# State Variable Analysis

 Question 1
Consider a state-variable model of a system
$\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} 0 &1 \\ -\alpha &-2\beta \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}+\begin{bmatrix} 0\\ \alpha \end{bmatrix}r$
$y=[1\;\;0]\begin{bmatrix} x_1\\ x_2 \end{bmatrix}$
where y is the output, and r is the input. The damping ratio $\xi$ and the undamped natural frequency $\omega _n$ (rad/sec) of the system are given by
 A $\xi =\frac{\beta }{\sqrt{\alpha }}; \; \omega _n=\sqrt{\alpha }$ B $\xi =\sqrt{\alpha }; \; \omega _n=\frac{\beta }{\sqrt{\alpha }}$ C $\xi =\frac{\sqrt{\alpha }}{\beta }; \; \omega _n=\sqrt{\beta }$ D $\xi =\sqrt{\beta }; \; \omega _n=\sqrt{\alpha }$
GATE EE 2019   Control Systems
Question 1 Explanation:
Characteristic equation is,
$|sI-A|=0$
$|sI-A|=\begin{vmatrix} s &-1 \\ \alpha &s+2\beta \end{vmatrix}$
$\;\;=s^2+2s\beta +\alpha =0$
$\therefore \;\; \omega _n^2=\alpha$
$\;\;\; \omega _n=\sqrt{\alpha }$
$\;\;\;2\xi \omega _n=2\beta$
$\;\;\;\xi =\frac{\beta }{\sqrt{\alpha }}$
 Question 2
Consider the system described by the following state space representation
$\begin{bmatrix} \dot{x_{1}(t)}\\ \dot{x_{2}(t)} \end{bmatrix}=\begin{bmatrix} 0 & 1\\ 0&-2 \end{bmatrix} \begin{bmatrix} {x_{1}(t)}\\ {x_{2}(t)} \end{bmatrix} +\begin{bmatrix} 0\\ 1 \end{bmatrix} u(t)$
$y(t)=\begin{bmatrix} 1 &0 \end{bmatrix} \begin{bmatrix} {x_{1}(t)}\\ {x_{2}(t)} \end{bmatrix}$
If u(t) is a unit step input and $\begin{bmatrix} {x_{1}(0)}\\ {x_{2}(0)} \end{bmatrix}=\begin{bmatrix} 1\\ 0 \end{bmatrix}$, the value of output y(t) at t = 1 sec (rounded off to three decimal places) is_________
 A 1.284 B 1.862 C 2.366 D 0.655
GATE EE 2017-SET-2   Control Systems
Question 2 Explanation:
$y(t)=\begin{bmatrix} 1 &0 \end{bmatrix}x(t)$
$x(t)=L^{-1}[(sI-A)^{-1}]x(0)+L^{-1}[(sI-A)^{-1}]BU(s)$
$L^{-1}[(sI-A)^{-1}]x(0)=\begin{bmatrix} 1\\ 0 \end{bmatrix}$
$L^{-1}[(sI-A)^{-1}]BU(s)=\begin{bmatrix} -0.25+0.5t+0.25e^{-2t}\\ 0.5-0.5e^{-2t} \end{bmatrix}$
$\therefore \;\; x(t)=\begin{bmatrix} 0.75+0.5t+0.25e^{-2t}\\ 0.5-0.5e^{-2t} \end{bmatrix}$
$\therefore \;\; y(t)=0.75+0.5t+0.25e^{-2t}$
$\therefore \;\;y(1)=1.284$
 Question 3
The transfer function of the system Y(s)/U(s) whose state-space equations are given below is: $\begin{bmatrix} \dot{x_{1}}(t)\\ \dot{x_{2}}(t) \end{bmatrix} =\begin{bmatrix} 1 & 2\\ 2&0 \end{bmatrix} \begin{bmatrix} {x_{1}}(t)\\ {x_{2}}(t) \end{bmatrix}+\begin{bmatrix} 1\\ 2 \end{bmatrix} u(t)$
$y(t)=[10] \begin{bmatrix} {x_{1}}(t)\\ {x_{2}}(t) \end{bmatrix}$
 A $\frac{(s+2)}{(s^{2}-2s-2)}$ B $\frac{(s-2)}{(s^{2}+s-4)}$ C $\frac{(s-4)}{(s^{2}+s-4)}$ D $\frac{(s+4)}{(s^{2}-s-4)}$
GATE EE 2017-SET-1   Control Systems
Question 3 Explanation:
\begin{aligned} \text{Given: } \dot{X}&=\begin{bmatrix} 1 & 2\\ 2 & 0 \end{bmatrix}X+\begin{bmatrix} 1\\ 2 \end{bmatrix}u \\ \text{and } Y&=\begin{bmatrix} 1 & 0 \end{bmatrix}X \\ \text{Transfer function} &=C[sI-A]^{-1}B+D \\ [sI-A]^{-1}&=\begin{bmatrix} \frac{s}{s^2-s-4} & \frac{2}{s^2-s-4}\\ \frac{2}{s^2-s-4} & \frac{s-1}{s^2-s-4} \end{bmatrix} \\ [sI-A]^{-1} \times B&=\begin{bmatrix} \frac{s+4}{s^2-s-4}\\ \frac{s-1}{s^2-s-4} \end{bmatrix} \\ \frac{Y(s)}{U(s)}&=C[sI-A]^{-1} \times B \\ &=\frac{s+4}{s^2-s-4} \end{aligned}
 Question 4
Consider a linear time invariant system $\dot{x}=Ax$, with initial condition $x(0)$ at t=0. Suppose $\alpha \; and \; \beta$ are eigenvectors of (2 x 2) matrix A corresponding to distinct eigenvalues $\lambda_{1} \; and \; \lambda _{2}$ respectively. Then the response $x(t)$ of the system due to initial condition $x(0)=\alpha$ is
 A $e^{\lambda _{1}t}\alpha$ B $e^{\lambda _{2}t}\beta$ C $e^{\lambda _{2}t}\alpha$ D $e^{\lambda _{1}t}\alpha + e^{\lambda _{2}t}\beta$
GATE EE 2016-SET-2   Control Systems
Question 4 Explanation:
$\dot{x}=Ax$
Eigen values are $\lambda _1$ and $\lambda _2$
we can write,
$\phi (t)=\begin{bmatrix} e^{\lambda _1 t} & 0\\ 0& e^{\lambda _2 t} \end{bmatrix}$
Response due to initial conditions,
$x(t)=\phi (t)x(0)$
$x(t)=\begin{bmatrix} e^{\lambda _1 t} & 0\\ 0& e^{\lambda _2 t} \end{bmatrix}\begin{bmatrix} \alpha \\ 0 \end{bmatrix}$
$=\alpha e^{\lambda _1 t}$
 Question 5
In the signal flow diagram given in the figure, $u_{1} \; and \; u_{2}$ are possible inputs whereas $y_{1} \; and \; y_{2}$ are possible outputs. When would the SISO system derived from this diagram be controllable and observable?
 A When $u_{1}$ is the only input and $y_{1}$ is the only output B When $u_{2}$ is the only input and $y_{1}$ is the only output C When $u_{1}$ is the only input and $y_{2}$ is the only output D When $u_{2}$ is the only input and $y_{2}$ is the only output
GATE EE 2015-SET-1   Control Systems
Question 5 Explanation:
Equations from the flow diagram,$\dot{x}_1=5x_1-2x_2+u_1$
$y_1=x_1$
$\dot{x}_2=5x_1+x_2+u_1+u_2$
$y_2=x_1-x_2$
$\dot{x}=\begin{bmatrix} 5 & -2\\ 2 & 1 \end{bmatrix}x+\begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}u$
$y=\begin{bmatrix} 1 & 0\\ 1 & -1 \end{bmatrix}x$
Considering the SISO cases:[/latex]

$1. \;\;u_1(input)$ and $y_1(output):$
$A\triangleq\begin{bmatrix} 5 & -2\\ 2 & 1 \end{bmatrix}$
$B\triangleq\begin{bmatrix} 1\\ 1 \end{bmatrix}$ and $C=\begin{bmatrix} 1 &0 \end{bmatrix}$
$O\triangleq\begin{bmatrix} C\\ CA \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 5 & -2 \end{bmatrix}$
$\Rightarrow \; Rank (O)=2$
$\Rightarrow \; Observable$
$C\triangleq\begin{bmatrix} B & AB \end{bmatrix}=\begin{bmatrix} 1 &3 \\ 1 & 3 \end{bmatrix}$
$\Rightarrow \; Rank(C)\neq 2$
$\Rightarrow \; non-controllable$

$2.\;\;u_2(input)$ and $y_1(output):$
$A\triangleq\begin{bmatrix} 5 & -2\\ 2 & 1 \end{bmatrix}$
$B\triangleq\begin{bmatrix} 0\\ 1 \end{bmatrix}$ and $C\triangleq \begin{bmatrix} 1 &-1 \end{bmatrix}$
$O\triangleq\begin{bmatrix} 1 & 0\\ 5 & -2 \end{bmatrix}$
$\Rightarrow \; Rank (O)=2$
$\Rightarrow \; Observable$
$C\triangleq\begin{bmatrix} 0 &-2 \\ 1 & 1 \end{bmatrix}$
$\Rightarrow \; Rank(C)=2$
$\Rightarrow \; controllable$

$3. \;\;u_2(input)$ and $y_2(output) :$
$Rank(O)\neq 2\Rightarrow$ non-observable
$Rank(C) =2 \Rightarrow$ controllable

$4. \; \; u_1 (input)$ and $y_2(output)$:
non-observale and non-controllable.
 Question 6
Consider the system described by following state space equations
$\begin{bmatrix} \dot{x_{1}}\\ \dot{x_{2}} \end{bmatrix}=\begin{bmatrix} 0 & 1\\ -1& -1 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}+\begin{bmatrix} 0\\ 1 \end{bmatrix}u;$ $y=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}$
If u is unit step input, then the steady state error of the system is
 A 0 B $1/2$ C $2/3$ D 1
GATE EE 2014-SET-3   Control Systems
Question 6 Explanation:
Given $A=\begin{bmatrix} 0 & 1\\ -1&-1 \end{bmatrix},$ $B=\begin{bmatrix} 0\\ 1 \end{bmatrix},$$C=\begin{bmatrix} 1 & 0 \end{bmatrix}$
Transfer function of the given system is given by
$\frac{Y(s)}{U(s)}=C(sI-A)^{-1}B$
Now, $(sI-A)=\begin{bmatrix} s & 0\\ 0 & s \end{bmatrix}-\begin{bmatrix} 0 & 1\\ -1&-1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} s & -1\\ 1&s+1 \end{bmatrix}$
$(sI-A)^{-1}=\frac{1}{|sI-A|}(Adj[A])$
$|sI-A|=s(s+1)+1=(s^2+s+1)$
and $Adj[A]=\begin{bmatrix} s+1 & 1\\ -1&s \end{bmatrix}$
$\therefore \;\;(sI-A)^{-1}=\frac{1}{(s^2+s+1)}\begin{bmatrix} s+1 & 1\\ -1&s \end{bmatrix}$
$\therefore$ Transfer function,
$\frac{Y(s)}{U(s)}=\frac{1}{(s^2+s+1)}\left \{ \begin{bmatrix} 1 &0 \end{bmatrix}\begin{bmatrix} s+1 & 1\\ -1&s \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix} \right \}$
$\;\;=\frac{1}{(s^2+s+1)}\left \{ \begin{bmatrix} (s+1) & 1 \end{bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}\right \}$
$\;\;=\frac{1}{(s^2+s+1)} \times 1$
$\frac{Y(s)}{U(s)}=\frac{1}{(s^2+s+1)}$
Given, input=unit step
$\therefore \;\; U(s)=\frac{1}{s}$
$\therefore \;\; Y(s)=\frac{1}{s}\frac{1}{(s^2+s+1)}=\frac{1}{s(s^2+s+1)}$
$\therefore$ Final Value
$\lim_{s\rightarrow 0}[sY(s)]$
$=\lim_{s\rightarrow 0}\left [ \frac{1}{(s^2+s+1)} \right ]=1$
$\therefore \;\;$ Error=Final value -Initial value
$e_{ss}=0$
 Question 7
The second order dynamic system
$\frac{dX}{dt}=PX+Qu$
$y=RX$
has the matrices P, Q and R as follows :
$P=\begin{bmatrix} -1 &1 \\ 0&-3 \end{bmatrix}\; Q=\begin{bmatrix} 0\\ 1 \end{bmatrix} \; R=\begin{bmatrix} 0 &1 \end{bmatrix}$
The system has the following controllability and observability properties:
 A Controllable and observable B Not controllable but observable C Controllable but not observable D Not controllable and not observable
GATE EE 2014-SET-2   Control Systems
Question 7 Explanation:
Given: $\dot{x}(t)=Px+Qu$
$y=Rx$
$P=\begin{bmatrix} -1 &1 \\ 0& -3 \end{bmatrix}$, $Q=\begin{bmatrix} 0\\ 1 \end{bmatrix}$, $R=\begin{bmatrix} 0 & 1 \end{bmatrix}$
For controllability,
$Q_c=\begin{bmatrix} Q &PQ \end{bmatrix}$
$[PQ]=\begin{bmatrix} -1 &1 \\ 0& -3 \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} 1\\ -3 \end{bmatrix}$
$\therefore \;\; Q_c=\begin{bmatrix} 0 &1 \\ 1& -3 \end{bmatrix}$
$\therefore \;\;|Q_c|=-1\neq 0\Rightarrow \;\; controllable$
Also, for observability,
$Q_0=\begin{bmatrix} R^T & P^TR^T \end{bmatrix}$
$\begin{bmatrix} P^T & R^T \end{bmatrix}=\begin{bmatrix} -1 & 0\\ 1 & -3 \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} 0\\ -3 \end{bmatrix}$
$\therefore \;\;Q_0=\begin{bmatrix} 0 & 0\\ 1 & -3 \end{bmatrix}$
$\therefore \;\; |Q_0|=(0 \times -3 -0 \times 1)=0\Rightarrow \;\; non-observable$
Since, $Q_c\neq 0$ and $Q_0=0$
$\therefore$ System is controllable but not observable.
 Question 8
The state transition matrix for the system
$\begin{bmatrix} \dot{x_{1}}\\ \dot{x_{2}} \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 1&1 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}+\begin{bmatrix} 1\\ 1 \end{bmatrix}u$
is
 A $\begin{bmatrix} e^{t} &0 \\ e^{t} & e^{t} \end{bmatrix}$ B $\begin{bmatrix} e^{t} &0 \\t^2 e^{t} & e^{t} \end{bmatrix}$ C $\begin{bmatrix} e^{t} &0 \\ te^{t} & e^{t} \end{bmatrix}$ D $\begin{bmatrix} e^{t} &t e^{t}\\ 0& e^{t} \end{bmatrix}$
GATE EE 2014-SET-2   Control Systems
Question 8 Explanation:
Given: $\begin{bmatrix} \dot{x}_1\\ \dot{x}_2 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 1&1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}+\begin{bmatrix} 1\\ 1 \end{bmatrix}u$
or, $\;\; \dot{x}(t)=Ax(t)+Bu$
Here, $[A]=\begin{bmatrix} 1 & 0\\ 1&1 \end{bmatrix}, \;\; B=\begin{bmatrix} 1\\ 1 \end{bmatrix}$
State transition matric is given by,
$e^{At}=L^{-1}[sI-A]^{-1}$
Now, $[sI-A]=\begin{bmatrix} s-1 & \\ 0 -1 & s-1 \end{bmatrix}$
$Adj[sI-A]=\begin{bmatrix} s-1 & \\ 1 & s-1 \end{bmatrix}$
and $|sI-A|=(s-1)^2$
$\therefore \;\; [sI-A]^{-1}=\frac{1}{|sI-A|}Adj[sI-A]$
$[sI-A]^{-1}=\frac{1}{(s-1)^2}\begin{bmatrix} s-1 & \\ 1 & s-1 \end{bmatrix}$
$=\begin{bmatrix} \frac{1}{s-1} & \\ \frac{1}{(s-1)^2} & \frac{1}{s-1} \end{bmatrix}$
$\therefore \;\; L^{-1}[sI-A]^{-1}=\begin{bmatrix} e^t & 0\\ te^t& e^t \end{bmatrix}$
$\therefore$ State transition matrix$=\begin{bmatrix} e^t & 0\\ te^t& e^t \end{bmatrix}$
 Question 9
The state variable formulation of a system is given as
$\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} -2 & 0\\ 0&-1 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}+\begin{bmatrix} 1\\1 \end{bmatrix}u$ $x_1(0)=0,x_2(0)=0\ and \; y=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}$
The response y(t) to the unit step input is
 A $\frac{1}{2}-\frac{1}{2}e^{-2t}$ B $1-\frac{1}{2}e^{-2t}-\frac{1}{2}e^{-2t}$ C $e^{-2t}-e^{-t}$ D $1-e^{-t}$
GATE EE 2013   Control Systems
Question 9 Explanation:
\begin{aligned} [sI-A]&=\begin{bmatrix} s & 0\\ 0 &s \end{bmatrix}-\begin{bmatrix} -2 & 0\\ 0& -1 \end{bmatrix} \\ &=\begin{bmatrix} s+2 & 0\\ 0 &s+1 \end{bmatrix} \\ [sI-A]^{-1}&=\begin{bmatrix} \frac{1}{s+2} & 0\\ 0 &\frac{1}{s+1} \end{bmatrix} \\ [sI-A]^{-1}[B]&=\begin{bmatrix} \frac{1}{s+2} & 0\\ 0 &\frac{1}{s+1} \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ &=\begin{bmatrix} \frac{1}{s+2}\\ \frac{1}{s+1} \end{bmatrix} \\ C[sI-A]^{-1}[B]&=\begin{bmatrix} 1 &0 \end{bmatrix}\begin{bmatrix} \frac{1}{s+2}\\ \frac{1}{s+1} \end{bmatrix}=\frac{1}{s+2} \\ G(s)&=\frac{1}{s+2} \\ \frac{Y(s)}{X(s)}&=\frac{1}{s+2} \\ Y(s)&=\frac{1}{s(s+2)}=\frac{1}{2}\left [ \frac{1}{s}-\frac{1}{s+2} \right ] \\ y(t)&=\frac{1}{2}(1-e^{-2t})=\frac{1}{2}-\frac{1}{2}e^{-2t} \end{aligned}
 Question 10
The state variable formulation of a system is given as
$\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} -2 & 0\\ 0&-1 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}+\begin{bmatrix} 1\\1 \end{bmatrix}u$ $x_1(0)=0,x_2(0)=0\ and \; y=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}$
The system is
 A controllable but not observable B not controllable but observable C both controllable and observable D both not controllable and not observable
GATE EE 2013   Control Systems
Question 10 Explanation:
$A=\begin{bmatrix} -2 &0 \\ 0& -1 \end{bmatrix},$ $B=\begin{bmatrix} 1\\ 1 \end{bmatrix}$
$AB=\begin{bmatrix} -2 &0 \\ 0& -1 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}=\begin{bmatrix} -2\\ -1 \end{bmatrix}$
For controllability, $|B \; : \;\;AB|\neq 0$
$\;\;\; \begin{bmatrix} 1 & -2\\ 1& -1 \end{bmatrix}=-1-(-2)=1 \neq 0$
The system is controllable.
$C^T=\begin{bmatrix} 1\\ 0 \end{bmatrix}$
$A^TC^T=\begin{bmatrix} -2 &0 \\ 0& -1 \end{bmatrix}\begin{bmatrix} 1\\ 0 \end{bmatrix}=\begin{bmatrix} -2\\ 0 \end{bmatrix}$
For observability, $[C^T\;: \; A^TC^T]\neq 0$
$\begin{bmatrix} 1 & -2\\ 0 & 0 \end{bmatrix}=0$
The system is not observable.
There are 10 questions to complete.