Chapter 2 : Fundamental of Mathematical Modeling

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:

– the Laplace Transform of time functions and the inverse

Laplace Transform.

– manually and with MatLab, the Transfer Function from

Mathematical Models

– Signal flow graphs from Block Diagram

– Transfer Function using Mason’s Rule

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Contents



1. Introduction

2. Laplace Transform

3. Transfer Function;- Numerical

- With MatLab

4. Block Diagram Models- Block Diagram Algebra

- Signal Flow Graphs and Mason’s Rule

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CHAP_2. 1: Introduction

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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 Newton’s laws of motion, Kirchhoff's laws

of electrical network, Ohm’s 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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Obtained by applying the physical laws of the process: Therefore:

i. Mechanical Systems Newton’s laws

ii. Electrical Systems *Kirchhoff's laws

* Ohm’s laws

* Mesh & Nodal Analysis

Example_1: Springer-mass-damper mechanical system.

Differential Equations

• Hence for viscous wall: 𝑓 𝑡 = 𝑏𝑣 𝑡

* Assumption: Wall friction force f 𝒕 is

linearly proportional to the velocity 𝒗 of

the mass 𝑀 .

• and for spring: 𝑓 𝑡 = 𝑘 0

𝑡𝑣 𝜏 𝑑𝜏

= 𝑘𝑦 𝑡

The time function

of 𝒓 𝒕 is called

forcing function

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Step_1: Free-Body Diagram (FBD): Step_2: Newton’s 2nd Law.

𝐹 = 𝑀𝑎 𝑡 : for translation motionOr:

• 𝑣 𝑡 = 𝑦 =𝑑𝑦 𝑡

𝑑𝑡

• 𝑎 𝑡 = 𝑦 =𝑑2𝑦 𝑡

𝑑𝑡2

Hence:

𝑀𝑎 𝑡 = −𝑏𝑣 𝑡 − 𝑘𝑦 𝑡 + 𝑟 𝑡

𝑀𝑑𝑣 𝑡

𝑑𝑡+ 𝑏𝑣 𝑡 + 𝑘𝑦 𝑡 = 𝑟 𝑡

Finally, the differential equation is:

𝑀𝑑2𝑦 𝑡

𝑑𝑡2 + 𝑏𝑑𝑦 𝑡

𝑑𝑡+ 𝑘𝑦 𝑡 = 𝑟 𝑡



Mechanical System Analysis:

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• For the following mechanism, define:

i. The input → 𝑭ii.The output → 𝒚.

Newton’s 2nd Law:

𝐹 = 𝑀𝑎 𝑡 : for translation motion

Hence:

𝑀𝑎 𝑡 = 𝐹 − 𝑘𝑦 𝑡 − 𝑓𝑣𝑣 𝑡

𝑀𝑑𝑣 𝑡

𝑑𝑡+ 𝑘𝑦 𝑡 + 𝑓𝑣𝑣 𝑡 = 𝐹 𝑡

Finally, the differential equation is:

𝑀𝑑2𝑦 𝑡

𝑑𝑡2 + 𝑓𝑣𝑑𝑦 𝑡

𝑑𝑡+ 𝑘𝑦 𝑡 = 𝐹 𝑡



Example_2: A Mechanical Mechanism.

• for viscous damper: 𝑓 𝑡 = 𝑓𝑣𝑣 𝑡

= 𝑓𝑣𝑑𝑦 𝑡

𝑑𝑡

y

k

f

F

mM

𝒇𝒗

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• For the following circuit, define:

i. The input → 𝒖𝒓ii.The output → 𝒖𝒄.

Kirchhoff voltage law (KVL):

𝑉𝑅 + 𝑉𝐿 + 𝑢𝑐 𝑡 = 𝑢𝑟 𝑡

Hence:

Ohm's laws:

𝐿𝑑𝑖 𝑡

𝑑𝑡+ 𝑅𝑖 𝑡 + 𝑢𝑐 𝑡 = 𝑢𝑟 𝑡

Or:

• 𝑖 𝑡 = 𝐶𝑑𝑢𝑐

𝑑𝑡(capacitor)

• 𝑣 𝑡 = 𝐿𝑑𝑖(𝑡)

𝑑𝑡(inductor)

Therefore, the differential equation is:

𝑑2𝑢𝑐 𝑡

𝑑𝑡2 +𝑅

𝐿

𝑑𝑢𝑐 𝑡

𝑑𝑡+

1

𝐿𝐶𝑢𝑐 𝑡 =

1

𝐿𝐶𝑢𝑟 𝑡



Example_3: A Passive Electrical Circuit.

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Initial conditions:

• 𝑖 0 = 𝐼0: current through inductor

• 𝑣 0 =1

𝐶 −∞

0𝑖 𝜏 𝑑𝜏 = 𝑉0:

voltage across capacitor

ur

uc

R L

Ci

• Define the differential equation for the following RLC electrical circuit:

Kirchhoff voltage law (KVL):

𝑉𝐿 + 𝑉𝑅 + 𝑉𝐶 = 0Hence:

Ohm’s lawsOr:

• 𝑣 𝑡 = 𝐿𝑑𝑖 𝑡

𝑑𝑡(inductor)

• 𝑣 𝑡 =1

𝐶 0

𝑡𝑖 𝜏 𝑑𝜏(capacitor)

Thus:

𝐿𝑑𝑖 𝑡

𝑑𝑡+ 𝑅𝑖 𝑡 +

1

𝐶 0

𝑡𝑖 𝜏 𝑑 𝑑𝜏 = 0

Therefore, the differential equation is:

𝑑2𝑖 𝑡

𝑑𝑡2 +𝑅

𝐿

𝑑𝑖 𝑡

𝑑𝑡+

1

𝐿𝐶𝑖 𝑡 = 0



Example_4: Source-Free Series RLC Circuit.

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Initial conditions:

• 𝑖 0 = 𝐼0: current through inductor

• 𝑣 0 =1

𝐶 −∞

0𝑖 𝜏 𝑑𝜏 = 𝑉0: voltage

across capacitor

+

−

1. Define the differential equation for the following Mechanical System:

2. Define the differential equation for the following RLC Electrical Network:



Try …

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𝐴𝑛𝑠: 𝑚 𝑥0 + 𝑏 𝑥0 + 𝑘𝑥0 = 𝑏 𝑥𝑖 + 𝑘𝑥𝑖

𝑘 𝑏

𝑥𝑖

𝑥0

𝑚



CHAP_2. 2: Laplace Transform

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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 –domain signals

ℒ 𝑓 𝑡 = 𝐹 𝑠 = 0

∞

𝑓 𝑡 𝑒−𝑠𝑡𝑑𝑡

ℒ−1 𝐹 𝑠 =1

2𝜋𝑗 𝜎−𝑗∞

𝜎+𝑗∞

𝐹 𝑠 𝑒𝑠𝑡𝑑𝑆 = 𝑓 𝑡 𝑢 𝑡

Transfer Function 𝐹 𝑠 is a function in Laplace domain 𝑠 where 𝒔 is a complex number: 𝒔 = 𝜶 + 𝒋𝝎

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A. Laplace Transform Table & Theorems

No. f(t) F(s)

1. δ 𝑡 1

2. u 𝑡1

𝑠

3. tu 𝑡1

𝑠2

4. 𝑡𝑛u 𝑡𝑛!

𝑠𝑛+1

5. 𝑒−𝑎𝑡u 𝑡1

𝑠 + 𝑎

6. 𝑡𝑒−𝑎𝑡u 𝑡1

𝑠 + 𝑎 2

7. 𝑡𝑛𝑒−𝑎𝑡u 𝑡𝑛!

𝑠 + 𝑎 𝑛+1

8. 1 − 𝑎𝑡 𝑒−𝑎𝑡u 𝑡𝑠

𝑠 + 𝑎 2

9.1

𝑎1 − 𝑒−𝑎𝑡u 𝑡

1

𝑠 𝑠 + 𝑎

⇐ Impulse function or Unit Impulse Function

⇐ Step function or Unit Step Function

⇐ Ramp function

1) Laplace Transform Table

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No. f(t) F(s)

10. 𝑠𝑖𝑛 𝜔𝑡 u 𝑡𝜔

𝑠2 + 𝜔2

11. 𝑐𝑜𝑠 𝜔𝑡 u 𝑡𝑠

𝑠2 + 𝜔2

12. 𝑒−𝑎𝑡 𝑠𝑖𝑛 𝜔𝑡 u 𝑡𝜔

𝑠 + 𝑎 2 + 𝜔2

13. 𝑒−𝑎𝑡 𝑐𝑜𝑠 𝜔𝑡 u 𝑡𝑠 + 𝑎

𝑠 + 𝑎 2 + 𝜔2

2) Laplace Transform TheoremsNo. Theorem Name

1. ℒ 𝑓 𝑡 = 𝐹 𝑠 = 0

∞

𝑓 𝑡 𝑒−𝑠𝑡𝑑𝑡 Laplace definition

2. ℒ 𝑘𝑓 𝑡 = 𝑘𝐹 𝑠 Linearity Theorem

3. ℒ 𝑓1 𝑡 + 𝑓2 𝑡 = 𝐹1 𝑠 + 𝐹2 𝑠 Linearity Theorem

4. ℒ 𝑒−𝑎𝑡𝑓 𝑡 = 𝐹 𝑠 + 𝑎 Frequency Shift Theorem

5. ℒ 𝑓 𝑡 − 𝑇 = 𝑒−𝑠𝑇𝐹 𝑠 Time Shift 𝑇 Theorem

6. ℒ 𝑓 𝑎𝑡 =1

𝑎𝐹

𝑠

𝑎Scaling Theorem

7. ℒ𝑑𝑓 𝑡

𝑑𝑡= 𝑠𝐹 𝑠 − 𝑓 0− Differentiation Theorem

8. ℒ𝑑2𝑓 𝑡

𝑑𝑡2= 𝑠2𝐹 𝑠 − 𝑠𝑓 0− − 𝑓′ 0− Differentiation Theorem

9. ℒ𝑑𝑛𝑓 𝑡

𝑑𝑡𝑛= 𝑠𝑛𝐹 𝑠 −

𝑘=1

𝑛

𝑠𝑛−𝑘 𝑓𝑘−1 0− Differentiation Theorem

(in general)

10. ℒ 0

𝑡

𝑓 𝜏 𝑑𝜏 =𝐹 𝑠

𝑠Integration Theorem

11. 𝑓 ∞ = 𝑙𝑖𝑚𝑠→0

𝑠𝐹 𝑠 Final Value Theorem

12. 𝑓 0+ = 𝑙𝑖𝑚𝑠→∞

𝑠𝐹 𝑠 Initial Value Theorem11/10/2021 15



ExampleWith 𝒖 𝒕 a unit step, find the Laplace Transform of the following function 𝑦 𝑡 , assuming zeros initial conditions:

𝑑2𝑦 𝑡

𝑑𝑡2 + 12𝑑𝑦 𝑡

𝑑𝑡+ 32𝑦 𝑡 = 32𝑢 𝑡

Solution:

• ℒ𝑑2𝑦 𝑡

𝑑𝑡2 = 𝑠2𝑌 𝑠 − 𝑠𝑦 0− − 𝑦′ 0−

• ℒ 12𝑑𝑦 𝑡

𝑑𝑡= 12 𝑠𝑌 𝑠 − 𝑦 0−

• ℒ 32𝑦 𝑡 = 32𝑌 𝑠

• ℒ 32𝑢 𝑡 = 32𝑈 𝑠

With zeros initial conditions, we finally obtain the Laplace Transform as:

𝑠2𝑌 𝑠 + 12𝑠𝑌 𝑠 + 32𝑌 𝑠 = 32𝑈 𝑠

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Zeros Initial Conditions:• 𝑦 0− = 0• 𝑦′ 0− = 0



Try…1) With 𝒇 𝒕 a unit step, find the Laplace Transform of the

following function:

𝑑2𝑦 𝑡


𝑑𝑡+ 10𝑦 𝑡 =

𝑑𝑓 𝑡

𝑑𝑡

Assume the initial conditions as:

• 𝑓 0− = 0, 𝑦 0− = 0 and𝑑𝑦 0−

𝑑𝑡= 1

2) Find the Laplace Transform of the following function:

𝑓 𝑡 = 𝑡𝑒−5𝑡

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𝐴𝑛𝑠: 𝑠2𝑌 𝑠 + 2𝑠𝑌 𝑠 + 10𝑌 𝑠 − 1 = 𝑠𝐹 𝑠

𝐴𝑛𝑠: 𝐹 𝑠 =1

𝑠 + 5 2



B. Inverse Laplace Transform

General form:

Three (3) possible cases:

i. Case_1: Roots of 𝐷 𝑠 are real & distinct;

ii.Case_2: Roots of 𝐷 𝑠 are real & repeated;

iii.Case_3: Roots of 𝐷 𝑠 are complex conjugate;

𝐹 𝑠 =𝑁 𝑠

𝐷 𝑠

Numerator

Denominator

o Hint: Use ‘Partial Fraction Expansion’

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• Example: 𝐹 𝑠 =2

𝑠+1 𝑠+2

Solution:

• 𝐹 𝑠 =2

𝑠+1 𝑠+2=

𝐴

𝑠+1+

𝐵

𝑠+2(Partial Fraction Expansion)

=2

𝑠+1−

2

𝑠+2

• The inverse Laplace Transform or Time Response is:

𝑓 𝑡 = 2𝑒−𝑡𝑢 𝑡 − 2𝑒−2𝑡𝑢 𝑡

As 𝑢 𝑡 is the unit step function, 𝑓 𝑡 can finally be expressed in another way as:

It is found that: 𝐴 = 2 and 𝐵 = −2

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i. Case_1: Roots of 𝐷 𝑠 are real & distinct

Item No.5 in the Laplace Transform Table

𝑓 𝑡 = 2𝑒−𝑡 − 2𝑒−2𝑡 = 2 𝑒−𝑡 − 𝑒−2𝑡



𝐹 𝑠 =2

𝑠+1 𝑠+2=

𝐴

𝑠+1+

𝐵

𝑠+2

=𝐴 𝑠+2 +𝐵 𝑠+1

𝑠+1 𝑠+2

Hence:

𝐴 𝑠 + 2 + 𝐵 𝑠 + 1 = 2

Solving steps:

Case_1: 𝑠 + 1 = 0 ⇒ 𝑠 = −1⇒ 𝐴 −1 + 2 = 2⇒ 𝐴 = 2

Case_2: 𝑠 + 2 = 0 ⇒ 𝑠 = −2⇒ 𝐵 −2 + 1 = 2⇒ 𝐵 = −2

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Work out Method:



Try…

- Find the Inverse Laplace Transform of the following function:

a) G 𝑠 =𝑠+3

𝑠+1 𝑠+2

Solution:

b) G 𝑠 =32

𝑠 𝑠2+12𝑠+32

Solution:

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𝐴𝑛𝑠: 𝑔 𝑡 = 2𝑒−𝑡 − 𝑒−2𝑡

𝐴𝑛𝑠: 𝑔 𝑡 = 𝑒−𝑡 − 2𝑒−4𝑡 + 𝑒−8𝑡



ii. Case_2: Roots of 𝐷 𝑠 are real & repeated

•Example: 𝐹 𝑠 =2

𝑠+1 𝑠+2 2

Solution:

• 𝐹 𝑠 =2

𝑠+1 𝑠+2 2 =𝐴

𝑠+1+

𝐵

𝑠+2+

𝐶

𝑠+2 2

=2

𝑠+1−

2

𝑠+2−

2

𝑠+2 2

• Finally,

Example:

• 𝐹 𝑠 =2

𝑠+1 𝑠+2 2 =𝐴

𝑠+1+

𝐵

𝑠+2+

𝐶

𝑠+2 2

=𝐴 𝑠+2 2+𝐵 𝑠+1 𝑠+2 +𝐶 𝑠+1

𝑠+1 𝑠+2 2

It is found that: 𝐴 = 2, 𝐵 = −2 and 𝐶 = −2

𝑓 𝑡 = 2𝑒−𝑡 − 2𝑒−2𝑡 − 2𝑡𝑒−2𝑡

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Work out Method:



Thus:

• 𝐹 𝑠 =2

𝑠+1 𝑠+2 2 =𝐴 𝑠+2 2+𝐵 𝑠+1 𝑠+2 +𝐶 𝑠+1

𝑠+1 𝑠+2 2

=𝐴 𝑠2+4𝑠+4 +𝐵 𝑠2+3𝑠+2 +𝐶 𝑠+1

𝑠+1 𝑠+2 2 =𝐴𝑠2+4𝐴𝑠+4𝐴+𝐵𝑠2+3𝐵𝑠+2𝐵+𝐶𝑠+𝐶

𝑠+1 𝑠+2 2

⇔ 𝐹 𝑠 =2

𝑠+1 𝑠+2 2 =𝑠2 𝐴+𝐵 +𝑠 4𝐴+3𝐵+𝐶 + 4𝐴+2𝐵+𝐶

𝑠+1 𝑠+2 2

Equating like powers of "𝒔" gives us a system of equations as:

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Powers of "𝒔" Equation

𝑠2 𝐴 + 𝐵 = 0

𝑠1 4𝐴 + 3𝐵 + 𝐶 = 0

𝑠0 4𝐴 + 2𝐵 + 𝐶 = 2

Solving the system of equations yields:

𝐴 = 2

B= −2

C= −2

Work out Method (Cont′d):



Try…

- Find the Inverse Laplace Transform of the following functions:

a) 𝐹 𝑠 =𝑠+3

𝑠+1 3 𝑠+2

Solution:

b) G 𝑠 =𝑠2+1

𝑠2 𝑠+2

Solution:

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𝐴𝑛𝑠: 𝑓 𝑡 = 𝑡2𝑒−𝑡 − 𝑡𝑒−𝑡 + 𝑒−𝑡 − 𝑒−2𝑡

= 𝑒−𝑡 𝑡2 − 𝑡 + 1 − 𝑒−2𝑡

𝐴𝑛𝑠: 𝑔 𝑡 =5

4𝑒−2𝑡 +

1

2𝑡 −

1

4



iii.Case_3: Roots of 𝐷 𝑠 are complex conjugate

•Example: 𝐹 𝑠 =3

𝑠 𝑠2+2𝑠+5

Solution:

• 𝐹 𝑠 =3

𝑠 𝑠2+2𝑠+5=

𝐴

𝑠+

𝐵𝑠+𝐶

𝑠2+2𝑠+5

= 35

𝑠−

3

5

𝑠+2

𝑠2+2𝑠+5

= 35

𝑠−

3

5

𝑠+1+1

𝑠2+2𝑠+5=

35

𝑠−

3

5

𝑠+1 + 1 2 2

𝑠+1 2+22

= 35

𝑠−

3

5

𝑠+1

𝑠+1 2+22 + 1 2 2

𝑠+1 2+22

• Finally,

It is found that: 𝐴 = 3 5, 𝐵 = −3 5 and 𝐶 = −6 5

𝑓 𝑡 =3

5−

3

5𝑒−𝑡 𝑐𝑜𝑠 2𝑡 +

1

2𝑒−𝑡𝑠𝑖𝑛2𝑡 =

3

5−

3

5𝑒−𝑡 𝑐𝑜𝑠 2𝑡 +

1

2𝑠𝑖𝑛 2𝑡

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•Example:

𝐹 𝑠 =3

𝑠 𝑠2+2𝑠+5=

𝐴

𝑠+

𝐵𝑠+𝐶

𝑠2+2𝑠+5=

𝐴 𝑠2+2𝑠+5 + 𝐵𝑠+𝐶 𝑠

𝑠 𝑠2+2𝑠+5

=𝐴𝑠2+2𝐴𝑠+5𝐴+𝐵𝑠2+𝐶𝑠

𝑠 𝑠2+2𝑠+5=

𝑠2 𝐴+𝐵 +𝑠 2𝐴+𝐶 +5𝐴

𝑠 𝑠2+2𝑠+5

⇔ 𝐹 𝑠 =3

𝑠 𝑠2+2𝑠+5=

𝑠2 𝐴+𝐵 +𝑠 2𝐴+𝐶 +5𝐴

𝑠 𝑠2+2𝑠+5

Equating like powers of "𝒔" gives us a system of equations as:

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Work out Method:

Powers of "𝒔" Equation

𝑠2 𝐴 + 𝐵 = 0

𝑠1 2𝐴 + 𝐶 = 0

𝑠0 5𝐴 = 3

Solving the system of equations yields:

𝐴 = 3 5

B= − 3 5

C= − 6 5



Try…

- Find the Inverse Laplace Transform of the following functions:

a) 𝐹 𝑠 =𝑠+3

𝑠2+4𝑠+5 𝑠+5

Solution:

b) G 𝑠 =10

𝑠2+9 𝑠+1

Solution:

11/10/2021 27

𝐴𝑛𝑠: 𝑓 𝑡 = −1

5𝑒−5𝑡 +

1

5𝑐𝑜𝑠 𝑡 𝑒−2𝑡 +

2

5𝑠𝑖𝑛 𝑡 𝑒−2𝑡

𝐴𝑛𝑠: 𝑔 𝑡 = 𝑒−𝑡 − 𝑐𝑜𝑠 3𝑡 −1

3𝑠𝑖𝑛 3𝑡



Partial-Fraction Expansion with MATLAB Three (3) steps to obtain the Partial-Fraction Expansion :

1. Define the coefficients in descending powers of 𝑠 as follow:o Coefficients of numerator polynomial as numerator coefficients ,o Coefficients of denominator polynomial as denominator

coefficients

2. Use the ′𝒓𝒆𝒔𝒊𝒅𝒖𝒆′ command to get the residues 𝒓 , poles 𝒑and direct terms 𝒌 of a Partial-Fraction Expansion of the ratio of the two polynomials 𝐵 𝑠 and 𝐴 𝑠 ;

3. The partial-fraction expansion of 𝐵 𝑠 𝐴 𝑠 is given by:

𝐵 𝑠

𝐴 𝑠= 𝑘 𝑠 +

𝑟 1

𝑠 − 𝑝 1+

𝑟 2

𝑠 − 𝑝 2+ ⋯ +

𝑟 𝑛

𝑠 − 𝑝 𝑛

Syntax:

𝑟, 𝑝, 𝑘 = 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 , 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠

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Example Get the partial-fraction expansion for the following expression:

𝐵 𝑠

𝐴 𝑠=

𝑠4+8𝑠3+16𝑠2+9𝑠+6

𝑠3+6𝑠2+11𝑠+6

Solution- Step_1: Coefficient definition:

o Numerator: 𝑁 = 1 8 16 9 6o Denominator: 𝐷 = 1 6 11 6

- Step_2: ‘residue’ command in Matlab

o 𝑟, 𝑝, 𝑘 = 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 1 8 16 9 6 , 1 6 11 6= 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 𝑁, 𝐷

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Solution (Cont’d) MatLab Results:

For direct terms 𝒌 :

𝒌 𝒔 = 𝟏𝒔𝟏 + 𝟐𝒔𝟎 = 𝒔 + 𝟐

- Step_3: Write the partial-fraction expansion as:

𝐵 𝑠

𝐴 𝑠=

𝑠4+8𝑠3+16𝑠2+9𝑠+6

𝑠3+6𝑠2+11𝑠+6⇔

𝐵 𝑠

𝐴 𝑠= 𝑠 + 2 −

6

𝑠+3−

4

𝑠+2+

3

𝑠+1

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Try… With MatLab, get the partial-fraction expansion for the following

expressions:

1) 𝐹 𝑠 =−4𝑠+8

𝑠2+6𝑠+8

2) 𝐺 𝑠 =2𝑠4+𝑠

𝑠2+1

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𝐴𝑛𝑠: 𝐹 𝑠 =−12

𝑠 + 4+

8

𝑠 + 2

𝐴𝑛𝑠: 𝐺 𝑠 = 2𝑠2 − 2 +0.5 − 𝑖

𝑠 − 𝑖+

0.5 + 𝑖

𝑠 + 𝑖



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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Examples

1) Find the response 𝑦 𝑡 for the following differential equation 𝑑𝑦 𝑡

𝑑𝑡+ 2𝑦 𝑡 = 𝑓 𝑡 , with input 𝑓 𝑡 and output 𝑦 𝑡 ;

if 𝑓 𝑡 = 𝑢 𝑡 − 𝑢 𝑡 − 1 i.e. a Unit Step Function, and the initial condition 𝑦 0− = −2.

Solution:- Step_1: Take the Laplace Transform of the differential equation.

• ℒ𝑑𝑦 𝑡

𝑑𝑡= 𝑠𝑌 𝑠 − 𝑦 0−

• ℒ 𝑦 𝑡 = 𝑌 𝑠

• ℒ 𝑓 𝑡 = 𝐹 𝑠 =1

𝑠− 𝑒−𝑠𝐹 𝑠 =

1

𝑠− 𝑒−𝑠 1

𝑠

So,

𝑑𝑦 𝑡

𝑑𝑡+ 2𝑦 𝑡 = 𝑢 𝑡 − 𝑢 𝑡 − 1 →

ℒ𝑠𝑌 𝑠 − 𝑦 0− + 2𝑌 𝑠 =

1

𝑠− 𝑒−𝑠 1

𝑠

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Solution (Cont’d)

- Step_2: Put initial conditions into the resulting equation.

• 𝑠𝑌 𝑠 − 𝑦 0− + 2𝑌 𝑠 =1

𝑠− 𝑒−𝑠 1

𝑠(The resulting equation)

• 𝑠𝑌 𝑠 + 2 + 2𝑌 𝑠 =1

𝑠− 𝑒−𝑠 1

𝑠(After putting the initial condition)

- Step_3: Solve for the output variable 𝑌 𝑠 .

• 𝑌 𝑠 =1

𝑠 𝑠+2− 𝑒−𝑠 1

𝑠 𝑠+2− 2

1

𝑠+2

- Step_4: Get result from the Laplace Transform Tables.

• ℒ−1 1

𝑠 𝑠+2=

1

21 − 𝑒−2𝑡

• ℒ−1 −𝑒−𝑠 1

𝑠 𝑠+2=

1

21 − 𝑒−2 𝑡−1 𝑢 𝑡 − 1

• ℒ−1 −21

𝑠+2= −2𝑒−2𝑡

Finally: 𝑦 𝑡 =12 1 − 𝑒−2𝑡 − 1

21 − 𝑒−2 𝑡−1 𝑢 𝑡 − 1 − 2𝑒−2𝑡

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Examples (Cont’d)

2) Find the response 𝑦 𝑡 for the following differential equation: 𝑑2𝑦 𝑡


𝑑𝑡+ 10𝑦 𝑡 =

𝑑𝑓 𝑡

𝑑𝑡, with input 𝑓 𝑡 and output 𝑦 𝑡 ;

if 𝑓 𝑡 = 𝑢 𝑡 i.e. a unit step input, and

the initial conditions: 𝑓 0− = 0, 𝑦 0− = 0 and𝑑𝑦 0−

𝑑𝑡= 1.

Solution:- Step_1: Take the Laplace Transform of the differential equation.

• ℒ𝑑2𝑦 𝑡

𝑑𝑡2 = 𝑠2𝑌 𝑠 − 𝑠𝑦 0− − 𝑦′ 0−

• ℒ 2𝑑𝑦 𝑡

𝑑𝑡= 2 𝑠𝑌 𝑠 − 𝑦 0−

• ℒ 10𝑦 𝑡 = 10𝑌 𝑠

• ℒ𝑑𝑓 𝑡

𝑑𝑡= 𝑠𝐹 𝑠 − 𝑓 0− = 𝑠

1

𝑠= 1

So,𝑑2𝑦 𝑡

𝑑𝑡2+ 2

𝑑𝑦 𝑡

𝑑𝑡+ 10𝑦 𝑡 =

𝑑𝑓 𝑡

𝑑𝑡→ℒ

𝑠2𝑌 𝑠 − 𝑠𝑦 0− − 𝑦′ 0− + 2 𝑠𝑌 𝑠 − 𝑦 0− + 10𝑌 𝑠 = 1

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Solution (Cont’d)

- Step_2: Put initial conditions into the resulting equation.

• 𝑠2𝑌 𝑠 − 𝑠𝑦 0− − 𝑦′ 0− + 2 𝑠𝑌 𝑠 − 𝑦 0− + 10𝑌 𝑠 = 1 (The resulting equation)

• 𝑠2𝑌 𝑠 − 𝑠. 0 − 1 + 2 𝑠𝑌 𝑠 − 0 + 10𝑌 𝑠 = 1 (After putting the initial condition)

⇔ 𝑠2𝑌 𝑠 − 1 + 2𝑠𝑌 𝑠 + 10𝑌 𝑠 = 1

- Step_3: Solve for the output variable 𝑌 𝑠 .

• 𝑌 𝑠 𝑠2 + 2𝑠 + 10 = 2

⇒ 𝑌 𝑠 =2

𝑠2+2𝑠+10

- Step_4: Get result from the Laplace Transform Tables.

• As the above result is not in the tables, let use the Inverse Laplace Transform.

Hence:

𝑌 𝑠 =2

𝑠2+2𝑠+10⇔ 𝑌 𝑠 =

2

𝑠+1 2+9=

2

𝑠+1 2+32

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From : 𝑌 𝑠 =2

𝑠+1 2+9=

2

𝑠+1 2+32

⇒ 𝑌 𝑠 =2

3

3

𝑠+1 2+32

We then use the table to determine the solution and finally get:

𝑦 𝑡 =2

3𝑒−𝑡𝑠𝑖𝑛 3𝑡

Try…- Find the response 𝑥 𝑡 for the following differential equations:

1) 𝑥 + 3 𝑥 + 2𝑥 = 0 with the following initial conditions: 𝑥 0 = 𝑎, 𝑥 0 = 𝑏

2) 𝑥 + 2 𝑥 + 5𝑥 = 3 with the following initial conditions: 𝑥 0 = 𝑎, 𝑥 0 = 𝑏

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CHAP_2. 3: Transfer Function

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2.3. Transfer Function (TF)A. Overview

• Commonly used to characterize the input—output relationships of components or systems that can be described by linear and time-invariant differential equations.

• It is defined as “the 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 zero”

𝑺𝒚𝒔𝒕𝒆𝒎Input Output

𝑟 𝑡 𝑐 𝑡𝒂

𝑺𝒖𝒃𝒚𝒔𝒕𝒆𝒎Input Output

𝑟 𝑡 𝑐 𝑡𝒃

𝑺𝒖𝒃𝒚𝒔𝒕𝒆𝒎 𝑺𝒖𝒃𝒚𝒔𝒕𝒆𝒎

With:• 𝑟 𝑡 ≡ 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛𝑝𝑢𝑡• 𝑐 𝑡 ≡ 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑙𝑒𝑑 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

11/10/2021 39



Overview (Cont’d)• A general form of the differential equation for Linear and Time

Invariant System (LTI-System) with an input 𝑟 𝑡 and output 𝑐 𝑡 is given by:

𝑎𝑛𝑑𝑛𝑐 𝑡

𝑑𝑡𝑛 + 𝑎𝑛−1𝑑𝑛−1𝑐 𝑡

𝑑𝑡𝑛−1 + ⋯ + 𝑎0𝑐 𝑡 = 𝑏𝑚𝑑𝑚𝑟 𝑡

𝑑𝑡𝑚 + 𝑏𝑚−1𝑑𝑚−1𝑟 𝑡

𝑑𝑡𝑚−1 + ⋯ + 𝑏0𝑟 𝑡

• The Transfer Function of this system is obtained by taking the Laplace transforms of both sides of Equation (assuming zero initial conditions),

𝑎𝑛𝑠𝑛𝐶 𝑠 + 𝑎𝑛−1𝑠𝑛−1𝐶 𝑠 + ⋯ + 𝑎0𝐶 𝑠 = 𝑏𝑚𝑠𝑚𝑅 𝑠 + 𝑏𝑚−1𝑠𝑚−1𝑅 𝑠 + ⋯ + 𝑏0𝑅 𝑠

𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯ + 𝑎0 𝐶 𝑠 = 𝑏𝑚𝑠𝑚 + 𝑏𝑚−1𝑠𝑚−1 + ⋯ + 𝑏0 𝑅 𝑠

𝐶 𝑠

𝑅 𝑠=

𝑏𝑚𝑠𝑚+𝑏𝑚−1𝑠𝑚−1+⋯+𝑏0

𝑎𝑛𝑠𝑛+𝑎𝑛−1𝑠𝑛−1+⋯+𝑎0= 𝐺 𝑠 •The equation 𝑮 𝒔 is called

Transfer Function (TF).

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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 𝐶 𝑠 to the input 𝑅 𝑠 .

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 𝑐 𝑡 , to an input 𝑟 𝑡 :

a)𝑑𝑐 𝑡

𝑑𝑡+ 2𝑐 𝑡 = 𝑟 𝑡

b) 𝑐 + 3 𝑐 + 𝑐 = 𝑟 𝑡

Solution:

a)𝑑𝑐 𝑡


• Taking the Laplace Transform:

𝑠𝐶 𝑠 + 2𝐶 𝑠 = 𝑅 𝑠

• The Transfer Function 𝐺 𝑠 , is:

𝐺 𝑠 =𝐶 𝑠

𝑅 𝑠=

1

𝑠+2

Example

11/10/2021 42

b) 𝑐 + 3 𝑐 + 2𝑐 = 𝑟 𝑡

• Taking the Laplace Transform:

𝑠2𝐶 𝑠 + 3s𝐶 𝑠 + 𝐶 𝑠 = 𝑅 𝑠

• The Transfer Function 𝐺 𝑠 , is:

𝐺 𝑠 =𝐶 𝑠

𝑅 𝑠=

1

𝑠2+3𝑠+1



From the given differential equations, find the Transfer Function represented by 𝑐 𝑡 , to an input 𝑟 𝑡 :

a) 5𝑑2𝑐 𝑡

𝑑𝑡2 − 3𝑐 𝑡 = 2𝑟 𝑡

b) 𝑥 + 2 𝑥 + 5𝑥 = 3𝑟 𝑡

Try…

11/10/2021 43



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 ′𝒕𝒇′ command to generate the Transfer Function (TF);

Syntax:

𝐺 𝑆 = 𝑡𝑓 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 , 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠

Example: 5𝑑4𝑦

𝑑𝑡4 − 4𝑑2𝑦

𝑑𝑡2 + 10𝑦 = 20𝑑𝑢

𝑑𝑡+ 4𝑢

Solution: -Step_1: Coefficient definition.

o Numerator: 𝑁 = 20 4o Denominator: 𝐷 = 5 0 − 4 10

-Step_2: ′𝒕𝒇′ command in Matlab.

o 𝐺 = 𝑡𝑓 20 4 , 5 0 − 4 10= 𝑡𝑓 𝑁, 𝐷

o Answer: 𝐺 =20𝑠+4

5𝑠^3−4𝑠+10

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• Use MATLAB to extract the Transfer Function represented by the following equations:

1.𝑑𝑐 𝑡


2. 5𝑑2𝑐 𝑡

𝑑𝑡2 − 3𝑐 𝑡 = 2𝑟 𝑡

Try…

11/10/2021 45



B. Open-Loop Transfer Function (OLTF)

OLTF can be represented using a block diagram:

𝑅 𝑠 𝐶 𝑠

𝑅 𝑠 𝐶 𝑠

Where:

𝐶 𝑠 = 𝐺 𝑠 𝑅 𝑠 or G 𝑠 =𝐶 𝑠

𝑅 𝑠

NOTE: G 𝑠 =𝐶 𝑠

𝑅 𝑠≡ Open-Loop Transfer Function.

𝐺 𝑠

𝑏𝑚𝑠𝑚 + 𝑏𝑚−1𝑠𝑚−1 + ⋯ + 𝑏0

𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯ + 𝑎0

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B. Closed-Loop Transfer Function (CLTF)

CLTF can be represented using a block diagram:

𝑅 𝑠 + E 𝑠 𝐶 𝑠

∓

Thus: 𝐸 𝑠 = R 𝑠 − 𝐻 𝑠 𝐶 𝑠 (eq.1) and 𝐶 𝑠 = G 𝑠 𝐸 𝑠 (eq.2)

eq.1 into eq.2: 𝐶 𝑠 = G 𝑠 [R 𝑠 − 𝐻 𝑠 𝐶 𝑠 ] (eq.3)

Re-arranging the eq.3 yields to:

𝐶 𝑠

𝑅 𝑠=

𝐺 𝑠

1±𝐺 𝑠 𝐻 𝑠≡ Closed-Loop Transfer Function.

𝐺 𝑠

𝐻 𝑠

Plant and Controller

Output

Feedback

Input Actuating signal (error)

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Feedback and its Effects Why Feedback?

• Feedback is a key tool that can be used to modify the behaviorof a system.

• 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.

• To reduce the error between the input and output of the system.

• It effects the system performance characteristics such as stability, overall system gain, sensitivity and bandwidth.

11/10/2021 48



o Feedback Effects on:

i. system sensitivity:

• It can make the system’s response less sensitive to externaldisturbances, parameter changes and noise

ii. system stability:

• An unstable system can be stabilized using feedback

• Stability refers to the ability of a system to follow its input signal

• A system that can’t control its output, or its output increases infinitely is an unstable system

• Adding feedback may also cause instability to an already stable system

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o Feedback Effects on (Cont’d)

iii. system gain:

• Feedback influences the system’s Transfer Function and consequently, the overall system gain:

• Thus, the system’s gain can be obtained by finding the magnitude of the system’s Transfer Function:

a) The system gain for OLCS: 𝐶 𝑠

𝑅 𝑠= 𝐺 𝑠

b) The system gain for CLCS: 𝐶 𝑠

𝑅 𝑠=

𝐺 𝑠

1+𝐺 𝑠 𝐻 𝑠

Effects of the product 𝐺 𝑠 𝐻 𝑠 :

– If 𝐺 𝑠 𝐻 𝑠 is negative;

𝐶 𝑠

𝑅 𝑠> 𝐺 𝑠 and we have positive feedback.

11/10/2021 50



Effects of the product 𝐺 𝑠 𝐻 𝑠 (Cont’d):

– If 𝐺 𝑠 𝐻 𝑠 is positive;

𝐶 𝑠

𝑅 𝑠< 𝐺 𝑠 and we have negative feedback.

o If 𝐺 𝑠 𝐻 𝑠 is positive and 𝐺 𝑠 𝐻 𝑠 ≫ 1;

𝐶 𝑠

𝑅 𝑠≈

𝐺 𝑠

𝐺 𝑠 𝐻 𝑠=

1

𝐻 𝑠

The gain is independent of the gain of the forward path 𝐺 𝑠 .

Depending on whether the feedback is positive + or negative − , the system gain of a Closed-Loop Control System (CLCS) can be increased or decreased.

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o Feedback Effects on (cont’d)

iv. system bandwidth:

• The system bandwidth increases as the gain is reduced with feedback;

• in some cases the gain × bandwidth = constant

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CHAP_2. 4: Block Diagram Models

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2.4 Block Diagram ModelsA. Overview

• A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals.

– Such diagram depicts the interrelationships that exist among the various components.

– Differing from a purely abstract mathematical representation, a block diagram has the advantage of indicating more realistically the signal flows of the actual system.

• Transfer function can be represented as a block diagram:

𝑏𝑚𝑠𝑚 + 𝑏𝑚−1𝑠𝑚−1 + ⋯ + 𝑏0

𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯ + 𝑎0

𝑅 𝑠 𝐶 𝑠

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Components of a Block Diagram for a LTI system

a) Signals

𝑅 𝑠 𝐶 𝑠

b) System

𝐺 𝑠𝑅 𝑠 𝐶 𝑠

Input Output

−

c) Summing Junction

𝑅1 𝑠 +

𝑅2 𝑠𝑅3 𝑠

𝐶 𝑠 = 𝑅1 𝑠 + 𝑅2 𝑠 − 𝑅3 𝑠

+

d) Pickoff point

𝑅 𝑠 𝑅 𝑠

𝑅 𝑠

𝑅 𝑠

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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

𝐺2 𝑠𝐺1 𝑠 𝐺3 𝑠𝑅 𝑠 𝑋2 𝑠𝑋1 𝑠 𝐶 𝑠

𝐺1 𝑠𝑋1 𝑠

𝑅 𝑠𝐺2 𝑠

𝐺3 𝑠

𝑋2 𝑠

𝑋3 𝑠

𝐶 𝑠±±

±

𝑅 𝑠𝐺 𝑠

𝐻 𝑠

𝐸 𝑠 𝐶 𝑠

𝑋 𝑠∓

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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.

𝐺1 𝑠𝑅 𝑠

𝐺2 𝑠 𝐺3 𝑠

𝑋1 𝑠= 𝐺1 𝑆 𝑅 𝑠

𝑋2 𝑠= 𝐺2 𝑠 𝐺1 𝑆 𝑅 𝑠

𝐶 𝑠= 𝐺3 𝑠 𝐺2 𝑠 𝐺1 𝑆 𝑅 𝑠

𝒂

𝑮𝟑 𝒔 𝑮𝟐 𝒔 𝑮𝟏 𝑺𝑅 𝑠 𝐶 𝑠

𝒃

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•Rules for reduction of the block diagram (Cont’d):

2. Equivalent Transfer Function for parallel substations.

𝐺1 𝑠𝑋1 𝑠 = 𝑅 𝑠 𝐺1 𝑠

𝑅 𝑠𝐺2 𝑠

𝐺3 𝑠

𝑋2 𝑠 = 𝑅 𝑠 𝐺2 𝑠

𝑋3 𝑠 = 𝑅 𝑠 𝐺3 𝑠

𝐶 𝑠= ±𝐺1 𝑠 ±𝐺2 𝑠 ± 𝐺3 𝑆 𝑅 𝑠

±±

±

±𝑮𝟏 𝒔 ±𝑮𝟐 𝒔 ± 𝑮𝟑 𝑺𝑅 𝑠 𝐶 𝑠

𝒃

𝒂

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3. Equivalent Transfer Function for substations with feedback.

𝐺 𝑠

1 ± 𝐺 𝑠 𝐻 𝑠

𝑅 𝑠 𝐶 𝑠

𝒃

𝒂

𝑅 𝑠𝐺 𝑠

𝐻 𝑠

𝐸 𝑠 𝐶 𝑠

𝑋 𝑠

∓

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•Rules for Reduction of Block Diagrams (Cont’d):

4. Moving Blocks/a Summing Junction.

𝑋 𝑠

∓

+ 𝐶 𝑠≡𝐺 𝑠

𝑅 𝑠

𝑋 𝑠

∓

+𝐺 𝑠

𝑅 𝑠

𝐺 𝑠

𝐺 𝑠𝐶 𝑠 𝑅 𝑠 𝐶 𝑠

𝒂 Back Past a Summing Junction/Behind the Block

𝒃 Forward Past a Summing Junction/Ahead of the Block

𝑋 𝑠

≡∓

+

∓

+

1

𝐺 𝑠

𝐺 𝑠𝐶 𝑠𝑅 𝑠

𝑋 𝑠

11/10/2021 62




5. Moving Blocks/a Pickoff Point.

𝑅 𝑠

𝑅 𝑠

𝑅 𝑠 𝐺 𝑠

≡𝐺 𝑠𝑅 𝑠

𝐺 𝑠

𝑅 𝑠 1

𝐺 𝑠𝐺 𝑠

𝑅 𝑠 𝐺 𝑠

𝑅 𝑠 𝑅 𝑠

𝒂 Backward Past Pickoff Point/Behind the Block

𝒃 Forward Past Pickoff Point/Ahead of the Block

≡

𝐺 𝑠𝑅 𝑠 𝐺 𝑠

𝑅 𝑠

𝑅 𝑠 𝐺 𝑠

1

𝐺 𝑠

𝑅 𝑠

𝑅 𝑠 𝐺 𝑠

𝑅 𝑠 𝐺 𝑠

𝐺 𝑠

𝐺 𝑠

𝑅 𝑠 𝐺 𝑠

𝑅 𝑠 𝐺 𝑠

11/10/2021 63



Three (3) tips for calculating the Equivalent Transfer Function of reduced Block Diagrams:

1) The Numerator of the closed-loop Transfer Function 𝐶 𝑠 𝑅 𝑠 is the product of the transfer functions of the

feed forward path.

2) The denominator of the closed-loop transfer function 𝐶 𝑠 𝑅 𝑠 is equal to:

1 − 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑜𝑢𝑛𝑑 𝑒𝑎𝑐ℎ 𝑙𝑜𝑜𝑝

3) The positive feedback loop yields a negative term in the denominator.

Conclusion of Block Diagram Reduction Techniques

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Example_1• Reduce the following block diagram

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Working…

1) Moving the summing point ahead of 𝐺1, we have:

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Working…

2) Combining 𝐺1 and 𝐺2 in Cascade, we get:

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Working…

3) Eliminating the feedback loop 𝐺1, 𝐺2 and 𝐻1, we get:

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Working…

4) Combing the two blocks in Cascade, we get

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Working…

5) Similarly eliminating the second feedback loop, we get:

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Working…

6) Similarly eliminating the third feedback loop, we get:

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Working…

7) Finally, the system is reduced to the following block diagram:

11/10/2021 72



Example_2• Deduce the response 𝒀 𝒔 of the following control system with two

inputs 𝑅 𝑠 and 𝐷 𝑠 .

• Solution:

a) Find:

- Step_1: 𝑌1 𝑠 𝐷 𝑠 when 𝑅 𝑠 = 0

𝑌1 𝑠

𝐷 𝑠 𝑅 𝑠 =0=

𝐺2

1+𝐺2 𝐻1−𝐻2

- Step_2: 𝑌2 𝑠 𝑅 𝑠 when 𝐷 𝑠 = 0

𝑌2 𝑠

𝑅 𝑠 𝐷 𝑠 =0=

𝐺1𝐺2

1+𝐺2 𝐻1−𝐻2

b) The total response 𝑌 𝑠 when 𝑅 𝑠and 𝐷 𝑠 are zeros is:

𝑌 𝑠 = 𝑌1 𝑠 + 𝑌2 𝑠

= 𝑌1 𝑠

𝐷 𝑠 𝑅 𝑠 =0∙ 𝐷 𝑠 +

𝑌2 𝑠

𝑅 𝑠 𝐷 𝑠 =0∙ 𝑅 𝑠

=𝐺2

1+𝐺2 𝐻1−𝐻2𝐷 𝑠 +

𝐺1𝐺2

1+𝐺2 𝐻1−𝐻2𝑅 𝑠

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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 calledMason’s 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 –represent signals

Nodes (sometimes) labeled with signal names

2) Branches –represent 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_1• Convert the following block diagram to a signal flow graph

• Solution:

Step_1:

- Identify and Label any unlabeled signals.

Step_2:

- Place a node for each signal.

𝑹 𝒔 𝑿𝟏 𝒔 𝑬 𝒔 𝑼 𝒔𝒀 𝒔

𝑿𝟐 𝒔

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Solution (Cont’d)Step_3:

- Connect nodes with branches, each representing a system block

Note: the −1 to provide negative feedback of 𝑿𝟐 𝒔

Step_4:

- To simplify, intermediate nodes with a single input and single output can be eliminated, if desired:

This makes sense for 𝑋1 𝑠 and 𝑋2 𝑠

Leave 𝑈 𝑠 to indicate separation between controller and plant.

𝑹 𝒔 𝑿𝟏 𝒔 𝑬 𝒔 𝑼 𝒔 𝒀 𝒔

𝑿𝟐 𝒔−𝟏

𝑯𝟏 𝒔 𝟏 𝑫 𝒔 𝑮 𝒔

𝑯𝟐 𝒔

𝑹 𝒔 𝑬 𝒔 𝑼 𝒔 𝒀 𝒔𝑯𝟏 𝒔 𝑫 𝒔 𝑮 𝒔

−𝑯𝟐 𝒔

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Example_2• Convert the following block diagram to a signal flow graph

• Solution:

Step_1: Label all signals, then place a node for each

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Solution (Cont’d)Step_3: Connect nodes with branches, each representing a system block

Step_4: Simplify – eliminate 𝑋5 𝑠 , 𝑋6 𝑠 and 𝑋7 𝑠

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Try…• Convert the following block diagram to a signal flow graph

2

1 2

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D. Mason’s 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, Mason’s 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 Mason’s rule formula, we need to define some terminology:

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Mason’s 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.

Non‐touching loop gains: the product of loop gains from non‐touching loops, taken two, three, four, or more at a time.

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Mason’s Rule: Illustration of Terminology From the following signal flow graph, let get:

1) The loop gain:

3 loops with gains: −𝐺1 𝑠 𝐻3 𝑠 ,

𝐺2 𝑠 𝐻1 𝑠 , and

−𝐺2 𝑠 𝐺3 𝑠 𝐻2 𝑠

2) Forward path gain:

2 forward paths with gains: 𝐺1 𝑠 𝐺2 𝑠 𝐺3 𝑠 𝐺4 𝑠

𝐺1 𝑠 𝐺2 𝑠 𝐺5 𝑠

3) Non-touching loop gains:

2 pairs of non-touching loops with gains: −𝐺1 𝑠 𝐻3 𝑠 ∙ 𝐺2 𝑠 𝐻1 𝑠

−𝐺1 𝑠 𝐻3 𝑠 ∙ −𝐺2 𝑠 𝐺3 𝑠 𝐻2 𝑠

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The Transfer Function, 𝐶 𝑠 𝑅 𝑠 , of a system representedby a signal-flow graph is:

𝐶 𝑠

𝑅 𝑠=

𝑖=1𝑛 𝑃𝑖∆𝑖

∆

Where: • 𝑛 = number of forward paths• 𝑃𝑖 = the 𝑖𝑡ℎ forward path gain• ∆= determinant of the System• ∆𝑖= determinant of the 𝑖𝑡ℎ forward path

∆ is called the signal flow graph determinant or characteristic function since ∆= 0 is the system characteristic equation.

Mason’s Rule

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• ∆= 1 − 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠

+ 𝑛𝑜𝑛 − 𝑡𝑜𝑢𝑐ℎ𝑖𝑛𝑔 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠 𝑡𝑎𝑘𝑒𝑛 2 𝑎𝑡 𝑎 𝑡𝑖𝑚𝑒

− 𝑛𝑜𝑛 − 𝑡𝑜𝑢𝑐ℎ𝑖𝑛𝑔 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠 𝑡𝑎𝑘𝑒𝑛 3 𝑎𝑡 𝑎 𝑡𝑖𝑚𝑒

+ 𝑛𝑜𝑛 − 𝑡𝑜𝑢𝑐ℎ𝑖𝑛𝑔 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠 𝑡𝑎𝑘𝑒𝑛 4 𝑎𝑡 𝑎 𝑡𝑖𝑚𝑒

− ⋯

• ∆𝑖= ∆ − 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑖𝑛 ∆ 𝒕𝒉𝒂𝒕 𝒕𝒐𝒖𝒄𝒉 𝑡ℎ𝑒 𝑖𝑡ℎ 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑝𝑎𝑡ℎ

= 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑖𝑛 ∆ 𝒕𝒉𝒂𝒕 𝒏𝒐𝒕 𝒕𝒐𝒖𝒄𝒉𝒊𝒏𝒈 𝑡ℎ𝑒 𝑖𝑡ℎ 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑝𝑎𝑡ℎ

Note: ∆𝑖= 1 if there are no non-touching loops to the 𝑖𝑡ℎ path.

Mason’s Rule (Cont’d)

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Seven (7) steps:

1) Calculate forward path gain 𝑃𝑖 for each forward path 𝑖.

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 ∆ from steps 2, 3, 4 and 5.

7) Calculate ∆𝒊 as portion of ∆ that not touching the forward path 𝒊.

Mason’s Rule Systematic Approach

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Example Apply Mason’s 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 𝑃𝑖 for

each forward path 𝒊:

•Number of forward paths is 2

•Forward path gains: 𝑃1 = 𝐺1𝐺2𝐺3𝐺4

𝑃2 = 𝐺1𝐺2𝐺5

Step_2: Get the 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠

•Number of loops with gains is 3

• 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛𝑠 = −𝐺1𝐻3 + 𝐺2𝐻1 − 𝐺2𝐺3𝐻2

Step_3: Get the 𝑁𝑇𝐿𝐺𝑠 2 @ 𝑡𝑖𝑚𝑒

each forward path 𝒊:

•Number of NTLs is 2 pairs

• 𝑁𝑇𝐿𝐺𝑠 = −𝐺1𝐻3𝐺2𝐻1

+ 𝐺1𝐻3𝐺2𝐺3𝐻2

Step_4: Get ∆:

•∆= 1 − −𝐺1𝐻3 + 𝐺2𝐻1 − 𝐺2𝐺3𝐻2

+ −𝐺1𝐻3𝐺2𝐻1 + 𝐺1𝐻3𝐺2𝐺3𝐻2



Solution (Cont’d)Step_5: Get ∆𝑖:

• ∆𝑖= ∆ − 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑖𝑛 ∆ 𝒕𝒉𝒂𝒕 𝒕𝒐𝒖𝒄𝒉 𝑡ℎ𝑒 𝑖𝑡ℎ 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑝𝑎𝑡ℎ

= 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑖𝑛 ∆ 𝒕𝒉𝒂𝒕 𝒏𝒐𝒕 𝒕𝒐𝒖𝒄𝒉𝒊𝒏𝒈 𝑡ℎ𝑒 𝑖𝑡ℎ 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑝𝑎𝑡ℎ

Thus:

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Note: Simplest way to find ∆𝑖 terms is to calculate ∆ with the 𝒊𝒕𝒉 path removed;

• Must remove nodes as well !!!

For 𝑖 = 1:

With forward path 1 removed, there are no loops, so:

∆1= ∆= 1 − 0 = 1

For 𝑖 = 2:

Similarly, removing forward path 2 leaves no loops, so:

∆2= ∆= 1 − 0 = 1

Finally:

𝑌 𝑠

𝑅 𝑠=

𝑖=1𝑛 𝑃𝑖∆𝑖

∆=

𝑃1∆1 + 𝑃2∆2

∆=

𝐺1𝐺2𝐺3𝐺4 + 𝐺1𝐺2𝐺5

1 + 𝐺1𝐻3 − 𝐺2𝐻1 + 𝐺2𝐺3𝐻2 − 𝐺1𝐻3𝐺2𝐻1 + 𝐺1𝐻3𝐺2𝐺3𝐻2



Try…• Apply Mason’s 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?

10/11/2021 91Control System for Industrial Automation, Dept of Electrical Eng, Faculty of


Chapter 2 : Fundamental of Mathematical Modeling

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