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(0)^k & =\beta_0 ,\; k \to \infty \quad \beta_0=0\;\\ | (0)^k & =\beta_0 ,\; k \to \infty \quad \beta_0=0\;\\ | ||
\end{alignat}</math> | \end{alignat}</math> | ||
| + | |||
| + | <math>\begin{bmatrix} | ||
| + | 1 & 1 \\ | ||
| + | -1 & -1 | ||
| + | \end{bmatrix}\begin{bmatrix} | ||
| + | 1 & 1 \\ | ||
| + | -1 & -1 | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 0 & 0 \\ | ||
| + | 0 & 0 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math>A^k=\begin{cases} | ||
| + | I_2, k=0 \\ | ||
| + | A, & k=1 \\ | ||
| + | 0, & k>1 | ||
| + | \end{cases}</math> | ||
| + | |||
| + | <math>\mathbf{(b)} \quad C_3=\begin{bmatrix} | ||
| + | B & AB & A^2B | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 1 & 1 & 0 \\ | ||
| + | 1 & -1 & 0 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math> X_{[3]}=A^3 X_{[0]}+\begin{bmatrix} | ||
| + | A^2 B & AB & B | ||
| + | \end{bmatrix}\begin{bmatrix} | ||
| + | u_{[0]} \\ | ||
| + | u_{[1]} \\ | ||
| + | u_{[2]} | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math> \begin{bmatrix} | ||
| + | 0 & 1 & 1 \\ | ||
| + | 0 & -1 & 1 | ||
| + | \end{bmatrix}\begin{bmatrix} | ||
| + | u_{[0]} \\ | ||
| + | u_{[1]} \\ | ||
| + | u_{[2]} | ||
| + | \end{bmatrix}= \begin{bmatrix} | ||
| + | 1 \\ | ||
| + | 1 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math> u_1=0, \quad u_2=1, \quad u_0=0</math> | ||
Latest revision as of 23:10, 20 May 2017
AC-2 P1.
$ \mathbf{a)} \quad C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 2 & 2 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $ $ \quad rank=2\ne \mbox 3 $
$ \quad Not\quad controllable $
$ \mathbf{b)} \quad Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} ,& \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix}. $
$ \mathbf{c)} \quad 0=\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \quad Not \quad observable $
$ \mathbf{d)} \quad X_1=r,X_2=-s,X_3=s \quad X=r\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+s\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\\ $
$ Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} ,& \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \end{Bmatrix}.\\ $
$ \mathbf{e)} \quad \lambda I-A=\begin{bmatrix} \lambda-1 & -1 & -1 \\ 0 & \lambda & -1 \\ 0 & 0 & \lambda+1 \end{bmatrix}\quad \lambda_1=1,\lambda_2=0,\lambda_3=-1\\ $
$ \qquad for \;\lambda_1=1 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} 0 & -1 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 0 \end{bmatrix}\\ $
$ \qquad for \;\lambda_2=0 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 & 1 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}\\ $
$ \qquad for \;\lambda_3=-1 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} -2 & -1 & -1 & 1 \\ 0 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \qquad rank<3 \qquad must\;contain\;\lambda=-1 \qquad \therefore\; No. $
$ \mathbf{f)} \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} \lambda-1 & -1 & -1 \\ 0 & \lambda & -1 \\ 0 & 0 & \lambda+1 \\ 0 & 1 & 1 \end{bmatrix} $
$ for\; \lambda_1=1 \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} 0 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_1=1 $
$ for\; \lambda_2=0 \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 \\ 0 & 0 & -1 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}\\ $
$ for \lambda-3=-1\quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} -2 & -1 & -1 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_3=-1 $
$ \because \;eigenvalues \left\{1,-1,2 \right\}\quad \therefore\;Yes. $
$ A-LC=<math>\begin{bmatrix} 1 & 1-L_1 & 1-L_1 \\ 0 & -L_2 & 1-L_2 \\ 0 & -L_3 & -1-L_3 \end{bmatrix}\quad LC=\begin{bmatrix} 0 & L_1 & L_1 \\ 0 & L_2 & L_2 \\ 0 & L_3 & L_3 \end{bmatrix} $
$ \lambda I-\left(A-LC \right)=\begin{bmatrix} \lambda-1 & L_1-1 & L_1-1 \\ 0 & \lambda+L_2 & L_2-1 \\ 0 & L_3 & \lambda+1+L_3 \end{bmatrix} $
For conditions $ \quad\lambda_1=1,\quad\lambda_2=-1,\quad\lambda_3=-2 $
$ \begin{cases} 3L_2 + 6L_3 = 9 \\ L_2 + 2L_3 = 3 \\ L_2=1 \\ L_3=1 \end{cases} \quad L=\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} $
$ \mathbf{g)} \quad \because\;\lambda_1=1,\quad Not\;stable. $
$ \mathbf{h)} \quad AU_1=\lambda_1 U_1 \quad AU_2=\lambda_2 U_2 \quad AU_3= \lambda_3 U_3\\ $
$ \begin{alignat}{2} y & = C X_(t) =C[U_1 e^t (\omega_1^T X_{(0)})+U_2 e^0 (\omega_2^T X_{(0)})+U_3 e^{-t} (\omega_3^T X_{(0)})]\\ & = -\omega_2^T \omega_{(0)}\\ \end{alignat} $
$ \therefore\;bounded $
$ :)\;\because \; \frac{1}{s}\; has\; pole=0 \quad\therefore\;Not \; BIBO \;Stable. $
P2.
$ \mathbf{(a)} \quad A=\frac{1}{2}\begin{bmatrix} 1 & -1 \\ -1 & -1 \end{bmatrix},\quad \lambda=0 $
$ \begin{alignat}{1} (0)^k & =\beta_0 ,\; k \to \infty \quad \beta_0=0\;\\ \end{alignat} $
$ \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $
$ A^k=\begin{cases} I_2, k=0 \\ A, & k=1 \\ 0, & k>1 \end{cases} $
$ \mathbf{(b)} \quad C_3=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix} $
$ X_{[3]}=A^3 X_{[0]}+\begin{bmatrix} A^2 B & AB & B \end{bmatrix}\begin{bmatrix} u_{[0]} \\ u_{[1]} \\ u_{[2]} \end{bmatrix} $
$ \begin{bmatrix} 0 & 1 & 1 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix} u_{[0]} \\ u_{[1]} \\ u_{[2]} \end{bmatrix}= \begin{bmatrix} 1 \\ 1 \end{bmatrix} $
$ u_1=0, \quad u_2=1, \quad u_0=0 $
