sistem kontrol i kuliah ii : transformasi laplace imron rosyadi, st email: pak.imron@gmail.com 1

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Sistem Kontrol IKuliah II : Transformasi Laplace

Imron Rosyadi, STEmail: pak.imron@gmail.com

1

From Lecture #1

2

3

Control System Design Process

• Diagram on the next page gives a flowchart of the control system design process

4Original System

- Plant - Sensors

- Actuators

New System

Math Model of Controller

Key Activities of the“MAD” Control Engineer:

- Modeling - Analysis - Design

- ImplementationMath Model

of Plant

MeasurementModeling

Implementation - Physical controller - Coupling controller

with plant

Desired Performance

DevelopPerformance Specifications

Analysis

Design

Simulation

5

Control System Design Process

• Hidden in this chart are three important elements :

1. Modeling the system (using mathematics)2. Analysis techniques for describing and

understanding the system’s behavior3. Design techniques for developing control

algorithms to modify the system’s behavior

• Modeling, analysis, and design = the MAD control theorist

• A fourth key element is Implementation

6

Modeling is the key!

• The single most important element in a control system design and development process is the formulation of a model of the system.

• A framework for describing a system in a precise way makes it possible to develop rigorous techniques for analyzing the system and designing controllers for the system

Modeling

• Key Point: most systems of interesting in engineering can be described (approximately) by▫Linear▫Ordinary▫Constant-coefficient▫Differential equations

• Call these LODEs• Where we are going looks like this:

PhysicalReality

LODELaplace

Transform

Requirescalculus to solve

Requiresalgebra to solve

7

Review of Complex Number

8

Complex Numbers: Notation and Properties (1)• A complex number:

• The complex plane

Rectangular (Cartesian) coordinates

Polar coordinates

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• Transformation between coordinates

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Complex Numbers: Notation and Properties (2)

10

Complex Numbers: Notation and Properties (3)• Euler’s Formula:

• Note differentiation property

)sin()cos( xjxe jx

jx

jx

je

xjxj

xjx

xdx

djx

dx

de

dx

d

)sin()cos(

)cos()sin(

)sin()cos(

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Exercise

• Show how Euler’s Formula is a parameterization of the unit circle

sincos je j

)tan(tantan

1sincos1

cossin1

22r

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Complex Numbers: Notation and Properties (4)

• Alternate notation for polar coordinates using Euler’s Formula

Compare to

Note: keep track of degrees and radians!

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Complex Math – Review

• Complex multiplication and division: the hard way

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Complex Math – Review• Complex multiplication and division: the easy

way Given:

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Exercise

7.1099.0

0.767.331.4

6.30.761.4

7.336.3

41

23

j

js

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Complex Math – Review• Complex conjugate:

• Some key results:

Given:

Define complex conjugate as

jbas

jbas

22 bass

22

222

)(

)(2

))(())((

bax

baaxx

jbaxjbaxsxsx

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Complex Math – Review• A function of a complex number is also a complex

number

• Example

Given: )}(Im{)}(Re{)( sGjsGsG

10

10)(

s

sG

45 js

415

10

1045

10)45(

jjjG

2606.06442.0

1660.06224.0 j

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Complex Math – Review• Derivatives of a function of complex numbers, G(s),

can be computed in the usual way

• Poles/Zeros

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Complex Math – Review• Poles/Zeros at infinity

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Laplace Transform

21

Laplace Transform Motivation

• Differential equations model dynamic systems

• Control system design requires simple methods for solving these equations!

• Laplace Transforms allow us to▫systematically solve linear time invariant (LTI)

differential equations for arbitrary inputs.▫easily combine coupled differential equations

into one equation.▫use with block diagrams to find

representations for systems that are made up of smaller subsystems.

uxbxm

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22

The Laplace Transform Definition

• Laplace Transform exists if integral converges for any value of s▫Region of convergence is not as important

for inverting “one-sided” transforms

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Laplace Transform Example (1)

• Example:

• Show that

Notation for “unit step”

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Laplace Transform Example (2)

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Laplace Transform of a Unit Step

• Find the Laplace Transform for the following function

otherwise0

01)(

ttus

s

s

es

dtesF stst

1

101

11)(

00

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Exercise

• Find the Laplace Transform for the following function

otherwise0

103)(

ttf

s

s

stst

es

es

es

dtesF

13

13

33)(

1

0

1

0

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The Laplace Transform Definition (Review)• Recall:

• The easiest way to use the Laplace Transform is by creating a table of Laplace Transform pairs. We can use several Laplace Transform properties to build the table.

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The function with the simplest Laplace Transform (1)

• A special input (class) has a very simple Laplace Transform

• The impulse function:▫Has unit “energy”

▫Is zero except at t=0

Think of pulse in the limit

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The function with the simplest Laplace Transform (2)

1t

sFtf

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LT Properties: Scaling and Linearity

• Proof: Both properties inherited from linearity of integration and the Laplace Transform definition

sFsFtftf

saFtaf

sFtf

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Example 1• Find the following Laplace Transforms

▫Hint: Use Euler’s Formula

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Example 1 (2) tjtj eej

t 2

1sin

22

11

2

1sin

s

jsjsjtL

tjtj eet 2

1cos

22

11

2

1cos

s

s

jsjstL

22

22

cos

sin

s

st

st

sFtf

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33

LT Properties: Time and Frequency Shift

• Proof of frequency shift: Combine exponentials

sFtfe

sFetutf

sFtf

t

s

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Example 2

• Find the following Laplace Transforms

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Example 2 (2)

22

22

)(

cos

as

as

s

ste

ass

atL

22

22

)(

sin

as

ste

ass

atL

22

22

cos

sin

as

aste

aste

sFtf

at

at

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LT Properties: Integration & Differentiation

• Proof of Differentiation Theorem: Integration by parts

vduuvudv

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LT Properties: Integration & Differentiation (2)

01

0

dfss

sFdf

fssFtfdt

dsFtf

t

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Example 3

• Find Laplace Transform for

• What is the Laplace Transform of▫Derivative of a step?▫Derivative of sine?

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Example 3 (2)

2

111)(

ssstdtut LL

1)0(1)(

u

ss

dt

tduL Impulse!

10

1

1)sin(22

s

s

ss

dt

tdL Cosine!

22

11

asste

ass

at

L

1

sin

1

1

1

2

2

2

s

st

dt

d

tudt

das

te

st

sFtf

at

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Exercise• What is the Laplace Transform of

tdt

dcos

1

1

1

1

11

1

)cos(22

2

2

2

2

ss

s

s

s

s

ss

dt

tdL -Sine!

1

1cos

2 s

tdt

dsFtf

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41

Initial Value Theorem

42

Final Value Theorem

43

Inverse Laplace and LODE solution

44

Inverse Laplace Transform

45

Partial Fraction Idea -1

46

Partial Fraction Idea -2

47

Partial Fraction Idea -3

48

Recall: Laplace differentiation theorem (1)

• The differentiation theorem

• Higher order derivatives

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Differentiation Theorem (revisited)

• Differentiation Theorem when initial conditions are zero

000

1

121 f

dt

df

dt

dsfssFstf

dt

dn

nnnn

n

n

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50

Solving differential equations: a simple example (1)• Consider

0,1 tdt

dx

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Solving differential equations: a simple example (2)

• Solution Summary▫Use differentiation theorem to take Laplace

Transform of differential equation▫Solve for the unknown Laplace Transform

Function▫Find the inverse Laplace Transform

0,0 txttx

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Example 1• Find the Laplace Transform for the solution to

120300

00...0

2

121

sXxssXxsxsXs

fsffssFstfdt

dL nnnn

n

n

Notation:

30,100,123 xxtxxx

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- Partial Fraction Expansions• In general, LODEs can be transformed into a function

that is expressed as a ratio of polynomials• In a partial fraction expansion we try to break it into its

parts, so we can use a table to go back to the time domain:

• Three ways of finding coefficients▫ Put partial fraction expansion over common

denominator and equate coefficients of s (Example 1)▫ Residue formula ▫ Equate both sides for several values of s (not

covered)

54

- Partial Fraction Expansions• Have to consider that in general we can

encounter:

▫Real, distinct roots▫Real repeated roots▫Complex conjugate pair roots (2nd order

terms)▫Repeated complex conjugate roots

222

2

22211 ))(()()()(

)()(

bas

GFsEs

bas

DCs

ps

B

ps

AK

sD

sNsX

Example 1, Part 2• Given X(s), find x(t).

• This Laplace Transform function is not immediately familiar, but it is made up of parts that are.

• Factor denominator, then use partial fraction expansion:

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Finding A, B, and C

• To solve, re-combine RHS and equate numerator coefficients (“Equate coefficients” method)

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Final Step• Example 1 completed:

• Since

• By inspection,

s

KtKe t 0,

0,2

52

2

1 2 teetx tt

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Residue Formula (1)• The residue formula allows us to find one

coefficient at a time by multiplying both sides of the equation by the appropriate factor.

• Returning to Example 1:

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Residue Formula (2)• For Laplace Transform with non-repeating roots,

• The general residue formula is:

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Example 2• Find the solution to the following differential

equation:

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Example 2 (2)

)2(

1

)1(

2)(

ss

sX

0,2)( 2 teetx tt

0)(21)(3)(

0)(20)(300)(2

2

sXssXssXs

sXxssXxsxsXs

3)()23( 2 ssXss

)2()1()2)(1(

3)(

s

B

s

A

ss

ssX

221

311

1

s

sXsA 112

322

2

s

sXsB

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Inverse Laplace Transform with Repeated Roots• We have discussed taking the inverse Laplace

transform of functions with non-repeated, real roots using partial fraction expansion.

• Now we will consider partial fraction expansion rules for functions with repeated (real) roots:▫# of constants = order of repeated roots

• Example:

23223

4

)3()3(3)3(

)1(

s

E

s

D

s

C

s

B

s

A

ss

s

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Repeated real roots in Laplace transform table• The easiest way to take an inverse Laplace

transform is to use a table of Laplace transform pairs.

222

222

1

2

)(

)(2)sin(

2)sin(

)(

1

!

)(

1)()(

as

astte

s

stt

asn

etas

te

sFtf

at

n

atn

at

Repeated Real Roots

Repeated Imaginary Roots(also use cosine term)

Repeated Complex Roots(also use cosine term)

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Example with repeated roots• Example: find x(t)

• Take Laplace Transform of both sides:

tef 2

21

1x

10,00

2 2

xx

exxx t

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Example with repeated roots (2)

•Terms with repeated roots:

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Example with repeated roots (3) C = 1B = 2

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Exercise 1• Find the solution to the following differential

equation 0)0(1)0(044 xxxxx

tt

ss

teetx

Ass

sA

s

s

ssXsB

s

B

s

A

s

ssX

ssssX

sXssXssXs

sXxssXxsxsXs

22

22

22

2

22

2

2

2

2

1:2

22

2

4

242

222

4

444

04440

040400

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Above …

Inverse Laplace and LODE solutions

- Partial fraction expansions - LODE solution examples

* Real roots* Real, repeated roots

Next:* Complex roots

69

NOTE:

A complex conjugate pair is actually two distinct, simple first order poles, so can find residues and combine in the usual way:

71

Inverse Laplace Transform with Complex Roots

• To simplify your algebra, don’t use first-order denominators such as

• Instead, rename variables

• So that

21 KKB 21 11 KjKjC

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Laplace Transform Pairs for Complex Roots

• More Laplace transform pairs (complex roots):

• Also, see the table in your textbook and most other control systems textbooks.

22

22

)()sin(

)()cos(

)()(

ste

s

ste

sFtf

t

t

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Return to example from above:74

75

Example with complex roots• Example: find x(t)

• Laplace Transform

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Example with complex roots (2)

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Example with complex roots (3)

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Example with complex roots (5)

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Exercise 2• Find solution to the following differential

equation

0)0(1)0(084 xxxxx

tetetx

ss

s

s

sss

ssX

sXssXssXs

sXxssXxsxsXs

tt 2sin2cos

22

2

22

2

42

484

4

0844

080400

22

2222

2

2

2

2

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Resume Lecture #2

81

•Review of Complex Number•Laplace Transform•Inverse Laplace Transform•Solving LODE

82

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