ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
FEEDBACK CONTROL SYSTEMS
1. Control System Design
2. Open and Closed-Loop Control Systems
3. Why Closed-Loop Control?
4. Case Study --- Speed Control of a DC Motor
5. Steady-State Errors in Unity Feedback Control Systems
1. Control System Design
1. Establishing system goals (objectives), e.g. to control the velocity of a motor accurately
−
What do you want to control? e.g. speed, position, level, temperature, ...
−
What do you want to achieve? e.g. fast response, less vibration, ...
−
How do you translate them into control terms? e.g. overshoot, rising time, ...
2. Obtaining a model of the process, the actuator, and the sensor. E.g. G(s) = k/(Ts+1)
−
Differential equations
−
Transfer functions
−
Block diagrams
−
State space equations
3. Designing a controller (PID, root locus, frequency response, state space)
−
Proportional (P) control
−
Integral (I) control
−
Proportional-Derivative (PD) control
−
Proportional-Integral (PI) control
−
Proportional-Derivative-Integral (PID) control
−
State space based control
4. Simulation (MATLAB) or experimental test.
2. Open and Closed-Loop Control Systems
Design requirements and specification:
−
−
−
−
−
An excellent transient response in terms of rise-time, size of overshoot and settling-time;
An excellent steady-state response in terms of steady-state error;
Acceptable stability margins;
Robustness in terms of disturbance rejection; and
Robustness in terms of sensitivity to parameter changes.
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
Open-loop control system: utilises an actuating device to control the process (plant) directly
without using feedback.
Input
Actuating
device
Process or
plant
Output
Closed-loop control system: uses a measurement of the output and feedback of this signal to
compare it with the desired input (reference or command).
R(s)
E(s)
+_
U(s)
Gc(s)
H(s)
Notation:
R(s): Laplace transform of the input signal;
U(s): Laplace transform of the control or actuating signal;
Y(s): Laplace transform of the output signal;
E(s): Laplace transform of the error signal;
Gc(s): Transfer function of the controller;
G(s): Transfer function of the process or plant;
H(s): Transfer function of the sensor.
2
G(s)
Y(s)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
3. Why Closed-Loop Control?
An advantage of the closed-loop control system is the fact that the use of feedback makes the
system response relatively insensitive to external disturbances (e.g. temperature and pressure) and
internal variations in system parameters (e.g. component tolerances) which are not known or
predicted.
3.1. Sensitivity of control systems to parameter variations
Suppose that G(s) changes to G(s)+∆G(s) due to the environment, ageing, etc. For small ∆G(s) <<
G(s), Y(s) changes to Y(s)+∆Y(s). This can be illustrated by expanding Y(s) to a first-order Taylor
series:
dY(s)
Y(G (s) + ∆G (s)) = Y(G (s)) +
∆G (s)
1
424
3 dG (s)
142
4 43
4
Y (s )
∆Y (s)
Open-loop:
R(s)
Y(s) = G(s)R(s)
∆Y(s) = ∆G(s)R(s)
G(s)
R(s)
Y(s)
+
Closed-loop: Y(s) = G(s)R(s)/[1+G(s)H(s)]
∆Y(s) = ∆G(s)R(s)/[1+G(s)H(s)]2
Y(s)
G(s)
_
H(s)
The system sensitivity is defined as: the ratio of the percentage change in the system transfer
function T(s) to the percentage change of the process transfer function G(s):
S TG =
Open-loop:
dT(s) / T(s) G (s) dT(s)
=
dG (s) / G (s) T(s) dG (s)
S TG = 1
Closed-loop: S TG =
1
1 + G (s)H(s)
A system with zero sensitivity is ideal (i.e. changes in the system parameters have no effect on the
transfer function).
K
Y(s)
Exercise: A temperature control system of a chemical R(s)
fluid in a tank is shown, where R is the resired
+
s+a
−
temperature, Y the actual temperature and 'a' is a constant
related to the mass in the tank. Determine the open-loop
and closed-loop sensitivity of this sytem to changes in
parameter 'a' due to chemical reactions an ageing. How would you reduce the sensitivity at low
frequencies (i.e. s→0)?
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
3.2. Disturbance rejection
Suppose that there is an input disturbance D(s) applied between the controller and plant:
D(s)
Open-loop:
U(s) = Gc(s)R(s)
Y(s) = Gc(s)G(s)R(s)+G(s)D(s)
R(s)
+
Gc(s)
Y(s)
G(s)
D(s)
Closed-loop: U(s) = Gc(s)[R(s) – H(s)Y(s)]
G c (s)G (s)
Y(s) =
R (s)
1 + G c (s)G (s)H(s)
U(s) +
+
R(s)
+
G (s)
+
D(s)
1 + G c (s)G (s)H(s)
Gc(s)
U(s) +
Y(s)
G(s)
–
H(s)
Exercise: For the systems shown above, the process transfer function G(s) = 1/(0.5s+1), the
controller Gc(s) = 99, R = 0, a step-unit disturbance D(s) = 1/s, and H(s) = 1. Determine the
response to the disturbance for both the open-loop and closed-loop systems and sketch the outputs.
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
3.3. Transient response
The transient response is the response of a system as a function of time.
A motor speed control system:
−
Objective: the actual speed y approaches the desired speed r quickly
−
Open-loop: Gc(s) = k, G(s) = k1/(Ts+1)
Assume, for example, kk1 = 1, T = 10
−
Closed-loop with a proportional control term k, H(s) = 1
Assume, for example, kk1 = 10, T = 10.
Unit-step response of closed-loop control
1
0.9
0.9
0.8
0.8
0.7
0.7
Amplitude
Amplitude
Unit-step response of open-loop control
1
0.6
0.5
0.6
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
5
10
0
0
15
5
10
15
Time (sec.)
Time (sec.)
3.4. Steady-state error
The steady-state error is the error after the transient response has decayed, leaving only the
continuous response, that is for a unity-feedback system:
e(∞) = lim[r ( t ) − y( t )]
t →∞
E(s) = R(s) - Y(s). Considering a unit-step input, R(s) = 1/s:
− Open-loop: Y(s) = G(s)R(s), E(s) = [1-G(s)]R(s).
By the final-value theorem,
e(∞) = lim e( t ) = lim sE (s) = 1 − G (0)
t →∞
s →0
− Closed-loop: Y(s) = G(s)R(s)/[1+G(s)H(s)]. For H(s) = 1, E(s) = R(s)/[1+G(s)]
By the final-value theorem,
e(∞) = lim e( t ) = lim sE (s) = 1 /[1 + G (0)]
t →∞
s →0
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
Exercise: Find the open and closed-loop steady-state errors for the following system for unit-step
and unit-ramp inputs.
R(s)
E(s)
+
−
25
s+5
Y(s)
3.5. Summary
The advantages of closed-loop control:
Reduce effects of process (plant) disturbances
Make system insensitive to process variations
Stabilise an unstable system
Create well-defined relations between output and reference
The disadvantages:
Risk of instability
Complexity in analysis and implementation, and expensive
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
4. Case Study --- Speed Control of a DC Motor
The purpose of this case study is:
to compare the open-loop and closed-loop control, and
to provide a guide to the solution of a first-order control design problem using a proportional (P)
controller.
Techniques used in the solution include:
Mathematical modelling of physical systems
Transfer function analysis
Laplace transforms
Feedback control
4.1. DC motor --- problem description
R
L
ia(t)
+
va
_
+
Armature circuit
vb = Kbω
_
J
θ Tm Bω
Td
For the armature controlled DC motor, we will use the following parameter values:
R = armature resistance = 1 Ω
L = armature inductance = 0.5 H
Km = motor-torque constant = 10 N-m/A
Kb = back emf constant = 0.1 V-s/rad
J = moment of inertia of the motor = 2 kg-m2
B = viscous-friction coefficient of the motor = 0.5 N-m-s/rad
The motor torque, Tm, is related to the armature current, ia, by a constant factor Km:
Tm(t) = Kmia(t)
The back emf, vb, is related to the rotational velocity, ω = dθ/dt, by the following equation:
vb(t) = Kbω(t)
4.2. Design objectives (goals)
The objective is to design a simple proportional controller (selecting the gain K) to achieve a fast
response to the step input, a small steady-state error, and a reduction of the effect of disturbance.
These requirements can be represented, for a unit-step input and unit-step disturbance, as:
Steady-state error ess ≤ 0.01 to a unit-step input
The effect of a unit-step disturbance < 0.005 rad/s
Time constant ≤ 0.2 s.
Can we meet these requirements using an open-loop controller? If not, we will try the closed-loop
controller.
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
4.3. Mathematical model of the DC motor
In this study, zero initial conditions are used.
Time relation
Laplace transform
The motor torque:
Tm(t) = Kmia(t)
Tm(s) = KmIa(s)
The back emf
voltage:
vb = Kbω(t)
Vb(s) = KbΩ(s)
Electrical circuit
equation:
v a ( t ) − K b ω( t ) = Ri a ( t ) + L
di a ( t )
dt
The torque relation: T1(t) = Tm(t) – Td(t)
Mechanical system
(rotor):
J
Va(s) – KbΩ(s) = (sL+R)Ia(s)
T1(s) = Tm(s) – Td(s)
dω( t )
+ Bω( t ) = Tm ( t ) − Td ( t )
dt
(Js+B)Ω(s) = KmIa(s)–Td(s)
Equations (1) and (2) are used to obtain the following block diagram:
Td(s)
Va(s)
+_
Km
Ls + R
(1)
Tm (s)
+
−
Tl(s)
1
Js + B
Ω(s)
Kb
4.4. Simplified model
Km
(Ls + R )(Js + B) + K m K b
Transfer function from Va to Ω: G (s) =
Transfer function from Td to Ω: G d (s) =
Neglecting L (assumed very small), we have the simplified first-order model:
Km
−R
G (s) =
, G d (s) =
RJs + RB + K m K b
RJs + RB + K m K b
Substituting the numerical values into G(s) and Gd(s): G (s) =
− (Ls + R )
(Ls + R )(Js + B) + K m K b
8
10
−1
, G d (s) =
2s + 1.5
2s + 1.5
(2)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
Td(s)
Va(s)
Tm (s)
10
+_
+
−
Tl(s)
1
2s + 0.5
Ω(s)
0.1
The block diagram can be simplified by moving the summing point ahead of the block with the
constant gain of 10:
Td(s)
1/10
Va(s)
+
−
10
2s + 0.5
_
Ω(s)
0.1
This can be further reduced by representing the feedback loop with an equivalent block:
Td(s)
1/10
Va(s)
+
−
Ω(s)
10
2s + 1.5
4.5. Simplicity versus accuracy
We must make a compromise between the simplicity of the model and the accuracy of the results of
the analysis. In deriving a reasonably simplified mathematical model, we frequently find it
necessary to ignore certain parameters.
4.6. Open-loop control
Introducing the constant gain (proportional), K, block into the system (R – desired velocity):
Td(s)
Motor dynamics
1/10
R(s)
K
Va(s)
+
−
9
10
2s + 1.5
Ω(s)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
Steady-state error: Td = 0, R(s) = 1/s,
E(s) = R(s) - Ω(s) =
e ss = lim sE(s) = 1 −
s→0
1
10K 
1 −

s  2s + 1.5 
10K
. By choosing K = 3/20, ess = 0, OK.
1 .5
Disturbance rejection: R(s) = 0, Td(s) = 1/s,
−1
s(2s + 1.5)
Taking the inverse Laplace transform: ω(t)=(-2/3)[1–e-0.75t]
→ ω(t→∞) = -2/3 rad/s, not OK!
Ω(s) =
Time constant: 2/1.5 = 4/3 s, again not OK!
A simple open-loop controller can not achieve the required goals.
4.7. Position and velocity sensors in control systems
4.7.1. Potentiometers
Converts mechanical energy to electrical energy. Input
displacement can either be linear or rotational.
Three terminal device: two fixed where voltage is
applied, third is variable with the movable shaft to
provide a voltage (with respect to ground) proportional to
the displacement:
e( t ) = K L y ( t )
e( t ) = K R θ( t )
+
Fixed
terminals Vs
y(t)
Variable
terminal
e(t) ∝ y(t)
Linear
Rotational
−
Rotary type available for single and multi-turn rotational motion.
Linear potentiometer
Rotary potentiometer
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
4.7.2. Incremental optical encoder
Converts linear or rotary displacement of shaft into a digital pulse signal for position and
velocity sensing.
Consists of a rotary glass or plastic disk with evenly spaced opaque and transparent region –
each region represents a displacement increment. For example, the angular resolution of 1000
lines disk is ∆θ = 2π/1000 = 0.006 radians per increment.
Light passes through transparent lines as the disk rotates with the movement of the shaft, giving
a sinusoidal or triangular waveform at the photo-detector, followed by an amplifier/comparator
to produce the pulse signal. Position information is obtained from counting the number of
pulses using a counter; speed information is obtained from the frequency of the pulse train.
Light
source(s)
Photodetector(s)
Squaring
Shaft
Transparent
Opaque
For direction of rotation information (clockwise or anti-clockwise), a second track of
opaque/transparent lines at a different radius of the disk, that is offset 90o with respect to the
first track (quadrature tracks), is included. The pulse trains from the two channels are then
compared for lead/lag to determine the direction of rotation.
4.7.3. Tachometer
Works as a voltage generator with a DC voltage proportional to the angular velocity of the input
shaft.
Used in velocity indicators, velocity feedback control and stabilisation.
Mathematical model:
e( t ) = K t
dθ( t )
= K t ω( t )
dt
Kt is given as a parameter by the manufacturer in volts/1000 rpm.
The transfer function is:
E(s)
= Kt
Ω(s)
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
4.8. Closed-loop control
A tachometer is attached directly to the motor that measures the angular velocity and produces a
proportional voltage that is fed-back and compared with the reference voltage corresponding to the
desired speed.
DC motor
+
eref
+
Controller
amplifier
M
Load
−
−
Ktω
T
Tachometer
-15V
Gain
control
VREF
ω
Driver
R2
ω
R1
M
KVREF − Ktω
R1
Summing
amplifier
Ktω
12
T
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
Assuming that Kt = 1 for simplicity:
Td(s)
1/10
R(s)
+
K
Va(s)
+
−
10
2s + 1.5
−
Ω(s)
Kt = 1
Steady-state error: Td = 0, R(s) = 1/s,
E(s) = R(s) - Ω(s) =
e ss = lim sE ( s ) =
s →0
1
2 s + 1 .5



s  2 s + 1.5 + 10 K 
1 .5
. By choosing K = 15, ess < 0.01, OK.
1.5 + 10 K
Disturbance rejection: R(s) = 0, Td(s) = 1/s,
Ω(s) =
−1
s(2s + 1.5 + 10K )
Taking the inverse Laplace transform (or from final-value theorem): ω(∞) = -1/(1.5+10K),
by choosing K = 20, |ω(∞)| < 0.005, OK.
Time constant: 2/(1.5+10K), by choosing K = 1, T = 0.17 s, OK.
From the above, we choose a suitable value K = 20.
From the above study, it can be seen that the larger K, the better performance for a simple firstorder system.
Can we choose K very large?
4.9. Problems with a very large gain K
Va(s) = K[R(s) - Ω(s)] = KE(s), in the time domain va(0) = Ke(0), the small error e(0) at the
beginning will lead to a large va(0) since K is large.
When K is big, the inductance L (which we ignored for simplicity) can not be neglected any
more. It will lead to a problem. Assume L = 1 mH, then the transfer function from Va to Ω is:
G (s) =
Km
10
=
2
(Ls + R )(Js + B) + K m K b 0.002s + 2s + 1.5
The closed-loop transfer function with the proportional term K is:
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
T(s) =
10K
0.002s + 2s + 1.5 + 10K
2
For K = 20, two real poles: s = -114 and s = -886 (ζ > 1)
For K = 500, two complex poles: s = -500+j1500 and s = -500-j1500 (ζ < 1)
A compromise should be found between the need to increase K and the need to avoid the above
problems.
4.10. Summary
Feedback (closed-loop) control can be used to stabilise systems, speed up the transient response,
improve the steady-state characteristics, provide disturbance rejection, and decrease the
sensitivity to parameter variations.
Proportional feedback control reduces errors and improves transient responses, but the high gain
(large K) may lead to many problems.
14
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
5. Steady-State Errors in Unity Feedback Control Systems
When a command input (desired output) is applied to a control system it is generally hoped that
after any transient effects have died away the system output will settle down to the command value.
The error with any system is the difference between the required (desired) output signal, i.e. the
reference input signal which specifies what is required, and the actual output signal. For the unityfeedback control system shown in the figure, the error is:
R(s)
E(s) = R (s) − Y(s) =
R (s)
1 + G (s)
E(s)
Y(s)
G(s)
+ _
Using the final-value theorem (assuming the system is stable), the steady-state error is:
sR (s)
s → 0 1 + G (s )
e ss = e(∞) = lim e( t ) = lim sE(s) = lim
t →∞
s →0
A general representation of G(s) is:
G (s) =
K (s m + b m −1s m −1 + L + b1s + b 0 )
s N (s n + a n −1s n −1 + L + a 1s + a 0 )
where K is a constant and m and n are integers and neither a0 nor b0 is zero. N is an integer, the
value of which is called the type of the system. Thus if N = 0 then the system is said to be type 0, if
N = 1 then type 1, if N = 2 then type 2 and so on. The type number is thus the number of 1/s factors
in the open-loop transfer function G(s). Since 1/s represents integration, the type number is the
number of integrators in the open-loop transfer function.
Exercise: What are the type numbers for the systems shown in the following figures?
(s + 1)
(s − 1)(s + 6)
Y(s)
K
(s + 1)
s(s − 1)(s + 6)
Y(s)
K
R(s)
+
−
R(s)
+
−
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
5.1. Static error constants
Just as we defined the damping ratio, the natural frequency, settling time, percentage overshoot etc.
as performance specifications for the transient response of a system, we can use static error
constants to specify the steady-state error characteristics of control systems.
5.1.1. Static position error constant Kpo
The steady-state error of the unity-feedback control system for a unit-step input is:
s
1
=
s →0 s(1 + G (s ))
1 + G (0)
e ss = e(∞) = lim e( t ) = lim sE(s) = lim
t →∞
s→0
The static position error constant Kpo is defined by:
K po = lim G (s) = G (0)
s →0
Thus, the steady-state error in terms of the static position error constant Kpo is;
1
e ss =
1 + K po
For a type 0 system, Kpo = C where C is a constant. For a type 1 system, Kpo = ∞. Hence, for a type
0 system, the static position error constant Kpo is finite, while for a type 1 or higher system, Kpo is
infinite. This is, ess = 1/(1+C) for a type 0 system, and ess = 0 for a type 1 or higher system.
Exercise: Find the static position error constants and steady-state errors, respectively, of the
systems shown in the following figures for a unit-step input.
R(s)
+
−
s +1
s+2
R(s)
Y(s)
+
16
−
s +1
s(s + 2)
Y(s)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
5.1.2. Static velocity error constant Kv
The steady-state error of a system with a unit-ramp (velocity) input is:
s
1
= lim
s →0 s (1 + G (s ))
s →0 sG (s )
e ss = e(∞) = lim e( t ) = lim sE(s) = lim
t →∞
s →0
2
The static velocity error constant Kv is defined by:
K v = lim sG (s) =
s →0
1
e ss
For a type 0 system, K v = lim sG (s) = 0 . For a type 1 system, K v = lim sG (s) = C . For a type 2 or
s →0
s →0
higher system, K v = lim sG (s) = ∞ .
s →0
5.1.3. Static acceleration error constant Ka
The steady-state error of a system with a unit-parabolic (acceleration) input, r(t) = t2/2, is:
e ss = e(∞) = lim e( t ) = lim sE(s) = lim
t →∞
s →0
s →0
s
1
= lim 2
s
→
0
s (1 + G (s))
s G (s)
3
The static acceleration error constant Ka is defined by:
K a = lim s 2 G (s) =
s →0
1
e ss
For a type 0 system, K a = lim s 2 G (s) = 0 . For a type 1 system, K a = lim s 2 G (s) = 0 . For a type 2
s →0
s →0
system, K a = lim s G (s) = C . For a type 3 or higher system, K a = lim s 2 G (s) = ∞ .
2
s →0
s →0
Exercise: A unity feedback system has the following forward transfer function:
G (s) =
1000(s + 8)
(s + 7)(s + 9)
(a) Evaluate the system type, Kpo, Kv, and Ka.
(b) Use your answers to (a) to find the steady-state errors for the unit-step, unit-ramp, and unitparabolic inputs.
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ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
Exercise: For the unity feedback system shown, determine the value of K to yield a 10% error in
the steady-state when the input is r(t) = 1.
s + 12
(s + 14)(s + 18)
R(s)
+
K
−
18
Y(s)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
5.2. Steady-State Error for Disturbances
D(s)
R(s)
E(s)
+
−
+
Gc(s)
+
Y(s)
G(s)
The error component when D(s) = 0 is: E(s) = R(s) - YR(s)
where the output due to the reference input, YR(s), is given by: YR (s) =
G c (s)G (s)
R (s)
1 + G c (s)G (s)
The steady-state error value due to R(s) is given by the final value theorem:
sR (s)
s →0 1 + G (s )G (s)
c
e R (∞) = lim sE(s) = lim
s →0
When R(s) = 0, the error in this case is given by:
E(s) = 0 - YD(s)
where the output due to the disturbance, YD(s), is:
YD (s) =
G (s)
D(s)
1 + G c (s)G (s)
The steady-state error value due to D(s) is found using final value theorem:
− sG (s)D(s)
s →0 1 + G (s )G (s )
c
e D (∞) = lim sE(s) = lim
s →0
The steady-state error due to the reference and disturbance inputs is:
e(∞) = eR(∞) + eD(∞)
Exercise: For the block diagram shown below, determine the stead-state error component due to a
unit-step disturbance.
D(s)
R(s)
+
100
+
−
19
+
1
s(s + 25)
Y(s)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
TUTORIAL PROBLEM SHEET 5
1. List the major advantages and disadvantages of closed-loop control systems.
2. For the system shown in the figure, what are the steady-state errors when a unit-step input is
applied to the following open-loop transfer functions G(s):
10
(s + 1)(s + 2)
6(s + 3)
(b) G (s) =
(s + 6)(s + 2)
10
(c) G (s) =
s(s + 1)(s + 2)
(a) G (s) =
R(s)
E(s)
+
G(s)
Y(s)
−
3. What are the type numbers for the systems shown in the following figures?
R(s)
+
K
−
(s + 1)
(s − 1)(s + 6)
Y(s)
(s + 1)
s(s − 1)(s + 6)
Y(s)
(a)
R(s)
+
K
−
(b)
4. Determine the steady-state error for the system shown in question (1) above with:
2(s + 1)
G (s) = 2
s (s + 2)
2
when subjected to the input r(t) = 1 + t + t . Plot the time response y(t) with MATLAB.
5. Find the static position error constants and steady-state errors, respectively, of the systems
shown in the following figures for a unit-step input.
R(s)
+
−
s +1
s+2
R(s)
Y(s)
+
−
(a)
s +1
s(s + 2)
(b)
Plot the time response y(t) for a unit-step input using MATLAB.
20
Y(s)
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2013)
________________________________________________________________________________
6. For the system shown in the following figure:
R(s)
+
Y(s)
1
s(s + 5)(s + 10)
K
−
(a) What value of K will yield a steady-state error in position of 0.01 for an input of r(t) = t/10?
(b) What is the value of Kv for the value of K found in (a)?
(c) Plot the time response using MATLAB.
7. Find the total steady-state error due to a unit-step input and a unit-step disturbance in the system
shown below.
D(s)
R(s)
+
1
s+5
−
+
+
100
s+2
Y(s)
8. Consider the system shown in the following figure with G(s) = 5/(s + 2). What are the values of
the gain K which will achieve the following design specifications?
D(s)
R(s)
+
K
+
+
Y(s)
G(s)
−
(a) Steady-state error due to a unit-step input R(s) to be less than 0.1.
(b) Steady-state error due to a unit-step change in disturbance D(s) to be less than 0.1.
(c) Validate the above results using MATLAB, that is to plot the time response for a unit-step
input and a unit-step disturbance, respectively.
9. Redo question (8) for G (s) =
10
.
s + 10s + 50
2
21