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ACS6001 Systems Modelling
Dr George PanoutsosDepartment of Automatic Control and Systems Engineering
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Welcome!
Dr George Panoutsos
Amy Johnson Building (ACSE), Room D8a
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MOLE
Lecture Slides
code, examples
Notes
Extra Material
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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)
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Module Structure Syllabus
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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
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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
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Systems Engineering Approach
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Systems Modelling
Performance
Dynamic AnalysisMathematical Analysis
ModellingAbstraction
Design
System Testing
Computer Simulation
Systems Engineering Approach
Computing
Control
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Systems Modelling
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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
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Systems Engineering
What is systems engineering?
Aslaksen & Belcher, Sys. Eng. Chapter 22.1.1
2.1.2
2.1.3
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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
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?
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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
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y=f(x)
yx
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Systems Modelling
2. Modelling the system Solving the model
The model is only an approximation of the real system!
Validation
Testing
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y=f(x)
yx
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Systems Modelling
3. Computer aided simulation Easier to analyse
complex and/or large scale systems
MATLAB/Simulink
(Dr Z Q Lang)
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Y1=f(x1)Y2=f(x1,x2)Y3=g(x3)Y4=g(x1,x3)Y5= Y3-Y1
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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?
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timeam
plitu
de
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Systems Modelling
Example of the systems engineering approach
Student attention level in lectures
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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
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Example:
The human cardiovascular system
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High abstraction level
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Example:
The human cardiovascular system
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Low abstraction level
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Example:
The human
cardiovascular system
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Very low abstraction level!
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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?)
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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!
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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
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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
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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!)
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but there are limitations:
Validation can be difficult Example?
(People dont trust computers!)
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Mathematical Models
can be applied to a variety of systems:
Engineering
Biomedical
Financial
Weather
+ many more!
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Mathematical Models
For this course:
Model is a set of mathematical equations describing the behaviour of a system
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Modelling errors
Parameters difficult to measure i.e. friction
Time varying parameters Due to wear, aging
Environmental errors Temperature, pressure, humidity
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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
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Example
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Modelling Techniques
Theoretical equations System behaviour is known from theory
Example: electrical circuit
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Modelling Techniques
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)
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Modelling Techniques
3. Fit a mathematical relationship curve fitting, regression or
Neural Networks, Fuzzy Modelling
Example: CV system
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Example:
Mathematical model of CV system (simplified form) Blood pressure (systolic mmHg)
Cortisol level (total metabolites)
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Blood pressure Cortisol
100 5000
120 6000
140 7000
160 8000
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In practice a combination of theoretical/empirical models may be used as necessary
Example: electrical circuit
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Independent study:
Complexity Generality!
regarding systems modelling
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Module Structure Syllabus
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1.Introduction
2.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
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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
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BXAYdtdY
+=
CXBYdtdYA
dtYd
++=22
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Physical System Equation Models
Note that these are both linear equations that can only be applied if a system is approximately linear.
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INPUT XSYSTEM
COMPONENT
OUTPUT Y
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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
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In general:
Electrical, mechanical systems: Approximately linear over certain operating range
Thermal, fluid systems: non-linear
Superposition principle
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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
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What if the system component is non-linear? Can we still apply linear equations?
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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 =
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Multiple inputs
with one output
Suppose
Expanding about and neglecting high order terms:
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),,( 21 nxxxgy =
)()(),,( 20202
10101
02010 xxxgxx
xgxxxgy
xxxxn
+
+===
)( 010
nxxn
xxxg
+=
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Example:
Pendulum
Torque is a non-linear function of
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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
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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 +
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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
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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
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Grossly non-linear systems
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Other non-linearities: delay, backlash A look-up table may be used to describe the system
Saturation
Dead-zone
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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
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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=
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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
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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
=+
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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 ==
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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:
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Module Structure Syllabus
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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
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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
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MS System
Newtons second law of motion states:
Consider a small displacement of mass y downwards:
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= onacceleratibodyofmassbodyonactingforces
KyspringbyexertedForce =yMKyThus =,
0=+KyyMor
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D-MS System
Forces are as before, but we also have the dampers force:
Newtons 2nd law of motion
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yCforceDamper =
yMKyyC =
0=++ KyyCyMor
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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
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Laplace Transform
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Time domain, differential equation
Laplace variable S, algebraic equation
Laplace
Transform
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D-MS System
Solution of:
Laplace transform:
Initial conditions?
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0=++ KyyCyM
[ ] 0)()0()()]0()0()([ 2 =++ sKYyssYCysysYsM
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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!
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[ ] 0)()0()()]0()0()([ 2 =++ sKYyssYCysysYsM
00;)0( 0 === tatyyyLet
0y
(Eq. 1)
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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:
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1;6;5 0 === yMK
MCLet
655)(: 2 ++
+=
ssssYThen
( )( )325)(++
+=
ssssY
-
Design example
Expanding as partial fractions:
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( ) ( )32
23)(
+
+=
sssY ( ) ( )32)(
21
++
+=
sk
sksY
( )( )( )( )
( )( ) 31
335
3252
221 ==+
+=
++++
=== ss s
sssssk
( )( )( )( )
( )( ) 21
225
3253
332 =
=++
=++++
=== ss s
sssssk
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Design example
Performance evaluation
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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
Performance evaluation
c/m=5 -> 3
(damper)
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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
Performance evaluation
k/m=6 -> 10
(spring)
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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
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Design example
Have we finished our design?
No! we cannot apply S domain equations to the system!
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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)( =
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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.
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Transfer function
The relationship between the input and output of a system component is often expressed in terms of the TRANSFER FUNCTION (TF).
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])([])([
txLtyLTF = with all initial conditions zero
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Transfer function
LTI only!
Cannot describe internal system structure (i.e. in/out only)
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Module Structure Syllabus
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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
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DC Motors
Widely applied in industry Robot manipulators
Disc drives
Machine tools
Home appliances
etc.
2
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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
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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
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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)
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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
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DC Motors
Creation of torque: Lorentz Force
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BlidFd
=
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DC Motors
Torque:
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T=I A B sin
where A is the area of the current loop (coil)
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DC Motors
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The DC motor so far works as a compass so something else is needed
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DC Motors
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Brushes Commutator
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DC Motors
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As an approximation
IKT a=
In practice: Several coils are added into to armature to increase torque.
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DC Motors
Back EMF (electro-motive force)
Faradays law of induction (one coil)
(many coils)
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sinsin
)cos(
ABdtdAB
dtBdA
dtde cb
==
=
=
+-
+-
ebbb Ke =
ab KK =
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DC Motors
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baaa edtdILIRe ++= T
dtdF
dtdJ =+ 2
2
Electrical and mechanical equations:
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DC Motors
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Differential equation model of armature controlled DC motors
=
==
=+
=
++=
ab
bb
a
baaa
KKdtdKe
TFJ
IKT
edtdILIRe
/ ,
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DC Motors
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Laplace transform
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DC Motors
Solve ia as a function of ea, then use the result in the torque equation
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DC Motors
Use the new torque result in the mechanical equation
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DC Motors
Use the back EMF equation and solve for theta
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DC Motors
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Armature Motor Load
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DC Motors
Frequently the armature inductance is neglected
La=0
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Differential equation
TF Model
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DC Motors
Frequently the armature inductance is neglected
La=0
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TF Model
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DC Motors
State-space representation: Define state variables:
Define control and input variables:
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DC Motors
DC Motor with a gearbox Inertias, frictions: common reference
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LNMN
nNN ML =
+
-Motor Load
Inertia Jm JLFriction Fm FLAngle m L
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DC Motors
DC Motor with a gearbox Load to motor gear ratio: NL/NM=n
Mechanical equation:
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DC Motors DC Motor with a gearbox
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LM NrNr 21 2 2 = 21 rrNNn MLML ===
mmcmmm FfrTJ = 1
LLcLL FfrJ = 2Tm=
=m
L=
2rfc=
cf
cf
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DC Motors
DC Motor with a gearbox Mechanical equation
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( )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
+ + + =
+ =
= + = +
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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
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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
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DC Motors
7. Use a software package to further investigate/analyse the system (i.e. simulink)
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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
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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
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Module Structure Syllabus
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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
-
What weve seen so far
Dynamic systems modelling: ODEs
Laplace transform
Transfer Function
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TF and matrix form representation For a single equation:
Or:
For a system of multiple simultaneous equations
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)()()( sRsGsY =
)()()()()( 2211 sRsGsRsGsY +=
-
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)()()()()()()()()()(
2221212
2121111
sRsGsRsGsYsRsGsRsGsY
+=
+=Two TFs
or in Matrix Form:
=
)()(
)()()()(
)()(
2
1
2221
1211
2
1
sRsR
sGsGsGsG
sYsY
-
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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
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k
M1c2
c1
M2
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Example:
Complex D-MS system
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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
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Example:
Difficulty: to get the damper forces right
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k
M1c2
c1
M2
x1
x2
w(t)
-
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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
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=
++
++0
)s(W)s(X)s(X
ksCMssCsC)CC(sMs
12
112
11212
2
Transfer function matrix
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Block Diagram
The facility to represent the relationship of a systems variables using diagrammatic means Block Diagram representation
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Modern Control Systems, Dorf & Bishop
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Block Diagram representation
Block diagram: core element in the systems engineering approach
Unidirectional blocks that represent the TF of the variables of interest
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G(s)X Y
Cause and effect relationship (TF)
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Example:
DC Motor TF model:
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G(s) (s)ea(s)
G(s) =
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The gain block
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AX(s) Y(s)
Y(s)=AX(s)
Or A=-5Or k/MOr 1/t
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The transfer function block
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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
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Block diagrams in series
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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
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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:
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24269193512
23
23
++++++
ssssss
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Negative feedback
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G(s)
H(s)
+
-
Y(s)U(s)V(s)
Z(s)
)()(1)()(
:)()()()]()(1[
)]()()()[()()...(),( eliminate
)()()()()()(
)()()(
sHsGsGsT
orsUsGsYsHsG
sYsHsUsGsYsZsV
sYsHsZsVsGsY
sZsUsV
+=
=+
=
=
=
=
-
Negative feedback
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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
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G(s)
H(s)
+
+
Y(s)U(s)V(s)
Z(s)
Equivalent diagram:U(s) Y(s)
)()(1)(
sHsGsG
-
Example:
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012
1asas ++
-
Revisiting the D-MS system
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( ) ( )32
23)(
+
+=
sssY
-
DC Motor
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Armature Motor Load
-
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System representation
We can also use signal flow graphs(independent study)
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5s1
431
23 ++ ss
31
2 +s
+
+
sss
32
4 ++5 8
+
-
Y(s)X(s)
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Module Structure Syllabus
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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
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State space models
Classical Modern Control
The time-domain approach Non-linear
Time-varying
Multi-variable
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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.
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(Dorf & Bishop, Modern Control Sys.)
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State variables For a dynamic system, the state of the
system is described using the state variables:[x1(t), x2(t),, xn(t)]
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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.
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State equations
The differential equations in the state variables relating the system inputs to the system outputs are known as the state equations.
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The D-MS system(with an external force)
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D-MSk
M
cy(t)
U(t))()()()(
:) (tUtKytyCtyM
lectureearlierseeissystemforEquation=++ (1)
-
We can represent this system using two state variables: Mass position
Mass speed
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)(ty)(ty
)()()()(
2
1
tytxtytx
=
=The state variables:
-
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)()()()( 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)
-
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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
-
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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!
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)()()()(
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)
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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
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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
-
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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
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Using state space models
Can use Matlab/Simulink to work directly with state space models
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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