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.