State Space Analysis

Question 1
For the given circuit, which one of the following is the correct state equation?
A
A
B
B
C
C
D
D
GATE EC 2020   Control Systems
Question 1 Explanation: 
From source transformation,

KVL in loop 1,
2i_{1}=2i+0.5\frac{\mathrm{d} i}{\mathrm{d} t}+V
\frac{\mathrm{d} i}{\mathrm{d} t}=-2V-4i+4i_{1}
KCL at node (a),
i=0.25\frac{\mathrm{d} V}{\mathrm{d} t}+\frac{V-i_{2}}{1}
\frac{\mathrm{d} v}{\mathrm{d} t}=-4V+4i+4i_{2}
\begin{bmatrix} v\\ i \end{bmatrix}=\begin{bmatrix} -4 & 4\\ -2 & -4 \end{bmatrix}\begin{bmatrix} v\\ i \end{bmatrix}+\begin{bmatrix} 0 &4 \\ 4 &0 \end{bmatrix}\begin{bmatrix} i_{1}\\ i_{2} \end{bmatrix}
Question 2
Let the state-space representation of an LTI system be \dot{x}(t)=Ax(t)+Bu(t), \dot{y}(t)=Cx(t)+du(t) where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let B=[0\;0\;1]^T and d=0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is
H(s)=\frac{1}{s^3+3s^2+2s+1}?

A
A
B
B
C
C
D
D
GATE EC 2019   Control Systems
Question 2 Explanation: 
\begin{aligned} X(t) &=A x(t)+B u(t) \\ y(t) &=C x(t) \\ B &=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \\ \frac{\gamma(s)}{U(s)} &=\\ \frac{\gamma(s)}{X_{1}(s)} \times \frac{X_{1}(s)}{U(s)}=& 1 \times \frac{1}{s^{3}+3 s^{2}+2 s+1} \\ x_{1}(s)\left(s^{3}+3 s^{2}+2 s+1\right]&=U(s) \\ x_{2} &=\dot{x}_{1}(t) \\ x_{2}(s) &=s x_{1}(s) \\ x_{3} &=\dot{x}_{2}(t) \\ \Rightarrow \quad x_{3}(s) &=s x_{2}(s)=s^{2} X_{1}(s) \\ \text { So, } \quad s x_{3}(s) &=-x_{1}(s)-2 x_{2}(s)-3 x_{3}(s)+u(s) \\ \dot{x}_{3}(t) &=-x_{1}(t)-2 r_{2}(t)-3 x_{3}(t)+u(t) \\ y(t) &=x_{1}(t) \\ \dot{x}(t) &=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3\end{array}\right] x(t)+\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] u(t) \\ y(t) &=[1 \quad 0 \quad 0] x(t) \end{aligned}
Question 3
The state equation and the output equation of a control system are given below:

\dot{x}=\begin{bmatrix} -4 & -1.5\\ 4 & 0 \end{bmatrix}x+\begin{bmatrix} 2\\ 0\end{bmatrix}u,
y=\begin{bmatrix} 1.5 & 0.625 \end{bmatrix}x.

The transfer function representation of the system is
A
\frac{3s+5}{s^{2}+4s+6}
B
\frac{3s-1.875}{s^{2}+4s+6}
C
\frac{4s+1.5}{s^{2}+4s+6}
D
\frac{4s+1.5}{s^{2}+4s+6}
GATE EC 2018   Control Systems
Question 3 Explanation: 
Transfer function
\begin{aligned} T(s) &=\frac{Y(s)}{U(s)}=\mathrm{C}(s I-A]^{-1} \mathrm{B} \\ A &=\left[\begin{array}{cc} -4 & -1.5 \\ 4 & 0 \end{array}\right] \\ B &=\left[\begin{array}{c} 2 \\ 0 \end{array}\right] \\ C&=[1.5\quad0.625]\\ {[s I-A]} & =\left[\begin{array}{cc}s+4 & 1.5 \\-4 & s\end{array}\right] \\ {[s I-A]^{-1}} & =\frac{1}{\left(s^{2}+4 s+6\right)}\left[\begin{array}{cc}s & -1.5 \\4 & s+4 \end{array}\right] \\ {[s I-A]^{-1} B} & =\frac{1}{s^{2}+4 s+6}\left[\begin{array}{c}2 s \\8 \end{array}\right] & \\ C[s I-A]^{-1} B & =\frac{1}{s^{2}+4 s+6}[1.5 \quad 0.625]\left[\begin{array}{c}2 s \\ 8\end{array}\right] \\ T(s) & =\frac{3 s+5}{s^{2}+4 s+6} \end{aligned}
Question 4
A second - order LTI system is described by the following state equations,
\frac{d}{dt}x_{1}(t)- x_{2}(t)=0
\frac{d}{dt}x_{2}(t)+2x_{1}(t)+3x_{2}(t)=r(t)
Where x_{1}(t) and x_{2}(t) are the two state variables and r(t) denotes the input. The output c(t) =x_{1}(t). The system is.
A
Undamped (oscillatory)
B
Under damped
C
Critically damped
D
Over damped
GATE EC 2017-SET-2   Control Systems
Question 4 Explanation: 
Given state variable equations are as follows:
\begin{aligned} &\frac{d}{d t} x_{1}(t)-x_{2}(t)=0 &\ldots(i)\\ &\frac{d}{d t} x_{2}(t)+2 x_{2}(t)+3 x_{2}(t)=r(t) &\ldots(ii) \end{aligned}
Also given that, input =r(t) and output =x_{1}(t)
By applying Laplace transform to equation (i), we
get,
s X_{1}(s)=X_{2}(s) \qquad \ldots(iii)
By applying Laplace transform to equation (ii).
we get,
\begin{aligned} s X_{2}(s)+2 X_{1}(s)+3 X_{2}(s)&=R(s) \\ x_{2}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3} &\ldots(iv) \end{aligned}
By substituting equation (iv) in equation (iii), we get,
\begin{aligned} s X_{\mathrm{i}}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3}\\ \left(s^{2}+3 s+2\right) X_{1}(s)&=R(s) \end{aligned}
So, the transfer function of the given system is,
\begin{aligned} \frac{X_{1}(s)}{R(s)} &=\frac{1}{s^{2}+3 s+2} \\ v r^{2} &=2 \\ 2 \zeta_{n} &=3 \\ \xi &=\frac{3}{2 \omega_{n}}=\frac{3}{2 \sqrt{2}}=1.06 \end{aligned}
As \xi \gt 1, the given system is overdamped.
Question 5
Consider the state space realization
\begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0&-9 \end{bmatrix} \begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}+ \begin{bmatrix} 0\\ 45 \end{bmatrix}u(t) , with the initial condition \begin{bmatrix} x_{1}(0)\\ x_{2}(0) \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}
where u(t) denotes the unit step function. The value of \lim_{t\rightarrow \infty }|\sqrt{x_{1}^{2}(t)+x_{2}^{2}(t)}| is _______.
A
4
B
5
C
6
D
7
GATE EC 2017-SET-2   Control Systems
Question 5 Explanation: 
\left[\begin{array}{l} \dot{x}_{1}(t) \\ \dot{x}_{2}(t) \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 0 & -9 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right]+\left[\begin{array}{c} 0 \\ 45 \end{array}\right] u(t)
By applying Laplace transform on both sides. get
\begin{aligned} s x_{1}(s)-x_{1}(0)&=0\\ X_{1}(s)&=\frac{x_{1}(0)}{s}=0 \quad \because x_{1}(0)=0 \\ \text{So,} x_{1}(t)&=0 \\ \text{and }\quad s X_{\lambda}(s)-x_{2}(0)&=-9 X_{1}(s)+\frac{45}{s} \\ X_{2}(s) &=\frac{45}{s(s+9)} \quad \because x_{2}(0)=0 \\ \text { Required value } &=\lim _{t \rightarrow \infty}|\sqrt{x_{1}^{2}(t)+x_{2}^{2}(t)}| \\ &=\left|\lim _{t \rightarrow \infty} x_{2}(t)\right| \quad \because x_{1}(t)=0 \\ \lim _{t \rightarrow \infty} x_{2}(t) &=\lim _{s \rightarrow 0} s X_{2}(s)=\frac{45}{9}=5 \end{aligned}
So,
Required value = |5| =5
Question 6
A second-order linear time-invariant system is described by the following state equations

\frac{d}{dt}x_{1}(t)+2x_{1}(t)=3u(t)
\frac{d}{dt}x_{2}(t)+x_{2}(t)=u(t)

where x_{1}(t) \; and \; x_{2}(t) are the two state variables and u(t) denotes the input. If the output c(t) = x_{1}(t), then the system is
A
controllable but not observable
B
observable but not controllable
C
both controllable and observable
D
neither controllable nor observable
GATE EC 2016-SET-3   Control Systems
Question 7
A network is described by the state model as
\dot{x_1}=2x_{1}-x_{2}+3u
\dot{x_2}=-4x_{2}-u
y=3x_{1}-2x_{2}
The transfer function H(s)(=\frac{Y(s)}{U(s)}) is
A
\frac{11s+35}{(s-2)(s+4)}
B
\frac{11s-35}{(s-2)(s+4)}
C
\frac{11s+38}{(s-2)(s+4)}
D
\frac{11s-38}{(s-2)(s+4)}
GATE EC 2015-SET-3   Control Systems
Question 7 Explanation: 
State matrix form:
\begin{array}{l} \dot{x}=\left[\begin{array}{cc} 2 & -1 \\ 0 & -4 \end{array}\right] x+\left[\begin{array}{c} 3 \\ -1 \end{array}\right] u \\ y=\left[\begin{array}{ll} 3 & -2 \end{array}\right] x \end{array}
The transfer function from tate matrix can be calculated as:
T(S)=C(s l-A]^{-1} B+D
Here,
[s l-A]^{-1}=\left[\begin{array}{cc} s-2 & 1 \\ 0 & s+4 \end{array}\right]^{-1}
\begin{aligned} &=\frac{1}{(s-2)(s+4)}\left[\begin{array}{cc} (s+4) & -1 \\ 0 & (s-2) \end{array}\right] \\ \Rightarrow &C[s /-A]^{-1} B \\ =& \frac{1}{(s-2)(s+4)}[3-2]\left[\begin{array}{cc} (s+4) & -1 \\ 0 & (s-2) \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \end{array}\right] \\ =& \frac{1}{(s-2)(s+4)}[3-2]\left[\begin{array}{c} 3 s+12+1 \\ -s+2 \end{array}\right] \\ =& \frac{1}{(s-2)(s+4)}[9 s+39+2 s-4] \\ =& \frac{11 s+35}{(s-2)(s+4)} \end{aligned}
Question 8
The state variable representation of a system is given as x=\begin{bmatrix} 0 & 1\\ 0&-1 \end{bmatrix}x; \; x(0)=\begin{bmatrix} 1\\ 0 \end{bmatrix} \; y= \begin{bmatrix} 0 &1 \end{bmatrix}x
The response y(t) is
A
sin(t)
B
1-e^{t}
C
1-cos(t)
D
0
GATE EC 2015-SET-2   Control Systems
Question 8 Explanation: 
The response of the system is given by
x(t)=L^{-1}\left[(s l-A)^{-1} x(0)+(s /-A)^{-1} B \cdot U(s)\right]
and the complete response
y(t)=[C] x(t)
From the given state model,
\begin{array}{l} A=\left[\begin{array}{ll}0 & 1 \\ 0 & -1\end{array}\right] ; B=0: C=[0 \quad 1] \; and \; x(0)=\left[\begin{array}{c}1 \\ 0\end{array}\right] \\ \therefore \quad(s l-A)^{-1}=\left[\begin{array}{cc}s & -1 \\ 0 & s+1\end{array}\right]^{-1} \\ =\frac{1}{s(s+1)}\left[\begin{array}{cc}s+1 & 1 \\ 0 & s\end{array}\right] \\ (s l-A)^{-1} x(0)=\frac{1}{s(s+1)}\left[\begin{array}{cc} s+1 & 1 \\ 0 & s \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \end{array}\right] \\ =\frac{1}{s(s+1)}\left[\begin{array}{c} s+1 \\ 0 \end{array}\right]=\left[\begin{array}{c} 1 / s \\ 0 \end{array}\right] \\ \Rightarrow x(t)=L^{-1}\left[(s I-A)^{-1} x(0)\right] \\ =\left[\begin{array}{c} 1 \\ 0 \end{array}\right] u(t) \end{array}
and the complete response
y(t)=\left[\begin{array}{lll}0 & 1\end{array}\right] x(t)=\left[\begin{array}{ll}0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 0\end{array}\right]=0
Question 9
The state transition matrix \phi (t) of a system \begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2} \end{bmatrix}=\begin{bmatrix} 0 & 1\\ 0&0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} is
A
\begin{bmatrix} t & 1\\ 1&0 \end{bmatrix}
B
\begin{bmatrix} 1 & 0\\ t&1 \end{bmatrix}
C
\begin{bmatrix} 0 & 1\\ 1&t \end{bmatrix}
D
\begin{bmatrix} 1 & t\\ 0&1 \end{bmatrix}
GATE EC 2014-SET-4   Control Systems
Question 9 Explanation: 
\begin{aligned} \text { Given } &\left[\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] \\ A &=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right] \\ [s I-A] &=\left[\begin{array}{ll} s & 0 \\ 0 & s \end{array}\right]-\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} s & -1 \\ 0 & s \end{array}\right] \\ [s I-A]=s^{2} & \\ \phi(t) &=\mathcal{L}^{-1}[s I-A]^{-1} \\ &=\frac{1}{s^{2}}\left[\begin{array}{ll} s & 1 \\ 0 & s \end{array}\right] \\ &=L^{-1}\left[\begin{array}{cc} \frac{1}{s} & \frac{1}{s^{2}} \\ 0 & \frac{1}{s} \end{array}\right]=\left[\begin{array}{ll} 1 & t \\ 0 & 1 \end{array}\right] \end{aligned}
Question 10
The state equation of a second-order linear system is given by

\dot{x}(t) = Ax(t), \;\; x(0) = x_{0}

For x_{0}=\begin{bmatrix} 1\\ -1 \end{bmatrix}, \; x(t)=\begin{bmatrix} e^{-t}\\ -e^{-t} \end{bmatrix} and for x_{0}=\begin{bmatrix} 0\\ 1 \end{bmatrix}, x(t)=\begin{bmatrix} e^{-t}-e^{-2t}\\ -e^{-t}+2e^{-2t} \end{bmatrix}

when x_{0}=\begin{bmatrix} 3\\ 5 \end{bmatrix}, x(t) is
A
\begin{bmatrix} -8e^{-t} +11e^{-2t}\\ 8e^{-t} -22e^{-2t} \end{bmatrix}
B
\begin{bmatrix} 11e^{-t} -8e^{-2t}\\ -11e^{-t} +16e^{-2t} \end{bmatrix}
C
\begin{bmatrix} 3e^{-t} -5e^{-2t}\\ -3e^{-t} +10e^{-2t} \end{bmatrix}
D
\begin{bmatrix} 5e^{-t} -3e^{-2t}\\ -5e^{-t} +6e^{-2t} \end{bmatrix}
GATE EC 2014-SET-3   Control Systems
Question 10 Explanation: 
Given \dot{x}(t)=A x(t), \quad x(0)=x_{0}
Taking the Laplace transform
\begin{aligned} s X(s)-x(0) &=A X(s) \\ [s I-A] X(s) &=x(0) \\ X(s) &=[s I-A]^{-1} x(0) \\ x(t) &=\mathcal{L}^{-1}[s I-A]^{-1} x(0) &\ldots(i) \end{aligned}
Condition given are
\begin{aligned} For x_{0}&=\left[\begin{array}{c}1 \\ -1\end{array}\right], x(t)=\left[\begin{array}{c}e^{-t} \\ -e^{-t}\end{array}\right] \\ For x_{0}&=\left[\begin{array}{l}0 \\ 1\end{array}\right], x(t)=\left[\begin{array}{l}e^{-t}-e^{-2 t} \\ -e^{-t}+2 e^{-2 t}\end{array}\right] \end{aligned}
Using the linearity property in equation (i)
K_{1} x_{1}(t)=\mathcal{L}^{-1}[s I-A]^{-1} x_{1}(0) K_{1}
K_{2} x_{2}(t)=\mathcal{L}^{-1}[s I-A]^{-1} x_{2}(0) K_{2}
Using the linearity property as
\begin{aligned} K_{1} x_{1}(t)+K_{2} x_{2}(t)&=\mathcal{L}^{-1}[s I-A]^{-1} \\ \left[K_{1} x_{1}(0)+K_{2} x_{2}(0)\right] & \quad \ldots(ii)\\ \text { Also } \quad X_{3}(s)&=[s I-A]^{-1} x_{3}(0) \\ \text{So, } \quad K_{1}\left[\begin{array}{r}1 \\ -1\end{array}\right]+&K_{2}\left[\begin{array}{l}0 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 5\end{array}\right] \\ \left[\begin{array}{l}K_{1}+O K_{2} \\ -K_{1}+K_{2}\end{array}\right]&=\left[\begin{array}{l}3 \\ 5\end{array}\right] \\ \Rightarrow \quad K_{1}&=3 \\ K_{2}&=8 \end{aligned}
So, from equation (ii), we get x(t)
\begin{aligned} x(t) &=K_{1} x_{1}(t)+K_{2} x_{2}(t) \\ &=3\left[\begin{array}{c} e^{-t} \\ -e^{-t} \end{array}\right]+8\left[\begin{array}{c} e^{-t}-e^{-2 t} \\ -e^{-t}+2 e^{-2 t} \end{array}\right] \\ &=\left[\begin{array}{c} 11 e^{-t}-8 e^{-2 t} \\ -1 t e^{-t}+16 e^{-2 t} \end{array}\right] \end{aligned}
There are 10 questions to complete.
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