Control 1 2014

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Dublin Institute of Technology
College of Engineering and Built Environment
Control Engineering 1
Mech 3019
Dr. Nigel Kent
Rm: 387
Email:nigel.kent@dit.ie
Ext: 3885
Dublin, 2013
Introduction
Aims
The aim of this module is to; Introduce the concept of control in an engi-
neering context and to indicate the wide variety of control tasks in engineer-
ing systems. Describe common control strategies; open loop and feedback.
Present a description of closed loop control systems and analyse the main
properties of feedback. Introduce the concepts of steady state error, dis-
turbances, disturbance rejection and stability. Present time and frequency
domain models as equivalent descriptions of system behaviour. Provide an
understanding of the need for compensation and to illustrate the proper-
ties of proportional, integral and derivative controllers. Understand that
proper control system design leads to systems that are efficiently and ad-
equately controlled, and can have a direct impact on energy consumption,
the environment and society.
Content
ˆ Concept of control systems, types of control, open and closed loop
systems, effects of feedback.
ˆ Differential equations and transfer functions of physical systems. Block
diagrams and their reduction.
ii
ˆ Analytical treatment of first and second order systems. Effect of dis-
turbances, steady state errors. Standard test inputs and system re-
sponse. Stability of linear systems, the Routh-Hurwitz Criterion.
ˆ Process control systems, 2 step, proportional, integral, and derivative
control actions. Generation of control actions.
ˆ Frequency response. Nyquist and Bode plots of the open loop system,
relative stability, the closed loop frequency response.
Learning Outcomes
On completion of the module you should be able to:
ˆ Derive the differential equations, transfer functions and block diagram
models of physical systems.
ˆ Apply transfer function and block diagram modelling in the analysis
and performance of control systems.
ˆ Evaluate the dynamic and steady state performance of first and second
order systems in response to standard test inputs.
ˆ Predict, analyse and evaluate control system performance using com-
puter simulation.
ˆ Explain the operation and application of standard controller modes.
ˆ Apply frequency response methods to predict system response.
ˆ Evaluate the performance and stability of open and closed loop sys-
tems.
ˆ Communicate and work effectively as part of a team
ˆ Recognise that good control system design has a direct impact on the
environment and society.
Assessment
End of semester written examination 60%
Continuous assessment & Laboratory Work 40%
Essential Reading
ˆ Class Notes
Recommended Reading
ˆ Control engineering an introductory course Wilkie, Johnson, Katebi
Palgrave Macmillian (2002)
ˆ Control engineering Bolton, W. Newes (2003)
ˆ Control systems engineering Nise N. S. Wiley (2003)
ˆ Integrated Electrical and Electronic Engineering Frazer and Milne
McGraw-Hill (1994)
ˆ The art of control engineering Dutton, K, Thompson, S. Barraclough
B. Addison-Wesley (1997)
ˆ Control Engineering Bissell, C. C. Stanley Thornes Pub. Ltd (1994)
ˆ Introduction to control system technology Bateson R. N. Prentice
Hall (1998)
ˆ Mechatronics Bolton W. Longman (1999)
Contents
Contents i
1 Control Basics 1
1.0.1 Proportional Control . . . . . . . . . . . . . . . . . . 10
1.0.2 Integral Control . . . . . . . . . . . . . . . . . . . . . 17
1.0.3 Derivative Control . . . . . . . . . . . . . . . . . . . 22
1.0.4 PID Control . . . . . . . . . . . . . . . . . . . . . . . 25
2 Models of Physical Systems 28
2.1 The Laplace Transform . . . . . . . . . . . . . . . . . . . . . 28
2.2 Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . 31
2.2.1 Method of Residues . . . . . . . . . . . . . . . . . . . 32
2.3 Mathematical models of physical systems . . . . . . . . . . . 34
3 Block Diagrams and their Reduction 62
3.1 Block diagram reductions . . . . . . . . . . . . . . . . . . . 63
4 The Performance of Control Systems 67
4.1 First Order Open Loop Systems . . . . . . . . . . . . . . . . 67
4.2 First Order Closed Loop Systems . . . . . . . . . . . . . . . 71
4.3 Second Order Systems . . . . . . . . . . . . . . . . . . . . . 85
4.3.1 Velocity (Rate) Feedback . . . . . . . . . . . . . . . . 107
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CONTENTS N.Kent
5 Frequency Response 113
5.1 Frequency Response of a First Order System . . . . . . . . . 114
5.1.1 Nyquist Stability Criterion . . . . . . . . . . . . . . . 127
5.2 The Bode Diagram . . . . . . . . . . . . . . . . . . . . . . . 134
5.3 The Routh Stability Criterion . . . . . . . . . . . . . . . . . 153
Appendices 169
ii
Chapter 1
Control Basics
For this course we will define control as the use of algorithms and feedback
for engineering systems. These systems can include mechanical, fluidic,
thermal, electrical and electronic systems. Examples include, feedback loops
in electronic amplifiers, setpoint control of chemical processing plants, 'fly
by wire' systems on aircrafts, cruise control for cars.
Control can be thought of as two (or more) systems linked together such
that one systems influences/controls the other. The terms open loop and
closed loop are used to describe such system configurations. Open loop
systems, as shown in, Fig 1.1 (a) are characterised by a user defined input
( 'r' in Fig 1.1 (a)) and this value is continually input into System 1. The
output of System 1 in turn affects the performance of System 2.
Closed loop systems or feedback systems, shown in Fig 1.1 (b), are
characterised by an interconnected cycle, where by the output of System
2 is 'fed back' to the input of System 1. This feature allows System 1 to
adapt based on the output of the overall system thus varying the effect on
System 2.
In general a controller senses the operating conditions of a system, com-
pares this condition against the desired behavior, calculates what, if any,
changes are required based on a model of the system's response to exter-
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CHAPTER 1. CONTROL BASICS N.Kent
nal inputs and then actuates the system to implement the desired change.
This basic feedback loop of sensing, calculation and actuation is the central
concept in control.
System 1 System 2 System 1 System 2
r u y u y
(a) Open Loop (b) Closed Loop
Fig 1.1: Feedback configurations. (a) Shows Open loop feedback where a predefined
value 'r' is input into System 1, the output of System 1 is directly inputted into System
2. (b) Shows Closed loop feedback where the output of System 2 is looped around and
inputted into System 1.
Open Loop Control
Open loop control does not compare the actual system output to the desired
system output to determine the action the controller should take. Instead,
a calibrated setting is used to obtain the desired result.
Example
Consider a manually operated valve with a dial indicator as shown in Fig
1.2. The dial indicator can tell the user if the valve is fully open or fully
closed or set at some location between both situations. To ensure appro-
priate operation the valve must be calibrated by measuring the flow rate at
different dial positions from open to closed.
Fig 1.3 shows a calibration curve obtained for the valve for different
pressure drops across the valve. Lets assume that in normal operation with
a pressure drop of ∆p2 we require a volume flow rate of q˙2. In this case
we can set our valve to the 'Setpoint' as indicated in Fig 1.3 and every-
thing will work fine. If for whatever reason the pressure drop in the pipe
increases to ∆p3 due perhaps to a blockage in another part of the pipe
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CHAPTER 1. CONTROL
BASICS N.Kent
Flow Rate (q˙)
Pressure Drop(∆p) or (
dp
dx
)
DialSetting
Fully Open
Fully Closed
Fig 1.2: Manual Valve
network forcing more fluid down the pipe we are interested in or a surge
in power causing the pump to run faster then the output flow rate will
increase to q˙3. Of course if for whatever reason the pressure drop in the
pipe decreases to ∆p1 due perhaps to a slow leak in a pipe then the out-
put flow rate will decrease to q˙1. The point is open-loop control cannot
correct for unexpected changes in pressure drop. More generally we say
open loop systems cannot correct for external disturbances in the system.
F
lo
w
R
at
e
(q˙
)
Dial Setting
∆p1
∆p2
∆p3
q˙3
q˙2
q˙1
Closed OpenSet Point
Fig 1.3: Graph of various flow rates for different pressure drops
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CHAPTER 1. CONTROL BASICS N.Kent
The primary advantage of open loop systems is that it is less expensive
than closed loop control, mainly because it is not necessary to measure the
actual result. The disadvantage of open loop control is that errors caused
by external disturbances are not corrected. Often a human operator must
correct slowly changing disturbances by manual adjustment. In this case
however, the human operator is providing feedback signal and as such is is
closing the loop.
Closed Loop Control
As we have seen a closed loop system is characterised by feedback. The
closed loop system uses the output from the 'Process' and feeds it back to
the 'Controller'. The controller then acts based on this revised input and
adjusts the manipulating element in order to adjust the process towards a
user defined setpoint.
Process
Measuring
Transmitter
Control Mode
Controlled Variable
Disturbance
Variables
Manipulated
Variable
Controller
error
Setpoint
Manipulating Element
Measured Value of
Controlled Variable
+
-
Fig 1.4: Closed loop Process Control
Take for example, Fig 1.4. Fig 1.4 shows a block diagram of a process
control system. Process control is concerned with maintaining variables such
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CHAPTER 1. CONTROL BASICS N.Kent
as temperature, liquid level in a tank, pH and other physical quantities. To
analyse this diagram we first need to outline a few definitions and discuss
the meaning of each block:
ˆ Process Block - is the representation of the physical variable to be
controlled. For the purposes of explanation lets take the physical
variable to be liquid level in a tank.
ˆ Measuring Transmitter - is the sensor configuration that deter-
mines the value of the controlled variable. The measuring transmitter
then converts the value of the controlled variable to a signal suitable
for transmission to the controller location, usually a location away
from the Process. A measuring transmitter will usually consist of a
sensing element and signal converter. In our case, the controlled vari-
able is any variable that will change relative to the level of liquid in
the tank and can be measured, say for example the height of liquid in
the tank. The sensing element might be a variable resistance pressure
sensor (P = ρgh) the output signal from the pressure sensor will need
to be converted from resistance to a signal the controller can read
usually current/voltage and then transmitted to the controller.
ˆ Controller is the unit that calculates the error in the system and
decides how the manipulating element should be changed to eliminate
this error. The error detector computes the difference between the de-
sired value (how much liquid do we want in the tank) called setpoint
and the measured value (how much fluid is in the tank.
error = setpoint - measured value of controlled variable
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CHAPTER 1. CONTROL BASICS N.Kent
The control block converts this error into a 'control action' that will
try eliminate this error i.e
measured value = desired value
The four most common control modes are on/off, proportional mode
(P), integral mode (I) and derivative mode (D). We will look at each
of these modes later in this section. Since we will deal with the control
modes later in this section, for our liquid level example, lets say for
now that the error detector reads a user defined setpoint value say 50%
i.e we want the tank half full of liquid. This setpoint is converted to an
appropiate signal current/voltage and compared to the measured value
converted from the pressure sensor - the difference in these signals will
tell the controller how far the level of liquid that is actually in the tank
is from the desired level.
ˆ Manipulating Element uses the controller output to regulate the
manipulated variable and usually consists of two parts, the actuator
and the final controlling element.
� The actuator converts the the controller output into an action
on the final controlling element
� The final controlling element then changes the value of the ma-
nipulated variable
The fill level of the tank can be controlled by means of a valve. A
pneumatic control valve may be used to control the flow of liquid into
the tank. A schematic of such a valve is shown in Fig 1.5. As the air
enters it pushes on a diaphragm opening the valve if no air is applied
the valve is closed by means of a spring. If such a device was used the
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CHAPTER 1. CONTROL BASICS N.Kent
controller would 'tell' the actuator in this case a pump, what pres-
sure to apply to the valve or final controlling element. The valve then
opens by an amount determined by the air pressure and changes the
flow rate into the tank.
Controlled
Variable
Spring
Diaphragm
Input air
pressure
signal
Fig 1.5: Pneumatically Operated Valve
ˆ Manipulated Variable - is the variable regulated by the final con-
trol element to achieve the desired value of the controlled variable.
The manipulated variable is capable of affecting change in the control
variable. In our example the manipulated variable is the flow rate into
that tank - this has the ability to affect the controlled variable i.e the
height of liquid in the tank.
ˆ Disturbances are process input variable but are not controlled by
the control system. Disturbance variables are capable of changing the
load on a system and are the main reason for using a closed loop
control system. Disturbances can be in the form of a slow leak in the
tank itself or is often an additional valve on the tank, since it is fair to
assume that the liquid stored will be used subsequently in the process
there may be an outlet valve on the tank to facilitate the use of liquid.
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CHAPTER 1. CONTROL BASICS N.Kent
This valve is not controlled by the controller but the controller must
react if this valve is open because if the valve is open the level of liquid
in the tank will drop.
The primary advantage of closed loop control is the potential for more
accurate control of the process. There are two disadvantages of closed loop
control.
1. Closed loop control is more expensive than open loop control
2. The feedback feature of closed loop control can, in certain circum-
stances, result in the system becoming unstable. An unstable system
produces an oscillation of the controlled variable
Simple forms of Feedback Control
Two position or On/Off Control
Two position or 'On Off' control is the simplest and least expensive mode
of feedback control. The controller has only two possible values depending
on the sign of the error the controller value changes. Consider the situation
shown in Figure 1.6.
The controller is configured such that if the thermostat reading is lower
than the required temperature the heater power is fully on and if the thermo-
stat temperature is higher than the required temperature the temperature
is fully off. This is represented
in Fig 1.7. The graph to the left of Fig 1.7
shows the measured error over time while the graph on the right shows the
controller response to this error.
It should be noted that the error can either be positive or negative
representing the measured temperature either being too hot or too cold. It
should also be noted that regardless of the magnitude or duration of the
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CHAPTER 1. CONTROL BASICS N.Kent
Power
Supply
Tank
Heating
Element
Thermostat
Feedback to On/Off Switch
Fig 1.6: Schematic On/Off control system showing tank of liquid to be heated and
controlled via a thermostat
Time Time
-ve
+ve
E
rr
or
0
C
on
tr
ol
le
r
O
u
tp
u
t
Min
Max
Fig 1.7: Graphs showing On/Off controller response(right) to measured system error
(right). Note that the response only depends on the sign of the error. If the error is
positive the controller output is maximum, if the error is negative the controller output
is minimum.
error the controller output can be in only one of two positions - either fully
on or fully off.
Mathematically this relationship can be represented as:
u =
 umax if e > 0umin if e < 0
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CHAPTER 1. CONTROL BASICS N.Kent
Where u is the controller output and e is the error. Most on off con-
trollers incorporate a 'dead-band' around e = 0 where no control action
occurs this is to limit the frequency of switching between the ON and OFF
states.
Generally the two position mode of control is best adapted to large
systems with relatively slow process rates so in the example of the heated
tank the the overall effect of the heater is relatively slow and fluctuations
in temperature are gradual i.e the temperature rises slowly and falls slowly,
thus, for such systems on/off control is satisfactory.
However, due to the nature of the switching of the output the temper-
ature in the tank is never held constant is is constantly fluctuating about
the setpoint as shown in Fig 1.8. For many processes this fluctuations may
be acceptable however for more sophisticated control operations additional
methods of control are required.
Time Time
C
on
tr
ol
le
r
O
u
tp
u
t
Process Variable
Setpoint
SP
On
100%
Off
0%
Fig 1.8: Graphs showing the result of On Off control. The graph on the right shows
how the process variable reacts based on the 'on/off' controller input shown on the right.
Notice how the process variable never settles at the setpoint - it continually oscillates
around the setpoint
1.0.1 Proportional Control
One of the biggest problems with On/Off control is that the control action
is exactly the same for small deviations from the setpoint (small errors) as
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CHAPTER 1. CONTROL BASICS N.Kent
it is for large deviations from the setpoint (large errors). With this in mind,
a form of control known as Proportional Control was developed. At its
most basic, proportional control changes the controller action based on the
magnitude of the error. Small errors in the system are met with a propor-
tionally small output response from the controller while large errors result
in a large output response from the controller. Taking our heated cylinder
from the On/Off control example previously, we can represent proportional
control by removing the switch and replacing it by a current limiting device
as shown in Fig 1.9.
Power
Supply
Tank
Heating
Element
Thermostat
Feedback to Current ControlCurrent
Control
Fig 1.9: Schematic control system showing tank of liquid to be heated and controlled
via a thermostat. Note the switch from the previous section has been replaced by a
current limiting device
In this case of the heated cylinder shown in Fig 1.9 if the temperature
change is relatively small then the corresponding controller action is rela-
tively small. i.e. if the temperature changes by only a small amount then
the current is changed by only a small amount. This relationship is shown
graphically in Fig 1.10. Notice the difference in controller output between
this graph and the one shown in Fig 1.8.
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CHAPTER 1. CONTROL BASICS N.Kent
Time Time
-ve
+ve
E
rr
or
0
C
on
tr
ol
le
r
O
u
tp
u
t
Min
Max
1 2 3 4 5 6 1 2 3 4 5 6
Fig 1.10: Proportional Control
Up to now we have described proportional control in rather vague terms
i.e 'small' changes in error lead to 'small' changes in controller output. As
engineers we need to build up more robust definitions of relations between
the input and output for any given system. The relationship between rela-
tive input and relative output for a given system can be defined in terms of
whats called proportional band or as the proportional gain KP of the
system.
ˆ Proportional Band is defined as the range over which proportional
control is implemented.
ˆ Proportional Gain or KP is the amount the error is multiplied by in
order to achieve the required controller output over the proportional
band.
Mathematically Proportional Control can be described as follows:
u =

umax if e ≥ emax
KP e if emin < e < emax
umin if e ≤ emin
Where u is the controller action Kp is the proportional gain and e is the
system error. If e ≥ emax or ≤ emin then the control action is as it would
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CHAPTER 1. CONTROL BASICS N.Kent
be in the case of On/Off control i.e umax or umin. If the error lies within
the emax and emin band, then the controller output is the error times the
proportional gain KP . This band or region is the proportional band. The
proportional band is usually expressed as a percentage of the scale range.
Example 1
In the case of the heated tank shown in Figure 1.9. Lets say the range
of temperatures possible in the tank, assuming the liquid is water, is from
room temperature, 22 degrees, to boiling at 100 degrees. If the heater spec-
ification is such that it can only heat the water by 10 degrees from fully on
to fully off then the proportional band is:
P.B =
10
100− 22 × 100 = 12.8%
That is a 12.8% change in temperature causes the heater to go from
fully hot to fully cold.
Example 2:
If the full scale range of the level of liquid in a tank is 1 metre (tank
empty to tank full) and the fill valve for the tank is set such that a change
in level of 0.1 metre will cause the valve to go from fully open to fully closed
then the proportional band setting is:
P.B =
0.1
1
× 100 = 10%
That is a 0.1m change in liquid level causes the valve to go from fully
open to fully closed.
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CHAPTER 1. CONTROL BASICS N.Kent
The relationship between KP and P.B is given by
P.B =
1
KP
× 100
It is usually arranged to have the manipulating variable, e.g the current
in the heater or the valve position from the above examples, giving the
desired value of the controlled position at the centre of the proportional
band. Taking the valve example this means the valve is half open when the
liquid is at the desired level so if the tank becomes too full we can close the
valve if that tank becomes too empty we can open the valve and allow more
liquid in.
Exercise:
A proportional type pneumatic controller is used to control the liquid
level in a tank at a desired level of 7.5 meters. The range of the instrument is
0-10 meters and the proportional band is set at 15%. The controller has an
output pressure range of 20-100 kN/m
2
and the output pressure increases
with increasing level. If the output pressure is 60 kN/m
2
when the tank
level is at the desired value, find:
a The liquid levels corresponding to output pressure of 20 kN/m
2
and 100
kN/m
2
respectively.
b The value of the output pressure for an error of 0.15 meters below the
desired value.
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CHAPTER 1. CONTROL BASICS N.Kent
Solution:
Problems with Proportional Control
Consider the situation from the previous exercise. The instrument range,
proportional band and output pressure can be graphed as shown in Fig 1.11.
In the case of the initial setpoint of the system, liquid level 7.5, can be seen
in point 'A' in Fig 1.11.
0 1 2 3 4 5 6 7 8 9 10
Liquid level (m)
20
30
40
50
60
70
80
90
100
P-Band 15%
Offset
A
B
P
re
ss
u
re
(2
0-
10
0)
k
N
/m
2
Fig 1.11: Caption
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CHAPTER 1. CONTROL BASICS N.Kent
If however the load of the system is changed in order to counter act this
the controller output must also change. In the case of Fig 1.11 we assume
an increase in load on the system causing the liquid level to drop to a value
of 7.125m. This results in a change in output pressure from 50% to 25%
of the pressure range, shown in point 'B' in Fig 1.11. In this case the level
stabilises and does not oscillate around a setpoint as in the case of on off
control. However the new steady state condition is less than the required
setpoint and the difference between original setpoint and new steady state
position is called the offset as shown in Fig 1.11.
The offset is proportional to the load change and inversely proportional
to the gain KP of the system. If the proportional band (p-band)is widened
as shown in Fig 1.12 then under the same loading conditions i.e the tank
level dropping to 7.125m the percentage change in controller output is 12.5
% as opposed to 25% in the case of a narrower p-band.
0 1 2 3 4 5 6 7 8 9 10
Liquid level (m)
20
30
40
50
60
70
80
90
100
P-Band 30%
Offset
A
B
P
re
ss
u
re
(2
0-
10
0)
k
N
/m
2
Fig 1.12: Caption
This may seem desireable however the case of the wider proportional
band it can be seen that the controller is less sensitive to changes in tank
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CHAPTER 1. CONTROL BASICS N.Kent
level i.e. the for the same change in tank level the controller response is less
for the wide proportional band than it is for the narrower (15%) band.
Obviously narrowing the proportional band will have the opposite effect,
and increase sensitivity however reducing the proportional band to zero will
in effect return us to On/Off control. Since we originally looked at getting
away from on/off control due to the oscillation problems encountered setting
the proportional band to zero is not desirable, however having a non zero
proportional band introduces the offset problem so we are left with the
following situation:
ˆ A wide proportional band (small gain) will provide a less sensitive
response, but a greater stability.
ˆ A narrow proportional band (large gain) will provide a more sensitive
response, but there is a practical limit to how narrow the proportional
band can be set.
ˆ Too narrow a proportional band (too much gain) will result in oscil-
lation and unstable control.
The proportional mode is used when the gain can be made large enough
to reduce the proportional offset to an acceptable level for the largest ex-
pected load change. If proportional control cannot reduce the proportional
offset to an acceptable level then Integral control can also be introduced in
combination with proportional control to eliminate the offset error entirely.
1.0.2 Integral Control
The integral control mode changes the output of the controller by an amount
proportional to the integral of the error signal. The proportion by which
the controller is changed based on the error is called the integral gain KI .
This relationship is shown in Fig 1.13.
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CHAPTER 1. CONTROL BASICS N.Kent
Time Time
-ve
+ve
E
rr
or
0
C
on
tr
ol
le
r
O
u
tp
u
t
Min
Max
1 2 3 4 5 6 1 2 3 4 5 6
Fig 1.13: Caption
Fig 1.13 shows the error signal as used previously, to describe both on
off and proportional control, on the left hand side and the corresponding
integral response on the right hand side. The figure can be described by
studying each graph over time.
Time 0-1 There is no error signal therefore the output of the integral con-
troller is zero.
Time 1-2 The error is relatively low and is a positive error therefore the
output from the integral controller increases over time - i.e as the area
under the error curve increases
Time 2-3 The error switches to negative and has a greater magnitude than
occurred between times 1-2. The resultant controller output decreases
at a sharper rate than between times 1-2. At time point 2 the net
overall error is negative.
Time 3-5 Between times 3-5 the error is large, sustained and positive in
sign. This results in a sharp increase in the integral controller response
and remains at a constant slope over times 3-5 since he overall area
increases between 3-5.
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CHAPTER 1. CONTROL BASICS N.Kent
Time 5-6 The error drops significantly although remains positive - this
results in the output integral controller remaining on an upward curve
however at a slower rate than between times 3-5 due to the lower error
magnitude.
Mathematically integral Control can be described as follows:
u(t) = KI
∫ t
0
e(τ).dτ
Where u is the integral output KI is the integral gain and e is the error
signal and τ is called the dummy variable of integration. Up to now you
would have seen integral of the form
∫ b
a
f(x)dx or
∫ 2pi
0
sinθ + cos2θdθ.
These functions describe curves i.e f(x) is a curve that can be plotted for
any value of x and can then be integrated over a range. In our case of error
e we have no such description of the curve just a series of values of error over
time. For this reason we introduce τ the dummy variable of integration. It
can simply be thought of as a mathematical tool to facilitate integration.
For our purposes we can just think of integral control as the sum of the
error over time.
Ok, now we have an integral controller - so what? If this integral action
was added to the control output from the proportional controller the net
effect would be to reduce the steady error problem that occurs with the
proportional controller. Since the integral controller output depends on the
integral of any error over time, a sustained steady state error would result in
a large response from the integral controller thus reducing the steady state
error and over time eliminating it entirely.
This type of combined control is called PI control and is demonstrated
graphically in Fig 1.14. The graph to the left of Fig 1.14 is the error signal
we are by now familiar with the graph to the right show the PI response
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CHAPTER 1. CONTROL BASICS N.Kent
between time 1 and time 2. The purely proportional response is shown by
the dashed line and is labeled 'P' while the purely integral response is shown
by the dashed-dot line indicated 'I'. The net response indicated P+I in Fig
1.14shows the overall response of a PI controller to a constant error between
time points 1 and 2.
Time Time
-ve
+ve
E
rr
or
0
C
on
tr
ol
le
r
O
u
tp
u
t
Ti
OI
OP
P+I
I
P
1 4 5 632 1 2
Fig 1.14: Caption
The time Ti as indicated in Fig 1.14 is called the 'Integral action time'
and is defined as the the time taken for the output from the integral control
(OI) mode to reach the same output as the proportional mode controller
output (OP ) for a constant error. The relationship between the integral
gain and the integral action time is shown as follows:
KI =
KP
Ti
Where KI is the integral gain KP is the proportional gain and Ti is the
integral action time. Different controller manufacturers use different types
of algorithms in effect the net result
is the same - where might might describe
changing the integral control action in terms of Integral gain others might
describe changing the integral control action in terms of integral action time.
Changing the integral action time on a controller has the following effects
20
CHAPTER 1. CONTROL BASICS N.Kent
ˆ If it is too short (large integral gain), over-reaction and instability will
result.
ˆ If it is too long (low integral gain), the integral action will be very
slow to take effect.
Exercise:
A step input of magnitude 4% is applied to the input of a proportional
plus integral controller. At a time t=0 the output undergoes a sudden
step change of 10% and after a lapse of 3 minutes the total change is 25%.
Calculate
a The proportional bandwidth
b Integral action time
Solution:
Mathematically the PI controller can be described as the sum of both
the proportional control and the integral control.
21
CHAPTER 1. CONTROL BASICS N.Kent
u(t) = KP .e(t) +KI
∫ t
0
e(τ).dτ
or
u(t) = KP
[
e(t) +
1
Ti
∫ t
0
e(τ).dτ
]
Ok, so now we have a controller that gives a response based on the error
at any given time (the proportional part) it also gives a response based
on what the error has been in the past (the integral response). We have
got rid of our oscillations due to on/off control and eliminated steady state
error due to changes in load. We could satisfy ourselves and say our job
is complete and in fact the PI controller is used in many industrial control
problems but lets go one step further. What if we could predict what the
error was going be in the future, surely that would be of benefit?
1.0.3 Derivative Control
Derivative control does exactly that, it predicts what the error signal will
be in the future and changes the control action based on that prediction. In
essence the process is very simple, derivative control measures the slope of
the error signal (or how fast the error is changing) at any given point then
predicts what the error will be at a time t + ∆t and decides what control
action should be implemented based on this prediction. It can be shown
graphically in Fig 1.15. In the case of Fig 1.15, since the controller response
is based on the slope of the error signal, we have changed our error signal
from before to allow visualisation of the response.
Lets break down the derivative controller response as we did in the case
of the integral response
22
CHAPTER 1. CONTROL BASICS N.Kent
Time Time
-ve
+ve
E
rr
or
0
C
on
tr
ol
le
r
O
u
tp
u
t
1 4 5 632 1 4 5 632
Fig 1.15: Caption
Time 0-1 There is no error signal, so there is no slope of the error signal,
therefore the output of the integral controller is zero.
Time 1-2 The error increases sharply at time point 1 and increases slowly
to time point 2. The resultant derivative controller action is an in-
crease in controller output between point 1 and point 2. Note we are
ignoring the sudden sharp change in slope that occurs at a single time
point i.e the initial jump at point 1.
Time 2-3 The error switches to negative and remains constant. The re-
sultant controller output is zero since the slope of the line between
points 2-3 is zero. Note in this case there is a system error however
the controller output is zero.
Time 3-5 The error increases significantly at time point 3 and continually
increases, at a constant rate up to time point 5. The resultant con-
troller action is larger than the controller action between point 1-2
due to the fact the slope of the error signal is greater.
Time 5-5.5 The error drops significantly although remains positive. How-
ever the controller response drops into the negative region due to the
fact the slope of the error output is now negative. The magnitude of
23
CHAPTER 1. CONTROL BASICS N.Kent
the negative controller response is directly related to the slope of the
error response
Time 5.5-6 The error response is constant again therefore the resultant
controller response is zero.
Mathematically derivative control can be described as follows:
u(t) = KD
de(t)
dt
Where u(t) is the derivative response, KD is called the derivative gain
and
de(t)
dt
is the slope of the error signal at time t.
Since derivative control action only occurs with changing error, and
will not respond to constant error as shown between point 2-3 in Fig 1.15.
Derivative control is never used alone but is used in combination with pro-
portional control or proportional and integral control. Fig 1.16 shows the
controller response from proportional and derivative (PD) controller. The
graph to the left of Fig 1.16 shows the error signal as before and the graph
on the right of Fig 1.16 shows the PD controller response over time points
0-2.
Time Time
-ve
+ve
E
rr
or
0
C
on
tr
ol
le
r
O
u
tp
u
t
1 4 5 632 21
P+D
P
D
OD
OP
Td
Fig 1.16: Caption
24
CHAPTER 1. CONTROL BASICS N.Kent
The purely derivative response labelled D in Fig 1.16 can be seen to be
constant between time 1 and time 2, since the slope of the error is constant
the derivative response is constant. However, the proportional response can
be seen to increase between time 1 and time 2. This is due to to the fact
the error itself is increasing between time 1 and time 2. Since proportional
controller output is dependent only on the error at any given time it also
increases between point 1 and point 2.
The overall response from a PD controller can be seen, labelled P+D,
in Fig 1.16. In a similar manner to the integral controller the derivative
controller can be described in terms of derivative gain KD or Derivative
Action Time (Td in Fig 1.16). Derivative action time is defined as the time
taken for the output from proportional control mode OP to equal the the
output from the derivative control mode OD. The relationship between KD
and Td is as follows
KD = KP .Td
The proportional plus derivative mode is especially useful when the pro-
cess under goes sudden load changes i.e. the error jumps rapidly. The
predictive nature of the derivative aspect ensures the controller response
acts rapidly.
1.0.4 PID Control
A combination of proportional integral and derivative (PID) controllers are
used in the majority of industrial control processes. A PID controller is
also called a three term or three mode controller. A graph summarising the
process is shown in Fig 1.17. For any given error signal the proportional
control takes into account the error at that time and eliminates oscillations,
25
CHAPTER 1. CONTROL BASICS N.Kent
the integral control aspect considers the error in the past and eliminates
steady state error while the derivative control component examines how
fast the error is changing into the future and acts to reduce the error before
is occurs.
Time
Past Present Future
E
rr
or
Integral Proportional Derivative
Fig 1.17: Caption
Mathematically the PID controller is described as
u(t) = KP .e(t) +KI
∫ t
0
e(τ).dτ +KD
de(t)
dt
or
u(t) = KP
[
e(t) +
1
Ti
∫ t
0
e(τ).dτ + Td
de(t)
dt
]
PID controller can, and are used in a range if industrial process con-
trol systems from keeping liquid at a constant level in a tank, to mixing
chemicals, to power consumption. In all cases the gains/action times of the
controller must be set to give appropriate control response for a given sys-
tem error. The process of setting these controllers to appropriate gains is
called PID tuning. Obviously having a knowledge of how a system behaves
to external loads and disturbances would of benefit when choosing these
gains. Although many systems
may on the surface seem very different, the
26
CHAPTER 1. CONTROL BASICS N.Kent
underlying mathematics behind each system can be remarkably similar. In
the next section we will look at how we can model real life systems and
investigate system response based on these models.
27
Chapter 2
Models of Physical Systems
2.1 The Laplace Transform
There has been lots written on the 'power' of the Laplace transform. At this
stage all you need to really know is that the Laplace transform is a method
of changing differential equations into algebraic equations that are easier
to solve. For example the differential equations describing how a system
behaves with time are changed into algebraic equations not involving time.
System behaviour in the 'time domain' is transformed into the 's-domain'
in which algebraic manipulations can be carried out. On completion of the
manipulation we can then perform an 'inverse Laplace transform' to convert
from the 's-domain' to the 'time domain' so that the behaviour over time
can be observed.
The Laplace transform can be written in the form:
F (s) = L{f(t)} =
∫ ∞
0
e−stf(t).dt
Where:
ˆ F (s) is our new domain it is customary to use a capital letter for a
28
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Laplace transform. In the case of Laplace transforms the new domain
is called the 'Frequency' domain and the operator s = σ+ jω is called
the complex frequency and units are (1/time).
ˆ L{} is the Laplace transform notation
ˆ f(t) is our original function plotted in the time domain 'signified by
the letter t.
ˆ
∫ ∞
0
e−stf(t).dt is the transform itself. We multiply each term of f(t)
by e−st and integrate wrt time from 0→∞
Fortunately there are tables available where the transform for many
different time based functions are available. However lets look at a few ex-
amples.
Step Input
f(t) = A
F (s) = L{A} =
∫ ∞
0
e−stA.dt
F (s) = A
[
− 1
s
.e−st
]∞
0
=
A
s
Exponential
f(t) = e−at
F (s) = L{e−at} =
∫ ∞
0
e−st.e−at.dt
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
F (s) =
∫ ∞
0
e−(s+a)t.dt
F (s) =
[
− 1
s+ a
.e−(s+a)t
]∞
0
F (s) = [0]−
[
− 1
s+ a
]
F (s) =
1
s+ a
Ramp input
f(t) = t
F (s) = L{t} =
∫ ∞
0
t.e−st.dt
F (s) =
1
s2
t
f(t)
0
σ
jω
F (s) = L{t} = 1
s2
Double ’x’
Double Pole
Time Domain s-Domain
Parabolic or Acceleration Signal
f(t) = t2
F (s) =
2
s3
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
t
f(t)
0
σ
jω
F (s) = L{t2} = 2
s3
Triple ’x’
Multiple pole of order 3
Time Domain s-Domain
Sinusoid
f(t) = sinωt
F (s) =
ω
s2 + ω2
t
f(t)
0
σ
jω
F (s) = L{sinωt} = ω
s2 + ω2
Time Domain s-Domain
Poles
A full table of required Laplace transforms will be available on the day
of the exam
2.2 Inverse Laplace Transform
On converting our system transfer function (the relationship between output
and input) from the time domain to the s domain and solving the required
algebraic equations we must now convert from the s domain to the time
domain. This can be done using the inverse Laplace transform.
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
The inverse Laplace transform of F (s) is defined as
f(t) =
1
2pij
∫ α+j∞
α−j∞
F (s).estds
Fortunately in most cases the order (value of the highest power) of the
numerator is less than that of the denominator. If this is the case then the
methos of partial fractions or method of residues can be used to find f(t) if
F (s) is known.
2.2.1 Method of Residues
Steps in implementing the Residue theroem
1. Write down F (s)
2. Factorise the denominator
3. Take each pole in turn and find the residue for that pole. To find the
residue for a given pole multiply F (s) by that pole and by est and set
s = pole value. The result is the reside for that pole.
4. Repeat this process for all poles.
The overall time based response f(t) =
∑
Residues
Example 1
F (s) =
1
s+ a
. What is the corresponding f(t)?
32
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
f(t) =
[ 1
(s+ a)︸ ︷︷ ︸
F (s)
× (s+ a)︸ ︷︷ ︸
Pole
× est︸︷︷︸
est
]
s→−a
f(t) =
[ 1
���
�(s+ a)
×����(s+ a)× est
]
s→−a
Setting s = −a
f(t) = e−at
Example 2
F (s) =
1
s(s+ 1)
. What is the corresponding f(t)?
f(t) =
[ 1
s(s+ 1)
.s.est
]
s→0
+
[ 1
s(s+ 1)
.(s+ 1).est
]
s→−1
f(t) = 1 + (−e−t)
f(t) = 1− e−t)
Exercise
F (s) =
1
(s+ 2)(s+ 3)(s+ 4)
. What is the corresponding f(t)?
Answer:
f(t) =
1
2
e−2t − e−3t + 1
2
e−4t
Multiple Poles
33
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
In the previous examples s is not raised to any power greater than one.
If it was there would be whats calle a multiple pole and the factors would
be in the form sn or (s + a)n These are considered multiple poles of order
n.
To find the residue of a multiple pole of order n we use:
R =
1
n− 1!
dn−1
dsn−1
[F (s).est.(s+ a)n]s→−a
Example 1
If F (s) =
1
s2
find f(t)
In this case n = 2
f(t) =
d
ds
[ 1
s2
.est.s2
]
s→0
f(t) = [t.est]s→0
f(t) = t.1
f(t) = t
2.3 Mathematical models of physical systems
A critical problem in engineering design and analysis is the determination
of a mathematical model. To be useful the model must not be too com-
plicated. Therefore simplifying assumptions are often made. The system
we are concerned with are dynamic in nature and their behaviour will be
described by differential equations.
34
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Although they will often be non linear systems it is customary to lin-
earise them about an operating point to obtain linear differential equations.
Therefore it should be borne in mind that such linear models are useful for
preliminary design and analysis only over a limited range.
Nonetheless these types of models are widely used in engineering utiliz-
ing, Laplace transforms frequency response and steady state models, (all of
which will cover in due course).
The components of all control system are diverse in nature and may
include electrical, electronic, mechanical, thermal and fluidic devices.
The differential equations for these devices are obtained using the basic
laws of physics eg Ohm's Law, Lenz Law, Newton's Laws, Bernoulli's Eqn,
Boyles Law, Kirchoffs Laws etc.
Laplace transformation of these differential equations yields an algebraic
equations in terms of complex frequency or what we have been calling s.
These algebraic equations can be easily manipulated to obtain the trans-
fer function of the system defined as the ratio of the output Laplace trans-
form and the input Laplace transform./
This transfer function represents a linear model of the system and is
usually shown in the form of a block diagram as shown below.
X(s) Y (s)
G(s)
Fig 2.1: Block diagram representation of a Transfer Function
Y (s)
X(s)
= G(s)
Y (s) = X(s).G(s)
35
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Recall
For a Capacitor
i(t) = C.
dv
dt
Taking the Laplace transform of both sides
I(s) = C.sV (s)
From Ohms Law we have V = I.Z
∴ V (s) = I(s).Z(s)
V (s)
I(s)
= Z(s)
From the Laplace transform of a capacitor
V (s)
I(s)
=
1
sC
Therefore Z(s) =
1
sC
i.e. in terms of Laplace the reactance of a Capacitor is
1
sC
Similarly for an inductor
v = L
di
dt
36
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Taking Laplace
V (s) = L.sI(s)
From Ohms Law
V (s)
I(s)
= Z(s)
Therefore for an inductor Z(s) = sL
1
jωC
1
sC
For sin waves In general
Capacitors
Inductors
jωL
sL
Example 1
Find the transfer function of the following system.
Solution:
vin(t) = i(t).R + vo(t)
37
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
R
Cvin(t) vout(t)
Fig 2.2: First order system - RC circuit
i(t) = C.
dvo(t)
dt
vin(t) = RC.
dvo(t)
dt
+ vo(t)
This is a first order linear differential equation.
Taking the Laplace transform the functions of time become functions of
s so we have
Vin(s) = RCsVo(s) + Vo(s)
Vin(s) = Vo(s)(1 + sRC)
Vo(s)
Vin(s)
=
1
1 + sRC
It is standard practice to let RC = τ . This value of τ is called the 'time
constant'of the system and can be considered a measure of how quickly
the system responds. We will look at this concept a little later but for now
the transfer function
Vo(s)
Vin(s)
can be written as
Vo(s)
Vin(s)
=
1
1 + sτ
This is the general form of a transfer function for a first order system
38
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
and is represented in block diagram format as shown below.
Vin(s) Vout(s)1
1 + sτ
Fig 2.3: First order transfer function
A common test signal in control engineering is the step input. A step
input can be thought of as turning a switch on, or more accurately increas-
ing the input signal from one level to another over a short period of time
such that in signal looks like a'step'.
f(t) = A where A is the magnitude of the input signal i.e. A Volts if
the input signal is a voltage.
In Laplace the step input of magnitude A is written as F (s) =
A
s
For the RC circuit we have just analysed lets apply a DC voltage of V
Volts at time t = 0 and calculate the output response.
vin(t) = V
Vin(s) =
V
s
The transfer function of the circuit is
Vo(s)
Vin(s)
=
1
1 + sτ
Rearranging we have
39
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Vo(s) = Vin(s)× 1
1 + sτ
Vo(s) =
V
s
× 1
1 + sτ
Important!!. Up to now at this point we have used the residue
method to find the inverse Laplace transform and find the time based re-
sponse of the system. We are still going to perform this step but we must
modify the transfer function such that the coefficient of s is 1.
Consider the transfer function
1
1 + sτ
Divide above and below by τ
We now have
Vo(s)
Vin(s)
=
1
τ
s+
1
τ
We now perform the residue method on this version of the transfer func-
tion so we have.
vo(t) =
V
s
.
1
τ
s+
1
τ
.s.ests→0 +
V
s
.
1
τ
s+
1
τ
.(s+
1
τ
).est
s→− 1
τ
Solving we have
vo(t) = V − V e− tτ
40
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
vo(t) = V (1− e− tτ )
Exercise:
What is the measure of vo(t) when t = τ?
vo(t) = V (1− e− tτ )
vo(t) = V (1− e− ττ )
when t = τ
vo(t) = V (1− e−1)
vo(t) = V (1− 0.367)
vo(t) = 0.632V
Now consider the above equation. The equation is telling us that after
a time t = τ seconds the output of the circuit will be 0.632 or 63.2% of the
input signal V . Note that the voltage V is the final value of the output (as
t → ∞) the output goes to V . Also remember that in this case τ = RC
where R and C are physical components. So by changing the values of R
and C we can determine how quickly the curcuit reaches 63.2% of its final
value and as a result ives an indication of how quickly the system will re-
spond.
The time constant τ of a system can be defined as the length of time
41
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
it takes the system to reach 63.2% of its final value for a step input into the
system.
Another way to look at the time constant is to say the time constant
is the time it would take the system to reach its final value if it were to
maintain its initial slope.
Consider
vo(t) = V − V e− tτ
dvo
dt
=
V
τ
e−
t
τ︸ ︷︷ ︸
Slope
At time t = 0
dvo
dt
=
V
τ
To understand this a little more look at the response of the circuit as
defined by vo(t) = V (1− e− tτ )
t
V
τ
0.632V
Time to reach V had the
output continued at initial
slope
vo
Fig 2.4: Time constant for first order system
42
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Exercise:
How long does it take for the output to reach its final value?
In theory the answer is∞ however after only four time constants we have
vo(t) = V (1− e− tτ )
vo(t) = V (1− e− 4ττ )
vo(t) = V (1− e−4)
vo(t) = 0.98V
i.e the output will reach 98% of its final value after only 4 time constants.
Example 2 - Fluid Model
Air and Reciever
Initially, po is at atmospheric pressure. At time t = 0, pin is applied.
Po
Pin
Restriction e.g. Valve
Fig 2.5: Pressure vessel
43
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
dpo
dt
∝ [pin(t)− po(t)]
dpo
dt
= K[pin(t)− po(t)]
dpo
dt
+Kpo(t) = Kpin(t)
This is a first order linear differential equation
Taking the Laplace of both sides we have
sPo(s) +KPo(s) = KPin(s)
Po(s)[s+K] = KPin(s)
Po(s)
Pin(s)
=
K
s+K
Pin(s) Pout(s)K
s+K
Fig 2.6: Block diagram for first order system
Example - The Integrator
If the input to a system is constant and the the output of a system ramps
this is known as an integrator. For the system shown below we consider q
to be the input to the system and x to be the output.
To simplify the system we neglect friction, fluid compressibility, leakage
past the piston and piston inertia.
44
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
x
Flowrate
q˙
Fig 2.7: Hydraulic Cylinder
With these assumptions in mind we see the velocity of the piston is di-
rectly proportional to the flow rate of the hydraulic fluid q. The differential
equation relating the the flow rate q and the piston displacement is:
dx
dt
∝ q
dx
dt
= Kq(t)
dx = Kq(t).dt
Integrating both sides
∫
dx =
∫
Kq(t).dt
x = K
∫
q(t).dt
We can solve using the Laplace transform in the following manner
dx
dt
= Kq(t)
45
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Taking the Laplace and assuming zero initial conditions
sX(s) = KQ(s)
X(s)
Q(s)
=
K
s
K
s
Q(s) X(s)
Fig 2.8: Hydraulic Cylinder
Comparing the both the time based and Laplace solutions we can see
that multiplication by
1
s
in the Laplace domain is the same as integrating
in the time domain.
Further we can say that differentiating in the time domain is the same
as multiplying by s in the Laplace domain.
The output for the system for a step input is as follows
t t
Constant
Flowrate
q˙ x
Run to end position
Fig 2.9: Cylinder output for constant input
Exercise:
46
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Derive the time based output for the system for a constant (step) input
to the system of magnitude q.
Mechanical Systems
Spring
k
x
f(t) = k × x(t)
Damper
B
x
f(t) = B
dx
dt
Mass
x
f(t) = m
d2x
dt2m
f(t)
To derive the transfer function of a mechanical system draw a free body
diagram of the mass. Remember the object of these exercises is to derive
the transfer function i.e. to relate the input and output.
47
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Example:
Find the transfer function of the following system:
x(t)
m
k
Spring Constant
Viscous Friction
f(t)
Fig 2.10: Spring mass damper systems
Solution:
The free body diagram of the mass looks as follows:
m
kx(t)
f(t)
B
dx
dt
We are looking for
X(s)
F (s)
∑
F = m
d2x
dt2
f(t) = −Bdx
dt
− k.x(t) = md
2x
dt2
f(t) = m
d2x
dt2
+B
dx
dt
+ k.x(t)
This is a second order linear differential equation
48
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Assuming zero conditions and taking the Laplace transform of both sides
we get
ms2X(s) +BsX(s) + kX(s) = F (s)
X(s)[ms2 +Bs+ k] = F (s)
X(s)
F (s)
=
1
ms2 +Bs+ k
Or can be also written as
X(s)
F (s)
=
1
m
s2 + B
m
s+ k
m
1
m
s2 + b
m
s+ k
m
X(s)F (s)
Exercise:
Find the transfer function
Xo(s)
Xi(s)
for the following system
xo(t)
m
k
xi(t)
B
mB
dxo
dt
k(xi(t)− xo(t))
Answer:
k
m
s2 + b
m
s+ k
m
Xo(s)Xi(s)
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Solution:
∑
F = m
d2x
dt2
k[xi(t)− xo(t)]−Bdxo
dt
= m
d2xo
dt2
kxi(t)− kxo(t)−Bdxo
dt
= m
d2xo
dt2
kxi(t) = m
d2xo
dt2
+B
dxo
dt
+ kxo(t)
This is a second order differential equation. Taking the Laplace trans-
form and assuming zero initial conditions
kXi(s) = ms
2Xo(s) +BsXo(s) + kXo(s)
kXi(s) = Xo(s)[ms
2 +Bs+ k]
Xo(s)
Xi(s)
=
k
ms2 +Bs+ k
Or as the answer gave
Xo(s)
Xi(s)
=
k
m
s2 + B
m
s+ k
m
The system is described as a second order differential equation. This
results in s2 terms in the Laplace transform of the system.
Example
Determine the transfer function
X(s)
F (s)
for the following system
50
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
f(t)m1m2
k2
k1
x(t)y(t)
D3 D1D2
Solution:
First determine the free body diagrams for each mass. In this case we
are jumping straight to the Laplace transform on the diagram itself. If you
are looking at these notes for the first time now - this was explained in class!
m1m2
k2Y (s)
D3sY (s)
k1(X(s)− Y (s))
D2s(X(s)− Y (s))
F (s)
k1(X(s)− Y (s))
D2s(X(s)− Y (s))
D1sX(s)
For m1
F (s)− k1(X(s)− Y (s))−D2s(X(s)− Y (s))−D1sX(s) = m1s2X(s)
For m2
k1(X(s)− Y (s)) +D2s(X(s)− Y (s))−D3sY (s)− k2Y (s) = m2s2Y (s)
From equations from m2
k1X(s)−k1Y (s)+D2sX(s)−D2sY (s)−D3sY (s)−k2Y (s) = m2s2Y (s)
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
k1X(s)+D2sX(s) = m2s
2Y (s)+D2sY (s)+D3sY (s)+k2Y (s)+k1Y (s)
X(s)(k1 +D2s) = Y (s)(m2s
2 +D2s+D3s+ k2 + k1)
Y (s) =
X(s)(k1 +D2s)
(m2s2 +D2s+D3s+ k2 + k1)
From equation from m1
F (s) + Y (s)(D2s+K1) = X(s)(m1s
2 +D1s+D2s+ k1)
Now since we are looking for
X(s)
F (s)
as given in the question we must elim-
inate Y (s).
Since Y (s) =
X(s)(k1 +D2s)
(m2s2 +D2s+D3s+ k2 + k1)
we can substitute this value for Y (s) into
F (s) + Y (s)(D2s+K1) = X(s)(m1s
2 +D1s+D2s+ k1)
After some manipulation we can determine
X(s)
F (s)
=
ms2 +D1s+D2s+ k1 + k2
(m1s2 +D1s+D2s+ k1)(m2s2 +D2s+D3s+ k2 + k1)− (k1 +D2s)2
This then is a fourth order system since when we multiply the denomi-
nator out further we get a m1m2s
4
term.
Liquid systems
52
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
Consider the tank as as shown with an input qi and and output qo
discharging water under its own head.
h(m)
Area = A(m2)
qo(t)
Restriction
qi(t)
Assuming laminar flow
Inflow - Outflow = Rate of accumulation.
The flow through pipes is opposed by a property of pipes known as as
hydraulic resistance. It is defined as
R =
h(t)
qo(t)
It follows therefore that qo(t) =
h(t)
R
We require a transfer function relating Qi(s) and H(s)
So:
qi(t)− qo(t) = A.dh
dt
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
qi(t)− h(t)
R
= A.
dh
dt
qi(t) = A.
dh
dt
+
h(t)
R
Taking the Laplace transform of both sides and assuming zero initial
conditions
Qi(s) = AsH(s) +
H(s)
R
Qi(s) = H(s)
[
As+
1
R
]
Qi(s) = H(s)
[1 + ARs
R
]
H(s)
Qi(s)
=
R
1 + ARs
H(s)
Qi(s)
=
R
1 + τs
where τ = AR
If we require the transfer function
Qo(s)
Qi(s)
then
qi(t)− qo(t) = A.dh
dt
R =
h(t)
qo(t)
=
dh
dqo
Rdqo = dh
R
dqo
dt
=
dh
dt
So now we have
54
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
qi(t)− qo(t) = ARdqo
dt
Taking Laplace and assuming zero initial conditions
Qi(s)−Qo(s) = ARsQo(s)
Qi(s) = ARsQo(s) +Qo(s)
Qi(s) = Qo(s)[ARs+ 1]
Qo(s)
Qi(s)
=
1
ARs+ 1
Interaction and Loading
Consider the system as shown below
h1
A1
q2(t)Restriction
q1(t)
h2 A2
q3(t)
Restriction
We can model the entire system as two block diagrams as shown
55
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
1
A1R1s+ 1
1
A2R2s+ 1
q1 q2 q3
We note however that the lower tank has no effect on what happens in
the upper tank. Instead of the word 'effect' Control engineers use the word
'loading'. Technically the way we would say this is that the two systems
are non interacting and that the lower tank does not load the upper tank.
Consider now the system below
h1
A1 Restriction
q1(t)
A2
qo(t)
h2
In this case the 2nd tank does have an effect on the 1st tank. We say
in this case the second tank loads the first tank and the two systems are
interacting. If we require the transfer function
Q3(s)
Q1(s)
then the system must
be considered as one unit. This type of system is called a 'Lumped System'
Modelling electromechanical systems
Consider the armature controlled DC servo motor as shown below.
The field current is constant and the voltage v(t) is applied to the arma-
ture which has a resistance Ra and negligible inductance. The effect of the
application of the input voltage v(t) will be to cause the armature to rotate.
The transfer function relating the voltage applied V (s) to the position of
the shaft θ(s) is found as follows.
56
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
v(t) vb(t)
J
D
θ(t)
ia Constant Field
Ra
Armature
First lets consider the mechanical aspects of the motor
Applying a voltage to the motor will cause a torque to be produced at
the shaft.
Torque
J
D
ω
∑
torques = J
dω
dt
Tm(t)−Dω(t) = J dω
dt
Tm(t) = J
dω
dt
+Dω(t)
Taking the Laplace and assuming zero initial conditions we have
Tm(s) = Jsω(s) +Dω(s)
Tm(s) = ω(s)[Js+D]
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
ω(s)
Tm(s)
=
1
Js+D
This is a first order system. However we are interested in position θ(s)
not angular velocity as given by ω(s)
Remember in terms of shaft position ω =
dθ
dt
Taking the Laplace of both sides we have
sθ(s) = ω(s)
θ(s) =
1
s
ω(s)
In terms of our mechanical system we can now write
θ(s)
Tm(s)
=
1
Js2 +Ds
This result is now a second order system.
Tm ωm(s) θ(s)1
s
1
Js+D
Tm θ(s)1
Js2 +Ds
Now considering the electrical aspects of the motor
58
CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
The Back e.m.f is proportional to the motor speed. Loading the motor
causes the speed to drop thus reducing vb a reduction in vb causes the more
current to pass through the armature thus increasing the torque.
ia =
vt − vb
Ra
Back e.m.f vb = Kbω(t)
Tm(t) = Kmia
Km or Kt is the torque constant of the motor and can be found on a
specification sheet for a motor.
In terms of an equation relating the applied voltage to the output speed
we have.
ia =
vt − vb
Ra
=
Tm(t)
Km
Substituting
vb = Kbω(t)
vt −Kbω(t)
Ra
=
Tm(t)
Km
Tm(t) =
[vt −Kbω(t)]Km
Ra
Tm(t) = J
dω
dt
+Dω(t)
[vt −Kbω(t)]Km
Ra
= J
dω
dt
+Dω(t)
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
vtKm
Ra
− KbKmω(t)
Ra
= J
dω
dt
+Dω(t)
vtKm
JRa
− KbKmω(t)
JRa
=
dω
dt
+
D
J
ω(t)
vtKm
JRa
=
dω
dt
+
D
J
ω(t) +
KbKmω(t)
JRa
Km
JRa
v(t) =
dω
dt
+ ω(t)
[D
J
+
KbKm
JRa
]
Let α =
[D
J
+
KbKm
JRa
]
and K =
Km
JRa
dω
dt
+ αω(t) = Kv(t)
This is a first order differential equation
Taking the Laplace transform of both sides and assuming zero initial
conditions we have
sω(s) + αω(s) = KV (s)
ω(s)[s+ α] = KV (s)
ω(s)
V (s)
=
K
[s+ α]
As before we require position as opposed to angular velocity. The result
of which is shown below
Note: to achieve position from velocity we need to integrate. In Laplace
domain we have already seen that multiplying by
1
s
is the same as inte-
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CHAPTER 2. MODELS OF PHYSICAL SYSTEMS N.Kent
V (s) ωm(s) θ(s)1
s
K
s+ α
θ(s)V (s) K
s2 + αs
grating in the time domain. This can be seen again in the block diagram
above. In the next section we will look at the power of block diagrams for
visualising systems.
61
Chapter 3
Block Diagrams and their Reduction
In the analysis of control systems it is very convenient to obtain the block
diagrams of the different components and their connections. If the various
components are non interacting it is possible to obtain the overall transfer
function of the system through a suitable combination of the transfer func-
tion of the component blocks utilising some basic rules of block diagram
transformations to reduce the original diagrams.
Lets look at a block diagram representation of the DC motor from the
previous Chapter
+
−
V (s)
Kb
1
Ra
1
s
1
Js+D
Km θ(s)
ia Tm ω(s)
Kbω(s)
V (s)−Kbω(s)
Now lets look at some reduction techniques for different configurations
of blocks.
62
CHAPTER 3. BLOCK DIAGRAMS AND THEIR REDUCTION N.Kent
3.1 Block diagram reductions
a Combining blocks in cascade (series)
G1 G2
x1 x2 x3
x1 x3
G1.G2
b Combining blocks in parallel
G1
G2
X Y
X Y
G1 +G2
+
+
63
CHAPTER 3. BLOCK DIAGRAMS AND THEIR REDUCTION N.Kent
c Moving a summing point behind a block
G
X2
Y
+
+
X1
G
X2
Y
+
+
X1
G
d Moving a pickoff point behind a block
G
YX
X
G
YX
X1
G
e Elimination of a feedback loop
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CHAPTER 3. BLOCK DIAGRAMS AND THEIR REDUCTION N.Kent
G
Y
X
H
+
−
G
1 +GH
Y
X
Example: Process Control.
G1 YX
H
+
−
G2 G3
Controller Actuator Process
Measurement
Exercises: Find the transfer function
C(s)
R(s)
of the following systems
G1 C(s)R(s)
H3
+
−
G2 G3
+
−
+
−
G4
H2
H1
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CHAPTER 3. BLOCK DIAGRAMS AND THEIR REDUCTION N.Kent
G1 C(s)R(s)
+
−
G2 G3
+
−
G4
H2
H1
+
−
G1 C(s)R(s)
+
−
G2 G3
+
−
G4
H4
H2
+
−
H3
−
H1
66
Chapter 4
The Performance of Control Systems
In general we do not know the inputs to a system design. Hence, for the
purposes of analysis it is customary to consider the the effects of the ap-
plication of a certain number of standard test inputs to the system hich
subject the system to sudden changes. The most common ones are:
ˆ A step input
ˆ A ramp input
ˆ A sinusoid (we will look at this later)
4.1 First Order Open Loop Systems
Lets remind ourselves what the response of a first order, open loop (no feed-
back) system is to (a) a step input and (b) a ramp input.
Consider the first order system as shown below.
We already have derived the transfer function
67
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
R
Cvin(t) vout(t)
Vo(s)
Vin(s)
=
1
τ
s+ 1
τ
Therefore we can write
Vo(s) = Vin(s).
1
τ
s+ 1
τ
For a step input
For a step input of magnitude V volts and using the method of residues
we can find vo(t)
vo(t) = V (1− e− tτ )
V
t
vo(t)
For a ramp input
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
L[vint] = 1
s2
Vo(s) = Vin(s).
1
τ
s+ 1
τ
Vo(s) =
1
s2
.
1
τ
s+ 1
τ
In terms of poles we have a multiple pole of order 2 at s = 0 and a
simple pole at s = − 1
τ
.
The residue of a multiple pole is given by
R = 1
(n− 1)! .
dn−1
dsn−1
[
F (s).(s+ α)2.est
]
s→−α
In this case
ˆ Vo(s) = F (s)
ˆ α = 0
ˆ n = 2
So we have
R1 = 1
(2− 1)! .
d
ds
[ 1
s2
.
1
τ
s+ 1
τ
.s2.est
]
s→0
R1 = d
ds
[ 1
τ
s+ 1
τ
.est
]
s→0
R1 = 1
τ
d
ds
[ est
s+ 1
τ
]
s→0
R1 = 1
τ
[(s+ 1
τ
).t.est − est.1
(s+ 1
τ
)2
]
s→0
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
R1 = 1
τ
[ 1
τ
.t.1− 1
1
τ
2
]
R1 = (t− τ)
For the residue at the simple pole
R2 =
[ 1
s2
.
1
τ
s+ 1
τ
.(s+ 1
τ
).est
]
s→− 1
τ
R2 =
[ 1
(− 1
τ
)2
.
1
τ
.
e−
t
τ
]
R2 = τ 2.1
τ
.e−
t
τ
R2 = τe− tτ
Now using the two residues
vo(t) = R1 +R2
vo(t) = (t− τ)− τe− tτ
t
v(t)
vin(t)
vo(t)
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
4.2 First Order Closed Loop Systems
Now lets look at the closed loop response to (a) a step input and (b) a ramp
input.
To keep in line with the conventions for a first order system we will re
write the transfer function as
1
1 + sτ
The block diagram for the system in closed loop form with a propor-
tional controller gain of K looks like
K
1
1 + sτ
+
−
R(s) C(s)
Controller
Closed Loop i.e Feedback is applied
The response to an input is found as before. However, our closed loop
transfer function (cltf) will be different than the open loop transfer func-
tion because we need to take the feedback into account. We use the block
diagram reduction rules to find our transfer function
C(s)
R(s)
Note in this case H(s) = 1 this is often the case and is called 'unity
feedback'
C(s)
R(s)
=
K
1 + sτ
1 +
K
1 + sτ
=
K
1 + sτ +K
C(s)
R(s)
=
K
τ
s+ 1+K
τ
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
C(s) = R(s).
K
τ
s+ 1+K
τ
For a step input
C(s) =
1
s
.
K
τ
s+ 1+K
τ
c(t) =
1
s
.
K
τ
s+ 1+K
τ
.s.ests→0 +
1
s
.
K
τ
s+ 1+K
τ
.(s+ 1+K
τ
).est
s→− 1+K
τ
c(t) =
K
1 +K
− K
1 +K
e−
1+K
τ
.t
c(t) =
K
1 +K
[1− e− 1+Kτ .t]
The final value of c(t) is when t→∞
c(t) =
K
1 +K
[1− 0] = K
1 +K
The closed loop response can be shown on the graph below
1
t
K
1 +K
Steady State Error
First Order
Closed Loop
Response
Note: in contrast to the open loop response the closed loop response
introduces a 'steady state error' into the system.
72
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
Steady State Error = Steady Input - Steady Output
In the case shown
Errorss = 1− K
1 +K
Errorss =
1
1 +K
We can observe two (related) things from this equation. The first is
that as we increase K the steady state error becomes smaller. The second
is that no matter how large we increase K by there will be some steady state
error. We mentioned previously that using proportional control introduced
an error that could not be eliminated by proportional control alone. This
is that error.
It is also worth noting that K is adjustable by the user and is called
the gain of the system thinking back to the proportional control K is the
factor we multiply the error by to get our system to reach its steady state
in the desired time or with an appropriate response.
Closed Loop Time Constant
In our closed loop transfer function (cltf)
c(t) =
K
1 +K
− K
1 +K
e−
1+K
τ
.t
we note that the term e−
1+K
τ
.t
is of the form e
− t
τcl
where τcl is the closed loop time constant.
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
Therefore
τcl =
τ
K + 1
Where τ is the open loop time constant and K is the gain as described
above.
Note that increasing K reduces the the closed loop time constant and
thus increases the system response.
Therefore since increasing K results in a reduction
of the steady state
error and also reduces the closed loop time constant we say that a high gain
is beneficial for for both transient and steady state performance.
More practical diagram
+
−
Desired Speed
Actual Speed
Motor and Load
ω
ω
t
K
1
1 + sτ
+
−
R(s) C(s)
ω
Power
Amplifier
Tachometer
Motor and Load
74
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
Relationship between Error and Input
Often we need to find what the steady state error is in a closed loop sys-
tem due to a given input. On way to find this error is to set t to infinity
thus finding the value of the output at the max possible time i.e. infinity.
We have applied this method in the last section. Now lets look at another
method.
Consider the relatively simple block diagram as shown below
G
+
−
R(s) C(s)εs
ε(s) = R(s)− C(s)
C(s) = ε(s).G(s)
ε(s) = R(s)− ε(s).G(s)
ε(s) + ε(s).G(s) = R(s)
ε(s)[1 +G(s)] = R(s)
ε(s)
R(s)
=
1
1 +G(s)
ε(s) =
1
1 +G(s)
.R(s)
Applying this technique to the previous problem we have
75
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
K
1
1 + sτ
+
−
R(s) C(s)ε(s)
Note: ε(s) is the error in Laplace domain while εss is the steady state
error.
In general we require the steady state error εss in the time domain. From
before we know this error to be
1
1 +K
for a step input into the system
ε(s)
R(s)
=
1
1 +G(s)
ε(s)
R(s)
=
1
1 +
K
1 + sτ
=
1 + sτ
1 + sτ +K
ε(s) = R(s).
1 + sτ
1 + sτ +K
For a step input R(s) =
1
s
ε(s) =
1
s
.
1 + sτ
1 + sτ +K
Now lets introduce whats called the Final Value Theroem. This theo-
rem simply states that the final (time based) value (or final f(t))of a Laplace
based function F (s) is given by
f(t)︸︷︷︸
t→∞
= F (s).s︸ ︷︷ ︸
s→0
Since we need to find the final value of
1
s
.
1 + sτ
1 + sτ +K
in order to find
the steady state error we can apply the final value theorem. This gives us
76
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
[εss(t)]t→∞ =
[1
s
.
1 + sτ
1 + sτ +K
.s
]
s→0
[εss(t)]t→∞ =
1
1 +K
This solution is as we had before. While this example is relatively simple
the final value theorem can be applied to situations where the solution may
not be so obvious.
Removal of steady state error
If the steady state error must be removed we can add integral action
to the proportional control. Remember earlier we said that integral action
simply 'adds up' the steady state error over a defined time and adjusts the
controller appropriately to eliminate the error.
In Laplace integration is performed my multiplying by
1
s
It is common
to be able to change the integral gain (or integral action time) just as the
proportional gain can be changed for a given system.
Our integrator block diagram therefore looks as follows
Ki
s
ε(s)
Where Ki is the integral gain. The modified block diagram therefore
looks as follows
Which can be reduced using the block diagram reduction rules to
77
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
K
1
1 + sτ
+
−
R(s) C(s)ε(s)
Ki
s
ε(s)
+
+
K +
Ki
s
1
1 + sτ
+
−
R(s) C(s)ε(s)
Now to find the error for this system
ε(s)
R(s)
=
1
1 +G(s)
ε(s)
R(s)
=
1
1 +
(
K +
Ki
s
)( 1
1 + sτ
)
ε(s)
R(s)
=
s
s+
(
sK +Ki
)( 1
1 + sτ
)
ε(s)
R(s)
=
s(1 + sτ)
s(1 + sτ) + (sK +Ki)
ε(s) = R(s).
s(1 + sτ)
s(1 + sτ) + (sK +Ki)
For a step input
ε(s) =
1
s
.
s(1 + sτ)
s(1 + sτ) + (sK +Ki)
Applying the final value theorem
[εss(t)]t→∞ =
[1
s
.
s(1 + sτ)
s(1 + sτ) + (sK +Ki)
.s
]
s→0
78
CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
[εss(t)]t→∞ = 0
i.e. There is no steady state error.
Note: The system with proportional control was first order. The system
with integral control is second order. Second order system have different
dynamic behaviour than first order and we will look at that later.
Ramp Input
Now lets look at the response of a closed loop first order system to a ramp
input.
K
1
1 + sτ
+
−
R(s) C(s)
Controller
r(t) = t
R(s) =
1
s2
We have already determined the closed loop transfer function for a first
order system and have written it as follows
C(s)
R(s)
=
K
τ
s+ 1+K
τ
In terms of output we can write
C(s) = R(s).
K
τ
s+ 1+K
τ
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
In this case R(s) =
1
s2
so we can write
C(s) =
1
s2
.
K
τ
s+ 1+K
τ
We require c(t)
Using the residue method
R1 = d
ds
[ 1
s2
.
K
τ
s+ 1+K
τ
.s2.est
]
s→0
R1 = K
τ
.
d
ds
[ est
s+ 1+K
τ
]
s→0
R1 = K
τ
[(s+ 1+K
τ
).t.est − est.1
(s+ 1+K
τ
)2
]
s→0
R1 = K
τ
[(1+K
τ
).t− 1
(1+K
τ
)2
]
R1 = K.τ
2
(1 +K)2τ
[1 +K
τ
.t− 1
]
R1 = K
(1 +K)2
.τ
[1 +K
τ
.t− 1
]
R1 = K
(1 +K)2
.τ.
1 +K
τ
.t− K
(1 +K)2
.τ
R1 = K
1 +K
.t− Kτ
(1 +K)2
In the case of Residue 2
R2 =
[ 1
s2
.
K
τ
s+ 1+K
τ
.(s+ 1+K
τ
).est
]
s→− 1+K
τ
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
R2 =
[ 1
(−1+K
τ
)2
.
K
τ
.e−
1+K
τ
t
]
R2 =
[ τ 2
(1 +K)2
.
K
τ
.e−
1+K
τ
t
]
R2 =
[ K.τ
(1 +K)2
.e−
1+K
τ
t
]
c(t) = R1 +R2
c(t) =
K
1 +K
.t− Kτ
(1 +K)2
+
K.τ
(1 +K)2
.e−
1+K
τ
t
t
v(t)
vin(t)
vo(t)
As t increases the third term will go to zero
Since the error ε(t) = r(t)− c(t) we have
ε(t) = t− K
1 +K
.t− Kτ
(1 +K)2
The equation above shows that the error will increases with time and
approach infinity as t approaches infinity.
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
Effect of Disturbance Signals
Many control systems are subject to unwanted disturbance signals. For
example, gust of wind affecting radar, sudden load on a motor shaft, spike
in current affecting temperature control systems.
A block diagram for a disturbance is shown below. In this case the
disturbance is at the output.
K G(s)
+
−
R(s) C(s)+
+
N(s)
H(s)
In order to fins the output C(s) superposition is used to find the contri-
bution to inputs R(s) and N(s).
The theorem of superposition states that that each input can be taken
alone and the value of output can be evaluated due to this input. The total
output is is the sum of the outputs for each input acting alone.
Applying superposition to the system shown we have
Considering N(s) acting alone
+
−
C(s)N(s)
K G(s) H(s)
C ′(S)
N(s)
=
1
1 +KG(s)H(s)
C ′(S) = N(s).
1
1 +KG(s)H(s)
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
Now considering R(s) acting alone
K G(s)
+
−
R(s) C(s)
H(s)
C ′′(S)
R(s)
=
KG(s)
1 +KG(s)H(s)
C ′′(S) = R(s).
KG(s)
1 +KG(s)H(s)
Now the total C(s)
C(s) = N(s).
1
1 +KG(s)H(s)
+R(s).
KG(s)
1 +KG(s)H(s)
As can be seen from the above equation the effect of the noise is reduced
by the ratio
1
1 +KG(s)H(s)
over the range of frequencies of interest. Therefore by increasing the
open loop gain we can alleviate the effect of disturbances occurring at the
output level.
Example: Speed Control System
Open loop motor controller
K
1
1 + sτ +
−
ωd
ωin ωout
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CHAPTER 4. THE PERFORMANCE OF CONTROL SYSTEMS N.Kent
If a disturbance occurs then teh controller does not react and the new
output speed is ωo − ωd
Closed loop motor controller
K
1
1 + sτ
+
−
R(s) ωo(s)
+
−
ωd(s)
In terms of superposition and looking at the output from the noise only