Introduction to Control Systems - · PDF fileIntroduction to Control Systems In this lecture,...

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 n

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


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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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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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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o

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.


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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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.


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


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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1. Find the closed loop transfer function for the system shown below.

Figure 9


Figure 10


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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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


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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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



(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



(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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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.


• 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


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


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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