Test 2 - Control System | Electronics and Communication (ECE)
Topic wise test for Control System (ECE) of GATE Electronics and Communication
Questions
Group I lists a set of four transfer functions. Group II gives a list of possible step response y (t). Match the step responses with the corresponding transfer functions.

- P - 3, Q - 1, R - 4, S - 2
- P - 3, Q - 2, R - 4, S - 1
- P - 2, Q - 1, R - 4, S - 2
- P - 3, Q - 4, R - 1, S - 2
The transfer function of a compensator is given as
Gc(s) = (s +1)/(s +2)
The phase of the above lead compensator is maximum at
- $\sqrt2$rad/s
- $\sqrt3$rad/s
- $\sqrt6$rad/s
- 1/ $\sqrt3$rad/s
A linear system is described by the following state equation:
X(t) = AX (t) + BU (t), A =$\left[
\begin{array}
\ 0 & 1 \\
-1 & 0
\end{array}
\right]$
The state-transition matrix of the system is
- $\left[ \begin{array} \ cost & sint \\\\ -sint & cost \end{array} \right]$
- $\left[ \begin{array} \ -cost & sint \\\\ -sint & -cost \end{array} \right]$
- $\left[ \begin{array} \ -cost & -sint \\\\ -sint & cost \end{array} \right]$
- $\left[ \begin{array} \ cost & -sint \\\\ sint & cost \end{array} \right]$
The transfer function Y(s)/R(s) of the system shown is

- 0
- $\dfrac{1}{s+1}$
- $\dfrac{2}{s+1}$
- $\dfrac{2}{s+3}$
The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by
- G(s)H(s) = k $\dfrac{s(s+1)}{(s+2)(s+3)}$
- G(s)H(s) = k$\dfrac{(s+1)}{s(s+2)(s+3)^2}$
- G(s)H(s) = k$\dfrac{s(s+1)}{s(s+1)(s+2)(s+3)}$
- G(s)H(s) = k $\dfrac{(s+1)}{s(s+2)(s+3)}$
The signal flow graph of a system is shown below.
Which of the following is the state variable representation of the system?
- $ x = \left[ \begin{array} \ 1 & 1 \\\\ -1 & 0 \end{array} \right] x + \left[ \begin{array} \ 0 \\\\ 2 \end{array} \right] u $ y = [0 & 0.5] x
- $ x = \left[ \begin{array} \ -1 & 1 \\\\ -1 & 0 \end{array} \right] x + \left[ \begin{array} \ 0 \\\\ 2 \end{array} \right] u $ y = [0 & 0.5] x
- $ x = \left[ \begin{array} \ 1 & -1 \\\\ -1 & 0 \end{array} \right] x + \left[ \begin{array} \ 0 \\\\ 2 \end{array} \right] u $ y = [0.5 & 0.5] x
- $ x = \left[ \begin{array} \ -1 & 1 \\\\ -1 & 0 \end{array} \right] x + \left[ \begin{array} \ 0 \\\\ 2 \end{array} \right] u $ y = [0.5 & 0.5] x
If A = $
x = \left[
\begin{array}
\ -2 & 2 \\
1 & -3
\end{array}
\right]
$, then sin At is
- $\dfrac{1}{3}$$ x = \left[ \begin{array} \ sin(-4t) + 2sin(-t) - 2sin(-4t) + 2sin(-t) \\\\ -sin(-4t) + sin(-t)2sin(-4t) + sin(-t) \end{array} \right] $
- $\left[ \begin{array} \ sin(-2t) sin(2t) \\\\ sin(t) sin(-3t) \end{array} \right]$
- $\dfrac{1}{3}$$ x = \left[ \begin{array} \ sin(4t) + 2sin(t) 2sin(-4t) - 2sin(-t) \\\\ -sin(-4t) + sin(t)2sin(4t) + sin(t) \end{array} \right] $
- $\dfrac{1}{3}$$ x = \left[ \begin{array} \ cos(-t) + 2cos(t) 2cos(-4t) + 2cos(-t) \\\\ -cos(-4t) + cos(-t)-2cos(-4t) + cos(-t) \end{array} \right] $
The zero-input response of a system given by the state-space equation
is
- $\left[ \begin{array} \ te^t \\\\ t \end{array} \right]$
- $\left[ \begin{array} \ e^t \\\\ t \end{array} \right]$
- $\left[ \begin{array} \ e^t \\\\ te^t \end{array} \right]$
- $\left[ \begin{array} \ t \\\\ te^t \end{array} \right]$* 1/2
A linear system is equivalently represented by two sets of state equations; $\bar X = AX + BU$ and W = CW + DU. The eigen values of the representations are also computed as $[\lambda]$ and $[\mu]$. Which of the following statements is true?
- $[\lambda] = [\mu] \ and \ X =W $
- $[\lambda] = [\mu] \ and \ X \ne W $
- $[\lambda] \ne [\mu] \ and \ X = W $
- $[\lambda] \ne [\mu] \ and \ X \ne W $
The transfer function of a plant is T (s) = $
\dfrac{5}{(s+5)(s^2+s+1)}
$. The second - order approximation of T(s) using dominate pole concept is
- $ \dfrac{1}{(s+5)(s+1)} $
- $ \dfrac{5}{(s+5)(s+1)} $
- $ \dfrac{5}{(s^2+s+1)} $
- $ \dfrac{1}{(s^2+s+1)} $
The figure shows the Nyquist plot of the open-loop transfer function G(s)H(s) of a system. If G(s)H(s) has one right hand pole, the closed loop system is

- always stable
- unstable with one closed loop right hand pole
- unstable with two closed loop right hand poles
- unstable with three closed loop right hand poles
The Nyquist plot of G (j$\omega$) H (j$\omega$) for a closed loop control system, passes through (- 1, j0) point in the GH-plane. The gain margin of the system in dB is equal to
- infinite
- greater than zero
- less than zero
- zero
The signal flow graph of a system is shown below.

The transfer function of the system is
- $ \dfrac{s+1}{(s^2+1)} $
- $ \dfrac{s-1}{(s^2+1)} $
- $ \dfrac{s+1}{(s^2+s+1)} $
- $ \dfrac{s-1}{(s^2+s+1)} $
The unit step response of an under-damped second order system has steady state value of -2. Which one of the following transfer functions has theses properties?
- $ \dfrac{-2.24}{(s^2+2.59s+1.12)} $
- $ \dfrac{-3.82}{(s^2+1.91s+1.91)} $
- $ \dfrac{-2.24}{(s^2-2.59s+1.12)} $
- $ \dfrac{-3.82}{(s^2-1.91s+1.91)} $
The magnitude of frequency responses of an underdamped second order system is 5 at 0 rad/sec and peaks to $\dfrac{10}{\sqrt3}$ at 5 $\sqrt2$ rad/sec. The transfer function of the system is
- $ \dfrac{500}{(s^2+10s+100)} $
- $ \dfrac{375}{(s^2+5s+75)} $
- $ \dfrac{720}{(s^2+12s+144)} $
- $ \dfrac{1125}{(s^2+25s+225)} $
Consider the signal flow graph shown in figure. The gain $
\dfrac{x_5}{x_1}
$ is
- $ \dfrac{1 - (be+cf+dg)}{abcd} $
- $ \dfrac{bedg}{1 - (be+cf+dg)} $
- $ \dfrac{bedg}{1 - (be+cf+dg) + bedg} $
- $ \dfrac{1 - (be+cf+dg) + bedg}{abcd} $
The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

- $k<5\ and \ \dfrac{1}{2} < k < \dfrac{1}{8}$
- $k<\dfrac{1}{8}\ and \ \dfrac{1}{2} < k < 5$
- $k<\dfrac{1}{8}\ and \ 5 < k$
- $k>\dfrac{1}{8}\ and \ k < 5$
A second-order system has the transfer function $\dfrac{C(s)}{R(s)}$ = $\dfrac{4}{s^2+4s+4}$ with r(t) as the unit-step function, the response c(t) of the system is represented by
The block diagram of a system with one input it and two outputs y1 and y2 is given below:

A state space model of the above system in terms of the state vector x and the output vector y = [y1 y2]T is
- $\bar x$ = [2]x + [1]u; y = [1 2]x
- $\bar x$ = [- 2]x + [1]u; y = $ \left[ \begin{array} \ 1 \\\\ 2 \end{array} \right] $x
- $\bar x$ = $ \left[ \begin{array} \ -2 & 0 \\\\ 0 & -2 \end{array} \right] $x + $ \left[ \begin{array} \ 1 \\\\ 1 \end{array} \right] $u; y = $ \left[ \begin{array} \ 1 & 2 \end{array} \right] $x
- $\bar x$ = $ \left[ \begin{array} \ 2 & 0 \\\\ 0 & 2 \end{array} \right] $x + $ \left[ \begin{array} \ 1 \\\\ 2 \end{array} \right] $u; y = $ \left[ \begin{array} \ 1 \\\\ 2 \end{array} \right] $x
A system with transfer function $
\left[
\begin{array}
\ Y(s) \\
X(s)
\end{array}
\right]
$ = $\dfrac{s}{s+p}$has an output y(t) = cos $
\left(
\begin{array}
\ 2t - \dfrac{\pi}{3}
\end{array}
\right)
$for the input signal x(t) = p cos $
\left(
\begin{array}
\ 2t - \dfrac{\pi}{2}
\end{array}
\right)
$. Then, the system parameter ‘p’ is
- $\sqrt3$
- $\dfrac{2}{\sqrt3}$
- 1
- $\dfrac{\sqrt3}{2}$
















