State Variable Analysis


Question 1
Consider the state-space description of an LTI system with matrices

A=\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right], B=\left[\begin{array}{l}0 \\ 1\end{array}\right], C=[3-2], D=1

For the input, \sin (\omega t), \omega \gt 0, the value of \omega for which the steady-state output of the system will be zero, is ___ (Round off to the nearest integer).
A
0
B
1
C
2
D
3
GATE EE 2023   Control Systems
Question 1 Explanation: 
We have, transfer function
\mathrm{T}(\mathrm{s})=\mathrm{C}[\mathrm{SI}-\mathrm{A}]^{-1} \mathrm{~B}+\mathrm{D} \quad (1)
where, [S I-A]=\left[\begin{array}{ll}S & 0 \\ 0 & S\end{array}\right]-\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right]
\begin{aligned} & =\left[\begin{array}{cc} \mathrm{s} & -1 \\ 1 & \mathrm{~s}+2 \end{array}\right] \\ \therefore \quad[\mathrm{sl}-\mathrm{A}]^{-1} & =\frac{1}{\mathrm{~s}^{2}+2 \mathrm{~s}+1}\left[\begin{array}{cc} \mathrm{s}+2 & 1 \\ -1 & \mathrm{~s} \end{array}\right] \end{aligned}

From eqn. (1), we get
\begin{aligned} \mathrm{T}(\mathrm{s})&=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{cc} \frac{\mathrm{s}+2}{(\mathrm{~s}+1)^{2}} & \frac{1}{(s+1)^{2}} \\ -\frac{1}{(s+1)^{2}} & \frac{s}{(s+1)^{2}} \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \end{array}\right]+1 \\ &=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{c} \frac{1}{(s+1)^{2}} \\ \frac{s}{(s+1)^{2}} \end{array}\right]+1 \\ &=\frac{3}{(s+1)^{2}}-\frac{2 s}{(s+1)}+1 \\ &=\frac{3-2 s+s^{2}+2 s+1}{s^{2}+2 s+1} \\ &=\frac{s^{2}+4}{s^{2}+2 s+1} \end{aligned}

Put, s=j \omega,
T(j \omega)=\frac{-\omega^{2}+4}{-\omega^{2}+\mathrm{j} 2 \omega+1}

Now, condition for output is zero,
\begin{aligned} -\omega^{2}+4 & =0 \\ \Rightarrow \quad \omega & =2 \mathrm{rad} / \mathrm{sec} . \end{aligned}
Question 2
The state space representation of a first-order system is given as
\overset{\bullet }{x}=-x+u
y=x
where,x is the state variable, u is the control input and y is the controlled output. Let u=-Kx be the control law, where K is the controller gain. To place a closed-loop pole at -2, the value of K is ___________________.
A
1
B
2
C
4
D
6
GATE EE 2021   Control Systems
Question 2 Explanation: 
\dot{x}=-x-K x=x(-k I-I)
Characteristic equation,
\begin{aligned} |S I+K I+I| &=0 \\ |(S+1+K) I| &=0 \\ \therefore \qquad\qquad S+1+K &=0 \\ S &=-1-K \\ -2 &=-1-K \\ K &=1 \end{aligned}


Question 3
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 3 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 4
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 4 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 5
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 5 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}


There are 5 questions to complete.