Csl2 16 j15

ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
Control Systems
Lecture 2

Today’s class
• System
• Transfer Function
• Laplace transform
• Differential equations
• Modelling mechanical systems

System
SystemInput
x(t)
Output
(response)
y(t)
Rules of operations
By writing the rules of operation, we get a differential equation as a
combination of inputs and outputs
dny
𝑑𝑡 𝑛
+ 𝑎 𝑛−1
dn−1y
𝑑𝑡 𝑛−1
+ ⋯ + 𝑎0y
= 𝑏 𝑚
dm
x
𝑑𝑡 𝑚 + ⋯ + 𝑏0x

CLASSIFICATION OF SYSTEMS
• Linear and Non-linear
– Linear - Having properties of Additivity and
scalability
• Time invariant and time varying
– Time invariant – system parameters do not change
with time.
• Networks with RLC components
– Time varying – system parameters vary with time
• Space shuttle losing mass due to fuel

CLASSIFICATION OF SYSTEMS
• Controls are classified with respect to
– technique involved to perform control (i.e. human/machines):
manual/automatic control
– Time dependence of output variable (i.e. constant/changing):
regulator/servo,
(also known as regulating/tracking control)
– fundamental structure of the control (i.e. the information used
for computing the control):
Open-loop/feedback control,
(also known as open-loop/closed-loop control)

Mathematical Modelling of
physical systems

Transfer function
System g(t)Input
x(t)
Output
(response)
y(t)
𝑌(𝑠)
𝑋(𝑠)
= 𝐺(𝑠)
Transfer
function
How to write the transfer function?

Fundamentals to transfer function
• Laplace Transform
A technique to solve differential equation
Transforming time domain function to frequency domain
function
Laplace Transform Definition





0
)()( dtetfsF st
Solving differential equation is easy
that is through algebra. No need to
carry out differentiation or
integration.

Example of Laplace Transform
technique
  atatadtty
a
dt
dy
t
t


0
0)(
conditioninitialzero
Integral Approach

Example of Laplace Transform
technique
2
)(
)(
][
TransformLaplaceTaking
conditioninitialzerowith
s
a
sY
s
a
ssY
aL
dt
dy
L
a
dt
dy









Laplace Transform Approach
atty
s
a
LsYL







 
)(
)]([
TransformLaplaceInverseTaking
2
11

Partial Fraction Expansion
A mathematical technique to
help taking Inverse Laplace
Transform

tt
eKeKtfsFL
s
K
s
K
sF
ss
sF
2
21
1
21
)()]([
21
)(
Expansion,FractionPartialBy
)2)(1(
2
)(








Solve

Three cases:
1. Roots of the denominator of F(s) are real
number and distinct
2. Roots of the denominator of F(s) are real
number and repeated
3. Roots of the denominator of F(s) are complex or
imaginary

Partial Fraction Expansion – Case 1: Real and
distinct roots
5)5(
)2(
)(





s
B
s
A
ss
s
sY
5
2
)5(
)2(
0




s
s
s
A
5
3
)(
)2(
5



s
s
s
B
5
5
3
5
2
)(


ss
sY

Partial Fraction Expansion – Case 1: Real and
distinct roots
5
5
3
5
2
)(


ss
sY
ttt
eeety 550
5
3
5
2
5
3
5
2
)( 

Taking Inverse Laplace Transform

Partial Fraction Expansion - MATLAB
2
)2)(1(
2
)(


ss
sY
)52(
3
)( 2


sss
sY
Real and repeated roots
Imaginary and complex roots

What is Transfer Function?
Mathematical model that separates input
from output
)()(2
)(
trtc
dt
tdc

2
1
)(
)(


ssR
sC

What is Transfer Function?
2
1
)(
)(
)()(2)(
)()(2
)(




ssR
sC
sRsCssC
trtc
dt
tdc

Mathematical modelling of physical
systems
• Systems to be modelled
–Mechanical
–Electrical
–Electro mechanical
–Pneumatic
–Thermal
–Hydraulic

Mechanical system modelling
• Translational
• Rotational
Example
Automobile suspension system
Along the road
1. The vertical displacements at the tires act as the motion
excitation to the automobile suspension system
2. Motion consists of a translational motion of the center of
mass
3. Rotational motion about the center of mass

Translational system
Example 1

Element Law Expression Laplace
Spring Spring Force α
change in length
F(t) = K x(t)
K – Stiffness
constant in N/m
F(s) = K
X(s)
Viscous Damper
or Dashpot
Force α velocity F(t) = fv v(t)
F(t) =fv dx/dt
fv - Friction or
damping
coefficient ,Ns/m
F(s) = fvsX(s)
Mass Newton’s second
law
Force α
acceleration
F(t) = M
d2x/dt2
F(s) =
Ms2X(s)
Translational system

Step 1: Decide input and output
Input variable:
)(tf
Output variable:
)(txMass position
)(txMass velocity
)(txMass acceleration
Applied force

Translational systems
Newton’s second law
𝑚𝑎 = 𝐹

Step 2: The Time and frequency response
representation
dt
tdx
ftkxtf
dt
txd
M v
)(
)()(
)(
2
2

Taking Laplace Transform

Example 2
F(t)
x1(t)
x2(t)
x3(t)

Csl2 16 j15

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