State Vector and Matrix

Q. Consider a linear system whose state space representation is $\dot{x}\left(t\right)=Ax\left(t\right).$If the initial state vector of the system is $x\left(0\right)=\left[\begin{array}{c}1\\ -2\end{array}\right]$, then the system response is $x\left(t\right)=\left[\begin{array}{c}{e}^{-2t}\\ -2e^{-2t}\end{array}\right].$ If the initial state vector of the system changes to $x\left(0\right)=\left[\begin{array}{c}1\\ -1\end{array}\right],$ then the system response becomes $x\left(t\right)=\left[\begin{array}{c}{e}^{-t}\\ -e^{-t}\end{array}\right].$

The eigen-value and eigen-vector pairs $({\lambda }_i,v_i)$ for the system are

1. $(-1,\left[\begin{array}{c}1\\ 1\end{array}\right])\text{and}(-2,\left[\begin{array}{c}1\\ 2\end{array}\right])$
2. $(-2,\left[\begin{array}{c}1\\ -1\end{array}\right])\text{and}(-1,\left[\begin{array}{c}1\\ -2\end{array}\right])$
3. $(-1,\left[\begin{array}{c}1\\ -1\end{array}\right])\text{and}(-2,\left[\begin{array}{c}1\\ -2\end{array}\right])$
4. $(-2,\left[\begin{array}{c}1\\ -1\end{array}\right])\text{and}(1,\left[\begin{array}{c}1\\ -2\end{array}\right])$

Q. A state variable system $x\left(t\right)=\left[\begin{array}{cc}0& 1\\ 0& -3\end{array}\right]X\left(t\right)+\left[\begin{array}{c}1\\ 0\end{array}\right]U\left(t\right),$
with the initial condition $X\left(0\right)=[-13{]}^T$ and the unit step input $u\left(t\right)$ has

The state transition equation:

1. $X\left(t\right)=\left[\begin{array}{c}t-e^{-t}\\ e^{-t}\end{array}\right]$
2. $X\left(t\right)=\left[\begin{array}{c}t-e^{-t}\\ 3e^{-3t}\end{array}\right]$
3. $X\left(t\right)=\left[\begin{array}{c}t-e^{-3t}\\ 3e^{-3t}\end{array}\right]$
4. $X\left(t\right)=\left[\begin{array}{c}t-e^{-2t}\\ e^{-t}\end{array}\right]$

Q. Given
$A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$,
the state transition matrix $e^{At}$ is given by

1. $\left[\begin{array}{cc}0& e^{-t}\\ e^{-t}& 0\end{array}\right]$
2. $\left[\begin{array}{cc}{e}^t& 0\\ 0& e^t\end{array}\right]$
3. $\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-t}\end{array}\right]$
4. $\left[\begin{array}{cc}0& e^t\\ e^t& 0\end{array}\right]$

Q. The matrix of any state space equations for the transfer function $C\left(s\right)/R\left(s\right)$ of the system, shown below in figure is

1. $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$
2. $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$
3. $[-1]$
4. $\left[3\right]$

Q. A state variable system $x\left(t\right)=\left[\begin{array}{cc}0& 1\\ 0& -3\end{array}\right]X\left(t\right)+\left[\begin{array}{c}1\\ 0\end{array}\right]U\left(t\right),$ with the initial condition $X\left(0\right)=[-13{]}^T$ and the unit step input $u\left(t\right)$ has

The state transition matrix

1. $\left[\begin{array}{cc}1& \frac{1}{3}(1-e^{-3t})\\ 0& e^{-3t}\end{array}\right]$
2. $\left[\begin{array}{cc}1& \frac{1}{3}(e^{-t}-e^{-3t})\\ 0& e^{-t}\end{array}\right]$
3. $\left[\begin{array}{cc}1& \frac{1}{3}(e^{-t}-e^{-3t})\\ 0& e^{-3t}\end{array}\right]$
4. $\left[\begin{array}{cc}1& (1-e^{-t})\\ 0& e^{-t}\end{array}\right]$

Q. Given the matrix:
$A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -6& -11& -6\end{array}\right]$

Its eigen values are ______.

1. $-1,-2,-4$
2. $1,2,3$
3. $-1,-2,-3$
4. $1,2,6$

Q. The state equation of a second-order linear system is given by,
$\dot{x}\left(t\right)=Ax\left(t\right),x\left(0\right)=x_0$

$For{x}_0=\left[\begin{array}{c}1\\ -1\end{array}\right],x\left(t\right)=\left[\begin{array}{c}{e}^{-t}\\ -e^{-t}\end{array}\right]\text{and for}{x}_0=\left[\begin{array}{c}0\\ 1\end{array}\right]$

$x\left(t\right)=\left[\begin{array}{cc}{e}^{-t}& e^{-2t}\\ -e^{-t}& +2e^{-2t}\end{array}\right]$

$When{x}_0=\left[\begin{array}{c}3\\ 5\end{array}\right],x\left(t\right)is$

1. $\left[\begin{array}{ccc}-8e^{-t}& +& 11e^{-2t}\\ 8e^{-t}& -& 22e^{-2t}\end{array}\right]$
2. $\left[\begin{array}{ccc}11e^{-t}& -& 8e^{-2t}\\ -11e^{-t}& +& 16e^{-2t}\end{array}\right]$
3. $\left[\begin{array}{ccc}3e^{-t}& -& 5e^{-2t}\\ -3e^{-t}& +& 10e^{-2t}\end{array}\right]$
4. $\left[\begin{array}{ccc}-5e^{-t}& +& 6e^{-2t}\\ -5e^{-t}& +& 6e^{-2t}\end{array}\right]$

Q. A linear system is equivalently represented by two sets of state equations.
$X=AX+BU$ and $W=CW+DU$
The eigen values of the representations are also computed as $\left[\lambda \right]$ and $\left[\mu \right]$. Which one of the following statements is true ?

1. $\left[\lambda \right]=\left[\mu \right]$ and $X=W$
2. $\left[\lambda \right]=\left[\mu \right]$ and $X\ne W$
3. $\left[\lambda \right]\ne \left[\mu \right]$ and $X=W$
4. $\left[\lambda \right]\ne \left[\mu \right]$ and $X\ne W$

Q. A second order system starts with an initial condition of $\left[\begin{array}{c}2\\ 3\end{array}\right]$ without any external input. The state transition matrix for the system is given by $\left[\begin{array}{cc}{e}^{-2t}& 0\\ 0& e^{-t}\end{array}\right]$. The state of the system at the end of $1$ second is given by

1. $\left[\begin{array}{c}0.271\\ 1.100\end{array}\right]$
   
2. $\left[\begin{array}{c}0.135\\ 0.368\end{array}\right]$
3. $\left[\begin{array}{c}0.271\\ 0.736\end{array}\right]$
4. $\left[\begin{array}{c}0.135\\ 1.100\end{array}\right]$

Q. For the system described by the state equation
$\dot{x}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0.5& 1& 2\end{array}\right]x+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u$
If the control signal $u$ is given by $u=[-0.5-3-5]x+v$, then the eigen values of the closed-loop system will be

1. $0,-1,-2$
2. $0,-1,-3$
3. $-1,-1,-2$
4. $0,-1,-1$

Q. For the system
$\dot{X}=\left[\begin{array}{cc}2& 0\\ 0& 4\end{array}\right]X+\left[\begin{array}{c}1\\ 1\end{array}\right]u;y=\left[\begin{array}{cc}4& 0\end{array}\right]X,$
with $u$ as unit impulse and with zero initial state, the output $y$ becomes

1. $2e^{2t}$
2. $4e^{2t}$
3. $2e^{4t}$
4. $4e^{4t}$

Q. The state model of a system is given as,

$\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{c}{x}_2\\ x_1\end{array}\right]$

$x^T\left(0\right)=\left[10\right]$
The solution of homogenous equation is given by,

1. $x\left(t\right)=\left[\begin{array}{c}{e}^t\\ 1\end{array}\right]$
2. $x\left(t\right)=\left[\begin{array}{c}{e}^t\\ t{e}^t\end{array}\right]$
3. $x\left(t\right)=\left[\begin{array}{c}0\\ e^t\end{array}\right]$
4. $x\left(t\right)=\left[\begin{array}{c}{e}^t\\ e^t\end{array}\right]$