Revision as of 23:19, 20 May 2017 by Congs (Talk | contribs)

p1 a) $ A=\begin{bmatrix} -1 & 1 \\ 0 & -2 \end{bmatrix} $

$ X(t)=\begin{bmatrix} X_1(t) \\ X_2(t) \end{bmatrix} $

$ \begin{cases} \dot{x}_1(t)=-X_1(t)+X_2(t) \\ \dot{x}_2(t)=-2X_2(t) \end{cases} $

$ \Phi(t)=\begin{bmatrix} \Phi_1(t)  &  \Phi_2(t)      \\ \end{bmatrix}  $

For $ \Phi_1(t) assume X_{(0)} =\begin{bmatrix} 1 \\ 0 \end{bmatrix} $

$ \begin{cases} \dot{x}_1(t)=e^{\begin{matrix} \int_{t}^{0} -1\, \mathrm{d}t \end{matrix}} X_1(0)=e^{-t}\\ \dot{x}_2(t)=e^{\begin{matrix} \int_{t}^{0} -2\, \mathrm{d}t \end{matrix}} X_2(0)=0 \end{cases} $

$ \therefore\Phi_1(t)=\begin{bmatrix} e^{-t} \\ 0 \end{bmatrix} $

$ For \quad \Phi_2(t) assume X_(0)=\begin{bmatrix} 0 \\ 1 \end{bmatrix} $

$ X_2(t)=e^{\begin{matrix} \int_{t}^{0} -2\, \mathrm{d}t \end{matrix}} X_2(0)=e^{-2t} $

$ \Phi_(t)=e^{\begin{matrix} \int_{t}^{0} A\, \mathrm{d}t \end{matrix}}=e^{\begin{bmatrix} -t & t \\ 0 & -2t \end{bmatrix}}=\begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-2t} \end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-2t} \end{bmatrix} $

$ \Phi_(t, \iota)=\Phi_(t)\Phi_(\iota)^-1=\begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-2t} \end{bmatrix}\begin{bmatrix} e^{-\iota} & \iota e^{-\iota} \\ 0 & e^{-2\iota} \end{bmatrix} $

b) $ A=\begin{bmatrix} -\cos t & \cos t \\ 0 & -2\cos t \end{bmatrix} $

$ \Phi_(t)=e^{\begin{matrix} \int_{t}^{0} A\, \mathrm{d}t \end{matrix}} =\begin{bmatrix} e^{\sin t} & 0 \\ 0 & e^{2\sin t} \end{bmatrix}\begin{bmatrix} 1 & - \sin t \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} e^{\sin t} & - \sin t e^ {\sin t} \\ 0 & e^{2\sin t} \end{bmatrix} $

$ \Phi_(t, \iota)=\Phi_(t)\cdot \Phi_(\iota)^-1=\begin{bmatrix} e^{\sin t} & -sin t e^{\sin t } \\ 0 & e^{2\sin t } \end{bmatrix}{\begin{bmatrix} e^{\sin t} & -sin t e^{\sin t } \\ 0 & e^{2\sin t } \end{bmatrix}}^{-1} $

p2 a) $ \left| {\lambda\Iota-A} \right|=\begin{bmatrix} \lambda+2 & -2 \\ 1 & \lambda-1 \end{bmatrix} $

$ \lambda_1 =0 \lambda_2=-1 Marginally stable ,not asy ,stable b) <math>c=\begin{bmatrix} B & AB \end{bmatrix} $=$ \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} $ rank=1 not observable, unobservable subspace$ \left\{ {\begin{bmatrix} 1 \\ 1 \end{bmatrix} } \right\} c) <math>0=\begin{bmatrix} C \\ CA \end{bmatrix} $=$ \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $

rank=1 not observable, unobservable subspace$ \left\{ {\begin{bmatrix} 1 \\ -1 \end{bmatrix} } \right\} d) e) \therefore talse \therefore talse f) <math>A-BK=\begin{bmatrix} -2-k_1 & 2-k_2 \\ -1-k_1 & 1-k_2 \end{bmatrix} $ $ \left| {\lambda-A+BK} \right|=\lambda^2+\left( {a+b+1} \right)\lambda+3a+3-ab=0 \lambda_1 =-3 ang \lambda_2=-1 <math>\begin{cases} -3a + 9 -ab=0 \\ 2a - b+3-ab=0 \end{cases} $

a=0 ,b=3 $ k=\begin{bmatrix} 0 & 3 \\ \end{bmatrix} $


g)$ \begin{bmatrix} \lambda\Iota-A \\ C \end{bmatrix} $=$ \begin{bmatrix} \lambda+2 & -2 \\ 1 & \lambda-1 \\ 1 & -1 \end{bmatrix} $ must contain $ \lambda=0 , no $

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