| Line 3: | Line 3: | ||
0 & -2 | 0 & -2 | ||
\end{bmatrix}</math> | \end{bmatrix}</math> | ||
| + | |||
<math>X(t)=\begin{bmatrix} | <math>X(t)=\begin{bmatrix} | ||
X_1(t) \\ | X_1(t) \\ | ||
| Line 8: | Line 9: | ||
\end{bmatrix} </math> | \end{bmatrix} </math> | ||
| − | <math> \dot{x}_1(t)=-X_1(t)+X_2(t) | + | <math>\begin{cases} |
| − | + | \dot{x}_1(t)=-X_1(t)+X_2(t) \\ | |
| + | \dot{x}_2(t)=-2X_2(t) | ||
| + | \end{cases}</math> | ||
| − | + | <math>\Phi(t)=\begin{bmatrix} | |
\Phi_1(t) & \Phi_2(t) \\ | \Phi_1(t) & \Phi_2(t) \\ | ||
\end{bmatrix} </math> | \end{bmatrix} </math> | ||
| − | For <math> \Phi_1(t) assume | + | |
| + | For <math> \Phi_1(t) assume X_{(0)} =\begin{bmatrix} | ||
1 \\ | 1 \\ | ||
0 | 0 | ||
\end{bmatrix} </math> | \end{bmatrix} </math> | ||
| + | <math> | ||
| + | \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}</math> | ||
| − | + | <math>\therefore\Phi_1(t)=\begin{bmatrix} | |
| − | + | e^{-t} \\ | |
| − | e^-t \\ | + | |
0 | 0 | ||
\end{bmatrix} </math> | \end{bmatrix} </math> | ||
| − | + | <math>For \quad \Phi_2(t) assume X_(0)=\begin{bmatrix} | |
0 \\ | 0 \\ | ||
1 | 1 | ||
\end{bmatrix} </math> | \end{bmatrix} </math> | ||
| − | <math>X_2(t)= | + | <math>X_2(t)=e^{\begin{matrix} \int_{t}^{0} -2\, \mathrm{d}t \end{matrix}} X_2(0)=e^{-2t}</math> |
| − | <math>\Phi_(t)= | + | <math>\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}</math> | ||
| − | + | <math>\Phi_(t, \boldsymbol)=\Phi_(t)\Phi_(\boldsymbol)^-1=\begin{bmatrix} | |
| − | - | + | e^{-t} & te^{-t} \\ |
| − | 0 | + | 0 & e^{-2t} |
| + | \end{bmatrix}\begin{bmatrix} | ||
| + | e^{-\boldsymbol} & \boldsymbol e^{-\boldsymbol} \\ | ||
| + | 0 & e^{-2\boldsymbol} | ||
\end{bmatrix}</math> | \end{bmatrix}</math> | ||
| − | + | b) <math>A=\begin{bmatrix} | |
| − | + | -\cos t & \cos t \\ | |
| − | + | 0 & -2\cos t | |
| − | 0 & | + | |
\end{bmatrix}</math> | \end{bmatrix}</math> | ||
| − | \begin{bmatrix} | + | |
| + | <math>\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 \\ | 1 & - \sin t \\ | ||
0 & 1 | 0 & 1 | ||
| − | \end{bmatrix} | + | \end{bmatrix}=\begin{bmatrix} |
| − | 0 & e^2\sin t | + | e^{\sin t} & - \sin t e^ {\sin t} \\ |
| + | 0 & e^{2\sin t} | ||
\end{bmatrix}</math> | \end{bmatrix}</math> | ||
| − | <math>\Phi_( | + | <math>\Phi_(t, \boldsymbol)=\Phi_(t)\cdot \Phi_(\boldsymbol)^-1=\begin{bmatrix} |
| − | & | + | e^{\sin t} & -sin t e^{\sin t } \\ |
| − | 0 & | + | 0 & e^{2\sin t } |
| − | \end{bmatrix}</math> | + | \end{bmatrix}{\begin{bmatrix} |
| + | e^{\sin t} & -sin t e^{\sin t } \\ | ||
| + | 0 & e^{2\sin t } | ||
| + | \end{bmatrix}}^{-1}</math> | ||
p2 a) <math>\left| {\lambda\Iota-A} \right|=\begin{bmatrix} | p2 a) <math>\left| {\lambda\Iota-A} \right|=\begin{bmatrix} | ||
Revision as of 09:47, 20 May 2017
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, \boldsymbol)=\Phi_(t)\Phi_(\boldsymbol)^-1=\begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-2t} \end{bmatrix}\begin{bmatrix} e^{-\boldsymbol} & \boldsymbol e^{-\boldsymbol} \\ 0 & e^{-2\boldsymbol} \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, \boldsymbol)=\Phi_(t)\cdot \Phi_(\boldsymbol)^-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 $
