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Unit I
Introduction to Control Systems
In this lecture, we lead you through a study of the basics of control system.
After completing the chapter, you should be able to
Describe a general process for designing a control system.
Understand the purpose of control engineering
Examine examples of control systems
Understand the principles of modern control engineering.
Realize few design examples.
Textbook
1. Richard C. Dorf and Robert H. Bishop, Modern Control Systems, Prentice
Hall, 2001.
1.1 INTRODUCTION
Control engineering is based on the foundations of feedback theory and
linear system analysis, and it generates the concepts of network theory and
communication theory. Accordingly, control engineering is not limited to any
engineering discipline but is applicable to aeronautical, chemical, mechanical,
environmental, civil, and electrical engineering.
A control system is an interconnection of components forming a system
configuration that will provide a desired system response. The basis for analysis
of a system is the foundation provided by linear system, which assumes a cause-
effect relationship for the components of a system. A component or process to be
controlled can be represented by a block as shown in Figure 1.
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Input Process Output
Figure 1 Process under control
An open-loop control system utilizes a controller or control actuator to obtain
the desired response as shown in Figure 2. The open-loop control system utilizes
an actuating device to control the process directly without using device. An
example of an open-loop control system is an electric toaster.
Output
response
Actuating
device
Process
Output
Figure 2 Open-loop control system (no feedback)
A closed-loop control system (Figure 3) utilizes an additional measure of the
actual output to compare the actual output with the desired output response. The
measure of the output is called the feedback signal. A feedback control system is
a control system that tends to maintain a relationship of one system variable to
another by comparing functions of these variables and using the difference as a
means of control. As the system is becoming more complex, the interrelationship
of many controlled variables may be considered in the control scheme. An
example of closed-loop control system is a person steering an automobile by
looking at the auto’s location on the road and making the appropriate
adjustments.Sky
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Desired
Output
Response
Difference or Actuating Error
Controller Process
Actual
Output
Response
Measurement Device
Figure 3 Closed-loop feedback system.
1.2 TEMPERATURE CONTROL SYSTEMS
Figure 4 shows a diagram of temperature control of an electric furnace. The
temperature in the electric furnace is measured by a thermometer, which is an
analog device. The analog temperature is converted to a digital temperature by an
A/D converter. The digital temperature is fed to a controller through an interface.
This digital temperature is compared with the programmed input temperature,
and if there is any error, the controller sends out a signal to the heater, through an
interface, amplifier, and relay, to bring the furnace temperature to a desired value.
Thermometer
A/D
Converter Interface
I
N
P
U
TRelay Amplifier Interface
Figure 4 Temperature control system.
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1.3 CONTROL SYSTEM DESIGN
The following table shows the control system design process.
1. Establish control goals
2. Identify the variables to control
3. Write the specifications for the
variables
4. Establish the system configuration
and identify the actuators
5. Obtain a model of the process, the
actuator, and the sensor
6. Describe a controller and select key
parameters to be adjusted
7. Optimize the parameters and
analyze the performance
• Variables to control are the quantities or conditions that are measured
and controlled.
• Process is a natural, progressively continuing operation marked by a
series of gradual changes that succeed one another in a relatively
fixed way and lead toward certain result or end.
• A system is a combination of components that act together and
perform a certain objective.
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1.4 DESIGN EXAMPLE: TURNABLE SPEED CONTROL
Battery Turntable
DC Amplifier DC Motor
Desired
Speed
Control
Device
(Amplifier)
Actuator
(DC Motor)
Process
(Turntable)
Actual
Speed
Figure 5 Open-loop control of speed of a turntable and a block diagram model.
Battery Turntable
DC Amplifier DC Motor
Tachometer
Figure 6 Closed-loop control of the speed of a turntable.Skyup
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1.5 DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
Desired
Head
Position
Error
Control
Device
Actuator and
Read Arm
Actual
Head
Position
Sensor
Figure 7 Closed-loop control system for disk drive.
A hard disk uses round, flat disks called platters, coated on both sides with a
special media material designed to store information in the form of magnetic
patterns. The platters are mounted by cutting a hole in the center and stacking
them onto a spindle. The platters rotate at high speed, driven by a special spindle
motor connected to the spindle. Special electromagnetic read/write devices called
heads are mounted onto sliders and used to either record information onto the
disk or read information from it. The sliders are mounted onto arms, all of which
are mechanically connected into a single assembly and positioned over the
surface of the disk by a device called an actuator. A logic board controls the
activity of the other components and communicates with the rest of the computer.
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Figure 8 A hard disk.
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Figure 9 Components of a hard disk
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1.6 FEEDBACK CONTROL OF AN ANTIAIRCRAFT GUN
Gun Azimuth (Elevation)
Demanded Azimuth (Elevation)
Control
System
Gun
Dynamics
Figure 10 Feedback control of an antiaircraft system.
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Exercises
E1.1
A precise optical signal source can control the output power level to within 1%.
A laser is controlled by an input current to yield the output power. A
microprocessor controls the input current to the laser. The microprocessor
compares the desired power level with a measured signal proportional to the laser
power output obtained from a sensor. Draw the block diagram representing the
closed-loop control system.
Desired
Power output
Error Micro
processor
Current Laser
(Process)
Output power
Measured
power
Sensor
(Measurement)
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E1.6
Automated highways may be prevalent in the next decade. Consider two
automated highway lanes merging into a single lane, and describe a control
system that ensures that the vehicle merge with a prescribed gap between two
vehicles.
Desired
gap
Error
Computer (Controller)
Brakes
Steering
Active vehicle
Actual gap
Measured gap
Sensor (Radar)
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Problems
P1.1
Many luxury automobiles have thermostatically controlled air-conditioning
systems for the comfort of the passengers. Sketch a block diagram of an air-
conditioning system where the driver sets the desired interior temperature on a
dashboard panel.
Desired
temperature
Error Thermostat
(Controller) Automobile
cabin
Cabin
temperature
Measured
temperature
Sensor (Measurement)
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P1.10
The role of air traffic control systems is increasing as airplane traffic increases at
busy airports. Engineers are developing air traffic control systems and collision
avoidance systems using the Global Positioning System (GPS) navigation
satellites. GPS allows each aircraft to know its position in the airspace landing
corridor very precisely. Sketch a block diagram depicting how an air traffic
controller might utilize GPS for aircraft collision avoidance.
Desired flight
Path from
traffic
controller
Autopilot (controller)
Allerons,
elevators, rubber, and
Engine
power
Aircraft
(Process)
Flight path
Measured flight path Global Positioning System
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P1.21
The potential of employing two or more helicopters for transporting payloads that
are too heavy for a single helicopter is a well-addressed issue in the civil and
military rotorcraft design arenas. A case of a multilift arrangement wherein two
helicopters jointly transport payloads has been named twin lift as shown in the
following figure. Develop the block diagram describing the pilots’ action, the
position of each helicopter, and the position of the load.
1 2
Load
Desired separation
distance
Measured separation
distance
Radar
(Measurement)
Separation
distance
Pilot Helicopter
Desired altitude
Measured altitude
Altimeter (Measurement)
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Design Problems
DP1.2
Many cars are fitted with cruise control that, at the press of a button,
automatically maintains a set speed. In this way, the driver can cruise at a speed
limit or economic speed without continually checking the speedmeter. Design a
feedback control in block diagram for a cruise control system.
Controller
1/k
Desired Speed
of auto
set by
driver
Desired
Shaft
speed
Electric motor Valve
Auto/
Engine
k
Drive shaft
speed
Measured
Shaft speed
INSTRUMENTATION AND CONTROL
TUTORIAL 1 – CREATING MODELS OF ENGINEERING SYSTEMS
This tutorial is of interest to any student studying control systems and in particular the EC
module D227 – Control System Engineering. The purpose of this tutorial is to introduce
students to the basic elements of engineering systems and how to create a transfer function
for them. The tutorial is mainly informative and consists of examples showing the derivation
of models for real hardware systems. The self assessment material is based on basic general
engineering knowledge.
On completion of this tutorial, you should be able to do the following.
• Derive the mathematical models of basic mechanical systems.
• Derive the mathematical models of basic fluid power systems.
• Derive the mathematical models of basic thermal systems.
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• Derive the mathematical models of basic electrical systems.
• Recognise the similarity between models of different systems.
• Explain the standard first and second order transfer functions.
• Explain the link between open and closed loop transfer functions.
If you are not familiar with instrumentation used in control engineering, you should complete
the tutorials on Instrumentation Systems.
In order to complete this tutorial, you must be familiar with basic mechanical and electrical
science. You should also be familiar with the Laplace transform and a tutorial on this may be
found in the maths section. You can also find tutorials on fluid power on this site.
Tutorial 2 in this series gives a detailed account of electric motor models and you may wish
to study this first.
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1. INTRODUCTION
Engineering systems is a very broad area ranging from control of a power station to control of a motor’s
speed. The student needs to have a broad base knowledge of engineering science in order to understand the
various elements and see how many of them are mathematically the same (analogues of each other).
Different kinds of engineering systems often conform to similar laws and there are clear analogies between
electrical, mechanical, thermal and fluid systems. The basic laws which we use most often concern
Resistance R, Capacitance C, Inductance L and conservation laws. You do not need to study all these in
detail and the appropriate law will be explained as required.
Here is a table showing the main analogue components. It is useful to note that capacitance is a zero order
differential equation, resistance is a first order differential equation and Inductance/inertia/inertance is a
second order differential equation.
MECHANICAL ELECTRICAL THERMAL FLUID
Spring
x = C F = (1/k) F
Electrical Capacitor
Q = C V
Thermal capacitor
Q = C ∆T
Fluid Capacitor
M = C x ∆p
Damper
Force = kd x velocity F = kd dx/dt Torque = kd x Ang.vel
Ohm's Law
V = R I
V = R dQ/dt
Heat Transfer Laws
∆T = R Φ ∆T = R dQ/dt
Fluid friction Laws do
not conform to this
pattern.
Newton's 2nd Law of
motion
Force = Mass x acceleration
F = M d2x/dt2
Law of Inductors
V = L d2q/dt2
No equivalence Fluid inertance
∆p = Ld2v/dt2
D'Alembert's Principles
∑Force = 0 ∑Moment = 0
Kirchoff's Laws
∑current = 0
Law of Conservation
of Energy
∑Energy = constant
Law of Conservation
of Mass
∑Mass = constant
Let’s look at the similarity of the various quantities used in these systems.
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2. SIMILARITY OF ELEMENTS
CAPACITANCE
The symbol C will be used for electrical, thermal and fluid capacitance. Mechanical capacitance is equivalent
to 1/k for mechanical systems where k is the spring stiffness.
RESISTANCE
The symbol R will be used for electrical and thermal resistance. Mechanical/hydraulic resistance is called the
damping coefficient and has various symbols.
INDUCTANCE / INERTIA / INERTANCE
The symbol L will be used for electrical inductance and fluid inertance. In mechanical systems, mass M is
the equivalent property for linear motion and moment of inertia I for angular motion.
OTHER EQUIVALENT PROPERTIES
Q is the symbol for electric charge and quantity of heat. This is equivalent to displacement in mechanical
systems, these being distance (usually x) or angle (usually θ).
V is the symbol for electric voltage (potential difference or e.m.f) and is equivalent to temperature T for thermal systems, Force F for mechanical systems and pressure p for fluid systems.
v or u is the symbol for velocity in mechanical systems and this is equivalent to electric current (I or i) and
heat flow rate Φ.
3 LAPLACE TRANSFOR,M AND TRANSFORMATIONS
Laplace is covered in detail in later tutorials and in the maths section. The purpose of this transform is to
allow differential equations to be converted into a normal algebraic equation in which the quantity s is just a
normal algebraic quantity. In this tutorial we should simply regard it as a shorthand method of writing
differential coefficients such that:
dθ becomes s θ
d θ becomes s
2 θ d θ
becomes s n
θ
dt dt 2
dt n
4 TRANSFER FUNCTIONS
The models of systems are often written in the form of a ratio of Output/Input. If the models are turned into a
function of s it is called a transfer function and this is usually denoted as G(s).
G(s) = Output
Input
The output and input are functions of s.
Now let’s examine the mathematical models of some mechanical systems.
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5. BASIC MODELS OF MECHANICAL SYSTEMS
5.1 GENERAL PROCEDURE
The general procedure for mechanical systems is as follows.
i. Adopt a suitable co-ordinate system with an appropriate sign convention. For linear motion, up is
positive and left to right is positive. For rotation anticlockwise is positive and clockwise is negative.
These may be ignored when convenient.
ii. Identify any disturbing forces acting on the system (inputs to the system).
iii. Identify displacements and/or velocities (outputs from the system).
iv. Draw a free body diagram for each mass showing all the forces and moments acting on it.
v. Apply Newton's 2nd Law to each free body diagram (F = Mass x Acceleration).
vi. Rearrange the equation(s) into a suitable form for solution by a convenient method.
Note that unless otherwise specified, ignore gravitational effects.
Let’s now examine mechanical elements in detail.
5.2 LINEAR MECHANICAL SYSTEMS.
5.2.1 SPRING
The basic law of a mechanical spring is Force ∝ change in length. The diagram shows the model with
mechanical symbols and as a block diagram.
Figure 1
The relationship has no derivatives in it may be written as a function of t or s with no transform involved.
As a function of time we write F(t) = k x(t) where k is the spring stiffness.
As a function of s we write F(s) = kx(s)
This can be arranged as a transfer function such that x
(s) F
= 1/k = C
C is the reciprocal of stiffness and it is called mechanical capacitance. The use of k is usually preferred in
mechanics but C is used in systems as it is directly analogous to electrical capacitance.
5.2.2 DAMPER or DASHPOT
Figure 2
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A damper may be idealised as a loosely fitting piston moving in a viscous fluid such that the force is directly
proportional to velocity. F ∝ v. Velocity v is the first derivative of distance so F ∝ dx/dt
The basic law of a dashpot is: F(t) = k dx
d dt
kd is the damping coefficient.
Changed into Laplace form. F = kd s x
Rearranged into a transfer function x
(s) = F
1
k ds
kd is the damping coefficient with units of Force/Velocity or N s/m. The diagram shows the model with
mechanical symbols and the control block.
5.2.3 MASS
When a mass is accelerated, the inertia has to be overcome and the inertia force is given by Newton’s Second
Law of Motion Force = Mass x Acceleration. Acceleration is the second derivative of x with time.
d 2 x
Basic Law F(t) =
M dt
2
Changed into Laplace form. F = Ms2 x
x 1 Rearranged into a transfer function (s) =
F
Ms 2
5.2.4 MASS - SPRING SYSTEM
Figure 3
For this spring - mass system, motion only occurs in one direction so the system has a single degree of
freedom. It is normal for the direction of motion to be expressed as the x direction regardless of the actual
direction. The free body diagram is as shown. The input is a disturbing force F which is a function of time
F(t). This could, for example, be a sinusoidal force. The output is a motion x which is a function of time x(t).
Let x be positive upwards.
Figure 4
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2 =
( )
The input force is opposed by the spring force and the inertia force (which always opposes changes in the
motion as stated in Newton’s third law of motion).
Spring force = k x
Inertia force = M d2x/dt2
D'Alembert's Principle is that all the forces and moments on the body must add up to zero. In this case it
means
F(t) - kx(t) - M d2x/dt2(t) = 0
or F(t) = M d2x/dt2(t) + kx (t)
Changing to a function of s we have F(s) = Ms2 x + kx = x [Ms2 + k]
x(s) = F F(1/M )
2 Ms
This may be shown as a transfer function. G(s) =
+ k x(s) =
s +
1/M
k/M
The block diagram for use in systems is as shown.
F(s) s 2 + k/M
Figure 5
5.2.5 SPRING DAMPER
Force balance as a function of time. F(t) = k x + kd dx/dt Force balance as a function of s F(s) = k x + kd s x
Rearrange into a transfer function. x
(s) = F
1/k
k /k s + 1 d
The units of kd/k are seconds and this is the time constant for the system T = kd/k x
(s) = 1/k .This is the standard first order equation which we shall study many times in these tutorials.
F Ts + 1
Figure 6
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5.2.6 MASS -SPRING - DAMPER SYSTEM
The input is the force F and the output is the movement x, both being functions of time.
Spring force Fs = kx
Damping force Fd = kd dx/dt Inertia force Fi = Md2x/dt2
The three forces oppose motion so if the total force on the system is zero then F = Fi + Fd + Fs 2
F(t) =
d x M
dt 2
dx + k
d dt + kx
F(s) =
Ms
2 x + k
d sx
+ kx
G(s) =
x (s) =
F
1/k
s (M/k) + s(k d /k) + 1
If we examine the units of (M/k)1/2 we find it is seconds and this is the second order time constant also with
the symbol T. The transfer function may be written as
G(s) =
x (s) =
F
T 2s 2
1/k
+ 2 δ Ts + 1
δ is the damping ratio defined as δ = kd/Cc and Cc is the critical damping ratio defined as (4Mk)½. The
k 2k M/k 2k M k
term 2δT is hence 2 d T = d = d = d and so the forgoing is correct. Cc 4Mk 2 M k k k
Figure 7
This is the standard 2nd order transfer function which will be analysed in detail later.
WORKED EXAMPLE No.1
A mass – spring –system has the following parameters.
Stiffness K = 800 N/m Mass M = 3 kg Damping Coefficient kd = 20 Ns/m
i. Calculate the time constant, critical damping coefficient and the damping ratio.
ii. Derive the equation for the force required when the piston is accelerating.
iii. Use the equation to evaluate the static deflection when F = 12 N.
iv. Use the equation to evaluate the force needed to make the mass accelerate at 4 m/s2
at the
moment when the velocity is 0.5 m/s.
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SOLUTION
i. T = √(M/k) = √(3/800) = 0.0612 seconds
cc = √4MK = √(4 x 3 x 800) = 97.97 Ns/m
δ = kd/ cc = 20/97.97 = 0.204
ii. For a constant acceleration s
2x = a (acceleration) and sx = v (velocity)
x (s) =
F
T 2s
2
1/k
+ 2 δ Ts + 1 F = kx(T
2s
2
+ 2 δ Ts + 1)
F = 800x (0.06122
s 2 + 2 x 0.204 x 0.0612s + 1 )
F = x(3 s 2 + 20 s + 800) = 0.00374s 2 x + 20 sx + 800 x
F = 3 a + 20 v + 800 x
iii. For a constant force and a static position there will be neither velocity nor acceleration so the s
and s2
terms are zero. F
= 800 x
x = 12
800
= 0.015 m or 15mm
iv. For velocity = 0.5 m/s and a = 4 m/s2
F = 3 a + 20 v + 800 x = 12 + 10 + 800x = 22 + 800 x
The deflection x would need to be evaluated from other methods x = v2/2a = 0.031 m
F = 46.8 N
SELF ASSESSMENT EXERCISE No.1
1. A mass – spring –system has the following parameters.
Stiffness K = 1200 N/m Mass M = 15 kg Damping Coefficient kd = 120 Ns/m
i. Calculate the time constant, critical damping coefficient and the damping ratio.
(0.112 s, 268.3 Ns/m and 0.447)
ii. If a constant force of 22 N is applied, what will be the static position of the mass?
(18 mm)
iii. Calculate the force needed to make the mass move with a constant acceleration of 12 m/s2
at the
point where the velocity is 1.2 m/s.
(396 N)
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5.3 ROTARY MECHANICAL SYSTEMS
The following is the rotary equivalent of the previous work.
5.3.1 TORSION BAR
This is the equivalent of a mass and spring. A metal rod clamped at one end and twisted at the other end
produces a torque opposing the twisting directly proportional to the angle of twist. The ratio T/θ is the
torsion stiffness of the torsion spring and is denoted with a k. T is torque ( N m)
θ is the angle of twist (radian)
k is the torsional stiffness ( N m/rad )
Balancing the torques we have T(t) = kθ (t) Change to Laplace form. T(s) = kθ (s)
Write as a transfer function. θ
(s) = 1
T k
Figure 8
5.3.2 TORSION DAMPER
A torsion damper may be idealised as vanes rotating in a viscous fluid so that the torque required to rotate it
is directly proportional to the angular velocity. kd is the torsion damping coefficient in N m s/radian
T(t) dθ
= k d dt
T(s) = k d s θ
G(s) =
θ
T (s) =
1
k d s
Figure 9
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5.3.3 MOMENT OF INERTIA
Rotating masses oppose changes to the motion and Newton's 2nd law for rotating masses is T = I d2θ/dt2 I is the moment of inertia in kg m2.
2
T(t) = I d θ
T(s) = I s 2θ
G(s) = θ
(s) = 1 . Note many text books also use J for moment of inertia.
dt 2 T Is2
Figure 10
SELF ASSESSMENT EXERCISE No.2
Derive the transfer function for a mass on a torsion bar fitted with a damper and show it is another
example of the second order transfer function. T is torque and J is moment of inertia.
G(s) =
θ (s) =
T
(J/k)s2
1/k
+ (Jk d
= /k)s + 1
T 2s
2
1/k
+ 2 δ Ts + 1
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5.3.4 GEARED SYSTEMS
When a mass is rotated through a gear system, the affect of the inertia is dramatically altered. Consider a
motor coupled to a load through a speed changing device such as a gear box. There is damping (viscous
friction) on the two bearings.
Figure 11
θm is the motor rotation and θo the output rotation. The gear ratio is Gr = θo/θm
Since this is a fixed number and is not a function of time, the speed and acceleration are also in the same ratio.
dθm/dt = ωm dθo/dt = ωo
Gr = ωo/ωm ω is the angular velocity
d2θm/dt2= αm d2θo/dt2 = αo Gr = αo/αm α is the angular acceleration.
The power transmitted by a shaft is given by Power = ωT. If there is no power lost, the output and input
power must be equal so it follows that ωm Tm = ωo To hence
Tm = ωo To /ωm = GrTo
(In reality friction significantly affects the torque)
Consider the inertia torque due the inertia on the output shaft Io.
To = Ioαo = Io αm x Gr Tm = To x Gr = Io αm x Gr2
Now consider the damping torque on the output shaft.
To = kdo ωo = kdo ωmGr Tm = To x Gr = kdo ωm Gr2
Now consider that there is an inertia and damping torque on the motor shaft and on the output shaft. The total torque produced on the motor shaft is
Tm = Imαm + kdm ωm + Gr To
To = Ioαo + kdo ωo Tm = Imαm + kdm ωm + Gr Ioαo + kdo ωo
Tm = Imαm + kdm ωm + Gr
2 Io αm + Gr2 kdo ωm
Tm = αm (Im +Gr
2 Io) + ωm (kdm +Gr2 kdo)
(Im +Gr2 Io) is the effective moment of inertia Ie and (kdm +Gr
2 kdo) is the effective damping coefficient kde. The equation may be written as Tm = αm (Ie) + ωm (kde)
In calculus form this becomes Tm = d2θ/dt2 (Ie) + dθ/dt (kde)
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Changing this into a function of s we have Tm(s) = s2θ(Ie) + sθ (kde) = sθsIe + kde
The output is the motor angle and the input is the motor torque so the geared system may be presented as a transfer function thus. θ (s)/Tm (s) = (1/Ie)/ss + Kde/Ie
Figure 12
WORKED EXAMPLE No.2
A DC Servo motor has a moment of inertia of 0.5 kg m
2. It is coupled to an aerial rotator through a gear
reduction ratio of 10. The driven mass has a moment of inertia of 1.2 kg m2. The damping on the motor
is 0.1 N m s/rad and on the rotator bearings it is 0.05 N m s/rad. Write down the transfer function θ/Tm in the simplest form. Calculate the torque required from the motor to
i. Turn the aerial at a constant rate of 0.02 rad/s.
ii. Accelerate the rotator at 0.005 rad/s2
at the start when ω = 0
SOLUTION
i. Ie = (Im +Gr2 Io) = (0.5 + 10
2 x 1.2) = 120.5 kg m
2.
Kde = (kdm +Gr2 kdo) = (0.1 + 10
2 x 0.05) = 5.1 N m s/rad.
θ (s)/Tm (s) = (1/Ie)/ss + Kde/Ie θ 1/Ie
(s) = Tm s(s +
k de /Ie )
Tm = Ie α + K de
ω
Tm = 120.5α + 5.1ω
If the rotator is moving at constant speed α (acceleration) is zero. Hence: Tm = 5.1ω = 5.1 x 0.02 = 0.102 Nm
ii. When accelerating at 0.005 rad/s2
the motor acceleration is 10 times larger at 0.05 rad/s2.
Tm = 120.5α + 5.1ω = 60.25 Nm when ω = 0
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SELF ASSESSMENT EXERCISE No.3
A DC Servo motor has a moment of inertia of 12 kg m
2. It is coupled to an aerial rotator through a gear
reduction ratio of 4. The driven mass has a moment of inertia of 15 kg m2. The damping on the motor is
0.2 N m s/rad and on the rotator bearings it is 0.4 N m s/rad.
Calculate the torque required from the motor to
i. Turn the aerial at a constant rate of 0.5 rad/s. (3.3 N)
ii. Accelerate the rotator at 0.02 rad/s2
at the start when ω = 0 (20.16 Nm)
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6. THERMAL SYSTEM MODELS
6.1 HEATING and COOLING
Consider a mass M kg at temperature θ1. The mass is placed in a hot environment at temperature θ2 and heat
Q is transferred into the mass causing its temperature to rise. The system could be for example, a resistance thermometer, and we want to know how long it takes for the sensor to warm up to the same temperature as
the liquid.
Figure 13
The laws of heat transfer tell us that the temperature rise is directly proportional to the heat added so:
dQ = Mc dθ1 = C dθ1
c is the specific heat capacity. C = Mc is the thermal capacitance in Joules/Kelvin.
Divide both sides by dt and: dQ
= Φ dt
dθ = C 1
dt
The rate of heat transfer into the mass is Φ = C dθ1/dt and the rate is governed by the thermal resistance
between the liquid and the mass. This obeys a law similar to Ohm’s Law so that:
Φ = (θ2 - θ1)/R R is the thermal resistance in Kelvin per Watt.
dθ θ − θ dθ θ − θ
Equating for Φ we have
C 1 = dt
dθ θ
1 2
R
θ 2
1 = dt
1 2
RC
1 + 1 = dt RC RC
In all systems, the product of the resistance and capacitance is the time constant T so we have
dθ θ 1 + 1
dt T
= θ
2
T Changing from a function of time into a function of s we have
θ sθ + 1
1
= θ 2
θ (Ts 1
+ 1) = θ 2
θ 1
1 (s) = +
T T θ 2
(Ts 1)
Figure 14
Note that this transfer function is the same standard first order equations derived for the spring - damper
system and thermal capacitance C is equivalent to 1/k and resistance R is equivalent to kd.
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6.2 INDUSTRIAL HEATING SYSTEM
The diagram shows a schematic of an industrial process for controlling the temperature of a tank of liquid.
The pneumatic controller will not be explained here but it has an input temperature set by adjustment of the
control. The temperature of the liquid is measured with a suitable device and turned into a standard signal in
the range 0.2 – 1 bar. This is connected to the controller and pressure sensing devices produce another air
signal (0.2 – 1 bar) depending on the error. This is sent to a valve that is opened pneumatically working on
the standard range. The overall result is that if the liquid is too cool, steam is allowed through to heat the
liquid.
If the valve opened instead of closing and a cooling fluid was used instead of steam, the system control is by
cooling.
The control equipment could just as likely be all electronic. Pneumatics are used in dangerous environments
such as heating up oil tanks.
Figure 15
The model for the above system will not be derived here but it will be more complicated than simply θ
1 (s) = θ 2
(Ts
1
+ 1)
because the controller has the facility to do more than proportional control. (Three term
control is covered in later tutorials)
WORKED EXAMPLE No.3
A simple thermal heating system has a transfer function
θo (s) = θ
i
(Ts
1
+ 1)
The temperature of the system at any time is θo and this is at 20oC when the set temperature θi is
changed from 20 oC to 100
oC. The time constant ‘T’ is 4 seconds. Deduce the formulae for how the
system temperature changes with time and sketch the graph.
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SOLUTION
θo = θ
i
1
Ts + 1 θ
i =
Tsθo + θo
θi is a constant (100oC) at all values of time after t = 0 (the start of the change).
θi
− θo = Ts θ o
= T dθ o
dt
Let θi
− θ o = x
Differentiate and - dθ o = dx
The equation becomes x = T dθ o
dt
= −T dx
dt
Rearrange and − dt
= dx
T x
Integrate without limits − t
= T
ln(x) + A
Substitute for x − t
= T
ln(θi - θo ) + A
When t = 0, θo = starting temperature = θ1 Hence
− t
= 0 T
= ln(θi
- θ ) + A 1
A = - ln(θi
- θ ) 1
θi - θ1 = change in temperature ∆θ
Substitute for A and − t
T t
= ln(θi
(θ - θ
- θ o )
)
− ln(∆ θ ) = ln
(θi - θ o )
∆θ
− Take anti logs and e T
− t
= i o
∆θ
∆θ e T = (θi
- θo )
− t
θo = θi
− ∆θ e T
Put in the values T = 4 ∆θ = 100 – 20 = 80 θi = 100
θo = 100 – 80e-t/4
Evaluating and plotting produces the result below. It is an exponential growth.
Figure 16
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7. HYDRAULIC SYSTEM MODELS
The basic theory for hydraulic and pneumatic components may be found in the tutorials on fluid power.
7.1 HYDRAULIC MOTOR
The following is the derivation of a model for use in control
theory. The formula relating flow rate Q and speed of rotation ω
is Q = k q ω = k dθ
q dt
kq is a constant known as the nominal displacement with units
of m3
per radian. θ is the angle of rotation in radian. Written as
a function of s this becomes Q = kq sθ
If we take the flow rate as the input and the angle of rotation as
the output the transfer function is: G(s) = θ
Q =
1
k q s
The formula that relates system pressure p to the output torque
T is T = kq p
If pressure is the input and torque the output then
T Figure 17 G(s) =
p = k q
This is a further definition of the constant kq.
WORKED EXAMPLE No.4
A hydraulic motor has a nominal displacement of 8 cm3/radian. Calculate the torque produced at a
pressure of 90 bar.
SOLUTION
T = p kq = 90 x 105
(N/m2) x 8 x 10
-6 (m
3/rad) = 72 Nm
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7.2 HYDRAULIC CYLINDER
Figure 18
The flow rate and movement are related by the law Q = A dx/dt. Expressed as a transfer function with x
being the output and Q the input we have: G(s) = x
= 1
Q As
Force and pressure are related by the law F = pA. The transfer function with p as the input and F as the
output is: G(s) = F
= A p
7.3 MODEL FOR A FLOW METERING VALVE AND ACTUATOR
Figure 19
The input to the system is the movement of the valve xi. This allows a flow of oil into the cylinder of Q m3/s
which makes the cylinder move a distance xo.
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o
o
x
Making a big assumption that for a constant supply pressure the flow rate is directly proportional to the valve
position we may say Q = kv xi
kv is the valve constant and examining its units we find they are m2/s
The area of the piston is A m2.
The velocity of the actuator is v = dxo/dt and this is related to the flow and the piston area by the law of
continuity such that Q = k
dx v x
i = A
dt
Changing to a function of s this becomes kv xi = Asxo
Expressed as a transfer function we have G(s) =
o (s) = x
i
1
(A/k v )s
The units of A/kv are seconds and we deduce this is yet another time constant T.
G(s) =
x o (s) = 1
x i
Ts
Note that this is not quite the standard first order equation 1/Ts+1 and the difference is that the output will
keep changing for a given input, unlike the previous examples where a limit is imposed on the output.
If the actuator is a motor instead of a cylinder the equation is similar but the output is angle instead of linear
motion.
WORKED EXAMPLE No.5
A hydraulic cylinder has bore of 90 mm and is controlled with a valve with a constant kv = 0.2 m2/s
Calculate the time constant T. Given that xi and xo are zero when t = 0, calculate the velocity of the
piston and the output position after 0.1 seconds when the input is changed suddenly to 5 mm.
SOLUTION
A = πD2/4 = 6.362 x 10
-3 m
2
T = A/kv = 6.362 x 10-3 / 0.2 = 0.032 seconds
G(s)
Tsx o
= x o
x
i
= x i
(s) = 1
Ts
T dx
= x dt i
dx o
dt
x = velocity = i
T
= 0.005 m
0.032 s
= 0.156 m/s
Velocity = distance /time distance = xo = v t = 0.156 x 0.1 = 0.0156 or 15.6 mm assuming the
velocity is constant.
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SELF ASSESSMENT EXERCISE No.4
1. A hydraulic motor has a nominal displacement of 5 cm3/radian. Calculate the torque produced at a
pressure of 120 bar.
(60 N m)
2. A hydraulic cylinder has bore of 50 mm and is controlled with a valve with a constant kv = 0.05 m2/s
Calculate the time constant T. Given that xi and xo are zero when t = 0, calculate the velocity of the
piston and the output position after 0.2 seconds when the input is changed suddenly to 4 mm.
(0.039 s, 0.102 m/s and 20 mm)
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7.4 ADVANCED HYDRAULIC MODEL
Consider the same system but this time let the actuator move a mass M kg and have to overcome a damping
force. Further suppose that the valve now meters the pressure and not the flow rate such that the pressure
applied to the cylinder is p = kvxi. Consider the free body diagram of the actuator.
Figure 20
The applied force is due to pressure Fd and this is determined by the pressure acting on the area A such that:
Fp = pA.
The applied force is opposed by the inertia force Fi and the damping force Fd.
Fi = M d2xo/dt2 and Fd = kd dxo/dt.
Balancing forces gives pA = d 2 x M + k dx o
dt 2 d
d 2 x o
dt
dx o
Substituting p = kvxi. we have k v x i A = M
dt 2
2
+ k d dt
In Laplace form we have k v x i A = Ms x o + k
d sx o
Rearranging it into a transfer function. G(s) =
x o (s) = x
i
(M/Ak v )s 2
1
+ (k d /Ak v )s
If we examine the units we find M/Akv = T2 where T is a time constant. The critical damping coefficient is Cc = √(4 M A kv) and the damping ratio is δ kd/Cc
The transfer function becomes: G(s) =
x o (s) = x
i
T 2s
2
1
+ 2 δ Ts
Figure 21
Note the similarity with the standard 2nd order equation 1/T2s2 + 2δTs + 1. The difference is again due to there being no limitation on the output. If the actuator is a motor instead of a cylinder, the transfer function is
similar but the output is angle and angular quantities are used instead of linear quantities.
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WORKED EXAMPLE No.6
A hydraulic cylinder has bore of 90 mm and moves a mass of 80 kg. It is controlled with a valve with a
constant kv = 20000 Pa/m. The damping coefficient is 180 Ns/m.
Calculate the time constant T, Cc and δ.
Given that xi and xo are zero when t = 0, calculate the initial acceleration of the mass when the input is changed suddenly to 5 mm.
Calculate the acceleration when the velocity reaches 2 mm/s.
Calculate the velocity when the acceleration is zero.
SOLUTION
A = πD2/4 = 6.362 x 10
-3 m
2
T = √(M/Akv)= √80/(20000 x 6.362 x 10-3
) = 0.793 seconds
Cc = √(4 M A kv) = 201.78 Ns/m δ = kd/Cc = 0.892
G(s) =
x o (s) = x
i
T 2s
2
1
+ 2 δ Ts
or in terms of time
xi = (T2
x acceleration) + (2δT x velocity)
The initial velocity is zero. 0.005 = 0.793
2 a + 0 a = 7.952 x 10
-3 m/s
2
When v = 0.002
0.005 = 0.7932
a + (2 x 0.892 x 0.793 x 0.002)
a = 0.005 - (2 x 0.892 x 0.793 x 0.002)/0.7932
= 3.452 x 10-3
m/s2
The system initially accelerates and will eventually settle down to a constant velocity with no
acceleration. Put a = 0.
0.005 = 0 + (2 x 0.892 x 0.793) x velocity
velocity = 0.00353 m/s or 3.53 mm/s.
SELF ASSESSMENT EXERCISE No.5
A hydraulic cylinder has bore of 50 mm and moves a mass of 10 kg. It is controlled with a valve with a
constant kv = 80 Pa/m. The damping coefficient is 2 Ns/m.
Calculate the time constant T, Cc and δ. (7.98 s, 2.5.7 Ns/m and 0.798)
Given that xi and xo are zero when t = 0, calculate the initial acceleration of the mass when the input is changed suddenly to 10 mm. (0.157 mm/s
2)
Calculate the acceleration when the velocity reaches 0.1 mm/s. (0.137 mm/s2)
Calculate the velocity when the acceleration is zero. (0.785 mm/s)
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V
8. ELECTRIC SYSTEM ELEMENTS MODELS
8.1 RESISTANCE
Applying Ohm's Law we have V = I R V/I = R
This may be a function of time or of s.
The equation may be expressed in terms of charge Q.
Since I = dQ/dt I(s) = sQ hence G(s) = V
(s) Q
= sR
Figure 22
This is similar to the model for the damper.
8.2 CAPACITANCE The law of a capacitor is Q = C V V/Q = 1/C
This is similar to the model for spring.
Differentiating with respect to time we have dQ/dt = C dV/dt
dQ/dt is current I so the equation may be expressed as I = C dV/dt
As a function of s this becomes I (s) = C sV
The transfer function is G(s) =
Figure 23
V (s) =
1
I sC
8.3 INDUCTANCE
Faraday's Law gives us V = L dI dt
d 2Q = L
dt 2
This is similar to the model for a mass and can be either a first or 2nd order equation as
required.
Expressed as a function of s we have V (s) = L sI or Ls2Q
Figure 24
G(s) =
V (s)
I = sL or
s 2 Q
8.4 POTENTIOMETER
Figure 25
f the supply voltage is constant and the current is negligible, the output voltage V is directly proportional to
the position x or angle θ so a simple transfer function is obtained.
G(s) =
G(s) =
Vo (s) x
o (s) θ
= constant = k p
= constant = k p
(linear)
(angular)
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V
V
V
i
8.5 R -C SERIES CIRCUIT
The input voltage Vi is the sum of the voltage over the
resistor and the capacitor so
Vi = I R + I /Cs
Vi = I (R + 1/Cs)
The output is the voltage over the capacitor so
Vo = I/Cs
Figure 26
The transfer function is then G(s) = o (s) =
V i
I(R
I/Cs
+ 1/Cs) =
1 RCs + 1
The units of RC are seconds and this is another electrical time constant T. The transfer function may be
written as
G(s) =
Vo (s) = V
i
1
Ts + 1
This is the standard first order equation and is the same as both the spring and damper and the thermal
example.
8.6 L - C - R in SERIES
This is 3 sub-systems in series. In this case we will take the
output as the voltage on the capacitor and the input as the voltage
across the series circuit.
The input voltage is the sum of all three voltages and is found by
adding them up.
Vi = I R + I sL + I/sC
Figure 27
The output voltage is Vo = I/sC The transfer function is then
G(s) =
G(s) =
o (s) = V
i
o (s) =
I(R
I/Cs =
+ sL + 1/Cs)
1
RCs
1
+ CLs2 + 1
V s 2 CL + sRC + 1
If we examine the units of CL we find it is seconds2 and we have yet another time constant defined as
T2 = CL and the equation may be rewritten as:
G(s) =
Vo (s) V
i
= 1
T 2s 2 + 2 δ Ts + 1
The damping ratio δ is defined as δ =
R and
4 L
C
4 L
is called the critical damping value. C
Note this is the standard 2nd order equation identical to the mass-spring-damper system.
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o
WORKED EXAMPLE No.7
A Capacitance of 200 µF is connected in series with a resistor of 20 kΩ as shown in figure 26. The
transfer function is θo (s) = θ
i
(Ts
1
+ 1)
The voltage across the resistor is suddenly changed from 3V to 10V.
Calculate the time constant T and derive formulae for how the voltage across the capacitor varies with
time. Sketch the graph.
SOLUTION
T = RC = 20 x 103
x 200 x 10-6
= 4 seconds
The derivation is identical to that in example 3 simply changing θ to V we get the result.
− t
V = V i
− ∆V e T
Put in the values T = 4 ∆V = 10 – 3 = 8 Vi = 10
Vo = 10 – 7e-t/4
Evaluating and plotting produces the result below. It is an exponential growth.
Figure 28
SELF ASSESSMENT EXERCISE No.6 1. Calculate the time constant for an RC circuit with a resistance of 220 Ω and capacitance of 470 nF. (103 µs)
2. Calculate the second order time constant and the damping ratio for a R-L-C circuit with L = 5 µH, C= 60 µF and R=6.8 Ω. (17.3 µs and 11.8 )
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⎠
⎜
a
R
⎟ R
a
2
9 ELECTRIC MOTORS
This is covered in greater depth in tutorial 2. Here are the basic models.
9.1 FIELD CONTROLLED MOTOR,
The main theory of electric motors is covered in another tutorial. It can be shown that for a d.c. servo motor
with field control
T = kf if If the motor drives an inertial load and has damping the dynamic equation becomes
d 2θ T = I
dt 2
dθ + k
d dt
T = Is2θ + k
d sθ = θ(Is
2 + k
d s) = k
f if
G(s) =
θ (s) =
if
Is2
k f
+ sk d
Figure 29
This models the relationship between the shaft angle and the
control current.
9.2 ARMATURE CONTROLLED MOTOR
It can be shown that the torque is related to armature voltage
⎛ dθ ⎞ k and resistance by the formula T = ⎜ Va
⎝ − k
dt ⎟
R
The torque must overcome inertia and damping as before so
T = Is 2θ +
k d sθ
Equating we get
2
k k k 2sθ
T = θ(Is
⎛
+ k d s) =
2 ⎞
(Va
− ksθ) R a
= Va − a R a
Figure 30 θ⎜ Is2
⎝
+ k d s +
k s ⎟ = k
V
R a ⎠ a
With rearrangement we find G(s) = θ (s) = (k/R a )
2
Va Is + sk d − k s/R a
This models the relationship between the angle of the shaft and the control voltage.
Worked examples and self assessment exercise for this section may be found in tutorial 2.
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θ
10. CLOSED LOOP SYSTEMS TRANSFER FUNCTION WITH UNITARY FEED BACK.
Consider a simple system with an input θi and output θo related by the transfer function G(s). If the system is
to be a controlled system in which we require the output to change and match the value of the input (set value), we must make the input the error θe instead of the set value. The error is obtained by comparing the output value with the input value with the signal summing device. This produces the result θe = θi - θo and because θo is subtracted, this idea is called NEGATIVE FEED BACK. The block diagram shows that the signal passes around a closed loop hence the name CLOSED LOOP SYSTEM.
G(s) = θ o
θ e
substitute θ e = θ
i
- θ o
Figure 31
G(s) =
θ o
θi
- θ o
divide the bottom line by θ o
G(s) =
1
θ i - 1
θ o
rearrange θ
i - 1 = θ o
1
G(s)
θi =
1
θ o G(s) + 1 =
1 + G(s)
G(s)
invert
θ o = θ
i
G(s)
1 + G(s) or
1 1/G(s) + 1
The transfer function for the closed system is hence o
θi
G(s) is the transfer function of the open loop system.
= 1
1/G(s) + 1
Let’s revisit the hydraulic open loop transfer functions derived previously.
When the hydraulic valve and actuator is turned into a closed loop system, the two transfer functions
become:
G(c.l) =
1
Ts + 1
for the first order version and
G(c.l) = T 2s 2
1
+ 2 δ Ts
for the second order version. + 1
These models are mathematically identical to the transfer function of the mass-spring- damper and the L-C-R
circuits.
Note that for any system with an open loop transfer function G(s) the closed loop transfer function with unit
feedback is
G cl
= 1
1/G(s) + 1
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= = =
WORKED EXAMPLE No.8
An open loop system has a transfer function G(s) = 2/(s2
+ 2s + 1). Derive the closed loop function when
unit feedback is used.
SOLUTION
1 1 2 2 G
cl =
2 2 2
1/G(s) + 1
s + 2s
2
+ 1 + 1
s + 2s
+ 1 + 2
s + 2s + 3
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SELF ASSESSMENT EXERCISE No.7
1. An open loop system has a transfer function G(s) = 5/(4s2
+ 2s + 2). Derive the closed loop
function when unit feedback is used.
Gcl = 5/(4s2
+ 2s + 7)
2. An open loop system has a transfer function G(s) = 10/(s3
+5s). Derive the closed loop function
when unit feedback is used.
Gcl = 10/(s3
+ 5s + 10)
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Shaft speed meter INSTRUMENTATION AND
CONTROL
TUTORIAL 3 – TRANSFER FUNCTION MANIPULATION
This tutorial is of interest to any student studying control systems and in particular the EC
module D227 – Control System Engineering.
On completion of this tutorial, you should be able to do the following.
Explain a basic open loop system.
Explain a basic closed loop system.
Explain the use of negative feedback
Manipulate transfer functions.
Explain the use of velocity feedback.
Explain the affect of disturbances to the output of a system.
Explain the use of proportional and derivative feedback.
Reduce complex systems to a single transfer function.
If you are not familiar with instrumentation used in control engineering, you should
complete the tutorials on Instrumentation Systems.
In order to complete the theoretical part of this tutorial, you must be familiar with basic
mechanical and electrical science.
You must also be familiar with the use of transfer functions and the Laplace Transform
(see maths tutorials).
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D.J.DUNN 1
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1. INTRODUCTION
The function of any control system is to automatically regulate the output and keep it at the desired value.
The desired value is the input to the system. If the input is changed the output must respond and change to
the new set value. If something happens to disturb the output without a change to the input, the output
must return to the correct value. Here is a short list of some of the things we might be controlling.
The speed or angle of a motor (electric or hydraulic)
The speed or position of a linear actuator (e.g. in robotics, ship’s stabilisers, aircraft control etcetera)
The temperature of an oven (e.g. heat treatment)
The pressure of a vessel (e.g. a steam boiler)
The quantity inside a container (e.g. metering contents in a vessel)
The flow of solids, liquids and gases (e.g. controlling the steam flow to a turbine)
There are some basic properties and terminology used with systems which we should examine next.
2. TRANSFER FUNCTION
Figure 1
Any item in any system may be represented as a simple block with an
input and output as shown. In general terms, the input is designated θ i and the output θo.
The ratio of output over input is often shown as G = θo/θi. When the model is a differential equation the
Laplace transform is used and this introduces the complex operator s. In this case G is called the
TRANSFER FUNCTION and strictly we should write G(s) = θo(s)/θi(s)
If G is a simple ratio, it is still a Transfer function but if the model is not a simple ratio and cannot be
transformed, it should not strictly be called a transfer function. You should study the tutorial on Laplace
and Fourier transforms in the maths section in order to fully appreciate this tutorial.
3. OPEN LOOP SYSTEMS
A system with no regulation is called an open loop system. For example a typical instrument system (see
tutorials on instrumentation) is an open loop system with an input and output but no control action at all.
Let’s take a d.c. servo motor as an example (see the tutorial on electric actuators). The speed of the servo
motor depends on the voltage and current supplied to it. A typical system might use a potentiometer
which you turn to an angle i (the input) to produce a voltage V and this is amplified with a power
amplifier producing electric power P that drives the motor at speed N (the output).
Figure 2
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D.J.DUNN 2
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The block diagram looks like this.
Figure 3
The block diagram show that the signal path from input to output is a linear chain not forming any loop
so this why it is called an OPEN LOOP SYSTEM.
WORKED EXAMPLE No.1
The speed of an electric motor is directly proportional to voltage such that N = 20V where V is in
Volts and N in rev/min. The motor is controlled by a power supply which has an output voltage
related to the position of the control knob by V = 2 i where V is in Volts and i is in degrees.
Draw the block diagram and deduce the overall transfer function. Determine the output speed when
the knob is set to 60o.
SOLUTION
Figure 4
G = N/i = 2i x 20 = 40 rev/min per degree N = 40 x 60 = 2400 rev/min
SELF ASSESSMENT EXERCISE No.1
A simple control system consists of a potentiometer with a transfer function of 0.02V/mm in series
with an amplifier with a gain of 12, in series with a V/I converter with a transfer function I = 0.5V
where V is in volts and I in mA. The output current is amplified with a gain of 1200 and the output
current supplied to an electro-magnetic torque arm which produces 3 Nm per Amp.
Draw the block diagram and deduce the overall transfer function. (0.432 Nm/mm)
Determine the input position of the potentiometer in mm which produces a torque output of 60 Nm.
(138.9 mm)
Consider the example of the servo motor again. Suppose the motor drives a load and that the load
suddenly increases. This would make the motor slow down as there would not be enough power to keep it
at the original speed. We would now have an error between the speed selected with the potentiometer and
the actual speed of the motor. To bring the speed back to the correct value, we have to turn up the power
and do this automatically we need a closed loop system. Open loop systems are incapable of maintaining
a correct output in all but the simplest cases.
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4. SUMMING DEVICES
In order to regulate any control system we must determine the error between the output and the input.
This is done with a summing device and the symbol for this is shown in the left diagram. These devices
may be electrical (e.g. a simple differential amplifier), pneumatic (e.g. a differential pressure cell) or
mechanical. We can put a Plus (+) or minus (-) sign in the symbol to show if it is adding or subtracting
and the symbol can be used with more than one signal as shown in the right diagram. In modern digital
systems it is a simple case of adding or subtracting the numbers stored in registers.
Figure 5
5. BASIC CLOSED LOOP CONTROL
Consider the example of the servo motor again. This time suppose we wish to control the angle of the
shaft o. The input potentiometer produces a voltage Vi and the output potentiometer produces a voltage
Vo to represent the angle of the shaft. If the two voltages are the same, the shaft is at the correct angle. If
there is an error, the voltages are different. The differential amplifier acts as the summing device and
produces a voltage Ve representing the error. The error is supplied to the power amplifier and power is
sent to the motor to rotate it in the direction that corrects the angle. When the voltages are equal again, the
error is zero and no power is supplied to the motor so it stops. Error in either direction can be corrected if
the power amplifier is capable of producing positive and negative current.
Figure 6
This description is somewhat over simplified and does not explain an actual working system. The motor
would have difficulty staying at the correct angle if there is a load trying to turn it. Note how the voltage
from the output potentiometer is fed back to the summing device so that the error is Ve = Vi - Vo. This is
NEGATIVE FEEDBACK and this is essential to make the system respond to the error.
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6. CLOSED LOOP SYSTEM TRANSFER FUNCTIONS
The most basic block diagram for a closed loop system is shown below. The main block is an open loop
system with a transfer function Gol. This relates the error and the output so that Gol = o/e.
The transfer function for the closed loop system is Gc.l. This relates the input i and output o.
The error is obtained by comparing the output value with the input value in the signal summing device.
This produces the result e = i - o and because o is subtracted, this idea is called NEGATIVE FEED
BACK. The block diagram shows that the signal passes around a closed loop hence the name CLOSED
LOOP SYSTEM.
Figure 7
The system shown is said to have UNITY FEED BACK as there is no processing in the feed back path.
The following result will be used many times in later tutorials on system analysis.
G ol
θo
θe
θe θi θo
G ol
θo
θi θo
G ol
θi θo θo
G ol
θi G
olθo θo
G ol
θi θo G
olθo
G ol
θi θo 1 G
o1
G
c.l.
θo
θi
G
ol 1 G
ol
1
1 1
G ol
This is the transfer function for closed loop unity feed back system. In practice ,at the very least, we have
a transducer to measure the output so we should show a block in the path representing the transducer. In
addition we may put other signal conditioners and processors in the path such as an amplifier or
attenuator.
Figure 8
G1 is the OPEN LOOP transfer function and is in the forward path. G2 is in the feed back path.
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i 1
i 1
i 2
1
2 i
The open loop transfer function is related to G1 and G2 as follows. θo
G1
θ
θe
G
e
θi
G 2θo
θo
1 θ i
G 2θo
G1
θi G
2 θo θo
G θ 1
G G 2θo θo
G θ 1 G G
2θo θo
G θ 1
θo 1 G G
G θo
θi
G
1
1 G G 1 2
θ G The transfer function for the closed system is hence G o 1
θ 1 G G 1
WORKED EXAMPLE No.2
The transfer function for a hydraulic system comprising of a hydraulic valve and actuator is
G(s) =1/(Ts). Write down the closed loop transfer function.
SOLUTION
G(s) (closed loop)
1
1 Ts 1
1
1
G(s)
WORKED EXAMPLE No.3
The transfer function for a hydraulic system comprising of a hydraulic valve and actuator with inertia
attached is G(s) =1/(T2s2+ 2Ts). Write down the closed loop transfer function.
SOLUTION
G(s) (closed loop)
1
1
T 2s
2 1
1
2δ Ts
1 G(s)
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SELF ASSESSMENT EXERCISE No.2
1. Find the closed loop transfer function for the system shown below.
Figure 9
2. Find the closed loop transfer function for the system shown below.
Figure 10
3. Find the closed loop transfer function for the system shown below.
Figure 11
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7. FEEDBACK PROCESSING
7.1 VELOCITY FEEDBACK
Velocity feedback is widely used to stabilise a system which tends to oscillate. This is not the same as
derivative control covered later. If the output of a system is motion xo then the rate of change dxo/dt is a
true velocity but the idea can be used for any output. Consider the block diagram below.
Figure 12
A typical electrical servo system uses position sensing and velocity feedback. The velocity signal is
derived from a tacho-generator or some other suitable speed measuring device. The resulting signal is
compared with the input and the output as shown.
The error signal is xe = xi - k2xo - k1dxo/dt
When a sudden (step) change is made, the error is a maximum and so the output changes very rapidly.
The velocity feedback is hence greatest at the start. The effect of the feedback is to reduce the error in a
manner directly proportional to the velocity. When the output is static the feedback is zero and so no error
results from it. The feedback has the same affect as damping and if the complete analysis is made, we
would see that control over k1 gives control over damping. This is useful in stabilising an oscillatory
system.
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i
i
WORKED EXAMPLE No.4
The diagram shows a closed loop system with velocity and negative feed-back. The transfer k
functions for the system is G(s) T 2s
2 2δ Ts
Derive the closed loop transfer function.
SOLUTION
Figure 13
k
The open loop transfer function is G(s) T
2s
2 2 δT s
θe θi θo αsθ o θ
i θo (α s 1)
θo
Gθe
G θ i
θo (α s 1)θo Gθ
i Gθo (α s 1)
θo Gθo (α s 1) Gθi
θo 1 G(α s 1) Gθi
θo θ
i
θo
G
1 G(α s
1
1)
θi
1
G
(α s
1)
Now substitute G
θo
T
2s
2
1
k
2 δ Ts
θ T
2s
2
k
2 δ Ts
(α s
1)
θo θ
i
θo
T 2s
2
k
2 δ Ts
k
k(α s
1)
θ T 2s
2 s2 δ T kα k)
This is the closed loop transfer function
The term with s is the effective damping term and as can be seen this is affected by the value of k.
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SELF ASSESSMENT EXERCISE No.3
The diagram shows how the arm of a robot is controlled using a controller and motor with position
and velocity feed-back. Determine the closed loop transfer function for the system.
Figure 14
7.2 DISTURBANCES
The affect of a disturbance on the output may be idealised on a block diagram as follows.
Figure 15
The disturbance d is added to the output to produce a new output o. G is the forward path transfer function.
θe θi
θo θo
G θ i θo d
θo
θ d θo Gθ
i Gθo d
θo 1 G Gθi d
θ Gθe
Gθi d
θo Gθe d θo
1 G
WORKED EXAMPLE No.5
A simple closed loop system consists of an amplifier with a gain of 10. For an input of 4 mA, show
that the effect of a disturbance added to the output of magnitude i) 0 and ii) 2
SOLUTION
i) d = 0, G = 10 i = 4 o = (G i+ d)/(1 + G) = (10 x 4+ 0)/(1 + 10) = 40/11
i) d = 2, G = 10 i = 4 o = (G i+ d)/(1 + G) = (10x 4+ 2)/(1 + 10) = 42/11
This shows that a disturbance of 2 produces an output error of 2/11.
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7.3 ELIMINATING THE AFFECT OF A DISTURBANCE
A special feed-back path is used to reduce or eliminate the affect of a disturbance added to the output.
The idealised system is shown below.
Figure 16
The disturbance (D) is processed through a transfer function G2 and added to the input.
G1 is the forward path transfer function. G2 is the feed-back path transfer function.
= G1e= G1( i - G2D) o = + D o = G1( i - G2D) + D o = G1 i - G1G2D + D
From the last line it can be seen that if G1G2D = D then o = G1 i The affect of the disturbance is completely removed when G1 = 1/G2
WORKED EXAMPLE No.6
For the system described above, the forward path transfer function is
G1(s)= 4/(s + 1)
Determine the transfer function for the feedback path which eliminates the affect of a disturbance.
SOLUTION
G2(s) = 1/G1(s) = (s + 1)/4
SELF ASSESSMENT EXERCISE No.4
1. A simple closed loop system consists of two amplifiers in series one with a gain of 3 and one with a
gain of 2. For an input of 6 mA, determine the output when a disturbance added to the output of
magnitude i) 0 and ii) 3
2. The forward path transfer function for a controlled system is G1(s)= 2/(3s2+ 1)
Determine the transfer function for the feedback path which eliminates the affect of a disturbance.
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7.4 USING PROPORTIONAL AND DIFFERENTIAL CONTROL IN THE FEEDBACK PATH
Let us consider that the feed-back transfer function is a P + D transfer function (Proportional plus
differential).
Figure 17 – PROPORTIONAL
For a proportional system, output is directly proportional to input. In effect it is an amplifier or attenuator
and k1 is a simple ratio.
Figure 18 -DIFFERENTIAL
For a differential block, the output is directly proportional to the rate of change of the input with time. A
tachometer is an example of this. In Laplace form the output is k2 sD. The units of k2 must be seconds.
We may write k2 = T k1 where T is called the DERIVATIVE TIME.
Figure 19 PROPORTIONAL PLUS DIFFERENTIAL
For P + D it follows that the transfer function is k1(Ts + 1)
WORKED EXAMPLE No.7
Given that G1 = 5/(s + 2) find the value of derivative time T and the constant which will eliminate
the disturbance D.
SOLUTION
Figure 20
G2 = 1/G1 = (s + 2)/5 = k1(Ts + 1 )
Rearrange to make the forms match.
0.2(s + 2) = k1(Ts + 1 ) 0.4(0.5s + 1) = k1(Ts + 1 )
Hence by comparison k1 = 0.4 and T = 0.5
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SELF ASSESSMENT EXERCISE No.5
1. The forward path transfer function of a controlled system is G(s) = 6/(s + 3). In order to eliminate
disturbances added to the output, a P + D feedback path is used. Find the value of the derivative time
and the constant required.
(0.333 sec and 0.5)
2. The forward path transfer function of a controlled system is G(s) = 10/(2s + 5). In order to eliminate
disturbances added to the output, a P + D feedback path is used. Find the value of the derivative time
and the constant required.
(0.4 s and 0.5)
3. The forward path transfer function of a controlled system is G(s) = 8/(5s + 10). In order to eliminate
disturbances added to the output, a P + D feedback path is used. Find the value of the derivative time
and the constant required.
(0.5 s and 1.25)
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8. SIMPLIFYING COMPLEX SYSTEMS
The symbol H is often used for feed back transfer functions but we can use any appropriate symbol to
help us simplify complex circuits. The diagrams below show the stages in reducing a block diagram to
one block with one transfer function.
Here is a more complex one.
Figure 21
Figure 22
Here is an even more complex one.
Figure 23
The technique now is to find the transfer function for the inner loop and work outwards as follows.
Figure 24
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WORKED EXAMPLE No.8
Derive the transfer function for the system below.
Figure 25
G1 = 3/s G2 = 1/(4s+5) G3 = 4 G4 = 1/s H1 = 5 H2 = 0.5
SOLUTION
Put in the data and D1
4
4s 5
Figure 26
4
4
1
4 x 0.5 4s 51
2
4s 7 4s 5 4s 5
3 4 12
G (3/s)(1/s)(D1 )
s 2 4s 7
s 2 4s 7 12
1 (3/s)(5)(D1 ) 1
15 4
1 60 s 2 4s 7 s 2 4s 7
60
G 12
s 4s 7
12
4s 2 7s 4s 7
4s3 7s 2 60s 2 4s3 67s 2
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SELF ASSESSMENT EXERCISE No.6
Derive the overall transfer functions for the systems below.
1.
5/(0.2s2
+ s + 40)
2.
Figure 27
8/(0.2s2
+ 1.8s + 52) Figure 28
3.
8/(4s2
+ 113s + 52) Figure 29
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TUTORIAL – STABILITY ANALYSIS
This tutorial is specifically written for students studying the EC module D227 – Control
System Engineering but is also useful for any student studying control.
On completion of this tutorial, you should be able to do the following.
• Explain the basic definition of system instability.
• Explain and plot Nyquist diagrams.
• Explain and calculate gain and phase margins.
• Explain and produce Bode plots.
The next tutorial continues the study of stability.
If you are not familiar with instrumentation used in control engineering, you should
complete the tutorials on Instrumentation Systems.
In order to complete the theoretical part of this tutorial, you must be familiar with basic
mechanical and electrical science.
You must also be familiar with the use of transfer functions and the Laplace Transform
(see maths tutorials).
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θ
( i 2 o
i o i 2 )
1
1. INSTABILITY
Consider a system in which the signals are added instead of subtracted by the summer. This is positive
feed back. The electronic amplifier shown is an example of this.
Figure 1
We may derive the closed loop transfer function as follows.
G = θo
θ = θ
+ G θ
G = θo
1 e i 2 o e
1 θ + i
G 2θo
G1
θi + G
2 θo ) = θo G θ + 1
G G θ 1
= θo
G θ = 1
θ - G G 1
2θo G θ =
1 θo (1 − G G
G
cl =
θo
θi
= G
1
1 − G G 1 2
If G1 G2 = 1 then Gcl = ∞ and the system is unstable. If G1 G2 ⟨ 1 then Gcl has a finite value which is small
or large depending on the values. In the electronic amplifier, the gain can be controlled by adjusting the feed back resistor (attenuator).
Suppose G1 = 1 and G2 = 0.8 Gcl = 1/(1- 0.8) = 5
Suppose G1 = 1 and G2 = 0.99 Gcl = 1/(1- 0.99) = 100 The closer the value of G2 is to 1 the higher the overall gain. It is essential that G2 is an attenuator if the
system is not to be unstable.
A system designed for negative feed back with a summer that subtracts should be stable but when the
signals vary, such as with a sinusoidal signal, it is possible for them to become unstable.
Consider an automatic control system such as the stabilisers on a ship. When the ship rolls, the stabilisers
change angle to bring it back before the roll becomes uncomfortably large. If the stabilisers moved the
wrong way, the ship would roll further. This should not happen with a well designed system but there are
reasons and causes that can make such a thing happen. Suppose the ship was rolling back and forth at its
natural frequency. All ships should have a righting force when they roll because of the buoyancy and will
roll at a natural frequency defined by the weight, size and distribution of mass and so on. The sensor
detects the roll and the hydraulic system is activated to move the stabiliser fins. Suppose that the
hydraulics move too slow (perhaps a leak in the line) and by the time the stabilisers have responded, the
ship has already righted itself and started to roll the other way. The stabilisers will now be in the wrong
position and will make the ship roll even further. The time delay has made matters worse instead of better
and this is basically what instability is about.
Another example of positive feed back is when you place a microphone in front of a loud speaker and get
a loud oscillation. Another example is when you push a child on swing. If you give a small push at the
start of each swing, the swing will move higher and higher. You are adding energy to the system, the
opposite affect of damping. If you stop pushing, friction will slowly bring it back to rest.
With positive feedback, energy is added to the system making the output grow out of control. A system
would not be designed with positive feedback but when a sinusoidal signal is applied, it is possible for the
negative feedback to be converted into positive feedback. This occurs when the phase shift of the
feedback is 180o
to the input and the gain of the system is one or more. When this happens, the feedback
reinforces the error instead of reducing it.
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2. NYQUIST DIAGRAMS
A system with negative feed back becomes unstable if the signal arriving back at the summer is larger
than the input signal and has shifted 180o
relative to it. Consider the block diagram of the closed loop
system. A sinusoidal signal is put in and the feed back is subtracted with the summer to produce the error.
Due to time delay the feed back is 180o
out of phase with the input. When they are summed the result is
an error signal larger than the input signal. This will produce instability and the output will grow and
grow.
Figure 2
A method of checking if this is going to happen is to disconnect the feed back at the summer and measure
the feed back over a wide range of frequencies.
Figure 3
If it is found that there is a frequency that produces a phase shift of 180
o and there is a gain in the signal,
then instability will result. The Nyquist diagram is the locus of the open loop transfer function plotted on
the complex plane. If a system is inherently unstable, the Nyquist diagram will enclose the point -1 (the
point where the phase angle is 180o
and unity gain).
Consider the following system.
Figure 4
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The transfer function relating θi and θ is G(s) G1 x G2 x G3 = K/s(1+s)(1+2s). Converting this into a
complex number (s = jω) we find
K− 3ω2 K(ω − 2ω
3 ) G( jω ) =
9ω4
+ (ω − 2ω3 ) −
j 9ω
4 + (ω − 2ω
3 )
The polar plot below (Nyquist Diagram) is shown for K = 1 and K = 0.4. We can see that at the 180o
position the radius is less than 1 when K = 1 so the system will be stable. When K = 2 the radius is greater
than 1 so the system is unstable. We conclude that turning up the gain makes the system become unstable.
Figure 5
The plot will cross the real axis when ω = 2ω3 or ω = 0.707 and this is true for all frequencies. The plot
will enclose the -1 point if
K− 3ω2
2 4 3
9ω4 + (ω − 2ω
3 ) ≤ −1
so the limit is when − K 3ω
= 9ω
+ (ω − 2ω )
Putting ω = 0.707 the limiting value of K is 1.5
ALTERNATIVE METHOD
There is another way to solve this and similar problems. The transfer function is broken down into
separate components so in the above case we have:
G(s) =
K x
1 s (1 +
x 1
s) (1 +
2s)
Each is turned into polar co-ordinates (see previous tutorial).
K produces a radius of
s
K and an angle of - 90
o for all radii
ω
1
1 + s
produces
a radius of 1
1 + ω2
and angle tan -1
( ω)
1
1 + 2s
produces a radius of 1
1 + 4ω2
and angle tan -1
2ω
When we multiply polar coordinates remember that the resultant radius is the product of the individual
radii and the resultant angle is the sum of the individual angles. The polar coordinates of the transfer
function are then:
Radius is K
x ω
1 x
1 + ω2
1
1 + 4ω2
Angle is - 90o −
tan
-1( ω)
− tan
-1 2ω
Put ω = 0.707 Radius = 1.414 x 0.8165 x 0.577 = 0.667 K Angle = -90 - 35.26 - 54.74 = -180o
If K = 1.5 the radius is 1 as stated previously.
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3 PHASE MARGIN and GAIN MARGIN
3.1 PHASE MARGIN
This is the additional phase lag which is needed to bring the system to the limit of stability. In other words
it is the angle between the point -1 and the vector of magnitude 1.
3.2 GAIN MARGIN
This is the additional gain required to bring the system to the limit of stability.
Figure 6
WORKED EXAMPLE No.1
The open loop transfer function of a system is G(s) = 200/(1+2S)(3+S)(5+S). Produce a polar plot
for ω = 3 to ω = 10. Determine the phase and gain margin.
SOLUTION
Evaluate the polar coordinates for 200/(1 + 2s), then 1/(2+s) then 1/(5+s) (See previous tutorial)
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Now add the three sets of angles and multiply the three sets of radii and plot the results.
Figure 7
The region of interest is where the plot is -180o
and the radius is 1. This would require a much more
accurate plot around the region for ω = 3 to 5 as shown below.
Figure 8
The phase margin is 180 – 166 = 14o
The gain margin is 1 – 0.65 = 0.35
SELF ASSESSMENT EXERCISE No.1
1 Determine the steady state gain and primary time constant for G(s) = 10/(s + 5). Determine the polar
coordinates when ω = 1/T (Gain = 2 and Radius = 1.414 and angle = -45
o)
2. Determine the steady state gain for G(s) = 0.5/(s+2)(s+10). Determine the polar coordinates when
ω = 0.5
(Gain = 0.025 , R1 = 0.0243 φ1 = -16.9o)
3. The open loop transfer function of a system is G(s) = 80/(s+1)(s+2)(s+4). Produce a polar plot for
ω = 3 to ω = 10. Determine the phase and gain margin. (0.11 and 3.5o).
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V
4. BODE PLOTS
These are logarithmic plots of the magnitude (radius of the polar plot) and phase angle of the transfer
function. First consider how to express the gain in decibels.
Strictly G is a power gain and G = Power out/Power In
If the power in and out were electric then we may say G = Vout Iout
V I in in
Using Ohms Law this with the same value of Resistance at input and output this becomes 2
G = o
V 2 i
2
I 2
or o
I 2 i
Expressing G in decibels G(db) = 10 log Vo
V
2
i
= 20 log Vo
V
i
or 20 log Io
Ii
From this, it is usual to express the modulus of G as ⎟G⎟ = 20 log⎟ (θo/θi)⎟
Note that the gain in db is the 20 log R where R is the radius of the polar plot in previous examples.
Consider the transfer function G(s) = 1
s
G(jω ) = 1
= - j
G = - j
= 1
G(db) =
1
20log⎜ ⎟ = −20logω
⎛ ⎞ ⎛ ⎞
⎜ ⎟
jω ω ω ⎝ ω ⎠ ⎝ ω ⎠ Plotting this equation produces the following graph. The graph shows a straight line passing through 0 db
at ω=1 with a gradient of -20 db per decade. The phase angle is -90o
at all values of ω.
Figure 9
Now consider the following transfer function
G(s) = 1
Ts + 1 G =
1 =
jω T + 1
1
jω T + 1
G(jω )
gain.
= 1
1 + ω2T 2
− j ωT
1 + ω2T 2
The radius of the polar coordinate is 1
(T2ω
2
+ 1)
and this is the
The gain in db is then
⎡ ⎛ ⎞⎤
⎜ =
⎟
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⎠
1 ⎝ ⎣
⎦
⎡ 1 2 2 ⎤ 2 2
G(db) = 20⎢log
⎜
⎥
( ) ⎟
= −20⎢ 2
log(T ω + 1)⎥ = −10log(T ω + 1)
⎢ T 2ω
2 + 1 ⎥ ⎣ ⎦ The phase angle is –tan
-1(ωT)
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⎟
⎡ 2 )⎥
⎜
⎜ ⎟
If we put T = 1 as a convenient example, and plot the result, we get two distinct straight lines shown on the left graph. The horizontal line is produced by very small values of ω and so it is called the LOW FREQUENCY ASYMPTOTE. The sloping straight line occurs at high values of ω and is called the HIGH FREQUENCY ASYMPTOTE and has a gradient of -20 db per decade. The two lines meet at
the breakpoint frequency or natural frequency given by ωn=1/T = 1 in this case.
The graph on the right shows phase angle plotted against ω and it goes from 0 to -90
o. The 45
o point
occurs at the break point frequency.
Figure 10
Now consider the following transfer function. (Standard first order response to a step input)
G(s) = K
s(1 +
sT) G(jω ) =
K
jω (1 +
jω T)
Note there is an easier way to find ⎟G⎟ as follows. Separate the two parts and find the modulus of each separately.
G = K 1
jω (1 + 1
ω2T
2
1 =
1 =
1
jω jω ω
1 =
1 =
1
1 + jω T 1 + jω T 1 + ω2T
2
G = K 1 1
= K⎛ =1 ⎞⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ Taking logs we get
ω 1 +
ω
2T
2
⎝ ω ⎠⎝
1 + ω2T 2 ⎠
Log G = logK
+ log⎛ =1 ⎞
⎛ 1 ⎞ log
⎜ ⎟ + ⎜ ⎟
⎝ ω ⎠ ⎝ 1 + ω2T
2 ⎠
Log G
= logK
− logω −
1 log(1 +
2
ω2T 2 )
Log G db = 20⎢⎣logK
− logω −
1 log(1 +
2 ω2T
⎤
⎦ There are three components to this and we may plot all three separately as shown. The graph for the
complete equation is the sum of the three components. The result is that the graph has two distinctive
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slopes of -20 db per decade and -40 db per decade. (K and T were taken arbitrarily as 10 giving a
breakpoint of ω = 1/T = 0.1.
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Figure 11
The plot of phase angle against frequency on the logarithmic scale shows that the phase angle shifts by
90o every time it passes through a breakpoint frequency. The plot for the case under examination is
shown.
A reasonable result is obtained by sketching the asymptotes for each and adding them together.
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ω 0.001 0.01 0.1 1.0 10 100 1000
-log ω 3 2 1 0 -1 -2 -3
-½log(ω2T2+1) tiny tiny tiny -0.048 -0.707 -1.707 -2.707
Total Gain (units) 3 2 1 -.048 -1.707 -3.707 -5.707
Total Gain db 60 40 20 -0.96 -34.14 -74.14 -114.14
φ degrees -90 -90.3 -92.9 -117 -169 -179 - 180
⎜
⎝ ⎠
⎝ ⎠
⎡
+ 1)⎥ ⎛ ⎜
⎟
1 = ⎟
⎜ ⎠
WORKED EXAMPLE No. 2
A system has a transfer function G(s) = 1
Where the time constant T is 0.5 seconds.
s(Ts + 1)
Plot the Bode diagram for gain and phase angle. Find the low frequency gain per decade, the high
frequency gain per decade and the break point frequency.
SOLUTION
G(s) = 1
G(jω( = 1 ⎛ 1 ⎞
G = 1 1 ⎛ =1 ⎞⎛ ⎞
⎜ ⎟
s(Ts + 1) jω ⎜ jω T + 1 ⎟ jω jω T + 1 ⎝ ω ⎠⎜ ω2T
2 + 1
⎟ G (db) =
20⎢⎣
− logω −
1 log(ω2
T 2 ⎤
2 ⎦ φ = -90 −
tan − 1 1 ⎞
⎝ ωT ⎟
Examining the table we see that the gain drops by 20 db per decade at low frequencies and by 40 db
per decade for high frequencies. Plotting the graph on logarithmic paper reveals a breakpoint
frequency of 2 rad/s which is also found by ωn = 1/T = 1/0.5 = 2
A quick way of drawing an approximate Bode plot is to evaluate the gain in db at the breakpoint frequency and draw asymptotes with a slope of -20 db per decade prior to it and -40 db per decade
after it. The phase angle may be found by adding the two components.
G(s) = 1 G(jω ) =
1 +
1 s(Ts + 1) jω jω T + 1
The phase angle for the first part is the angle of a vector at position -1/ω on the j axis which
corresponds to -90o. The phase angle of the second part is the angle of a vector at -1 on the real axis and - ωT on the j axis. The two phase angles may be added to produce the overall result.
A quick way to draw the Bode phase plot is to note that the break point frequency occurs at the mid
point of the phase shift (-135o
in this case) so draw the asymptotes such that they change by -90o at
each breakpoint frequency.
Figure 12
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GAIN AND PHASE MARGINS FROM THE BODE PLOT
Gain and phase margins may be found from Bode plots as follows. Locate the point where the gain is zero
db (unity gain) and project down onto the phase diagram. The phase margin is the margin between the
phase plot and -180o.
Locate the point where the phase angle reaches ±180o. Project this back to the gain plot and the gain margin is the margin between this point and the zero db level. If the gain is increased until this is zero, the
system becomes unstable.
Figure 13
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( )
WORKED EXAMPLE No.3
Draw the asymptotes of the Bode plots for the systems having a transfer function
G(s) =
K
T s + 1 (T s 1 2
+ 1)
K is 10, T1 is 2 seconds and T2 is 0.2 seconds.
Find the value of K which makes the system stable.
SOLUTION
The two break point frequencies are 1/T1 = 1/2 = 0.5 rad/s and 1/T2 = 1/0.2 = 5 rad/s.
GAIN
Locate the two frequencies and draw the asymptotes. The first one is 20 db/decade up to 0.5 rad/s.
The second one is -20 db per decade until it intercepts 5 rad/s. From then on it is -40 db per decade.
PHASE
The phase angle diagram is no so easy to construct from asymptotes. Locate the break point
frequencies. These mark the mid points between 0 and 90o
(45o) for both functions. (K has zero
angle). The resultant phase angle varies from 0 to -180o
reaching -135o
half way between the break
points.
Figure 14
The phase angle reaches 180o
at around ω = 110 rad/s. The gain at 110 rad/s is about -60 db hence the gain margin is about 60 db. To make the gain unity (zero db) we need an extra gain of:
60 = 20 log G G = 1000
If the plot is repeated with K = 2 000 it will be seen that the gain margin is about zero.
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1 2
1 2
SELF ASSESSMENT EXERCISE No.2
1. A system has a transfer function
T is 0.1 seconds.
What is the steady state gain? (4)
G(s) =
4
s(T s + 1)
What is the low frequency gain per decade, the high frequency gain per decade and the break point
frequency? (-20 db, -40 db and 10 rad/s)
2. A system has a transfer function
G(s) = (T s
1
+ 1)(T s
+ 1)
T1 is 0.25 seconds and T2 is 0.15 seconds.
Find the low frequency gain per decade, the high frequency gain per decade and the break point
frequencies. (0 db, -40 db and 4 rad/s and 6.7 rad/s)
Find the gain margin and phase margin. (-80 db and 0o
approximately)
3. Draw the asymptotes of the Bode plots for the systems having a transfer function
G(s) = (T s
K
+ 1)(T s + 1)
K is 2, T1 is 0.1 seconds and T2 is 10 seconds.
Find the gain margin and the value of K which makes the system stable. (130 db and 3 x 106
approx)
4. The diagram shows the bode gain and phase plot for a system. Determine the gain margin and whether or not
the system is stable. (45 db and 10o
approx Unstable)
Figure 15
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