acs6001 - gp - l1-9 single file

ACS6001 Systems Modelling

Dr George PanoutsosDepartment of Automatic Control and Systems Engineering

Welcome!

Dr George Panoutsos

Amy Johnson Building (ACSE), Room D8a

[email protected]

ACS6001 2010-2011 Copyright

2

MOLE

Lecture Slides

code, examples

Notes

Extra Material


3

Module Structure

ACS6001 Systems Modelling 30 Credits for the whole of ACS6001

Assessment: Assignment 12.5% - ACS6001.02b GP Monday 29 Nov. 2010, 4pm

(barcode system)


4

Module Structure Syllabus


5

1.Introduction2.Theoretical & Empirical Equations

3. System Linearisation

4. First Order

Systems

5. Second Order

Systems

6. Transfer Function

Models

7. DC Motor Models

8. Block Diagrams

9. State Space Models

Recommended text (secondary reading)

Aslaksen E. and Belcher R., Systems Engineering, Prentice Hall, 1992

Close C. M., Frederick D. K and Newell J. C., Modelling and Analysis of Dynamic Systems, 3rd Ed., John Wiley & Sons, INC. 2002

Dorf R. C. and Bishop R. H., Modern Control Systems, 9th

Ed., Prentice Hall, 2001

Hung V. V. and Esfandiari R. S., Dynamic Systems Modelling and Analysis, McGraw-Hill, 1998


6

Systems Engineering Approach


7

Systems Modelling

Performance

Dynamic AnalysisMathematical Analysis

ModellingAbstraction

Design

System Testing

Computer Simulation

Systems Engineering Approach

Computing

Control

Systems Modelling


8

Systems Engineering

What is systems engineering?

Systems engineering is defined as the artof designing and optimising systemsstarting with an expressed need andending up with the complete set ofspecifications for all the system elements.

Aslaksen & Belcher, Sys. Eng. Ch.2


9

Systems Engineering

What is systems engineering?

Aslaksen & Belcher, Sys. Eng. Chapter 22.1.1

2.1.2

2.1.3


10

Systems Modelling

1. System description and abstraction We need to

have an understanding of the system be able identify the systems relevant

components be able to suggest an appropriate

abstraction level

Example


11

?

Systems Modelling

2. Modelling the system Solving the model

A mathematical model is a description of a system in terms of equationsClose, Modeling and Analysis of DynSys, pp2-4


12

y=f(x)

yx

Systems Modelling

2. Modelling the system Solving the model

The model is only an approximation of the real system!

Validation

Testing


13

y=f(x)

yx

Systems Modelling

3. Computer aided simulation Easier to analyse

complex and/or large scale systems

MATLAB/Simulink

(Dr Z Q Lang)


14

Y1=f(x1)Y2=f(x1,x2)Y3=g(x3)Y4=g(x1,x3)Y5= Y3-Y1

Systems Modelling

4. System analysis and performance evaluation

Performance related to design/specification

Did we achieve our targets?

If not, what we can do to improve?


15

timeam

plitu

de

Systems Modelling

Example of the systems engineering approach

Student attention level in lectures


16

System Abstraction

What is abstraction and why we need it? simplification

Partitioning the system allows for a reduced complexity

Easier to focus on specific system properties

Fundamental to the systems approach


17

Example:

The human cardiovascular system


18

High abstraction level

Example:

The human cardiovascular system


19

Low abstraction level

Example:

The human

cardiovascular system


20

Very low abstraction level!

Example:

The human cardiovascular system What is the correct abstraction level?

Problem specificIt needs to include the systems properties under

investigation

Sometimes its not possible to get to the right abstraction level! (why?)


21

Systems Modelling

What is a model?

A model is representation of something

The model should be as similar as possible to the

entity that it represents

but can never be exactly same as the real thing!


22

Systems Modelling

Types of model Physical toy model of a car

Pilot plant a small/large scale working model

of a system

Mathematical set of equations describing

the real system


23

If we have the model of a system we can Try out proposed changes,

experimentation

Reduce cost easier, cheaper, faster but not always!

Reduce risk

(People trust computers!)ACS6001 2010-2011 Copyright

24

but there are limitations:

The model is not 100% accurate Approximation (errors? Semester 2)

The system, or specific parameters, may not be possible to be modelled

Can also be time consuming, expensive to develop a model (contradicts advantages!)


25

but there are limitations:

Validation can be difficult Example?

(People dont trust computers!)


26

Mathematical Models

can be applied to a variety of systems:

Engineering

Biomedical

Financial

Weather

+ many more!


27

Mathematical Models

For this course:

Model is a set of mathematical equations describing the behaviour of a system


28

Modelling errors

Parameters difficult to measure i.e. friction

Time varying parameters Due to wear, aging

Environmental errors Temperature, pressure, humidity


29

Good modelling practice

As simple as possible!

Model each component in the system separately

Test each component model

Combine the components models into a model of the full system

Test the full system modelACS6001 2010-2011 Copyright

30

Example


31

Modelling Techniques

Theoretical equations System behaviour is known from theory

Example: electrical circuit


32


Empirical Equations System behaviour is not known from theory

or too complex!

Input-output relationship is measured via practical tests

1. Select and apply an appropriate input signal2. Measure the corresponding output (response)


33


3. Fit a mathematical relationship curve fitting, regression or

Neural Networks, Fuzzy Modelling

Example: CV system


34

Example:

Mathematical model of CV system (simplified form) Blood pressure (systolic mmHg)

Cortisol level (total metabolites)


35

Blood pressure Cortisol

100 5000

120 6000

140 7000

160 8000

In practice a combination of theoretical/empirical models may be used as necessary

Example: electrical circuit


36

Independent study:

Complexity Generality!

regarding systems modelling


1



2

1.Introduction

2.Theoretical & Empirical Equations


4. First Order

Systems

5. Second Order

Systems


Models

7. DC Motor Models

8. Block Diagrams


Physical System Equation Models

Physical system equation models usually take the form of first or second order ordinary differential equations (o.d.e. s)

1st

2nd


3

BXAYdtdY

+=

CXBYdtdYA

dtYd

++=22

Physical System Equation Models

Note that these are both linear equations that can only be applied if a system is approximately linear.


4

INPUT XSYSTEM

COMPONENT

OUTPUT Y


5

Test for linearity:Suppose: and

What does give?

If , system is linear.If not, system is non-linear.

Examples

)()( 11 TYOutputTXInput )()( 22 TYOutputTXInput

)()( 21 TXTXofinput +

)()()()( 2121 TYTYTXTX ++

Superposition principle

Close, Modeling and Analysis of Dyn Sys


6

In general:

Electrical, mechanical systems: Approximately linear over certain operating range

Thermal, fluid systems: non-linear

Superposition principle

Easiest if we apply a linear model to represent the system 1st 2nd order ode

Many system components are close to linear for a normal range of inputs Normal = range of inputs expected under

normal operating conditions


7

What if the system component is non-linear? Can we still apply linear equations?


8


9

Suppose the input (x)/ output (y) relationship is: Carry out Taylor Series expansion about

(1)

If the range is small, the slope at is a good approximation to the curve.

Hence, (1) approximates to:

Or:

where and

(m is the slope of the curve at the operating point).

This can be alternatively written as:

or as: y = m x

Linearisation[ ])()( txgty =

+

+

++====

!3)(

2)()()(

30

03

320

02

2

00

0xx

dxgdxx

dxgdxx

dxdgxgy

xxxxxx

0xx 0xx =

)()( 00

0 xxdxdgxgy

xx

+==

)( 00 xxmyy +=

)( 00 xgy =0xxdx

dgm=

=

)()( 00 xxmyy =

Multiple inputs

with one output

Suppose

Expanding about and neglecting high order terms:


10

),,( 21 nxxxgy =

)()(),,( 20202

10101

02010 xxxgxx

xgxxxgy

xxxxn

+

+===

)( 010

nxxn

xxxg

+=

Example:

Pendulum

Torque is a non-linear function of


11

M

L

sinMgLT =

-4 -3 -2 -1 0 1 2 3 4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-pipi

T

Angle

Torq

ue


12

To form a linear approximation:

Take first derivative and evaluate at equilibrium:

This is reasonably accurate over the range

( )00

sin

=

=dd

MgLT

( )ooMgL 00cos = MgL=

44 +


13

500 600 700 800 900 1000 1100 120040

50

60

70

80

90

100

110

120

130

140

150

Angle: 0.63, MSE : 19.34, STD Res : 4.32

Sampling Time

Blo

od P

ress

ure

Real BME system:Blood pressure monitoring in the ICU

Types of systems

Linear across the whole range

Linear for specific range

Linear only around a specific point

Other types of non-linearity Saturation

Dead-zone


14

Grossly non-linear systems


15

Other non-linearities: delay, backlash A look-up table may be used to describe the system

Saturation

Dead-zone


16

Modelled by a first order o.d.e.

Example 1 A cup of coffee cooling from to room temperature of .

Modelling

It is known from theoretical considerations that the rate of heat loss if proportional to the difference between the temperature of the coffee and room temperature (thus, the rate of heat loss decreases as the temperature of the coffee falls).

It is also known from theoretical considerations that the rate of fall in temperature is proportional to the rate of heat loss.

Thus: (1)

where T is the temperature of the coffee, is the room temperature, t is the time variable and k is a constant.

This will be developed further in a little while.

1st Order Systems

Co100 Co21

)( roomTTkdtdT

=

roomT


17

1st Order SystemsExample 2

A water tank with a steady flow of water into the top of the tank has a leak from the bottom of the tank.

The rate of leakage is proportional to the depth of the water in the tank, H.

Water flows into the tank at a constant rate, S.

Modelling

Rate of inflow = S

Rate of outflow = KH, where K is a constant

Hence, rate of change of water level in tank is:

(2)

This will be developed further in a little while.

KHSdt

dH=


18

Models are created in order to investigate how the real system that the model represents behaves. We may wish to determine the following from the models above:

Example 1 How long will it take for the coffee to cool to ?

Example 2

What level will the water settle down to in steady-state?

What is the depth of water at some general time t?

These questions can be answered by solving the relevant system model.

N.B. solving or simulating a model involves forming an expression for the output variable of the system as a function of time.

Use of models

Co90

Solving example 1

To predict the time t when the coffee has cooled to a temperature of :

Rearrange equation (1):

(3)

We need to find an expression for T in terms of the other system variables.

Unfortunately, it is impossible to get the required expression directly from equation (3) because this equation contains both T and dT/dt terms.

This is a well known problem with first order differential equations.

C90o

roomkTkTdtdT

=+


It was shown last week how to solve this sort of problem. For a system with input u and output x, represented by the equation:

(4)

a solution can be obtained as: (5)

The general solution (5) can be used as a solution to equation (3) if the following substitutions are made:

x = T, a = -k, b = k, u =

;

Thus, kt = 0.1353

We are trying to find a value for t but we have only got a value for kt

Hence, we need to know k.

k depends on the strength of the coffee, and the amount of sugar.

We can find an approximate vale for k but not its exact value.

Hence, although we know the general form of the model from the background theory, there is an error because we do not know the exact value of k.

buaxdtdx

=

abu)1e()0(xe)t(x atat +=

roomT

.

21e7990 kt +=

7969e kt = 1449.16979ekt ==

21)1e(e10090:C90Tat,Thus ktkto ==


Solving equation 2 (Leaking water tank)

What level (value of H) will the water settle down to in steady-state?

What is the level of the water (value of H) at some general time t?

When the water level reaches steady-state, the rate of change of level is zero, i.e. dH/dt=0.

Hence, from equation (2), S = KH [Equation (2) was dH/dt = S KH]

Thus, H = S/K

Note that we need to know the value of the parameter K in order to get H

To get an expression for the value of H at some general time t, we need to solve equation (2).

To do this, we can apply the general solution for a first order system given in equation (5) if we make the following substitutions:

X = H, a = K, b = 1, u = S.ACS6001 2010-2011 Copyright



1



4. First Order

Systems

5. Second Order

Systems


Models

7. DC Motor Models

8. Block Diagrams


2nd order systems

These are modelled by 2nd order ODEs

We will look at two examples: Mass-spring system (MS system)

Damped mass-spring system (D-MS System)

Note: 1)elastic deformation 2) linearity


2

MS System

Newtons second law of motion states:

Consider a small displacement of mass y downwards:


3

= onacceleratibodyofmassbodyonactingforces

KyspringbyexertedForce =yMKyThus =,

0=+KyyMor

D-MS System

Forces are as before, but we also have the dampers force:

Newtons 2nd law of motion


4

yCforceDamper =

yMKyyC =

0=++ KyyCyMor

Solution of 2nd order system equations Output = f(t)

Analytical type of solution (as the one developed for 1st order systems) is very difficult for 2nd order systems!

We can use Laplace transform instead


5

Laplace Transform


6

Time domain, differential equation

Laplace variable S, algebraic equation

Laplace

Transform

D-MS System

Solution of:

Laplace transform:

Initial conditions?


7

0=++ KyyCyM

[ ] 0)()0()()]0()0()([ 2 =++ sKYyssYCysysYsM

D-MS System

Initial conditions:

This means displacement of mass M at time zero is and initial velocity is zero.

Note: If we specify that initial conditions are zero, we must check that this is true ------ we cannot just assume that initial conditions are zero!


8

[ ] 0)()0()()]0()0()([ 2 =++ sKYyssYCysysYsM

00;)0( 0 === tatyyyLet

0y

(Eq. 1)


9

Inserting initial conditions in equation (1) gives:

i.e. (Eq.2)

Equation (2) can be written as:

where and (Eq.3)

The equation B(s) = 0 is known as the characteristic equation of the system.

The roots of the characteristic equation are the values of s which make B(s) = 0

The roots are also known as the poles of the system.

The poles determine the character of the response (i.e. whether it is oscillatory or not).

The roots of A(s) are known as the zeros of the system.

At the zeros, the function for Y(s) in equation (3) becomes equal to zero.

0)()()( 002 =++ sKYCysCsYMsysYMs

( )KCsMs

yCMssY++

+= 2

0)(

)()()(

2

0

sBsA

MKs

MCs

yMCs

sY =++

+

= 0)( yMCssA

+=

MK

MCsssB ++= 2)(

Design example

returning to equation 3:

Factorising the denominator:


10

1;6;5 0 === yMK

MCLet

655)(: 2 ++

+=

ssssYThen

( )( )325)(++

+=

ssssY

Design example

Expanding as partial fractions:


11

( ) ( )32

23)(

+

+=

sssY ( ) ( )32)(

21

++

+=

sk

sksY

( )( )( )( )

( )( ) 31

335

3252

221 ==+

+=

++++

=== ss s

sssssk

( )( )( )( )

( )( ) 21

225

3253

332 =

=++

=++++

=== ss s

sssssk

Design example

Performance evaluation


12

0 0.5 1 1.5 2 2.5 3 3.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9Step Response

Time (sec)

Ampl

itude

Design example


c/m=5 -> 3

(damper)


13

0 0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7Step Response

Time (sec)

Ampl

itude

Design example


k/m=6 -> 10

(spring)


14

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45Step Response

Time (sec)

Ampl

itude

Design example

Have we finished our design?

No! we cannot apply S domain equations to the system!


15


16

Finally, we require y(t).

Apply Inverse Laplace Transform:

(using standard Laplace transform tables)

Laplace transform table shows that:

Hence, (Eq.4)

For first term in Y(s), and for second term

Hence, applying inverse Laplace transform on Eq.4,

[ ] ( )aseLat

=

1

ateas

L =

11

( ) ( )32

23)(

+

+=

sssY

2=a 3=a

tt eety 32 23)( =

Simulation languages Clearly, the process of simulation by solving the model

equations is tedious. Because of this, simulation is usually carried out by using

purpose-designed simulation languages Different simulation languages require the system to be

described in different ways. One way is to express the system in terms of first and

second order ordinary differential equations (as just developed in recent lectures).

Another way is to express the system in terms of the transfer functions of its components.


17

Transfer function

The relationship between the input and output of a system component is often expressed in terms of the TRANSFER FUNCTION (TF).


18

])([])([

txLtyLTF = with all initial conditions zero

Transfer function

LTI only!

Cannot describe internal system structure (i.e. in/out only)


19



1



4. First Order

Systems

5. Second Order

Systems


Models

7. DC Motor Models

8. Block Diagrams


DC Motors

Widely applied in industry Robot manipulators

Disc drives

Machine tools

Home appliances

etc.

2


DC Motors

Benefits Provide accurate speed/position control

Portable

High torque

Drawbacks Not enough power for certain applications (use

hydraulics instead)

Difficult to use in gas explosion risk environments

3


DC Motors

Main components: A rotor (armature)

which rotates and therefore creates the torque for the load

A stator (does not move) which is used to create a magnetic field for the rotor


4

DC Motors

Electromagnetic phenomena: Torque creation due to a current in a

conductor (rotor) that is exposed to a magnetic field (created by the stator)

Induction of back EMF to the rotor due to a moving conductor (rotor) in a magnetic field (created by the stator)


5

DC Motors

Physics: Some units that well use: Force: N=kgm/s2

Torque: Nm

Current: A (Ampere)=C/s (Coulomb/s)

Magnetic Field: T (Tesla)=N/(Am)

Magnetic flux: Wb (Webber) = Tm2


6

DC Motors

Creation of torque: Lorentz Force


7

BlidFd

=

DC Motors

Torque:


8

T=I A B sin

where A is the area of the current loop (coil)

DC Motors


9

The DC motor so far works as a compass so something else is needed

DC Motors


10

Brushes Commutator

DC Motors


11

As an approximation

IKT a=

In practice: Several coils are added into to armature to increase torque.

DC Motors

Back EMF (electro-motive force)

Faradays law of induction (one coil)

(many coils)


12

sinsin

)cos(

ABdtdAB

dtBdA

dtde cb

==

=

=

+-

+-

ebbb Ke =

ab KK =

DC Motors


13

baaa edtdILIRe ++= T

dtdF

dtdJ =+ 2

2

Electrical and mechanical equations:

DC Motors


14

Differential equation model of armature controlled DC motors

=

==

=+

=

++=

ab

bb

a

baaa

KKdtdKe

TFJ

IKT

edtdILIRe

/ ,

DC Motors


15

Laplace transform

DC Motors

Solve ia as a function of ea, then use the result in the torque equation


16

DC Motors

Use the new torque result in the mechanical equation


17

DC Motors

Use the back EMF equation and solve for theta


18

DC Motors


19

Armature Motor Load

-

DC Motors

Frequently the armature inductance is neglected

La=0


20

Differential equation

TF Model

DC Motors

Frequently the armature inductance is neglected

La=0


21

TF Model

DC Motors

State-space representation: Define state variables:

Define control and input variables:


22

DC Motors

DC Motor with a gearbox Inertias, frictions: common reference


23

LNMN

nNN ML =

+

-Motor Load

Inertia Jm JLFriction Fm FLAngle m L

DC Motors

DC Motor with a gearbox Load to motor gear ratio: NL/NM=n

Mechanical equation:


24

DC Motors DC Motor with a gearbox


25

LM NrNr 21 2 2 = 21 rrNNn MLML ===

mmcmmm FfrTJ = 1

LLcLL FfrJ = 2Tm=

=m

L=

2rfc=

cf

cf

DC Motors

DC Motor with a gearbox Mechanical equation


26

( )2 2

2 2

( )

or

where:,

m L m m L m m

m m m

m L m L

J n J F n F T

J F T

J J n J F F n F

+ + + =

+ =

= + = +

DC Motors

The modelling process:

1. Schematic diagram of the system Real system -> diagram

2. Mathematically describe the physical laws governing the system

3. Apply the physics/equation to the system


27

DC Motors

4. Derive an output-input set of equations describing the process under investigation

5. Apply Laplace transform to bring the equations to a transfer function format

6. Draw a block diagram of the process/system in TF format


28

DC Motors

7. Use a software package to further investigate/analyse the system (i.e. simulink)


29

Conclusions: DC Motors

Transfer functions for systems under investigation are not typically available Control/Systems engineer to derive them!

Physical models are very useful Only if the underlying physics are

known/understood/straightforward


30

Conclusions: DC Motors

Physical models allow the engineer to: See the effect of various system parameters

Give physical meaning to the states of the system

Get a better understanding of the limitations of the model

modelling is the most time consuming stage of the control design workflowACS6001 2010-2011 Copyright

31



1



4. First Order

Systems

5. Second Order

Systems


Models

7. DC Motor Models

8. Block Diagrams


What weve seen so far

Dynamic systems modelling: ODEs

Laplace transform

Transfer Function


2

TF and matrix form representation For a single equation:

Or:

For a system of multiple simultaneous equations


3

)()()( sRsGsY =

)()()()()( 2211 sRsGsRsGsY +=


4

)()()()()()()()()()(

2221212

2121111

sRsGsRsGsYsRsGsRsGsY

+=

+=Two TFs

or in Matrix Form:

=

)()(

)()()()(

)()(

2

1

2221

1211

2

1

sRsR

sGsGsGsG

sYsY


5

In general, for J inputs and I outputs

=

)(

)()(

)(

)(

)(

)()()(

)(

)()(

2

1

2

1

21

111

2

1

sR

sRsR

sG

sG

sG

sGsGsG

sY

sYsY

JIJ

J

I

J

I

Why would we be interested in the matrix form?

Example:

Complex D-MS system


6

k

M1c2

c1

M2

Example:

Complex D-MS system


7

k

M1c2

c1

M2

Review of modelling strategy(1) Split system into its separate component parts

(2) Model each component separately

(3) Combine the component models into a model of the full system

Here, there are two main components, M1 and M2

Example:

Difficulty: to get the damper forces right


8

k

M1c2

c1

M2

x1

x2

w(t)


9

Equating forces:

Newtons 2nd law of motion: ODE (1)

Equating forces:

Newtons 2nd law of motion: ODE (2)

1MonForces)t(Kx:forceSpring 1

])t(x)t(x[C:forceDamper 211 )MtorespectwithMofvelocityrelative 21)t(xM:forcengAccelerati 11

2MonForces

)t(W:forceDisturbing

])t(x)t(x[C)t(xC:forcesDamper 12122 )t(xM:forcengAccelerati 22


10

=

++

++0

)s(W)s(X)s(X

ksCMssCsC)CC(sMs

12

112

11212

2

Transfer function matrix

Block Diagram

The facility to represent the relationship of a systems variables using diagrammatic means Block Diagram representation


11

Modern Control Systems, Dorf & Bishop

Block Diagram representation

Block diagram: core element in the systems engineering approach

Unidirectional blocks that represent the TF of the variables of interest


12

G(s)X Y

Cause and effect relationship (TF)

Example:

DC Motor TF model:


13

G(s) (s)ea(s)

G(s) =

The gain block


14

AX(s) Y(s)

Y(s)=AX(s)

Or A=-5Or k/MOr 1/t

The transfer function block


15

F(s)X(s) Y(s)

F(s): any transfer function

Example: first order system: F(s)=A/(s+5)

the integrator: F(s) = 1/s

Block diagrams in series


16

F(s)X(s) Y(s) G(s) V(s)

Equivalent diagram:

F(s)G(s)X(s) V(s)

Numerical example

Block diagrams in parallel


17

F(s)

X(s)

V1(s)

G(s) V2(s)

Y(s)+

Equivalent diagram:F(s)+G(s)X(s) Y(s)

Numerical example

Example:


18

24269193512

23

23

++++++

ssssss

Negative feedback


19

G(s)

H(s)

+

-

Y(s)U(s)V(s)

Z(s)

)()(1)()(

:)()()()]()(1[

)]()()()[()()...(),( eliminate

)()()()()()(

)()()(

sHsGsGsT

orsUsGsYsHsG

sYsHsUsGsYsZsV

sYsHsZsVsGsY

sZsUsV

+=

=+

=

=

=

=

Negative feedback


20

Equivalent diagram:U(s) Y(s)

Numerical example (H: unity)

)()(1)(

sHsGsG

+

G(s)

H(s)

+

-

Y(s)U(s)V(s)

Z(s)

Positive feedback


21

G(s)

H(s)

+

+

Y(s)U(s)V(s)

Z(s)

Equivalent diagram:U(s) Y(s)

)()(1)(

sHsGsG

Example:


22

012

1asas ++

Revisiting the D-MS system


23

( ) ( )32

23)(

+

+=

sssY

DC Motor


24

Armature Motor Load

-

System representation

We can also use signal flow graphs(independent study)


25


26

5s1

431

23 ++ ss

31

2 +s

+

+

sss

32

4 ++5 8

+

-

Y(s)X(s)



1



4. First Order

Systems

5. Second Order

Systems


Models

7. DC Motor Models

8. Block Diagrams


State space models

Classical Modern Control

The time-domain approach Non-linear

Time-varying

Multi-variable


2

State variables Definition: The state of a system is a set of

variables such that the knowledge of these variables and the input functions will, with the equations describing the dynamics, provide the future state and output of the system.


3

(Dorf & Bishop, Modern Control Sys.)

State variables For a dynamic system, the state of the

system is described using the state variables:[x1(t), x2(t),, xn(t)]


4

Simple Example:

ON/OFF switch!

The state variables allow the system outputs to be calculated for all time provided that the inputs are known for all time and the initial conditions are known.


5

State equations

The differential equations in the state variables relating the system inputs to the system outputs are known as the state equations.


6

The D-MS system(with an external force)


7

D-MSk

M

cy(t)

U(t))()()()(

:) (tUtKytyCtyM

lectureearlierseeissystemforEquation=++ (1)

We can represent this system using two state variables: Mass position

Mass speed


8

)(ty)(ty

)()()()(

2

1

tytxtytx

=

=The state variables:


9

)()()()( tUtKytyCtyM =++

Substituting in the state variables: )()()()( 122 tUtKxtCxtxM =++

What happened to the 2nd order dynamics? State space representation of systems: not unique!

=

=

MtCx

MtKx

MtUtx

txtx)()()()(

)()(

212

21

State space representation

(1)

(2)

(3)


10

We may also select the following state variables:

)()(

)()(2

2

1

tyKtxtKytx=

=

=

=

MtCx

MtxK

MtUKtx

Ktxtx)()()()(

/)()(

2122

2

21

=++ )()()()( tUtKytyCtyM )()()()( 12222 tUtxtxKCtx

KM

=++Using Eq.1 :

State space representation


11

UMx

x

MC

MKx

x

+

=

1010

2

1

2

1

=

=

MtCx

MtKx

MtUtx

txtx)()()()(

)()(

212

21

This is usually expressed in matrix form:

State space representation (Eq.3)

Linear Dependency

The state variables must be linearly independent!


12

)()()()(

2

1

tKytxtytx

=

=

this will not result in 1st order odes

Linear Dependency

In general: For a system of order n, if [x1(t), x2(t),, xn-1(t)]

are state variables then xn(t) is not a legitimate system variable if it can be expressed as a liner combination of the other state variables such as:

xn(t)=c1x1(t),+c2x2(t)+ + cn-1xn-1(t)


13


14

In general, for a system with n state variables and m inputs:

And in matrix form:

+

=

m

1

nm1n

m111

n

21

nn1n

2221n11211

n

21

u

u

bb

bb

x

xx

aa

aaaaa

x

xx


15

n

21

x

xx

is known as the state vector

Usually the state space models are expressed in the following form: uBxAx +=

,vectorstatetheisxwhere

m

1

u

u

vectorinputtheisu

=

nn1n

n111

aa

aa

A

=

nm1n

m111

bb

bb

B


16

xCy =The output y of the system can be expressed as:

=

n

1

y

y

ywhere

=

nn1n

n111

cc

cc

C

Using state space models

We can work directly with the governing differential equations of the system in the time domain

1st order ODEs: easier to solve

Treat multivariable systems as SISO

Directly design and analyse non-linear systems


17

Using state space models

Can use Matlab/Simulink to work directly with state space models


18

ACS6001 - GP - L1-2ACS6001 Systems ModellingWelcome!MOLEModule StructureModule Structure SyllabusRecommended text (secondary reading)Systems Engineering ApproachSystems ModellingSystems EngineeringSystems EngineeringSystems ModellingSystems ModellingSystems ModellingSystems ModellingSystems ModellingSystems ModellingSystem AbstractionExample:Example:Example:Example:Systems ModellingSystems ModellingIf we have the model of a system we canbut there are limitations:but there are limitations:Mathematical ModelsMathematical ModelsModelling errorsGood modelling practiceSlide Number 31Modelling TechniquesModelling TechniquesModelling TechniquesExample:Slide Number 36

ACS6001 - GP - L3-4Independent study:Module Structure SyllabusPhysical System Equation ModelsPhysical System Equation ModelsSlide Number 5Slide Number 6Slide Number 7What if the system component is non-linear?Slide Number 9Multiple inputsExample:Slide Number 12Slide Number 13Types of systemsGrossly non-linear systemsSlide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21

ACS6001 - GP - L5-6Module Structure Syllabus2nd order systemsMS SystemD-MS SystemSolution of 2nd order system equationsLaplace TransformD-MS SystemD-MS SystemSlide Number 9Design exampleDesign exampleDesign exampleDesign exampleDesign exampleDesign exampleSlide Number 16Simulation languagesTransfer functionTransfer function

ACS6001 - GP - L7Module Structure SyllabusDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsDC MotorsConclusions: DC MotorsConclusions: DC Motors

ACS6001 - GP - L8Module Structure SyllabusWhat weve seen so farTF and matrix form representationSlide Number 4Slide Number 5Example:Example:Example:Slide Number 9Slide Number 10Block DiagramBlock Diagram representationExample:The gain blockThe transfer function blockBlock diagrams in seriesBlock diagrams in parallelExample:Negative feedbackNegative feedbackPositive feedbackExample:Revisiting the D-MS systemDC MotorSystem representationSlide Number 26

ACS6001 - GP - L9Module Structure SyllabusState space modelsState variablesState variablesSimple Example:State equationsThe D-MS system(with an external force)Slide Number 8Slide Number 9Slide Number 10Slide Number 11Linear DependencyLinear DependencySlide Number 14Slide Number 15Slide Number 16Using state space modelsUsing state space models

acs6001 - gp - l1-9 single file

Documents

systems engineeringwelcome

systems modelling1

copyright 14y1

analysis of dynamic

system analysis

system abstraction

modern control systems

system description