chapter 2 : fundamental of mathematical modeling
TRANSCRIPT
Chapter_2 : Fundamental of Mathematical
Modeling
Objectives
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Upon completing this topic, students should be able
to find:
โ the Laplace Transform of time functions and the inverse
Laplace Transform.
โ manually and with MatLab, the Transfer Function from
Mathematical Models
โ Signal flow graphs from Block Diagram
โ Transfer Function using Masonโs Rule
11/10/2021 2
Contents
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
1. Introduction
2. Laplace Transform
3. Transfer Function;- Numerical
- With MatLab
4. Block Diagram Models- Block Diagram Algebra
- Signal Flow Graphs and Masonโs Rule
11/10/2021 3
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
CHAP_2. 1: Introduction
11/10/2021 4
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
To understand and control complex systems, one must obtain
quantitative mathematical models of these systems. A mathematical model is a set of equations (usually
differential equations) that represents the dynamics of systems.
differential equations are obtained by using physical laws of
engineering such as Newtonโs laws of motion, Kirchhoff's laws
of electrical network, Ohmโs laws, etc.
The equations of the mathematical model may be solved using
mathematical tools such as the Laplace Transform.
Before solving the equations, we usually need to linearize them.
In practice, the complexity of the system requires some
assumptions in the determination model.
2. 1 Introduction
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Obtained by applying the physical laws of the process: Therefore:
i. Mechanical Systems Newtonโs laws
ii. Electrical Systems *Kirchhoff's laws
* Ohmโs laws
* Mesh & Nodal Analysis
Example_1: Springer-mass-damper mechanical system.
Differential Equations
โข Hence for viscous wall: ๐ ๐ก = ๐๐ฃ ๐ก
* Assumption: Wall friction force f ๐ is
linearly proportional to the velocity ๐ of
the mass ๐ .
โข and for spring: ๐ ๐ก = ๐ 0
๐ก๐ฃ ๐ ๐๐
= ๐๐ฆ ๐ก
The time function
of ๐ ๐ is called
forcing function
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Step_1: Free-Body Diagram (FBD): Step_2: Newtonโs 2nd Law.
๐น = ๐๐ ๐ก : for translation motionOr:
โข ๐ฃ ๐ก = ๐ฆ =๐๐ฆ ๐ก
๐๐ก
โข ๐ ๐ก = ๐ฆ =๐2๐ฆ ๐ก
๐๐ก2
Hence:
๐๐ ๐ก = โ๐๐ฃ ๐ก โ ๐๐ฆ ๐ก + ๐ ๐ก
๐๐๐ฃ ๐ก
๐๐ก+ ๐๐ฃ ๐ก + ๐๐ฆ ๐ก = ๐ ๐ก
Finally, the differential equation is:
๐๐2๐ฆ ๐ก
๐๐ก2 + ๐๐๐ฆ ๐ก
๐๐ก+ ๐๐ฆ ๐ก = ๐ ๐ก
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Mechanical System Analysis:
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โข For the following mechanism, define:
i. The input โ ๐ญii.The output โ ๐.
Newtonโs 2nd Law:
๐น = ๐๐ ๐ก : for translation motion
Hence:
๐๐ ๐ก = ๐น โ ๐๐ฆ ๐ก โ ๐๐ฃ๐ฃ ๐ก
๐๐๐ฃ ๐ก
๐๐ก+ ๐๐ฆ ๐ก + ๐๐ฃ๐ฃ ๐ก = ๐น ๐ก
Finally, the differential equation is:
๐๐2๐ฆ ๐ก
๐๐ก2 + ๐๐ฃ๐๐ฆ ๐ก
๐๐ก+ ๐๐ฆ ๐ก = ๐น ๐ก
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Example_2: A Mechanical Mechanism.
โข for viscous damper: ๐ ๐ก = ๐๐ฃ๐ฃ ๐ก
= ๐๐ฃ๐๐ฆ ๐ก
๐๐ก
y
k
f
F
mM
๐๐
11/10/2021 8
โข For the following circuit, define:
i. The input โ ๐๐ii.The output โ ๐๐.
Kirchhoff voltage law (KVL):
๐๐ + ๐๐ฟ + ๐ข๐ ๐ก = ๐ข๐ ๐ก
Hence:
Ohm's laws:
๐ฟ๐๐ ๐ก
๐๐ก+ ๐ ๐ ๐ก + ๐ข๐ ๐ก = ๐ข๐ ๐ก
Or:
โข ๐ ๐ก = ๐ถ๐๐ข๐
๐๐ก(capacitor)
โข ๐ฃ ๐ก = ๐ฟ๐๐(๐ก)
๐๐ก(inductor)
Therefore, the differential equation is:
๐2๐ข๐ ๐ก
๐๐ก2 +๐
๐ฟ
๐๐ข๐ ๐ก
๐๐ก+
1
๐ฟ๐ถ๐ข๐ ๐ก =
1
๐ฟ๐ถ๐ข๐ ๐ก
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Example_3: A Passive Electrical Circuit.
11/10/2021 9
Initial conditions:
โข ๐ 0 = ๐ผ0: current through inductor
โข ๐ฃ 0 =1
๐ถ โโ
0๐ ๐ ๐๐ = ๐0:
voltage across capacitor
ur
uc
R L
Ci
โข Define the differential equation for the following RLC electrical circuit:
Kirchhoff voltage law (KVL):
๐๐ฟ + ๐๐ + ๐๐ถ = 0Hence:
Ohmโs lawsOr:
โข ๐ฃ ๐ก = ๐ฟ๐๐ ๐ก
๐๐ก(inductor)
โข ๐ฃ ๐ก =1
๐ถ 0
๐ก๐ ๐ ๐๐(capacitor)
Thus:
๐ฟ๐๐ ๐ก
๐๐ก+ ๐ ๐ ๐ก +
1
๐ถ 0
๐ก๐ ๐ ๐ ๐๐ = 0
Therefore, the differential equation is:
๐2๐ ๐ก
๐๐ก2 +๐
๐ฟ
๐๐ ๐ก
๐๐ก+
1
๐ฟ๐ถ๐ ๐ก = 0
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Example_4: Source-Free Series RLC Circuit.
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Initial conditions:
โข ๐ 0 = ๐ผ0: current through inductor
โข ๐ฃ 0 =1
๐ถ โโ
0๐ ๐ ๐๐ = ๐0: voltage
across capacitor
+
โ
1. Define the differential equation for the following Mechanical System:
2. Define the differential equation for the following RLC Electrical Network:
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Try โฆ
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๐ด๐๐ : ๐ ๐ฅ0 + ๐ ๐ฅ0 + ๐๐ฅ0 = ๐ ๐ฅ๐ + ๐๐ฅ๐
๐ ๐
๐ฅ๐
๐ฅ0
๐
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
CHAP_2. 2: Laplace Transform
11/10/2021 12
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
The differential equations (time-domain) are transformed by
the Laplace Transform into algebraic equations (frequency-
domain), which are easier to solve.
Equations:
1) Laplace Transform:
2) Inverse Laplace Transform:
2. 2 Laplace Transform
Time-domain signals Frequency โdomain signals
โ ๐ ๐ก = ๐น ๐ = 0
โ
๐ ๐ก ๐โ๐ ๐ก๐๐ก
โโ1 ๐น ๐ =1
2๐๐ ๐โ๐โ
๐+๐โ
๐น ๐ ๐๐ ๐ก๐๐ = ๐ ๐ก ๐ข ๐ก
Transfer Function ๐น ๐ is a function in Laplace domain ๐ where ๐ is a complex number: ๐ = ๐ถ + ๐๐
11/10/2021 13
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
A. Laplace Transform Table & Theorems
No. f(t) F(s)
1. ฮด ๐ก 1
2. u ๐ก1
๐
3. tu ๐ก1
๐ 2
4. ๐ก๐u ๐ก๐!
๐ ๐+1
5. ๐โ๐๐กu ๐ก1
๐ + ๐
6. ๐ก๐โ๐๐กu ๐ก1
๐ + ๐ 2
7. ๐ก๐๐โ๐๐กu ๐ก๐!
๐ + ๐ ๐+1
8. 1 โ ๐๐ก ๐โ๐๐กu ๐ก๐
๐ + ๐ 2
9.1
๐1 โ ๐โ๐๐กu ๐ก
1
๐ ๐ + ๐
โ Impulse function or Unit Impulse Function
โ Step function or Unit Step Function
โ Ramp function
1) Laplace Transform Table
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No. f(t) F(s)
10. ๐ ๐๐ ๐๐ก u ๐ก๐
๐ 2 + ๐2
11. ๐๐๐ ๐๐ก u ๐ก๐
๐ 2 + ๐2
12. ๐โ๐๐ก ๐ ๐๐ ๐๐ก u ๐ก๐
๐ + ๐ 2 + ๐2
13. ๐โ๐๐ก ๐๐๐ ๐๐ก u ๐ก๐ + ๐
๐ + ๐ 2 + ๐2
2) Laplace Transform TheoremsNo. Theorem Name
1. โ ๐ ๐ก = ๐น ๐ = 0
โ
๐ ๐ก ๐โ๐ ๐ก๐๐ก Laplace definition
2. โ ๐๐ ๐ก = ๐๐น ๐ Linearity Theorem
3. โ ๐1 ๐ก + ๐2 ๐ก = ๐น1 ๐ + ๐น2 ๐ Linearity Theorem
4. โ ๐โ๐๐ก๐ ๐ก = ๐น ๐ + ๐ Frequency Shift Theorem
5. โ ๐ ๐ก โ ๐ = ๐โ๐ ๐๐น ๐ Time Shift ๐ Theorem
6. โ ๐ ๐๐ก =1
๐๐น
๐
๐Scaling Theorem
7. โ๐๐ ๐ก
๐๐ก= ๐ ๐น ๐ โ ๐ 0โ Differentiation Theorem
8. โ๐2๐ ๐ก
๐๐ก2= ๐ 2๐น ๐ โ ๐ ๐ 0โ โ ๐โฒ 0โ Differentiation Theorem
9. โ๐๐๐ ๐ก
๐๐ก๐= ๐ ๐๐น ๐ โ
๐=1
๐
๐ ๐โ๐ ๐๐โ1 0โ Differentiation Theorem
(in general)
10. โ 0
๐ก
๐ ๐ ๐๐ =๐น ๐
๐ Integration Theorem
11. ๐ โ = ๐๐๐๐ โ0
๐ ๐น ๐ Final Value Theorem
12. ๐ 0+ = ๐๐๐๐ โโ
๐ ๐น ๐ Initial Value Theorem11/10/2021 15
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
ExampleWith ๐ ๐ a unit step, find the Laplace Transform of the following function ๐ฆ ๐ก , assuming zeros initial conditions:
๐2๐ฆ ๐ก
๐๐ก2 + 12๐๐ฆ ๐ก
๐๐ก+ 32๐ฆ ๐ก = 32๐ข ๐ก
Solution:
โข โ๐2๐ฆ ๐ก
๐๐ก2 = ๐ 2๐ ๐ โ ๐ ๐ฆ 0โ โ ๐ฆโฒ 0โ
โข โ 12๐๐ฆ ๐ก
๐๐ก= 12 ๐ ๐ ๐ โ ๐ฆ 0โ
โข โ 32๐ฆ ๐ก = 32๐ ๐
โข โ 32๐ข ๐ก = 32๐ ๐
With zeros initial conditions, we finally obtain the Laplace Transform as:
๐ 2๐ ๐ + 12๐ ๐ ๐ + 32๐ ๐ = 32๐ ๐
11/10/2021 16
Zeros Initial Conditions:โข ๐ฆ 0โ = 0โข ๐ฆโฒ 0โ = 0
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Tryโฆ1) With ๐ ๐ a unit step, find the Laplace Transform of the
following function:
๐2๐ฆ ๐ก
๐๐ก2 + 2๐๐ฆ ๐ก
๐๐ก+ 10๐ฆ ๐ก =
๐๐ ๐ก
๐๐ก
Assume the initial conditions as:
โข ๐ 0โ = 0, ๐ฆ 0โ = 0 and๐๐ฆ 0โ
๐๐ก= 1
2) Find the Laplace Transform of the following function:
๐ ๐ก = ๐ก๐โ5๐ก
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๐ด๐๐ : ๐ 2๐ ๐ + 2๐ ๐ ๐ + 10๐ ๐ โ 1 = ๐ ๐น ๐
๐ด๐๐ : ๐น ๐ =1
๐ + 5 2
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
B. Inverse Laplace Transform
General form:
Three (3) possible cases:
i. Case_1: Roots of ๐ท ๐ are real & distinct;
ii.Case_2: Roots of ๐ท ๐ are real & repeated;
iii.Case_3: Roots of ๐ท ๐ are complex conjugate;
๐น ๐ =๐ ๐
๐ท ๐
Numerator
Denominator
o Hint: Use โPartial Fraction Expansionโ
11/10/2021 18
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
โข Example: ๐น ๐ =2
๐ +1 ๐ +2
Solution:
โข ๐น ๐ =2
๐ +1 ๐ +2=
๐ด
๐ +1+
๐ต
๐ +2(Partial Fraction Expansion)
=2
๐ +1โ
2
๐ +2
โข The inverse Laplace Transform or Time Response is:
๐ ๐ก = 2๐โ๐ก๐ข ๐ก โ 2๐โ2๐ก๐ข ๐ก
As ๐ข ๐ก is the unit step function, ๐ ๐ก can finally be expressed in another way as:
It is found that: ๐ด = 2 and ๐ต = โ2
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i. Case_1: Roots of ๐ท ๐ are real & distinct
Item No.5 in the Laplace Transform Table
๐ ๐ก = 2๐โ๐ก โ 2๐โ2๐ก = 2 ๐โ๐ก โ ๐โ2๐ก
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
๐น ๐ =2
๐ +1 ๐ +2=
๐ด
๐ +1+
๐ต
๐ +2
=๐ด ๐ +2 +๐ต ๐ +1
๐ +1 ๐ +2
Hence:
๐ด ๐ + 2 + ๐ต ๐ + 1 = 2
Solving steps:
Case_1: ๐ + 1 = 0 โ ๐ = โ1โ ๐ด โ1 + 2 = 2โ ๐ด = 2
Case_2: ๐ + 2 = 0 โ ๐ = โ2โ ๐ต โ2 + 1 = 2โ ๐ต = โ2
11/10/2021 20
Work out Method:
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Tryโฆ
- Find the Inverse Laplace Transform of the following function:
a) G ๐ =๐ +3
๐ +1 ๐ +2
Solution:
b) G ๐ =32
๐ ๐ 2+12๐ +32
Solution:
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๐ด๐๐ : ๐ ๐ก = 2๐โ๐ก โ ๐โ2๐ก
๐ด๐๐ : ๐ ๐ก = ๐โ๐ก โ 2๐โ4๐ก + ๐โ8๐ก
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
ii. Case_2: Roots of ๐ท ๐ are real & repeated
โขExample: ๐น ๐ =2
๐ +1 ๐ +2 2
Solution:
โข ๐น ๐ =2
๐ +1 ๐ +2 2 =๐ด
๐ +1+
๐ต
๐ +2+
๐ถ
๐ +2 2
=2
๐ +1โ
2
๐ +2โ
2
๐ +2 2
โข Finally,
Example:
โข ๐น ๐ =2
๐ +1 ๐ +2 2 =๐ด
๐ +1+
๐ต
๐ +2+
๐ถ
๐ +2 2
=๐ด ๐ +2 2+๐ต ๐ +1 ๐ +2 +๐ถ ๐ +1
๐ +1 ๐ +2 2
It is found that: ๐ด = 2, ๐ต = โ2 and ๐ถ = โ2
๐ ๐ก = 2๐โ๐ก โ 2๐โ2๐ก โ 2๐ก๐โ2๐ก
11/10/2021 22
Work out Method:
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Thus:
โข ๐น ๐ =2
๐ +1 ๐ +2 2 =๐ด ๐ +2 2+๐ต ๐ +1 ๐ +2 +๐ถ ๐ +1
๐ +1 ๐ +2 2
=๐ด ๐ 2+4๐ +4 +๐ต ๐ 2+3๐ +2 +๐ถ ๐ +1
๐ +1 ๐ +2 2 =๐ด๐ 2+4๐ด๐ +4๐ด+๐ต๐ 2+3๐ต๐ +2๐ต+๐ถ๐ +๐ถ
๐ +1 ๐ +2 2
โ ๐น ๐ =2
๐ +1 ๐ +2 2 =๐ 2 ๐ด+๐ต +๐ 4๐ด+3๐ต+๐ถ + 4๐ด+2๐ต+๐ถ
๐ +1 ๐ +2 2
Equating like powers of "๐" gives us a system of equations as:
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Powers of "๐" Equation
๐ 2 ๐ด + ๐ต = 0
๐ 1 4๐ด + 3๐ต + ๐ถ = 0
๐ 0 4๐ด + 2๐ต + ๐ถ = 2
Solving the system of equations yields:
๐ด = 2
B= โ2
C= โ2
Work out Method (Contโฒd):
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Tryโฆ
- Find the Inverse Laplace Transform of the following functions:
a) ๐น ๐ =๐ +3
๐ +1 3 ๐ +2
Solution:
b) G ๐ =๐ 2+1
๐ 2 ๐ +2
Solution:
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๐ด๐๐ : ๐ ๐ก = ๐ก2๐โ๐ก โ ๐ก๐โ๐ก + ๐โ๐ก โ ๐โ2๐ก
= ๐โ๐ก ๐ก2 โ ๐ก + 1 โ ๐โ2๐ก
๐ด๐๐ : ๐ ๐ก =5
4๐โ2๐ก +
1
2๐ก โ
1
4
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
iii.Case_3: Roots of ๐ท ๐ are complex conjugate
โขExample: ๐น ๐ =3
๐ ๐ 2+2๐ +5
Solution:
โข ๐น ๐ =3
๐ ๐ 2+2๐ +5=
๐ด
๐ +
๐ต๐ +๐ถ
๐ 2+2๐ +5
= 35
๐ โ
3
5
๐ +2
๐ 2+2๐ +5
= 35
๐ โ
3
5
๐ +1+1
๐ 2+2๐ +5=
35
๐ โ
3
5
๐ +1 + 1 2 2
๐ +1 2+22
= 35
๐ โ
3
5
๐ +1
๐ +1 2+22 + 1 2 2
๐ +1 2+22
โข Finally,
It is found that: ๐ด = 3 5, ๐ต = โ3 5 and ๐ถ = โ6 5
๐ ๐ก =3
5โ
3
5๐โ๐ก ๐๐๐ 2๐ก +
1
2๐โ๐ก๐ ๐๐2๐ก =
3
5โ
3
5๐โ๐ก ๐๐๐ 2๐ก +
1
2๐ ๐๐ 2๐ก
11/10/2021 25
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
โขExample:
๐น ๐ =3
๐ ๐ 2+2๐ +5=
๐ด
๐ +
๐ต๐ +๐ถ
๐ 2+2๐ +5=
๐ด ๐ 2+2๐ +5 + ๐ต๐ +๐ถ ๐
๐ ๐ 2+2๐ +5
=๐ด๐ 2+2๐ด๐ +5๐ด+๐ต๐ 2+๐ถ๐
๐ ๐ 2+2๐ +5=
๐ 2 ๐ด+๐ต +๐ 2๐ด+๐ถ +5๐ด
๐ ๐ 2+2๐ +5
โ ๐น ๐ =3
๐ ๐ 2+2๐ +5=
๐ 2 ๐ด+๐ต +๐ 2๐ด+๐ถ +5๐ด
๐ ๐ 2+2๐ +5
Equating like powers of "๐" gives us a system of equations as:
11/10/2021 26
Work out Method:
Powers of "๐" Equation
๐ 2 ๐ด + ๐ต = 0
๐ 1 2๐ด + ๐ถ = 0
๐ 0 5๐ด = 3
Solving the system of equations yields:
๐ด = 3 5
B= โ 3 5
C= โ 6 5
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Tryโฆ
- Find the Inverse Laplace Transform of the following functions:
a) ๐น ๐ =๐ +3
๐ 2+4๐ +5 ๐ +5
Solution:
b) G ๐ =10
๐ 2+9 ๐ +1
Solution:
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๐ด๐๐ : ๐ ๐ก = โ1
5๐โ5๐ก +
1
5๐๐๐ ๐ก ๐โ2๐ก +
2
5๐ ๐๐ ๐ก ๐โ2๐ก
๐ด๐๐ : ๐ ๐ก = ๐โ๐ก โ ๐๐๐ 3๐ก โ1
3๐ ๐๐ 3๐ก
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Partial-Fraction Expansion with MATLAB Three (3) steps to obtain the Partial-Fraction Expansion :
1. Define the coefficients in descending powers of ๐ as follow:o Coefficients of numerator polynomial as numerator coefficients ,o Coefficients of denominator polynomial as denominator
coefficients
2. Use the โฒ๐๐๐๐๐ ๐๐โฒ command to get the residues ๐ , poles ๐and direct terms ๐ of a Partial-Fraction Expansion of the ratio of the two polynomials ๐ต ๐ and ๐ด ๐ ;
3. The partial-fraction expansion of ๐ต ๐ ๐ด ๐ is given by:
๐ต ๐
๐ด ๐ = ๐ ๐ +
๐ 1
๐ โ ๐ 1+
๐ 2
๐ โ ๐ 2+ โฏ +
๐ ๐
๐ โ ๐ ๐
Syntax:
๐, ๐, ๐ = ๐๐๐ ๐๐๐ข๐ ๐๐ข๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐๐๐๐๐ก๐ , ๐๐๐๐๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐๐๐๐๐๐ก๐
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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Example Get the partial-fraction expansion for the following expression:
๐ต ๐
๐ด ๐ =
๐ 4+8๐ 3+16๐ 2+9๐ +6
๐ 3+6๐ 2+11๐ +6
Solution- Step_1: Coefficient definition:
o Numerator: ๐ = 1 8 16 9 6o Denominator: ๐ท = 1 6 11 6
- Step_2: โresidueโ command in Matlab
o ๐, ๐, ๐ = ๐๐๐ ๐๐๐ข๐ 1 8 16 9 6 , 1 6 11 6= ๐๐๐ ๐๐๐ข๐ ๐, ๐ท
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Solution (Contโd) MatLab Results:
For direct terms ๐ :
๐ ๐ = ๐๐๐ + ๐๐๐ = ๐ + ๐
- Step_3: Write the partial-fraction expansion as:
๐ต ๐
๐ด ๐ =
๐ 4+8๐ 3+16๐ 2+9๐ +6
๐ 3+6๐ 2+11๐ +6โ
๐ต ๐
๐ด ๐ = ๐ + 2 โ
6
๐ +3โ
4
๐ +2+
3
๐ +1
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Tryโฆ With MatLab, get the partial-fraction expansion for the following
expressions:
1) ๐น ๐ =โ4๐ +8
๐ 2+6๐ +8
2) ๐บ ๐ =2๐ 4+๐
๐ 2+1
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๐ด๐๐ : ๐น ๐ =โ12
๐ + 4+
8
๐ + 2
๐ด๐๐ : ๐บ ๐ = 2๐ 2 โ 2 +0.5 โ ๐
๐ โ ๐+
0.5 + ๐
๐ + ๐
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
The Laplace Transform can be used to solve differential
equations using a four step process:
1. Take the Laplace Transform of the differential equationusing the derivative property (and, perhaps, others) as necessary.
2. Put initial conditions into the resulting equation.
3. Solve for the output variable.
4. Get result from the Laplace Transform Tables. (look up the terms individually):
- If the result is in a form that is not in the tables, you'll need to use the Inverse Laplace Transform.
C. Steps to Solving Differential Equations
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Examples
1) Find the response ๐ฆ ๐ก for the following differential equation ๐๐ฆ ๐ก
๐๐ก+ 2๐ฆ ๐ก = ๐ ๐ก , with input ๐ ๐ก and output ๐ฆ ๐ก ;
if ๐ ๐ก = ๐ข ๐ก โ ๐ข ๐ก โ 1 i.e. a Unit Step Function, and the initial condition ๐ฆ 0โ = โ2.
Solution:- Step_1: Take the Laplace Transform of the differential equation.
โข โ๐๐ฆ ๐ก
๐๐ก= ๐ ๐ ๐ โ ๐ฆ 0โ
โข โ ๐ฆ ๐ก = ๐ ๐
โข โ ๐ ๐ก = ๐น ๐ =1
๐ โ ๐โ๐ ๐น ๐ =
1
๐ โ ๐โ๐ 1
๐
So,
๐๐ฆ ๐ก
๐๐ก+ 2๐ฆ ๐ก = ๐ข ๐ก โ ๐ข ๐ก โ 1 โ
โ๐ ๐ ๐ โ ๐ฆ 0โ + 2๐ ๐ =
1
๐ โ ๐โ๐ 1
๐
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Solution (Contโd)
- Step_2: Put initial conditions into the resulting equation.
โข ๐ ๐ ๐ โ ๐ฆ 0โ + 2๐ ๐ =1
๐ โ ๐โ๐ 1
๐ (The resulting equation)
โข ๐ ๐ ๐ + 2 + 2๐ ๐ =1
๐ โ ๐โ๐ 1
๐ (After putting the initial condition)
- Step_3: Solve for the output variable ๐ ๐ .
โข ๐ ๐ =1
๐ ๐ +2โ ๐โ๐ 1
๐ ๐ +2โ 2
1
๐ +2
- Step_4: Get result from the Laplace Transform Tables.
โข โโ1 1
๐ ๐ +2=
1
21 โ ๐โ2๐ก
โข โโ1 โ๐โ๐ 1
๐ ๐ +2=
1
21 โ ๐โ2 ๐กโ1 ๐ข ๐ก โ 1
โข โโ1 โ21
๐ +2= โ2๐โ2๐ก
Finally: ๐ฆ ๐ก =12 1 โ ๐โ2๐ก โ 1
21 โ ๐โ2 ๐กโ1 ๐ข ๐ก โ 1 โ 2๐โ2๐ก
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Examples (Contโd)
2) Find the response ๐ฆ ๐ก for the following differential equation: ๐2๐ฆ ๐ก
๐๐ก2 + 2๐๐ฆ ๐ก
๐๐ก+ 10๐ฆ ๐ก =
๐๐ ๐ก
๐๐ก, with input ๐ ๐ก and output ๐ฆ ๐ก ;
if ๐ ๐ก = ๐ข ๐ก i.e. a unit step input, and
the initial conditions: ๐ 0โ = 0, ๐ฆ 0โ = 0 and๐๐ฆ 0โ
๐๐ก= 1.
Solution:- Step_1: Take the Laplace Transform of the differential equation.
โข โ๐2๐ฆ ๐ก
๐๐ก2 = ๐ 2๐ ๐ โ ๐ ๐ฆ 0โ โ ๐ฆโฒ 0โ
โข โ 2๐๐ฆ ๐ก
๐๐ก= 2 ๐ ๐ ๐ โ ๐ฆ 0โ
โข โ 10๐ฆ ๐ก = 10๐ ๐
โข โ๐๐ ๐ก
๐๐ก= ๐ ๐น ๐ โ ๐ 0โ = ๐
1
๐ = 1
So,๐2๐ฆ ๐ก
๐๐ก2+ 2
๐๐ฆ ๐ก
๐๐ก+ 10๐ฆ ๐ก =
๐๐ ๐ก
๐๐กโโ
๐ 2๐ ๐ โ ๐ ๐ฆ 0โ โ ๐ฆโฒ 0โ + 2 ๐ ๐ ๐ โ ๐ฆ 0โ + 10๐ ๐ = 1
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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Solution (Contโd)
- Step_2: Put initial conditions into the resulting equation.
โข ๐ 2๐ ๐ โ ๐ ๐ฆ 0โ โ ๐ฆโฒ 0โ + 2 ๐ ๐ ๐ โ ๐ฆ 0โ + 10๐ ๐ = 1 (The resulting equation)
โข ๐ 2๐ ๐ โ ๐ . 0 โ 1 + 2 ๐ ๐ ๐ โ 0 + 10๐ ๐ = 1 (After putting the initial condition)
โ ๐ 2๐ ๐ โ 1 + 2๐ ๐ ๐ + 10๐ ๐ = 1
- Step_3: Solve for the output variable ๐ ๐ .
โข ๐ ๐ ๐ 2 + 2๐ + 10 = 2
โ ๐ ๐ =2
๐ 2+2๐ +10
- Step_4: Get result from the Laplace Transform Tables.
โข As the above result is not in the tables, let use the Inverse Laplace Transform.
Hence:
๐ ๐ =2
๐ 2+2๐ +10โ ๐ ๐ =
2
๐ +1 2+9=
2
๐ +1 2+32
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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From : ๐ ๐ =2
๐ +1 2+9=
2
๐ +1 2+32
โ ๐ ๐ =2
3
3
๐ +1 2+32
We then use the table to determine the solution and finally get:
๐ฆ ๐ก =2
3๐โ๐ก๐ ๐๐ 3๐ก
Tryโฆ- Find the response ๐ฅ ๐ก for the following differential equations:
1) ๐ฅ + 3 ๐ฅ + 2๐ฅ = 0 with the following initial conditions: ๐ฅ 0 = ๐, ๐ฅ 0 = ๐
2) ๐ฅ + 2 ๐ฅ + 5๐ฅ = 3 with the following initial conditions: ๐ฅ 0 = ๐, ๐ฅ 0 = ๐
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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CHAP_2. 3: Transfer Function
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
2.3. Transfer Function (TF)A. Overview
โข Commonly used to characterize the inputโoutput relationships of components or systems that can be described by linear and time-invariant differential equations.
โข It is defined as โthe ratio of the Laplace Transform of the output(response function) to the Laplace Transform of the input (driving function) under the assumption that all initial conditions are zeroโ
๐บ๐๐๐๐๐Input Output
๐ ๐ก ๐ ๐ก๐
๐บ๐๐๐๐๐๐๐Input Output
๐ ๐ก ๐ ๐ก๐
๐บ๐๐๐๐๐๐๐ ๐บ๐๐๐๐๐๐๐
With:โข ๐ ๐ก โก ๐๐๐๐๐๐๐๐๐ ๐๐๐๐ข๐กโข ๐ ๐ก โก ๐๐๐๐ก๐๐๐๐๐๐ ๐ฃ๐๐๐๐๐๐๐
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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Overview (Contโd)โข A general form of the differential equation for Linear and Time
Invariant System (LTI-System) with an input ๐ ๐ก and output ๐ ๐ก is given by:
๐๐๐๐๐ ๐ก
๐๐ก๐ + ๐๐โ1๐๐โ1๐ ๐ก
๐๐ก๐โ1 + โฏ + ๐0๐ ๐ก = ๐๐๐๐๐ ๐ก
๐๐ก๐ + ๐๐โ1๐๐โ1๐ ๐ก
๐๐ก๐โ1 + โฏ + ๐0๐ ๐ก
โข The Transfer Function of this system is obtained by taking the Laplace transforms of both sides of Equation (assuming zero initial conditions),
๐๐๐ ๐๐ถ ๐ + ๐๐โ1๐ ๐โ1๐ถ ๐ + โฏ + ๐0๐ถ ๐ = ๐๐๐ ๐๐ ๐ + ๐๐โ1๐ ๐โ1๐ ๐ + โฏ + ๐0๐ ๐
๐๐๐ ๐ + ๐๐โ1๐ ๐โ1 + โฏ + ๐0 ๐ถ ๐ = ๐๐๐ ๐ + ๐๐โ1๐ ๐โ1 + โฏ + ๐0 ๐ ๐
๐ถ ๐
๐ ๐ =
๐๐๐ ๐+๐๐โ1๐ ๐โ1+โฏ+๐0
๐๐๐ ๐+๐๐โ1๐ ๐โ1+โฏ+๐0= ๐บ ๐ โขThe equation ๐ฎ ๐ is called
Transfer Function (TF).
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Three (3) steps to obtain manually the Transfer Function:1. Write the differential equation for the system.2. Take the Laplace transform of the differential equation,
assuming all initial conditions are zeros.3. Take the ratio of the output ๐ถ ๐ to the input ๐ ๐ .
This ratio is the transfer function.
The key advantage of Transfer Functions is that: they allow engineers to use simple algebraic equations instead of
complex differential equations for analyzing and designing systems.
A. 1. To obtain Transfer Function Manually
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From the given differential equations, find the Transfer Function represented by ๐ ๐ก , to an input ๐ ๐ก :
a)๐๐ ๐ก
๐๐ก+ 2๐ ๐ก = ๐ ๐ก
b) ๐ + 3 ๐ + ๐ = ๐ ๐ก
Solution:
a)๐๐ ๐ก
๐๐ก+ 2๐ ๐ก = ๐ ๐ก
โข Taking the Laplace Transform:
๐ ๐ถ ๐ + 2๐ถ ๐ = ๐ ๐
โข The Transfer Function ๐บ ๐ , is:
๐บ ๐ =๐ถ ๐
๐ ๐ =
1
๐ +2
Example
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b) ๐ + 3 ๐ + 2๐ = ๐ ๐ก
โข Taking the Laplace Transform:
๐ 2๐ถ ๐ + 3s๐ถ ๐ + ๐ถ ๐ = ๐ ๐
โข The Transfer Function ๐บ ๐ , is:
๐บ ๐ =๐ถ ๐
๐ ๐ =
1
๐ 2+3๐ +1
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
From the given differential equations, find the Transfer Function represented by ๐ ๐ก , to an input ๐ ๐ก :
a) 5๐2๐ ๐ก
๐๐ก2 โ 3๐ ๐ก = 2๐ ๐ก
b) ๐ฅ + 2 ๐ฅ + 5๐ฅ = 3๐ ๐ก
Tryโฆ
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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A. 2. To obtain Transfer Function with MatLab Two (2) steps to obtain the Transfer Function:
1. Define the coefficients of the differential equation in descending power as :o input signal coefficients as numerator coefficients ,o output signal coefficients as denominator coefficients
2. Use the โฒ๐๐โฒ command to generate the Transfer Function (TF);
Syntax:
๐บ ๐ = ๐ก๐ ๐๐ข๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐๐๐๐๐ก๐ , ๐๐๐๐๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐๐๐๐๐๐ก๐
Example: 5๐4๐ฆ
๐๐ก4 โ 4๐2๐ฆ
๐๐ก2 + 10๐ฆ = 20๐๐ข
๐๐ก+ 4๐ข
Solution: -Step_1: Coefficient definition.
o Numerator: ๐ = 20 4o Denominator: ๐ท = 5 0 โ 4 10
-Step_2: โฒ๐๐โฒ command in Matlab.
o ๐บ = ๐ก๐ 20 4 , 5 0 โ 4 10= ๐ก๐ ๐, ๐ท
o Answer: ๐บ =20๐ +4
5๐ ^3โ4๐ +10
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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โข Use MATLAB to extract the Transfer Function represented by the following equations:
1.๐๐ ๐ก
๐๐ก+ 2๐ ๐ก = ๐ ๐ก
2. 5๐2๐ ๐ก
๐๐ก2 โ 3๐ ๐ก = 2๐ ๐ก
Tryโฆ
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
B. Open-Loop Transfer Function (OLTF)
OLTF can be represented using a block diagram:
๐ ๐ ๐ถ ๐
๐ ๐ ๐ถ ๐
Where:
๐ถ ๐ = ๐บ ๐ ๐ ๐ or G ๐ =๐ถ ๐
๐ ๐
NOTE: G ๐ =๐ถ ๐
๐ ๐ โก Open-Loop Transfer Function.
๐บ ๐
๐๐๐ ๐ + ๐๐โ1๐ ๐โ1 + โฏ + ๐0
๐๐๐ ๐ + ๐๐โ1๐ ๐โ1 + โฏ + ๐0
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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B. Closed-Loop Transfer Function (CLTF)
CLTF can be represented using a block diagram:
๐ ๐ + E ๐ ๐ถ ๐
โ
Thus: ๐ธ ๐ = R ๐ โ ๐ป ๐ ๐ถ ๐ (eq.1) and ๐ถ ๐ = G ๐ ๐ธ ๐ (eq.2)
eq.1 into eq.2: ๐ถ ๐ = G ๐ [R ๐ โ ๐ป ๐ ๐ถ ๐ ] (eq.3)
Re-arranging the eq.3 yields to:
๐ถ ๐
๐ ๐ =
๐บ ๐
1ยฑ๐บ ๐ ๐ป ๐ โก Closed-Loop Transfer Function.
๐บ ๐
๐ป ๐
Plant and Controller
Output
Feedback
Input Actuating signal (error)
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Feedback and its Effects Why Feedback?
โข Feedback is a key tool that can be used to modify the behaviorof a system.
โข This behavior altering effect of feedback is a key mechanism that control engineers exploit deliberately to achieve the objective of acting on a system to ensure that the desired performance specifications are achieved.
Feedback Effects.
โข To reduce the error between the input and output of the system.
โข It effects the system performance characteristics such as stability, overall system gain, sensitivity and bandwidth.
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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o Feedback Effects on:
i. system sensitivity:
โข It can make the systemโs response less sensitive to externaldisturbances, parameter changes and noise
ii. system stability:
โข An unstable system can be stabilized using feedback
โข Stability refers to the ability of a system to follow its input signal
โข A system that canโt control its output, or its output increases infinitely is an unstable system
โข Adding feedback may also cause instability to an already stable system
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
o Feedback Effects on (Contโd)
iii. system gain:
โข Feedback influences the systemโs Transfer Function and consequently, the overall system gain:
โข Thus, the systemโs gain can be obtained by finding the magnitude of the systemโs Transfer Function:
a) The system gain for OLCS: ๐ถ ๐
๐ ๐ = ๐บ ๐
b) The system gain for CLCS: ๐ถ ๐
๐ ๐ =
๐บ ๐
1+๐บ ๐ ๐ป ๐
Effects of the product ๐บ ๐ ๐ป ๐ :
โ If ๐บ ๐ ๐ป ๐ is negative;
๐ถ ๐
๐ ๐ > ๐บ ๐ and we have positive feedback.
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Effects of the product ๐บ ๐ ๐ป ๐ (Contโd):
โ If ๐บ ๐ ๐ป ๐ is positive;
๐ถ ๐
๐ ๐ < ๐บ ๐ and we have negative feedback.
o If ๐บ ๐ ๐ป ๐ is positive and ๐บ ๐ ๐ป ๐ โซ 1;
๐ถ ๐
๐ ๐ โ
๐บ ๐
๐บ ๐ ๐ป ๐ =
1
๐ป ๐
The gain is independent of the gain of the forward path ๐บ ๐ .
Depending on whether the feedback is positive + or negative โ , the system gain of a Closed-Loop Control System (CLCS) can be increased or decreased.
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o Feedback Effects on (contโd)
iv. system bandwidth:
โข The system bandwidth increases as the gain is reduced with feedback;
โข in some cases the gain ร bandwidth = constant
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
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CHAP_2. 4: Block Diagram Models
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2.4 Block Diagram ModelsA. Overview
โข A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals.
โ Such diagram depicts the interrelationships that exist among the various components.
โ Differing from a purely abstract mathematical representation, a block diagram has the advantage of indicating more realistically the signal flows of the actual system.
โข Transfer function can be represented as a block diagram:
๐๐๐ ๐ + ๐๐โ1๐ ๐โ1 + โฏ + ๐0
๐๐๐ ๐ + ๐๐โ1๐ ๐โ1 + โฏ + ๐0
๐ ๐ ๐ถ ๐
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Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Components of a Block Diagram for a LTI system
a) Signals
๐ ๐ ๐ถ ๐
b) System
๐บ ๐ ๐ ๐ ๐ถ ๐
Input Output
โ
c) Summing Junction
๐ 1 ๐ +
๐ 2 ๐ ๐ 3 ๐
๐ถ ๐ = ๐ 1 ๐ + ๐ 2 ๐ โ ๐ 3 ๐
+
d) Pickoff point
๐ ๐ ๐ ๐
๐ ๐
๐ ๐
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Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Four (4) steps for drawing Block Diagram:
1) Write the equations that describe the dynamic behavior for each component.
2) Take Laplace Transform of these equations, assuming zero initial conditions.
3) Represent each Laplace-transformed equation individually in block form.
4) Assembly the elements into a complete block diagram.
Procedures for drawing Block Diagram
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Blocks of Block diagram can be connected in three basic forms:
1) Cascade:
2) Parallel
3) Feedback
Basic Forms of Block Diagram Blocks
๐บ2 ๐ ๐บ1 ๐ ๐บ3 ๐ ๐ ๐ ๐2 ๐ ๐1 ๐ ๐ถ ๐
๐บ1 ๐ ๐1 ๐
๐ ๐ ๐บ2 ๐
๐บ3 ๐
๐2 ๐
๐3 ๐
๐ถ ๐ ยฑยฑ
ยฑ
๐ ๐ ๐บ ๐
๐ป ๐
๐ธ ๐ ๐ถ ๐
๐ ๐ โ
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B. Block Diagram Algebra
Simplify block diagrams into simpler, recognizable forms;
In order to determine the equivalent transfer function.
1. Simplify to instances of the three standard forms;
then simplify those forms
2. Move blocks around relative to summing junctions and pickoff points;
simplify to a standard form:
i. Move blocks forward/backward past summing junctions
ii. Move blocks forward/backward past pickoff points
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Rules for Reduction of Block Diagrams
1. Any number of cascaded blocks can be reduced by a single block representing transfer function being a product of transfer functions of all cascaded blocks.
๐บ1 ๐ ๐ ๐
๐บ2 ๐ ๐บ3 ๐
๐1 ๐ = ๐บ1 ๐ ๐ ๐
๐2 ๐ = ๐บ2 ๐ ๐บ1 ๐ ๐ ๐
๐ถ ๐ = ๐บ3 ๐ ๐บ2 ๐ ๐บ1 ๐ ๐ ๐
๐
๐ฎ๐ ๐ ๐ฎ๐ ๐ ๐ฎ๐ ๐บ๐ ๐ ๐ถ ๐
๐
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โขRules for reduction of the block diagram (Contโd):
2. Equivalent Transfer Function for parallel substations.
๐บ1 ๐ ๐1 ๐ = ๐ ๐ ๐บ1 ๐
๐ ๐ ๐บ2 ๐
๐บ3 ๐
๐2 ๐ = ๐ ๐ ๐บ2 ๐
๐3 ๐ = ๐ ๐ ๐บ3 ๐
๐ถ ๐ = ยฑ๐บ1 ๐ ยฑ๐บ2 ๐ ยฑ ๐บ3 ๐ ๐ ๐
ยฑยฑ
ยฑ
ยฑ๐ฎ๐ ๐ ยฑ๐ฎ๐ ๐ ยฑ ๐ฎ๐ ๐บ๐ ๐ ๐ถ ๐
๐
๐
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โขRules for reduction of the block diagram (Contโd):
3. Equivalent Transfer Function for substations with feedback.
๐บ ๐
1 ยฑ ๐บ ๐ ๐ป ๐
๐ ๐ ๐ถ ๐
๐
๐
๐ ๐ ๐บ ๐
๐ป ๐
๐ธ ๐ ๐ถ ๐
๐ ๐
โ
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โขRules for Reduction of Block Diagrams (Contโd):
4. Moving Blocks/a Summing Junction.
๐ ๐
โ
+ ๐ถ ๐ โก๐บ ๐
๐ ๐
๐ ๐
โ
+๐บ ๐
๐ ๐
๐บ ๐
๐บ ๐ ๐ถ ๐ ๐ ๐ ๐ถ ๐
๐ Back Past a Summing Junction/Behind the Block
๐ Forward Past a Summing Junction/Ahead of the Block
๐ ๐
โกโ
+
โ
+
1
๐บ ๐
๐บ ๐ ๐ถ ๐ ๐ ๐
๐ ๐
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โขRules for reduction of the block diagram (Contโd):
5. Moving Blocks/a Pickoff Point.
๐ ๐
๐ ๐
๐ ๐ ๐บ ๐
โก๐บ ๐ ๐ ๐
๐บ ๐
๐ ๐ 1
๐บ ๐ ๐บ ๐
๐ ๐ ๐บ ๐
๐ ๐ ๐ ๐
๐ Backward Past Pickoff Point/Behind the Block
๐ Forward Past Pickoff Point/Ahead of the Block
โก
๐บ ๐ ๐ ๐ ๐บ ๐
๐ ๐
๐ ๐ ๐บ ๐
1
๐บ ๐
๐ ๐
๐ ๐ ๐บ ๐
๐ ๐ ๐บ ๐
๐บ ๐
๐บ ๐
๐ ๐ ๐บ ๐
๐ ๐ ๐บ ๐
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Three (3) tips for calculating the Equivalent Transfer Function of reduced Block Diagrams:
1) The Numerator of the closed-loop Transfer Function ๐ถ ๐ ๐ ๐ is the product of the transfer functions of the
feed forward path.
2) The denominator of the closed-loop transfer function ๐ถ ๐ ๐ ๐ is equal to:
1 โ ๐๐๐๐๐ข๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ ๐๐๐ ๐น๐ข๐๐๐ก๐๐๐๐ ๐๐๐๐ข๐๐ ๐๐๐โ ๐๐๐๐
3) The positive feedback loop yields a negative term in the denominator.
Conclusion of Block Diagram Reduction Techniques
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Example_1โข Reduce the following block diagram
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Workingโฆ
1) Moving the summing point ahead of ๐บ1, we have:
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Workingโฆ
2) Combining ๐บ1 and ๐บ2 in Cascade, we get:
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Workingโฆ
3) Eliminating the feedback loop ๐บ1, ๐บ2 and ๐ป1, we get:
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Workingโฆ
4) Combing the two blocks in Cascade, we get
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Workingโฆ
5) Similarly eliminating the second feedback loop, we get:
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Workingโฆ
6) Similarly eliminating the third feedback loop, we get:
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Workingโฆ
7) Finally, the system is reduced to the following block diagram:
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Example_2โข Deduce the response ๐ ๐ of the following control system with two
inputs ๐ ๐ and ๐ท ๐ .
โข Solution:
a) Find:
- Step_1: ๐1 ๐ ๐ท ๐ when ๐ ๐ = 0
๐1 ๐
๐ท ๐ ๐ ๐ =0=
๐บ2
1+๐บ2 ๐ป1โ๐ป2
- Step_2: ๐2 ๐ ๐ ๐ when ๐ท ๐ = 0
๐2 ๐
๐ ๐ ๐ท ๐ =0=
๐บ1๐บ2
1+๐บ2 ๐ป1โ๐ป2
b) The total response ๐ ๐ when ๐ ๐ and ๐ท ๐ are zeros is:
๐ ๐ = ๐1 ๐ + ๐2 ๐
= ๐1 ๐
๐ท ๐ ๐ ๐ =0โ ๐ท ๐ +
๐2 ๐
๐ ๐ ๐ท ๐ =0โ ๐ ๐
=๐บ2
1+๐บ2 ๐ป1โ๐ป2๐ท ๐ +
๐บ1๐บ2
1+๐บ2 ๐ป1โ๐ป2๐ ๐
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C. Signal Flow Graphs
Alternative method to block diagram representation, developed by Samuel Jefferson Mason.
They provide graphical description of systems.
Advantage:
the availability of a flow graph gain formula, also calledMasonโs gain formula.
A signal flow graph consists of a network in which nodesare connected by directed branches.
It depicts the flow of signals from one point of a system to another and gives the relationships among the signals.
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Components of Signal Flow Graphs
Signal flow graphs consist of:
1) Nodes โrepresent signals
Nodes (sometimes) labeled with signal names
2) Branches โrepresent system blocks
Branches labeled with system transfer functions
Unidirectional line segment joining two nodes.
4) Path: a branch or continuous sequence of branches.
5) Arrows indicate signal flow direction
6) Implicit summation at nodes:
Always a positive sum
Negative signs associated with branch transfer functions
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Four (4) steps:
1) Identify and label all signals on the block diagram
2) Place a node for each signal
3) Connect nodes with branches in place of the blocks;
Maintain correct direction
Label branches with corresponding transfer functions
Negate transfer functions as necessary to provide negativefeedback
4) If desired, simplify where possible.
Conversion Procedures from a Block
Diagram to a Signal Flow Graph
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Example_1โข Convert the following block diagram to a signal flow graph
โข Solution:
Step_1:
- Identify and Label any unlabeled signals.
Step_2:
- Place a node for each signal.
๐น ๐ ๐ฟ๐ ๐ ๐ฌ ๐ ๐ผ ๐๐ ๐
๐ฟ๐ ๐
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Solution (Contโd)Step_3:
- Connect nodes with branches, each representing a system block
Note: the โ1 to provide negative feedback of ๐ฟ๐ ๐
Step_4:
- To simplify, intermediate nodes with a single input and single output can be eliminated, if desired:
This makes sense for ๐1 ๐ and ๐2 ๐
Leave ๐ ๐ to indicate separation between controller and plant.
๐น ๐ ๐ฟ๐ ๐ ๐ฌ ๐ ๐ผ ๐ ๐ ๐
๐ฟ๐ ๐โ๐
๐ฏ๐ ๐ ๐ ๐ซ ๐ ๐ฎ ๐
๐ฏ๐ ๐
๐น ๐ ๐ฌ ๐ ๐ผ ๐ ๐ ๐๐ฏ๐ ๐ ๐ซ ๐ ๐ฎ ๐
โ๐ฏ๐ ๐
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Example_2โข Convert the following block diagram to a signal flow graph
โข Solution:
Step_1: Label all signals, then place a node for each
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Solution (Contโd)Step_3: Connect nodes with branches, each representing a system block
Step_4: Simplify โ eliminate ๐5 ๐ , ๐6 ๐ and ๐7 ๐
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Tryโฆโข Convert the following block diagram to a signal flow graph
2
1 2
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D. Masonโs Rule (Mason, 1953)
The block diagram reduction technique requiressuccessive application of fundamental relationships in order to arrive at the system transfer function.
On the other hand, Masonโs rule for reducing a signal-flow graph to a single transfer function requires the application of one formula.
The formula was derived by S. J. Mason when he related the signal-flow graph to the simultaneous equations that can be written from the graph.
Before presenting the Masonโs rule formula, we need to define some terminology:
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Masonโs Rule: Terminology Loop: a closed path that originates and terminates on the
same node.
Non-touching loops: loops that do not have any nodes in common.
Loop gain: Total gain (product of individual gains) around any path in the signal flow graph.
Beginning and ending at the same node.
Not passing through any node more than once.
Forward path gain: gain along any path from the input to the output.
Not passing through any node more than once.
Nonโtouching loop gains: the product of loop gains from nonโtouching loops, taken two, three, four, or more at a time.
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Masonโs Rule: Illustration of Terminology From the following signal flow graph, let get:
1) The loop gain:
3 loops with gains: โ๐บ1 ๐ ๐ป3 ๐ ,
๐บ2 ๐ ๐ป1 ๐ , and
โ๐บ2 ๐ ๐บ3 ๐ ๐ป2 ๐
2) Forward path gain:
2 forward paths with gains: ๐บ1 ๐ ๐บ2 ๐ ๐บ3 ๐ ๐บ4 ๐
๐บ1 ๐ ๐บ2 ๐ ๐บ5 ๐
3) Non-touching loop gains:
2 pairs of non-touching loops with gains: โ๐บ1 ๐ ๐ป3 ๐ โ ๐บ2 ๐ ๐ป1 ๐
โ๐บ1 ๐ ๐ป3 ๐ โ โ๐บ2 ๐ ๐บ3 ๐ ๐ป2 ๐
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The Transfer Function, ๐ถ ๐ ๐ ๐ , of a system representedby a signal-flow graph is:
๐ถ ๐
๐ ๐ =
๐=1๐ ๐๐โ๐
โ
Where: โข ๐ = number of forward pathsโข ๐๐ = the ๐๐กโ forward path gainโข โ= determinant of the Systemโข โ๐= determinant of the ๐๐กโ forward path
โ is called the signal flow graph determinant or characteristic function since โ= 0 is the system characteristic equation.
Masonโs Rule
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โข โ= 1 โ ๐๐๐๐ ๐๐๐๐๐
+ ๐๐๐ โ ๐ก๐๐ข๐โ๐๐๐ ๐๐๐๐ ๐๐๐๐๐ ๐ก๐๐๐๐ 2 ๐๐ก ๐ ๐ก๐๐๐
โ ๐๐๐ โ ๐ก๐๐ข๐โ๐๐๐ ๐๐๐๐ ๐๐๐๐๐ ๐ก๐๐๐๐ 3 ๐๐ก ๐ ๐ก๐๐๐
+ ๐๐๐ โ ๐ก๐๐ข๐โ๐๐๐ ๐๐๐๐ ๐๐๐๐๐ ๐ก๐๐๐๐ 4 ๐๐ก ๐ ๐ก๐๐๐
โ โฏ
โข โ๐= โ โ ๐๐๐๐ ๐๐๐๐ ๐ก๐๐๐๐ ๐๐ โ ๐๐๐๐ ๐๐๐๐๐ ๐กโ๐ ๐๐กโ ๐๐๐๐ค๐๐๐ ๐๐๐กโ
= ๐๐๐๐ ๐๐๐๐ ๐ก๐๐๐๐ ๐๐ โ ๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐ ๐กโ๐ ๐๐กโ ๐๐๐๐ค๐๐๐ ๐๐๐กโ
Note: โ๐= 1 if there are no non-touching loops to the ๐๐กโ path.
Masonโs Rule (Contโd)
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Seven (7) steps:
1) Calculate forward path gain ๐๐ for each forward path ๐.
2) Calculate all loop Transfer Functions
3) Consider non-touching loop gains (NTLGs) taken 2 at a time
4) Consider non-touching loop gains (NTLGs) taken 3 at a time
5) Etc.
6) Calculate โ from steps 2, 3, 4 and 5.
7) Calculate โ๐ as portion of โ that not touching the forward path ๐.
Masonโs Rule Systematic Approach
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Example Apply Masonโs Rule to calculate the transfer function of the system
represented by the following Signal Flow Graph.
Solution:
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Step_1: Get the forward path gain ๐๐ for
each forward path ๐:
โขNumber of forward paths is 2
โขForward path gains: ๐1 = ๐บ1๐บ2๐บ3๐บ4
๐2 = ๐บ1๐บ2๐บ5
Step_2: Get the ๐๐๐๐ ๐๐๐๐๐
โขNumber of loops with gains is 3
โข ๐๐๐๐ ๐๐๐๐๐ = โ๐บ1๐ป3 + ๐บ2๐ป1 โ ๐บ2๐บ3๐ป2
Step_3: Get the ๐๐๐ฟ๐บ๐ 2 @ ๐ก๐๐๐
each forward path ๐:
โขNumber of NTLs is 2 pairs
โข ๐๐๐ฟ๐บ๐ = โ๐บ1๐ป3๐บ2๐ป1
+ ๐บ1๐ป3๐บ2๐บ3๐ป2
Step_4: Get โ:
โขโ= 1 โ โ๐บ1๐ป3 + ๐บ2๐ป1 โ ๐บ2๐บ3๐ป2
+ โ๐บ1๐ป3๐บ2๐ป1 + ๐บ1๐ป3๐บ2๐บ3๐ป2
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Solution (Contโd)Step_5: Get โ๐:
โข โ๐= โ โ ๐๐๐๐ ๐๐๐๐ ๐ก๐๐๐๐ ๐๐ โ ๐๐๐๐ ๐๐๐๐๐ ๐กโ๐ ๐๐กโ ๐๐๐๐ค๐๐๐ ๐๐๐กโ
= ๐๐๐๐ ๐๐๐๐ ๐ก๐๐๐๐ ๐๐ โ ๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐ ๐กโ๐ ๐๐กโ ๐๐๐๐ค๐๐๐ ๐๐๐กโ
Thus:
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Note: Simplest way to find โ๐ terms is to calculate โ with the ๐๐๐ path removed;
โข Must remove nodes as well !!!
For ๐ = 1:
With forward path 1 removed, there are no loops, so:
โ1= โ= 1 โ 0 = 1
For ๐ = 2:
Similarly, removing forward path 2 leaves no loops, so:
โ2= โ= 1 โ 0 = 1
Finally:
๐ ๐
๐ ๐ =
๐=1๐ ๐๐โ๐
โ=
๐1โ1 + ๐2โ2
โ=
๐บ1๐บ2๐บ3๐บ4 + ๐บ1๐บ2๐บ5
1 + ๐บ1๐ป3 โ ๐บ2๐ป1 + ๐บ2๐บ3๐ป2 โ ๐บ1๐ป3๐บ2๐ป1 + ๐บ1๐ป3๐บ2๐บ3๐ป2
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Tryโฆโข Apply Masonโs Rule to calculate the transfer function of the
system represented by the following Signal Flow Graph.
Solution:
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Thank You!!!
Any Questions?
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