Introduction to Control Systems Chapter 2.pptx

Department of Electrical & Computer Engineering
Introduction to Control Systems
Addis Ababa Science &
Technology University
College of Electrical & Mechanical
Engineering
Talegeta M.

2
Chapter 2:
Modelling and Representation of Physical Systems
Contents
i. Introduction
ii. Review on Laplace transform
iii. Control system models
a) Differential equation
b) Transfer function
c) State space
iv. System representation
a) Block diagrams
b) Signal flow graph
v. System mathematical modelling
a) Mathematical modeling of electrical networks
b) Mathematical modeling of mechanical systems
o Transitional
o Rotational
o Mechanical system with gears
c) Mathematical modelling of electromechanical system
vi. Block diagram algebra
vii. Signal flow graph (reading assignment)
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2.1 Introduction
- Mathematical models of physical systems are key
elements in the design and analysis of control systems
- Transfer functions: a mathematical representation
to describe relationship between inputs and outputs of
the physics of a system, i.e., of the differential
equations that govern the motion of bodies.
- Desired outcomes :
- Understand the application of Laplace transforms and
their role in obtaining transfer functions of physical
system.

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• Approaches for solving dynamic system
problem
oDefine the system and its components
oFormulate the mathematical model and list
the necessary assumptions.
oWrite the differential equations describing the
model.
oSolve the equations for the desired output
variables
oExamine the solutions and the assumptions
oIf necessary, reanalyse or redesign the system
2.1 Introduction

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Modeling of dynamic systems
Model: A representation of a system.
Types of Models:
a) Physical models (prototypes)
b) Mathematical models (e.g., input-output relationships)
o Analytical models (using physical laws)
o Computer (numerical) models
o Experimental models (using input/output experimental data)
Models for physical dynamic systems:
o Lumped-parameter models
o Continuous-parameter models.
Example: Spring element (flexibility, inertia, damping)
2.1 Introduction

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Mathematical model comparison between continuous and
discrete control systems
2.1 Introduction

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NB:
o Even if most physical systems are nonlinear, we
will consider as they are linear or we will take
linearization approximations, which allow us
to use Laplace transform methods.
o In this course only Continuous-time LTI system
is considered.
2.1 Introduction

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Laplace & inverse Laplace transform
Let f(t) be a function of time t such that f(t) = 0 for t < 0 and s is
a complex variable. We use L as a symbolic operator to stand
for Laplace integral on the quantity that is prefixed by it.
The Laplace transform of f(t) is defined as
L [f(t)]=F(s)=
F(s) denotes the Laplace transform of f(t).
The function f(t) in time domain can be found from the Laplace
transform F(s) by the reverse process known as the inverse
Laplace transformation
Where;
Where;
2.2 Review on Laplace Transform
Laplace Transform: takes a function of t (time) to a function of a complex
variable s (frequency)

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Laplace transform of a common function.
i) Exponential function for otherwise.
F(s)=L []=
F(s)=L []=
F(s)=L []=
F(s)=L []=
ii) Step function
for
for
-where A is a constant and us(t) is an unit step function.
-The unit-step function has a height of unity
F(s)=L []=
F(s)=L []=
F(s)=L []=

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iii) Ramp function
for
for
-where R is the slope of a linear function and us(t) is an unit step function.
-The unit-step function has a height of unity
F(s)=L []=R
F(s)=L []=
F(s)=L []=
iv) Sinusoidal function
for for
Where
F(s)=L []=
F(s)=L []=

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F(s)=L []=
F(s)=L []=
F(s)=L []=
F(s)=L []=
Performing the same calculation, one can obtain
F(s)=L []=
Properties of Laplace Transform
1- Multiplication by a constant, L [
L [
2- Multiplication by . L [
3-Translation in time , , L [

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4- Change in time scale, , L [
5- Real differentiation , L
L
6-Initial value theorem,
7-Final value theorem
8-Real integration
L
9-Complex differentiation , L
10- Convolution integral, L

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Laplace transform table of some basic signals
2.2 Review on Laplace Transform

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Laplace transform tables

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Example:
Find the transfer function represented by
 Taking the Laplace transform of the differential at zero
initial conditions
 Then the transfer function becomes:

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Control system model & representation
oDifferential equation
oTransfer function
oState space
oBlock diagrams
oSignal flow graph
2.3 Control System Representation

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Example : Continuous-time LTI system
)
(
)
(
)
(
)
(
2
2
t
u
t
y
dt
t
dy
RC
dt
t
y
d
LC 


constants
2nd
order ordinary
differential equation
1
1
)
(
)
(
)
(
)
(
)
(
)
(
2
2






RCs
LCs
s
U
s
Y
s
U
s
Y
s
RCsY
s
Y
LCs
Transfer function
)
(s
U )
(s
Y
1
1
2

 RCs
LCs Block diagram
Y(s)
U(s) 1
1
2

 RCs
LCs
Signal flow graph

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Transfer function & Block diagrams from
differential equations
• In control system differential equations can
describe the relation ship b/n the input and
output of the system.
• Which relates the output c(t) to the input
r(t) by taking system parameters and
• Using Laplace transform we transfer this
differential equation into transfer function of the
system.

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• Taking all initial conditions zero, the Laplace transform
becomes:
• Then, the input output ratio becomes;
• We call it this ratio of output signal to input signal,
G(s), as the Transfer Function of the system.
• Block diagrams

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2.4 System Mathematical Modelling
Mathematical Modelling of Electrical Networks
oElectrical circuits with passive elements:
Example:
Determine the mathematical modelling and the transfer
function for the following electrical system considering
the voltage source as an input & voltage across the
capacitor as an output.

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Solution
Substituting current
Voltage-charge relation
Apply KVL
around the loop

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• Taking the Laplace transform assuming all initial
conditions zero:
• Solving the transfer function
Vc(s)/V(s)-Output/Input, we obtain:
• Block diagram representation:

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• To simplify solving our problem, we can take the
Laplace transform equation of voltage-current
relation for all the three electrical elements as
follows(assuming zero initial conditions):
oFor the Capacitor:
oFor Resistor:
oFor Inductor:
oIn general for any impedance combination:

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• Hence, the above problem can be easily solved as:
• Solving V/I
• Calculating voltage across the capacitor in terms of
loop current I:
• Then the mathematical modelling of the system in
the form of TF will be:

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oElectrical circuits with active elements:
• Operational amplifier
It is an electronic amplifier used as a basic
building block for implementing transfer functions:
Circuit diagram for operational amplifier

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Ideal operational Amplifier

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• Characteristics of operational Amplifier:
o Differential input : V2(t) – V1(t)
o High input impedance : Zi = ∞ (ideal)
o Low output impedance: Zo = 0 (ideal)
o High constant gain amplification A= ∞ (ideal)
o Hence, the output Vo(t) = A(V2(t) – V1(t))
• Operational amplifiers:
o Inverting operational amplifier
o Non-inverting operational amplifier

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• Inverting operational amplifier
• From the characteristics of operational amplifier,
since the input impedance is high =0 and from
KCL:
If V2(t) is grounded, the
amplifier is called an
inverting operational
amplifier.
I1(s) = -I2(s)

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• Considering the gain A is large VI(t) = 0,
thus
oI1(s) = Vi(s)/Z1(s) and
oI2(s) = Vo(s)/Z2(s)
• Equating the above two equation the
transfer function for an inverting
amplifier configured as above becomes:

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Example:
Represent the following circuit with its
equivalent mathematical modelling in
terms of transfer function:

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Solution
What will be the mathematical modelling in terms
of differential equation ?

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• Non- Inverting Operational Amplifier:
General non inverting
operational amplifier
Using voltage division
Substituting the above
equations:
For large gain A:

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Example:
Find the Transfer function for the network
below(determine the mathematical modelling in
frequency domain)

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Tip:
• Transfer function & connection diagram for
analogue integrator and differentiator operational
amplifiers.
V2 s
( )
V1 s
( )
1

RCs
V2 s
( )
V1 s
( )
RCs


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Mathematical Modelling for Mechanical
Systems
oTransitional mechanical systems

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Example
Determine the transfer function for the mechanical system taking
the force on the mass as an input and the displacement as output.
f(t) control force
fe (t)= k x(t) linear spring, elastic force,
fa (t) = µv(t) linear viscous friction, friction force

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Solution
• Free body diagram in time and frequency domain
• Applying newton’s law: ∑ f = 0
In deferential equation form
Taking Laplace transform
The transfer function becomes:
m a(t) = f(t) - fe (t) - fa (t)

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• In Mechanical Systems, the number of equations of motion
required is equal to the number of linearly independent
motion which implies a point of motion in a system can still
move if other points of motion are held still or the vice versa.
• It is also called Degree of Freedom.
Exercise: Find the transfer function,
Answer:

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oRotational Mechanical Systems:

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Example:
Find the transfer function for the following
rotational mechanical system ϴ2(s)/T(s):

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Solution
Torques on J1
due only to the
motion of J1
Torques on J1
due only to the
motion of J2
Final free-body
diagram for J1
For the first momentum

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• For the second momentum
Torques on J2
due only to the
motion of J1
Torques on J2 due
only to the motion
ofJ1
Final free-body
diagram for J 2

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• From the above two free body diagram:
Block diagram

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oMechanical Systems with Gears:
• Most of the systems which are driven by motors
are associated with gear trains for driving the
load, which provides mechanical advantage for
rotational systems.
o An input gear with radius r1
and N1 teeth is rotated through
angle 1(t) due to a torque,
ϴ
T1(t).
o An output gear with radius r2
and N2 teeth responds by
rotating through angle 2(t) and
ϴ
delivering a torque, T2(t)-.
o Let us now find the relation-ship
between the rotation of
Gear1, 1(t) and Gear2, 2(t)
ϴ ϴ

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• The distance travelled along each gear
circumstance is equal:
• The number of teeth along the circumstance is
proportional with the radius:
• Torque relation for input and output gear:
• From the above equations we get:

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Example:
Mechanical impedances (spring, damper,
inertia) which are driven by gears:

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Solution
a) Equivalent
system at the
output after
reflection of input
torque
b) Equivalent
system at the input
after reflection of
impedances

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• Hence the equation of motion for the first case:
• Substituting 2 with 1,
ϴ ϴ 2= 1
ϴ ϴ
• After simplification
Which gives representation in the form of part b

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Exercise:
Find the transfer function for the following system
taking the angular displacement of the second
inertia as an output when we apply an input torque
on the first inertia:

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oMathematical Modelling for
Electromechanical Systems:
• Most systems in Mechatronics are of the mixed type,
e.g., electromechanical, hydromechanical, etc
• Electromechanical systems are systems which have
both electrical and mechanical variable.
• It has different application areas: Robot control,
Trackers (ex. Sun & star tracking), Computer tape
& Disk drives, different servo applications.
• Motor is one of the electromechanical system which
yields a displacement output taking voltage as
an input:

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Example:
• Mathematical modelling for DC motor
(Electromechanical system)
Input: Voltage u
Output: Angular velocity, 
or angular position, ϴ
ea
ia
dc
Ra La
J
m
Dm
eb

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Solution
voltage
emf
-
back
e
,
e
dt
di
L
i
R
e b
b
a
a
a
a
a 



m
m
m Dω
ω
J
T 
 
Tm = Te
Mechanical torque = Electrical torque

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• Power transformation
• Combing previous equations results in the
following mathematical model:
• Taking Laplace transform of the system’s
differential equations with zero initial conditions
gives:
a
t
m i
K
T 
m
b
b ω
K
e 










0
a
a
m
b
i
e
t
m
m
a
a
a
a
K
-
Dω
ω
J
K
i
R
dt
di
L



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• Eliminating Ia yields the input-output transfer
function
• Most of the time Ra >>La for DC motors. Hence, we
can take La = 0.
• Assuming small inductance, La  0
{¿(La 𝑠+Ra)I𝑎(𝑠)+K𝑏 Ω𝑚 (𝑠)=𝐸𝑎(𝑠)
¿(Js+D)Ω𝑚 (s)− Kt 𝐼𝑎(𝑠)=0
Ωm(s)
Ea(s)
=
Kt
La J s
2
+(J Ra+D La) 𝑠+ D Ra+K t 𝐾𝑏
Kt/Ra
𝐽𝑠+(𝐷+𝐾𝑡𝐾𝑏/𝑅𝑎)

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Reading assignment 1
• Mathematical Modelling for fluid Systems:
o Hydraulic and
o Pneumatic systems
• Mathematical Modelling for thermal Systems:

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Block diagram elements
oSumming Junction
oPickoff Points
oSignals
oSystem/subsystem
Pickoff Point
2.5 Block Diagram Algebra

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Block diagram reduction
• Block arrangements
oCascade/series
oParallel
oFeedback
Cascaded/series blocks
Considering that the
interconnected subsystem has
no loading effect on the
adjacent subsystem.

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Parallel blocks
Blocks with their input signals has the same takeoff
point and their output signal sinks at the same summing
junction.

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Feedback blocks
Can be negative feedback or Positive feedback
C(s) , G (s )=
𝐶(𝑠)
𝑅 (𝑠 )

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Rules
Moving blocks
If necessary for block reduction move blocks before and
after a summing points, pickoff points.
1. Moving block before a summing point
2. Moving block ahead a summing point

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3. Moving block before a pickoff point
4. Moving block ahead a pickoff point

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• Block diagram algebra table

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• Block diagram algebra table

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Example
Calculate the transfer function for the following system
in terms of subsystems transfer functions.

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• Solution
Moving takeoff point
beyond a block
Cascading +
positive feedback
Cascading + negative feedback
Cascading +
negative feedback

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Exercise:
Find the transfer function G(s)= C(s)/R(s)
Answer:
2
2
1
)
( 3




S
S
s
G
3
s

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2.6 Signal flow Graph
• Signal-flow graphs are an alternative to block diagrams.
• Unlike block diagrams, which consist of blocks, signals,
summing junctions, and pickoff points,
• A signal-flow graph consists only of branches which
represent systems, and nodes which represent signals.
• A signal is a node with the signal's name written
adjacent to it.
R(s) C(s)
)
(s
G
)
(s
R )
(s
C

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oTerminologies on Signal flow graphs
• Node: It is a point representing a variable.
o Input Node : Node which has only outgoing branches.
X1 is input node.
o Output node/ sink node: Only incoming branches,
X2.
o Mixed nodes: Has both incoming and outgoing
branches.
• Branch : A line joining two nodes.
• Transmittance : It is the gain between two
nodes. It is generally written on the branch near
the arrow i.e. G.
branch node
X2
X1 G
node

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Block diagram and signal flow graphs comparison

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NB:
oIf a summing point is placed before a take off point in
the direction of signal flow, in such a case the summing
point and take off point shall be represented by a single
node.
oIf a summing point is placed after a take off point in
the direction of signal flow, in such a case the summing
point and take off point shall be represented by separate
nodes connected by a branch having transmittance unity.

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• Example
Converting complicated Block diagram to signal flow
graph.

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oSignal flow graph reduction
• It is a technique for reducing signal-flow graphs to
single transfer functions that relate the output of a
system to its input.
• To reduce signal flow graphs we will follow
Mason’s Rule.
• First let’s define terminologies which helps to
evaluate Mason’s Rule.
• Path: It is the traversal of connected branches in
the direction of branch arrows, such that no node is
traversed more than once.

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• Forward path : A path which originates from the
input node and terminates at the output node and
along which no node is traversed more than once.
• Forward Path gain : It is the product of branch
transmittances of a forward path.
• Loop : Path that originates and terminates at the
same node and along which no other node is
traversed more than once.
• Self loop: Path that originates and terminates at
the same node.
o Loop gain: it is the product of branch transmittances
of a loop.

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• Non-touching loops: Loops that don’t have any
common node or branch.
Example 1
o Forward Path: direct path from R(s) to C(s)
o Forward path gain 1: G1G2G3G4
o Forward path gain 2: G5G6G7G8
L 1 = G2 H2 L 2 = H3
L3= G7 H7
Non-touching loops are L1 & L2, L1 & L3, L2 &L3
Self loop: L2 =H3

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Example 2:
It has four loop gain

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• For the above flow graph: we have two forward paths
and their corresponding gain becomes:
• Non touching loops:
• Non touching loop gain
o The product of loop gains from non-touching loops taken
two, three, four, or more at a time
• All three of the non-touching-loop gains taken two at
a time are

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• In our example there are no non-touching-loop
gains taken three at a time since three non-
touching loops do not exist.
At this point we can drive Mason’s Equation:
• The Transfer function C(s)/R(s):
Where;
k= number of
forward paths
Tk= the k forward-
path gain
th

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 ∆ = 1 – (Σ loop gains) + (Σ non-touching loop
gains taken two at a time) – (Σ non-touching loop
gains taken three at a time)+ so on.
 ∆ k = 1 – (loop-gain which does not touch the Kth
forward path)
Example:
Find the transfer function C(s)/R(s) from the signal flow
graph.

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Solution:
• It has only one forward path gain:
• Identifying loop gains: there are fours
• Identifying non-touching loop and calculating
gains:
• Taken two at a time:
T1 =

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Taken three at a time:
Calculating
(Since we have only one forward path)
Substituting the above equation in to the Mason’s rule:

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Exercise:
Reduce the signal flow graph

Introduction to Control Systems Chapter 2.pptx

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