Difference between revisions of "ECE-QE AC2-2015" - Rhea

Revision as of 01:57, 17 May 2017

AC-2 P1.

$ \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 1 \quad and \quad Model 3 \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}\quad X_[0]=0 $

$ \quad\therefore\quad X_1=-X_3 , X_2=X_3 $

$ \quad \therefore \quad X_[0]=\quad 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 2 \quad and \quad Model 3 \quad are \quad already \quad bounded \quad when \quad t>0 $

$ \quad \therefore\omega_1^T X_[0]=0 $

$ \begin{bmatrix} -2 & -\frac{1}{2} &-\frac{3}{2}\\ \end{bmatrix} \quad X_[0]=0 $

$ 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} $

$ \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} $