28/02/2014

DAE32103_Control System 1

Muhammad Faizal bin Ismail

Dept. of Electrical Engineering

PPD, UTHM

ifaizal@uthm.edu.my

013-7143106 1

1. Introduction

2. Laplace Transform Theorem

3. Common Time Domain Input Function

4. Transfer Function

Outline:

1. Introduction

2. Laplace Transform – Table/ Theorem/ Eg.

3. Common Time Domain Input Function

4. Transfer Function – Open/ Closed Loop & Eg.

5. Electrical Elements Modelling – Table & Eg.

6. Mechanical Elements Modelling - Table & Eg.

7. Block Diagram Reduction - Table & Eg.

8. System Response – Poles/ Zeros, Second Order, Steady State Error, Stability Analysis

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OUTLINE

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DAE32103_Control System 2

1. Intro - Objective of this chapter

After completing this chapter you will be able to:

• Describe the fundamental of Laplace transforms.

• Apply the Laplace transform to solve linear ordinary differential equations.

• Apply Mathematical model, called a transfer function for linear time-invariant electrical, mechanical and electromechanical systems.

3

2. What is Laplace Transform?

• Laplace transform is a method or techniques used to transform the time (t) domain to the Laplace/frequency (s) domain

• What is algebra & calculus?

4

Time Domain Frequency Domain

Differential equations

Input q(t)

Output h(t)

Algebraic equations

Input Q(s)

Output H(s)

Calculus Algebra

Laplace Transformation

Inverse Laplace

Transformation

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DAE32103_Control System 3

Laplace Transform (cont.)

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The Laplace transform solution

consists of the following three

steps:

(1) the Laplace transformation of

q1(t) and (r dhldt + h = Gq) to

frequency domain

(2) the algebraic solution for H(s)

(3) the inverse Laplace

transformation of H(s) to time

domain h(t).

(4) The calculus solution is shown as

step 4.

Definition of the Laplace Transform

• Laplace transform is defined as

• Inverse Laplace transform is defined as

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[ ])(tf )()(

0

sFdtetfst

== ∫∞

−

L

L-1 ∫∞+

∞−

==

j

j

sttfdsesF

jsF

σ

σπ

)()(2

1)]([

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Laplace Theorem

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Laplace Table

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Example 1

Find the Laplace transform for

9

1)( =tf

Solution:

Example 2

Find the Laplace transform for

10

atetf

−=)(

Solution:

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DAE32103_Control System 6

Example 3

Find the inverse Laplace transform of( )( )21

32

10)(

++=

ssssF

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Solution:

Expanding F(s) by partial fraction:

Where,

Then, taking the inverse Laplace transform

)(9

40

3

105

9

5)(

332tueteetf

ttt

++−=

−−

Example 4

Given the ,solve for y(t) if all initial conditions are

zero. Use the Laplace transform method.

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Solution:

Substitute the corresponding F(s) for each term:

Solving for the response:

Where, K1= 1 when s=0

K2=-2 when s=-4

K3= 1 when s=-8Hence

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3. Common Time Domain Input

Functions

• Unit Step Function

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cont.

• Unit Ramp Function

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DAE32103_Control System 8

cont.

• Unit Impulse Function

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4. Transfer Function

• Definition:

Ratio of the output to the input; with all initial conditions are zero

• If the transformed input signal is X(s) and the transformed output signal is Y(s), then the transfer function M(s) is define as;

• From this,

• Therefore the output is

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DAE32103_Control System 9

TF of Linear Time Invariant Systems

• In practice, the input-output relation of lines time-invariant system with

continuous-data input is often described by a differential equation

• The linear time-invariant system is described by the following nth-order

differential equation with constant real coefficients;

).()(

....)()(

)()(

.....)(

)(

)(

011

1

1

011

1

1

trbdt

tdrb

dt

trdb

dt

trdb

tcadt

tdca

dt

tcda

td

tcda

m

m

mm

m

m

n

n

nn

n

n

+++=

++++

−

−

−

−

−

−

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c(t) is output

r(t) is input

cont.

• Taking the Laplace transform of both sides,

• If we assume that all initial conditions are zero, hence

• Now, form the ratio of output transform, C(s) divided by input

transform. The ratio, G(s) is called transfer function.

).(___)(....)()(

)(___)(.....)()(

0

1

1

0

1

1

trofconditioninitialsRbsRsbsRsb

tcofconditioninitialsCasCsasCsa

m

m

m

m

n

n

n

n

++++=

++++

−

−

−

−

)().....()().....(01

1

101

1

1sRbsbsbsbsCasasasa m

m

m

m

n

n

n

n ++++=++++−

−

−

−

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)....(

)....(

)(

)()(

01

1

1

01

1

1

asasasa

bsbsbsb

sR

sCsG

n

n

n

n

m

m

m

m

++++

++++==

−

−

−

−

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DAE32103_Control System 10

cont.

• The transfer function can be represented as a block diagram

• General block diagram

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Block Diagram of Open Loop

System

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Block Diagram of Closed Loop

System

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Example 1

• Problem: Find the transfer function represented by

• Solution:

Taking the Laplace transform of both sides, assuming zero initial

conditions, we have

The transfer function, G(s) is

)()(2)(

trtcdt

tdc=+

)()(2)( sRsCssC =+

22

2

1

)(

)()(

+==

ssR

sCsG

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Example 2

• Problem: Use the result of Example 1 to find the response, c(t), to an input, r(t)=u(t), a unit step and assuming zero initial conditions.

• Solution:

Since r(t)=u(t), R(s)=1/s, hence

Expanding by partial fractions, we get

Finally, taking the inverse Laplace transform of each term yields

)2(

1)()()(

+==

sssGsRsC

2

2/12/1)(

+−=

sssC

23

)(2

1

2

1)(

2tuetc

t

−=

−

Example 3

• Problem: Find the transfer function, G(s)=C(s)/R(s), corresponding to the

differential equation

• Solution:

rdt

dr

dt

rdc

dt

dc

dt

cd

dt

cd34573

2

2

2

2

3

3

++=+++

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Example 4

• Problem: Find the differential equation corresponding to the transfer function,

• Solution:26

12)(

2++

+=

ss

ssG

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Example 5

• Problem: Find the ramp response for a system whose transfer function is,

• Solution:

)8)(4()(

++=

ss

ssG

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