Test 1 - Control System | Electronics and Communication (ECE)
Topic wise test 1 for Control System (ECE) of GATE Electronics and Communication
Questions
A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has
- a higher type number
- reduced damping
- higher noise amplification
- larger transient overshoot
If the closed-loop transfer function of a control system is given as T(s) = $$\dfrac{s-5}{(s+2)(s+3)}$$, then it is
- an unstable system
- an uncontrollable system
- a minimum phase system
- a non-minimum phase system
The open-loop transfer function of a unity-gain feedback control system is given by
G (s) = $$\dfrac{k}{(s+1)(s+2)}$$
The gain margin of the system in dB is given by
- 0
- 1
- 20
- $\infty$
Despite the presence of negative feedback, control systems still have problems of instability because the
- used components have nonlinearities
- dynamic equations of the subsystems are not known
- mathematical analysis involves approximations
- system has large negative phase angle at high frequencies
A system with transfer function
G (s) = $$\dfrac{(s^2+9)(s+2)}{(s+1)(s+3)(s+4)}$$
Is excited by sin($\omega$t). The steady-state output of the system is zero at
- $\omega$= 1 rad/s
- $\omega$= 2 rad/s
- $\omega$= 3 rad/s
- $\omega$= 4 rad/s
The input-output transfer function of a plant H(x) = $$\dfrac{100}{s(s+10)^2}$$. The plant is placed in a unity negative feedback configuration as shown in the figure below.

The gain margin of the system under closed loop unity negative feedback is
- 0dB
- 20dB
- 26 dB
- 46 dB
Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems?
For the asymptotic Bode magnitude plot shown below, the system transfer function can be

- $$ \dfrac{10s+1}{0.1s + 1} $$
- $$ \dfrac{100s+1}{0.1s + 1} $$
- $$ \dfrac{100s}{10s + 1} $$
- $$ \dfrac{0.1s+1}{10s + 1} $$
The number of open right half plane of
G (s) = $
\dfrac{10}{s^5+2s^4+3s^3+6s^2+5s+3}
$
- 0
- 1
- 2
- 3
The state variable description of an LTI system is given by
$
\left(
\begin{array}
x_1 \\ x_2 \\ x_3
\end{array}
\right)
$= $
\left(
\begin{array}
\ 0&a_1&0\\
0&0&a_2\\
a_3&0&0
\end{array}
\right)
$$
\left(
\begin{array}
\ x_1\\
x_2\\
x_3
\end{array}
\right)
$+ $
\left(
\begin{array}
\ 0\\
0\\
1
\end{array}
\right)
u
$
y = (1 0 0) $
\left(
\begin{array}
\ x_1\\
x_2\\
x_3
\end{array}
\right)
$
Where y is output and u is the input. The system is controllable for
- a1 $\ne$0, a2 = 0, a3 $\ne$0
- a1 = 0, a2 $\ne$0, a3 $\ne$0
- a1 = 0, a2 $\ne$0, a3 = 0
- a1 $\ne$ 0, a2 $\ne$0, a3 = 0
The root locus of the system G (s) H(s) = $\dfrac{k}{s(s+2)(s+3)}$ has the break - away point located at
- (- 0.5,0)
- (- 2.548,0)
- (- 4,0)
- (- 0.784,0)
The input-output transfer function of a plant H(x) = $\dfrac{100}{s(s+10)^2}$. The plant is placed in a unity negative feedback configuration as shown in the figure below.

The signal flow graph that DOES NOT model the plant transfer function H(s) is
The feedback configuration and the pole-zero locations of are shown below. The root locus for negative values of k , i.e. for -$\infty$< k < 0, has breakaway / break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

- $\pm \sqrt2 $ and 00
- $\pm \sqrt2 $ and 450
- $\pm \sqrt3 $ and 00
- $\pm \sqrt3 $ and 450
For the transfer function G (j$\omega$ ) = 5 + j$\omega$, the corresponding Nyquist plot for positive frequency has the form
A causal system having the transfer function H(s) = 1 /(s+2) is excited with 10u(t) . The time at which the output reaches 99% of its steady state value is
- 2.7 sec
- 2.5 sec
- 2.3 sec
- 2.1 sec
The transfer function of a compensator is given as
Gc(s) = $\dfrac{s+a}{s+b}$
Gc (s) is a lead compensator if
- a = 1, b = 2
- a = 3, b = 2
- a = – 3, b = – 1
- a = 3, b = 1
The gain margin for the system with open-loop transfer function
G(s) H(s) = $\dfrac{2(1+s)}{s^2}, \ is$
- $\infty$
- 0
- 1
- – $\infty$
A double integrator plant, $G(s) = \dfrac{k}{s^2} \ H(s) = 1$ is to be compensated to achieve the damping ratio $\xi$ = 0.5, and an undamped natural, $\omega_n = 5\ rad/s$. Which one of the following compensator Gc(S) will be suitable?
- $\dfrac{s+3}{s+9.9}$
- $\dfrac{s+9.9}{s+3}$
- $\dfrac{s-6}{s+8.33}$
- $\dfrac{s+6}{s}$
Consider the system
$\dfrac{dx}{dt}$= Ax + Bu with A = $
\left[
\begin{array}
\ 1&0\\
0&1
\end{array}
\right]
$ and B = $
\left[
\begin{array}
\ p\\
q
\end{array}
\right]
$
where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true?
- The system is completely state controllable for any non-zero values of p and q.
- Only p = 0 and q = 0 result in controllability.
- The system is uncontrollable for all values of p and q.
- We cannot conclude about controllability from the given data.






