Control Systems Test - 4 - PDF Flipbook

Control Systems Test - 4

205 Views
68 Downloads
PDF 4,122,886 Bytes

Download as PDF

REPORT DMCA


GATE
EEE

Control
Systems

Test-04Solutions


CONTROL SYSTEMS
1. The poles and zeroes of an all-pass network are located in which

part of the s-plane?
a) Poles and zeroes are in the right half of s-plane
b) Poles and zeroes are in the left half of s-plane
c) Poles in the right half and zeroes in the left half of s-plane
d) Poles in the left half and zeroes in the right half of s-plane
Answer: (d)
2. To detect the position error in a position control system, which
of the following may be used?
1) Potentiometers
2) Synchros
3) LVDT
Select the correct answer using the code given below:
a) 1 and 2
b) 1 and 3
c) 2 and 3
d) 1, 2 and 3
Answer: (d)
3. Which of the following is the transfer function of a system
having the Nyquist plot shown in Fig. below?

1


a)
( +2)2( +5)

b)
2( +2)( +5)

c) ( +1)
2( +2)( +5)

d) ( +1)( +3)
2( +2)( +5)

Answer: (b)

Solution:
At ω = 0 ∞∠-1800
At ω = ∞ 0∠-3600
Initial angle (at ω = 0) -1800 ∴ it is a type ‘2’ system ω → ∞
∠-3600 ⇒ the difference between poles and zeros must be ‘4’

[(P – Z) (–900) = -3600 P = 4 and Z = 0]

From the given options only (b) matches.

4. The transfer function 1+0.5s represents a
1+s

a) Lead network

b) Lag network

c) Lag-lead network

d) PID controller

Answer: (b)

Solution:

T. F = 1+0.5s = T(s)
1+s

∠T(jω) = tan−1 0.5 ω − tan−1 ω < 0

It is a lag network.

2


In s-plane

5. What is the effect of lag compensator on the system bandwidth

(BW) and the signal to noise ratio (SNR)?

a) BW is reduced and SNR is improved

b) BW is reduced and SNR is deteriorated

c) BW is increased and SNR is improved

d) BW is increased and SNR is deteriorated

Answer: (a)

6. Consider the unit step response of a unity feedback control

system whose open loop transfer function is given by G(s) =
s(s1+1). What are the values of oscillation and the damping ratio?

a) ωn = 1, ζ = 1

b) ωn = 1, ζ = 0.5

c) ωn = 0.5, ζ = 1

d) ωn = 0.5, ζ = 0.5

Answer: (b)

Solution:

Characteristic equation: s2 + s + 1

T. F = G(s) = 1
1+G(s) s2+s+1

3


Comparing with s2 + 2ξωn + ωn2
ω2n = 1, ωn = 1

2ξωn = 1 ⇒ ξ = 0.5
7. What is the effect of phase lead compensator on gain cross-over

frequency (ωgc) and on the bandwidth (ωb)?
a) Both are increased
b) ωgc is increased but ωb is decreased
c) ωgc is decreased but ωb is increased
d) Both are decreased
Answer: (a)
8. Consider the following statements:
1. If the input is a sine wave of radian frequency ω, the output in

general is non-sinusoidal containing frequencies which are
multiple of ω.
2. The jump resonance may occur
3. The system exhibits self-sustained oscillation of fixed
frequency and amplitude
4. The response to a particular test signal is a guide to the
behaviour to other inputs.
Which of the above statements are correct and peculiar to
nonlinear system?
a) 1, 3 and 4
b) 2, 3 and 4
c) 1, 2 and 3
d) 1, 2 and 4

4


Answer: (c)

Solution:

Refer the peculiar characteristics shown by a nonlinear system.
9. If a system has an open-loop transfer function 11−+ss, then the gain

of the system at frequency 1 rad/sec will be

a) 1

b) 1/2

c) Zero

d) -1

Answer: (a)

Solution:

G(jω) = 1−jω
1+jω

Gain, |G(jω)| = �11+−jjωω� = 1 at all frequencies

10. The asymptotic approximation of the log-magnitude versus

frequency plot of a minimum phase system with real poles and

one zero is shown in Fig. Its transfer functions is

a) 20( +5)
( +2)( +25)

b) 10( +5)
( +2)2( +25)

5


c) 20( +5)
2( +2)( +25)

d) 50( +5)
2( +2)( +25)

Answer: (d)

Solution:

|TF| = [20logk − 2(2logω)]

= 54dB|ω=0.1
20logk = 54dB + 2(2log 0.1)

20logk = 14dB

k=5

TF = 5(s+5)(2)(25)
5(s)2(s+2)(s+25)

TF = 50(s+5)
s2(s+2)(s+25)

11. In the control system shown below the controller which can

give zero steady-state error to a ramp input is of

a) Proportional type
b) Integral type
c) Derivative type
d) Proportional plus derivative type
Answer: (b)

6


Solution:

Let, controller be G(s)

For ramp input error = A/KV

Where, KV = sli→m0 sG(s) × 9
s(s+2)

KV = sli→m0 9G(s)
s+2

For error to be zero G(s) should be of type 1.

Hence, option (b) is correct.

12. Loop gain vs phase plot is known as

a) Nyquist plot

b) Bode Plot

c) Nichol's chart

d) Inverse Nyquist plot

Answer: (c)

Solution:
Constant M-circle in log-magnitude vs phase angle plane →

Nichol's charts

13. The effect of integral controller on the steady-state error ess and

that on the relative stability Rs of the system is
a) Both are increased

b) ess is increased but Rs is reduced

c) ess is reduced but Rs is increased

d) Both are reduced

Answer: (d)

7


Solution:

With the effect of integral controller, the steady state error as

relative stability reduces, because integral controller will add

one pole in the system which will the settling time results in

reduction in relative stability.

14. The Z-transform of x(K) is given by

X(Z) = �1−e−T�Z−1
(1−Z−1)�1−e−TZ−1�

The initial value x(0) is

a) zero

b) 1

c) 2

d) 3

Answer: (a)

Solution:

X(Z) = �1−e−T�Z−1
(1−Z−1)�1−e−TZ−1�

x(Z) = Z�1−e−T�
(Z−1)�Z−e−T�

so, x(0) = 0

15. For a tachometer if θ(t) is the rotor displacement e(t) is the

output voltage and Kt is the tachometer constant, then the

transfer function is defined as

a) Kt · s2

b) Kt · s

c) Kt / s

8


d) Kt
Answer: (b)

16. The frequency response of a linear system G (jω) is provided

in the tabular form below.
| ( )| 1.3 1.2 1.0 0.8 0.5 0.3
∠ ( ) −1300 −1400 −1500 −1600 −1800 −2000

The gain margin and phase margin of the system are

a) 6 dB and 300
b) 6 dB and -300

c) -6 dB and 300

d) -6 dB and -300

Answer: (a)

Solution:

From the given table

�G�jωpc�H�jωpc�� = 0.5

∴ GM = 1 = 2 (or)
0.5

GM = 20 log 2 = 6 dB

From the given table
∠G�jωgc�H�jωgc� = −1500
PM = 1800 − 1500 = 300

GM = 6 dB

PM = 300

9


17. If D is the rotor diameter and L, the axial length, then a high
performance a.c servomotor is characterized by which one of the
following?
a) Large D and Large L
b) Large D and Small L
c) Small D and Small L
d) Small D and Large L
Answer: (d)

18. For the following network to work as lag compensator, the
value of R2 should be

a) R2 ≥ 20Ω

b) R2 ≤ 10Ω

c) R2C ≤ R21C
2

d) Any value of R2

Answer: (d)

Solution:

E2(s) = R2+S1C
E1(s) R1+R2+S1C

= 1+sR2C
1+sC(R1+R2)

For lag compensator,

10


tan−1 �ωC(R11+R2)� ≥ tan−1 �ωC1R2�
ωC(R1 + R2) > ωCR2
⇒ R1 > 0
Which is already given

19. Which one of the following statements is correct for the system

represented by the block diagram shown in the below figure?

a) Open loop is stable but closed loop unstable.

b) Open loop and closed loop both are unstable

c) Open loop and closed loop both are stable.

d) Open loop is unstable but closed loop stable.

Answer: (d)

Solution:

Open loop function,

G(s) = 5
(s−1)(s+2)

One pole lies on R.H.S of s-plane (s = 1)

So, open loop system is unstable.

Closed loop transfer function,

5 5
= =C(s) s2+s+3

R(s)
(s−1)(s+2)

1+(s−1)5(s+2)

Roots are complex conjugate but lies on L.H.S of s-plane.

Closed loop system is stable.

11


20. The phase-lead compensation is used to
a) increase rise time and decrease overshoot.
b) decrease both rise time and overshoot.
c) increase both rise time and overshoot.
d) decrease rise time and increase overshoot.
Answer: (b)
Solution:
Phase-lead compensation is used to decrease rise time and to
decrease overshoot.

21. Assertion (A): The stator windings of a control transformer
have higher impedance per phase.
Reason (R): The rotor of a control transformer is a cylindrical
in shape.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of
A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (d)

22. The figure below gives two equivalent block diagrams

12


The value of transfer function of block marked 'X' is given by

a) G(s)

b) 1
( )

c) 1

d) none of these

Answer: (b)

23. Match List-I (Transfer Function of the System) with List-II

(Type and Order of the System) and select the correct answer

using the codes given below the lists:

List – I

A. 2(s+2)
s(s+5)

B. (s+2)
(s+3)(s+5)

C. 2(s+2)
s2(s+5)

D. 5(s+2)
(s+1)(s+3)(s+5)

List – II

1. Type 0, second order

2. Type 1, second order

3. Type 0, third order

4. Type 2, third order

Codes:

ABC D

a) 2 1 4 3

b) 4 3 2 1

13


c) 2 3 4 1
d) 4 1 2 3
Answer: (a)
Solution:
Type of system: Number of poles at origin
Order of system: Degree of denominator polynomial or number
of total poles
A: Type-1, order 2
B: Type-0, order 2
C: Type-2, order 3
D: Type-0, order 3
24. The Bode plot of a transfer function G(s) is shown in the figure
below

The gain (20log| ( )|) is 32 dB and -8 dB at 1 rad/s and 10

rad/s respectively. The phase is negative for all ω. Then G(s) is

a) 39.8


b) 39.8
2

c) 32


d) 32
2

14


Answer: (b)

Solution:

Slope = −8−32 = −40dB/dec
log10−log1

= −2 × 20dB/dec

i.e., two poles are at origin. So the transfer function of the bode

plot is G(s) = K
s2

Finding K:

|G(s)| = K
ω2

|G(s)|dB = 20 log K − 40 logω

Given |G(s)|dB = −8 dB at ω = 10 rad/s

∴ −8 = 20 log k – 40 log 10

⇒ 20 log k = 32

⇒ log k = 32 = 1.6
20

⇒ K =101.6 = 39.81

∴ G(s) = 39.81
s2

25. Consider the single input single output system with its state

variable representation:

−1 0 0 1
̇ = � 0 −2 0 � + �1�
0 0 −3
0

= [1 0 2]

The system is

a) neither controllable nor observable

b) controllable but not observable

15


c) uncontrollable but observable

d) both controllable and observable

Answer: (a)

26. The characteristic equation of a feedback control system is

given by

s3 + 5s2 + (K + 6)s + K = 0

There K > 0 is a scalar variable parameter. In the root locus

diagram of the system the asymptotes of the root loci for large

values of K meet at a point in the s-plane, whose coordinates are

a) (–3, 0)

b) (–2, 0)

c) (–1, 0)

d) (2, 0)

Answer: (b)

Solution:

s3 + 5s2 + (K + 6)s + K = 0

⇒ 1 + K(s+1) = 0
s3+5s2+6s

⇒ 1 + K(s+1) = 0
s(s2+5s+6s)

⇒ 1 + K(s+1) = 0
s(s+2)(s+3)

⇒ σA = (−2)+(−3)+(−1) = −4 = −2
3−1 2

Hence, point is (-2, 0)

16


27. The state variable description of a linear autonomous system is

̇ = Ax, where x is the two-dimensional state vector and A is

given by

= �−02 −02�

The poles of the system are located at

a) –2 and +2

b) –2j and +2j

c) –2 and –2

d) +2 and +2

Answer: (a)

Solution:

[sI − A] = �0s 80� − �−02 −02�

= �82 2s�

[sI − A]−1 = �−s2 −s2�
2−4

Poles of the system s = ±2

Hence, option (a) is correct

28. Consider the block diagram:

17


Which of the block diagrams given above can be reduced to

transfer function? ( ) = 1
a) 1 and 3 ( ) 1+ 1 2

b) 2 and 4

c) 1 and 2

d) 2 and 3

Answer: (a)

Solution:

1. ( ) = 1
( ) 1+ 1 2

2. ( ) = 1
( ) 1+ 1

3. ( ) = 1 . 1 2 = 1
( ) 2 1+ 1 2 1 2+1

4. ( ) = 1 . 1 2 = 1
( ) 2 1− 1 2 1− 1 2

29. The gain margin of a system is –10dB. It is increased by 5dB.

Then the system is

a) stable

b) unstable

18


c) marginally stable

d) unstable for –10dB itself

Answer: (a)

Solution:

Gain margin = -10 + 5 = - 5 dB

We have, -5 = 20 log (GM)

GM = 0.5623

Since gain margin is positive, hence the system is stable

30. In the following pick out the time-invariant systems

i) dy(t) + 5y(t) = u(t)
dt

ii) t d2y(t) + dy(t) + 2y(t) = v(t)
dt2 dt

iii) d2y(t) + 5 dy(t) + 6y(t) = w(t)
dt2 dt

a) (i) and (ii)

b) (i) and (iii)

c) (ii) and (iii)

d) (i) only

Answer: (a)

19


Data Loading...