped between the matrices \(\mathbf{\tilde{\Phi}}_{RY0}\) and \(\mathbf{\tilde{\Phi}}_{RY1}\), we draw the following conclusion. If the product of any two poles in a given spectrum (15) (we denote these poles by the symbols \(\phi_{y1}\) and \(\phi_{y2}\), and the other two poles by the symbols \(\phi_{x1}\) and \(\phi_{x2}\)) is equal to a positive number \(a_{52}\), then the output regulator matrix  
\[\left. \begin{array}{ll}\mathbf{F}_{RY}\right|_{m_{y}=a_{5,2}}=\mathbf{J}_{RY} \begin{bmatrix}m_{x}&s_{x}&0\\ 0&0&s_{y}\end{bmatrix},\end{array}