(Created page with "2016 AC-2 P1. (a) <math> X=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} y \\ \dot{y} \end{bmatrix}</math> <math>\begin{cases} \dot{x}=\begin{bmatrix} \do...") |
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<math>y[k]=\begin{bmatrix} | <math>y[k]=\begin{bmatrix} | ||
| + | -1 & 1 | ||
| + | \end{bmatrix}x[k]</math> | ||
| + | |||
| + | let <math>x_[k]=\begin{bmatrix} | ||
| + | a\\ | ||
| + | b | ||
| + | \end{bmatrix} \quad y[0]=\begin{bmatrix} | ||
| + | -1 & 1 | ||
| + | \end{bmatrix}\begin{bmatrix} | ||
| + | a\\ | ||
| + | b | ||
| + | \end{bmatrix}=1</math> | ||
| + | |||
| + | <math> -a+b=1</math> | ||
| + | |||
| + | <math>y[1]=\begin{bmatrix} | ||
-1 & 1 | -1 & 1 | ||
\end{bmatrix}x[1]=\begin{bmatrix} | \end{bmatrix}x[1]=\begin{bmatrix} | ||
| Line 75: | Line 91: | ||
-1 & 2 | -1 & 2 | ||
\end{bmatrix}x[0]=0</math> | \end{bmatrix}x[0]=0</math> | ||
| + | |||
| + | <math>x[1]=Ax[0]</math> | ||
| + | |||
| + | <math>3a+2b=0</math> | ||
| + | |||
| + | <math>\therefore a=-\frac{2}{5} \quad b=\frac{3}{5}</math> | ||
| + | |||
| + | <math>x[0]=\begin{bmatrix} | ||
| + | -\frac{2}{5}\\ | ||
| + | \frac{3}{5} | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | ii)<math>x[0]=\begin{bmatrix} | ||
| + | a\\ | ||
| + | b | ||
| + | \end{bmatrix} \quad x[2]=Ax[1]=A^2x[0]</math> | ||
| + | |||
| + | <math>A^2=0 \quad x[2]=0</math> | ||
| + | |||
| + | <math>y[2]=[-1 \quad 1]\quad x[2]=0 \quad y[1]=[-1 \quad 1] \quad x[1]=1</math> | ||
| + | |||
| + | <math>[-1\quad 1]\quad \begin{bmatrix} | ||
| + | 2 &4\\ | ||
| + | -1&2 | ||
| + | \end{bmatrix}\quad\begin{bmatrix} | ||
| + | a\\ | ||
| + | b | ||
| + | \end{bmatrix}=1</math> | ||
| + | |||
| + | we only have -3a-2b=1,so we can't uniquely determine a,b. | ||
| + | |||
| + | P3 (a)<math>\lambda I-A=\begin{bmatrix} | ||
| + | \lambda+2&-4\\ | ||
| + | 1&\lambda-2 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math>\lambda_1=\lambda_2=0</math> | ||
| + | |||
| + | <math>\begin{bmatrix} | ||
| + | -2-\lambda_1 & 4\\ | ||
| + | -1 & 2-\lambda_1 | ||
| + | \end{bmatrix}\begin{bmatrix} | ||
| + | u_1 \\ | ||
| + | u_2 | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 0\\ | ||
| + | 0 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math>\begin{cases} | ||
| + | -2u_1-\lambda_1u_1+4u_2=0\\ | ||
| + | -u_1+2u_2-\lambda_1u_2=0 | ||
| + | \end{cases}</math> | ||
| + | |||
| + | <math>u_1=2u_2</math> | ||
| + | |||
| + | <math> \therefore eigenvector \begin{bmatrix} | ||
| + | u_1\\ | ||
| + | u_2 | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 2\\ | ||
| + | 1 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math>J=MAM^{-1}</math> | ||
| + | |||
| + | (b)<math>e^{At}=L^{-1}\begin{bmatrix} | ||
| + | (SI-A)^{-1} | ||
| + | \end{bmatrix}=L^{-1}\begin{bmatrix} | ||
| + | \frac{s-2}{s^2} & \frac{4}{s^2} \\ | ||
| + | \frac{-1}{s^2} & \frac{s+2}{s^2} | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math>L^{-1}\begin{bmatrix} | ||
| + | \frac{s-2}{s^2} | ||
| + | \end{bmatrix}= L^{-1}\begin{bmatrix} | ||
| + | \frac{1}{s}-\frac{2}{s^2} | ||
| + | \end{bmatrix}=1-2t</math> | ||
| + | |||
| + | <math>L^{-1}\begin{bmatrix} | ||
| + | \frac{4}{s^2} | ||
| + | \end{bmatrix}=4t</math> | ||
| + | |||
| + | <math>L^{-1}\begin{bmatrix} | ||
| + | \frac{-1}{s^2} | ||
| + | \end{bmatrix}=-t</math> | ||
| + | |||
| + | <math>L^{-1}\begin{bmatrix} | ||
| + | \frac{s+2}{s^2} | ||
| + | \end{bmatrix}=L^{-1}\begin{bmatrix} | ||
| + | \frac{1}{s}+\frac{2}{s^2} | ||
| + | \end{bmatrix}=1+2t</math> | ||
| + | |||
| + | <math>e^{At}=\begin{bmatrix} | ||
| + | 1-2t & 4t\\ | ||
| + | -t & 1+2t | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | (c)T(s)=<math>C(SI-A)^{-1}B</math> | ||
| + | |||
| + | =<math>[-1 \quad 1] \quad | ||
| + | \begin{bmatrix} | ||
| + | \frac{s-2}{s^2} & \frac{4}{s^2} \\ | ||
| + | \frac{-1}{s^2} & \frac{s+2}{s^2} | ||
| + | \end{bmatrix} \quad | ||
| + | \begin{bmatrix} | ||
| + | 2\\ | ||
| + | 1 | ||
| + | \end{bmatrix}</math> | ||
Revision as of 05:41, 22 May 2017
2016 AC-2 P1. (a) $ X=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} y \\ \dot{y} \end{bmatrix} $
$ \begin{cases} \dot{x}=\begin{bmatrix} \dot{x_1} \\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} x_2 \\ -2x_1 x_2-u x_1+2u \end{bmatrix}\\ y=x_1 \end{cases} $
(b) $ u \equiv 2 $
$ \dot{x} =\begin{bmatrix} x_2 \\ -2x_1 x_2-2x_1+4 \end{bmatrix}=\begin{bmatrix} x_2 \\ -2x_1 (x_2+1)+4 \end{bmatrix} $
let $ \begin{cases} -2x_1 (x_2+1)+4=0 \\ x_2=0 \end{cases} \Rightarrow \begin{cases} -2x_1 +4=0 \\ x_2=0 \end{cases} \Rightarrow \begin{cases} x_1=2 \\ x_2=0 \end{cases} $
$ \therefore The \; equilibrum\; point\; is \;x_e=\begin{bmatrix} 2 \\ 0 \end{bmatrix} $
(c) $ u \equiv 2 \quad x_e=\begin{bmatrix} 2 \\ 0 \end{bmatrix}, \quad let \;x=f(x) $
The Jacobin of $ \dot{x} $ is: $ \begin{align} Df(x)= \begin{bmatrix} 0 & 1 \\ -2x_1-2 & -2x_1 \end{bmatrix} \end{align} $
The linear dynamics around $ x_e $ is $ \frac{d}{dt}f(x)=\begin{bmatrix} 0 & 1 \\ -2 & -4 \end{bmatrix} f(x) $
which is stable, locally stable at $ x_e $.
P2. i) $ x[k+1]=A x[k] $
$ y[k]=\begin{bmatrix} -1 & 1 \end{bmatrix}x[k] $
let $ x_[k]=\begin{bmatrix} a\\ b \end{bmatrix} \quad y[0]=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=1 $
$ -a+b=1 $
$ y[1]=\begin{bmatrix} -1 & 1 \end{bmatrix}x[1]=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix}x[0]=0 $
$ x[1]=Ax[0] $
$ 3a+2b=0 $
$ \therefore a=-\frac{2}{5} \quad b=\frac{3}{5} $
$ x[0]=\begin{bmatrix} -\frac{2}{5}\\ \frac{3}{5} \end{bmatrix} $
ii)$ x[0]=\begin{bmatrix} a\\ b \end{bmatrix} \quad x[2]=Ax[1]=A^2x[0] $
$ A^2=0 \quad x[2]=0 $
$ y[2]=[-1 \quad 1]\quad x[2]=0 \quad y[1]=[-1 \quad 1] \quad x[1]=1 $
$ [-1\quad 1]\quad \begin{bmatrix} 2 &4\\ -1&2 \end{bmatrix}\quad\begin{bmatrix} a\\ b \end{bmatrix}=1 $
we only have -3a-2b=1,so we can't uniquely determine a,b.
P3 (a)$ \lambda I-A=\begin{bmatrix} \lambda+2&-4\\ 1&\lambda-2 \end{bmatrix} $
$ \lambda_1=\lambda_2=0 $
$ \begin{bmatrix} -2-\lambda_1 & 4\\ -1 & 2-\lambda_1 \end{bmatrix}\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix} $
$ \begin{cases} -2u_1-\lambda_1u_1+4u_2=0\\ -u_1+2u_2-\lambda_1u_2=0 \end{cases} $
$ u_1=2u_2 $
$ \therefore eigenvector \begin{bmatrix} u_1\\ u_2 \end{bmatrix}=\begin{bmatrix} 2\\ 1 \end{bmatrix} $
$ J=MAM^{-1} $
(b)$ e^{At}=L^{-1}\begin{bmatrix} (SI-A)^{-1} \end{bmatrix}=L^{-1}\begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix} $
$ L^{-1}\begin{bmatrix} \frac{s-2}{s^2} \end{bmatrix}= L^{-1}\begin{bmatrix} \frac{1}{s}-\frac{2}{s^2} \end{bmatrix}=1-2t $
$ L^{-1}\begin{bmatrix} \frac{4}{s^2} \end{bmatrix}=4t $
$ L^{-1}\begin{bmatrix} \frac{-1}{s^2} \end{bmatrix}=-t $
$ L^{-1}\begin{bmatrix} \frac{s+2}{s^2} \end{bmatrix}=L^{-1}\begin{bmatrix} \frac{1}{s}+\frac{2}{s^2} \end{bmatrix}=1+2t $
$ e^{At}=\begin{bmatrix} 1-2t & 4t\\ -t & 1+2t \end{bmatrix} $
(c)T(s)=$ C(SI-A)^{-1}B $
=$ [-1 \quad 1] \quad \begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix} \quad \begin{bmatrix} 2\\ 1 \end{bmatrix} $
