# Control Systems Test - 1 - PDF Flipbook

Control Systems Test - 1

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GATE
EEE

Control
Systems

Test-01Solutions

CONTROL SYSTEMS

1.The transfer function of the system Y(s)/U(s) whose state-space

equations are given below is:

� ̇̇12(( ))� = �21 02� � 12(( ))� + �21� ( )

( ) = [1 0] � 12(( ))�

a) ( + 2)
( 2 − 2 − 2)

b) ( −2) 4)
( 2 + −

c) ( − 4)
( 2 + − 4)

d) ( + 4)
( 2 − − 4)

Solution:

= [ − ]−1

− = � −−21 − 2 �

= [1 0]� −−21 − 2 ��12�
( −1)( )−4

= + 4
2 − − 4

2. The open-loop transfer function of a unity feedback system is

given by

( ) = ( +2)
+1)(

The breakaway point of the root locus plot is given by

a) -0.423

1

b) -0.523

c) -0.700

d) none of these

Solution:

( ) = ( +2)
+1)(

Its characteristic equation is,

1 + ( +2) = 0
+1)(

= − ( + 1)( + 2)

= − ( 2 + 3 + 2)

= − [ 3 + 3 2 + 2 ]

Now = 0

⇒ 3 3 + 6 + 2 = 0

∴ −1.577 cannot be breakaway point, the breakaway point is

−0.423

3. The open-loop transfer function of a feedback system is

( ). ( ) = ( + 4 + 20)
4)( 2 +

The four branches of root locus originate at

a) –2, –3, –1 + j4 – 1 – j4

b) –1, –2, –3 + j4 – 3 – j4

c) 0, –4, –3 + j4 – 2 – j4

d) none of these

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Solution:

All the branches of root locus originate from poles

4. Unit step input, gives a response y(t) = te-t, t > 0. The transfer

function of the system is

a) ( 1
+1)2

b) ( 1
+1)2

c) (
+1)2

d) 1
( +1)

Solution:

Step input = u(t)
Output y(t) = − ( )

∴ = [ ] = � − ( )�
[ ] [ ( )]

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

5. A linear discrete-time system has the characteristic equation z3 –

0.81 z = 0. The system

a) is stable

b) is marginally, stable

c) is unstable

d) stability cannot be assessed from the given information

Solution:

3

Put = 1+
1−

The characteristic equation is

�11−+ �3 − 0.81 �11−+ � = 0
⇒ 2 + 3 2 + 3 + 1 − 0.81 − 0.81 = 0

Since all the coefficient of the first is positive, hence the system
is stable.
6. In position control systems, the device used for providing rate-
feedback voltage is called
a) potentiometer
b) synchro transmitter
c) synchro transformer
d) tachogenerator
7. A minimum-phase system with no zeros has a phase angle of -
2700 at gain crossover frequency. The system is
a) stable
b) unstable
c) marginally stable
d) conditionally stable
Solution:

4

Phase margin = 1800 + ϕ

= 1800 + (- 2700)

= -900

For a stable system phase margin should be positive.

Hence the system is unstable.

8. The transfer function for a system is

( ) = ( +2) , > 0
( +1)( +3)( +4)

Two branches (loci) of the plot directed along asymptotes are

centered at a point

a) -3

b) -4

c) -2

d) -1

Solution:

( ) = ( +2) , > 0
( +1)( +3)( +4)

Point of intersection of asymptotes with real axis is called the

centroid ( ).

= ∑ −∑

= −1− 3 − 4 − (−2) = −3
3−1

9. A temperature control system is

a) a servomechanism

b) a digital system

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d) an analogue process control system
10. A position control system is
a) a process control system
b) an automatic regulatory system
c) a servomechanism
d) a stochastic control system
11. The figure shown below represents root locus of a unity
feedback control system. Which one of the following is the
correct open loop transfer function of the system?

a) ( +5)
( +1)( +2)

b) ( +1)
( +2)( +5)

c)
( +1)( +5)

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

Solution:

Root locus starts from poles and end to zeros.

6

Here root locus is starting from -1 and -2 and one branch is

ending at -5 and other to -∞.

So, Poles at -1, -2

Zero → -5 and ∞

= ( +5)
( +1)( +2)

12. The transfer function V2(s)/V1(s) of the circuit shown in below

is

a) 0.5 +1
+1

b) 3 +6
+2

c) +2
+1

d) +1
+2

Solution:

Take general RC circuit as shown below.

Transfer function 2( ) = + 1 = 1( 2+1)
1( ) 2 ( 1 2+ 1+ 2)
1 1
+ 2 + 1

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, 1 = 2 = 100 ; = 10

∴ 2( ) = 100×10−6� ×10×103×100×10−6+1�
1( ) ( ×10×103×100×10−6+100×10−6)

= +1
+2

13. Traffic light on roads are examples of

a) closed-loop system

b) open-loop system

c) both open-loop and closed-loop

d) open-loop but can be made closed-loop

14. The frequency at which the Nyquist diagram cuts (-1, 0) circle

is known as

a) gain crossover frequency

b) phase crossover frequency

c) damping frequency

d) natural frequency

15. GC(s) is a lead compensator if

a) a = 1, b = 2

b) a = 3, b = 2

c) a = -3, b = -1

d) a = 3, b = 1

Solution:

In lead compensator zero is near to origin.

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That means a = 1, b = 2 whose pole zero plot is given below.

16. A unity feedback system has OLTF.

( ) = 9
( +3)

a) Natural frequency = 9

b) Natural frequency = 3

c) Damping ratio = 1/2

d) Damping ratio = 1/6

Solution:

( ) = 9
( +3)

Characteristic equation is

1 + 9 = 0
( +3)

⇒ 2 + 3 + 9 = 0

∴ 2 = 9
⇒ = 3
2 = 3
⇒ = 0.5

17. Which one of the following is correct?

A system with gain margin close to zero dB or a phase margin

close to zero degree is

a) Highly stable

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b) Oscillatory
c) Relatively stable
d) Unstable
Solution:

GM = 0 dB
-20 log a = 0

⇒ = 1
For a < 1 system stable
For a > 1 unstable
For a = 1 oscillatory system
18. The unit circle of the Nyquist plot transforms into 0 dB line of
the amplitude plot of the Bode diagram at
a) zero frequency
b) low frequency
c) high frequency
d) any frequency
19. The signal flow graph of a system is shown below. U(s) is the
input and C(s) is the output.

10

Assuming, h1 = b1 and h0 = b0 – b1a1, the input-output transfer

function, G(s) = C(s)/U(s) of the system is given by

a) ( ) = 0 + 1
2+ 0 + 1

b) ( ) = 1 + 0
2+ 1+ 0

c) ( ) = 1 + 0
2+ 1 + 0

d) ( ) = 0 + 1
2+ 0 + 1

Solution:

Number of forward paths = 2

Number of poles = 2

Applying Mason’s gain formula

= ( )
( )

= ℎ0(1)�1 �(1)�1 �(1)+ℎ11 ∙1�1−(1)1 (− 1)�
1−�(1)1 (− 1)+(1)�1 �(1)�1 �(− 0)�

= ℎ1 +ℎ1 1+ℎ0
2+ 1 + 0

Given ℎ1 = 1, ℎ0 = 0 − 1 1

= 1 + 1 1+ 0− 1 1
2+ 1 + 0

= 1 + 0
2+ 1 + 0

20. ( ) = This system is operated in closed- loop with
( + )

unity feedback. The closed-loop system is

a) stable

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b) unstable
c) marginally stable
d) conditionally stable
21. Routh's array for a system is given below

The system is

a) stable

b) unstable

c) marginally stable

d) conditionally stable

22. A second order system has a transfer function given by

( ) = 25
2+8 +25

If the system, initially at rest, in subjected to a unit input at t = 0,

the second peak in a response will occur

a) π sec

b) π/3 sec

c) 2π/3 sec

d) π/2 sec

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Solution:

( ) = 25
2+8 +25

∴ 2 = 25 ⇒ = 5

2 = 8

⇒ = 8 = 0.8
2×5

∴ = = = =
�1−(0.8)2 5√0.36 5×0.6 3

Hence, time for second peak = 3 =

23. In order to reduce the steady-state error of a control system,

which one of the following control scheme will be employed?

c) Rate feedback

d) Integral feedback

Solution:

PI control increases the system type by one, thus improves the

24. If the root locus lies only on the negative real axis then the

time response is

a) over damped

b) under damped

c) oscillation

d) sustains oscillations

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25. The Laplace transformation of f (t) is F(s). Given ( ) =

, the final value of f (t) is
2+ 2

a) infinity

b) zero

c) one

d) none of these

Solution:

SF(s) poles are at s = ± jω i.e., on the imaginary axis.

∴ System/function oscillates with a fixed amplitude of frequency

ω rad/sec; hence final value can’t be determined.

26. Assertion (A): Lag compensation makes the closed loop

system more sluggish.

Reason (R): Lag compensation decreases the gain crossover

frequency and the bandwidth.

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 a correct explanation of

A.

c) A is true, but R is false.

d) A is false, but R is true.

Solution:

The rise time and settling time of lag compensated system are

longer ⇒ Sluggish system. The gain cross over frequency is

decreased and bandwidth of system is reduced

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27. A system with a large gain margin or a large phase margin is

usually a/an

a) unstable system

b) stable system

c) very stable but very sluggish system

d) highly oscillatory system

28. The Laplace Transform of the function ( ) = − is given

by

a)
+

b)

c)

d) none of these

29. Given the matrix

01 0
= � 0 0 1�
−6
−6 −11
What are the Eigen values of A?

a) -1, -2, -3

b) -1, 2, 3

c) 0, 0, -6

d) -6, -11, -6

15

Solution:
010

= � 0 0 1 �
−6 −11 −6

To find eigenvalues, | − | = 0
−1 0
�0 −1 �
6 11 + 6

( 2 + 6 + 11) + 6 = 0
3 + 6 2 + 11 + 6 = 0
( + 1)( + 2)( + 3) = 0

= −1, −2, −3
30. Match List-I with List-II and select the correct answer using

the codes given below the Lists:
List-I (Properties of frictional force)

List-II (Frictional force)
1. Coulomb friction y
2. Viscous friction

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3. Static friction
Codes:

AB C
a) 1 2 3
b) 2 1 3
c) 3 2 1
d) 2 3 1