Control Systems Test - 5 - PDF Flipbook
Control Systems Test - 5
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GATE
EEE
Control
Systems
Test-05Solutions
CONTROL SYSTEMS
1. The state equations of a system are given by
−3 1 0 0
ẋ = � 0 −3 0 � x + �0�
0 0 −1
1
y = [1 0 1]x
The system is
a) controllable and observable
b) controllable but not completely observable
c) neither controllable nor completely observable
d) not completely controllable but observable
Answer: (d)
2. Match List-I (Time Domain Specification) with List-II
(Equation for Finding its Value) and select the correct answer:
List – I
1. Peak overshoot
2. Peak time
3. Rise time
4. Settling time (2%)
List – II
1. π
�ωn�1−ξ2�
2. 4
ξ(ωn)
3. exp ��−1π−ξξ2�
4. π − �cos−1 �ωn�ξ1−ξ2��
1
Codes:
A BC D
a) 3 2 4 1
b) 4 1 3 2
c) 3 1 4 2
d) 4 2 3 1
Answer: (c)
Solution:
Settling time → 4
ξ(ωn)
Peak time → tp = π
ωn�1−ξ2
Rise time → tr = π − �cos−1 �ωn�ξ1−ξ2��
− πξ
Peak overshoot → MD = e �1−ξ2
3. Assertion (A): Sampled-data system requires hold circuit.
Reason (R): Hold circuit converts the signal to analog form.
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: (a)
Solution:
Hold circuit convert signal to analog form.
2
4. Match List-I (Nature of Eigen value) with List-II (Nature of
Singular Point) and select the correct answer using the codes
given the below the lists:
List-I
A. Real negative and distinct
B. Real equal but opposite in sign
C. Purely imaginary pair
D. Complex conjugate pair
List-II
1. Centre
2. Focus point
3. Saddle point
4. Stable node
5. Unstable node
Codes:
ABC D
a) 1 2 5 3
b) 4 3 1 2
c) 1 3 5 2
d) 4 2 1 3
Answer: (b)
3
5. A Bode magnitude plot for the transfer function G(s) of a plant
is shown in the figure. Which one of the following transfer
functions best describes the plant?
a) 1000(s+10)
s+1000
b) 10(s+10)
s(s+1000)
c) s+1000
10s(s+10)
d) s+1000
10(s+10)
Answer: (d)
Solution:
Corner frequencies,
ω1 = 2πf1 = 2π(10)
ω2 = 2πf2 = 2π × 103
∴ G(S) = K�11++ωωs1s2�
20logK = 20dB
K = 10
G(S) = 10�11++22ππs1s1003�
4
G(S) = (2π×10×10)�s+2π103�
(2π×1000)(s+2π10)
G(S) = (s+2π1000)
10×(s+2π10)
G(S) = (j2πf+2π1000)
10(j2πf+2π10)
G(S) = (jf+1000)
10(jf+10)
Let us assume s = jf
G(S) = (s+1000)
10(s+10)
6. Consider the following statements:
1. Bandwidth is increased.
2. Peak overshoot in the step response is increased.
Which of these are the effects of using lead compensation in a
feedback system?
a) 1 only
b) 2 only
c) Both 1 and 2
d) Neither 1 nor 2
Answer: (a)
Solution:
By using lead compensator rise time decreases
↓ tr ∝ 1
B.W.↑
Hence B.W increases.
5
7. Consider the system shown in the figure below. What is the
transfer function relating C to N?
a) KG
1+KG
b) 1
1+KG
c) G
1+KG
d) 1
KG
Answer: (b)
Solution:
( )
( ) =?
Removing input R(s), the block diagram reduces to
N − C. K. G = C
N = C(1 + KG)
= �1+1 �
6
8. For the feedback system shown in the given figure, the forward
path does not affect the system output when KG is
a) small
b) negative
c) one
d) very large
Answer: (d)
Solution:
Transfer function, C = KG
R 1+KGH
Output, C = KG . R
1+KGH
When, >> 1 then,
1 + KGH ≃ KGH
C = KG . R = R
1+KGH H
(Independent of KG)
9. Let ẋ = �01 2b� x + �01� u, where b is an unknown constant. This
system is
a) Uncontrollable for b = 1
b) Uncontrollable for b = 0
c) Uncontrollable for all values of b
d) Controllable for all values of b
7
Answer: (d)
Solution:
QC = [B AB]
B = �01� A = �10 2b�
AB = �01 2b� �10� = �2b�
QC = �10 2b�
For controlling
|QC| ≠ 0
|QC| = 2 ≠ 0
10. The pole-zero plot shown below in the figure is that of which
one of the following?
a) Integrator
b) PD controller
c) PID controller
d) Lag-lead compensator
Answer: (d)
8
11. The impulse response of a linear system is given by g(t) –
Ke−t�τ tangent is drawn to the impulse response current time I =
0. The tangent will intersect the time axis at
a) 1 τ
2
b) τ
c) 2τ
d) Kτ
Answer: (b)
Solution:
y(t) = ke−t/τ
Slope at t = 0,
ddyt �t=0 = − k
τ
Equation of line,
y − k = − k (t)
τ
When line intersects t-axis, y = 0
⇒ −k = − k t1 ⇒ t1 = τ
τ
9
12. Compared to continuous time system, the discrete system is
a) More accurate but less stable
b) Less accurate but more stable
c) More accurate and more stable
d) Less accurate and less stable
Answer: (c)
13. The transfer function of a system is Y(s) = s+s 2. The steady state
R(s)
output y (t) is Acos(2t + ∅) for the input cos (2t). The values of
A and ∅, respectively are
a) 1 , −450
√2
b) 1 , +450
√2
c) √2, −450
d) √2, +450
Answer: (b)
Solution:
A = �jωjω+2�ω=2 = 2 = 2 = 1
√22+22 2√2 √2
ϕ = ∠ jωjω+2�ω=2 = 900 − tan−1 2 = 450
2
14. The system matrix of a continuous time system is given by
= �−03 −15�
What is the characteristic equation?
a) s2 + 5s + 3 = 0
b) s2 – 3s – 5 = 0
10
c) s2 + 3s + 5 = 0
d) s2 + s + 2 = 0
Answer: (a)
Solution:
Characteristic equation |sI − A| = 0
�3s s−+15� = 0
s(s + 5) + 3 = 0
s2 + 5s + 3 = 0
15. The state - transition matrix of the above system is
a) �e−2te−+3te−3t e−02t�
b) �e−03t e−2te−−2te−3t�
c) �e−03t e−2te−+2te−3t�
d) �e3t e−2te−−2te−3t�
0
Answer: (b)
Solution:
eAt = L−1[(SI − A)−1]
1 1
= L−1 �s+3 (s+3)1(s+2)�
0
s+2
eAt = �e−02t e−2te−−2te−2t�
11
16. What is represented by state transition matrix of a system?
a) Free response
b) Impulse response
c) Step response
d) Forced response
Answer: (a)
17. Match List-I (Element of Mechanical System) with List-II
(Analogous Element in Electrical System) and select the
correct answer using the code given below the lists:
List – I
A. Mass
B. Spring
C. Damper
D. Displacement
List – II
1. Inductance
2. Capacitance
3. Resistance
4. Charge current
Codes:
ABC D
a) 1 2 3 4
b) 3 4 5 2
c) 1 4 3 2
d) 3 2 5 4
12
Answer: (a)
Solution:
Mechanical system Electrical Electrical
(Force-voltage) (Force-current)
Force Voltage Current
Mass Inductance Capacitance
Damper
Spring Resistance 1
Resistance
1
Capacitance 1
Inductance
Displacement Charge Magnetic flux
linkage
18. A lead compensator is described by the function ss++αβ. What are
the relative values of α and β?
a) α > β
b) α < β
c) α ≥ β
d) α ≤ β
Answer: (b)
Solution:
For phase lead compensator
∠T(s) > 0
tan−1 ω − tan−1 ω > 0
α β
⇒α 100
Answer: (a)
18
Solution:
Pole zero plot is given below
Lead G(s) = s+0.1
s+1
∠G(s) = ∠ tan−1 ω − tan−1 ω
0.1 1
Phase lead occur from ω = 0.1 to ω = 1
Range 0.1 < ω < 1
28. A unity feedback system has an open-loop transfer function
G(jω) = K
jω(j0.2ω+1)(j0.05ω+1)
The phase cross-over frequency of the Nyquist plot is given by
a) 1 rad/sec
b) 100 rad/sec
c) 10 rad/sec
d) none of these
Answer: (c)
29. A lead compensating network
A. improves response time
B. increases resonant frequency
C. stabilizes the system with low phase margin
D. enables moderate increase in gain without affecting stability
a) A, B, C, D
b) A, D
19
c) B, C
d) B, D
Answer: (a)
30. Three blocks G1, G2 and G3 are connected in some fashion
such that overall transfer function is
1+ 2(1+ 1 2)
1+ 1 2
The blocks are connected in the following manner:
a) G1, G2 with negative feedback and combination in parallel
with G3.
b) G1, G3 with negative feedback and in parallel G2.
c) G1, G2 is cascade and combination in parallel with G3.
d) G1, G3 in cascade and combination in parallel with G2.
Answer: (a)
Solution:
C = G1+G3(G1G2+1)
R G1G2+1
= G1 + G3
1+G1G2
So, G1, G2 constitutes negative feedback and combination in
parallel to G3.
20
RG3 + R G1 = C
1+G1G2
Transfer function,
C = G3 + G1
R 1+G1G2
21
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