Control Systems (ME338)
s
School of Computing, Engineering and Mathematics
Division of Engineering and Product Design
Semester One Examinations, January 2012
B.ENG. HONOURS DEGREE COURSE
CONTROL SYSTEMS (ME338)
EXAMINER: DR. K. SAEED
Instructions to Candidates:
 The time allowed is TWO hours
 Attempt any FOUR questions only
 The total number of questions is SIX
 Each question carries 25 marks
 This is a CLOSED-BOOK examination
Special Requirements:
 Formula sheet (pages 7-8)
 Graph paper
Page 1 of 8
Control Systems (ME 338)
Question 1
(a) Describe briefly, by drawing an appropriate diagram, the transient and steadystate response of a control system.
(7 marks)
(b) A liquid system is shown in Figure Q1. In this system, liquid passes through the
load valve in the side of the tank.
Q
i
G(s) 
25
ss  3s
Q
0
Figure Q1
If the flow through this restriction is laminar, determine for the system:
(i)
the closed-loop transfer function and the characteristic equation of the
system.
(6 marks)
(ii) the damping ratio (𝜉), the peak time (𝑡𝑝 ), the natural frequency (𝜔𝑛 ), the
phase angle (𝜙) and the damping frequency (𝜔𝑑 ).
(7 marks)
(iii) The percentage of maximum overshoot and the settling time for the 2%
criteria.
(5 marks)
QUESTION 2 COMMENCES ON THE NEXT PAGE
Page 2 of 8
Control Systems (ME 338)
Question 2
A thermal system used for heating flowing water is shown in Figure Q2 below:
G(s) 
R
sRCs  1
Figure Q2
An electric heating element is provided in the storage tank to heat the flow of water.
The inlet water temperature is θi and the outlet water temperature is θ0. The storage
tank is insulated from the surrounding atmosphere to reduce heat loss. The thermal
resistance of the insulation ‘R’ ( 0C/Joules/sec) is given as 2.58. The energy input to
the system ‘q’ (Joules/sec) obtained from the electric heater is 0.0969. Determine
for the system:
(a) the natural frequency and the damping ratio of the system.
(5 marks)
(b) the time constant (Td) if the damping is to be made critical for the system using a
derivative error control.
(20 marks)
QUESTION 3 COMMENCES ON THE NEXT PAGE
Page 3 of 8
Control Systems (ME 338)
Question 3
A moment of inertia system is shown in Figure Q3 below:
Figure Q3
A moment of inertia of the system (J) is given as 1 kg-m2. The inertia torque
developed in the system is Jα, where ‘α’ is the angular acceleration produced due to
the applied torque ‘T’ (Nm). The viscous damping is offered by the damper. The
damping torque varies in direct proportion to the angular velocity of the shaft. The
coefficient of viscous damping ‘f’ of the damper is 3 Nm/rad/sec. The moment of
inertial system has a tachometer to measure its rotational speed. The open-loop
transfer function of the moment of inertial system and tachometer are given as:
Moment of Inertia System:
𝐺1 (𝑠) =
Tachometer:
𝐺2 (𝑠) =
1
𝑠(𝐽𝑠+𝑓)
𝐾
𝑠+1
Determine the following:
(a) Draw the block-diagram representation of the system.
(3 marks)
(b) Find the open-loop and closed-loop transfer functions of the system.
(3 marks)
(c) Draw the root-locus plot of the system.
(15 marks)
(d) The conditions when the system becomes unstable.
(4 marks)
Page 4 of 8
Control Systems (ME 338)
Question 4
(a) Describe the effects of the addition of zeros and poles on a control system’s
behaviour.
(5 marks)
(b) The model for a space vehicle control system is shown in Figure Q4 below:
R(s)
Gc(s)
G (s ) 
1
s
Lead
Compensator
2
Space
Vehicle
G (s) 
10
s  10
C(s)
Sensor
Figure Q4
Design a lead compensator Gc(s) such that the damping ratio (𝜉) and the
undamped natural frequency (𝜔𝑛 ) of the dominant closed-loop pole are 0.5 and
2 rad/sec respectively.
(20 marks)
Question 5
(a) Describe briefly lead and lag compensators.
(5 marks)
(b) Consider a unity-feedback control system having the following feed-forward
transfer function:
𝐺(𝑠) =
10
𝑠(𝑠 + 2)(𝑠 + 8)
Design a compensator such that the dominant closed-loop poles are located at
𝑠 = −2 ± 𝑗2√3 and the static velocity error constant Kv is equal to 80 sec-1.
(20 marks)
Page 5 of 8
Control Systems (ME 338)
Question 6
A mass-spring damper system is shown in Figure Q6 below:
k
u(t)
m
y (t)
b
Figure Q6
It has a spring stiffness of 0.04 N-m, a damping coefficient (b) of 0.75 and a mass of
100 kg. Determine for the system:
(a) the mathematical model and the transfer function of the system.
(10 marks)
(b) the state-space model of the system.
(15 marks)
Page 6 of 8
Control Systems (ME 338)
APPENDIX I
Formula Sheet

The output time response equation for the underdamped system is given
below:
e  nt
c(t )  1 
sin( d t   )
1  2

Rise time (tr):t r 
 
n 1   2
1  2
  tan 1

Where,
tp 


Peak time:

% of Maximum Peak (Mp) :
%
of
n 1   2
overshoot  e
max imum
 


2
 1

Settling time (ts) :t s  4 1  4T (2%criteria )
 n

Roots finding equation: p1 , p 2  n  j n 1   2

The generalised 2nd order system equation




x100
ts  3
1
 n
 3T (5%criteria )
  cos 
s 2  2n s   n2  0

Damping ratio with derivative control
'  
 nTd
2
Page 7 of 8
Control Systems (ME 338)
For Root Locus Plot:

The angle of asymptotes is given by
2k  1180 0

where: k = 0, 1 and 2.
PZ
The three asymptotes intersect on the real axis at :
x

 Poles   Zeros
PZ
Compensator Design
1

s
T1
 (T1 s  1)(T2 s  1)
Gc ( s )  K c
 Kc


 T1
( s  1)( T2 s  1)
s
T

1


1

 s 
T2


1
 s 

T2







Where, ‘γ ‘ is not equal to β. Then the compensated system will have the
transfer function as:
1

s
T1
G( s)Gc ( s)  K c 


s
T1

1

 s 
T2


1
 s 
T2



G ( s)



Error equation:
K v  lim sGc ( s )G ( s )
s 0
Conditions for Lag compensator:
1
T2
 50 
1
s
T2
s
 00
s s1
Page 8 of 8