(Created page with "AC-2 2014 P1. （a）i） <math> \begin{bmatrix} \dot{x}_1(t) \\ \dot{x}_2(t) \end{bmatrix}=\begin{bmatrix} -1 & -\frac{1}{2} \\ \frac{1}{2} & -1 \end{bmatrix}\begin{bmatr...") |
|||
| Line 80: | Line 80: | ||
as \\ | as \\ | ||
t\rightarrow \infty</math> | t\rightarrow \infty</math> | ||
| + | |||
<math>e^{At}=\frac{1}{2}\begin{bmatrix} | <math>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} \\ | ||
| Line 90: | Line 91: | ||
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| + | |||
<math>X(t)=e^{At}X(0)+\begin{matrix} \int_{0}^{t}e^{A(t-I)}BU dI | <math>X(t)=e^{At}X(0)+\begin{matrix} \int_{0}^{t}e^{A(t-I)}BU dI | ||
\end{matrix} | \end{matrix} | ||
=e^{At}X(0)+\begin{matrix} \int_{0}^{t}e^{A(t-I)}dI BU | =e^{At}X(0)+\begin{matrix} \int_{0}^{t}e^{A(t-I)}dI BU | ||
\end{matrix}</math> | \end{matrix}</math> | ||
| + | |||
<math>X(\infty)=e^(Atrightarrow\infty)X(0)+0Bu=X(0)=\begin{bmatrix} | <math>X(\infty)=e^(Atrightarrow\infty)X(0)+0Bu=X(0)=\begin{bmatrix} | ||
4 \\ | 4 \\ | ||
| Line 114: | Line 117: | ||
| − | + | ii）Can’t resolve the rest of questions | |
| − | + | ||
| Line 132: | Line 134: | ||
\end{bmatrix} \\ | \end{bmatrix} \\ | ||
rank=2 \\ | rank=2 \\ | ||
| − | not controllable | + | not \quad controllable |
0=\begin{bmatrix} | 0=\begin{bmatrix} | ||
C \\ | C \\ | ||
| Line 203: | Line 205: | ||
1 & 0 & 0 | 1 & 0 & 0 | ||
\end{bmatrix}=3 \\ | \end{bmatrix}=3 \\ | ||
| − | The only uncontrollable and unobservable \lambda has negative real part \\ | + | The \quad only \quad uncontrollable \quad and \quad unobservable \quad\lambda \quad has\quad negative \quad real \quad part \\ |
| − | Stablizable and | + | Stablizable \quad and \quad detectable |
</math> | </math> | ||
<math>H(S)=C(SI-A)^{-1}B+D=(\frac{1}{S-1}-\frac{2}{(S+1)(S-1)}-\frac{8}{(S-1)(S+1)(S-2)})=\frac{S^2-3S-6}{(S-1)(S+1)(S-2)} \\ | <math>H(S)=C(SI-A)^{-1}B+D=(\frac{1}{S-1}-\frac{2}{(S+1)(S-1)}-\frac{8}{(S-1)(S+1)(S-2)})=\frac{S^2-3S-6}{(S-1)(S+1)(S-2)} \\ | ||
</math> | </math> | ||
| + | |||
<math> | <math> | ||
P_1=1 \\ | P_1=1 \\ | ||
P_2=-1 \\ | P_2=-1 \\ | ||
P_3=2 \\ | P_3=2 \\ | ||
| − | not all poles have neqative real parts \\ | + | not \;all\; poles \;have \;neqative\; real\; parts \\ |
| − | not BIBO stable | + | not \;BIBO\; stable\; |
</math> | </math> | ||
| Line 225: | Line 228: | ||
\dot{x}_2(t)=-costx_1(t)-tx_2(t) | \dot{x}_2(t)=-costx_1(t)-tx_2(t) | ||
</math> | </math> | ||
| + | |||
<math>\Phi(t)=\begin{bmatrix} | <math>\Phi(t)=\begin{bmatrix} | ||
\Phi_1(t) & \Phi_2(t) | \Phi_1(t) & \Phi_2(t) | ||
| Line 236: | Line 240: | ||
X_1(0)=1 X_2(0)=0 | X_1(0)=1 X_2(0)=0 | ||
</math> | </math> | ||
| + | |||
<math>X_1(t)=e^{-\int_{0}^{t}IdI}\\ | <math>X_1(t)=e^{-\int_{0}^{t}IdI}\\ | ||
X_1(0)=e^{-\frac{1}{2}t^2} | X_1(0)=e^{-\frac{1}{2}t^2} | ||
| Line 278: | Line 283: | ||
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| + | |||
<math>\Phi(tI)=\Phi(t)\Phi^{-1}(I)=\begin{bmatrix} | <math>\Phi(tI)=\Phi(t)\Phi^{-1}(I)=\begin{bmatrix} | ||
e^{-\frac{1}{2}t^2} & 0 \\ | e^{-\frac{1}{2}t^2} & 0 \\ | ||
Latest revision as of 04:34, 21 May 2017
AC-2 2014
P1. （a）i） $ \begin{bmatrix} \dot{x}_1(t) \\ \dot{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 \quad 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 \\ rank\begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=rank\begin{bmatrix} 0 & 0 & 0 \\ -2 & 2 & 0 \\ -5 & 4 & -1 \\ 1 & 0 & 0 \end{bmatrix}=3 \\ $
For$ \lambda _2=-1 \\ rank\begin{bmatrix} \lambda I-A & B \end{bmatrix}=rank\begin{bmatrix} -2 & 0 & 0 & 1 \\ -2 & 0 & 0 & 1 \\ -5 & 4 & -3 & 2 \end{bmatrix}=2 \\ uncontrollable \\ rank\begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=rank\begin{bmatrix} -2 & 0 & 0 \\ -2 & 0 & 0 \\ -5 & 4 & -3 \\ 1 & 0 & 0 \end{bmatrix}=2 \\ unobservable $
For$ \lambda _3=2 \\ rank\begin{bmatrix} \lambda I-A & B \end{bmatrix}=rank\begin{bmatrix} 1 & 0 & 0 & 1 \\ -2 & 3 & 0 & 1 \\ -5 & 4 & 0 & 2 \end{bmatrix}=3 \\ rank\begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=rank\begin{bmatrix} 1 & 0 & 0 \\ -2 & 3 & 0 \\ -5 & 4 & 0 \\ 1 & 0 & 0 \end{bmatrix}=3 \\ The \quad only \quad uncontrollable \quad and \quad unobservable \quad\lambda \quad has\quad negative \quad real \quad part \\ Stablizable \quad and \quad detectable $
$ H(S)=C(SI-A)^{-1}B+D=(\frac{1}{S-1}-\frac{2}{(S+1)(S-1)}-\frac{8}{(S-1)(S+1)(S-2)})=\frac{S^2-3S-6}{(S-1)(S+1)(S-2)} \\ $
$ P_1=1 \\ P_2=-1 \\ P_3=2 \\ not \;all\; poles \;have \;neqative\; real\; parts \\ not \;BIBO\; stable\; $
P3 $ X(t)=\begin{bmatrix} X_1(t) \\ X_2(t) \end{bmatrix} \dot{x}_1(t)=-tx_1(t) \dot{x}_2(t)=-costx_1(t)-tx_2(t) $
$ \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} \\ X_1(0)=1 X_2(0)=0 $
$ X_1(t)=e^{-\int_{0}^{t}IdI}\\ X_1(0)=e^{-\frac{1}{2}t^2} \dot{x}_2(t)=-coste^{-\frac{1}{2}t^2}-tx_2(t) for \Phi_2(t) \\ assume \\ X(0)=\begin{bmatrix} 0 \\ 1 \end{bmatrix} \\ X_1(t)=0 \dot{x}_2(t)=-tx_2(t) x_2(t)=e^{-\frac{1}{2}t^2}x_2(0)=e^{-\frac{1}{2}t^2} \Phi_2(t)=\begin{bmatrix} 0 \\ e^{-\frac{1}{2}t^2} \end{bmatrix} \\ $
$ \Phi(t)=e^{-\int_{0}^{t}AdI}=e^{{-\int_{0}^{t}}\begin{bmatrix} -t & 0 \\ -cost & -t \end{bmatrix}}dI=e^{\begin{bmatrix} -\frac{1}{2}t^2 & 0 \\ sint & -\frac{1}{2}t^2 \end{bmatrix}}=e^{\begin{bmatrix} -\frac{1}{2}t^2 & 0 \\ 0 &-\frac{1}{2}t^2 \end{bmatrix}}e^{\begin{bmatrix} 0 & 0 \\ sint & 0 \end{bmatrix}}=\begin{bmatrix} e^{-\frac{1}{2}t^2} & 0 \\ 0 & e^{-\frac{1}{2}t^2} \end{bmatrix}\begin{bmatrix} 1 & 0 \\ sint & 1 \end{bmatrix}=\begin{bmatrix} e^{-\frac{1}{2}t^2} & 0 \\ sinte^{-\frac{1}{2}t^2} & e^{-\frac{1}{2}t^2} \end{bmatrix} $
$ \Phi(tI)=\Phi(t)\Phi^{-1}(I)=\begin{bmatrix} e^{-\frac{1}{2}t^2} & 0 \\ sinte^{-\frac{1}{2}t^2} & e^{-\frac{1}{2}t^2} \end{bmatrix}\begin{bmatrix} e^{-\frac{1}{2}I^2} & 0 \\ sinIe^{-\frac{1}{2}I^2} & e^{-\frac{1}{2}I^2} \end{bmatrix}=\begin{bmatrix} e^{-\frac{1}{2}(t^2-I^2)} & 0 \\ (sint-sinI)e^{-\frac{1}{2}(t^2-I^2)} & e^{-\frac{1}{2}(t^2-I^2)} \end{bmatrix} $
