Review of Laplace Transformsguppy.mpe.nus.edu.sg/anpoo/FBControla/2Laplace Transforms...Solution: From Table L-1 L-1 5L-1 L-1 2L-1 ME2142/ME2142E Feedback Control Systems 15 Using

ME2142/ME2142E Feedback Control Systems1

Review of Laplace TransformsReview of Laplace Transforms

ME2142/ME242E

Feedback Control Systems

ME2142/ME242E

Feedback Control Systems


Complex variableComplex variable

js

Real part Imaginary part

1j

Re

Im

s

s-plane

Mathematical preliminariesMathematical preliminaries


Complex functionComplex function

Complex functionComplex function

Real part Imaginary part

yx jGGsG )(Im

G(s)-plane

xG

yG )(sG

G22yx GGG

x

y

GG1tan

jeGsG )(

yx jGGsG )(

Complex conjugate


Euler’s TheoremEuler’s Theorem

sincos je j

sincos je j Its Complex conjugates

Euler’s Theorem

jj eej

21sin

jj ee 21cos

Some useful formulas:


Laplace TransformsLaplace Transforms

A mathematical tool that transforms difficult differential equations into simple algebraic equations where solutions can be easily obtained. A mathematical tool that transforms difficult differential equations into simple algebraic equations where solutions can be easily obtained.

Inverse Transform:

jc

jc

st dsesFj

)(21

Definition:

0for0)( ttf

Normally Tables of Laplace Transform pairs are used for taking the Laplace Transfrom and the Inverse Transforms.


Properties of Laplace TransformsProperties of Laplace Transforms

Linearity:

2

1s

t


So L and LSo L and L


Use of Linearity Property:

ExampleExample

2

1s

t

4

2)2sin( 2

st

4

103)2sin(53 22

sstt

From Tables, we have:

Then L

2

1s

t



Translation:

If , then

L )()( sFtf Notation:



If , then

GivenGiven

43 6)()(

ssFttf

Example on use of Translation Property

?)(,)2()( 13

1 sGttg ?)(,)( 232

2 sGtetg t

From Tables:From Tables:

SinceSince 42

116)(),2()(s

esGtftg s

AlsoAlso 422

2 )2(6)(),()(

s

sGtfetg t



Variable transform:

FindFind

43 6)()(

ssFttf

)2()2(8)( 33 tftttg

44

48)2/(

621)(

sssG

Then, from Linearity propertiesThen, from Linearity properties

38)(for)( ttgsG

SinceSinceApproach 1Approach 1

44

486)8()(ss

sG

Approach 2Approach 2

AndAnd



Derivatives:

Useful

Not so useful



Final-Value Theorem:

)(lim)(lim0

ssFtfst

Initial-Value Theorem:

)(lim)0( ssFfs

Useful

Not so useful


Finding the Inverse Finding the Inverse

1. Using L-1

2. Using the Table of Transform Pairs

3. Using properties

4. Using Partial Fraction Expansion(We do not normally use Approach 1. We use a combination of 2 to 4.)

1. Using L-1

2. Using the Table of Transform Pairs

3. Using properties

4. Using Partial Fraction Expansion(We do not normally use Approach 1. We use a combination of 2 to 4.)

jc

jc

st dsesFj

sF )(21)(

Not normally

used


Using Table and Properties Using Table and Properties

ExampleExample

Thus L-1

ts

21

2)3(5

stte

s3

2 5)3(

1

168

)3(51)( 222

sss

sF

168

2s )4sin(24

422 t

s

)4sin(25)( 3 ttetsF t

Lookup from TableLookup from Table

Lookup from Table; Use Translation propertyLookup from Table; Use Translation property

Lookup from TableLookup from Table atas

a sin22

L

Find the inverse of

Solution:

From Table L-1

L-1 5L-1

L-1 2L-1


Using Table and Properties Using Table and Properties

ExampleExample52

2)( 2

8

ssesF

s

Solution: To make use of

228

2

8

2)1(2

522)(

se

ssesF s

s

)2sin(2)1(

222 te

st

From Table L-1 )2sin(2

222 t

s

8for0

8for)8(2sin)( )8(

tttesF t

otherwise0andfor)()( atatfsFe asAlso Property L-1

Thus L-1

Find the inverse of

12)1( 22 sss

atas

a sin22

LWe write

Note that

Using translation property L-1

ME2142/ME2142E Feedback Control Systems

16

Using Partial Fractions Using Partial Fractions

Frequently we can write

with N(s) and D(s) being polynomials in s

Example

or Eqn(1-1)

are the zeros and poles of F(s) respectively. They can either be real or complex. If they are complex, they always occur in conjugate pairs.

)()()(

sDsNsF

nm pppzzz ,,,and,,, 2121

mnasasasabsbsbsbsF n

nn

n

mm

mm

with)(

011

1

011

1

)())(()())(()(

21

21

n

m

pspspszszszsKsF



n

n

psa

psa

psa

sDsNsF

2

2

1

1

)()()(

Eqn(1-3)kps

kk sDsNpsa

)()()(

Of the form (s+p)

If Eqn(1-1) has distinct (not repeated) real poles, then F(s) can always be expanded into a sum of partial fractions:

Eqn(1-2)

where ak are constants.

To determine the value of ak, multiply both sides of Eqn(1-2) by (s+pk) and let s = -pk.

n

kn

k

kkkkk ps

psapspsa

pspsa

pspsa

sDsNps

)()()()(

)()()(

2

2

1

1



ExampleExampleFind the inverse of

Solution:We let

then

Thus and

)2)(1(3)(

ssssF

21)2)(1(3)( 21

sa

sa

ssssF

22)(1(

)3()1(1

1

sssssa

12)(1(

)3()2(2

2

sssssa

21

12)(

sssF 0for2)( 2 teetf tt



If Eqn(1-1) has multiple real poles, each multiple pole pr of order q will be equivalent to partial fractions of the form:

where bk are constants.

qr

q

rr psb

psb

psb

)()()( 221

If Eqn(1-1) has denominators of the form each of these will be equivalent to partial fractions of the form:

where a, b, c and d are constants.

)( 2 bass

bassdcs

basssNsF 22 )()()(

More than one real pole (s+p)with the same value

Indicates a pair of complex conjugate

poles



Given , find f(t).

Solution:

Let

Multiplying both sides by (s+1)3

Letting we have

Comparing terms in s2, we have

Comparing constant terms, we have giving

Thus and

ExampleExample 3

2

)1(32)(

s

sssF

322

12 )1()1(32 bsbsbss

1s 23 b

11 b

33

221

3

2

)1()1()1()1(32)(

s

bs

bsb

ssssF

3213 bbb 02 b

3)1(2

11)(

sssF tt etetf 2)(


Solving differential equations Solving differential equations

1. Take the Laplace transform of the differential equation to convert it into an algebraic equation. The Laplace transform of the dependent variable is then obtained using simple Algebra.

2. The time function of the dependent variable is then obtained by taking the inverse of the Laplace transform.

1. Take the Laplace transform of the differential equation to convert it into an algebraic equation. The Laplace transform of the dependent variable is then obtained using simple Algebra.

2. The time function of the dependent variable is then obtained by taking the inverse of the Laplace transform.



ExampleExampleSolve for y(t) given

Solution:

Transforming

or

giving

0)0(,0)0(,352 yyyyy

ssYsssYssYsYs 3)()52()(5)(2)( 22

)52(3)( 2

sss

sY522

sscbs

sa

s

sYyssYysysYs 3)(5)0()(2)0()0()(2



ExampleExample

We have

Multiplying both sides by s and letting s=0, we get

Multiplying both sides by the denominators

we have

Comparing terms in s,

Comparing terms in s2,

Thus

And

)52(3)( 2

sss

sY522

sscbs

sa

53

a

)52( 2 sss

)()52(3 22 csbsssa

5/65/60 cc

5/35/30 bb

522

53

53)( 2

sss

ssY

0for2sin1032cos

53

53)( ttetety tt

2222 2)1(2

103

2)1(1

53

53

ss

ss


Laplace TransformsLaplace Transforms

END

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Review of Laplace Transformsguppy.mpe.nus.edu.sg/anpoo/FBControla/2Laplace Transforms...Solution: From Table L-1 L-1 5L-1 L-1 2L-1 ME2142/ME2142E Feedback Control Systems 15 Using