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1 \\ \end{bmatrix}\end{Bmatrix}</math> | 1 \\ \end{bmatrix}\end{Bmatrix}</math> | ||
| − | <math>(f | + | <math>(f \sim j)\quad Can't \quad resolve.</math> |
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<math>\quad Energy \quad is \quad U^{2}[0] +U^{2}[1] +U^{2} [2]= \frac{6}{9}= \frac{2}{3}</math> | <math>\quad Energy \quad is \quad U^{2}[0] +U^{2}[1] +U^{2} [2]= \frac{6}{9}= \frac{2}{3}</math> | ||
| − | P4. <math> ( X_1^{2}-1)( X_2-2)=0</math> | + | P4. <math>\quad ( X_1^{2}-1)( X_2-2)=0</math> |
| − | + | ||
| − | + | <math>\quad -X_2( X_1^{2}+1)=0</math> | |
| + | |||
| + | <math>\quad soll:\quad X_1=\pm1 , \quad X_2=0</math> | ||
<math>\quad X_{e1}=\begin{bmatrix} | <math>\quad X_{e1}=\begin{bmatrix} | ||
| Line 222: | Line 224: | ||
0 & -2\\ | 0 & -2\\ | ||
\end{bmatrix} \quad all \quad\lambda_i\quad negative \quad asy \quad stable</math> | \end{bmatrix} \quad all \quad\lambda_i\quad negative \quad asy \quad stable</math> | ||
| − | + | ||
| + | <math> D f( X_{e2})=\begin{bmatrix} | ||
4 & 0 \\ | 4 & 0 \\ | ||
0 & -2\\ | 0 & -2\\ | ||
\end{bmatrix} \quad has \quad a\quad positive \quad \lambda,\quad unstable</math> | \end{bmatrix} \quad has \quad a\quad positive \quad \lambda,\quad unstable</math> | ||
Latest revision as of 06:16, 17 May 2017
AC-2 P1.
$ \mathbf{a)} \quad e^{At}=\begin{bmatrix} -1 & -1 & 1\\ -1 & 1 & 2\\ 1 & 1 & -2\\ \end{bmatrix}\begin{bmatrix} e^{t} & 0 & 0\\ 0 & e^{-t} & 0\\ 0 & 0 & e^{0}\\ \end{bmatrix}\begin{bmatrix} -2 & -\frac{1}{2} & -\frac{3}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ -1 & 0 & -1\\ \end{bmatrix} $
=$ \begin{bmatrix} -e^{t} & -e^{-t} & 1\\ -e^{t} & e^{-t} & 2\\ e^{t} & e^{-t} & -2\\ \end{bmatrix}\begin{bmatrix} -2 & -\frac{1}{2} & -\frac{3}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ -1 & 0 & -1\\ \end{bmatrix} $
=$ \begin{bmatrix} 2e^{t}-1 & \frac{1}{2}e^{t} -\frac{1}{2}e^{-t} & \frac{3}{2}e^{t}-\frac{1}{2}e^{-t}-1\\ 2e^{t}-2 & \frac{1}{2}e^{t}+\frac{1}{2}e^{-t} & \frac{1}{2}e^{t}+\frac{1}{2}e^{-t}-2\\ -2e^{t}+2 & -\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} & -\frac{3}{2}e^{t}+\frac{1}{2}e^{-t}+2\\ \end{bmatrix} $
$ \mathbf{b)}\quad\lambda_1=1,\quad\lambda_2=-1,\quad\lambda_3=0 $, $ \because\quad\lambda_1>0 \quad\therefore\quad unstable. $
$ \mathbf{c)}\quad If \quad we \quad want \quad t \to \infty, X(t) \to0 $
$ \quad Model \quad 1 \quad and \quad Model \quad3 \quad need \quad to \quad be \quad zero. $
$ \omega_1^T X[0]=0 $
$ \omega_3^T X[0]=0 $
$ \begin{bmatrix} -2 & -\frac{1}{2} &-\frac{3}{2}\\ -1 & 0 & -1 \\\end{bmatrix} X[0]=0 $
$ \quad\therefore\quad X_1=-X_3 , X_2=X_3 $
$ \quad \therefore \quad X[0]=X_3\begin{bmatrix} -1 \\ 1 \\ 1 \\ \end{bmatrix} $
$ \quad The \quad set\quad is\begin{Bmatrix} \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \end{Bmatrix} $
$ \quad If \quad we \quad want \quad to\quad remain\quad bounded $
$ \quad Model \quad 2 \quad and \quad 3 \quad are \quad already \quad bounded \quad when \quad t \geqslant 0 $
$ \quad \therefore\omega_1^T X[0]=0 $
$ \begin{bmatrix} -2 & -\frac{1}{2} &-\frac{3}{2}\\ \end{bmatrix} X[0]=0 $
$ \quad X_1=-\frac{1}{4} X_2-\frac{3}{4}X_3 $
$ \quad X=\begin{bmatrix} -\frac{1}{4} X_2-\frac{3}{4}X_3\\ X_2\\ X_3\\ \end{bmatrix}=X_2\begin{bmatrix} -\frac{1}{4}\\ 1\\ 0\\ \end{bmatrix} +X_3\begin{bmatrix} -\frac{3}{4}\\ 0\\ 1\\ \end{bmatrix} $
$ \quad \therefore \quad The \quad set\quad is\begin{Bmatrix} \begin{bmatrix} -\frac{1}{4} \\ 1 \\ 0\\ \end{bmatrix}，\begin{bmatrix} -\frac{3}{4}\\ 0\\ 1\\ \end{bmatrix}\end{Bmatrix} $
$ \mathbf{d)} \quad C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -1 & -1 & -1 \end{bmatrix} $
$ \quad rank=1\ne \mbox 3,\quad not\quad controllable $
$ \quad The\quad reachable\quad subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \end{Bmatrix} $
$ \mathbf{e)} \quad O=\begin{bmatrix} \quad C\\ \quad CA\\ \quad CA^2 \\ \end{bmatrix} =\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $
$ \quad rank=1\ne \mbox 3,\quad not \quad observable $
$ \quad unobservable \quad subspace\quad is \begin{Bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix},\begin{bmatrix} -1\\ 0 \\ 1 \\ \end{bmatrix}\end{Bmatrix} $
$ (f \sim j)\quad Can't \quad resolve. $
P2.$ \Phi（t）=e^{\zeta_0^t\begin{bmatrix} -t & 1\\ 0 & -1\\ \end{bmatrix}dt}=e^{\begin{bmatrix} -\frac{1}{2}t^{2} & 0\\ 0 & -t\\ \end{bmatrix}}e^{\begin{bmatrix} 0 & t \\ 0 & 0\\ \end{bmatrix}}=\begin{bmatrix} e^{-\frac{1}{2}t^{2}} & 0 \\ 0 & e^{-t}\\ \end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1\\ \end{bmatrix} $
$ \Phi（t）=\begin{bmatrix} e^{-\frac{1}{2}t^{2}} & te^{-\frac{1}{2}t^{2}} \\ 0 & e^{-t}\\ \end{bmatrix} $
P3.$ \quad Let \quad X[3]= \begin{bmatrix} 1 & 1\\ \end{bmatrix}^{t} ,\quad X[0]=\begin{bmatrix} 0 & 0\\ \end{bmatrix}^{t} $
$ \quad C_3 =\begin{bmatrix} B & AB & A^2B\\ \end{bmatrix} =\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ \end{bmatrix}^{t} $
$ \quad U =\begin{bmatrix} u(2)\\ u(1)\\ u(0)\\ \end{bmatrix} =\quad C_3^{t} ( C_3C_3^{t})^{-1} X[3]=\begin{bmatrix} \frac{2}{3} & -\frac{1}{3}\\ \frac{1}{3} & \frac{1}{3}\\ - \frac{1}{3} & \frac{2}{3}\\ \end{bmatrix}\begin{bmatrix} 1\\ 1\\ \end{bmatrix}=\begin{bmatrix} \frac{1}{3}\\ \frac{2}{3}\\ \frac{1}{3}\\ \end{bmatrix} $
$ \quad \therefore \quad U[0]=\frac{1}{3}, \quad U[1]=\frac{2}{3}, \quad U[2]=\frac{1}{3} $
$ \quad Energy \quad is \quad U^{2}[0] +U^{2}[1] +U^{2} [2]= \frac{6}{9}= \frac{2}{3} $
P4. $ \quad ( X_1^{2}-1)( X_2-2)=0 $
$ \quad -X_2( X_1^{2}+1)=0 $
$ \quad soll:\quad X_1=\pm1 , \quad X_2=0 $
$ \quad X_{e1}=\begin{bmatrix} 1\\ 0\\ \end{bmatrix},\quad X_{e2}=\begin{bmatrix} -1\\ 0\\ \end{bmatrix} $
$ D f(X)=\begin{bmatrix} 2 X_1 X_2-4 X_1 & X_1^{2}-1\\ -2 X_1X_2 & -X_1^{2}-1\\ \end{bmatrix} $
$ D f( X_{e1})=\begin{bmatrix} -4 & 0 \\ 0 & -2\\ \end{bmatrix} \quad all \quad\lambda_i\quad negative \quad asy \quad stable $
$ D f( X_{e2})=\begin{bmatrix} 4 & 0 \\ 0 & -2\\ \end{bmatrix} \quad has \quad a\quad positive \quad \lambda,\quad unstable $
