| Line 4: | Line 4: | ||
（a）i） | （a）i） | ||
<math>\begin{bmatrix} | <math>\begin{bmatrix} | ||
| − | x_1(t)\\ | + | x_1(t) \\ |
x_2(t) | x_2(t) | ||
\end{bmatrix}=\begin{bmatrix} | \end{bmatrix}=\begin{bmatrix} | ||
| − | -1 &-\frac{1}{2}\\ | + | -1 &-\frac{1}{2} \\ |
\frac{1}{2} & -1 | \frac{1}{2} & -1 | ||
\end{bmatrix}\begin{bmatrix} | \end{bmatrix}\begin{bmatrix} | ||
| Line 13: | Line 13: | ||
x_2(t) | x_2(t) | ||
\end{bmatrix}+\begin{bmatrix} | \end{bmatrix}+\begin{bmatrix} | ||
| − | \frac{x_0(t)}{2}\\ | + | \frac{x_0(t)}{2} \\ |
\frac{x_3(t)}{2} | \frac{x_3(t)}{2} | ||
\end{bmatrix}=\begin{bmatrix} | \end{bmatrix}=\begin{bmatrix} | ||
| − | -1 &-\frac{1}{2}\\ | + | -1 &-\frac{1}{2} \\ |
\frac{1}{2} & -1 | \frac{1}{2} & -1 | ||
\end{bmatrix}\begin{bmatrix} | \end{bmatrix}\begin{bmatrix} | ||
| − | x_1(t) \\ | + | x_1(t) \\ |
x_2(t) | x_2(t) | ||
\end{bmatrix}+begin{bmatrix} | \end{bmatrix}+begin{bmatrix} | ||
| − | \frac{1}{2}&0 \\ | + | \frac{1}{2}&0 \\ |
0& \frac{1}{2} | 0& \frac{1}{2} | ||
\end{bmatrix}\begin{bmatrix} | \end{bmatrix}\begin{bmatrix} | ||
| − | x_0(t) \\ | + | x_0(t) \\ |
| − | + | x_3(t) | |
\end{bmatrix}</math> | \end{bmatrix}</math> | ||
| Line 114: | Line 114: | ||
ii） | ii） | ||
Can’t resolve the rest of questions | Can’t resolve the rest of questions | ||
| + | |||
| + | |||
| + | P2 | ||
| + | <math>\lambda _1=1 \\ | ||
| + | \lambda _2=-1 \\ | ||
| + | \lambda _3=2 \\ | ||
| + | not stable </math> | ||
| + | |||
| + | <math>C=\begin{bmatrix} | ||
| + | B & AB & A^2B | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 1 & 1 & 1 \\ | ||
| + | 1 & 1 & 1 \\ | ||
| + | 2 & 5 & 11 | ||
| + | \end{bmatrix} \\ | ||
| + | rank=2 \\ | ||
| + | not controllable | ||
| + | 0=\begin{bmatrix} | ||
| + | C \\ | ||
| + | CA \\ | ||
| + | CA^2 | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 1 & 0 & 0 \\ | ||
| + | 1 & 0 & 0 \\ | ||
| + | 1 & 0 & 0 | ||
| + | \end{bmatrix}\\ | ||
| + | rank=1 \\ | ||
| + | not observable | ||
| + | </math> | ||
| + | |||
| + | For<math>\lambda _1=1 \\ | ||
| + | rank\begin{bmatrix} | ||
| + | \lambda I-A & B | ||
| + | \end{bmatrix}=rank\begin{bmatrix} | ||
| + | 0 & 0 & 0 & 1 \\ | ||
| + | -2 & 2 & 0 & 1 \\ | ||
| + | -5 & 4 & -1 & 2 | ||
| + | \end{bmatrix}=3 \\ | ||
| + | </math> | ||
Revision as of 03:13, 21 May 2017
AC-2 2014
P1. （a）i） $ \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}=\begin{bmatrix} -1 &-\frac{1}{2} \\ \frac{1}{2} & -1 \end{bmatrix}\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}+\begin{bmatrix} \frac{x_0(t)}{2} \\ \frac{x_3(t)}{2} \end{bmatrix}=\begin{bmatrix} -1 &-\frac{1}{2} \\ \frac{1}{2} & -1 \end{bmatrix}\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}+begin{bmatrix} \frac{1}{2}&0 \\ 0& \frac{1}{2} \end{bmatrix}\begin{bmatrix} x_0(t) \\ x_3(t) \end{bmatrix} $
ii) $ A=\begin{bmatrix} -1 & \frac{1}{2} \\ \frac{1}{2} & -1 \end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}+\begin{bmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}+\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} $
$ e^A=\begin{bmatrix} e^{-1} & 0 \\ 0 & e^{-1} \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} e^{-\frac{1}{2}} & 0 \\ 0 & e^\frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}=\frac{1}{2}\begin{bmatrix} e^{-\frac{3}{2}}+e^{-\frac{1}{2}} & -e^{-\frac{3}{2}}+e^{-\frac{1}{2}} \\ -e^{-\frac{3}{2}}+e^{-\frac{1}{2}} & e^{-\frac{3}{2}}+e^{-\frac{1}{2}} \end{bmatrix} $
iii) $ \lambda _1=-\frac{1}{2} \lambda _2=-\frac{3}{2}\\ stable \\ X(t)\rightarrow X(\infty) \\ as \\ t\rightarrow \infty $ $ e^{At}=\frac{1}{2}\begin{bmatrix} e^{-\frac{3}{2}t}+e^{-\frac{1}{2}t} & e^{-\frac{3}{2}t}+e^{-\frac{1}{2}t} \\ e^{-\frac{3}{2}t}+e^{-\frac{1}{2}t} & e^{-\frac{3}{2}t}+e^{-\frac{1}{2}t} \end{bmatrix} t\rightarrow \infty e^{At}\rightarrow\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $ $ X(t)=e^{At}X(0)+\begin{matrix} \int_{0}^{t}e^{A(t-I)}BU dI \end{matrix} =e^{At}X(0)+\begin{matrix} \int_{0}^{t}e^{A(t-I)}dI BU \end{matrix} $ $ X(\infty)=e^(Atrightarrow\infty)X(0)+0Bu=X(0)=\begin{bmatrix} 4 \\ 1 \end{bmatrix} $
(b)$ X(t)=\begin{bmatrix} 0 & 0 &0 \\ \frac{1}{2} & -1 & \frac{1}{2}\\ 0 & \frac{1}{2} & -1 \end{bmatrix}\begin{bmatrix} x_0 \\ x_1 \\ x_2 \end{bmatrix}+\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}U(t) $
ii）
Can’t resolve the rest of questions
P2
$ \lambda _1=1 \\ \lambda _2=-1 \\ \lambda _3=2 \\ not stable $
$ C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 5 & 11 \end{bmatrix} \\ rank=2 \\ not controllable 0=\begin{bmatrix} C \\ CA \\ CA^2 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}\\ rank=1 \\ not observable $
For$ \lambda _1=1 \\ rank\begin{bmatrix} \lambda I-A & B \end{bmatrix}=rank\begin{bmatrix} 0 & 0 & 0 & 1 \\ -2 & 2 & 0 & 1 \\ -5 & 4 & -1 & 2 \end{bmatrix}=3 \\ $
