Difference between revisions of "ECE-QE AC2-2011" - Rhea

Revision as of 21:45, 19 May 2017

AC-2 2011

P1.

$ \mathbf{a)} \qquad C=\begin{bmatrix} B & AB\end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -3 & 3 \end{bmatrix} $

$ \Rightarrow \qquad Not \quad controllable. \qquad Subspace \begin{bmatrix} 1 \\ -3 \end{bmatrix} $

$ \mathbf{b)} \qquad 0=\begin{bmatrix} C \\ CA \end{bmatrix}=\begin{bmatrix} 3 & 1 \\ 6 & 2 \end{bmatrix} $

Not observable.

$ 3x_1+r=0 \qquad x_1=-\frac{1}{3}r \qquad span \begin{bmatrix} 1 \\ -3 \end{bmatrix} $

$ \mathbf{c)} \qquad \lambda I-A=\begin{vmatrix} \lambda-2 & -1 \\ 0 & \lambda+1 \end{vmatrix} \qquad (\lambda-2)(\lambda+1)=0 \qquad \lambda_1=2 \qquad \lambda_2=-1 \\ $

$ \lambda=2 \qquad \begin{bmatrix} 0 & -1 & 1 \\ 0 & 3 &-3 \end{bmatrix} \\ rank<2 \qquad associated \qquad with \\ \qquad \qquad \lambda=2>0 \\ $

$ \Rightarrow \qquad not \qquad stablizable \\ $

$ \mathbf{d)} \qquad \lambda=-1 \qquad \begin{vmatrix} -3 & -1 & 1 \\ 0 & 0 & -3 \end{vmatrix} \qquad rank=2 \\ $

$ \Rightarrow \qquad \lambda=2 \qquad has \qquad to \qquad be \qquad eigeavalue $

  $ \Rightarrow \qquad \begin{bmatrix} 1  &  -1 \end{bmatrix}\qquad can \qquad not \qquad be \qquad eigeavalues   $

$ \mathbf{e)} \qquad \lambda=2 \qquad \begin{bmatrix} 0 & -1 \\ 0 & 3 \\ 3 & 1 \end{bmatrix} \qquad \lambda=-1 \qquad \begin{bmatrix} -3 & -1 \\ 0 & 0 \\ 3 & 1 \end{bmatrix} \\ unobservable \qquad modes \qquad associated \qquad to \qquad \lambda=-1 \\ which \qquad is \qquad stable \qquad \longrightarrow \qquad detectable \\ $

$ \mathbf{f)} \qquad \begin{bmatrix} \frac{\lambda I-A}{C} \end{bmatrix}=\begin{bmatrix} -3 & -1 \\ 0 & 0 \\ 3 & 1 \end{bmatrix} \qquad \lambda=-1 \qquad rank<2 \\ must \qquad contains \qquad eigenvalue \qquad =-1 \\ let\qquad l=\begin{bmatrix} l_1 \\ l_2 \end{bmatrix} \qquad A-lC=\begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix} - \begin{bmatrix} 3l_1 & l_1 \\ 3l_2 & l_2 \end{bmatrix}\\ \qquad \qquad \qquad Ā =\begin{bmatrix} 2-3l_1 & 1-l_1 \\ -3l_2 & -1-l_2 \end{bmatrix} \\ \lambda I-Ā=\begin{bmatrix} -3+3l_1 & l_1-1 \\ 3l_2 & l_2 \end{bmatrix} \qquad \lambda=-1 \\ \qquad \qquad (-3+3l_1)l_2-3l_2(l_1-1)=0 \\ \qquad \qquad $