Test 1 - Control System | Electronics and Communication (ECE)

Description: Topic wise test 1 for Control System (ECE) of GATE Electronics and Communication
Number of Questions: 20
Created by:
Tags: Steady State Transfer Function Pole-zero Plot Pole-Zero Plot Gains Phase Margin Stability
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A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has

  1. a higher type number

  2. reduced damping

  3. higher noise amplification

  4. larger transient overshoot


Correct Option: C
Explanation:

PD controller may accentuate noise at higher frequency. It does not affect the type of system and it increases the damping. It also reduces the maximum 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

  1. an unstable system

  2. an uncontrollable system

  3. a minimum phase system

  4. a non-minimum phase system


Correct Option: D
Explanation:

In a minimum phase system, all the poles as well as zeros are on the left half of the s-plane. In given system as there is right half zero (s = 5), the system is 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

  1. 0

  2. 1

  3. 20

  4. $\infty$


Correct Option: D
Explanation:

Given system is 2nd order and for 2nd order system, G.M. is infinite.

Despite the presence of negative feedback, control systems still have problems of instability because the

  1. used components have nonlinearities

  2. dynamic equations of the subsystems are not known

  3. mathematical analysis involves approximations

  4. system has large negative phase angle at high frequencies


Correct Option: A
Explanation:

Despite the presence of negative feedback, control systems still have problems of instability because used components have nonlinearity. There are always some variations as compared to ideal characteristics.

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

  1. $\omega$= 1 rad/s

  2. $\omega$= 2 rad/s

  3. $\omega$= 3 rad/s

  4. $\omega$= 4 rad/s


Correct Option: C
Explanation:

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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

  1. 0dB

  2. 20dB

  3. 26 dB

  4. 46 dB


Correct Option: C
Explanation:

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?


Correct Option: C
Explanation:

Transfer function for the given pole zero plot is: $$ \dfrac{(s+Z_1)(s+Z_2)}{(s+P_1)(s+P_2)} $$ From the plot $R_e \ (P_1 \ and \ P_2) > (Z_1 \ and Z_2)$

So, these are two lead compensator. Hence both high pass filters and the system is high pass filter.

For the asymptotic Bode magnitude plot shown below, the system transfer function can be

  1. $$ \dfrac{10s+1}{0.1s + 1} $$

  2. $$ \dfrac{100s+1}{0.1s + 1} $$

  3. $$ \dfrac{100s}{10s + 1} $$

  4. $$ \dfrac{0.1s+1}{10s + 1} $$


Correct Option: A
Explanation:

Here by using figure

K = 1, w1 = .1 rad/s w2 = 10 rad/s

H(s) = K (1 + s/w1)/(1 + s/w2) = 1(1 + s/.1)/(1 + s/10) = (1 + 10s)/(1 + .1s)

The number of open right half plane of G (s) = $ \dfrac{10}{s^5+2s^4+3s^3+6s^2+5s+3} $

  1. 0

  2. 1

  3. 2

  4. 3


Correct Option: C
Explanation:

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

  1. a1 $\ne$0, a2 = 0, a3 $\ne$0

  2. a1 = 0, a2 $\ne$0, a3 $\ne$0

  3. a1 = 0, a2 $\ne$0, a3 = 0

  4. a1 $\ne$ 0, a2 $\ne$0, a3 = 0


Correct Option: D
Explanation:

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The root locus of the system G (s) H(s) = $\dfrac{k}{s(s+2)(s+3)}$ has the break - away point located at

  1. (- 0.5,0)

  2. (- 2.548,0)

  3. (- 4,0)

  4. (- 0.784,0)


Correct Option: D
Explanation:

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


Correct Option: D
Explanation:

$$ TF = H(s) \\ = \dfrac{100}{s(s^2+100)} \ne \dfrac{100}{s(s+10)^2} $$

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

  1. $\pm \sqrt2 $ and 00

  2. $\pm \sqrt2 $ and 450

  3. $\pm \sqrt3 $ and 00

  4. $\pm \sqrt3 $ and 450


Correct Option: B
Explanation:

For the transfer function G (j$\omega$ ) = 5 + j$\omega$, the corresponding Nyquist plot for positive frequency has the form


Correct Option: A
Explanation:

We have $G(j\omega) = 5+j\omega \\ Here \ \sigma = 5. \ Thus\ G(j\omega)\ is\ a\ straight\ line\ parallel\ to\ j\omega\ axis. $

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

  1. 2.7 sec

  2. 2.5 sec

  3. 2.3 sec

  4. 2.1 sec


Correct Option: C
Explanation:

The transfer function of a compensator is given as

Gc(s) = $\dfrac{s+a}{s+b}$

Gc (s) is a lead compensator if

  1. a = 1, b = 2

  2. a = 3, b = 2

  3. a = – 3, b = – 1

  4. a = 3, b = 1


Correct Option: A
Explanation:

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The gain margin for the system with open-loop transfer function G(s) H(s) = $\dfrac{2(1+s)}{s^2}, \ is$

  1. $\infty$

  2. 0

  3. 1

  4. $\infty$


Correct Option: D
Explanation:

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?

  1. $\dfrac{s+3}{s+9.9}$

  2. $\dfrac{s+9.9}{s+3}$

  3. $\dfrac{s-6}{s+8.33}$

  4. $\dfrac{s+6}{s}$


Correct Option: A
Explanation:

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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?

  1. The system is completely state controllable for any non-zero values of p and q.

  2. Only p = 0 and q = 0 result in controllability.

  3. The system is uncontrollable for all values of p and q.

  4. We cannot conclude about controllability from the given data.


Correct Option: C
Explanation:

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